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Removing DESpecs directory which deserted to git
| author | Klaus Thoden <kthoden@mpiwg-berlin.mpg.de> |
|---|---|
| date | Wed, 29 Nov 2017 16:55:37 +0100 |
| parents | 22d6a63640c6 |
| children |
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<?xml version="1.0" encoding="utf-8"?><echo xmlns="http://www.mpiwg-berlin.mpg.de/ns/echo/1.0/" xmlns:de="http://www.mpiwg-berlin.mpg.de/ns/de/1.0/" xmlns:dcterms="http://purl.org/dc/terms" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:echo="http://www.mpiwg-berlin.mpg.de/ns/echo/1.0/" xmlns:xhtml="http://www.w3.org/1999/xhtml" xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" version="1.0RC"> <metadata> <dcterms:identifier>ECHO:R04RNX9Y.xml</dcterms:identifier> <dcterms:creator identifier="GND:118201913">Bélidor, Bernard Forest de</dcterms:creator> <dcterms:title xml:lang="fr">Nouveau cours de mathématique à l' usage de l' artillerie et du génie : où l' on applique les parties les plus utiles de cette science à la théorie et à la pratique des différens sujets qui peuvent avoir rapport à la guerre</dcterms:title> <dcterms:date xsi:type="dcterms:W3CDTF">1757</dcterms:date> <dcterms:language xsi:type="dcterms:ISO639-3">fra</dcterms:language> <dcterms:rights>CC-BY-SA</dcterms:rights> <dcterms:license xlink:href="http://creativecommons.org/licenses/by-sa/3.0/">CC-BY-SA</dcterms:license> <dcterms:rightsHolder xlink:href="http://www.mpiwg-berlin.mpg.de">Max Planck Institute for the History of Science, Library</dcterms:rightsHolder> <parameters>despecs=1.1.2</parameters> </metadata> <text xml:lang="fr" type="free"> <div xml:id="echoid-div1" type="section" level="1" n="1"><pb file="0001" n="1"/> <pb file="0002" n="2"/> <pb file="0003" n="3"/> <pb file="0004" n="4"/> <pb file="0005" n="5"/> <pb file="0006" n="6"/> <pb file="0007" n="7"/> </div> <div xml:id="echoid-div2" type="section" level="1" n="2"> <head xml:id="echoid-head1" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHEMATIQUE, <lb/>A L’USAGE <lb/>DE L’ARTILLERIE ET DU GENIE,</head> <p> <s xml:id="echoid-s1" xml:space="preserve">Où l’on applique les parties les plus utiles de cette ſcience <lb/>à la théorie & </s> <s xml:id="echoid-s2" xml:space="preserve">à la pratique des différens ſujets qui peuvent <lb/>avoir rapport à la guerre.</s> <s xml:id="echoid-s3" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div3" type="section" level="1" n="3"> <head xml:id="echoid-head2" xml:space="preserve">NOUVELLE EDITION, <lb/>Corrigée & conſidérablement augmentée.</head> <p style="it"> <s xml:id="echoid-s4" xml:space="preserve">Par M. </s> <s xml:id="echoid-s5" xml:space="preserve"><emph style="sc">Belidor</emph>, Colonel d’Infanterie, Chevalier de l’Ordre <lb/>Royal & </s> <s xml:id="echoid-s6" xml:space="preserve">Militaire de Saint Louis, Membre des Académies Royales <lb/>des Sciences de France, d’Angleterre & </s> <s xml:id="echoid-s7" xml:space="preserve">de Pruſſe.</s> <s xml:id="echoid-s8" xml:space="preserve"/> </p> <figure> <image file="0007-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0007-01"/> </figure> </div> <div xml:id="echoid-div4" type="section" level="1" n="4"> <head xml:id="echoid-head3" xml:space="preserve">A PARIS, <lb/>Chez <emph style="sc">Nyon</emph>, Quai des Auguſtins, près le Pont S. Michel, <lb/>à l’Occaſion.</head> <head xml:id="echoid-head4" xml:space="preserve">M. DCC. LVII.</head> <head xml:id="echoid-head5" style="it" xml:space="preserve">AVEC APPROBATION ET PRIVILEGE DU ROI.</head> <pb file="0008" n="8"/> <handwritten/> <handwritten/> <handwritten/> <pb file="0009" n="9"/> <figure> <image file="0009-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0009-01"/> </figure> </div> <div xml:id="echoid-div5" type="section" level="1" n="5"> <head xml:id="echoid-head6" style="it" xml:space="preserve">PRÉFACE.</head> <p> <s xml:id="echoid-s9" xml:space="preserve"><emph style="sc">QUoique</emph> le titre de cet Ouvrage me paroiſſe an-<lb/>noncer ſuffiſamment ce que l’on s’y eſt propoſé, & </s> <s xml:id="echoid-s10" xml:space="preserve">qu’il <lb/>ſoit connu par pluſieurs éditions, je ne me crois pas <lb/>pour cela diſpenſé de rendre compte ici du Livre en gé-<lb/>néral d’une maniere plus détaillée, & </s> <s xml:id="echoid-s11" xml:space="preserve">des additions con-<lb/>ſidérables que j’y ai faites. </s> <s xml:id="echoid-s12" xml:space="preserve">Perſuadé & </s> <s xml:id="echoid-s13" xml:space="preserve">convaincu, par <lb/>une longue expérience, que les Officiers & </s> <s xml:id="echoid-s14" xml:space="preserve">les Ingé-<lb/>nieurs militaires ne doivent pas étudier les Mathéma-<lb/>tiques de la même maniere qu’une perſonne qui vou-<lb/>droit s’y livrer entiérement, & </s> <s xml:id="echoid-s15" xml:space="preserve">en faire ſon étude prin-<lb/>cipale, j’ai tâché de réunir dans un ſeul volume tout ce <lb/>qui leur eſt abſolument néceſſaire, en joignant, autant <lb/>qu’il m’a été poſſible, les applications des principes que <lb/>je donne, à des exemples ſenſibles, & </s> <s xml:id="echoid-s16" xml:space="preserve">qui ont un rap-<lb/>port direct aux opérations qu’ils ſont obligés de faire <lb/>dans les places qu’ils ont à remplir. </s> <s xml:id="echoid-s17" xml:space="preserve">C’eſt ſans doute à <lb/>cela que je puis attribuer le ſuccès qu’il a eu, & </s> <s xml:id="echoid-s18" xml:space="preserve">c’eſt <lb/>pour le rendre encore plus intéreſſant que j’ai toujours <lb/>travaillé ſur le même plan. </s> <s xml:id="echoid-s19" xml:space="preserve">Preſque toutes les additions <lb/>que j’ai faites ont pour objet des queſtions ou des mé-<lb/>thodes utiles dans la pratique, dont la préciſion doit <lb/>être le but de toutes les études d’un Ingénieur. </s> <s xml:id="echoid-s20" xml:space="preserve">Si le <lb/>goût des Mathématiques n’avoit pas fait des progrès <lb/>auſſi ſurprenans depuis une quarantaine d’année, j’aurois <lb/>pu me contenter dans cette nouvelle édition de corri- <pb o="iv" file="0010" n="10" rhead="PREÉFACE."/> ger les fautes qui s’étoient gliſſées dans la premiere, & </s> <s xml:id="echoid-s21" xml:space="preserve"><lb/>de donner des démonſtrations plus rigoureuſes & </s> <s xml:id="echoid-s22" xml:space="preserve">plus <lb/>élégantes de certaines propoſitions, ſans rien ajouter de <lb/>nouveau; </s> <s xml:id="echoid-s23" xml:space="preserve">mais eu égard aux connoiſſances que l’on <lb/>exige actuellement des Ingénieurs, j’ai fait toutes les <lb/>additions qui m’ont paru abſolument néceſſaires pour <lb/>rendre cet Ouvrage complet dans ſon genre. </s> <s xml:id="echoid-s24" xml:space="preserve">Une <lb/>théorie abrégée, mais rigoureuſement démontrée d’un <lb/>petit nombre de principes & </s> <s xml:id="echoid-s25" xml:space="preserve">des premiers élémens de <lb/>chaque partie des Mathématiques, analogue à l’art de <lb/>la guerre, & </s> <s xml:id="echoid-s26" xml:space="preserve">à tout ce qui en dépend; </s> <s xml:id="echoid-s27" xml:space="preserve">c’eſt à quoi doi-<lb/>vent ſe borner les études d’un habile Militaire. </s> <s xml:id="echoid-s28" xml:space="preserve">S’il veut <lb/>après cela donner dans toutes les autres ſciences étran-<lb/>geres à ſa profeſſion, quoique dépendantes des Mathé-<lb/>matiques, il ne fait que décorer ſon eſprit, ſans ſe ren-<lb/>dre plus utile à l’état, qui ne peut tirer aucun ſecours <lb/>de ces vérités ſublimes, deſtinées plutôt à faire briller le <lb/>génie dans une aſſemblée de Sçavans, qu’à rendre des <lb/>ſervices importans au Prince dans des occaſions dange-<lb/>reuſes.</s> <s xml:id="echoid-s29" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s30" xml:space="preserve">Cet Ouvrage eſt diviſé en ſeize Livres. </s> <s xml:id="echoid-s31" xml:space="preserve">Dans le pre-<lb/>mier, je donne les premiers élémens d’Algebre, après <lb/>avoir donné les définitions des propoſitions dont on ſe <lb/>ſert en Géométrie, & </s> <s xml:id="echoid-s32" xml:space="preserve">des termes le plus en uſage dans <lb/>cette ſcience. </s> <s xml:id="echoid-s33" xml:space="preserve">On y traite d’abord du calcul arithmé-<lb/>tique, par rapport à la Multiplication & </s> <s xml:id="echoid-s34" xml:space="preserve">à la Diviſion, <lb/>en ſe ſervant de ce que l’on appelle communément par-<lb/>ties aliquotes: </s> <s xml:id="echoid-s35" xml:space="preserve">c’eſt une des premieres additions qui <lb/>m’a paru néceſſaire pour montrer aux Commençans <lb/>des manieres abrégées de faire ces opérations, qui de-<lb/>viennent fort longues en ſuivant les regles générales, <lb/>dans les cas où le multiplicande & </s> <s xml:id="echoid-s36" xml:space="preserve">le multiplicateur <pb o="v" file="0011" n="11" rhead="PRÉFACE."/> ſont tous deux des nombres complexes. </s> <s xml:id="echoid-s37" xml:space="preserve">Delà je paſſe <lb/>au calcul des fractions numériques & </s> <s xml:id="echoid-s38" xml:space="preserve">algébriques, aux-<lb/>quelles j’ai ajouté la théorie & </s> <s xml:id="echoid-s39" xml:space="preserve">la pratique des fractions <lb/>décimales, que je démontre par le principe de la nu-<lb/>mération: </s> <s xml:id="echoid-s40" xml:space="preserve">cette partie m’a paru indiſpenſablement né-<lb/>ceſſaire pour mettre un Ingénieur au fait des Livres <lb/>dont il eſt obligé de faire uſage. </s> <s xml:id="echoid-s41" xml:space="preserve">Tout le monde ſçait <lb/>que les Tables des ſinus, dont on ſe ſert ſi fréquem-<lb/>ment dans la Trigonométrie, ſont conſtruites par le <lb/>moyen des décimales. </s> <s xml:id="echoid-s42" xml:space="preserve">On opere toujours avec plus de <lb/>ſûreté quand on connoît la nature des nombres ſur <lb/>leſquels on opere. </s> <s xml:id="echoid-s43" xml:space="preserve">On voit encore dans le même Livre <lb/>un uſage important des décimales dans l’approxima-<lb/>tion des racines quarrées & </s> <s xml:id="echoid-s44" xml:space="preserve">cubiques qu’il faut déter-<lb/>miner avec tout le ſoin poſſible dans certaines occa-<lb/>ſions. </s> <s xml:id="echoid-s45" xml:space="preserve">J’ai encore ajouté un Traité complet du calcul <lb/>des Expoſans, que j’ai mis devant le chapitre de la for-<lb/>mation des puiſſances auxquelles ce calcul a un rapport <lb/>direct.</s> <s xml:id="echoid-s46" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s47" xml:space="preserve">Dans le ſecond Livre, je traite des raiſons ou rapports <lb/>arithmétiques & </s> <s xml:id="echoid-s48" xml:space="preserve">géométriques, des progreſſions & </s> <s xml:id="echoid-s49" xml:space="preserve"><lb/>proportions qui en réſultent, dont je démontre les prin-<lb/>cipales propriétés. </s> <s xml:id="echoid-s50" xml:space="preserve">De la comparaiſon de la progreſſion <lb/>arithmétique des expoſans d’une même lettre à la pro-<lb/>greſſion géométrique des puiſſances de cette même <lb/>lettre, je déduis la nature & </s> <s xml:id="echoid-s51" xml:space="preserve">les principales propriétés <lb/>des logarithmes, dont on eſt obligé de faire uſage <lb/>dans un grand nombre de queſtions, & </s> <s xml:id="echoid-s52" xml:space="preserve">dont les Ingé-<lb/>nieurs doivent néceſſairement ſe ſervir dans les calculs <lb/>trigonométriques, pour déterminer avec préciſion des <lb/>diſtances inacceſſibles. </s> <s xml:id="echoid-s53" xml:space="preserve">Cette partie, dont je n’avois point <lb/>parlé dans l’ancienne édition, ſe trouve démontrée avec <pb o="vj" file="0012" n="12" rhead="PRÉFACE."/> toute la briéveté poſſible; </s> <s xml:id="echoid-s54" xml:space="preserve">j’eſpere qu’elle n’en ſera pas <lb/>plus difficile à concevoir. </s> <s xml:id="echoid-s55" xml:space="preserve">Je paſſe delà aux regles gé-<lb/>nérales de la méthode analytique dans la recherche de <lb/>la vérité. </s> <s xml:id="echoid-s56" xml:space="preserve">Je montre l’uſage & </s> <s xml:id="echoid-s57" xml:space="preserve">l’application de tout ce <lb/>qui précede, ſoit en Arithmétique, ſoit en Algebre dans <lb/>cette partie, qui eſt la plus importante des Mathéma-<lb/>tiques, & </s> <s xml:id="echoid-s58" xml:space="preserve">quieſt eſſentiellement attachée à cette ſcience. <lb/></s> <s xml:id="echoid-s59" xml:space="preserve">Je donne enſuite un grand nombre d’exemples ſur des <lb/>problêmes, donton peut avoir beſoin dans les différentes <lb/>opérations militaires, qui ſont du détail d’un Ingénieur. </s> <s xml:id="echoid-s60" xml:space="preserve"><lb/>J’ai auſſi ajouté quelques ſolutions générales pour ac-<lb/>coutumer les Commençans à ces expreſſions indéter-<lb/>minées & </s> <s xml:id="echoid-s61" xml:space="preserve">aux idées abſtraites, afin de leur faire mieux <lb/>ſentir l’avantage que l’on peut retirer de l’étude de l’ Al-<lb/>gebre. </s> <s xml:id="echoid-s62" xml:space="preserve">Enfin je termine ce Livre par un Traité complet <lb/>des Equations du ſecond degré, dont je n’avois dit que <lb/>deux mots dans la premiere édition. </s> <s xml:id="echoid-s63" xml:space="preserve">Dans cette partie <lb/>je diſcute la nature des racines poſitives & </s> <s xml:id="echoid-s64" xml:space="preserve">négatives; </s> <s xml:id="echoid-s65" xml:space="preserve"><lb/>je fais voir la différence des unes aux autres; </s> <s xml:id="echoid-s66" xml:space="preserve">les cas où <lb/>ce ſont les racines négatives qui réſolvent le problême <lb/>dans le ſens qu’on s’étoit propoſé: </s> <s xml:id="echoid-s67" xml:space="preserve">d’où il ſuit qu’on ne <lb/>doit point confondre les racines négatives avec celles <lb/>qui ne réſolvent pas le problême comme on le de-<lb/>mande. </s> <s xml:id="echoid-s68" xml:space="preserve">Comme dans la ſolution des équations du ſe-<lb/>cond degré on arrive quelquefois à des radicaux aſſez <lb/>compliqués, j’ai encore ajouté un petit Traité du calcul <lb/>des Incommenſurables.</s> <s xml:id="echoid-s69" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s70" xml:space="preserve">Dans le troiſieme Livre, je commence à traiter de la <lb/>Géométrie, & </s> <s xml:id="echoid-s71" xml:space="preserve">j’examine d’abord les différentes poſitions <lb/>des lignes droites les unes à l’égard des autres; </s> <s xml:id="echoid-s72" xml:space="preserve">ce qui <lb/>me conduit à examiner les propriétés des angles & </s> <s xml:id="echoid-s73" xml:space="preserve">des <lb/>lignes paralleles. </s> <s xml:id="echoid-s74" xml:space="preserve">J’ai ajouté dans ce Livre quelques pro- <pb o="vij" file="0013" n="13" rhead="PRÉFACE."/> blêmes qui m’ont paru néceſſaires pour faire mieux en-<lb/>tendre ce que j’ai à dire dans les Livres ſuivans.</s> <s xml:id="echoid-s75" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s76" xml:space="preserve">Le quatrieme Livre traite des propriétés des ſurfaces <lb/>en général, & </s> <s xml:id="echoid-s77" xml:space="preserve">comme il n’y a point de ſurfaces qu’on <lb/>ne puiſſe réduire en triangles, je commence par expli-<lb/>quer aſſez au long tout ce qui a rapport aux triangles <lb/>& </s> <s xml:id="echoid-s78" xml:space="preserve">aux parallélogrammes. </s> <s xml:id="echoid-s79" xml:space="preserve">J’ai auſſi ajouté dans cette <lb/>partie pluſieurs propoſitions ſur les rapports des trian-<lb/>gles comparés entr’eux, ſoit qu’il s’agiſſe d’une ſimple <lb/>ſimilitude, ou d’une égalité parfaite.</s> <s xml:id="echoid-s80" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s81" xml:space="preserve">Dans le cinquieme Livre, j’examine les propriétés <lb/>du cercle, principalement par rapport à la meſure des <lb/>angles, & </s> <s xml:id="echoid-s82" xml:space="preserve">delà je déduis celles des ſécantes intérieures <lb/>ou extérieures, & </s> <s xml:id="echoid-s83" xml:space="preserve">celles des tangentes; </s> <s xml:id="echoid-s84" xml:space="preserve">j’en fais l’appli-<lb/>cation ſur quelques problêmes, dont la ſolution dépend <lb/>de ces mêmes propriétés.</s> <s xml:id="echoid-s85" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s86" xml:space="preserve">Le ſixieme Livre eſt un Traité de l’inſcription & </s> <s xml:id="echoid-s87" xml:space="preserve">de <lb/>la circonſcription des figures régulieres au cercle. </s> <s xml:id="echoid-s88" xml:space="preserve">J’exa-<lb/>mine enſuite, relativement à cet objet, les propriétés de <lb/>la quadratrice, dont je donne la conſtruction, & </s> <s xml:id="echoid-s89" xml:space="preserve">par <lb/>le moyen de laquelle je réſous d’une maniere aiſée les <lb/>problêmes que l’on peut propoſer ſur la diviſion des arcs <lb/>de cercle, ou des différens ſecteurs en pluſieurs parties <lb/>égales.</s> <s xml:id="echoid-s90" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s91" xml:space="preserve">Dans le ſeptieme Livre, on applique la doctrine des <lb/>proportions aux figures planes: </s> <s xml:id="echoid-s92" xml:space="preserve">on y explique les rap-<lb/>ports des périmetres des figures ſemblables, & </s> <s xml:id="echoid-s93" xml:space="preserve">celui de <lb/>leurs ſurfaces. </s> <s xml:id="echoid-s94" xml:space="preserve">On donne enſuite la maniere de les ajou-<lb/>ter, ſouſtraire, multiplier, & </s> <s xml:id="echoid-s95" xml:space="preserve">diviſer, ſuivant une raiſon <lb/>donnée quelconque; </s> <s xml:id="echoid-s96" xml:space="preserve">ce que l’on fait par l’invention <lb/>des lignes proportionnelles à d’autres lignes données de <lb/>grandeur. </s> <s xml:id="echoid-s97" xml:space="preserve">J’ai ajouté dans cette partie deux théorêmes <pb o="viij" file="0014" n="14" rhead="PRÉFACE."/> extrêmement curieux, l’un ſur le rapport de deux trian-<lb/>gles qui ont un angle égal, compris entre côtés iné-<lb/>gaux, qui eſt d’un grand uſage dans la Géodéſie, & </s> <s xml:id="echoid-s98" xml:space="preserve"><lb/>l’autre ſur la maniere de trouver l’aire d’un triangle, <lb/>dont on connoît les trois côtés. </s> <s xml:id="echoid-s99" xml:space="preserve">La démonſtration que <lb/>j’en donne eſt une des plus ſimples que l’on puiſſe <lb/>trouver: </s> <s xml:id="echoid-s100" xml:space="preserve">le lecteur en jugera par la comparaiſon avec <lb/>celles de la même propoſition qui ſe trouvent dans les <lb/>autres Livres.</s> <s xml:id="echoid-s101" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s102" xml:space="preserve">Après avoir examiné les principales propriétés des <lb/>lignes & </s> <s xml:id="echoid-s103" xml:space="preserve">des ſurfaces, je paſſe, dans le huitieme Livre, à <lb/>la théorie des ſolides ou corps, dont je recherche les <lb/>propriétés par rapport à leurs ſuperficies & </s> <s xml:id="echoid-s104" xml:space="preserve">à leurs ſoli-<lb/>dités. </s> <s xml:id="echoid-s105" xml:space="preserve">J’enſeigne la maniere de toiſer, non ſeulement les <lb/>priſmes, les pyramides, les cônes, les ſpheres, mais <lb/>encore les différentes parties de ces corps. </s> <s xml:id="echoid-s106" xml:space="preserve">A l’occaſion <lb/>de la pyramide tronquée, je donne une méthode gé-<lb/>nérale pour trouver une ſurface plane ſemblable à deux <lb/>autres propoſées, & </s> <s xml:id="echoid-s107" xml:space="preserve">moyenne géométrique entre ces <lb/>deux, ſans être obligé d’extraire de racines quarrées. </s> <s xml:id="echoid-s108" xml:space="preserve">Je <lb/>donne enſuite la maniere de trouver des ſolides qui <lb/>aient entr’eux une raiſon donnée, & </s> <s xml:id="echoid-s109" xml:space="preserve">je fais voir d’où <lb/>dépend la ſolution des problêmes de ce genre, qui ont <lb/>tous rapport à la duplication du cube. </s> <s xml:id="echoid-s110" xml:space="preserve">La méthode que <lb/>j’ai ſuivie dans ce Livre eſt entiérement différente de <lb/>celle qui ſe trouve dans les autres Elémens; </s> <s xml:id="echoid-s111" xml:space="preserve">elle eſt ſi <lb/>ſimple, qu’en moins de ſeize propoſitions, on voit tout <lb/>ce qu’Archimede a découvert de plus beau ſur la ſphere, <lb/>& </s> <s xml:id="echoid-s112" xml:space="preserve">de ma théorie, je laiſſe entrevoir celle de toiſer toutes <lb/>ſortes de voûtes en plein ceintre, qui auroient pour baſe <lb/>des polygones réguliers quelconques.</s> <s xml:id="echoid-s113" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s114" xml:space="preserve">Ces huit premiers Livres font comme une premiere <pb o="ix" file="0015" n="15" rhead="PRÉFACE."/> partie du Cours de Mathématique. </s> <s xml:id="echoid-s115" xml:space="preserve">Afin d’en faire voir <lb/>l’utilité, on a mis après chaque propoſition des corol-<lb/>laires qui en montrent la fécondité & </s> <s xml:id="echoid-s116" xml:space="preserve">l’on voit avec <lb/>admiration l’étendue de la Géométrie dont il ſuffit de <lb/>ſçavoir les premiers élémens pour découvrir les mêmes <lb/>vérités qui ſemblent ſe préſenter d’elles-mêmes à notre <lb/>eſprit, pour établir davantage l’utilité & </s> <s xml:id="echoid-s117" xml:space="preserve">l’importance <lb/>des premieres, & </s> <s xml:id="echoid-s118" xml:space="preserve">qui ſemblent par-là s’empreſſer de <lb/>nous dédommager des premiers ſoins que nous avons <lb/>pris pour arriver à la connoiſſance de ces premieres vé-<lb/>rités.</s> <s xml:id="echoid-s119" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s120" xml:space="preserve">Comme les ſimples élémens renfermés dans les huit <lb/>premiers Livres ne ſont pas ſuffiſans pour entendre beau-<lb/>coup de choſes intéreſſantes, qui ſont traitées dans les <lb/>ſuivans, principalement la théorie du jet des bombes, <lb/>& </s> <s xml:id="echoid-s121" xml:space="preserve">le toiſé des voûtes qui demande une connoiſſance au <lb/>moins élémentaire des propriétés des ſections coniques, <lb/>je donne dans le neuvieme Livre un petit Traité, où <lb/>j’explique les principales propriétés de ces courbes par <lb/>rapport à leurs axes & </s> <s xml:id="echoid-s122" xml:space="preserve">à leurs diametres, dont je recher-<lb/>che les tangentes, & </s> <s xml:id="echoid-s123" xml:space="preserve">ſur leſquelles je donne quelques <lb/>problêmes.</s> <s xml:id="echoid-s124" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s125" xml:space="preserve">Le dixieme Livre qui comprend la Trigonométrie & </s> <s xml:id="echoid-s126" xml:space="preserve"><lb/>le nivellement, peut encore être regardé comme un <lb/>des plus néceſſaires à un Ingénieur, dont tout l’Art dé-<lb/>pend de ces deux parties; </s> <s xml:id="echoid-s127" xml:space="preserve">la premiere dans la guerre, <lb/>& </s> <s xml:id="echoid-s128" xml:space="preserve">la ſeconde dans la paix, où il peut être chargé de <lb/>l’exécution des projets les plus importans, & </s> <s xml:id="echoid-s129" xml:space="preserve">qui ont <lb/>abſolument beſoin de la ſcience du nivellement. </s> <s xml:id="echoid-s130" xml:space="preserve">On <lb/>enſeigne dans ce Livre l’uſage des Tables des Sinus, Tan-<lb/>gentes, Sécantes, & </s> <s xml:id="echoid-s131" xml:space="preserve">de leurs Logarithmes; </s> <s xml:id="echoid-s132" xml:space="preserve">la théorie <lb/>du calcul des triangles, que l’on applique enſuite à me-<lb/> <pb o="x" file="0016" n="16" rhead="PRÉFACE."/> ſurer les hauteurs & </s> <s xml:id="echoid-s133" xml:space="preserve">les diſtances inacceſſibles ou acceſ-<lb/>ſibles; </s> <s xml:id="echoid-s134" xml:space="preserve">à la maniere de calculer les parties d’une forti-<lb/>fication, pour la tracer enſuite ſur le terrein. </s> <s xml:id="echoid-s135" xml:space="preserve">Comme <lb/>la meſure des diſtances inacceſſibles eſt de la derniere <lb/>importance dans les travaux militaires, je donne des pro-<lb/>blêmes nouveaux ſur la maniere de les déterminer, par <lb/>le moyen de certaines lignes connues qui ſe trouvent <lb/>déja déterminées. </s> <s xml:id="echoid-s136" xml:space="preserve">Ces problêmes, dont la ſolution dé-<lb/>pend des principes précédens, méritent l’attention de <lb/>ceux que j’ai eu en vue: </s> <s xml:id="echoid-s137" xml:space="preserve">ainſi ils ne peuvent mieux faire <lb/>que de les étudier avec ſoin.</s> <s xml:id="echoid-s138" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s139" xml:space="preserve">Le onzieme Livre eſt un Traité du calcul ordinaire <lb/>des ouvrages de maçonnerie, où j’explique en même <lb/>tems le toiſé des bois. </s> <s xml:id="echoid-s140" xml:space="preserve">Cette partie eſt encore néceſſaire <lb/>aux Ingénieurs, qui ſont quelquefois obligés de faire <lb/>les devis & </s> <s xml:id="echoid-s141" xml:space="preserve">détails de tout ce qui doit entrer dans l’exé-<lb/>cution des ouvrages néceſſaires dans une fortification. <lb/></s> <s xml:id="echoid-s142" xml:space="preserve">On l’a traité d’une maniere ſi claire & </s> <s xml:id="echoid-s143" xml:space="preserve">ſi facile, que les <lb/>Commençans pourront en peu de jours ſe rendre fa-<lb/>miliers ces ſortes de calculs.</s> <s xml:id="echoid-s144" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s145" xml:space="preserve">Dans le douzieme Livre, on fait une application <lb/>générale de la Géométrie à la meſure des ſolides régu-<lb/>liers & </s> <s xml:id="echoid-s146" xml:space="preserve">irréguliers, qui peuvent ſe rencontrer dans la <lb/>pratique: </s> <s xml:id="echoid-s147" xml:space="preserve">par exemple, on y enſeigne la maniere de <lb/>toiſer la ſolidité des voûtes en plein ceintre, ou en tiers <lb/>point; </s> <s xml:id="echoid-s148" xml:space="preserve">celles des voûtes elliptiques ſurbaiſſées, ou ſur-<lb/>montées ſur des plans circulaires ou rectilignes. </s> <s xml:id="echoid-s149" xml:space="preserve">J’ai ajouté <lb/>auſſi dans cet endroit un Traité du Toiſé des ſurfaces <lb/>des voûtes à pans en plein ceintre, & </s> <s xml:id="echoid-s150" xml:space="preserve">des voûtes en lu-<lb/>nettes, ſans autre ſecours que les propriétés du cercle. <lb/></s> <s xml:id="echoid-s151" xml:space="preserve">Je donne auſſi le Toiſé du ſolide de ces mêmes voûtes. </s> <s xml:id="echoid-s152" xml:space="preserve"><lb/>Enſuite on applique les mêmes principes à t oiſer les <pb o="xj" file="0017" n="17" rhead="PRÉFACE."/> revêtemens d’une fortification, par exemple, les oril-<lb/>lons & </s> <s xml:id="echoid-s153" xml:space="preserve">les flancs concaves, les arrondiſſemens des con-<lb/>tre-forts, les pyramides tronquées qui ſe trouvent aux <lb/>angles des mêmes ouvrages, & </s> <s xml:id="echoid-s154" xml:space="preserve">l’onglet d’un batardeau. <lb/></s> <s xml:id="echoid-s155" xml:space="preserve">Enfin je termine cette partie par l’expoſition d’un prin-<lb/>cipe général pour trouver les ſurfaces & </s> <s xml:id="echoid-s156" xml:space="preserve">les ſolides en-<lb/>gendrés par les mouvemens d’une ligne droite ou courbe, <lb/>& </s> <s xml:id="echoid-s157" xml:space="preserve">une ſurface rectiligne ou curviligne autour d’un axe de <lb/>révolution, par le moyen du centre de gravité de ces <lb/>lignes ou ſurfaces génératrices. </s> <s xml:id="echoid-s158" xml:space="preserve">Cette découverte peut <lb/>être regardée comme une des plus importantes que l’on <lb/>ait faite en Géométrie. </s> <s xml:id="echoid-s159" xml:space="preserve">Tout le monde convient que <lb/>l’on en eſt redevable au P. </s> <s xml:id="echoid-s160" xml:space="preserve">Guildin: </s> <s xml:id="echoid-s161" xml:space="preserve">enſorte que l’on ap-<lb/>pelle ce principe communément la Regle du P. </s> <s xml:id="echoid-s162" xml:space="preserve">Guildin.</s> <s xml:id="echoid-s163" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s164" xml:space="preserve">Le treizieme Livre eſt encore une application des <lb/>mêmes principes à la Géodéſie ou diviſion des champs <lb/>en parties qui aient entr’elles des rapports déterminés, <lb/>quelle que ſoit la figure du terrein que l’on veut partager, <lb/>& </s> <s xml:id="echoid-s165" xml:space="preserve">en commençant la diviſion par des lignes tirées d’un <lb/>point donné. </s> <s xml:id="echoid-s166" xml:space="preserve">Delà je paſſe à l’explication d’une ma-<lb/>chine connue de tout le monde, ſous le nom de compas <lb/>de proportion, parce que cet inſtrument eſt réellement <lb/>fondé ſur la nature & </s> <s xml:id="echoid-s167" xml:space="preserve">les propriétés des proportions. </s> <s xml:id="echoid-s168" xml:space="preserve">Il <lb/>peut être d’un grand uſage pour abréger les opérations <lb/>dans un grand nombre de cas, comme pour trouver <lb/>des lignes proportionnelles à des lignes données, pour <lb/>couper des lignes données en parties égales, pour con-<lb/>noître les degrés d’un arc dont on a la corde, ou bien <lb/>pour diviſer un angle propoſé en pluſieurs parties égales, <lb/>enfin pour trouver des ſurfaces ou des ſolides qui aient <lb/>des raiſons données avec d’autres ſurfaces ou d’autres <lb/>ſolides propoſés; </s> <s xml:id="echoid-s169" xml:space="preserve">ce qui peut avoir une application, <pb o="xij" file="0018" n="18" rhead="PRÉFACE."/> lorſqu’il faut déterminer le calibre des boulets par leurs <lb/>peſanteurs & </s> <s xml:id="echoid-s170" xml:space="preserve">réciproquement. </s> <s xml:id="echoid-s171" xml:space="preserve">Je donne enſuite un pro-<lb/>blême fort curieux ſur la maniere de faire l’analyſe de <lb/>la fonte de chaque eſpece de métal, dont le canon eſt <lb/>compoſé: </s> <s xml:id="echoid-s172" xml:space="preserve">j’ai fait voir par-là comment on pouvoit ap-<lb/>pliquer à l’Artillerie des queſtions qui lui paroiſſent <lb/>étrangeres, comme le problême d’Hieron, qui ne differe <lb/>que de nom de celui-ci. </s> <s xml:id="echoid-s173" xml:space="preserve">Enfin je termine ce Livre par <lb/>une diſſertation, où je recherche la longueur que doivent <lb/>avoir les boulets relativement à leur calibre, pour que <lb/>la force du boulet ſoit la plus grande qu’il eſt poſſible; <lb/></s> <s xml:id="echoid-s174" xml:space="preserve">& </s> <s xml:id="echoid-s175" xml:space="preserve">je rapporte un précis des expériences que j’ai faites <lb/>depuis par ordre du Roi, pour reconnoître ſi cette <lb/>théorie étoit bien fondée; </s> <s xml:id="echoid-s176" xml:space="preserve">j’ai auſſi ajouté une for-<lb/>mule fort curieuſe à ce que j’avois dit dans l’ancienne <lb/>édition ſur la maniere de nombrer les boulets en pile <lb/>dans les Arcenaux: </s> <s xml:id="echoid-s177" xml:space="preserve">ſur quoi l’on pourra remarquer une <lb/>propriété des nombres triangulaires qui m’a paru mé-<lb/>riter attention pour la ſommation des nombres quarrés.</s> <s xml:id="echoid-s178" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s179" xml:space="preserve">Le quatorzieme Livre eſt entiérement deſtiné à ex-<lb/>pliquer les regles du jet des bombes. </s> <s xml:id="echoid-s180" xml:space="preserve">Comme cette <lb/>théorie a un rapport direct avec le mouvement des <lb/>corps, j’explique d’abord les plus belles découvertes de <lb/>Galilée ſur les corps qui tombent, en vertu de la peſan-<lb/>teur, après avoir expliqué les regles principales du choc <lb/>des corps durs, parce que cette partie a auſſi un rap-<lb/>port direct au jet des bombes, où il faut eſtimer la <lb/>force que la bombe acquiert par la vîteſſe que ſa chûte <lb/>lui communique, afin de connoître les effets qu’elle <lb/>peut produire pour proportionner les ouvrages qui doi-<lb/>vent être à l’épreuve de la bombe à la force du choc. <lb/></s> <s xml:id="echoid-s181" xml:space="preserve">Je donne auſſi des ſolutions géométriques & </s> <s xml:id="echoid-s182" xml:space="preserve">algébri- <pb o="xiij" file="0019" n="19" rhead="PRÉFACE."/> ques des différens problêmes qui ont rapport au jet <lb/>des bombes, pour faire voir l’accord de l’analyſe avec <lb/>la conſtruction géométrique, & </s> <s xml:id="echoid-s183" xml:space="preserve">pour initier les Com-<lb/>mençans à l’application de l’Algebre à la Géométrie.</s> <s xml:id="echoid-s184" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s185" xml:space="preserve">Dans le quinzieme Livre, j’explique les principales <lb/>propriétés des machines, en faiſant uſage du principe <lb/>de M. </s> <s xml:id="echoid-s186" xml:space="preserve">Varignon, & </s> <s xml:id="echoid-s187" xml:space="preserve">quelquefois auſſi de celui de M. </s> <s xml:id="echoid-s188" xml:space="preserve">Deſ-<lb/>cartes, quoique le premier ſoit plus géométrique. </s> <s xml:id="echoid-s189" xml:space="preserve">Après <lb/>avoir examiné les machines ſimples, qui font l’objet <lb/>de la méchanique en général, après avoir donné la <lb/>maniere d’en calculer les forces, on fait voir les diffé-<lb/>rens uſages auxquels elles ſont propres, ſoit pour les <lb/>manœuvres de l’Artillerie, ou pour la pratique des Arts. <lb/></s> <s xml:id="echoid-s190" xml:space="preserve">Ces mêmes principes généraux ſont enſuite appliqués <lb/>à la conſtruction des magaſins à poudre, ou de tout <lb/>autre édifice, où l’on examine la différence des pouſ-<lb/>ſées des voûtes en plein ceintre, avec celle des voûtes <lb/>ſurbaiſſées, ou des voûtes en tiers point. </s> <s xml:id="echoid-s191" xml:space="preserve">On détermine <lb/>enſuite quel eſt le choc des bombes & </s> <s xml:id="echoid-s192" xml:space="preserve">des boulets de <lb/>canon qui viennent rencontrer des ſurfaces horizontales <lb/>ou inclinées, & </s> <s xml:id="echoid-s193" xml:space="preserve">quelle élévation il faut donner à un <lb/>mortier, pour qu’une bombe venant à tomber ſur un <lb/>magaſin à poudre, choque la voûte avec toute ſa peſan-<lb/>teur abſolue.</s> <s xml:id="echoid-s194" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s195" xml:space="preserve">Enfin le ſeizieme & </s> <s xml:id="echoid-s196" xml:space="preserve">dernier Livre eſt une ſuite du <lb/>précédent. </s> <s xml:id="echoid-s197" xml:space="preserve">On y examine l’équilibre des fluides en-<lb/>tr’eux, ou avec les ſolides qui y ſont plongés. </s> <s xml:id="echoid-s198" xml:space="preserve">Les <lb/>vîteſſes des eaux qui s’écoulent par différentes ouver-<lb/>tures; </s> <s xml:id="echoid-s199" xml:space="preserve">les chocs des mêmes fluides contre des ſurfaces <lb/>en repos ou en mouvement, ſelon les vîteſſes, les den-<lb/>ſités, & </s> <s xml:id="echoid-s200" xml:space="preserve">la ſituation des corps expoſés au courant. </s> <s xml:id="echoid-s201" xml:space="preserve">J’y <lb/>ai ajouté une théorie abrégée du choc d’un fluide contre <pb o="xiv" file="0020" n="20" rhead="PRÉFACE."/> une ſurface quelconque, & </s> <s xml:id="echoid-s202" xml:space="preserve">diſpoſée comme on voudra, <lb/>en ſuppoſant que les tranches horizontales de ce fluide <lb/>ont des vîteſſes qui ſuivent la raiſon des racines quarrées <lb/>des hauteurs. </s> <s xml:id="echoid-s203" xml:space="preserve">Enfin je termine ce Livre par un diſcours <lb/>ſur la nature & </s> <s xml:id="echoid-s204" xml:space="preserve">les propriétés de l’air, où l’on fait voir <lb/>comment la peſanteur de ce fluide produit tous les ef-<lb/>fets qu’on attribuoit autrefois à l’horreur du vuide. </s> <s xml:id="echoid-s205" xml:space="preserve">On <lb/>peut après cela voir dans notre Architecture Hydrau-<lb/>lique ce qui a rapport au reſſort de l’air, & </s> <s xml:id="echoid-s206" xml:space="preserve">à la force <lb/>prodigieuſe de ſa dilatation, confirmé par pluſieurs <lb/>expériences qui ſe trouvent détaillées dans le même <lb/>Ouvrage.</s> <s xml:id="echoid-s207" xml:space="preserve"/> </p> <figure> <image file="0020-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0020-01"/> </figure> <pb o="xv" file="0021" n="21"/> </div> <div xml:id="echoid-div6" type="section" level="1" n="6"> <head xml:id="echoid-head7" xml:space="preserve">TABLE <lb/>DES MATIERES <lb/>Contenues dans cet Ouvrage. <lb/>LIVRE PREMIER.</head> <note style="it" position="right" xml:space="preserve"> <lb/>Introduction à la Géométrie. # Page 1 <lb/>Définitions des termes dont on fait uſage. # ibid. <lb/>Réduction des quantités algébriques à leurs moindres termes. # 11 <lb/>Additions des quantités algébriques complexes & incomplexes. # 12 <lb/>Souſtraction des quantités algébriques incomplexes & complexes. # 13 <lb/>Multiplication des quantités incomplexes. # 14 <lb/>Multiplication des quantités complexes. # 15 <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Le quarré d’une grandeur quelconque, exprimée par deux <lb/># lettres poſitiyes, eſt égal au quarré de chacune de ces lettres, plus à deux <lb/># rectangles compris ſous les mêmes lettres. # 19 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Le cube d’une grandeur quelconque, exprimée par deux <lb/># lettres, eſt égal au cube de la premiere, plus au cube de la ſeconde, plus à <lb/># trois parallélepipedes du quarré de la premiere par la ſeconde, plus enfin <lb/># à trois autres parallélepipedes du quarré de la ſeconde par la premiere. # 20 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si on a une ligne droite diviſée en deux également dans <lb/># un point, & en deux parties inégales dans un autre point, le rectangle des <lb/># parties inégales, plus le quarré de la partie moyenne eſt égal au quarré de <lb/># la moitié de la ligne. # ibid. <lb/><emph style="sc">Proposit</emph>. IV. <emph style="sc">Theor</emph>. Si l’on a une ligne droite, diviſée en deux égale-<lb/># ment, & qu’on lui ajoute une autre ligne quelconque; le rectangle de la <lb/># ſomme de ces deux lignes par la ligne ajoutée, avec le quarré de la demi-<lb/># propoſée, eſt égal au quarré de la ligne égale à la moitié de la propoſée, <lb/># plus la ligne ajoutée. # 21 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Si l’on a deux lignes, dont l’une ſoit double de l’autre, le <lb/># quarré de la premiere ſera quadruple du quarré de la ſeconde. # 22 <lb/>De la diviſion des quantités algébriques incomplexes & complexes. # ibid. <lb/>Définitions des parties aliquotes. # 28 <lb/>Multiplication des quantités complexes, par le moyen des parties aliquotes. # ibid. & ſuiv. <lb/>Traité des fractions numériques & algébriques. # 37 <lb/>Définitions des fractions, & des parties dont elles ſont compoſées. # ibid. <lb/><emph style="sc">Probl</emph>. I. Evaluer une fraction. # 39 <lb/><emph style="sc">Probl</emph>. II. Trouver le plus grand commun diviſeur de deux nombres. # 40 <lb/><emph style="sc">Probl</emph>. III. Réduire pluſieurs fractions données au même dénominateur. # 42 <lb/>De l’addition, ſouſtraction, multiplication, & diviſion des fractions. # 43 & ſuiv. <pb o="xvj" file="0022" n="22" rhead="TABLE"/> Des fractions décimales, & des quatre opérations de l’Arithmétique ſur ces <lb/># ſortes de fractions. # Pages 54 & ſuiv. <lb/>Uſages des fractions décimales. # 63 & ſuiv. <lb/>Du calcul des expoſans, de la formation des puiſſances, & de l’extraction <lb/># des racines. # 68 & ſuiv. <lb/>De la formation des puiſſances des quantités exponentielles, & de l’extrac-<lb/># tion de leurs racines. # 70 <lb/>De la formation des puiſſances des polynomes, & de l’extraction de leurs <lb/># racines. # 74 <lb/>De l’extraction de la racine quarrée des quantités algébriques complexes. # 75 <lb/>De la formation du quarré du nombre quelconque, & de l’extraction de ſes <lb/># racines. # 78 <lb/>De la formation du cube d’une quantité complexe, & de l’extraction de la <lb/># racine cube des quantités algébriques & numériques. # 91 <lb/>De la formation algébrique du cube d’un nombre quelconque, & de l’extrac-<lb/># tion de ſa racine cube. # 95 <lb/>Reglé génér ale de l’extraction des racines cubiques des quantités numériques. # 97 <lb/>Maniere d’approcher le plus près qu’il eſt poſſible de la racine cube d’un <lb/># nombre donnè par le moyen des décimales. # 100 <lb/></note> </div> <div xml:id="echoid-div7" type="section" level="1" n="7"> <head xml:id="echoid-head8" xml:space="preserve">LIVRE II,</head> <p> <s xml:id="echoid-s208" xml:space="preserve">Qui traite des rapports, proportions, progreſſions arithmétiques & </s> <s xml:id="echoid-s209" xml:space="preserve">géo-<lb/>métriques, des logarithmes, de la réſolution analytique des problêmes <lb/>du premier & </s> <s xml:id="echoid-s210" xml:space="preserve">du ſecond degré.</s> <s xml:id="echoid-s211" xml:space="preserve"/> </p> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Si quatre grandeurs ſont en proportion géométrique, le pro-<lb/># duit des extrêmes eſt égal à celui des moyens. # 110 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si quatre grandeurs ſont tellement diſpoſées que le pro-<lb/># duit des extrémes ſoit égal au produit des moyens, ces quatre grandeurs <lb/># ſeront en proportion géométrique. # 113 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si deux raiſons ont un même rapport à une troiſieme, <lb/># elles ſont égales entr’elles. # 115 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Lorſque pluſieurs grandeurs ſont en proportion géométri-<lb/># que, la ſomme des antécédens eſt à celle des conſéquens, comme un ſeul <lb/># antécédent à ſon conſéquent. # 116 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Deux grandeurs demeurent toujours dans le même rapport, <lb/># quoique l’on leur ajoute, pourvu que les ajoutées ſoient proportionnelles. # ibid. <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Deux grandeurs gardent toujours le même rapport, quoi-<lb/># que l’on en retranche, pourvu que les parties retranchées ſoient proportion-<lb/># nelles. # 117 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Si on multiplie les deux termes d’une raiſon par une <lb/># même quantité, les produits ſont dans la même raiſon des quantités non <lb/># multipliées. # ibid. <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Si on diviſe les deux termes d’un rapport par une même <lb/># grandeur, il reſte toujours le même. # ibid. <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Theor</emph>. Si l’on multiplie deux proportions termes par termes, <pb o="xvij" file="0023" n="23" rhead="DES MATIERES."/> # les produits ſeront encore en proportion. # 118 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Theor</emph>. Dans une proportion continue, le quarré du premier terme <lb/># eſt à celui du ſecond, comme le premier au troiſieme. # ibid. <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Theor</emph>. Lorſque quatre grandeurs ſont en proportion arithméti-<lb/># que, la ſomme des extrêmes eſt égale à celle des moyens. # 119 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Theor</emph>. Lorſque quatre grandeurs ſont tellement diſpoſées que <lb/># la ſomme des extrêmes eſt égale à celle des moyens, elles ſont en propor-<lb/># tion arithmétique. # 121 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Theor</emph>. Dans une progreſſion arithmétique, la ſomme de deux <lb/># termes également éloignés des extrêmes eſt égale à celle des mêmes ex-<lb/># trêmes. # 122 <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Theor</emph>. Toute progreſſion géométrique croiſſante ou décroiſſante <lb/># peut être repréſentée par la ſuite, a : aq : aq<emph style="sub">2</emph>, &c. ou aq<emph style="sub">3</emph>: aq<emph style="sub">2</emph>: aq: a, &c. # 125 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Theor</emph>. Dans une progreſſion géométrique quelconque, la ſomme <lb/># des antécédens eſt à celle des conſéquens, comme un antécédent à ſon con-<lb/># ſéquent. # 127 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Theor</emph>. Dans une progreſſion géométrique, le produit de deux <lb/># termes également éloignés des extrêmes eſt égal à celui des extrêmes. # 128 <lb/><emph style="sc">Probl</emph>. Inſérer pluſieurs moyens proportionnels entre deux nombres donnés. # 129 <lb/>Des logarithmes, de leur nature, & de leurs uſages. # 130 <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Theor</emph>. Dans la ſuite des puiſſances d’une quantité quelconque, <lb/># dont les termes forment une progreſſion géométrique, les expoſans ſont en <lb/># progreſſion arithmétique. # 131 <lb/><emph style="sc">Prop</emph>. XVIII. <emph style="sc">Theor</emph>. L’expoſant des termes d’une raiſon doublée ou triplée <lb/># eſt égal au quarré ou au cube de celui des raiſons ſimples dont elle eſt dou-<lb/># blée ou triplée. # 140 <lb/>Regles générales pour la réſolution des problêmes, ou application du calcul <lb/># algébrique à la maniere de dégager les inconnues. # 141 <lb/>Uſages de l’Addition & de la Souſtraction, Multiplication & Diviſion, & <lb/># extraction des racines pour dégager les inconnues. # 142 <lb/>Maniere de ſubſtituer dans une équation la valeur des inconnues. # 146 <lb/>Maniere de réduire toutes les inconnues à une ſeule, lorſqu’on a autant d’é-<lb/># quations que d’inconnues. # 148 <lb/>Application des Regles précédentes à pluſieurs problêmes curieux & utiles. <lb/># 149 & ſuiv. <lb/>De la réſolution des équations du ſecond degré. # 158 <lb/>Remarque générale & importante ſur la nature des équations du ſecond degré. <lb/># 161 <lb/></note> </div> <div xml:id="echoid-div8" type="section" level="1" n="8"> <head xml:id="echoid-head9" xml:space="preserve">LIVRE III,</head> <head xml:id="echoid-head10" xml:space="preserve">Où l’on conſidere les différentes poſitions des lignes droites les unes à <lb/>l’égard des autres.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Probl</emph>. D’un point donné hors d’une ligne, mener une perpendicu-<lb/># laire à cette ligne. # 180 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Probl</emph>. D’un point donné ſur une ligne, élever une perpendiculaire <lb/># à cette ligne. # ibid. <pb o="xviij" file="0024" n="24" rhead="TABLE"/> <emph style="sc">Prop</emph>. III. <emph style="sc">Probl</emph>. Diviſer une ligne donnée en parties égales. # 181 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. D’un même point ſur une ligne donnée, on ne peut élever <lb/># qu’une perpendiculaire. # ibid. <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. D’un point donné hors d’une ligne, on ne peut abaiſſer à <lb/># cette ligne qu’une perpendiculaire. # 182 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Une perpendiculaire eſt la plus courte de toutes les lignes <lb/># que l’on peut mener d’un point à une ligne. # ibid. <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Lorſque deux lignes ſe coupent, elles forment des angles <lb/># oppoſés au ſommet qui ſont égaux. # 183 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Si deux lignes paralleles en rencontrent une troiſieme, <lb/># elles font des angles égaux du même côté. # 184 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Theor</emph>. Si deux lignes paralleles ſont coupées par une troiſieme, <lb/># les angles alternes internes ſont égaux, les angles internes ou externes <lb/># d’un même côté, pris enſemble, valent deux droits. # 185 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Theor</emph>. Suppoſant qu’une ligne coupe deux autres lignes, ces der-<lb/># nieres ſeront paralleles, 1°. ſi les angles alternes internes, ou alternes ex-<lb/># ternes ſont égaux, 2°. ſi les angles internes ou externes d’un même côté, <lb/># pris enſemble, valent deux droits. # ibid. <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Une ligne quelconque, & un point étant donné ſur le même <lb/># plan, mener par ce point une parallele à la propoſée. # 186 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Trouver le rayon d’un cercle qui paſſe par trois points <lb/># donnés. # 187 <lb/></note> </div> <div xml:id="echoid-div9" type="section" level="1" n="9"> <head xml:id="echoid-head11" xml:space="preserve">LIVRE IV,</head> <head xml:id="echoid-head12" xml:space="preserve">Qui traite des propriétés des triangles & des Parallélogrammes.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. L’angle extérieur d’un triangle eſt égal aux deux intérieurs <lb/># oppoſés, & les trois enſemble valent deux droits. # 189 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Deux triangles ſont parfaitement égaux, lorſque les trois <lb/># côtés de l’un ſont égaux aux trois côtés de l’autre. # 191 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Deux triangles ſont égaux en tout, lorſqu’ils ont un angle <lb/># égal compris entre deux côtés égaux chacun à chacun. # 192 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Deux triangles ſont parfaitement égaux, lorſqu’ils ont <lb/># deux angles égaux ſur un côté égal. # 193 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Deux parallélogrammes ſont égaux, lorſqu’ayant même <lb/># baſe ils ſont compris entre paralleles. # ibid. <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Deux triangles ſont égaux, lorſqu’ayant même baſe ils <lb/># ſont compris entre paralleles. # 194 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Les complémens des parallélogrammes ſont égaux. # 195 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Les parallélogrammes qui ont même baſe ſont comme leurs <lb/># hauteurs. # ibid. <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Theor</emph>. Si l’on coupe les deux côtés d’un triangle par une ligne <lb/># parallele à la baſe, ils ſeront coupés en parties proportionnelles. # 197 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Theor</emph>. Deux triangles ſont ſemblables, lorſqu’ils ont tous leurs <lb/># côtés proportionnels. # 199 <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Theor</emph>. Deux triangles ſont ſemblables, lorſqu’ils ont un angle <lb/># égal compris entre côtés proportionnels. # 200 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Theor</emph>. Deux triangles ſont ſemblables, lorſqu’ils ont deux angles <pb o="xix" file="0025" n="25" rhead="DES MATIERES."/> # égaux chacun à chacun. # ibid. <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Theor</emph>. Si de l’angle droit d’un triangle rectangle on abaiſſe une <lb/># perpendiculaire ſur l’hypoténuſe, elle diviſera ce triangle en deux autres <lb/># ſemblables entr’eux & au propoſé. # 202 <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Theor</emph>. Le quarré de l’hypoténuſe eſt égal au quarré des deux <lb/># autres côtés. # ibid. <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Theor</emph>. Dans tout triangle obtuſangle, le quarré du côté oppoſé à <lb/># l’angle obtus eſt égal au quarré des deux autres côtés, plus à deux rectangles <lb/># compris ſous un des côtés, & la partie de ce même côté, compriſe entre ſon <lb/># prolongement, & la rencontre d’une perpendiculaire abaiſſée de l’angle oppoſé <lb/># à ce côté ſur ce même côté. # 205 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Theor</emph>. Dans tout triangle, le quarré d’un côté oppoſé à un angle <lb/># aigu, eſt égal à la ſomme des quarrés des deux autres côtés, moins deux <lb/># rectangles compris ſous le plus grand côté, & la partie de ce grand côté, <lb/># compriſe entre l’angle, auquel le premier eſt oppoſé, & la rencontre de ce <lb/># grand côté par la perpendiculaire abaiſſée du plus grand angle ſur ce côté. # 207 <lb/></note> </div> <div xml:id="echoid-div10" type="section" level="1" n="10"> <head xml:id="echoid-head13" xml:space="preserve">LIVRE V,</head> <head xml:id="echoid-head14" xml:space="preserve">Où l’on traite des propriétés du cercle.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Une perpendiculaire abaiſſée du centre d’un cercle ſur une <lb/># corde, diviſe cette corde & ſon arc en deux parties egales. # 210 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si une droite paſſe par le centre, & diviſe une corde en deux <lb/># parties égales, elle lui ſera perpendiculaire. # 211 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si une droite eſt perpendiculaire ſur le milieu d’une corde, <lb/># elle paſſe néceſſairement par le centre. # ibid. <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Une droite menée du centre au point de contingence eſt per-<lb/># pendiculaire à la tangente. # 212 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Un angle à la circonférence a pour meſure la moitié de l’arc <lb/># compris entre ſes côtés. # 213 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Un angle formé par une tangente & par une corde, a pour <lb/># meſure la moitié de l’arc compris entre ſes côtés. # 214 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Un angle qui a ſon ſommet au dedans du cercle entre le <lb/># centre & la circonférence, a pour meſure la moitié de l’arc ſur lequel il eſt <lb/># appuyé, plus la moitié de l’arc compris entre ſes côtés prolongés. # ibid. <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Un angle, dont le ſommet eſt hors de la circonférence, a <lb/># pour meſure la moitié de l’arc concave, moins la moitié de l’arc convexe, <lb/># compris entre ſes côtés. # 215 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Theor</emph>. Si deux droites ſe coupent au dedans d’un cercle, les rec-<lb/># tangles des ſegmens ſont égaux. # 216 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Theor</emph>. Si d’un point, hors d’un cercle, on mene deux ſécantes <lb/># terminées à la partie concave de la circonférence, le produit des ſécantes <lb/># par leurs parties extérieures ſont égaux. # ibid. <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Theor</emph>. Le quarré d’une ordonnée eſt égal au produit de ſes abſ-<lb/># ciſſes. # 217 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. D’un point donnè, mener une tangente à un cercle ſur le <lb/># même plan. # 218 <lb/> <pb o="xx" file="0026" n="26" rhead="TABLE"/> <emph style="sc">Prop</emph>. XIII. <emph style="sc">Theor</emph>. Le quarré d’une tangente eſt égal au rectangle d’une ſé-<lb/># cante entiere par ſa partie extérieure. # 219 <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Theor</emph>. Si l’on a une tangente perpendiculaire à l’extrêmité <lb/># d’un diametre, & que de l’autre extrêmité du même diametre on mene tant <lb/># de lignes que l’on voudra, le quarré du diametre eſt toujours égal au quarre <lb/># de chaque ligne par la partie intérieure. # 220 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Diviſer une ligne donnée en moyenne & extrême raiſon. <lb/># ibid. <lb/></note> </div> <div xml:id="echoid-div11" type="section" level="1" n="11"> <head xml:id="echoid-head15" xml:space="preserve">LIVRE VI, <lb/>Qui traite des Polygones réguliers, inſcrits & circonſcrits au cercle.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Probl</emph>. Inſcrire un héxagone dans un cercle. # 223 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Probl</emph>. Décrire un dodécagone dans un cercle. # 224 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Probl</emph>. Inſcrire un décagone dans un cercle. # 225 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Une ligne égale à la ſomme des côtés d’un héxagone & d’un <lb/># décagone inſcrits au même cercle, eſt diviſée en moyenne & extrême raiſon <lb/># au point de jonction. # 226 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Le quarré du côté d’un pentagone régulier inſcrit au cercle, <lb/># eſt égal à la ſomme des quarrés des côtés de l’exagone & du décagone <lb/># inſcrits au même cercle. # ibid. <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Probl</emph>. Inſcrire un pentagone dans un cercle. # 227 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Probl</emph>. Inſcrire un quarré dans un cercle. # 228 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Probl</emph>. Inſcrire un octogone dans un cercle. # ibid. <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Diviſer un angle quelconque en trois parties égales par le <lb/># moyen de la quadratrice. # 231 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Décrire un ennéagone régulier dans un cercle. # 232 <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Décrire un eptagone régulier dans un cercle. # ibid. <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Décrire un décagone dans un cercle. # ibid. <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Probl</emph>. Circonſcrire un polygone quelconque autour d’un cercle. # 233 <lb/></note> </div> <div xml:id="echoid-div12" type="section" level="1" n="12"> <head xml:id="echoid-head16" xml:space="preserve">LIVRE VII, <lb/>Où l’on conſidere les rapports qu’ont entr’eux les circuits des figures ſem-<lb/>blables, & les proportions de leurs ſuperficies.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Les circuits des polygones ſemblables ſont comme les rayons <lb/># des cercles auxquels ils ſont inſcrits. # 234 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. La ſurface d’un poligone régulier quelconque eſt égale à <lb/># celle d’un triangle qui auroit une baſe égale au contour du poligone, & pour <lb/># hauteur une ligne égale à la perpendiculaire abaiſſée du centre de ce poligone <lb/># ſur un de ſes côtés. # 235 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. La ſurface d’un cercle eſt égale à celle d’un triangle qui <lb/># auroit pour baſe la circonférence du cercle, & pour hauteur le rayon du <lb/># même cercle. # 236 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Les ſurfaces des deux polygones ſemblables ſont entr’elles <lb/># comme les quarré des rayons ou lignes homologues. # 240 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Les ſurfaces des cercles ſont les quarrés de leurs rayons. # 241 <pb o="xxj" file="0027" n="27" rhead="DES MATIERES"/> <emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Deux triangles ſemblables ſont entr’eux comme les quarrés <lb/># des côtés homologues. # 242 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Les parallélogrammes ſont comme les produits des baſes <lb/># par leurs hauteurs. # 243 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Si trois lignes ſont en proportion continue, le quarré de <lb/># la premiere eſt au quarré de la ſeconde, comme la premiere à la troi-<lb/># ſieme. # 244 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Theor</emph>. Le rectangle de deux lignes quelconques eſt moyen pro-<lb/># portionnel entre les quarrés des mêmes lignes. # ibid. <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Trouver une moyenne proportionnelle entre deux lignes <lb/># données. # 245 <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Trouver une troiſieme proportionnelle à deux lignes don-<lb/># nées. # 246 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Trouver une quatrieme proportionnelle à trois lignes don-<lb/># nées. # 248 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Probl</emph>. Faire un quarré égal à un rectangle. # ibid. <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Probl</emph>. Trouver un quarré qui ſoit à un autre dans une raiſon <lb/># donnée. # 249 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Trouver le rapport des figures ſemblables. # 250 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Probl</emph>. Sur une ligne donnée, faire un rectangle égal à un <lb/># autre. # ibid. <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Theor</emph>. Deux triangles qui ont un angle égal, ſont entr’eux <lb/># comme les produits des côtés qui contiennent l’angle égal. # 252 <lb/><emph style="sc">Prop</emph>. XVIII. <emph style="sc">Theor</emph>. La ſurface d’un triangle eſt égale à la racine quarrée <lb/># d’un produit de quatre dimenſions, fait de la demi-ſomme des trois côtés, <lb/># multipliée par la différence de chacun de ces côtés à la même demi-ſomme. <lb/># 253 <lb/></note> </div> <div xml:id="echoid-div13" type="section" level="1" n="13"> <head xml:id="echoid-head17" xml:space="preserve">LIVRE VIII, <lb/>Qui traite des propriétés des corps, de leurs ſurfaces, & de leurs ſolidités.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. La ſurface d’un priſme droit, ſans y comprendre les baſes, <lb/># eſt égale à celle d’un rectangle qui auroit même hauteur, & pour baſe une <lb/># ligne égale au contour du polygone. # 261 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. La ſurface d’une pyramide droite eſt égale à celle d’un <lb/># triangle qui auroit pour baſe une ligne égale à la ſomme des côtés, & pour <lb/># hauteur la moitié de la perpendiculaire abaiſſée du ſommet de la pyramide <lb/># ſur l’un des côtés de la baſe. # 262 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Les parallélepipedes & les priſmes droits ſont comme les <lb/># produits de leurs trois dimenſions. # 263 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Toute pyramide eſt le tiers d’un priſme de même baſe & <lb/># même hauteur. # 264 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Deux pyramides de même hauteur ſont entr’elles comme <lb/># leurs baſes. # 267 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Deux priſmes ſont égaux, lorſqu’ils ont des baſes réci-<lb/># proques à leurs hauteurs. # ibid. <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Une pyramide tronquée quelconque eſt égale à une autre <lb/># pyramide de même hauteur, qui auroit une baſe égale à la ſomme des baſes, <pb o="xxij" file="0028" n="28" rhead="TABLE"/> # inférieure & ſupérieure, plus une baſe moyenne géométrique entre ces deux <lb/># baſes. # 268 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Une demi-ſphere eſt les deux tiers du cylindre circonſcrit <lb/># de même baſe & de même hauteur. # 271 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Theor</emph>. Les ſolidités des ſpheres ſont comme les cubes de leurs dia-<lb/># metres. # 273 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Theor</emph>. La ſurface de ſa demi-ſphere eſt égale à la ſurface convexe <lb/># du cylindre auquel elle eſt inſcrite. # 275 <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Theor</emph>. La ſolidité d’une zone eſt égale aux deux tiers du cylindre <lb/># du grand cercle, plus au tiers du cylindre du plus petit cercle. # 277 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Theor</emph>. Si l’on coupe une demi-ſphere inſcrite dans un cylindre <lb/># par un plan parallele à la baſe, la ſurface de la zone eſt égale à celle du cy-<lb/># lindre correſpondant. # 279 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Theor</emph>. Si trois lignes ſont en proportion continue, le ſolide fait <lb/># ſur ces trois lignes, eſt égal au cube de la moyenne. # 280 <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Theor</emph>. Lorſque quatre lignes ſont en progreſſion geométrique, <lb/># le cube fait ſur la premiere, eſt au cube ſur la ſeconde, comme la premiere <lb/># à la quatrieme. # ibid. <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Entre deux lignes données, trouver deux moyennes pro-<lb/># portionnelles. # 281 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Probl</emph>. Entre deux nombres donnés, trouver deux moyens pro-<lb/># portionnels. # 282 <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Probl</emph>. Faire un cube qui ſoit à un autre dans une raiſon don-<lb/># née. # 283 <lb/><emph style="sc">Prop</emph>. XVIII. <emph style="sc">Probl</emph>. Faire un cube égal à un parallélepipede propoſé. # 284 <lb/></note> </div> <div xml:id="echoid-div14" type="section" level="1" n="14"> <head xml:id="echoid-head18" xml:space="preserve">LIVRE IX, <lb/>Qui traite des Sections coniques. <lb/>CHAPITRE PREMIER. <lb/>Des propriétés de la Parabole.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Dans la parabole, le quarré d’une ordonnée quelconque eſt <lb/># égal au produit de ſon abſciſſe par le parametre. # 288 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Les quarrés des ordonnées à l’axe ſont comme leurs abſ-<lb/># ciſſes. # 289 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Probl</emph>. Par un point donné, mener une tangente à la parabole. # 290 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. La ſounormale eſt toujours égale à la moitié du para-<lb/># metre. # 292 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. La ſoutangente eſt double de l’abſciſſe. # ibid. <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Une parallele à une tangente eſt coupée en deux également <lb/># par le diametre qui paſſe par le point touchant. # 293 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Le quarré d’une ordonnée à un diametre eſt égal au pro-<lb/># duit de ſon abſciſſe par le parametre de ce diametre. # 295 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. Si l’on coupe un cône par un plan parallele à un de ſes <lb/># côtés, la ſection ſera une parabole. # 297 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Décrire une parabole, le parametre étant donné. # 298 <pb o="xxiij" file="0029" n="29" rhead="DES MATIERES."/> <emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Trouver l’axe d’une parabole donnée. # ibid. <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Trouver le parametre d’un diametre quelconque. # 299 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Trouver le foyer d’une parabole. # ibid. <lb/></note> </div> <div xml:id="echoid-div15" type="section" level="1" n="15"> <head xml:id="echoid-head19" xml:space="preserve">CHAPITRE II, <lb/>Qui traite de l’Ellipſe.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Dans l’ellipſe, le quarré d’une ordonnée à l’axe eſt au rec-<lb/># tangle de ſes abſciſſes, comme le quarré du petit axe au quarré du grand <lb/># axe. # 301 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si des extrêmités de deux diametres conjugués on mene à <lb/># un même axe deux ordonnées, le quarré d’une des abſciſſes correſpondantes, <lb/># à partir du centre, eſt égal au rectangle des parties du même axe, faites <lb/># par l’autre ordonnée. # 304 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Le quarré d’une ordonnée à un diametre quelconque eſt <lb/># au produit de ſes abſciſſes, comme le quarré du diametre parallele aux <lb/># ordonnées, eſt à celui du diametre des abſciſſes. # 305 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. La ſomme des quarrés de deux diametres conjugués eſt <lb/># égale à celle des quarrés des deux axes. # 308 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Si par l’extrêmité de l’axe on mene une tangente qui aille <lb/># rencontrer deux diametres conjugués, prolongés autant qu’il ſera néceſ-<lb/># ſaire, le rectangle des parties de cette tangente eſt égal au quarré de la <lb/># moitié de l’axe qui lui eſt parallele. # 310 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Si l’on coupe un cône par un plan oblique à la baſe, de <lb/># maniere que les deux côtés du cône ſoient coupés entre le ſommet & la baſe, <lb/># la ſection eſt une ellipſe. # 311 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Si l’on coupe un cylindre par un plan oblique à la baſe, <lb/># la ſection ſera une ellipſe. # 312 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Theor</emph>. La ſomme des diſtances d’un point de l’ellipſe aux foyers <lb/># eſt égale au grand axe de cette courbe. # ibid. <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Les deux axes d’une ellipſe étant donnés, la décrire par <lb/># un mouvement continu. # 314 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Trouver le centre & les axes d’une ellipſe donnée. # 315 <lb/></note> </div> <div xml:id="echoid-div16" type="section" level="1" n="16"> <head xml:id="echoid-head20" xml:space="preserve">CHAPITRE III, <lb/>Qui traite de l’Hyperbole.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Dans l’hyperbole, le quarré d’une ordonnée à l’axe eſt au <lb/># rectangle de ſes abſciſſes, comme le quarré de l’axe parallele aux ordonnées <lb/># eſt au quarré de l’axe ſur lequel on prend les abſciſſes. # 316 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si une droite parallele au ſecond axe coupe l’hyperbole en <lb/># deux points, le quarré du ſecond axe eſt égal au rectangle des parties de <lb/># cette ligne, terminée aux aſymptotes. # 318 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si l’on a deux lignes paralleles & terminées aux aſymp-<lb/># totes, les rectangles de leurs parties ſont égaux. # 319 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Si par deux points quelconques d’une hyperbole ou de deux <lb/># hyperboles oppoſées, on mene quatre lignes paralleles entr’elles deux à <lb/># deux terminées aux aſymptotes, les rectangles des parties de ces lignes <pb o="xxiv" file="0030" n="30" rhead="TABLE"/> # ſeront reſpectivement égaux. # ibid. <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Probl</emph>. Par un point donné, mener une tangente à une hyper-<lb/># bole. # 320 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Le quarré d’une ordonnée à un diametre quelconque eſt au <lb/># produit de ſes abſciſſes, comme le quarré du diametre parallele à cette <lb/># ordonnée, eſt au quarré du diametre ſur lequel on prend les abſciſſes. # 321 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Si l’on coupe un cône par un plan parallele à l’axe, la <lb/># courbe ſera une hyperbole. # 322 <lb/></note> </div> <div xml:id="echoid-div17" type="section" level="1" n="17"> <head xml:id="echoid-head21" xml:space="preserve">LIVRE X, <lb/>Qui traite de la Trigonométrie rectiligne & du Nivellement. <lb/>Du calcul des triangles rectangles.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Probl</emph>. Connoiſſant dans un triangle rectangle un côté & un angle, <lb/># trouver le côté oppoſé à l’angle aigu. # 332 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Probl</emph>. Connoiſſant dans un triangle un angle & un côté, trouver <lb/># l’hypoténuſe. # 333 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Probl</emph>. Dans un triangle rectangle, dont on connoît un angle & le <lb/># côté oppoſé, trouver le côté oppoſé à l’autre angle. # ibid. <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Probl</emph>. Connoiſſant les deux côtés qui contiennent l’angle droit <lb/># dans un triangle rectangle, trouver un des angles de la baſe. # 334 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Probl</emph>. Connoiſſant dans un triangle rectangle les deux côtés qui <lb/># contiennent un angle aigu, trouver la valeur de cet angle. # ibid. <lb/></note> </div> <div xml:id="echoid-div18" type="section" level="1" n="18"> <head xml:id="echoid-head22" xml:space="preserve">De la réſolution des triangles obtuſangles ou acutangles.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Dans tous triangles, les ſinus des angles ſont comme les <lb/># côtés oppoſés. # 335 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Dans un triangle obtuſangle, le ſinus de l’angle obtus eſt <lb/># le même que celui de ſon ſupplément. # ibid. <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Probl</emph>. Connoiſſant deux angles & un côté dans un triangle, on <lb/># demande les autres côtés. # 336 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Connoiſſant dans un triangle deux côtés & un angle oppoſé <lb/># à l’un de ces côtés, trouver les deux autres angles. # 337 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Theor</emph>. Dans un triangle quelconque, dont on connoît deux côtés & <lb/># l’angle compris entre ces côtés, la ſomme des deux côtés connus eſt à leur <lb/># différence, comme la tangente de la moitié de la ſomme des deux angles in-<lb/># connus eſt à la tangente de la moitié de leur différence. # ibid. <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Connoiſſant dans un triangle deux côtés & l’angle compris, <lb/># trouver les deux autres angles. # 338 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Theor</emph>. Dans tout triangle, dont on connoît les trois côtés, le <lb/># plus grand côté eſt à la ſomme des deux autres, comme la différence des <lb/># deux mêmes côtés eſt à la différence des ſegmens de la baſe. # 340 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Probl</emph>. Connoiſſant les trois côtés d’un triangle, trouver les <lb/># ſegmens de la baſe. # ibid. <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Probl</emph>. Trouver une diſtance inacceſſible. # 343 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Trouver la diſtance de deux objets inacceſſibles. # 345 <lb/> <pb o="xxv" file="0031" n="31" rhead="DES MATIERES."/> <emph style="sc">Prop</emph>. XVI. <emph style="sc">Probl</emph>. Tirer une ligne parallele à une ligne inacceſſible. # 347 <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Probl</emph>. Meſurer une hauteur acceſſible ou inacceſſible. # 348 <lb/></note> </div> <div xml:id="echoid-div19" type="section" level="1" n="19"> <head xml:id="echoid-head23" xml:space="preserve">Problêmes de Trigonométrie applicables à la fortification.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Probl</emph>. I. Connoiſſant la longueur d’une ligne, dont on ne peut approcher, <lb/># & les angles de deux ſtations, dont la diſtance eſt inconnue, trouver les <lb/># angles & les lignes de cette figure. # 359 <lb/><emph style="sc">Probl</emph>. II. Connoiſſant une ligne & ſes parties avec les angles obſervés d’un ſeul <lb/># point, trouver les diſtances de ce point aux extrêmités de la même ligne. # 360 <lb/></note> </div> <div xml:id="echoid-div20" type="section" level="1" n="20"> <head xml:id="echoid-head24" xml:space="preserve">Théorie & pratique du Nivellement.</head> <note style="it" position="right" xml:space="preserve"> <lb/>CHAP. I. Où l’on donne l’uſage du niveau d’eau. # 364 <lb/>CHAP. II. Où l’on donne la maniere de faire un nivellement compoſé. # 367 <lb/>CHAP. III. Où l’on donne la maniere de niveler entre deux termes, où <lb/># ſe trouvent des hauteurs & des fonds. # 369 <lb/>CHAP. IV. Où l’on donne la maniere de calculer la différence du niveau <lb/># vrai au niveau apparent pour une ligne quelconque. # 372 <lb/>CHAP. V. Où l’on donne la deſcription du niveau de M. Huyghens. # 374 <lb/>CHAP. VI. Où l’on donne la maniere de ſe ſervir du niveau de M. Huy-<lb/># ghens. # 377 <lb/>CHAP. VII. Où l’on donne la maniere de ſe ſervir du niveau de M. Huy-<lb/># ghens dans le nivellement compoſé. # 379 <lb/></note> </div> <div xml:id="echoid-div21" type="section" level="1" n="21"> <head xml:id="echoid-head25" xml:space="preserve">LIVRE XI.</head> <head xml:id="echoid-head26" xml:space="preserve">Du Toiſé en général, où l’on donne la maniere de faire le toiſé des plans, <lb/># des ſolides, & de la charpente.</head> <note style="it" position="right" xml:space="preserve"> <lb/>CHAP. I. Maniere de multiplier deux dimenſions, dont l’une contient des <lb/># toiſes & des parties de toiſe, & la ſeconde des toiſes ſeulement. # 387 <lb/>CHAP. II. Maniere de multiplier deux dimenſions qui contiennent cha-<lb/># cune des toiſes, des pieds, des pouces, &c. # 392 <lb/>CHAP. III. Maniere de multiplier trois dimenſions exprimées en toiſes, <lb/># pieds, pouces, &c. # 397 <lb/>CHAP. IV. Maniere de toiſer les bois de charpente. # 402 <lb/></note> </div> <div xml:id="echoid-div22" type="section" level="1" n="22"> <head xml:id="echoid-head27" xml:space="preserve">LIVRE XII,</head> <head xml:id="echoid-head28" xml:space="preserve">Où l’on applique la Géométrie à la meſure des ſuperficies & des ſolides.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Probl</emph>. Meſurer les figures triangulaires. # 409 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Probl</emph>. Meſurer les quadrilateres quelconques. # 410 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Pro</emph>. Meſurer la ſurface des polygones réguliers & irréguliers. # 411 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Probl</emph>. Meſurer la ſuperficie des cercles & de leurs parties. # 412 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Probl</emph>. Trouver la ſurface d’une ellipſe. # 413 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Probl</emph>. Trouver l’aire d’une parabole. # 414 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Probl</emph>. Meſurer les ſurfaces des priſmes & des cylindres. # 415 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Probl</emph>. Trouver les ſurfaces des pyramides & des cônes. # ibid. <pb o="xxvj" file="0032" n="32" rhead="TABLE"/> <emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Trouver les ſurfaces des ſpheres, de leurs ſegmens, & de <lb/># leurs zones. # 416 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Trouver la ſolidité des cubes, des parallélepipedes, des <lb/># priſmes, & des cylindres. # 417 <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Cuber les pyramides & les cônes. # 418 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Trouver la ſolidité des pyramides & des cônes tron-<lb/># qués. # 419 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Probl</emph>. Trouver la ſolidité des ſecteurs de cylindre & de cône <lb/># tronqué. # 420 <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Probl</emph>. Trouver la ſolidité d’une ſphere. # 422 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Cuber un paraboloïde. # 424 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Probl</emph>. Cuber un ſphéroïde elliptique. # 425 <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Probl</emph>. Cuber un hyperboloïde. # 426 <lb/><emph style="sc">Prop</emph>. XVIII. <emph style="sc">Probl</emph>. Trouver la ſolidité de la maçonnerie de toutes ſortes <lb/># de voûtes. # 427 <lb/><emph style="sc">Prop</emph>. XIX. <emph style="sc">Theor</emph>. La ſurface d’un pande voûte en plein ceintre eſt double <lb/># du triangle correſpondant de la baſe. # 432 <lb/>Application de la Géométrie au toiſé des parties d’une fortification. # 436 <lb/><emph style="sc">Prop</emph>. XX. <emph style="sc">Probl</emph>. Maniere de cuber l’onglet d’un bâtardeau. # 442 <lb/><emph style="sc">Prop</emph>. XXI. <emph style="sc">Probl</emph>. Connoiſſant le centre de gravité d’une ligne droite, trou-<lb/># ver la valeur de la ſurface qu’elle décrira dans ſa révolution autour d’un <lb/># axe. # 446 <lb/><emph style="sc">Prop</emph>. XXII. <emph style="sc">Probl</emph>. Trouver la ſurface d’une demi-ſphere, connoiſſant le <lb/># centre de gravité de la demi-circonférence du cercle générateur. # 447 <lb/><emph style="sc">Prop</emph>. XXIII. <emph style="sc">Probl</emph>. Connoiſſant le centre de gravité d’un rectangle qui <lb/># tourne autour d’un axe, trouver le ſolide qu’il décrit dans ce mouve-<lb/># ment. # 448 <lb/><emph style="sc">Prop</emph>. XXIV. <emph style="sc">Probl</emph>. Connoiſſant le centre de gravitè d’un triangle iſoſcele <lb/># qui tourne autour de ſon axe, trouver le ſolide du corps qu’il décrira. # 449 <lb/><emph style="sc">Prop</emph>. XXV. <emph style="sc">Probl</emph>. Connoiſſant le centre de gravité d’un cercle, trouver la <lb/># ſolidité de la ſphere engendrée par la révolution de ce cercle, autour de ſon <lb/># diametre. # 451 <lb/></note> </div> <div xml:id="echoid-div23" type="section" level="1" n="23"> <head xml:id="echoid-head29" xml:space="preserve">LIVRE XIII,</head> <head xml:id="echoid-head30" xml:space="preserve">Où l’on applique la Géométrie à la diviſion des champs, & à l’uſage du <lb/># compas de proportion.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Probl</emph>. Diviſer un triangle en autant de parties égales qu’on voudra <lb/># par des lignes tirées de l’angle oppoſé à la baſe. # 454 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Probl</emph>. Diviſer un triangle en deux parties égales par une ligne tirée <lb/># d’un point donné ſur un des côtés du triangle. # ibid. <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Probl</emph>. Diviſer un triangle en trois parties égales par des lignes <lb/># tirées d’un point pris ſur un de ſes côtés. # 455 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Probl</emph>. Diviſer un triangle en trois parties égales par des lignes <lb/># qui partent des trois angles. # ibid. <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Probl</emph>. Diviſer un triangle en deux parties égales par des lignes <lb/># tirées d’un point donné à volonté dans la ſurface du triangle. # 456 <pb o="xxvij" file="0033" n="33" rhead="DES MATIERES."/> <emph style="sc">Prop</emph>. VI. <emph style="sc">Probl</emph>. Diviſer un triangle en deux parties égales par une ligne <lb/># parallele à la baſe. # ibid. <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Probl</emph>. Diviſer un trapézoïde en deux parties égales par une ligne <lb/># parallele à la baſe. # 457 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Probl</emph>. Diviſer un trapeze en deux parties égales par une ligne <lb/># parallele à l’un des côtés. # 458 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Diviſer un trapézoïde en trois parties égales. # 459 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Diviſer un trapeze en deux parties égales. # ibid. <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Diviſer un trapeze en deux parties égales par une ligne <lb/># tirée d’un de ſes angles. # ibid. <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Diviſer un trapézoïde en deux parties égales par une ligne <lb/># tirée d’un point pris ſur l’un de ſes côtés. # 460 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Probl</emph>. Diviſer un pentagone en trois parties égales par des <lb/># lignes tirées d’un de ſes angles. # ibid. <lb/></note> </div> <div xml:id="echoid-div24" type="section" level="1" n="24"> <head xml:id="echoid-head31" xml:space="preserve">Uſages du compas de proportion.</head> <note style="it" position="right" xml:space="preserve"> <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Probl</emph>. Diviſer une ligne droite en autant de parties égales que <lb/># l’on voudra, par le moyen du compas de proportion. # 461 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Trouver une troiſieme proportionnelle à deux lignes don-<lb/># nées. # 462 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Probl</emph>. Trouver une quatrieme proportionnelle à trois lignes <lb/># données. # 463 <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Probl</emph>. Inſcrire un polygone quelconque dans un cercle. # ibid. <lb/><emph style="sc">Prop</emph>. XVIII. <emph style="sc">Probl</emph>. Décrire un polygone régulier quelconque ſur une ligne <lb/># donnée. # 464 <lb/><emph style="sc">Prop</emph>. XIX. <emph style="sc">Probl</emph>. Faire un angle de tant de degrés que l’on voudra. # ibid. <lb/><emph style="sc">Prop</emph>. XX. <emph style="sc">Probl</emph>. Un angle étant donné ſur le papier, en trouver la valeur <lb/># par la ligne des cordes. # 465 <lb/><emph style="sc">Prop</emph>. XXI. <emph style="sc">Probl</emph>. Connoiſſant le nombre des degrés d’un arc de cercle, <lb/># trouver ſon rayon. # ibid. <lb/><emph style="sc">Prop</emph>. XXII. <emph style="sc">Probl</emph>. Ouvrir le compas de proportion de maniere que les <lb/># lignes des cordes faſſent tel angle que l’on voudra. # ibid. <lb/><emph style="sc">Prop</emph>. XXIII. <emph style="sc">Probl</emph>. Le compas de proportion étant ouvert d’une grandeur <lb/># quelconque, connoître la valeur de l’angle, formé par la ligne des cor-<lb/># des. # 466 <lb/><emph style="sc">Prop</emph>. XXIV. <emph style="sc">Probl</emph>. Faire un quarré qui ſoit à un autre dans une raiſon <lb/># donnée. # ibid. <lb/><emph style="sc">Prop</emph>. XXV. <emph style="sc">Probl</emph>. Trouver le rapport d’un quarré à un autre. # 467 <lb/><emph style="sc">Prop</emph>. XXVI. <emph style="sc">Probl</emph>. Ouvrir le compas de proportion de maniere que les <lb/># lignes des plans faſſent un angle droit. # ibid. <lb/><emph style="sc">Prop</emph>. XXVII. <emph style="sc">Probl</emph>. Faire un quarré égal à deux autres donnés. # 468 <lb/><emph style="sc">Prop</emph>. XXVIII. <emph style="sc">Probl</emph>. Faire un cube qui ſoit à un autre dans une raiſon <lb/># donnée. # ibid. <lb/><emph style="sc">Prop</emph>. XXIX. <emph style="sc">Probl</emph>. Trouver le rapport qui eſt entre deux cubes. # ibid. <lb/><emph style="sc">Prop</emph>. XXX. <emph style="sc">Probl</emph>. Faire l’analyſe du métal dont on fait les pieces de <lb/># canon. # 469 <lb/><emph style="sc">Prop</emph>. XXXI. <emph style="sc">Pro</emph>. Trouver le calibre des boulets, & des pieces de canon. # 472 <pb o="xxviij" file="0034" n="34" rhead="TABLE"/> <emph style="sc">Prop</emph>. XXXII. <emph style="sc">Probl</emph>. Trouver le diametre des cylindres qui ſervent à me-<lb/># ſurer la poudre. # 473 <lb/><emph style="sc">Prop</emph>. XXXIII. <emph style="sc">Probl</emph>. Trouver la longueur des pieces de canon relative-<lb/># ment à leur calibre. # 475 <lb/><emph style="sc">Prop</emph>. XXXIV. <emph style="sc">Probl</emph>. Trouver le nombre des boulets qui ſont en piles. # 485 <lb/></note> </div> <div xml:id="echoid-div25" type="section" level="1" n="25"> <head xml:id="echoid-head32" xml:space="preserve">LIVRE XIV.</head> <head xml:id="echoid-head33" xml:space="preserve">Du mouvement des corps, & du jet des bombes.</head> <note style="it" position="right" xml:space="preserve"> <lb/>CHAP. I. Du choc des corps. # 493 <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Si deux corps égaux en maſſe ſont mus avec des vîteſſes <lb/># inégales, les forces de leur choc ſont comme leurs vîteſſes. # 497 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si deux corps inégaux & de même matiere ſont pouſſés <lb/># avec des vîteſſes égales, les forces de leurs chocs ſont comme leurs maſſes. # ibid. <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si les maſſes & les vîteſſes de deux corps ſont réciproques, <lb/># leurs forces ſont égales. # 499 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Si deux corps, ſans reſſort, ſe meuvent dans la même di-<lb/># rection avec des vîteſſes inégales, & vers un même point, la quantité de <lb/># mouvement, après le choc, ſera égale à celle qu’ils avoient avant le <lb/># choc. # 500 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Si deux corps ſe meuvent avec des vîteſſes inégales dans des <lb/># ſens directement oppoſés, la quantité de mouvement, après le choc, eſt <lb/># égale à la différence des quantités de mouvement avant le choc. # 501 <lb/>CHAP. II. Du mouvement des corps jettés. # 502 <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Si rien ne s’oppoſoit au mouvement des corps, ils ſeroient <lb/># toujours en mouvement avec la même vîteſſe, & ſuivant la même direc-<lb/># tion. # 504 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Un corps qui tombe reçoit à chaque inſtant des degrés <lb/># égaux de vîteſſe. # 505 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si deux corps égaux ſe meuvent pendant le même tems, l’un <lb/># avec une vîteſſe uniformément accélérée, l’autre avec une vîteſſe uniforme, <lb/># égale au dernier degré de vîteſſe acquiſe par le premier, l’eſpace parcouru <lb/># par le ſecond ſera double de l’eſpace parcouru par le premier. # 506 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Probl</emph>. Un corps eſt tombé perpendiculairement pendant quatre ſe-<lb/># condes, on demande l’eſpace que la peſanteur lui a fait parcourir. # 512 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Probl</emph>. Un corps a parcouru en tombant, par la force de la pe-<lb/># ſanteur, un certain eſpace; on demande le tems qu’il lui a fallu pour le <lb/># parcourir. # 513 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Si un corps eſt pouſſé à la fois par deux forces motrices, <lb/># capables de lui faire parcourir chacune une ligne donnée de grandeur & <lb/># de poſition, par l’effort compoſé de ces deux forces, il parcourra la dia-<lb/># gonale du parallélogramme formé ſur les directions de ces forces dans le <lb/># même tems qu’il eût décrit l’un des côtés de ce même parallélogramme, <lb/># par l’action d’une ſeule force. # 514 <lb/>CHAP. III. Théorie & pratique du jet des bombes; conſtruction & uſage <lb/># de l’inſtrument univerſel. # 518 <lb/><emph style="sc">Prop</emph>. VII. <emph style="sc">Theor</emph>. Si un corps eſt pouſſé par une force motrice ſuivant une <pb o="xxix" file="0035" n="35" rhead="DES MATIERES."/> # ligne parallele ou oblique à l’horizon, en vertu de cette force & de celle <lb/># de la peſanteur, il décrit une parabole. # 519 <lb/><emph style="sc">Prop</emph>. VIII. <emph style="sc">Probl</emph>. Connoiſſant la ligne de projection ſuppoſée horizontale, <lb/># & la ligne de chûte d’une parabole décrite par un mobile quelconque, trou-<lb/># ver de quel hauteur ce mobile doit tomber pour avoir à la fin de ſa chûte <lb/># une vîteſſe avec laquelle il puiſſe parcourir la même ligne de projection d’un <lb/># mouvement uniforme, dans le même tems que la peſanteur lui fait parcourir <lb/># la ligne de chûte. # 521 <lb/><emph style="sc">Prop</emph>. IX. <emph style="sc">Probl</emph>. Le parametre d’une parabole décrite par un mobile eſt qua-<lb/># druple de la ligne de hauteur. # 523 <lb/><emph style="sc">Prop</emph>. X. <emph style="sc">Probl</emph>. Connoiſſant la ligne de but, l’angle formé par la verti-<lb/># cale & la direction du mortier, l’angle formé par la même direction & la <lb/># ligne de but, trouver le parametre, la ligne de projection, & la ligne de <lb/># chûte. # 525 <lb/><emph style="sc">Prop</emph>. XI. <emph style="sc">Probl</emph>. Trouver quelle élévation il faut donner à un mortier pour <lb/># jetter une bombe à tel endroit que l’on voudra, pourvu que cet endroit ſoit <lb/># de niveau avec la batterie. # 526 <lb/><emph style="sc">Prop</emph>. XII. <emph style="sc">Probl</emph>. Trouver quelle élévation il faut donner à un mortier <lb/># pour chaſſer une bombe à une diſtance donnée, en ſuppoſant que la batterie <lb/># n’eſt pas de niveau avec l’endroit où l’on veut jetter la bombe. # 528 <lb/><emph style="sc">Prop</emph>. XIII. <emph style="sc">Probl</emph>. La ligne de but, l’angle qu’elle fait avec la verticale, <lb/># & la charge du mortier étant donnée, trouver l’angle d’élévation, ſous <lb/># lequel il faut pointer le mortier pour qu’elle tombe à un point donné. # 530 <lb/><emph style="sc">Prop</emph>. XIV. <emph style="sc">Probl</emph>. Conſtruire un inſtrument univerſel pour jetter des bom-<lb/># bes ſur toutes ſortes de plans. # 532 <lb/><emph style="sc">Prop</emph>. XV. <emph style="sc">Probl</emph>. Trouver par l’inſtrument univerſel, quelle hauteur il <lb/># faut donner à un mortier pour jetter une bombe à une diſtance donnée de <lb/># niveau avec la batterie. # 533 <lb/><emph style="sc">Prop</emph>. XVI. <emph style="sc">Probl</emph>. Trouver par l’inſtrument univerſel quelle élévation il <lb/># faut donner à un mortier pour jetter une bombe à une diſtance donnée ſur <lb/># un objet qui n’eſt pas de niveau avec la batterie. # 535 <lb/><emph style="sc">Prop</emph>. XVII. <emph style="sc">Theor</emph>. Si l’on tire deux bombes avec une même charge ſous <lb/># différens angles d’élévation, la portée de la premiere eſt à celle de la ſe-<lb/># conde, comme le ſinus d’un angle double de l’angle d’élévation du mortier <lb/># pour la premiere bombe, eſt au ſinus d’un angle double de l’élévation pour <lb/># la ſeconde. # 536 <lb/><emph style="sc">Prop</emph>. XVIII. <emph style="sc">Theor</emph>. Si l’on tire deux bombes à différens degrés d’éléva-<lb/># tion avec une même charge, il y aura même raiſon du ſinus de l’angle <lb/># double de la premiere élévation au ſinus double de la ſeconde, que de la <lb/># portée de la premiere élévation à la portée de la ſeconde. # 538 <lb/><emph style="sc">Prop</emph>. XIX. <emph style="sc">Probl</emph>. Connoiſſant l’amplitude d’une parabole décrite par une <lb/># bombe, connoître la hauteur à laquelle elle s’eſt élevée au deſſus de l’ho-<lb/># riſon. # 539 <lb/><emph style="sc">Prop</emph>. XX. <emph style="sc">Probl</emph>. Connoiſſant où une bombe s’eſt élevée, trouver la force <lb/># qu’elle a acquiſe en tombant de cette hauteur d’un mouvement accéléré. # ibid. <lb/></note> <pb o="xxx" file="0036" n="36" rhead="TABLE"/> </div> <div xml:id="echoid-div26" type="section" level="1" n="26"> <head xml:id="echoid-head34" xml:space="preserve">LIVRE XV, <lb/>Qui traite de la méchanique ſtatique.</head> <note style="it" position="right" xml:space="preserve"> <lb/>CHAP. I. Introduction à la méchanique. # 543 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Si un corps eſt pouſſé à la fois par deux puiſſances repré-<lb/># ſentées par les côtés d’un quarré, & dirigées ſuivant ces mêmes côtés, il <lb/># décrira la diagonale du quarré dans le même tems qu’il eût décrit le côté, <lb/># s’il n’avoit été pouſſé que par une ſeule force. # 546 <lb/>CHAP. II. Où l’on fait voir le rapport des puiſſances qui ſoutiennent des <lb/># poids avec des cordes. # 553 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Si deux puiſſances ſoutiennent un poids, tendant à ſuivre <lb/># une direction verticale, ces puiſſances ſont en équilibre, ſi elles ſont en <lb/># raiſon réciproque des perpendiculaires abaiſſées d’un point de cette verti-<lb/># cale ſur leur direction. # 554 <lb/>CHAP. III. Du plan incliné. # 557 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Si une puiſſance ſoutient un poids, 1°. par une ligne de di-<lb/># rection parallele au plan incliné, la puiſſance eſt au poids, comme ſa hau-<lb/># teur eſt à ſa longueur, 2°. ſi la direction de la puiſſance eſt parallele à la <lb/># baſe du plan incliné, la puiſſance eſt au poids, comme la hauteur du plan <lb/># eſt à la baſe. # ibid. <lb/>CHAP. IV. <emph style="sc">Theor</emph>. Du levier. # 560 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Deux puiſſances ſont en équilibre ſur un levier, ſi elles ſont <lb/># en raiſon réciproque des perpendiculaires tirées du point d’appui ſur leurs <lb/># directions. # ibid. <lb/>CHAP. V. De la roue dans ſon aiſſieu. # 566 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Siune puiſſance ſoutient un poids à l’aide d’une roue, & que <lb/># la direction de la puiſſance ſoit tangente à la roue, la puiſſance eſt au poids <lb/># comme le rayon du treuil à celui de la roue. # ibid. <lb/>CHAP. VI. De la poulie. # 567 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Si une puiſſance ſoutient un poids à l’aide d’une poulie, dont <lb/># la chape ſoit mobile, la puiſſance eſt égale au poids. 2°. Si la chape eſt <lb/># mobile, & ſi le poids y eſt attaché de maniere qu’il ſoit enlevé par la puiſ-<lb/># ſance, ſuivant des directions paralleles, la puiſſance ſera moitié du <lb/># poids. # 568 <lb/>CHAP. VII. Du coin. # 570 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. La force qui chaſſe le coin eſt à la réſiſtance, comme la moitié <lb/># de la tête du coin eſt à la longueur d’un de ſes côtés. # 572 <lb/>CHAP. VIII. De la vis. # 573 <lb/><emph style="sc">Prop</emph>. <emph style="sc">Theor</emph>. Si une puiſſance enleve un poids à l’aide d’une vis, la puiſ-<lb/># ſance ſera au poids, comme la hauteur d’un des pas de la vis eſt à la cir-<lb/># conférence du cercle que décrira la puiſſance appliquée au levier, par le <lb/># moyen duquel on meut la vis. # 574 <lb/>CHAP. IX. Des machines compoſées. # 575 <lb/>Analogie des poulies moufiées. Si une puiſſance ſoutient un poids à l’aide de <lb/># pluſieurs poulies mouflées, la puiſſance eſt au poids comme l’unité au dou-<lb/># ble du nombre des poulies mobiles. # 576 <pb o="xxxj" file="0037" n="37" rhead="DES MATIERES."/> Analogie des roues dentées. La puiſſance eſt au poids comme le produit des <lb/># rayons des pignons eſt au produit des rayons des roues. # 579 <lb/></note> </div> <div xml:id="echoid-div27" type="section" level="1" n="27"> <head xml:id="echoid-head35" xml:space="preserve">LIVRE XVI, <lb/>Qui traite de l’Hydroſtatique & de l’Hydraulique.</head> <note style="it" position="right" xml:space="preserve"> <lb/>CHAP. I. De l’équilique, & du mouvement des liqueurs. # 602 <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Si on met une liqueur dans un vaſe, ſa ſurface ſera de ni-<lb/># veau, & toutes ſes parties en équilibre. # 605 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si on verſe une liqueur dans un ſiphon, elle ſe mettra de <lb/># niveau dans les deux branches. # 609 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si l’on met dans les deux branches du ſiphon des liqueurs <lb/># de peſanteurs différentes, les hauteurs de ces liqueurs ſeront dans la raiſon <lb/># réciproque des peſanteurs ſpécifiques, ſi les diametres ſont égaux. # 610 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Si un corps eſt d’une denſité égale, plus petite ou plus <lb/># grande, que celle du fluide dans lequel il eſt plongé, 1°. il demeurera en équi-<lb/># libre dans tel endroit qu’il ſoit plongé; 2°. il ſurnagera; 3°. il deſcendra <lb/># au fond avec une vîteſſe égale à celle qu’il reçoit des differences des pe-<lb/># ſanteurs ſpécifiques. # 611 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Theor</emph>. Si l’on a un vaſe plus gros par en bas que par en haut, <lb/># rempli d’une liqueur quelconque, cette liqueur aura autant de force pour <lb/># ſortir par une ouverture égale à ſa baſe, que ſi cette ouverture étoit égale <lb/># à celle d’en haut. # 616 <lb/>CHAP. II. De la vîteſſe des fluides qui ſortent par des ouvertures faites <lb/># aux vaſes qui les contiennent. <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Si l’on a un tuy au vertical, & rempli d’une liqueur quel-<lb/># conque, la vîteſſe de cette liqueur, à l’ouverture de la baſe, eſt expri-<lb/># mée par la racine quarrée de la hauteur. # 619 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si le trou n’eſt pas égal à la baſe, la vîteſſe eſt encore <lb/># exprimée par la racine quarrée de la hauteur. # ibid. <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Trouver la dépenſe d’un jet d’eau pendant une minute par <lb/># un ajutage de quatre lignes de diametre, & une hauteur de 40 pieds. # 625 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Si un vaſe ſe déſemplit par une ouverture plus petite que <lb/># la baſe, les quantités d’eau qui s’écouleront dans des tems égaux, ſeront <lb/># comme les nombres impairs, pris dans un ordre renverſé. <lb/>CHAP. III. Du cours des rivieres, & du choc des fluides contre les ſur-<lb/># faces des corps qu’elles rencontrent. <lb/><emph style="sc">Prop</emph>. I. <emph style="sc">Theor</emph>. Toute riviere ou fleuve qui n’eſt point arrêté dans ſon mou-<lb/># vement, eſt mu d’une vîteſſe accélérée. # 629 <lb/><emph style="sc">Prop</emph>. II. <emph style="sc">Theor</emph>. Si un fluide choque avec différentes vîteſſes des ſur-<lb/># faces égales, expoſées perpendiculairement à ſon courant, les forces du <lb/># choc ſeront comme les quarrés des vîteſſes. # 633 <lb/><emph style="sc">Prop</emph>. III. <emph style="sc">Theor</emph>. Si deux ſurfaces égales ſont expoſées au même fluide, <lb/># l’une perpendiculairement, l’autre obliquement, les forces du choc ſeront <lb/># comme le quarré du ſinus total au quarré de celui de l’angle d’incli-<lb/># naiſon. # 635 <lb/><emph style="sc">Prop</emph>. IV. <emph style="sc">Theor</emph>. Si deux ſurfaces égales ſont expoſées, l’une perpendicu- <pb o="xxxij" file="0038" n="38" rhead="TABLE"/> # lairement, l’autre obliquement à un fluide, dont toutes les tranches on <lb/># des vîteſſes qui croiſſent comme les racines quarrées des hauteurs, les chocs <lb/># ſont comme les cubes du ſinus total & du ſinus d’inclinaiſon. # 538 <lb/><emph style="sc">Prop</emph>. V. <emph style="sc">Probl</emph>. Connoiſſant la vîteſſe de l’eau, trouver le choc de cette eau <lb/># contre une ſurface donnée. # 641 <lb/><emph style="sc">Prop</emph>. VI. <emph style="sc">Theor</emph>. Si l’on a un vaiſſeau rempli d’eau, toujours entretenu <lb/># à la même hauteur, les chocs à la ſortie de deux ajutages égaux ſeront dans <lb/># la raiſon des hauteurs d’eau au deſſus des ajutages. # 642 <lb/>Diſcours ſur la nature & les propriétés de l’air. # 643 <lb/></note> </div> <div xml:id="echoid-div28" type="section" level="1" n="28"> <head xml:id="echoid-head36" xml:space="preserve">Fin de la Table.</head> <pb file="0039" n="39"/> <figure> <image file="0039-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0039-01"/> </figure> </div> <div xml:id="echoid-div29" type="section" level="1" n="29"> <head xml:id="echoid-head37" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE, <lb/>A L’USAGE <lb/>DES INGÉNIEURS <lb/>ET OFFICIERS D’ARTILLERIE.</head> <head xml:id="echoid-head38" xml:space="preserve">LIVRE PREMIER, <lb/>Où l’on donne l’Introduction à la Géométrie. <lb/><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head39" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s212" xml:space="preserve">1. </s> <s xml:id="echoid-s213" xml:space="preserve">LA Géométrie eſt une ſcience qui ne conſidere pas tant <lb/>la grandeur en elle-même, que le rapport qu’elle peut avoii <lb/>avec une grandeur de même nature qu’elle.</s> <s xml:id="echoid-s214" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div30" type="section" level="1" n="30"> <head xml:id="echoid-head40" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s215" xml:space="preserve">2. </s> <s xml:id="echoid-s216" xml:space="preserve">Tout ce qui peut tomber en queſtion, s’appelle propoſi- <pb o="2" file="0040" n="40" rhead="NOUVEAU COURS"/> tion. </s> <s xml:id="echoid-s217" xml:space="preserve">Il y en a de différentes ſortes, & </s> <s xml:id="echoid-s218" xml:space="preserve">elles changent de nom <lb/>ſuivant leur objet. </s> <s xml:id="echoid-s219" xml:space="preserve">Par exemple,</s> </p> </div> <div xml:id="echoid-div31" type="section" level="1" n="31"> <head xml:id="echoid-head41" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s220" xml:space="preserve">3. </s> <s xml:id="echoid-s221" xml:space="preserve">Axiome eſt une propoſition ſi claire, qu’elle n’a pas <lb/>beſoin de démonſtration pour qu’on en voie la vérité. </s> <s xml:id="echoid-s222" xml:space="preserve">De <lb/>ces propoſitions ſont les ſuivantes. </s> <s xml:id="echoid-s223" xml:space="preserve">Le tout eſt plus grand qu’une <lb/>de ſes parties; </s> <s xml:id="echoid-s224" xml:space="preserve">deux choſes égales à une même troiſieme, ſont égales <lb/>entr’elles; </s> <s xml:id="echoid-s225" xml:space="preserve">ſi à des quantités égales on ajoute des quantités égales, <lb/>les quantités qui en réſulteront ſeront encore égales, &</s> <s xml:id="echoid-s226" xml:space="preserve">c. </s> <s xml:id="echoid-s227" xml:space="preserve">On fait <lb/>un grand uſage de ces propoſitions dans la Géométrie, ſi ſim-<lb/>ples qu’elles paroiſſent.</s> <s xml:id="echoid-s228" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div32" type="section" level="1" n="32"> <head xml:id="echoid-head42" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s229" xml:space="preserve">4. </s> <s xml:id="echoid-s230" xml:space="preserve">Théorême eſt une propoſition dont il faut démontrer <lb/>la vérité.</s> <s xml:id="echoid-s231" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div33" type="section" level="1" n="33"> <head xml:id="echoid-head43" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s232" xml:space="preserve">5. </s> <s xml:id="echoid-s233" xml:space="preserve">Problême eſt une propoſition dans laquelle il s’agit d’exé-<lb/>cuter quelqu’opération, ſuivant certaines conditions, & </s> <s xml:id="echoid-s234" xml:space="preserve">de <lb/>prouver enſuite que l’on a réellement fait ce qui étoit en <lb/>queſtion.</s> <s xml:id="echoid-s235" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div34" type="section" level="1" n="34"> <head xml:id="echoid-head44" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s236" xml:space="preserve">6. </s> <s xml:id="echoid-s237" xml:space="preserve">Lemme eſt une propoſition qui en précéde une autre, <lb/>pour en faciliter l’intelligence & </s> <s xml:id="echoid-s238" xml:space="preserve">la démonſtration.</s> <s xml:id="echoid-s239" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div35" type="section" level="1" n="35"> <head xml:id="echoid-head45" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s240" xml:space="preserve">7. </s> <s xml:id="echoid-s241" xml:space="preserve">Corollaire eſt une propoſition qui n’eſt qu’une ſuite ou <lb/>une conſéquence de la propoſition précédente. </s> <s xml:id="echoid-s242" xml:space="preserve">Comme toutes <lb/>ces propoſitions ont pour objet la grandeur; </s> <s xml:id="echoid-s243" xml:space="preserve">voici l’idée qu’il <lb/>faut s’en former.</s> <s xml:id="echoid-s244" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div36" type="section" level="1" n="36"> <head xml:id="echoid-head46" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s245" xml:space="preserve">8. </s> <s xml:id="echoid-s246" xml:space="preserve">On appelle grandeur tout ce qui eſt ſuſceptible d’aug-<lb/>mentation ou de diminution. </s> <s xml:id="echoid-s247" xml:space="preserve">On conſidére en Géométrie <lb/>trois ſortes de grandeurs ou dimenſions; </s> <s xml:id="echoid-s248" xml:space="preserve">longueur, largeur, & </s> <s xml:id="echoid-s249" xml:space="preserve"><lb/>profondeur.</s> <s xml:id="echoid-s250" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div37" type="section" level="1" n="37"> <head xml:id="echoid-head47" xml:space="preserve">IX.</head> <p> <s xml:id="echoid-s251" xml:space="preserve">9. </s> <s xml:id="echoid-s252" xml:space="preserve">La longueur conſidérée ſans largeur & </s> <s xml:id="echoid-s253" xml:space="preserve">ſans profondeur, <lb/>ſe nomme ligne.</s> <s xml:id="echoid-s254" xml:space="preserve"/> </p> <pb o="3" file="0041" n="41" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div38" type="section" level="1" n="38"> <head xml:id="echoid-head48" xml:space="preserve">X.</head> <p> <s xml:id="echoid-s255" xml:space="preserve">10. </s> <s xml:id="echoid-s256" xml:space="preserve">La longueur & </s> <s xml:id="echoid-s257" xml:space="preserve">la largeur conſidérées enſemble, ſans <lb/>avoir égard à l’épaiſſeur ou profondeur, ſe nomme ſurface. <lb/></s> <s xml:id="echoid-s258" xml:space="preserve">On l’appelle ſurface plane, lorſque tous ſes points ne ſont pas <lb/>plus élevés les uns que les autres, comme cela arrive dans les <lb/>ſurfaces plates & </s> <s xml:id="echoid-s259" xml:space="preserve">unies, telles que ſont celles des glaces ou <lb/>miroirs.</s> <s xml:id="echoid-s260" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div39" type="section" level="1" n="39"> <head xml:id="echoid-head49" xml:space="preserve">XI.</head> <p> <s xml:id="echoid-s261" xml:space="preserve">11. </s> <s xml:id="echoid-s262" xml:space="preserve">La longueur, la largeur, & </s> <s xml:id="echoid-s263" xml:space="preserve">la profondeur conſidérées <lb/>enſemble, ſe nomment corps ou ſolide. </s> <s xml:id="echoid-s264" xml:space="preserve">La longueur, la lar-<lb/>geur, & </s> <s xml:id="echoid-s265" xml:space="preserve">la profondeur ſont toutes des grandeurs de même <lb/>nature: </s> <s xml:id="echoid-s266" xml:space="preserve">on ne leur a donné différens noms que relativement <lb/>à la maniere dont on les conçoit placées dans les corps.</s> <s xml:id="echoid-s267" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div40" type="section" level="1" n="40"> <head xml:id="echoid-head50" xml:space="preserve">XII.</head> <p> <s xml:id="echoid-s268" xml:space="preserve">12. </s> <s xml:id="echoid-s269" xml:space="preserve">Le point eſt l’extrêmité d’un corps ou d’une ſurface, <lb/>ou bien d’une ligne; </s> <s xml:id="echoid-s270" xml:space="preserve">on le conçoit comme indiviſible, ou <lb/>ſans dimenſion, c’eſt-à-dire qu’on ne lui attribue ni longueur, <lb/>ni largeur, ni profondeur. </s> <s xml:id="echoid-s271" xml:space="preserve">Ainſi le point ne peut être l’objet <lb/>de la Géométrie, qui ne conſidere que l’étendue avec laquelle <lb/>il n’a aucun rapport.</s> <s xml:id="echoid-s272" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div41" type="section" level="1" n="41"> <head xml:id="echoid-head51" xml:space="preserve">XIII.</head> <p> <s xml:id="echoid-s273" xml:space="preserve">13. </s> <s xml:id="echoid-s274" xml:space="preserve">La ligne droite eſt la plus courte de toutes celles que <lb/>l’on peut mener d’un point A à un autre point B, comme <lb/>A B. </s> <s xml:id="echoid-s275" xml:space="preserve">D’où il ſuit, 1°. </s> <s xml:id="echoid-s276" xml:space="preserve">qu’il n’y a qu’un ſeul chemin qui ſoit <lb/>le plus court d’un point à un autre. </s> <s xml:id="echoid-s277" xml:space="preserve">2°. </s> <s xml:id="echoid-s278" xml:space="preserve">Que deux points <lb/>ſuffiſent pour déterminer la poſition d’une ligne droite. </s> <s xml:id="echoid-s279" xml:space="preserve">3°. <lb/></s> <s xml:id="echoid-s280" xml:space="preserve">Que ſi une ligne droite a deux points communs avec une <lb/>autre ligne, elle ſe confond entiérement avec elle.</s> <s xml:id="echoid-s281" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div42" type="section" level="1" n="42"> <head xml:id="echoid-head52" xml:space="preserve">XIV.</head> <p> <s xml:id="echoid-s282" xml:space="preserve">14. </s> <s xml:id="echoid-s283" xml:space="preserve">La ligne courbe eſt celle qui n’eſt pas la plus courte <lb/>d’un point à un autre, comme C D. </s> <s xml:id="echoid-s284" xml:space="preserve">Il y a donc une infinité <lb/>de lignes courbes qui peuvent paſſer par deux points, puiſqu’il <lb/>y a une infinité de chemins qui ne ſont pas les plus courts.</s> <s xml:id="echoid-s285" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div43" type="section" level="1" n="43"> <head xml:id="echoid-head53" xml:space="preserve">XV.</head> <p> <s xml:id="echoid-s286" xml:space="preserve">15. </s> <s xml:id="echoid-s287" xml:space="preserve">La ligne mixte eſt celle qui eſt en partie courbe, & </s> <s xml:id="echoid-s288" xml:space="preserve"><lb/>en partie droite, comme E F.</s> <s xml:id="echoid-s289" xml:space="preserve"/> </p> <pb o="4" file="0042" n="42" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div44" type="section" level="1" n="44"> <head xml:id="echoid-head54" xml:space="preserve">XVI.</head> <p> <s xml:id="echoid-s290" xml:space="preserve">16. </s> <s xml:id="echoid-s291" xml:space="preserve">Une ligne perpendiculaire eſt une ligne droite C D, qui <lb/> <anchor type="note" xlink:label="note-0042-01a" xlink:href="note-0042-01"/> aboutiſſant ſur une autre A B, ne penche pas plus d’un côté <lb/>que de l’autre.</s> <s xml:id="echoid-s292" xml:space="preserve"/> </p> <div xml:id="echoid-div44" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-01" xlink:href="note-0042-01a" xml:space="preserve">Figure 4.</note> </div> </div> <div xml:id="echoid-div46" type="section" level="1" n="45"> <head xml:id="echoid-head55" xml:space="preserve">XVII.</head> <p> <s xml:id="echoid-s293" xml:space="preserve">17. </s> <s xml:id="echoid-s294" xml:space="preserve">Le Quarré eſt une figure rectiligne, formée par quatre <lb/> <anchor type="note" xlink:label="note-0042-02a" xlink:href="note-0042-02"/> côtés égaux, qui aboutiſſent perpendiculairement les uns ſur <lb/>les autres.</s> <s xml:id="echoid-s295" xml:space="preserve"/> </p> <div xml:id="echoid-div46" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-02" xlink:href="note-0042-02a" xml:space="preserve">Figure 1.</note> </div> </div> <div xml:id="echoid-div48" type="section" level="1" n="46"> <head xml:id="echoid-head56" xml:space="preserve">XVIII.</head> <p> <s xml:id="echoid-s296" xml:space="preserve">18. </s> <s xml:id="echoid-s297" xml:space="preserve">Le Rectangle eſt un quadrilatere, dont tous les côtés ne <lb/> <anchor type="note" xlink:label="note-0042-03a" xlink:href="note-0042-03"/> ſont pas égaux entr’eux, mais ſeulement deux à deux, & </s> <s xml:id="echoid-s298" xml:space="preserve">qui <lb/>aboutiſſent perpendiculairement les uns aux autres.</s> <s xml:id="echoid-s299" xml:space="preserve"/> </p> <div xml:id="echoid-div48" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-03" xlink:href="note-0042-03a" xml:space="preserve">Figure 2.</note> </div> </div> <div xml:id="echoid-div50" type="section" level="1" n="47"> <head xml:id="echoid-head57" xml:space="preserve">XIX.</head> <p> <s xml:id="echoid-s300" xml:space="preserve">19. </s> <s xml:id="echoid-s301" xml:space="preserve">Le Cube eſt un corps qui a la figure d’un dez à jouer, <lb/> <anchor type="note" xlink:label="note-0042-04a" xlink:href="note-0042-04"/> renfermé par ſix quarrés égaux, & </s> <s xml:id="echoid-s302" xml:space="preserve">dont toutes les dimenſions <lb/>ſont égales entr’elles; </s> <s xml:id="echoid-s303" xml:space="preserve">cette figure étant la plus ſimple de <lb/>toutes, on y ramene tous les ſolides: </s> <s xml:id="echoid-s304" xml:space="preserve">c’eſt pourquoi lorſqu’on <lb/>propoſe de trouver la ſolidité d’un corps on ſe ſert du mot <lb/>cuber, qui ſignifie la même choſe.</s> <s xml:id="echoid-s305" xml:space="preserve"/> </p> <div xml:id="echoid-div50" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-04" xlink:href="note-0042-04a" xml:space="preserve">Figure 3.</note> </div> </div> <div xml:id="echoid-div52" type="section" level="1" n="48"> <head xml:id="echoid-head58" xml:space="preserve">XX.</head> <p> <s xml:id="echoid-s306" xml:space="preserve">20. </s> <s xml:id="echoid-s307" xml:space="preserve">Le Parallelepipede eſt un ſolide renfermé par ſix rectan-<lb/>gles, dont les côtés oppoſés ſont égaux, & </s> <s xml:id="echoid-s308" xml:space="preserve">qui n’a pas ſes <lb/>trois dimenſions égales.</s> <s xml:id="echoid-s309" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s310" xml:space="preserve">21. </s> <s xml:id="echoid-s311" xml:space="preserve">Il y a une maniere de conſidérer les trois eſpeces de <lb/> <anchor type="note" xlink:label="note-0042-05a" xlink:href="note-0042-05"/> l’étendue, c’eſt-à-dire la ligne, la ſurface, & </s> <s xml:id="echoid-s312" xml:space="preserve">le ſolide ou corps, <lb/>qui eſt très-propre à expliquer beaucoup de choſes en Géo-<lb/>métrie; </s> <s xml:id="echoid-s313" xml:space="preserve">c’eſt d’imaginer la ligne compoſée d’une infinité de <lb/>points, la ſurface compoſée d’une infinité de lignes, & </s> <s xml:id="echoid-s314" xml:space="preserve">le <lb/>corps compoſé d’une infinité de plans. </s> <s xml:id="echoid-s315" xml:space="preserve">Pour faire entendre <lb/>ceci, conſidérez deux points, comme A B éloignés l’un <lb/>de l’autre d’une diſtance quelconque; </s> <s xml:id="echoid-s316" xml:space="preserve">ſi l’on ſuppoſe que le <lb/>point A ſe meut pour aller vers le point B, ſans s’écarter <lb/>ni à droite ni à gauche, & </s> <s xml:id="echoid-s317" xml:space="preserve">qu’il laiſſe ſur ſon chemin une <lb/>trace d’autres points, la ſomme de tous ces points formera <lb/>une ligne droite A B, puiſqu’il n’y aura point d’eſpace dans <lb/>la longueur A B, ſi petit qu’il ſoit, que le point A n’ait par- <pb o="5" file="0043" n="43" rhead="DE MATHÉMATIQUE. Liv. I."/> couru. </s> <s xml:id="echoid-s318" xml:space="preserve">Ainſi toute ligne droite peut être conſiderée, comme <lb/>formée par une multitude de points, dont la quantité eſt ex-<lb/>primée par la longueur de la même ligne.</s> <s xml:id="echoid-s319" xml:space="preserve"/> </p> <div xml:id="echoid-div52" type="float" level="2" n="1"> <note position="left" xlink:label="note-0042-05" xlink:href="note-0042-05a" xml:space="preserve">Figure 5.</note> </div> <p> <s xml:id="echoid-s320" xml:space="preserve">22. </s> <s xml:id="echoid-s321" xml:space="preserve">L’on concevra de même que le plan eſt compoſé d’une <lb/> <anchor type="note" xlink:label="note-0043-01a" xlink:href="note-0043-01"/> infinité de lignes; </s> <s xml:id="echoid-s322" xml:space="preserve">car ſuppoſant que la ligne A C ſe meut <lb/>le long de la ligne C D, en demeurant toujours également <lb/>inclinée, il eſt viſible que ſi elle laiſſe après elle autant de <lb/>lignes qu’il y a de points dans C D, que lorſqu’elle ſera par-<lb/>venue au point D, toutes les lignes compoſeront enſemble la <lb/>ſurface B C.</s> <s xml:id="echoid-s323" xml:space="preserve"/> </p> <div xml:id="echoid-div53" type="float" level="2" n="2"> <note position="right" xlink:label="note-0043-01" xlink:href="note-0043-01a" xml:space="preserve">Figure 2.</note> </div> <p> <s xml:id="echoid-s324" xml:space="preserve">23. </s> <s xml:id="echoid-s325" xml:space="preserve">Enſin ſi l’on a un plan A B qui ſe meuve le long de <lb/> <anchor type="note" xlink:label="note-0043-02a" xlink:href="note-0043-02"/> la ligne B C, & </s> <s xml:id="echoid-s326" xml:space="preserve">qu’il laiſſe autant de plans après lui qu’il y <lb/>a de points dans cette ligne, l’on voit que lorſqu’il ſera ar-<lb/>rivé à l’extrêmité C, il aura formé un corps tel que D B qui <lb/>ſera compoſé d’une infinité de plans, dont la ſomme ſera ex-<lb/>primée par la ligne B C, pourvu que cette ligne ſoit perpen-<lb/>diculaire au plan générateur.</s> <s xml:id="echoid-s327" xml:space="preserve"/> </p> <div xml:id="echoid-div54" type="float" level="2" n="3"> <note position="right" xlink:label="note-0043-02" xlink:href="note-0043-02a" xml:space="preserve">Figure 5 & 6.</note> </div> <p> <s xml:id="echoid-s328" xml:space="preserve">24. </s> <s xml:id="echoid-s329" xml:space="preserve">Comme on entend par la génération d’une choſe les <lb/>parties qui l’ont formée, il s’enſuit, ſelon ce qui vient d’être <lb/>dit, que le point eſt le générateur de la ligne; </s> <s xml:id="echoid-s330" xml:space="preserve">la ligne eſt la <lb/>génératrice de la ſurface, & </s> <s xml:id="echoid-s331" xml:space="preserve">la ſurface génératrice du corps; <lb/></s> <s xml:id="echoid-s332" xml:space="preserve">& </s> <s xml:id="echoid-s333" xml:space="preserve">par conſéquent le point peut être lui-même conſidéré com-<lb/>me le principe générateur de toute ſorte de grandeur.</s> <s xml:id="echoid-s334" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s335" xml:space="preserve">25. </s> <s xml:id="echoid-s336" xml:space="preserve">Si l’on ſuppoſe que la ligne A C ſoit de huit pieds, & </s> <s xml:id="echoid-s337" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0043-03a" xlink:href="note-0043-03"/> la ligne C D de ſix, & </s> <s xml:id="echoid-s338" xml:space="preserve">que l’on conſidere ces nombres com-<lb/>me exprimant la quantité de points qui ſe trouve dans ces <lb/>lignes, l’on verra qu’en multipliant 8 par 6, le produit 48 <lb/>ſera la ſurface A D; </s> <s xml:id="echoid-s339" xml:space="preserve">car cette ſurface étant compoſée d’une <lb/>infinité de lignes, & </s> <s xml:id="echoid-s340" xml:space="preserve">chacune de ces lignes étant compoſée <lb/>d’une infinité de points, il s’enſuit que la ſurface eſt com-<lb/>poſée d’une infinité de points, dont la quantité eſt repré-<lb/>ſentée par le produit de tous les points de la ligne C D, par <lb/>rous les points de la ligne A C, c’eſt-à-dire de ſa longueur <lb/>A C, par ſa largeur C D, qui donne 48 pieds, qu’il faut bien <lb/>ſe garder de confondre avec le pied courant; </s> <s xml:id="echoid-s341" xml:space="preserve">car le pied cou-<lb/>tant n’eſt qu’une longueur ſans largeur, au lieu que ceux qui <lb/>ſont formés par le produit de deux dimenſions, ſont autant <lb/>de ſurfaces quarrées, qui ſervent à meſurer toutes les ſu-<lb/>perficies.</s> <s xml:id="echoid-s342" xml:space="preserve"/> </p> <div xml:id="echoid-div55" type="float" level="2" n="4"> <note position="right" xlink:label="note-0043-03" xlink:href="note-0043-03a" xml:space="preserve">Figure 2.</note> </div> <p> <s xml:id="echoid-s343" xml:space="preserve">26. </s> <s xml:id="echoid-s344" xml:space="preserve">Or comme le ſolide eſt compoſé d’autant de plans qu’il <lb/> <anchor type="note" xlink:label="note-0043-04a" xlink:href="note-0043-04"/> <pb o="6" file="0044" n="44" rhead="NOUVEAU COURS"/> y a de points dans la ligne C B, il faut donc multiplier le <lb/>plan A B par la ligne C B pour avoir le contenu de ce ſolide; <lb/></s> <s xml:id="echoid-s345" xml:space="preserve">ainſi ſuppoſant que le plan A B vaut 48 pieds quarrés, & </s> <s xml:id="echoid-s346" xml:space="preserve"><lb/>que le nombre des points de la ligne B C ſoit exprimé par <lb/>4 pieds courans, multipliant 48 par 4, on aura 192 pieds pour <lb/>la valeur du ſolide A C. </s> <s xml:id="echoid-s347" xml:space="preserve">Il faut encore faire attention que <lb/>ces pieds ſont différens du pied courant, & </s> <s xml:id="echoid-s348" xml:space="preserve">du pied quarré, <lb/>car ce ſont autant de ſolides qui ont un pied de long, un de <lb/>large, & </s> <s xml:id="echoid-s349" xml:space="preserve">un de haut, qui ſont par conſéquent des cubes, <lb/>puiſqu’ils ont les trois dimenſions égales: </s> <s xml:id="echoid-s350" xml:space="preserve">ainſi il faut remar-<lb/>quer que les lignes meſurent les lignes, les ſurfaces meſurent <lb/>les ſurfaces, & </s> <s xml:id="echoid-s351" xml:space="preserve">les ſolides meſurent les ſolides; </s> <s xml:id="echoid-s352" xml:space="preserve">car la raiſon <lb/>ſeule nous démontre que la meſure doit être de même na-<lb/>ture que la choſe, ou la grandeur meſurée.</s> <s xml:id="echoid-s353" xml:space="preserve"/> </p> <div xml:id="echoid-div56" type="float" level="2" n="5"> <note position="right" xlink:label="note-0043-04" xlink:href="note-0043-04a" xml:space="preserve">Figure 5.</note> </div> <p> <s xml:id="echoid-s354" xml:space="preserve">27. </s> <s xml:id="echoid-s355" xml:space="preserve">Mais comme il s’agit beaucoup moins ici de chercher <lb/>la valeur abſolue des grandeurs, que le rapport qu’elles ont <lb/>entr’elles, nous nous ſervirons de lettres de l’alphabet pour <lb/>exprimer les grandeurs, afin de rendre générales les démonſ-<lb/>trations des propoſitions que nous établirons. </s> <s xml:id="echoid-s356" xml:space="preserve">Pour concevoir <lb/>la raiſon de cette généralité, on fera attention que la géné-<lb/>ralité d’un ſigne dépend de ſon indétermination; </s> <s xml:id="echoid-s357" xml:space="preserve">car dès-lors <lb/>qu’une grandeur eſt indéterminée, on peut l’appliquer à telle <lb/>eſpece de choſes que l’on voudra. </s> <s xml:id="echoid-s358" xml:space="preserve">Ainſi, par exemple, le <lb/>nombre 7 étant indéterminé par rapport à l’eſpece de ſes <lb/>unités, puiſqu’il ne ſignifie pas plus ſept hommes que ſept <lb/>chevaux, on peut l’employer pour marquer telle eſpece d’u-<lb/>nités que l’on voudra, d’hommes ou de chevaux, &</s> <s xml:id="echoid-s359" xml:space="preserve">c. </s> <s xml:id="echoid-s360" xml:space="preserve">ainſi <lb/>ſon indétermination le rend d’autant plus général, & </s> <s xml:id="echoid-s361" xml:space="preserve">propre à <lb/>déſigner telle ſorte d’unité que l’on jugera à propos. </s> <s xml:id="echoid-s362" xml:space="preserve">Si donc l’in-<lb/>détermination d’un ſigne eſt la plus grande poſſible, ſa gé-<lb/>néralité ſera auſſi la plus grande qu’on puiſſe imaginer. </s> <s xml:id="echoid-s363" xml:space="preserve">Pour <lb/>arriver à ce dernier degré de généralité, on remarquera en-<lb/>core qu’une grandeur ne peut être indéterminée qu’en deux <lb/>manieres; </s> <s xml:id="echoid-s364" xml:space="preserve">ſçavoir, la premiere par rapport à l’eſpece ſeule-<lb/>ment, & </s> <s xml:id="echoid-s365" xml:space="preserve">non pas à l’égard du nombre des unités, & </s> <s xml:id="echoid-s366" xml:space="preserve">la <lb/>ſeconde par rapport au nombre & </s> <s xml:id="echoid-s367" xml:space="preserve">à l’eſpece tout enſemble. <lb/></s> <s xml:id="echoid-s368" xml:space="preserve">De cette premiere claſſe ſont les ſignes de l’Arithmétique, qui <lb/>ſont toujours indéterminés par rapport aux différentes ſortes <lb/>d’unités, & </s> <s xml:id="echoid-s369" xml:space="preserve">jamais à l’égard du nombre de ces unités; </s> <s xml:id="echoid-s370" xml:space="preserve">& </s> <s xml:id="echoid-s371" xml:space="preserve"><lb/>de la ſeconde claſſe ſont les ſignes de l’alphabet ou les lettres, <pb o="7" file="0045" n="45" rhead="DE MATHÉMATIQUE. Liv. I."/> qui ne déſignant aucune eſpece en particulier, peuvent être <lb/>employées pour les déſigner toutes, & </s> <s xml:id="echoid-s372" xml:space="preserve">n’étant point fixées <lb/>pour aucun nombre, peuvent auſſi être employées à les re-<lb/>préſenter en général tous. </s> <s xml:id="echoid-s373" xml:space="preserve">Donc puiſque l’indétermination <lb/>des lettres eſt la plus grande poſſible, leur généralité eſt auſſi <lb/>la plus grande, & </s> <s xml:id="echoid-s374" xml:space="preserve">par conſéquent tout ce qu’on démontre <lb/>par leur ſecours, eſt démontré généralement. </s> <s xml:id="echoid-s375" xml:space="preserve">On remar-<lb/>quera encore que l’on auroit pu prendre tout autre caractere <lb/>que ceux de l’alphabet, mais ces caracteres étant déja con-<lb/>nus, il étoit naturel de s’en ſervir, & </s> <s xml:id="echoid-s376" xml:space="preserve">c’eſt ce qui les a fait <lb/>préférer à tout autre.</s> <s xml:id="echoid-s377" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s378" xml:space="preserve">28. </s> <s xml:id="echoid-s379" xml:space="preserve">Pour exprimer une ligne, on ſe ſervira d’une des let-<lb/>tres de l’alphabet, a, b, c, d, e, &</s> <s xml:id="echoid-s380" xml:space="preserve">c; </s> <s xml:id="echoid-s381" xml:space="preserve">& </s> <s xml:id="echoid-s382" xml:space="preserve">pour exprimer un <lb/>plan, on en mettra deux l’une contre l’autre pour marquer <lb/>les deux dimenſions de ce plan; </s> <s xml:id="echoid-s383" xml:space="preserve">& </s> <s xml:id="echoid-s384" xml:space="preserve">pour marquer un ſolide, <lb/>on en mettra trois de ſuite, parce qu’un ſolide quelconque <lb/>a trois dimenſions, & </s> <s xml:id="echoid-s385" xml:space="preserve">de plus, parce que l’on eſt convenu <lb/>de repréſenter la multiplication de deux grandeurs, en met-<lb/>tant ces grandeurs les unes auprès des autres. </s> <s xml:id="echoid-s386" xml:space="preserve">Par exemple, <lb/>ab repréſente un plan, dont les deux dimenſions ſont a & </s> <s xml:id="echoid-s387" xml:space="preserve">b, <lb/>& </s> <s xml:id="echoid-s388" xml:space="preserve">ſe multiplient l’une par l’autre; </s> <s xml:id="echoid-s389" xml:space="preserve">de même b c d repréſente <lb/>un ſolide, dont les trois dimenſions ſont b, c, d, dont le <lb/>produit a donné ce ſolide.</s> <s xml:id="echoid-s390" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s391" xml:space="preserve">29. </s> <s xml:id="echoid-s392" xml:space="preserve">Comme dans une même propoſition on nomme tou-<lb/>jours les lignes égales par les mêmes lettres, & </s> <s xml:id="echoid-s393" xml:space="preserve">les lignes <lb/>inégales par des lettres différentes; </s> <s xml:id="echoid-s394" xml:space="preserve">dès que l’on verra a b, c d, <lb/>on jugera que ce ſont des rectangles, parce que leurs dimen-<lb/>ſions ſont inégales, au lieu que a a ſignifie un quarré, parce <lb/>que les deux dimenſions ſont égales.</s> <s xml:id="echoid-s395" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s396" xml:space="preserve">30. </s> <s xml:id="echoid-s397" xml:space="preserve">De même quand on verra a a a, l’on jugera que c’eſt <lb/>un cube, parce que les trois dimenſions ſont égales; </s> <s xml:id="echoid-s398" xml:space="preserve">& </s> <s xml:id="echoid-s399" xml:space="preserve">quand <lb/>on verra a b c, on jugera que c’eſt un parallelepipede, puiſ-<lb/>que ſes trois dimenſions ſont inégales.</s> <s xml:id="echoid-s400" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s401" xml:space="preserve">31. </s> <s xml:id="echoid-s402" xml:space="preserve">Les caracteres de l’alphabet ſont bien plus propres à <lb/>exprimer les grandeurs que les nombres; </s> <s xml:id="echoid-s403" xml:space="preserve">car quand je vois le <lb/>nombre 8, je ne ſçais s’il repréſente une ligne de huit pieds <lb/>courans, ou un plan de huit pieds quarrés, ou un ſolide de <lb/>huit pieds cubes; </s> <s xml:id="echoid-s404" xml:space="preserve">car un plan qui auroit quatre pieds de long <lb/>ſur deux pieds de large, auroit huit pour ſa ſuperficie; </s> <s xml:id="echoid-s405" xml:space="preserve">& </s> <s xml:id="echoid-s406" xml:space="preserve">un <lb/>ſolide qui auroit ſes dimenſions exprimées par une ligne <lb/> <pb o="8" file="0046" n="46" rhead="NOUVEAU COURS"/> de deux pieds, auroit auſſi huit pieds cubes pour ſa ſolidité <lb/>Ainſi dans les opérations que l’on fait avec les chiffres, il faut <lb/>que la mémoire ſoit aſſujettie à retenir ce qu’ils ſignifient, <lb/>au lieu que celles qui ſe font avec les lettres, ne la fatiguent <lb/>aucunement, puiſque la nature des grandeurs eſt repréſentée <lb/>par les lettres mêmes; </s> <s xml:id="echoid-s407" xml:space="preserve">car dés que je vois a a, b c d, j’ap-<lb/>perçois auſſitôt que a a eſt un quarré, & </s> <s xml:id="echoid-s408" xml:space="preserve">que b c d eſt un ſo-<lb/>lide; </s> <s xml:id="echoid-s409" xml:space="preserve">au lieu que ſi ces grandeurs étoient repréſentées par des <lb/>nombres, je ne ſçaurois ce qu’elles ſignifient.</s> <s xml:id="echoid-s410" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s411" xml:space="preserve">32. </s> <s xml:id="echoid-s412" xml:space="preserve">Comme on fait avec les lettres de l’alphabet les opéra-<lb/>tions que l’on fait avec les nombres, c’eſt-à-dire l’Addition, <lb/>la Souſtraction, la Multiplication, la Diviſion, & </s> <s xml:id="echoid-s413" xml:space="preserve">l’Extraction <lb/>des racines, & </s> <s xml:id="echoid-s414" xml:space="preserve">que de plus on opére ſur les quantités incon-<lb/>nues, de même que ſur les quantités connues (& </s> <s xml:id="echoid-s415" xml:space="preserve">c’eſt encore <lb/>un des grands avantages du calcul algébrique ſur le numéri-<lb/>que), on eſt convenu de repréſenter les quantités connues <lb/>par les premieres lettres de l’alphabet a, b, c, d, e, &</s> <s xml:id="echoid-s416" xml:space="preserve">c. </s> <s xml:id="echoid-s417" xml:space="preserve">& </s> <s xml:id="echoid-s418" xml:space="preserve">les <lb/>quantités inconnues par les dernieres u, x, y, z, afin de les <lb/>diſtinguer des premieres.</s> <s xml:id="echoid-s419" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s420" xml:space="preserve">33. </s> <s xml:id="echoid-s421" xml:space="preserve">L’on ſe ſert en Algebre de quelques ſignes pour indi-<lb/>quer les opérations que l’on fait ſur les lettres: </s> <s xml:id="echoid-s422" xml:space="preserve">par exemple, <lb/>ce ſigne + ſignifie plus, & </s> <s xml:id="echoid-s423" xml:space="preserve">déſigne l’addition de la quantité <lb/>qui le précéde à celle qui le ſuit. </s> <s xml:id="echoid-s424" xml:space="preserve">Ainſi a + b marque que la <lb/>grandeur b eſt ajoutée à la grandeur a; </s> <s xml:id="echoid-s425" xml:space="preserve">on ſe ſert même quel-<lb/>quefois de ces ſignes dans les calculs numériques, & </s> <s xml:id="echoid-s426" xml:space="preserve">il y a <lb/>des occaſions où il vaut mieux dire 5 + 3 que 8, quoique l’un <lb/>foit égal à l’autre.</s> <s xml:id="echoid-s427" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s428" xml:space="preserve">34. </s> <s xml:id="echoid-s429" xml:space="preserve">Ce ſigne - ſignifie moins, & </s> <s xml:id="echoid-s430" xml:space="preserve">déſigne la ſouſtraction <lb/>de la grandeur qui le ſuit de celle qui le précéde. </s> <s xml:id="echoid-s431" xml:space="preserve">a - b, <lb/>marque la différence de la grandeur a à la grandeur b.</s> <s xml:id="echoid-s432" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s433" xml:space="preserve">35. </s> <s xml:id="echoid-s434" xml:space="preserve">Si l’on veut marquer le produit d’une grandeur par <lb/>une autre, ou le faire en deux manieres, 1°. </s> <s xml:id="echoid-s435" xml:space="preserve">en mettant le <lb/>multiplicateur à côté du multiplicande, comme nous l’avons <lb/>déja dit, n°. </s> <s xml:id="echoid-s436" xml:space="preserve">28. </s> <s xml:id="echoid-s437" xml:space="preserve">Ainſi a b repréſente le produit de a par b, b c d <lb/>repréſente le produit des trois grandeurs b, c, d, les unes par <lb/>les autres. </s> <s xml:id="echoid-s438" xml:space="preserve">2°. </s> <s xml:id="echoid-s439" xml:space="preserve">On déſigne encore la multiplication de deux <lb/>ou de pluſieurs grandeurs, en mettant ce ſigne x entre deux, <lb/>ainſi a x b déſigne le produit de a par b, de même a x b x c <lb/>déſigne celui des trois grandeurs a b c, 2 x 3 x 4 déſigne celui <lb/>des trois nombres 2, 3, 4 qui vaut 24. </s> <s xml:id="echoid-s440" xml:space="preserve">Il eſt même quelque- <pb o="9" file="0047" n="47" rhead="DE MATHÉMATIQUE. Liv. I."/> fois à propos en Arithmétique de ſe contenter d’indiquer la <lb/>multiplication pour reconnoître plus aiſément les facteurs du <lb/>produit. </s> <s xml:id="echoid-s441" xml:space="preserve">On appelle facteurs les nombres ou quantités algébri-<lb/>ques, de la multiplication deſquels réſulte le produit dont il <lb/>s’agit.</s> <s xml:id="echoid-s442" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s443" xml:space="preserve">36. </s> <s xml:id="echoid-s444" xml:space="preserve">Quand on veut marquer qu’une grandeur eſt diviſée <lb/>par une autre, on met celle que l’on regarde comme divi-<lb/>dende au deſſus d’une petite barre horizontale, & </s> <s xml:id="echoid-s445" xml:space="preserve">celle que <lb/>l’on regarde comme diviſeur au deſſous de la même barre. <lb/></s> <s xml:id="echoid-s446" xml:space="preserve">Par exemple, {ab/c} déſigne que la grandeur a b eſt diviſée par <lb/>la quantité c; </s> <s xml:id="echoid-s447" xml:space="preserve">de même {bcd/gf} marque le quotient de b c d diviſé <lb/>par gf.</s> <s xml:id="echoid-s448" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s449" xml:space="preserve">37. </s> <s xml:id="echoid-s450" xml:space="preserve">Lorſqu’on verra ce ſigne = précédé d’une quantité, <lb/>& </s> <s xml:id="echoid-s451" xml:space="preserve">ſuivi d’une autre, cela voudra dire que ces quantités ſont <lb/>égales; </s> <s xml:id="echoid-s452" xml:space="preserve">c’eſt pourquoi on le nomme ſigne d’égalité: </s> <s xml:id="echoid-s453" xml:space="preserve">ainſi a b <lb/>= c d, ſignifie que le produit a b eſt égal au produit c d.</s> <s xml:id="echoid-s454" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s455" xml:space="preserve">38. </s> <s xml:id="echoid-s456" xml:space="preserve">Les deux quantités algébriques différentes, entre leſ-<lb/>quelles ſe trouve le ſigne d’égalité, ſont nommées enſemble <lb/>Equation; </s> <s xml:id="echoid-s457" xml:space="preserve">ainſi a = b, ay = bx, cd + xx = bb, y = {ab/c} <lb/>ſont des équations.</s> <s xml:id="echoid-s458" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s459" xml:space="preserve">L’on appelle membres de l’équation, les quantités qui ſe <lb/>trouvent de part & </s> <s xml:id="echoid-s460" xml:space="preserve">d’autre du ſigne d’égalité. </s> <s xml:id="echoid-s461" xml:space="preserve">Ainſi les quan-<lb/>tités a b c, d f x ſont les membres de l’équation a b c = d f x, <lb/>dont a b c eſt le premier membre, & </s> <s xml:id="echoid-s462" xml:space="preserve">d f x le ſecond.</s> <s xml:id="echoid-s463" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s464" xml:space="preserve">39. </s> <s xml:id="echoid-s465" xml:space="preserve">Si l’on a un produit qui réſulte de la multiplication <lb/>d’une même lettre pluſieurs fois par elle-même, comme a a a, <lb/>a a a b b b, on peut abréger cette expreſſion en écrivant cette <lb/>lettre une ſeule fois, & </s> <s xml:id="echoid-s466" xml:space="preserve">mettant un peu au deſſus, vers la <lb/>droite, un nombre qui marque combien de fois cette lettre <lb/>ſe multiplie par elle-même, ou, ce qui revient au même, com-<lb/>bien de fois on auroit dû l’écrire: </s> <s xml:id="echoid-s467" xml:space="preserve">ainſi au lieu de a a a on écrit <lb/>a<emph style="sub">3</emph>; </s> <s xml:id="echoid-s468" xml:space="preserve">au lieu de a a b b on écrit a<emph style="sub">2</emph>b<emph style="sub">2</emph>; </s> <s xml:id="echoid-s469" xml:space="preserve">au lieu de {aaabb/ccdd} on écrit <lb/>{a<emph style="sub">3</emph>b<emph style="sub">2</emph>/c<emph style="sub">2</emph>d<emph style="sub">2</emph>}. </s> <s xml:id="echoid-s470" xml:space="preserve">Ce nombre eſt appellé expoſant.</s> <s xml:id="echoid-s471" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s472" xml:space="preserve">40. </s> <s xml:id="echoid-s473" xml:space="preserve">Si un même produit doit être pris un certain nombre <lb/>de fois, on écrit au devant le nombre qui déſigne combien de <lb/>fois il le faut prendre. </s> <s xml:id="echoid-s474" xml:space="preserve">Ainſi 3ab marque que l’on prend trois <lb/>fois le produit a b, 5a<emph style="sub">3</emph> b<emph style="sub">2</emph> déſigne que l’on prend cinq fois la <pb o="10" file="0048" n="48" rhead="NOUVEAU COURS"/> grandeur a<emph style="sub">3</emph>b<emph style="sub">2</emph>. </s> <s xml:id="echoid-s475" xml:space="preserve">Ce nombre eſt appellé coefficient; </s> <s xml:id="echoid-s476" xml:space="preserve">il faut bien <lb/>ſe garder de le confondre avec celui que nous appellons expo-<lb/>ſant. </s> <s xml:id="echoid-s477" xml:space="preserve">b<emph style="sub">3</emph> eſt totalement différent de 3b, & </s> <s xml:id="echoid-s478" xml:space="preserve">ne peut jamais lui <lb/>être égal. </s> <s xml:id="echoid-s479" xml:space="preserve">Un exemple en nombre ſuffit pour en voir la diffé-<lb/>rence. </s> <s xml:id="echoid-s480" xml:space="preserve">Suppoſons que b = 5, on aura 3b = 3 x 5 = 15, & </s> <s xml:id="echoid-s481" xml:space="preserve"><lb/>b<emph style="sub">3</emph> = 5 x 5 x 5 = 125.</s> <s xml:id="echoid-s482" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s483" xml:space="preserve">41. </s> <s xml:id="echoid-s484" xml:space="preserve">On ſe ſert quelquefois des expoſans pour marquer le <lb/>quarré ou le cube d’une ligne déſignée dans une figure. </s> <s xml:id="echoid-s485" xml:space="preserve">A B<emph style="sub">2</emph> <lb/>marque le quarré de A B, A B<emph style="sub">3</emph> marque le cube de la même <lb/>ligne.</s> <s xml:id="echoid-s486" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s487" xml:space="preserve">42. </s> <s xml:id="echoid-s488" xml:space="preserve">Quand une quantité algébrique a été multipliée une <lb/>fois, deux fois, trois ou quatre fois par elle-même, &</s> <s xml:id="echoid-s489" xml:space="preserve">c, le pro-<lb/>duit qui en réſulte eſt appellé puiſſance ou degré; </s> <s xml:id="echoid-s490" xml:space="preserve">ainſi a ou <lb/>a<emph style="sub">1</emph> eſt nommé premiere puiſſance ou premier degré de la gran-<lb/>deur a; </s> <s xml:id="echoid-s491" xml:space="preserve">aa ou a<emph style="sub">2</emph> ſeconde puiſſance, ou ſecond degré, & </s> <s xml:id="echoid-s492" xml:space="preserve">ſou-<lb/>vent le quarré de a; </s> <s xml:id="echoid-s493" xml:space="preserve">de même aaa ou a<emph style="sub">3</emph> eſt le troiſieme degré, <lb/>la troiſieme puiſſance, & </s> <s xml:id="echoid-s494" xml:space="preserve">quelquefois le cube de a; </s> <s xml:id="echoid-s495" xml:space="preserve">enfin aaaa <lb/>ou a<emph style="sub">4</emph> le quatrieme degré, la quatrieme puiſſance de a, ou bien <lb/>le quarré-quarré de la même grandeur, puiſqu’il réſulte de <lb/>la multiplication du quarré a<emph style="sub">2</emph> par lui-même. </s> <s xml:id="echoid-s496" xml:space="preserve">Il en eſt ainſi des <lb/>autres.</s> <s xml:id="echoid-s497" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s498" xml:space="preserve">43. </s> <s xml:id="echoid-s499" xml:space="preserve">Une puiſſance peut être regardée comme le produit <lb/>d’une certaine puiſſance par une autre puiſſance; </s> <s xml:id="echoid-s500" xml:space="preserve">ainſi a<emph style="sub">5</emph> eſt <lb/>le produit de a<emph style="sub">3</emph> par a<emph style="sub">2</emph>, ou de la troiſieme puiſſance de a par <lb/>la ſeconde.</s> <s xml:id="echoid-s501" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s502" xml:space="preserve">44. </s> <s xml:id="echoid-s503" xml:space="preserve">Il peut auſſi y avoir des puiſſances faites du produit <lb/>de deux ou pluſieurs lettres multipliées l’une par l’autre; </s> <s xml:id="echoid-s504" xml:space="preserve">car <lb/>ſi l’on multiplie a b par lui-même une fois, le produit a a b b <lb/>ſera la ſeconde puiſſance de la quantité a b: </s> <s xml:id="echoid-s505" xml:space="preserve">de même a<emph style="sub">3</emph>b<emph style="sub">3</emph> eſt <lb/>le cube de la même grandeur.</s> <s xml:id="echoid-s506" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s507" xml:space="preserve">45. </s> <s xml:id="echoid-s508" xml:space="preserve">Le nombre ou la grandeur algébrique de la multipli-<lb/>cation, de laquelle réſulte une puiſſance, eſt appellé racine, <lb/>& </s> <s xml:id="echoid-s509" xml:space="preserve">il y a autant de racines que de puiſſances; </s> <s xml:id="echoid-s510" xml:space="preserve">ainſi a eſt la <lb/>racine quarrée de a<emph style="sub">2</emph>, la racine cube de a<emph style="sub">3</emph>, la racine cin-<lb/>quieme de a<emph style="sub">5</emph>, &</s> <s xml:id="echoid-s511" xml:space="preserve">c; </s> <s xml:id="echoid-s512" xml:space="preserve">de même ab<emph style="sub">2</emph> eſt la racine cube de a<emph style="sub">3</emph>b<emph style="sub">6</emph>; <lb/></s> <s xml:id="echoid-s513" xml:space="preserve">abc eſt la racine quatrieme de a<emph style="sub">4</emph>b<emph style="sub">4</emph>c<emph style="sub">4</emph>.</s> <s xml:id="echoid-s514" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s515" xml:space="preserve">46. </s> <s xml:id="echoid-s516" xml:space="preserve">Les quantités algébriques ſont appellées incomplexes <lb/>ou monomes, lorſqu’elles ne ſont pas jointes enſemble par les <pb o="11" file="0049" n="49" rhead="DE MATHÉMATIQUE. Liv. I."/> ſignes + & </s> <s xml:id="echoid-s517" xml:space="preserve">-; </s> <s xml:id="echoid-s518" xml:space="preserve">ainſi ab, cd, {bb/a}, {ff/g} ſont des quantités in-<lb/>complexes ou monomes. </s> <s xml:id="echoid-s519" xml:space="preserve">Monome ſignifie qui n’eſt compoſé <lb/>que d’un ſeul terme; </s> <s xml:id="echoid-s520" xml:space="preserve">au contraire lorſqu’elles ſont liées en-<lb/>ſemble par les ſignes + & </s> <s xml:id="echoid-s521" xml:space="preserve">-, on les appelle complexes ou <lb/>polynomes, c’eſt-à-dire qui ont pluſieurs termes. </s> <s xml:id="echoid-s522" xml:space="preserve">Ainſi b c + <lb/>a d, e f + g h, a a b - b c d, {aa + cc/a}, a b + c d - a c ſont <lb/>des quantités complexes ou polynomes. </s> <s xml:id="echoid-s523" xml:space="preserve">Si les quantités algébri-<lb/>ques n’ont que deux termes, on les appelle quelquefois bi-<lb/>nomes, & </s> <s xml:id="echoid-s524" xml:space="preserve">trinomes lorſqu’elles en ont trois; </s> <s xml:id="echoid-s525" xml:space="preserve">mais au delà elles <lb/>retiennent le nom général de polynomes; </s> <s xml:id="echoid-s526" xml:space="preserve">dans le dernier exem-<lb/>ple, a b, c d, a c ſont les termes de la quantité complexe a b <lb/>+ c d - a c.</s> <s xml:id="echoid-s527" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s528" xml:space="preserve">47. </s> <s xml:id="echoid-s529" xml:space="preserve">Lorſqu’une quantité algébrique n’eſt précédée d’aucun <lb/>ſigne, on ſuppoſe toujours qu’elle a le ſigne +, & </s> <s xml:id="echoid-s530" xml:space="preserve">alors on <lb/>l’appelle quantité poſitive, pour la diſtinguer de celles qui ſont <lb/>précédées du ſigne -, & </s> <s xml:id="echoid-s531" xml:space="preserve">ab que l’on appelle quantités néga-<lb/>tives: </s> <s xml:id="echoid-s532" xml:space="preserve">+ a b eſt la même choſe que a b, & </s> <s xml:id="echoid-s533" xml:space="preserve">eſt cenſé poſitif: <lb/></s> <s xml:id="echoid-s534" xml:space="preserve">- a c, - b c, ſont des quantités négatives.</s> <s xml:id="echoid-s535" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s536" xml:space="preserve">48. </s> <s xml:id="echoid-s537" xml:space="preserve">Lorſqu’une quantité n’a point de coefficient, ni d’expo-<lb/>ſant particulier, on lui ſuppoſe toujours l’unité pour coeffi-<lb/>cient & </s> <s xml:id="echoid-s538" xml:space="preserve">pour expoſant. </s> <s xml:id="echoid-s539" xml:space="preserve">Ainſi a b eſt la même choſe que 1a<emph style="sub">1</emph>b<emph style="sub">1</emph>, <lb/>a b c eſt le même que 1a<emph style="sub">1</emph>b<emph style="sub">1</emph>c<emph style="sub">1</emph>, & </s> <s xml:id="echoid-s540" xml:space="preserve">ainſi de toutes les autres.</s> <s xml:id="echoid-s541" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s542" xml:space="preserve">49. </s> <s xml:id="echoid-s543" xml:space="preserve">Lorſque des quantités incomplexes ou les termes d’une <lb/>quantité complexe contiennent préciſément les mêmes let-<lb/>tres, on les appelle des quantités ſemblables: </s> <s xml:id="echoid-s544" xml:space="preserve">ainſi 3ab & </s> <s xml:id="echoid-s545" xml:space="preserve"><lb/>2ab, 5ac & </s> <s xml:id="echoid-s546" xml:space="preserve">2ac ſont des quantités ſemblables. </s> <s xml:id="echoid-s547" xml:space="preserve">Il faut bien <lb/>remarquer que la ſimilitude des quantités algébriques ne dé-<lb/>pend ni des ſignes, ni des coefficiens, comme on le voit par <lb/>ces exemples, mais ſeulement des lettres & </s> <s xml:id="echoid-s548" xml:space="preserve">du nombre de <lb/>fois qu’elles ſont écrites. </s> <s xml:id="echoid-s549" xml:space="preserve">Pour reconnoître plus aiſément la <lb/>ſimilitude de pluſieurs termes, on obſervera dans les produits <lb/>de mettre les lettres dans leur ordre naturel ou alphabéti-<lb/>que; </s> <s xml:id="echoid-s550" xml:space="preserve">ainſi l’on écrira a b c, & </s> <s xml:id="echoid-s551" xml:space="preserve">non pas c a b, ni b c a.</s> <s xml:id="echoid-s552" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div58" type="section" level="1" n="49"> <head xml:id="echoid-head59" xml:space="preserve"><emph style="sc">Premiere</emph> <emph style="sc">Regle</emph></head> <head xml:id="echoid-head60" style="it" xml:space="preserve">Pour réduire les Quantités algébriques à leurs moindres termes.</head> <p> <s xml:id="echoid-s553" xml:space="preserve">50. </s> <s xml:id="echoid-s554" xml:space="preserve">Quand on a des quantités algébriques complexes, qui <lb/>renferment des termes ſemblables, il faut ajouter les coeffi- <pb o="12" file="0050" n="50" rhead="NOUVEAU COURS"/> ciens de ceux qui ont le même ſigne, & </s> <s xml:id="echoid-s555" xml:space="preserve">donner le même <lb/>ſigne à leur ſomme, afin de réduire la quantité propoſée; <lb/></s> <s xml:id="echoid-s556" xml:space="preserve">ainſi 4ab - 2ac + 2ab - 3ac ſe réduit à 6ab - 5ac, 28abd + <lb/>15acf + 8abd + 7acf = 36abd + 22acf.</s> <s xml:id="echoid-s557" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s558" xml:space="preserve">51. </s> <s xml:id="echoid-s559" xml:space="preserve">Quand les quantités ſemblables ont des ſignes diffé-<lb/>rens, il faut ſouſtraire le plus petit coefficient du plus grand, <lb/>& </s> <s xml:id="echoid-s560" xml:space="preserve">donner á la différence le ſigne du plus grand. </s> <s xml:id="echoid-s561" xml:space="preserve">Par exem-<lb/>ple, pour réduire cd + 6ab + 4aa - 4ab, on écrira cd + 4aa <lb/>+ 2ab en ôtant 4ab de 6ab; </s> <s xml:id="echoid-s562" xml:space="preserve">de même 2ab + 5cd + 3ab - 7cd <lb/>ſe réduit à 5ab - 2cd.</s> <s xml:id="echoid-s563" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s564" xml:space="preserve">52. </s> <s xml:id="echoid-s565" xml:space="preserve">Enfin lorſque deux termes ſont égaux, & </s> <s xml:id="echoid-s566" xml:space="preserve">qu’ils ont <lb/>des ſignes différens, ils ſe réduiſent à rien; </s> <s xml:id="echoid-s567" xml:space="preserve">ainſi a<emph style="sub">2</emph>b + 2cd <lb/>- a<emph style="sub">2</emph>b = 2cd, puiſque - a<emph style="sub">2</emph>b ſouſtrait de + a<emph style="sub">2</emph>b donne o pour <lb/>différence.</s> <s xml:id="echoid-s568" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div59" type="section" level="1" n="50"> <head xml:id="echoid-head61" xml:space="preserve"><emph style="sc">Seconde</emph> <emph style="sc">Regle</emph>.</head> <head xml:id="echoid-head62" style="it" xml:space="preserve"><emph style="sc">Addition</emph> des Quantités algébriques incomplexes & <lb/>complexes.</head> <p> <s xml:id="echoid-s569" xml:space="preserve">53. </s> <s xml:id="echoid-s570" xml:space="preserve">Pour ajouter enſemble des quantités algébriques, qui <lb/>ne ſont précédées d’aucuns ſignes, il faut les écrire de ſuite, <lb/>& </s> <s xml:id="echoid-s571" xml:space="preserve">les lier avec le ſigne +: </s> <s xml:id="echoid-s572" xml:space="preserve">ainſi pour ajouter les quantités <lb/>a b, a c, a d, on écrira a b + a c + a d; </s> <s xml:id="echoid-s573" xml:space="preserve">de même la ſomme <lb/>des quantités e f, g h, m n eſt égale à e f + g h + m n.</s> <s xml:id="echoid-s574" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s575" xml:space="preserve">54. </s> <s xml:id="echoid-s576" xml:space="preserve">Si les quantités que l’on veut ajouter ſont complexes, <lb/>on les écrira de ſuite avec leurs ſignes, & </s> <s xml:id="echoid-s577" xml:space="preserve">après avoir réduit <lb/>les termes ſemblables, on aura la ſomme de ces quantités. <lb/></s> <s xml:id="echoid-s578" xml:space="preserve">Par exemple, pour ajouter 2aab - 3acd avec acc + 5acd <lb/>- 6aab, on écrira 2aab - 3acd + acc + 5acd - 6aab, ce <lb/>qui ſe réduit à acc + 2acd - 4aab. </s> <s xml:id="echoid-s579" xml:space="preserve">Pour ajouter 6add + 5aac <lb/>- 4abb avec 2aac - 2abb, l’on écrira 6add + 5aac - 4abb <lb/>+ 2aac - 2abb qui ſe réduit à 6add - 6abb + 7aac. </s> <s xml:id="echoid-s580" xml:space="preserve">Enfin <lb/>pour ajouter abc - ddc - dcc avec dcc - abc + 3ddc, on écrira <lb/>abc - ddc - ddc + dcc - abc + 3ddc qui ſe réduit à 2ddc. </s> <s xml:id="echoid-s581" xml:space="preserve">En <lb/>général dans l’Addition algébrique, ſoit des monomes, ſoit <lb/>des polynomes, on écrit les quantités à la ſuite les unes des <lb/>autres avec leurs ſignes, & </s> <s xml:id="echoid-s582" xml:space="preserve">l’on fait aprés la réduction des <lb/>quantités ſemblables, s’il y en a.</s> <s xml:id="echoid-s583" xml:space="preserve"/> </p> <pb o="13" file="0051" n="51" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div60" type="section" level="1" n="51"> <head xml:id="echoid-head63" style="it" xml:space="preserve"><emph style="sc">Soustraction</emph> des Quantités algébriques incomplexes & <lb/>complexes.</head> <p> <s xml:id="echoid-s584" xml:space="preserve">55. </s> <s xml:id="echoid-s585" xml:space="preserve">Pour ſouſtraire une quantité algébrique d’une autre, <lb/>il faut l’écrire à la ſuite de celle dont on l’a ſouſtrait, en chan-<lb/>geant les ſignes de cette quantité, c’eſt-à-dire en mettant + <lb/>où il y a -, & </s> <s xml:id="echoid-s586" xml:space="preserve">- où il y a +: </s> <s xml:id="echoid-s587" xml:space="preserve">il faut enſuite faire la ré-<lb/>duction des quantités ſemblables, s’il y en a.</s> <s xml:id="echoid-s588" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s589" xml:space="preserve">Par exemple, pour ſouſtraire b b de a a, je l’écris à la ſuite <lb/>de a a avec le ſigne -, parce qu’il eſt cenſé avoir le ſigne +, <lb/>& </s> <s xml:id="echoid-s590" xml:space="preserve">la différence eſt a a - b b. </s> <s xml:id="echoid-s591" xml:space="preserve">De même pour ſouſtraire c + d <lb/>de a + b, il faut changer les ſignes de c + d, & </s> <s xml:id="echoid-s592" xml:space="preserve">écrire a + b <lb/>- c - d qui ſera la différence demandée. </s> <s xml:id="echoid-s593" xml:space="preserve">Pour ſouſtraire <lb/>b - d de a + c, on écrira a + c - b + d. </s> <s xml:id="echoid-s594" xml:space="preserve">Pour ſouſtraire <lb/>2bb - 3cc de aa + bb, on écrira aa + bb - 2bb + 3cc, & </s> <s xml:id="echoid-s595" xml:space="preserve"><lb/>réduiſant, on aura aa - bb + 3cc. </s> <s xml:id="echoid-s596" xml:space="preserve">Enfin pour ſouſtraire ab <lb/>- dc + bb - 3aa de aa - dc + 3bc - bb, on écrira aa - dc <lb/>+ 3bc - bb - ab + dc - bb + 3aa, ce qui donne, en rédui-<lb/>ſant, 4aa + 3bc - 2bb - ab, il en ſeroit de même des autres.</s> <s xml:id="echoid-s597" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div61" type="section" level="1" n="52"> <head xml:id="echoid-head64" style="it" xml:space="preserve">Eclairciſſement ſur la Souſtraction littérale.</head> <p> <s xml:id="echoid-s598" xml:space="preserve">Il n’eſt pas difficile de concevoir pourquoi on change le <lb/>ſigne +, exprimé ou ſous-entendu en -, car c’eſt en cela <lb/>préciſément que conſiſte la Souſtraction <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s599" xml:space="preserve">mais preſque <anchor type="note" xlink:label="note-0051-01a" xlink:href="note-0051-01"/> tous les Commençans ſont ſurpris de voir qu’il faut changer <lb/>les ſignes de - en +, cependant cela eſt facile à compren-<lb/>dre, ſi l’on fait attention que pour ôter b - d d’une quan-<lb/>tité quelconque a + c, il ne faut pas ôter b tout ſeul, puiſ-<lb/>que ce ſeroit trop ôter de toute la quantité d, b étant plus <lb/>grand que b - d de la même quantité d; </s> <s xml:id="echoid-s600" xml:space="preserve">donc puiſque l’on <lb/>auroit réellement ôté d, en écrivant - b, il faut le remettre <lb/>en écrivant + d.</s> <s xml:id="echoid-s601" xml:space="preserve"/> </p> <div xml:id="echoid-div61" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0051-01" xlink:href="note-0051-01a" xml:space="preserve">Art. 34.</note> </div> <p> <s xml:id="echoid-s602" xml:space="preserve">Mais comme on entendra mieux ceci par les nombres, <lb/>ſuppoſons qu’il faille retrancher du nombre 12 la quantité <lb/>6 - 2. </s> <s xml:id="echoid-s603" xml:space="preserve">Selon la regle, il faut écrire 12 - 6 + 2, dont la <lb/>différence eſt 8; </s> <s xml:id="echoid-s604" xml:space="preserve">car comme 6 - 2 eſt égale à 4, l’on voit <lb/>qu’on ne peut retrancher que 4 de 12, & </s> <s xml:id="echoid-s605" xml:space="preserve">que par conſéquent <lb/>ſi au lieu de 4 on en retranche 6, il faut rendre à 12 la quan- <pb o="14" file="0052" n="52" rhead="NOUVEAU COURS"/> tité 2 que l’on avoit ôté de trop. </s> <s xml:id="echoid-s606" xml:space="preserve">Enfin pour expliquer ceci <lb/>d’une autre façon, ſuppoſons deux perſonnes, dont l’une a <lb/>cent écus & </s> <s xml:id="echoid-s607" xml:space="preserve">ne doit rien, & </s> <s xml:id="echoid-s608" xml:space="preserve">l’autre au contraire n’a rien & </s> <s xml:id="echoid-s609" xml:space="preserve"><lb/>doit cent écus, il eſt certain que la premiere eſt plus riche <lb/>que la ſeconde de deux cens écus; </s> <s xml:id="echoid-s610" xml:space="preserve">par conſéquent ſi l’on <lb/>retranche - de +, la différence ſera +.</s> <s xml:id="echoid-s611" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div63" type="section" level="1" n="53"> <head xml:id="echoid-head65" style="it" xml:space="preserve"><emph style="sc">Multiplication</emph> des Quantités incomplexes.</head> <p> <s xml:id="echoid-s612" xml:space="preserve">56. </s> <s xml:id="echoid-s613" xml:space="preserve">Pour multiplier deux quantités quelconques incom-<lb/>plexes l’une par l’autre, il faut avoir égard aux ſignes, aux <lb/>coefficiens & </s> <s xml:id="echoid-s614" xml:space="preserve">aux lettres: </s> <s xml:id="echoid-s615" xml:space="preserve">ainſi la Multiplication renferme <lb/>trois parties.</s> <s xml:id="echoid-s616" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s617" xml:space="preserve">1°. </s> <s xml:id="echoid-s618" xml:space="preserve">Si le multiplicande & </s> <s xml:id="echoid-s619" xml:space="preserve">le multiplicateur ont le ſigne +, <lb/>on donnera le ſigne + au produit, & </s> <s xml:id="echoid-s620" xml:space="preserve">c’eſt ce que l’on expri-<lb/>me, en diſant que + par + donne +.</s> <s xml:id="echoid-s621" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s622" xml:space="preserve">Si le multiplicande a le ſigne +, & </s> <s xml:id="echoid-s623" xml:space="preserve">le multiplicateur le <lb/>ſigne -, le produit aura le ſigne -, & </s> <s xml:id="echoid-s624" xml:space="preserve">c’eſt ce que l’on ex-<lb/>prime, en diſant + par - donne -.</s> <s xml:id="echoid-s625" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s626" xml:space="preserve">Si le multiplicande a le ſigne -, & </s> <s xml:id="echoid-s627" xml:space="preserve">le multiplicateur le <lb/>ſigne +, le produit aura le ſigne -, ou bien - par + <lb/>donne -.</s> <s xml:id="echoid-s628" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s629" xml:space="preserve">Enfin ſi le multiplicande & </s> <s xml:id="echoid-s630" xml:space="preserve">le multiplicateur ont le ſigne <lb/>-, le produit aura le ſigne +, c’eſt-à-dire que - par -<lb/>donne +. </s> <s xml:id="echoid-s631" xml:space="preserve">Regle général, toutes les fois que le multiplicande <lb/>& </s> <s xml:id="echoid-s632" xml:space="preserve">le multiplicateur ont le même ſigne, le produit eſt poſitif <lb/>ou précédé du ſigne +, & </s> <s xml:id="echoid-s633" xml:space="preserve">il eſt négatif ou précédé du ſigne <lb/>- toutes les fois que le multiplicande & </s> <s xml:id="echoid-s634" xml:space="preserve">le multiplicateur <lb/>ſont des ſignes différens.</s> <s xml:id="echoid-s635" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s636" xml:space="preserve">2°. </s> <s xml:id="echoid-s637" xml:space="preserve">Si le multiplicande & </s> <s xml:id="echoid-s638" xml:space="preserve">le multiplicateur enſemble, ou <lb/>ſéparément, ont des coefficiens différens de l’unité, on les <lb/>multipliera l’un par l’autre, & </s> <s xml:id="echoid-s639" xml:space="preserve">le produit ſervira de coefficient <lb/>au produit que l’on cherche.</s> <s xml:id="echoid-s640" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s641" xml:space="preserve">3°. </s> <s xml:id="echoid-s642" xml:space="preserve">Enfin pour multiplier les lettres les unes par les autres, <lb/>on les poſera de ſuite les unes auprès des autres pour indi-<lb/>quer la multiplication des grandeurs qu’elles déſignent; </s> <s xml:id="echoid-s643" xml:space="preserve">car <lb/>on a vu (n°. </s> <s xml:id="echoid-s644" xml:space="preserve">35.) </s> <s xml:id="echoid-s645" xml:space="preserve">que cette maniere de les diſpoſer a été choiſie <lb/>pour la marque de la multiplication. </s> <s xml:id="echoid-s646" xml:space="preserve">Tout ceci deviendra ſen-<lb/>ſible par des exemples.</s> <s xml:id="echoid-s647" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s648" xml:space="preserve">Soit propoſé de multiplier la quantité 3ab ou + 3ab par <pb o="15" file="0053" n="53" rhead="DE MATHÉMATIQUE. Liv. I."/> 2ac ou + 2ac; </s> <s xml:id="echoid-s649" xml:space="preserve">je dis par le premier article de la Regle, + par <lb/>+ donne +: </s> <s xml:id="echoid-s650" xml:space="preserve">paſſant enſuite aux coefficiens, je dis, 3 fois 2, <lb/>font 6; </s> <s xml:id="echoid-s651" xml:space="preserve">& </s> <s xml:id="echoid-s652" xml:space="preserve">enfin aux lettres, ab par ac donne a<emph style="sub">2</emph>bc: </s> <s xml:id="echoid-s653" xml:space="preserve">on aura donc <lb/>au produit 6a<emph style="sub">2</emph>bc ou + 6a<emph style="sub">2</emph>bc. </s> <s xml:id="echoid-s654" xml:space="preserve">De même 4ac, multiplié par <lb/>- 5ab<emph style="sub">2</emph> donne - 20a<emph style="sub">2</emph>b<emph style="sub">2</emph>c; </s> <s xml:id="echoid-s655" xml:space="preserve">en diſant + par - donne -, <lb/>5 fois 4, font 20, ac par ab<emph style="sub">2</emph> donne a<emph style="sub">2</emph>b<emph style="sub">2</emph>c. </s> <s xml:id="echoid-s656" xml:space="preserve">De même - 6a<emph style="sub">3</emph>b<emph style="sub">2</emph>, <lb/>multiplié par 4a<emph style="sub">2</emph>bc<emph style="sub">2</emph>, donne - 24a<emph style="sub">5</emph>b<emph style="sub">3</emph>c<emph style="sub">2</emph>: </s> <s xml:id="echoid-s657" xml:space="preserve">enfin - 8abc par <lb/>- 5bcd, donne + 40ab<emph style="sub">2</emph>c<emph style="sub">2</emph>d.</s> <s xml:id="echoid-s658" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s659" xml:space="preserve">57. </s> <s xml:id="echoid-s660" xml:space="preserve">Pour multiplier deux ou pluſieurs quantités qui ont des <lb/>expoſans, & </s> <s xml:id="echoid-s661" xml:space="preserve">qui ſont compoſées des mêmes lettres, il faut <lb/>ajouter les expoſans des mêmes lettres, & </s> <s xml:id="echoid-s662" xml:space="preserve">leur ſomme ſera <lb/>les expoſans des lettres du produit: </s> <s xml:id="echoid-s663" xml:space="preserve">ainſi 3a<emph style="sub">2</emph>b<emph style="sub">3</emph> x 5a<emph style="sub">3</emph>b<emph style="sub">2</emph> = 15a<emph style="sub">5</emph>b<emph style="sub">5</emph>. <lb/></s> <s xml:id="echoid-s664" xml:space="preserve">De même a<emph style="sub">2</emph>b<emph style="sub">2</emph>c<emph style="sub">3</emph>, multiplié par ab<emph style="sub">3</emph>c<emph style="sub">2</emph>, donne a<emph style="sub">3</emph>b<emph style="sub">5</emph>c<emph style="sub">5</emph>; </s> <s xml:id="echoid-s665" xml:space="preserve">car il eſt <lb/>évident que a<emph style="sub">2</emph>b<emph style="sub">2</emph>c<emph style="sub">3</emph> = aabbccc, & </s> <s xml:id="echoid-s666" xml:space="preserve">ab<emph style="sub">3</emph>c<emph style="sub">2</emph> = abbbcc; </s> <s xml:id="echoid-s667" xml:space="preserve">donc le pro-<lb/>duit de ces quantités ſe trouvera, en plaçant toutes ces lettres <lb/>les unes auprès des autres, & </s> <s xml:id="echoid-s668" xml:space="preserve">ſera aaabbbbbccccc, ou a<emph style="sub">3</emph>b<emph style="sub">5</emph>c<emph style="sub">5</emph>, <lb/>en ſubſtituant les expoſans qui marquent combien de fois cha-<lb/>que lettre doit être écrite. </s> <s xml:id="echoid-s669" xml:space="preserve">Ceci eſt ſuffiſant pour la Multipli-<lb/>cation des quantités incomplexes.</s> <s xml:id="echoid-s670" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div64" type="section" level="1" n="54"> <head xml:id="echoid-head66" style="it" xml:space="preserve"><emph style="sc">Multiplication</emph> des Quantités complexes.</head> <p> <s xml:id="echoid-s671" xml:space="preserve">58. </s> <s xml:id="echoid-s672" xml:space="preserve">La Multiplication des quantités complexes ſe réduit à <lb/>celle des quantités incomplexes, en obſervant de faire au-<lb/>tant de multiplications particulieres qu’il y a de termes au <lb/>multiplicande & </s> <s xml:id="echoid-s673" xml:space="preserve">au multiplicateur, en ſuivant préciſément <lb/>les mêmes regles pour les ſignes, les coefficiens, & </s> <s xml:id="echoid-s674" xml:space="preserve">pour les <lb/>lettres. </s> <s xml:id="echoid-s675" xml:space="preserve">Si le multiplicateur n’a qu’un terme, il y aura autant <lb/>de multiplications particulieres par ce terme, qu’il y aura de <lb/>termes au multiplicande. </s> <s xml:id="echoid-s676" xml:space="preserve">Lorſqu’on aura trouvé tous les ter-<lb/>mes du produit, on obſervera d’en faire la réduction, s’il <lb/>s’en trouve de ſemblables: </s> <s xml:id="echoid-s677" xml:space="preserve">par exemple, pour multiplier 2a <lb/>+ b par 3c, l’on dira + par + donne +; </s> <s xml:id="echoid-s678" xml:space="preserve">2 fois 3 font 6, a par <lb/>c donne ac, le premier terme du produit ſera 6ac: </s> <s xml:id="echoid-s679" xml:space="preserve">de même <lb/>on dira + par + donne +, 3 fois 1 c’eſt 3, b par c donne <lb/>bc, & </s> <s xml:id="echoid-s680" xml:space="preserve">le ſecond terme du produit ſera bc; </s> <s xml:id="echoid-s681" xml:space="preserve">les ajoutant enſem-<lb/>ble, le produit total ſera 6ac + 3bc. </s> <s xml:id="echoid-s682" xml:space="preserve">Pour multiplier a - b <lb/>par d, l’on dira + par + donne +; </s> <s xml:id="echoid-s683" xml:space="preserve">1 par 1 donne 1, a par d <lb/>donne a d, & </s> <s xml:id="echoid-s684" xml:space="preserve">le premier terme ſera + 1ad, ou ſimplement <lb/>ad: </s> <s xml:id="echoid-s685" xml:space="preserve">paſſant au ſecond, on dira - par + donne -; </s> <s xml:id="echoid-s686" xml:space="preserve">1 par 1 <pb o="16" file="0054" n="54" rhead="NOUVEAU COURS"/> donne 1, b par d donne bd, & </s> <s xml:id="echoid-s687" xml:space="preserve">le ſecond terme ſera - 1bd, <lb/>ou ſimplement - bd; </s> <s xml:id="echoid-s688" xml:space="preserve">les ajoutant enſemble, on aura ab - bd <lb/>pour le produit total.</s> <s xml:id="echoid-s689" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s690" xml:space="preserve">Si le multiplicateur eſt auſſi complexe, ou compoſé de plu-<lb/>ſieurs termes, pour établir un certain ordre dans la maniere <lb/>de faire la multiplication, on met le multiplicande & </s> <s xml:id="echoid-s691" xml:space="preserve">le mul-<lb/>tiplicateur l’un au deſſous de l’autre, on multiplie tous les ter-<lb/>mes du multiplicande par tous les termes du multiplicateur; <lb/></s> <s xml:id="echoid-s692" xml:space="preserve">ce qui donne autant de produits particuliers qu’il y a de ter-<lb/>mes au multiplicateur, & </s> <s xml:id="echoid-s693" xml:space="preserve">dont chacun contient autant de <lb/>termes qu’il y en a au multiplicande. </s> <s xml:id="echoid-s694" xml:space="preserve">Ainſi pour multiplier <lb/>a + c par a + c, je mets une de ces quantités ſous l’autre, <lb/>& </s> <s xml:id="echoid-s695" xml:space="preserve">commençant à multiplier par la gauche, je dis a par a donne <lb/>aa, a par + c donne + ac; </s> <s xml:id="echoid-s696" xml:space="preserve">multipliant enſuite par le ſecond <lb/>terme c du multiplicateur, je dis + c par a donne + ac, & </s> <s xml:id="echoid-s697" xml:space="preserve"><lb/>+ c par + c donne + cc; </s> <s xml:id="echoid-s698" xml:space="preserve">additionnant le tout, le produit eſt <lb/>aa + ac + ac + cc; </s> <s xml:id="echoid-s699" xml:space="preserve">& </s> <s xml:id="echoid-s700" xml:space="preserve">pour abréger, au lieu d’écrire deux <lb/>fois la même quantité ac, je marque ſeulement 2ac <anchor type="note" xlink:href="" symbol="*"/>, ce qui <anchor type="note" xlink:label="note-0054-01a" xlink:href="note-0054-01"/> donne aa + 2ac + cc.</s> <s xml:id="echoid-s701" xml:space="preserve"/> </p> <div xml:id="echoid-div64" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0054-01" xlink:href="note-0054-01a" xml:space="preserve">Art. 50.</note> </div> <p> <s xml:id="echoid-s702" xml:space="preserve">59. </s> <s xml:id="echoid-s703" xml:space="preserve">Pour multiplier a - b par a - b, je poſe encore une <lb/>de ces quantités ſous l’autre, & </s> <s xml:id="echoid-s704" xml:space="preserve">je dis a par a donne aa, & </s> <s xml:id="echoid-s705" xml:space="preserve"><lb/>puis a par - b donne - ab (car on ſous-entend toujours que <lb/>a a le ſigne +). </s> <s xml:id="echoid-s706" xml:space="preserve">Multipliant enſuite par la ſeconde lettre <lb/>du multiplicateur, je dis - b par a donne - ab, & </s> <s xml:id="echoid-s707" xml:space="preserve">- b par <lb/>- b donne + bb; </s> <s xml:id="echoid-s708" xml:space="preserve">après avoir fait l’addition je trouve au pro-<lb/>duit aa - 2ab + bb. </s> <s xml:id="echoid-s709" xml:space="preserve">Tout ceci eſt évident par le premier ar-<lb/>ticle du n°. </s> <s xml:id="echoid-s710" xml:space="preserve">56; </s> <s xml:id="echoid-s711" xml:space="preserve">ce ſeroit toujours la même choſe pour des <lb/>opérations plus compliquées, comme on peut le voir dans les <lb/>exemples qui ſuivent.</s> <s xml:id="echoid-s712" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Multiplicande # 2a + b # a - b <lb/>Multiplicateur # 3c # d <lb/>Produit # 6ac + 3bc # ad - bd <lb/></note> <note position="right" xml:space="preserve"># # a + c # # a - b <lb/># # a + c # # a - b <lb/>Premier produit # aa + ac # 1<emph style="sub">er</emph> produit # aa - ab <lb/>Second produit # ac + cc # 2<emph style="sub">e</emph> produit # - ab + bb <lb/>Produit total. # aa + 2ac + cc # Prod. total. # aa - 2ab + bb. <lb/></note> <pb o="17" file="0055" n="55" rhead="DE MATHÉMATIQUE. Liv. I."/> <note position="right" xml:space="preserve">Multiplicande # aa + bb - ad - xx <lb/>Multiplicateur # aa + bc <lb/>Premier produit # a<emph style="sub">4</emph> + a<emph style="sub">2</emph>b<emph style="sub">2</emph> - a<emph style="sub">3</emph>d - a<emph style="sub">2</emph>x<emph style="sub">2</emph> <lb/>Second produit # + a<emph style="sub">2</emph>bc + b<emph style="sub">3</emph>c - abcd - bcxx <lb/>Prod. total # a<emph style="sub">4</emph> + a<emph style="sub">2</emph>b<emph style="sub">2</emph> + a<emph style="sub">2</emph>bc - a<emph style="sub">3</emph>d + b<emph style="sub">3</emph>c - a<emph style="sub">2</emph>x<emph style="sub">2</emph> - abcd - bcx<emph style="sub">2</emph> <lb/>Multiplicande # a<emph style="sub">3</emph> + a<emph style="sub">2</emph>b + ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>Multiplicateur # a - b <lb/>Premier produit # a<emph style="sub">4</emph> + a<emph style="sub">3</emph>b + a<emph style="sub">2</emph>b<emph style="sub">2</emph> + ab<emph style="sub">3</emph> \\ - a<emph style="sub">3</emph>b - a<emph style="sub">2</emph>b<emph style="sub">2</emph> - ab<emph style="sub">3</emph> - b<emph style="sub">4</emph> <lb/>Produit total # a<emph style="sub">4</emph> - b<emph style="sub">4</emph>. <lb/></note> <p> <s xml:id="echoid-s713" xml:space="preserve">Car il eſt viſible que tous les termes intermédiaires ſe détrui-<lb/>ſent par la réduction, puiſqu’ils ont des ſignes différens, & </s> <s xml:id="echoid-s714" xml:space="preserve"><lb/>qu’ils ſont ſemblables avec les mêmes coefficiens.</s> <s xml:id="echoid-s715" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div66" type="section" level="1" n="55"> <head xml:id="echoid-head67" xml:space="preserve"><emph style="sc">Démonstration des</emph> <emph style="sc">Regles</emph></head> <head xml:id="echoid-head68" style="it" xml:space="preserve">De la Multiplication des quantités complexes ou incomplexes <lb/>données au n°. 57.</head> <p> <s xml:id="echoid-s716" xml:space="preserve">Il n’eſt pas difficile de concevoir pourquoi + multiplié par <lb/>+ donne +; </s> <s xml:id="echoid-s717" xml:space="preserve">mais on n’apperçoit pas avec la même facilité <lb/>pourquoi + multiplié par -, ou - par + donne -, & </s> <s xml:id="echoid-s718" xml:space="preserve">l’on <lb/>conçoit encore moins comment - multiplié par - donne +; <lb/></s> <s xml:id="echoid-s719" xml:space="preserve">c’eſt pourquoi nous nous arrêterons principalement à expli-<lb/>quer ces derniers cas.</s> <s xml:id="echoid-s720" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s721" xml:space="preserve">La raiſon du premier cas eſt, que multipliant par exemple <lb/>a - b par d, l’on ne peut multiplier a par d ſans que le pro-<lb/>duit a d ne ſoit plus grand qu’il n’étoit, parce que a eſt <lb/>plus grand que a - b, & </s> <s xml:id="echoid-s722" xml:space="preserve">par conſéquent pour ôter ce qu’il y <lb/>a de trop dans le produit a d, il faut multiplier b par d, & </s> <s xml:id="echoid-s723" xml:space="preserve"><lb/>ôter le produit b d de a d pour avoir a d - b d; </s> <s xml:id="echoid-s724" xml:space="preserve">ce qui fait <lb/>voir que + par - doit donner -.</s> <s xml:id="echoid-s725" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s726" xml:space="preserve">Et pour le faire voir en nombres, multiplions 15 - 5 par <lb/>6: </s> <s xml:id="echoid-s727" xml:space="preserve">or comme 15 - 5 eſt égal à 10, c’eſt proprement 10 qu’il <lb/>faut multiplier par 6, & </s> <s xml:id="echoid-s728" xml:space="preserve">non pas 15 entiers, à moins que ſelon <lb/>la regle on ne multiplie auſſi 5 par 6 pour en ôter le produit <pb o="18" file="0056" n="56" rhead="NOUVEAU COURS"/> 30 de 90, produit de 15 par 6; </s> <s xml:id="echoid-s729" xml:space="preserve">ce qui donne 60, de même <lb/>qu’on l’auroit eu en multipliant 10 par 6.</s> <s xml:id="echoid-s730" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s731" xml:space="preserve">A l’égard du dernier cas, il paroît bien étrange que -<lb/>par - donne +; </s> <s xml:id="echoid-s732" xml:space="preserve">mais ce qui fait qu’on met +, c’eſt que <lb/>les deux termes, qui ſont précédés du ſigne -, donnant deux <lb/>multiplications négatives, par leſquelles on ôte plus qu’il ne <lb/>faut, l’on eſt obligé de mettre + au produit des deux termes <lb/>qui ont le ſigne -, pour remplacer ce que l’on avoit ôté de <lb/>trop. </s> <s xml:id="echoid-s733" xml:space="preserve">Par exemple, pour multiplier a - b par a - b, je vois, <lb/>aprés avoir fait la regle, que du produit aa il faut retrancher <lb/>- 2ab, & </s> <s xml:id="echoid-s734" xml:space="preserve">que retranchant plus qu’il ne faut de la quantité <lb/>bb, il faut rendre cette même quantité en la mettant avec <lb/>le ſigne +; </s> <s xml:id="echoid-s735" xml:space="preserve">ce qui remet toutes choſes dans l’état où elles <lb/>doivent être.</s> <s xml:id="echoid-s736" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s737" xml:space="preserve">Comme cette regle eſt abſolument indiſpenſable pour la <lb/>pratique des opérations algébriques, on ne ſçauroit trop ſe <lb/>convaincre de ſa vérité & </s> <s xml:id="echoid-s738" xml:space="preserve">de la certitude des principes ſur leſ-<lb/>quels elle eſt appuyée. </s> <s xml:id="echoid-s739" xml:space="preserve">Pour cela, il ſuffit de faire attention <lb/>à la nature de la multiplication. </s> <s xml:id="echoid-s740" xml:space="preserve">En général, multiplier un <lb/>nombre par un autre, c’eſt prendre le premier autant de fois <lb/>qu’il eſt marqué par l’autre, & </s> <s xml:id="echoid-s741" xml:space="preserve">de la même maniere qu’il eſt <lb/>marqué par l’autre. </s> <s xml:id="echoid-s742" xml:space="preserve">On ſçait que l’on appelle multiplicande <lb/>celui que l’on doit prendre pluſieurs fois, & </s> <s xml:id="echoid-s743" xml:space="preserve">multiplicateur <lb/>celui qui marque combien de fois on doit prendre le premier.</s> <s xml:id="echoid-s744" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s745" xml:space="preserve">Les unités du multiplicateur marquent combien de fois il <lb/>faut répéter le multiplicande, & </s> <s xml:id="echoid-s746" xml:space="preserve">le ſigne du même multi-<lb/>plicateur déſigne de quelle maniere il faut prendre le même <lb/>multiplicande. </s> <s xml:id="echoid-s747" xml:space="preserve">Si donc le multiplicateur a le ſigne +, la <lb/>multiplication ſe fait par addition, & </s> <s xml:id="echoid-s748" xml:space="preserve">ſi au contraire il a le <lb/>ſigne -, elle ſe fait par ſouſtraction, & </s> <s xml:id="echoid-s749" xml:space="preserve">le produit réſulte <lb/>d’une ſouſtraction répétée pluſieurs fois. </s> <s xml:id="echoid-s750" xml:space="preserve">Il faut encore con-<lb/>cevoir comment la multiplication ſe fait par ſouſtraction: <lb/></s> <s xml:id="echoid-s751" xml:space="preserve">pour cela on fera attention que les quantités négatives ne <lb/>ſont pas moins réelles que les quantités poſitives; </s> <s xml:id="echoid-s752" xml:space="preserve">mais elles <lb/>leurs ſont ſeulement oppoſées: </s> <s xml:id="echoid-s753" xml:space="preserve">on peut donc les multiplier <lb/>comme les autres. </s> <s xml:id="echoid-s754" xml:space="preserve">Ainſi ſi l’on regarde le bien que l’on poſſede <lb/>comme quelque choſe de poſitif, les dettes que l’on fait, ſeront <lb/>des grandeurs négatives, & </s> <s xml:id="echoid-s755" xml:space="preserve">l’on ſçait aſſez par expérience <lb/>qu’elles peuvent ſe multiplier, ainſi que les biens, quoique <lb/>bien plus facilement. </s> <s xml:id="echoid-s756" xml:space="preserve">Un homme qui accumule ſes dettes, <pb o="19" file="0057" n="57" rhead="DE MATHÉMATIQUE. Liv. I."/> multiplie par moins, & </s> <s xml:id="echoid-s757" xml:space="preserve">c’eſt ainſi qu’il faut entendre toutes <lb/>ces expreſſions. </s> <s xml:id="echoid-s758" xml:space="preserve">Tout cela poſé, + a x - b doit donner - ab; <lb/></s> <s xml:id="echoid-s759" xml:space="preserve">car le multiplicande ayant le ſigne +, & </s> <s xml:id="echoid-s760" xml:space="preserve">le multiplicateur le <lb/>ſigne -, indique qu’il faut ſouſtraire a autant de fois qu’il <lb/>eſt marqué par b. </s> <s xml:id="echoid-s761" xml:space="preserve">De même - a x + b doit donner - ab; </s> <s xml:id="echoid-s762" xml:space="preserve"><lb/>car le multiplicateur b étant poſitif, indique qu’il faut répéter <lb/>pluſieurs fois la quantité négative - a. </s> <s xml:id="echoid-s763" xml:space="preserve">Le réſultat de toutes <lb/>ces quantités négatives égales ne pourra jamais donner que <lb/>du négatif: </s> <s xml:id="echoid-s764" xml:space="preserve">ainſi - a x + b donne - ab: </s> <s xml:id="echoid-s765" xml:space="preserve">enfin - a x - b <lb/>doit donner + ab; </s> <s xml:id="echoid-s766" xml:space="preserve">car le multiplicande ayant le ſigne - eſt <lb/>négatif, & </s> <s xml:id="echoid-s767" xml:space="preserve">le multiplicateur ayant auſſi le même ſigne, fait <lb/>voir que la multiplication ſe fait par ſouſtraction, c’eſt-à-dire <lb/>qu’il faut ſouſtraire la quantité négative - a autant de fois qu’il <lb/>eſt marqué par les unités de b, & </s> <s xml:id="echoid-s768" xml:space="preserve">par conſéquent c’eſt mettre <lb/>a autant de fois poſitif, par la même raiſon que pour ſouſ-<lb/>traire une quantité négative une fois, il faut la mettre une <lb/>fois poſitive. </s> <s xml:id="echoid-s769" xml:space="preserve">Enfin cette derniere partie de la regle des ſignes <lb/>répond parfaitement à ce que l’on dit ordinairement d’un <lb/>homme qui acquitte ſes dettes.</s> <s xml:id="echoid-s770" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s771" xml:space="preserve">Les deux dernieres parties de la regle n’ont pas beſoin de <lb/>démonſtration; </s> <s xml:id="echoid-s772" xml:space="preserve">car il eſt évident que puiſque les coefficiens <lb/>ſont des nombres, ils doivent ſe multiplier comme des nom-<lb/>bres, & </s> <s xml:id="echoid-s773" xml:space="preserve">la maniere dont on indique la multiplication des let-<lb/>tres eſt de pure convention: </s> <s xml:id="echoid-s774" xml:space="preserve">ainſi elle ne peut être conteſtée.</s> <s xml:id="echoid-s775" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div67" type="section" level="1" n="56"> <head xml:id="echoid-head69" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s776" xml:space="preserve">Pour donner une idée de la facilité que l’on a de démon-<lb/>trer les propoſitions de Géométrie par le moyen du calcul <lb/>algébrique, j’ai cru qu’il étoit à propos, avant d’aller plus <lb/>loin, de faire une application de la multiplication à la dé-<lb/>monſtration des propoſitions ſuivantes.</s> <s xml:id="echoid-s777" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div68" type="section" level="1" n="57"> <head xml:id="echoid-head70" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s778" xml:space="preserve">60. </s> <s xml:id="echoid-s779" xml:space="preserve">Le quarré d’une grandeur quelconque, exprimée par deux <lb/>lettres poſitives, eſt égale au quarré de chacune de ces lettres, plus <lb/>à deux rectangles compris ſous les mêmes lettres.</s> <s xml:id="echoid-s780" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s781" xml:space="preserve">Car ſi l’on multiplie a + b par a + b, l’on aura au produit <pb o="20" file="0058" n="58" rhead="NOUVEAU COURS"/> aa + 2ab + bb, qui eſt compoſé des quarrés a a & </s> <s xml:id="echoid-s782" xml:space="preserve">b b, & </s> <s xml:id="echoid-s783" xml:space="preserve"><lb/>de deux rectangles compris ſous les mêmes lettres a & </s> <s xml:id="echoid-s784" xml:space="preserve">b, qui <lb/>ſont 2ab.</s> <s xml:id="echoid-s785" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div69" type="section" level="1" n="58"> <head xml:id="echoid-head71" xml:space="preserve">PROPOSITION II <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s786" xml:space="preserve">61. </s> <s xml:id="echoid-s787" xml:space="preserve">Le cube d’une grandeur quelconque exprimée par deux let-<lb/>tres, eſt égal au cube de la premiere, plus au cube de la ſeconde, <lb/>plus à trois parallelepipedes du quarré de la premiere par la ſe-<lb/>conde, plus enfin à trois autres parallelepipedes du quarré de la <lb/>ſeconde par la premiere.</s> <s xml:id="echoid-s788" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s789" xml:space="preserve">Car le quarré de a + b étant (n°. </s> <s xml:id="echoid-s790" xml:space="preserve">60.) </s> <s xml:id="echoid-s791" xml:space="preserve">aa + 2ab + bb, <lb/>ſi on le multiplie encore par a + b, l’on aura le cube a<emph style="sub">3</emph> + 3a<emph style="sub">2</emph>b <lb/>+ 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>, qui renferme a<emph style="sub">3</emph> & </s> <s xml:id="echoid-s792" xml:space="preserve">b<emph style="sub">3</emph>, cubes des deux lettres a <lb/>& </s> <s xml:id="echoid-s793" xml:space="preserve">b, plus trois parallelepipedes 3a<emph style="sub">2</emph>b du quarré aa par b; </s> <s xml:id="echoid-s794" xml:space="preserve">plus <lb/>enfin trois autres parallelepipedes du quarré bb par a, 3abb.</s> <s xml:id="echoid-s795" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s796" xml:space="preserve">Nous nous ſervirons de ceci dans la ſuite pour démontrer <lb/>les opérations de la racine quarrée & </s> <s xml:id="echoid-s797" xml:space="preserve">cubique.</s> <s xml:id="echoid-s798" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Racine # a + b <lb/>par # a + b <lb/># aa + ab <lb/># ab + bb <lb/>Quarré # aa + 2ab + bb <lb/></note> <note position="right" xml:space="preserve">Quarré # aa + 2ab + bb <lb/>par # a + b <lb/># a<emph style="sub">3</emph> + 2a<emph style="sub">2</emph>b + ab<emph style="sub">2</emph> <lb/># + a<emph style="sub">2</emph>b + 2ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>Cube # a<emph style="sub">3</emph> + 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/></note> </div> <div xml:id="echoid-div70" type="section" level="1" n="59"> <head xml:id="echoid-head72" xml:space="preserve">PROPOSITION II <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s799" xml:space="preserve">62. </s> <s xml:id="echoid-s800" xml:space="preserve">Si l’on a une ligne A B diviſée en deux également au <lb/> <anchor type="note" xlink:label="note-0058-03a" xlink:href="note-0058-03"/> point C, & </s> <s xml:id="echoid-s801" xml:space="preserve">en deux inégalement au point D, je dis que le rec-<lb/>tangle A D x D B, compris ſous les parties inégales A D & </s> <s xml:id="echoid-s802" xml:space="preserve"><lb/>D B, plus le quarré de la moyenne partie C D, eſt égal au quarré <lb/>de la moitié de la ligne, c’eſt-à-dire à <emph style="ol">A C</emph><emph style="sub">2</emph> ou <emph style="ol">C B</emph><emph style="sub">2</emph>.</s> <s xml:id="echoid-s803" xml:space="preserve"/> </p> <div xml:id="echoid-div70" type="float" level="2" n="1"> <note position="left" xlink:label="note-0058-03" xlink:href="note-0058-03a" xml:space="preserve">Figure 7.</note> </div> <p> <s xml:id="echoid-s804" xml:space="preserve">Nous nommerons A C ou C B a, C D x, ainſi D B ſera <lb/>a - x, & </s> <s xml:id="echoid-s805" xml:space="preserve">A D a + x.</s> <s xml:id="echoid-s806" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div72" type="section" level="1" n="60"> <head xml:id="echoid-head73" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s807" xml:space="preserve">Si l’on ajoute à A D x D B (aa - xx) le quarré de C D <pb o="21" file="0059" n="59" rhead="DE MATHÉMATIQUE. Liv. I."/> (xx), l’on pourra former cette équation A D x D B + C D<emph style="sub">2</emph> <lb/>(aa - xx + xx) = A C<emph style="sub">2</emph> (aa), puiſqu’en effaçant ce qui <lb/>ſe détruit dans le premier membre, on auroit aa = aa; </s> <s xml:id="echoid-s808" xml:space="preserve">ce <lb/>qu’il falloit démontrer.</s> <s xml:id="echoid-s809" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div73" type="section" level="1" n="61"> <head xml:id="echoid-head74" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s810" xml:space="preserve">63. </s> <s xml:id="echoid-s811" xml:space="preserve">Il ſuit de cette propoſition, que ſi une ligne eſt coupée <lb/>en deux également en C, & </s> <s xml:id="echoid-s812" xml:space="preserve">en deux inégalement en D, le <lb/>quarré A C<emph style="sub">2</emph> de la moitié de la ligne, moins le quarré C D<emph style="sub">2</emph> <lb/>de la moyenne partie C D, eſt égal au rectangle A D x D B, <lb/>compris ſous les parties inégales A D, D B; </s> <s xml:id="echoid-s813" xml:space="preserve">ce qui eſt évident, <lb/>puiſque A C<emph style="sub">2</emph> - C D<emph style="sub">2</emph> (aa - xx) = A D x D B (aa - xx).</s> <s xml:id="echoid-s814" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div74" type="section" level="1" n="62"> <head xml:id="echoid-head75" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s815" xml:space="preserve">64. </s> <s xml:id="echoid-s816" xml:space="preserve">Si l’on a une ligne droite A B diviſée en deux également <lb/> <anchor type="note" xlink:label="note-0059-01a" xlink:href="note-0059-01"/> en C, & </s> <s xml:id="echoid-s817" xml:space="preserve">qu’on lui ajoute une droite B E, je dis que le rectangle <lb/>de la droite A E, ſomme de ces deux lignes par la droite B E que <lb/>l’on a ajoutée, avec le quarré de la moyenne C B, ſera égal au quarré <lb/>de la ligne C E, compoſée de la moitié C B, & </s> <s xml:id="echoid-s818" xml:space="preserve">de l’ ajoutée B E.</s> <s xml:id="echoid-s819" xml:space="preserve"/> </p> <div xml:id="echoid-div74" type="float" level="2" n="1"> <note position="right" xlink:label="note-0059-01" xlink:href="note-0059-01a" xml:space="preserve">Figure 7.</note> </div> <p> <s xml:id="echoid-s820" xml:space="preserve">Nous nommerons A C ou C B a, C E x, ainſi B E ſera <lb/>x - a, & </s> <s xml:id="echoid-s821" xml:space="preserve">A E x + a.</s> <s xml:id="echoid-s822" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div76" type="section" level="1" n="63"> <head xml:id="echoid-head76" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s823" xml:space="preserve">Il eſt évident que ſi l’on ajoute au rectangle de A E x B E <lb/>(xx - aa) le quarré de C B (aa), l’on pourra former cette <lb/>équation A E x B E + <emph style="ol">C B</emph><emph style="sub">2</emph> (xx - aa + aa) = C E<emph style="sub">2</emph> (xx), <lb/>puiſqu’en effaçant tout ce qui ſe détruit, il vient xx = xx; <lb/></s> <s xml:id="echoid-s824" xml:space="preserve">C. </s> <s xml:id="echoid-s825" xml:space="preserve">Q. </s> <s xml:id="echoid-s826" xml:space="preserve">F. </s> <s xml:id="echoid-s827" xml:space="preserve">D.</s> <s xml:id="echoid-s828" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div77" type="section" level="1" n="64"> <head xml:id="echoid-head77" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s829" xml:space="preserve">65. </s> <s xml:id="echoid-s830" xml:space="preserve">Il ſuit de cette propoſition, que ſi à une ligne diviſée <lb/>en deux également l’on en ajoute une autre, le quarré de la <lb/>ligne C E, compoſé de la moitié de la ligne & </s> <s xml:id="echoid-s831" xml:space="preserve">de l’ajoutée, <lb/>moins le quarré de la moyenne C B, ſera égal au rectangle <lb/>compris ſous toute la ligne A E, & </s> <s xml:id="echoid-s832" xml:space="preserve">la partie ajoutée B E; </s> <s xml:id="echoid-s833" xml:space="preserve">ce <lb/>qui eſt évident, puiſque C E<emph style="sub">2</emph> - C B<emph style="sub">2</emph> = A E x B E (xx - aa).</s> <s xml:id="echoid-s834" xml:space="preserve"/> </p> <pb o="22" file="0060" n="60" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div78" type="section" level="1" n="65"> <head xml:id="echoid-head78" xml:space="preserve">PROPOSITION V.</head> <p style="it"> <s xml:id="echoid-s835" xml:space="preserve">66. </s> <s xml:id="echoid-s836" xml:space="preserve">Si l’on a deux lignes, dont la premiere ſoit double de la <lb/>ſeconde, je dis que le quarré de la premiere ſera quadruple du quarré <lb/>de la ſeconde.</s> <s xml:id="echoid-s837" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div79" type="section" level="1" n="66"> <head xml:id="echoid-head79" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s838" xml:space="preserve">Si de ces deux lignes la ſeconde ſe nomme a, la premiere <lb/>ſera 2a: </s> <s xml:id="echoid-s839" xml:space="preserve">or multipliant 2a par 2a, l’on aura 4aa pour le quarré <lb/>de la premiere; </s> <s xml:id="echoid-s840" xml:space="preserve">& </s> <s xml:id="echoid-s841" xml:space="preserve">ſi l’on multiplie a par lui-même, l’on aura <lb/>aa pour le quarré de la ſeconde, & </s> <s xml:id="echoid-s842" xml:space="preserve">par conſéquent le quarré <lb/>de la premiere eſt quadruple du quarré de la ſeconde.</s> <s xml:id="echoid-s843" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div80" type="section" level="1" n="67"> <head xml:id="echoid-head80" style="it" xml:space="preserve">De la Diviſion des Quantités algébriques incomplexes & <lb/>complexes.</head> <p> <s xml:id="echoid-s844" xml:space="preserve">67. </s> <s xml:id="echoid-s845" xml:space="preserve">Pour diviſer une quantité algébrique par une autre, <lb/>on met celle que l’on doit diviſer au deſſus d’une barre ho-<lb/>rizontale, & </s> <s xml:id="echoid-s846" xml:space="preserve">celle par laquelle on diviſe au deſſous de la même <lb/>barre (n°. </s> <s xml:id="echoid-s847" xml:space="preserve">38.)</s> <s xml:id="echoid-s848" xml:space="preserve">, en obſervant d’effacer les lettres communes <lb/>au dividende & </s> <s xml:id="echoid-s849" xml:space="preserve">au diviſeur, s’il y en a quelques-unes, & </s> <s xml:id="echoid-s850" xml:space="preserve">ce <lb/>qui reſte marque le quotient. </s> <s xml:id="echoid-s851" xml:space="preserve">Ainſi pour diviſer a par b, j’écris <lb/>{a/b}, ce qui ſigniſie a diviſé par b; </s> <s xml:id="echoid-s852" xml:space="preserve">pour diviſer a b c par fg, j’é-<lb/>cris {abc/fg}; </s> <s xml:id="echoid-s853" xml:space="preserve">pour diviſer ab<emph style="sub">2</emph>c<emph style="sub">3</emph> par abc<emph style="sub">2</emph>, ou abbccc par abcc, j’écris <lb/>{aabbccc/abcc}, ce qui ſe réduit à abc, en effaçant les lettres com-<lb/>munes au dividende & </s> <s xml:id="echoid-s854" xml:space="preserve">au diviſeur. </s> <s xml:id="echoid-s855" xml:space="preserve">Si l’on multiplie le quo-<lb/>tient abc par le diviſeur abcc, l’on aura a<emph style="sub">2</emph>b<emph style="sub">2</emph>c<emph style="sub">3</emph>; </s> <s xml:id="echoid-s856" xml:space="preserve">ce qui prouve <lb/>que la Diviſion eſt bien faite, puiſque le produit du diviſeur <lb/>par le quotient eſt égal au dividende.</s> <s xml:id="echoid-s857" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s858" xml:space="preserve">68. </s> <s xml:id="echoid-s859" xml:space="preserve">Si le dividende & </s> <s xml:id="echoid-s860" xml:space="preserve">le diviſeur ſont chacun précédés de <lb/>coefficiens, il faudra les diviſer l’un par l’autre, ſelon les regles <lb/>de la diviſion des nombres, & </s> <s xml:id="echoid-s861" xml:space="preserve">le quotient ſera le coefficient <lb/>du quotient. </s> <s xml:id="echoid-s862" xml:space="preserve">Ainſi 21ab<emph style="sub">2</emph> diviſé par 7ab = 3b; </s> <s xml:id="echoid-s863" xml:space="preserve">{28abc<emph style="sub">3</emph>/4a<emph style="sub">2</emph>bc} = {7c<emph style="sub">2</emph>/a}; <lb/></s> <s xml:id="echoid-s864" xml:space="preserve">{36a<emph style="sub">2</emph>b<emph style="sub">4</emph>/9a<emph style="sub">3</emph>bc<emph style="sub">2</emph>} = {4b<emph style="sub">3</emph>/ac<emph style="sub">2</emph>}. </s> <s xml:id="echoid-s865" xml:space="preserve">L’on peut remarquer que lorſque le dividende <lb/>& </s> <s xml:id="echoid-s866" xml:space="preserve">le diviſeur ont chacun des lettres ſemblables avec des ex-<lb/>poſans, la diviſion de ces lettres ſe fait par la ſouſtraction des <lb/>expoſans: </s> <s xml:id="echoid-s867" xml:space="preserve">ainſi {a<emph style="sub">3</emph>/a<emph style="sub">2</emph>} = a = a<emph style="sub">3-2</emph>{a<emph style="sub">5</emph>b<emph style="sub">4</emph>/a<emph style="sub">2</emph>b<emph style="sub">3</emph>} = a<emph style="sub">3</emph>b = a<emph style="sub">5 - 2</emph> b<emph style="sub">4 - 3</emph>, <pb o="23" file="0061" n="61" rhead="DE MATHÉMATIQUE. Liv. I."/> {36ac<emph style="sub">2</emph>f<emph style="sub">3</emph>/4a<emph style="sub">3</emph>cf<emph style="sub">2</emph>} = {9c<emph style="sub">2 - 1</emph>f<emph style="sub">3 - 2</emph>/a<emph style="sub">3 - 1</emph>} = {9cf/a<emph style="sub">2</emph>}, & </s> <s xml:id="echoid-s868" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s869" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s870" xml:space="preserve">69. </s> <s xml:id="echoid-s871" xml:space="preserve">A l’égard des ſignes, ſi le dividende & </s> <s xml:id="echoid-s872" xml:space="preserve">le diviſeur ont <lb/>chacun le même ſigne + ou -, il faut que le quotient ait le <lb/>ſigne +: </s> <s xml:id="echoid-s873" xml:space="preserve">la raiſon en eſt, qu’une quantité négative eſt con-<lb/>tenue dans une quantité négative, de la même maniere qu’une <lb/>quantité poſitive eſt contenue dans une quantité poſitive. </s> <s xml:id="echoid-s874" xml:space="preserve">Mais <lb/>s’ils avoient différens ſignes, le quotient auroit le ſigne -, <lb/>parce que les quantités poſitives & </s> <s xml:id="echoid-s875" xml:space="preserve">négatives étant des quan-<lb/>tités oppoſées les unes aux autres, ſe contiennent négative-<lb/>ment, & </s> <s xml:id="echoid-s876" xml:space="preserve">par conſéquent le quotient doit avoir le ſigne -. <lb/></s> <s xml:id="echoid-s877" xml:space="preserve">Par exemple, + a<emph style="sub">2</emph> b diviſé par + a = + ab; </s> <s xml:id="echoid-s878" xml:space="preserve">de même <lb/>- ab diviſé par - b donne + a; </s> <s xml:id="echoid-s879" xml:space="preserve">ce qui ſe peut encore dé-<lb/>montrer par la preuve de la Diviſion, par laquelle le pro-<lb/>duit du diviſeur par le quotient doit redonner le dividende. </s> <s xml:id="echoid-s880" xml:space="preserve"><lb/>Multipliant donc le quotient + a par le diviſeur - b, on <lb/>aura - ab, puiſque - par + donne - (n°. </s> <s xml:id="echoid-s881" xml:space="preserve">57). </s> <s xml:id="echoid-s882" xml:space="preserve">Si l’on diviſe <lb/>+ ab par - a, le quotient ſera - b; </s> <s xml:id="echoid-s883" xml:space="preserve">car multipliant le quo-<lb/>tient - b par le diviſeur - a, on aura + ab, puiſque - par <lb/>- donne + (n°. </s> <s xml:id="echoid-s884" xml:space="preserve">57). </s> <s xml:id="echoid-s885" xml:space="preserve">Enfin ſi l’on diviſe - ab par + a, le <lb/>quotient ſera - b; </s> <s xml:id="echoid-s886" xml:space="preserve">car multipliant le quotient - b par le divi-<lb/>ſeur + a, on aura - ab, puiſque - par + donne -.</s> <s xml:id="echoid-s887" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s888" xml:space="preserve">70. </s> <s xml:id="echoid-s889" xml:space="preserve">Si le dividende eſt complexe, & </s> <s xml:id="echoid-s890" xml:space="preserve">le diviſeur toujours <lb/>incomplexe, on fera ſur chaque terme les mêmes opérations <lb/>que nous venons d’expliquer, & </s> <s xml:id="echoid-s891" xml:space="preserve">la ſomme des quotiens par-<lb/>ticuliers ſera le quotient total. </s> <s xml:id="echoid-s892" xml:space="preserve">Ainſi pour diviſer ab + ad par <lb/>a, je dis ab diviſé par a donne b, que j’écris au quotient. </s> <s xml:id="echoid-s893" xml:space="preserve">Je <lb/>dis enſuite ab diviſé par a donne d au quotient, qui étant <lb/>ajouté au premier b, donne pour le quotient total b + d; </s> <s xml:id="echoid-s894" xml:space="preserve">ce <lb/>qui eſt encore évident, puiſqu’en multipliant le quotient b + d <lb/>par le diviſeur a, on aura ab + ad égal au dividende.</s> <s xml:id="echoid-s895" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s896" xml:space="preserve">71. </s> <s xml:id="echoid-s897" xml:space="preserve">Quand le dividende & </s> <s xml:id="echoid-s898" xml:space="preserve">le diviſeur ſont chacun des <lb/>quantités algébriques complexes, on ſuit à peu près le même <lb/>procédé que dans la diviſion des nombres. </s> <s xml:id="echoid-s899" xml:space="preserve">Par exemple, pour <lb/>diviſer aa + 2ab + bb par a + b, je poſe les premiers termes <lb/>du diviſeur ſous les premiers termes du dividende, & </s> <s xml:id="echoid-s900" xml:space="preserve">je com-<lb/>mence par chercher combien de fois le premier terme a du <lb/>diviſeur eſt contenu dans le premier terme a<emph style="sub">2</emph> du dividende, <lb/>en diſant, en a<emph style="sub">2</emph> combien de fois a, ou a<emph style="sub">2</emph> diviſé par a donne a <lb/>au quotient: </s> <s xml:id="echoid-s901" xml:space="preserve">je multiplie le diviſeur entier a + b par a, &</s> <s xml:id="echoid-s902" xml:space="preserve"> <pb o="24" file="0062" n="62" rhead="NOUVEAU COURS"/> je retranche le produit aa + ab du dividende <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s903" xml:space="preserve">ce que je fais <anchor type="note" xlink:label="note-0062-01a" xlink:href="note-0062-01"/> en l’écrivant à la ſuite de cette même quantité avec des ſignes <lb/>contraires, & </s> <s xml:id="echoid-s904" xml:space="preserve">j’ai aa + 2ab + bb - aa - ab; </s> <s xml:id="echoid-s905" xml:space="preserve">ce qui ſe <lb/>réduit à ab + bb. </s> <s xml:id="echoid-s906" xml:space="preserve">Je fais ſur le reſte la même opération, en <lb/>diſant ab diviſé par a, donne b au quotient, que je mets à <lb/>côté du premier terme que j’ai déja trouvé: </s> <s xml:id="echoid-s907" xml:space="preserve">je multiplie pa-<lb/>reillement le diviſeur entier a + b par b, ce qui me donne <lb/>pour produit ab + bb, qu’il faut encore retrancher du reſte <lb/>ab + bb, ce que je fais en le mettant à la ſuite de cette quan-<lb/>tité avec des ſignes contraires: </s> <s xml:id="echoid-s908" xml:space="preserve">j’ai donc ab + bb - ab - bb, <lb/>ce qui ſe réduit à zero par la regle de la réduction des quan-<lb/>tités ſemblables, d’où je conclus que le quotient eſt a + b, <lb/>puiſqu’il ne reſte rien.</s> <s xml:id="echoid-s909" xml:space="preserve"/> </p> <div xml:id="echoid-div80" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0062-01" xlink:href="note-0062-01a" xml:space="preserve">Art. 55.</note> </div> <p> <s xml:id="echoid-s910" xml:space="preserve">72. </s> <s xml:id="echoid-s911" xml:space="preserve">Pour diviſer a<emph style="sub">2</emph> - 2ab + bb par a - b, je dis comme <lb/>ci-deſſus, a<emph style="sub">2</emph> diviſé par a donne a au quotient: </s> <s xml:id="echoid-s912" xml:space="preserve">je multiplie le <lb/>diyiſeur entier a - b par le quotient a, dont le produit eſt <lb/>aa - ab, que je retranche du dividende, en le mettant après <lb/>avec des ſignes contraires pour avoir le reſte aa - 2ab + bb <lb/>- aa + ab, ce qui ſe réduit à - ab + bb. </s> <s xml:id="echoid-s913" xml:space="preserve">Je fais ſur le reſte <lb/>la même opération, & </s> <s xml:id="echoid-s914" xml:space="preserve">je dis - ab diviſé par a, donne - b, <lb/>que j’écris à la ſuite du premier terme du quotient: </s> <s xml:id="echoid-s915" xml:space="preserve">je mul-<lb/>tiplie le diviſeur a - b par - b, & </s> <s xml:id="echoid-s916" xml:space="preserve">j’ôte le produit - ab + bb <lb/>du reſte qui m’a ſervi de dividende pour avoir - ab + bb <lb/>+ ab - bb, qui ſe réduit à zero par la réduction des quan-<lb/>tités ſemblables, d’où je conclus encore que a - b eſt le quo-<lb/>tient.</s> <s xml:id="echoid-s917" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s918" xml:space="preserve">73. </s> <s xml:id="echoid-s919" xml:space="preserve">Pour diviſer aa - bb par a + b, je dis aa diviſé par a <lb/>donne a, qui étant multiplié par le diviſeur, donne pour pro-<lb/>duit aa + ab; </s> <s xml:id="echoid-s920" xml:space="preserve">le retranchant du dividende, il reſte aa - bb <lb/>- aa - ab; </s> <s xml:id="echoid-s921" xml:space="preserve">qui étant réduit, donne - bb - ab, ou - ab <lb/>- bb, que je diviſe encore par a + b, en diſant - ab diviſé <lb/>par + a donne - b. </s> <s xml:id="echoid-s922" xml:space="preserve">Multipliant le diviſeur par - b, il vient <lb/>- ab - bb, qui étant retranché du dividende partiel, donne <lb/>- ab - bb + ab + bb ou zero, en effaçant ce qui ſe dé-<lb/>truit; </s> <s xml:id="echoid-s923" xml:space="preserve">d’où il ſuit quele quotient eſt a - b, ce qui eſt évident, <lb/>puiſqu’en multipliant ce quotient par le diviſeur, on retrouve <lb/>le dividende.</s> <s xml:id="echoid-s924" xml:space="preserve"/> </p> <pb o="25" file="0063" n="63" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div82" type="section" level="1" n="68"> <head xml:id="echoid-head81" xml:space="preserve"><emph style="sc">Exemples de</emph> <emph style="sc">Division</emph>.</head> <p> <s xml:id="echoid-s925" xml:space="preserve">1<emph style="sub">er</emph> {Dividende aa + 2ab + bb/Diviſeur a + b} {Quotient total a + b.</s> <s xml:id="echoid-s926" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s927" xml:space="preserve">Produit aa + ab (a, premier quotient.</s> <s xml:id="echoid-s928" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s929" xml:space="preserve">Souſtraction aa + 2ab + bb - aa - ab.</s> <s xml:id="echoid-s930" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s931" xml:space="preserve">{Réduction ou nou- \\ veau dividende { ab + bb/Diviſeur a + b} (b, ſecond quotient.</s> <s xml:id="echoid-s932" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s933" xml:space="preserve">Produit ab + bb</s> </p> <p> <s xml:id="echoid-s934" xml:space="preserve">Souſtraction ab + bb - ab - bb = o.</s> <s xml:id="echoid-s935" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s936" xml:space="preserve">2<emph style="sub">e</emph> Dividende aa - 2ab + bb (a - b, quotient total.</s> <s xml:id="echoid-s937" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s938" xml:space="preserve">Diviſeur a - b</s> </p> <p> <s xml:id="echoid-s939" xml:space="preserve">Produit aa - ab</s> </p> <p> <s xml:id="echoid-s940" xml:space="preserve">Souſtraction aa - 2ab + bb - aa + ab (a, I<emph style="sub">er</emph> quot.</s> <s xml:id="echoid-s941" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s942" xml:space="preserve">{Réduction ou nou- \\ veau dividende { - ab + bb/Diviſeur a - b} (- b, ſecond quotient.</s> <s xml:id="echoid-s943" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s944" xml:space="preserve">Produit - ab - bb</s> </p> <p> <s xml:id="echoid-s945" xml:space="preserve">Souſtraction - ab + bb + ab + bb = o.</s> <s xml:id="echoid-s946" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s947" xml:space="preserve">3<emph style="sub">e</emph> {Dividende aa - bb (quotient total (a - b)/Diviſeur a + b (a, premier quotient.</s> <s xml:id="echoid-s948" xml:space="preserve">}</s> </p> <p> <s xml:id="echoid-s949" xml:space="preserve">Produit aa + bb</s> </p> <p> <s xml:id="echoid-s950" xml:space="preserve">Souſtraction aa - bb - aa - ab</s> </p> <p> <s xml:id="echoid-s951" xml:space="preserve">Réduction ou nouveau dividende. </s> <s xml:id="echoid-s952" xml:space="preserve">{- ab - bb</s> </p> <p> <s xml:id="echoid-s953" xml:space="preserve">Diviſeur a + b (- b, ſecond quotient.</s> <s xml:id="echoid-s954" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s955" xml:space="preserve">Produit - ab - bb</s> </p> <p> <s xml:id="echoid-s956" xml:space="preserve">Souſtraction - ab - bb + ab + bb = o.</s> <s xml:id="echoid-s957" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s958" xml:space="preserve">4<emph style="sub">e</emph> {Dividende a<emph style="sub">4</emph> x x x - b<emph style="sub">4</emph> (quot. </s> <s xml:id="echoid-s959" xml:space="preserve">total a<emph style="sub">3</emph> + a<emph style="sub">2</emph>b + ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>/Diviſeur a - b (premier quotient a<emph style="sub">3</emph>}</s> </p> <p> <s xml:id="echoid-s960" xml:space="preserve">Produit a<emph style="sub">4</emph> - a<emph style="sub">3</emph>b</s> </p> <p> <s xml:id="echoid-s961" xml:space="preserve">Souſtraction a<emph style="sub">4</emph> - a<emph style="sub">4</emph> + a<emph style="sub">3</emph>b x x - b<emph style="sub">4</emph></s> </p> <p> <s xml:id="echoid-s962" xml:space="preserve">{Réduction ou di- \\ vidende partiel { a<emph style="sub">3</emph>b x x - b<emph style="sub">4</emph>/Diviſeur a - b (ſecond quotient a<emph style="sub">2</emph>b}</s> </p> <p> <s xml:id="echoid-s963" xml:space="preserve">Produit a<emph style="sub">3</emph>b - a<emph style="sub">2</emph>b<emph style="sub">2</emph></s> </p> <p> <s xml:id="echoid-s964" xml:space="preserve">Souſtraction a<emph style="sub">3</emph>b - a<emph style="sub">3</emph>b + a<emph style="sub">2</emph>b<emph style="sub">2</emph> - b<emph style="sub">4</emph></s> </p> <pb o="26" file="0064" n="64" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s965" xml:space="preserve">{Réduction ou nou- \\ veau dividende { a<emph style="sub">2</emph>b<emph style="sub">2</emph> - b<emph style="sub">4</emph>/Diviſeur a - b (troiſieme quotient ab<emph style="sub">2</emph>}</s> </p> <p> <s xml:id="echoid-s966" xml:space="preserve">Produit a<emph style="sub">2</emph>b<emph style="sub">2</emph> - ab<emph style="sub">3</emph></s> </p> <p> <s xml:id="echoid-s967" xml:space="preserve">Souſtraction a<emph style="sub">2</emph>b<emph style="sub">2</emph> - a<emph style="sub">2</emph>b<emph style="sub">2</emph> + ab<emph style="sub">3</emph> - b<emph style="sub">4</emph></s> </p> <p> <s xml:id="echoid-s968" xml:space="preserve">{Réduction ou nou- \\ veau dividende {ab<emph style="sub">3</emph> - b<emph style="sub">4</emph>/Diviſeur a - b} (quatrieme quotient + b<emph style="sub">3</emph></s> </p> <p> <s xml:id="echoid-s969" xml:space="preserve">Produit ab<emph style="sub">3</emph> - b<emph style="sub">4</emph></s> </p> <p> <s xml:id="echoid-s970" xml:space="preserve">Souſtraction ab<emph style="sub">3</emph> - b<emph style="sub">4</emph> - ab<emph style="sub">3</emph> + b<emph style="sub">4</emph> = o.</s> <s xml:id="echoid-s971" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div83" type="section" level="1" n="69"> <head xml:id="echoid-head82" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s972" xml:space="preserve">Quoique le quotient ait plus de termes que le dividende, <lb/>il ne faut pas croire pour cela que le dividende ſoit plus petit <lb/>que le quotient; </s> <s xml:id="echoid-s973" xml:space="preserve">car tant que le diviſeur a - b ſera quelque <lb/>choſe de poſitif, le produit du quotient poſitif a<emph style="sub">3</emph> + a<emph style="sub">2</emph>b + ab<emph style="sub">2</emph> <lb/>+ b par la quantité poſitive a - b, donnera certainement au <lb/>produit quelque choſe de plus grand que ce même quotient: <lb/></s> <s xml:id="echoid-s974" xml:space="preserve">donc a<emph style="sub">4</emph> - b<emph style="sub">4</emph>, qui eſt le produit, eſt plus grand que a<emph style="sub">3</emph> + a<emph style="sub">2</emph>b <lb/>+ ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>. </s> <s xml:id="echoid-s975" xml:space="preserve">D’ailleurs en Algebre une quantité qui a plus <lb/>de dimenſion qu’une autre, eſt toujours regardée comme la <lb/>plus grande.</s> <s xml:id="echoid-s976" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s977" xml:space="preserve">Si l’on avoit des quantités plus compoſées que les précé-<lb/>dentes, on ſuivroit le même procédé dans l’opération, comme <lb/>ſi l’on propoſoit de diviſer la quantité 6a<emph style="sub">2</emph> + 10ab + 17ac <lb/>+ 15bc + 12c<emph style="sub">2</emph> par 2a + 3c, on écriroit le dividende au deſſus <lb/>du diviſeur, & </s> <s xml:id="echoid-s978" xml:space="preserve">le reſte ſe feroit comme on le voit ci-deſſous.</s> <s xml:id="echoid-s979" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s980" xml:space="preserve">{Dividende 6a<emph style="sub">2</emph> + 10ab + 17ac + 15bc + 12c<emph style="sub">2</emph> { 3a + 5b + 4c, quot. </s> <s xml:id="echoid-s981" xml:space="preserve">total.</s> <s xml:id="echoid-s982" xml:space="preserve">/Diviſeur 2a + 3c (3a, premier quotient.</s> <s xml:id="echoid-s983" xml:space="preserve">}</s> </p> <p> <s xml:id="echoid-s984" xml:space="preserve">Produit 6a<emph style="sub">2</emph> + 9ac</s> </p> <p> <s xml:id="echoid-s985" xml:space="preserve">Souſtraction 6a<emph style="sub">2</emph> + 10ab + 17ac - 6a<emph style="sub">2</emph> - 9ac + 15bc + 12c<emph style="sub">2</emph></s> </p> <p> <s xml:id="echoid-s986" xml:space="preserve">{Réduction ou nou- \\ veau dividende { 10ab + 8ac + 15bc + 12cc/Diviſeur 2a + 3c (5b, ſecond quotient.</s> <s xml:id="echoid-s987" xml:space="preserve">}</s> </p> <p> <s xml:id="echoid-s988" xml:space="preserve">Produit 10ab + 15bc</s> </p> <p> <s xml:id="echoid-s989" xml:space="preserve">Souſtraction 10ab + 8ac + 15bc + 12cc - 10ab - 15bc.</s> <s xml:id="echoid-s990" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s991" xml:space="preserve">{Réduction ou nou- \\ veau dividende {8ac + 12cc/Diviſeur 2a + 3c} (4c, troiſieme quotient.</s> <s xml:id="echoid-s992" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s993" xml:space="preserve">Produit 8ac + 12cc</s> </p> <p> <s xml:id="echoid-s994" xml:space="preserve">Souſtraction 8ac + 12cc - 8ac - 12cc = o.</s> <s xml:id="echoid-s995" xml:space="preserve"/> </p> <pb o="27" file="0065" n="65" rhead="DE MATHÉMATIQUE. Liv. I."/> <p> <s xml:id="echoid-s996" xml:space="preserve">Si le dividende & </s> <s xml:id="echoid-s997" xml:space="preserve">le diviſeur contenoient pluſieurs puiſ-<lb/>ſances d’une même lettre, il faudroit diſpoſer les termes du <lb/>dividende par rapport aux différentes puiſſances d’une même <lb/>lettre, en regardant comme premier terme celui dans lequel <lb/>cette puiſſance ſeroit la plus élevée, comme ſecond celui où <lb/>elle ſe trouveroit d’un degré moins élevée, & </s> <s xml:id="echoid-s998" xml:space="preserve">ainſi des autres. <lb/></s> <s xml:id="echoid-s999" xml:space="preserve">Ayant fait la même opération ſur le diviſeur, il faudroit faire <lb/>la Diviſion ſelon les regles précédentes; </s> <s xml:id="echoid-s1000" xml:space="preserve">c’eſt ce que l’on ap-<lb/>pelle ordonner une quantité par rapport à une lettre. </s> <s xml:id="echoid-s1001" xml:space="preserve">Par <lb/>exemple, ſi l’on propoſe de diviſer 22a<emph style="sub">4</emph>b + 9ab<emph style="sub">4</emph> + 12a<emph style="sub">2</emph>b<emph style="sub">3</emph> <lb/>19a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 8a<emph style="sub">5</emph>, par 4a<emph style="sub">3</emph> + 2ab<emph style="sub">2</emph> + 3b<emph style="sub">3</emph> + 5a<emph style="sub">2</emph>b, on commencera <lb/>par ordonner le dividende par rapport à la lettre a, en regar-<lb/>dant le terme 8a<emph style="sub">5</emph> comme le premier, parce qu’il contient la <lb/>plus haute puiſſance de la lettre a; </s> <s xml:id="echoid-s1002" xml:space="preserve">& </s> <s xml:id="echoid-s1003" xml:space="preserve">en ſuivant le même <lb/>principe, on aura le dividende ordonné, 8a<emph style="sub">5</emph> + 22a<emph style="sub">4</emph>b + 19a<emph style="sub">3</emph>b<emph style="sub">2</emph> <lb/>+ 12a<emph style="sub">2</emph>b<emph style="sub">3</emph> + 9ab<emph style="sub">4</emph>, on fera de même pour le diviſeur, & </s> <s xml:id="echoid-s1004" xml:space="preserve">l’on <lb/>aura le diviſeur ordonné, 4a<emph style="sub">3</emph> + 5a<emph style="sub">2</emph>b + 2ab<emph style="sub">2</emph> + 3b<emph style="sub">3</emph>. </s> <s xml:id="echoid-s1005" xml:space="preserve">Le reſte <lb/>de la Diviſion ſe fera préciſément comme les précédentes.</s> <s xml:id="echoid-s1006" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1007" xml:space="preserve">{Dividende 8a<emph style="sub">5</emph> + 22a<emph style="sub">4</emph>b + 19a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 12a<emph style="sub">2</emph>b<emph style="sub">3</emph> + 9ab<emph style="sub">4</emph>/Diviſeur 4a<emph style="sub">3</emph> + 5a<emph style="sub">2</emph>b + 2ab<emph style="sub">2</emph> + 3b<emph style="sub">3</emph> (2a<emph style="sub">2</emph> + 3ab, quot. </s> <s xml:id="echoid-s1008" xml:space="preserve">total.</s> <s xml:id="echoid-s1009" xml:space="preserve">}</s> </p> <p> <s xml:id="echoid-s1010" xml:space="preserve">Produit 8a<emph style="sub">5</emph> + 10a<emph style="sub">4</emph>b + 4a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 6a<emph style="sub">2</emph>b<emph style="sub">3</emph> (1<emph style="sub">er</emph> quotient 2a<emph style="sub">2</emph>.</s> <s xml:id="echoid-s1011" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1012" xml:space="preserve">Souſtraction 8a<emph style="sub">5</emph> + 22a<emph style="sub">4</emph>b + 19a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 12ab<emph style="sub">3</emph> + 9ab<emph style="sub">4</emph> - 8a<emph style="sub">5</emph> <lb/>- 10a<emph style="sub">4</emph>b - 4a<emph style="sub">3</emph>b<emph style="sub">2</emph> - 6a<emph style="sub">2</emph>b<emph style="sub">3</emph>.</s> <s xml:id="echoid-s1013" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1014" xml:space="preserve">{Réduction ou nou- \\ veau dividende { 12a<emph style="sub">4</emph>b + 15a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 6a<emph style="sub">2</emph>b<emph style="sub">3</emph> + 9ab<emph style="sub">4</emph>/Diviſeur 4a<emph style="sub">3</emph> + 5a<emph style="sub">2</emph>b + 2ab<emph style="sub">2</emph> + 3b<emph style="sub">3</emph> (2<emph style="sub">e</emph> quotient 3ab.</s> <s xml:id="echoid-s1015" xml:space="preserve">}</s> </p> <p> <s xml:id="echoid-s1016" xml:space="preserve">Produit. </s> <s xml:id="echoid-s1017" xml:space="preserve">12a<emph style="sub">4</emph> + 15a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 6a<emph style="sub">2</emph>b<emph style="sub">3</emph> + 9ab<emph style="sub">4</emph></s> </p> <p> <s xml:id="echoid-s1018" xml:space="preserve">Souſtraction 12a<emph style="sub">4</emph>b + 15a<emph style="sub">3</emph>b<emph style="sub">2</emph> + 6a<emph style="sub">2</emph>b<emph style="sub">3</emph> + 9ab<emph style="sub">4</emph> - 12a<emph style="sub">4</emph>b<emph style="sub">2</emph> <lb/>- 15a<emph style="sub">3</emph>b<emph style="sub">2</emph> - 6a<emph style="sub">2</emph>b<emph style="sub">3</emph> - 9ab<emph style="sub">4</emph> = o.</s> <s xml:id="echoid-s1019" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div84" type="section" level="1" n="70"> <head xml:id="echoid-head83" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s1020" xml:space="preserve">Nous n’avons point parlé des quatre Regles ordinaires d’A-<lb/>rithmétique, parce que nous avons ſuppoſé que ceux qui étu-<lb/>dieront ce Traité, ſçauront au moins l’Addition, la Souſtrac-<lb/>tion, la Multiplication & </s> <s xml:id="echoid-s1021" xml:space="preserve">la Diviſion; </s> <s xml:id="echoid-s1022" xml:space="preserve">mais comme pluſieurs <lb/>pourroient n’avoir aucune connoiſſance des parties plus rele-<lb/>vées, & </s> <s xml:id="echoid-s1023" xml:space="preserve">même ignorer la maniere dont on doit pratiquer la <lb/>Multiplication dans certain cas, lorſque le multiplicateur &</s> <s xml:id="echoid-s1024" xml:space="preserve"> <pb o="28" file="0066" n="66" rhead="NOUVEAU COURS"/> le multiplicande ſont chacun des nombres complexes; </s> <s xml:id="echoid-s1025" xml:space="preserve">nous <lb/>allons commencer par expliquer la méthode de faire cette opé-<lb/>ration par le ſecours des parties aliquotes, que nous applique-<lb/>rons ſur le champ à des exemples. </s> <s xml:id="echoid-s1026" xml:space="preserve">Cette partie eſt d’autant <lb/>plus néceſſaire, qu’elle ſervira beaucoup pour l’intelligence <lb/>du toiſé, que nous donnerons dans la ſuite.</s> <s xml:id="echoid-s1027" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div85" type="section" level="1" n="71"> <head xml:id="echoid-head84" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s1028" xml:space="preserve">74. </s> <s xml:id="echoid-s1029" xml:space="preserve">On dit qu’une grandeur eſt partie aliquote d’un tout <lb/>ou d’une autre grandeur, lorſqu’elle eſt contenue un nombre <lb/>de fois juſte dans cette autre. </s> <s xml:id="echoid-s1030" xml:space="preserve">Ainſi le pied eſt partie aliquote <lb/>de la toiſe, parce qu’il y eſt contenu ſix fois juſte; </s> <s xml:id="echoid-s1031" xml:space="preserve">le ſol eſt <lb/>une partie aliquote de la livre, parce que la livre vaut vingt <lb/>ſols: </s> <s xml:id="echoid-s1032" xml:space="preserve">de même ces autres nombres, 2, 4, 5, 10 ſols ſont des <lb/>parties aliquotes de la livre, parce que chacun d’eux eſt con-<lb/>tenue exactement un certain nombre de fois dans la livre.</s> <s xml:id="echoid-s1033" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1034" xml:space="preserve">Lorſqu’une grandeur n’eſt pas contenue exactement dans <lb/>une autre, & </s> <s xml:id="echoid-s1035" xml:space="preserve">ſans reſte, elle eſt appellée partie aliquante <lb/>de cette grandeur: </s> <s xml:id="echoid-s1036" xml:space="preserve">ainſi 9 ſols eſt une partie aliquante de la <lb/>livre, parce que cette grandeur eſt contenue deux fois dans la <lb/>livre, avec un reſte 2; </s> <s xml:id="echoid-s1037" xml:space="preserve">de même 17 ſols, 15 ſols ſont des par-<lb/>ties aliquantes de la livre pour la même raiſon: </s> <s xml:id="echoid-s1038" xml:space="preserve">5 pouces, <lb/>7 pouces, 8 pouces ſont des parties aliquantes du pied, parce <lb/>que chacune de ces grandeurs ſont contenues dans le pied, <lb/>avec des reſtes.</s> <s xml:id="echoid-s1039" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div86" type="section" level="1" n="72"> <head xml:id="echoid-head85" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s1040" xml:space="preserve">75. </s> <s xml:id="echoid-s1041" xml:space="preserve">Quoique, ſelon les définitions précédentes, une partie <lb/>aliquante ne puiſſe pas être partie aliquote d’un même tout, <lb/>néanmoins on peut décompoſer cette quantité en d’autres, <lb/>qui ſoient parties aliquotes du tout, & </s> <s xml:id="echoid-s1042" xml:space="preserve">dont la ſomme ſoit <lb/>égale à la partie aliquante propoſée; </s> <s xml:id="echoid-s1043" xml:space="preserve">ainſi ce nombre 17 ſols <lb/>eſt égal à 10 + 5 + 2, qui ſont chacun des parties aliquotes <lb/>de la livre, dont il n’eſt qu’une partie aliquante. </s> <s xml:id="echoid-s1044" xml:space="preserve">Tout l’art <lb/>des opérations que nous allons faire conſiſte à décompoſer les <lb/>parties aliquantes en parties aliquotes, en faiſant enſorte, au-<lb/>tant qu’il eſt poſſible, que ces parties ſoient non ſeulement par-<lb/>ties aliquotes de ce tout ou de l’unité principale, mais encore <lb/>les unes des autres.</s> <s xml:id="echoid-s1045" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1046" xml:space="preserve">76. </s> <s xml:id="echoid-s1047" xml:space="preserve">On appelle multiplication complexe celle dans laquelle <pb o="29" file="0067" n="67" rhead="DE MATHÉMATIQUE. Liv. I."/> le multiplicateur ou le multiplicande, ou tous les deux en-<lb/>ſemble, contiennent chacun des unités de différentes eſpeces, <lb/>quoique réductibles à la même, ainſi que dans la queſtion <lb/>ſuivante.</s> <s xml:id="echoid-s1048" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div87" type="section" level="1" n="73"> <head xml:id="echoid-head86" xml:space="preserve"><emph style="sc">Exemple</emph> I.</head> <p> <s xml:id="echoid-s1049" xml:space="preserve">On demande le prix de 45 toiſes 3 pieds de maçonnerie, à <lb/>9 liv. </s> <s xml:id="echoid-s1050" xml:space="preserve">la toiſe.</s> <s xml:id="echoid-s1051" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"># 45 toiſ. # 3 pieds. <lb/># 9 liv. <lb/># 405 <lb/>Pour 3 toiſes # 4 # 10 <lb/>Total # 409 # 10 <lb/></note> <p> <s xml:id="echoid-s1052" xml:space="preserve">Pour avoir le prix que l’on cher-<lb/>che, il faudra multiplier 45 toiſes <lb/>3 pieds par 9 liv. </s> <s xml:id="echoid-s1053" xml:space="preserve">ou, pour mieux <lb/>dire, il faudra prendre 45 fois 9l. <lb/></s> <s xml:id="echoid-s1054" xml:space="preserve">& </s> <s xml:id="echoid-s1055" xml:space="preserve">la moitié de 9 livres, parce que <lb/>3 pieds ſont la moitié d’une toiſe, <lb/>dont le prix doit auſſi être moitié <lb/>du prix de la toiſe: </s> <s xml:id="echoid-s1056" xml:space="preserve">car en général il eſt ridicule de dire que <lb/>l’on multiplie des livres, des ſols & </s> <s xml:id="echoid-s1057" xml:space="preserve">des deniers par des toiſes, <lb/>des pieds, des pouces, &</s> <s xml:id="echoid-s1058" xml:space="preserve">c. </s> <s xml:id="echoid-s1059" xml:space="preserve">D’ailleurs, ſuivant un tel énoncé, <lb/>il eſt impoſſible de déterminer la nature des unités du pro-<lb/>duit: </s> <s xml:id="echoid-s1060" xml:space="preserve">mais il faut regarder un des nombres comme un nom-<lb/>bre abſtrait, c’eſt-à-dire dont les unités ne marquent que des <lb/>nombres de fois, & </s> <s xml:id="echoid-s1061" xml:space="preserve">dont les parties marquent des parties cor-<lb/>reſpondantes d’une fois. </s> <s xml:id="echoid-s1062" xml:space="preserve">Ainſi dans notre exemple, comme <lb/>on cherche le prix de 45 toiſes 3 pieds, à 9 liv. </s> <s xml:id="echoid-s1063" xml:space="preserve">la toiſe, puiſque <lb/>pour une toiſe il faut prendre une fois 9 liv. </s> <s xml:id="echoid-s1064" xml:space="preserve">pour 45 toiſes, il <lb/>faudra prendre 45 fois 9 livres, & </s> <s xml:id="echoid-s1065" xml:space="preserve">pour 3 pieds, moitié d’une <lb/>toiſe, il faudra prendre une moitié de fois 9 liv. </s> <s xml:id="echoid-s1066" xml:space="preserve">ou la moitié <lb/>de 9 liv. </s> <s xml:id="echoid-s1067" xml:space="preserve">Le produit de 9 liv. </s> <s xml:id="echoid-s1068" xml:space="preserve">par 45 liv. </s> <s xml:id="echoid-s1069" xml:space="preserve">eſt 405 livres, la moitié <lb/>de 9 liv. </s> <s xml:id="echoid-s1070" xml:space="preserve">eſt 4 liv. </s> <s xml:id="echoid-s1071" xml:space="preserve">10 ſols; </s> <s xml:id="echoid-s1072" xml:space="preserve">ainſi la ſomme 409 liv. </s> <s xml:id="echoid-s1073" xml:space="preserve">10 ſols eſt le <lb/>prix demandé.</s> <s xml:id="echoid-s1074" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div88" type="section" level="1" n="74"> <head xml:id="echoid-head87" xml:space="preserve"><emph style="sc">Exemple</emph> II.</head> <p> <s xml:id="echoid-s1075" xml:space="preserve">On demande le prix de 3 toiſes 2 pieds 6 pouces, à 5 liv. </s> <s xml:id="echoid-s1076" xml:space="preserve">4 ſ. <lb/></s> <s xml:id="echoid-s1077" xml:space="preserve">6 den. </s> <s xml:id="echoid-s1078" xml:space="preserve">la toiſe courante.</s> <s xml:id="echoid-s1079" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">5 liv. # 4 ſols # 6 den. <lb/>3 toiſ. # 2 pi. # 6 pouces. <lb/>15 liv. # 13 ſols # 6 den., prix de 3 toiſes. <lb/>1 . . # 14 . . # 10, prix de 2 pieds. <lb/>0 . . # 8 . . # 8 {1/2}, prix de 6 pouces. <lb/>17 . . # 17 . . # 0 {1/2}, prix total. <lb/></note> <p> <s xml:id="echoid-s1080" xml:space="preserve">Pour avoir le prix de-<lb/>mandé, il faudra multi-<lb/>plier 5 liv. </s> <s xml:id="echoid-s1081" xml:space="preserve">4 ſ. </s> <s xml:id="echoid-s1082" xml:space="preserve">6 den. </s> <s xml:id="echoid-s1083" xml:space="preserve">par <lb/>3 toiſes 2 pieds 6 pouces, <lb/>ou, pour mieux dire, il fau-<lb/>dra chercher le prix de 3 <emph style="sub">t.</emph> <lb/>à 5 liv. </s> <s xml:id="echoid-s1084" xml:space="preserve">4 ſols 6 den. </s> <s xml:id="echoid-s1085" xml:space="preserve">le prix <pb o="30" file="0068" n="68" rhead="NOUVEAU COURS"/> de deux pieds & </s> <s xml:id="echoid-s1086" xml:space="preserve">celui de ſix pouces, en conſidérant ces nom-<lb/>bres comme des parties de la toiſe, & </s> <s xml:id="echoid-s1087" xml:space="preserve">prenant pour leur prix <lb/>les mêmes parties du prix de la toiſe. </s> <s xml:id="echoid-s1088" xml:space="preserve">Ayant diſpoſé ces nom-<lb/>bres l’un au deſſus de l’autre, comme on voit ici, on commen-<lb/>cera la Multiplication par les plus petites eſpeces, parce qu’il <lb/>n’y a qu’un chiffre au rang des livres, & </s> <s xml:id="echoid-s1089" xml:space="preserve">l’on dira: </s> <s xml:id="echoid-s1090" xml:space="preserve">3 fois 6 font <lb/>18, poſe 6 d. </s> <s xml:id="echoid-s1091" xml:space="preserve">& </s> <s xml:id="echoid-s1092" xml:space="preserve">retiens 1 pour 12. </s> <s xml:id="echoid-s1093" xml:space="preserve">On paſſera delà aux ſols, <lb/>en diſant, 3 fois 4 font 12, & </s> <s xml:id="echoid-s1094" xml:space="preserve">1 que j’ai retenu c’eſt 13, que <lb/>je poſe au rang des ſols. </s> <s xml:id="echoid-s1095" xml:space="preserve">On paſſera de même aux livres, & </s> <s xml:id="echoid-s1096" xml:space="preserve"><lb/>l’on dira, 3 fois 5 font 15 l.</s> <s xml:id="echoid-s1097" xml:space="preserve">, que je mets au rang des livres. <lb/></s> <s xml:id="echoid-s1098" xml:space="preserve">Pour avoir le prix de deux pieds, on fera attention que deux <lb/>pieds étant le tiers de la toiſe, il faudra auſſi que le prix de <lb/>deux pieds ſoit le tiers du prix de la toiſe: </s> <s xml:id="echoid-s1099" xml:space="preserve">par conſéquent il <lb/>faudra diviſer le prix de la toiſe par 3, en diſant, le tiers de <lb/>5 l. </s> <s xml:id="echoid-s1100" xml:space="preserve">eſt 1 pour 3, reſte 2 l. </s> <s xml:id="echoid-s1101" xml:space="preserve">ou 40 ſols, qui joints avec les 4 ſols <lb/>ſuivans, font 44, dont le tiers eſt 14 pour 42, reſte 2 ſols ou <lb/>24 den.</s> <s xml:id="echoid-s1102" xml:space="preserve">, leſquels joints avec les 6 den. </s> <s xml:id="echoid-s1103" xml:space="preserve">ſuivans, font 30 den.</s> <s xml:id="echoid-s1104" xml:space="preserve">, <lb/>dont le tiers eſt 10, que l’on poſera au rang des deniers. </s> <s xml:id="echoid-s1105" xml:space="preserve">Enfin <lb/>pour avoir le prix de 6 pouces, on remarquera que 6 pouces <lb/>étant le quart de 2 pieds ou 24 pouces, le prix de 6 pouces <lb/>doit être le quart du prix de deux pieds, & </s> <s xml:id="echoid-s1106" xml:space="preserve">l’on prendra le <lb/>quart d’une liv. </s> <s xml:id="echoid-s1107" xml:space="preserve">14 ſ. </s> <s xml:id="echoid-s1108" xml:space="preserve">10 den.</s> <s xml:id="echoid-s1109" xml:space="preserve">, en diſant, le quart d’une livre <lb/>n’eſt point, je poſe zero au rang des livres; </s> <s xml:id="echoid-s1110" xml:space="preserve">je réduis la livre en <lb/>ſols, ce qui me donne 20 ſols, leſquels ajoutés à 14, font 34, <lb/>dont le quart eſt 8 pour 32, reſte 2 ſ. </s> <s xml:id="echoid-s1111" xml:space="preserve">ou 24 den.</s> <s xml:id="echoid-s1112" xml:space="preserve">, leſquels <lb/>ajoutés aux dix ſuivans, font 34 den.</s> <s xml:id="echoid-s1113" xml:space="preserve">, dont le quart eſt 8 {1/2}, <lb/>que je poſe au rang des deniers. </s> <s xml:id="echoid-s1114" xml:space="preserve">Faiſant l’addition de ces dif-<lb/>férens produits, on aura pour le prix total de 3 toiſes 2 pieds <lb/>6 pouces, à 5 liv. </s> <s xml:id="echoid-s1115" xml:space="preserve">4 ſ. </s> <s xml:id="echoid-s1116" xml:space="preserve">4 den. </s> <s xml:id="echoid-s1117" xml:space="preserve">la toiſe, 17 liv. </s> <s xml:id="echoid-s1118" xml:space="preserve">17 ſols 0 {1/2} den.</s> <s xml:id="echoid-s1119" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div89" type="section" level="1" n="75"> <head xml:id="echoid-head88" xml:space="preserve"><emph style="sc">Exemple</emph> III.</head> <p> <s xml:id="echoid-s1120" xml:space="preserve">On demande le prix de 43 aunes deux tiers d’étoffe, à 12 l. <lb/></s> <s xml:id="echoid-s1121" xml:space="preserve">10ſ. </s> <s xml:id="echoid-s1122" xml:space="preserve">8 den. </s> <s xml:id="echoid-s1123" xml:space="preserve">l’aune.</s> <s xml:id="echoid-s1124" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1125" xml:space="preserve">Comme dans cet exemple la premiere partie 43 du multi-<lb/>plicateur eſt compoſée de deux chiffres, & </s> <s xml:id="echoid-s1126" xml:space="preserve">que l’on ne verroit <lb/>pas tout d’un coup la valeur de 43 fois 8 deniers, on commen-<lb/>cera la Multiplication par les plus hautes eſpeces. </s> <s xml:id="echoid-s1127" xml:space="preserve">On cher-<lb/>chera donc d’abord le prix de 43 aunes à 12 livres, le prix <lb/>de 43 aunes à 10 ſols, & </s> <s xml:id="echoid-s1128" xml:space="preserve">le prix de 43 aunes à 8 deniers.</s> <s xml:id="echoid-s1129" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1130" xml:space="preserve">On trouvera le prix de 43 aunes, à 12 liv. </s> <s xml:id="echoid-s1131" xml:space="preserve">l’aune, en multi- <pb o="31" file="0069" n="69" rhead="DE MATHÉMATIQUE. Liv. I."/> pliant 43 par 12. </s> <s xml:id="echoid-s1132" xml:space="preserve">Pour <lb/>avoir enſuite le prix de <lb/>43 aunes, à 10 ſols, on <lb/>remarquera que le prix <lb/>de 43 aunes, à une livre, <lb/>ſeroit 43 liv.</s> <s xml:id="echoid-s1133" xml:space="preserve">; donc puiſ-<lb/>que 10 ſols ſont la moitié <lb/>d’une livre, le prix de <lb/>43 aunes, à 10 ſols, ſera <lb/>la moitié de 43 liv. </s> <s xml:id="echoid-s1134" xml:space="preserve">On <lb/>en prendra donc la moi-<lb/>tié, en diſant: </s> <s xml:id="echoid-s1135" xml:space="preserve">la moitié <lb/>de 4 eſt 2, que l’on poſera <lb/>au deſſous des dixaines <lb/>de livres; </s> <s xml:id="echoid-s1136" xml:space="preserve">la moitié de 3 eſt 1, que l’on poſera ſous les unités <lb/>des livres, reſte une livre, dont la moitié eſt 10 ſols, que l’on <lb/>poſera au rang des ſols. </s> <s xml:id="echoid-s1137" xml:space="preserve">Pour avoir le prix de 43 aunes, à 8 d. <lb/></s> <s xml:id="echoid-s1138" xml:space="preserve">on remarquera que 8 den. </s> <s xml:id="echoid-s1139" xml:space="preserve">ſont le tiers de 2 ſols: </s> <s xml:id="echoid-s1140" xml:space="preserve">on commen-<lb/>cera donc par chercher le produit de 43 aunes, à 2 ſols, que <lb/>l’on barrera, parce qu’il ne doit point entrer dans la ſomme. </s> <s xml:id="echoid-s1141" xml:space="preserve"><lb/>Pour avoir le faux produit, on prendra le cinquieme de celui <lb/>que l’on vient de trouver pour 10 ſols, en diſant: </s> <s xml:id="echoid-s1142" xml:space="preserve">le cin-<lb/>quieme de 21 liv. </s> <s xml:id="echoid-s1143" xml:space="preserve">eſt 4, que je poſe au rang des livres, reſte <lb/>une livre, laquelle jointe avec 10 ſols, donne 30 ſols, dont le <lb/>cinquieme eſt 6. </s> <s xml:id="echoid-s1144" xml:space="preserve">Je prends le tiers de ce produit, en diſant: </s> <s xml:id="echoid-s1145" xml:space="preserve"><lb/>le tiers de 4 liv. </s> <s xml:id="echoid-s1146" xml:space="preserve">eſt une liv. </s> <s xml:id="echoid-s1147" xml:space="preserve">pour 3, reſte une livre, qui jointe <lb/>avec les 6 ſols ſuivans, donne 26 ſols, dont le tiers eſt 8 pour <lb/>24, reſte 2 ſols ou 24 deniers, dont le tiers eſt 8, que je poſe <lb/>au rang des deniers, & </s> <s xml:id="echoid-s1148" xml:space="preserve">j’ai le prix de 43 aunes, à 8 deniers <lb/>l’aune. </s> <s xml:id="echoid-s1149" xml:space="preserve">Enfin pour avoir le prix des deux tiers d’aunes, je <lb/>prends deux fois le tiers du prix d’une aune, en diſant: </s> <s xml:id="echoid-s1150" xml:space="preserve">le tiers <lb/>de 12 liv. </s> <s xml:id="echoid-s1151" xml:space="preserve">eſt 4 livres, que je poſe au rang des livres. </s> <s xml:id="echoid-s1152" xml:space="preserve">Le tiers <lb/>de 10 ſols eſt 3 pour 9, reſte un ſol ou 12 deniers, qui joints <lb/>aux 8 ſuivans, font 20, dont le tiers eſt 6 {2/3}; </s> <s xml:id="echoid-s1153" xml:space="preserve">j’écris deux fois <lb/>le produit, puis faiſant l’addition des produits particuliers, je <lb/>trouve pour le prix total 547 liv. </s> <s xml:id="echoid-s1154" xml:space="preserve">5 ſols 9 den.</s> <s xml:id="echoid-s1155" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"># 12 liv. # 10 ſols # 8 den. <lb/># 43 {2/3} <lb/>Prix de 43 aunes. # 36 <lb/>à 12 liv. # 48 <lb/>à 10 ſols . . # 21 liv. # 10 ſols # 0 den. <lb/>Faux prod. de 2 ſ. <lb/>à 8 den. . . # liv. <lb/># 1 # 8 # 8 <lb/>Prix d’un tiers. # 4 # 3 # 6 {2/3} <lb/># 4 # 3 # 6 {2/3} <lb/><emph style="sc">Total</emph> # 547 # 5 # 8 {1/3} <lb/></note> </div> <div xml:id="echoid-div90" type="section" level="1" n="76"> <head xml:id="echoid-head89" xml:space="preserve"><emph style="sc">Exemple</emph> IV.</head> <p> <s xml:id="echoid-s1156" xml:space="preserve">On demande le prix de 5 marcs 6 onces 2 gros de cuivre, à <lb/>4 liv. </s> <s xml:id="echoid-s1157" xml:space="preserve">7 ſols 8 den. </s> <s xml:id="echoid-s1158" xml:space="preserve">le marc.</s> <s xml:id="echoid-s1159" xml:space="preserve"/> </p> <pb o="32" file="0070" n="70" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s1160" xml:space="preserve">Tout le monde ſçait que la livre vaut deux marcs, le marc <lb/>8 onces, l’once 8 gros, le gros 3 deniers, le denier 24 grains, <lb/>ce qui donne 9216 grains pour la livre. </s> <s xml:id="echoid-s1161" xml:space="preserve">Cela poſé,</s> </p> <p> <s xml:id="echoid-s1162" xml:space="preserve">Ayant diſpoſé ces deux <lb/> <anchor type="note" xlink:label="note-0070-01a" xlink:href="note-0070-01"/> nombres, comme on voit <lb/>ici, en regardant 4 liv. </s> <s xml:id="echoid-s1163" xml:space="preserve">7 ſ. <lb/></s> <s xml:id="echoid-s1164" xml:space="preserve">8 den. </s> <s xml:id="echoid-s1165" xml:space="preserve">comme le multipli-<lb/>cande, & </s> <s xml:id="echoid-s1166" xml:space="preserve">5 marcs 6 onces <lb/>2 gros comme le multipli-<lb/>cateur: </s> <s xml:id="echoid-s1167" xml:space="preserve">comme la partie de <lb/>ce même multiplicateur, qui <lb/>contient les marcs, n’eſt <lb/>compoſée que d’un ſeul chiffre, on cherchera d’abord le prix <lb/>de 5 marcs, à 4 liv. </s> <s xml:id="echoid-s1168" xml:space="preserve">7 ſols 8 den. </s> <s xml:id="echoid-s1169" xml:space="preserve">le marc, que l’on trouvera <lb/>en multipliant 4 liv. </s> <s xml:id="echoid-s1170" xml:space="preserve">5 ſols 8 den. </s> <s xml:id="echoid-s1171" xml:space="preserve">par 5, à commencer par les <lb/>deniers, en diſant, cinq fois 8 font 40 deniers, je poſe 4, & </s> <s xml:id="echoid-s1172" xml:space="preserve"><lb/>retiens 3 pour 36; </s> <s xml:id="echoid-s1173" xml:space="preserve">paſſant enſuite aux ſols, 7 fois 5 font 35, <lb/>& </s> <s xml:id="echoid-s1174" xml:space="preserve">3 que j’ai retenue font 38, poſe 18, & </s> <s xml:id="echoid-s1175" xml:space="preserve">retiens une livre; </s> <s xml:id="echoid-s1176" xml:space="preserve"><lb/>paſſant de même aux livres, 5 fois 4 font 20, & </s> <s xml:id="echoid-s1177" xml:space="preserve">une que j’ai <lb/>retenue font 21. </s> <s xml:id="echoid-s1178" xml:space="preserve">Pour avoir après cela le prix de 6 onces, <lb/>qui eſt une partie aliquante du marc, on les diviſera en ces <lb/>deux parties, 4 & </s> <s xml:id="echoid-s1179" xml:space="preserve">2, qui ſont chacune partie aliquote du <lb/>marc, & </s> <s xml:id="echoid-s1180" xml:space="preserve">partie aliquote l’une de l’autre; </s> <s xml:id="echoid-s1181" xml:space="preserve">& </s> <s xml:id="echoid-s1182" xml:space="preserve">comme 4 onces <lb/>ſont la moitié du marc, on prendra la moitié du prix d’un <lb/>marc, en diſant, la moitié de 4 liv. </s> <s xml:id="echoid-s1183" xml:space="preserve">eſt 2, la moitié de 7 ſols <lb/>eſt 3 pour 6, reſte un ſol ou 12 deniers, qui joints avec les 8 <lb/>ſuivans, font 20, dont la moitié eſt 10; </s> <s xml:id="echoid-s1184" xml:space="preserve">on prendra de même <lb/>la moitié de ce dernier produit pour avoir le prix de deux <lb/>onces, que l’on trouvera d’une livre 1 ſol 11 den. </s> <s xml:id="echoid-s1185" xml:space="preserve">Enfin pour <lb/>avoir le prix de deux gros, on remarquera que le gros étant la <lb/>8<emph style="sub">e</emph> partie de l’once, deux gros ſeront la 8<emph style="sub">e</emph> partie de deux onces, <lb/>& </s> <s xml:id="echoid-s1186" xml:space="preserve">par conſéquent le prix de deux gros ſera auſſi la huitieme <lb/>partie de celui de deux onces, que l’on vient d’écrire. </s> <s xml:id="echoid-s1187" xml:space="preserve">On dira <lb/>donc, la huitieme partie d’une livre n’eſt point, je poſe o au <lb/>rang des livres; </s> <s xml:id="echoid-s1188" xml:space="preserve">la huitieme partie de 21 ſols eſt 2 pour 16, <lb/>reſte 5 ſols, qui valent 6 deniers, leſquels joints avec les 11 d. </s> <s xml:id="echoid-s1189" xml:space="preserve"><lb/>ſuivans, donnent 71, dont la huitieme partie eſt 8 pour 64, <lb/>avec un reſte 7; </s> <s xml:id="echoid-s1190" xml:space="preserve">ce qui donne en tout pour le prix de deux gros, <lb/>o liv. </s> <s xml:id="echoid-s1191" xml:space="preserve">2 ſols 8 den, {7/8}. </s> <s xml:id="echoid-s1192" xml:space="preserve">Ajoutant ces différens produits, on aura <lb/>le prix total de 25 liv. </s> <s xml:id="echoid-s1193" xml:space="preserve">6 ſols 9 den. </s> <s xml:id="echoid-s1194" xml:space="preserve">{7/8}.</s> <s xml:id="echoid-s1195" xml:space="preserve"/> </p> <div xml:id="echoid-div90" type="float" level="2" n="1"> <note position="right" xlink:label="note-0070-01" xlink:href="note-0070-01a" xml:space="preserve"># 4 liv. # 7 ſols # 8 den. <lb/># 5 m. # 6 on. # 2 gros <lb/>Prix de 5 marcs. # 21 liv. # 18 ſols # 4 den. <lb/>de 4 onc. # 2 # 3 # 10 <lb/>de 2 onc. # 1 # 1 # 11 <lb/>de 2 gros # 0 # 2 # 8 {7/8} <lb/><emph style="sc">Total</emph> # 25 # 6 # 9 {7/8} <lb/></note> </div> <pb o="33" file="0071" n="71" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div92" type="section" level="1" n="77"> <head xml:id="echoid-head90" xml:space="preserve"><emph style="sc">Exemple</emph> V.</head> <p> <s xml:id="echoid-s1196" xml:space="preserve">On demande le prix de 325 marcs 7 onces 5 gros 2 deniers <lb/>16 grains d’un certain métal, à 54 liv. </s> <s xml:id="echoid-s1197" xml:space="preserve">18 ſols 9 den. </s> <s xml:id="echoid-s1198" xml:space="preserve">le marc.</s> <s xml:id="echoid-s1199" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"># 325 <emph style="sub">marcs</emph> \\ 54 <emph style="sub">liv.</emph> # 7 <emph style="sub">onces</emph> \\ 18 <emph style="sub">ſols</emph> # 5<emph style="sub">gros</emph> \\ 9<emph style="sub">den.</emph> # 2 <emph style="sub">den.</emph> # 16 <emph style="sub">grains.</emph> <lb/>Pour 325 <emph style="sub">marcs</emph> # 1300 <lb/>à 54 <emph style="sub">liv.</emph> # 1625 # 0 # 0 <lb/>à 10 ſols # 162 # 10 # 0 <lb/>à 4 ſols # 65 # 0 # 0 <lb/>à 4 ſols # 65 # 0 # 0 <lb/>à 6 den. # 8 # 2 # 6 <lb/>à 3 den. # 4 # 1 # 3 <lb/>Prix de 325 <emph style="sub">marcs</emph> à \\ 54 liv. 18 ſ. 9 den. # 17854 <emph style="sub">liv.</emph> # 13 <emph style="sub">ſols</emph> # 9 <emph style="sub">den.</emph> <lb/></note> <note style="it" position="right" xml:space="preserve">Prix de # { # 4 onces # 27 # 9 # 4 <lb/># # 2 onces # 13 # 14 # 8 <lb/># # 1 once # 6 # 17 # 4 <lb/># # 4 gros # 3 # 8 # 8 <lb/># # 1 gros # 0 # 17 # 2 <lb/># # 1 denier # 0 # 5 # 8 <lb/># # 1 denier # 0 # 5 # 8 <lb/># # 8 grains # 0 # 1 # 10 <lb/># # 8 grains # 0 # 1 # 10 <lb/># # # 17907 # 15 # 11 <lb/></note> <p> <s xml:id="echoid-s1200" xml:space="preserve">Comme le premier terme du multiplicande, & </s> <s xml:id="echoid-s1201" xml:space="preserve">celui du mul-<lb/>tiplicateur ſont nombres compoſés de pluſieurs chiffres, on <lb/>cherchera d’abord le prix de 325 marcs, à 54 liv. </s> <s xml:id="echoid-s1202" xml:space="preserve">le marc, ce <lb/>qui ſe fera en multipliant 325 par 54; </s> <s xml:id="echoid-s1203" xml:space="preserve">on cherchera enſuite le <lb/>prix de 325 marcs, à 18 ſols le marc, ce qui ſe fera en divi-<lb/>ſant 18 ſols en ſes parties, 10 + 4 + 4, qui ſont chacune des <lb/>parties aliquotes de la livre, & </s> <s xml:id="echoid-s1204" xml:space="preserve">prenant pour 10 ſols la moitié <lb/>de 325, en diſant, la moitié de 3 eſt 1 pour 2, reſte 1, qui joint <lb/>avec le 2 ſuivant fait 12, dont la moitié eſt 6; </s> <s xml:id="echoid-s1205" xml:space="preserve">la moitié de <lb/>5 eſt 2 pour 4, reſte une livre ou 20 ſols, dont la moitié eſt <lb/>10 ſols, que je poſe au rang des ſols. </s> <s xml:id="echoid-s1206" xml:space="preserve">Pour 4 ſols on cherchera <lb/>le cinquieme de 325, parce que 4 ſols fait la cinquieme partie <pb o="34" file="0072" n="72" rhead="NOUVEAU COURS"/> de la livre, & </s> <s xml:id="echoid-s1207" xml:space="preserve">l’on dira la cinquieme partie de 32 eſt 6 pour <lb/>30, reſte 2, qui joints avec le 5 ſuivant, font 25, dont la cin-<lb/>quieme partie eſt 5, ainſi l’on écrira deux fois 65, qui eſt le <lb/>cinquieme de 325, & </s> <s xml:id="echoid-s1208" xml:space="preserve">l’on aura le prix de 325 marcs à 18 ſols.</s> <s xml:id="echoid-s1209" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1210" xml:space="preserve">On paſſera enſuite aux deniers 9, que l’on diviſera en deux <lb/>parties, 6, 3, dont la premiere 6 eſt la huitieme partie de 4 ſols, <lb/>& </s> <s xml:id="echoid-s1211" xml:space="preserve">la ſeconde 3 eſt moitié de la premiere 6; </s> <s xml:id="echoid-s1212" xml:space="preserve">on prendra donc <lb/>la huitieme partie du prix que l’on vient de trouver pour 4 ſ. <lb/></s> <s xml:id="echoid-s1213" xml:space="preserve">en diſant la huitieme partie de 65 eſt 8 pour 64, poſe 8 au rang <lb/>des livres, reſte 1 liv. </s> <s xml:id="echoid-s1214" xml:space="preserve">ou 20 ſols, dont la huitieme partie eſt 2 ſ. </s> <s xml:id="echoid-s1215" xml:space="preserve"><lb/>pour 16, reſte 4 ſols ou 48 deniers, dont la huitieme partie eſt <lb/>6 deniers; </s> <s xml:id="echoid-s1216" xml:space="preserve">pour 3 den. </s> <s xml:id="echoid-s1217" xml:space="preserve">on prendra la moitié de ce que l’on vient <lb/>de trouver pour 6, & </s> <s xml:id="echoid-s1218" xml:space="preserve">l’on aura évidemment 4 liv. </s> <s xml:id="echoid-s1219" xml:space="preserve">1 ſol 3 den. </s> <s xml:id="echoid-s1220" xml:space="preserve"><lb/>Toutes ces opérations achevées, on aura le prix de 225 marcs, <lb/>à 54 liv. </s> <s xml:id="echoid-s1221" xml:space="preserve">18 ſ. </s> <s xml:id="echoid-s1222" xml:space="preserve">9 den. </s> <s xml:id="echoid-s1223" xml:space="preserve">le marc; </s> <s xml:id="echoid-s1224" xml:space="preserve">& </s> <s xml:id="echoid-s1225" xml:space="preserve">comme cette partie devient <lb/>déja un peu compliquée, on pourra d’abord prendre la ſomme <lb/>de ces produits particuliers, pour être moins expoſé à ſe tromper <lb/>dans l’addition totale. </s> <s xml:id="echoid-s1226" xml:space="preserve">On paſſera enſuite aux onces, & </s> <s xml:id="echoid-s1227" xml:space="preserve">l’on <lb/>diviſera lenombre 7, qui marque combien il y en a en 4, 2, 1, <lb/>qui ſont chacune partie aliquote du marc, & </s> <s xml:id="echoid-s1228" xml:space="preserve">partie aliquote <lb/>l’une de l’autre; </s> <s xml:id="echoid-s1229" xml:space="preserve">pour 4 onces on prendra la moitié de 54 liv. </s> <s xml:id="echoid-s1230" xml:space="preserve"><lb/>18 ſols 9 den. </s> <s xml:id="echoid-s1231" xml:space="preserve">en diſant, la moitié de 54 livres eſt 27 livres, la <lb/>moitié de 18 ſols eſt 9 ſols, la moitié de 9 den. </s> <s xml:id="echoid-s1232" xml:space="preserve">eſt 4 den. </s> <s xml:id="echoid-s1233" xml:space="preserve">{1/2}; </s> <s xml:id="echoid-s1234" xml:space="preserve"><lb/>pour 2 onces on prendra la moitié de ce que l’on vient de trou-<lb/>ver, en diſant, la moitié de 27 eſt 13 pour 26, je poſe 13 au <lb/>rang des livres, reſte 1 liv. </s> <s xml:id="echoid-s1235" xml:space="preserve">ou 20 ſols, qui joint avec les 9 qui <lb/>ſont après, donne 29 ſols, dont la moitié eſt; </s> <s xml:id="echoid-s1236" xml:space="preserve">14 pour 28, <lb/>reſte un ſol ou 12 deniers, qui joints avec le 4 ſuivant, font <lb/>16 deniers, dont la moitié eſt 8 (on négligera ici toutes les <lb/>fractions, parce qu’elles ne pourroient monter qu’à 3 ou 4 d. </s> <s xml:id="echoid-s1237" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s1238" xml:space="preserve">que d’ailleurs, pour en avoir exactement la ſomme, cela ſup-<lb/>poſeroit le calcul de ces nombres, que nous n’avons pas encore <lb/>donné). </s> <s xml:id="echoid-s1239" xml:space="preserve">Pour une once on prendra encore la moitié de ce <lb/>que l’on vient de trouver, en diſant, la moitié de 13 eſt 6 <lb/>pour 12, reſte 1 liv. </s> <s xml:id="echoid-s1240" xml:space="preserve">ou 20 ſols, qui joints avec les 14 ſuivans, <lb/>font 34, dont la moitié eſt 17; </s> <s xml:id="echoid-s1241" xml:space="preserve">la moitié de 8 deniers eſt 4 <lb/>On paſſera des onces au gros, & </s> <s xml:id="echoid-s1242" xml:space="preserve">l’on diviſera 5 en deux par-<lb/>ties, 4 & </s> <s xml:id="echoid-s1243" xml:space="preserve">1, pour 4 gros on prendra la moitié du prix d’une <lb/>once, parce que l’once vaut 8 gros, & </s> <s xml:id="echoid-s1244" xml:space="preserve">l’on aura 3 liv. </s> <s xml:id="echoid-s1245" xml:space="preserve">8 ſ. </s> <s xml:id="echoid-s1246" xml:space="preserve">8 d. </s> <s xml:id="echoid-s1247" xml:space="preserve"><lb/>Pour un gros on prendra le quart du prix de 4 gros, en diſant, <pb o="35" file="0073" n="73" rhead="DE MATHEMATIQUE. Liv. I."/> le quart de 3 liv. </s> <s xml:id="echoid-s1248" xml:space="preserve">n’eſt point, je poſe zero au rang des livres, <lb/>le quart de 68 eſt 17, le quart de 8 eſt 2.</s> <s xml:id="echoid-s1249" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1250" xml:space="preserve">On paſſera pareillement aux deniers, & </s> <s xml:id="echoid-s1251" xml:space="preserve">pour 2 den. </s> <s xml:id="echoid-s1252" xml:space="preserve">on pren-<lb/>dra deux fois le tiers de 17 ſols 2 den. </s> <s xml:id="echoid-s1253" xml:space="preserve">que l’on vient de trouver <lb/>pour le prix du gros, qui vaut 3 deniers, & </s> <s xml:id="echoid-s1254" xml:space="preserve">l’on aura 5 ſ. </s> <s xml:id="echoid-s1255" xml:space="preserve">8 den. <lb/></s> <s xml:id="echoid-s1256" xml:space="preserve">que l’on écrira deux fois. </s> <s xml:id="echoid-s1257" xml:space="preserve">Enfin pour avoir le prix de 16 grains, <lb/>on prendra encore deux fois le tiers de 5 ſols 8 den. </s> <s xml:id="echoid-s1258" xml:space="preserve">que l’on <lb/>vient de trouver pour le prix d’un denier, qui vaut 24 grains, <lb/>dont 16 grains ſont les deux tiers, & </s> <s xml:id="echoid-s1259" xml:space="preserve">l’on aura un ſol 10 den. </s> <s xml:id="echoid-s1260" xml:space="preserve"><lb/>que l’on écrira deux fois; </s> <s xml:id="echoid-s1261" xml:space="preserve">ajoutant tous ces prix particuliers, <lb/>on aura le prix total de 325 marcs 7 onces 5 gros 2 den. </s> <s xml:id="echoid-s1262" xml:space="preserve">16 gr. </s> <s xml:id="echoid-s1263" xml:space="preserve"><lb/>que l’on trouvera, par l’addition, de 17907 liv. </s> <s xml:id="echoid-s1264" xml:space="preserve">15 ſols 11 den.</s> <s xml:id="echoid-s1265" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div93" type="section" level="1" n="78"> <head xml:id="echoid-head91" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s1266" xml:space="preserve">On pourroit, ſans ſçavoir le calcul des fractions, opérer ſur <lb/>les plus petites parties des deniers, en imaginant le denier diviſé <lb/>en douze parties, & </s> <s xml:id="echoid-s1267" xml:space="preserve">chaque partie diviſée encore en douze au-<lb/>tres parties, ainſi pour {1/2} on prendroit 6, pour {1/3} on prendroit 4, <lb/>& </s> <s xml:id="echoid-s1268" xml:space="preserve">ainſi de ſuite, & </s> <s xml:id="echoid-s1269" xml:space="preserve">dans l’addition de ces parties, on retiendroit <lb/>autant de deniers que l’on auroit trouvé de fois douze. </s> <s xml:id="echoid-s1270" xml:space="preserve">Nous <lb/>allons appliquer cette méthode à l’exemple ſuivant.</s> <s xml:id="echoid-s1271" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div94" type="section" level="1" n="79"> <head xml:id="echoid-head92" xml:space="preserve"><emph style="sc">Exemple</emph> VI.</head> <p> <s xml:id="echoid-s1272" xml:space="preserve">On demande le prix de 247 toiſes 5 pieds 9 pouces de maçon-<lb/>nerie, à 25 liv. </s> <s xml:id="echoid-s1273" xml:space="preserve">19 ſols 11 den. </s> <s xml:id="echoid-s1274" xml:space="preserve">la toiſe.</s> <s xml:id="echoid-s1275" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1276" xml:space="preserve">Après avoir diſpoſé le mul-<lb/> <anchor type="note" xlink:label="note-0073-01a" xlink:href="note-0073-01"/> tiplicande & </s> <s xml:id="echoid-s1277" xml:space="preserve">le multiplica-<lb/>teur, comme on le voit ici, <lb/>on multipliera d’abord 247 <lb/>par 25 pour avoir le prix de <lb/>247 toiſes, à 25 liv. </s> <s xml:id="echoid-s1278" xml:space="preserve">la toiſe. <lb/></s> <s xml:id="echoid-s1279" xml:space="preserve">On cherchera enſuite le prix <lb/>de 247 toiſes, à 19 ſ. </s> <s xml:id="echoid-s1280" xml:space="preserve">en pre-<lb/>nant d’abord pour 10 ſols la <lb/>moitié du nombre 247, regar-<lb/>dé comme 247 livres, & </s> <s xml:id="echoid-s1281" xml:space="preserve">l’on <lb/>dira, la moitié de 2 eſt 1, la <lb/>moitié de 4 eſt 2, la moitié de <lb/>7 eſt 3 pour 6, reſte une livre <lb/>ou 20 ſols, dont la moitié eſt <lb/>10. </s> <s xml:id="echoid-s1282" xml:space="preserve">On cherchera pareille- <pb o="36" file="0074" n="74" rhead="NOUVEAU COURS"/> ment le prix de 247 toiſes à 5 ſols, & </s> <s xml:id="echoid-s1283" xml:space="preserve">l’on prendra la moitié du <lb/>prix que l’on vient de trouver pour 10, en diſant, la moitié de <lb/>12 eſt 6, la moitié de 3 eſt 1, reſte 1 liv. </s> <s xml:id="echoid-s1284" xml:space="preserve">qui joint avec les 10 ſ. <lb/></s> <s xml:id="echoid-s1285" xml:space="preserve">ſuivant fait 30 ſols, dont la moitié eſt 15. </s> <s xml:id="echoid-s1286" xml:space="preserve">On prendra encore <lb/>le prix de 247 toiſes, à 4 ſ. </s> <s xml:id="echoid-s1287" xml:space="preserve">en prenant le cinquieme de 247, & </s> <s xml:id="echoid-s1288" xml:space="preserve"><lb/>l’on dira le cinquieme de 24 eſt 4 pour 20, le cinquieme de 47 <lb/>eſt 9 pour 45, reſte 2 liv. </s> <s xml:id="echoid-s1289" xml:space="preserve">ou 40 ſols, dont le cinquieme eſt 8, <lb/>que l’on poſera au rang des ſols. </s> <s xml:id="echoid-s1290" xml:space="preserve">Ces opérations faites, on aura <lb/>le prix de 247 toiſes, à 19 ſols: </s> <s xml:id="echoid-s1291" xml:space="preserve">caril eſt évident que 10 + 5 + 4 <lb/>eſt égal à 19: </s> <s xml:id="echoid-s1292" xml:space="preserve">on cherchera enſuite le prix de 247 toiſ. </s> <s xml:id="echoid-s1293" xml:space="preserve">à 11 den. </s> <s xml:id="echoid-s1294" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s1295" xml:space="preserve">pour ce, l’on partagera les 11 den. </s> <s xml:id="echoid-s1296" xml:space="preserve">en parties aliquotes de <lb/>4 ſols, 6 + 3 + 2, & </s> <s xml:id="echoid-s1297" xml:space="preserve">comme 6 eſt le 8<emph style="sub">e</emph> de 4 ſols ou de 48 den. </s> <s xml:id="echoid-s1298" xml:space="preserve"><lb/>on prendra le huitieme du prix que l’on vient de trouver pour <lb/>4 ſols, en diſant, la huitieme partie de 49 eſt 6 pour 48, reſte <lb/>une livre ou 20 ſols, qui joints avec les 8 ſuivans, fait 28 ſols, <lb/>dont le huitieme eſt 3 pour 24, reſte 4 ſ. </s> <s xml:id="echoid-s1299" xml:space="preserve">ou 48 deniers, dont <lb/>le huitieme eſt 6. </s> <s xml:id="echoid-s1300" xml:space="preserve">Pour 3 den. </s> <s xml:id="echoid-s1301" xml:space="preserve">on prendra la moitié du der-<lb/>nier prix que l’on trouvera de 3 liv. </s> <s xml:id="echoid-s1302" xml:space="preserve">1 ſol 9 den. </s> <s xml:id="echoid-s1303" xml:space="preserve">Enfin pour <lb/>2 den. </s> <s xml:id="echoid-s1304" xml:space="preserve">on prendra le tiers de ce même nombre, que l’on trou-<lb/>vera de 2 liv. </s> <s xml:id="echoid-s1305" xml:space="preserve">1 ſol 2 deniers: </s> <s xml:id="echoid-s1306" xml:space="preserve">on cherchera enſuite le prix de <lb/>5 pieds, que l’on diviſera en deux parties 3. </s> <s xml:id="echoid-s1307" xml:space="preserve">2, pour 3 pieds, <lb/>on prendra la moitié du prix de la toiſe, en diſant, la moitié <lb/>de 25 eſt 12 pour 24, reſte 1 ou 20, qui joints à 19, font 39; </s> <s xml:id="echoid-s1308" xml:space="preserve"><lb/>la moitié de 39 ſols eſt 19 pour 38, reſte 1 ſol ou 12 den. </s> <s xml:id="echoid-s1309" xml:space="preserve">qui <lb/>joints aux 11 ſuivans, font 23, dont la moitié eſt 11 den. </s> <s xml:id="echoid-s1310" xml:space="preserve">& </s> <s xml:id="echoid-s1311" xml:space="preserve"><lb/>ſuivant la remarque précédente, la moitié de 12 eſt 6. </s> <s xml:id="echoid-s1312" xml:space="preserve">Pour <lb/>2 pieds on prendra le tiers du même prix, en diſant: </s> <s xml:id="echoid-s1313" xml:space="preserve">le tiers <lb/>de 25 eſt 8 pour 24, reſte 1 liv. </s> <s xml:id="echoid-s1314" xml:space="preserve">ou 20 ſols, leſquels joints avec <lb/>les 19 ſuivans, font 39, dont le tiers eſt 13; </s> <s xml:id="echoid-s1315" xml:space="preserve">le tiers de 11 eſt <lb/>3, reſte 2 ou 24, dont le tiers eſt 8. </s> <s xml:id="echoid-s1316" xml:space="preserve">Enfin pour avoir le prix <lb/>de 9 pouces, je les regarde comme 6 + 3: </s> <s xml:id="echoid-s1317" xml:space="preserve">pour 6 pouces, je <lb/>prends le quart du prix de deux pieds, en diſant, le quart de <lb/>8 eſt 2, le quart de 13 eſt 3 pour 12, reſte 1 ſol ou 12 den. </s> <s xml:id="echoid-s1318" xml:space="preserve">qui <lb/>joints avec les 3 ſuivans, font 15, dont le quart eſt 3, reſte 3 <lb/>ou 36, qui joints aux 8 ſuivans, font 44, dont le quart eſt 11. </s> <s xml:id="echoid-s1319" xml:space="preserve"><lb/>Enfin pour 3 pouces je prends la moitié de 2 liv. </s> <s xml:id="echoid-s1320" xml:space="preserve">3 ſ. </s> <s xml:id="echoid-s1321" xml:space="preserve">3 den. </s> <s xml:id="echoid-s1322" xml:space="preserve">{11/12} <lb/>que je trouve d’une livre 1 ſol 7 den. </s> <s xml:id="echoid-s1323" xml:space="preserve">{11/12}. </s> <s xml:id="echoid-s1324" xml:space="preserve">A joutant tous ces pro-<lb/>duits particuliers, on aura pour le prix total de 247 toiſes <lb/>5 pieds 9 pouces, à 25 liv. </s> <s xml:id="echoid-s1325" xml:space="preserve">19 ſols 11 den. </s> <s xml:id="echoid-s1326" xml:space="preserve">la toiſe; </s> <s xml:id="echoid-s1327" xml:space="preserve">6445 liv. </s> <s xml:id="echoid-s1328" xml:space="preserve"><lb/>@7 ſols 8 deniers.</s> <s xml:id="echoid-s1329" xml:space="preserve"/> </p> <div xml:id="echoid-div94" type="float" level="2" n="1"> <note position="right" xlink:label="note-0073-01" xlink:href="note-0073-01a" xml:space="preserve">Prix de # 247 <emph style="sub">toiſ.</emph> # 5 <emph style="sub">pi.</emph> # 9 <emph style="sub">pou.</emph> <lb/>247 <emph style="sub">toiſ.</emph> # 25 <emph style="sub">liv.</emph> # 19 <emph style="sub">ſ.</emph> # 11 <emph style="sub">den.</emph> <lb/>à 25 liv. # {1235 \\ 494 <lb/>à 10 ſols # 123 # 10 # 0 <lb/>à 5 ſols # 61 # 15 # 0 <lb/>à 4 ſols # 49 # 8 # 0 <lb/>à 6 den. # 6 # 3 # 6 <lb/>à 3 den. # 3 # 1 # 9 <lb/>à 2 den. # 2 # 1 # 2 <lb/>Prix de 3<emph style="sub">pi.</emph> # 12 # 19 # 11 # 6 <lb/>de 2 pieds # 8 # 13 # 3 # 8 <lb/>de 6 pouces # 2 # 3 # 3 # 11 <lb/>de 3 pouces # 1 # 1 # 7 # 11 <lb/>Prix total # 6445 # 17 \\ E ij # 8 # 0 <lb/></note> </div> <pb o="37" file="0075" n="75" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div96" type="section" level="1" n="80"> <head xml:id="echoid-head93" xml:space="preserve">TRAITÉ</head> <head xml:id="echoid-head94" style="it" xml:space="preserve">DES FRACTIONS NUMÉRIQUES ET ALGÉBRIQUES.</head> <head xml:id="echoid-head95" xml:space="preserve"><emph style="sc">Définition</emph> I.</head> <p> <s xml:id="echoid-s1330" xml:space="preserve">76. </s> <s xml:id="echoid-s1331" xml:space="preserve">SI l’on diviſe une unité quelconque, que nous appelle-<lb/>rons unité principale, comme une toiſe, un pied, une livre, &</s> <s xml:id="echoid-s1332" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s1333" xml:space="preserve">en un certain nombre de parties égales, chacune de ces parties <lb/>ſera appellée unité fractionnaire, pour la diſtinguer de l’unité <lb/>principale que l’on diviſe, & </s> <s xml:id="echoid-s1334" xml:space="preserve">le nombre qui marquera com-<lb/>bien on prend de ces parties égales, ſera appellé une fraction, <lb/>que l’on exprime ainſi, {2/3}, {5/6}, & </s> <s xml:id="echoid-s1335" xml:space="preserve">que l’on prononce deux tiers, <lb/>cinq ſixiemes. </s> <s xml:id="echoid-s1336" xml:space="preserve">On a déja vu qu’une barre placée entre deux <lb/>grandeurs, indique la diviſion de la grandeur ſupérieure, & </s> <s xml:id="echoid-s1337" xml:space="preserve"><lb/>c’eſt encore ce qui arrive ici.</s> <s xml:id="echoid-s1338" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div97" type="section" level="1" n="81"> <head xml:id="echoid-head96" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s1339" xml:space="preserve">77. </s> <s xml:id="echoid-s1340" xml:space="preserve">Le nombre que l’on met au deſſous de la barre s’appelle <lb/>dénominateur, parce qu’il fait voir en combien de parties égales <lb/>on a partagé ou diviſé l’unité principale. </s> <s xml:id="echoid-s1341" xml:space="preserve">Dans les fractions pré-<lb/>cédentes, les nombres 3 & </s> <s xml:id="echoid-s1342" xml:space="preserve">6 ſont les dénominateurs de ces <lb/>fractions, parce qu’ils déſignent que les unités principales ont <lb/>été diviſées en trois ou en ſix parties égales.</s> <s xml:id="echoid-s1343" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div98" type="section" level="1" n="82"> <head xml:id="echoid-head97" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s1344" xml:space="preserve">78. </s> <s xml:id="echoid-s1345" xml:space="preserve">Le nombre que l’on met au deſſus de la barre horizon-<lb/>tale s’appelle numérateur, parce qu’il compte effectivement <lb/>combien on prend de parties égales: </s> <s xml:id="echoid-s1346" xml:space="preserve">ainſi 2 & </s> <s xml:id="echoid-s1347" xml:space="preserve">5 ſont les nu-<lb/>mérateurs des fractions {2/3} & </s> <s xml:id="echoid-s1348" xml:space="preserve">{5/6}. </s> <s xml:id="echoid-s1349" xml:space="preserve">Les fractions algébriques ſe <lb/>marquent préciſément de la même maniere; </s> <s xml:id="echoid-s1350" xml:space="preserve">ainſi {a/b}, {c/d}, {f/g} ſont <lb/>des fractions algébriques, dont les numérateurs ſont a, c, f, <lb/>& </s> <s xml:id="echoid-s1351" xml:space="preserve">les dénominateurs b, d, g.</s> <s xml:id="echoid-s1352" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div99" type="section" level="1" n="83"> <head xml:id="echoid-head98" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s1353" xml:space="preserve">79. </s> <s xml:id="echoid-s1354" xml:space="preserve">Si le numérateur eſt égal, plus petit ou plus grand que <lb/>le dénominateur, la fraction ſera auſſi égale à l’unité, ou plus <lb/>petite ou plus grande que l’unité; </s> <s xml:id="echoid-s1355" xml:space="preserve">car un tout eſt égal à toutes <lb/>ſes parties priſes enſemble, & </s> <s xml:id="echoid-s1356" xml:space="preserve">plus grand qu’une de ſes parties, <pb o="38" file="0076" n="76" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s1357" xml:space="preserve">plus petit que toutes ſes parties priſes enſemble, ajoutées à <lb/>quelqu’une de ſes parties.</s> <s xml:id="echoid-s1358" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div100" type="section" level="1" n="84"> <head xml:id="echoid-head99" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s1359" xml:space="preserve">80. </s> <s xml:id="echoid-s1360" xml:space="preserve">La grandeur d’une fraction dépend de la grandeur du <lb/>numérateur de cette fraction; </s> <s xml:id="echoid-s1361" xml:space="preserve">enſorte que de deux fractions <lb/>qui ont même dénominateur, la plus grande eſt celle qui a le <lb/>plus grand numérateur, & </s> <s xml:id="echoid-s1362" xml:space="preserve">la plus petite, celle qui a le plus petit <lb/>numérateur; </s> <s xml:id="echoid-s1363" xml:space="preserve">car il eſt évident que la fraction {5/6} eſt plus grande <lb/>que la fraction {3/6}, par la même raiſon que 5 eſt plus grand que <lb/>3, quelle que ſoit la nature des unités du 6 & </s> <s xml:id="echoid-s1364" xml:space="preserve">du 3, pourvu <lb/>qu’elle ſoit la même pour l’un & </s> <s xml:id="echoid-s1365" xml:space="preserve">pour l’autre.</s> <s xml:id="echoid-s1366" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div101" type="section" level="1" n="85"> <head xml:id="echoid-head100" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s1367" xml:space="preserve">81. </s> <s xml:id="echoid-s1368" xml:space="preserve">Plus le nombre dans lequel on diviſe un même tout eſt <lb/>grand, plus chaque partie eſt petite, & </s> <s xml:id="echoid-s1369" xml:space="preserve">par conſéquent plus le <lb/>dénominateur d’une fraction eſt grand, le numérateur reſtant <lb/>le même, plus auſſi la fraction eſt petite; </s> <s xml:id="echoid-s1370" xml:space="preserve">c’eſt ce que les Géo-<lb/>metres expriment, en diſant que deux fractions qui ont un <lb/>même numérateur ſont entr’elles réciproquement comme leurs <lb/>dénominateurs; </s> <s xml:id="echoid-s1371" xml:space="preserve">car il eſt évident que la fraction {2/3} eſt plus <lb/>grande que la fraction {2/5}, pourvu qu’elles ſoient chacune frac-<lb/>tion d’une même unité principale, d’une toiſe par exemple, <lb/>d’un pied, &</s> <s xml:id="echoid-s1372" xml:space="preserve">c.</s> <s xml:id="echoid-s1373" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div102" type="section" level="1" n="86"> <head xml:id="echoid-head101" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s1374" xml:space="preserve">82. </s> <s xml:id="echoid-s1375" xml:space="preserve">Les fractions étant des parties de certaines grandeurs ou <lb/>unités principales, ſont de même nature qu’elles, & </s> <s xml:id="echoid-s1376" xml:space="preserve">par con-<lb/>ſéquent ſont ſuſceptibles comme elles d’augmentation ou de <lb/>diminution. </s> <s xml:id="echoid-s1377" xml:space="preserve">Donc on peut faire ſur les fractions les mêmes <lb/>opérations que l’on fait ſur les entiers, c’eſt-à-dire qu’on peut <lb/>les ajouter, les ſouſtraire, les multiplier, ou les diviſer les unes <lb/>par les autres.</s> <s xml:id="echoid-s1378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1379" xml:space="preserve">Outre les quatre opérations qui leur ſont communes avec <lb/>les nombres entiers, il y en a trois autres qui leur ſont parti-<lb/>culieres, & </s> <s xml:id="echoid-s1380" xml:space="preserve">dont les premieres dépendent. </s> <s xml:id="echoid-s1381" xml:space="preserve">La premiere de ces <lb/>trois eſt d’évaluer une fraction, ou de déterminer ſa valeur en <lb/>quantités connues; </s> <s xml:id="echoid-s1382" xml:space="preserve">la ſeconde eſt de réduire les fractions à <lb/>leurs moindres termes, & </s> <s xml:id="echoid-s1383" xml:space="preserve">la troiſieme eſt de les réduire au <lb/>même dénominateur. </s> <s xml:id="echoid-s1384" xml:space="preserve">Nous allons commencer par expliquer <lb/>ces opérations, par le ſecours deſquelles on pourra faire aiſé-<lb/>ment toutes les autres.</s> <s xml:id="echoid-s1385" xml:space="preserve"/> </p> <pb o="39" file="0077" n="77" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div103" type="section" level="1" n="87"> <head xml:id="echoid-head102" xml:space="preserve"><emph style="sc">Probleme</emph> I.</head> <p style="it"> <s xml:id="echoid-s1386" xml:space="preserve">83. </s> <s xml:id="echoid-s1387" xml:space="preserve">Evaluer une fraction, ou, ce qui eſt la même choſe, trouver <lb/>en valeurs connues, moindre que l’unité principale, une quantité <lb/>égale à une fraction propoſée.</s> <s xml:id="echoid-s1388" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1389" xml:space="preserve">On diviſera l’unité principale en autant de parties égales <lb/>qu’il y a d’unités au dénominateur; </s> <s xml:id="echoid-s1390" xml:space="preserve">on multipliera enſuite le <lb/>quotient par le numérateur, & </s> <s xml:id="echoid-s1391" xml:space="preserve">le produit ſera la valeur de la <lb/>fraction propoſée. </s> <s xml:id="echoid-s1392" xml:space="preserve">Comme ſi l’on propoſoit d’évaluer cette <lb/>fraction {2/5} de liv. </s> <s xml:id="echoid-s1393" xml:space="preserve">je diviſe la livre, qui eſt ici l’unité principale, <lb/>& </s> <s xml:id="echoid-s1394" xml:space="preserve">qui vaut 20 ſols, en cinq parties égales, dont chacune eſt <lb/>4 ſols, leſquels multipliés par le numérateur 2, font connoître <lb/>que la fraction {2/5} de liv. </s> <s xml:id="echoid-s1395" xml:space="preserve">vaut 5 ſols. </s> <s xml:id="echoid-s1396" xml:space="preserve">De même ſi l’on propoſe <lb/>d’évaluer cette fraction {5/6} de pied, je diviſe le pied ou 12 pouces <lb/>en ſix parties égales, leſquelles ſont chacune de deux pouces, <lb/>je multiplie ce quotient 2 par le numérateur 5; </s> <s xml:id="echoid-s1397" xml:space="preserve">le produit 10 <lb/>me marque que la fraction {5/6} de pied vaut 10 pouces. </s> <s xml:id="echoid-s1398" xml:space="preserve">Cette <lb/>premiere opération n’a pas lieu dans les fractions algébriques, <lb/>{a/b} eſt {a/b}, & </s> <s xml:id="echoid-s1399" xml:space="preserve">l’on ne pourroit l’évaluer qu’après avoir ſubſtitué à <lb/>la place de a & </s> <s xml:id="echoid-s1400" xml:space="preserve">de b les grandeurs qu’elles expriment.</s> <s xml:id="echoid-s1401" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div104" type="section" level="1" n="88"> <head xml:id="echoid-head103" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s1402" xml:space="preserve">84. </s> <s xml:id="echoid-s1403" xml:space="preserve">On dit qu’une fraction eſt réduite à ſes moindres termes, <lb/>ou à ſa plus ſimple expreſſion, lorſque le numérateur & </s> <s xml:id="echoid-s1404" xml:space="preserve">le dé-<lb/>nominateur de cette fraction n’ont pas d’autres diviſeurs com-<lb/>muns que l’unité: </s> <s xml:id="echoid-s1405" xml:space="preserve">ainſi ces fractions {2/3}, {7/5}, {8/9} ſont des fractions <lb/>réduites à leurs moindres termes. </s> <s xml:id="echoid-s1406" xml:space="preserve">Il n’en eſt pas de même des <lb/>fractions {3/9}, {4/16}, qui ſont telles, qu’on en peut trouver d’autres <lb/>qui leur ſoient égales, & </s> <s xml:id="echoid-s1407" xml:space="preserve">dont les termes ſoient plus petit, <lb/>comme {1/3} pour la premiere, & </s> <s xml:id="echoid-s1408" xml:space="preserve">{2/8} ou {1/4} pour la ſeconde, que l’on <lb/>trouve en diviſant les deux termes de la premiere par 3, & </s> <s xml:id="echoid-s1409" xml:space="preserve">les <lb/>deux termes de la ſeconde par 2 ou par 4.</s> <s xml:id="echoid-s1410" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1411" xml:space="preserve">85. </s> <s xml:id="echoid-s1412" xml:space="preserve">Si le nombre par lequel on diviſe les deux termes d’une <lb/>fraction eſt le plus grand diviſeur poſſible, commun au nu-<lb/>mérateur & </s> <s xml:id="echoid-s1413" xml:space="preserve">au dénominateur, la fraction qui réſultera des <lb/>deux quotiens, diviſés l’un par l’autre, ſera auſſi la plus ſimple <lb/>fraction poſſible, & </s> <s xml:id="echoid-s1414" xml:space="preserve">égale à la premiere.</s> <s xml:id="echoid-s1415" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1416" xml:space="preserve">86. </s> <s xml:id="echoid-s1417" xml:space="preserve">En Algebre une fraction eſt réduite à ſes moindres ter-<lb/>mes, lorſqu’elle n’a point de lettre commune au numérateur <pb o="40" file="0078" n="78" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s1418" xml:space="preserve">au dénominateur. </s> <s xml:id="echoid-s1419" xml:space="preserve">Ainſi {a/b}, {c/d}, {gf/mn} ſont des fractions algé-<lb/>briques irréductibles.</s> <s xml:id="echoid-s1420" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div105" type="section" level="1" n="89"> <head xml:id="echoid-head104" xml:space="preserve"><emph style="sc">Probleme</emph> II.</head> <p style="it"> <s xml:id="echoid-s1421" xml:space="preserve">87. </s> <s xml:id="echoid-s1422" xml:space="preserve">Trouver le plus grand commun diviſeur de deux nombres, <lb/>360 & </s> <s xml:id="echoid-s1423" xml:space="preserve">792, ou, ce qui eſt la même choſe, réduire la fraction {360/792} à <lb/>ſes moindres termes.</s> <s xml:id="echoid-s1424" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div106" type="section" level="1" n="90"> <head xml:id="echoid-head105" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s1425" xml:space="preserve">On diviſera le plus grand nombre 792 par le plus petit 360, <lb/>& </s> <s xml:id="echoid-s1426" xml:space="preserve">négligeant le quotient 2, on diviſera de nouveau le plus petit <lb/>360 par le reſte 72; </s> <s xml:id="echoid-s1427" xml:space="preserve">& </s> <s xml:id="echoid-s1428" xml:space="preserve">comme la diviſion de ces deux nombres <lb/>ſe fait exactement, on en conclura que 72 eſt le plus grand <lb/>diviſeur poſſible, commun aux deux nombres 792 & </s> <s xml:id="echoid-s1429" xml:space="preserve">360. <lb/></s> <s xml:id="echoid-s1430" xml:space="preserve">De même ſoit propoſé de trouver le plus grand commun di-<lb/>viſeur des deux nombres 91 & </s> <s xml:id="echoid-s1431" xml:space="preserve">294, ou, ce qui eſt la même <lb/>choſe, de réduire la fraction {91/294} à ſes moindres termes; </s> <s xml:id="echoid-s1432" xml:space="preserve">je <lb/>diviſe le plus grand nombre 294 par le plus petit 91, il vient <lb/>3 au quotient, que je néglige, avec un reſte 21; </s> <s xml:id="echoid-s1433" xml:space="preserve">je diviſe le <lb/>plus petit nombre 91 par le reſte 21, il vient encore 3 au quo-<lb/>tient, que je néglige pareillement, avec un reſte 7: </s> <s xml:id="echoid-s1434" xml:space="preserve">je diviſe <lb/>le premier reſte 21 par le ſecond 7, & </s> <s xml:id="echoid-s1435" xml:space="preserve">comme la diviſion ſe <lb/>fait exactement & </s> <s xml:id="echoid-s1436" xml:space="preserve">ſans reſte, je conclus que le nombre 7 eſt <lb/>le plus grand commun diviſeur aux deux nombres 294 & </s> <s xml:id="echoid-s1437" xml:space="preserve">91. </s> <s xml:id="echoid-s1438" xml:space="preserve"><lb/>En général le reſte qui diviſe exactement le reſte précédent, eſt <lb/>toujours le plus grand commun diviſeur que l’on cherche; </s> <s xml:id="echoid-s1439" xml:space="preserve">di-<lb/>viſant donc le numérateur & </s> <s xml:id="echoid-s1440" xml:space="preserve">le dénominateur de la 1<emph style="sub">re</emph> fraction <lb/>{360/792} par le plus grand diviſeur commun 72, on aura la frac-<lb/>tion {5/11}, qui eſt irréductible, & </s> <s xml:id="echoid-s1441" xml:space="preserve">égale à la propoſée. </s> <s xml:id="echoid-s1442" xml:space="preserve">Diviſant <lb/>de même le numérateur & </s> <s xml:id="echoid-s1443" xml:space="preserve">le dénominateur de la ſeconde <lb/>fraction {91/294} par le plus grand commun diviſeur 7, on aura la <lb/>nouvelle fraction {13/42} égale à la précédente, & </s> <s xml:id="echoid-s1444" xml:space="preserve">réduite à ſa plus <lb/>ſimple expreſſion.</s> <s xml:id="echoid-s1445" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div107" type="section" level="1" n="91"> <head xml:id="echoid-head106" style="it" xml:space="preserve">Démonſtration de cette pratique.</head> <p> <s xml:id="echoid-s1446" xml:space="preserve">Pour concevoir la raiſon de ces opérations, on fera atten-<lb/>tion, 1°. </s> <s xml:id="echoid-s1447" xml:space="preserve">qu’un nombre qui diviſe exactement une grandeur, <lb/>eſt auſſi diviſeur exact de ſes multiples, ou des nombres qui <lb/>réſultent du produit de cette grandeur par une autre quelcon-<lb/>que. </s> <s xml:id="echoid-s1448" xml:space="preserve">Par exemple, ſi 3 eſt diviſeur de 6, il ſera auſſi divi- <pb o="41" file="0079" n="79" rhead="DE MATHÉMATIQUE. Liv. I."/> ſeur de 6 x 4, de 6 x 5, ou des nombres 24 & </s> <s xml:id="echoid-s1449" xml:space="preserve">30, &</s> <s xml:id="echoid-s1450" xml:space="preserve">c.</s> <s xml:id="echoid-s1451" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1452" xml:space="preserve">2°. </s> <s xml:id="echoid-s1453" xml:space="preserve">Qu’un nombre qui diviſe les deux parties d’un tout, <lb/>ſera auſſi diviſeur du tout, parce qu’un nombre eſt égal à <lb/>toutes ſes parties priſes enſemble; </s> <s xml:id="echoid-s1454" xml:space="preserve">ainſi le nombre 3 étant di-<lb/>viſeur des nombres 9 & </s> <s xml:id="echoid-s1455" xml:space="preserve">6, eſt auſſi diviſeur de leur ſomme 15.</s> <s xml:id="echoid-s1456" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1457" xml:space="preserve">3°. </s> <s xml:id="echoid-s1458" xml:space="preserve">Que ſi un nombre eſt diviſeur d’un tout & </s> <s xml:id="echoid-s1459" xml:space="preserve">d’une de ſes <lb/>parties, il ſera auſſi diviſeur de l’autre partie; </s> <s xml:id="echoid-s1460" xml:space="preserve">car s’il ne la di-<lb/>viſoit pas, il ne ſeroit pas diviſeur du tout, ce qui eſt contre <lb/>l’hypotheſe: </s> <s xml:id="echoid-s1461" xml:space="preserve">ainſi le nombre 3 étant diviſeur du tout 15, & </s> <s xml:id="echoid-s1462" xml:space="preserve"><lb/>d’une de ſes parties 9, eſt auſſi diviſeur de l’autre 6.</s> <s xml:id="echoid-s1463" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1464" xml:space="preserve">Cela poſé, que a & </s> <s xml:id="echoid-s1465" xml:space="preserve">b repréſentent les deux nombres, dont <lb/>on demande le plus grand commun diviſeur, que a diviſé par <lb/>b donne un quotient f avec le reſte d, on aura a = bf + d; <lb/></s> <s xml:id="echoid-s1466" xml:space="preserve">car un dividende quelconque eſt égal au produit du diviſeur <lb/>par le quotient joint au reſte de la diviſion. </s> <s xml:id="echoid-s1467" xml:space="preserve">Que b, diviſé par <lb/>le premier reſte d, donne un quotient g avec le reſte c, on aura <lb/>par la même raiſon b = dg + c: </s> <s xml:id="echoid-s1468" xml:space="preserve">enfin que le dernier reſte c <lb/>diviſe exactement le premier d, en donnant h au quotient, on <lb/>aura encore d = ch; </s> <s xml:id="echoid-s1469" xml:space="preserve">& </s> <s xml:id="echoid-s1470" xml:space="preserve">raſſemblant toutes ces égalités, on <lb/>aura a = bf + d, b = dg + c, & </s> <s xml:id="echoid-s1471" xml:space="preserve">d = ch. </s> <s xml:id="echoid-s1472" xml:space="preserve">Or il eſt évident <lb/>que c eſt diviſeur des quantités a & </s> <s xml:id="echoid-s1473" xml:space="preserve">b, car puiſque c eſt di-<lb/>viſeur de d, il eſt auſſi diviſeur de ſon multiple dg; </s> <s xml:id="echoid-s1474" xml:space="preserve">d’ailleurs <lb/>il eſt diviſeur de lui-même; </s> <s xml:id="echoid-s1475" xml:space="preserve">donc il diviſe dg + c; </s> <s xml:id="echoid-s1476" xml:space="preserve">donc il eſt <lb/>diviſeur de b, à cauſe de l’équation b = dg + c. </s> <s xml:id="echoid-s1477" xml:space="preserve">Puiſque c eſt <lb/>diviſeur de d & </s> <s xml:id="echoid-s1478" xml:space="preserve">de b, il eſt auſſi diviſeur des multiples de b; </s> <s xml:id="echoid-s1479" xml:space="preserve"><lb/>donc il diviſe bf + d; </s> <s xml:id="echoid-s1480" xml:space="preserve">donc il eſt diviſeur de a, à cauſe de <lb/>l’égalité a = bf + d.</s> <s xml:id="echoid-s1481" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1482" xml:space="preserve">Si l’on met dans l’équation b = dg + c la quantité ch à la <lb/>place de d qui lui eſt égale, on aura b = cgh + c; </s> <s xml:id="echoid-s1483" xml:space="preserve">ſubſtituant <lb/>pareillement cette valeur de b dans celle de a, ainſi que celle <lb/>de d, on aura a = cfgh + cf + ch; </s> <s xml:id="echoid-s1484" xml:space="preserve">donc au lieu de la frac-<lb/>tion {a/b} on auroit, ſuivant les ſuppoſitions que nous avons faites, <lb/>{cfgh + cf + ch/cgh + c}, dans laquelle fraction il eſt aiſé de voir qu’il n’y <lb/>a que la quantité c qui ſoit un diviſeur commun au numéra-<lb/>teur & </s> <s xml:id="echoid-s1485" xml:space="preserve">au dénominateur, & </s> <s xml:id="echoid-s1486" xml:space="preserve">que cette lettre eſt en même tems <lb/>le plus grand commun diviſeur. </s> <s xml:id="echoid-s1487" xml:space="preserve">Comme le procédé numé-<lb/>rique eſt préciſément le même, il faut auſſi qu’il faſſe trouver <lb/>le commun diviſeur que l’on cherche; </s> <s xml:id="echoid-s1488" xml:space="preserve">ainſi l’on pourra tou- <pb o="42" file="0080" n="80" rhead="NOUVEAU COURS"/> jours réduire une fraction quelconque à ſes moindres termes.</s> <s xml:id="echoid-s1489" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div108" type="section" level="1" n="92"> <head xml:id="echoid-head107" xml:space="preserve"><emph style="sc">Probleme</emph> III.</head> <p style="it"> <s xml:id="echoid-s1490" xml:space="preserve">88. </s> <s xml:id="echoid-s1491" xml:space="preserve">Réduire deux ou pluſieurs fractions à un même dénomina-<lb/>teur, de maniere qu’elles ſoient toujours égales aux fractions pro-<lb/>poſées.</s> <s xml:id="echoid-s1492" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div109" type="section" level="1" n="93"> <head xml:id="echoid-head108" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s1493" xml:space="preserve">S’il n’y a que deux fractions, on multipliera le numérateur <lb/>& </s> <s xml:id="echoid-s1494" xml:space="preserve">le dénominateur de chacune par le dénominateur de l’autre; <lb/></s> <s xml:id="echoid-s1495" xml:space="preserve">& </s> <s xml:id="echoid-s1496" xml:space="preserve">s’il y en a pluſieurs, on multipliera le numérateur & </s> <s xml:id="echoid-s1497" xml:space="preserve">le dé-<lb/>nominateur de chacune par le produit des dénominateurs des <lb/>autres fractions.</s> <s xml:id="echoid-s1498" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1499" xml:space="preserve">Dans l’un & </s> <s xml:id="echoid-s1500" xml:space="preserve">dans l’autre cas les fractions auront même dé-<lb/>nominateur; </s> <s xml:id="echoid-s1501" xml:space="preserve">car le produit de tant de nombres que l’on vou-<lb/>dra, multipliés les uns par les autres, ſera toujours le même. <lb/></s> <s xml:id="echoid-s1502" xml:space="preserve">De plus, chacune ſera égale à la premiere fraction propoſée, <lb/>puiſque le numérateur augmente par la multiplication dans <lb/>la même proportion que les parties du dénominateur dimi-<lb/>nuent. </s> <s xml:id="echoid-s1503" xml:space="preserve">La regle eſt précìſément la même pour les fractions <lb/>algébriques, & </s> <s xml:id="echoid-s1504" xml:space="preserve">ſe démontre de la même maniere, comme on <lb/>le va voir dans les exemples ſuivans.</s> <s xml:id="echoid-s1505" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1506" xml:space="preserve">Soient propoſées les fractions {2/3} & </s> <s xml:id="echoid-s1507" xml:space="preserve">{4/5}, pour être réduites au <lb/>même dénominateur, on multipliera les deux termes 2 & </s> <s xml:id="echoid-s1508" xml:space="preserve">3 <lb/>de la premiere par le dénominateur 5 de la ſeconde, & </s> <s xml:id="echoid-s1509" xml:space="preserve">réci-<lb/>proquement les deux termes 4 & </s> <s xml:id="echoid-s1510" xml:space="preserve">5 de la ſeconde par le déno-<lb/>minateur 3 de la premiere, & </s> <s xml:id="echoid-s1511" xml:space="preserve">l’on aura les deux nouvelles frac-<lb/>tions {10/15} & </s> <s xml:id="echoid-s1512" xml:space="preserve">{12/15} égales aux précédentes, & </s> <s xml:id="echoid-s1513" xml:space="preserve">réduites en même dé-<lb/>nomination. </s> <s xml:id="echoid-s1514" xml:space="preserve">De même pour réduire les fractions algébriques <lb/>{a/b} & </s> <s xml:id="echoid-s1515" xml:space="preserve">{c/d} à la même dénomination, je multiplie a & </s> <s xml:id="echoid-s1516" xml:space="preserve">b par d, & </s> <s xml:id="echoid-s1517" xml:space="preserve"><lb/>les termes c & </s> <s xml:id="echoid-s1518" xml:space="preserve">d de la ſeconde par b, pour avoir les fractions <lb/>{ad/bd}, {cb/bd} qui ſont égales aux précédentes, & </s> <s xml:id="echoid-s1519" xml:space="preserve">ont même déno-<lb/>minateur bd.</s> <s xml:id="echoid-s1520" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1521" xml:space="preserve">Si l’on a pluſieurs fractions, comme {2/3}, {3/4}, {5/6} à réduire, on <lb/>multipliera les termes 2 & </s> <s xml:id="echoid-s1522" xml:space="preserve">3 de la fraction {2/3} par 24, produit <lb/>des deux autres dénominateurs 6 & </s> <s xml:id="echoid-s1523" xml:space="preserve">4; </s> <s xml:id="echoid-s1524" xml:space="preserve">de même les termes 3 <lb/>& </s> <s xml:id="echoid-s1525" xml:space="preserve">4 de la fraction {3/4} par le nombre 18, produit des dénomina-<lb/>teurs 3 & </s> <s xml:id="echoid-s1526" xml:space="preserve">6 des deux autres; </s> <s xml:id="echoid-s1527" xml:space="preserve">& </s> <s xml:id="echoid-s1528" xml:space="preserve">enfin les termes 5 & </s> <s xml:id="echoid-s1529" xml:space="preserve">6 de la frac-<lb/>tion {5/6} par 12, produit des dénominateurs 3 & </s> <s xml:id="echoid-s1530" xml:space="preserve">4 des deux <pb o="43" file="0081" n="81" rhead="DE MATHÉMATIQUE. Liv. I."/> premieres, & </s> <s xml:id="echoid-s1531" xml:space="preserve">l’on aura les trois nouvelles fractions {48/72}, {54/72}, {60/72} <lb/>égales aux précédentes, & </s> <s xml:id="echoid-s1532" xml:space="preserve">qui ont même dénominateur.</s> <s xml:id="echoid-s1533" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1534" xml:space="preserve">En agiſſant de même, on verra que les fractions {a/b}, {c/d}, {f/g} <lb/>deviendront celles-ci {adg/bdg}, {cbg/bdg}, {bdf/bdg}, qui ont évidemment même <lb/>dénominateur.</s> <s xml:id="echoid-s1535" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div110" type="section" level="1" n="94"> <head xml:id="echoid-head109" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s1536" xml:space="preserve">89. </s> <s xml:id="echoid-s1537" xml:space="preserve">Après avoir réduit les fractions propoſées en même dé-<lb/>nomination, il eſt à propos de voir ſi le dénominateur n’a pas <lb/>quelque diviſeur par lequel on puiſſe diviſer tous les numéra-<lb/>teurs, afin de ſimplifier les nouvelles fractions, ainſi que dans <lb/>l’exemple précédent, où l’on peut diviſer tous les numéra-<lb/>teurs & </s> <s xml:id="echoid-s1538" xml:space="preserve">le dénominateur commun par 6, ce qui réduit les frac-<lb/>tions à celles-ci {8/12}, {9/12}, {10/12} égales aux premieres, ayant même <lb/>dénomination, & </s> <s xml:id="echoid-s1539" xml:space="preserve">les plus ſimples que l’on puiſſe trouver, qui <lb/>rempliſſent ces conditions.</s> <s xml:id="echoid-s1540" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1541" xml:space="preserve">90. </s> <s xml:id="echoid-s1542" xml:space="preserve">S’il y a pluſieurs dénominateurs parmi les fractions à <lb/>réduire, qui ayent entr’eux un diviſeur commun, deux par <lb/>exemple, on pourra diviſer une fois par ce diviſeur chaque ter-<lb/>me des nouvelles fractions réduites; </s> <s xml:id="echoid-s1543" xml:space="preserve">s’il y en a trois qui ayent <lb/>un diviſeur commun, on pourra diviſer toutes les nouvelles <lb/>fractions deux fois de ſuite par le même diviſeur, ou bien, ſi <lb/>l’on veut, une fois par le quarré de ce diviſeur commun. </s> <s xml:id="echoid-s1544" xml:space="preserve">Dans <lb/>l’exemple propoſé ci-deſſus; </s> <s xml:id="echoid-s1545" xml:space="preserve">on a diviſé toutes les nouvelles <lb/>fractions par 6, parce que deux d’entr’elles avoient un même <lb/>diviſeur 3, ſçavoir, la fraction {2/3} & </s> <s xml:id="echoid-s1546" xml:space="preserve">fraction {5/6}, & </s> <s xml:id="echoid-s1547" xml:space="preserve">deux autres <lb/>des mêmes fractions avoient à leurs dénominateurs un divi-<lb/>ſeur commun 2, ſçavoir, la fraction {3/4} & </s> <s xml:id="echoid-s1548" xml:space="preserve">la fraction {5/6}, c’eſt <lb/>pourquoi l’on diviſe par 2 x 3 ou par 6. </s> <s xml:id="echoid-s1549" xml:space="preserve">On trouvera aiſément <lb/>la raiſon de ces opérations, ſi l’on décompoſe les dénomina-<lb/>teurs de ces fractions dans leurs facteurs.</s> <s xml:id="echoid-s1550" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div111" type="section" level="1" n="95"> <head xml:id="echoid-head110" style="it" xml:space="preserve">De l’Addition des Fractions.</head> <p> <s xml:id="echoid-s1551" xml:space="preserve">91 Si les fractions que l’on veut ajouter enſemble n’ont pas <lb/>un même dénominateur, on commencera par les y réduire: <lb/></s> <s xml:id="echoid-s1552" xml:space="preserve">ainſi ſi l’on propoſe d’ajouter enſemble les fractions {2/3}, {4/5}, {5/6}, on <lb/>les réduira au même dénominateur, ſuivant l’art. </s> <s xml:id="echoid-s1553" xml:space="preserve">88, & </s> <s xml:id="echoid-s1554" xml:space="preserve">l’on <lb/>aura à la place de ces fractions {60/90}, {72/90}, {75/90}, ou plus ſimplement <lb/>(article 89.) </s> <s xml:id="echoid-s1555" xml:space="preserve">{20/30}, {24/30}, {25/30}, qui ſont égales aux précédentes. </s> <s xml:id="echoid-s1556" xml:space="preserve">On <pb o="44" file="0082" n="82" rhead="NOUVEAU COURS"/> prendra la ſomme de leurs numérateurs, pour en faire celui <lb/>d’une nouvelle fraction, qui conſervera le même dénomina-<lb/>teur commun, & </s> <s xml:id="echoid-s1557" xml:space="preserve">qui ſera la ſomme des fractions propoſées; <lb/></s> <s xml:id="echoid-s1558" xml:space="preserve">cette ſomme ſe trouvera {69/30} ou {23/10}, qui eſt irréductible. </s> <s xml:id="echoid-s1559" xml:space="preserve">On opé-<lb/>reroit de même ſur des fractions littérales; </s> <s xml:id="echoid-s1560" xml:space="preserve">ainſi {a/b} + {c/d} + {f/g} <lb/>= {adg + bcg + bdf/bdg}.</s> <s xml:id="echoid-s1561" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1562" xml:space="preserve">Si les fractions ont déja même dénomination, on n’aura <lb/>pas la peine de les y réduire, le reſte de l’opération s’achevera <lb/>comme dans le cas précédent. </s> <s xml:id="echoid-s1563" xml:space="preserve">La raiſon de cette opération eſt <lb/>évidente, car puiſque les fractions comptent des unités de <lb/>même eſpece, étant réduites au même dénominateur, la ſomme <lb/>de ces fractions ne differe pas de celle des numérateurs, par la <lb/>même raiſon que la ſomme de ces différens nombres, 10 écus, <lb/>20 écus, 15 écus eſt égale à la ſomme des nombres 10 + 20 <lb/>+ 15 = 45 écus.</s> <s xml:id="echoid-s1564" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div112" type="section" level="1" n="96"> <head xml:id="echoid-head111" style="it" xml:space="preserve">De la Souſtraction des Fractions.</head> <p> <s xml:id="echoid-s1565" xml:space="preserve">92. </s> <s xml:id="echoid-s1566" xml:space="preserve">Si les fractions ont un même dénominateur, on fera <lb/>une nouvelle fraction, dont le numérateur ſoit égal à la dif-<lb/>férence des numérateurs des fractions propoſées, & </s> <s xml:id="echoid-s1567" xml:space="preserve">qui retien-<lb/>dra le même dénominateur. </s> <s xml:id="echoid-s1568" xml:space="preserve">Par exemple, ſi l’on veut ôter <lb/>{4/6} de {5/6}, on ôtera le numérateur 4 du numérateur 5, & </s> <s xml:id="echoid-s1569" xml:space="preserve">l’on <lb/>écrira le reſte 1 au deſſus de la barre de diviſion, en mettant <lb/>au deſſous le dénominateur pour avoir la fraction {1/6} égale à la <lb/>différence des fractions propoſées. </s> <s xml:id="echoid-s1570" xml:space="preserve">De même {5/9} - {3/9} = {2/9}, & </s> <s xml:id="echoid-s1571" xml:space="preserve"><lb/>en Algebre {a/b} - {c/b} = {a - c/b}, {d/f} - {g/f} = {d - g/f}.</s> <s xml:id="echoid-s1572" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1573" xml:space="preserve">Si les fractions n’ont pas un même dénominateur, on com-<lb/>mencera par les y réduire (n°. </s> <s xml:id="echoid-s1574" xml:space="preserve">88), & </s> <s xml:id="echoid-s1575" xml:space="preserve">le reſte ſe fera comme <lb/>dans le premier cas, ſoit ſur les fractions numériques, ſoit ſur <lb/>les fractions algébriques. </s> <s xml:id="echoid-s1576" xml:space="preserve">Par exemple, ſi l’on propoſe d’ôter <lb/>la fraction {3/7} de la fraction {2/3}, on les réduira d’abord en celles-ci <lb/>qui leur ſont égales, {14/21} & </s> <s xml:id="echoid-s1577" xml:space="preserve">{9/21}, dont la différence eſt {5/21} égale à <lb/>celle des fractions primitives {3/7} & </s> <s xml:id="echoid-s1578" xml:space="preserve">{2/3}; </s> <s xml:id="echoid-s1579" xml:space="preserve">de même {5/8} - {4/9} = {45/72} - {32/72} <lb/>= {13/72}. </s> <s xml:id="echoid-s1580" xml:space="preserve">De même pour ôter de la fraction {a/b} celle-ci {c/d}, on les <lb/>réduira d’abord au même dénominateur, & </s> <s xml:id="echoid-s1581" xml:space="preserve">prenant la diffé-<lb/>rence des numérateurs des nouvelles fractions, on aura pour <pb o="45" file="0083" n="83" rhead="DE MATHÉMATIQUE. Liv. I."/> celle des fractions propoſées {ad - bc/bd}; </s> <s xml:id="echoid-s1582" xml:space="preserve">de même encore {f/g} - {d/h} <lb/>= {fh - dg/gh}, {r/s} - {x/z} = {rz - sx/sz}, &</s> <s xml:id="echoid-s1583" xml:space="preserve">c.</s> <s xml:id="echoid-s1584" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1585" xml:space="preserve">93. </s> <s xml:id="echoid-s1586" xml:space="preserve">Si l’on avoit pluſieurs fractions à ôter de pluſieurs au-<lb/>tres fractions, on commenceroit par réduire celles que l’on <lb/>doit ôter en même dénomination (ſelon l’art. </s> <s xml:id="echoid-s1587" xml:space="preserve">88.) </s> <s xml:id="echoid-s1588" xml:space="preserve">pour avoir <lb/>une ſeule fraction égale à leur ſomme; </s> <s xml:id="echoid-s1589" xml:space="preserve">on feroit la même choſe <lb/>pour les fractions dont on doit ſouſtraire les premieres: </s> <s xml:id="echoid-s1590" xml:space="preserve">enfin <lb/>on prendra la différence de ces nouvelles fractions, & </s> <s xml:id="echoid-s1591" xml:space="preserve">l’on <lb/>aura celle des fractions propoſées. </s> <s xml:id="echoid-s1592" xml:space="preserve">Par exemple, ſi l’on veut <lb/>ôter les fractions {1/2}, {2/9}, {4/5} des fractions {2/3}, {3/4}, {5/6}, je réduis les <lb/>premieres en même dénomination, pour avoir à leur place les <lb/>fractions {45/90}, {20/90}, {72/90}, dont la ſomme eſt {137/90}. </s> <s xml:id="echoid-s1593" xml:space="preserve">Je réduis pareille-<lb/>ment les fractions {2/3}, {3/4}, {5/6} en même dénomination pour avoir à <lb/>leur place les fractions {48/72}, {54/72}, {60/72}, ou plus ſimplement {8/12}, {9/12}, <lb/>{10/12}, dont la ſomme eſt {27/12}; </s> <s xml:id="echoid-s1594" xml:space="preserve">réduiſant donc les deux fractions {137/90} <lb/>& </s> <s xml:id="echoid-s1595" xml:space="preserve">{27/12} en même dénomination, la premiere deviendra {1644/1080}, & </s> <s xml:id="echoid-s1596" xml:space="preserve"><lb/>la ſeconde {2430/1080}; </s> <s xml:id="echoid-s1597" xml:space="preserve">prenant la différence de ces fractions, on aura <lb/>celle des fractions propoſées de {786/1080}. </s> <s xml:id="echoid-s1598" xml:space="preserve">On voit par cet exemple <lb/>comment on peut déterminer laquelle de deux fractions eſt la <lb/>plus grande, & </s> <s xml:id="echoid-s1599" xml:space="preserve">de combien l’une ſurpaſſe l’autre, ce qui dans <lb/>certains cas ne s’apperçoit pas tout d’un coup comme dans ces <lb/>deux-ci, {48/55} & </s> <s xml:id="echoid-s1600" xml:space="preserve">{27/32}, à moins que l’on n’ait beaucoup d’habitude <lb/>au calcul.</s> <s xml:id="echoid-s1601" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1602" xml:space="preserve">94. </s> <s xml:id="echoid-s1603" xml:space="preserve">Si l’on vouloit ſouſtraire un entier & </s> <s xml:id="echoid-s1604" xml:space="preserve">une fraction d’un <lb/>autre entier, & </s> <s xml:id="echoid-s1605" xml:space="preserve">d’une autre fraction, il faudroit d’abord <lb/>réduire l’entier en fraction, ce qui ſe feroit en le multipliant <lb/>par le dénominateur de la fraction qui lui eſt jointe: </s> <s xml:id="echoid-s1606" xml:space="preserve">ainſi pour <lb/>que a - {cx/d} ſoit tout en fraction, il faut multiplier a par d, & </s> <s xml:id="echoid-s1607" xml:space="preserve"><lb/>écrire {ad - cx/d}; </s> <s xml:id="echoid-s1608" xml:space="preserve">de même pour ne faire qu’une ſeule fraction <lb/>de l’entier 2y + {bb/f}, l’on multipliera 2y par f pour avoir la frac-<lb/>tion {2fy + bb/f}; </s> <s xml:id="echoid-s1609" xml:space="preserve">enſuite pour ſouſtraire ces deux fractions l’une <lb/>de l’autre, par exemple, {ad-cx/d} de {2fy + bb/f}, je les réduis au <lb/>même dénominateur, & </s> <s xml:id="echoid-s1610" xml:space="preserve">j’ai pour la ſeconde {2dfy + bbd/df}, & </s> <s xml:id="echoid-s1611" xml:space="preserve">pour <lb/>la premiere {adf - cfx/df}, dont la différence eſt {2dfy + bbd - adf + cfx/df}.</s> <s xml:id="echoid-s1612" xml:space="preserve"> <pb o="46" file="0084" n="84" rhead="NOUVEAU COURS"/> Pour concevoir aiſément la raiſon de toutes ces opérations, il <lb/>ſuffit de faire attention que les fractions ayant même dénomi-<lb/>nateur, leur différence eſt préciſément celle des numérateurs; <lb/></s> <s xml:id="echoid-s1613" xml:space="preserve">car il eſt évident que la différence de {3/5} & </s> <s xml:id="echoid-s1614" xml:space="preserve">{2/5} eſt {1/5}, par la même <lb/>raiſon que la différence de 3 à 2 eſt 1.</s> <s xml:id="echoid-s1615" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div113" type="section" level="1" n="97"> <head xml:id="echoid-head112" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s1616" xml:space="preserve">95. </s> <s xml:id="echoid-s1617" xml:space="preserve">Une fraction n’eſt plus que la moitié, le tiers ou le quart <lb/>de ce qu’elle étoit, ſi on multiplie ſon dénominateur par 2, <lb/>par 3 ou par 4, puiſque le nombre des parties dans leſquelles <lb/>on diviſe l’unitéprincipale devenant double, triple ou qua-<lb/>druple, chaque partie diminue dans la même proportion, & </s> <s xml:id="echoid-s1618" xml:space="preserve"><lb/>que d’ailleurs on n’en prend que le même nombre, puiſque le <lb/>numérateur ne change pas.</s> <s xml:id="echoid-s1619" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div114" type="section" level="1" n="98"> <head xml:id="echoid-head113" style="it" xml:space="preserve">De la Multiplication des Fractions.</head> <p> <s xml:id="echoid-s1620" xml:space="preserve">96. </s> <s xml:id="echoid-s1621" xml:space="preserve">On peut multiplier une fraction par un entier ou par <lb/>une autre fraction. </s> <s xml:id="echoid-s1622" xml:space="preserve">Si le multiplicateur eſt un entier, on mul-<lb/>tipliera le numérateur de la fraction par l’entier donné, le pro-<lb/>duit ſera le numérateur d’une nouvelle fraction, qui conſer-<lb/>vera le même dénominateur que la fraction multiplicande, & </s> <s xml:id="echoid-s1623" xml:space="preserve"><lb/>cette nouvelle fraction ſera le produit cherché. </s> <s xml:id="echoid-s1624" xml:space="preserve">Par exemple, <lb/>ſi l’on veut multiplier la fraction {2/3} par l’entier 4, je multiplie <lb/>le numérateur 2 par l’entier 4, & </s> <s xml:id="echoid-s1625" xml:space="preserve">le produit 8 ſera le numéra-<lb/>teur de la fraction {8/3} égale au produit cherché. </s> <s xml:id="echoid-s1626" xml:space="preserve">De même la <lb/>fraction {4/5} x 6 = {24/5} la fraction {27/32} x 3 = {81/32}; </s> <s xml:id="echoid-s1627" xml:space="preserve">il en eſt de même <lb/>pour les fractions algébriques. </s> <s xml:id="echoid-s1628" xml:space="preserve">Le produit de {a/b} x c = {ac/b}, {a/g} x <lb/>√c+d\x{0020}={ac + ad/g}, {fg/a} x √a - b\x{0020} = {afg - bfg/a}.</s> <s xml:id="echoid-s1629" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1630" xml:space="preserve">97. </s> <s xml:id="echoid-s1631" xml:space="preserve">Si le multiplicateur eſt auſſi une fraction, on multi-<lb/>pliera les deux numérateurs l’un par l’autre, & </s> <s xml:id="echoid-s1632" xml:space="preserve">les deux déno-<lb/>minateurs de même, le produit des numérateurs ſera le nu-<lb/>mérateur d’une nouvelle fraction, dont le produit des déno-<lb/>minateurs ſera le dénominateur, laquelle fraction ſera le pro-<lb/>duit cherché: </s> <s xml:id="echoid-s1633" xml:space="preserve">ainſi {2/3} x {4/5} = {8/15}, {3/5} x {8/9} = {24/45} ou {8/15}, en les rédui-<lb/>ſant à leur plus ſimple expreſſion: </s> <s xml:id="echoid-s1634" xml:space="preserve">il en ſeroit de même ſi les <lb/>fractions étoient algébriques, {a/b} x {c/d} = {ac/bd}, {fg/a} x {bd/gh} = {bdfg/agh} = <lb/>{bfd/ah}; </s> <s xml:id="echoid-s1635" xml:space="preserve">de même {a + b/c}x{a - b/g} = {aa - bb/cg}, {a-b/f}x{c-d/g} = {ac-bc-ad+bd/fg}.</s> <s xml:id="echoid-s1636" xml:space="preserve"/> </p> <pb o="47" file="0085" n="85" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div115" type="section" level="1" n="99"> <head xml:id="echoid-head114" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s1637" xml:space="preserve">Pour entendre la raiſon de ces opérations, on fera atten-<lb/>tion qu’une fraction devient d’autant plus grande, que ſon nu-<lb/>mérateur augmente, le dénominateur reſtant le même; </s> <s xml:id="echoid-s1638" xml:space="preserve">donc <lb/>pour avoir une fraction deux ou trois fois plus grande, il ſuffit <lb/>de multiplier le numérateur par 2 ou par 3: </s> <s xml:id="echoid-s1639" xml:space="preserve">donc pour le pre-<lb/>mier cas, pour multiplier une fraction par un entier, il ſuffit <lb/>de multiplier le numérateur de la fraction par l’entier.</s> <s xml:id="echoid-s1640" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1641" xml:space="preserve">Pour le ſecond cas, lorſque le multiplicateur eſt auſſi une <lb/>fraction, on remarquera que lorſque je multiplie une fraction <lb/>{2/3}, par exemple par {4/5}, & </s> <s xml:id="echoid-s1642" xml:space="preserve">que je multiplie d’abord le numéra-<lb/>teur 2 de la premiere par le numérateur 4 de la ſeconde, je <lb/>multiplie par un nombre cinq fois trop grand, puiſque je ne <lb/>me propoſe pas de multiplier cette fraction par l’entier 4, mais <lb/>ſeulement par la cinquieme partie de 4; </s> <s xml:id="echoid-s1643" xml:space="preserve">& </s> <s xml:id="echoid-s1644" xml:space="preserve">c’eſt ce que je fais <lb/>effectivement en multipliant le dénominateur 3 par le déno-<lb/>minateur 5 (art. </s> <s xml:id="echoid-s1645" xml:space="preserve">95); </s> <s xml:id="echoid-s1646" xml:space="preserve">car après cette multiplication, les par-<lb/>ties ne ſont plus que la cinquieme partie de ce qu’elles étoient <lb/>avant.</s> <s xml:id="echoid-s1647" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1648" xml:space="preserve">98. </s> <s xml:id="echoid-s1649" xml:space="preserve">Si l’on avoit un entier & </s> <s xml:id="echoid-s1650" xml:space="preserve">une fraction à multiplier par <lb/>un entier & </s> <s xml:id="echoid-s1651" xml:space="preserve">une fraction, on donneroit à chaque entier le <lb/>même dénominateur que la fraction qui l’accompagne, en le <lb/>multipliant par le dénominateur, & </s> <s xml:id="echoid-s1652" xml:space="preserve">le diviſant par le même; <lb/></s> <s xml:id="echoid-s1653" xml:space="preserve">on multiplieroit les deux nouvelles fractions qui en réſulte-<lb/>roient l’une par l’autre, & </s> <s xml:id="echoid-s1654" xml:space="preserve">le produit ſeroit le produit que l’on <lb/>demande. </s> <s xml:id="echoid-s1655" xml:space="preserve">Par exemple, (3 + {5/6}) x (4 + {8/9}) = ({18 + 5/6}) x ({36+8/9}) <lb/>= {23/6} x {44/9} = {1012/54}; </s> <s xml:id="echoid-s1656" xml:space="preserve">de même pour multiplier {bx/a} - y par {bx/a} +y <lb/>je réduis les entiers en fractions, en le multipliant par le dé-<lb/>nominateur de la fraction, à laquelle ils ſont liés par les ſignes <lb/>+ ou -, & </s> <s xml:id="echoid-s1657" xml:space="preserve">il vient {bx - ay/a} & </s> <s xml:id="echoid-s1658" xml:space="preserve">{bx + ay/a}, & </s> <s xml:id="echoid-s1659" xml:space="preserve">multipliant les deux <lb/>numérateurs l’un par l’autre, c’eſt-à-dire bx - ay par bx + ay, <lb/>il vient bbxx - abxy + abxy - aayy ou bbxx - aayy, à qui <lb/>il faut donner pour dénominateur le produit des dénomina-<lb/>teurs des deux fractions, qui ſera aa, & </s> <s xml:id="echoid-s1660" xml:space="preserve">l’on écrira {bbxx - aayy/aa} <lb/>pour le produit de la multiplication, ou bien {bbxx/aa} - yy.</s> <s xml:id="echoid-s1661" xml:space="preserve"/> </p> <pb o="48" file="0086" n="86" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div116" type="section" level="1" n="100"> <head xml:id="echoid-head115" xml:space="preserve"><emph style="sc">Remarque</emph></head> <p> <s xml:id="echoid-s1662" xml:space="preserve">99. </s> <s xml:id="echoid-s1663" xml:space="preserve">Si dans le premier cas le multiplicateur étoit égal au <lb/>dénominateur de la fraction propoſée, le produit ſeroit égal <lb/>au numérateur, & </s> <s xml:id="echoid-s1664" xml:space="preserve">alors la multiplication ſe fait, en ôtant le <lb/>dénominateur, ainſi {2/3} x 3 = 2, {a/b} x b = b.</s> <s xml:id="echoid-s1665" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1666" xml:space="preserve">Si dans le même cas le dénominateur étoit diviſible par <lb/>l’entier propoſé, il faudroit faire la diviſion, & </s> <s xml:id="echoid-s1667" xml:space="preserve">du quotient <lb/>faire le dénominateur d’une nouvelle fraction qui auroit même <lb/>numérateur, & </s> <s xml:id="echoid-s1668" xml:space="preserve">ſeroit le produit demandé. </s> <s xml:id="echoid-s1669" xml:space="preserve">Ainſi pour multi-<lb/>plier {5/12} par 3, on diviſera le dénominateur 12 par 3, & </s> <s xml:id="echoid-s1670" xml:space="preserve">le quo-<lb/>tient 4 ſera le dénominateur d’une nouvelle fraction {5/4}, qui <lb/>conſervera le même numérateur, & </s> <s xml:id="echoid-s1671" xml:space="preserve">ſera égale au produit cher-<lb/>ché. </s> <s xml:id="echoid-s1672" xml:space="preserve">En opérant de cette maniere, la fraction qui viendra ſera <lb/>tout d’un coup réduite à ſa plus ſimple expreſſion, & </s> <s xml:id="echoid-s1673" xml:space="preserve">l’on n’a <lb/>pas deux opérations à faire. </s> <s xml:id="echoid-s1674" xml:space="preserve">Il eſt de plus évident que la frac-<lb/>tion {5/4} eſt le produit de la fraction {5/12} par 3, puiſque les par-<lb/>ties dans leſquelles on diviſe l’unité principale ſont devenues <lb/>trois fois plus grandes qu’elles n’étoient, & </s> <s xml:id="echoid-s1675" xml:space="preserve">que l’on en prend <lb/>toujours le même nombre.</s> <s xml:id="echoid-s1676" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1677" xml:space="preserve">100. </s> <s xml:id="echoid-s1678" xml:space="preserve">Dans le ſecond cas, c’eſt-à-dire lorſque le multiplica-<lb/>teur eſt auſſi une fraction, ſi le numérateur de la fraction mul-<lb/>tiplicande eſt diviſible par le dénominateur de la fraction <lb/>multiplicateur, & </s> <s xml:id="echoid-s1679" xml:space="preserve">réciproquement le dénominateur de la pre-<lb/>miere diviſible par le numérateur de la ſeconde, on fera les <lb/>diviſions, le premier quotient ſera le numérateur d’une frac-<lb/>tion, & </s> <s xml:id="echoid-s1680" xml:space="preserve">le ſecond le dénominateur de la même fraction, la-<lb/>quelle ſera le produit que l’on cherche. </s> <s xml:id="echoid-s1681" xml:space="preserve">Par exemple, ſi l’on <lb/>propoſe de multiplier la fraction {8/9} par la fraction {3/4}, dans leſ-<lb/>quelles le numérateur 8 de la premiere eſt diviſible par le dé-<lb/>nominateur 4 de la ſeconde, & </s> <s xml:id="echoid-s1682" xml:space="preserve">réciproquement le dénomina-<lb/>teur 9 de la premiere diviſible par le numérateur 3 de la ſe-<lb/>conde. </s> <s xml:id="echoid-s1683" xml:space="preserve">Je diviſe donc 8 par 4, & </s> <s xml:id="echoid-s1684" xml:space="preserve">9 par 3; </s> <s xml:id="echoid-s1685" xml:space="preserve">des quotiens 2 & </s> <s xml:id="echoid-s1686" xml:space="preserve">3, <lb/>je fais la fraction {2/3}, qui eſt le produit demandé: </s> <s xml:id="echoid-s1687" xml:space="preserve">en opérant <lb/>de cette maniere, la fraction qui vient au produit eſt tout d’un <lb/>coup réduite à ſa plus ſimple expreſſion, au lieu qu’il auroit <lb/>fallu réduire la fraction {24/36} que l’on eût trouvée, en ſuivant le <lb/>procédé ordinaire. </s> <s xml:id="echoid-s1688" xml:space="preserve">On doit faire attention à cette remarque, <lb/>lorſque les fractions que l’on veut multiplier les unes par les <lb/>autres ſont des nombres un peu conſidérables.</s> <s xml:id="echoid-s1689" xml:space="preserve"/> </p> <pb o="49" file="0087" n="87" rhead="DE MATHÉMATIQUE. Liv. I."/> <p> <s xml:id="echoid-s1690" xml:space="preserve">101. </s> <s xml:id="echoid-s1691" xml:space="preserve">Il arrive quelquefois dans ce ſecond cas, que le pro-<lb/>duit eſt plus petit que le multiplicande, ce qui paroît d’abord <lb/>ſurprenant; </s> <s xml:id="echoid-s1692" xml:space="preserve">mais on ne ſera pas long-temps embarraſſé par <lb/>cette difficulté apparente, ſi l’on fait attention à la nature de <lb/>la Multiplication, qui eſt une opération, par laquelle on cher-<lb/>che un nombre qui ſoit au multiplicande, comme le multi-<lb/>plicateur eſt à l’unité. </s> <s xml:id="echoid-s1693" xml:space="preserve">Si donc le multiplicateur eſt plus petit <lb/>que l’unité, il faut que le produit ſoit auſſi plus petit que le <lb/>multiplicande; </s> <s xml:id="echoid-s1694" xml:space="preserve">ce qui arrivera néceſſairement toutes les fois <lb/>que la fraction propoſée pour multiplicateur ne vaudra pas un <lb/>entier. </s> <s xml:id="echoid-s1695" xml:space="preserve">D’ailleurs, quand je multiplie une fraction {8/9} par une <lb/>autre {3/4}, c’eſt-à-dire que j’en prends les trois quarts, qui ſeront <lb/>certainement plus petits que cette fraction.</s> <s xml:id="echoid-s1696" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1697" xml:space="preserve">102. </s> <s xml:id="echoid-s1698" xml:space="preserve">La Multiplication des fractions ſert à faire connoître <lb/>ce que c’eſt qu’une fraction de fraction, qui paroît d’abord <lb/>quelque choſe de bien compliqué. </s> <s xml:id="echoid-s1699" xml:space="preserve">Si l’on demande, par exem-<lb/>ple, ce que vaut la moitié des trois quarts des quatre cin-<lb/>quiemes d’un écu, on multipliera, les unes par les autres, les <lb/>fractions {1/2}, {3/4}, {4/5}; </s> <s xml:id="echoid-s1700" xml:space="preserve">ce qui donnera au produit {12/40} ou {3/10}: </s> <s xml:id="echoid-s1701" xml:space="preserve">je diviſe <lb/>l’écu en dix parties pour en avoir le dixieme, il me vient 6 ſols: <lb/></s> <s xml:id="echoid-s1702" xml:space="preserve">donc {3/10} valent 18 ſols; </s> <s xml:id="echoid-s1703" xml:space="preserve">& </s> <s xml:id="echoid-s1704" xml:space="preserve">par conſéquent 18 ſols ſont la moitié <lb/>des trois quarts des quatre cinquiemes d’un écu. </s> <s xml:id="echoid-s1705" xml:space="preserve">Enfin on re-<lb/>marquera encore que l’on peut énoncer une même fraction de <lb/>pluſieurs manieres. </s> <s xml:id="echoid-s1706" xml:space="preserve">On peut dire que la fraction {3/10} d’écu vaut <lb/>les trois dixiemes d’un écu, ou la dixieme partie de trois écus. </s> <s xml:id="echoid-s1707" xml:space="preserve"><lb/>Toutes ces expreſſions reviennent abſolument au même; </s> <s xml:id="echoid-s1708" xml:space="preserve">car <lb/>ſi trois écus ſont triples d’un écu, en prenant la dixieme partie <lb/>de trois écus, on ne prend qu’un dixieme; </s> <s xml:id="echoid-s1709" xml:space="preserve">& </s> <s xml:id="echoid-s1710" xml:space="preserve">prenant les trois <lb/>dixiemes d’un écu, on en prend trois fois plus, ce qui fait <lb/>une compenſation parfaite.</s> <s xml:id="echoid-s1711" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div117" type="section" level="1" n="101"> <head xml:id="echoid-head116" style="it" xml:space="preserve">De la Diviſion des Fractions.</head> <p> <s xml:id="echoid-s1712" xml:space="preserve">103. </s> <s xml:id="echoid-s1713" xml:space="preserve">On peut diviſer une fraction par un entier, ou par <lb/>une autre fraction. </s> <s xml:id="echoid-s1714" xml:space="preserve">Si le diviſeur eſt un entier, on multipliera <lb/>le dénominateur de la fraction dividende par cet entier, & </s> <s xml:id="echoid-s1715" xml:space="preserve">le <lb/>produit ſera le dénominateur d’une nouvelle fraction, qui <lb/>ayant même numérateur, ſera le quotient demandé. </s> <s xml:id="echoid-s1716" xml:space="preserve">Pour <lb/>diviſer la fraction {3/4} par 5, on multipliera le dénominateur 4 <lb/>par l’entier 5, & </s> <s xml:id="echoid-s1717" xml:space="preserve">la fraction {3/20} eſt le quotient cherché: </s> <s xml:id="echoid-s1718" xml:space="preserve">de même <pb o="50" file="0088" n="88" rhead="NOUVEAU COURS"/> {5/6} diviſé par 3 = {5/18}, {7/5} diviſé par 6 = {7/30}. </s> <s xml:id="echoid-s1719" xml:space="preserve">La regle eſt la même <lb/>pour les quantités algébriques: </s> <s xml:id="echoid-s1720" xml:space="preserve">{a/b} diviſé par c = {a/bc}; </s> <s xml:id="echoid-s1721" xml:space="preserve">la frac-<lb/>tion {fg + gh/c} diviſée par d = {fg + gh/cd}, {aa - bb/a} diviſé par a + b <lb/>= {aa - bb/c x c + b} = {a - b/c}, car aa - bb eſt le produit de a + b par <lb/>a - b; </s> <s xml:id="echoid-s1722" xml:space="preserve">donc a + b ſe trouve un diviſeur commun au numé-<lb/>rateur & </s> <s xml:id="echoid-s1723" xml:space="preserve">au dénominateur, & </s> <s xml:id="echoid-s1724" xml:space="preserve">par conſéquent la fraction eſt <lb/>réductible.</s> <s xml:id="echoid-s1725" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1726" xml:space="preserve">104. </s> <s xml:id="echoid-s1727" xml:space="preserve">Si le numérateur de la fraction dividende étoit divi-<lb/>ſible par l’entier donné, on feroit la diviſion, afin de n’être <lb/>point obligé de réduire la fraction qui viendroit au quotient, <lb/>& </s> <s xml:id="echoid-s1728" xml:space="preserve">qui ſeroit néceſſairement réductible ſi l’on multiplioit le dé-<lb/>nominateur par l’entier propoſé pour diviſeur: </s> <s xml:id="echoid-s1729" xml:space="preserve">ainſi la frac-<lb/>tion {8/9} diviſée par 4 = {2/9}, {35/48} diviſé par 7 = {5/48}, en général {ab/c} <lb/>diviſé par b = {a/c}, {fgh/cd} diviſé par gh = {f/cd}. </s> <s xml:id="echoid-s1730" xml:space="preserve">La raiſon de toutes <lb/>ces opérations ſe tire toujours du même principe; </s> <s xml:id="echoid-s1731" xml:space="preserve">car diviſer <lb/>une fraction par un entier, comme 2, 3 ou 4, c’eſt en cher-<lb/>cher une qui ne ſoit que la moitié, le tiers ou le quart de la <lb/>fraction propoſée, & </s> <s xml:id="echoid-s1732" xml:space="preserve">c’eſt ce que l’on exécute effectivement, <lb/>en ſuivant l’une ou l’autre méthode. </s> <s xml:id="echoid-s1733" xml:space="preserve">Dans la premiere, lorſ-<lb/>qu’on multiplie le dénominateur, les parties dans leſquelles on <lb/>diviſe l’unité principale, ne ſont plus que la moitié, le tiers ou <lb/>le quart de ce qu’elles étoient, puiſque leur nombre devient <lb/>double ou triple, ou quadruple: </s> <s xml:id="echoid-s1734" xml:space="preserve">donc la fraction n’eſt plus <lb/>auſſi que la moitié, le tiers ou le quart de ce qu’elle étoit, <lb/>puiſque l’on ne touche pas au numérateur. </s> <s xml:id="echoid-s1735" xml:space="preserve">Dans la ſeconde <lb/>pratique, les parties reſtent bien les mêmes, puiſque l’on ne <lb/>touche pas au dénominateur; </s> <s xml:id="echoid-s1736" xml:space="preserve">mais la fraction diminue par la <lb/>diviſion du numérateur, qui n’eſt plus que la moitié, le tiers <lb/>ou le quart de ce qu’il étoit, ſuivant qu’il a été diviſé par 2 ou <lb/>par 3, ou par 4. </s> <s xml:id="echoid-s1737" xml:space="preserve">Seulement il eſt à remarquer que l’une de ces <lb/>deux méthodes peut toujours avoir lieu, puiſqu’il eſt toujours <lb/>poſſible de multiplier un nombre par un autre, & </s> <s xml:id="echoid-s1738" xml:space="preserve">que la ſe-<lb/>conde n’eſt d’uſage que lorſque le numérateur eſt diviſible par <lb/>l’entier donné; </s> <s xml:id="echoid-s1739" xml:space="preserve">auquel cas on doit préférer cette méthode à la <lb/>plus générale, pour que la fraction ſoit réduite à ſes moindres <lb/>termes dès la premiere opération.</s> <s xml:id="echoid-s1740" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1741" xml:space="preserve">105. </s> <s xml:id="echoid-s1742" xml:space="preserve">Si le diviſeur eſt auſſi une fraction, on multipliera le <pb o="51" file="0089" n="89" rhead="DE MATHÉMATIQUE. Liv. I."/> numérateur de la fraction dividende par le dénominateur de <lb/>la fraction diviſeur, & </s> <s xml:id="echoid-s1743" xml:space="preserve">le dénominateur de la même fraction <lb/>dividende par le numérateur de l’autre, c’eſt ce qu’on appelle <lb/>ordinairement multiplier en croix. </s> <s xml:id="echoid-s1744" xml:space="preserve">Cette regle eſt générale <lb/>pour les fractions numériques & </s> <s xml:id="echoid-s1745" xml:space="preserve">algébriques; </s> <s xml:id="echoid-s1746" xml:space="preserve">ainſi pour di-<lb/>viſer la fraction {2/3} par la fraction {4/5}, je multiplie le numérateur <lb/>2 de la premiere fraction dividende par le dénominateur 5 <lb/>de la fraction diviſeur; </s> <s xml:id="echoid-s1747" xml:space="preserve">je multiplie de même le dénomi-<lb/>nateur 3 de la premiere par le numérateur 4 de la ſeconde, & </s> <s xml:id="echoid-s1748" xml:space="preserve"><lb/>mettant les deux produits 10, 12 en fraction, j’ai pour quo-<lb/>tient des fractions données, diviſées l’une par l’autre, la frac-<lb/>tion {10/12} ou {5/6} qui lui eſt égale. </s> <s xml:id="echoid-s1749" xml:space="preserve">De même la fraction {15/17} diviſée <lb/>par {3/4} = {15 x 4/17 x 3} = {60/51} = {20/17}, en réduiſant le produit. </s> <s xml:id="echoid-s1750" xml:space="preserve">En général <lb/>une fraction {a/b} diviſée par {c/d} = {a x d/b x c} = {ad/bc}, une fraction {df + gh/b} <lb/>diviſée par la fraction {a/c} = {cdf + cgh/ab}, {a + b/c} x {df/a - b} = {√a + b\x{0020} x √a - b\x{0020}/cdf} <lb/>= {aa - bb/cdf}, & </s> <s xml:id="echoid-s1751" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s1752" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div118" type="section" level="1" n="102"> <head xml:id="echoid-head117" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s1753" xml:space="preserve">La raiſon de cette opération eſt toujours déduite des mê-<lb/>mes principes que les précédentes. </s> <s xml:id="echoid-s1754" xml:space="preserve">Quand je multiplie le dé-<lb/>nominateur 3 de la fraction {2/3} par le numérateur 4 de la frac-<lb/>tion {4/5}, je rends la fraction propoſée cinq fois plus petite <lb/>(art. </s> <s xml:id="echoid-s1755" xml:space="preserve">103.) </s> <s xml:id="echoid-s1756" xml:space="preserve">que je ne me propoſe de le faire, puiſque je ne veux <lb/>pas la diviſer par quatre entier, mais ſeulement par la cinquieme <lb/>partie de 4, puiſque la fraction {4/5} ne vaut que cela (art. </s> <s xml:id="echoid-s1757" xml:space="preserve">102); <lb/></s> <s xml:id="echoid-s1758" xml:space="preserve">donc il faut la rendre cinq fois plus grande pour la remettre dans <lb/>l’état où elle doit être; </s> <s xml:id="echoid-s1759" xml:space="preserve">c’eſt ce que je fais en multipliant en-<lb/>ſuite le numérateur de la fraction dividende par le dénomina-<lb/>teur 5 de la fraction diviſeur. </s> <s xml:id="echoid-s1760" xml:space="preserve">La démonſtration ſubſiſte tou-<lb/>jours dans toute ſa force pour les fractions algébriques, cepen-<lb/>dant on peut la prouver directement comme il ſuit.</s> <s xml:id="echoid-s1761" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1762" xml:space="preserve">Pour prouver que la fraction {a/b}, diviſée par la fraction {c/d} <lb/>donne au quotient {ad/bc}, nous ſuppoſerons que {a/b} = f, & </s> <s xml:id="echoid-s1763" xml:space="preserve">que <lb/>{c/d} = g, & </s> <s xml:id="echoid-s1764" xml:space="preserve">nous ferons voir que {ad/bc} = {f/g}: </s> <s xml:id="echoid-s1765" xml:space="preserve">pour cela, faites at-<lb/>tention que puiſque l’on a par hypotheſe {a/b} = f, & </s> <s xml:id="echoid-s1766" xml:space="preserve">{c/d} = g, on <pb o="52" file="0090" n="90" rhead="NOUVEAU COURS"/> aura a = bf, & </s> <s xml:id="echoid-s1767" xml:space="preserve">c = dg. </s> <s xml:id="echoid-s1768" xml:space="preserve">Mettant donc ces valeurs de a & </s> <s xml:id="echoid-s1769" xml:space="preserve"><lb/>de c dans la fraction {ad/bc}, on aura la nouvelle fraction {bfd/bdg}, qui <lb/>étant réduite à ſa plus ſimple expreſſion, devient {f/g}; </s> <s xml:id="echoid-s1770" xml:space="preserve">donc {ad/bc} <lb/>= {f/g}: </s> <s xml:id="echoid-s1771" xml:space="preserve">C. </s> <s xml:id="echoid-s1772" xml:space="preserve">Q. </s> <s xml:id="echoid-s1773" xml:space="preserve">F. </s> <s xml:id="echoid-s1774" xml:space="preserve">D.</s> <s xml:id="echoid-s1775" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1776" xml:space="preserve">106. </s> <s xml:id="echoid-s1777" xml:space="preserve">Si le numérateur de la fraction dividende eſt diviſible <lb/>par le numérateur du diviſeur, & </s> <s xml:id="echoid-s1778" xml:space="preserve">le dénominateur de la même <lb/>fraction diviſible par celui du diviſeur, il faudra faire les di-<lb/>viſions, & </s> <s xml:id="echoid-s1779" xml:space="preserve">les quotiens mis en fraction, ſeront le quotient de-<lb/>mandé, qui ſe trouvera de cette maniere réduit à ſa plus ſimple <lb/>expreſſion. </s> <s xml:id="echoid-s1780" xml:space="preserve">Par exemple, pour diviſer la fraction {8/9} par la frac-<lb/>tion {2/3}, je diviſe le numérateur 8 par le numérateur 2, & </s> <s xml:id="echoid-s1781" xml:space="preserve">le <lb/>dénominateur 9 par le dénominateur 3, avec les quotiens 4 <lb/>& </s> <s xml:id="echoid-s1782" xml:space="preserve">3, je fais la fraction {4/3}, qui eſt le quotient que l’on demande. <lb/></s> <s xml:id="echoid-s1783" xml:space="preserve">En ſuivant la regle générale, on auroit multiplié 8 par 3, & </s> <s xml:id="echoid-s1784" xml:space="preserve"><lb/>9 par 2, ce qui auroit donné la fraction {24/18}, qui ne vaut en <lb/>effet que {4/3}, en diviſant ſes deux termes par 6, qui leur eſt com-<lb/>mun. </s> <s xml:id="echoid-s1785" xml:space="preserve">Il ſera toujours poſſible de faire la diviſion, en ſuivant <lb/>la regle générale, mais il faut préférer cette derniere à la pre-<lb/>miere, lorſque la diviſion peut ſe faire.</s> <s xml:id="echoid-s1786" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1787" xml:space="preserve">107. </s> <s xml:id="echoid-s1788" xml:space="preserve">Si l’on avoit un entier & </s> <s xml:id="echoid-s1789" xml:space="preserve">une fraction à diviſer par un <lb/>entier & </s> <s xml:id="echoid-s1790" xml:space="preserve">une fraction, on réduiroit chaque entier en fraction, <lb/>qui auroit même dénominateur que la fraction à laquelle il eſt <lb/>uni par les ſignes + & </s> <s xml:id="echoid-s1791" xml:space="preserve">-, & </s> <s xml:id="echoid-s1792" xml:space="preserve">l’on feroit la diviſion de ces <lb/>fractions, ſuivant l’une des regles précédentes. </s> <s xml:id="echoid-s1793" xml:space="preserve">Ainſi pour di-<lb/>viſer 6 + {3/4} par 2 + {5/6}, je change la premiere en {27/4}, & </s> <s xml:id="echoid-s1794" xml:space="preserve">la ſe-<lb/>conde en {17/6}, je multiplie ces deux fractions en croix, & </s> <s xml:id="echoid-s1795" xml:space="preserve">j’ai <lb/>pour le quotient {162/68} ou {81/34}, qui eſt irréductible.</s> <s xml:id="echoid-s1796" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1797" xml:space="preserve">108. </s> <s xml:id="echoid-s1798" xml:space="preserve">Il y a encore une autre maniere de diviſer une fraction <lb/>par une autre fraction, en opérant ſur le numérateur ou ſur <lb/>le dénominateur ſeulement. </s> <s xml:id="echoid-s1799" xml:space="preserve">On opére ſur le numérateur ſeu-<lb/>lement, lorſque le numérateur du dividende eſt diviſible par <lb/>celui du diviſeur; </s> <s xml:id="echoid-s1800" xml:space="preserve">& </s> <s xml:id="echoid-s1801" xml:space="preserve">voici ce qu’on fait en ce cas: </s> <s xml:id="echoid-s1802" xml:space="preserve">on diviſe <lb/>le numérateur du dividende par celui du diviſeur, & </s> <s xml:id="echoid-s1803" xml:space="preserve">enſuite <lb/>on multiplie le quotient par le dénominateur du même divi-<lb/>ſeur, le produit étant diviſé par le dénominateur du dividende, <lb/>donne le quotient des deux fractions. </s> <s xml:id="echoid-s1804" xml:space="preserve">Par exemple, ſi l’on <lb/>propoſe de diviſer la fraction {18/49} par la fraction {3/5}, je diviſe le <lb/>numérateur 18 du dividende par le numérateur 3 du diviſeur, <pb o="53" file="0091" n="91" rhead="DE MATHÉMATIQUE. Liv. I."/> le quotient eſt 6, que je multiplie par 5, dénominateur du di-<lb/>viſeur, le produit 30 diviſé par 49 me donne une fraction {30/49} <lb/>égale au quotient que je cherche: </s> <s xml:id="echoid-s1805" xml:space="preserve">cette pratique ſe déduit tou-<lb/>jours des mêmes principes. </s> <s xml:id="echoid-s1806" xml:space="preserve">Quand je diviſe 18 par 3, j’ai une <lb/>fraction cinq fois plus petite que celle que je cherche, car ce <lb/>n’eſt pas par 3 que je veux la diviſer, mais par {3/5}, ou la cin-<lb/>quieme partie de 3; </s> <s xml:id="echoid-s1807" xml:space="preserve">c’eſt donc pour rétablir cette trop grande <lb/>diminution, que je multiplie par 5 le quotient que j’ai trouvé.</s> <s xml:id="echoid-s1808" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1809" xml:space="preserve">On opére ſur le dénominateur ſeulement, lorſque le déno-<lb/>minateur du dividende eſt diviſible par le dénominateur du <lb/>diviſeur, & </s> <s xml:id="echoid-s1810" xml:space="preserve">voici ce que l’on fait: </s> <s xml:id="echoid-s1811" xml:space="preserve">On diviſe le dénomina-<lb/>teur du dividende par celui du diviſeur, & </s> <s xml:id="echoid-s1812" xml:space="preserve">on multiplie le <lb/>quotient par le numérateur du diviſeur; </s> <s xml:id="echoid-s1813" xml:space="preserve">ce nouveau produit <lb/>ſert de dénominateur à une fraction qui retient toujours le <lb/>même numérateur que la fraction dividende, & </s> <s xml:id="echoid-s1814" xml:space="preserve">cette fraction <lb/>eſt le quotient cherché. </s> <s xml:id="echoid-s1815" xml:space="preserve">Par exemple, pour diviſer la fraction <lb/>{18/49} par la fraction {5/7}, je diviſe le dénominateur 49 par 7; </s> <s xml:id="echoid-s1816" xml:space="preserve">je <lb/>multiplie le quotient 7 par le numérateur 5 du diviſeur, le pro-<lb/>duit eſt 35, que je fais ſervir de dénominateur à une nouvelle <lb/>fraction, dont le numérateur eſt toujours 18, & </s> <s xml:id="echoid-s1817" xml:space="preserve">j’ai {18/35} pour <lb/>le quotient demandé. </s> <s xml:id="echoid-s1818" xml:space="preserve">La raiſon de cette méthode eſt encore <lb/>aiſée à déduire des principes que l’on a donnés. </s> <s xml:id="echoid-s1819" xml:space="preserve">Quand je diviſe <lb/>le dénominateur du dividende par le dénominateur 7 du di-<lb/>viſeur, j’ai une fraction ſept fois plus grande que la précé-<lb/>dente, maisje veux qu’elle ſoit ſeulement {5/7} de fois plus grande <lb/>que la propoſée; </s> <s xml:id="echoid-s1820" xml:space="preserve">donc il faut multiplier le nouveau quotient, <lb/>afin que par la multiplication du dénominateur il y ait une com-<lb/>penſation de ce que l’on avoit fait de trop. </s> <s xml:id="echoid-s1821" xml:space="preserve">En général on doit <lb/>encore préférer ces méthodes à la méthode générale, lorſ-<lb/>qu’elles peuvent avoir lieu; </s> <s xml:id="echoid-s1822" xml:space="preserve">car en opérant ainſi, les quotiens <lb/>ſeront irréductibles ſi le dividende avoit été réduit à ſa plus <lb/>ſimple expreſſion avant de commencer la Diviſion: </s> <s xml:id="echoid-s1823" xml:space="preserve">dans les <lb/>exemples précédens, ſi l’on eût ſuivi la regle générale, on eût <lb/>trouvé pour le premier {90/147}, pour le ſecond {126/245}, au lieu des frac-<lb/>tions {30/49} & </s> <s xml:id="echoid-s1824" xml:space="preserve">{18/35}.</s> <s xml:id="echoid-s1825" xml:space="preserve"/> </p> <figure> <image file="0091-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0091-01"/> </figure> <pb o="54" file="0092" n="92" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div119" type="section" level="1" n="103"> <head xml:id="echoid-head118" xml:space="preserve">TRAITÉ</head> <head xml:id="echoid-head119" style="it" xml:space="preserve">DES FRACTIONS DÉCIMALES.</head> <p> <s xml:id="echoid-s1826" xml:space="preserve">109. </s> <s xml:id="echoid-s1827" xml:space="preserve">OUtre les fractions dont nous venons de parler, il y en <lb/>a encore d’autres qui ſont d’un grand uſage en Mathématique, <lb/>& </s> <s xml:id="echoid-s1828" xml:space="preserve">dont la connoiſſance eſt abſolument néceſſaire pour avoir <lb/>dans certaines occaſions les grandeurs dont on a beſoin avec <lb/>toute la préciſion poſſible.</s> <s xml:id="echoid-s1829" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div120" type="section" level="1" n="104"> <head xml:id="echoid-head120" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s1830" xml:space="preserve">110. </s> <s xml:id="echoid-s1831" xml:space="preserve">Si on diviſe un tout ou unité principale par l’unité, <lb/>ſuivie d’un ou de pluſieurs zero, par les nombres 10, 100, <lb/>1000, 10000, &</s> <s xml:id="echoid-s1832" xml:space="preserve">c, qui ſont les puiſſances ſucceſſives de 10, <lb/>& </s> <s xml:id="echoid-s1833" xml:space="preserve">que l’on prenne pluſieurs de ces parties égales, la fraction <lb/>qui marque combien on prend de ces parties égales, eſt ap-<lb/>pellée fraction décimale, & </s> <s xml:id="echoid-s1834" xml:space="preserve">ſe marque ainſi: </s> <s xml:id="echoid-s1835" xml:space="preserve">{3/10}, {7/100}, {48/1000}, & </s> <s xml:id="echoid-s1836" xml:space="preserve"><lb/>ainſi des autres.</s> <s xml:id="echoid-s1837" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1838" xml:space="preserve">On a trouvé le ſecret d’opérer ſur ces ſortes de fractions, <lb/>préciſément de la même maniere que l’on opére ſur les nom-<lb/>bres naturels; </s> <s xml:id="echoid-s1839" xml:space="preserve">& </s> <s xml:id="echoid-s1840" xml:space="preserve">de plus, de réduire toute fraction donnée à <lb/>une fraction décimale qui lui ſoit égale, ou qui n’en différe <lb/>que d’une quantité infiniment petite, & </s> <s xml:id="echoid-s1841" xml:space="preserve">c’eſt ce qui a rendu <lb/>leur uſage ſi fréquent dans les Mathématiques.</s> <s xml:id="echoid-s1842" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div121" type="section" level="1" n="105"> <head xml:id="echoid-head121" xml:space="preserve"><emph style="sc">Premier principe</emph>.</head> <p> <s xml:id="echoid-s1843" xml:space="preserve">111. </s> <s xml:id="echoid-s1844" xml:space="preserve">Puiſque les fractions décimales ſont des fractions, on <lb/>peut les exprimer comme les autres fractions; </s> <s xml:id="echoid-s1845" xml:space="preserve">ainſi pour mar-<lb/>quer 3 dixiemes, 58 centiemes on peut écrire {3/10}, {58/100}: </s> <s xml:id="echoid-s1846" xml:space="preserve">mais il <lb/>y a une autre maniere de les marquer, c’eſt d’écrire le numé-<lb/>rateur ſeulement, en ſous-entendant le dénominateur. </s> <s xml:id="echoid-s1847" xml:space="preserve">Par <lb/>exemple, au lieu d’écrire {3/10}, {58/100}, on écrit. </s> <s xml:id="echoid-s1848" xml:space="preserve">3. </s> <s xml:id="echoid-s1849" xml:space="preserve">58, en mettant <lb/>un point ſur la gauche du numérateur, de maniere qu’il y ait <lb/>après ce point autant de chiffres qu’il y auroit de zero au dé-<lb/>nominateur après l’unité; </s> <s xml:id="echoid-s1850" xml:space="preserve">de même s’il y avoit des entiers <lb/>joints aux fractions, comme 15 {25/100}, 38 {245/1000}, on pourroit écrire <lb/>15. </s> <s xml:id="echoid-s1851" xml:space="preserve">25 & </s> <s xml:id="echoid-s1852" xml:space="preserve">38. </s> <s xml:id="echoid-s1853" xml:space="preserve">245. </s> <s xml:id="echoid-s1854" xml:space="preserve">De cette maniere, quoique le dénomina-<lb/>teur ne ſoit pas exprimé, on peut cependant toujours le con- <pb o="55" file="0093" n="93" rhead="DE MATHÉMATIQUE. Liv. I."/> noître: </s> <s xml:id="echoid-s1855" xml:space="preserve">car s’il y a deux chiffres après le point, on concluera <lb/>que le dénominateur eſt 100, s’il y en a trois, on concluera <lb/>que le dénominateur eſt 1000, & </s> <s xml:id="echoid-s1856" xml:space="preserve">ainſi de ſuite.</s> <s xml:id="echoid-s1857" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1858" xml:space="preserve">112. </s> <s xml:id="echoid-s1859" xml:space="preserve">Il ſuit delà que ſi l’on a des expreſſions, comme 253. <lb/></s> <s xml:id="echoid-s1860" xml:space="preserve">27, cela ſignifie 253 {27/100}, de même que 483. </s> <s xml:id="echoid-s1861" xml:space="preserve">547 ſignifie 483 <lb/>entiers {547/1000}. </s> <s xml:id="echoid-s1862" xml:space="preserve">Il ſuit encore delà que ſi l’on veut mettre ſous <lb/>cette forme la quantité 28 {3/100}, il faudra l’écrire ainſi, 28. </s> <s xml:id="echoid-s1863" xml:space="preserve">03, <lb/>en mettant un zero devant le 3, afin qu’il y ait deux chiffres <lb/>après le point, pour que l’on connoiſſe que le dénominateur <lb/>eſt l’unité ſuivie de deux zero ou 100. </s> <s xml:id="echoid-s1864" xml:space="preserve">De même pour mettre <lb/>ſous cette forme 53 {48/10000}, on écrira 53. </s> <s xml:id="echoid-s1865" xml:space="preserve">0048, en mettant <lb/>deux zero avant les chiffres 48, pour marquer que le déno-<lb/>minateur a quatre zero après l’unité, & </s> <s xml:id="echoid-s1866" xml:space="preserve">compte des dix mil-<lb/>liemes.</s> <s xml:id="echoid-s1867" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1868" xml:space="preserve">113. </s> <s xml:id="echoid-s1869" xml:space="preserve">S’il n’y avoit point d’entiers avec la fraction, mais ſeu-<lb/>lement {325/1000}, on écriroit ainſi: </s> <s xml:id="echoid-s1870" xml:space="preserve">0.</s> <s xml:id="echoid-s1871" xml:space="preserve">325, en faiſant voir par le <lb/>zero mis avant le point, qu’il n’y a pas d’entier. </s> <s xml:id="echoid-s1872" xml:space="preserve">Si l’on fait <lb/>bien attention, on verra que cette expreſſion 0.</s> <s xml:id="echoid-s1873" xml:space="preserve">325 eſt égale à <lb/>{3/10} + {2/100} + {5/1000}; </s> <s xml:id="echoid-s1874" xml:space="preserve">car {3/10} eſt égale à {30/100}, à {300/1000}, & </s> <s xml:id="echoid-s1875" xml:space="preserve">{2/100} = {20/1000}, <lb/>puiſqu’une fraction ne change pas de valeur lorſqu’on multiplie <lb/>ſon numérateur & </s> <s xml:id="echoid-s1876" xml:space="preserve">ſon dénominateur par un même nombre: <lb/></s> <s xml:id="echoid-s1877" xml:space="preserve">donc au lieu d’exprimer la fraction 0.</s> <s xml:id="echoid-s1878" xml:space="preserve">325 en diſant 325 cen-<lb/>tiemes, on auroit pu l’énoncer ainſi: </s> <s xml:id="echoid-s1879" xml:space="preserve">3 dixiemes, 2 centiemes, <lb/>5 milliemes; </s> <s xml:id="echoid-s1880" xml:space="preserve">ce qui fait voir que les chiffres de cette quantité <lb/>0.</s> <s xml:id="echoid-s1881" xml:space="preserve">325 vont en augmentant en proportion décuple, en allant <lb/>de droite à gauche, & </s> <s xml:id="echoid-s1882" xml:space="preserve">diminuent dans la même proportion, <lb/>en allant de gauche à droite: </s> <s xml:id="echoid-s1883" xml:space="preserve">car il eſt évident qu’un centieme <lb/>eſt dix fois plus grand qu’un millieme, & </s> <s xml:id="echoid-s1884" xml:space="preserve">qu’un dixieme eſt <lb/>dix fois plus grand qu’un centieme. </s> <s xml:id="echoid-s1885" xml:space="preserve">En conſidérant les frac-<lb/>tions décimales ſous ce point de vue, on peut les définir en <lb/>diſant que ce ſont des nombres moindres que les entiers qui <lb/>ſuivent la proportion des différens ordres de la numération.</s> <s xml:id="echoid-s1886" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1887" xml:space="preserve">En effet, après avoir fixé le terme des unités ou nombres en-<lb/>tiers, rien n’empêche d’imaginer d’autres nombres, dont les <lb/>unités ſuivent toujours la même progreſſion, ainſi que dans ce <lb/>nombre 6325.</s> <s xml:id="echoid-s1888" xml:space="preserve">489, dans lequel les unités du premier chiffre 2, <lb/>qui eſt à la gauche du 5, où ſe terminent les entiers, ſont dix fois <lb/>plus grandes que les unités du même 5, & </s> <s xml:id="echoid-s1889" xml:space="preserve">les unités du 4 qui <lb/>eſt immédiatement à la droite du même 5, ſont dix fois plus <lb/>petites que les unités du 5, ou les unités du 3 qui occupe le <pb o="56" file="0094" n="94" rhead="NOUVEAU COURS"/> ſecond rang à la gauche du chiffre 5 des unités, ſont cent fois <lb/>plus grandes que celles du même 5, & </s> <s xml:id="echoid-s1890" xml:space="preserve">les unités du 8 qui tient <lb/>le ſecond rang vers la droite, après le 5, ſont cent fois plus <lb/>petites que les unités du même 5, & </s> <s xml:id="echoid-s1891" xml:space="preserve">ainſi des autres qui pour-<lb/>roient occuper des rangs égaux, tant vers la droite, que vers <lb/>la gauche du chiffre des unités: </s> <s xml:id="echoid-s1892" xml:space="preserve">enſorte que l’on peut dire, en <lb/>partant de ce chiffre vers la droite, unités, dixiemes, cen-<lb/>tiemes, milliemes, &</s> <s xml:id="echoid-s1893" xml:space="preserve">c, de même que l’on dit, en partant de <lb/>ce même chiffre vers la gauche, unités, dixaines, centaines, <lb/>mille, &</s> <s xml:id="echoid-s1894" xml:space="preserve">c. </s> <s xml:id="echoid-s1895" xml:space="preserve">Cette maniere d’enviſager les fractions décimales <lb/>jette un grand jour dans toutes les opérations que l’on fait ſur <lb/>elles, & </s> <s xml:id="echoid-s1896" xml:space="preserve">l’on ne peut ſe la rendre trop familiere.</s> <s xml:id="echoid-s1897" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div122" type="section" level="1" n="106"> <head xml:id="echoid-head122" xml:space="preserve"><emph style="sc">Second principe</emph>.</head> <p> <s xml:id="echoid-s1898" xml:space="preserve">114. </s> <s xml:id="echoid-s1899" xml:space="preserve">Pluſieurs fractions décimales, comme 0 3, 0.</s> <s xml:id="echoid-s1900" xml:space="preserve">54, 0.</s> <s xml:id="echoid-s1901" xml:space="preserve">008, <lb/>ou leurs égales, {3/10}, {54/100}, {8/100}, étant ſous leur premiere forme, <lb/>pourront aiſément ſe réduire à la même dénomination; </s> <s xml:id="echoid-s1902" xml:space="preserve">car <lb/>{3/10}, comme on l’a déja dit, eſt égal à {30/100}, à {300/1000}, & </s> <s xml:id="echoid-s1903" xml:space="preserve">{54/100} eſt égal <lb/>à {540/1000}: </s> <s xml:id="echoid-s1904" xml:space="preserve">donc les fractions propoſées pourront auſſi s’écrire ſous <lb/>cette forme, 0.</s> <s xml:id="echoid-s1905" xml:space="preserve">300, 0.</s> <s xml:id="echoid-s1906" xml:space="preserve">540, 0.</s> <s xml:id="echoid-s1907" xml:space="preserve">008. </s> <s xml:id="echoid-s1908" xml:space="preserve">Il eſt évident que ces chan-<lb/>gemens ne font point changer la valeur des fractions, puiſque <lb/>l’on ne fait par cette opération que multiplier les numérateurs <lb/>& </s> <s xml:id="echoid-s1909" xml:space="preserve">dénominateurs par les mêmes nombres. </s> <s xml:id="echoid-s1910" xml:space="preserve">Ces principes une <lb/>fois bien compris, il eſt aiſé de voir que l’on peut opérer ſur <lb/>les fractions comme ſur les nombres entiers; </s> <s xml:id="echoid-s1911" xml:space="preserve">& </s> <s xml:id="echoid-s1912" xml:space="preserve">comme l’on <lb/>peut réduire toute fraction en fraction décimale qui lui ſoit <lb/>égale, ou qui n’en différe pas ſenſiblement, il ſuit auſſi que <lb/>l’on peut rappeller toutes les opérations des fractions à celles <lb/>des nombres entiers: </s> <s xml:id="echoid-s1913" xml:space="preserve">c’eſt pourquoi nous n’entrerons pas dans <lb/>un grand détail d’exemples. </s> <s xml:id="echoid-s1914" xml:space="preserve">Nous allons commencer par ex-<lb/>pliquer l’art de faire ſur ces quantités les quatre Regles prin-<lb/>cipales de l’Arithmétique; </s> <s xml:id="echoid-s1915" xml:space="preserve">nous donnerons enſuite la maniere <lb/>de réduire une fraction quelconque en décimales, & </s> <s xml:id="echoid-s1916" xml:space="preserve">les diffé-<lb/>rentes applications que l’on peut faire de ces opérations aux <lb/>calculs qui ſont le plus en uſage.</s> <s xml:id="echoid-s1917" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div123" type="section" level="1" n="107"> <head xml:id="echoid-head123" style="it" xml:space="preserve">De l’Addition des Fractions décimales.</head> <p> <s xml:id="echoid-s1918" xml:space="preserve">115. </s> <s xml:id="echoid-s1919" xml:space="preserve">Si les fractions propoſées ne ſont pas réduites à la même <lb/>dénomination, on commencera par les y réduire (art. </s> <s xml:id="echoid-s1920" xml:space="preserve">113):</s> <s xml:id="echoid-s1921" xml:space="preserve"> <pb o="57" file="0095" n="95" rhead="DE MATHÉMATIQUE. Liv. I."/> cette préparation faite, on les rangera les unes ſous les autres, <lb/>de maniere que les dixiemes ſoient ſous les dixiemes, les cen-<lb/>tiemes ſous les centiemes, les milliemes ſous les milliemes, <lb/>formant chacun une colonne verticale, & </s> <s xml:id="echoid-s1922" xml:space="preserve">l’on fera l’Addition <lb/>ſuivant les regles que l’on a données pour l’Addition des nom-<lb/>bres entiers. </s> <s xml:id="echoid-s1923" xml:space="preserve">Par exemple, ſi l’on veut avoir la ſomme des <lb/>fractions 0.</s> <s xml:id="echoid-s1924" xml:space="preserve">3, 0.</s> <s xml:id="echoid-s1925" xml:space="preserve">25, 0.</s> <s xml:id="echoid-s1926" xml:space="preserve">489, 0.</s> <s xml:id="echoid-s1927" xml:space="preserve">056, on les réduira en même dé-<lb/>nomination que le nombre 0.</s> <s xml:id="echoid-s1928" xml:space="preserve">489, ou 0.</s> <s xml:id="echoid-s1929" xml:space="preserve">056, dont chacun a <lb/>des milliemes, & </s> <s xml:id="echoid-s1930" xml:space="preserve">l’on aura, en les diſpoſant par ordre comme <lb/>il convient, <lb/> <anchor type="note" xlink:label="note-0095-01a" xlink:href="note-0095-01"/> dont la ſomme ſe trouvera être de # 1.</s> <s xml:id="echoid-s1931" xml:space="preserve">095, # c’eſt-à-dire l’entier, <lb/>& </s> <s xml:id="echoid-s1932" xml:space="preserve">{95/1000}.</s> <s xml:id="echoid-s1933" xml:space="preserve"/> </p> <div xml:id="echoid-div123" type="float" level="2" n="1"> <note position="right" xlink:label="note-0095-01" xlink:href="note-0095-01a" xml:space="preserve"># 0.300 <lb/># 0.250 <lb/># 0.489 <lb/># 0.056 <lb/></note> </div> <p> <s xml:id="echoid-s1934" xml:space="preserve">116. </s> <s xml:id="echoid-s1935" xml:space="preserve">S’il y avoit des entiers joints aux <lb/> <anchor type="note" xlink:label="note-0095-02a" xlink:href="note-0095-02"/> fractions, comme dans les nombres ſui-<lb/>vans, 25.</s> <s xml:id="echoid-s1936" xml:space="preserve">43, 3.</s> <s xml:id="echoid-s1937" xml:space="preserve">054, 69.</s> <s xml:id="echoid-s1938" xml:space="preserve">067, 36.</s> <s xml:id="echoid-s1939" xml:space="preserve">48, ce <lb/>ſeroit préciſément la même opération, & </s> <s xml:id="echoid-s1940" xml:space="preserve"><lb/>l’on auroit, en les ajoutant comme on voit <lb/>ici, après les avoir réduit à la dénomination <lb/>de 3.</s> <s xml:id="echoid-s1941" xml:space="preserve">054, 134.</s> <s xml:id="echoid-s1942" xml:space="preserve">031, c’eſt-à-dire 134 entiers, plus lafraction {3@/1000}.</s> <s xml:id="echoid-s1943" xml:space="preserve"/> </p> <div xml:id="echoid-div124" type="float" level="2" n="2"> <note position="right" xlink:label="note-0095-02" xlink:href="note-0095-02a" xml:space="preserve"># 25.430 <lb/># 3.054 <lb/># 69.067 <lb/># 36.480 <lb/># 134.031 <lb/></note> </div> <p> <s xml:id="echoid-s1944" xml:space="preserve">On peut même ſe diſpenſer de réduire les frac-<lb/> <anchor type="note" xlink:label="note-0095-03a" xlink:href="note-0095-03"/> tions propoſées à la même dénomination, en ob-<lb/>ſervant tout le reſte, comme on l’a expliqué au <lb/>commencement de l’art. </s> <s xml:id="echoid-s1945" xml:space="preserve">114. </s> <s xml:id="echoid-s1946" xml:space="preserve">Ainſi ſi l’on veut <lb/>ajouter les fractions ſuivantes, 0.</s> <s xml:id="echoid-s1947" xml:space="preserve">35, 0.</s> <s xml:id="echoid-s1948" xml:space="preserve">48, 054, <lb/>0.</s> <s xml:id="echoid-s1949" xml:space="preserve">345, 0.</s> <s xml:id="echoid-s1950" xml:space="preserve">0048, on les diſpoſera comme on le voit <lb/>ici, & </s> <s xml:id="echoid-s1951" xml:space="preserve">l’on aura pour la ſomme que l’on a demandée</s> </p> <div xml:id="echoid-div125" type="float" level="2" n="3"> <note position="right" xlink:label="note-0095-03" xlink:href="note-0095-03a" xml:space="preserve"># 0.35 <lb/># 0.48 <lb/># 0.54 <lb/># 0.345 <lb/># 0.0048 <lb/># 1.7198 <lb/></note> </div> </div> <div xml:id="echoid-div127" type="section" level="1" n="108"> <head xml:id="echoid-head124" style="it" xml:space="preserve">De la Souſtraction des Fractions décimales.</head> <p> <s xml:id="echoid-s1952" xml:space="preserve">117. </s> <s xml:id="echoid-s1953" xml:space="preserve">Si les fractions n’ont pas même dénomination, pour <lb/>plus grande facilité, on commencera par les réduire à celle du <lb/>plus grand dénominateur, ſuivant la méthode de l’art. </s> <s xml:id="echoid-s1954" xml:space="preserve">113; <lb/></s> <s xml:id="echoid-s1955" xml:space="preserve">enſuite on les diſpoſera de maniere que les dixiemes ſoient au <lb/>deſſous des dixiemes, les centiemes ſous les centiemes, & </s> <s xml:id="echoid-s1956" xml:space="preserve"><lb/>ainſi des autres nombres: </s> <s xml:id="echoid-s1957" xml:space="preserve">cela fait, on fera la Souſtraction <pb o="58" file="0096" n="96" rhead="NOUVEAU COURS"/> comme elle ſe pratique ſur les nombres entiers. <lb/></s> <s xml:id="echoid-s1958" xml:space="preserve"> <anchor type="note" xlink:label="note-0096-01a" xlink:href="note-0096-01"/> Ainſi pour ôter la fraction décimale 0.</s> <s xml:id="echoid-s1959" xml:space="preserve">025 de <lb/>0.</s> <s xml:id="echoid-s1960" xml:space="preserve">5894, on écrira comme on voit ici, <lb/>& </s> <s xml:id="echoid-s1961" xml:space="preserve">faiſant la Souſtraction, le reſte ſera . </s> <s xml:id="echoid-s1962" xml:space="preserve">. </s> <s xml:id="echoid-s1963" xml:space="preserve">.</s> <s xml:id="echoid-s1964" xml:space="preserve"/> </p> <div xml:id="echoid-div127" type="float" level="2" n="1"> <note position="right" xlink:label="note-0096-01" xlink:href="note-0096-01a" xml:space="preserve"># 0.5894 <lb/># 0.0250 <lb/># 0.5644 <lb/></note> </div> <p> <s xml:id="echoid-s1965" xml:space="preserve">Si l’on avoit des entiers & </s> <s xml:id="echoid-s1966" xml:space="preserve">des fractions à ſouſtraire d’un <lb/>entier & </s> <s xml:id="echoid-s1967" xml:space="preserve">d’une fraction, la méthode ſeroit toujours la même:</s> <s xml:id="echoid-s1968" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"># 68.05489 <lb/># 47.9453 <lb/># 20.10959 <lb/></note> <p> <s xml:id="echoid-s1969" xml:space="preserve">ainſi pour ôter 47.</s> <s xml:id="echoid-s1970" xml:space="preserve">9453 de 68.</s> <s xml:id="echoid-s1971" xml:space="preserve">05489, on écrira, <lb/>fans même ſe donner la peine de réduire le pre-<lb/>mier à la domination du ſecond, & </s> <s xml:id="echoid-s1972" xml:space="preserve">le reſte ſera</s> </p> <p> <s xml:id="echoid-s1973" xml:space="preserve">La démonſtration de ces deux opérations eſt la même que <lb/>celle des mêmes opérations ſur les nombres entiers; </s> <s xml:id="echoid-s1974" xml:space="preserve">car puiſ-<lb/>que l’on prend la ſomme ou la différence des dixiemes, des cen-<lb/>tiemes, des milliemes, on a auſſi la ſomme ou la différence de <lb/>ces fractions, puiſqu’elles ne contiennent que des dixiemes, <lb/>des centiemes, & </s> <s xml:id="echoid-s1975" xml:space="preserve">des milliemes, &</s> <s xml:id="echoid-s1976" xml:space="preserve">c. </s> <s xml:id="echoid-s1977" xml:space="preserve">La preuve de ces deux <lb/>opérations ſe fait auſſi comme dans les autres par l’opération <lb/>contraire; </s> <s xml:id="echoid-s1978" xml:space="preserve">ainſi il n’eſt pas néceſſaire d’inſiſter davantage.</s> <s xml:id="echoid-s1979" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div129" type="section" level="1" n="109"> <head xml:id="echoid-head125" style="it" xml:space="preserve">De la Multiplication des Fractions décimales.</head> <p> <s xml:id="echoid-s1980" xml:space="preserve">118. </s> <s xml:id="echoid-s1981" xml:space="preserve">Pour multiplier deux nombres l’un par l’autre, donc <lb/>un ſeul, ou tous les deux enſemble, renferment des parties <lb/>décimales, on fera la Multiplication comme ſi ces nombres <lb/>étoient tous nombres entiers; </s> <s xml:id="echoid-s1982" xml:space="preserve">& </s> <s xml:id="echoid-s1983" xml:space="preserve">lorſqu’on aura trouvé le pro-<lb/>duit, on ſéparera vers la droite autant de chiffres qu’il y a de <lb/>décimales, tant au multiplicande qu’au multiplicateur. </s> <s xml:id="echoid-s1984" xml:space="preserve">Les <lb/>chiffres qui ſeront à la gauche du point marqueront les en-<lb/>tiers, & </s> <s xml:id="echoid-s1985" xml:space="preserve">ceux qui ſeront à la droite marqueront les décimales. <lb/></s> <s xml:id="echoid-s1986" xml:space="preserve">Par exemple, pour multiplier 24.</s> <s xml:id="echoid-s1987" xml:space="preserve">35 par 2.</s> <s xml:id="echoid-s1988" xml:space="preserve">3, on écrira</s> </p> <note position="right" xml:space="preserve"># 24.35 <lb/># 2.3 <lb/># 7305 <lb/># 4870 <lb/># 56.005 <lb/></note> <p> <s xml:id="echoid-s1989" xml:space="preserve">Ayant fait la Multiplication comme s’il <lb/>n’y avoit point de décimales, & </s> <s xml:id="echoid-s1990" xml:space="preserve">trouvé <lb/>le produit 56005, on écrira 56.</s> <s xml:id="echoid-s1991" xml:space="preserve">005, fai-<lb/>ſant enſorte qu’il ſe trouve trois chiffres à <lb/>la droite du point, parce qu’il y avoit trois <lb/>rangs de décimales, tant au multipli-<lb/>cande qu’au multiplicateur, ſçavoir, 2 à l’un, & </s> <s xml:id="echoid-s1992" xml:space="preserve">1 à l’autre. <lb/></s> <s xml:id="echoid-s1993" xml:space="preserve">De même pour multiplier 4.</s> <s xml:id="echoid-s1994" xml:space="preserve">35 par 6.</s> <s xml:id="echoid-s1995" xml:space="preserve">7, j’écris</s> </p> <pb o="59" file="0097" n="97" rhead="DE MATHÉMATIQUE. Liv. I."/> <note position="right" xml:space="preserve"># 4.35 <lb/># 6.7 <lb/># 3045 <lb/># 2610 <lb/># 29.145 <lb/></note> <p> <s xml:id="echoid-s1996" xml:space="preserve">Faiſant la Multiplication comme s’il n’y <lb/>avoit point de décimales, je trouve le produit <lb/>29145, & </s> <s xml:id="echoid-s1997" xml:space="preserve">j’écris 29.</s> <s xml:id="echoid-s1998" xml:space="preserve">145, faiſant enſorte qu’il <lb/>y ait trois rangs de décimales aprés le point, <lb/>parce qu’il y en a deux au multiplicande, & </s> <s xml:id="echoid-s1999" xml:space="preserve"><lb/>un au multiplicateur.</s> <s xml:id="echoid-s2000" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2001" xml:space="preserve">119. </s> <s xml:id="echoid-s2002" xml:space="preserve">Il pourroit arriver que le nombre des rangs de déci-<lb/>males du multiplicande & </s> <s xml:id="echoid-s2003" xml:space="preserve">du multiplicateur fût plus grand <lb/>que le nombre des chiffres du produit; </s> <s xml:id="echoid-s2004" xml:space="preserve">ce qui arrive lorſqu’il <lb/>n’y a point d’entiers joints aux fractions décimales, & </s> <s xml:id="echoid-s2005" xml:space="preserve">qu’elles <lb/>ſont d’un certain ordre; </s> <s xml:id="echoid-s2006" xml:space="preserve">en ce cas on mettroit vers la gauche <lb/>autant de zero qu’il ſeroit néceſſaire, pour qu’il y ait après le <lb/>point autant de rangs de chiffres qu’il y en a au multiplicande <lb/>& </s> <s xml:id="echoid-s2007" xml:space="preserve">au multiplicateur. </s> <s xml:id="echoid-s2008" xml:space="preserve">Par exemple, ſi l’on propoſe de multi-<lb/>plier ces deux nombres, qui ne contiennent</s> </p> <note position="right" xml:space="preserve"># 0.0054 <lb/># 0.012 <lb/># 108 <lb/># 54 <lb/># 0.0000648 <lb/></note> <p> <s xml:id="echoid-s2009" xml:space="preserve">que des décimales, 0.</s> <s xml:id="echoid-s2010" xml:space="preserve">0054 par 0.</s> <s xml:id="echoid-s2011" xml:space="preserve">012, les <lb/>ayant diſpoſés comme on voit ici, fait la <lb/>multiplication comme à l’ordinaire, & </s> <s xml:id="echoid-s2012" xml:space="preserve">trouvé <lb/>le produit 648 des chiffres ſignificatifs, multi-<lb/>pliés les uns par les autres, on écrira 0.</s> <s xml:id="echoid-s2013" xml:space="preserve">0000648, <lb/>en faiſant enſorte, par l’addition de quatre <lb/>zero, qu’il y ait après le point autant de rangs qu’il y en a, <lb/>tant au multiplicande qu’au multiplicateur.</s> <s xml:id="echoid-s2014" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2015" xml:space="preserve">De même 0.</s> <s xml:id="echoid-s2016" xml:space="preserve">0048, multiplié par 0.</s> <s xml:id="echoid-s2017" xml:space="preserve">027,</s> </p> <note position="right" xml:space="preserve"># 0.0048 <lb/># 0.027 <lb/># 336 <lb/># 96 <lb/># 0.0001296 <lb/></note> <p> <s xml:id="echoid-s2018" xml:space="preserve">donne au produit, en multipliant les chiffres <lb/>ſignificatifs les uns par les autres, 1296, & </s> <s xml:id="echoid-s2019" xml:space="preserve"><lb/>j’ajoute à ce produit, vers la gauche, trois zero, <lb/>afin qu’il y ait autant de rangs de décimales <lb/>après le point qu’il y en a, tant au multipli-<lb/>cande qu’au multiplicateur.</s> <s xml:id="echoid-s2020" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div130" type="section" level="1" n="110"> <head xml:id="echoid-head126" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s2021" xml:space="preserve">Pour entendre plus aiſément la raiſon par laquelle on dé-<lb/>montre l’opération précédente, nous l’appliquerons au pre-<lb/>mier exemple, dans lequel il s’agiſſoit de multiplier 24.</s> <s xml:id="echoid-s2022" xml:space="preserve">35 par <lb/>2.</s> <s xml:id="echoid-s2023" xml:space="preserve">3. </s> <s xml:id="echoid-s2024" xml:space="preserve">Lorſque je multiplie ces nombres l’un par l’autre, com-<lb/>me s’ils n’avoient point de décimales, je rends le multiplicande <lb/>cent fois plus grand qu’il n’eſt, puiſque les unités du 4 qui ſe <pb o="60" file="0098" n="98" rhead="NOUVEAU COURS"/> trouvoient par le point au rang des unités ſimples, ſe trou-<lb/>vent par la ſuppreſſion du même point au rang des centaines. <lb/></s> <s xml:id="echoid-s2025" xml:space="preserve">De même je rends le multiplicateur 2.</s> <s xml:id="echoid-s2026" xml:space="preserve">3 dix fois plus grand <lb/>qu’il n’eſt effectivement, en le conſidérant comme 23: </s> <s xml:id="echoid-s2027" xml:space="preserve">le pro-<lb/>duit qui réſulte de ces deux nombres ſera donc dixfois cent fois <lb/>plus grand qu’il ne doit être, ou mille fois plus grand: </s> <s xml:id="echoid-s2028" xml:space="preserve">donc pour <lb/>le réduire à ſa juſte valeur, il faudra le rendre mille fois plus petit; </s> <s xml:id="echoid-s2029" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s2030" xml:space="preserve">c’eſt ce que l’on fait en retranchant vers la droite autant de <lb/>rangs de décimales qu’il y en a, tant au multiplicande qu’au <lb/>multiplicateur. </s> <s xml:id="echoid-s2031" xml:space="preserve">Dans notre exemple, on en a retranché 3, ce <lb/>qui a fait que le chiffre 6 du produit 56005, qui étoit au rang <lb/>des mille, s’eſt trouvé au rang des unités, en écrivant 56.</s> <s xml:id="echoid-s2032" xml:space="preserve">005. </s> <s xml:id="echoid-s2033" xml:space="preserve"><lb/>On appliquera le même raiſonnement à tout autre exemple.</s> <s xml:id="echoid-s2034" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div131" type="section" level="1" n="111"> <head xml:id="echoid-head127" style="it" xml:space="preserve">De la Diviſion des Fractions décimales.</head> <p> <s xml:id="echoid-s2035" xml:space="preserve">120. </s> <s xml:id="echoid-s2036" xml:space="preserve">Pour diviſer un nombre décimal par un autre, ſoit <lb/>qu’ils ne contiennent l’un & </s> <s xml:id="echoid-s2037" xml:space="preserve">l’autre que des décimales, ſoit <lb/>que le dividende & </s> <s xml:id="echoid-s2038" xml:space="preserve">le diviſeur ayent encore, outre ces déci-<lb/>males, des nombres entiers, ou ſeulement l’un des deux, regle <lb/>générale, on regardera ces nombres comme s’ils étoient tous <lb/>nombres entiers: </s> <s xml:id="echoid-s2039" xml:space="preserve">on les diviſera l’un par l’autre, ſuivant la <lb/>méthode de la Diviſion des nombres entiers; </s> <s xml:id="echoid-s2040" xml:space="preserve">& </s> <s xml:id="echoid-s2041" xml:space="preserve">lorſqu’on aura <lb/>trouvé le quotient, on fera enſorte qu’il y ait après le point <lb/>un nombre de décimales égal à celui du dividende, moins <lb/>celui du diviſeur.</s> <s xml:id="echoid-s2042" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2043" xml:space="preserve">Soit, par exemple, propoſé de diviſer 88.</s> <s xml:id="echoid-s2044" xml:space="preserve">392 par 254.</s> <s xml:id="echoid-s2045" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2046" xml:space="preserve">Je diviſe ces deux nombres comme s’ils</s> </p> <note position="right" xml:space="preserve"># 88.392 # { # 2.54 <lb/># 762 # # 34.8 <lb/># 1219 <lb/># 1016 <lb/># 2032 <lb/># 2032 <lb/></note> <p> <s xml:id="echoid-s2047" xml:space="preserve">étoient 88392 & </s> <s xml:id="echoid-s2048" xml:space="preserve">254, ayant trouvé le quo-<lb/>tient 348, j’écris 34.</s> <s xml:id="echoid-s2049" xml:space="preserve">8, de maniere qu’il y ait <lb/>après le point un rang de décimales, parce <lb/>qu’il y en a trois au dividende, & </s> <s xml:id="echoid-s2050" xml:space="preserve">deux au di-<lb/>viſeur, dont la différence eſt 1.</s> <s xml:id="echoid-s2051" xml:space="preserve"/> </p> <pb o="61" file="0099" n="99" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div132" type="section" level="1" n="112"> <head xml:id="echoid-head128" xml:space="preserve"><emph style="sc">Exemple</emph> II.</head> <p> <s xml:id="echoid-s2052" xml:space="preserve">Soit propoſé de diviſer 158.</s> <s xml:id="echoid-s2053" xml:space="preserve">0802 par 32.</s> <s xml:id="echoid-s2054" xml:space="preserve">46.</s> <s xml:id="echoid-s2055" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2056" xml:space="preserve">Je diviſe ces deux nombres comme <lb/> <anchor type="note" xlink:label="note-0099-01a" xlink:href="note-0099-01"/> s’ils ne contenoient point de décimales, <lb/>& </s> <s xml:id="echoid-s2057" xml:space="preserve">ayant trouvé le quoitient 487, je l’é-<lb/>cris ainſi, 4.</s> <s xml:id="echoid-s2058" xml:space="preserve">87, c’eſt-à-dire quatre en-<lb/>tiers {87/100}, en faiſant enſorte qu’il y ait <lb/>deux chiffres de décimales, parce que <lb/>la différence de 2 à 4 eſt 2.</s> <s xml:id="echoid-s2059" xml:space="preserve"/> </p> <div xml:id="echoid-div132" type="float" level="2" n="1"> <note position="right" xlink:label="note-0099-01" xlink:href="note-0099-01a" xml:space="preserve"># 158.0802 # { # 32.46 <lb/># 12984 # # 4.87 <lb/># 28240 <lb/># 25968 <lb/># 22722 <lb/># 22722 <lb/># 00000 <lb/></note> </div> <p> <s xml:id="echoid-s2060" xml:space="preserve">121. </s> <s xml:id="echoid-s2061" xml:space="preserve">Il ſuit de cette Regle générale, que s’il y a autant <lb/>de décimales au diviſeur qu’au dividende, le quotient ſera des <lb/>entiers; </s> <s xml:id="echoid-s2062" xml:space="preserve">car puiſque (hyp.) </s> <s xml:id="echoid-s2063" xml:space="preserve">le diviſeur a autant de rangs de <lb/>décimales que le dividende, la différence ſera 0, & </s> <s xml:id="echoid-s2064" xml:space="preserve">par con-<lb/>ſéquent il n’y aura point de décimales au’quotient. </s> <s xml:id="echoid-s2065" xml:space="preserve">Il ſuit en-<lb/>core delà, que s’il n’y a point de décimales au diviſeur, il y en <lb/>aura autant au quotient qu’au dividende. </s> <s xml:id="echoid-s2066" xml:space="preserve">Si le dividende n’a-<lb/>voit point de parties décimales, ou en avoit moins que le di-<lb/>viſeur, on lui ajouteroit autant de zero qu’il ſeroit néceſſaire, <lb/>pour que le nombre de ſes décimales fût égal à celui des déci-<lb/>males du diviſeur, & </s> <s xml:id="echoid-s2067" xml:space="preserve">dans ce cas le quotient aura toujours <lb/>des entiers, à moins que le nombre des entiers du diviſeur ne <lb/>fût plus grand que celui des entiers du dividende. </s> <s xml:id="echoid-s2068" xml:space="preserve">Par exem-<lb/>ple, ſi l’on propoſe de diviſer 883.</s> <s xml:id="echoid-s2069" xml:space="preserve">92 par 2.</s> <s xml:id="echoid-s2070" xml:space="preserve">54, le quotient <lb/>ſera 348, parce que la différence des décimales du dividende à <lb/>celles du diviſeur eſt zero.</s> <s xml:id="echoid-s2071" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2072" xml:space="preserve">De même ſi l’on veut diviſer 5952 par <lb/> <anchor type="note" xlink:label="note-0099-02a" xlink:href="note-0099-02"/> 1.</s> <s xml:id="echoid-s2073" xml:space="preserve">24, on ajoutera deux zero au dividen-<lb/>de, parce qu’il y a deux rangs de déci-<lb/>males au diviſeur: </s> <s xml:id="echoid-s2074" xml:space="preserve">puis faiſant la diviſion <lb/>des nombres 5952.</s> <s xml:id="echoid-s2075" xml:space="preserve">00, 1.</s> <s xml:id="echoid-s2076" xml:space="preserve">24 comme s’ils <lb/>étoient 595200, 124, on trouvera le quo-<lb/>tient de 4800 entiers.</s> <s xml:id="echoid-s2077" xml:space="preserve"/> </p> <div xml:id="echoid-div133" type="float" level="2" n="2"> <note position="right" xlink:label="note-0099-02" xlink:href="note-0099-02a" xml:space="preserve"># 5952.00 # { # 1.24 <lb/># 496 # # 4800 <lb/># 992 <lb/># 992 <lb/># 000 <lb/></note> </div> <p> <s xml:id="echoid-s2078" xml:space="preserve">Pour entendre plus aiſément la démonſtration de cette Regle <lb/>générale, nous allons établir pluſieurs principes.</s> <s xml:id="echoid-s2079" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div135" type="section" level="1" n="113"> <head xml:id="echoid-head129" xml:space="preserve"><emph style="sc">Premier principe</emph>.</head> <p> <s xml:id="echoid-s2080" xml:space="preserve">122. </s> <s xml:id="echoid-s2081" xml:space="preserve">Une fraction décimale qui contient des entiers & </s> <s xml:id="echoid-s2082" xml:space="preserve">des <pb o="62" file="0100" n="100" rhead="NOUVEAU COURS"/> décimales, peut être énoncée comme ſi elle ne contenoit que <lb/>des décimales: </s> <s xml:id="echoid-s2083" xml:space="preserve">ainſi la fraction 24.</s> <s xml:id="echoid-s2084" xml:space="preserve">32, qui vaut 24 entiers & </s> <s xml:id="echoid-s2085" xml:space="preserve"><lb/>32 centiemes, peut être énoncée ainſi, deux mille quatre cens <lb/>trente-deux centiemes; </s> <s xml:id="echoid-s2086" xml:space="preserve">car {2400/100} ou deux mille quatre cens cen-<lb/>tiemes, valent 24 entiers, puiſque le numérateur eſt 24 fois <lb/>plus grand que le dénominateur.</s> <s xml:id="echoid-s2087" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div136" type="section" level="1" n="114"> <head xml:id="echoid-head130" xml:space="preserve"><emph style="sc">Second principe</emph>.</head> <p> <s xml:id="echoid-s2088" xml:space="preserve">123. </s> <s xml:id="echoid-s2089" xml:space="preserve">Les unités du quotient doivent toujours être de même <lb/>nature que celles du dividende, lorſque le diviſeur eſt un nom-<lb/>bre entier qui marque des nombres de fois: </s> <s xml:id="echoid-s2090" xml:space="preserve">ainſi ſi le divi-<lb/>dende a pour unités des milliemes, & </s> <s xml:id="echoid-s2091" xml:space="preserve">que le diviſeur ſoit un <lb/>nombre entier abſtrait, comme 3 ou 4, le quotient vaudra le <lb/>tiers ou le quart des milliemes du dividende, & </s> <s xml:id="echoid-s2092" xml:space="preserve">aura par con-<lb/>ſéquent des unités de même nature.</s> <s xml:id="echoid-s2093" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div137" type="section" level="1" n="115"> <head xml:id="echoid-head131" xml:space="preserve"><emph style="sc">Troisieme principe</emph>.</head> <p> <s xml:id="echoid-s2094" xml:space="preserve">124. </s> <s xml:id="echoid-s2095" xml:space="preserve">Plus un diviſeur eſt grand, le dividende reſtant le <lb/>même, plus le quotient eſt petie; </s> <s xml:id="echoid-s2096" xml:space="preserve">& </s> <s xml:id="echoid-s2097" xml:space="preserve">réciproquement plus le <lb/>diviſeur eſt petit, le dividende étant toujours ſuppoſé le mê-<lb/>me, plus le quotient eſt grand: </s> <s xml:id="echoid-s2098" xml:space="preserve">car il eſt viſible que plus un <lb/>nombre eſt petit, plus il eſt contenu de fois dans un autre.</s> <s xml:id="echoid-s2099" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div138" type="section" level="1" n="116"> <head xml:id="echoid-head132" style="it" xml:space="preserve">Démonſtration de la Regle générale.</head> <p> <s xml:id="echoid-s2100" xml:space="preserve">Pour rendre cette démonſtration plus intelligible, nous al-<lb/>lons appliquer les raiſonnemens au premier exemple. </s> <s xml:id="echoid-s2101" xml:space="preserve">Quand <lb/>je diviſe ce nombre 88.</s> <s xml:id="echoid-s2102" xml:space="preserve">392 par celui-ci, 2.</s> <s xml:id="echoid-s2103" xml:space="preserve">54, comme s’ils <lb/>étoient des nombres entiers, le quotient 348 que je trouve, ne <lb/>doit avoir que des nombres entiers par le ſecond principe: <lb/></s> <s xml:id="echoid-s2104" xml:space="preserve">mais puiſque le dividende eſt 88.</s> <s xml:id="echoid-s2105" xml:space="preserve">392, & </s> <s xml:id="echoid-s2106" xml:space="preserve">non pas 88392, c’eſt-<lb/>à-dire 88 milles 392 milliemes, les unités du quotient, par le <lb/>ſecond principe, doivent être des milliemes: </s> <s xml:id="echoid-s2107" xml:space="preserve">doncle quotient <lb/>348 eſt mille fois plus grand qu’il ne doit être, en ſuppoſant <lb/>le diviſeur toujours entier, & </s> <s xml:id="echoid-s2108" xml:space="preserve">que les unités du dividende ſont <lb/>des milliemes: </s> <s xml:id="echoid-s2109" xml:space="preserve">il faudroit donc en ce cas l’écrire ainſi, 0.</s> <s xml:id="echoid-s2110" xml:space="preserve">348. </s> <s xml:id="echoid-s2111" xml:space="preserve"><lb/>Préſentement ſi l’on ſuppoſe que le diviſeur devienne ce qu’il <lb/>eſt réellement, c’eſt-à-dire 2.</s> <s xml:id="echoid-s2112" xml:space="preserve">54, ou deux cens cinquante-quatre <lb/>centiemes, puiſque les centiemes ſont cent fois plus petits que <lb/>les unités, le nombre 2.</s> <s xml:id="echoid-s2113" xml:space="preserve">54 ſera auſſi cent fois plus petit que <pb o="63" file="0101" n="101" rhead="DE MATHÉMATIQUE. Liv. I."/> 254: </s> <s xml:id="echoid-s2114" xml:space="preserve">donc le quotient doit devenir cent fois plus grand; </s> <s xml:id="echoid-s2115" xml:space="preserve">& </s> <s xml:id="echoid-s2116" xml:space="preserve"><lb/>c’eſt ce que l’on fait en retranchant un rang de décimales vers <lb/>la droite dans le quotient 0.</s> <s xml:id="echoid-s2117" xml:space="preserve">348, qui devient 34.</s> <s xml:id="echoid-s2118" xml:space="preserve">8: </s> <s xml:id="echoid-s2119" xml:space="preserve">car en opé-<lb/>rant ainſi, les unités du 3 qui étoient des dixiemes, ſont deve-<lb/>nues des dixaines, les unités du 4 qui étoient des centiemes, <lb/>ſont devenues des unités ſimples, & </s> <s xml:id="echoid-s2120" xml:space="preserve">par conſéquent le quo-<lb/>tient 0.</s> <s xml:id="echoid-s2121" xml:space="preserve">348 étant écrit ainſi 34.</s> <s xml:id="echoid-s2122" xml:space="preserve">8, eſt devenu cent fois plus <lb/>grand; </s> <s xml:id="echoid-s2123" xml:space="preserve">d’où il ſuit que la méthode que l’on a donnée met les <lb/>choſes dans l’état où elles doivent être.</s> <s xml:id="echoid-s2124" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2125" xml:space="preserve">Pour entendre la raiſon des opérations énoncées dans l’ar-<lb/>ticle 120, on fera attention que le quotient d’une diviſion ne <lb/>change pas, lorſqu’on multiplie le diviſeur & </s> <s xml:id="echoid-s2126" xml:space="preserve">le dividende <lb/>par un même nombre. </s> <s xml:id="echoid-s2127" xml:space="preserve">Ainſi 12 diviſé par 4, donne 3 au quo-<lb/>tient: </s> <s xml:id="echoid-s2128" xml:space="preserve">que je multiplie 12 & </s> <s xml:id="echoid-s2129" xml:space="preserve">4 par 5, le quotient des produits <lb/>60 & </s> <s xml:id="echoid-s2130" xml:space="preserve">20, diviſés l’un par l’autre, ſera toujours 3. </s> <s xml:id="echoid-s2131" xml:space="preserve">Cela poſé, <lb/>lorſque je diviſe deux nombres, qui ont chacun même nom-<lb/>bre de décimales, comme s’ils n’en avoient point, je ne fais <lb/>que multiplier le dividende & </s> <s xml:id="echoid-s2132" xml:space="preserve">le diviſeur par un même nom-<lb/>bre; </s> <s xml:id="echoid-s2133" xml:space="preserve">ce qui ne doit point changer le quotient: </s> <s xml:id="echoid-s2134" xml:space="preserve">ainſi quand je <lb/>diviſe 88 3.</s> <s xml:id="echoid-s2135" xml:space="preserve">92 par 2.</s> <s xml:id="echoid-s2136" xml:space="preserve">54, comme s’ils étoient 88392 & </s> <s xml:id="echoid-s2137" xml:space="preserve">254, <lb/>je multiplie le dividende & </s> <s xml:id="echoid-s2138" xml:space="preserve">le diviſeur par 100; </s> <s xml:id="echoid-s2139" xml:space="preserve">le quotient ne <lb/>doit donc pas changer: </s> <s xml:id="echoid-s2140" xml:space="preserve">mais le quotient de 88392 diviſé par <lb/>254, eſt 348: </s> <s xml:id="echoid-s2141" xml:space="preserve">donc ce même nombre eſt auſſi le quotient de <lb/>883.</s> <s xml:id="echoid-s2142" xml:space="preserve">92 diviſé par 2.</s> <s xml:id="echoid-s2143" xml:space="preserve">54. </s> <s xml:id="echoid-s2144" xml:space="preserve">Cette raiſon peut donner la démonſ-<lb/>tration de tous les cas imaginables, c’eſt pourquoi on fera <lb/>très-bien de l’étudier avec attention.</s> <s xml:id="echoid-s2145" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div139" type="section" level="1" n="117"> <head xml:id="echoid-head133" style="it" xml:space="preserve">Uſages des Fractions décimales.</head> <p> <s xml:id="echoid-s2146" xml:space="preserve">125. </s> <s xml:id="echoid-s2147" xml:space="preserve">Premier uſage. </s> <s xml:id="echoid-s2148" xml:space="preserve">Approcher ſi près que l’on voudra du <lb/>quotient d’une diviſion qui ne peut pas ſe faire ſans reſte.</s> <s xml:id="echoid-s2149" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2150" xml:space="preserve">On cherchera d’abord le quotient du dividende, diviſé par <lb/>le diviſeur, & </s> <s xml:id="echoid-s2151" xml:space="preserve">l’on mettra à la ſuite du reſte autant de zero <lb/>que l’on voudra avoir de décimales au quotient: </s> <s xml:id="echoid-s2152" xml:space="preserve">ſi l’on veut <lb/>avoir le quotient à un millieme ou un dix millieme près, on <lb/>ajoutera trois ou quatre zero à la ſuite du reſte, & </s> <s xml:id="echoid-s2153" xml:space="preserve">l’on conti-<lb/>nuera la diviſion comme à l’ordinaire, en mettant les chiffres <lb/>à la ſuite du quotient comme ils viendront, après les avoir <lb/>ſéparés des entiers du quotient par un point, comme on va <lb/>voir dans l’exemple ſuivant.</s> <s xml:id="echoid-s2154" xml:space="preserve"/> </p> <pb o="64" file="0102" n="102" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s2155" xml:space="preserve">Soit propoſé de diviſer 353 par 15, & </s> <s xml:id="echoid-s2156" xml:space="preserve">de trouver un quo-<lb/>tient qui ne differe pas du vrai quotient de la dix millieme <lb/>partie de l’unité.</s> <s xml:id="echoid-s2157" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2158" xml:space="preserve">Après avoir diviſé 353 par 15, & </s> <s xml:id="echoid-s2159" xml:space="preserve">trouvé <lb/> <anchor type="note" xlink:label="note-0102-01a" xlink:href="note-0102-01"/> le quotient 23 avec le reſte 8, j’ajoute à <lb/>ce reſte quatre zero, parce que je veux <lb/>pouſſer les décimales juſqu’aux dix mil-<lb/>liemes, & </s> <s xml:id="echoid-s2160" xml:space="preserve">continuant la Diviſion comme <lb/>à l’ordinaire, je dis, en 80 combien de fois <lb/>15, cinq fois; </s> <s xml:id="echoid-s2161" xml:space="preserve">je poſe 5 au quotient (après <lb/>avoir mis un point à la ſuite du 3 pour ſé-<lb/>parer les entiers des décimales); </s> <s xml:id="echoid-s2162" xml:space="preserve">cinq fois <lb/>15 font 75, que je poſé ſous 80. </s> <s xml:id="echoid-s2163" xml:space="preserve">75 de 80, <lb/>reſte 5; </s> <s xml:id="echoid-s2164" xml:space="preserve">j’abaiſſe un zero à côté du 5, & </s> <s xml:id="echoid-s2165" xml:space="preserve">je <lb/>dis, en 50 combien de fois 15, trois fois; <lb/></s> <s xml:id="echoid-s2166" xml:space="preserve">je poſe 3 au quotient; </s> <s xml:id="echoid-s2167" xml:space="preserve">trois fois 15 font <lb/>45, de 50, reſte 5; </s> <s xml:id="echoid-s2168" xml:space="preserve">j’abaiſſe encore un zero, & </s> <s xml:id="echoid-s2169" xml:space="preserve">diviſant 50 <lb/>par 15, je trouve encore 3 au quotient; </s> <s xml:id="echoid-s2170" xml:space="preserve">& </s> <s xml:id="echoid-s2171" xml:space="preserve">comme l’on eſt ar-<lb/>rivé à un reſte 5, qui ſera toujours le même, les quotiens qui <lb/>ſuivront ſeront auſſi toujours les mêmes, & </s> <s xml:id="echoid-s2172" xml:space="preserve">l’on aura tout d’un <lb/>coup un quotient qui ne différera pas, ſi l’on veut, du vrai <lb/>quotient de la cent millionieme partie de l’unité.</s> <s xml:id="echoid-s2173" xml:space="preserve"/> </p> <div xml:id="echoid-div139" type="float" level="2" n="1"> <note position="right" xlink:label="note-0102-01" xlink:href="note-0102-01a" xml:space="preserve">353 # {15 <lb/>30 # 23.5333 <lb/>53 <lb/>45 <lb/>8.0000 <lb/>7.5 <lb/>50 <lb/>450 <lb/>500 <lb/>45 <lb/>50 <lb/></note> </div> <p> <s xml:id="echoid-s2174" xml:space="preserve">126. </s> <s xml:id="echoid-s2175" xml:space="preserve">Second uſage. </s> <s xml:id="echoid-s2176" xml:space="preserve">Trouver une fraction décimale égale <lb/>à une fraction donnée moindre que l’unité, ou bien, ce qui <lb/>revient au même, faire la diviſion d’un nombre par un nom-<lb/>bre plus grand que lui.</s> <s xml:id="echoid-s2177" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2178" xml:space="preserve">Soit propoſé de réduire la frac-<lb/> <anchor type="note" xlink:label="note-0102-02a" xlink:href="note-0102-02"/> tion {5/7} en décimale, ou, ce qui eſt <lb/>la même choſe, de trouver le quo-<lb/>tient approché de 5 diviſé par 7, <lb/>juſqu’à ce qu’il ne differe pas de la <lb/>millieme partie de l’unité. </s> <s xml:id="echoid-s2179" xml:space="preserve">On ajou-<lb/>tera à la ſuite du numérateur 5 ſix <lb/>zero, & </s> <s xml:id="echoid-s2180" xml:space="preserve">faiſant la diviſion comme <lb/>à l’ordinaire, on trouvera au quo-<lb/>tient 0.</s> <s xml:id="echoid-s2181" xml:space="preserve">714285, ou 174 mille 285 <lb/>millioniemes pour le quotient de 5 <lb/>diviſé par 7, ou pour la valeur ap-<lb/>prochée de la fraction {5/7}, avec un <lb/>reſte cinq, ou cinq millioniemes, <pb o="65" file="0103" n="103" rhead="DE MATHÉMATIQUE. Liv. I."/> dont il faudroit encore prendre la ſeptieme partie. </s> <s xml:id="echoid-s2182" xml:space="preserve">Si l’on vou-<lb/>loit continuer plus loin la Diviſion, on trouveroit une ſuite <lb/>infinie de périodes égales à 714285; </s> <s xml:id="echoid-s2183" xml:space="preserve">car il eſt évident qu’en <lb/>mettant un zero à la ſuite du 5, on auroit encore 50 à diviſer <lb/>par 7, & </s> <s xml:id="echoid-s2184" xml:space="preserve">les mêmes quotients reparoîtroient avec les mêmes <lb/>reſtes, ce qui donneroit en un inſtant une approximation pro-<lb/>digieuſe, mais cependant toujours telle, qu’il y manqueroit <lb/>quelque choſe. </s> <s xml:id="echoid-s2185" xml:space="preserve">Dans la pratique la plus rigoureuſe, on ſe con-<lb/>tente ordinairement de ſix chiffres décimaux, ou tout au plus <lb/>de huit.</s> <s xml:id="echoid-s2186" xml:space="preserve"/> </p> <div xml:id="echoid-div140" type="float" level="2" n="2"> <note position="right" xlink:label="note-0102-02" xlink:href="note-0102-02a" xml:space="preserve">5.000000 # {7 <lb/>49 # 0.714285 <lb/>10 <lb/>7 <lb/>30 <lb/>28 <lb/>20 <lb/>14 <lb/>60 <lb/>56 <lb/>40 <lb/>35 <lb/>5 <lb/></note> </div> <p> <s xml:id="echoid-s2187" xml:space="preserve">127. </s> <s xml:id="echoid-s2188" xml:space="preserve">Il y a des fractions qui peuvent ſe réduire en fractions <lb/>décimales, & </s> <s xml:id="echoid-s2189" xml:space="preserve">d’autres qui ne peuvent jamais s’y réduire, com-<lb/>me la fraction {5/7} & </s> <s xml:id="echoid-s2190" xml:space="preserve">la fraction {6/7}, pour laquelle on trouveroit <lb/>0.</s> <s xml:id="echoid-s2191" xml:space="preserve">857142, 857142, &</s> <s xml:id="echoid-s2192" xml:space="preserve">c. </s> <s xml:id="echoid-s2193" xml:space="preserve">en ſuivant le même procédé que nous <lb/>avons ſuivi pour la premiere. </s> <s xml:id="echoid-s2194" xml:space="preserve">Il n’en eſt pas de même des frac-<lb/>tions {4/5}, {5/16}, pour leſquelles on trouve des fractions décimales <lb/>complettes & </s> <s xml:id="echoid-s2195" xml:space="preserve">ſans reſte, 0.</s> <s xml:id="echoid-s2196" xml:space="preserve">8, 0.</s> <s xml:id="echoid-s2197" xml:space="preserve">3125, en ſuivant toujours le <lb/>même procédé.</s> <s xml:id="echoid-s2198" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2199" xml:space="preserve">128. </s> <s xml:id="echoid-s2200" xml:space="preserve">Troiſieme uſage. </s> <s xml:id="echoid-s2201" xml:space="preserve">Réduire en fraction décimale les par-<lb/>ties connues d’une certaine meſure, comme de la toiſe, du <lb/>pied, & </s> <s xml:id="echoid-s2202" xml:space="preserve">de la livre, &</s> <s xml:id="echoid-s2203" xml:space="preserve">c. </s> <s xml:id="echoid-s2204" xml:space="preserve">On fera d’abord une fraction qui <lb/>aura pour numérateur le nombre des parties que l’on veut ré-<lb/>duire en décimales, & </s> <s xml:id="echoid-s2205" xml:space="preserve">pour dénominateur, le nombre qui <lb/>marque combien de fois cette partie eſt contenue dans la me-<lb/>ſure dont il s’agit. </s> <s xml:id="echoid-s2206" xml:space="preserve">On réduira cette fraction en décimales par <lb/>l’article précédent, & </s> <s xml:id="echoid-s2207" xml:space="preserve">l’on aura la fraction décimalc deman-<lb/>dée. </s> <s xml:id="echoid-s2208" xml:space="preserve">Par exemple, ſi je veux avoir une fraction décimale de <lb/>la toiſe, qui vale 5 pieds, ou bien réduire 5 pieds en parties <lb/>décimales de la toiſe, je prends cette fraction {5/6}, dont le nu-<lb/>mérateur 5 exprime le nombre de pieds, dont je veux avoir la <lb/>valeur en décimales, & </s> <s xml:id="echoid-s2209" xml:space="preserve">le dénominateur 6 marque combien <lb/>de fois le pied eſt contenu dans la toiſe: </s> <s xml:id="echoid-s2210" xml:space="preserve">jeréduis cette fraction <lb/>en décimale, ſuivant l’art. </s> <s xml:id="echoid-s2211" xml:space="preserve">125, & </s> <s xml:id="echoid-s2212" xml:space="preserve">j’ai pour la valeur de 3 pieds <lb/>en décimale, 0.</s> <s xml:id="echoid-s2213" xml:space="preserve">8333, qui n’en differe pas de la dixmillieme <lb/>partie de la toiſe. </s> <s xml:id="echoid-s2214" xml:space="preserve">De même ſi je veux réduire 9 pouces en dé-<lb/>cimales de la toiſe, ou, ce qui revient même, avoir une partie <lb/>décimale de la toiſe égale à 9 pouces, je prends la fraction {9/72}, <lb/>dont le numérateur ſoit 9, & </s> <s xml:id="echoid-s2215" xml:space="preserve">le dénominateur le nombre 72, qui <lb/>me marque combien de fois le pouce eſt contenu dans la toiſe, <lb/>& </s> <s xml:id="echoid-s2216" xml:space="preserve">diviſant 9 par 72, ſelon la méthode de l’art. </s> <s xml:id="echoid-s2217" xml:space="preserve">125, je trouve <pb o="66" file="0104" n="104" rhead="NOUVEAU COURS"/> pour la valeur de 9 pouces en parties décimales de la toiſe, 0.</s> <s xml:id="echoid-s2218" xml:space="preserve">125. <lb/></s> <s xml:id="echoid-s2219" xml:space="preserve">Si l’on vouloit avoir une partie décimale de la toiſe égale à <lb/>5 pieds 9 pouces, il n’y auroit qu’à prendre la ſomme des deux <lb/>nombres 0.</s> <s xml:id="echoid-s2220" xml:space="preserve">8333, & </s> <s xml:id="echoid-s2221" xml:space="preserve">0.</s> <s xml:id="echoid-s2222" xml:space="preserve">125, ou 0.</s> <s xml:id="echoid-s2223" xml:space="preserve">1250 que l’on trouveroit de <lb/>0.</s> <s xml:id="echoid-s2224" xml:space="preserve">9583.</s> <s xml:id="echoid-s2225" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2226" xml:space="preserve">129. </s> <s xml:id="echoid-s2227" xml:space="preserve">Si l’on vouloit réduire en parties décimales de la livre, <lb/>des nombres de ſols & </s> <s xml:id="echoid-s2228" xml:space="preserve">de deniers, on s’y prendroit de la même <lb/>maniere. </s> <s xml:id="echoid-s2229" xml:space="preserve">Par exemple, ſi l’on me demande une partie déci-<lb/>male de la livre égale à 7 ſols 8 den. </s> <s xml:id="echoid-s2230" xml:space="preserve">je cherche d’abord une <lb/>partie décimale de la livre, égale à 7 ſols; </s> <s xml:id="echoid-s2231" xml:space="preserve">ce que je fais en di-<lb/>viſant 7 par 20, ou en cherchant une fraction décimale égale <lb/>à la fraction {7/20}, que l’on trouvera de 0.</s> <s xml:id="echoid-s2232" xml:space="preserve">35={35/100}. </s> <s xml:id="echoid-s2233" xml:space="preserve">Je cherche <lb/>enſuite une fraction décimale de même valeur que la fraction <lb/>{8/240}, qui vaut 8 deniers, puiſque le denier eſt la 240<emph style="sub">e</emph> partie de la <lb/>livre, que je trouve de 0.</s> <s xml:id="echoid-s2234" xml:space="preserve">0333, & </s> <s xml:id="echoid-s2235" xml:space="preserve">prenant la ſomme de ces <lb/>deux fractions, j’ai pour la valeur de 7 ſols 8 den. </s> <s xml:id="echoid-s2236" xml:space="preserve">en fractions <lb/>décimales de la livre, 0.</s> <s xml:id="echoid-s2237" xml:space="preserve">3833.</s> <s xml:id="echoid-s2238" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2239" xml:space="preserve">130. </s> <s xml:id="echoid-s2240" xml:space="preserve">Quatrieme uſage des fractions décimales. </s> <s xml:id="echoid-s2241" xml:space="preserve">Faire la multi-<lb/>plication des nombres complexes par le moyen des décimales.</s> <s xml:id="echoid-s2242" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2243" xml:space="preserve">Soit propoſé de trouver le prix de 27 toiſes 5 pieds 9 pouces, <lb/>à 4 liv. </s> <s xml:id="echoid-s2244" xml:space="preserve">7 ſols 8 den. </s> <s xml:id="echoid-s2245" xml:space="preserve">la toiſe.</s> <s xml:id="echoid-s2246" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2247" xml:space="preserve">Je réduis les 5 pieds 9 pouces en parties <lb/> <anchor type="note" xlink:label="note-0104-01a" xlink:href="note-0104-01"/> décimales de la toiſe, ſuivant l’art. </s> <s xml:id="echoid-s2248" xml:space="preserve">127, & </s> <s xml:id="echoid-s2249" xml:space="preserve"><lb/>j’ai 27<emph style="sub">t</emph>. </s> <s xml:id="echoid-s2250" xml:space="preserve">9583 de toiſe; </s> <s xml:id="echoid-s2251" xml:space="preserve">de même je réduis <lb/>les 7 ſols 8 den. </s> <s xml:id="echoid-s2252" xml:space="preserve">en parties décimales de la <lb/>livre, & </s> <s xml:id="echoid-s2253" xml:space="preserve">j’ai par l’art. </s> <s xml:id="echoid-s2254" xml:space="preserve">128, 4<emph style="sub">1</emph>.</s> <s xml:id="echoid-s2255" xml:space="preserve">3833, je mul-<lb/>tiplie ces deux nombres l’un par l’autre, au <lb/>lieu de multiplier 27 toiſes 5 pieds 9 pouces <lb/>par 4 liv. </s> <s xml:id="echoid-s2256" xml:space="preserve">7 ſols 8 deniers, & </s> <s xml:id="echoid-s2257" xml:space="preserve">ayant trouvé <lb/>le produit 122.</s> <s xml:id="echoid-s2258" xml:space="preserve">54961639, je fais enſorte <lb/>qu’il y aìt huit rangs de décimales après le <lb/>point (art. </s> <s xml:id="echoid-s2259" xml:space="preserve">117), parce qu’il y en a quatre au multiplicande <lb/>& </s> <s xml:id="echoid-s2260" xml:space="preserve">au multiplicateur, & </s> <s xml:id="echoid-s2261" xml:space="preserve">j’ai 122 au rang des livres, reſte à <lb/>ſçavoir ce que vaut la fraction décimale 0.</s> <s xml:id="echoid-s2262" xml:space="preserve">54961639, expri-<lb/>mée en ſols & </s> <s xml:id="echoid-s2263" xml:space="preserve">en deniers: </s> <s xml:id="echoid-s2264" xml:space="preserve">c’eſt ce que nous allons expliquer <lb/>dans l’article ſuivant.</s> <s xml:id="echoid-s2265" xml:space="preserve"/> </p> <div xml:id="echoid-div141" type="float" level="2" n="3"> <note position="right" xlink:label="note-0104-01" xlink:href="note-0104-01a" xml:space="preserve">27.9583 <lb/>4.3833 <lb/>838749 <lb/>838749 <lb/>2236664 <lb/>838749 <lb/>1118332 <lb/>122.54961639 <lb/></note> </div> <p> <s xml:id="echoid-s2266" xml:space="preserve">131. </s> <s xml:id="echoid-s2267" xml:space="preserve">Réduire une fraction décimale en parties connues, <lb/>ou, ce qui eſt la même choſe, évaluer une fraction décimale.</s> <s xml:id="echoid-s2268" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2269" xml:space="preserve">On multipliera la fraction décimale par le nombre qui mar-<lb/>que combien de fois la quantité à laquelle on veut réduire, eſt <pb o="67" file="0105" n="105" rhead="DE MATHÉMATIQUE. Liv. I."/> contenue dans le tout, dont la fraction eſt décimale; </s> <s xml:id="echoid-s2270" xml:space="preserve">& </s> <s xml:id="echoid-s2271" xml:space="preserve">lorſ-<lb/>qu’on aura trouvé le produit, le nombre qui ſe trouvera hors <lb/>du rang des décimales, ſera celui que l’on demande: </s> <s xml:id="echoid-s2272" xml:space="preserve">par exem-<lb/>ple, ſi l’on me propoſe d’évaluer cette fraction de livre 0.</s> <s xml:id="echoid-s2273" xml:space="preserve">35 <lb/>en ſols, je multiplie 0.</s> <s xml:id="echoid-s2274" xml:space="preserve">35 par 20, le produit eſt 7.</s> <s xml:id="echoid-s2275" xml:space="preserve">00; </s> <s xml:id="echoid-s2276" xml:space="preserve">d’où je <lb/>conclus que cette fraction vaut 7 ſols, puiſque le nombre 7 ſe <lb/>trouve hors du rang des décimales. </s> <s xml:id="echoid-s2277" xml:space="preserve">De même ſi l’on me propoſe <lb/>d’évaluer en pieds & </s> <s xml:id="echoid-s2278" xml:space="preserve">pouces cette fraction de toiſe 0.</s> <s xml:id="echoid-s2279" xml:space="preserve">9583, je <lb/>multiplie d’abord ce nombre par 6, qui marque combien de <lb/>fois la toiſe contient le pied, je trouve au produit 5.</s> <s xml:id="echoid-s2280" xml:space="preserve">7498, <lb/>d’où je conclus d’abord que cette fraction vaut cinq pieds; </s> <s xml:id="echoid-s2281" xml:space="preserve">la <lb/>partie décimale 7498 exprime donc des parties décimales de <lb/>pieds. </s> <s xml:id="echoid-s2282" xml:space="preserve">Pour ſçavoir ce qu’elle vaut de pouces, je multiplie ce <lb/>nombre par 12, qui marque combien de fois le pouce eſt con-<lb/>tenu dans le pied, le produit eſt 8.</s> <s xml:id="echoid-s2283" xml:space="preserve">9976; </s> <s xml:id="echoid-s2284" xml:space="preserve">d’où je conclus en-<lb/>core que 0.</s> <s xml:id="echoid-s2285" xml:space="preserve">7498 de pied valent 8 pouces, puiſque 8 ſe trouve <lb/>au rang des entiers; </s> <s xml:id="echoid-s2286" xml:space="preserve">& </s> <s xml:id="echoid-s2287" xml:space="preserve">de plus je prends 9, à cauſe que la frac-<lb/>tion reſtante {9976/10000} eſt preſque égale à l’unité: </s> <s xml:id="echoid-s2288" xml:space="preserve">donc la fraction <lb/>décimale de toiſe 0.</s> <s xml:id="echoid-s2289" xml:space="preserve">9583 vaut 5 pieds 9 pouces, comme on le <lb/>ſçait par l’art. </s> <s xml:id="echoid-s2290" xml:space="preserve">127.</s> <s xml:id="echoid-s2291" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2292" xml:space="preserve">La raiſon de cette pratique eſt aiſé à déduire de la nature <lb/>des décimales; </s> <s xml:id="echoid-s2293" xml:space="preserve">car lorſque je multiplie la fraction 0.</s> <s xml:id="echoid-s2294" xml:space="preserve">35 de liv. <lb/></s> <s xml:id="echoid-s2295" xml:space="preserve">par 20, le produit 7.</s> <s xml:id="echoid-s2296" xml:space="preserve">00 eſt bien vingt fois plus grand, mais il <lb/>n’exprime plus que des parties décimales de ſols, au lieu qu’au-<lb/>paravant il exprimoit des parties décimales de livres, qui ſont <lb/>vingt fois plus grandes.</s> <s xml:id="echoid-s2297" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2298" xml:space="preserve">En ſuivant ce procédé, on trouvera que la fraction de livre <lb/>0.</s> <s xml:id="echoid-s2299" xml:space="preserve">54961639, vaut 10 ſols 11 den. </s> <s xml:id="echoid-s2300" xml:space="preserve">+ {907840/1000000}; </s> <s xml:id="echoid-s2301" xml:space="preserve">& </s> <s xml:id="echoid-s2302" xml:space="preserve">ſi l’on eût <lb/>cherché le même prix, ſuivant les regles des parties aliquotes, <lb/>on auroit trouvé 122 liv. </s> <s xml:id="echoid-s2303" xml:space="preserve">11 ſols o d.</s> <s xml:id="echoid-s2304" xml:space="preserve">; ce qui montre la préci-<lb/>ſion de chacune de ces méthodes. </s> <s xml:id="echoid-s2305" xml:space="preserve">Cependant il faut avouer <lb/>que celle des parties aliquotes a quelque choſe de plus expé-<lb/>ditif dans la pratique, quoique les principes ſur leſquels cha-<lb/>que méthode eſt fondée ſoient fort ſimples, & </s> <s xml:id="echoid-s2306" xml:space="preserve">à la portée de <lb/>tout le monde.</s> <s xml:id="echoid-s2307" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div143" type="section" level="1" n="118"> <head xml:id="echoid-head134" style="it" xml:space="preserve">Remarque générale ſur les Fractions décimales.</head> <p> <s xml:id="echoid-s2308" xml:space="preserve">132. </s> <s xml:id="echoid-s2309" xml:space="preserve">Lorſque les fractions décimales contiennent beaucoup <lb/>de chiffres, on retranche ordinairement les deux derniers, vu <pb o="68" file="0106" n="106" rhead="NOUVEAU COURS"/> l’erreur inſenſible qui en réſulte. </s> <s xml:id="echoid-s2310" xml:space="preserve">Par exemple, pour évaluer <lb/>cette derniere fraction de livres 54961639, on aura à peu près <lb/>la même préciſion en évaluant celle-ci, 5497, que l’on a, en <lb/>retranchant les quatre derniers chiffres de la précédente, & </s> <s xml:id="echoid-s2311" xml:space="preserve"><lb/>mettant une unité au 6 pour compenſer ce retranchement. <lb/></s> <s xml:id="echoid-s2312" xml:space="preserve">On verra encore d’autres uſages des fractions décimales dans <lb/>l’extraction des racines quarrées & </s> <s xml:id="echoid-s2313" xml:space="preserve">cubiques, qui ſont encore <lb/>plus importans que les précédens.</s> <s xml:id="echoid-s2314" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div144" type="section" level="1" n="119"> <head xml:id="echoid-head135" xml:space="preserve">DU CALCUL DES EXPOSANS, <lb/>DE LA FORMATION DES PUISSANCES, <lb/><emph style="sc">ET DE L’</emph><emph style="sc">Extraction des</emph> <emph style="sc">Racines</emph>. <lb/>Du Calcul des Expoſans.</head> <p> <s xml:id="echoid-s2315" xml:space="preserve">133. </s> <s xml:id="echoid-s2316" xml:space="preserve"><emph style="sc">NOus</emph> avons déja vu (art. </s> <s xml:id="echoid-s2317" xml:space="preserve">39.) </s> <s xml:id="echoid-s2318" xml:space="preserve">qu’un expoſant eſt un <lb/>nombre que l’on met vers la droite d’une lettre, un peu plus <lb/>élevée qu’elle, & </s> <s xml:id="echoid-s2319" xml:space="preserve">qui marque combien de fois on auroit dû <lb/>écrire cette lettre, ou combien de fois elle eſt multipliée par <lb/>elle - même. </s> <s xml:id="echoid-s2320" xml:space="preserve">a<emph style="sub">3</emph>, a<emph style="sub">5</emph>, a<emph style="sub">2</emph>b<emph style="sub">3</emph>, a<emph style="sub">4</emph>b<emph style="sub">8</emph> ſont des quantités exponan-<lb/>tielles. </s> <s xml:id="echoid-s2321" xml:space="preserve">Mais on trouve ſouvent en Algebre des quantités qui <lb/>ont des expoſans poſitifs & </s> <s xml:id="echoid-s2322" xml:space="preserve">négatifs, poſitifs entiers, & </s> <s xml:id="echoid-s2323" xml:space="preserve">poſitifs <lb/>fractionnaires, négatifs entiers, & </s> <s xml:id="echoid-s2324" xml:space="preserve">négatifs fractionnaires, <lb/>comme a<emph style="sub">3</emph>, a<emph style="sub">-3</emph>, a<emph style="sub">{3/2}</emph>, a<emph style="sub">-{3/2}</emph>; </s> <s xml:id="echoid-s2325" xml:space="preserve">on trouve même quelquefois des <lb/>lettres qui ont zero pour expoſant, comme a°, b°, &</s> <s xml:id="echoid-s2326" xml:space="preserve">c. </s> <s xml:id="echoid-s2327" xml:space="preserve">Il <lb/>s’agit de ſçavoir ce que peuvent ſignifier ces expreſſions, c’eſt <lb/>ce que nous allons démontrer, & </s> <s xml:id="echoid-s2328" xml:space="preserve">c’eſt à quoi ſe réduit ce qu’on <lb/>appelle calcul des expoſans.</s> <s xml:id="echoid-s2329" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2330" xml:space="preserve">Comme ce calcul eſt fondé ſur ces deux ſuppoſitions, 1°. </s> <s xml:id="echoid-s2331" xml:space="preserve">que <lb/>deux lettres ou pluſieurs, miſes à côté l’une de l’autre, déſi-<lb/>gneront toujours le produit des grandeurs qu’elles expriment; <lb/></s> <s xml:id="echoid-s2332" xml:space="preserve">2°. </s> <s xml:id="echoid-s2333" xml:space="preserve">Que pour diviſer deux grandeurs algébriques l’une par <lb/>l’autre, il faut les poſer en fraction, & </s> <s xml:id="echoid-s2334" xml:space="preserve">effacer les lettres com-<lb/>munes au diviſeur & </s> <s xml:id="echoid-s2335" xml:space="preserve">au dividende, ou communes au numéra-<lb/>teur & </s> <s xml:id="echoid-s2336" xml:space="preserve">au dénominateur, il faut ſe rendre attentif à tout ce <lb/>qui eſt renfermé dans ces deux hypotheſes, & </s> <s xml:id="echoid-s2337" xml:space="preserve">l’on en déduira <lb/>aifément tout ce que nous allons voir.</s> <s xml:id="echoid-s2338" xml:space="preserve"/> </p> <pb o="69" file="0107" n="107" rhead="DE MATHÉMATIQUE. Liv. I."/> <p> <s xml:id="echoid-s2339" xml:space="preserve">134. </s> <s xml:id="echoid-s2340" xml:space="preserve">Pour multiplier deux grandeurs qui ont les mêmes <lb/>lettres avec différens expoſans l’une par l’autre, il faut écrire <lb/>ces lettres les unes à côté des autres, & </s> <s xml:id="echoid-s2341" xml:space="preserve">leur donner la ſomme <lb/>des expoſans des deux facteurs: </s> <s xml:id="echoid-s2342" xml:space="preserve">ainſi a<emph style="sub">3</emph> x a<emph style="sub">2</emph> = a<emph style="sub">3 + 2</emph> = a<emph style="sub">5</emph>; <lb/></s> <s xml:id="echoid-s2343" xml:space="preserve">a<emph style="sub">2</emph>b<emph style="sub">3</emph> x a<emph style="sub">4</emph>b<emph style="sub">2</emph> = a<emph style="sub">2 + 4</emph>b<emph style="sub">3 + 2</emph> = a<emph style="sub">6</emph>b<emph style="sub">5</emph>; </s> <s xml:id="echoid-s2344" xml:space="preserve">car a<emph style="sub">3</emph> = aaa, & </s> <s xml:id="echoid-s2345" xml:space="preserve">a<emph style="sub">2</emph> = aa: </s> <s xml:id="echoid-s2346" xml:space="preserve"><lb/>donc a<emph style="sub">3</emph> x a<emph style="sub">2</emph> = aaa x aa = a<emph style="sub">5</emph>; </s> <s xml:id="echoid-s2347" xml:space="preserve">de même a<emph style="sub">2</emph>b<emph style="sub">3</emph> = aabbb, & </s> <s xml:id="echoid-s2348" xml:space="preserve">a<emph style="sub">4</emph>b<emph style="sub">2</emph> <lb/>= aaaabb : </s> <s xml:id="echoid-s2349" xml:space="preserve">donc a<emph style="sub">2</emph>b<emph style="sub">3</emph> x a<emph style="sub">4</emph>b<emph style="sub">2</emph> = aabbb x aaaabb = aaaaaabbbbb.</s> <s xml:id="echoid-s2350" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2351" xml:space="preserve">135. </s> <s xml:id="echoid-s2352" xml:space="preserve">Comme la Diviſion fait toujours le contraire de la <lb/>Multiplication, elle doit auſſi ſe faire par une voie oppoſée: <lb/></s> <s xml:id="echoid-s2353" xml:space="preserve">donc puiſque la multiplication des quantités qui ont les mêmes <lb/>lettres, avec différens expoſans, ſe fait par l’Addition de ces <lb/>mêmes expoſans, la Diviſion doit ſe faire par la Souſtraction <lb/>des expoſans des lettres communes au dividende & </s> <s xml:id="echoid-s2354" xml:space="preserve">au diviſeur: </s> <s xml:id="echoid-s2355" xml:space="preserve"><lb/>ainſi {a<emph style="sub">3</emph>/a<emph style="sub">2</emph>} = a<emph style="sub">3 - 2</emph> = a, & </s> <s xml:id="echoid-s2356" xml:space="preserve">c’eſt ce que l’on fait, lorſqu’après les <lb/>avoir mis en fraction, on efface les lettres communes au nu-<lb/>mérateur & </s> <s xml:id="echoid-s2357" xml:space="preserve">au dénominateur; </s> <s xml:id="echoid-s2358" xml:space="preserve">car{a<emph style="sub">3</emph>/a<emph style="sub">2</emph>} = {aaa/aa} effaçant aa au nu-<lb/>mérateur & </s> <s xml:id="echoid-s2359" xml:space="preserve">au dénominateur, il vient a au quotient, de même <lb/>que par la Souſtraction des expoſans. </s> <s xml:id="echoid-s2360" xml:space="preserve">Tout de même {a<emph style="sub">4</emph>b<emph style="sub">2</emph>c<emph style="sub">5</emph>/a<emph style="sub">3</emph>bc<emph style="sub">2</emph>} = <lb/>{aaaabbccccc/aaabcc} = abccc = abc<emph style="sub">3</emph>, ce que l’on eût auſſi trouvé par la <lb/>Souſtraction des expoſans, en faiſant {a<emph style="sub">4</emph>b<emph style="sub">2</emph>c<emph style="sub">5</emph>/a<emph style="sub">3</emph>bc<emph style="sub">2</emph>} = a<emph style="sub">4 - 3</emph>b<emph style="sub">2 - 1</emph>c<emph style="sub">5 - 2</emph> <lb/>= abc<emph style="sub">3</emph>. </s> <s xml:id="echoid-s2361" xml:space="preserve">De même {d<emph style="sub">2</emph>f<emph style="sub">3</emph>g<emph style="sub">4</emph>/dfg<emph style="sub">2</emph>} = d<emph style="sub">2 - 1</emph>f<emph style="sub">3 - 1</emph>g<emph style="sub">4 - 2</emph> = df<emph style="sub">2</emph>g<emph style="sub">2</emph>; </s> <s xml:id="echoid-s2362" xml:space="preserve">demê-<lb/>me encore {a<emph style="sub">2</emph>b<emph style="sub">5</emph>/a<emph style="sub">3</emph>b<emph style="sub">2</emph>} = {b<emph style="sub">3</emph>/a} en effaçant les lettres communes au nu-<lb/>mérateur & </s> <s xml:id="echoid-s2363" xml:space="preserve">au dénominateur, ou bien en faiſant la ſouſtrac-<lb/>tion des expoſans a<emph style="sub">2 - 3</emph>b<emph style="sub">5 - 2</emph> = a<emph style="sub">-1</emph>b<emph style="sub">3</emph>. </s> <s xml:id="echoid-s2364" xml:space="preserve">On voit à préſent ce <lb/>que c’eſt qu’un expoſant négatif; </s> <s xml:id="echoid-s2365" xml:space="preserve">car il eſt évident que le né-<lb/>gatif vient de la ſouſtraction d’un nombre plus grand, ôté d’un <lb/>plus petit que lui: </s> <s xml:id="echoid-s2366" xml:space="preserve">donc une quantité qui a un expoſant né-<lb/>gatif eſt le quotient d’une certaine puiſſance d’une lettre di-<lb/>viſée par une plus haute puiſſance de la même lettre; </s> <s xml:id="echoid-s2367" xml:space="preserve">ainſi <lb/>a<emph style="sub">-2</emph> peut venir de {a<emph style="sub">3</emph>/a<emph style="sub">5</emph>}, ou bien de {a<emph style="sub">5</emph>/a<emph style="sub">7</emph>} ou de {a/a<emph style="sub">3</emph>}, &</s> <s xml:id="echoid-s2368" xml:space="preserve">c, car {a<emph style="sub">3</emph>/a<emph style="sub">5</emph>} = <lb/>{a<emph style="sub">3</emph> x 1/a<emph style="sub">3</emph> x a<emph style="sub">2</emph>}; </s> <s xml:id="echoid-s2369" xml:space="preserve">donc en diviſant le numérateur & </s> <s xml:id="echoid-s2370" xml:space="preserve">le dénominateur de <lb/>la fraction par une même grandeur a<emph style="sub">3</emph>, il vient au quotient <lb/>{1/a<emph style="sub">2</emph>}: </s> <s xml:id="echoid-s2371" xml:space="preserve">mais on trouve auſſi le quotient de {a<emph style="sub">3</emph>/a<emph style="sub">5</emph>} en ôtant l’expoſant <lb/>5 du diviſeur de l’expoſant 3 du dividende, & </s> <s xml:id="echoid-s2372" xml:space="preserve">le quotient eſt <pb o="70" file="0108" n="108" rhead="NOUVEAU COURS"/> a<emph style="sub">3 - 5</emph> = a<emph style="sub">-2</emph>; </s> <s xml:id="echoid-s2373" xml:space="preserve">donc a<emph style="sub">-2</emph> = {1/a<emph style="sub">2</emph>}: </s> <s xml:id="echoid-s2374" xml:space="preserve">en général une lettre élevée à <lb/>une puiſſance négative eſt égale au quotient de l’unité, diviſée <lb/>par la même puiſſance poſitive de la même lettre. </s> <s xml:id="echoid-s2375" xml:space="preserve">a<emph style="sub">-2</emph>b<emph style="sub">3</emph> = b<emph style="sub">3</emph> x <lb/>{1/a<emph style="sub">2</emph>} = {b<emph style="sub">3</emph>/a<emph style="sub">2</emph>}, a<emph style="sub">-4</emph> = {1/a<emph style="sub">4</emph>}, a<emph style="sub">-m</emph> = {1/a<emph style="sub">m</emph>}, & </s> <s xml:id="echoid-s2376" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s2377" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2378" xml:space="preserve">136. </s> <s xml:id="echoid-s2379" xml:space="preserve">Si l’expoſant du diviſeur eſt égal à l’expoſant du divi-<lb/>dende, la différence de l’expoſant ſera zero: </s> <s xml:id="echoid-s2380" xml:space="preserve">donc a° = {a<emph style="sub">2</emph>/a<emph style="sub">2</emph>} <lb/>= {a<emph style="sub">3</emph>/a<emph style="sub">3</emph>} = a<emph style="sub">2 - 2</emph> = a<emph style="sub">3 - 3</emph>, &</s> <s xml:id="echoid-s2381" xml:space="preserve">c. </s> <s xml:id="echoid-s2382" xml:space="preserve">Donc en général a° = 1; </s> <s xml:id="echoid-s2383" xml:space="preserve">car une <lb/>grandeur, diviſée par elle-même, donne toujours 1 au quo-<lb/>tient, puiſqu’elle ſe contient une fois.</s> <s xml:id="echoid-s2384" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div145" type="section" level="1" n="120"> <head xml:id="echoid-head136" style="it" xml:space="preserve">De la formation des Puiſſances, des Quantités exponentielles, <lb/>& de l’extraction de leurs racines.</head> <p> <s xml:id="echoid-s2385" xml:space="preserve">137. </s> <s xml:id="echoid-s2386" xml:space="preserve">On appelle puiſſance en général, tout produit qui ré-<lb/>ſulte de la multiplication d’une quantité multipliée pluſieurs <lb/>fois par elle-même. </s> <s xml:id="echoid-s2387" xml:space="preserve">a, a<emph style="sub">2</emph>, a<emph style="sub">5</emph> ſont des puiſſances de a, parce <lb/>que ces produits réſultent de a, multiplié pluſieurs fois par <lb/>lui-même: </s> <s xml:id="echoid-s2388" xml:space="preserve">dans ces exemples il a été multiplié trois fois, <lb/>deux fois, cinq fois, parce que les expoſans ſont 3, 2 & </s> <s xml:id="echoid-s2389" xml:space="preserve">5.</s> <s xml:id="echoid-s2390" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2391" xml:space="preserve">138. </s> <s xml:id="echoid-s2392" xml:space="preserve">Comme la multiplication d’une même lettre, qui a dif-<lb/>férens expoſans, ſe fait par l’addition des expoſans (art. </s> <s xml:id="echoid-s2393" xml:space="preserve">133), <lb/>les puiſſances d’une quantité algébrique, qui a déja un ex-<lb/>poſant, ou les produits de cette quantité, multipliée pluſieurs <lb/>fois par elle-même, ſe trouveront par l’addition de cet expo-<lb/>ſant, répétés autant de fois qu’il y a d’unité dans la puiſſance <lb/>à laquelle on veut élever cette quantité; </s> <s xml:id="echoid-s2394" xml:space="preserve">mais l’addition ré-<lb/>pétée d’un même nombre ſe fait par la multiplication: </s> <s xml:id="echoid-s2395" xml:space="preserve">donc <lb/>la formation des puiſſances d’une quantité exponentielle ſe <lb/>fera en multipliant ſon expoſant par le nombre qui marque <lb/>la puiſſance à laquelle on veut l’élever: </s> <s xml:id="echoid-s2396" xml:space="preserve">ainſi pour élever a<emph style="sub">2</emph> à <lb/>la 3<emph style="sub">e</emph>, 4<emph style="sub">e</emph> ou 5<emph style="sub">e</emph> puiſſance, il faudra ajouter l’expoſant 2 à lui-<lb/>même trois fois, quatre fois, ou cinq fois, ou, ce qui eſt la <lb/>même choſe, le multiplier par 3, par 4, ou par 5, & </s> <s xml:id="echoid-s2397" xml:space="preserve">l’on aura <lb/>pour la 3<emph style="sub">e</emph>, 4<emph style="sub">e</emph>, ou 5<emph style="sub">e</emph> puiſſance de a<emph style="sub">2</emph>; </s> <s xml:id="echoid-s2398" xml:space="preserve">a<emph style="sub">6</emph>, a<emph style="sub">8</emph>, a<emph style="sub">10</emph>. </s> <s xml:id="echoid-s2399" xml:space="preserve">De même pour <lb/>élever a<emph style="sub">2</emph> b<emph style="sub">3</emph> c<emph style="sub">4</emph> à la cinquieme puiſſance, il faudra multiplier <lb/>les expoſans des lettres a, b, c, qui ſont 2, 3, 4 par 5, & </s> <s xml:id="echoid-s2400" xml:space="preserve">les <lb/>produits, mis à côtés des mêmes lettres, donneront la puiſ- <pb o="71" file="0109" n="109" rhead="DE MATHÉMATIQUE. Liv. I."/> ſance demandée égale à a<emph style="sub">10</emph>, b<emph style="sub">15</emph>, c<emph style="sub">20</emph>. </s> <s xml:id="echoid-s2401" xml:space="preserve">De même la quatrieme <lb/>puiſſance de c<emph style="sub">2</emph> b<emph style="sub">3</emph> f<emph style="sub">6</emph> eſt c<emph style="sub">8</emph> b<emph style="sub">12</emph> f<emph style="sub">24</emph>, & </s> <s xml:id="echoid-s2402" xml:space="preserve">ainſi du reſte.</s> <s xml:id="echoid-s2403" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2404" xml:space="preserve">139. </s> <s xml:id="echoid-s2405" xml:space="preserve">Si l’on avoit une fraction que l’on voulût élever à une <lb/>puiſſance, & </s> <s xml:id="echoid-s2406" xml:space="preserve">dont le numérateur & </s> <s xml:id="echoid-s2407" xml:space="preserve">le dénominateur fuſſent <lb/>chacuns des quantités exponentielles, on l’éleveroit à cette <lb/>puiſſance en multipliant les expoſans du numérateur & </s> <s xml:id="echoid-s2408" xml:space="preserve">du dé-<lb/>nominateur par l’expoſant de la puiſſance; </s> <s xml:id="echoid-s2409" xml:space="preserve">car une fraction <lb/>multipliée par une fraction eſt égale au produit des numéra-<lb/>teurs, diviſé par celui des dénominateurs. </s> <s xml:id="echoid-s2410" xml:space="preserve">Ainſi pour élever <lb/>la fraction {a<emph style="sub">2</emph>b<emph style="sub">3</emph>/c<emph style="sub">4</emph>} à la ſeconde puiſſance, on écrira {a<emph style="sub">2</emph> x <emph style="sub">2</emph>b<emph style="sub">3</emph> x <emph style="sub">2</emph>/c<emph style="sub">4</emph> x <emph style="sub">2</emph>} = <lb/>{a<emph style="sub">4</emph>b<emph style="sub">6</emph>/c<emph style="sub">8</emph>}; </s> <s xml:id="echoid-s2411" xml:space="preserve">de même la 3<emph style="sub">e</emph> puiſſance de la fraction {a<emph style="sub">3</emph>f<emph style="sub">2</emph>c<emph style="sub">4</emph>/b<emph style="sub">2</emph>g<emph style="sub">2</emph>} = {a<emph style="sub">9</emph>f<emph style="sub">6</emph>c<emph style="sub">12</emph>/b<emph style="sub">6</emph>g<emph style="sub">6</emph>}, <lb/>& </s> <s xml:id="echoid-s2412" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s2413" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2414" xml:space="preserve">140. </s> <s xml:id="echoid-s2415" xml:space="preserve">L’extraction des racines fait préciſément le contraire <lb/>de la formation des puiſſances. </s> <s xml:id="echoid-s2416" xml:space="preserve">Extraire la racine d’une quan-<lb/>tité algébrique, c’eſt chercher la quantité qui, multipliée par <lb/>elle-même, a donné la quantité dont on cherche la racine. <lb/></s> <s xml:id="echoid-s2417" xml:space="preserve">Comme il y a différentes puiſſances, il y a auſſi différentes <lb/>racines: </s> <s xml:id="echoid-s2418" xml:space="preserve">la racine quarrée d’une quantité algébrique eſt la <lb/>lettre ou quantité, qui multipliée une fois par elle-même, a <lb/>donné le quarré propoſé; </s> <s xml:id="echoid-s2419" xml:space="preserve">la racine cube eſt celle qui, multi-<lb/>pliée deux fois par elle-même, a donné le cube propoſé, ou <lb/>bien dont l’expoſant, multiplié par 3, a donné ce même cube. </s> <s xml:id="echoid-s2420" xml:space="preserve"><lb/>Si l’on veut indiquer cette racine, on ſe ſert du ſigne √\x{0020}, que <lb/>l’on appelle ſigne radical, & </s> <s xml:id="echoid-s2421" xml:space="preserve">qui ſert pour marquer toutes les <lb/>racines, en mettant au deſſus un chiffre qui marque la racine <lb/>que l’on veut prendre. </s> <s xml:id="echoid-s2422" xml:space="preserve">Ainſi <emph style="sub">2</emph>√\x{0020}, <emph style="sub">3</emph>√\x{0020}, <emph style="sub">4</emph>√\x{0020}, <emph style="sub">5</emph>√\x{0020} ſont des ſignes qui <lb/>indiquent les racines ſeconde, troiſieme, quatrieme ou cin-<lb/>quieme; </s> <s xml:id="echoid-s2423" xml:space="preserve">quand on veut marquer une racine quarrée, on <lb/>ſous-entend preſque toujours le 2, & </s> <s xml:id="echoid-s2424" xml:space="preserve">l’on marque ainſi √\x{0020}: </s> <s xml:id="echoid-s2425" xml:space="preserve"><lb/>par exemple, √a<emph style="sub">2</emph>\x{0020} indique qu’il faut prendre la racine quarrée <lb/>de la quantité a<emph style="sub">2</emph>, <emph style="sub">3</emph>√a<emph style="sub">3</emph>\x{0020} indique que l’on prend la racine cube <lb/>de a<emph style="sub">3</emph>. </s> <s xml:id="echoid-s2426" xml:space="preserve">La racine quarrée de a<emph style="sub">2</emph> eſt a, car a x a donne a<emph style="sub">2</emph>; </s> <s xml:id="echoid-s2427" xml:space="preserve">la <lb/>racine cube de a<emph style="sub">3</emph> eſt a, car a x a x a donne a<emph style="sub">3</emph>: </s> <s xml:id="echoid-s2428" xml:space="preserve">de même la <lb/>racine quatrieme de a<emph style="sub">4</emph> eſt a, car a x a x a x a donne a<emph style="sub">4</emph>, & </s> <s xml:id="echoid-s2429" xml:space="preserve"><lb/>ainſi de ſuite.</s> <s xml:id="echoid-s2430" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2431" xml:space="preserve">141. </s> <s xml:id="echoid-s2432" xml:space="preserve">Comme l’extraction des racines eſt une opération di-<lb/>rectement oppoſée à la formation des puiſſances, que celle-ci <pb o="72" file="0110" n="110" rhead="NOUVEAU COURS"/> décompoſe les quantités que l’autre avoit compoſées, la ma-<lb/>niere dont on doit la pratiquer doit auſſi être directement op-<lb/>poſée à celle dont on ſe ſert pour l’élévation des puiſſances. <lb/></s> <s xml:id="echoid-s2433" xml:space="preserve">Mais (n°. </s> <s xml:id="echoid-s2434" xml:space="preserve">136.) </s> <s xml:id="echoid-s2435" xml:space="preserve">la formation des puiſſances ſe fait, en mul-<lb/>tipliant l’expoſant de la quantité que l’on veut élever par <lb/>l’expoſant de la puiſſance à laquelle on veut élever cette quan-<lb/>tité; </s> <s xml:id="echoid-s2436" xml:space="preserve">donc l’extraction des racines ſe fera en diviſant l’expo-<lb/>ſant de la quantité donnée par l’expoſant de la racine que l’on <lb/>demande. </s> <s xml:id="echoid-s2437" xml:space="preserve">Si l’expoſant de la grandeur donnée eſt diviſible <lb/>par l’expoſant de la racine, on aura la racine exacte, ſinon on <lb/>aura pour la racine cherchée une quantité, dont l’expoſant ſera <lb/>une fraction, ou bien on ſe contentera d’indiquer la racine, <lb/>en la mettant ſous le ſigne √\x{0020}, au deſſus duquel on mettra un <lb/>nombre qui marque la racine que l’on demande. </s> <s xml:id="echoid-s2438" xml:space="preserve">Tout ceci <lb/>s’entendra aiſément par des exemples. </s> <s xml:id="echoid-s2439" xml:space="preserve">Pour avoir la racine <lb/>quarrée ou 2<emph style="sub">e</emph> de a<emph style="sub">2</emph>b<emph style="sub">6</emph>, je diviſe les expoſans 2 & </s> <s xml:id="echoid-s2440" xml:space="preserve">6 par 2, ex-<lb/>poſant de la racine; </s> <s xml:id="echoid-s2441" xml:space="preserve">je mets les quotiens 1 & </s> <s xml:id="echoid-s2442" xml:space="preserve">3 en expoſant <lb/>à côté des lettres ab, & </s> <s xml:id="echoid-s2443" xml:space="preserve">j’ai pour la racine demandée a<emph style="sub">1</emph>b<emph style="sub">3</emph>; </s> <s xml:id="echoid-s2444" xml:space="preserve"><lb/>(car lorſqu’une lettre n’a pas d’expoſant, on lui ſuppoſe tou-<lb/>jours l’unité pour expoſant). </s> <s xml:id="echoid-s2445" xml:space="preserve">Si l’on multiplie ab<emph style="sub">3</emph> ou abbb par <lb/>lui-même une fois, on aura a<emph style="sub">2</emph>b<emph style="sub">6</emph> ou aabbbbbb; </s> <s xml:id="echoid-s2446" xml:space="preserve">donc ab<emph style="sub">3</emph> eſt la <lb/>racine quarrée de a<emph style="sub">2</emph>b<emph style="sub">6</emph>: </s> <s xml:id="echoid-s2447" xml:space="preserve">pour avoir la racine cinquieme de <lb/>a<emph style="sub">10</emph>b<emph style="sub">15</emph>c<emph style="sub">20</emph>, j’écris d’abord a<emph style="sub">{10/5}</emph>b<emph style="sub">{15/5}</emph>c<emph style="sub">{20/5}</emph>, & </s> <s xml:id="echoid-s2448" xml:space="preserve">faiſant la diviſion des ex-<lb/>poſans par l’expoſant 5 de la racine cinquieme, j’ai a<emph style="sub">2</emph>b<emph style="sub">3</emph>c<emph style="sub">4</emph> = <lb/><emph style="sub">5</emph>√a<emph style="sub">10</emph>b<emph style="sub">15</emph>c<emph style="sub">20</emph>\x{0020}: </s> <s xml:id="echoid-s2449" xml:space="preserve">de même <emph style="sub">3</emph>√8a<emph style="sub">3</emph>b<emph style="sub">9</emph>c<emph style="sub">12</emph>\x{0020} = 2a<emph style="sub">{3/3}</emph>b<emph style="sub">{9/3}</emph>c<emph style="sub">{12/3}</emph> = 2ab<emph style="sub">3</emph>c<emph style="sub">4</emph>, car <lb/>le cube de 2 eſt 8, celui de a eſt a<emph style="sub">3</emph>, de b<emph style="sub">3</emph> eſt b<emph style="sub">3 x 3</emph> ou b<emph style="sub">9</emph>, celui <lb/>de c<emph style="sub">4</emph> eſt c<emph style="sub">4 x 3</emph> ou c<emph style="sub">12</emph>.</s> <s xml:id="echoid-s2450" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2451" xml:space="preserve">Si l’on me demande la racine cinquieme de a<emph style="sub">6</emph>b<emph style="sub">8</emph>, comme <lb/>les expoſans 6 & </s> <s xml:id="echoid-s2452" xml:space="preserve">8 ne ſont pas diviſibles par 5, expoſant de la <lb/>racine, je puis indiquer cette racine en deux manieres, ou <lb/>bien en mettant le ſigne √\x{0020} avec un 5 au deſſus devant la quan-<lb/>tité a<emph style="sub">6</emph>b<emph style="sub">8</emph>, de cette maniere: </s> <s xml:id="echoid-s2453" xml:space="preserve"><emph style="sub">5</emph>√a<emph style="sub">6</emph>b<emph style="sub">8</emph>\x{0020}, ou bien en mettant aux <lb/>lettres ab les expoſans fractionnaires {6/5}, {8/5}, en cette maniere: <lb/></s> <s xml:id="echoid-s2454" xml:space="preserve">a<emph style="sub">{6/5}</emph>b<emph style="sub">{8/5}</emph>, & </s> <s xml:id="echoid-s2455" xml:space="preserve">ces deux expreſſions <emph style="sub">5</emph>√a<emph style="sub">6</emph>b<emph style="sub">8</emph>\x{0020}, ou a<emph style="sub">{6/5}</emph>b<emph style="sub">{8/5}</emph> ſont égales, car <lb/>elles déſignent chacune la racine cinquieme d’une même <lb/>grandeur.</s> <s xml:id="echoid-s2456" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2457" xml:space="preserve">142. </s> <s xml:id="echoid-s2458" xml:space="preserve">Il ſuit delà, que lorſqu’on trouvera une quantité avec <lb/>un expoſant fractionnaire, on en pourra conclure que l’on <pb o="73" file="0111" n="111" rhead="DE MATHÉMATIQUE. Liv. I."/> prend la racine marquée par le dénominateur de cette même <lb/>quantité élevée à une puiſſance égale au numérateur de la frac-<lb/>tion: </s> <s xml:id="echoid-s2459" xml:space="preserve">ainſi a<emph style="sub">{3/2}</emph>=<emph style="sub">2</emph>√a<emph style="sub">3</emph>\x{0020}, a<emph style="sub">{5/6}</emph>=<emph style="sub">6</emph>√a<emph style="sub">5</emph>\x{0020}, a<emph style="sub">{2/3}</emph>b<emph style="sub">{4/3}</emph>=<emph style="sub">3</emph>√a<emph style="sub">2</emph>b<emph style="sub">4</emph>\x{0020}, a<emph style="sub">{1/2}</emph>b<emph style="sub">{4/5}</emph>=<emph style="sub">2</emph>√a\x{0020} <lb/>X <emph style="sub">5</emph>√b<emph style="sub">4</emph>\x{0020}, &</s> <s xml:id="echoid-s2460" xml:space="preserve">c.</s> <s xml:id="echoid-s2461" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2462" xml:space="preserve">143. </s> <s xml:id="echoid-s2463" xml:space="preserve">Il ſuit encore des mêmes principes, que a<emph style="sub">-{3/2}</emph>={1/a<emph style="sub">{3/2}</emph>}= <lb/>{1/√a<emph style="sub">3</emph>\x{0020}}; </s> <s xml:id="echoid-s2464" xml:space="preserve">car par la fin de l’art. </s> <s xml:id="echoid-s2465" xml:space="preserve">134. </s> <s xml:id="echoid-s2466" xml:space="preserve">a<emph style="sub">-3</emph>={1/a<emph style="sub">3</emph>}, & </s> <s xml:id="echoid-s2467" xml:space="preserve">par la même <lb/>raiſon a<emph style="sub">-{3/2}</emph>={1/a<emph style="sub">{3/2}</emph>}. </s> <s xml:id="echoid-s2468" xml:space="preserve">Mais par l’article précédent a<emph style="sub">{3/2}</emph>=√a<emph style="sub">3</emph>\x{0020}; </s> <s xml:id="echoid-s2469" xml:space="preserve">donc <lb/>a<emph style="sub">-{3/2}</emph>={1/√a<emph style="sub">3</emph>\x{0020}}: </s> <s xml:id="echoid-s2470" xml:space="preserve">de même a<emph style="sub">-{3/2}</emph>b<emph style="sub">{5/6}</emph>={b<emph style="sub">{5/6}</emph>/a<emph style="sub">{3/2}</emph>={<emph style="sub">6</emph>√b<emph style="sub">5</emph>\x{0020}/√a<emph style="sub">3</emph>\x{0020}}; </s> <s xml:id="echoid-s2471" xml:space="preserve">de même encore <lb/>a<emph style="sub">-3</emph>b<emph style="sub">-{4/5}</emph>={1/a<emph style="sub">3</emph>b<emph style="sub">{4/5}</emph>}={1/a<emph style="sub">3</emph><emph style="sub">5</emph>√b<emph style="sub">4</emph>\x{0020}}, ou {a<emph style="sub">-3</emph>/√b<emph style="sub">4</emph>\x{0020}}, & </s> <s xml:id="echoid-s2472" xml:space="preserve">ainſi desautres. </s> <s xml:id="echoid-s2473" xml:space="preserve">On voit <lb/>par tout ce que nous venons de dire ce que ſignifie un expo-<lb/>ſant poſitif ou négatif entier, ce que ſignifie un expoſant en-<lb/>tier, fractionnaire poſitif ou fractionnaire négatif, & </s> <s xml:id="echoid-s2474" xml:space="preserve">enfin ce <lb/>que c’eſt qu’un expoſant zero.</s> <s xml:id="echoid-s2475" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2476" xml:space="preserve">144. </s> <s xml:id="echoid-s2477" xml:space="preserve">Lorſqu’on aura une des expreſſions précédentes, com-<lb/>me a<emph style="sub">-3</emph>, a<emph style="sub">-{3/2}</emph>, a<emph style="sub">{4/5}</emph>, a<emph style="sub">0</emph>, & </s> <s xml:id="echoid-s2478" xml:space="preserve">autres ſemblables, on pourra pren-<lb/>dre en leurs places leurs égales, {1/a<emph style="sub">3</emph>}, {1/a<emph style="sub">{3/2}</emph>} ou {1/√a<emph style="sub">3</emph>\x{0020}}, <emph style="sub">5</emph>√a<emph style="sub">4</emph>\x{0020}, & </s> <s xml:id="echoid-s2479" xml:space="preserve">1 à <lb/>la place de a<emph style="sub">0</emph>, ſi cela eſt à propos, & </s> <s xml:id="echoid-s2480" xml:space="preserve">réciproquement ſubſti-<lb/>tuer les premieres expreſſions à la place des ſecondes, ſi le <lb/>calcul le demande ainſi. </s> <s xml:id="echoid-s2481" xml:space="preserve">Voici les formules générales de toutes <lb/>ces expreſſions: </s> <s xml:id="echoid-s2482" xml:space="preserve">a<emph style="sub">-m</emph>={1/a<emph style="sub">m</emph>}, a<emph style="sub">{m/n}</emph>=<emph style="sub">n</emph>√a<emph style="sub">m</emph>\x{0020}, a<emph style="sub">-{m/n}</emph>={1/<emph style="sub">n</emph>√a<emph style="sub">m</emph>\x{0020}}, a<emph style="sub">0</emph>, b<emph style="sub">0</emph>, q<emph style="sub">0</emph>=1.</s> <s xml:id="echoid-s2483" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2484" xml:space="preserve">Si l’on avoit des fractions algébriques, dont on voulût <lb/>extraire les racines, on extrairoit celle du numérateur & </s> <s xml:id="echoid-s2485" xml:space="preserve">celle <lb/>du dénominateur, ſuivant les regles précédentes, en ſuppo-<lb/>ſant que les deux termes ſont des quantités incomplexes: </s> <s xml:id="echoid-s2486" xml:space="preserve">car <lb/>puiſque l’on éleve les fractions à des puiſſances propoſées, en <lb/>y élevant le numérateur & </s> <s xml:id="echoid-s2487" xml:space="preserve">le dénominateur (art. </s> <s xml:id="echoid-s2488" xml:space="preserve">139), il <lb/>faut, par la raiſon contraire, extraire les racines, en prenant <lb/>celle du numérateur & </s> <s xml:id="echoid-s2489" xml:space="preserve">celle du dénominateur. </s> <s xml:id="echoid-s2490" xml:space="preserve">Ainſi la racine <lb/>ſeconde de {a<emph style="sub">6</emph>b<emph style="sub">8</emph>/c<emph style="sub">4</emph>}={a<emph style="sub">{6/2}</emph>b<emph style="sub">{8/2}</emph>/c<emph style="sub">{4/2}</emph>}={a<emph style="sub">3</emph>b<emph style="sub">4</emph>/c<emph style="sub">2</emph>}, la racine 3<emph style="sub">e</emph> ou cubique de <lb/>{a<emph style="sub">9</emph>f<emph style="sub">6</emph>c<emph style="sub">12</emph>/b<emph style="sub">6</emph>g<emph style="sub">6</emph>}={a<emph style="sub">{9/3}</emph>f<emph style="sub">{6/3}</emph>c<emph style="sub">{12/3}</emph>/b<emph style="sub">{6/3}</emph>g<emph style="sub">{6/3}</emph>}={a<emph style="sub">3</emph>f<emph style="sub">2</emph>c<emph style="sub">4</emph>/b<emph style="sub">2</emph>g<emph style="sub">2</emph>}, & </s> <s xml:id="echoid-s2491" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s2492" xml:space="preserve"/> </p> <pb o="74" file="0112" n="112" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div146" type="section" level="1" n="121"> <head xml:id="echoid-head137" style="it" xml:space="preserve">De la formation des Puiſſances, des Polinomes, & de l’extrac-<lb/>tion de leurs racines.</head> <p> <s xml:id="echoid-s2493" xml:space="preserve">145. </s> <s xml:id="echoid-s2494" xml:space="preserve">On trouve les puiſſances des quantités algébriques <lb/>complexes de la même maniere que celles des quantités al-<lb/>gébriques incomplexes, c’eſt-à-dire en multipliant ces quan-<lb/>tités par elles-mêmes autant de fois moins une qu’il y a d’unités <lb/>dans l’expoſant de la puiſſance à laquelle on veut élever cette <lb/>quantité. </s> <s xml:id="echoid-s2495" xml:space="preserve">Pour avoir le quarré de a + b, je multiplie a + b <lb/>par a + b, & </s> <s xml:id="echoid-s2496" xml:space="preserve">j’ai (art. </s> <s xml:id="echoid-s2497" xml:space="preserve">60.) </s> <s xml:id="echoid-s2498" xml:space="preserve">a<emph style="sub">2</emph> + 2ab + bb. </s> <s xml:id="echoid-s2499" xml:space="preserve">Demême le quarré <lb/>ou la ſeconde puiſſance de a - b eſt a<emph style="sub">2</emph> - 2ab + bb: </s> <s xml:id="echoid-s2500" xml:space="preserve">d’où il <lb/>ſuit que généralement le quarré d’un binome contient tou-<lb/>jours les quarrés des deux termes, plus ou moins deux rectan-<lb/>gles du premier par le ſecond; </s> <s xml:id="echoid-s2501" xml:space="preserve">plus, lorſque les deux termes <lb/>ſont poſitifs ou négatifs, & </s> <s xml:id="echoid-s2502" xml:space="preserve">moins lorſque l’un ou l’autre eſt <lb/>négatif: </s> <s xml:id="echoid-s2503" xml:space="preserve">car il eſt clair que - <emph style="ol">a - b</emph> x - <emph style="ol">a - b</emph> donne a<emph style="sub">2</emph> + 2ab <lb/>+ bb, de même que <emph style="ol">a + b</emph> x <emph style="ol">a + b</emph>, & </s> <s xml:id="echoid-s2504" xml:space="preserve">que - <emph style="ol">a + b</emph> x - <emph style="ol">a + b</emph> <lb/>donne <emph style="ol">a<emph style="sub">2</emph> - 2ab + b<emph style="sub">2</emph></emph>, auſſi-bien que <emph style="ol">a - b</emph> x <emph style="ol">a - b</emph>.</s> <s xml:id="echoid-s2505" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2506" xml:space="preserve">Si la quantité que l’on veut élever au quarré avoit plus de <lb/>deux termes, 4 ou 5, par exemple, comme a + b + c + d, <lb/>on trouveroit toujours le quarré de cette quantité, en la mul-<lb/>tipliant une fois par elle-même: </s> <s xml:id="echoid-s2507" xml:space="preserve">mais on peut le trouver d’une <lb/>maniere beaucoup plus expéditive. </s> <s xml:id="echoid-s2508" xml:space="preserve">Je prends d’abord les quar-<lb/>rés de tous les termes qui compoſent cette quantité, ſoit que <lb/>tous ces termes ſoient poſitifs, ſoit que tous ſoient négatifs, <lb/>ou qu’il y en ait de poſitifs & </s> <s xml:id="echoid-s2509" xml:space="preserve">de négatifs. </s> <s xml:id="echoid-s2510" xml:space="preserve">Je prends enſuite le <lb/>double du premier terme, que je multiplie par tous les ſui-<lb/>vans, en donnant au produit le ſigne du premier terme, ſi <lb/>chacun des ſuivans a le même ſigne que ce premier terme, & </s> <s xml:id="echoid-s2511" xml:space="preserve"><lb/>un ſigne différent ſi celui du terme par lequel je multiplie le <lb/>double du premier eſt différent de celui du même premier. </s> <s xml:id="echoid-s2512" xml:space="preserve">Je <lb/>prends pareillement le double du ſecond, que je combine avec <lb/>les ſuivans par multiplication, en ſuivant la même regle; </s> <s xml:id="echoid-s2513" xml:space="preserve">je <lb/>prends de même le double du troiſieme, que je combine en-<lb/>core de même avec les autres, juſqu’à ce que je ſois arrivé à <lb/>l’avant dernier, que je combine avec le dernier de la même <lb/>maniere, & </s> <s xml:id="echoid-s2514" xml:space="preserve">l’opération eſt achevée.</s> <s xml:id="echoid-s2515" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2516" xml:space="preserve">Ainſi pour élever a + b - c + d - f + g à la ſeconde puiſſance, <pb o="75" file="0113" n="113" rhead="DE MATHÉMATIQUE. Liv. I."/> je prends d’abord tous les quarrés poſitifs des termes qui com-<lb/>poſent cette quantité, & </s> <s xml:id="echoid-s2517" xml:space="preserve">j’ai pour une premiere partie du <lb/>quarré que je cherche a<emph style="sub">2</emph> + b<emph style="sub">2</emph> + c<emph style="sub">2</emph> + d<emph style="sub">2</emph> + f<emph style="sub">2</emph> + g<emph style="sub">2</emph>: </s> <s xml:id="echoid-s2518" xml:space="preserve">je prends <lb/>enſuite le double du premier terme a, qui eſt 2a, que je com-<lb/>bine par multiplication avec tous les ſuivans, & </s> <s xml:id="echoid-s2519" xml:space="preserve">j’ai pour une <lb/>ſeconde partie du quarré que je cherche 2ab - 2ac + 2ad -<lb/>2af + 2ag, en donnant le ſigne + aux termes qui ont le même <lb/>ſigne + que 2a, & </s> <s xml:id="echoid-s2520" xml:space="preserve">le ſigne à ceux qui ont le ſigne -. </s> <s xml:id="echoid-s2521" xml:space="preserve">Je <lb/>prends pareillement le double de b, qui eſt 2b, & </s> <s xml:id="echoid-s2522" xml:space="preserve">le combi-<lb/>nant, ainſi que j’ai fait pour a, avec ceux qui le ſuivent, j’ai <lb/>- 2bc + 2bd - 2bf + 2bg pour la troiſieme partie du quarré <lb/>que je cherche. </s> <s xml:id="echoid-s2523" xml:space="preserve">Je prends encore le double de - c, qui eſt - 2c, <lb/>& </s> <s xml:id="echoid-s2524" xml:space="preserve">j’ai - 2cd + 2cf - 2cg, en mettant + aux termes qui ont <lb/>le ſigne -, & </s> <s xml:id="echoid-s2525" xml:space="preserve">- à ceux qui ont le ſigne +. </s> <s xml:id="echoid-s2526" xml:space="preserve">Je trouve de <lb/>même, en prenant le double du quatrieme terme d, qui eſt 2d, <lb/>- 2df + 2dg; </s> <s xml:id="echoid-s2527" xml:space="preserve">& </s> <s xml:id="echoid-s2528" xml:space="preserve">enfin prenant le double de - f, qui eſt <lb/>- 2f, je trouve - 2fg pour la derniere partie du quarré que <lb/>je cherche. </s> <s xml:id="echoid-s2529" xml:space="preserve">Ajoutant toutes ces parties, j’ai pour le quarré de-<lb/>mandé a<emph style="sub">2</emph> + b<emph style="sub">2</emph> + c<emph style="sub">2</emph> + d<emph style="sub">2</emph> + f<emph style="sub">2</emph> + g<emph style="sub">2</emph> + 2ab - 2ac + 2ad <lb/>- 2af + 2ag - 2bc + 2bd - 2bf + 2bg - 2cd + 2cf - 2cg <lb/>- 2df + 2dg - 2fg. </s> <s xml:id="echoid-s2530" xml:space="preserve">La preuve de cette pratique ſe fera en <lb/>multipliant cette quantité par elle-même, & </s> <s xml:id="echoid-s2531" xml:space="preserve">l’on trouvera les <lb/>mêmes quantités, quoique dans un ordre différent. </s> <s xml:id="echoid-s2532" xml:space="preserve">Mais la <lb/>valeur du quarré ne dépend pas de l’ordre dans lequel les <lb/>termes ſont diſpoſés: </s> <s xml:id="echoid-s2533" xml:space="preserve">il ſera toujours le même, pourvu qu’il <lb/>y ait autant de termes qu’il doit y en avoir, & </s> <s xml:id="echoid-s2534" xml:space="preserve">que chacun <lb/>d’eux ait le ſigne qu’il doit avoir. </s> <s xml:id="echoid-s2535" xml:space="preserve">On pourroit encore ſe ſervir <lb/>du même abrégé, ſi les termes avoient des coefficiens diffé-<lb/>rens de l’unité. </s> <s xml:id="echoid-s2536" xml:space="preserve">Par exemple, le quarré de 3a - 2b + 4c ſe <lb/>trouvera en ſuivant cette méthode, 9a<emph style="sub">2</emph> + 4b<emph style="sub">2</emph> + 16c<emph style="sub">2</emph> - 12ab <lb/>+ 24ac - 16bc.</s> <s xml:id="echoid-s2537" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div147" type="section" level="1" n="122"> <head xml:id="echoid-head138" style="it" xml:space="preserve">De l’Extraction de la Racine quarrée, des Quantités algébriques <lb/>complexes.</head> <p> <s xml:id="echoid-s2538" xml:space="preserve">146. </s> <s xml:id="echoid-s2539" xml:space="preserve">Pour extraire la racine quarrée d’une quantité algé-<lb/>brique complexe, par exemple, celle de la quantité a<emph style="sub">2</emph> + 2ab <lb/>+ bb, il faut dire, la racine de aa eſt a, qu’il faut poſer à la <lb/>racine: </s> <s xml:id="echoid-s2540" xml:space="preserve">ayant multipliée cette racine par elle-même, il faut <lb/>ôter le produit aa de la quantité propoſée pour avoir le reſte <pb o="76" file="0114" n="114" rhead="NOUVEAU COURS"/> 2ab + bb: </s> <s xml:id="echoid-s2541" xml:space="preserve">enſuite il faut doubler a, & </s> <s xml:id="echoid-s2542" xml:space="preserve">diviſer le reſte 2ab <lb/>+ bb par ce diviſeur 2a. </s> <s xml:id="echoid-s2543" xml:space="preserve">Faiſant la diviſion de 2ab par 2a, il <lb/>vient b, qu’il faut mettre à la ſuite de la racine, & </s> <s xml:id="echoid-s2544" xml:space="preserve">à la ſuite <lb/>du diviſeur 2a. </s> <s xml:id="echoid-s2545" xml:space="preserve">Enfin il faut multiplier par ce quotient b le <lb/>diviſeur devenu 2a + b, & </s> <s xml:id="echoid-s2546" xml:space="preserve">ſouſtraire le produit 2ab + bb du <lb/>reſte 2ab + bb; </s> <s xml:id="echoid-s2547" xml:space="preserve">& </s> <s xml:id="echoid-s2548" xml:space="preserve">comme il ne reſte rien après la ſouſtraction, <lb/>l’on conclura que la racine de a<emph style="sub">2</emph> + 2ab + bb, eſt a + b.</s> <s xml:id="echoid-s2549" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2550" xml:space="preserve">Pour voir ſi l’on a bien fait l’opération, on multipliera la <lb/>racine a + b par elle-même, & </s> <s xml:id="echoid-s2551" xml:space="preserve">comme le produit eſt a<emph style="sub">2</emph> + 2ab <lb/>+ bb égal à la quantité propoſée, on ſera ſûr que l’on a bien <lb/>opéré.</s> <s xml:id="echoid-s2552" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2553" xml:space="preserve">147. </s> <s xml:id="echoid-s2554" xml:space="preserve">Pour extraire la racine quarrée de a<emph style="sub">2</emph> - 2ab + bb, il <lb/>faut dire, la racine quarrée de a<emph style="sub">2</emph> eſt a, qu’il faut mettre à la <lb/>racine; </s> <s xml:id="echoid-s2555" xml:space="preserve">enſuite ôter le quarré aa de cette racine de la quantité <lb/>propoſée pour avoir le reſte, - 2ab + bb, qu’il faut pareille-<lb/>ment diviſer par + 2a, le quotient eſt - b, que je poſe à la <lb/>racine, & </s> <s xml:id="echoid-s2556" xml:space="preserve">à côté du diviſeur. </s> <s xml:id="echoid-s2557" xml:space="preserve">Je multiplie 2a - b par - b, le <lb/>produit eſt 2ab + bb; </s> <s xml:id="echoid-s2558" xml:space="preserve">ôtant ce produit de - 2ab + bb, comme <lb/>il ne reſte rien, je conclus que a - b eſt la racine.</s> <s xml:id="echoid-s2559" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div148" type="section" level="1" n="123"> <head xml:id="echoid-head139" xml:space="preserve"><emph style="sc">Article</emph> 146.</head> <p> <s xml:id="echoid-s2560" xml:space="preserve">a<emph style="sub">2</emph> + 2ab + b<emph style="sub">2</emph> <lb/>Reſte 2ab + b<emph style="sub">2</emph> <lb/>Souſtract. </s> <s xml:id="echoid-s2561" xml:space="preserve">2ab - bb <lb/>0 0</s> </p> <p> <s xml:id="echoid-s2562" xml:space="preserve">{a + b, racine. <lb/></s> <s xml:id="echoid-s2563" xml:space="preserve">2a diviſeur. </s> <s xml:id="echoid-s2564" xml:space="preserve"><lb/>2a + b <lb/>b <lb/>2ab + bb</s> </p> </div> <div xml:id="echoid-div149" type="section" level="1" n="124"> <head xml:id="echoid-head140" xml:space="preserve"><emph style="sc">Article</emph> 147.</head> <p> <s xml:id="echoid-s2565" xml:space="preserve">a<emph style="sub">2</emph> - 2ab + b<emph style="sub">2</emph> <lb/>Reſte - 2ab + bb <lb/>Souſtraction + 2ab - bb <lb/>0 0</s> </p> <p> <s xml:id="echoid-s2566" xml:space="preserve">{a - b, racine. <lb/></s> <s xml:id="echoid-s2567" xml:space="preserve">2a diviſeur. </s> <s xml:id="echoid-s2568" xml:space="preserve"><lb/>2a - b <lb/>- b <lb/>- 2ab + bb</s> </p> <p> <s xml:id="echoid-s2569" xml:space="preserve">148. </s> <s xml:id="echoid-s2570" xml:space="preserve">De même pour avoir la racine quarrée de la quantité <lb/>9a<emph style="sub">2</emph> - 12ab + 4b<emph style="sub">2</emph> + 24ac - 16bc + 16c<emph style="sub">2</emph> + 24ac - 16bc, <lb/>je dis, la racine quarrée de 9a<emph style="sub">2</emph> eſt 3a, que je poſe à la racine, <lb/>& </s> <s xml:id="echoid-s2571" xml:space="preserve">j’ôte le quarré de cette racine de la quantité propoſée: </s> <s xml:id="echoid-s2572" xml:space="preserve">pour <lb/>avoir le reſte - 12ab + 4b<emph style="sub">2</emph> + 24ac - 16bc + 16c<emph style="sub">2</emph>, je double <lb/>cette racine 3a, & </s> <s xml:id="echoid-s2573" xml:space="preserve">j’ai 6a pour diviſeur. </s> <s xml:id="echoid-s2574" xml:space="preserve">Je diviſe - 12ab par <pb o="77" file="0115" n="115" rhead="DE MATHÉMATIQUE. Liv. I."/> 6a, le quotient eſt - 2b, que j’écris à la racine, à côté de 3a, <lb/>& </s> <s xml:id="echoid-s2575" xml:space="preserve">à côté du diviſeur 6a; </s> <s xml:id="echoid-s2576" xml:space="preserve">ce qui me donne 6a - 2b, que je <lb/>multiplie par - 2b, pour avoir le produit - 12ab + 4bb, <lb/>que j’écris au deſſous du premier reſte avec des ſignes con-<lb/>traires pour avoir un ſecond reſte, en effaçant ce qui ſe détruit, <lb/>que je trouve être 24ac - 16bc + 16c<emph style="sub">2</emph>: </s> <s xml:id="echoid-s2577" xml:space="preserve">je double encore ce <lb/>que j’ai trouvé à la racine pour avoir le nouveau diviſeur 6a <lb/>- 4b, par lequel je diviſe le premier terme 24ac du ſecond <lb/>reſte; </s> <s xml:id="echoid-s2578" xml:space="preserve">ce qui me donne au quotient 4c, que j’écris à la ſuite <lb/>de la racine, & </s> <s xml:id="echoid-s2579" xml:space="preserve">à côté du diviſeur 6a - 4b: </s> <s xml:id="echoid-s2580" xml:space="preserve">je multiplie cette <lb/>ſomme par le même quotient 4c, & </s> <s xml:id="echoid-s2581" xml:space="preserve">j’en ôte le produit 24ac <lb/>- 16bc + 16c<emph style="sub">2</emph> du dernier reſte; </s> <s xml:id="echoid-s2582" xml:space="preserve">& </s> <s xml:id="echoid-s2583" xml:space="preserve">comme la Souſtraction ſe <lb/>fait ſans reſte, je conclus que 3a - 2b + 4c eſt la racine du <lb/>quarré propoſé: </s> <s xml:id="echoid-s2584" xml:space="preserve">je leve cette quantité au quarré, & </s> <s xml:id="echoid-s2585" xml:space="preserve">je trouve <lb/>qu’elle donne effectivement une quantité égale à celle que <lb/>l’on avoit donnée pour en extraire la racine.</s> <s xml:id="echoid-s2586" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div150" type="section" level="1" n="125"> <head xml:id="echoid-head141" xml:space="preserve"><emph style="sc">Article</emph> 148.</head> <p> <s xml:id="echoid-s2587" xml:space="preserve">9a<emph style="sub">2</emph> - 12ab + 4b<emph style="sub">2</emph> + 24ac - 16bc <lb/>-9a<emph style="sub">2</emph> {+ 16cc <lb/>1<emph style="sub">er</emph> reſte - 12ab + 4b<emph style="sub">2</emph> + 24ac <lb/>- 16bc + 16cc <lb/>+ 12ab - 4bb <lb/>Second reſte 24ac - 16bc + 16c<emph style="sub">2</emph> <lb/>- 24ac + 16bc - 16c<emph style="sub">2</emph> <lb/>0 0 0}</s> </p> <p> <s xml:id="echoid-s2588" xml:space="preserve">{3a - 2b + 4c, racine. <lb/></s> <s xml:id="echoid-s2589" xml:space="preserve">6a premier diviſeur. </s> <s xml:id="echoid-s2590" xml:space="preserve"><lb/>6a - 2b <lb/>- 2b <lb/>- 12ab + 4bb <lb/>6a - 4b, 2<emph style="sub">e</emph> diviſ. </s> <s xml:id="echoid-s2591" xml:space="preserve"><lb/>6a - 4b + 4c <lb/>+ 4c <lb/>24ac - 16bc + 16cc</s> </p> <p> <s xml:id="echoid-s2592" xml:space="preserve">Il eſt évident que la méthode dont on ſe ſert pour extraire <lb/>la racine doit la faire trouver néceſſairement, ſi la quantité <lb/>propoſée en a une: </s> <s xml:id="echoid-s2593" xml:space="preserve">car nous avons déja vu pluſieurs fois que <lb/>le quarré d’une quantité complexe contient le quarré du pre-<lb/>mier terme, le double du premier par le ſecond, & </s> <s xml:id="echoid-s2594" xml:space="preserve">le quarré <lb/>du ſecond. </s> <s xml:id="echoid-s2595" xml:space="preserve">Lorſque l’on a pris la racine quarrée du premier <lb/>terme, on a celui de la racine: </s> <s xml:id="echoid-s2596" xml:space="preserve">ainſi pour avoir le ſecond de la <lb/>même racine, il n’y a qu’à doubler ce premier, & </s> <s xml:id="echoid-s2597" xml:space="preserve">diviſer par <lb/>ce double un terme qui renferme deux lettres; </s> <s xml:id="echoid-s2598" xml:space="preserve">& </s> <s xml:id="echoid-s2599" xml:space="preserve">ſi l’on a un <lb/>quotient, ce ſera le ſecond terme de la racine, pourvu que le <lb/>quarré de ce ſecond terme ſoit encore contenu dans la quan-<lb/>tité propoſée. </s> <s xml:id="echoid-s2600" xml:space="preserve">Or par notre méthode on prend le quarré de <pb o="78" file="0116" n="116" rhead="NOUVEAU COURS"/> ce terme, puiſque l’on ajoute ce nombre au diviſeur pour mul-<lb/>tiplier le tout par ce ſecond terme; </s> <s xml:id="echoid-s2601" xml:space="preserve">& </s> <s xml:id="echoid-s2602" xml:space="preserve">s’il ne reſte rien, on ſera <lb/>ſûr que la quantité eſt un quarré parfait, & </s> <s xml:id="echoid-s2603" xml:space="preserve">de plus celui des <lb/>deux termes que l’on a trouvés, puiſque l’on a pu en ſouſtraire <lb/>le quarré du premier, le double rectangle du même premier <lb/>par le ſecond, & </s> <s xml:id="echoid-s2604" xml:space="preserve">le quarré du ſecond. </s> <s xml:id="echoid-s2605" xml:space="preserve">Le raiſonnement eſt <lb/>toujours le même, quelque ſoit le nombre des termes de la <lb/>racine; </s> <s xml:id="echoid-s2606" xml:space="preserve">car on peut toujours regarder ce que l’on a trouvé <lb/>comme le premier, & </s> <s xml:id="echoid-s2607" xml:space="preserve">ce que l’on cherche comme le ſecond <lb/>d’une quantité de deux termes, puiſque l’on peut toujours <lb/>réduire un polinome quelconque, comme a + b + c + d à <lb/>un binome, en ſuppoſant a + b + c = f; </s> <s xml:id="echoid-s2608" xml:space="preserve">ce qui donne <lb/>a + b + c + d = f + d.</s> <s xml:id="echoid-s2609" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2610" xml:space="preserve">149. </s> <s xml:id="echoid-s2611" xml:space="preserve">Si la quantité propoſée pour en extraire la racine n’eſt <lb/>pas un quarré parfait, on ſe contentera d’indiquer que l’on <lb/>en prend la racine, en la mettant ſous le ſigne √, que l’on <lb/>appelle radical, comme nous avons déja vu: </s> <s xml:id="echoid-s2612" xml:space="preserve">ainſi la racine de <lb/>aa-bb eſt √aa - bb\x{0020}, la racine de a<emph style="sub">2</emph> - 2bc = ac eſt <lb/>√a<emph style="sub">2</emph> - 2bc + ac\x{0020}, & </s> <s xml:id="echoid-s2613" xml:space="preserve">l’on appelle ces quantités, des quantités <lb/>radicales ou irrationnelles, quelquefois incommenſurables.</s> <s xml:id="echoid-s2614" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div151" type="section" level="1" n="126"> <head xml:id="echoid-head142" style="it" xml:space="preserve">De la formation du quarré d’un nombre quelconque, & de l’ex-<lb/>traction des racines ſur les grandeurs numériques.</head> <p> <s xml:id="echoid-s2615" xml:space="preserve">150. </s> <s xml:id="echoid-s2616" xml:space="preserve">Le quarré d’un nombre quelconque ſe trouve en mul-<lb/>tipliant ce nombre par lui-même: </s> <s xml:id="echoid-s2617" xml:space="preserve">ainſi le quarré de 3247 ſe <lb/>trouveroit en multipliant ce nombre une fois par lui-même, <lb/>ſuivant les regles de la Multiplication. </s> <s xml:id="echoid-s2618" xml:space="preserve">Mais pour déterminer <lb/>avec plus de préciſion les différentes parties qui compoſent ce <lb/>quarré, & </s> <s xml:id="echoid-s2619" xml:space="preserve">faire entendre plus aiſément ce que nous avons à <lb/>dire ſur l’extraction des racines, nous rapporterons la forma-<lb/>tion du quarré de ce nombre à celle du quarré d’une quantité <lb/>algébrique complexe, en le regardant lui-même comme une <lb/>quantité de cette nature, & </s> <s xml:id="echoid-s2620" xml:space="preserve">le décompoſant en ſes parties <lb/>3000 + 200 + 40 + 7, & </s> <s xml:id="echoid-s2621" xml:space="preserve">faiſant 3000 = a, 200 = b, 40 = c, <lb/>7 = d: </s> <s xml:id="echoid-s2622" xml:space="preserve">donc le quarré 3247, ou de 3000 + 200 + 40 + 7 <lb/>ſera repréſenté par celui de la quantité algébrique a + b + c + d, <lb/>qui eſt a<emph style="sub">2</emph> + 2ab + b<emph style="sub">2</emph> + 2ac + 2bc + c<emph style="sub">2</emph> + 2ad + 2bd + 2cd <lb/>+ dd, ou a<emph style="sub">2</emph> + 2ab + b<emph style="sub">2</emph> + √2a + 2b\x{0020} x c + c<emph style="sub">2</emph> + √2a + 2b + 2c\x{0020} x d + d<emph style="sub">2</emph></s> </p> <pb o="79" file="0117" n="117" rhead="DE MATHÉMATIQUE. Liv. I."/> <p> <s xml:id="echoid-s2623" xml:space="preserve">En faiſant toutes les Multiplications néceſſaires, on trouvera <lb/> <anchor type="note" xlink:label="note-0117-01a" xlink:href="note-0117-01"/> 10,54,30,09 = a<emph style="sub">2</emph> + 2ab + b<emph style="sub">2</emph> + √2a + 2b\x{0020} x c + c<emph style="sub">2</emph> + √2a + 2b + 2c\x{0020} <lb/>x d + dd.</s> <s xml:id="echoid-s2624" xml:space="preserve"/> </p> <div xml:id="echoid-div151" type="float" level="2" n="1"> <note position="right" xlink:label="note-0117-01" xlink:href="note-0117-01a" xml:space="preserve">9 # 00 # 00 # 00 # = # a<emph style="sub">2</emph> <lb/>1 # 20 # 00 # 00 # = # 2ab <lb/># 4 # 00 # 00 # = # bb <lb/># 25 # 60 # 00 # = # √2a + 2b\x{0020} x c <lb/># # 16 # 00 # = # c<emph style="sub">2</emph> <lb/># 4 # 53 # 60 # = # √2a + 2b + 2c\x{0020} x d <lb/># # # 49 # = # d<emph style="sub">2</emph> <lb/></note> </div> <p> <s xml:id="echoid-s2625" xml:space="preserve">Sur quoi l’on remarquera qu’en partageant ces produits par-<lb/>tiels en tranches de deux chiffres chacunes, excepté la der-<lb/>niere à gauche, qui peut n’en contenir qu’un,</s> </p> <p> <s xml:id="echoid-s2626" xml:space="preserve">1°. </s> <s xml:id="echoid-s2627" xml:space="preserve">Le quarré du chiffre ſignificatif 3 du premier terme 3000, <lb/>aura après lui autant de tranches de deux chiffres chacunes, <lb/>qu’il a de chiffres après lui dans le nombre 3247, & </s> <s xml:id="echoid-s2628" xml:space="preserve">qu’ainſi <lb/>ce quarré doit ſe trouver au produit total dans la premiere <lb/>tranche à gauche 10.</s> <s xml:id="echoid-s2629" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2630" xml:space="preserve">2°. </s> <s xml:id="echoid-s2631" xml:space="preserve">Que le produit 1200000 fait du double 6000 du pre-<lb/>mier terme 3000, multiplié par le ſecond 200, ſera renfermé <lb/>dans le premier chiffre de la ſeconde tranche, joint au reſte <lb/>que l’on aura eu, en ôtant le quarré de 9000000 de la premiere.</s> <s xml:id="echoid-s2632" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2633" xml:space="preserve">3°. </s> <s xml:id="echoid-s2634" xml:space="preserve">Que le quarré du même ſecond terme 40000 ſera en-<lb/>core contenu dans le dernier chiffre de la ſeconde tranche, <lb/>ayant après lui autant de tranches qu’il y a de chiffres dans le <lb/>nombre 3247; </s> <s xml:id="echoid-s2635" xml:space="preserve">d’où il ſuit, en réſumant ces trois articles, que <lb/>les deux premieres tranches contiennent le quarré 9000000 du <lb/>premier terme 3000, le double du produit du premier terme <lb/>3000, par le ſecond 200, & </s> <s xml:id="echoid-s2636" xml:space="preserve">enfin le quarré du ſecond.</s> <s xml:id="echoid-s2637" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2638" xml:space="preserve">4°. </s> <s xml:id="echoid-s2639" xml:space="preserve">Que le produit du double des deux premiers termes par <lb/>le ſecond 40, repréſenté par √2a + 2b\x{0020} x c, lequel eſt 256000, <lb/>ſe trouvera renfermé dans le premier chiffre 3 de la troiſieme <lb/>tranche, puiſque le 6 eſt directement au deſſus du 3 de cette <lb/>tranche, & </s> <s xml:id="echoid-s2640" xml:space="preserve">que le quarré 1600 du même troiſieme terme ſera <lb/>encore contenu dans la troiſieme tranche, jointe à ce qu’il <lb/>précéde. </s> <s xml:id="echoid-s2641" xml:space="preserve">Ainſi les trois premieres tranches contiennent le <lb/>quarré des trois premiers termes, le double du premier, mul-<lb/>tiplié par le ſecond, & </s> <s xml:id="echoid-s2642" xml:space="preserve">le double des deux premiers par le troi- <pb o="80" file="0118" n="118" rhead="NOUVEAU COURS"/> ſreme, leſquels produits ſont repréſentés par a<emph style="sub">2</emph> + b<emph style="sub">2</emph> + c<emph style="sub">2</emph> + 2ab <lb/>+ √2a + 2b\x{0020} x c.</s> <s xml:id="echoid-s2643" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2644" xml:space="preserve">5°. </s> <s xml:id="echoid-s2645" xml:space="preserve">Enfin l’on verra que le double du produit des trois pre-<lb/>miers termes 3000 + 200 + 40, multipliés par le ſecond, eſt <lb/>renfermé dans le premier chiffre de la derniere tranche, & </s> <s xml:id="echoid-s2646" xml:space="preserve"><lb/>que le quarré de ce dernier terme 7 eſt renfermé dans le der-<lb/>nier chiffre de la derniere tranche; </s> <s xml:id="echoid-s2647" xml:space="preserve">& </s> <s xml:id="echoid-s2648" xml:space="preserve">qu’ainſi le quarré de la <lb/>grandeur complexe 3000 + 200 + 40 + 7, ou du nombre <lb/>3247, eſt renfermé dans le nombre 10443009, puiſque ce <lb/>nombre renferme tous les produits dont il peut être compoſé.</s> <s xml:id="echoid-s2649" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2650" xml:space="preserve">Tout cela poſé, il ſera facile d’entendre ce que nous allons <lb/>dire ſur l’extraction des racines.</s> <s xml:id="echoid-s2651" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2652" xml:space="preserve">151. </s> <s xml:id="echoid-s2653" xml:space="preserve">Extraire la racine quarrée d’un nombre, c’eſt cher-<lb/>cher un nombre qui, multiplié par lui-même, donne au pro-<lb/>duit un nombre égal au nombre propoſé: </s> <s xml:id="echoid-s2654" xml:space="preserve">extraire la racine de <lb/>25, c’eſt chercher le nombre 5, qui multiplié par lui-même <lb/>une fois, donne 25 au produit. </s> <s xml:id="echoid-s2655" xml:space="preserve">Toutes les fois qu’un nombre <lb/>propoſé, pour en extraire la racine, ne contiendra que deux <lb/>chiffres, ou ſera moindre que 100, on pourra, ſans aucune <lb/>opération, trouver ſa racine vraie ou la plus proche, par le <lb/>moyen de la Table ſuivante.</s> <s xml:id="echoid-s2656" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">1 # 2 # 3 # 4 # 5 # 6 # 7 # 8 # 9 <lb/>1 # 4 # 9 # 16 # 25 # 36 # 49 # 64 # 81 <lb/></note> <p> <s xml:id="echoid-s2657" xml:space="preserve">Mais lorſque les nombres ſeront plus conſidérables, l’opé-<lb/>ration devient plus compliquée, & </s> <s xml:id="echoid-s2658" xml:space="preserve">c’eſt ce que nous allons <lb/>détailler, après que nous aurons donné les réflexions ſui-<lb/>vantes, qui ſont néceſſaires pour une exacte démonſtration <lb/>de la regle générale de l’extraction des racines quarrées.</s> <s xml:id="echoid-s2659" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2660" xml:space="preserve">152. </s> <s xml:id="echoid-s2661" xml:space="preserve">Le plus grand nombre poſſible de deux chiffres 99 ne <lb/>peut avoir plus d’un chiffre à ſa racine: </s> <s xml:id="echoid-s2662" xml:space="preserve">car ſuppoſons qu’il <lb/>puiſſe en avoir deux, & </s> <s xml:id="echoid-s2663" xml:space="preserve">que ce nombre de deux chiffres ſoit <lb/>le plus petit poſſible, qui eſt 10, ou élevant 10 au quarré, on <lb/>verra que ce quarré 100 eſt plus grand que 99: </s> <s xml:id="echoid-s2664" xml:space="preserve">donc un nom-<lb/>bre de deux chiffres quelconque ne peut en avoir plus d’un à <lb/>ſa racine; </s> <s xml:id="echoid-s2665" xml:space="preserve">ce qui eſt viſible auſſi par la Table précédente. </s> <s xml:id="echoid-s2666" xml:space="preserve">Ainſi <lb/>toutes les racines d’un chiffre ſont compriſes, depuis 1 juſqu’à <lb/>99 incluſivement.</s> <s xml:id="echoid-s2667" xml:space="preserve"/> </p> <pb o="81" file="0119" n="119" rhead="DE MATHÉMATIQUE. Liv. I."/> <p> <s xml:id="echoid-s2668" xml:space="preserve">153. </s> <s xml:id="echoid-s2669" xml:space="preserve">Le plus grand nombre poſſible de quatre chiffres ne <lb/>peut en avoir plus de deux à ſa racine. </s> <s xml:id="echoid-s2670" xml:space="preserve">Prenons le plus grand <lb/>nombre poſſible de quatre chiffres, qui eſt 9999, puiſque ſi on <lb/>lui ajoute l’unité, il devient 10000, qui en a 5; </s> <s xml:id="echoid-s2671" xml:space="preserve">& </s> <s xml:id="echoid-s2672" xml:space="preserve">ſuppoſons <lb/>que ce nombre puiſſe avoir à ſa racine le plus petit nombre <lb/>compoſé detrois chiffres, qui eſt 100, j’éleve 100 à ſon quarré, <lb/>& </s> <s xml:id="echoid-s2673" xml:space="preserve">il me vient 10000, qui eſt plus grand que le nombre 9999: <lb/></s> <s xml:id="echoid-s2674" xml:space="preserve">donc il ne peut pas avoir à ſa racine aucun nombre de trois <lb/>chiffres. </s> <s xml:id="echoid-s2675" xml:space="preserve">D’où il ſuit que toutes les racines de deux chiffres <lb/>ſont renfermées, depuis 100 juſqu’à 9999 incluſivement.</s> <s xml:id="echoid-s2676" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2677" xml:space="preserve">154. </s> <s xml:id="echoid-s2678" xml:space="preserve">Le plus grand nombre de ſix chiffres ne peut en avoir <lb/>plus de trois à ſa racine. </s> <s xml:id="echoid-s2679" xml:space="preserve">Prenons le plus grand nombre de ſix <lb/>chiffres, qui eſt 999999, & </s> <s xml:id="echoid-s2680" xml:space="preserve">ſuppoſons qu’il puiſſe avoir pour <lb/>racine le plus petit nombre de quatre chiffres, qui eſt 1000, <lb/>j’éléve 1000 à ſon quarré, & </s> <s xml:id="echoid-s2681" xml:space="preserve">j’ai 1000000, qui a ſept chiffres, <lb/>& </s> <s xml:id="echoid-s2682" xml:space="preserve">eſt plus grand que le nombre 999999, & </s> <s xml:id="echoid-s2683" xml:space="preserve">par conſéquent <lb/>ce nombre ne peut donner que trois chiffres à la racine. </s> <s xml:id="echoid-s2684" xml:space="preserve">D’où <lb/>il ſuit que les racines de trois chiffres ſont renfermées, depuis <lb/>10000 juſqu’à 999999 incluſivement.</s> <s xml:id="echoid-s2685" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2686" xml:space="preserve">155. </s> <s xml:id="echoid-s2687" xml:space="preserve">En continuant le même raiſonnement, on verra que <lb/>toutes les racines de quatre chiffres ſont compriſes, depuis <lb/>1000000 juſqu’à 99999999; </s> <s xml:id="echoid-s2688" xml:space="preserve">que les nombres ou racines de 5 <lb/>chiffres ſont contenues depuis 100000000 juſqu’à 9999999999 <lb/>incluſivement, &</s> <s xml:id="echoid-s2689" xml:space="preserve">c.</s> <s xml:id="echoid-s2690" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div153" type="section" level="1" n="127"> <head xml:id="echoid-head143" xml:space="preserve"><emph style="sc">Remarque</emph> <emph style="sc">Génerale</emph>.</head> <p> <s xml:id="echoid-s2691" xml:space="preserve">156. </s> <s xml:id="echoid-s2692" xml:space="preserve">Il ſuit de tout ce que nous venons de dire, qu’en gé-<lb/>néral un nombre aura toujours à ſa racine autant de chiffres <lb/>qu’on aura de tranches de deux chiffres, en le partageant de <lb/>deux en deux de droite à gauche, excepté la derniere tranche, <lb/>qui peut n’en contenir qu’un.</s> <s xml:id="echoid-s2693" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2694" xml:space="preserve">Ainſi 99 ne peut avoir qu’un chiffre, parce qu’il n’a qu’une <lb/>tranche de deux chiffres. </s> <s xml:id="echoid-s2695" xml:space="preserve">100 & </s> <s xml:id="echoid-s2696" xml:space="preserve">9999 auront deux chiffres à <lb/>leurs racines, parce qu’en les diviſant en tranches de cette <lb/>maniere: </s> <s xml:id="echoid-s2697" xml:space="preserve">1,00; </s> <s xml:id="echoid-s2698" xml:space="preserve">99,99, chacun en contient deux. </s> <s xml:id="echoid-s2699" xml:space="preserve">De même <lb/>10000 & </s> <s xml:id="echoid-s2700" xml:space="preserve">999999 auront auſſi trois chiffres à leurs racines, <lb/>ainſi que tous les nombres qui leur ſont intermédiaires, parce <lb/>qu’en partageant chacun en tranches, on a 1,00,00 & </s> <s xml:id="echoid-s2701" xml:space="preserve">99,99,99; <lb/></s> <s xml:id="echoid-s2702" xml:space="preserve">ils contiennent chacun trois tranches. </s> <s xml:id="echoid-s2703" xml:space="preserve">De plus, le quarré du <lb/>premier chiffre ſe trouvera dans la premiere tranche, le quarré <pb o="82" file="0120" n="120" rhead="NOUVEAU COURS"/> du ſecond ſe trouvera dans la ſeconde tranche, le quarré du <lb/>troiſieme ſe trouvera dans la troiſieme tranche; </s> <s xml:id="echoid-s2704" xml:space="preserve">ce qui revient <lb/>encore à ce que nous avons vu précédemment dans la forma-<lb/>tion algébrique du quarré du nombre (art. </s> <s xml:id="echoid-s2705" xml:space="preserve">150). </s> <s xml:id="echoid-s2706" xml:space="preserve">Cela poſé, <lb/>nous allons donner la regle générale, & </s> <s xml:id="echoid-s2707" xml:space="preserve">nous en ferons l’ap-<lb/>plication à quelques exemples.</s> <s xml:id="echoid-s2708" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div154" type="section" level="1" n="128"> <head xml:id="echoid-head144" style="it" xml:space="preserve">Regle générale pour l’extraction des Racines quarrées.</head> <p> <s xml:id="echoid-s2709" xml:space="preserve">157. </s> <s xml:id="echoid-s2710" xml:space="preserve">Un nombre étant donné pour en extraire la racine, <lb/>on partagera ce nombre en tranches de deux chiffres chacunes, <lb/>excepté la premiere à gauche, qui peut n’en contenir qu’un <lb/>ſeul; </s> <s xml:id="echoid-s2711" xml:space="preserve">on cherchera quel eſt le plus grand quarré contenu dans <lb/>la premiere tranche, on en prendra la racine, que l’on poſera <lb/>à la racine, à côté d’un nombre propoſé, après l’avoir ſéparé <lb/>par une petite barre verticale; </s> <s xml:id="echoid-s2712" xml:space="preserve">on élevera cette racine à ſon <lb/>quarré pour l’ôter de la premiere tranche, & </s> <s xml:id="echoid-s2713" xml:space="preserve">l’on écrira le reſte <lb/>au deſſous, & </s> <s xml:id="echoid-s2714" xml:space="preserve">l’opération ſera faite ſur la premiere tranche. <lb/></s> <s xml:id="echoid-s2715" xml:space="preserve">A côté de ce reſte, on abaiſſera la ſeconde tranche, en met-<lb/>tant un point au deſſous du premier chiffre de cette ſeconde <lb/>tranche; </s> <s xml:id="echoid-s2716" xml:space="preserve">on doublera ce que l’on a trouvé à la racine, & </s> <s xml:id="echoid-s2717" xml:space="preserve">par <lb/>ce nombre on diviſera les chiffres qui ſont terminés au premier <lb/>chiffre de la ſeconde tranche; </s> <s xml:id="echoid-s2718" xml:space="preserve">on mettra le quotient à la ſuite <lb/>du diviſeur, & </s> <s xml:id="echoid-s2719" xml:space="preserve">on multipliera le diviſeur ainſi augmenté par <lb/>ce même quotient. </s> <s xml:id="echoid-s2720" xml:space="preserve">Si le produit peut être ôté des chiffres du <lb/>reſte & </s> <s xml:id="echoid-s2721" xml:space="preserve">de la ſeconde tranche, le quotient ſera le ſecond chiffre <lb/>de la racine, & </s> <s xml:id="echoid-s2722" xml:space="preserve">on le poſera à côté du premier chiffre. </s> <s xml:id="echoid-s2723" xml:space="preserve">Si ce <lb/>produit étoit plus grand, on diminueroit le quotient d’une <lb/>unité, & </s> <s xml:id="echoid-s2724" xml:space="preserve">l’on feroit l’opération de même, juſqu’à ce qu’on <lb/>eût un nombre que l’on pût retrancher des chiffres ſur leſ-<lb/>quels on opére; </s> <s xml:id="echoid-s2725" xml:space="preserve">& </s> <s xml:id="echoid-s2726" xml:space="preserve">l’ayant trouvé, on cherchera le reſte qui <lb/>doit venir par la ſouſtraction de ce produit.</s> <s xml:id="echoid-s2727" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2728" xml:space="preserve">On abaiſſera la troiſieme tranche à côté de ce reſte, en met-<lb/>tant un point ſous le premier chiffre de la troiſieme tranche, <lb/>& </s> <s xml:id="echoid-s2729" xml:space="preserve">l’on diviſeroit les chiffres terminés au premier de la troi-<lb/>ſieme, par le double de ce qu’on auroit trouvé à la racine: </s> <s xml:id="echoid-s2730" xml:space="preserve">on <lb/>poſera de même ce quotient à côté du diviſeur; </s> <s xml:id="echoid-s2731" xml:space="preserve">& </s> <s xml:id="echoid-s2732" xml:space="preserve">ſi le pro-<lb/>duit du divifeur ainſi augmenté, multiplié par le quotient, <lb/>donne un produit moindre que le ſecond reſte, joint à la troi-<lb/>fieme tranche, ce quotient ſera le troiſieme chiffre de la racine;</s> <s xml:id="echoid-s2733" xml:space="preserve"> <pb o="83" file="0121" n="121" rhead="DE MATHÉMATIQUE. Liv. I."/> ſinon il faudra diminuer ce quotient de l’unité, juſqu’à ce que <lb/>ce quotient, poſé à côté du diviſeur, multipliant le tout, donne <lb/>un produit moindre, ou au moins égal aux chiffres ſur leſquels <lb/>on opére.</s> <s xml:id="echoid-s2734" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2735" xml:space="preserve">S’il n’y a que trois tranches, & </s> <s xml:id="echoid-s2736" xml:space="preserve">qu’après avoir retranché ce <lb/>produit il ne reſte rien, on aura la racine demandée; </s> <s xml:id="echoid-s2737" xml:space="preserve">s’il y en <lb/>davantage, on abaiſſera la tranche ſuivante à côté du reſte, <lb/>& </s> <s xml:id="echoid-s2738" xml:space="preserve">l’on fera toujours la même opération, en prenant toujours <lb/>pour diviſeur le double des chiffres que l’on a trouvés à la ra-<lb/>cine, & </s> <s xml:id="echoid-s2739" xml:space="preserve">prenant pour les termes de la racine ceux qui auront <lb/>les conditions expliquées ci-devant; </s> <s xml:id="echoid-s2740" xml:space="preserve">ſçavoir, que le produit <lb/>de ce nombre par lui-même, plus le produit du même nom-<lb/>bre par le diviſeur ſoit plus petit, ou au moins égal aux chiffres <lb/>ſupérieurs.</s> <s xml:id="echoid-s2741" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div155" type="section" level="1" n="129"> <head xml:id="echoid-head145" xml:space="preserve"><emph style="sc">Exemple</emph> I.</head> <p> <s xml:id="echoid-s2742" xml:space="preserve">158. </s> <s xml:id="echoid-s2743" xml:space="preserve">Soit propoſé d’extraire la racine quarrée du nombre <lb/>1936, je partage ce nombre en tranches de deux chiffres cha-<lb/>cune, en l’écrivant ainſi, 19,36; </s> <s xml:id="echoid-s2744" xml:space="preserve">& </s> <s xml:id="echoid-s2745" xml:space="preserve">je dis, en 19 quel eſt le <lb/>plus grand quarré qui y ſoit contenu, ce quarré eſt 16, dont <lb/>la racine eſt 4, je poſe 4 à la racine, après avoir ſéparé le <lb/>nombre 1936 de ſa racine par une petite barre verticale. </s> <s xml:id="echoid-s2746" xml:space="preserve">Je <lb/>dis enſuite, quatre fois 4 font 16 de 19, reſte 3: </s> <s xml:id="echoid-s2747" xml:space="preserve">j’abaiſſe la <lb/>ſeconde tranche 36 à côté du reſte 3, en mettant un point ſous <lb/>le premier chiffre 3 de cette tranche: </s> <s xml:id="echoid-s2748" xml:space="preserve">je double le chiffre 4 <lb/>que j’ai trouvé à la racine pour avoir le diviſeur 8, par lequel <lb/>je diviſe les deux premiers chiffres 33 du reſte, & </s> <s xml:id="echoid-s2749" xml:space="preserve">de la ſeconde <lb/>tranche; </s> <s xml:id="echoid-s2750" xml:space="preserve">& </s> <s xml:id="echoid-s2751" xml:space="preserve">je dis en 33 combien de fois 8, quatre fois: </s> <s xml:id="echoid-s2752" xml:space="preserve">je <lb/>poſe 4 à côté du diviſeur 8; </s> <s xml:id="echoid-s2753" xml:space="preserve">ce qui me donne 84, que je mul-<lb/>tiplie par le même quotient 4, en diſant, quatre fois 4 font <lb/>16, poſe 6 & </s> <s xml:id="echoid-s2754" xml:space="preserve">retiens 1: </s> <s xml:id="echoid-s2755" xml:space="preserve">quatre fois 8 font 32, & </s> <s xml:id="echoid-s2756" xml:space="preserve">1 que j’ai <lb/>retenu font 33: </s> <s xml:id="echoid-s2757" xml:space="preserve">le produit eſt 336, qui ôté de 336, reſte o; <lb/></s> <s xml:id="echoid-s2758" xml:space="preserve">d’où je conclus que 44 eſt la racine du quarré propoſé. </s> <s xml:id="echoid-s2759" xml:space="preserve">Pour <lb/>voir ſi je ne me ſuis pas trompé, j’éleve 44 à ſon quarré, il <lb/>me vient 1936, qui eſt le nombre propoſé.</s> <s xml:id="echoid-s2760" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div156" type="section" level="1" n="130"> <head xml:id="echoid-head146" xml:space="preserve"><emph style="sc">Article</emph> 158.</head> <note position="right" xml:space="preserve"># 19,36 # { # 44, racine. # 44 # Preuve. <lb/># 16 # # 8, divifeur. # 44 <lb/># 336 # # 84, diviſ. augmenté. # 176 <lb/># 336 # # 4, quotient. # 176 <lb/>Différence # 000 # 336, produit. # 1936 <lb/></note> <pb o="84" file="0122" n="122" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div157" type="section" level="1" n="131"> <head xml:id="echoid-head147" xml:space="preserve"><emph style="sc">Exemple</emph> II.</head> <p> <s xml:id="echoid-s2761" xml:space="preserve">159. </s> <s xml:id="echoid-s2762" xml:space="preserve">Soit propoſé d’extraire la racine du nombre 10543009, <lb/>après l’avoir partagé en tranches de deux chiffres chacune, & </s> <s xml:id="echoid-s2763" xml:space="preserve"><lb/>placé comme on voit ci-après à la gauche d’une barre verti-<lb/>cale, à côté de laquelle je dois mettre la racine: </s> <s xml:id="echoid-s2764" xml:space="preserve">je dis, en 10 <lb/>quel eſt le plus grand quarré qui y ſoit contenu, ce quarré eſt <lb/>9, dont la racine eſt 3, que je poſe à la racine: </s> <s xml:id="echoid-s2765" xml:space="preserve">j’éleve 3 à ſon <lb/>quarré, il me vient 9, que je retranche de 10, reſte 1. </s> <s xml:id="echoid-s2766" xml:space="preserve">J’abaiſſe <lb/>la ſeconde tranche 54 à côté du reſte 1, en obſervant de mettre <lb/>un point ſous le premier chiffre 5; </s> <s xml:id="echoid-s2767" xml:space="preserve">& </s> <s xml:id="echoid-s2768" xml:space="preserve">doublant ce que j’ai <lb/>trouvé à la racine, il me vient 6 pour diviſeur: </s> <s xml:id="echoid-s2769" xml:space="preserve">je dis en 15 <lb/>combien de fois 6, deux fois; </s> <s xml:id="echoid-s2770" xml:space="preserve">j’écris 6 au deſſous du diviſeur, <lb/>& </s> <s xml:id="echoid-s2771" xml:space="preserve">je mets à côté le quotient 2, & </s> <s xml:id="echoid-s2772" xml:space="preserve">je multiplie 62 par 2, le <lb/>produit eſt 124, lequel retranché de 154, donne 30 pour ſe-<lb/>cond reſte: </s> <s xml:id="echoid-s2773" xml:space="preserve">j’abaiſſe la ſeconde tranche, qui eſt 30, à côté du <lb/>reſte 30, en mettant un point ſous le premier chiffre 3 de cette <lb/>ſeconde tranche; </s> <s xml:id="echoid-s2774" xml:space="preserve">je double ce que j’ai à la racine pour avoir <lb/>le ſecond diviſeur 64, par lequel je diviſe les chiffres 303, & </s> <s xml:id="echoid-s2775" xml:space="preserve"><lb/>je dis en 30 combien de ſois 6, cinq ſois, je poſe le 5 à la ſuite <lb/>de 64, en écrivant 645. </s> <s xml:id="echoid-s2776" xml:space="preserve">Je multiplie ce nombre par 5, & </s> <s xml:id="echoid-s2777" xml:space="preserve"><lb/>comme le produit 3225 ne peut pas être ôté du 3030, j’eſſaie <lb/>le 4; </s> <s xml:id="echoid-s2778" xml:space="preserve">j’écris donc 644, & </s> <s xml:id="echoid-s2779" xml:space="preserve">multipliant ce nombre par 4, le pro-<lb/>duit eſt 2576, qui pouvant être ôté de 3030, m’indique que <lb/>ce 4 eſt bon, & </s> <s xml:id="echoid-s2780" xml:space="preserve">je le poſe à la racine. </s> <s xml:id="echoid-s2781" xml:space="preserve">J’ôte le nombre 2576 <lb/>de 3030, le reſte eſt 454, à côté duquel j’abaiſſe la quatrieme <lb/>tranche, en mettant un point ſous le premier chiffre 0 de cette <lb/>tranche 09 pour diviſer les chiffres 4540 par le double de ce <lb/>qui eſt à la racine, qui eſt. </s> <s xml:id="echoid-s2782" xml:space="preserve">648. </s> <s xml:id="echoid-s2783" xml:space="preserve">Je dis donc en 45 combien de <lb/>fois 6, ſept fois, je poſe le 7 à côté du diviſeur 648, en écri-<lb/>vant 6487, & </s> <s xml:id="echoid-s2784" xml:space="preserve">je multiplie ce nombre par 7, le produit eſt <lb/>45409, lequel étant préciſément égal au nombre 45409, in-<lb/>dique que le 7 eſt bon: </s> <s xml:id="echoid-s2785" xml:space="preserve">je le poſe à la racine qui ſe trouve de <lb/>3247, comme on le ſçait d’ailleurs par l’article 150.</s> <s xml:id="echoid-s2786" xml:space="preserve"/> </p> <pb o="85" file="0123" n="123" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div158" type="section" level="1" n="132"> <head xml:id="echoid-head148" xml:space="preserve"><emph style="sc">Article</emph> 159.</head> <p> <s xml:id="echoid-s2787" xml:space="preserve">10,54,30,09 <lb/>1 54 <lb/>1 24 <lb/>3030 <lb/>2576 <lb/>45409 <lb/>45409 <lb/>00000</s> </p> <p> <s xml:id="echoid-s2788" xml:space="preserve">{ 3247 <lb/>6, I<emph style="sub">er</emph> diviſeur. <lb/></s> <s xml:id="echoid-s2789" xml:space="preserve">62 <lb/>2 <lb/>124, produit. </s> <s xml:id="echoid-s2790" xml:space="preserve"><lb/>64, 2<emph style="sub">e</emph> diviſeur. </s> <s xml:id="echoid-s2791" xml:space="preserve"><lb/>645 <lb/>5 <lb/>prod. </s> <s xml:id="echoid-s2792" xml:space="preserve">d’épreuve <lb/>644 <lb/>4 <lb/>2576, produit. </s> <s xml:id="echoid-s2793" xml:space="preserve"><lb/>648, 3<emph style="sub">e</emph> diviſeur. </s> <s xml:id="echoid-s2794" xml:space="preserve"><lb/>6487 <lb/>7 <lb/>45409, produit.</s> <s xml:id="echoid-s2795" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2796" xml:space="preserve">Preuve de l’opération. <lb/></s> <s xml:id="echoid-s2797" xml:space="preserve">3247 <lb/>3247 <lb/>22729 <lb/>12988 <lb/>6494 <lb/>9741 <lb/>10543009</s> </p> </div> <div xml:id="echoid-div159" type="section" level="1" n="133"> <head xml:id="echoid-head149" xml:space="preserve"><emph style="sc">Exemple</emph> III.</head> <p> <s xml:id="echoid-s2798" xml:space="preserve">160. </s> <s xml:id="echoid-s2799" xml:space="preserve">Soit propoſé d’extraire la racine du nombre 867972, <lb/>je diviſe ce nombre en tranches de deux chiffres chacune, en <lb/>l’écrivant ainſi: </s> <s xml:id="echoid-s2800" xml:space="preserve">86,79,72; </s> <s xml:id="echoid-s2801" xml:space="preserve">puis je dis, en 86 quel eſt le plus <lb/>grand quarré qui y ſoit contenu, ce quarré eſt 81, dont la ra-<lb/>cine eſt 9, que je poſe à la racine; </s> <s xml:id="echoid-s2802" xml:space="preserve">j’éléve 9 à ſon quarré, & </s> <s xml:id="echoid-s2803" xml:space="preserve"><lb/>j’ôte ce quarré 81 de 86, reſte 5: </s> <s xml:id="echoid-s2804" xml:space="preserve">j’abaiſſe à côté du 5 la ſe-<lb/>conde tranche 79, en mettant un point ſous le premier chiffre <lb/>7, ce qui me donne 57 à diviſer par 18, double de 9, qui eſt <lb/>à la racine. </s> <s xml:id="echoid-s2805" xml:space="preserve">Je fais la diviſion, & </s> <s xml:id="echoid-s2806" xml:space="preserve">je dis, en 5 combien de fois <lb/>1, il y eſt cinq fois; </s> <s xml:id="echoid-s2807" xml:space="preserve">mais comme je vois aiſément qu’il n’eſt <lb/>pas bon, parce qu’il me donneroit un produit trop fort, j’eſ-<lb/>ſaie tout d’un coup le 4, en le mettant à la ſuite du diviſeur <lb/>84: </s> <s xml:id="echoid-s2808" xml:space="preserve">je multiplie 184 par 4, le produit eſt 736, qui étant plus <lb/>grand que 579, me marque que le 4 n’eſt pas encore bon, ainſi <lb/>je ne le mets point à la racine. </s> <s xml:id="echoid-s2809" xml:space="preserve">J’éprouve le 3, en mettant 183, <lb/>& </s> <s xml:id="echoid-s2810" xml:space="preserve">multipliant ce nombre par 3, le produit eſt 549; </s> <s xml:id="echoid-s2811" xml:space="preserve">comme <lb/>ce produit eſt moindre que 579, je mets le 3 à la racine. </s> <s xml:id="echoid-s2812" xml:space="preserve">J’ôte <lb/>enſuite 549 de 579, le reſte eſt 30; </s> <s xml:id="echoid-s2813" xml:space="preserve">j’abaiſſe à côté de ce reſte <lb/>la troiſieme tranche 72, en mettant le point ſous le premier <pb o="86" file="0124" n="124" rhead="NOUVEAU COURS"/> chiffre 7, afin de diviſer le nombre 3072 par 186, double de <lb/>ce qui eſt à la racine: </s> <s xml:id="echoid-s2814" xml:space="preserve">je dis donc en 3 combien de fois 1, il <lb/>y eſt trois fois: </s> <s xml:id="echoid-s2815" xml:space="preserve">mais comme je vois que le 3 eſt trop fort, <lb/>j’eſſaie le 2 en le mettant à côté du diviſeur 186; </s> <s xml:id="echoid-s2816" xml:space="preserve">ce qui me <lb/>donne 1862, que je multiplie par 2; </s> <s xml:id="echoid-s2817" xml:space="preserve">le produit eſt 3724: <lb/></s> <s xml:id="echoid-s2818" xml:space="preserve">comme ce produit eſt plus grand que 3072, je conclus que le <lb/>2 n’eſt pas encore bon; </s> <s xml:id="echoid-s2819" xml:space="preserve">je mets 1 à la racine, qui ſera cer-<lb/>tainement bon, puiſqu’en mettant 1 à la ſuite du diviſeur, & </s> <s xml:id="echoid-s2820" xml:space="preserve"><lb/>multipliant par 1, le produit eſt 1861, moindre que 3072: </s> <s xml:id="echoid-s2821" xml:space="preserve"><lb/>j’ôte ce nombre 1861 de 3072, le reſte eſt 1211.</s> <s xml:id="echoid-s2822" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2823" xml:space="preserve">Sil’on veut faire la preuve de cette opération, il faudra élever <lb/>la racine 931 que l’on a trouvée à ſon quarré, lui ajouter le reſte <lb/>1211, & </s> <s xml:id="echoid-s2824" xml:space="preserve">l’on doit trouver un nombre égal au nombre propoſé.</s> <s xml:id="echoid-s2825" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div160" type="section" level="1" n="134"> <head xml:id="echoid-head150" xml:space="preserve"><emph style="sc">Article</emph> 160.</head> <p> <s xml:id="echoid-s2826" xml:space="preserve">86,79,72 <lb/>579 <lb/>549 <lb/>30772 <lb/>1861 <lb/>Reſte 1211</s> </p> <p> <s xml:id="echoid-s2827" xml:space="preserve">{ 931 <lb/>18, I<emph style="sub">er</emph> diviſeur. <lb/></s> <s xml:id="echoid-s2828" xml:space="preserve">184 <lb/>4 <lb/>produit d’épreuve. </s> <s xml:id="echoid-s2829" xml:space="preserve"><lb/>183 <lb/>3 <lb/>549, bon produit. </s> <s xml:id="echoid-s2830" xml:space="preserve"><lb/>186, ſecond diviſeur. </s> <s xml:id="echoid-s2831" xml:space="preserve"><lb/>1862 <lb/>2 <lb/>produit d’épreuve. </s> <s xml:id="echoid-s2832" xml:space="preserve"><lb/>1861 <lb/>1 <lb/>1861</s> </p> <p> <s xml:id="echoid-s2833" xml:space="preserve">Preuve de l’opération. <lb/></s> <s xml:id="echoid-s2834" xml:space="preserve">931 <lb/>931 <lb/>931 <lb/>2793 <lb/>8379 <lb/>866761 <lb/>1211 <lb/>867972</s> </p> <p style="it"> <s xml:id="echoid-s2835" xml:space="preserve">Maniere d’approcher le plus près qu’il eſt poſſible de la racine <lb/>quarrée d’un nombre, dont on ne peut avoir la racine ſans reſte, <lb/>par le moyen des décimales.</s> <s xml:id="echoid-s2836" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2837" xml:space="preserve">161. </s> <s xml:id="echoid-s2838" xml:space="preserve">Comme le principal uſage de la racine quarrée dans la <lb/>Géométrie, & </s> <s xml:id="echoid-s2839" xml:space="preserve">ſurtout dans la Géométrie pratique, eſt de <lb/>trouver en nombre le côté d’un quarré égal à une quantité de <lb/>toiſes, ou de pieds quarrés, il eſt néceſſaire, pour agir avec <lb/>plus de préciſion, d’approcher le plus près qu’il eſt poſſible de <pb o="87" file="0125" n="125" rhead="DE MATHÉMATIQUE. Liv. I."/> la racine qu’on cherche, en faiſant enſorte que les reſtes que <lb/>l’on néglige ſoient de ſi petite valeur, qu’on puiſſe les regarder <lb/>comme de nulle conſéquence. </s> <s xml:id="echoid-s2840" xml:space="preserve">Pour cela, voici ce qu’il faut <lb/>faire.</s> <s xml:id="echoid-s2841" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div161" type="section" level="1" n="135"> <head xml:id="echoid-head151" style="it" xml:space="preserve">Regle générale d’approximation.</head> <p> <s xml:id="echoid-s2842" xml:space="preserve">162. </s> <s xml:id="echoid-s2843" xml:space="preserve">On ajoutera au nombre propoſé, pour en extraire la <lb/>racine, autant de tranches de deux zero chacune, que l’on vou-<lb/>dra avoir de décimales à la racine; </s> <s xml:id="echoid-s2844" xml:space="preserve">& </s> <s xml:id="echoid-s2845" xml:space="preserve">aprés avoir ſéparé les <lb/>entiers de la racine d’avec les décimales qui doivent ſuivre, <lb/>on continuera le procédé de l’extraction des racines, préciſé-<lb/>ment de la même maniere qu’il ſe pratique ſur les nombres <lb/>entiers, comme on le verra dans l’exemple ſuivant.</s> <s xml:id="echoid-s2846" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2847" xml:space="preserve">8,69,00,00,00, <lb/>469 <lb/>441 <lb/>2800 <lb/>2336 <lb/>46400 <lb/>41209 <lb/>519100 <lb/>471584 <lb/>Reſte 47516</s> </p> <p> <s xml:id="echoid-s2848" xml:space="preserve">{ 29,478 <lb/>4, premier diviſeur. <lb/></s> <s xml:id="echoid-s2849" xml:space="preserve">49 <lb/>9 <lb/>441 <lb/>58, ſecond diviſeur. </s> <s xml:id="echoid-s2850" xml:space="preserve"><lb/>585 <lb/>5 <lb/>produit d’épreuve. </s> <s xml:id="echoid-s2851" xml:space="preserve"><lb/>584 <lb/>4 <lb/>2336, bon produit. </s> <s xml:id="echoid-s2852" xml:space="preserve"><lb/>588, troiſieme diviſeur. </s> <s xml:id="echoid-s2853" xml:space="preserve"><lb/>5888 <lb/>8 <lb/>produit d’épreuve. </s> <s xml:id="echoid-s2854" xml:space="preserve"><lb/>5887 <lb/>7 <lb/>41209, bon produit. </s> <s xml:id="echoid-s2855" xml:space="preserve"><lb/>5894, 4<emph style="sub">me</emph> diviſeur. </s> <s xml:id="echoid-s2856" xml:space="preserve"><lb/>58948 <lb/>8 <lb/>471584</s> </p> <pb o="88" file="0126" n="126" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s2857" xml:space="preserve">163. </s> <s xml:id="echoid-s2858" xml:space="preserve">Soit propoſé d’extraire la racine de 869, juſqu’à ce que <lb/>la racine ne différe pas d’un millieme de la vraie valeur. </s> <s xml:id="echoid-s2859" xml:space="preserve">On <lb/>ajoutera au nombre propoſé ſix zero, parce que l’on veut avoir <lb/>des milliemes, en écrivant au lieu de 869, 8,69.</s> <s xml:id="echoid-s2860" xml:space="preserve">00,00,00, & </s> <s xml:id="echoid-s2861" xml:space="preserve"><lb/>après les avoir partagé en tranches de deux en deux, on dira <lb/>en 8 quel eſt le plus grand quarré qui y ſoit contenu; </s> <s xml:id="echoid-s2862" xml:space="preserve">ce quarré <lb/>eſt 4, dont la racine eſt 2, que je poſe à la racine. </s> <s xml:id="echoid-s2863" xml:space="preserve">J’ôte ce <lb/>quarré 4 du premier chiffre 8, il me reſte 4, à côté duquel <lb/>j’abaiſſe la ſeconde tranche 69, en mettant un point ſous le 6, <lb/>afin de faire voir que c’eſt 46 que je diviſe par le diviſeur 4, <lb/>double de la racine. </s> <s xml:id="echoid-s2864" xml:space="preserve">Je dis donc, en 46 combien de fois 4, <lb/>neuf fois, je poſe 9 à côté du diviſeur 4, & </s> <s xml:id="echoid-s2865" xml:space="preserve">au deſſous, & </s> <s xml:id="echoid-s2866" xml:space="preserve">je <lb/>multiplie 49 par 9, le produit eſt 441, moindre que 469; </s> <s xml:id="echoid-s2867" xml:space="preserve">ce <lb/>qui me montre que le 9 eſt un des chiffres de la racine. </s> <s xml:id="echoid-s2868" xml:space="preserve">J’ôte <lb/>441 de 469, le reſte eſt 28. </s> <s xml:id="echoid-s2869" xml:space="preserve">J’abaiſſe à côté de ce reſte la pre-<lb/>miere tranche de décimales, en mettant un point ſous le pre-<lb/>mier zero, pour marquer que c’eſt 280 que je veux diviſer par <lb/>le nombre 58, double de ce qui eſt à la racine. </s> <s xml:id="echoid-s2870" xml:space="preserve">Je fais la Divi-<lb/>ſion, & </s> <s xml:id="echoid-s2871" xml:space="preserve">je dis, en 28 combien de fois 5, il y eſt cinq fois: </s> <s xml:id="echoid-s2872" xml:space="preserve">je <lb/>poſe le 5 à côté du diviſeur 58, & </s> <s xml:id="echoid-s2873" xml:space="preserve">au deſſous, & </s> <s xml:id="echoid-s2874" xml:space="preserve">je multiplie <lb/>585 par 5, le produit eſt 2925, qui étant plus grand que 28,00, <lb/>me marque que l’on ne peut pas mettre 5 à la racine: </s> <s xml:id="echoid-s2875" xml:space="preserve">je prends <lb/>le 4, & </s> <s xml:id="echoid-s2876" xml:space="preserve">j’écris 584, queje multiplie par 4, le produit eſt 2336, <lb/>lequel étant moindre que 2800, me marque que le 4 eſt bon, <lb/>& </s> <s xml:id="echoid-s2877" xml:space="preserve">je le poſe à la racine, après avoir ſéparé les entiers 29 des <lb/>décimales par un point. </s> <s xml:id="echoid-s2878" xml:space="preserve">J’ôte ce produit 2336 de 2800, le reſte <lb/>eſt 464, à côté duquel j’abaiſſe la ſeconde tranche de déci-<lb/>males, en mettant un point ſous le premier zero. </s> <s xml:id="echoid-s2879" xml:space="preserve">Je diviſe <lb/>4640 par 588, double de ce qui eſt à la racine; </s> <s xml:id="echoid-s2880" xml:space="preserve">& </s> <s xml:id="echoid-s2881" xml:space="preserve">je dis, en 46 <lb/>combien de fois 5, il y eſt 8, je poſe le 8 à côté du diviſeur <lb/>588; </s> <s xml:id="echoid-s2882" xml:space="preserve">& </s> <s xml:id="echoid-s2883" xml:space="preserve">au deſſous de ce même diviſeur, je multiplie 5888 par <lb/>8, le produit eſt 47104, qui étant plus grand que 46400, fait <lb/>voir que je ne puis pas mettre 8 à la racine; </s> <s xml:id="echoid-s2884" xml:space="preserve">je prends le 7, que <lb/>je mets à côté du diviſeur 588, & </s> <s xml:id="echoid-s2885" xml:space="preserve">au deſſous, puis multipliant <lb/>5887 par 7, le produit eſt 41209; </s> <s xml:id="echoid-s2886" xml:space="preserve">& </s> <s xml:id="echoid-s2887" xml:space="preserve">comme ce produit eſt <lb/>moindre que 46400, je poſe le 7 à la racine. </s> <s xml:id="echoid-s2888" xml:space="preserve">Je retranche le <lb/>produit 41209 de 46400, le reſte eſt 5191, à côté duquel j’a-<lb/>baiſſe la troiſieme tranche, en mettant un point ſous le pre-<lb/>mier zero, pour diviſer le nombre 51910 par 5894, double de <lb/>ce que j’ai trouvé à la racine. </s> <s xml:id="echoid-s2889" xml:space="preserve">Je fais la Diviſion, en diſant, <pb o="89" file="0127" n="127" rhead="DE MATHÉMATIQUE. Liv. I."/> en 51 combien de fois 5, il y eſt neuf fois: </s> <s xml:id="echoid-s2890" xml:space="preserve">mais comme je vois <lb/>que le 9 eſt trop fort, je paſſe au 8; </s> <s xml:id="echoid-s2891" xml:space="preserve">je poſe 8 à côté du divi-<lb/>ſeur & </s> <s xml:id="echoid-s2892" xml:space="preserve">au deſſous, & </s> <s xml:id="echoid-s2893" xml:space="preserve">je multiplie 58948 par 8: </s> <s xml:id="echoid-s2894" xml:space="preserve">le produit eſt <lb/>471584; </s> <s xml:id="echoid-s2895" xml:space="preserve">& </s> <s xml:id="echoid-s2896" xml:space="preserve">comme il eſt moindre que le nombre 519184, je <lb/>poſe le 8 à la racine, qui ſe trouve de 29.</s> <s xml:id="echoid-s2897" xml:space="preserve">478, ou, ce qui re-<lb/>vient au même, 29 entiers, plus {478/1000}.</s> <s xml:id="echoid-s2898" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2899" xml:space="preserve">164. </s> <s xml:id="echoid-s2900" xml:space="preserve">Si l’on ſuppoſe que le nombre 869 ſoit un nombre <lb/>de toiſes quarrées, ce que l’on trouve à la racine au rang <lb/>des entiers, marque des toiſes linéaires, & </s> <s xml:id="echoid-s2901" xml:space="preserve">le nombre que l’on <lb/>trouve au rang des décimales, marque des parties de toiſes <lb/>linéaires, comme des pieds, des pouces, & </s> <s xml:id="echoid-s2902" xml:space="preserve">des lignes. </s> <s xml:id="echoid-s2903" xml:space="preserve">Pour <lb/>ſçavoir ce que vaut de pieds la partie décimale {478/1000} ou 0.</s> <s xml:id="echoid-s2904" xml:space="preserve">478, <lb/>on multipliera, ſuivant la regle (art. </s> <s xml:id="echoid-s2905" xml:space="preserve">131.) </s> <s xml:id="echoid-s2906" xml:space="preserve">le nombre 478 par <lb/>6, qui marque combien la toiſe contient de pieds; </s> <s xml:id="echoid-s2907" xml:space="preserve">& </s> <s xml:id="echoid-s2908" xml:space="preserve">le reſte <lb/>par 12, qui marque combien le pied vaut de pouces, & </s> <s xml:id="echoid-s2909" xml:space="preserve">le reſte <lb/>encore par 12, qui marque combien le pouce vaut de lignes. <lb/></s> <s xml:id="echoid-s2910" xml:space="preserve">En ſuivant ce procédé, tous les nombres qui ſe trouveront <lb/>hors les décimales, marqueront les parties de la toiſe que l’on <lb/>demande, qui ſont 2 pieds 10 pouces 4 lignes 11 points, ou <lb/>ſi l’on veut, à cauſe du reſte que l’on a encore négligé dans <lb/>les décimales, 2 pieds 10 pouces 5 lignes. </s> <s xml:id="echoid-s2911" xml:space="preserve">La racine du nom-<lb/>bre 869 toiſes eſt donc 29 toiſes 2 pieds 10 pouces 5 lignes.</s> <s xml:id="echoid-s2912" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2913" xml:space="preserve">165. </s> <s xml:id="echoid-s2914" xml:space="preserve">Si l’on a un nombre compoſé de toiſes, pieds & </s> <s xml:id="echoid-s2915" xml:space="preserve">pouces <lb/>propoſé pour en extraire la racine, comme ſi l’on demandoit <lb/>celle du nombre 24 toiſes 3 pieds 9 pouces, il faudroit réduire <lb/>les fractions {1/2} & </s> <s xml:id="echoid-s2916" xml:space="preserve">{9/72} de toiſes, qui ſont la même choſe que <lb/>3 pieds 9 pouces, en fractions décimales de la toiſe, ſuivantla <lb/>méthode de l’art. </s> <s xml:id="echoid-s2917" xml:space="preserve">128; </s> <s xml:id="echoid-s2918" xml:space="preserve">de la ſomme de ces deux fractions dé-<lb/>cimales, & </s> <s xml:id="echoid-s2919" xml:space="preserve">du nombre propoſé faire un ſeul nombre, que l’on <lb/>trouvera de 24.</s> <s xml:id="echoid-s2920" xml:space="preserve">625000, dont on extraira la racine, ſuivant les <lb/>méthodes données ci - devant; </s> <s xml:id="echoid-s2921" xml:space="preserve">cette racine ſe trouvera de <lb/>4,962, c’eſt-à-dire de 4 toiſes, plus {962/1000} de toiſes que l’on éva-<lb/>luera, ſuivant la méthode de l’art. </s> <s xml:id="echoid-s2922" xml:space="preserve">131, & </s> <s xml:id="echoid-s2923" xml:space="preserve">que l’on trouvera <lb/>de 5 pieds 9 pouces 3 lignes 2 points.</s> <s xml:id="echoid-s2924" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div162" type="section" level="1" n="136"> <head xml:id="echoid-head152" style="it" xml:space="preserve">Démonſtration de la Racine quarrée.</head> <p> <s xml:id="echoid-s2925" xml:space="preserve">166. </s> <s xml:id="echoid-s2926" xml:space="preserve">Pour démontrer les opérations précédentes, nous ex-<lb/>trairons encore la racine quarrée du nombre 676, & </s> <s xml:id="echoid-s2927" xml:space="preserve">nous fe-<lb/>rons voir la raiſon de chaque opération.</s> <s xml:id="echoid-s2928" xml:space="preserve"/> </p> <pb o="90" file="0128" n="128" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s2929" xml:space="preserve">On voit par les articles 152, 153, 154 & </s> <s xml:id="echoid-s2930" xml:space="preserve">155 pour quelle <lb/>raiſon on diviſe le nombre donné en tranches de deux chiffres <lb/>chacune, & </s> <s xml:id="echoid-s2931" xml:space="preserve">comment chaque tranche doit donner un chiffre <lb/>à la racine. </s> <s xml:id="echoid-s2932" xml:space="preserve">Cela poſé, pour extraire la racine de 676, après <lb/>avoir partagé le nombre en tranches de deux chiffres chacune, <lb/>excepté la premiere qui n’en contient qu’un, je dis, la racine <lb/>quarrée de 6 eſt 2, que je poſe à la racine, & </s> <s xml:id="echoid-s2933" xml:space="preserve">qui vaut 20, puiſ-<lb/>qu’il doit y avoir deux chiffres à la racine, dont il eſt le pre-<lb/>mier. </s> <s xml:id="echoid-s2934" xml:space="preserve">Lors donc que j’éleve 2 au quarré, & </s> <s xml:id="echoid-s2935" xml:space="preserve">que je retranche 4 <lb/>de 6, c’eſt comme ſi j’élevois 40 au quarré, & </s> <s xml:id="echoid-s2936" xml:space="preserve">que je retranchaſſe <lb/>400 de 600, puiſque le 6 vaut réellement 600. </s> <s xml:id="echoid-s2937" xml:space="preserve">Selon la regle, <lb/>j’abiſſe la ſeconde tranche à côté du reſte 2, & </s> <s xml:id="echoid-s2938" xml:space="preserve">j’ai 276: </s> <s xml:id="echoid-s2939" xml:space="preserve">je <lb/>mets un point ſous le 7, parce que nous avons fait voir que le <lb/>double du premier terme, multiplié par le ſecond, doit ſe <lb/>trouver compris dans les deux premiers chiffres 27 (n°. </s> <s xml:id="echoid-s2940" xml:space="preserve">150); <lb/></s> <s xml:id="echoid-s2941" xml:space="preserve">mais j’ai le double du premier, & </s> <s xml:id="echoid-s2942" xml:space="preserve">ce nombre 27 contient le <lb/>double du premier, multiplié par le ſecond: </s> <s xml:id="echoid-s2943" xml:space="preserve">donc en diviſant <lb/>27 par le double du premier, je dois trouver le ſecond: </s> <s xml:id="echoid-s2944" xml:space="preserve">je fais <lb/>la Diviſion, & </s> <s xml:id="echoid-s2945" xml:space="preserve">je dis, en 27 combien de fois 4, il y eſt ſix fois: </s> <s xml:id="echoid-s2946" xml:space="preserve"><lb/>je mets le 6 à côté du diviſeur & </s> <s xml:id="echoid-s2947" xml:space="preserve">au deſſous, ſelon la regle; </s> <s xml:id="echoid-s2948" xml:space="preserve">ce <lb/>qui me donne néceſſairement par la Multiplication le quarré <lb/>de 6, lequel doit être contenu dans les deux derniers chiffres: </s> <s xml:id="echoid-s2949" xml:space="preserve"><lb/>je dis donc ſix fois 6 font 36, je poſe 6 & </s> <s xml:id="echoid-s2950" xml:space="preserve">retiens 3; </s> <s xml:id="echoid-s2951" xml:space="preserve">ſix fois<unsure/> <lb/>4 font 24, & </s> <s xml:id="echoid-s2952" xml:space="preserve">3 de retenus, font 27, le produit eſt 276: </s> <s xml:id="echoid-s2953" xml:space="preserve">donc <lb/>le 6 eſt le ſecond chiffre de la racine: </s> <s xml:id="echoid-s2954" xml:space="preserve">donc 26 eſt la racine du <lb/>nombre propoſé, puiſque ce nombre contient le quarré du <lb/>premier 2 ou 20, qui eſt 400, le double du premier 40, mul-<lb/>tiplié par 6 ou 240, & </s> <s xml:id="echoid-s2955" xml:space="preserve">enfin le quarré 36 du ſecond.</s> <s xml:id="echoid-s2956" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2957" xml:space="preserve">Le raiſonnement que nous faiſons pour une racine de deux <lb/>chiffres ſe peut appliquer à tout autre; </s> <s xml:id="echoid-s2958" xml:space="preserve">car on pourra toujours <lb/>partager un nombre quelconque de chiffres en deux parties, <lb/>dont la premiere contiendra tous les chiffres, excepté le der-<lb/>nier à droite, & </s> <s xml:id="echoid-s2959" xml:space="preserve">la ſeconde contiendra le dernier chiffre. </s> <s xml:id="echoid-s2960" xml:space="preserve">De <lb/>cette maniere, on verra que lorſqu’on aura trouvé la racine <lb/>des premiers chiffres, le reſte qui viendra, joint avec la der-<lb/>niere tranche, doit contenir le double des premiers chiffres <lb/>trouvés, multiplié par le dernier avec le quarré du dernier. <lb/></s> <s xml:id="echoid-s2961" xml:space="preserve">D’ailleurs ce double produit ſera toujours placé de maniere, <lb/>que les chiffres ſignificatifs de ce même produit ſeront tou-<lb/>jours terminés au premier chiffre de la derniere tranche: </s> <s xml:id="echoid-s2962" xml:space="preserve">donc <pb o="91" file="0129" n="129" rhead="DE MATHÉMATIQUE. Liv. I."/> en faiſant la Diviſion, on doit trouver le dernier chiffre. </s> <s xml:id="echoid-s2963" xml:space="preserve">Ceci <lb/>peut encore ſe démontrer indépendamment de cette ſuppoſi-<lb/>tion, par la formation du quarré, expliquée au n°. </s> <s xml:id="echoid-s2964" xml:space="preserve">150, & </s> <s xml:id="echoid-s2965" xml:space="preserve"><lb/>même on ne peut mieux faire que d’y recourir encore, pour <lb/>voir de quelle maniere on a déduit de cette formation la regle <lb/>que nous venons de voir; </s> <s xml:id="echoid-s2966" xml:space="preserve">c’eſt en cela que conſiſte l’eſprit <lb/>géométrique, & </s> <s xml:id="echoid-s2967" xml:space="preserve">c’eſt par l’étude de la compoſition des quan-<lb/>tités que l’on acquiert le grand art de les décompoſer; </s> <s xml:id="echoid-s2968" xml:space="preserve">je dis <lb/>le grand art, car c’eſt le plus difficile de toute la Géométrie, <lb/>& </s> <s xml:id="echoid-s2969" xml:space="preserve">la décompoſition des quantités eſt ſon objet dans toutes <lb/>les méthodes de calcul que l’on propoſe.</s> <s xml:id="echoid-s2970" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div163" type="section" level="1" n="137"> <head xml:id="echoid-head153" style="it" xml:space="preserve">De la formation du Cube d’une quantité complexe, & de l’extrac-<lb/>tion de la racine cube des quantités algébriques & numériques.</head> <p> <s xml:id="echoid-s2971" xml:space="preserve">167. </s> <s xml:id="echoid-s2972" xml:space="preserve">Nous avons déja vu, n°. </s> <s xml:id="echoid-s2973" xml:space="preserve">61, que le cube d’une quan-<lb/>tité, compoſée de deux termes, contient le cube du premier <lb/>terme, le cube du ſecond, plus deux parallelepipedes, dont <lb/>le premier a pour baſe le triple du quarré du premier, & </s> <s xml:id="echoid-s2974" xml:space="preserve">le ſe-<lb/>cond pour hauteur, & </s> <s xml:id="echoid-s2975" xml:space="preserve">l’autre a pour baſe le triple du quarré <lb/>du ſecond, & </s> <s xml:id="echoid-s2976" xml:space="preserve">pour hauteur le premier; </s> <s xml:id="echoid-s2977" xml:space="preserve">ce que nous avons <lb/>démontré généralement, en élevant a + b à ſon cube, que nous <lb/>avons trouvé a<emph style="sub">3</emph> + 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>.</s> <s xml:id="echoid-s2978" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2979" xml:space="preserve">168. </s> <s xml:id="echoid-s2980" xml:space="preserve">Le cube d’une quantité, compoſé de trois termes, ou de <lb/>quatre termes, ſe trouvera de même, en multipliant cette <lb/>quantité deux fois de ſuite par elle-même; </s> <s xml:id="echoid-s2981" xml:space="preserve">mais on peut la <lb/>trouver plus aiſément, en rapportant la quantité à l’expreſſion <lb/>générale a + b, qui peut repréſenter une quantité complexe <lb/>quelconque, en faiſant, par exemple dans celle-ci, c + d + f <lb/>+ g, c + d = a, & </s> <s xml:id="echoid-s2982" xml:space="preserve">f + g =b. </s> <s xml:id="echoid-s2983" xml:space="preserve">Voici de quelle maniere on <lb/>s’y prendroit pour élever tout d’un coup c + d + f + g au cube. <lb/></s> <s xml:id="echoid-s2984" xml:space="preserve">On prendroit d’abord le cube de c + d, qui eſt c<emph style="sub">3</emph> + 3c<emph style="sub">2</emph>d + <lb/>3cd<emph style="sub">2</emph> + d<emph style="sub">3</emph>, & </s> <s xml:id="echoid-s2985" xml:space="preserve">de même le cube de f + g, qui eſt f<emph style="sub">3</emph> + 3f<emph style="sub">2</emph>g <lb/>+ 3fg<emph style="sub">2</emph> + g<emph style="sub">3</emph>; </s> <s xml:id="echoid-s2986" xml:space="preserve">on prendroit enſuite le triple du quarré de c + d <lb/>que l’on trouvera de 3c<emph style="sub">2</emph> + 6cd + 3d<emph style="sub">2</emph>, que l’on multipliera <lb/>par f + g, ce qui donnera 3c<emph style="sub">2</emph>f + 6cdf + 3d<emph style="sub">2</emph>f + 3c<emph style="sub">2</emph>g + 6cdg <lb/>+ 3d<emph style="sub">2</emph>g. </s> <s xml:id="echoid-s2987" xml:space="preserve">De même on prendra le triple du quarré de f + g, <lb/>qui ſera 3ff + 6fg + 3g<emph style="sub">2</emph>, que l’on multipliera par c + d, & </s> <s xml:id="echoid-s2988" xml:space="preserve"><lb/>l’on aura 3cff + 6cfg + 3cg<emph style="sub">2</emph> + 3dff + 6dfg + 3dg<emph style="sub">2</emph>; </s> <s xml:id="echoid-s2989" xml:space="preserve">ajou-<lb/>tant tous ces produits enſemble, on aura pour le cube total de <pb o="92" file="0130" n="130" rhead="NOUVEAU COURS"/> la grandeur c + d + f + g, la quantité c<emph style="sub">2</emph> + 3c<emph style="sub">2</emph>d + 3cd<emph style="sub">2</emph> <lb/>+ d<emph style="sub">3</emph> + f<emph style="sub">3</emph> + 3f<emph style="sub">2</emph>g + 3fg<emph style="sub">2</emph> + g<emph style="sub">3</emph> + 3c<emph style="sub">2</emph>f + 6cdf + 3d<emph style="sub">2</emph>f + 3c<emph style="sub">2</emph>g <lb/>+ 6cdg + 3d<emph style="sub">2</emph>g + 3cff + 6cfg + 3cg<emph style="sub">2</emph> + 3dff + 6dfg + 3dg<emph style="sub">2</emph>.</s> <s xml:id="echoid-s2990" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2991" xml:space="preserve">169. </s> <s xml:id="echoid-s2992" xml:space="preserve">Quand cette méthode n’auroit pas l’avantage d’être <lb/>plus expéditive, & </s> <s xml:id="echoid-s2993" xml:space="preserve">moins ſujette à jetter dans l’erreur, elle <lb/>devient ici néceſſaire, pour faire connoître comment on peut <lb/>ramener la formation du cube d’une quantité complexe de <lb/>tant de termes que l’on voudra, à la formation du cube, <lb/>du binome a + b; </s> <s xml:id="echoid-s2994" xml:space="preserve">& </s> <s xml:id="echoid-s2995" xml:space="preserve">pour montrer pareillement comment <lb/>l’extraction des racines cubes des mêmes polinomes ſe rappelle <lb/>à l’extraction de la racine cube de a + b.</s> <s xml:id="echoid-s2996" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2997" xml:space="preserve">De même ſi l’on vouloit élever au cube la quantité complexe <lb/>3c + 2d + 5f, on feroit 3c + 2d = a, & </s> <s xml:id="echoid-s2998" xml:space="preserve">5f = b. </s> <s xml:id="echoid-s2999" xml:space="preserve">Cela poſé, <lb/>on chercheroit d’abord a<emph style="sub">3</emph>, que l’on trouveroit en élevant le <lb/>binome 3c + 2d au cube, ſuivant la regle du binome a + b, & </s> <s xml:id="echoid-s3000" xml:space="preserve"><lb/>qui eſt 27c<emph style="sub">3</emph> + 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph>. </s> <s xml:id="echoid-s3001" xml:space="preserve">On chercheroit enſuite <lb/>le triple du quarré du premier terme, multiplié par le ſecond, <lb/>ou 3a<emph style="sub">2</emph>b qui eſt 135cf<emph style="sub">2</emph> + 180cdf + 60d<emph style="sub">2</emph>f. </s> <s xml:id="echoid-s3002" xml:space="preserve">On prendroit de <lb/>même le triple du quarré du ſecond, multiplié par le premier, <lb/>ou 3ab<emph style="sub">2</emph> qui ſe trouveroit être 225cf<emph style="sub">2</emph> + 150df<emph style="sub">2</emph>: </s> <s xml:id="echoid-s3003" xml:space="preserve">enfin on auroit <lb/>pour b<emph style="sub">3</emph> ou le cube du ſecond terme, 125f<emph style="sub">3</emph>. </s> <s xml:id="echoid-s3004" xml:space="preserve">En aſſemblant toutes <lb/>ces quantités, on auroit pour le cube du trinome 3c + 2d + 5f, <lb/>27c<emph style="sub">3</emph> + 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph> + 135c<emph style="sub">2</emph>f + 180cdf + 60d<emph style="sub">2</emph>f + <lb/>225cf<emph style="sub">2</emph> + 150df<emph style="sub">2</emph> + 125f<emph style="sub">3</emph>.</s> <s xml:id="echoid-s3005" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div164" type="section" level="1" n="138"> <head xml:id="echoid-head154" style="it" xml:space="preserve">De l’Extraction des Racines Cubes des quantités algébriques.</head> <head xml:id="echoid-head155" xml:space="preserve"><emph style="sc">Regle generale</emph>.</head> <p> <s xml:id="echoid-s3006" xml:space="preserve">170. </s> <s xml:id="echoid-s3007" xml:space="preserve">Pour extraire la racine cube d’une quantité algébri-<lb/>que, il faudra prendre d’abord la racine cube d’un des termes <lb/>de cette quantité, qui ſera un cube parfait, & </s> <s xml:id="echoid-s3008" xml:space="preserve">l’écrire à la <lb/>racine: </s> <s xml:id="echoid-s3009" xml:space="preserve">pour avoir le ſecond terme de la racine, il faudra <lb/>prendre le triple du quarré du premier terme que l’on vient de <lb/>mettre à la racine, & </s> <s xml:id="echoid-s3010" xml:space="preserve">par cette quantité diviſer un terme du <lb/>polinome propoſé qui puiſſe donner un quotient exact; </s> <s xml:id="echoid-s3011" xml:space="preserve">il fau-<lb/>dra ajouter à côté du diviſeur le triple du premier terme, mul-<lb/>tiplié par ce quotient, le quarré du même quotient, & </s> <s xml:id="echoid-s3012" xml:space="preserve">multi-<lb/>plier le tout par le même quotient; </s> <s xml:id="echoid-s3013" xml:space="preserve">& </s> <s xml:id="echoid-s3014" xml:space="preserve">ſi le polinome propoſé <lb/>eſt un cube parfait, & </s> <s xml:id="echoid-s3015" xml:space="preserve">n’a que quatre termes, il faut que le <lb/>produit qui viendra, ſoit égal à ce qui reſte de la même quan- <pb o="93" file="0131" n="131" rhead="DE MATHÉMATIQUE. Liv. I."/> citè, après en avoir retranché le cube du premier terme.</s> <s xml:id="echoid-s3016" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div165" type="section" level="1" n="139"> <head xml:id="echoid-head156" xml:space="preserve"><emph style="sc">Exemple</emph> I.</head> <p> <s xml:id="echoid-s3017" xml:space="preserve">171. </s> <s xml:id="echoid-s3018" xml:space="preserve">Soit propoſé d’extraire la racine cube du polinome <lb/>a<emph style="sub">3</emph> + 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>. </s> <s xml:id="echoid-s3019" xml:space="preserve">Ayant diſpoſé cette quantité à la gau-<lb/>che d’une barre verticale, comme on le voit ci-après, je dis, <lb/>la racine cube de a<emph style="sub">3</emph> eſt a, que je poſe à la racine: </s> <s xml:id="echoid-s3020" xml:space="preserve">j’éleve cette <lb/>racine à ſon cube, & </s> <s xml:id="echoid-s3021" xml:space="preserve">ôtant a<emph style="sub">3</emph> de la quantité propoſée, il me <lb/>vient pour reſte 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>. </s> <s xml:id="echoid-s3022" xml:space="preserve">Pour avoir le ſecond terme <lb/>de la racine, j’éleve la grandeur a à ſon quarré, dont le triple <lb/>3a<emph style="sub">2</emph> me ſert de diviſeur, que je place au deſſous de la racine. <lb/></s> <s xml:id="echoid-s3023" xml:space="preserve">Je cherche dans le reſte un terme diviſible par 3a<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s3024" xml:space="preserve">je vois <lb/>que le premier de ce reſte 3a<emph style="sub">2</emph>b eſt effectivement diviſible par <lb/>3a<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s3025" xml:space="preserve">me donne au quotient b. </s> <s xml:id="echoid-s3026" xml:space="preserve">J’écris au deſſous du diviſeur <lb/>3a<emph style="sub">2</emph> la quantité ſuivante, 3a<emph style="sub">2</emph> + 3ab + b<emph style="sub">2</emph>, qui contient le triple <lb/>du quarré du premier terme, le triple du premier par le ſecond, <lb/>& </s> <s xml:id="echoid-s3027" xml:space="preserve">le quarré du ſecond ou du quotient b: </s> <s xml:id="echoid-s3028" xml:space="preserve">je multiplie cette <lb/>quantité par le même quotient b, & </s> <s xml:id="echoid-s3029" xml:space="preserve">j’ai 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>, <lb/>qui eſt égal au reſte, & </s> <s xml:id="echoid-s3030" xml:space="preserve">me fait voir que b eſt le ſecond terme <lb/>de la racine. </s> <s xml:id="echoid-s3031" xml:space="preserve">Je le mets donc à la ſuite de a, ce qui me donne <lb/>a + b pour la racine cube demandée.</s> <s xml:id="echoid-s3032" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div166" type="section" level="1" n="140"> <head xml:id="echoid-head157" xml:space="preserve"><emph style="sc">Article</emph> 171.</head> <p> <s xml:id="echoid-s3033" xml:space="preserve">a<emph style="sub">3</emph> + 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>- a<emph style="sub">3</emph> <lb/>Reſte 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>Souſtract. </s> <s xml:id="echoid-s3034" xml:space="preserve">- 3a<emph style="sub">2</emph>b - 3ab<emph style="sub">2</emph> - b<emph style="sub">3</emph> <lb/>0 0 0</s> </p> <p> <s xml:id="echoid-s3035" xml:space="preserve">{a + b, racine. <lb/></s> <s xml:id="echoid-s3036" xml:space="preserve">3a<emph style="sub">2</emph>, diviſeur. </s> <s xml:id="echoid-s3037" xml:space="preserve"><lb/>3a<emph style="sub">2</emph> + 3ab + b<emph style="sub">2</emph> <lb/>b <lb/>3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>, produit.</s> <s xml:id="echoid-s3038" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div167" type="section" level="1" n="141"> <head xml:id="echoid-head158" xml:space="preserve"><emph style="sc">Exemple</emph> II.</head> <p> <s xml:id="echoid-s3039" xml:space="preserve">172. </s> <s xml:id="echoid-s3040" xml:space="preserve">Soit encore propoſé d’extraire la racine cube de la quan-<lb/>tité 27c<emph style="sub">3</emph> + 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph>. </s> <s xml:id="echoid-s3041" xml:space="preserve">Ayant écrit cette quantité <lb/>à la gauche d’une ligne verticale, de l’autre côté de laquelle je <lb/>dois mettre la racine, je dis, la racine cube de 27c<emph style="sub">3</emph> eſt 3c, <lb/>puiſqu’en élevant 3c au cube, j’ai 27c<emph style="sub">3</emph>: </s> <s xml:id="echoid-s3042" xml:space="preserve">j’ôte ce cube de la <lb/>quantité propoſée, le reſte eſt 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d. </s> <s xml:id="echoid-s3043" xml:space="preserve">Je triple <lb/>le quarré de ce qui eſt à la racine, & </s> <s xml:id="echoid-s3044" xml:space="preserve">j’ai pour diviſeur 27c<emph style="sub">2</emph>. <lb/></s> <s xml:id="echoid-s3045" xml:space="preserve">Je cherche dans le reſte un terme qui ſoit diviſible par 27c<emph style="sub">2</emph>, ce <lb/>terme eſt 54c<emph style="sub">2</emph>d, qui me donne au quotient 2d: </s> <s xml:id="echoid-s3046" xml:space="preserve">j’écris au <pb o="94" file="0132" n="132" rhead="NOUVEAU COURS"/> deſſous du diviſeur le même diviſeur 27c<emph style="sub">2</emph>, avec les termes ſui-<lb/>vans, + 18cd + 4dd, je multiplie toute cette quantité par le <lb/>quotient 2d; </s> <s xml:id="echoid-s3047" xml:space="preserve">& </s> <s xml:id="echoid-s3048" xml:space="preserve">comme le produit eſt 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph> <lb/>égal au reſte, je conclus que 2d eſt le ſecond terme que je <lb/>cherche, & </s> <s xml:id="echoid-s3049" xml:space="preserve">je le mets à la racine, qui eſt 3c + 2d.</s> <s xml:id="echoid-s3050" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div168" type="section" level="1" n="142"> <head xml:id="echoid-head159" xml:space="preserve"><emph style="sc">Article</emph> 172.</head> <p> <s xml:id="echoid-s3051" xml:space="preserve">27c<emph style="sub">3</emph> + 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph> <lb/>- 27c<emph style="sub">3</emph> <lb/>Reſte 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph> <lb/>- 54c<emph style="sub">2</emph>d - 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph> <lb/>0 0 0</s> </p> <p> <s xml:id="echoid-s3052" xml:space="preserve">{ 3c + 2d, racine. <lb/></s> <s xml:id="echoid-s3053" xml:space="preserve">27c<emph style="sub">2</emph>, diviſeur. </s> <s xml:id="echoid-s3054" xml:space="preserve"><lb/>27c<emph style="sub">2</emph> + 18cd + 4dd <lb/>2d <lb/>54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph>, produit.</s> <s xml:id="echoid-s3055" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3056" xml:space="preserve">173. </s> <s xml:id="echoid-s3057" xml:space="preserve">Si la quantité devoit avoir plus de deux termes à la <lb/>racine, on ſuivroit toujours le même procédé, c’eſt-à-dire <lb/>que l’on prendroit pour diviſeur le triple du quarré de ce que <lb/>l’on auroit trouvé pour diviſer le reſte par cette quantité, & </s> <s xml:id="echoid-s3058" xml:space="preserve">le <lb/>quotient qui viendroit ſe détermineroit de la même maniere <lb/>que l’on a déterminé le ſecond terme de la racine. </s> <s xml:id="echoid-s3059" xml:space="preserve">Par exem-<lb/>ple, ſi l’on me propoſe d’extraire la racine cube de la quantité <lb/>27c<emph style="sub">3</emph> + 54c<emph style="sub">2</emph>d + 36cd<emph style="sub">2</emph> + 8d<emph style="sub">3</emph> + 135c<emph style="sub">2</emph> f + 180dcf + 60d<emph style="sub">2</emph>f <lb/>+ 225cf<emph style="sub">2</emph> + 150df<emph style="sub">2</emph> + 125f<emph style="sub">3</emph>. </s> <s xml:id="echoid-s3060" xml:space="preserve">Après avoir trouvé les deux <lb/>premiers termes de la racine 3c + 2d, avec le reſte 135c<emph style="sub">2</emph>f, &</s> <s xml:id="echoid-s3061" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s3062" xml:space="preserve">comme il eſt marqué ci-après, pour avoir le troiſieme terme <lb/>de la racine, il faudra prendre pour diviſeur le triple du quarré <lb/>de ce qui eſt à la racine, que l’on trouvera être 27c<emph style="sub">2</emph> + 46cd <lb/>+ 12dd: </s> <s xml:id="echoid-s3063" xml:space="preserve">on cherchera donc un terme qui ſoit diviſible par <lb/>27c<emph style="sub">2</emph>: </s> <s xml:id="echoid-s3064" xml:space="preserve">ce terme eſt le premier du dernier reſte 135c<emph style="sub">2</emph>f, lequel <lb/>diviſé par 27c<emph style="sub">2</emph>, donne 5f au quotient: </s> <s xml:id="echoid-s3065" xml:space="preserve">j’écris au deſſous du <lb/>diviſeur ce même diviſeur, avec les quantités ſuivantes, 45cf <lb/>+ 30df + 25ff, dont les deux premiers termes ſont le triple de <lb/>ce qui eſt à la racine, multiplié par le quotient 5f; </s> <s xml:id="echoid-s3066" xml:space="preserve">le troi-<lb/>ſieme, le quarré du même quotient 5f: </s> <s xml:id="echoid-s3067" xml:space="preserve">je multiplie toute cette <lb/>quantité par 5f, & </s> <s xml:id="echoid-s3068" xml:space="preserve">comme le produit qui en réſulte détruit <lb/>tous les termes du dernier reſte, étant pris en moins, je con-<lb/>clus que 5f eſt le troiſieme terme de la racine, & </s> <s xml:id="echoid-s3069" xml:space="preserve">je le poſe à <lb/>la ſuite des autres.</s> <s xml:id="echoid-s3070" xml:space="preserve"/> </p> <pb o="95" file="0133" n="133" rhead="DE MATHÉMATIQUE. Liv. I."/> </div> <div xml:id="echoid-div169" type="section" level="1" n="143"> <head xml:id="echoid-head160" xml:space="preserve"><emph style="sc">Article</emph> 173.</head> <p> <s xml:id="echoid-s3071" xml:space="preserve">Reſte { 135c<emph style="sub">2</emph>f + 180dcf + 60d<emph style="sub">2</emph>f + <lb/>225cf<emph style="sub">2</emph> + 150df<emph style="sub">2</emph> + 125f<emph style="sub">3</emph> <lb/>- 135c<emph style="sub">2</emph>f - 180dcf - 60d<emph style="sub">2</emph>f <lb/>- 225cf<emph style="sub">2</emph> - 150df - 125f<emph style="sub">3</emph> <lb/>Produit <lb/>négatif. </s> <s xml:id="echoid-s3072" xml:space="preserve">0 0 0 0 0 0</s> </p> <p> <s xml:id="echoid-s3073" xml:space="preserve">{ 3c + 2d + 5f, racine. <lb/></s> <s xml:id="echoid-s3074" xml:space="preserve">27c<emph style="sub">2</emph> + 36c<emph style="sub">2</emph>d + 12dd, div. </s> <s xml:id="echoid-s3075" xml:space="preserve"><lb/>27c<emph style="sub">2</emph> + 36cd + 12dd<emph style="sub">2</emph> + 45cf <lb/>+ 30df + 25ff x 5f</s> </p> </div> <div xml:id="echoid-div170" type="section" level="1" n="144"> <head xml:id="echoid-head161" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s3076" xml:space="preserve">Cette pratique porte ſa démonſtration avec elle; </s> <s xml:id="echoid-s3077" xml:space="preserve">car il eſt <lb/>évident qu’en la ſuivant, on doit reconnoître ſi la quantité <lb/>propoſée eſt un cube, puiſque l’on ôte de cette quantité toutes <lb/>les parties qui forment le cube d’une quantité complexe. <lb/></s> <s xml:id="echoid-s3078" xml:space="preserve">Quand on a un peu d’habitude au calcul, on voit tout d’un <lb/>coup ſi une quantité propoſée eſt un cube parfait; </s> <s xml:id="echoid-s3079" xml:space="preserve">car ſi elle <lb/>ne contient que deux termes, trois termes, ou cinq, ſix, ſept, <lb/>huit, neuf, & </s> <s xml:id="echoid-s3080" xml:space="preserve">non pas dix termes, on ſera ſûre qu’elle n’eſt <lb/>point un cube parfait; </s> <s xml:id="echoid-s3081" xml:space="preserve">car elle ne peut être cube parfait que <lb/>d’un binome ou d’un trinome, ou d’une quantité plus com-<lb/>pliquée, & </s> <s xml:id="echoid-s3082" xml:space="preserve">le binome ne donne que quatre termes à ſon cube, <lb/>& </s> <s xml:id="echoid-s3083" xml:space="preserve">le trinome en donne dix: </s> <s xml:id="echoid-s3084" xml:space="preserve">donc les intermédiaires ne ſont <lb/>pas des cubes.</s> <s xml:id="echoid-s3085" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div171" type="section" level="1" n="145"> <head xml:id="echoid-head162" style="it" xml:space="preserve">De la formation algébrique du Cube d’un nombre quelconque, & <lb/>de l’extraction de racine cube de quantités numériques.</head> <p> <s xml:id="echoid-s3086" xml:space="preserve">174. </s> <s xml:id="echoid-s3087" xml:space="preserve">Pour élever un nombre comme celui-ci, 47 à ſon cube, <lb/>on peut le faire en deux manieres, ou en multipliant 47 par <lb/>lui-même pour avoir ſon quarré 2209, & </s> <s xml:id="echoid-s3088" xml:space="preserve">multipliant encore <lb/>ce quarré par 47, ce qui donnera 103823, ou bien en ſe ſervant <lb/>de la formule a<emph style="sub">3</emph> + 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>: </s> <s xml:id="echoid-s3089" xml:space="preserve">pour cela, je regarde <lb/>le nombre 47 comme une quantité complexe, que je repré-<lb/>ſente par a + b; </s> <s xml:id="echoid-s3090" xml:space="preserve">ſçavoir 40 par a, & </s> <s xml:id="echoid-s3091" xml:space="preserve">7 par b, ce qui me donne <lb/>40 + 7 = a + b. </s> <s xml:id="echoid-s3092" xml:space="preserve">Je cherche d’abord a<emph style="sub">3</emph> en élevant 40 au cube, <lb/>& </s> <s xml:id="echoid-s3093" xml:space="preserve">j’ai a<emph style="sub">3</emph> = 64000: </s> <s xml:id="echoid-s3094" xml:space="preserve">je prends enſuite le triple du quarré de <lb/>40, que je multiplie par 7, pour avoir 3a<emph style="sub">2</emph>b, ce qui me donne <lb/>3a<emph style="sub">2</emph>b = 33600. </s> <s xml:id="echoid-s3095" xml:space="preserve">Je cherche pareillement 3ab<emph style="sub">2</emph>, ou le triple du <lb/>premier, multiplié par le quarré du ſecond, ce qui donne <lb/>3ab<emph style="sub">2</emph> = 5880: </s> <s xml:id="echoid-s3096" xml:space="preserve">enfin pour b<emph style="sub">3</emph>, j’ai b<emph style="sub">3</emph> = 343. </s> <s xml:id="echoid-s3097" xml:space="preserve">Raſſemblant toutes <pb o="96" file="0134" n="134" rhead="NOUVEAU COURS"/> <anchor type="note" xlink:label="note-0134-01a" xlink:href="note-0134-01"/> Sur quoi l’on remarquera, 1°. </s> <s xml:id="echoid-s3098" xml:space="preserve">Qu’en diviſant les produits par-<lb/>ticuliers & </s> <s xml:id="echoid-s3099" xml:space="preserve">le cube total en tranches de trois chiffres chacune, <lb/>excepté la derniere à gauche, qui peut n’en contenir que deux <lb/>ou un; </s> <s xml:id="echoid-s3100" xml:space="preserve">que le nombre 64, cube du premier chiffre 4 de la <lb/>quantité 47, a après lui autant de tranches de trois qu’il y a de <lb/>rangs de chiffres à ſa racine; </s> <s xml:id="echoid-s3101" xml:space="preserve">ſçavoir une tranche de, 000 après <lb/>64, & </s> <s xml:id="echoid-s3102" xml:space="preserve">un chiffre 7 après 4 dans 47.</s> <s xml:id="echoid-s3103" xml:space="preserve"/> </p> <div xml:id="echoid-div171" type="float" level="2" n="1"> <note position="right" xlink:label="note-0134-01" xlink:href="note-0134-01a" xml:space="preserve"># { # 64,000 = a<emph style="sub">3</emph> <lb/># # 33,600 = 3a<emph style="sub">2</emph>b <lb/>ces égalités, on aura # # 5,880 = 3ab<emph style="sub">2</emph> <lb/># # 343 = b<emph style="sub">3</emph> <lb/># # 103,823 = √a + b\x{0020}<emph style="sub">3</emph> <lb/></note> </div> <p> <s xml:id="echoid-s3104" xml:space="preserve">2°. </s> <s xml:id="echoid-s3105" xml:space="preserve">Que le produit repréſenté par 3a<emph style="sub">2</emph>b eſt placé de maniere <lb/>que le triple du quarré de 4 ou 16, qui eſt 48, multiplié par <lb/>7 ou 336, a deux zero après lui: </s> <s xml:id="echoid-s3106" xml:space="preserve">donc il aura auſſi deux chif-<lb/>fres après lui dans le cube total, & </s> <s xml:id="echoid-s3107" xml:space="preserve">ſera contenu dans les chif-<lb/>fres qui ſe terminent au premier 8 de la ſeconde tranche.</s> <s xml:id="echoid-s3108" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3109" xml:space="preserve">3°. </s> <s xml:id="echoid-s3110" xml:space="preserve">Que le produit repréſenté par 3ab<emph style="sub">2</emph>, ou le triple 12 du <lb/>premier chiffre 4, multiplié par 49, quarré du ſecond, a un <lb/>rang de chiffres après lui, puiſqu’il eſt 5580; </s> <s xml:id="echoid-s3111" xml:space="preserve">& </s> <s xml:id="echoid-s3112" xml:space="preserve">qu’enfin le cube <lb/>du ſecond chiffre 7 eſt renſermé tout entier dans la ſeconde <lb/>tranche.</s> <s xml:id="echoid-s3113" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3114" xml:space="preserve">Ceci ſuppoſé, il ſera facile d’entendre la méthode de l’ex-<lb/>traction de la racíne cube que nous allons donner, après quel-<lb/>ques remarques, qui ſont abſolument néceſſaires, pour qu’il <lb/>n’y ait rien à déſirer ſur cette partie.</s> <s xml:id="echoid-s3115" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3116" xml:space="preserve">Pour extraire la racine cube d’une quantité quelconque, il <lb/>faut d’abord connoître les cubes des neuf premiers chiffres; <lb/></s> <s xml:id="echoid-s3117" xml:space="preserve">ce que l’on connoîtra par le moyen de la Table ſuivante, qui <lb/>ſuffit, lorſque les nombres propoſés n’ont que trois chiffres.</s> <s xml:id="echoid-s3118" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">1 # 2 # 3 # 4 # 5 # 6 # 7 # 8 # 9 # 10 <lb/>1 # 8 # 27 # 64 # 125 # 216 # 343 # 512 # 729 # 1000 <lb/></note> <p> <s xml:id="echoid-s3119" xml:space="preserve">175. </s> <s xml:id="echoid-s3120" xml:space="preserve">On remarquera d’abord que le plus grand nombre de <lb/>trois chiffres ne peut avoir qu’un chiffre à ſa racine cube, car <lb/>le plus grand nombre de trois chiffres eſt 999, & </s> <s xml:id="echoid-s3121" xml:space="preserve">le plus petit <lb/>de deux chiffres eſt 10, dont le cube 1000 eſt de quatre chif- <pb o="97" file="0135" n="135" rhead="DE MATHÉMATIQUE. Liv. I."/> fres; </s> <s xml:id="echoid-s3122" xml:space="preserve">ainſi toutes les racines cubes d’un chiffre ſont compriſes <lb/>incluſivement depuis 1 juſqu’à 999.</s> <s xml:id="echoid-s3123" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3124" xml:space="preserve">176. </s> <s xml:id="echoid-s3125" xml:space="preserve">Le plus grand nombre de ſix chiffres ne peut en avoir <lb/>que deux à ſa racine; </s> <s xml:id="echoid-s3126" xml:space="preserve">le plus grand nombre de ſix chiffres eſt <lb/>999999, & </s> <s xml:id="echoid-s3127" xml:space="preserve">le plus petit de trois chiffres eſt 100, dont le cube <lb/>eſt 1000000, qui a ſept chiffres, & </s> <s xml:id="echoid-s3128" xml:space="preserve">eſt plus grand que 999999. <lb/></s> <s xml:id="echoid-s3129" xml:space="preserve">Ainſi toutes les racines cubes de deux chiffres ſont compriſes <lb/>depuis 1000 juſqu’à 999,999 incluſivement.</s> <s xml:id="echoid-s3130" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3131" xml:space="preserve">177. </s> <s xml:id="echoid-s3132" xml:space="preserve">Le plus grand nombre de neuf chiffres ne peut en avoir <lb/>que trois à ſa racine; </s> <s xml:id="echoid-s3133" xml:space="preserve">car le plus grand nombre de neuf chif-<lb/>fres eſt 999999999, & </s> <s xml:id="echoid-s3134" xml:space="preserve">le plus petit nombre de quatre chiffres <lb/>eſt 1000, dont le cube eſt 1000000000 qui contient dix chif-<lb/>fres, & </s> <s xml:id="echoid-s3135" xml:space="preserve">eſt néceſſairement plus grand que 999999999; </s> <s xml:id="echoid-s3136" xml:space="preserve">d’où il <lb/>ſuit que les racines cubes de trois chiffres ſont compriſes, de-<lb/>puis 1000000 juſqu’à 999,999,999 incluſivement.</s> <s xml:id="echoid-s3137" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3138" xml:space="preserve">178. </s> <s xml:id="echoid-s3139" xml:space="preserve">En continuant toujours le même raiſonnement, on <lb/>verra qu’en général un nombre propoſé doit avoir autant de <lb/>chiffres à ſa racine cube qu’il aura de tranches de trois chiffres <lb/>chacune, excepté la premiere à gauche, qui peut n’en con-<lb/>tenir que deux ou même un, mais que l’on regarde toujours <lb/>comme une tranche; </s> <s xml:id="echoid-s3140" xml:space="preserve">car 999 ne donne qu’un chiffre à la ra-<lb/>cine, comme on l’a démontré, art. </s> <s xml:id="echoid-s3141" xml:space="preserve">175, & </s> <s xml:id="echoid-s3142" xml:space="preserve">ce nombre ne <lb/>contient qu’une tranche de trois chiffres. </s> <s xml:id="echoid-s3143" xml:space="preserve">1000 donne deux <lb/>chiffres à la racine cube, parce que, outre la tranche des trois <lb/>zero, il contient encore une tranche d’un chiffre. </s> <s xml:id="echoid-s3144" xml:space="preserve">De même <lb/>999999 ne peut donner que deux chiffres à la racine, ainſi <lb/>que tous les intermédiaires entre lui & </s> <s xml:id="echoid-s3145" xml:space="preserve">1000, parce qu’ils ne <lb/>contiennent que deux tranches, & </s> <s xml:id="echoid-s3146" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s3147" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3148" xml:space="preserve">Tout cela poſé, nous allons donner la regle générale, & </s> <s xml:id="echoid-s3149" xml:space="preserve"><lb/>l’appliquer à quelques exemples.</s> <s xml:id="echoid-s3150" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div173" type="section" level="1" n="146"> <head xml:id="echoid-head163" style="it" xml:space="preserve">Regle générale pour l’extraction de la Racine cube des quantités <lb/>numériques.</head> <p> <s xml:id="echoid-s3151" xml:space="preserve">179. </s> <s xml:id="echoid-s3152" xml:space="preserve">1°. </s> <s xml:id="echoid-s3153" xml:space="preserve">On commencera par partager le nombre donné en <lb/>tranches de trois chiffres chacune, en comptant pour une <lb/>tranche la premiere à gauche, qui peut ne contenir que deux <lb/>chiffres, ou même un ſeul.</s> <s xml:id="echoid-s3154" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3155" xml:space="preserve">2°. </s> <s xml:id="echoid-s3156" xml:space="preserve">On cherchera le plus grand cube contenu dans la pre-<lb/>miere tranche à gauche, on en prendra la racine, que l’on <pb o="98" file="0136" n="136" rhead="NOUVEAU COURS"/> poſera à la droite du nombre propoſé, après en avoir ſéparé <lb/>la racine par une barre verticale. </s> <s xml:id="echoid-s3157" xml:space="preserve">On élevera cette racine à ſon <lb/>cube, que l’on ôtera de la premiere tranche, & </s> <s xml:id="echoid-s3158" xml:space="preserve">l’opération ſera <lb/>achevée pour cette tranche.</s> <s xml:id="echoid-s3159" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3160" xml:space="preserve">3°. </s> <s xml:id="echoid-s3161" xml:space="preserve">A côté du reſte que l’on aura trouvé, en ôtant le cube <lb/>du premier chiffre de la racine de la premiere tranche, on <lb/>abaiſſera la ſeconde tranche, en obſervant de mettre un point <lb/>ſous le premier chiffre de cette ſeconde tranche: </s> <s xml:id="echoid-s3162" xml:space="preserve">pour avoir le <lb/>ſecond chiffre de la racine, on élevera le premier au quarré, <lb/>dont on prendra le triple, qui ſera le diviſeur dont il faudra <lb/>ſe ſervir pour trouver le ſecond chiffre de la racine.</s> <s xml:id="echoid-s3163" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3164" xml:space="preserve">4°. </s> <s xml:id="echoid-s3165" xml:space="preserve">On diviſera les chiffres terminés à celui ſous lequel on <lb/>a mis un point, par le diviſeur trouvé, & </s> <s xml:id="echoid-s3166" xml:space="preserve">l’on aura un quotient, <lb/>que l’on éprouvera comme il ſuit, avant que de le poſer à la <lb/>racine. </s> <s xml:id="echoid-s3167" xml:space="preserve">Il faudra ajouter enſemble les produits repréſentés par <lb/>3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>, c’eſt-à-dire le produit du diviſeur par le <lb/>chiffre que l’on éprouve, le triple du premier terme de la ra-<lb/>cine par le quarré du même chiffre, & </s> <s xml:id="echoid-s3168" xml:space="preserve">enfin le cube de ce <lb/>même chiffre, en obſervant de les placer avant l’addition, de <lb/>maniere qu’ils ſe paſſent tous d’un chiffre en avant. </s> <s xml:id="echoid-s3169" xml:space="preserve">Il faudra <lb/>ôter la ſomme de la ſeconde tranche, jointe au reſte que l’on <lb/>a trouvé, & </s> <s xml:id="echoid-s3170" xml:space="preserve">ſi la ſouſtraction ſe peut faire, on mettra le chiffre <lb/>à la racine, ſinon il faudra diminuer d’une unité, juſqu’à @e <lb/>que la ſomme de ces produits ſoit moindre, ou tout au moins <lb/>égale aux chiffres ſur leſquels on opere. </s> <s xml:id="echoid-s3171" xml:space="preserve">Si le nombre propoſé <lb/>n’a que deux tranches, l’extraction ſera faite, & </s> <s xml:id="echoid-s3172" xml:space="preserve">la racine ſera <lb/>la racine exacte que l’on cherche, ſi la ſouſtraction n’a pas <lb/>donné de reſte. </s> <s xml:id="echoid-s3173" xml:space="preserve">Si le nombre avoit encore d’autres tranches, <lb/>on les abaiſſeroit l’une aprés l’autre à côté du dernier reſte, en <lb/>déterminant les diviſeurs, & </s> <s xml:id="echoid-s3174" xml:space="preserve">les chiffres que l’on doit mettre à <lb/>la racine, comme on a fait pour le ſecond chiffre de la même <lb/>racine.</s> <s xml:id="echoid-s3175" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div174" type="section" level="1" n="147"> <head xml:id="echoid-head164" xml:space="preserve"><emph style="sc">Exemple</emph> I.</head> <p> <s xml:id="echoid-s3176" xml:space="preserve">180. </s> <s xml:id="echoid-s3177" xml:space="preserve">Soit propoſé d’extraire la racine cube du nombre <lb/>103823. </s> <s xml:id="echoid-s3178" xml:space="preserve">Aprés avoir partagé ce nombre en tranches de trois <lb/>chiffres chacune, je dis, en 103 quel eſt le plus grand cube <lb/>qui y ſoit contenu? </s> <s xml:id="echoid-s3179" xml:space="preserve">Ce cube eſt 64 (comme on le peut voir <lb/>aiſément par la Table des cubes), dont la racine cube eſt 4, <lb/>Je poſe 4 à la racine, à la droite du nombre propoſé, après <pb o="99" file="0137" n="137" rhead="DE MATHÉMATIQUE. Liv. I."/> l’avoir ſéparée par une barre verticale, & </s> <s xml:id="echoid-s3180" xml:space="preserve">je ſouſtrais le cube 64 <lb/>de cette racine 4 de 103, le reſte eſt 39. </s> <s xml:id="echoid-s3181" xml:space="preserve">J’abaiſſe enſuite la ſe-<lb/>conde tranche 823 à côté du reſte 39, en mettant un point <lb/>ſous le premier chiffre 8, pour marquer que 398, eſt le <lb/>dividende ſur lequel il faut opérer, & </s> <s xml:id="echoid-s3182" xml:space="preserve">qui contient le triple <lb/>du quarré du premier terme, multiplié par le ſecond: </s> <s xml:id="echoid-s3183" xml:space="preserve">pour <lb/>avoir ce ſecond terme, je triple le quarré de 4, & </s> <s xml:id="echoid-s3184" xml:space="preserve">j’ai 48 pour <lb/>diviſeur, par lequel je diviſe 398, en imaginant le chiffre 8 <lb/>du diviſeur ſous le chiffre 8 du dividende partiel, & </s> <s xml:id="echoid-s3185" xml:space="preserve">je dis, <lb/>en 39 combien de fois 4, il y eſt neuf fois; </s> <s xml:id="echoid-s3186" xml:space="preserve">mais comme je <lb/>prévois que le 9 n’eſt pas bon, j’eſſaie le 8, quoique je ſçache <lb/>bien qu’il n’eſt pas non plus celui que je demande, mais pour <lb/>montrer la maniere dont on fait l’épreuve. </s> <s xml:id="echoid-s3187" xml:space="preserve">Je multiplie d’abord <lb/>le diviſeur par 8, & </s> <s xml:id="echoid-s3188" xml:space="preserve">j’ai 384 qui me repréſente le produit dé-<lb/>ſigné par 3a<emph style="sub">2</emph>b. </s> <s xml:id="echoid-s3189" xml:space="preserve">Je multiplie enſuite le nombre 12, triple de ce <lb/>qui eſt à la racine, par 64, quarré de 8, le produit eſt 768, <lb/>que j’écris au deſſous du premier 384, de maniere qu’il dé-<lb/>borde le dernier chiffre 4 d’un rang vers la droite, & </s> <s xml:id="echoid-s3190" xml:space="preserve">ce nom-<lb/>bre me repréſente 3ab<emph style="sub">2</emph>. </s> <s xml:id="echoid-s3191" xml:space="preserve">Enfin je prends le cube de 8, qui eſt <lb/>512, que j’écris au deſſous du ſecond produit, de maniere que <lb/>le 2 déborde d’un rang le dernier chiffre 7 de ce ſecond pro-<lb/>duit. </s> <s xml:id="echoid-s3192" xml:space="preserve">J’ajoute ces trois nombres enſemble, & </s> <s xml:id="echoid-s3193" xml:space="preserve">je trouve la <lb/>ſomme 46582. </s> <s xml:id="echoid-s3194" xml:space="preserve">Comme ce produit eſt plus grand que le reſte, <lb/>joint avec la ſeconde tranche 39823, je conclus que le 8 n’eſt <lb/>pas encore bon; </s> <s xml:id="echoid-s3195" xml:space="preserve">je diminue d’une unité, & </s> <s xml:id="echoid-s3196" xml:space="preserve">j’éprouve le 7 de <lb/>la même maniere: </s> <s xml:id="echoid-s3197" xml:space="preserve">je multiplie le diviſeur 48 par 7, & </s> <s xml:id="echoid-s3198" xml:space="preserve">j’ai 336, <lb/>qui me repréſente le produit déſigné par 3a<emph style="sub">2</emph>b; </s> <s xml:id="echoid-s3199" xml:space="preserve">je multiplie <lb/>enſuite 12, triple de ce qui eſt à la racine, par 49, quarré de <lb/>7, & </s> <s xml:id="echoid-s3200" xml:space="preserve">j’ai au produit 588, que je place de maniere, que le der-<lb/>nier chiffre 8 déborde d’un rang le dernier chiffre du produit <lb/>ſupérieur, & </s> <s xml:id="echoid-s3201" xml:space="preserve">ce produit me déſigne 3ab<emph style="sub">2</emph>. </s> <s xml:id="echoid-s3202" xml:space="preserve">Enfin j’éleve 7 au <lb/>cube, qui eſt 343, que j’écris encore au deſſous du ſecond <lb/>produit, de maniere que le dernier chiffre 3 paſſe le dernier <lb/>du produit ſupérieur d’un rang vers la droite: </s> <s xml:id="echoid-s3203" xml:space="preserve">j’ajoute enſem-<lb/>ble ces produits, & </s> <s xml:id="echoid-s3204" xml:space="preserve">je trouve que leur ſomme eſt 39823, égale <lb/>au nombre ſur lequel j’opere; </s> <s xml:id="echoid-s3205" xml:space="preserve">d’où je conclus que le 7 eſt bon, <lb/>je le poſe à la racine, que je trouve être 47, comme on le ſçait <lb/>déja par l’article 174.</s> <s xml:id="echoid-s3206" xml:space="preserve"/> </p> <pb o="100" file="0138" n="138" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div175" type="section" level="1" n="148"> <head xml:id="echoid-head165" xml:space="preserve"><emph style="sc">Article</emph> 180.</head> <p> <s xml:id="echoid-s3207" xml:space="preserve">103,823 <lb/>64 <lb/>39823 <lb/>39823 <lb/>00000</s> </p> <p> <s xml:id="echoid-s3208" xml:space="preserve">{47, racine. <lb/></s> <s xml:id="echoid-s3209" xml:space="preserve">48 = 3a<emph style="sub">2</emph>, diviſeur. </s> <s xml:id="echoid-s3210" xml:space="preserve"><lb/>384 = 3a<emph style="sub">2</emph>b <lb/>768 = 3ab<emph style="sub">2</emph> <lb/>512 = b<emph style="sub">3</emph> <lb/>46592 = 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>336 = 3a<emph style="sub">2</emph>b <lb/>588 = 3ab<emph style="sub">2</emph> <lb/>343 = b<emph style="sub">3</emph> <lb/>39823 = 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph></s> </p> <p> <s xml:id="echoid-s3211" xml:space="preserve">}Epreuve du 8. <lb/></s> <s xml:id="echoid-s3212" xml:space="preserve">}Epreuve du 7.</s> <s xml:id="echoid-s3213" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div176" type="section" level="1" n="149"> <head xml:id="echoid-head166" xml:space="preserve"><emph style="sc">Exemple</emph> II.</head> <p> <s xml:id="echoid-s3214" xml:space="preserve">181. </s> <s xml:id="echoid-s3215" xml:space="preserve">Soit propoſé d’extraire la racine cube du nombre <lb/>99865243. </s> <s xml:id="echoid-s3216" xml:space="preserve">Aprés avoir partagé ce nombre en tranches de <lb/>trois chiffres en trois chiffres, à commencer par la droite, je <lb/>cherche d’abord la racine cube de 99,865, préciſément de la <lb/>même maniere que dans l’exemple précédent, en faiſant abſ-<lb/>traction pour un moment de la troiſieme tranche 243. </s> <s xml:id="echoid-s3217" xml:space="preserve">Je dis <lb/>donc en 99 quel eſt le plus grand cube qui y ſoit contenu? </s> <s xml:id="echoid-s3218" xml:space="preserve">Ce <lb/>cube eſt 64, dont la racine eſt 4, que je poſe à la racine, à la <lb/>droite du nombre propoſé: </s> <s xml:id="echoid-s3219" xml:space="preserve">je cube 4, & </s> <s xml:id="echoid-s3220" xml:space="preserve">j’ôte le produit 64 de <lb/>99, le reſte eſt 35, & </s> <s xml:id="echoid-s3221" xml:space="preserve">toute l’opération eſt faite pour la pre-<lb/>miere tranche. </s> <s xml:id="echoid-s3222" xml:space="preserve">J’abaiſſe la ſeconde tranche 865, en mettant <lb/>un point ſous le premier chiffre de cette tranche, pour mar-<lb/>quer que le nombre 358 contient le triple du quarré du pre-<lb/>mier terme, multiplié par le ſecond. </s> <s xml:id="echoid-s3223" xml:space="preserve">Je triple le quarré de ce <lb/>qui eſt à la racine, & </s> <s xml:id="echoid-s3224" xml:space="preserve">j’ai le diviſeur 48, par lequel il faut di-<lb/>viſer 358 pour avoir le ſecond chiffre de la racine. </s> <s xml:id="echoid-s3225" xml:space="preserve">Je diviſe <lb/>donc 358 par 48, & </s> <s xml:id="echoid-s3226" xml:space="preserve">je dis, en 35 combien de fois 4, il y eſt <lb/>huit fois; </s> <s xml:id="echoid-s3227" xml:space="preserve">mais ni le 8 ni le 7 ne peuvent être mis à la racine, <lb/>car en faiſant l’épreuve du 7, comme dans l’exemple précé-<lb/>dent, on verra que les produits déſignés par 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>, <lb/>qu’il faut retrancher du reſte, joint à la ſeconde tranche, don-<lb/>nent un nombre trop grand 39823. </s> <s xml:id="echoid-s3228" xml:space="preserve">Ainſi j’éprouve le 6; </s> <s xml:id="echoid-s3229" xml:space="preserve">pour <lb/>cela je multiplie le diviſeur 48 par 6 pour avoir le produit 288, <lb/>déſigné par 3a<emph style="sub">2</emph>b. </s> <s xml:id="echoid-s3230" xml:space="preserve">Je multiplie enſuite le triple de ce qui eſt à <pb o="101" file="0139" n="139" rhead="DE MATHÉMATIQUE. Liv. I."/> la racine, ou 12 par le quarré de 6, qui eſt 36, & </s> <s xml:id="echoid-s3231" xml:space="preserve">j’ai 432, qui <lb/>me repréſente 3ab<emph style="sub">2</emph>, que j’écris au deſſous du premier produit, <lb/>de maniere que le dernier chiffre 2 ſurpaſſe d’un rang vers la <lb/>droite le chiffre ſupérieur. </s> <s xml:id="echoid-s3232" xml:space="preserve">Enfin j’écris le cube de 6, qui eſt <lb/>216, de maniere que le 6 déborde encore d’un rang; </s> <s xml:id="echoid-s3233" xml:space="preserve">je prends <lb/>la ſomme de ces trois produits, que je trouve être 33336. <lb/></s> <s xml:id="echoid-s3234" xml:space="preserve">Comme ce nombre eſt moindre que 35865, je conclus que le <lb/>6 eſt bon, & </s> <s xml:id="echoid-s3235" xml:space="preserve">je le poſe à la racine; </s> <s xml:id="echoid-s3236" xml:space="preserve">je ſouſtrais 33336 de 35865, <lb/>& </s> <s xml:id="echoid-s3237" xml:space="preserve">le reſte eſt 2529. </s> <s xml:id="echoid-s3238" xml:space="preserve">Si l’on n’avoit pas encore la troiſieme <lb/>tranche 243, l’opération ſeroit achevée, & </s> <s xml:id="echoid-s3239" xml:space="preserve">la racine ſeroit <lb/>46, avec le reſte 2529, qui ne pourroit pas donner une <lb/>unité mais puiſqu’elle s’y trouve, il faut encore déterminer <lb/>le troiſieme chiffre de cette racine: </s> <s xml:id="echoid-s3240" xml:space="preserve">pour cela, je quarre 46 à <lb/>part, & </s> <s xml:id="echoid-s3241" xml:space="preserve">je trouve pour ſon quarré 2116, dont je prends le <lb/>triple, qui eſt 6348, par lequel je dois diviſer le nombre qui <lb/>contient le troiſieme chiffre, multiplié par le triple du quarré <lb/>du premier terme, que je regarde comme 46; </s> <s xml:id="echoid-s3242" xml:space="preserve">j’abaiſſe la troi-<lb/>ſieme tranche 243 à côté du reſte 2529, en mettant un point <lb/>ſous le premier chiffre 2 de cette tranche, & </s> <s xml:id="echoid-s3243" xml:space="preserve">je diviſe 25292 <lb/>par 6348, en diſant, en 25 combien de fois 6, il y eſt quatre <lb/>fois; </s> <s xml:id="echoid-s3244" xml:space="preserve">mais en faiſant l’épreuve comme ci-devant, on verroit <lb/>que le 4 ne peut pas être mis à la racine, ainſi j’éprouve le 3. </s> <s xml:id="echoid-s3245" xml:space="preserve"><lb/>Je prends d’abord le produit du diviſeur par 3, que je trouve <lb/>19044, qui me repréſente 3a<emph style="sub">2</emph>b, je prends enſuite le triple de <lb/>ce qui eſt à la racine, que je multiplie par 9, quarré du chiffre <lb/>3, que j’éprouve, & </s> <s xml:id="echoid-s3246" xml:space="preserve">j’ai 1242 que je place au deſſous du pre-<lb/>mier produit, de maniere que le 2 déborde d’un chiffre, & </s> <s xml:id="echoid-s3247" xml:space="preserve"><lb/>ce produit me repréſente 3ab<emph style="sub">2</emph>. </s> <s xml:id="echoid-s3248" xml:space="preserve">Enfin j’écris au deſſous de ce <lb/>ſecond produit 27, cube de 3, de maniere que le 7 déborde <lb/>d’un rang les chiffres ſupérieurs: </s> <s xml:id="echoid-s3249" xml:space="preserve">j’ajoute ces trois grandeurs <lb/>enſemble, & </s> <s xml:id="echoid-s3250" xml:space="preserve">j’ai pour leur ſomme 1916847. </s> <s xml:id="echoid-s3251" xml:space="preserve">Comme ce pro-<lb/>duit eſt moindre que 2529243, je conclus que le 3 eſt bon, & </s> <s xml:id="echoid-s3252" xml:space="preserve"><lb/>je le poſe à la racine. </s> <s xml:id="echoid-s3253" xml:space="preserve">J’ôte ce dernier produit du nombre <lb/>2529243, le reſte eſt 612396, qui ne pouvoit donner une <lb/>unité de plus à la racine, & </s> <s xml:id="echoid-s3254" xml:space="preserve">de cette maniere l’opération ſe <lb/>trouve achevée.</s> <s xml:id="echoid-s3255" xml:space="preserve"/> </p> <pb o="102" file="0140" n="140" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div177" type="section" level="1" n="150"> <head xml:id="echoid-head167" xml:space="preserve"><emph style="sc">Article</emph> 181.</head> <p> <s xml:id="echoid-s3256" xml:space="preserve">99865243 <lb/>64 <lb/>35865 <lb/>33336 <lb/>2529243 <lb/>1916847 <lb/>612396</s> </p> <p> <s xml:id="echoid-s3257" xml:space="preserve">{463 <lb/>48 = 3a<emph style="sub">2</emph>, I<emph style="sub">er</emph> diviſeur. <lb/></s> <s xml:id="echoid-s3258" xml:space="preserve">288 = 3a<emph style="sub">2</emph>b <lb/>432 = 3ab<emph style="sub">2</emph> <lb/>216 = b<emph style="sub">3</emph> <lb/>33336 = 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>6348 = 3a<emph style="sub">2</emph>, ſecond diviſ. </s> <s xml:id="echoid-s3259" xml:space="preserve"><lb/>19044 = 3a<emph style="sub">2</emph>b <lb/>1242 = 3ab<emph style="sub">2</emph> <lb/>27 = b<emph style="sub">3</emph> <lb/>1916847 = 3a<emph style="sub">2</emph>b + 3ab + b<emph style="sub">3</emph></s> </p> <p> <s xml:id="echoid-s3260" xml:space="preserve">} Epreuve du 6. <lb/></s> <s xml:id="echoid-s3261" xml:space="preserve">} Epreuve du 3.</s> <s xml:id="echoid-s3262" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div178" type="section" level="1" n="151"> <head xml:id="echoid-head168" style="it" xml:space="preserve">Maniere d’approcher le plus prés qu’il eſt poſſible de la racine cube <lb/>d’un nombre donné, par le moyen des décimales.</head> <p> <s xml:id="echoid-s3263" xml:space="preserve">182. </s> <s xml:id="echoid-s3264" xml:space="preserve">On ajoutera au nombre propoſé, pour en extraire la <lb/>racine, autant de tranches de trois zero chacune que l’on vou-<lb/>dra avoir de décimales à la racine: </s> <s xml:id="echoid-s3265" xml:space="preserve">on extraira d’abord la ra-<lb/>cine du nombre propoſé, comme on a fait ci-devant, & </s> <s xml:id="echoid-s3266" xml:space="preserve">aprés <lb/>avoir trouvé le reſte, puiſque la racine n’eſt pas complette, on <lb/>abaiſſera auprés de ce reſte la premiere tranche, & </s> <s xml:id="echoid-s3267" xml:space="preserve">l’on opérera ſur <lb/>cette partie comme ſur des nombres entiers; </s> <s xml:id="echoid-s3268" xml:space="preserve">on fera l’épreuve <lb/>des chiffres qu’il faudra mettre à la racine, préciſément de la <lb/>même maniere, comme on verra ſuſaffimment dans l’exemple <lb/>ſuivant, dans lequel on ſe contentera d’indiquer les opérations <lb/>ſans s’arrêter à les détailler.</s> <s xml:id="echoid-s3269" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3270" xml:space="preserve">183. </s> <s xml:id="echoid-s3271" xml:space="preserve">Si l’on ſuppoſe que 694 ſoit un nombre de toiſes, dont <lb/>on demande la racine en toiſes, pieds, pouces, il faudra ré-<lb/>duire les décimales 853 en valeur connue, ſuivant la méthode <lb/>de l’article 131, en multipliant ce nombre 0.</s> <s xml:id="echoid-s3272" xml:space="preserve">853 par 6, pre-<lb/>nant les entiers pour les pieds, & </s> <s xml:id="echoid-s3273" xml:space="preserve">multipliant encore le reſte <lb/>par 12 pour avoir les pouces, & </s> <s xml:id="echoid-s3274" xml:space="preserve">ainſi de ſuite pour les lignes <lb/>& </s> <s xml:id="echoid-s3275" xml:space="preserve">les points. </s> <s xml:id="echoid-s3276" xml:space="preserve">En opérant de cette maniere, on verra que la <lb/>racine cube de 694 toiſes cubes eſt 8 toiſ. </s> <s xml:id="echoid-s3277" xml:space="preserve">5 pieds 1 pouce 5 lig.</s> <s xml:id="echoid-s3278" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3279" xml:space="preserve">184. </s> <s xml:id="echoid-s3280" xml:space="preserve">Si au contraire on propoſoit un nombre qui contînt <lb/>des toiſes, des pieds, des pouces pour en extraire la racine, <pb o="103" file="0141" n="141" rhead="DE MATHÉMATIQUE. Liv. I."/> il faudroit chercher une fraction décimale de la toiſe égale <lb/>aux pieds, pouces, lignes, qui ſont joints au nombre entier, <lb/>en chercher la racine, ſuivant les regles précédentes; </s> <s xml:id="echoid-s3281" xml:space="preserve">& </s> <s xml:id="echoid-s3282" xml:space="preserve">la ra-<lb/>cine que l’on trouvera ſera celle que l’on demande, exprimée <lb/>en toiſes & </s> <s xml:id="echoid-s3283" xml:space="preserve">parties décimales de toiſes, que l’on réduira en <lb/>pieds, pouces, lignes & </s> <s xml:id="echoid-s3284" xml:space="preserve">points, ſuivant la méthode de l’ar-<lb/>ticle 131.</s> <s xml:id="echoid-s3285" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div179" type="section" level="1" n="152"> <head xml:id="echoid-head169" xml:space="preserve"><emph style="sc">Article</emph> 182.</head> <p> <s xml:id="echoid-s3286" xml:space="preserve">694,000,000,000 <lb/>512 <lb/>182000 <lb/>169 472 <lb/>12 528000 <lb/>11 682125 <lb/>845875000 <lb/>705142479</s> </p> <p> <s xml:id="echoid-s3287" xml:space="preserve">{8.</s> <s xml:id="echoid-s3288" xml:space="preserve">853 <lb/>192 = 3a<emph style="sub">2</emph>, premier diviſeur. <lb/></s> <s xml:id="echoid-s3289" xml:space="preserve">1536 = 3a<emph style="sub">2</emph>b <lb/>1536 = 3ab<emph style="sub">2</emph> <lb/>512 = b<emph style="sub">3</emph> <lb/>169472 = 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>23232 = 3a<emph style="sub">2</emph>, ſecond diviſeur. </s> <s xml:id="echoid-s3290" xml:space="preserve"><lb/>116160 = 3a<emph style="sub">2</emph>b <lb/>6600 = 3ab<emph style="sub">2</emph> <lb/>125 = b<emph style="sub">3</emph> <lb/>116821 <lb/>25 = 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph> <lb/>2349675 = 3a<emph style="sub">2</emph>, troiſieme diviſeur. </s> <s xml:id="echoid-s3291" xml:space="preserve"><lb/>7049025 = 3a<emph style="sub">2</emph>b <lb/>23895 = 3ab<emph style="sub">2</emph> <lb/>27= b<emph style="sub">3</emph> <lb/>705141477= 3a<emph style="sub">2</emph>b + 3ab<emph style="sub">2</emph> + b<emph style="sub">3</emph>.</s> <s xml:id="echoid-s3292" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div180" type="section" level="1" n="153"> <head xml:id="echoid-head170" style="it" xml:space="preserve">Démonſtration de la Racine Cube.</head> <p> <s xml:id="echoid-s3293" xml:space="preserve">185. </s> <s xml:id="echoid-s3294" xml:space="preserve">Le cube d’un nombre quelconque peut être regardé <lb/>comme celui d’un binome, dont le premier terme repréſente <lb/>cous les chiffres, excepté le premier à droite, & </s> <s xml:id="echoid-s3295" xml:space="preserve">le ſecond <lb/>repréſente ce dernier. </s> <s xml:id="echoid-s3296" xml:space="preserve">Or le cube d’un binome contient le <lb/>cube du premier terme, le triple du quarré du premier par le <lb/>ſecond, le triple du premier par le quarré du ſecond, & </s> <s xml:id="echoid-s3297" xml:space="preserve">le <lb/>cube du ſecond: </s> <s xml:id="echoid-s3298" xml:space="preserve">ainſi il n’y a qu’à faire voir que par la mé-<lb/>thode propoſée on détermine toutes ces parties, dont le cube <lb/>eſt compoſé; </s> <s xml:id="echoid-s3299" xml:space="preserve">c’eſt ce qu’il eſt aiſé de reconnoître: </s> <s xml:id="echoid-s3300" xml:space="preserve">car dans le <lb/>premier exemple, lorſque je poſe 4 à la racine cube, comme <pb o="104" file="0142" n="142" rhead="NOUVEAU COURS"/> je ſçais qu’il doit y avoir deux chiffres, c’eſt réellement 40 que <lb/>je poſe, dont le cube eſt 64000, que je retranche de 10383, <lb/>& </s> <s xml:id="echoid-s3301" xml:space="preserve">le reſte eſt 39823. </s> <s xml:id="echoid-s3302" xml:space="preserve">Je triple enſuite le quarré de 4, & </s> <s xml:id="echoid-s3303" xml:space="preserve">je di-<lb/>viſe 398 par 48, comme ſi je diviſois 39823 par 4800, puiſ-<lb/>que le 8 eſt poſé ſous le premier chiffre de la ſeconde tranche. <lb/></s> <s xml:id="echoid-s3304" xml:space="preserve">Or il eſt certain que le quotient qui doit me venir eſt le ſecond <lb/>terme de la racine, puiſque le triple du quarré du premier ter-<lb/>me par le ſecond doit avoir deux chiffres après lui: </s> <s xml:id="echoid-s3305" xml:space="preserve">d’ailleurs <lb/>j’ôte encore le triple du quarré du ſecond par le premier, par <lb/>la maniere dont je poſe le produit du triple du premier terme <lb/>par le quarré du ſecond, en l’avançant d’un rang vers la droite, <lb/>puiſque ce produit ne doit avoir qu’un chiffre après lui, & </s> <s xml:id="echoid-s3306" xml:space="preserve"><lb/>enfin j’ôte le cube du ſecond terme. </s> <s xml:id="echoid-s3307" xml:space="preserve">D’où il ſuit que j’ai ôté <lb/>du nombre propoſé toutes les parties qui forment un cube, & </s> <s xml:id="echoid-s3308" xml:space="preserve"><lb/>ſi le cube eſt parfait, il ne doit rien reſter après la ſouſtraction <lb/>de la ſomme de ces trois produits. </s> <s xml:id="echoid-s3309" xml:space="preserve">Si le cube eſt imparfait, <lb/>on prend toujours le plus approchant, à quelque défaut près, <lb/>mais on eſt aſſuré qu’il ne s’en faut pas d’une unité que la ra-<lb/>cine ne ſoit celle qu’on cherche par l’épreuve que l’on fait, <lb/>puiſque ſi l’on augmentoit d’une unité, le cube de la racine <lb/>ſeroit plus grand que le nombre propoſé.</s> <s xml:id="echoid-s3310" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3311" xml:space="preserve">On appliquera le même raiſonnement à une racine de tant <lb/>de chiffres que l’on voudra, puiſque l’on peut regarder les chif-<lb/>fres trouvés comme le premier terme de la racine, & </s> <s xml:id="echoid-s3312" xml:space="preserve">celui qui <lb/>reſte à trouver comme le ſecond, en regardant le nombre <lb/>propoſé comme s’il ne contenoit que deux tranches.</s> <s xml:id="echoid-s3313" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3314" xml:space="preserve">La preuve de l’extraction des racines quarrées & </s> <s xml:id="echoid-s3315" xml:space="preserve">cubiques <lb/>ſe fait en élevant les racines trouvées au quarré ou au cube: <lb/></s> <s xml:id="echoid-s3316" xml:space="preserve">ſi le nombre propoſé étoit un quarré ou un cube parfait, on <lb/>doit trouver en multipliant la racine une ou deux fois par elle-<lb/>même un nombre égal au premier; </s> <s xml:id="echoid-s3317" xml:space="preserve">ſi les nombres ne ſont pas <lb/>des quarrés ou des cubes parfaits, en ajoutant le reſte avec la <lb/>même puiſſance de la racine, on doit retrouver le nombre <lb/>propoſé.</s> <s xml:id="echoid-s3318" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div181" type="section" level="1" n="154"> <head xml:id="echoid-head171" style="it" xml:space="preserve">De l’Extraction des Racines quarrées & cubiques, des Fractions <lb/>numériques.</head> <p> <s xml:id="echoid-s3319" xml:space="preserve">186. </s> <s xml:id="echoid-s3320" xml:space="preserve">Pour extraire la racine quarrée d’une fraction numé-<lb/>rique, il faut extraire la racine du numérateur & </s> <s xml:id="echoid-s3321" xml:space="preserve">du dénomi- <pb o="105" file="0143" n="143" rhead="DE MATHÉMATIQUE. Liv. II."/> nateur, & </s> <s xml:id="echoid-s3322" xml:space="preserve">des deux racines en faire une nouvelle fraction, <lb/>qui ſera la fraction demandée: </s> <s xml:id="echoid-s3323" xml:space="preserve">ainſi la racine de {16/25} eſt {4/5}, & </s> <s xml:id="echoid-s3324" xml:space="preserve"><lb/>ainſi des autres. </s> <s xml:id="echoid-s3325" xml:space="preserve">La raiſon eſt, que l’on éleve une fraction au <lb/>quarré, en multipliant le numérateur par lu-même, ainſi que <lb/>le dénominateur. </s> <s xml:id="echoid-s3326" xml:space="preserve">Ainſi pour en extraire la racine, il faut <lb/>prendre celle du numérateur & </s> <s xml:id="echoid-s3327" xml:space="preserve">du dénominateur.</s> <s xml:id="echoid-s3328" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3329" xml:space="preserve">187. </s> <s xml:id="echoid-s3330" xml:space="preserve">Quand le dénominateur de la fraction n’eſt pas un <lb/>quarré, on multiplie le numérateur & </s> <s xml:id="echoid-s3331" xml:space="preserve">le dénominateur par ce <lb/>même dénominateur: </s> <s xml:id="echoid-s3332" xml:space="preserve">de cette maniere la fraction n’a pas <lb/>changé de valeur, & </s> <s xml:id="echoid-s3333" xml:space="preserve">de plus le dénominateur eſt un quarré <lb/>parfait, ce qui contribue beaucoup à déterminer exactement <lb/>la valeur de la racine fractionnaire. </s> <s xml:id="echoid-s3334" xml:space="preserve">Ainſi pour extraire la ra-<lb/>cine quarrée de {3/8}, je multiplie 3, & </s> <s xml:id="echoid-s3335" xml:space="preserve">8 par 8 pour avoir la frac-<lb/>tion {24/64}, dont la racine eſt à peu près {5/8}, puiſqu’en l’élevant au <lb/>quarré il vient {25/64}, qui ne differe de la fraction {3/8} que de {1/64}. </s> <s xml:id="echoid-s3336" xml:space="preserve">De <lb/>même la racine de {3/5} ou de {15/25} eſt {4/5}, ou à peu près: </s> <s xml:id="echoid-s3337" xml:space="preserve">quand on <lb/>veut les avoir encore plus exactement, il faut chercher une <lb/>fraction décimale égale à la fraction propoſée, & </s> <s xml:id="echoid-s3338" xml:space="preserve">en extraire <lb/>la racine, ſuivant les regles ordinaires.</s> <s xml:id="echoid-s3339" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3340" xml:space="preserve">188. </s> <s xml:id="echoid-s3341" xml:space="preserve">De même pour extraire la racine cube d’une fraction <lb/>numérique, il faudra chercher celle du numérateur & </s> <s xml:id="echoid-s3342" xml:space="preserve">celle du <lb/>dénominateur: </s> <s xml:id="echoid-s3343" xml:space="preserve">par exemple, la racine de {216/64} eſt {6/4} ou {3/2}; </s> <s xml:id="echoid-s3344" xml:space="preserve">de <lb/>même celle de {512/729} eſt {8/9}. </s> <s xml:id="echoid-s3345" xml:space="preserve">Si le dénominateur n’étoit pas un cube <lb/>parfait, on multiplieroit les deux termes de la fraction par le <lb/>quarré du même dénominateur pour avoir la racine cube que <lb/>l’on demande avec plus de préciſion; </s> <s xml:id="echoid-s3346" xml:space="preserve">tout ceci eſt évident par <lb/>la formation des puiſſances des fractions.</s> <s xml:id="echoid-s3347" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div182" type="section" level="1" n="155"> <head xml:id="echoid-head172" style="it" xml:space="preserve">Fin du premier Livre.</head> <figure> <image file="0143-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0143-01"/> </figure> <pb o="106" file="0144" n="144"/> </div> <div xml:id="echoid-div183" type="section" level="1" n="156"> <head xml:id="echoid-head173" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head174" xml:space="preserve">LIVRE SECOND,</head> <p style="it"> <s xml:id="echoid-s3348" xml:space="preserve">Où l’on traite des raiſons ou rapports, proportions & </s> <s xml:id="echoid-s3349" xml:space="preserve">pro-<lb/>greſſions géométriques & </s> <s xml:id="echoid-s3350" xml:space="preserve">arithmétiques, des Logarithmes, <lb/>de la réſolution analytique des Problêmes du premier & </s> <s xml:id="echoid-s3351" xml:space="preserve"><lb/>ſecond degré, & </s> <s xml:id="echoid-s3352" xml:space="preserve">de leurs opérations.</s> <s xml:id="echoid-s3353" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div184" type="section" level="1" n="157"> <head xml:id="echoid-head175" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s3354" xml:space="preserve">189. </s> <s xml:id="echoid-s3355" xml:space="preserve">ON appelle homogenes les grandeurs de même nature, <lb/>comme deux lignes, deux ſurfaces ou deux ſolides, deux eſpaces <lb/>ou deux tems, &</s> <s xml:id="echoid-s3356" xml:space="preserve">c.</s> <s xml:id="echoid-s3357" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3358" xml:space="preserve">190. </s> <s xml:id="echoid-s3359" xml:space="preserve">Les grandeurs qui ne ſont pas de même nature, ſont <lb/>appellées grandeurs hétérogenes: </s> <s xml:id="echoid-s3360" xml:space="preserve">ainſi une toiſe & </s> <s xml:id="echoid-s3361" xml:space="preserve">une livre de <lb/>monnoie ſont des grandeurs hétérogenes: </s> <s xml:id="echoid-s3362" xml:space="preserve">ainſi qu’une ligne <lb/>& </s> <s xml:id="echoid-s3363" xml:space="preserve">une ſurface, ou bien un ſolide & </s> <s xml:id="echoid-s3364" xml:space="preserve">un tems, parce ces gran-<lb/>deurs ne peuvent pas ſe contenir l’une l’autre, n’étant pas de <lb/>même nature.</s> <s xml:id="echoid-s3365" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3366" xml:space="preserve">191. </s> <s xml:id="echoid-s3367" xml:space="preserve">On appelle raiſon ou rapport de deux ou de pluſieurs <lb/>grandeurs, la comparaiſon que l’on peut faire de ces grandeurs <lb/>entr’elles. </s> <s xml:id="echoid-s3368" xml:space="preserve">Ainſi pour déterminer combien il peut y avoir de <lb/>ſortes de raiſons ou de rapports, il faut examiner en combien <lb/>de manieres on peut comparer une grandeur à une autre.</s> <s xml:id="echoid-s3369" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3370" xml:space="preserve">192. </s> <s xml:id="echoid-s3371" xml:space="preserve">1°. </s> <s xml:id="echoid-s3372" xml:space="preserve">On peut comparer une grandeur à une autre, en <lb/>examinant combien cette grandeur ſurpeſſe celle à laquelle on <lb/>la compare, ou de combien elle en eſt ſurpaſſée, & </s> <s xml:id="echoid-s3373" xml:space="preserve">cette com-<lb/>paraiſon eſt appellée raiſon ou rapport arithmétique. </s> <s xml:id="echoid-s3374" xml:space="preserve">Ainſi ſi je <pb o="107" file="0145" n="145" rhead="NOUVEAU COURS DE MATHÉM. Liv. II."/> conſidere de combien 15 eſt plus grand que 5, le nombre 10 <lb/>que je trouve, en retranchant 5 de 15, eſt le rapport arith-<lb/>métique de 15 à 5, que l’on marque ordinairement ainſi, <lb/>15 - 5; </s> <s xml:id="echoid-s3375" xml:space="preserve">& </s> <s xml:id="echoid-s3376" xml:space="preserve">de même en Algebre a - b eſt le rapport arith-<lb/>métique de a à b. </s> <s xml:id="echoid-s3377" xml:space="preserve">D’où il ſuit qu’en général on peut toujours <lb/>connoître le rapport arithmétique de deux grandeurs par la <lb/>Souſtraction, puiſque c’eſt par cette opération que l’on peut <lb/>connoître de combien l’une ſurpaſſe l’autre.</s> <s xml:id="echoid-s3378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3379" xml:space="preserve">193. </s> <s xml:id="echoid-s3380" xml:space="preserve">2°. </s> <s xml:id="echoid-s3381" xml:space="preserve">On peut comparer une grandeur à une autre, en <lb/>examinant combien l’une contient l’autre, ou y eſt contenue, <lb/>& </s> <s xml:id="echoid-s3382" xml:space="preserve">cette comparaiſon eſt appellée rapport géométrique. </s> <s xml:id="echoid-s3383" xml:space="preserve">Ainſi <lb/>dans la comparaiſon que je fais de 12 à 4, je puis examiner <lb/>combien de fois 12 contient 4; </s> <s xml:id="echoid-s3384" xml:space="preserve">& </s> <s xml:id="echoid-s3385" xml:space="preserve">dans celle de a à b, je puis <lb/>examiner combien de fois a contient b, & </s> <s xml:id="echoid-s3386" xml:space="preserve">comme on ne le peut <lb/>ſçavoir que par la Diviſion, ce rapport ſe marque ainſi, {12/4}, <lb/>{a/b}; </s> <s xml:id="echoid-s3387" xml:space="preserve">car on peut prendre une diviſion indiquée pour la diviſion <lb/>même, ou pour le quotient qui réſulte de leur diviſion. </s> <s xml:id="echoid-s3388" xml:space="preserve">Ainſi <lb/>lorſqu’il eſt beſoin, on peut ſe ſervir de ces termes, diviſiòn <lb/>indiquée, quotient, fraction, raiſon ou rapport géométrique, puiſ-<lb/>que tous ſignifient la même choſe ou le même nombre. </s> <s xml:id="echoid-s3389" xml:space="preserve">Le <lb/>quotient de 12 diviſé par 4 eſt 3; </s> <s xml:id="echoid-s3390" xml:space="preserve">la fraction {12/4} eſt 3, le rap-<lb/>port géométrique de 12 à 4 eſt encore 3. </s> <s xml:id="echoid-s3391" xml:space="preserve">Il faut remarquer <lb/>encore que comme l’on ſe ſert plus communément dans les <lb/>Mathématiques de rapport géométrique, on dit tout ſimple-<lb/>ment rapport, pour exprimer le rapport géométrique de deux <lb/>grandeurs.</s> <s xml:id="echoid-s3392" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3393" xml:space="preserve">194. </s> <s xml:id="echoid-s3394" xml:space="preserve">Les grandeurs qui ont entr’elles un rapport de nom-<lb/>bre à nombre, ſont appellées commenſurables, parce qu’elles <lb/>ont au moins l’unité pour commune meſure: </s> <s xml:id="echoid-s3395" xml:space="preserve">par exemple, <lb/>une ligne de quatre pieds eſt dite commenſurable avec une <lb/>ligne de neuf pieds, parce que le rapport de ces deux lignes <lb/>eſt celui des deux nombres 4 & </s> <s xml:id="echoid-s3396" xml:space="preserve">9.</s> <s xml:id="echoid-s3397" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3398" xml:space="preserve">195. </s> <s xml:id="echoid-s3399" xml:space="preserve">Les grandeurs qui n’ont point un rapport de nombre <lb/>à nombre, ou qui ne peuvent avoir de meſures communes, ſi <lb/>petites qu’elles ſoient, ſont nommées incommenſurables. </s> <s xml:id="echoid-s3400" xml:space="preserve">Par <lb/>exemple, ſi l’on a un quarré de 16 pieds, & </s> <s xml:id="echoid-s3401" xml:space="preserve">un autre de 32 <lb/>pieds, la racine du premier quarré ſera incommenſurable avec <lb/>celle du ſecond: </s> <s xml:id="echoid-s3402" xml:space="preserve">car comme 32 n’eſt point un quarré parfait, <lb/>ſi près que l’on puiſſe approcher de ce nombre, il y aura tou- <pb o="108" file="0146" n="146" rhead="NOUVEAU COURS"/> jours quelque reſte; </s> <s xml:id="echoid-s3403" xml:space="preserve">& </s> <s xml:id="echoid-s3404" xml:space="preserve">cette racine ſera incommenſurable avec <lb/>celle de 16, puiſque l’on ne pourra jamais la déterminer exac-<lb/>tement.</s> <s xml:id="echoid-s3405" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3406" xml:space="preserve">196. </s> <s xml:id="echoid-s3407" xml:space="preserve">Dans un rapport quelconque arithmétique ou géomé-<lb/>trique, il y a toujours deux termes, le premier eſt appellé anté-<lb/>cédent, & </s> <s xml:id="echoid-s3408" xml:space="preserve">le ſecond conſéquent; </s> <s xml:id="echoid-s3409" xml:space="preserve">dans le rapport de 12 à 4, 12 <lb/>eſt l’antécédent, & </s> <s xml:id="echoid-s3410" xml:space="preserve">4 eſt le conſéquent; </s> <s xml:id="echoid-s3411" xml:space="preserve">dans celui de a à b, <lb/>a eſt antécédent, & </s> <s xml:id="echoid-s3412" xml:space="preserve">b conſéquent.</s> <s xml:id="echoid-s3413" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3414" xml:space="preserve">197. </s> <s xml:id="echoid-s3415" xml:space="preserve">Une raiſon eſt égale à une autre, quand l’antécédent <lb/>de l’une contient autant de fois ſon conſéquent que l’antécé-<lb/>dent de l’autre contient le ſien. </s> <s xml:id="echoid-s3416" xml:space="preserve">Par exemple, la raiſon de 12 <lb/>à 4 eſt égale à celle de 15 à 5, parce que 12 contient 4 autant <lb/>de fois que 15 contient 5, ſçavoir trois fois. </s> <s xml:id="echoid-s3417" xml:space="preserve">Cette égalité de <lb/>raiſon ſe marque quelquefois ainſi, {12/4} = {15/5}; </s> <s xml:id="echoid-s3418" xml:space="preserve">& </s> <s xml:id="echoid-s3419" xml:space="preserve">ſi a a même <lb/>rapport avec b que c avec d, l’on peut encore exprimer cette <lb/>égalité de rapport, en mettant {a/b} = {c/d}, qui fait voir que les <lb/>quatre grandeurs a b & </s> <s xml:id="echoid-s3420" xml:space="preserve">c d forment deux rapports géométri-<lb/>ques égaux.</s> <s xml:id="echoid-s3421" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3422" xml:space="preserve">198. </s> <s xml:id="echoid-s3423" xml:space="preserve">Comme cette expreſſion {12/4} ou {a/b} repréſentent égale-<lb/>ment des rapports géométriques des diviſions & </s> <s xml:id="echoid-s3424" xml:space="preserve">des fractions: <lb/></s> <s xml:id="echoid-s3425" xml:space="preserve">on remarquera que lorſqu’il s’agira de rapport, on appellera le <lb/>terme qui eſt au deſſus de la ligne, antécédent, & </s> <s xml:id="echoid-s3426" xml:space="preserve">le terme qui <lb/>eſt au deſſous, conſéquent; </s> <s xml:id="echoid-s3427" xml:space="preserve">& </s> <s xml:id="echoid-s3428" xml:space="preserve">que quand il s’agira de diviſion, <lb/>le premier ſera appellé dividende, & </s> <s xml:id="echoid-s3429" xml:space="preserve">le ſecond diviſeur; </s> <s xml:id="echoid-s3430" xml:space="preserve">& </s> <s xml:id="echoid-s3431" xml:space="preserve"><lb/>qu’enfin lorſqu’il s’agira de fraction, le premier ſera appellé <lb/>numérateur, & </s> <s xml:id="echoid-s3432" xml:space="preserve">le ſecond dénominateur.</s> <s xml:id="echoid-s3433" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3434" xml:space="preserve">199. </s> <s xml:id="echoid-s3435" xml:space="preserve">On appelle raiſon d’égalité celle où l’antécédent eſt <lb/>égal au conſéquent, & </s> <s xml:id="echoid-s3436" xml:space="preserve">raiſon d’inégalité, lorſque les deux <lb/>termes ſont inégaux; </s> <s xml:id="echoid-s3437" xml:space="preserve">ce qui peut arriver de deux manieres: <lb/></s> <s xml:id="echoid-s3438" xml:space="preserve">la premiere, quand l’antécédent eſt plus grand que le conſé-<lb/>quent, & </s> <s xml:id="echoid-s3439" xml:space="preserve">pour lors on nomme cette raiſon, raiſon de plus <lb/>grande inégalité; </s> <s xml:id="echoid-s3440" xml:space="preserve">& </s> <s xml:id="echoid-s3441" xml:space="preserve">lorſque l’antécédent eſt plus petit que le <lb/>conſéquent, on l’appelle raiſon de moindre inégalité.</s> <s xml:id="echoid-s3442" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3443" xml:space="preserve">200. </s> <s xml:id="echoid-s3444" xml:space="preserve">Deux rapports égaux forment ce que l’on appelle une <lb/>proportion; </s> <s xml:id="echoid-s3445" xml:space="preserve">ſi les deux rapports égaux ſont arithmétiques, la <lb/>proportion eſt arithmétique; </s> <s xml:id="echoid-s3446" xml:space="preserve">ſi les deux rapports égaux ſont <lb/>géométriques, la proportion eſt géométrique. </s> <s xml:id="echoid-s3447" xml:space="preserve">Ainſi dans <lb/>toute proportion il y a quatre termes, puiſque chacun des <lb/>deux rapports en a deux. </s> <s xml:id="echoid-s3448" xml:space="preserve">Il y a proportion arithmétique entre <pb o="109" file="0147" n="147" rhead="DE MATHÉMATIQUE. Liv. II."/> quatre grandes, lorſque la premiere ſurpaſſe la ſeconde autant <lb/>que la troiſieme ſurpaſſe la quatrieme, ou bien lorſque la ſe-<lb/>conde ſurpaſſe la premiere autant que la quatrieme ſurpaſſe la <lb/>troiſieme. </s> <s xml:id="echoid-s3449" xml:space="preserve">Ainſi ces quatre nombre 9 7, 5 3 forment une <lb/>proportion arithmétique, que l’on peut marquer ainſi, 9 - 7 <lb/>= 5 - 3, ou 2 = 2. </s> <s xml:id="echoid-s3450" xml:space="preserve">Mais on la marque plus communément <lb/>de cette maniere, 9. </s> <s xml:id="echoid-s3451" xml:space="preserve">7: </s> <s xml:id="echoid-s3452" xml:space="preserve">5. </s> <s xml:id="echoid-s3453" xml:space="preserve">3, que l’on prononce ainſi, 9 eſt à <lb/>7, comme 5 eſt à 3. </s> <s xml:id="echoid-s3454" xml:space="preserve">Le point qui eſt entre le 9 & </s> <s xml:id="echoid-s3455" xml:space="preserve">le 6 ſignifie <lb/>eſt à, & </s> <s xml:id="echoid-s3456" xml:space="preserve">les deux points qui ſont entre chaque rapport, ſigni-<lb/>fient comme. </s> <s xml:id="echoid-s3457" xml:space="preserve">Le point qui ſépare les deux termes du ſecond <lb/>rapport, ſignifie la même choſe que celui qui eſt entre les deux <lb/>premiers termes 9 & </s> <s xml:id="echoid-s3458" xml:space="preserve">7. </s> <s xml:id="echoid-s3459" xml:space="preserve">La proportion arithmétique ſe mar-<lb/>que de même en Algebre. </s> <s xml:id="echoid-s3460" xml:space="preserve">Si a - b = c - d, on écrit ſi a. </s> <s xml:id="echoid-s3461" xml:space="preserve">b: </s> <s xml:id="echoid-s3462" xml:space="preserve">c. </s> <s xml:id="echoid-s3463" xml:space="preserve">d <lb/>que l’on exprime, en diſant, a eſt à b arithmétiquement, <lb/>comme c eſt à d. </s> <s xml:id="echoid-s3464" xml:space="preserve">Il y a proportion géométrique entre quatre <lb/>nombres, lorſque le premier contient le ſecond, ou y eſt con-<lb/>tenu autant de fois que le troiſieme contient le quatrieme, ou <lb/>y eſt contenu. </s> <s xml:id="echoid-s3465" xml:space="preserve">Ainſi ces quatre nombres 12, 4, 15 & </s> <s xml:id="echoid-s3466" xml:space="preserve">5, ſont <lb/>en proportion géométrique, puiſque 12 contient 4 autant de <lb/>fois que 15 contient 5: </s> <s xml:id="echoid-s3467" xml:space="preserve">cette proportion peut ſe marquer ainſi, <lb/>{12/4} = {15/5}, & </s> <s xml:id="echoid-s3468" xml:space="preserve">cette maniere eſt peut-être la plus naturelle; </s> <s xml:id="echoid-s3469" xml:space="preserve">mais <lb/>le plus communément on la marque ainſi, 12. </s> <s xml:id="echoid-s3470" xml:space="preserve">4 :</s> <s xml:id="echoid-s3471" xml:space="preserve">: 15. </s> <s xml:id="echoid-s3472" xml:space="preserve">5, <lb/>c’eſt-à-dire que 12 eſt à 4 géométriquement, comme 15 eſt à <lb/>5. </s> <s xml:id="echoid-s3473" xml:space="preserve">La proportion géométrique ſe marque de même en Al-<lb/>gebre: </s> <s xml:id="echoid-s3474" xml:space="preserve">ainſi ſi a contient b autant de fois que c contient d, <lb/>on écrit a. </s> <s xml:id="echoid-s3475" xml:space="preserve">b :</s> <s xml:id="echoid-s3476" xml:space="preserve">: c. </s> <s xml:id="echoid-s3477" xml:space="preserve">d.</s> <s xml:id="echoid-s3478" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3479" xml:space="preserve">201. </s> <s xml:id="echoid-s3480" xml:space="preserve">Une proportion arithmétique ou géométrique eſt ap-<lb/>pellée diſcrete, lorſque les quatre termes ſont quatre gran-<lb/>deurs différentes; </s> <s xml:id="echoid-s3481" xml:space="preserve">& </s> <s xml:id="echoid-s3482" xml:space="preserve">lorſque dans l’une ou l’autre le même <lb/>nombre eſt conſéquent d’un rapport, & </s> <s xml:id="echoid-s3483" xml:space="preserve">antécédent de l’autre, <lb/>la proportion eſt appellée continue; </s> <s xml:id="echoid-s3484" xml:space="preserve">ainſi ces trois grandeurs <lb/>3, 5, 7 ſont en proportion arithmétique continue, parce que <lb/>l’on a 3. </s> <s xml:id="echoid-s3485" xml:space="preserve">5 : </s> <s xml:id="echoid-s3486" xml:space="preserve">5. </s> <s xml:id="echoid-s3487" xml:space="preserve">7, & </s> <s xml:id="echoid-s3488" xml:space="preserve">cette proportion ſe marque ainſi · 3.</s> <s xml:id="echoid-s3489" xml:space="preserve">5.</s> <s xml:id="echoid-s3490" xml:space="preserve">7 <lb/>que l’on exprime, en diſant, 3 eſt à 5, comme 5 eſt à 7 arith-<lb/>métiquement, afin de la diſtinguer de la proportion diſcrete <lb/>arithmétique, comme celle-ci, 2.</s> <s xml:id="echoid-s3491" xml:space="preserve">4:</s> <s xml:id="echoid-s3492" xml:space="preserve">8.</s> <s xml:id="echoid-s3493" xml:space="preserve">10, & </s> <s xml:id="echoid-s3494" xml:space="preserve">autres ſembla-<lb/>bles. </s> <s xml:id="echoid-s3495" xml:space="preserve">De même ces trois grandeurs 18, 6, 2 forment une pro-<lb/>portion géométrique continue, parce que 18. </s> <s xml:id="echoid-s3496" xml:space="preserve">6 :</s> <s xml:id="echoid-s3497" xml:space="preserve">: 6. </s> <s xml:id="echoid-s3498" xml:space="preserve">2, où <lb/>l’on voit que 6 eſt conſéquent dans le premier rapport, & </s> <s xml:id="echoid-s3499" xml:space="preserve">an-<lb/>récédent dans le ſecond. </s> <s xml:id="echoid-s3500" xml:space="preserve">Pour diſtinguer cette eſpece de pro- <pb o="110" file="0148" n="148" rhead="NOUVEAU COURS"/> portion des autres, on eſt convenu de la marquer ainſi {.</s> <s xml:id="echoid-s3501" xml:space="preserve">./.</s> <s xml:id="echoid-s3502" xml:space="preserve">.} <lb/>18.</s> <s xml:id="echoid-s3503" xml:space="preserve">6.</s> <s xml:id="echoid-s3504" xml:space="preserve">2, de même en Algebre {.</s> <s xml:id="echoid-s3505" xml:space="preserve">./.</s> <s xml:id="echoid-s3506" xml:space="preserve">.} a. </s> <s xml:id="echoid-s3507" xml:space="preserve">b. </s> <s xml:id="echoid-s3508" xml:space="preserve">c marque que les trois <lb/>grandeurs a, b, c forment une progreſſion géométrique.</s> <s xml:id="echoid-s3509" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3510" xml:space="preserve">201. </s> <s xml:id="echoid-s3511" xml:space="preserve">Les quantités qui forment une proportion arithméti-<lb/>que ou géométrique ſont appellées proportionnelles. </s> <s xml:id="echoid-s3512" xml:space="preserve">Le premier <lb/>& </s> <s xml:id="echoid-s3513" xml:space="preserve">le dernier terme d’une proportion quelconque ſont appellés <lb/>extrêmes, & </s> <s xml:id="echoid-s3514" xml:space="preserve">le ſecond & </s> <s xml:id="echoid-s3515" xml:space="preserve">le troiſieme ſont appellés moyens. <lb/></s> <s xml:id="echoid-s3516" xml:space="preserve">Dans les proportions continues arithmétiques ou géométri-<lb/>ques, le terme qui ſert de conſéquent & </s> <s xml:id="echoid-s3517" xml:space="preserve">d’antécédent eſt ap-<lb/>pellé moyen arithmétique ou géométrique.</s> <s xml:id="echoid-s3518" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div185" type="section" level="1" n="158"> <head xml:id="echoid-head176" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s3519" xml:space="preserve">Je crois devoir avertir ici ceux qui commencent la Géo-<lb/>métrie, qu’il eſt de la derniere importance de bien ſçavoir les <lb/>propoſitions de ce ſecond Livre, particuliérement la premiere <lb/>& </s> <s xml:id="echoid-s3520" xml:space="preserve">ſes corollaires, puiſque c’eſt preſque par elle ſeule que ſont <lb/>démontrées toutes les propoſitions où il s’agit de rapport & </s> <s xml:id="echoid-s3521" xml:space="preserve">de <lb/>proportion. </s> <s xml:id="echoid-s3522" xml:space="preserve">Pour leur en faciliter l’intelligence, nous leur <lb/>donnerons pluſieurs démonſtrations de cette propoſition, & </s> <s xml:id="echoid-s3523" xml:space="preserve"><lb/>nous nous arrêterons principalement à celles qui ſont démon-<lb/>trées par des raiſons métaphyſiques.</s> <s xml:id="echoid-s3524" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div186" type="section" level="1" n="159"> <head xml:id="echoid-head177" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head178" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3525" xml:space="preserve">Si quatre grandeurs ſont en proportion géométrique, le produit <lb/>des extrêmes ſera égal à celui des moyens, c’eſt-à-dire que ſi l’on a <lb/>a. </s> <s xml:id="echoid-s3526" xml:space="preserve">b :</s> <s xml:id="echoid-s3527" xml:space="preserve">: c. </s> <s xml:id="echoid-s3528" xml:space="preserve">d, on aura ad = bc.</s> <s xml:id="echoid-s3529" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div187" type="section" level="1" n="160"> <head xml:id="echoid-head179" xml:space="preserve"><emph style="sc">Premiere démonstration</emph>.</head> <p> <s xml:id="echoid-s3530" xml:space="preserve">202. </s> <s xml:id="echoid-s3531" xml:space="preserve">Puiſqu’une proportion n’eſt autre choſe que l’égalité <lb/>de deux rapports, au lieu de l’exprimer ainſi, a. </s> <s xml:id="echoid-s3532" xml:space="preserve">b :</s> <s xml:id="echoid-s3533" xml:space="preserve">: c. </s> <s xml:id="echoid-s3534" xml:space="preserve">d, on <lb/>peut la marquer de cette maniere, {a/b} = {c/d}. </s> <s xml:id="echoid-s3535" xml:space="preserve">Si je multiplie les <lb/>deux termes de cette égalité par une même grandeur bd, je ne <lb/>troublerai point l’égalité; </s> <s xml:id="echoid-s3536" xml:space="preserve">ainſi j’aurai {abd/b} = {cbd/d@}: </s> <s xml:id="echoid-s3537" xml:space="preserve">mais {abd/b} = <lb/>ad, en effaçant la lettre b, commune au numérateur & </s> <s xml:id="echoid-s3538" xml:space="preserve">au dé-<lb/>nominateur; </s> <s xml:id="echoid-s3539" xml:space="preserve">& </s> <s xml:id="echoid-s3540" xml:space="preserve">de même {cbd/d}=6c: </s> <s xml:id="echoid-s3541" xml:space="preserve">donc on aura ad = bc. </s> <s xml:id="echoid-s3542" xml:space="preserve">Ce qui <lb/>prouve que le produit des extrêmes eſt égal au produit des <lb/>moyens. </s> <s xml:id="echoid-s3543" xml:space="preserve">C. </s> <s xml:id="echoid-s3544" xml:space="preserve">Q. </s> <s xml:id="echoid-s3545" xml:space="preserve">F. </s> <s xml:id="echoid-s3546" xml:space="preserve">D.</s> <s xml:id="echoid-s3547" xml:space="preserve"/> </p> <pb o="111" file="0149" n="149" rhead="DE MATHÉMATIQUE. Liv. II."/> </div> <div xml:id="echoid-div188" type="section" level="1" n="161"> <head xml:id="echoid-head180" xml:space="preserve"><emph style="sc">Seconde démonstration</emph>.</head> <p> <s xml:id="echoid-s3548" xml:space="preserve">203. </s> <s xml:id="echoid-s3549" xml:space="preserve">Puiſque l’on a a. </s> <s xml:id="echoid-s3550" xml:space="preserve">b :</s> <s xml:id="echoid-s3551" xml:space="preserve">: c. </s> <s xml:id="echoid-s3552" xml:space="preserve">d, à cauſe de l’égalité des rap-<lb/>ports {a/b}, {c/d}; </s> <s xml:id="echoid-s3553" xml:space="preserve">ſi l’on ſuppoſe que {a/b} = f, on aura auſſi {c/d}=f. </s> <s xml:id="echoid-s3554" xml:space="preserve">Mul-<lb/>tipliant chaque membre de la premiere égalité par b, on aura <lb/>{ab/b} = bf, ou a = bf; </s> <s xml:id="echoid-s3555" xml:space="preserve">multipliant chaque membre de la ſe-<lb/>conde égalité par d, on aura {cd/d} = df, ou c = df: </s> <s xml:id="echoid-s3556" xml:space="preserve">donc en <lb/>mettant dans la proportion a. </s> <s xml:id="echoid-s3557" xml:space="preserve">b :</s> <s xml:id="echoid-s3558" xml:space="preserve">: c. </s> <s xml:id="echoid-s3559" xml:space="preserve">d à la place de a & </s> <s xml:id="echoid-s3560" xml:space="preserve">de c <lb/>ſur valeurs bf & </s> <s xml:id="echoid-s3561" xml:space="preserve">df, on aura bf : </s> <s xml:id="echoid-s3562" xml:space="preserve">b :</s> <s xml:id="echoid-s3563" xml:space="preserve">: df : </s> <s xml:id="echoid-s3564" xml:space="preserve">d, ou le produit des <lb/>extrêmes eſt égal à celui des moyens, puiſque l’un & </s> <s xml:id="echoid-s3565" xml:space="preserve">l’autre <lb/>donne également bdf.</s> <s xml:id="echoid-s3566" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div189" type="section" level="1" n="162"> <head xml:id="echoid-head181" xml:space="preserve"><emph style="sc">Troisieme démonstration</emph>.</head> <p> <s xml:id="echoid-s3567" xml:space="preserve">204. </s> <s xml:id="echoid-s3568" xml:space="preserve">Suppoſons qu’au lieu de la proportion a. </s> <s xml:id="echoid-s3569" xml:space="preserve">b :</s> <s xml:id="echoid-s3570" xml:space="preserve">: c. </s> <s xml:id="echoid-s3571" xml:space="preserve">d on <lb/>me donne celle-ci 12. </s> <s xml:id="echoid-s3572" xml:space="preserve">6 :</s> <s xml:id="echoid-s3573" xml:space="preserve">: 4. </s> <s xml:id="echoid-s3574" xml:space="preserve">2; </s> <s xml:id="echoid-s3575" xml:space="preserve">il faut démontrer pour quelle <lb/>raiſon le produit des moyens 6 x 4 eſt égal au produit des ex-<lb/>trêmes 12 x 2. </s> <s xml:id="echoid-s3576" xml:space="preserve">Pour cela je fais attention que 12 étant double <lb/>de 6; </s> <s xml:id="echoid-s3577" xml:space="preserve">ſi je viens à multiplier 12 & </s> <s xml:id="echoid-s3578" xml:space="preserve">6 par le même nombre 4, <lb/>le produit de 12 par 4 ſera double du produit de 6 par le même <lb/>nombre 4; </s> <s xml:id="echoid-s3579" xml:space="preserve">mais ſi au lieu de multiplier 2 par 4, je multiplie <lb/>ce nombre par un autre, qui ne ſoit que la moitié de 4, il eſt <lb/>néceſſaire que le produit devienne la moitié de celui de 12 par <lb/>4: </s> <s xml:id="echoid-s3580" xml:space="preserve">donc il ſera égal à celui de 6 par 4, puiſqu’il perd autant <lb/>du côté du multiplicateur 2, que le nombre 6 gagne par ſon <lb/>multiplicateur 4. </s> <s xml:id="echoid-s3581" xml:space="preserve">En un mot, 6 n’eſt que la moitié de 12; </s> <s xml:id="echoid-s3582" xml:space="preserve">mais <lb/>par la nature de la proportion, il a un multiplicateur double <lb/>de celui de 12, ce qui fait une compenſation parfaite. </s> <s xml:id="echoid-s3583" xml:space="preserve">On peut <lb/>appliquer ce raiſonnement à tel autre rapport que ce ſoit, ſoit <lb/>numérique, ſoit algébrique. </s> <s xml:id="echoid-s3584" xml:space="preserve">Ainſi notre démonſtration eſt <lb/>générale, parce qu’elle ne dépend pas de l’exemple auquel elle <lb/>eſt appliquée, mais de l’univerſalité des principes ſur leſquels <lb/>elle eſt fondée.</s> <s xml:id="echoid-s3585" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div190" type="section" level="1" n="163"> <head xml:id="echoid-head182" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s3586" xml:space="preserve">205. </s> <s xml:id="echoid-s3587" xml:space="preserve">Il ſuit de cette propoſition, que dans une proportion <lb/>géométrique continue, le produit des extrêmes eſt égalau quarré <lb/>du terme moyen: </s> <s xml:id="echoid-s3588" xml:space="preserve">car ſi l’on a {.</s> <s xml:id="echoid-s3589" xml:space="preserve">./.</s> <s xml:id="echoid-s3590" xml:space="preserve">.} a. </s> <s xml:id="echoid-s3591" xml:space="preserve">b. </s> <s xml:id="echoid-s3592" xml:space="preserve">c, ou bien a. </s> <s xml:id="echoid-s3593" xml:space="preserve">b :</s> <s xml:id="echoid-s3594" xml:space="preserve">: b: </s> <s xml:id="echoid-s3595" xml:space="preserve">c, <lb/>on aura ac = bb.</s> <s xml:id="echoid-s3596" xml:space="preserve"/> </p> <pb o="112" file="0150" n="150" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div191" type="section" level="1" n="164"> <head xml:id="echoid-head183" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s3597" xml:space="preserve">206. </s> <s xml:id="echoid-s3598" xml:space="preserve">Il ſuit encore que connoiſſant les trois termes a, b, c <lb/>d’une proportion, on pourra connoître le quatrieme; </s> <s xml:id="echoid-s3599" xml:space="preserve">car ſi <lb/>l’on nomme x ce quatrieme, l’on aura a. </s> <s xml:id="echoid-s3600" xml:space="preserve">b :</s> <s xml:id="echoid-s3601" xml:space="preserve">: c. </s> <s xml:id="echoid-s3602" xml:space="preserve">x; </s> <s xml:id="echoid-s3603" xml:space="preserve">par con-<lb/>ſéquent ax = bc, ou bien en diviſant chaque membre de l’é-<lb/>galité par a, {ax/a}, ou x = {bc/a}, qui fait voir que pour trouver ce <lb/>quatrieme terme, il faut multiplier le ſecond par le ſecond par le troiſieme, <lb/>& </s> <s xml:id="echoid-s3604" xml:space="preserve">diviſer le produit par le premier.</s> <s xml:id="echoid-s3605" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div192" type="section" level="1" n="165"> <head xml:id="echoid-head184" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s3606" xml:space="preserve">207. </s> <s xml:id="echoid-s3607" xml:space="preserve">Il ſuit encore qu’on peut prendre le produit du ſecond <lb/>& </s> <s xml:id="echoid-s3608" xml:space="preserve">du troiſieme terme d’une proportion diviſé par le premier, <lb/>pour le quatrieme terme de la même proportion: </s> <s xml:id="echoid-s3609" xml:space="preserve">car comme <lb/>x eſt égal à {bc/a}, on pourra avec les trois termes a, b, c écrire <lb/>a. </s> <s xml:id="echoid-s3610" xml:space="preserve">b :</s> <s xml:id="echoid-s3611" xml:space="preserve">: c, {bc/a}, & </s> <s xml:id="echoid-s3612" xml:space="preserve">c’eſt ſur cette proportion qu’eſt fondée la regle, <lb/>appellée Regle de Trois, qui fait trouver le quatrieme terme <lb/>d’une proportion, dont les trois autres ſont connus. </s> <s xml:id="echoid-s3613" xml:space="preserve">Si dans <lb/>une proportion quelconque on connoît trois termes, on pourra <lb/>toujours connoître le quatrieme, de quelque maniere qu’ils <lb/>ſoient diſpoſés.</s> <s xml:id="echoid-s3614" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3615" xml:space="preserve">208. </s> <s xml:id="echoid-s3616" xml:space="preserve">De même dans la proportion continue, connoiſſant <lb/>les deux premiers termes, on pourra connoître le troiſieme, <lb/>en diviſant le quarré du moyen par le premier. </s> <s xml:id="echoid-s3617" xml:space="preserve">Ainſi ayant les <lb/>deux premiers termes a,b de la proportion continue, on aura <lb/>x = {bb/a}, puiſque a. </s> <s xml:id="echoid-s3618" xml:space="preserve">b :</s> <s xml:id="echoid-s3619" xml:space="preserve">: b. </s> <s xml:id="echoid-s3620" xml:space="preserve">{bb/a}.</s> <s xml:id="echoid-s3621" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3622" xml:space="preserve">209. </s> <s xml:id="echoid-s3623" xml:space="preserve">Mais ſi l’on avoit le premier terme a & </s> <s xml:id="echoid-s3624" xml:space="preserve">le troiſieme c, <lb/>& </s> <s xml:id="echoid-s3625" xml:space="preserve">qu’on voulût avoir le terme moyen, que nous appellerons x, <lb/>on multipliera le premier & </s> <s xml:id="echoid-s3626" xml:space="preserve">le troiſieme l’un par l’autre, & </s> <s xml:id="echoid-s3627" xml:space="preserve"><lb/>l’on prendra la racine du produit; </s> <s xml:id="echoid-s3628" xml:space="preserve">cette racine ſera la moyenne <lb/>proportionnelle demandée: </s> <s xml:id="echoid-s3629" xml:space="preserve">car ayant a : </s> <s xml:id="echoid-s3630" xml:space="preserve">x :</s> <s xml:id="echoid-s3631" xml:space="preserve">: x : </s> <s xml:id="echoid-s3632" xml:space="preserve">c, on aura <lb/>xx = ac, & </s> <s xml:id="echoid-s3633" xml:space="preserve">par conſéquent x = √ac\x{0020}</s> </p> <pb o="113" file="0151" n="151" rhead="DE MATHÉMATIQUE. Liv. II."/> </div> <div xml:id="echoid-div193" type="section" level="1" n="166"> <head xml:id="echoid-head185" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head186" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3634" xml:space="preserve">210. </s> <s xml:id="echoid-s3635" xml:space="preserve">Si quatre grandeurs ſont diſpoſées de telle ſorte que le pro-<lb/>duit des extrêmes ſoit égal au produit des moyens, ces quatre gran-<lb/>deurs ſeront proportionnelles.</s> <s xml:id="echoid-s3636" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div194" type="section" level="1" n="167"> <head xml:id="echoid-head187" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3637" xml:space="preserve">Si quatre grandeurs a, b, c, d donnent ad = bc, je dis que <lb/>l’on aura a. </s> <s xml:id="echoid-s3638" xml:space="preserve">b :</s> <s xml:id="echoid-s3639" xml:space="preserve">: c. </s> <s xml:id="echoid-s3640" xml:space="preserve">d, ou bien que {a/b} = {c/d}. </s> <s xml:id="echoid-s3641" xml:space="preserve">Pour le prouver il <lb/>n’y a qu’à diviſer les deux membres de l’équation ad=bc, par <lb/>une même grandeur bd, on aura {ad/bd}={bc/bd}, ou en effaçant les <lb/>lettres communes pour faire la diviſion {a/b}={c/d}. </s> <s xml:id="echoid-s3642" xml:space="preserve">Or comme on <lb/>a diviſé des grandeurs égales par d’autres grandeurs égales, on <lb/>aura des quotients égaux {a/b} & </s> <s xml:id="echoid-s3643" xml:space="preserve">{c/d} qui donnent a. </s> <s xml:id="echoid-s3644" xml:space="preserve">b :</s> <s xml:id="echoid-s3645" xml:space="preserve">: c. </s> <s xml:id="echoid-s3646" xml:space="preserve">d. </s> <s xml:id="echoid-s3647" xml:space="preserve">C.</s> <s xml:id="echoid-s3648" xml:space="preserve">Q.</s> <s xml:id="echoid-s3649" xml:space="preserve">F.</s> <s xml:id="echoid-s3650" xml:space="preserve">D.</s> <s xml:id="echoid-s3651" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3652" xml:space="preserve">211. </s> <s xml:id="echoid-s3653" xml:space="preserve">Ce théorême, qui eſt l’inverſe du précédent, ſert à <lb/>faire voir que quatre grandeurs ſont proportionnelles, en fai-<lb/>ſant voir que le produit des extrêmes eſt égal à celui des moyens: <lb/></s> <s xml:id="echoid-s3654" xml:space="preserve">c’eſt pourquoi il eſt à propos d’être bien prévenu de ce prin-<lb/>cipe, qui ſera le fondement de toutes les démonſtrations al-<lb/>gébriques que nous allons donner.</s> <s xml:id="echoid-s3655" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div195" type="section" level="1" n="168"> <head xml:id="echoid-head188" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s3656" xml:space="preserve">212. </s> <s xml:id="echoid-s3657" xml:space="preserve">Il ſuit de cette propoſition, qu’une équation peut tou-<lb/>jours être regardée comme ayant un de ſes membres formé du <lb/>produit des extrêmes, & </s> <s xml:id="echoid-s3658" xml:space="preserve">l’autre de celui des moyens d’une <lb/>proportion; </s> <s xml:id="echoid-s3659" xml:space="preserve">& </s> <s xml:id="echoid-s3660" xml:space="preserve">que l’on peut même faire une proportion avec <lb/>les racines des produits qui forment chaque membre de l’é-<lb/>quation, comme on le verra ailleurs.</s> <s xml:id="echoid-s3661" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div196" type="section" level="1" n="169"> <head xml:id="echoid-head189" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s3662" xml:space="preserve">213. </s> <s xml:id="echoid-s3663" xml:space="preserve">Il ſuit encore du théorême précédent, que ſi quatre <lb/>grandeurs ſont en proportion géométrique, elles le ſeront <lb/>encore dans les quatre changemens ſuivans, que l’on déſigne <lb/>par ces mots invertendo, alternando, componendo, dividendo, <lb/>& </s> <s xml:id="echoid-s3664" xml:space="preserve">que d’autres appellent en raiſon inverſe, en raiſon alterne, <lb/>compoſition & </s> <s xml:id="echoid-s3665" xml:space="preserve">diviſion.</s> <s xml:id="echoid-s3666" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3667" xml:space="preserve">214. </s> <s xml:id="echoid-s3668" xml:space="preserve">Pour changer une proportion donnée en raiſon in- <pb o="114" file="0152" n="152" rhead="NOUVEAU COURS"/> verſe, l’on met les antécédens à la place des conſéquens, & </s> <s xml:id="echoid-s3669" xml:space="preserve"><lb/>les conſéquens à celle des antécédens, c’eſt-à-dire que ſi <lb/>a. </s> <s xml:id="echoid-s3670" xml:space="preserve">b :</s> <s xml:id="echoid-s3671" xml:space="preserve">: c. </s> <s xml:id="echoid-s3672" xml:space="preserve">d, on aura auſſi b. </s> <s xml:id="echoid-s3673" xml:space="preserve">a :</s> <s xml:id="echoid-s3674" xml:space="preserve">: d. </s> <s xml:id="echoid-s3675" xml:space="preserve">c; </s> <s xml:id="echoid-s3676" xml:space="preserve">ce qui eſt bien évident, <lb/>puiſqu’on vient de voir que les quatre termes d’une propor-<lb/>tion peuvent toujours former une équation; </s> <s xml:id="echoid-s3677" xml:space="preserve">& </s> <s xml:id="echoid-s3678" xml:space="preserve">comme la <lb/>proportion inverſe, auſſi-bien que la directe donne b c = a d; <lb/></s> <s xml:id="echoid-s3679" xml:space="preserve">il s’enſuit qu’en renverſant les termes, cela n’empêche pas <lb/>qu’ils ne ſoient en proportion.</s> <s xml:id="echoid-s3680" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3681" xml:space="preserve">215. </s> <s xml:id="echoid-s3682" xml:space="preserve">Pour changer une proportion en raiſon alterne ou al-<lb/>ternando, on met les moyens à la place les uns des autres ſans <lb/>changer les extrêmes, c’eſt-à-dire que ſi l’on a a. </s> <s xml:id="echoid-s3683" xml:space="preserve">b :</s> <s xml:id="echoid-s3684" xml:space="preserve">: c. </s> <s xml:id="echoid-s3685" xml:space="preserve">d, on <lb/>aura auſſi a. </s> <s xml:id="echoid-s3686" xml:space="preserve">c :</s> <s xml:id="echoid-s3687" xml:space="preserve">: b. </s> <s xml:id="echoid-s3688" xml:space="preserve">d; </s> <s xml:id="echoid-s3689" xml:space="preserve">ce qui eſt bien évident, puiſqu’on a tou-<lb/>jours a d pour le produit des extrêmes, & </s> <s xml:id="echoid-s3690" xml:space="preserve">b c pour le produit <lb/>des moyens; </s> <s xml:id="echoid-s3691" xml:space="preserve">& </s> <s xml:id="echoid-s3692" xml:space="preserve">que ces produits ſont égaux, à cauſe de la pre-<lb/>miere proportion a. </s> <s xml:id="echoid-s3693" xml:space="preserve">b :</s> <s xml:id="echoid-s3694" xml:space="preserve">: c. </s> <s xml:id="echoid-s3695" xml:space="preserve">d qui donne a d = b c.</s> <s xml:id="echoid-s3696" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3697" xml:space="preserve">216. </s> <s xml:id="echoid-s3698" xml:space="preserve">Pour changer une proportion en compoſant ou com-<lb/>ponendo, on ajoute le conſéquent à l’antécédent, & </s> <s xml:id="echoid-s3699" xml:space="preserve">l’on com-<lb/>pare la ſomme au conſéquent ou à l’antécédent: </s> <s xml:id="echoid-s3700" xml:space="preserve">on fait la <lb/>même opération pour chaque rapport, c’eſt-à-dire que ſi l’on <lb/>a a. </s> <s xml:id="echoid-s3701" xml:space="preserve">b :</s> <s xml:id="echoid-s3702" xml:space="preserve">: c. </s> <s xml:id="echoid-s3703" xml:space="preserve">d, on aura auſſi a + b. </s> <s xml:id="echoid-s3704" xml:space="preserve">b :</s> <s xml:id="echoid-s3705" xml:space="preserve">: c + d. </s> <s xml:id="echoid-s3706" xml:space="preserve">d; </s> <s xml:id="echoid-s3707" xml:space="preserve">ce <lb/>qui ſera évident, ſi l’on fait voir que ces quatre termes don-<lb/>nent un produit des extrêmes égal au produit des moyens. </s> <s xml:id="echoid-s3708" xml:space="preserve">Le <lb/>produit des extrêmes eſt a d + b d, & </s> <s xml:id="echoid-s3709" xml:space="preserve">celui des moyens eſt <lb/>b c + b d, évidemment égal au premier, puiſque la proportion <lb/>primitive donne a d = b c, & </s> <s xml:id="echoid-s3710" xml:space="preserve">que b d eſt égal dans l’un & </s> <s xml:id="echoid-s3711" xml:space="preserve">dans <lb/>l’autre.</s> <s xml:id="echoid-s3712" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3713" xml:space="preserve">217. </s> <s xml:id="echoid-s3714" xml:space="preserve">Le changement appellé dividendo, que l’on pourroit <lb/>nommer avec plus de raiſon detrahendo ou de ſouſtraction, ſe <lb/>fait en ôtant le conſéquent de l’antécédent, dans chaque rap-<lb/>port, & </s> <s xml:id="echoid-s3715" xml:space="preserve">en comparant chaque différence à l’antécédent, ou au <lb/>conſéquent: </s> <s xml:id="echoid-s3716" xml:space="preserve">par exemple, ſi l’on a a. </s> <s xml:id="echoid-s3717" xml:space="preserve">b :</s> <s xml:id="echoid-s3718" xml:space="preserve">: c. </s> <s xml:id="echoid-s3719" xml:space="preserve">d, on aura auſſi <lb/>a - b. </s> <s xml:id="echoid-s3720" xml:space="preserve">b :</s> <s xml:id="echoid-s3721" xml:space="preserve">: c - d. </s> <s xml:id="echoid-s3722" xml:space="preserve">d, ou a. </s> <s xml:id="echoid-s3723" xml:space="preserve">a - b :</s> <s xml:id="echoid-s3724" xml:space="preserve">: c. </s> <s xml:id="echoid-s3725" xml:space="preserve">c - d: </s> <s xml:id="echoid-s3726" xml:space="preserve">car dans l’un <lb/>& </s> <s xml:id="echoid-s3727" xml:space="preserve">dans l’autre, le produit des moyens eſt égal au produit des <lb/>extrêmes. </s> <s xml:id="echoid-s3728" xml:space="preserve">Dans le premier cas, le produit des moyens eſt <lb/>b c - b d, & </s> <s xml:id="echoid-s3729" xml:space="preserve">celui des extrêmes eſt a d - b d égal au premier: <lb/></s> <s xml:id="echoid-s3730" xml:space="preserve">dans le ſecond, le produit des moyens eſt a c - b c, & </s> <s xml:id="echoid-s3731" xml:space="preserve">celui <lb/>des extrêmes a c - a d évidemment égal à l’autre, puiſque les <lb/>termes de l’un ſont égaux aux termes de l’autre; </s> <s xml:id="echoid-s3732" xml:space="preserve">car a c = a c, <lb/>& </s> <s xml:id="echoid-s3733" xml:space="preserve">a d = b c par la proportion a. </s> <s xml:id="echoid-s3734" xml:space="preserve">b :</s> <s xml:id="echoid-s3735" xml:space="preserve">: c. </s> <s xml:id="echoid-s3736" xml:space="preserve">d.</s> <s xml:id="echoid-s3737" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3738" xml:space="preserve">218. </s> <s xml:id="echoid-s3739" xml:space="preserve">Il y a encore beaucoup d’autres changemens différens <pb o="115" file="0153" n="153" rhead="DE MATHÉMATIQUE. Liv. II."/> de ceux-ci, que l’on peut faire dans une proportion ſans la dé-<lb/>truire, mais qui réſultent de la combinaiſon de ces premiers, & </s> <s xml:id="echoid-s3740" xml:space="preserve"><lb/>dont l’uſage eſt moins fréquent dans les Mathématiques: </s> <s xml:id="echoid-s3741" xml:space="preserve">il ſuffit <lb/>d’avoir la regle générale pour reconnoître ſi les changemens <lb/>que l’on fait ne détruiſent point la proportion; </s> <s xml:id="echoid-s3742" xml:space="preserve">& </s> <s xml:id="echoid-s3743" xml:space="preserve">pour cela <lb/>il n’y a qu’à examiner dans tous les cas ſi le produit des extrê-<lb/>mes eſt égal à celui des moyens.</s> <s xml:id="echoid-s3744" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3745" xml:space="preserve">Nous allons donner un eſpece de tableau de ces change-<lb/>mens, en nombres & </s> <s xml:id="echoid-s3746" xml:space="preserve">en lettres, pour que l’on puiſſe plus aiſé-<lb/>ment ſe les graver dans la mémoire.</s> <s xml:id="echoid-s3747" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3748" xml:space="preserve">Si l’on a a. </s> <s xml:id="echoid-s3749" xml:space="preserve">b :</s> <s xml:id="echoid-s3750" xml:space="preserve">: c. </s> <s xml:id="echoid-s3751" xml:space="preserve">d, on aura</s> </p> <note position="right" xml:space="preserve">Invertendo # b. a :: d. c, ou d. c :: b. a. <lb/>Alternando # a. c :: b. d. <lb/>Componendo # a + b. a :: c + d. d, ou a. a + b :: c. c+ d. <lb/>Dividendo # a - b. a :: c - d. d, ou a. a - b ::c. c - d. <lb/></note> </div> <div xml:id="echoid-div197" type="section" level="1" n="170"> <head xml:id="echoid-head190" xml:space="preserve">En nombres.</head> <p> <s xml:id="echoid-s3752" xml:space="preserve">Si 3. </s> <s xml:id="echoid-s3753" xml:space="preserve">4 :</s> <s xml:id="echoid-s3754" xml:space="preserve">: 6. </s> <s xml:id="echoid-s3755" xml:space="preserve">8, on aura</s> </p> <note position="right" xml:space="preserve">Invertendo # 4. 3 :: 8. 6, ou 8. 6 :: 4. 3. <lb/>Alternando # 3. 6 :: 4. 8. <lb/>Componendo # 3. 7 :: 6. 14, ou 7. 4 :: 14. 8. <lb/>Dividendo # 3. 4-3 :: 8. 8-6, ou 3.1 :: 6. 2. <lb/></note> <p> <s xml:id="echoid-s3756" xml:space="preserve">Dans les deux premiers changemens, le produit des extrê-<lb/>mes & </s> <s xml:id="echoid-s3757" xml:space="preserve">des moyens ſont les mêmes que ceux que donnent la <lb/>proportion; </s> <s xml:id="echoid-s3758" xml:space="preserve">& </s> <s xml:id="echoid-s3759" xml:space="preserve">dans les autres, les produits des extrêmes & </s> <s xml:id="echoid-s3760" xml:space="preserve"><lb/>des moyens ſont ſimplement égaux, ſans être les mêmes que <lb/>ceux de la proportion primitive.</s> <s xml:id="echoid-s3761" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div198" type="section" level="1" n="171"> <head xml:id="echoid-head191" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head192" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3762" xml:space="preserve">219. </s> <s xml:id="echoid-s3763" xml:space="preserve">Lorſque deux raiſons ont un même rapport à une troiſieme, <lb/>ces deux raiſons ſont égales entr’elles, c’eſt-à-dire que ſi l’on a <lb/>a. </s> <s xml:id="echoid-s3764" xml:space="preserve">b :</s> <s xml:id="echoid-s3765" xml:space="preserve">: e. </s> <s xml:id="echoid-s3766" xml:space="preserve">f, & </s> <s xml:id="echoid-s3767" xml:space="preserve">c. </s> <s xml:id="echoid-s3768" xml:space="preserve">d :</s> <s xml:id="echoid-s3769" xml:space="preserve">: e. </s> <s xml:id="echoid-s3770" xml:space="preserve">f, on aura a. </s> <s xml:id="echoid-s3771" xml:space="preserve">b :</s> <s xml:id="echoid-s3772" xml:space="preserve">: c. </s> <s xml:id="echoid-s3773" xml:space="preserve">d.</s> <s xml:id="echoid-s3774" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div199" type="section" level="1" n="172"> <head xml:id="echoid-head193" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3775" xml:space="preserve">Si l’on diviſe l’antécédent a par ſon conſéquent, & </s> <s xml:id="echoid-s3776" xml:space="preserve">que le <lb/>quotient ſoit g; </s> <s xml:id="echoid-s3777" xml:space="preserve">en diviſant de même c par d, & </s> <s xml:id="echoid-s3778" xml:space="preserve">e par f, les <lb/>quotients ſeront auſſi g & </s> <s xml:id="echoid-s3779" xml:space="preserve">g; </s> <s xml:id="echoid-s3780" xml:space="preserve">ce qui donnera a = bg, c = dg, <lb/>& </s> <s xml:id="echoid-s3781" xml:space="preserve">e = fg: </s> <s xml:id="echoid-s3782" xml:space="preserve">pour faire voir que a. </s> <s xml:id="echoid-s3783" xml:space="preserve">b :</s> <s xml:id="echoid-s3784" xml:space="preserve">: c : </s> <s xml:id="echoid-s3785" xml:space="preserve">d, il n’y a qu’à mettre <pb o="116" file="0154" n="154" rhead="NOUVEAU COURS"/> à la place de a ſa valeur b g, & </s> <s xml:id="echoid-s3786" xml:space="preserve">à la place de c ſa valeur d g, on <lb/>aura bg. </s> <s xml:id="echoid-s3787" xml:space="preserve">b :</s> <s xml:id="echoid-s3788" xml:space="preserve">: dg. </s> <s xml:id="echoid-s3789" xml:space="preserve">d. </s> <s xml:id="echoid-s3790" xml:space="preserve">Le produit des extrêmes ſera bdg = bdg, <lb/>produit des moyens. </s> <s xml:id="echoid-s3791" xml:space="preserve">Plus ſimplement, puiſque a. </s> <s xml:id="echoid-s3792" xml:space="preserve">b :</s> <s xml:id="echoid-s3793" xml:space="preserve">: e. </s> <s xml:id="echoid-s3794" xml:space="preserve">f, & </s> <s xml:id="echoid-s3795" xml:space="preserve"><lb/>que c. </s> <s xml:id="echoid-s3796" xml:space="preserve">d :</s> <s xml:id="echoid-s3797" xml:space="preserve">: e. </s> <s xml:id="echoid-s3798" xml:space="preserve">f, on aura {a/b} = {e/f}, & </s> <s xml:id="echoid-s3799" xml:space="preserve">{c/d} = {e/f}: </s> <s xml:id="echoid-s3800" xml:space="preserve">donc {a/b} = {c/d}: </s> <s xml:id="echoid-s3801" xml:space="preserve">donc <lb/>a. </s> <s xml:id="echoid-s3802" xml:space="preserve">b :</s> <s xml:id="echoid-s3803" xml:space="preserve">: c. </s> <s xml:id="echoid-s3804" xml:space="preserve">d. </s> <s xml:id="echoid-s3805" xml:space="preserve">C. </s> <s xml:id="echoid-s3806" xml:space="preserve">Q. </s> <s xml:id="echoid-s3807" xml:space="preserve">F. </s> <s xml:id="echoid-s3808" xml:space="preserve">D.</s> <s xml:id="echoid-s3809" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div200" type="section" level="1" n="173"> <head xml:id="echoid-head194" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head195" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3810" xml:space="preserve">220. </s> <s xml:id="echoid-s3811" xml:space="preserve">Lorſque pluſieurs grandeurs ſont en proportion géomé-<lb/>trique, ou qu’elles forment des rapports égaux, la ſomme des an-<lb/>técédens eſt à la ſomme des conſéquens, comme un ſeul antécédent <lb/>eſt à ſon conſéquent; </s> <s xml:id="echoid-s3812" xml:space="preserve">c’eſt-à-dire que ſi des grandeurs, comme <lb/>a, b, c, d forment les rapports égaux {a/b}={c/d}={e/f}, l’on aura <lb/>a + c + e. </s> <s xml:id="echoid-s3813" xml:space="preserve">b + d + f :</s> <s xml:id="echoid-s3814" xml:space="preserve">: a. </s> <s xml:id="echoid-s3815" xml:space="preserve">b, ou comme c. </s> <s xml:id="echoid-s3816" xml:space="preserve">d.</s> <s xml:id="echoid-s3817" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div201" type="section" level="1" n="174"> <head xml:id="echoid-head196" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3818" xml:space="preserve">Pour le prouver, nous ferons voir que le produit des moyens <lb/>eſt égal au produit des extrêmes, ou, ce qui eſt la même choſe, <lb/>que a b + b c + b e = a b + a d + a f; </s> <s xml:id="echoid-s3819" xml:space="preserve">ce qui eſt bien évident: <lb/></s> <s xml:id="echoid-s3820" xml:space="preserve">car 1°. </s> <s xml:id="echoid-s3821" xml:space="preserve">a b=a b, 2°. </s> <s xml:id="echoid-s3822" xml:space="preserve">Puiſque {a/b}={c/d}, ou que a. </s> <s xml:id="echoid-s3823" xml:space="preserve">b :</s> <s xml:id="echoid-s3824" xml:space="preserve">: c. </s> <s xml:id="echoid-s3825" xml:space="preserve">d, on <lb/>@ ad = bc. </s> <s xml:id="echoid-s3826" xml:space="preserve">3° Puiſque {a/b}={e/f}, ou que a. </s> <s xml:id="echoid-s3827" xml:space="preserve">b :</s> <s xml:id="echoid-s3828" xml:space="preserve">: e. </s> <s xml:id="echoid-s3829" xml:space="preserve">f, on aura <lb/>a f = b e. </s> <s xml:id="echoid-s3830" xml:space="preserve">Donc toutes les parties qui compoſent le produit <lb/>des extrêmes ſont égales à celles qui forment le produit des <lb/>moyens, & </s> <s xml:id="echoid-s3831" xml:space="preserve">partant il y a proportion. </s> <s xml:id="echoid-s3832" xml:space="preserve">C. </s> <s xml:id="echoid-s3833" xml:space="preserve">Q. </s> <s xml:id="echoid-s3834" xml:space="preserve">F. </s> <s xml:id="echoid-s3835" xml:space="preserve">D.</s> <s xml:id="echoid-s3836" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div202" type="section" level="1" n="175"> <head xml:id="echoid-head197" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head198" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3837" xml:space="preserve">221. </s> <s xml:id="echoid-s3838" xml:space="preserve">Deux grandeurs demeurent en même raiſon, quoi que l’on <lb/>leur ajoute, pourvu que ce que l’on ajoute à la premiere, ſoit à ce <lb/>que l’on ajoute à la ſeconde, comme la premiere eſt à la ſeconde.</s> <s xml:id="echoid-s3839" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div203" type="section" level="1" n="176"> <head xml:id="echoid-head199" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3840" xml:space="preserve">Si aux deux grandeurs a & </s> <s xml:id="echoid-s3841" xml:space="preserve">b l’on ajoute les deux grandeurs <lb/>c & </s> <s xml:id="echoid-s3842" xml:space="preserve">d, & </s> <s xml:id="echoid-s3843" xml:space="preserve">que a ſoit à b, comme c & </s> <s xml:id="echoid-s3844" xml:space="preserve">à d, je dis que a + c. <lb/></s> <s xml:id="echoid-s3845" xml:space="preserve">b+d :</s> <s xml:id="echoid-s3846" xml:space="preserve">: a. </s> <s xml:id="echoid-s3847" xml:space="preserve">b: </s> <s xml:id="echoid-s3848" xml:space="preserve">car puiſque a. </s> <s xml:id="echoid-s3849" xml:space="preserve">b :</s> <s xml:id="echoid-s3850" xml:space="preserve">: c. </s> <s xml:id="echoid-s3851" xml:space="preserve">d: </s> <s xml:id="echoid-s3852" xml:space="preserve">donc alternando (n°.</s> <s xml:id="echoid-s3853" xml:space="preserve">215.) </s> <s xml:id="echoid-s3854" xml:space="preserve"><lb/>a. </s> <s xml:id="echoid-s3855" xml:space="preserve">c :</s> <s xml:id="echoid-s3856" xml:space="preserve">: b. </s> <s xml:id="echoid-s3857" xml:space="preserve">d: </s> <s xml:id="echoid-s3858" xml:space="preserve">donc componendo (n°. </s> <s xml:id="echoid-s3859" xml:space="preserve">216.) </s> <s xml:id="echoid-s3860" xml:space="preserve">a + c. </s> <s xml:id="echoid-s3861" xml:space="preserve">a :</s> <s xml:id="echoid-s3862" xml:space="preserve">: b + d. </s> <s xml:id="echoid-s3863" xml:space="preserve">b, <lb/>& </s> <s xml:id="echoid-s3864" xml:space="preserve">alternando. </s> <s xml:id="echoid-s3865" xml:space="preserve">a + c. </s> <s xml:id="echoid-s3866" xml:space="preserve">b + d :</s> <s xml:id="echoid-s3867" xml:space="preserve">: a. </s> <s xml:id="echoid-s3868" xml:space="preserve">d. </s> <s xml:id="echoid-s3869" xml:space="preserve">C. </s> <s xml:id="echoid-s3870" xml:space="preserve">Q. </s> <s xml:id="echoid-s3871" xml:space="preserve">F. </s> <s xml:id="echoid-s3872" xml:space="preserve">D.</s> <s xml:id="echoid-s3873" xml:space="preserve"/> </p> <pb o="117" file="0155" n="155" rhead="DE MATHÉMATIQUE. Liv. II."/> </div> <div xml:id="echoid-div204" type="section" level="1" n="177"> <head xml:id="echoid-head200" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head201" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3874" xml:space="preserve">222. </s> <s xml:id="echoid-s3875" xml:space="preserve">Deux grandeurs demeurent toujours en même rapport, <lb/>quoique l’on retranche de l’une ou de l’autre, pourvu que ce que <lb/>l’on retranche de la premiere, ſoit à ce que l’on retranche de la ſe-<lb/>conde, comme la premiere eſt à la ſeconde.</s> <s xml:id="echoid-s3876" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div205" type="section" level="1" n="178"> <head xml:id="echoid-head202" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3877" xml:space="preserve">Si l’on a deux grandeurs a & </s> <s xml:id="echoid-s3878" xml:space="preserve">b, & </s> <s xml:id="echoid-s3879" xml:space="preserve">deux autres c & </s> <s xml:id="echoid-s3880" xml:space="preserve">d, <lb/>telles que a ſoit à b, comme c à d, je dis que a - c. </s> <s xml:id="echoid-s3881" xml:space="preserve">b - d :</s> <s xml:id="echoid-s3882" xml:space="preserve">: <lb/>a. </s> <s xml:id="echoid-s3883" xml:space="preserve">b : </s> <s xml:id="echoid-s3884" xml:space="preserve">car puiſque a. </s> <s xml:id="echoid-s3885" xml:space="preserve">b :</s> <s xml:id="echoid-s3886" xml:space="preserve">: c. </s> <s xml:id="echoid-s3887" xml:space="preserve">d: </s> <s xml:id="echoid-s3888" xml:space="preserve">donc alternando (art. </s> <s xml:id="echoid-s3889" xml:space="preserve">215.) <lb/></s> <s xml:id="echoid-s3890" xml:space="preserve">a. </s> <s xml:id="echoid-s3891" xml:space="preserve">c :</s> <s xml:id="echoid-s3892" xml:space="preserve">: b. </s> <s xml:id="echoid-s3893" xml:space="preserve">d, & </s> <s xml:id="echoid-s3894" xml:space="preserve">dividendo (art. </s> <s xml:id="echoid-s3895" xml:space="preserve">217.) </s> <s xml:id="echoid-s3896" xml:space="preserve">a - c. </s> <s xml:id="echoid-s3897" xml:space="preserve">a :</s> <s xml:id="echoid-s3898" xml:space="preserve">: b - d. </s> <s xml:id="echoid-s3899" xml:space="preserve">d, & </s> <s xml:id="echoid-s3900" xml:space="preserve"><lb/>@ncore alternando, a - c. </s> <s xml:id="echoid-s3901" xml:space="preserve">b - d :</s> <s xml:id="echoid-s3902" xml:space="preserve">: a. </s> <s xml:id="echoid-s3903" xml:space="preserve">b. </s> <s xml:id="echoid-s3904" xml:space="preserve">C. </s> <s xml:id="echoid-s3905" xml:space="preserve">Q. </s> <s xml:id="echoid-s3906" xml:space="preserve">F. </s> <s xml:id="echoid-s3907" xml:space="preserve">D.</s> <s xml:id="echoid-s3908" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div206" type="section" level="1" n="179"> <head xml:id="echoid-head203" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head204" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3909" xml:space="preserve">223. </s> <s xml:id="echoid-s3910" xml:space="preserve">Si l’on multiplie les deux termes d’une raiſon par une même <lb/>quantité, les produits ſeront dans la même raiſon que ces termes <lb/>avant d’être multipliés.</s> <s xml:id="echoid-s3911" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div207" type="section" level="1" n="180"> <head xml:id="echoid-head205" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3912" xml:space="preserve">Pour prouver que ſi l’on multiplie deux grandeurs, comme <lb/>a & </s> <s xml:id="echoid-s3913" xml:space="preserve">b par une autre grandeur c, l’on a ac. </s> <s xml:id="echoid-s3914" xml:space="preserve">bc :</s> <s xml:id="echoid-s3915" xml:space="preserve">: a. </s> <s xml:id="echoid-s3916" xml:space="preserve">b, conſidérés <lb/>que le produit des extrêmes & </s> <s xml:id="echoid-s3917" xml:space="preserve">celui des moyens donnent <lb/>abc = abc. </s> <s xml:id="echoid-s3918" xml:space="preserve">C. </s> <s xml:id="echoid-s3919" xml:space="preserve">Q. </s> <s xml:id="echoid-s3920" xml:space="preserve">F. </s> <s xml:id="echoid-s3921" xml:space="preserve">D.</s> <s xml:id="echoid-s3922" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div208" type="section" level="1" n="181"> <head xml:id="echoid-head206" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head207" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3923" xml:space="preserve">224. </s> <s xml:id="echoid-s3924" xml:space="preserve">Si l’on diviſe les deux termes d’une raiſon par une même <lb/>quantité, les quotients ſeront dans la même raiſon que les grandeurs <lb/>que l’on a diviſées.</s> <s xml:id="echoid-s3925" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div209" type="section" level="1" n="182"> <head xml:id="echoid-head208" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3926" xml:space="preserve">Pour démontrer que ſi l’on diviſe deux grandeurs a & </s> <s xml:id="echoid-s3927" xml:space="preserve">b <lb/>par une même grandeur c, les quotients ſeront dans la même <lb/>raiſon que les grandeurs, nous ſuppoſerons que {a/c} = d, & </s> <s xml:id="echoid-s3928" xml:space="preserve">que <lb/>{b/c} = f. </s> <s xml:id="echoid-s3929" xml:space="preserve">Cela poſé, on aura a = c d, & </s> <s xml:id="echoid-s3930" xml:space="preserve">b = c f, ainſi pour <pb o="118" file="0156" n="156" rhead="NOUVEAU COURS"/> prouver que a.</s> <s xml:id="echoid-s3931" xml:space="preserve">b:</s> <s xml:id="echoid-s3932" xml:space="preserve">:d.</s> <s xml:id="echoid-s3933" xml:space="preserve">f, on n’a qu’à mettre à la place de a & </s> <s xml:id="echoid-s3934" xml:space="preserve"><lb/>de b dans la proportion leurs valeurs cd & </s> <s xml:id="echoid-s3935" xml:space="preserve">cf pour avoir cd. <lb/></s> <s xml:id="echoid-s3936" xml:space="preserve">cf:</s> <s xml:id="echoid-s3937" xml:space="preserve">:d.</s> <s xml:id="echoid-s3938" xml:space="preserve">f, qui donnera cdf = cdf pour le produit des ex -<lb/>trêmes & </s> <s xml:id="echoid-s3939" xml:space="preserve">des moyens.</s> <s xml:id="echoid-s3940" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div210" type="section" level="1" n="183"> <head xml:id="echoid-head209" xml:space="preserve">PROPOSITION IX.</head> <head xml:id="echoid-head210" xml:space="preserve"><emph style="sc">Ttheoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3941" xml:space="preserve">225. </s> <s xml:id="echoid-s3942" xml:space="preserve">Si l’on multiplie deux proportions, termes par termes, <lb/>les produits qui en réſulteront ſeront encore en proportion.</s> <s xml:id="echoid-s3943" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div211" type="section" level="1" n="184"> <head xml:id="echoid-head211" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3944" xml:space="preserve">Soient les deux proportions a.</s> <s xml:id="echoid-s3945" xml:space="preserve">b:</s> <s xml:id="echoid-s3946" xml:space="preserve">:c.</s> <s xml:id="echoid-s3947" xml:space="preserve">d, & </s> <s xml:id="echoid-s3948" xml:space="preserve">l’autre f. </s> <s xml:id="echoid-s3949" xml:space="preserve">g:</s> <s xml:id="echoid-s3950" xml:space="preserve">: m. </s> <s xml:id="echoid-s3951" xml:space="preserve">n, <lb/>il faut prouver que af.</s> <s xml:id="echoid-s3952" xml:space="preserve">bg:</s> <s xml:id="echoid-s3953" xml:space="preserve">:cm.</s> <s xml:id="echoid-s3954" xml:space="preserve">dn, ou que bgcm = afdn, c’eſt -<lb/>à - dire que le produit des extrêmes eſt égal à celui des moyens. <lb/></s> <s xml:id="echoid-s3955" xml:space="preserve">Pour cela, conſidérez que bgcm = bcgm = bc x gm, & </s> <s xml:id="echoid-s3956" xml:space="preserve">que <lb/>afdn = adfn = ad x fn: </s> <s xml:id="echoid-s3957" xml:space="preserve">mais ad = bc, puiſque a.</s> <s xml:id="echoid-s3958" xml:space="preserve">b:</s> <s xml:id="echoid-s3959" xml:space="preserve">:c.</s> <s xml:id="echoid-s3960" xml:space="preserve">d, & </s> <s xml:id="echoid-s3961" xml:space="preserve"><lb/>gm = fn, puiſque f.</s> <s xml:id="echoid-s3962" xml:space="preserve">g:</s> <s xml:id="echoid-s3963" xml:space="preserve">:m.</s> <s xml:id="echoid-s3964" xml:space="preserve">n. </s> <s xml:id="echoid-s3965" xml:space="preserve">Donc bgcm = afdn, c’eſt-à-<lb/>dire qu’il y a proportion, puiſque le produit des extrêmes eſt <lb/>égal à celui des moyens.</s> <s xml:id="echoid-s3966" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div212" type="section" level="1" n="185"> <head xml:id="echoid-head212" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s3967" xml:space="preserve">226. </s> <s xml:id="echoid-s3968" xml:space="preserve">Il ſuit de cette propoſition, que ſi quatre grandeurs ſont <lb/>en proportion géométrique, leurs quarrés, leurs cubes, ou en <lb/>général les mêmes puiſſances de ces grandeurs y ſeront auſſi, <lb/>c’eſt - à - dire que ſi l’on a a.</s> <s xml:id="echoid-s3969" xml:space="preserve">b:</s> <s xml:id="echoid-s3970" xml:space="preserve">:c.</s> <s xml:id="echoid-s3971" xml:space="preserve">d, on aura a<emph style="sub">2</emph>.</s> <s xml:id="echoid-s3972" xml:space="preserve">b<emph style="sub">2</emph>:</s> <s xml:id="echoid-s3973" xml:space="preserve">:c<emph style="sub">2</emph>.</s> <s xml:id="echoid-s3974" xml:space="preserve">d<emph style="sub">2</emph>, <lb/>ou a<emph style="sub">3</emph>.</s> <s xml:id="echoid-s3975" xml:space="preserve">b<emph style="sub">3</emph>:</s> <s xml:id="echoid-s3976" xml:space="preserve">:c<emph style="sub">3</emph>.</s> <s xml:id="echoid-s3977" xml:space="preserve">d<emph style="sub">3</emph>: </s> <s xml:id="echoid-s3978" xml:space="preserve">car en multipliant la proportion a.</s> <s xml:id="echoid-s3979" xml:space="preserve">b:</s> <s xml:id="echoid-s3980" xml:space="preserve">:c.</s> <s xml:id="echoid-s3981" xml:space="preserve">d <lb/>par elle - même une ou pluſieurs fois, on retombe dans le cas <lb/>de la propoſition préſente. </s> <s xml:id="echoid-s3982" xml:space="preserve">D’ailleurs il eſt aiſé de voir que <lb/>dans tous ces cas le produit des extrêmes eſt égal à celui des <lb/>moyens.</s> <s xml:id="echoid-s3983" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div213" type="section" level="1" n="186"> <head xml:id="echoid-head213" xml:space="preserve">PROPOSITION X.</head> <head xml:id="echoid-head214" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s3984" xml:space="preserve">227. </s> <s xml:id="echoid-s3985" xml:space="preserve">Dans une proportion continue, le quarré du premier terme <lb/>eſt au quarré du ſecond, comme le premier au troiſieme; </s> <s xml:id="echoid-s3986" xml:space="preserve">c’eſt-à-<lb/>dire que ſi l’on a la proportion continue {.</s> <s xml:id="echoid-s3987" xml:space="preserve">./.</s> <s xml:id="echoid-s3988" xml:space="preserve">.} a.</s> <s xml:id="echoid-s3989" xml:space="preserve">b.</s> <s xml:id="echoid-s3990" xml:space="preserve">c, ou a.</s> <s xml:id="echoid-s3991" xml:space="preserve">b:</s> <s xml:id="echoid-s3992" xml:space="preserve">:b.</s> <s xml:id="echoid-s3993" xml:space="preserve">c, <lb/>on aura auſſi a<emph style="sub">2</emph>.</s> <s xml:id="echoid-s3994" xml:space="preserve">b<emph style="sub">2</emph>:</s> <s xml:id="echoid-s3995" xml:space="preserve">:a.</s> <s xml:id="echoid-s3996" xml:space="preserve">c.</s> <s xml:id="echoid-s3997" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div214" type="section" level="1" n="187"> <head xml:id="echoid-head215" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s3998" xml:space="preserve">Puiſque a.</s> <s xml:id="echoid-s3999" xml:space="preserve">b:</s> <s xml:id="echoid-s4000" xml:space="preserve">:b.</s> <s xml:id="echoid-s4001" xml:space="preserve">c, on aura bb = ac, & </s> <s xml:id="echoid-s4002" xml:space="preserve">multipliant chaque <lb/>membre de cette égalité par a, on aura abb = a<emph style="sub">2</emph>c; </s> <s xml:id="echoid-s4003" xml:space="preserve">d’où l’on <pb o="119" file="0157" n="157" rhead="DE MATHÉMATIQUE. Liv. II."/> tire la proportion a<emph style="sub">2</emph>.</s> <s xml:id="echoid-s4004" xml:space="preserve">b<emph style="sub">2</emph>:</s> <s xml:id="echoid-s4005" xml:space="preserve">:a.</s> <s xml:id="echoid-s4006" xml:space="preserve">c; </s> <s xml:id="echoid-s4007" xml:space="preserve">car nous avons déja vu que <lb/>lorſque l’on a une équation on en peut tirer une proportion, <lb/>& </s> <s xml:id="echoid-s4008" xml:space="preserve">réciproquement d’une proportion, on en peut toujours tirer <lb/>une équation (art. </s> <s xml:id="echoid-s4009" xml:space="preserve">212). </s> <s xml:id="echoid-s4010" xml:space="preserve">C. </s> <s xml:id="echoid-s4011" xml:space="preserve">Q. </s> <s xml:id="echoid-s4012" xml:space="preserve">F. </s> <s xml:id="echoid-s4013" xml:space="preserve">D.</s> <s xml:id="echoid-s4014" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div215" type="section" level="1" n="188"> <head xml:id="echoid-head216" style="it" xml:space="preserve">Des Proportions & Progreſſions arithmétiques.</head> <p> <s xml:id="echoid-s4015" xml:space="preserve">228. </s> <s xml:id="echoid-s4016" xml:space="preserve">Nous avons déja dit qu’une proportion arithmétique <lb/>eſt l’égalité de deux rapports arithmétiques, & </s> <s xml:id="echoid-s4017" xml:space="preserve">qu’elle réſulte <lb/>de quatre nombres, tels que le premier ſurpaſſe le ſecond, <lb/>d’autant que le troiſieme ſurpaſſe le quatrieme, comme dans <lb/>les nombres ſuivans, 2.</s> <s xml:id="echoid-s4018" xml:space="preserve">5:</s> <s xml:id="echoid-s4019" xml:space="preserve">6.</s> <s xml:id="echoid-s4020" xml:space="preserve">9′, qui ſont en proportion arith -<lb/>métique.</s> <s xml:id="echoid-s4021" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div216" type="section" level="1" n="189"> <head xml:id="echoid-head217" xml:space="preserve">PROPOSITION XI.</head> <head xml:id="echoid-head218" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s4022" xml:space="preserve">229. </s> <s xml:id="echoid-s4023" xml:space="preserve">Lorſque quatre grandeurs ſont en proportion arithmétique, <lb/>la ſomme des extrêmes eſt égale à celle des moyens; </s> <s xml:id="echoid-s4024" xml:space="preserve">c’eſt-à-dire que <lb/>ſi l’on a a.</s> <s xml:id="echoid-s4025" xml:space="preserve">b:</s> <s xml:id="echoid-s4026" xml:space="preserve">c.</s> <s xml:id="echoid-s4027" xml:space="preserve">d, on aura a + d = b + c.</s> <s xml:id="echoid-s4028" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div217" type="section" level="1" n="190"> <head xml:id="echoid-head219" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4029" xml:space="preserve">Puiſqu’il y a proportion entre les quatre grandeurs a,b,c,d, <lb/>& </s> <s xml:id="echoid-s4030" xml:space="preserve">qu’une proportion n’eſt que l’égalité de rapports, l’excès <lb/>de b ſur a ſera égal à celui de d ſur c: </s> <s xml:id="echoid-s4031" xml:space="preserve">ſuppoſant que cet excès <lb/>ſoit une quantité f, on aura b = a + f; </s> <s xml:id="echoid-s4032" xml:space="preserve">& </s> <s xml:id="echoid-s4033" xml:space="preserve">de même d = c <lb/>+ f. </s> <s xml:id="echoid-s4034" xml:space="preserve">Donc au lieu de la proportion a. </s> <s xml:id="echoid-s4035" xml:space="preserve">b: </s> <s xml:id="echoid-s4036" xml:space="preserve">c. </s> <s xml:id="echoid-s4037" xml:space="preserve">d, on aura celle -<lb/>ci, a.</s> <s xml:id="echoid-s4038" xml:space="preserve">a + f:</s> <s xml:id="echoid-s4039" xml:space="preserve">c.</s> <s xml:id="echoid-s4040" xml:space="preserve">c + f: </s> <s xml:id="echoid-s4041" xml:space="preserve">prenant la ſomme des extrêmes & </s> <s xml:id="echoid-s4042" xml:space="preserve"><lb/>des moyens de cette nouvelle proportion, égale à la premiere, <lb/>on aura a + c + f = a + f + c; </s> <s xml:id="echoid-s4043" xml:space="preserve">ce qui eſt bien évident, <lb/>puiſque tout eſt égal de part & </s> <s xml:id="echoid-s4044" xml:space="preserve">d’autre. </s> <s xml:id="echoid-s4045" xml:space="preserve">C. </s> <s xml:id="echoid-s4046" xml:space="preserve">Q. </s> <s xml:id="echoid-s4047" xml:space="preserve">F. </s> <s xml:id="echoid-s4048" xml:space="preserve">D.</s> <s xml:id="echoid-s4049" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div218" type="section" level="1" n="191"> <head xml:id="echoid-head220" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s4050" xml:space="preserve">230. </s> <s xml:id="echoid-s4051" xml:space="preserve">Il ſuit delà, que ſi l’on connoît trois termes quelcon -<lb/>ques d’une proportion arithmétique, on connoîtra auſſi le qua -<lb/>trieme: </s> <s xml:id="echoid-s4052" xml:space="preserve">par exemple, ſi l’on donne ces trois nombres 2, 5, 7 <lb/>pour les trois premiers termes d’une proportion arithmétique, <lb/>dont on demande le quatrieme, ſoit x ce quatrieme terme, <lb/>on aura 2.</s> <s xml:id="echoid-s4053" xml:space="preserve">5:</s> <s xml:id="echoid-s4054" xml:space="preserve">7.</s> <s xml:id="echoid-s4055" xml:space="preserve">x: </s> <s xml:id="echoid-s4056" xml:space="preserve">donc 2 + x = 5 + 7; </s> <s xml:id="echoid-s4057" xml:space="preserve">& </s> <s xml:id="echoid-s4058" xml:space="preserve">ôtant de chaque <lb/>membre le même nombre 2, on aura 2 + x - 2, ou x = 5 <lb/>+ 7 - 2 = 10; </s> <s xml:id="echoid-s4059" xml:space="preserve">ce qui eſt bien évident, puiſque l’excés de <pb o="120" file="0158" n="158" rhead="NOUVEAU COURS"/> 10 ſur 7 eſt 3, comme l’excès de 5 ſur 2 eſt 3. </s> <s xml:id="echoid-s4060" xml:space="preserve">D’où l’on dé -<lb/>duit généralement que le quatrieme terme d’une proportion <lb/>arithmétique ſe trouve en prenant la ſomme des moyens, & </s> <s xml:id="echoid-s4061" xml:space="preserve"><lb/>ôtant le premier extrême de cette ſomme.</s> <s xml:id="echoid-s4062" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div219" type="section" level="1" n="192"> <head xml:id="echoid-head221" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s4063" xml:space="preserve">231. </s> <s xml:id="echoid-s4064" xml:space="preserve">Si la proportion eſt continue, c’eſt - à - dire ſi un terme <lb/>eſt à la fois antécédent du ſecond rapport, & </s> <s xml:id="echoid-s4065" xml:space="preserve">conſéquent du <lb/>premier, on aura la ſomme des extrêmes égale au double du <lb/>terme moyen. </s> <s xml:id="echoid-s4066" xml:space="preserve">Ainſi ſi l’on a cette proportion continue arith -<lb/>métique a. </s> <s xml:id="echoid-s4067" xml:space="preserve">b: </s> <s xml:id="echoid-s4068" xml:space="preserve">b. </s> <s xml:id="echoid-s4069" xml:space="preserve">c, on aura a + c = b + b = 2b: </s> <s xml:id="echoid-s4070" xml:space="preserve">car puiſ -<lb/>que ces trois grandeurs ſont en proportion arithmétique, la <lb/>premiere ſurpaſſe la ſeconde, autant que la même ſeconde ſur -<lb/>paſſe la troiſieme, & </s> <s xml:id="echoid-s4071" xml:space="preserve">appellant d l’excès de la premiere ſur la <lb/>ſeconde, on aura a = b + d, & </s> <s xml:id="echoid-s4072" xml:space="preserve">b = a - d: </s> <s xml:id="echoid-s4073" xml:space="preserve">donc puiſque <lb/>l’excès de b ſur c eſt encore le même, on aura b = c + d, ou <lb/>b - d = c; </s> <s xml:id="echoid-s4074" xml:space="preserve">mais nous avons b = a - d: </s> <s xml:id="echoid-s4075" xml:space="preserve">donc b - d = a - d <lb/>- d = a - 2d = c. </s> <s xml:id="echoid-s4076" xml:space="preserve">Ainſi au lieu de la proportion continue <lb/>a. </s> <s xml:id="echoid-s4077" xml:space="preserve">b: </s> <s xml:id="echoid-s4078" xml:space="preserve">b. </s> <s xml:id="echoid-s4079" xml:space="preserve">c, on aura celle - ci a. </s> <s xml:id="echoid-s4080" xml:space="preserve">a - d: </s> <s xml:id="echoid-s4081" xml:space="preserve">a - d. </s> <s xml:id="echoid-s4082" xml:space="preserve">a - 2d, dans <lb/>laquelle il eſt évident que la ſomme des extrêmes a + a - 2d <lb/>eſt égale à celle des moyens a + a - d - d, ou au double du <lb/>moyen a - d; </s> <s xml:id="echoid-s4083" xml:space="preserve">ce qui eſt encore une autre démonſtration de <lb/>la même propriété.</s> <s xml:id="echoid-s4084" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div220" type="section" level="1" n="193"> <head xml:id="echoid-head222" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s4085" xml:space="preserve">232. </s> <s xml:id="echoid-s4086" xml:space="preserve">Connoiſſant les deux extrêmes d’une proportion con -<lb/>tinue arithmétique, il ſera facile de trouver le moyen terme, <lb/>en prenant la moitié de la ſomme des deux termes donnés: <lb/></s> <s xml:id="echoid-s4087" xml:space="preserve">ainſi ſi l’on demande un terme moyen arithmétique entre 3 <lb/>& </s> <s xml:id="echoid-s4088" xml:space="preserve">5, on prendra la moitié de la ſomme de ces deux nombres <lb/>8, qui eſt 4, & </s> <s xml:id="echoid-s4089" xml:space="preserve">ce nombre ſera le moyen que l’on cherche: </s> <s xml:id="echoid-s4090" xml:space="preserve"><lb/>car il eſt évident que l’on a 3. </s> <s xml:id="echoid-s4091" xml:space="preserve">4: </s> <s xml:id="echoid-s4092" xml:space="preserve">4. </s> <s xml:id="echoid-s4093" xml:space="preserve">5. </s> <s xml:id="echoid-s4094" xml:space="preserve">En Algebre c’eſt la <lb/>même choſe, pour trouver un moyen arithmétique entre les <lb/>deux grandeurs a & </s> <s xml:id="echoid-s4095" xml:space="preserve">b, j’ajoute ces deux nombres enſemble <lb/>pour avoir a + b, dont la moitié {a + b/2} eſt le moyen demandé; </s> <s xml:id="echoid-s4096" xml:space="preserve"><lb/>en effet a. </s> <s xml:id="echoid-s4097" xml:space="preserve">{a + b/2}:</s> <s xml:id="echoid-s4098" xml:space="preserve">{a + b/2}. </s> <s xml:id="echoid-s4099" xml:space="preserve">b, puiſque la différence du premier <lb/>terme au ſecond eſt égale à celle du méme ſecond au troiſieme.</s> <s xml:id="echoid-s4100" xml:space="preserve"/> </p> <pb o="121" file="0159" n="159" rhead="DE MATHÉMATIQUE. Liv. II."/> </div> <div xml:id="echoid-div221" type="section" level="1" n="194"> <head xml:id="echoid-head223" xml:space="preserve">PROPOSITION XII.</head> <head xml:id="echoid-head224" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s4101" xml:space="preserve">233. </s> <s xml:id="echoid-s4102" xml:space="preserve">Si quatre grandeurs ſont telles que la ſomme des extrêmes, <lb/>ſoit égale à celle des moyens, ces quatre grandeurs ſont en pro -<lb/>portion arithmétique; </s> <s xml:id="echoid-s4103" xml:space="preserve">c’eſt - à - dire que ſi les quatre grandeurs <lb/>a, b, c, d ſont telles que a + d, ſomme des extrêmes, ſoit égale à <lb/>c + d, ſomme des moyens, on aura a. </s> <s xml:id="echoid-s4104" xml:space="preserve">b: </s> <s xml:id="echoid-s4105" xml:space="preserve">c. </s> <s xml:id="echoid-s4106" xml:space="preserve">d.</s> <s xml:id="echoid-s4107" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div222" type="section" level="1" n="195"> <head xml:id="echoid-head225" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4108" xml:space="preserve">Tout ſe réduit à prouver que l’excès de a ſur b eſt égal à <lb/>celui de c par d, ou réciproquement que l’excès de b ſur a eſt <lb/>égal à celui de d ſur c; </s> <s xml:id="echoid-s4109" xml:space="preserve">puiſque a + d = b + c, en ajoutant <lb/>de part & </s> <s xml:id="echoid-s4110" xml:space="preserve">d’autre de cette égalité la même quantité, on ne <lb/>changera pas l’égalité. </s> <s xml:id="echoid-s4111" xml:space="preserve">Ajoutons dans chaque membre la <lb/>quantité négative - b - d, on aura a + d - b - d = c + d <lb/>- b - d, ou a - b = c - d, puiſque + d - d ſe détruiſent <lb/>dans le premier membre; </s> <s xml:id="echoid-s4112" xml:space="preserve">& </s> <s xml:id="echoid-s4113" xml:space="preserve">que - b + b ſe détruiſent dans <lb/>le ſecond: </s> <s xml:id="echoid-s4114" xml:space="preserve">donc l’excès de a ſur b eſt égal à celui de c ſur d, <lb/>on prouveroit avec la même facilité que l’excès de b ſur a eſt <lb/>égal à celui de d ſur c: </s> <s xml:id="echoid-s4115" xml:space="preserve">donc ſi quatre grandeurs ſont telles, <lb/>que la ſomme des extrêmes ſoit égale à celle des moyens, ces <lb/>quatre grandeurs ſont en proportion arithmétique. </s> <s xml:id="echoid-s4116" xml:space="preserve">C. </s> <s xml:id="echoid-s4117" xml:space="preserve">Q. </s> <s xml:id="echoid-s4118" xml:space="preserve">F. </s> <s xml:id="echoid-s4119" xml:space="preserve">D.</s> <s xml:id="echoid-s4120" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div223" type="section" level="1" n="196"> <head xml:id="echoid-head226" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s4121" xml:space="preserve">234. </s> <s xml:id="echoid-s4122" xml:space="preserve">Il ſuit delà, que l’on aura toujours prouvé que quatre <lb/>grandeurs ſont en proportion arithmétique, dès qu’on aura <lb/>démontré que la ſomme des extrêmcs eſt égale à celle des <lb/>moyens. </s> <s xml:id="echoid-s4123" xml:space="preserve">Il ſuit encore de cette propoſition, que l’on peut faire <lb/>ſur cette proportion les changemens appellés alternando & </s> <s xml:id="echoid-s4124" xml:space="preserve">in -<lb/>vertendo ſans la détruire: </s> <s xml:id="echoid-s4125" xml:space="preserve">car il eſt évident que ſi l’on a 3. </s> <s xml:id="echoid-s4126" xml:space="preserve">5: </s> <s xml:id="echoid-s4127" xml:space="preserve">7. </s> <s xml:id="echoid-s4128" xml:space="preserve">9, <lb/>on aura auſſi 3. </s> <s xml:id="echoid-s4129" xml:space="preserve">7: </s> <s xml:id="echoid-s4130" xml:space="preserve">5. </s> <s xml:id="echoid-s4131" xml:space="preserve">9, & </s> <s xml:id="echoid-s4132" xml:space="preserve">5. </s> <s xml:id="echoid-s4133" xml:space="preserve">3: </s> <s xml:id="echoid-s4134" xml:space="preserve">9. </s> <s xml:id="echoid-s4135" xml:space="preserve">7.</s> <s xml:id="echoid-s4136" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div224" type="section" level="1" n="197"> <head xml:id="echoid-head227" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s4137" xml:space="preserve">235. </s> <s xml:id="echoid-s4138" xml:space="preserve">Si pluſieurs grandeurs ſont telles, que toutes ſe ſur -<lb/>paſſent également les unes les autres, on appelle progreſſion <lb/>arithmétique, la ſuite de rapports égaux qui en réſulte. </s> <s xml:id="echoid-s4139" xml:space="preserve">La <lb/>progreſſion arithmétique ſe marque de la même maniere que <lb/>la proportion continue: </s> <s xml:id="echoid-s4140" xml:space="preserve">ainſi {.</s> <s xml:id="echoid-s4141" xml:space="preserve">/.</s> <s xml:id="echoid-s4142" xml:space="preserve">} a. </s> <s xml:id="echoid-s4143" xml:space="preserve">b. </s> <s xml:id="echoid-s4144" xml:space="preserve">c. </s> <s xml:id="echoid-s4145" xml:space="preserve">d. </s> <s xml:id="echoid-s4146" xml:space="preserve">f marque que les <lb/>grandeurs a, b, c, d ſont en progreſſion arithmétique.</s> <s xml:id="echoid-s4147" xml:space="preserve"/> </p> <pb o="122" file="0160" n="160" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s4148" xml:space="preserve">236. </s> <s xml:id="echoid-s4149" xml:space="preserve">On diſtingue deux principales ſortes de progreſſions <lb/>arithmétiques; </s> <s xml:id="echoid-s4150" xml:space="preserve">progreſſion arithmétique croiſſante, & </s> <s xml:id="echoid-s4151" xml:space="preserve">progreſ -<lb/>ſion arithmétique décroiſſante. </s> <s xml:id="echoid-s4152" xml:space="preserve">La premiere eſt celle où les ter -<lb/>mes vont en augmentant, & </s> <s xml:id="echoid-s4153" xml:space="preserve">dans laquelle chaque terme eſt <lb/>moindre que celui qui le ſuit; </s> <s xml:id="echoid-s4154" xml:space="preserve">la ſeconde eſt celle où les ter -<lb/>mes vont en diminuant, ou, ce qui revient au même, dans <lb/>laquelle chacun eſt plus grand que celui qui le ſuit, comme <lb/>dans les deux progreſſions ſuivantes, dont la premiere eſt <lb/>croiſſante, & </s> <s xml:id="echoid-s4155" xml:space="preserve">la ſeconde décroiſſante. </s> <s xml:id="echoid-s4156" xml:space="preserve">{.</s> <s xml:id="echoid-s4157" xml:space="preserve">/.</s> <s xml:id="echoid-s4158" xml:space="preserve">} 2. </s> <s xml:id="echoid-s4159" xml:space="preserve">5. </s> <s xml:id="echoid-s4160" xml:space="preserve">7. </s> <s xml:id="echoid-s4161" xml:space="preserve">9. </s> <s xml:id="echoid-s4162" xml:space="preserve">11. </s> <s xml:id="echoid-s4163" xml:space="preserve">13, & </s> <s xml:id="echoid-s4164" xml:space="preserve"><lb/>{.</s> <s xml:id="echoid-s4165" xml:space="preserve">/.</s> <s xml:id="echoid-s4166" xml:space="preserve">} 15. </s> <s xml:id="echoid-s4167" xml:space="preserve">12. </s> <s xml:id="echoid-s4168" xml:space="preserve">9. </s> <s xml:id="echoid-s4169" xml:space="preserve">6. </s> <s xml:id="echoid-s4170" xml:space="preserve">3. </s> <s xml:id="echoid-s4171" xml:space="preserve">1. </s> <s xml:id="echoid-s4172" xml:space="preserve">Chacune de ces deux ſortes de pro -<lb/>greſſions, en contiennent une infinité de différentes, ſelon <lb/>les différens rapports qui régnent dans chaque progreſſion en <lb/>particulier.</s> <s xml:id="echoid-s4173" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div225" type="section" level="1" n="198"> <head xml:id="echoid-head228" xml:space="preserve">PROPOSITION XIII.</head> <head xml:id="echoid-head229" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s4174" xml:space="preserve">237. </s> <s xml:id="echoid-s4175" xml:space="preserve">Dans une progreſſion arithmétique quelconque, la ſomme <lb/>de deux termes également éloignés des extrêmes, eſt égale à celle <lb/>des mêmes extrêmes.</s> <s xml:id="echoid-s4176" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div226" type="section" level="1" n="199"> <head xml:id="echoid-head230" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4177" xml:space="preserve">Soit {.</s> <s xml:id="echoid-s4178" xml:space="preserve">/.</s> <s xml:id="echoid-s4179" xml:space="preserve">} a.</s> <s xml:id="echoid-s4180" xml:space="preserve">b.</s> <s xml:id="echoid-s4181" xml:space="preserve">e.</s> <s xml:id="echoid-s4182" xml:space="preserve">d.</s> <s xml:id="echoid-s4183" xml:space="preserve">f.</s> <s xml:id="echoid-s4184" xml:space="preserve">g.</s> <s xml:id="echoid-s4185" xml:space="preserve">h une progreſſion arithmétique croiſſante, <lb/>je dis que e + f, ſomme de deux termes également éloignée <lb/>des extrêmes, eſt égale à la ſomme des mêmes extrêmes a + h. <lb/></s> <s xml:id="echoid-s4186" xml:space="preserve">Puiſqu’une progreſſion n’eſt qu’une ſuite de rapports égaux, <lb/>ſuppoſons que le rapport arithmétique de a à b ſoit c, c’eſt - à -<lb/>dire que b ſurpaſſe a de la quantité c, on aura b = a + c, par <lb/>la même raiſon b ſera ſurpaſſé par e de la même grandeur c: </s> <s xml:id="echoid-s4187" xml:space="preserve"><lb/>donc e = b + c, ou a + c + c = a + 2c. </s> <s xml:id="echoid-s4188" xml:space="preserve">En continuant le <lb/>même raiſonnement, on verra que d = a + 3c, que f = <lb/>a + 4c, que g = a + 5c, & </s> <s xml:id="echoid-s4189" xml:space="preserve">h = a + 6c: </s> <s xml:id="echoid-s4190" xml:space="preserve">donc au lieu de la <lb/>premiere, on aura celle - ci {.</s> <s xml:id="echoid-s4191" xml:space="preserve">/.</s> <s xml:id="echoid-s4192" xml:space="preserve">} a. </s> <s xml:id="echoid-s4193" xml:space="preserve">a + c. </s> <s xml:id="echoid-s4194" xml:space="preserve">a + 2c. </s> <s xml:id="echoid-s4195" xml:space="preserve">a + 3c. </s> <s xml:id="echoid-s4196" xml:space="preserve">a + 4c. </s> <s xml:id="echoid-s4197" xml:space="preserve"><lb/>a + 5c. </s> <s xml:id="echoid-s4198" xml:space="preserve">a + 6c, dans laquelle il eſt évident que la ſomme de <lb/>deux termes quelconques, également éloignés des extrêmes, <lb/>eſt égale à celle des extrêmes. </s> <s xml:id="echoid-s4199" xml:space="preserve">Ainſi la ſomme du troiſieme & </s> <s xml:id="echoid-s4200" xml:space="preserve"><lb/>du cinquieme terme eſt 2a + 6c, & </s> <s xml:id="echoid-s4201" xml:space="preserve">la ſomme des extrêmes <lb/>eſt auſſi 2a + 6c, c’eſt - à - dire que e + f = a + h. </s> <s xml:id="echoid-s4202" xml:space="preserve">C. </s> <s xml:id="echoid-s4203" xml:space="preserve">Q. </s> <s xml:id="echoid-s4204" xml:space="preserve">F. </s> <s xml:id="echoid-s4205" xml:space="preserve">D.</s> <s xml:id="echoid-s4206" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div227" type="section" level="1" n="200"> <head xml:id="echoid-head231" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s4207" xml:space="preserve">238. </s> <s xml:id="echoid-s4208" xml:space="preserve">Si le nombre des termes de la progreſſion arithmétique <pb o="123" file="0161" n="161" rhead="DE MATHÉMATIQUE. Liv. II."/> eſt impair, la ſomme des extrêmes ſera égale au double du <lb/>terme moyen; </s> <s xml:id="echoid-s4209" xml:space="preserve">& </s> <s xml:id="echoid-s4210" xml:space="preserve">la ſomme de tous les termes d’une progreſſion <lb/>arithmétique ſera égale au produit de la ſomme des extrê-<lb/>mes, multipliée par la moitié du nombre des termes: </s> <s xml:id="echoid-s4211" xml:space="preserve">car ſi <lb/>l’on multiplioit la ſomme des extrêmes par le nombre des ter-<lb/>mes, le produit ſeroit double de la ſomme de tous les termes, <lb/>puiſque la ſomme des extrêmes ne vaut pas un terme tout ſeul, <lb/>mais deux termes enſemble également éloignés des extrêmes.</s> <s xml:id="echoid-s4212" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div228" type="section" level="1" n="201"> <head xml:id="echoid-head232" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s4213" xml:space="preserve">239. </s> <s xml:id="echoid-s4214" xml:space="preserve">Si l’on prend deux termes quelconques, & </s> <s xml:id="echoid-s4215" xml:space="preserve">deux autres <lb/>termes également éloignés du terme moyen, ſi le nombre des <lb/>termes eſt impair, ou des moyens ſi le nombre des termes eſt <lb/>pair, ces quatre termes ſeront en proportion arithmétique: <lb/></s> <s xml:id="echoid-s4216" xml:space="preserve">par exemple, dans la progreſſion {.</s> <s xml:id="echoid-s4217" xml:space="preserve">/.</s> <s xml:id="echoid-s4218" xml:space="preserve">} a. </s> <s xml:id="echoid-s4219" xml:space="preserve">a + c. </s> <s xml:id="echoid-s4220" xml:space="preserve">a + 2c. </s> <s xml:id="echoid-s4221" xml:space="preserve">a + 3c. </s> <s xml:id="echoid-s4222" xml:space="preserve"><lb/>a + 4c. </s> <s xml:id="echoid-s4223" xml:space="preserve">a + 5c. </s> <s xml:id="echoid-s4224" xml:space="preserve">a + 6c; </s> <s xml:id="echoid-s4225" xml:space="preserve">les deux premiers termes a & </s> <s xml:id="echoid-s4226" xml:space="preserve">a + c, <lb/>& </s> <s xml:id="echoid-s4227" xml:space="preserve">les deux derniers a + 5c & </s> <s xml:id="echoid-s4228" xml:space="preserve">a + 6c forment une proportion <lb/>arithmétique a. </s> <s xml:id="echoid-s4229" xml:space="preserve">a + c: </s> <s xml:id="echoid-s4230" xml:space="preserve">a + 5c. </s> <s xml:id="echoid-s4231" xml:space="preserve">a + 6c: </s> <s xml:id="echoid-s4232" xml:space="preserve">car il eſt évident que <lb/>le ſecond ſurpaſſe le premier, d’autant que le quatrieme ſur-<lb/>paſſe le troiſieme.</s> <s xml:id="echoid-s4233" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div229" type="section" level="1" n="202"> <head xml:id="echoid-head233" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s4234" xml:space="preserve">240. </s> <s xml:id="echoid-s4235" xml:space="preserve">Il ſuit encore de cette propoſition, & </s> <s xml:id="echoid-s4236" xml:space="preserve">de l’expreſſion <lb/>générale, qu’un terme quelconque d’une progreſſion arithmé-<lb/>tique croiſſante eſt égal au premier terme, plus au produit de <lb/>la différence du ſecond au premier, multipliée par le nombre <lb/>des termes qui le précéde: </s> <s xml:id="echoid-s4237" xml:space="preserve">ainſi le cinquieme terme a + 4c <lb/>de la progreſſion, citée dans ces corollaires, eſt égal au pre-<lb/>mier terme a, plus quatre fois l’excès c du ſecond ſur le pre-<lb/>mier, parce qu’il a quatre termes avant lui. </s> <s xml:id="echoid-s4238" xml:space="preserve">Ainſi l’on voit ce <lb/>qu’il faut faire pour trouver un terme quelconque, lorſque <lb/>l’on connoît le premier & </s> <s xml:id="echoid-s4239" xml:space="preserve">la différence du ſecond au premier. <lb/></s> <s xml:id="echoid-s4240" xml:space="preserve">Par exemple, ſi l’on me demande le ſixieme terme d’une pro-<lb/>greſſion arithmétique croiſſante, dont le premier terme eſt 2, <lb/>& </s> <s xml:id="echoid-s4241" xml:space="preserve">la différence du ſecond au premier eſt 3; </s> <s xml:id="echoid-s4242" xml:space="preserve">je multiplie cette <lb/>différence 3 par 5, parce qu’il y a cinq termes devant le 6<emph style="sub">e</emph>, <lb/>& </s> <s xml:id="echoid-s4243" xml:space="preserve">j’ajoute au produit 15 le premier terme 2, ce qui me donne <lb/>17 pour le ſixieme terme.</s> <s xml:id="echoid-s4244" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div230" type="section" level="1" n="203"> <head xml:id="echoid-head234" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s4245" xml:space="preserve">241. </s> <s xml:id="echoid-s4246" xml:space="preserve">Réciproquement étant donnés le premier & </s> <s xml:id="echoid-s4247" xml:space="preserve">le ſixieme <pb o="124" file="0162" n="162" rhead="NOUVEAU COURS"/> termes d’une progreſſion, on pourra trouver la différence de <lb/>cette progreſſion, & </s> <s xml:id="echoid-s4248" xml:space="preserve">tous les termes intermédiaires. </s> <s xml:id="echoid-s4249" xml:space="preserve">Ainſi ſi le <lb/>premier terme eſt 2, & </s> <s xml:id="echoid-s4250" xml:space="preserve">le ſixieme eſt 17, j’ôte le premier du <lb/>dernier, & </s> <s xml:id="echoid-s4251" xml:space="preserve">je diviſe le reſte 15 par 5, qui marque le nombre <lb/>des termes qui précédent le ſixieme; </s> <s xml:id="echoid-s4252" xml:space="preserve">le quotient 3 eſt la dif-<lb/>férence; </s> <s xml:id="echoid-s4253" xml:space="preserve">de même en Algebre ſi un terme eſt a, & </s> <s xml:id="echoid-s4254" xml:space="preserve">le ſixieme <lb/>a + 5c, j’ôte a de a + 5c, & </s> <s xml:id="echoid-s4255" xml:space="preserve">je diviſe 5c par 5 pour avoir l’ex-<lb/>cès c du ſecond terme ſur le premier.</s> <s xml:id="echoid-s4256" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div231" type="section" level="1" n="204"> <head xml:id="echoid-head235" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s4257" xml:space="preserve">242. </s> <s xml:id="echoid-s4258" xml:space="preserve">On voit encore comment il faudroit s’y prendre pour <lb/>trouver tous les termes d’une progreſſion arithmétique, dont <lb/>on connoîtroit le premier & </s> <s xml:id="echoid-s4259" xml:space="preserve">le ſecond: </s> <s xml:id="echoid-s4260" xml:space="preserve">car puiſque trois ter-<lb/>mes de ſuite forment une proportion continue arithmétique, <lb/>il n’y a qu’à ôter le premier du double du ſecond pour avoir <lb/>le troiſieme terme.</s> <s xml:id="echoid-s4261" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div232" type="section" level="1" n="205"> <head xml:id="echoid-head236" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s4262" xml:space="preserve">243. </s> <s xml:id="echoid-s4263" xml:space="preserve">On tire encore de cette propoſition la méthode d’in-<lb/>ſérer tant de moyens proportionnels arithmétiques que l’on <lb/>veut entre deux nombres donnés. </s> <s xml:id="echoid-s4264" xml:space="preserve">Pour cela, il faut ôter le <lb/>plus petit nombre du plus grand, & </s> <s xml:id="echoid-s4265" xml:space="preserve">diviſer le reſte par le <lb/>nombre qui exprime combien on veut avoir de moyens arith-<lb/>métiques, augmenté de l’unité. </s> <s xml:id="echoid-s4266" xml:space="preserve">Par exemple, ſi l’on me de-<lb/>mande quatre moyens arithmétiques entre 2 & </s> <s xml:id="echoid-s4267" xml:space="preserve">17, j’ôte 2 de <lb/>17, le reſte eſt 15, que je diviſe par 5, plus grand d’une unité <lb/>que le nombre des moyens arithmétiques que je demande. </s> <s xml:id="echoid-s4268" xml:space="preserve">Le <lb/>quotient 3 eſt la différence du ſecond terme au premier: </s> <s xml:id="echoid-s4269" xml:space="preserve">ainſi <lb/>en ajoutant cette différence au premier terme, le ſecond eſt <lb/>5, & </s> <s xml:id="echoid-s4270" xml:space="preserve">la progreſſion eſt {.</s> <s xml:id="echoid-s4271" xml:space="preserve">/.</s> <s xml:id="echoid-s4272" xml:space="preserve">} 2. </s> <s xml:id="echoid-s4273" xml:space="preserve">5. </s> <s xml:id="echoid-s4274" xml:space="preserve">8. </s> <s xml:id="echoid-s4275" xml:space="preserve">11. </s> <s xml:id="echoid-s4276" xml:space="preserve">14. </s> <s xml:id="echoid-s4277" xml:space="preserve">17, qui eſt telle qu’en-<lb/>tre 2 & </s> <s xml:id="echoid-s4278" xml:space="preserve">17 il y a quatre moyens arithmétiques.</s> <s xml:id="echoid-s4279" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div233" type="section" level="1" n="206"> <head xml:id="echoid-head237" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s4280" xml:space="preserve">244. </s> <s xml:id="echoid-s4281" xml:space="preserve">Tout ce que nous venons de dire ſur les progreſſions <lb/>arithmétiques croiſſantes ſe démontrera avec la même facilité, <lb/>& </s> <s xml:id="echoid-s4282" xml:space="preserve">à peu près de la même maniere ſur les progreſſions décroiſ-<lb/>ſantes. </s> <s xml:id="echoid-s4283" xml:space="preserve">Il faut encore remarquer qu’une progreſſion arithmé-<lb/>tique peut commencer par zero, & </s> <s xml:id="echoid-s4284" xml:space="preserve">qu’en ce cas la différence <lb/>eſt égale au ſecond terme; </s> <s xml:id="echoid-s4285" xml:space="preserve">c’eſt ce qui arrive dans la progreſ-<lb/>ſion des nombres naturels {.</s> <s xml:id="echoid-s4286" xml:space="preserve">/.</s> <s xml:id="echoid-s4287" xml:space="preserve">} 0. </s> <s xml:id="echoid-s4288" xml:space="preserve">1. </s> <s xml:id="echoid-s4289" xml:space="preserve">2. </s> <s xml:id="echoid-s4290" xml:space="preserve">3. </s> <s xml:id="echoid-s4291" xml:space="preserve">4, &</s> <s xml:id="echoid-s4292" xml:space="preserve">c. </s> <s xml:id="echoid-s4293" xml:space="preserve">Il faut encore <pb o="125" file="0163" n="163" rhead="DE MATHÉMATIQUE. Liv. II."/> remarquer que toute progreſſion, dont la différence ne ſera <lb/>pas égale au ſecond terme, ne pourra commencer par zero.</s> <s xml:id="echoid-s4294" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div234" type="section" level="1" n="207"> <head xml:id="echoid-head238" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s4295" xml:space="preserve">245. </s> <s xml:id="echoid-s4296" xml:space="preserve">Si l’on a pluſieurs termes de ſuite, tels que chacun, ex-<lb/>cepté le premier, ſoit antécédent & </s> <s xml:id="echoid-s4297" xml:space="preserve">conſéquent d’une ſuite de <lb/>rapports géométriques égaux, toutes ces quantités formeront <lb/>une progreſſion géométrique. </s> <s xml:id="echoid-s4298" xml:space="preserve">Par exemple, les nombres ſuivans <lb/>64, 32, 16, 8, 4, 2, 1 forment une progreſſion géométrique: </s> <s xml:id="echoid-s4299" xml:space="preserve">car <lb/>64. </s> <s xml:id="echoid-s4300" xml:space="preserve">32@: </s> <s xml:id="echoid-s4301" xml:space="preserve">32. </s> <s xml:id="echoid-s4302" xml:space="preserve">16, & </s> <s xml:id="echoid-s4303" xml:space="preserve">32. </s> <s xml:id="echoid-s4304" xml:space="preserve">16 :</s> <s xml:id="echoid-s4305" xml:space="preserve">: 16. </s> <s xml:id="echoid-s4306" xml:space="preserve">8; </s> <s xml:id="echoid-s4307" xml:space="preserve">ce qui montre évidemment <lb/>que chaque terme peut être conſéquent & </s> <s xml:id="echoid-s4308" xml:space="preserve">antécédent des <lb/>rapports égaux. </s> <s xml:id="echoid-s4309" xml:space="preserve">On marque ordinairement que des quantités <lb/>ſont en progreſſion géométrique, en mettant au devant vers <lb/>la gauche une petite barre entre quatre points de cette maniere: <lb/></s> <s xml:id="echoid-s4310" xml:space="preserve">{:</s> <s xml:id="echoid-s4311" xml:space="preserve">/:</s> <s xml:id="echoid-s4312" xml:space="preserve">} 64. </s> <s xml:id="echoid-s4313" xml:space="preserve">32. </s> <s xml:id="echoid-s4314" xml:space="preserve">16. </s> <s xml:id="echoid-s4315" xml:space="preserve">8. </s> <s xml:id="echoid-s4316" xml:space="preserve">4. </s> <s xml:id="echoid-s4317" xml:space="preserve">2, &</s> <s xml:id="echoid-s4318" xml:space="preserve">c.</s> <s xml:id="echoid-s4319" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4320" xml:space="preserve">On peut encore définir une progreſſion géométrique, en <lb/>diſant, que c’eſt une ſuite de nombres, tels que chacun, diviſé <lb/>par celui qui le ſuit, donne toujours le même quotient. </s> <s xml:id="echoid-s4321" xml:space="preserve">On <lb/>diſtingue deux principales ſortes de progreſſions géométriques: <lb/></s> <s xml:id="echoid-s4322" xml:space="preserve">l’une que l’on appelle croiſſante, c’eſt celle dans laquelle cha-<lb/>que terme eſt moindre que celui qui le ſuit, & </s> <s xml:id="echoid-s4323" xml:space="preserve">l’autre décroiſ-<lb/>ſante, c’eſt celle dans laquelle chaque terme eſt toujours plus <lb/>grand que celui qui le ſuit.</s> <s xml:id="echoid-s4324" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div235" type="section" level="1" n="208"> <head xml:id="echoid-head239" xml:space="preserve">PROPOSITION XIV.</head> <head xml:id="echoid-head240" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s4325" xml:space="preserve">246. </s> <s xml:id="echoid-s4326" xml:space="preserve">Toute progreſſion géométrique croiſſante peut être repréſenté <lb/>par celle-ci {:</s> <s xml:id="echoid-s4327" xml:space="preserve">/:</s> <s xml:id="echoid-s4328" xml:space="preserve">} a. </s> <s xml:id="echoid-s4329" xml:space="preserve">aq. </s> <s xml:id="echoid-s4330" xml:space="preserve">aq<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4331" xml:space="preserve">aq<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4332" xml:space="preserve">aq<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4333" xml:space="preserve">aq<emph style="sub">5</emph>, &</s> <s xml:id="echoid-s4334" xml:space="preserve">c. </s> <s xml:id="echoid-s4335" xml:space="preserve">Et toute progreſſion <lb/>géométrique décroiſſante par celle-ci, qui eſt l’inverſe de la précé-<lb/>dente {:</s> <s xml:id="echoid-s4336" xml:space="preserve">/:</s> <s xml:id="echoid-s4337" xml:space="preserve">} aq<emph style="sub">6</emph>. </s> <s xml:id="echoid-s4338" xml:space="preserve">aq<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4339" xml:space="preserve">aq<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4340" xml:space="preserve">aq<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4341" xml:space="preserve">aq<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4342" xml:space="preserve">aq<emph style="sub">1</emph> a.</s> <s xml:id="echoid-s4343" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div236" type="section" level="1" n="209"> <head xml:id="echoid-head241" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s4344" xml:space="preserve">Pour faire voir que ces quantités ſont en progreſſion géo-<lb/>métrique, il n’y a qu’à diviſer un terme quelconque par le ſui-<lb/>vant, & </s> <s xml:id="echoid-s4345" xml:space="preserve">ce même terme par celui qui le ſuit immédiatement, <lb/>& </s> <s xml:id="echoid-s4346" xml:space="preserve">voir ſi le quotient eſt le même. </s> <s xml:id="echoid-s4347" xml:space="preserve">Dans la premiere progreſ-<lb/>ſion, je diviſe aq<emph style="sub">3</emph> par aq<emph style="sub">2</emph>, le quotient eſt q. </s> <s xml:id="echoid-s4348" xml:space="preserve">Je diviſe enſuite <lb/>aq<emph style="sub">2</emph> par aq, & </s> <s xml:id="echoid-s4349" xml:space="preserve">le quotient eſt encore q: </s> <s xml:id="echoid-s4350" xml:space="preserve">donc il y a progreſ-<lb/>ſion, puiſque aq. </s> <s xml:id="echoid-s4351" xml:space="preserve">aq@: </s> <s xml:id="echoid-s4352" xml:space="preserve">aq<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4353" xml:space="preserve">aq<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4354" xml:space="preserve">De même pour la ſeconde, <lb/>je diviſe aq<emph style="sub">6</emph> par aq<emph style="sub">5</emph>, le quotient eſt q. </s> <s xml:id="echoid-s4355" xml:space="preserve">Je diviſe le même aq<emph style="sub">5</emph> <pb o="126" file="0164" n="164" rhead="NOUVEAU COURS"/> par aq<emph style="sub">4</emph>, le quotient eſt q, égal au premier: </s> <s xml:id="echoid-s4356" xml:space="preserve">donc ces termes <lb/>ſont en progreſſion géométrique, puiſqu’ils donnent un même <lb/>quotient. </s> <s xml:id="echoid-s4357" xml:space="preserve">C. </s> <s xml:id="echoid-s4358" xml:space="preserve">Q. </s> <s xml:id="echoid-s4359" xml:space="preserve">F. </s> <s xml:id="echoid-s4360" xml:space="preserve">D.</s> <s xml:id="echoid-s4361" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div237" type="section" level="1" n="210"> <head xml:id="echoid-head242" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s4362" xml:space="preserve">247. </s> <s xml:id="echoid-s4363" xml:space="preserve">Il ſuit delà, que dans une progreſſion géométrique <lb/>croiſſante, le quarré du premier terme eſt au quarré du ſecond, <lb/>comme le premier terme au troiſieme; </s> <s xml:id="echoid-s4364" xml:space="preserve">car dans la ſuite {:</s> <s xml:id="echoid-s4365" xml:space="preserve">/:</s> <s xml:id="echoid-s4366" xml:space="preserve">} a. <lb/></s> <s xml:id="echoid-s4367" xml:space="preserve">aq. </s> <s xml:id="echoid-s4368" xml:space="preserve">aq<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4369" xml:space="preserve">aq<emph style="sub">3</emph>, &</s> <s xml:id="echoid-s4370" xml:space="preserve">c. </s> <s xml:id="echoid-s4371" xml:space="preserve">on a a<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4372" xml:space="preserve">a<emph style="sub">2</emph>q<emph style="sub">2</emph>:</s> <s xml:id="echoid-s4373" xml:space="preserve">: a. </s> <s xml:id="echoid-s4374" xml:space="preserve">aq<emph style="sub">2</emph>: </s> <s xml:id="echoid-s4375" xml:space="preserve">puiſque le produit <lb/>des extrêmes eſt égal à celui des moyens, a<emph style="sub">3</emph>q<emph style="sub">2</emph> = a<emph style="sub">3</emph>q<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4376" xml:space="preserve">Il ſuit <lb/>encore de la même formation des progreſſions, que le cube du <lb/>premier terme eſt au cube du ſecond, comme le premier au <lb/>quatrieme: </s> <s xml:id="echoid-s4377" xml:space="preserve">car a<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4378" xml:space="preserve">a<emph style="sub">3</emph>q<emph style="sub">3</emph>:</s> <s xml:id="echoid-s4379" xml:space="preserve">: a. </s> <s xml:id="echoid-s4380" xml:space="preserve">aq<emph style="sub">3</emph>, puiſque a<emph style="sub">4</emph>q<emph style="sub">3</emph>, produit des ex-<lb/>trêmes eſt égal à a<emph style="sub">4</emph>q<emph style="sub">3</emph>, produit des moyens. </s> <s xml:id="echoid-s4381" xml:space="preserve">En général ſi l’on <lb/>appelle a le premier terme d’une progreſſion, & </s> <s xml:id="echoid-s4382" xml:space="preserve">b le ſecond; </s> <s xml:id="echoid-s4383" xml:space="preserve"><lb/>m la puiſſance quelconque à laquelle on éleve les deux pre-<lb/>miers termes, on aura a<emph style="sub">m</emph>. </s> <s xml:id="echoid-s4384" xml:space="preserve">b<emph style="sub">m</emph>:</s> <s xml:id="echoid-s4385" xml:space="preserve">: a eſt au terme, dont le rang <lb/>ſeroit déſigné par le nombre m + 1.</s> <s xml:id="echoid-s4386" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div238" type="section" level="1" n="211"> <head xml:id="echoid-head243" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s4387" xml:space="preserve">248. </s> <s xml:id="echoid-s4388" xml:space="preserve">Suppoſant toujours que la progreſſion va en croiſſant, <lb/>un terme quelconque eſt égal au produit du premier terme, <lb/>multiplié par le quotient du ſecond, diviſé par le premier, le-<lb/>quel quotient eſt élevé à la puiſſance, marquée par le nombre <lb/>des termes qui précédent. </s> <s xml:id="echoid-s4389" xml:space="preserve">Ainſi le quatrieme terme eſt égal <lb/>au premier a, multiplié par q, quotient du ſecond aq, diviſé <lb/>par le premier, élevé à la troiſieme puiſſance, parce qu’il y a <lb/>trois termes qui précédent le quatrieme; </s> <s xml:id="echoid-s4390" xml:space="preserve">ce terme eſt aq<emph style="sub">3</emph>: <lb/></s> <s xml:id="echoid-s4391" xml:space="preserve">ainſi connoiſſant les deux premiers termes d’une progreſſion <lb/>géométrique, on connoîtra aiſément un terme quelconque. </s> <s xml:id="echoid-s4392" xml:space="preserve"><lb/>Pour cela, il n’y aura qu’à diviſer le ſecond par le premier, <lb/>multiplier le premier terme par ce quotient, élevé à une puiſ-<lb/>ſance, marqué par le nombre des termes qui précédent celui <lb/>qu’on cherche. </s> <s xml:id="echoid-s4393" xml:space="preserve">Par exemple, ſi l’on me demande le ſixieme <lb/>terme d’une progreſſion géométrique croiſſante, dont le pre-<lb/>mier eſt a, & </s> <s xml:id="echoid-s4394" xml:space="preserve">le ſecond aq, je diviſe le ſecond par le premier a, <lb/>le quotient eſt q: </s> <s xml:id="echoid-s4395" xml:space="preserve">je multiplie a par ce quotient q, élevé à la <lb/>cinquieme puiſſance, & </s> <s xml:id="echoid-s4396" xml:space="preserve">le ſixieme terme eſt aq<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4397" xml:space="preserve">Il en ſeroit <lb/>de même en nombres. </s> <s xml:id="echoid-s4398" xml:space="preserve">Si le premier terme eſt a, & </s> <s xml:id="echoid-s4399" xml:space="preserve">le ſecond b; </s> <s xml:id="echoid-s4400" xml:space="preserve"><lb/>je diviſe b par a, le quotient eſt {b/a}, & </s> <s xml:id="echoid-s4401" xml:space="preserve">qu’on me demande le cin- <pb o="127" file="0165" n="165" rhead="DE MATHÉMATIQUE. Liv. II."/> quieme terme de la progreſſion croiſſante, dont a & </s> <s xml:id="echoid-s4402" xml:space="preserve">b ſeroient <lb/>les deux premiers termes: </s> <s xml:id="echoid-s4403" xml:space="preserve">je multiplie a par la quatrieme puiſ-<lb/>ſance de {b/a}, qui eſt {b<emph style="sub">4</emph>/a<emph style="sub">4</emph>}, & </s> <s xml:id="echoid-s4404" xml:space="preserve">appellant x ce cinquieme terme, j’ai <lb/>x = {ab<emph style="sub">4</emph>/a<emph style="sub">4</emph>} ou {b<emph style="sub">4</emph>/a<emph style="sub">3</emph>}. </s> <s xml:id="echoid-s4405" xml:space="preserve">D’où il ſuit encore qu’un terme quelconque <lb/>d’une progreſſion géométrique croiſſante eſt égal au ſecond <lb/>terme, élevé à une puiſſance moindre d’un degré que le nu-<lb/>méro de ce terme, diviſé par le premier terme, élevé à une <lb/>puiſſance moindre de deux degrés que le même numéro.</s> <s xml:id="echoid-s4406" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div239" type="section" level="1" n="212"> <head xml:id="echoid-head244" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s4407" xml:space="preserve">249. </s> <s xml:id="echoid-s4408" xml:space="preserve">Si l’on ſuppoſe a égal à l’unité la ſuite ou progreſſion <lb/>{:</s> <s xml:id="echoid-s4409" xml:space="preserve">/:</s> <s xml:id="echoid-s4410" xml:space="preserve">} a. </s> <s xml:id="echoid-s4411" xml:space="preserve">aq. </s> <s xml:id="echoid-s4412" xml:space="preserve">aq<emph style="sub">2</emph>, &</s> <s xml:id="echoid-s4413" xml:space="preserve">c. </s> <s xml:id="echoid-s4414" xml:space="preserve">deviendra {:</s> <s xml:id="echoid-s4415" xml:space="preserve">/:</s> <s xml:id="echoid-s4416" xml:space="preserve">} q<emph style="sub">1</emph>. </s> <s xml:id="echoid-s4417" xml:space="preserve">q<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4418" xml:space="preserve">q<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4419" xml:space="preserve">q<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4420" xml:space="preserve">q<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4421" xml:space="preserve">q<emph style="sub">6</emph>, &</s> <s xml:id="echoid-s4422" xml:space="preserve">c. </s> <s xml:id="echoid-s4423" xml:space="preserve">D’où <lb/>il ſuit que toutes les puiſſances d’un nombre forment une pro-<lb/>greſſion géométrique; </s> <s xml:id="echoid-s4424" xml:space="preserve">ce qui eſt d’ailleurs évident par l’idée <lb/>que l’on doit avoir des puiſſances ſucceſſives d’un nombre.</s> <s xml:id="echoid-s4425" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div240" type="section" level="1" n="213"> <head xml:id="echoid-head245" xml:space="preserve">PROPOSITION XV.</head> <head xml:id="echoid-head246" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s4426" xml:space="preserve">250. </s> <s xml:id="echoid-s4427" xml:space="preserve">Dans une progreſſion quelconque, la ſomme des antécé-<lb/>dens eſt à la ſomme des conſéquens, comme un ſeul antécédent eſt <lb/>à ſon conſéquent; </s> <s xml:id="echoid-s4428" xml:space="preserve">c’eſt-à-dire que ſi les grandeurs a, b, c, d, f, <lb/>font une progreſſion géométrique, on aura cette proportion, a + b <lb/>+ c + d + f. </s> <s xml:id="echoid-s4429" xml:space="preserve">b + c + d + f:</s> <s xml:id="echoid-s4430" xml:space="preserve">: a. </s> <s xml:id="echoid-s4431" xml:space="preserve">b.</s> <s xml:id="echoid-s4432" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div241" type="section" level="1" n="214"> <head xml:id="echoid-head247" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4433" xml:space="preserve">Il faut démontrer que le produit des extrêmes ab + bb + bc <lb/>+ bd eſt égal au produit des moyens. </s> <s xml:id="echoid-s4434" xml:space="preserve">ab + ac + ad + af. <lb/></s> <s xml:id="echoid-s4435" xml:space="preserve">1°. </s> <s xml:id="echoid-s4436" xml:space="preserve">ab = ab. </s> <s xml:id="echoid-s4437" xml:space="preserve">2°. </s> <s xml:id="echoid-s4438" xml:space="preserve">Puiſque par la nature de la progreſſion a. </s> <s xml:id="echoid-s4439" xml:space="preserve">b:</s> <s xml:id="echoid-s4440" xml:space="preserve">: <lb/>b. </s> <s xml:id="echoid-s4441" xml:space="preserve">c, bb = ac. </s> <s xml:id="echoid-s4442" xml:space="preserve">3°. </s> <s xml:id="echoid-s4443" xml:space="preserve">Par la même raiſon, puiſque a. </s> <s xml:id="echoid-s4444" xml:space="preserve">b:</s> <s xml:id="echoid-s4445" xml:space="preserve">: b. </s> <s xml:id="echoid-s4446" xml:space="preserve">c, & </s> <s xml:id="echoid-s4447" xml:space="preserve"><lb/>que b. </s> <s xml:id="echoid-s4448" xml:space="preserve">c :</s> <s xml:id="echoid-s4449" xml:space="preserve">: c. </s> <s xml:id="echoid-s4450" xml:space="preserve">d, on aura a. </s> <s xml:id="echoid-s4451" xml:space="preserve">b:</s> <s xml:id="echoid-s4452" xml:space="preserve">: c. </s> <s xml:id="echoid-s4453" xml:space="preserve">d; </s> <s xml:id="echoid-s4454" xml:space="preserve">donc ad = bc. </s> <s xml:id="echoid-s4455" xml:space="preserve">4°. </s> <s xml:id="echoid-s4456" xml:space="preserve">Puiſque <lb/>a. </s> <s xml:id="echoid-s4457" xml:space="preserve">b :</s> <s xml:id="echoid-s4458" xml:space="preserve">: c. </s> <s xml:id="echoid-s4459" xml:space="preserve">d:</s> <s xml:id="echoid-s4460" xml:space="preserve">: d. </s> <s xml:id="echoid-s4461" xml:space="preserve">f, on aura a. </s> <s xml:id="echoid-s4462" xml:space="preserve">b :</s> <s xml:id="echoid-s4463" xml:space="preserve">: d. </s> <s xml:id="echoid-s4464" xml:space="preserve">f; </s> <s xml:id="echoid-s4465" xml:space="preserve">donc af = bd. </s> <s xml:id="echoid-s4466" xml:space="preserve">Ainſi <lb/>toutes les parties du produit des extrêmes ſont égales à toutes <lb/>les parties du produit des moyens; </s> <s xml:id="echoid-s4467" xml:space="preserve">d’où il ſuit que la pro-<lb/>portion a lieu.</s> <s xml:id="echoid-s4468" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div242" type="section" level="1" n="215"> <head xml:id="echoid-head248" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s4469" xml:space="preserve">251. </s> <s xml:id="echoid-s4470" xml:space="preserve">Si la progreſſion eſt décroiſſante, & </s> <s xml:id="echoid-s4471" xml:space="preserve">décroît juſqu’à <lb/>l’infini, le dernier terme pourra être regardé comme zero: <lb/></s> <s xml:id="echoid-s4472" xml:space="preserve">ainſi la ſomme des antécédens, qui eſt tous les termes, excepté <pb o="128" file="0166" n="166" rhead="NOUVEAU COURS"/> le dernier, ſera la ſomme de tous les termes de la progreſſion; <lb/></s> <s xml:id="echoid-s4473" xml:space="preserve">& </s> <s xml:id="echoid-s4474" xml:space="preserve">la ſomme des conſéquens ſera la ſomme de tous les termes, <lb/>excepté le premier, ce qui ne détruira pas la proportion. </s> <s xml:id="echoid-s4475" xml:space="preserve">Cette <lb/>propoſition & </s> <s xml:id="echoid-s4476" xml:space="preserve">ſon corollaire donnent la ſolution des problêmes <lb/>que l’on peut propoſer pour la ſommation des ſuites des pro-<lb/>greſſions géométriques, comme on verra dans le Traité des <lb/>Equations. </s> <s xml:id="echoid-s4477" xml:space="preserve">On ne peut trop ſçavoir cette propoſition, & </s> <s xml:id="echoid-s4478" xml:space="preserve">ce <lb/>qui précéde, ſi l’on veut trouver la ſolution de ces ſortes de <lb/>problêmes.</s> <s xml:id="echoid-s4479" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div243" type="section" level="1" n="216"> <head xml:id="echoid-head249" xml:space="preserve">PROPOSITION XVI.</head> <head xml:id="echoid-head250" xml:space="preserve"><emph style="sc">Theoreme</emph></head> <p style="it"> <s xml:id="echoid-s4480" xml:space="preserve">252. </s> <s xml:id="echoid-s4481" xml:space="preserve">Dans une progreſſion géométrique, telle que {:</s> <s xml:id="echoid-s4482" xml:space="preserve">/:</s> <s xml:id="echoid-s4483" xml:space="preserve">} a.</s> <s xml:id="echoid-s4484" xml:space="preserve">b.</s> <s xml:id="echoid-s4485" xml:space="preserve">c.</s> <s xml:id="echoid-s4486" xml:space="preserve">d.</s> <s xml:id="echoid-s4487" xml:space="preserve">f.</s> <s xml:id="echoid-s4488" xml:space="preserve">g. <lb/></s> <s xml:id="echoid-s4489" xml:space="preserve">le produit de deux termes, également éloignés des extrêmes, eſt égal <lb/>au produit des mêmes extrémes.</s> <s xml:id="echoid-s4490" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div244" type="section" level="1" n="217"> <head xml:id="echoid-head251" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4491" xml:space="preserve">Prenons les termes c, d, qui ſont également éloignés des <lb/>extrêmes; </s> <s xml:id="echoid-s4492" xml:space="preserve">il faut prouver que c d eſt égal au produit des ex-<lb/>trêmes ag. </s> <s xml:id="echoid-s4493" xml:space="preserve">Pour cela, faites attention que la nature de la <lb/>progreſſion donne les proportions ſuivantes.</s> <s xml:id="echoid-s4494" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s4495" xml:space="preserve">a. </s> <s xml:id="echoid-s4496" xml:space="preserve">b :</s> <s xml:id="echoid-s4497" xml:space="preserve">: b. </s> <s xml:id="echoid-s4498" xml:space="preserve">c, b. </s> <s xml:id="echoid-s4499" xml:space="preserve">c :</s> <s xml:id="echoid-s4500" xml:space="preserve">: c. </s> <s xml:id="echoid-s4501" xml:space="preserve">d, c. </s> <s xml:id="echoid-s4502" xml:space="preserve">d :</s> <s xml:id="echoid-s4503" xml:space="preserve">: d. </s> <s xml:id="echoid-s4504" xml:space="preserve">f</s> </p> <p style="it"> <s xml:id="echoid-s4505" xml:space="preserve">b. </s> <s xml:id="echoid-s4506" xml:space="preserve">c :</s> <s xml:id="echoid-s4507" xml:space="preserve">: c. </s> <s xml:id="echoid-s4508" xml:space="preserve">d, c. </s> <s xml:id="echoid-s4509" xml:space="preserve">d :</s> <s xml:id="echoid-s4510" xml:space="preserve">: d. </s> <s xml:id="echoid-s4511" xml:space="preserve">f, d. </s> <s xml:id="echoid-s4512" xml:space="preserve">f :</s> <s xml:id="echoid-s4513" xml:space="preserve">: f. </s> <s xml:id="echoid-s4514" xml:space="preserve">g.</s> <s xml:id="echoid-s4515" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4516" xml:space="preserve">Multipliant deux à deux termes par termes, on aura</s> </p> <p style="it"> <s xml:id="echoid-s4517" xml:space="preserve">ab. </s> <s xml:id="echoid-s4518" xml:space="preserve">bc :</s> <s xml:id="echoid-s4519" xml:space="preserve">: bc. </s> <s xml:id="echoid-s4520" xml:space="preserve">cd, bc. </s> <s xml:id="echoid-s4521" xml:space="preserve">cd :</s> <s xml:id="echoid-s4522" xml:space="preserve">: cd. </s> <s xml:id="echoid-s4523" xml:space="preserve">df, cd. </s> <s xml:id="echoid-s4524" xml:space="preserve">df :</s> <s xml:id="echoid-s4525" xml:space="preserve">: df. </s> <s xml:id="echoid-s4526" xml:space="preserve">fg.</s> <s xml:id="echoid-s4527" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4528" xml:space="preserve">D’où l’on déduit celle-ci, en diviſant chaque terme des rap-<lb/>ports par les lettres communes à l’antécédent & </s> <s xml:id="echoid-s4529" xml:space="preserve">au conſéquent.</s> <s xml:id="echoid-s4530" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s4531" xml:space="preserve">a. </s> <s xml:id="echoid-s4532" xml:space="preserve">c :</s> <s xml:id="echoid-s4533" xml:space="preserve">: b. </s> <s xml:id="echoid-s4534" xml:space="preserve">d, b . </s> <s xml:id="echoid-s4535" xml:space="preserve">d :</s> <s xml:id="echoid-s4536" xml:space="preserve">: c. </s> <s xml:id="echoid-s4537" xml:space="preserve">f, c. </s> <s xml:id="echoid-s4538" xml:space="preserve">f :</s> <s xml:id="echoid-s4539" xml:space="preserve">: d. </s> <s xml:id="echoid-s4540" xml:space="preserve">g.</s> <s xml:id="echoid-s4541" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4542" xml:space="preserve">Et puiſque toutes ces raiſons ſont égales entr’elles, on aura <lb/>cette proportion a. </s> <s xml:id="echoid-s4543" xml:space="preserve">c :</s> <s xml:id="echoid-s4544" xml:space="preserve">: d. </s> <s xml:id="echoid-s4545" xml:space="preserve">g: </s> <s xml:id="echoid-s4546" xml:space="preserve">donc ag = dc, c’eſt-à-dire que <lb/>le produit des extrêmes de la progreſſion eſt égal à celui de <lb/>deux termes quel conques, également éloignés des mêmes ex-<lb/>trêmes. </s> <s xml:id="echoid-s4547" xml:space="preserve">C. </s> <s xml:id="echoid-s4548" xml:space="preserve">Q. </s> <s xml:id="echoid-s4549" xml:space="preserve">F. </s> <s xml:id="echoid-s4550" xml:space="preserve">D.</s> <s xml:id="echoid-s4551" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div245" type="section" level="1" n="218"> <head xml:id="echoid-head252" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s4552" xml:space="preserve">253. </s> <s xml:id="echoid-s4553" xml:space="preserve">Il ſuit de cette propoſition, que les deux extrêmes & </s> <s xml:id="echoid-s4554" xml:space="preserve"><lb/>deux termes quelconques qui en ſeront également éloignés, <lb/>formeront une proportion, dont les deux premiers ſeront les <pb o="129" file="0167" n="167" rhead="DE MATHÉMATIQUE. Liv. II."/> extrêmes, & </s> <s xml:id="echoid-s4555" xml:space="preserve">les deux autres les moyens. </s> <s xml:id="echoid-s4556" xml:space="preserve">Si le nombre des ter-<lb/>mes de la progreſſion eſt impair, le produit des extrêmes ou de <lb/>deux termes, qui en ſeront chacun également éloignés, ſera <lb/>égal à celui des moyens.</s> <s xml:id="echoid-s4557" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div246" type="section" level="1" n="219"> <head xml:id="echoid-head253" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s4558" xml:space="preserve">254. </s> <s xml:id="echoid-s4559" xml:space="preserve">Tout ce que nous avons dit ſur les progreſſions arith-<lb/>métiques croiſſantes ſe doit auſſi entendre des progreſſions <lb/>décroiſſantes, en faiſant les changemens néceſſaires. </s> <s xml:id="echoid-s4560" xml:space="preserve">Au reſte <lb/>toute progreſſion décroiſſante ſe peut rappeller à une progreſ-<lb/>ſion croiſſante, en allant de droite à gauche. </s> <s xml:id="echoid-s4561" xml:space="preserve">On remarquera <lb/>de plus, que les deux derniers théorêmes auroient pu ſe dé-<lb/>montrer bien facilement par la progreſſion générale {:</s> <s xml:id="echoid-s4562" xml:space="preserve">/:</s> <s xml:id="echoid-s4563" xml:space="preserve">} a. </s> <s xml:id="echoid-s4564" xml:space="preserve">aq. <lb/></s> <s xml:id="echoid-s4565" xml:space="preserve">aq<emph style="sub">2</emph>, &</s> <s xml:id="echoid-s4566" xml:space="preserve">c: </s> <s xml:id="echoid-s4567" xml:space="preserve">mais c’eſt préciſément à cauſe de cette facilité que <lb/>j’ai cru qu’il falloit les démontrer un peu autrement; </s> <s xml:id="echoid-s4568" xml:space="preserve">car cette <lb/>expreſſion ne vous laiſſe aucun raiſonnement à faire, en vous <lb/>donnant tout d’un coup ce que vous demandez, & </s> <s xml:id="echoid-s4569" xml:space="preserve">l’on court <lb/>ſouvent riſque de déraiſonner, ou au moins d’ignorer l’art de <lb/>raiſonner, lorſque l’on ne raiſonne que par formule, ſans ſe <lb/>mettre en peine de le faire par ſoi-même.</s> <s xml:id="echoid-s4570" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div247" type="section" level="1" n="220"> <head xml:id="echoid-head254" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s4571" xml:space="preserve">255. </s> <s xml:id="echoid-s4572" xml:space="preserve">Inſérer pluſieurs moyens proportionnels entre deux nom-<lb/>bres donnés.</s> <s xml:id="echoid-s4573" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div248" type="section" level="1" n="221"> <head xml:id="echoid-head255" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s4574" xml:space="preserve">Il faudra diviſer le plus grand par le plus petit; </s> <s xml:id="echoid-s4575" xml:space="preserve">& </s> <s xml:id="echoid-s4576" xml:space="preserve">pour <lb/>avoir la raiſon de la progreſſion, il faudra extraire la racine <lb/>du quotient, marquée par le nombre des moyens proportion-<lb/>nels, augmenté de l’unité. </s> <s xml:id="echoid-s4577" xml:space="preserve">Par exemple, ſi l’on me demande <lb/>trois moyens proportionnels géométriques entre 4 & </s> <s xml:id="echoid-s4578" xml:space="preserve">64, je <lb/>diviſe 64 par 4, le quotient eſt 16, dont j’extrais la racine <lb/>quatrieme, qui eſt 2, parce que l’on demande trois moyens <lb/>proportionnels, & </s> <s xml:id="echoid-s4579" xml:space="preserve">cette racine eſt la raiſon de la progreſſion, <lb/>c’eſt-à-dire que chaque terme eſt double de celui qui le ſuit: <lb/></s> <s xml:id="echoid-s4580" xml:space="preserve">ainſi le ſecond terme ſera 8, & </s> <s xml:id="echoid-s4581" xml:space="preserve">le troiſieme 16, le quatrieme <lb/>32, & </s> <s xml:id="echoid-s4582" xml:space="preserve">la progreſſion eſt {:</s> <s xml:id="echoid-s4583" xml:space="preserve">/:</s> <s xml:id="echoid-s4584" xml:space="preserve">} 4.</s> <s xml:id="echoid-s4585" xml:space="preserve">8. </s> <s xml:id="echoid-s4586" xml:space="preserve">16. </s> <s xml:id="echoid-s4587" xml:space="preserve">32. </s> <s xml:id="echoid-s4588" xml:space="preserve">64, où l’on voit qu’il ſe <lb/>trouve trois moyens entre 4 & </s> <s xml:id="echoid-s4589" xml:space="preserve">64. </s> <s xml:id="echoid-s4590" xml:space="preserve">Si l’on en avoit demandé <lb/>quatre, il auroit fallu extraire la racine cinquieme du quotient <lb/>du plus grand nombre, diviſé par le plus petit.</s> <s xml:id="echoid-s4591" xml:space="preserve"/> </p> <pb o="130" file="0168" n="168" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div249" type="section" level="1" n="222"> <head xml:id="echoid-head256" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4592" xml:space="preserve">La raiſon de cette opération ſe déduit immédiatement de <lb/>la formule ou expreſſion générale des progreſſions {:</s> <s xml:id="echoid-s4593" xml:space="preserve">/:</s> <s xml:id="echoid-s4594" xml:space="preserve">} a. </s> <s xml:id="echoid-s4595" xml:space="preserve">aq. <lb/></s> <s xml:id="echoid-s4596" xml:space="preserve">aq<emph style="sub">2</emph>.</s> <s xml:id="echoid-s4597" xml:space="preserve">aq<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4598" xml:space="preserve">aq<emph style="sub">4</emph>, &</s> <s xml:id="echoid-s4599" xml:space="preserve">c. </s> <s xml:id="echoid-s4600" xml:space="preserve">Je ſuppoſe que l’on me demande trois moyens <lb/>géométriques entre a & </s> <s xml:id="echoid-s4601" xml:space="preserve">aq<emph style="sub">4</emph>, je diviſe aq par a, le quotient eſt <lb/>q<emph style="sub">4</emph>, dont la racine quatrieme q eſt la raiſon de la progreſſion: </s> <s xml:id="echoid-s4602" xml:space="preserve"><lb/>ainſi aq ſera le ſecond terme, aq x q ſera le troiſieme, aq<emph style="sub">2</emph> x q <lb/>ou aq<emph style="sub">3</emph> ſera le quatrieme.</s> <s xml:id="echoid-s4603" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4604" xml:space="preserve">Il faut encore remarquer qu’une progreſſion géométrique <lb/>quelconque ne peut jamais avoir zero pour un de ſes termes, <lb/>à moins qu’il ne ſerve d’expoſant: </s> <s xml:id="echoid-s4605" xml:space="preserve">car une progreſſion quel-<lb/>conque peut commencer par l’unité, ou par une grandeur éle-<lb/>vée à la puiſſance zero, comme a°, q°, qui ne différe pas de <lb/>l’unité (art. </s> <s xml:id="echoid-s4606" xml:space="preserve">136).</s> <s xml:id="echoid-s4607" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s4608" xml:space="preserve">Des Logarithmes, de leur nature, & </s> <s xml:id="echoid-s4609" xml:space="preserve">de leurs uſages.</s> <s xml:id="echoid-s4610" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div250" type="section" level="1" n="223"> <head xml:id="echoid-head257" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s4611" xml:space="preserve">256. </s> <s xml:id="echoid-s4612" xml:space="preserve">Les logarithmes ſont des nombres en progreſſion arith-<lb/>métique, correſpondans à d’autres nombres en progreſſion <lb/>géométrique. </s> <s xml:id="echoid-s4613" xml:space="preserve">Par exemple, ſi l’on diſpoſe l’une au deſſous de <lb/>l’autre, ces deux ſuites 2, 4, 8, 16, 32; </s> <s xml:id="echoid-s4614" xml:space="preserve">& </s> <s xml:id="echoid-s4615" xml:space="preserve">35, 7, 9, 11, dont <lb/>la premiere eſt une progreſſion géométrique, & </s> <s xml:id="echoid-s4616" xml:space="preserve">la ſeconde <lb/>une progreſſion arithmétique, comme on le voit ici:</s> <s xml:id="echoid-s4617" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4618" xml:space="preserve">3, 5, 7, 9, 11</s> </p> <p> <s xml:id="echoid-s4619" xml:space="preserve">2, 4, 8, 16, 32.</s> <s xml:id="echoid-s4620" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4621" xml:space="preserve">Chaque terme inférieur de la progreſſion arithmétique eſt <lb/>appellé logarithme du terme inférieur correſpondant: </s> <s xml:id="echoid-s4622" xml:space="preserve">ainſi 3 <lb/>eſt le logarithme de 2, 5 celui de 4, & </s> <s xml:id="echoid-s4623" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s4624" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4625" xml:space="preserve">257. </s> <s xml:id="echoid-s4626" xml:space="preserve">De même ſi l’on prend ces deux autres ſuites,</s> </p> <note position="right" xml:space="preserve">0 # 1 # 2 # 3 # 4 # 5 <lb/>1, # 10, # 100, # 1000, # 10000, # 100000, <lb/></note> <p> <s xml:id="echoid-s4627" xml:space="preserve">dont l’une eſt une progreſſion arithmétique, dont la différence <lb/>eſt l’unité, & </s> <s xml:id="echoid-s4628" xml:space="preserve">l’autre eſt la progreſſion géométrique réſultante <lb/>des différentes puiſſances de 10: </s> <s xml:id="echoid-s4629" xml:space="preserve">chaque terme de la progreſ-<lb/>ſion arithmétique ſera le logarithme du terme de la progreſſion <lb/>géométrique auquel il répond: </s> <s xml:id="echoid-s4630" xml:space="preserve">ainſi 1 eſt le logarithme de 10, <lb/>3 eſt celui de 1000, & </s> <s xml:id="echoid-s4631" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s4632" xml:space="preserve"/> </p> <pb o="131" file="0169" n="169" rhead="DE MATHÉMATIQUE. Liv. II."/> </div> <div xml:id="echoid-div251" type="section" level="1" n="224"> <head xml:id="echoid-head258" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s4633" xml:space="preserve">258. </s> <s xml:id="echoid-s4634" xml:space="preserve">Comme on peut prendre une infinité de progreſſions <lb/>arithmétiques, dont les termes ſoient poſés au deſſus de ceux <lb/>d’une progreſſion géométrique, il ſuit delà que chaque terme <lb/>de cette progreſſion pourroit avoir une infinité de logarithmes: <lb/></s> <s xml:id="echoid-s4635" xml:space="preserve">mais on eſt convenu de donner à la progreſſion décuple les <lb/>logarithmes de la progreſſion arithmétique des nombres na-<lb/>turels, en donnant zero pour logarithme à l’unité.</s> <s xml:id="echoid-s4636" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div252" type="section" level="1" n="225"> <head xml:id="echoid-head259" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s4637" xml:space="preserve">Comme les propriétés des logarithmes dépendent des pro-<lb/>portions, progreſſions géométriques & </s> <s xml:id="echoid-s4638" xml:space="preserve">arithmétiques, & </s> <s xml:id="echoid-s4639" xml:space="preserve">de plus <lb/>de celles des expoſans, comme on le verra ci-après, il eſt de la <lb/>derniere importance d’avoir préſent à l’eſprit tout ce que <lb/>nous avons vu ſur ces différentes parties: </s> <s xml:id="echoid-s4640" xml:space="preserve">c’eſt pourquoi nous <lb/>allons reprendre la formule des progreſſions géométriques, & </s> <s xml:id="echoid-s4641" xml:space="preserve"><lb/>l’examiner par rapport aux logarithmes.</s> <s xml:id="echoid-s4642" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div253" type="section" level="1" n="226"> <head xml:id="echoid-head260" xml:space="preserve">PROPOSITION XVII.</head> <head xml:id="echoid-head261" xml:space="preserve"><emph style="sc">Theoreme fondamental</emph>.</head> <p style="it"> <s xml:id="echoid-s4643" xml:space="preserve">259. </s> <s xml:id="echoid-s4644" xml:space="preserve">Dans la ſuite des puiſſances d’une quantité quelconque, <lb/>dont les termes forment une progreſſion géométrique, les expoſans <lb/>ſont en progreſſion arithmétique.</s> <s xml:id="echoid-s4645" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div254" type="section" level="1" n="227"> <head xml:id="echoid-head262" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s4646" xml:space="preserve">Que cette ſuite ſoit repréſentée par celle des puiſſances ſuc-<lb/>ceſſives de q, qui eſt {:</s> <s xml:id="echoid-s4647" xml:space="preserve">/:</s> <s xml:id="echoid-s4648" xml:space="preserve">} q<emph style="sub">0</emph>. </s> <s xml:id="echoid-s4649" xml:space="preserve">q<emph style="sub">1</emph>. </s> <s xml:id="echoid-s4650" xml:space="preserve">q<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4651" xml:space="preserve">q<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4652" xml:space="preserve">q<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4653" xml:space="preserve">q<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4654" xml:space="preserve">q<emph style="sub">6</emph>. </s> <s xml:id="echoid-s4655" xml:space="preserve">q<emph style="sub">7</emph>. </s> <s xml:id="echoid-s4656" xml:space="preserve">q<emph style="sub">8</emph>. </s> <s xml:id="echoid-s4657" xml:space="preserve">q<emph style="sub">9</emph>. </s> <s xml:id="echoid-s4658" xml:space="preserve">q<emph style="sub">10</emph>, &</s> <s xml:id="echoid-s4659" xml:space="preserve">c, <lb/>il eſt évident que ces quantités forment une progreſſion géo-<lb/>métrique, comme nous l’avons déja dit, puiſque chaque ter-<lb/>me, diviſé par le précédent, donne toujours le même quotient <lb/>q. </s> <s xml:id="echoid-s4660" xml:space="preserve">De plus il eſt encore évident que les expoſans ſont en pro-<lb/>greſſion arithmétique, qui eſt celle des nombres naturels. <lb/></s> <s xml:id="echoid-s4661" xml:space="preserve">C. </s> <s xml:id="echoid-s4662" xml:space="preserve">Q. </s> <s xml:id="echoid-s4663" xml:space="preserve">F. </s> <s xml:id="echoid-s4664" xml:space="preserve">D.</s> <s xml:id="echoid-s4665" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div255" type="section" level="1" n="228"> <head xml:id="echoid-head263" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p><s xml:id="echoid-s4666" xml:space="preserve">60. </s> <s xml:id="echoid-s4667" xml:space="preserve">Donc ces expoſans peuvent être regardés comme les <lb/>logarithmes des termes auxquels ils répondent, ſuivant la dé-<lb/>finition des logarithmes: </s> <s xml:id="echoid-s4668" xml:space="preserve">ainſi le logarithme d’un nombre n’eſt <lb/>autre choſe que l’expoſant d’une puiſſance; </s> <s xml:id="echoid-s4669" xml:space="preserve">& </s> <s xml:id="echoid-s4670" xml:space="preserve">ce que nous diſons <pb o="132" file="0170" n="170" rhead="NOUVEAU COURS"/> ici des lettres, peut s’entendre des nombres, par exemple, la <lb/>progreſſion géométrique double, qui réſulte de toutes les puiſ <lb/>ſances ſucceſſives de 2, qui eſt # {:</s> <s xml:id="echoid-s4671" xml:space="preserve">/:</s> <s xml:id="echoid-s4672" xml:space="preserve">} 1. </s> <s xml:id="echoid-s4673" xml:space="preserve">2. </s> <s xml:id="echoid-s4674" xml:space="preserve">4. </s> <s xml:id="echoid-s4675" xml:space="preserve">8. </s> <s xml:id="echoid-s4676" xml:space="preserve">16. </s> <s xml:id="echoid-s4677" xml:space="preserve">32. </s> <s xml:id="echoid-s4678" xml:space="preserve">64, &</s> <s xml:id="echoid-s4679" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s4680" xml:space="preserve">auroit pu s’écrire ainſi # {:</s> <s xml:id="echoid-s4681" xml:space="preserve">/:</s> <s xml:id="echoid-s4682" xml:space="preserve">} 2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s4683" xml:space="preserve">2<emph style="sub">1</emph>. </s> <s xml:id="echoid-s4684" xml:space="preserve">2<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4685" xml:space="preserve">2<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4686" xml:space="preserve">2<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4687" xml:space="preserve">2<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4688" xml:space="preserve">2<emph style="sub">6</emph>, &</s> <s xml:id="echoid-s4689" xml:space="preserve">c. </s> <s xml:id="echoid-s4690" xml:space="preserve"><lb/>Et de même la progreſſion décuple, ou celle des puiſſances ſuc-<lb/>ceſſives de 10, qui eſt # {:</s> <s xml:id="echoid-s4691" xml:space="preserve">/:</s> <s xml:id="echoid-s4692" xml:space="preserve">} 1. </s> <s xml:id="echoid-s4693" xml:space="preserve">10. </s> <s xml:id="echoid-s4694" xml:space="preserve">100. </s> <s xml:id="echoid-s4695" xml:space="preserve">1000. </s> <s xml:id="echoid-s4696" xml:space="preserve">10000. </s> <s xml:id="echoid-s4697" xml:space="preserve">100000, <lb/>auroit pu s’écrire ainſi # {:</s> <s xml:id="echoid-s4698" xml:space="preserve">/:</s> <s xml:id="echoid-s4699" xml:space="preserve">} 10<emph style="sub">0</emph>. </s> <s xml:id="echoid-s4700" xml:space="preserve">10<emph style="sub">1</emph>. </s> <s xml:id="echoid-s4701" xml:space="preserve">10<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4702" xml:space="preserve">10<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4703" xml:space="preserve">10<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4704" xml:space="preserve">10<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4705" xml:space="preserve"><lb/>Dans l’une & </s> <s xml:id="echoid-s4706" xml:space="preserve">dans l’autre, les nombres 0, 1, 2, 3, 4, 5 ſont <lb/>les logarithmes des termes auxquels ils répondent, & </s> <s xml:id="echoid-s4707" xml:space="preserve">en même <lb/>tems les expoſans des puiſſances de 10. </s> <s xml:id="echoid-s4708" xml:space="preserve">Nous avons déja averti <lb/>que l’on s’en tenoit à la derniere ſuite pour calculer les loga-<lb/>rithmes des nombres naturels, comme nous le verrons dans la <lb/>ſuite.</s> <s xml:id="echoid-s4709" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div256" type="section" level="1" n="229"> <head xml:id="echoid-head264" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s4710" xml:space="preserve">261. </s> <s xml:id="echoid-s4711" xml:space="preserve">Donc ſi l’on prend quatre termes quelconques en pro-<lb/>portion géométrique, leurs expoſans ou leurs logarithmes for-<lb/>meront une proportion arithmétique. </s> <s xml:id="echoid-s4712" xml:space="preserve">Par exemple, ſi l’on <lb/>prend ces quatre termes q<emph style="sub">0</emph>, q<emph style="sub">1</emph>, q<emph style="sub">4</emph> & </s> <s xml:id="echoid-s4713" xml:space="preserve">q<emph style="sub">5</emph> qui ſont en proportion <lb/>géométrique, puiſque l’on a q<emph style="sub">0</emph>. </s> <s xml:id="echoid-s4714" xml:space="preserve">q<emph style="sub">1</emph>: </s> <s xml:id="echoid-s4715" xml:space="preserve">q<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4716" xml:space="preserve">q<emph style="sub">5</emph>, & </s> <s xml:id="echoid-s4717" xml:space="preserve">que d’ailleurs le <lb/>produit des extrêmes eſt égal à celui des moyens, il eſt viſible <lb/>que leurs expoſans ou leurs logarithmes ſont en proportion <lb/>arithmétique, puiſque 0. </s> <s xml:id="echoid-s4718" xml:space="preserve">1 : </s> <s xml:id="echoid-s4719" xml:space="preserve">4. </s> <s xml:id="echoid-s4720" xml:space="preserve">5.</s> <s xml:id="echoid-s4721" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div257" type="section" level="1" n="230"> <head xml:id="echoid-head265" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s4722" xml:space="preserve">262. </s> <s xml:id="echoid-s4723" xml:space="preserve">Pour trouver le produit d’un terme de cette ſuite par <lb/>un autre, il faut chercher un terme, dont l’expoſant ſoit égal <lb/>à la ſomme des expoſans des deux termes: </s> <s xml:id="echoid-s4724" xml:space="preserve">car on a vu dans le <lb/>calcul des expoſans (art: </s> <s xml:id="echoid-s4725" xml:space="preserve">134), que le produit des quantités <lb/>exponentielles ſe trouve par l’addition des expoſans. </s> <s xml:id="echoid-s4726" xml:space="preserve">Ainſi <lb/>pour multiplier q<emph style="sub">2</emph> par q<emph style="sub">3</emph>, je cherche le terme dont l’expoſant <lb/>ſoit 5, égal à la ſomme des expoſans 2 + 3, & </s> <s xml:id="echoid-s4727" xml:space="preserve">le terme q<emph style="sub">5</emph> eſt <lb/>le produit demandé. </s> <s xml:id="echoid-s4728" xml:space="preserve">Donc pour avoir le produit de deux <lb/>nombres par le moyen des logarithmes, il faut ajouter les lo-<lb/>garithmes de ces deux nombres, & </s> <s xml:id="echoid-s4729" xml:space="preserve">la ſomme ſera le logarithme <lb/>du produit, pourvu que la progreſſion arithmétique que l’on a <lb/>choiſie, ſoit telle que zero ſoit le logarithme de l’unité.</s> <s xml:id="echoid-s4730" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div258" type="section" level="1" n="231"> <head xml:id="echoid-head266" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s4731" xml:space="preserve">263. </s> <s xml:id="echoid-s4732" xml:space="preserve">Pour diviſer un terme quelconque de cette ſuite par <pb o="133" file="0171" n="171" rhead="DE MATHÉMATIQUE. Liv. II."/> un autre, il faut retrancher l’expoſant du diviſeur de celui du <lb/>dividende, & </s> <s xml:id="echoid-s4733" xml:space="preserve">la diſſérence ſera l’expoſant du quotient: </s> <s xml:id="echoid-s4734" xml:space="preserve">par <lb/>exemple, pour diviſer q<emph style="sub">9</emph> par q<emph style="sub">4</emph>, je retranche 4 de 9, le reſte 5 <lb/>eſt l’expoſant du quotient, qui eſt q<emph style="sub">5</emph>: </s> <s xml:id="echoid-s4735" xml:space="preserve">car on a vu dans le calcul <lb/>des expoſans (art. </s> <s xml:id="echoid-s4736" xml:space="preserve">135), que la diviſion ſe fait par la ſouſtraction <lb/>des expoſans de ces quantités. </s> <s xml:id="echoid-s4737" xml:space="preserve">Donc en général pour diviſer un <lb/>nombre par un autre, par le moyen des logarithmes, il ſaut <lb/>ſouſtraire le logarithme du diviſeur de celui du dividende, & </s> <s xml:id="echoid-s4738" xml:space="preserve"><lb/>chercher un nombre, dont le logarithme ſoit égal à la diffé-<lb/>rence des deux logarithmes des nombres donnés; </s> <s xml:id="echoid-s4739" xml:space="preserve">le nombre <lb/>correſpondant ſera le quotient que l’on demande, en ſuppo-<lb/>ſant toujours que zero ſoit le logarithme de l’unité.</s> <s xml:id="echoid-s4740" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div259" type="section" level="1" n="232"> <head xml:id="echoid-head267" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s4741" xml:space="preserve">264. </s> <s xml:id="echoid-s4742" xml:space="preserve">Pour faire une Regle de Trois par le moyen des loga-<lb/>rithmes, il faudra ajouter enſemble les logarithmes des deux <lb/>moyens, & </s> <s xml:id="echoid-s4743" xml:space="preserve">de la ſomme retrancher le logarithme du premier <lb/>extrême, le reſte ſera le logarithme du dernier extrême: </s> <s xml:id="echoid-s4744" xml:space="preserve">car <lb/>une regle de Trois ſe fait en multipliant ces deux moyens l’un <lb/>par l’autre, & </s> <s xml:id="echoid-s4745" xml:space="preserve">diviſant par le premier extrême. </s> <s xml:id="echoid-s4746" xml:space="preserve">Mais par le <lb/>corollaire 3<emph style="sub">e</emph>, la multiplication de deux termes de notre pro-<lb/>greſſion ſe fait par l’addition des logarithmes ou expoſans des <lb/>deux moyens, & </s> <s xml:id="echoid-s4747" xml:space="preserve">le terme qui a pour expoſant la ſomme de ces <lb/>expoſans, eſt le produit de ces deux termes. </s> <s xml:id="echoid-s4748" xml:space="preserve">Et par le corol-<lb/>laire 4<emph style="sub">e</emph>, la diviſion de ce produit par le premier terme ſe fait <lb/>par la ſouſtraction des expoſans: </s> <s xml:id="echoid-s4749" xml:space="preserve">donc en ôtant l’expoſant du <lb/>premier terme de la ſomme des expoſans des deux moyens, on <lb/>a l’expoſant ou le logarithme du quatrieme terme. </s> <s xml:id="echoid-s4750" xml:space="preserve">Ainſi pour <lb/>trouver un terme quatrieme proportionnel géométrique aux <lb/>trois termes q<emph style="sub">2</emph>, q<emph style="sub">3</emph>, 9<emph style="sub">5</emph>, je prends la ſomme 8 des expoſans 5.</s> <s xml:id="echoid-s4751" xml:space="preserve">3 <lb/>des termes moyens q<emph style="sub">3</emph>, q<emph style="sub">5</emph>; </s> <s xml:id="echoid-s4752" xml:space="preserve">de cette ſomme j’ôte 2, expoſant <lb/>du premier, & </s> <s xml:id="echoid-s4753" xml:space="preserve">le reſte 6 eſt le logarithme du quatrieme terme <lb/>que je cherche, qui eſt q<emph style="sub">6</emph>: </s> <s xml:id="echoid-s4754" xml:space="preserve">& </s> <s xml:id="echoid-s4755" xml:space="preserve">en effet, q<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4756" xml:space="preserve">q<emph style="sub">3</emph>: </s> <s xml:id="echoid-s4757" xml:space="preserve">q<emph style="sub">5</emph>. </s> <s xml:id="echoid-s4758" xml:space="preserve">q<emph style="sub">6</emph>, puiſ-<lb/>que le produit des extrêmes eſt égal à celui des moyens. </s> <s xml:id="echoid-s4759" xml:space="preserve">D’ail-<lb/>leurs, comme ces quatre termes ſont en proportion géomé-<lb/>trique, leurs expoſans ou logarithmes, par le corollaire 2, ſont <lb/>en proportion arithmétique: </s> <s xml:id="echoid-s4760" xml:space="preserve">ainſi le logarithme que l’on cher-<lb/>che eſt le quatrieme terme d’une proportion arithmétique, qui <lb/>ſe détermine en ôtant le premier terme de la ſomme des deux <lb/>moyens (art. </s> <s xml:id="echoid-s4761" xml:space="preserve">230). </s> <s xml:id="echoid-s4762" xml:space="preserve">Ainſi en général pour faire une Regle de <pb o="134" file="0172" n="172" rhead="NOUVEAU COURS"/> Trois par les logarithmes, il faut ajouter enſemble les loga-<lb/>rithmes des moyens, & </s> <s xml:id="echoid-s4763" xml:space="preserve">de la ſomme ôter celui du premier ex-<lb/>trême, le reſte eſt celui du quatrieme terme.</s> <s xml:id="echoid-s4764" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4765" xml:space="preserve">265. </s> <s xml:id="echoid-s4766" xml:space="preserve">Comme toute Multiplication renferme cette propor-<lb/>tion, l’unité eſt au multiplicateur, comme le multiplicande eſt au <lb/>produit; </s> <s xml:id="echoid-s4767" xml:space="preserve">il ſuit que faire une Multiplication ou une Regle de <lb/>Trois c’eſt la même choſe: </s> <s xml:id="echoid-s4768" xml:space="preserve">donc il faut ajouter le logarithme <lb/>du multiplicateur à celui du multiplicande, & </s> <s xml:id="echoid-s4769" xml:space="preserve">de la ſomme <lb/>ôter le logarithme de l’unité. </s> <s xml:id="echoid-s4770" xml:space="preserve">C’eſt pour cela que dans les pro-<lb/>greſſions arithmétiques que l’on a choiſies, pour déterminer les <lb/>logarithmes des nombres naturels, on a donné zero pour loga-<lb/>rithme à l’unité, afin que toute multiplication ſe réduisît à <lb/>l’addition de deux nombres.</s> <s xml:id="echoid-s4771" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4772" xml:space="preserve">266. </s> <s xml:id="echoid-s4773" xml:space="preserve">Comme toute Diviſion renferme cette proportion, <lb/>l’unité eſt au diviſeur, comme le quotient eſt au dividende: </s> <s xml:id="echoid-s4774" xml:space="preserve">il ſuit <lb/>qu’on ne peut faire une diviſion qu’on ne faſſe réellement une <lb/>regle de Trois; </s> <s xml:id="echoid-s4775" xml:space="preserve">& </s> <s xml:id="echoid-s4776" xml:space="preserve">comme dans cette regle de proportion, le <lb/>terme que l’on cherche eſt le troiſieme, il faut ajouter enſem-<lb/>ble les logarithmes ou expoſans des extrêmes, qui ſont l’unité <lb/>& </s> <s xml:id="echoid-s4777" xml:space="preserve">le dividende, & </s> <s xml:id="echoid-s4778" xml:space="preserve">de la ſomme ôter l’expoſant du diviſeur, <lb/>pour avoir le logarithme ou l’expoſant du quotient: </s> <s xml:id="echoid-s4779" xml:space="preserve">donc ſi le <lb/>logarithme de l’unité eſt zero, toute diviſion ſur les loga-<lb/>rithmes ſe réduira à la ſouſtraction de deux nombres; </s> <s xml:id="echoid-s4780" xml:space="preserve">c’eſt <lb/>encore pour cette raiſon que l’on a donné zero pour logarithme <lb/>à l’unité.</s> <s xml:id="echoid-s4781" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div260" type="section" level="1" n="233"> <head xml:id="echoid-head268" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s4782" xml:space="preserve">267. </s> <s xml:id="echoid-s4783" xml:space="preserve">Pour élever un terme quelconque à une puiſſance pro-<lb/>poſée, il ſuffit de multiplier ſon expoſant par celui de la puiſ-<lb/>ſance à la quelle on veut l’élever, & </s> <s xml:id="echoid-s4784" xml:space="preserve">faire du produit l’expoſant <lb/>de la même lettre, qui ſera la puiſſance demandée, comme on <lb/>l’a démontré dans la formation des puiſſances des quantités <lb/>exponentielles. </s> <s xml:id="echoid-s4785" xml:space="preserve">Par exemple, pour élever q<emph style="sub">2</emph> au cube, je mul-<lb/>tiplie ſon expoſant 2 par 3, expoſant de la puiſſance deman-<lb/>dée; </s> <s xml:id="echoid-s4786" xml:space="preserve">le produit 6 mis en expoſant au devant de la même quan-<lb/>tité, me donne q<emph style="sub">6</emph>, qui eſt le cube de q<emph style="sub">2</emph>: </s> <s xml:id="echoid-s4787" xml:space="preserve">donc en général pour <lb/>trouver la puiſſance d’un nombre, par le moyen des logarith-<lb/>mes, il faut multiplier le logarithme de ce nombre par l’ex-<lb/>poſant de la puiſſance, & </s> <s xml:id="echoid-s4788" xml:space="preserve">le produit ſera le logarithme de la <lb/>puiſſance que l’on demande, que l’on trouvera à côté de ce <lb/>même logarithme.</s> <s xml:id="echoid-s4789" xml:space="preserve"/> </p> <pb o="135" file="0173" n="173" rhead="DE MATHÉMATIQUE. Liv. II."/> </div> <div xml:id="echoid-div261" type="section" level="1" n="234"> <head xml:id="echoid-head269" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s4790" xml:space="preserve">268. </s> <s xml:id="echoid-s4791" xml:space="preserve">Pour extraire la racine d’un terme quelconque de <lb/>cette ſuite, il faut diviſer l’expoſant ou le logarithme de ce <lb/>terme par l’expoſant de la racine, par 2 ſi c’eſt la racine quar-<lb/>rée que l’on demande, par 3 ſi c’eſt la racine cubique, & </s> <s xml:id="echoid-s4792" xml:space="preserve">ainſi <lb/>des autres: </s> <s xml:id="echoid-s4793" xml:space="preserve">car on a vu dans le Traité du calcul des expo-<lb/>ſans, que la racine des quantités exponentielles ſe fait en di-<lb/>viſant leur expoſant par l’expoſant de la racine. </s> <s xml:id="echoid-s4794" xml:space="preserve">Ainſi pour <lb/>extraire la racine cubique de q<emph style="sub">9</emph>, je diviſe le logarithme ou ex-<lb/>poſant 9 par 3, le quotient eſt 3: </s> <s xml:id="echoid-s4795" xml:space="preserve">ainſi q<emph style="sub">3</emph> eſt la racine cubique <lb/>de cette quantité. </s> <s xml:id="echoid-s4796" xml:space="preserve">Donc en général, par le moyen des loga-<lb/>rithmes, l’extraction d’une racine quarrée ou cubique ſe ré-<lb/>duit à diviſer un nombre par 2 ou par 3; </s> <s xml:id="echoid-s4797" xml:space="preserve">& </s> <s xml:id="echoid-s4798" xml:space="preserve">c’eſt principale-<lb/>ment dans cette opération que l’on voit tout d’un coup l’im-<lb/>portance de cette découverte, dont on eſt redevable au Baron <lb/>de Neper, Ecoſſois, dont le nom ſera toujours reſpecté des <lb/>plus grands Calculateurs.</s> <s xml:id="echoid-s4799" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div262" type="section" level="1" n="235"> <head xml:id="echoid-head270" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s4800" xml:space="preserve">269. </s> <s xml:id="echoid-s4801" xml:space="preserve">Comme tout ceci eſt de la derniere importance, nous <lb/>allons en faire l’application ſur un ſyſtême de logarithme quel-<lb/>conque, différent de celui des Tables ordinaires, après quoi <lb/>nous expoſerons en peu de mots la maniere dont on a trouvé <lb/>les logarithmes des nombres naturels. </s> <s xml:id="echoid-s4802" xml:space="preserve">Nous ne pouvons trop <lb/>recommander aux Commençans de s’appliquer à généraliſer <lb/>les idées, en examinant particuliérement la poſſibilité d’une <lb/>infinité de ſyſtêmes de logarithmes, & </s> <s xml:id="echoid-s4803" xml:space="preserve">en tâchant de décou-<lb/>vrir les raiſons qui ont déterminé les premiers qui en ont cal-<lb/>culé des Tables, à ſe ſervir de la progreſſion décuple. </s> <s xml:id="echoid-s4804" xml:space="preserve">On verra <lb/>que cette raiſon eſt priſe de la nature des logarithmes conſi-<lb/>dérés comme expoſans des puiſſances de 10.</s> <s xml:id="echoid-s4805" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Logarithmes # {./.} # 0. # 1. # 2. # 3. # 4. # 5. # 6. # 7. # 8. # 9 <lb/>Progreſſion géométrique # {../..} # 1. # 2. # 4. # 8. # 16. # 32. # 64. # 128. # 256. # 512. <lb/></note> <p> <s xml:id="echoid-s4806" xml:space="preserve">1°. </s> <s xml:id="echoid-s4807" xml:space="preserve">Pour multiplier un terme quelconque de cette ſuite, 8, <lb/>par exemple par 16, j’ajoute enſemble leurs logarithmes 3 & </s> <s xml:id="echoid-s4808" xml:space="preserve">4, <lb/>la ſomme eſt 7; </s> <s xml:id="echoid-s4809" xml:space="preserve">& </s> <s xml:id="echoid-s4810" xml:space="preserve">le nombre 128 qui ſe trouve au deſſous eſt <lb/>le produit de 16 par 8. </s> <s xml:id="echoid-s4811" xml:space="preserve">De même pour multiplier le nombre 8 <lb/>de la progreſſion géométrique par 32, j’ajoute enſemble leurs <pb o="136" file="0174" n="174" rhead="NOUVEAU COURS"/> logarithmes 3 & </s> <s xml:id="echoid-s4812" xml:space="preserve">5; </s> <s xml:id="echoid-s4813" xml:space="preserve">la ſomme 8 eſt le logarithme du produit <lb/>244, comme on peut s’en convaincre aiſément, en faiſant la <lb/>multiplication.</s> <s xml:id="echoid-s4814" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4815" xml:space="preserve">2°. </s> <s xml:id="echoid-s4816" xml:space="preserve">Pour diviſer un nombre quelconque de la progreſſion <lb/>géométrique par un autre terme de la même progreſſion, 128 <lb/>par 4, j’ôte le logarithme de 4 du logarithme de 128; </s> <s xml:id="echoid-s4817" xml:space="preserve">ces deux <lb/>logarithmes ſont 2 & </s> <s xml:id="echoid-s4818" xml:space="preserve">7, dont la différence 5 eſt le logarithme <lb/>du quotient 32. </s> <s xml:id="echoid-s4819" xml:space="preserve">De même pour diviſer 512 par 64, j’ôte 6, <lb/>logarithme ou expoſant du diviſeur, de 9 expoſant du divi-<lb/>dende, la différence 3 eſt le logarithme du quotient 8. </s> <s xml:id="echoid-s4820" xml:space="preserve">En <lb/>effet 512, diviſé par 64, donne 8.</s> <s xml:id="echoid-s4821" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4822" xml:space="preserve">3°. </s> <s xml:id="echoid-s4823" xml:space="preserve">Pour trouver un quatrieme terme proportionnel aux trois <lb/>nombres 4, 32, 64, je prends la ſomme des logarithmes des <lb/>deux moyens, qui eſt 11, j’en ôte le logarithme 2 du premier <lb/>extrême 4, le reſte eſt 9, logarithme de 512 qui eſt le terme <lb/>que l’on demande.</s> <s xml:id="echoid-s4824" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4825" xml:space="preserve">4°. </s> <s xml:id="echoid-s4826" xml:space="preserve">Pour élever 8 au cube, je multiplie ſon expoſant ou ſon <lb/>logarithme, qui eſt 3 par 3, expoſant de la puiſſance, & </s> <s xml:id="echoid-s4827" xml:space="preserve">j’ai 9 <lb/>au produit, qui eſt le logarithme du cube de 8, qui eſt 512, <lb/>comme on l’a déja vu par la Table des Cubes.</s> <s xml:id="echoid-s4828" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4829" xml:space="preserve">5°. </s> <s xml:id="echoid-s4830" xml:space="preserve">Pour extraire la racine quarrée de 256, je diviſe ſon lo-<lb/>garithme 8 par 2, expoſant de la racine quarré le quotient <lb/>4 eſt le logarithme de la racine 16: </s> <s xml:id="echoid-s4831" xml:space="preserve">élevant 16 au quarré, on <lb/>aura effectivement 256, comme il eſt aiſé de le voir.</s> <s xml:id="echoid-s4832" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div263" type="section" level="1" n="236"> <head xml:id="echoid-head271" xml:space="preserve"><emph style="sc">Remarque</emph> <emph style="sc">Générale.</emph></head> <p> <s xml:id="echoid-s4833" xml:space="preserve">270. </s> <s xml:id="echoid-s4834" xml:space="preserve">On voit par-là que toute Multiplication ſe réduit à <lb/>l’Addition de deux nombres; </s> <s xml:id="echoid-s4835" xml:space="preserve">que toute Diviſion ſe fait par la <lb/>Souſtraction de deux nombres; </s> <s xml:id="echoid-s4836" xml:space="preserve">& </s> <s xml:id="echoid-s4837" xml:space="preserve">que toute Regle de Trois ſe <lb/>fait par l’Addition de deux nombres, & </s> <s xml:id="echoid-s4838" xml:space="preserve">par la Souſtraction <lb/>d’un troiſieme de la ſomme des deux premiers; </s> <s xml:id="echoid-s4839" xml:space="preserve">enfin que la <lb/>formation des puiſſances ſe fait en doublant ou triplant le lo-<lb/>garithme du nombre, dont on veut avoir le quarré ou le cube, <lb/>& </s> <s xml:id="echoid-s4840" xml:space="preserve">que l’extrction des racines ſe réduit à prendre la moitié, <lb/>le tiers, ou le quart du logarithme d’un nombre propoſé, pour <lb/>avoir la racine ſeconde, troiſieme, ou quatrieme. </s> <s xml:id="echoid-s4841" xml:space="preserve">Mais pour <lb/>cela, il faut que les nombres propoſés ſoient préciſément quel-<lb/>ques-uns des termes de la progreſſion, pour avoir leurs loga-<lb/>rithmes. </s> <s xml:id="echoid-s4842" xml:space="preserve">Ainſi afin de rendre un ſi grand avantage pratica-<lb/>ble ſur tous les nombres poſſibles, il a fallu trouver leurs lo- <pb o="137" file="0175" n="175" rhead="DE MATHÉMATIQUE. Liv. II."/> garithmes, ou, ce qui eſt la même choſe, l’expoſant du rang <lb/>que chacun occupe dans la progreſſion des nombres à laquelle <lb/>on s’eſt arrêté pour calculer les logarithmes. </s> <s xml:id="echoid-s4843" xml:space="preserve">C’eſt ce que nous <lb/>allons détailler dans les articles ſuivans.</s> <s xml:id="echoid-s4844" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4845" xml:space="preserve">271. </s> <s xml:id="echoid-s4846" xml:space="preserve">On a imaginé que tous les nombres naturels étoient <lb/>renfermés dans une ſeule progreſſion géométrique, dont cha-<lb/>queterme étoit des puiſſances différentes du nombre 10; </s> <s xml:id="echoid-s4847" xml:space="preserve">toutes <lb/>puiſſances fractionnaires, excepté les termes de la progreſſion <lb/>décuple, {.</s> <s xml:id="echoid-s4848" xml:space="preserve">./.</s> <s xml:id="echoid-s4849" xml:space="preserve">.}10. </s> <s xml:id="echoid-s4850" xml:space="preserve">100. </s> <s xml:id="echoid-s4851" xml:space="preserve">1000. </s> <s xml:id="echoid-s4852" xml:space="preserve">10000, &</s> <s xml:id="echoid-s4853" xml:space="preserve">c, qui ſont des puiſſances <lb/>complettes de 10. </s> <s xml:id="echoid-s4854" xml:space="preserve">Pour cela, on a inſéré entre 1 & </s> <s xml:id="echoid-s4855" xml:space="preserve">10 9999999 <lb/>moyens géométriques, & </s> <s xml:id="echoid-s4856" xml:space="preserve">entre chaque expoſant 0 & </s> <s xml:id="echoid-s4857" xml:space="preserve">1 de ces <lb/>nombres, autant de moyens arithmétiques correſpondans aux <lb/>premiers; </s> <s xml:id="echoid-s4858" xml:space="preserve">& </s> <s xml:id="echoid-s4859" xml:space="preserve">pour avoir plus commodément ces moyens arith-<lb/>métiques, on a ajouté ſept décimales à la ſuite de chaque ex-<lb/>poſant; </s> <s xml:id="echoid-s4860" xml:space="preserve">ce qui ne change pas la progreſſion arithmétique. </s> <s xml:id="echoid-s4861" xml:space="preserve">Ainſi <lb/>au lieu de la premiere ſuite {.</s> <s xml:id="echoid-s4862" xml:space="preserve">./.</s> <s xml:id="echoid-s4863" xml:space="preserve">.} 10<emph style="sub">0</emph>. </s> <s xml:id="echoid-s4864" xml:space="preserve">10<emph style="sub">1</emph>. </s> <s xml:id="echoid-s4865" xml:space="preserve">10<emph style="sub">2</emph>. </s> <s xml:id="echoid-s4866" xml:space="preserve">10<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4867" xml:space="preserve">10<emph style="sub">4</emph>. </s> <s xml:id="echoid-s4868" xml:space="preserve">10<emph style="sub">5</emph>, on <lb/>a celle-ci, {.</s> <s xml:id="echoid-s4869" xml:space="preserve">./.</s> <s xml:id="echoid-s4870" xml:space="preserve">.}10<emph style="sub">0.0000000</emph>. </s> <s xml:id="echoid-s4871" xml:space="preserve">10<emph style="sub">1.0000000</emph>. </s> <s xml:id="echoid-s4872" xml:space="preserve">10<emph style="sub">3.0000000</emph>, &</s> <s xml:id="echoid-s4873" xml:space="preserve">c. </s> <s xml:id="echoid-s4874" xml:space="preserve">tou-<lb/>jours telle que les expoſans ſont en progreſſion arithmétique, <lb/>& </s> <s xml:id="echoid-s4875" xml:space="preserve">que chaque terme eſt une puiſſance complette du nombre <lb/>10. </s> <s xml:id="echoid-s4876" xml:space="preserve">En ſuppoſant donc qu’entre les expoſans 0.</s> <s xml:id="echoid-s4877" xml:space="preserve">0000000, il y <lb/>ait 999,9999 moyens arithmétiques, on trouvera que le pre-<lb/>mier eſt 0.</s> <s xml:id="echoid-s4878" xml:space="preserve">0000001, & </s> <s xml:id="echoid-s4879" xml:space="preserve">que le terme de la progreſſion géomé-<lb/>trique qui lui répond, ou, ce qui eſt la même choſe que la <lb/>puiſſance de 10 correſpondante à ce logarithme, eſt 10<emph style="sub">0.0000001</emph>: <lb/></s> <s xml:id="echoid-s4880" xml:space="preserve">car, ſelon l’article 243, pour inſérer un nombre de moyens <lb/>arithmétiques entre deux nombres quelconques, il faut ôter <lb/>le plus petit du plus grand, & </s> <s xml:id="echoid-s4881" xml:space="preserve">diviſer le reſte par le nombre <lb/>des moyens que l’on demande, augmenté de l’unité. </s> <s xml:id="echoid-s4882" xml:space="preserve">Suivant <lb/>cette regle, j’ôte le plus petit terme 0.</s> <s xml:id="echoid-s4883" xml:space="preserve">0000000 de 1.</s> <s xml:id="echoid-s4884" xml:space="preserve">0000000, <lb/>ou, ce qui eſt la même choſe, 0 de 1, le reſte eſt 1, que je <lb/>diviſe par le nombre 9999999 des moyens arithmétiques pro-<lb/>portionnels, augmenté de l’unité, qui eſt 10000000. </s> <s xml:id="echoid-s4885" xml:space="preserve">Ce pre-<lb/>mier moyen arithmétique eſt donc {1/10000000}, ou en réduiſant <lb/>cette fraction en décimales 0.</s> <s xml:id="echoid-s4886" xml:space="preserve">00000001; </s> <s xml:id="echoid-s4887" xml:space="preserve">le ſecond moyen <lb/>arithmétique ſera 0.</s> <s xml:id="echoid-s4888" xml:space="preserve">00000002, & </s> <s xml:id="echoid-s4889" xml:space="preserve">le terme de la progreſſion <lb/>géométrique correſpondant à ce logarithme ſera 10<emph style="sub">0.00000002</emph>, <lb/>en continuant le même raiſonnement, on a conſtruit des Ta-<lb/>bles des Logarithmes de tous les nombres naturels, & </s> <s xml:id="echoid-s4890" xml:space="preserve">l’on a <lb/>trouvé que le nombre 2 eſt à peu près égal à 10, élevé à la <lb/>puiſſance 0.</s> <s xml:id="echoid-s4891" xml:space="preserve">3010300, ou 10<emph style="sub">0.3010300</emph>. </s> <s xml:id="echoid-s4892" xml:space="preserve">On a trouvé de même que <pb o="138" file="0176" n="176" rhead="NOUVEAU COURS"/> 3 étoit égal à 10, élevé à la puiſſance 0.</s> <s xml:id="echoid-s4893" xml:space="preserve">4771213, ou égal <lb/>10<emph style="sub">0.4771213</emph>, & </s> <s xml:id="echoid-s4894" xml:space="preserve">l’on a appellé ces nombres, logarithmes de 2 & </s> <s xml:id="echoid-s4895" xml:space="preserve"><lb/>de 3.</s> <s xml:id="echoid-s4896" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4897" xml:space="preserve">272. </s> <s xml:id="echoid-s4898" xml:space="preserve">On a inſéré le même nombre de moyens arithméti-<lb/>ques entre les expoſans 1.</s> <s xml:id="echoid-s4899" xml:space="preserve">0000000, & </s> <s xml:id="echoid-s4900" xml:space="preserve">2.</s> <s xml:id="echoid-s4901" xml:space="preserve">0000000, ou entre <lb/>les nombres 1 & </s> <s xml:id="echoid-s4902" xml:space="preserve">2, & </s> <s xml:id="echoid-s4903" xml:space="preserve">l’on a trouvé que 12, par exemple, étoit <lb/>égal à 10, élevé à la puiſſance 1.</s> <s xml:id="echoid-s4904" xml:space="preserve">0791812, ou que 12 = <lb/>10<emph style="sub">1.0791812</emph>. </s> <s xml:id="echoid-s4905" xml:space="preserve">Quand on a eu une fois trouvé les logarithmes des <lb/>nombres, appellés premiers, c’eſt-à-dire qui n’ont point de <lb/>diviſeur autre que l’unité, la plus grande partie du travail <lb/>s’eſt trouvée achevée, puiſque pour avoir les logarithmes des <lb/>nombres multiples ou ſous-multiples de ceux-ci, il n’a fallu <lb/>qu’ajouter à leurs logarithmes celui du multiplicateur, ou bien <lb/>en ſouſtraire celui du diviſeur. </s> <s xml:id="echoid-s4906" xml:space="preserve">Par exemple, lorſqu’on a trouvé <lb/>que le logarithme de 2 eſt 0.</s> <s xml:id="echoid-s4907" xml:space="preserve">3010300, on a découvert aiſé-<lb/>ment & </s> <s xml:id="echoid-s4908" xml:space="preserve">ſans calcul celui de 5, en ôtant 0.</s> <s xml:id="echoid-s4909" xml:space="preserve">3010300 de 1.</s> <s xml:id="echoid-s4910" xml:space="preserve">0000000, <lb/>logarithme de 10, & </s> <s xml:id="echoid-s4911" xml:space="preserve">ce logarithme eſt 6989700.</s> <s xml:id="echoid-s4912" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4913" xml:space="preserve">273. </s> <s xml:id="echoid-s4914" xml:space="preserve">Il faut bien prendre garde que lorſque nous diſons que <lb/>l’on a renfermé dans une ſeule progreſſion géométrique tous <lb/>les nombres naturels, on ne veut pas dire pour cela que les <lb/>nombres naturels ſont en progreſſion géométrique, mais ſeu-<lb/>lement que chacun d’eux en particulier eſt un terme de cette <lb/>progreſſion, dont le numéro ou le rang qu’il occupe eſt mar-<lb/>qué par ſon logarithme. </s> <s xml:id="echoid-s4915" xml:space="preserve">Auſſi les logarithmes de quatre nom-<lb/>bres, pris de ſuite dans les Tables des Logarithmes, ne ſont-<lb/>ils pas en progreſſion arithmétique, ce qui devroit arriver, ſi <lb/>les nombres auxquels ils répondent formoient une progreſſion <lb/>géométrique.</s> <s xml:id="echoid-s4916" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4917" xml:space="preserve">274. </s> <s xml:id="echoid-s4918" xml:space="preserve">On appelle caracteriſtique d’un logarithme le nombre <lb/>de ce logarithme qui eſt au rang des entiers: </s> <s xml:id="echoid-s4919" xml:space="preserve">ainſi pour peu <lb/>que l’on y faſſe attention, on verra que le caracteriſtique <lb/>des nombres moindres que 10, eſt 0; </s> <s xml:id="echoid-s4920" xml:space="preserve">que celui des nombres <lb/>moindres que 100, eſt 1; </s> <s xml:id="echoid-s4921" xml:space="preserve">que celui des nombres moindres que <lb/>1000, eſt 2, & </s> <s xml:id="echoid-s4922" xml:space="preserve">qu’en général le caractériſtique du logarithme <lb/>d’un nombre renferme autant d’unités que la plus proche puiſ-<lb/>ſance de 10, à laquelle un nombre eſt ſupérieur, contient de <lb/>zero. </s> <s xml:id="echoid-s4923" xml:space="preserve">Ainſi le logarithme de 99 ne peut avoir pour caracté-<lb/>riſtique que l’unité, parce que la plus proche puiſſance de 10, <lb/>à laquelle il eſt ſupérieur, qui eſt 10, n’a qu’un zero.</s> <s xml:id="echoid-s4924" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4925" xml:space="preserve">275. </s> <s xml:id="echoid-s4926" xml:space="preserve">Les nombres fractionnaires, moindres que l’unité, <pb o="139" file="0177" n="177" rhead="DE MATHÉMATIQUE. Liv. II."/> auront des expoſans ou des logarithmes négatifs: </s> <s xml:id="echoid-s4927" xml:space="preserve">car dans une <lb/>progreſſion arithmétique, les termes qui ſont avant le zero ſont <lb/>négatifs; </s> <s xml:id="echoid-s4928" xml:space="preserve">& </s> <s xml:id="echoid-s4929" xml:space="preserve">d’ailleurs l’unité a zero pour expoſant. </s> <s xml:id="echoid-s4930" xml:space="preserve">Donc, &</s> <s xml:id="echoid-s4931" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s4932" xml:space="preserve">De plus, les fractions {1/2}, {1/3}, {1/4}, {1/5}, &</s> <s xml:id="echoid-s4933" xml:space="preserve">c. </s> <s xml:id="echoid-s4934" xml:space="preserve">dont le numérateur eſt <lb/>l’unité, & </s> <s xml:id="echoid-s4935" xml:space="preserve">le dénominateur, quelques-uns des nombres natu-<lb/>rels, auront pour logarithmes ceux des nombres entiers qui <lb/>leur ſervent de dénominateurs, pris en moins ou négatifs. </s> <s xml:id="echoid-s4936" xml:space="preserve"><lb/>D’où il ſuit que l’on peut aiſément opérer ſur les fractions, <lb/>par le moyen des logarithmes.</s> <s xml:id="echoid-s4937" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4938" xml:space="preserve">Si l’on veut avoir un plus grand détail des logarithmes, & </s> <s xml:id="echoid-s4939" xml:space="preserve"><lb/>particuliérement ſur la conſtruction de leurs Tables, on peut <lb/>conſulter le Livre de Trigonométrie de M. </s> <s xml:id="echoid-s4940" xml:space="preserve">Rivard. </s> <s xml:id="echoid-s4941" xml:space="preserve">Cette <lb/>étude ne peut qu’être utile, & </s> <s xml:id="echoid-s4942" xml:space="preserve">d’ailleurs comme on eſt obligé <lb/>de ſe ſervir de ces nombres artificiels dans la pratique du <lb/>calcul des triangles, on agit toujours avec plus de ſûreté dans <lb/>ſes opérations, lorſque l’on connoît bien les propriétés des <lb/>nombres dont on ſe ſert.</s> <s xml:id="echoid-s4943" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div264" type="section" level="1" n="237"> <head xml:id="echoid-head272" style="it" xml:space="preserve">Des Raiſons compoſées.</head> <head xml:id="echoid-head273" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s4944" xml:space="preserve">276. </s> <s xml:id="echoid-s4945" xml:space="preserve">Une raiſon compoſée eſt le produit de deux rapports <lb/>multipliés les uns par les autres: </s> <s xml:id="echoid-s4946" xml:space="preserve">par exemple, la raiſon de a b <lb/>à c d eſt compoſée de la raiſon de a à b, & </s> <s xml:id="echoid-s4947" xml:space="preserve">de c à d. </s> <s xml:id="echoid-s4948" xml:space="preserve">Ainſi une <lb/>raiſon compoſée peut être regardée comme le produit de deux <lb/>fractions, puiſque chaque raiſon peut être regardée comme <lb/>une fraction. </s> <s xml:id="echoid-s4949" xml:space="preserve">Il en eſt de même dans les nombres: </s> <s xml:id="echoid-s4950" xml:space="preserve">la raiſon de <lb/>10 à 21 eſt compoſée de celle de 2 à 3, & </s> <s xml:id="echoid-s4951" xml:space="preserve">de celle de 5 à 7. <lb/></s> <s xml:id="echoid-s4952" xml:space="preserve">Les raiſons de la Multiplication, deſquelles réſulte la raiſon <lb/>compoſée, ſont appellées raiſons compoſantes.</s> <s xml:id="echoid-s4953" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4954" xml:space="preserve">277. </s> <s xml:id="echoid-s4955" xml:space="preserve">Si les raiſons compoſantes ſont égales, la raiſon com-<lb/>poſée qui en réſulte eſt appellée raiſon doublée, s’il y a deux rai-<lb/>ſons égales, raiſon triplée, ſi l’on a multipliée trois raiſons <lb/>égales l’une par l’autre. </s> <s xml:id="echoid-s4956" xml:space="preserve">Par exemple, ſi l’on a la proportion <lb/>a. </s> <s xml:id="echoid-s4957" xml:space="preserve">b:</s> <s xml:id="echoid-s4958" xml:space="preserve">: c. </s> <s xml:id="echoid-s4959" xml:space="preserve">d, ou, ce qui eſt la même choſe, {a/b} = {c/d}, la raiſon <lb/>de a c à b d eſt doublée de celle de a à b, ou de celle de c à d, <lb/>puiſque la proportion ſuppoſe qu’il y a égalité entre ces deux <lb/>raiſons. </s> <s xml:id="echoid-s4960" xml:space="preserve">Si l’on a a. </s> <s xml:id="echoid-s4961" xml:space="preserve">b:</s> <s xml:id="echoid-s4962" xml:space="preserve">: c. </s> <s xml:id="echoid-s4963" xml:space="preserve">d:</s> <s xml:id="echoid-s4964" xml:space="preserve">: f. </s> <s xml:id="echoid-s4965" xml:space="preserve">g, ou {a/b} = {c/d} = {f/g}, la raiſon <pb o="140" file="0178" n="178" rhead="NOUVEAU COURS"/> de a c f à b d g ſera triplée de celle de a à b, ou bien de celle <lb/>de c à d, puiſque ces trois raiſons ſont égales.</s> <s xml:id="echoid-s4966" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4967" xml:space="preserve">278. </s> <s xml:id="echoid-s4968" xml:space="preserve">Quand on dit que deux produits ſont entr’eux en rai-<lb/>ſon doublée de deux autres grandeurs, c’eſt comme ſi l’on di-<lb/>ſoit que le premier produit eſt au ſecond, comme le quarré <lb/>d’une grandeur eſt au quarré de l’autre: </s> <s xml:id="echoid-s4969" xml:space="preserve">ainſi ſuppoſant tou-<lb/>jours que a. </s> <s xml:id="echoid-s4970" xml:space="preserve">b:</s> <s xml:id="echoid-s4971" xml:space="preserve">: c. </s> <s xml:id="echoid-s4972" xml:space="preserve">d, lorſque je dis que la raiſon de a c à b d <lb/>eſt doublée de celle de a à b, c’eſt comme ſi je faiſois cette <lb/>proportion, ac. </s> <s xml:id="echoid-s4973" xml:space="preserve">bd:</s> <s xml:id="echoid-s4974" xml:space="preserve">: aa. </s> <s xml:id="echoid-s4975" xml:space="preserve">bb. </s> <s xml:id="echoid-s4976" xml:space="preserve">Pour démontrer cette propor-<lb/>tion, il n’y a qu’à faire voir que le produit des extrêmes eſt <lb/>égal à celui des moyens, ou que a a b d = a c b b; </s> <s xml:id="echoid-s4977" xml:space="preserve">ce qui eſt évi-<lb/>dent, ſi l’on diviſe chaque membre par a b, puiſque a d = b c.</s> <s xml:id="echoid-s4978" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4979" xml:space="preserve">279. </s> <s xml:id="echoid-s4980" xml:space="preserve">De même lorſqu’on dit que la raiſon d’un produit de <lb/>trois dimenſions à un autre produit de trois dimenſions, eſt <lb/>triplée de celle d’une grandeur linéaire à une autre, c’eſt comme <lb/>ſi l’on diſoit que le premier produit eſt au ſecond, comme le <lb/>cube de la premiere grandeur eſt au cube de la ſeconde. </s> <s xml:id="echoid-s4981" xml:space="preserve">Par <lb/>exemple, ſi l’on a a. </s> <s xml:id="echoid-s4982" xml:space="preserve">b:</s> <s xml:id="echoid-s4983" xml:space="preserve">: c. </s> <s xml:id="echoid-s4984" xml:space="preserve">d:</s> <s xml:id="echoid-s4985" xml:space="preserve">: f. </s> <s xml:id="echoid-s4986" xml:space="preserve">g, quand on dit que la <lb/>raiſon de a c f à b d g eſt triplée de celle de a à b, c’eſt comme <lb/>ſi l’on faiſoit cette proportion, a c f. </s> <s xml:id="echoid-s4987" xml:space="preserve">b d g:</s> <s xml:id="echoid-s4988" xml:space="preserve">: a<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4989" xml:space="preserve">b<emph style="sub">3</emph>. </s> <s xml:id="echoid-s4990" xml:space="preserve">Pour prou-<lb/>ver cette proportion, il n’y a qu’à faire voir que le produit des <lb/>extrêmes eſt égal à celui des moyens, ou que a c f b<emph style="sub">3</emph> = a<emph style="sub">3</emph>b d g; <lb/></s> <s xml:id="echoid-s4991" xml:space="preserve">ce qui eſt aiſé à faire, car a b = a b: </s> <s xml:id="echoid-s4992" xml:space="preserve">donc en diviſant chaque <lb/>membre par cette même quantité, on aura c f b<emph style="sub">2</emph> = a<emph style="sub">2</emph> dg; </s> <s xml:id="echoid-s4993" xml:space="preserve">mais <lb/>puiſque a. </s> <s xml:id="echoid-s4994" xml:space="preserve">b:</s> <s xml:id="echoid-s4995" xml:space="preserve">: c. </s> <s xml:id="echoid-s4996" xml:space="preserve">d, b c = a d: </s> <s xml:id="echoid-s4997" xml:space="preserve">donc diviſant encore le premier <lb/>membre par b c, & </s> <s xml:id="echoid-s4998" xml:space="preserve">le ſecond par a d, on aura b f = a g; </s> <s xml:id="echoid-s4999" xml:space="preserve">ce <lb/>qui eſt encore vrai, puiſque a. </s> <s xml:id="echoid-s5000" xml:space="preserve">b:</s> <s xml:id="echoid-s5001" xml:space="preserve">: f. </s> <s xml:id="echoid-s5002" xml:space="preserve">g.</s> <s xml:id="echoid-s5003" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div265" type="section" level="1" n="238"> <head xml:id="echoid-head274" xml:space="preserve">PROPOSITION XVIII.</head> <head xml:id="echoid-head275" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s5004" xml:space="preserve">280. </s> <s xml:id="echoid-s5005" xml:space="preserve">L’expoſant des deux termes d’une raiſon doublée eſt égal <lb/>au quarré de celui qui eſt entre les deux termes de la raiſon ſimple, <lb/>& </s> <s xml:id="echoid-s5006" xml:space="preserve">l’expoſant des deux termes d’une raiſon triplée eſt égal au cube <lb/>de celui des deux termes de la raiſon ſimple.</s> <s xml:id="echoid-s5007" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div266" type="section" level="1" n="239"> <head xml:id="echoid-head276" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s5008" xml:space="preserve">On entend ici par l’expoſant d’une raiſon, le quotient qui <lb/>réſulte de la diviſion des deux termes l’un par l’autre. </s> <s xml:id="echoid-s5009" xml:space="preserve">Cela <lb/>poſé, ſi l’on imagine que le quotient de a, diviſé par b, ſoit f, <lb/>& </s> <s xml:id="echoid-s5010" xml:space="preserve">que celui de c, diviſé par d, ſoit auſſi f, ce qui donnera <pb o="141" file="0179" n="179" rhead="DE MATHÉMATIQUE. Liv. II."/> a. </s> <s xml:id="echoid-s5011" xml:space="preserve">b:</s> <s xml:id="echoid-s5012" xml:space="preserve">: c. </s> <s xml:id="echoid-s5013" xml:space="preserve">d, il faut démontrer que {a c/b d} = f f; </s> <s xml:id="echoid-s5014" xml:space="preserve">ce qui eſt évident, <lb/>car {a/b} = f, & </s> <s xml:id="echoid-s5015" xml:space="preserve">{c/d} = f: </s> <s xml:id="echoid-s5016" xml:space="preserve">donc {a/b} X {c/d} = f f. </s> <s xml:id="echoid-s5017" xml:space="preserve">De même ſi a. </s> <s xml:id="echoid-s5018" xml:space="preserve">b:</s> <s xml:id="echoid-s5019" xml:space="preserve">: c. <lb/></s> <s xml:id="echoid-s5020" xml:space="preserve">d:</s> <s xml:id="echoid-s5021" xml:space="preserve">: f. </s> <s xml:id="echoid-s5022" xml:space="preserve">g, & </s> <s xml:id="echoid-s5023" xml:space="preserve">que le quotient de a, diviſé par b, ſoit q, ainſi <lb/>que celui de c, diviſé par d, & </s> <s xml:id="echoid-s5024" xml:space="preserve">de f par g, on aura {a c f/b d d} = q<emph style="sub">3</emph>; </s> <s xml:id="echoid-s5025" xml:space="preserve"><lb/>car (hypoth.) </s> <s xml:id="echoid-s5026" xml:space="preserve">{a/b} = q, {c/d} = q, {f/g} = q: </s> <s xml:id="echoid-s5027" xml:space="preserve">donc {a c f/b d g} = q<emph style="sub">3</emph>. </s> <s xml:id="echoid-s5028" xml:space="preserve">Il en <lb/>eſt de même en nombres, la raiſon de 12 à 3 eſt 4, celle de <lb/>20 à 5 eſt 4, & </s> <s xml:id="echoid-s5029" xml:space="preserve">celle de 12 X 20, ou de 240 à 5 X 4 & </s> <s xml:id="echoid-s5030" xml:space="preserve">20, & </s> <s xml:id="echoid-s5031" xml:space="preserve"><lb/>16, quarré de 4.</s> <s xml:id="echoid-s5032" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div267" type="section" level="1" n="240"> <head xml:id="echoid-head277" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s5033" xml:space="preserve">281. </s> <s xml:id="echoid-s5034" xml:space="preserve">La raiſon qui eſt entre les quarrés de deux nombres <lb/>eſt doublée de celle qui eſt entre les racines; </s> <s xml:id="echoid-s5035" xml:space="preserve">la raiſon qui eſt <lb/>entre les cubes de deux nombres eſt triplée de celle qui eſt en-<lb/>tre les racines, & </s> <s xml:id="echoid-s5036" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s5037" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5038" xml:space="preserve">Il faut bien prendre garde de confondre la raiſon double <lb/>avec la raiſon doublée, & </s> <s xml:id="echoid-s5039" xml:space="preserve">de même la raiſon triple avec la rai-<lb/>ſon triplée. </s> <s xml:id="echoid-s5040" xml:space="preserve">Une raiſon double ou triple n’eſt qu’une raiſon <lb/>ſimple, dans laquelle l’antécédent eſt double ou triple du con-<lb/>ſéquent; </s> <s xml:id="echoid-s5041" xml:space="preserve">mais une raiſon doublée eſt une raiſon compoſée de <lb/>deux raiſons égales, & </s> <s xml:id="echoid-s5042" xml:space="preserve">une raiſon triplée eſt une raiſon com-<lb/>poſée du produit de trois raiſons égales.</s> <s xml:id="echoid-s5043" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s5044" xml:space="preserve">Regles générales pour la réſolution des Problêmes ou application <lb/>du calcul analytique à la méthode de dégager les inconnues.</s> <s xml:id="echoid-s5045" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div268" type="section" level="1" n="241"> <head xml:id="echoid-head278" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s5046" xml:space="preserve">282. </s> <s xml:id="echoid-s5047" xml:space="preserve">Lorſqu’une quantité eſt poſitive, & </s> <s xml:id="echoid-s5048" xml:space="preserve">qu’elle ne ſe trouve <lb/>qu’une ſeule fois dans un ſeul membre d’une équation, on <lb/>l’appelle quantité dégagée: </s> <s xml:id="echoid-s5049" xml:space="preserve">par exemple, dans l’équation a + b <lb/>= x, la quantité x eſt une quantité dégagée.</s> <s xml:id="echoid-s5050" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div269" type="section" level="1" n="242"> <head xml:id="echoid-head279" xml:space="preserve"><emph style="sc">Axiome</emph> I.</head> <p> <s xml:id="echoid-s5051" xml:space="preserve">283. </s> <s xml:id="echoid-s5052" xml:space="preserve">Si à des grandeurs égales on ajoute des grandeurs éga-<lb/>les, les tous ſeront égaux.</s> <s xml:id="echoid-s5053" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div270" type="section" level="1" n="243"> <head xml:id="echoid-head280" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s5054" xml:space="preserve">284. </s> <s xml:id="echoid-s5055" xml:space="preserve">Si de grandeurs égales on ôte des grandeurs égales, les <lb/>reſtes ſeront égaux.</s> <s xml:id="echoid-s5056" xml:space="preserve"/> </p> <pb o="142" file="0180" n="180" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div271" type="section" level="1" n="244"> <head xml:id="echoid-head281" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s5057" xml:space="preserve">285. </s> <s xml:id="echoid-s5058" xml:space="preserve">Si on multiplie des grandeurs égales par une même <lb/>grandeur, les produits ſeront égaux.</s> <s xml:id="echoid-s5059" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div272" type="section" level="1" n="245"> <head xml:id="echoid-head282" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s5060" xml:space="preserve">286. </s> <s xml:id="echoid-s5061" xml:space="preserve">Si l’on diviſe des grandeurs égales par une même <lb/>grandeur, les quotiens ſeront égaux.</s> <s xml:id="echoid-s5062" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div273" type="section" level="1" n="246"> <head xml:id="echoid-head283" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s5063" xml:space="preserve">287. </s> <s xml:id="echoid-s5064" xml:space="preserve">Si l’on extrait la racine de quantités égales, les racines <lb/>ſeront égales.</s> <s xml:id="echoid-s5065" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div274" type="section" level="1" n="247"> <head xml:id="echoid-head284" xml:space="preserve"><emph style="sc">Premiere</emph> <emph style="sc">Regle</emph>,</head> <p style="it"> <s xml:id="echoid-s5066" xml:space="preserve">Où l’on fait voir l’uſage de l’ Addition & </s> <s xml:id="echoid-s5067" xml:space="preserve">de la Souſtr action pour <lb/>le dégagement des inconnues.</s> <s xml:id="echoid-s5068" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5069" xml:space="preserve">288. </s> <s xml:id="echoid-s5070" xml:space="preserve">Pour dégager une quantité, il faut faire paſſer les <lb/>grandeurs qui l’accompagnent dans l’autre membre avec des <lb/>ſignes contraires, & </s> <s xml:id="echoid-s5071" xml:space="preserve">les effacer dans le membre où elles ſont. <lb/></s> <s xml:id="echoid-s5072" xml:space="preserve">Par exemple, ſi l’on a cette équation a + c = x - d, pour <lb/>dégager x, il faut faire paſſer - d du ſecond membre dans <lb/>le premier avec le ſigne +, & </s> <s xml:id="echoid-s5073" xml:space="preserve">l’on aura a + c + d = x, où <lb/>la quantité x eſt dégagée, puiſque ſa valeur eſt a + c + d: </s> <s xml:id="echoid-s5074" xml:space="preserve"><lb/>car comme on n’a fait qu’ajouter d à chaque membre de l’équa-<lb/>tion, il s’enſuit par l’axiome premier, que l’on n’a point changé <lb/>l’égalité.</s> <s xml:id="echoid-s5075" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5076" xml:space="preserve">De même pour dégager y dans l’équation y + a = b + c, <lb/>l’on fera paſſer a du premier membre dans le ſecond avec le <lb/>ſigne -, pour avoir y = b + c - a, qui donne la valeur de <lb/>y, puiſque par le ſecond axiome on n’a fait que retrancher la <lb/>même grandeur de deux grandeurs égales.</s> <s xml:id="echoid-s5077" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div275" type="section" level="1" n="248"> <head xml:id="echoid-head285" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s5078" xml:space="preserve">289. </s> <s xml:id="echoid-s5079" xml:space="preserve">Il ſuit de la regle précédente, premiérement, que l’on <lb/>peut rendre tous les termes d’une équation poſitifs, en tranſ-<lb/>poſant ceux qui ont le ſigne - d’un membre de l’équation dans <lb/>l’autre, & </s> <s xml:id="echoid-s5080" xml:space="preserve">leur donnant le ſigne +. </s> <s xml:id="echoid-s5081" xml:space="preserve">Par exemple, pour ren-<lb/>dre poſitifs tous les termes de l’équation a b - c c + c d - d d <lb/>= a a + b b, il n’y a qu’à faire paſſer les termes c c & </s> <s xml:id="echoid-s5082" xml:space="preserve">d d, qui <lb/>ont le ſigne - du premier membre dans le ſecond, en leur <lb/>donnant le ſigne +; </s> <s xml:id="echoid-s5083" xml:space="preserve">& </s> <s xml:id="echoid-s5084" xml:space="preserve">après les avoir effacés du premier <pb o="143" file="0181" n="181" rhead="DE MATHÉMATIQUE. Liv. II."/> membre, on aura ab + cd = aa + bb + cc + dd, où il n’y <lb/>a plus de quantités négatives. </s> <s xml:id="echoid-s5085" xml:space="preserve">De même ſi l’on a aa - dd + cd <lb/>- ab = ac + cc - ad, l’on n’a qu’à faire paſſer dd & </s> <s xml:id="echoid-s5086" xml:space="preserve">ab du <lb/>premier membre dans le ſecond, & </s> <s xml:id="echoid-s5087" xml:space="preserve">aa du ſecond dans le pre-<lb/>mier, avec des ſignes contraires, & </s> <s xml:id="echoid-s5088" xml:space="preserve">l’on aura aa + cd + ad <lb/>= ac + cc + dd + ab, où il n’y a plus de termes négatiſs.</s> <s xml:id="echoid-s5089" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5090" xml:space="preserve">290. </s> <s xml:id="echoid-s5091" xml:space="preserve">L’on peut encore par la même regle faire paſſer tous <lb/>les termes d’un des membres d’une équation dans l’autre, en <lb/>réduiſant l’égalité à zero: </s> <s xml:id="echoid-s5092" xml:space="preserve">car pour faire paſſer, par exemple, <lb/>les termes du ſecond membre de cette équation aa + bb = cd <lb/>+ bc - dd; </s> <s xml:id="echoid-s5093" xml:space="preserve">dans le premier, l’on n’a qu’à tranſpoſer les ter-<lb/>mes, en leur donnant des ſignes contraires, & </s> <s xml:id="echoid-s5094" xml:space="preserve">l’on aura aa + bb <lb/>- cd + bb + dd = o.</s> <s xml:id="echoid-s5095" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div276" type="section" level="1" n="249"> <head xml:id="echoid-head286" xml:space="preserve"><emph style="sc">Seconde</emph> <emph style="sc">Regle</emph>,</head> <p style="it"> <s xml:id="echoid-s5096" xml:space="preserve">Où l’on fait voir l’uſage de la Multiplication pour dégager les <lb/>inconnues, & </s> <s xml:id="echoid-s5097" xml:space="preserve">pour délivrer les équations des fractions qu’elles <lb/>contiennent.</s> <s xml:id="echoid-s5098" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5099" xml:space="preserve">291. </s> <s xml:id="echoid-s5100" xml:space="preserve">Pour dégager une quantité qui ſe trouve diviſée par <lb/>quelque nombre, ou par quelque lettre, il faut multiplier les <lb/>autres termes de l’équation par le diviſeur de cette quantité, <lb/>ſans toucher à cette quantité, que pour en effacer le diviſeur: <lb/></s> <s xml:id="echoid-s5101" xml:space="preserve">ainſi pour dégager {xx/c} dans l’équation a + b = {xx/c}, il faut mul-<lb/>tiplier le membre a + b par le diviſeur c, & </s> <s xml:id="echoid-s5102" xml:space="preserve">l’on aura ac + bc <lb/>= xx, ou xx eſt dégagée. </s> <s xml:id="echoid-s5103" xml:space="preserve">De même ſi l’on avoit c + b = z, <lb/>il faut pour dégager z, multiplier les termes c + b par le divi-<lb/>ſeur 2, & </s> <s xml:id="echoid-s5104" xml:space="preserve">l’on aura 2c + 2b = z; </s> <s xml:id="echoid-s5105" xml:space="preserve">ce qui eſt évident par le <lb/>3<emph style="sub">e</emph> axiome, puiſqu’ayant multiplié les deux membres de cette <lb/>équation par une même quantité, on n’a rien changé à l’é-<lb/>galité.</s> <s xml:id="echoid-s5106" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div277" type="section" level="1" n="250"> <head xml:id="echoid-head287" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s5107" xml:space="preserve">292. </s> <s xml:id="echoid-s5108" xml:space="preserve">Comme la diviſion indiquée, ou autrement {a/b} n’eſt <lb/>qu’une fraction; </s> <s xml:id="echoid-s5109" xml:space="preserve">il ſuit de la regle précédente, que l’on peut <lb/>non ſeulement dégager les quantités inconnues qui ſont divi-<lb/>ſées, mais que l’on peut encore délivrer de fractions les ter-<lb/>mes d’une équation, en multipliant tous les autres termes de <lb/>l’équation par les dénominateurs des fractions: </s> <s xml:id="echoid-s5110" xml:space="preserve">par exemple, <pb o="144" file="0182" n="182" rhead="NOUVEAU COURS"/> pour ôter la fraction qui ſe trouve dans l’équation a + {dd/e} <lb/>+ b = d + c, je multiplie tous ces termes par le dénomina-<lb/>teur c de la fraction {dd/c}, & </s> <s xml:id="echoid-s5111" xml:space="preserve">il vient ac + dd + bc = dc + cc, <lb/>où il n’y a plus de fractions. </s> <s xml:id="echoid-s5112" xml:space="preserve">Pour ôter les fractions de l’équa-<lb/>tion xd + {bbc/a} - cc = dd - {aad/c} + bc, je commence par mul-<lb/>tiplier tous les termes de l’équation par le dénominateur a de <lb/>la premiere fraction, pour avoir adx + bbc - acc = add -<lb/>{aaad/c} + abc, où il n’y a plus de fractions dans le premier mem-<lb/>bre; </s> <s xml:id="echoid-s5113" xml:space="preserve">enſuite je multiplie tous les termes de cette nouvelle <lb/>équation par le dénominateur de la ſeconde fraction, pour <lb/>avoir adcx + bbcc - cccc = acdd - a<emph style="sub">3</emph>d + abcc, où il n’y a <lb/>plus de fractions. </s> <s xml:id="echoid-s5114" xml:space="preserve">Enſin ſi l’on avoit une équation, comme <lb/>{a/b} = {c/d} + {x/a} = {b/c} + {y/c}, l’on en feroit évanouir toutes les frac-<lb/>tions, en multipliant chaque numérateur par les dénomina-<lb/>teurs de toutes les autres fractions, & </s> <s xml:id="echoid-s5115" xml:space="preserve">l’on aura aacdc + abccc <lb/>+ bcdcx = abbdc + abcdy.</s> <s xml:id="echoid-s5116" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5117" xml:space="preserve">293. </s> <s xml:id="echoid-s5118" xml:space="preserve">Mais au lieu de multiplier l’un après l’autre chaque <lb/>numérateur par tous les dénominateurs des autres fractions, <lb/>on peut tout d’un coup ôter les fractions d’une équation, en <lb/>multipliant chaque terme par le produit de tous les dénomina-<lb/>nateurs, & </s> <s xml:id="echoid-s5119" xml:space="preserve">en effaçant dans les numérateurs & </s> <s xml:id="echoid-s5120" xml:space="preserve">dénomina-<lb/>teurs de chaque nouvelle fraction les lettres ſemblables.</s> <s xml:id="echoid-s5121" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div278" type="section" level="1" n="251"> <head xml:id="echoid-head288" xml:space="preserve"><emph style="sc">Troisieme</emph> <emph style="sc">Regle</emph>,</head> <head xml:id="echoid-head289" style="it" xml:space="preserve">Où l’on fait voir l’uſage de la Diviſion pour dégager les <lb/>inconnues.</head> <p> <s xml:id="echoid-s5122" xml:space="preserve">294. </s> <s xml:id="echoid-s5123" xml:space="preserve">Lorſqu’une quantité inconnue, que l’on veut dégager, <lb/>eſt multipliée par une grandeur connue, on dégagera l’incon-<lb/>nue, en diviſant chaque membre de l’équation par cette gran-<lb/>deur connue. </s> <s xml:id="echoid-s5124" xml:space="preserve">Ainſi pour dégager l’inconnue dans l’équation <lb/>ax = bb - cc, l’on diviſera chaque membre par a, & </s> <s xml:id="echoid-s5125" xml:space="preserve">l’on <lb/>aura x = {bb - cc/a}. </s> <s xml:id="echoid-s5126" xml:space="preserve">De même ſi l’on a cz = dd + az, on dé-<lb/>gagera l’inconnue z, en faiſant paſſer az du ſecond membre <lb/>dans le premier, avec un ſigne contraire, pour avoir cz - az <lb/>= dd, & </s> <s xml:id="echoid-s5127" xml:space="preserve">diviſant chaque membre par c-a, l’on aura z = {dd/c-a};</s> <s xml:id="echoid-s5128" xml:space="preserve"> <pb o="145" file="0183" n="183" rhead="DE MATHÉMATIQUE. Liv. II."/> ce qui eſt bien évident, par l’axiome 4<emph style="sub">e</emph>, puiſqu’ayant diviſé <lb/>chaque membre de l’équation par la même grandeur, les quo-<lb/>tients doivent être égaux.</s> <s xml:id="echoid-s5129" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div279" type="section" level="1" n="252"> <head xml:id="echoid-head290" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s5130" xml:space="preserve">295. </s> <s xml:id="echoid-s5131" xml:space="preserve">Il ſuit de cette regle, que lorſque tous les termes d’une <lb/>équation ſont multipliés par une même lettre, ou par une <lb/>même grandeur, on peut rendre l’équation plus ſimple, en <lb/>diviſant tous les termes par cette grandeur. </s> <s xml:id="echoid-s5132" xml:space="preserve">Par exemple, ſi <lb/>l’on a aa + ab = ac - ad, où tous les termes ſont multipliés <lb/>par a, l’on n’a qu’à diviſer les deux membres de cette équa-<lb/>tion par la même lettre a, il viendra l’équation a + b = c - d, <lb/>qui eſt plus ſimple que la précédente: </s> <s xml:id="echoid-s5133" xml:space="preserve">mais s’il ſe trouvoit quel-<lb/>que terme qui ne pût pas être diviſé comme les autres, ne con-<lb/>tenant pas de lettres ſemblables au diviſeur; </s> <s xml:id="echoid-s5134" xml:space="preserve">cela n’empêche <lb/>pas que la diviſion ne ſe faſſe toujours, parce que quand on ne <lb/>peut pas la faire effectivement ſur quelque terme, on la fait <lb/>par indiction. </s> <s xml:id="echoid-s5135" xml:space="preserve">Par exemple, pour diviſer cette équation abb <lb/>- cbb = cdx + bbc par bb, dans laquelle le terme cdx n’a <lb/>point de lettres ſemblables au diviſeur, l’on efface bb des au-<lb/>tres termes, & </s> <s xml:id="echoid-s5136" xml:space="preserve">l’on marque pour celui-ci {cdx/bb}: </s> <s xml:id="echoid-s5137" xml:space="preserve">ainſi l’on a a - c <lb/>= {cdx/bb} + c.</s> <s xml:id="echoid-s5138" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5139" xml:space="preserve">Enſin lorſque les deux membres d’une équation ont <lb/>un diviſeur commun, on pourra les réduire à une équation <lb/>plus ſimple, en diviſant chaque membre par le diviſeur qui <lb/>eſt commun. </s> <s xml:id="echoid-s5140" xml:space="preserve">Par exemple, ſi l’on a une équation comme <lb/>bbx - bxx = abb - abx, dont les membres ont pour divi-<lb/>ſeur commun bb - bx, on fera la diviſion qui donnera cette <lb/>autre équation, qui donnera x = a.</s> <s xml:id="echoid-s5141" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div280" type="section" level="1" n="253"> <head xml:id="echoid-head291" xml:space="preserve"><emph style="sc">Quatrieme</emph> <emph style="sc">Regle</emph>,</head> <head xml:id="echoid-head292" style="it" xml:space="preserve">Où l’on fait voir l’uſage de l’extraction des racines pour dégager <lb/>les inconnues.</head> <p> <s xml:id="echoid-s5142" xml:space="preserve">296. </s> <s xml:id="echoid-s5143" xml:space="preserve">Quand on a une équation, où l’un des membres ne <lb/>contient que des grandeurs connues, & </s> <s xml:id="echoid-s5144" xml:space="preserve">que l’autre où eſt l’in-<lb/>connue eſt un quarré ou un cube parfait, il faut extraire la <lb/>racine de ces deux membres pour avoir une nouvelle équation, <lb/>dans laquelle on pourra dégager l’inconnue. </s> <s xml:id="echoid-s5145" xml:space="preserve">Par exemple, ſi <lb/>l’on a xx + 2ax + aa = bc + dd, où le premier membre de <pb o="146" file="0184" n="184" rhead="NOUVEAU COURS"/> cette équation eſt un quarré parfait, on extraira la racine de <lb/>chaque membre. </s> <s xml:id="echoid-s5146" xml:space="preserve">Celle du premier membre, ſuivant la mé-<lb/>thode de l’article 147, eſt x + a, & </s> <s xml:id="echoid-s5147" xml:space="preserve">celle du ſecond, par l’ar-<lb/>ticle 149, eſt √bc + dd\x{0020}: </s> <s xml:id="echoid-s5148" xml:space="preserve">donc l’équation devient x + a = <lb/>√bc + dd\x{0020}; </s> <s xml:id="echoid-s5149" xml:space="preserve">& </s> <s xml:id="echoid-s5150" xml:space="preserve">faiſant paſſer a du premier membre dans le <lb/>ſecond (art. </s> <s xml:id="echoid-s5151" xml:space="preserve">288), on aura x = √bc + dd\x{0020}-a, qui fait voir que <lb/>ſi l’on extrait la racine de bc + dd, & </s> <s xml:id="echoid-s5152" xml:space="preserve">que l’on ôte de cette <lb/>racine la grandeur a, la différence ſera la valeur de x.</s> <s xml:id="echoid-s5153" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5154" xml:space="preserve">De même pour dégager x dans l’équation xx - 2ax + aa <lb/>= bb, j’extrais la racine de chaque membre, & </s> <s xml:id="echoid-s5155" xml:space="preserve">j’ai x - a = b; <lb/></s> <s xml:id="echoid-s5156" xml:space="preserve">d’où l’on déduit en tranſpoſant x = b + a.</s> <s xml:id="echoid-s5157" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5158" xml:space="preserve">297. </s> <s xml:id="echoid-s5159" xml:space="preserve">Comme le premier membre de cette équation eſt un <lb/>cube parfait, x<emph style="sub">3</emph> + 3ax<emph style="sub">2</emph> + 3a<emph style="sub">2</emph>x + a<emph style="sub">3</emph> = aab, en tirant la ra-<lb/>cine cube de chaque membre, on aura l’équation plus ſimple <lb/>x + a = <emph style="sub">3</emph>√aab\x{0020}; </s> <s xml:id="echoid-s5160" xml:space="preserve">& </s> <s xml:id="echoid-s5161" xml:space="preserve">en tranſpoſant, l’on aura x = <emph style="sub">3</emph>√aab\x{0020} - a, <lb/>qui fait voir que ſi l’on extrait la racine cubique de aab, & </s> <s xml:id="echoid-s5162" xml:space="preserve">que <lb/>l’on ôte de cette racine la grandeur a, le reſte ſera la valeur <lb/>de x. </s> <s xml:id="echoid-s5163" xml:space="preserve">De même le premier membre de cette équation x<emph style="sub">3</emph> - 3axx <lb/>+ 3a<emph style="sub">2</emph>x - a<emph style="sub">3</emph> = bdd, étant encore un cube parfait, ſi l’on ex-<lb/>trait la racine cube de chaque membre, l’on aura x - a = <lb/><emph style="sub">3</emph>√bdd\x{0020}, ou x = a + <emph style="sub">3</emph>√bdd\x{0020}, qui fait voir que la grandeur a, <lb/>plus la racine cube de bdd eſt égale à x.</s> <s xml:id="echoid-s5164" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div281" type="section" level="1" n="254"> <head xml:id="echoid-head293" xml:space="preserve"><emph style="sc">Cinquieme</emph> <emph style="sc">Regle</emph>,</head> <head xml:id="echoid-head294" style="it" xml:space="preserve">Où l’on donne la maniere de ſubſtituer dans une équation la valeur <lb/>des inconnues.</head> <p> <s xml:id="echoid-s5165" xml:space="preserve">298. </s> <s xml:id="echoid-s5166" xml:space="preserve">Quand on connoît la valeur de quelques lettres que l’on <lb/>veut faire évanouir dans une équation, on ſubſtitue à leur <lb/>place les quantités qui leur ſont égales avec le même ſigne. <lb/></s> <s xml:id="echoid-s5167" xml:space="preserve">Par exemple, ſi l’on a l’équation a + z = y + b - c, où l’on <lb/>veut faire évanouir z, & </s> <s xml:id="echoid-s5168" xml:space="preserve">que l’on ſuppoſe z = d + e, on effa-<lb/>cera z dans l’équation, & </s> <s xml:id="echoid-s5169" xml:space="preserve">l’on mettra à ſa place ſa valeur d + e; </s> <s xml:id="echoid-s5170" xml:space="preserve"><lb/>ce qui donnera a + d + e = y + b - c, où z ne ſe trouve plus. </s> <s xml:id="echoid-s5171" xml:space="preserve"><lb/>Si l’on a cette équation b + d - x = c + z, dans laquelle on <lb/>veut faire évanouir x, ſuppoſant que x = a - e, l’on effacera <lb/>x, & </s> <s xml:id="echoid-s5172" xml:space="preserve">l’on mettra à ſa place - a + e, à cauſe que x a le ſigne <lb/>-, & </s> <s xml:id="echoid-s5173" xml:space="preserve">l’on aura b + d - a + e = c + z, où x ne ſe trouve <lb/>plus.</s> <s xml:id="echoid-s5174" xml:space="preserve"/> </p> <pb o="147" file="0185" n="185" rhead="DE MATHÉMATIQUE. Liv. II."/> <p> <s xml:id="echoid-s5175" xml:space="preserve">299. </s> <s xml:id="echoid-s5176" xml:space="preserve">Si la lettre qu’on veut faire évanouir eſt multipliée ou <lb/>diviſée dans l’équation par quelqu’autre grandeur, il faut mul-<lb/>tiplier ou diviſer ſa valeur par cette même grandeur, & </s> <s xml:id="echoid-s5177" xml:space="preserve">l’écrire <lb/>dans l’équation avec le même ſigne. </s> <s xml:id="echoid-s5178" xml:space="preserve">Par exemple, ſi de l’é-<lb/>quation bb + ax - cc = ad + aa - yy, on veut faire éva-<lb/>nouir x, ſuppoſant que x = e + f, comme x eſt multipliée <lb/>par a dans l’équation, il faut multiplier ſa valeur e + f, par la <lb/>même lettre a, pour avoir ax = ac + af, & </s> <s xml:id="echoid-s5179" xml:space="preserve">mettant ac + af à <lb/>la place de ax, l’on aura bb + ac + af - cc = ad + aa - yy, <lb/>où x ne ſe trouve plus.</s> <s xml:id="echoid-s5180" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5181" xml:space="preserve">300. </s> <s xml:id="echoid-s5182" xml:space="preserve">Pour faire évanouir de l’équation cc + yy - 2bd = aa <lb/>- bz la lettre z, ſuppoſant que z = d - e + g, il faut mul-<lb/>tiplier la valeur de z par b, pour avoir bz = bd - be + bg; </s> <s xml:id="echoid-s5183" xml:space="preserve">& </s> <s xml:id="echoid-s5184" xml:space="preserve"><lb/>comme bz a le ſigne - dans l’équation, il faut changer les <lb/>ſignes de bd - be + bg, & </s> <s xml:id="echoid-s5185" xml:space="preserve">mettre dans l’équation - bd + be <lb/>- bg; </s> <s xml:id="echoid-s5186" xml:space="preserve">ce qui donnera cc + yy = aa - bd + be - bg, où <lb/>z ne ſe trouve plus.</s> <s xml:id="echoid-s5187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5188" xml:space="preserve">301. </s> <s xml:id="echoid-s5189" xml:space="preserve">Pour faire évanouir y de l’équation 2ab + ez = be + <lb/>{ddy/a - f}, ſuppoſant que l’on a y = e - g, il faut multiplier e - g <lb/>par dd, pour avoir ddy = dde - ddg; </s> <s xml:id="echoid-s5190" xml:space="preserve">mais comme ddy eſt <lb/>diviſé par a - f dans l’équation, il faut pour y ſubſtituer dde <lb/>- ddg le diviſer auſſi par a - f, & </s> <s xml:id="echoid-s5191" xml:space="preserve">alors on aura 2ab + ez = <lb/>{dde - ddg/a - f}, où y ne ſe trouve plus.</s> <s xml:id="echoid-s5192" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5193" xml:space="preserve">302. </s> <s xml:id="echoid-s5194" xml:space="preserve">Pour faire évanouir u de l’équation aa + dd = au + bd, <lb/>ſuppoſant que l’on a u = {aa - cc + fg/b + d}, il faut, à cauſe que u eſt <lb/>égal à une fraction, multiplier le numérateur de cette fraction <lb/>par a, pour avoir a u = {a<emph style="sub">3</emph> - acc + afg/b + d}, & </s> <s xml:id="echoid-s5195" xml:space="preserve">puis mettre à la place <lb/>de au dans la premiere équation, la fraction qui lui eſt égale, <lb/>& </s> <s xml:id="echoid-s5196" xml:space="preserve">l’on aura aa + dd = {a<emph style="sub">3</emph> + afg - acc/b + d} + bd, dans laquelle u ne <lb/>ſe trouve plus. </s> <s xml:id="echoid-s5197" xml:space="preserve">Si l’on veut ôter la fraction de cette équation, <lb/>l’on n’aura qu’à multiplier les autres termes par le dénomina-<lb/>teur b + d (art. </s> <s xml:id="echoid-s5198" xml:space="preserve">283), & </s> <s xml:id="echoid-s5199" xml:space="preserve">l’équation ſera transformée en celle-<lb/>ci, aab + aad + d<emph style="sub">3</emph> = a<emph style="sub">3</emph> + acc + afg + bbd, après avoir <lb/>effacé les termes b d d, qui ſe trouvent dans chaque membre <lb/>avec le même ſigne.</s> <s xml:id="echoid-s5200" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5201" xml:space="preserve">303. </s> <s xml:id="echoid-s5202" xml:space="preserve">Si la lettre qu’on veut faire évanouir eſt le côté d’un <lb/>quarré ou d’un cube, il faut quarrer ou cuber ſa valeur, & </s> <s xml:id="echoid-s5203" xml:space="preserve"><lb/> <pb o="148" file="0186" n="186" rhead="NOUVEAU COURS"/> mettre ſon quarré ou ſon cube dans l’équation à la place du <lb/>quarré ou du cube de la lettre qu’on veut faire évanouir. </s> <s xml:id="echoid-s5204" xml:space="preserve">Par <lb/>exemple, ſi l’on veut faire évanouir y de l’équation yy - 2bd <lb/>= 2ax + dd, ſuppoſant que y = b + d, il faut quarrer la va-<lb/>leur de y pour avoir yy = bb + 2bd + dd, & </s> <s xml:id="echoid-s5205" xml:space="preserve">mettre la va-<lb/>leur du quarré de y à la place de yy, & </s> <s xml:id="echoid-s5206" xml:space="preserve">l’on aura cette équa-<lb/>tion, bb + 2bd + dd - 2bd = 2ax + dd, & </s> <s xml:id="echoid-s5207" xml:space="preserve">effaçant + 2bd <lb/>& </s> <s xml:id="echoid-s5208" xml:space="preserve">- 2bd, qui ſe détruiſent, & </s> <s xml:id="echoid-s5209" xml:space="preserve">dd qui eſt commun au premier <lb/>& </s> <s xml:id="echoid-s5210" xml:space="preserve">au ſecond membre avec le même ſigne, l’équation deviendra <lb/>bb = 2ax, d’où dégageant x, il vient x = {bb/2a}, qui eſt la va-<lb/>leur de x. </s> <s xml:id="echoid-s5211" xml:space="preserve">L’on pourra de même ſubſtituer dans une équation <lb/>la valeur d’un cube, quand on connoîtra celle de ſa racine.</s> <s xml:id="echoid-s5212" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5213" xml:space="preserve">Comme l’on ne fait par la ſubſtitution que mettre une gran-<lb/>deur égale à la place d’une autre dans une équation, il s’enſuit <lb/>que les deux membres de cette équation demeurent toujours <lb/>égaux.</s> <s xml:id="echoid-s5214" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div282" type="section" level="1" n="255"> <head xml:id="echoid-head295" xml:space="preserve"><emph style="sc">Sixieme</emph> <emph style="sc">Regle</emph>,</head> <head xml:id="echoid-head296" style="it" xml:space="preserve">Où l’on fait voir comment on peut faire évanouir toutes les incon-<lb/>nues d’une équation.</head> <p> <s xml:id="echoid-s5215" xml:space="preserve">304. </s> <s xml:id="echoid-s5216" xml:space="preserve">Pour réſoudre un problême par Algebre, il faut com-<lb/>mencer par conſidérer attentivement l’état de la queſtion, & </s> <s xml:id="echoid-s5217" xml:space="preserve"><lb/>toutes les conditions qu’elle renferme; </s> <s xml:id="echoid-s5218" xml:space="preserve">enſuite marquer ce que <lb/>l’on connoît avec les premieres lettres de l’alphabet, & </s> <s xml:id="echoid-s5219" xml:space="preserve">ce que <lb/>l’on ne connoît pas avec les dernieres: </s> <s xml:id="echoid-s5220" xml:space="preserve">conſidérant après cela <lb/>le problême comme réſolu, on tâchera de trouver autant d’é-<lb/>quations que l’on a employé de lettres inconnues, que nous <lb/>appellerons premieres équations.</s> <s xml:id="echoid-s5221" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5222" xml:space="preserve">On choiſira la plus ſimple de toutes ces équations, pour dé-<lb/>gager une des inconnues qu’elle renferme; </s> <s xml:id="echoid-s5223" xml:space="preserve">& </s> <s xml:id="echoid-s5224" xml:space="preserve">ayant trouvé la <lb/>valeur de cette inconnue, on la ſubſtituera dans les autres <lb/>équations aux endroits où cette inconnue ſe trouvera.</s> <s xml:id="echoid-s5225" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5226" xml:space="preserve">On recommencera de nouveau à choiſir la plus ſimple des <lb/>autres équations pour y dégager une ſeconde inconnue, dont <lb/>on ſubſtituera, comme auparavant, la valeur dans les autres <lb/>équations, & </s> <s xml:id="echoid-s5227" xml:space="preserve">l’on réïtérera la même choſe pour faire évanouir <lb/>l’une après l’autre toutes les lettres inconnues; </s> <s xml:id="echoid-s5228" xml:space="preserve">& </s> <s xml:id="echoid-s5229" xml:space="preserve">de cette <lb/>maniere on trouvera la valeur connue de toutes les inconnues; <lb/></s> <s xml:id="echoid-s5230" xml:space="preserve">ce qui donnera la ſolution du problême.</s> <s xml:id="echoid-s5231" xml:space="preserve"/> </p> <pb o="149" file="0187" n="187" rhead="DE MATHEMATIQUE. Liv. II."/> <p> <s xml:id="echoid-s5232" xml:space="preserve">Pour rendre ceci plus ſenſible, nous allons faire évanouir <lb/>toutes les inconnues des trois équations x + y = z + a, y + z <lb/>= b + x, & </s> <s xml:id="echoid-s5233" xml:space="preserve">x + z = c + y. </s> <s xml:id="echoid-s5234" xml:space="preserve">Pour cela, je commence par <lb/>chercher la valeur de z dans la premiere équation, en la dé-<lb/>gageant de a, que je fais paſſer dans l’autre membre avec le <lb/>ſigne contraire, afin d’avoir x + y - a = z, qui me donne <lb/>la valeur de z; </s> <s xml:id="echoid-s5235" xml:space="preserve">enſuite je mets cette valeur à la place de z dans <lb/>les autres équations (art. </s> <s xml:id="echoid-s5236" xml:space="preserve">298.) </s> <s xml:id="echoid-s5237" xml:space="preserve">qui ſe trouvent changées en <lb/>celles-ci, 2y + x - a = b + x, & </s> <s xml:id="echoid-s5238" xml:space="preserve">2x + y - a = c + y, & </s> <s xml:id="echoid-s5239" xml:space="preserve"><lb/>comme x ſe trouve dans le premier & </s> <s xml:id="echoid-s5240" xml:space="preserve">le ſecond membre de <lb/>la premiere équation avec le ſigne +, de même y dans la <lb/>ſeconde; </s> <s xml:id="echoid-s5241" xml:space="preserve">je les efface, & </s> <s xml:id="echoid-s5242" xml:space="preserve">en dégageant les inconnues qui <lb/>reſtent, il vient 2y = b + a, & </s> <s xml:id="echoid-s5243" xml:space="preserve">2x = c + a, ou bien y = <lb/>{b + a/2}, & </s> <s xml:id="echoid-s5244" xml:space="preserve">x = {c + a/2}, où les valeurs de x & </s> <s xml:id="echoid-s5245" xml:space="preserve">de y ſe trouvent <lb/>tout d’un coup, ſans avoir été obligé de faire une ſeconde <lb/>ſubſtitution. </s> <s xml:id="echoid-s5246" xml:space="preserve">Si préſentement on met dans la premiere équa-<lb/>tion, où l’inconnue a été dégagée, la valeur de x & </s> <s xml:id="echoid-s5247" xml:space="preserve">de y, on <lb/>aura {b + a + c + a/2} - a = z, ou {b + c/2} = z. </s> <s xml:id="echoid-s5248" xml:space="preserve">Par conſéquent <lb/>on a trouvé la valeur des inconnues x, y & </s> <s xml:id="echoid-s5249" xml:space="preserve">z en lettres connues.</s> <s xml:id="echoid-s5250" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div283" type="section" level="1" n="256"> <head xml:id="echoid-head297" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s5251" xml:space="preserve">On s’eſt contenté de donner ſeulement un petit exemple <lb/>de cette regle, parce qu’on en va voir l’application, auſſi-bien <lb/>que des précédentes, dans tout ce qui ſuit, où l’on va réſoudre <lb/>pluſieurs problêmes curieux, que l’on a rapportés exprès pour <lb/>familiariſer les Commençans avec le calcul algébrique, & </s> <s xml:id="echoid-s5252" xml:space="preserve"><lb/>pour rendre intéreſſant ce que l’on a vu juſqu’ici, qu’il eſt à <lb/>propos d’entendre parfaitement, pour avoir le plaiſir de com-<lb/>prendre ſans peine tout ce qui compoſe la ſuite de cet ouvrage.</s> <s xml:id="echoid-s5253" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div284" type="section" level="1" n="257"> <head xml:id="echoid-head298" style="it" xml:space="preserve">Application des Regles précédentes à la réſolution de pluſieurs <lb/>Problêmes curieux.</head> <head xml:id="echoid-head299" xml:space="preserve"><emph style="sc">Premiere question</emph>.</head> <p> <s xml:id="echoid-s5254" xml:space="preserve">Trois perſonnes ont gagné enſemble au jeu 875 livres, la <lb/>ſeconde perſonne a gagné deux fois autant que la premiere, & </s> <s xml:id="echoid-s5255" xml:space="preserve"><lb/>10 liv. </s> <s xml:id="echoid-s5256" xml:space="preserve">de plus, la troiſieme a gagné autant que la premiere & </s> <s xml:id="echoid-s5257" xml:space="preserve"><lb/>la ſeconde, & </s> <s xml:id="echoid-s5258" xml:space="preserve">15 liv. </s> <s xml:id="echoid-s5259" xml:space="preserve">de plus. </s> <s xml:id="echoid-s5260" xml:space="preserve">On demande combien chaque <lb/>perſonne a gagné.</s> <s xml:id="echoid-s5261" xml:space="preserve"/> </p> <pb o="150" file="0188" n="188" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s5262" xml:space="preserve">Pour réſoudre cette queſtion, j’appelle x le gain de la pre-<lb/>miere perſonne; </s> <s xml:id="echoid-s5263" xml:space="preserve">par conſéquent celui de la ſeconde ſera 2x, <lb/>parce qu’elle a gagné le double de la premiere; </s> <s xml:id="echoid-s5264" xml:space="preserve">& </s> <s xml:id="echoid-s5265" xml:space="preserve">comme elle <lb/>a encore gagné 10 livres de plus, ſon gain ſera 2x + 20. </s> <s xml:id="echoid-s5266" xml:space="preserve">Or <lb/>comme la troiſieme perſonne a gagné autant que la premiere <lb/>& </s> <s xml:id="echoid-s5267" xml:space="preserve">la ſeconde, & </s> <s xml:id="echoid-s5268" xml:space="preserve">même 15 liv. </s> <s xml:id="echoid-s5269" xml:space="preserve">de plus, j’ajoute enſemble le <lb/>gain des deux premieres perſonnes, c’eſt-à-dire x & </s> <s xml:id="echoid-s5270" xml:space="preserve">2x + 10, <lb/>à quoi ajoutant 15, le gain de la troiſieme perſonne ſera <lb/>3x + 25; </s> <s xml:id="echoid-s5271" xml:space="preserve">& </s> <s xml:id="echoid-s5272" xml:space="preserve">comme le gain des trois perſonnes eſt égal à <lb/>875, je forme cette équation x + 2x + 10 + 3x + 25 = <lb/>875; </s> <s xml:id="echoid-s5273" xml:space="preserve">d’où je dégage la quantité inconnue, en faiſant paſſer la <lb/>ſomme des nombres que je connois du premier membre dans <lb/>le ſecond (art. </s> <s xml:id="echoid-s5274" xml:space="preserve">288.) </s> <s xml:id="echoid-s5275" xml:space="preserve">avec le ſigne -, & </s> <s xml:id="echoid-s5276" xml:space="preserve">réduiſant le tout en <lb/>un ſeul terme; </s> <s xml:id="echoid-s5277" xml:space="preserve">ce qui donne cette nouvelle équation 6x = 875 <lb/>- 25, ou 6x = 840, que je diviſe par 6 (art. </s> <s xml:id="echoid-s5278" xml:space="preserve">294.) </s> <s xml:id="echoid-s5279" xml:space="preserve">pour avoir <lb/>x = 140, qui me fait voir que la premiere perſonne a gagné <lb/>140 livres. </s> <s xml:id="echoid-s5280" xml:space="preserve">Pour avoir le gain de la ſeconde perſonne, je dou-<lb/>ble 140, & </s> <s xml:id="echoid-s5281" xml:space="preserve">j’ajoute 10 au produit, qui donne 2x + 10 = 290: <lb/></s> <s xml:id="echoid-s5282" xml:space="preserve">enfin ſi j’ajoute cette équation à la précédente, & </s> <s xml:id="echoid-s5283" xml:space="preserve">15 à la ſom-<lb/>me, j’aurai le gain de la troiſieme perſonne, c’eſt - à - dire <lb/>3x + 25 = 445; </s> <s xml:id="echoid-s5284" xml:space="preserve">par conſéquent la premiere perſonne a ga-<lb/>gné 140 livres, la ſeconde 290 livres, & </s> <s xml:id="echoid-s5285" xml:space="preserve">la troiſieme 445; </s> <s xml:id="echoid-s5286" xml:space="preserve">ce <lb/>qui eſt bien évident, puiſque ces trois ſommes font enſemble <lb/>875 livres, & </s> <s xml:id="echoid-s5287" xml:space="preserve">qu’elles rempliſſent toutes les conditions du <lb/>problême.</s> <s xml:id="echoid-s5288" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div285" type="section" level="1" n="258"> <head xml:id="echoid-head300" xml:space="preserve"><emph style="sc">Seconde question</emph>.</head> <p> <s xml:id="echoid-s5289" xml:space="preserve">Quatre Sappeurs ont fait chacun une quantité de toiſes de <lb/>ſappe, & </s> <s xml:id="echoid-s5290" xml:space="preserve">ils ont gagné enſemble 140 livres; </s> <s xml:id="echoid-s5291" xml:space="preserve">le ſecond Sappeur <lb/>a gagné trois fois plus que le premier, moins 8 livres; </s> <s xml:id="echoid-s5292" xml:space="preserve">le troi-<lb/>ſieme a gagné la moitié de ce qu’ont gagné enſemble le pre-<lb/>mier & </s> <s xml:id="echoid-s5293" xml:space="preserve">le ſecond, moins 12 livres; </s> <s xml:id="echoid-s5294" xml:space="preserve">& </s> <s xml:id="echoid-s5295" xml:space="preserve">le quatrieme a gagné <lb/>autant que le premier & </s> <s xml:id="echoid-s5296" xml:space="preserve">le troiſieme: </s> <s xml:id="echoid-s5297" xml:space="preserve">l’on demande combien <lb/>ils ont gagné chacun.</s> <s xml:id="echoid-s5298" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5299" xml:space="preserve">Pour réſoudre cette queſtion, j’appelle x le gain du pre-<lb/>mier Sappeur; </s> <s xml:id="echoid-s5300" xml:space="preserve">ainſi 3x - 8 ſera le gain du ſecond Sappeur; <lb/></s> <s xml:id="echoid-s5301" xml:space="preserve">2x - 16 le gain du troiſieme; </s> <s xml:id="echoid-s5302" xml:space="preserve">& </s> <s xml:id="echoid-s5303" xml:space="preserve">3x - 16 le gain du qua-<lb/>trieme: </s> <s xml:id="echoid-s5304" xml:space="preserve">& </s> <s xml:id="echoid-s5305" xml:space="preserve">comme toutes ces quantités, priſes enſemble, ſont <lb/>égales à 240 livres, je forme cette équation x + 3x - 8 + 2x <lb/>- 16 + 3x - 16 = 140, que je réduis à ſa plus ſimple expreſ- <pb o="151" file="0189" n="189" rhead="DE MATHEMATIQUE. Liv. II."/> ſion, en ajoutant enſemble toutes les quantités ſemblables, <lb/>& </s> <s xml:id="echoid-s5306" xml:space="preserve">il vient 9x - 40 = 140, ou bien 9x = 180, en faiſant <lb/>paſſer 140 du premier membre dans le ſecond. </s> <s xml:id="echoid-s5307" xml:space="preserve">Or ſi l’on diviſe <lb/>les membres de cette équation par 9 (art. </s> <s xml:id="echoid-s5308" xml:space="preserve">294.) </s> <s xml:id="echoid-s5309" xml:space="preserve">pour dégager <lb/>l’inconnue, l’on trouvera x = 20, qui montre que le gain du <lb/>premier Sappeur eſt 20 livres: </s> <s xml:id="echoid-s5310" xml:space="preserve">ainſi le gain du ſecond, qui eſt <lb/>3x - 8, ſera 52 livres; </s> <s xml:id="echoid-s5311" xml:space="preserve">celui du troiſieme, qui eſt 2x - 16, <lb/>ſera 24 livres; </s> <s xml:id="echoid-s5312" xml:space="preserve">& </s> <s xml:id="echoid-s5313" xml:space="preserve">celui du quatrieme, qui eſt 3x - 16, ſera <lb/>44 livres; </s> <s xml:id="echoid-s5314" xml:space="preserve">ce qui eſt évident, puiſque ces quatre nombres, <lb/>pris enſemble, font 140 livres, & </s> <s xml:id="echoid-s5315" xml:space="preserve">rempliſſent les autres con-<lb/>ditions du problême.</s> <s xml:id="echoid-s5316" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div286" type="section" level="1" n="259"> <head xml:id="echoid-head301" xml:space="preserve"><emph style="sc">Troisieme question</emph>.</head> <p> <s xml:id="echoid-s5317" xml:space="preserve">Cinq Canonniers ont tiré dans une après midi 96 coups de <lb/>canon; </s> <s xml:id="echoid-s5318" xml:space="preserve">le ſecond a tiré le double du premier, & </s> <s xml:id="echoid-s5319" xml:space="preserve">deux coups <lb/>de plus; </s> <s xml:id="echoid-s5320" xml:space="preserve">le troiſieme a tiré autant que le premier & </s> <s xml:id="echoid-s5321" xml:space="preserve">le ſecond, <lb/>moins ſix coups; </s> <s xml:id="echoid-s5322" xml:space="preserve">le quatrieme autant que le ſecond & </s> <s xml:id="echoid-s5323" xml:space="preserve">le troi-<lb/>ſieme, plus dix coups; </s> <s xml:id="echoid-s5324" xml:space="preserve">le cinquieme a tiré autant que le pre-<lb/>mier & </s> <s xml:id="echoid-s5325" xml:space="preserve">le quatrieme, moins vingt coups: </s> <s xml:id="echoid-s5326" xml:space="preserve">on demande com-<lb/>bien de coups de canon ils ont tiré chacun.</s> <s xml:id="echoid-s5327" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5328" xml:space="preserve">Ayant nommé x le nombre de coups que le premier a tiré, <lb/>je trouverai pour le ſecond 2x+2; </s> <s xml:id="echoid-s5329" xml:space="preserve">pour le troiſieme 3x + 2 - 6, <lb/>ou, ce qui eſt la même choſe, 3x - 4; </s> <s xml:id="echoid-s5330" xml:space="preserve">pour le quatrieme <lb/>5x + 2 - 4 + 10, ou bien 5x + 8; </s> <s xml:id="echoid-s5331" xml:space="preserve">enfin pour le cinquieme <lb/>6x + 8 - 20, ou bien 6x - 12. </s> <s xml:id="echoid-s5332" xml:space="preserve">Or comme toutes ces quan-<lb/>tités priſes enſemble doivent être égales à 96, je forme cette <lb/>équation x + 2x + 2 + 3x - 4 + 5x + 8 + 6x - 12 = 96, <lb/>que je réduis à ſa plus ſimple expreſſion, en ajoutant dans une <lb/>ſomme les quantités connues, qui ont le ſigne + ou -, & </s> <s xml:id="echoid-s5333" xml:space="preserve">il <lb/>vient 17x - 6 = 96, ou bien 17x = 102, en faiſant paſſer <lb/>- 6 du premier membre dans le ſecond: </s> <s xml:id="echoid-s5334" xml:space="preserve">pour avoir préſente-<lb/>ment la valeur de x, je diviſe cette équation par 17, & </s> <s xml:id="echoid-s5335" xml:space="preserve">je <lb/>trouve x = 6; </s> <s xml:id="echoid-s5336" xml:space="preserve">ce qui fait voir que le premier Canonnier a tiré <lb/>ſix coups; </s> <s xml:id="echoid-s5337" xml:space="preserve">ainſi le ſecond, qui eſt 2x + 2, en a tiré 14; </s> <s xml:id="echoid-s5338" xml:space="preserve">le troi-<lb/>ſieme, qui eſt 3x - 4, en a auſſi tiré 14; </s> <s xml:id="echoid-s5339" xml:space="preserve">le quatrieme, qui <lb/>eſt 5x + 8, en aura tiré 38; </s> <s xml:id="echoid-s5340" xml:space="preserve">& </s> <s xml:id="echoid-s5341" xml:space="preserve">le cinquieme, qui eſt 6x - 12, <lb/>en aura tiré 24; </s> <s xml:id="echoid-s5342" xml:space="preserve">ce qui eſt évident, puiſque tous ces nombres, <lb/>pris enſemble, font 96.</s> <s xml:id="echoid-s5343" xml:space="preserve"/> </p> <pb o="152" file="0190" n="190" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div287" type="section" level="1" n="260"> <head xml:id="echoid-head302" xml:space="preserve"><emph style="sc">Quatrieme question</emph>.</head> <p> <s xml:id="echoid-s5344" xml:space="preserve">Un Officier de Mineurs a fait faire en trois mois mille toiſes <lb/>courantes de galeries de mines; </s> <s xml:id="echoid-s5345" xml:space="preserve">il a fait dans le ſecond mois <lb/>le double de l’ouvrage du premier, & </s> <s xml:id="echoid-s5346" xml:space="preserve">50 toiſes de plus, parce <lb/>qu’il a reçu un renfort de Mineurs; </s> <s xml:id="echoid-s5347" xml:space="preserve">le troiſieme mois il a fait <lb/>200 toiſes d’ouvrage de moins que le ſecond, parce qu’une <lb/>partie de ſon monde eſt tombée malade: </s> <s xml:id="echoid-s5348" xml:space="preserve">on demande combien <lb/>il a fait de toiſes de galeries dans le premier mois, dans le <lb/>ſecond, & </s> <s xml:id="echoid-s5349" xml:space="preserve">dans le troiſieme?</s> <s xml:id="echoid-s5350" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5351" xml:space="preserve">Pour réſoudre cette queſtion, je nomme x la quantité de <lb/>toiſes de galeries de mines qui s’eſt faite le premier mois, <lb/>2x+50 pour ce qui s’eſt fait le ſecond mois, & </s> <s xml:id="echoid-s5352" xml:space="preserve">2x+50-200, <lb/>ou bien 2x - 150 pour la quantité qui s’eſt faite le troiſieme <lb/>mois; </s> <s xml:id="echoid-s5353" xml:space="preserve">& </s> <s xml:id="echoid-s5354" xml:space="preserve">comme la ſomme de ces quantités doit être égale à <lb/>1000 toiſes, je forme cette équation x + 2x + 50 + 2x - 150 <lb/>= 1000, qui étant réduite à ſa plus ſimple expreſſion (art. </s> <s xml:id="echoid-s5355" xml:space="preserve">50.) <lb/></s> <s xml:id="echoid-s5356" xml:space="preserve">donne 5x - 100 = 1000, ou bien x = 1100, & </s> <s xml:id="echoid-s5357" xml:space="preserve">diviſant <lb/>chaque membre de cette équation par 5, l’on aura x = 220; </s> <s xml:id="echoid-s5358" xml:space="preserve"><lb/>ce qui fait voir que dans le premier mois on a fait 220 toiſes <lb/>courantes de galeries de mines: </s> <s xml:id="echoid-s5359" xml:space="preserve">par conſéquent on en a fait <lb/>400 le ſecond mois, & </s> <s xml:id="echoid-s5360" xml:space="preserve">290 le troiſieme; </s> <s xml:id="echoid-s5361" xml:space="preserve">ce qui eſt évident, <lb/>puiſque ces quantités font enſemble 1000 toiſes.</s> <s xml:id="echoid-s5362" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div288" type="section" level="1" n="261"> <head xml:id="echoid-head303" xml:space="preserve"><emph style="sc">Cinquieme question</emph>.</head> <p> <s xml:id="echoid-s5363" xml:space="preserve">On a fait un détachement de Grenadiers pour attaquer un <lb/>poſte, parmi leſquels il s’en trouve deux qui raiſonnant en-<lb/>ſemble ſur les grenades qu’ils ont dans leurs gibernes, le pre-<lb/>mier dit au ſecond: </s> <s xml:id="echoid-s5364" xml:space="preserve">Si tu m’avois donné une de tes grenades, <lb/>j’en aurois autant que toi, & </s> <s xml:id="echoid-s5365" xml:space="preserve">le ſecond lui répond: </s> <s xml:id="echoid-s5366" xml:space="preserve">ſi tu m’en <lb/>avois donné une des tiennes, j’en aurois le double de celles <lb/>que tu as: </s> <s xml:id="echoid-s5367" xml:space="preserve">on demande combien ils avoient de grenades <lb/>chacun?</s> <s xml:id="echoid-s5368" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5369" xml:space="preserve">Comme cette queſtion renferme deux inconnues, je nomme <lb/>y le nombre des grenades qu’a le premier Grenadier, & </s> <s xml:id="echoid-s5370" xml:space="preserve">z le <lb/>nombre de celles qu’a le ſecond; </s> <s xml:id="echoid-s5371" xml:space="preserve">& </s> <s xml:id="echoid-s5372" xml:space="preserve">je fais autant d’équations <lb/>qu’il y a d’inconnues, ſelon l’article 304. </s> <s xml:id="echoid-s5373" xml:space="preserve">Pour former la pre-<lb/>miere équation, je dis, ſi y avoit une grenade de plus, & </s> <s xml:id="echoid-s5374" xml:space="preserve">z une <lb/>grenade de moins, ces deux quantités ſeroient égales, ce qui <pb o="153" file="0191" n="191" rhead="DE MATHÉMATIQUE. Liv. II."/> donne y + 1 = z - 1. </s> <s xml:id="echoid-s5375" xml:space="preserve">Pour avoir la ſeconde équation, <lb/>je fais encore ce raiſonnement, ſi z avoit une grenade de <lb/>plus, & </s> <s xml:id="echoid-s5376" xml:space="preserve">y une de moins, la premiere quantité ſeroit dou-<lb/>ble de la ſeconde; </s> <s xml:id="echoid-s5377" xml:space="preserve">ce qui donne cette égalité z + 1 = 2y <lb/>- 2. </s> <s xml:id="echoid-s5378" xml:space="preserve">Préſentement que j’ai autant d’équations que d’incon-<lb/>nues, je dégage l’inconnue z de la premiere équation, en <lb/>faiſant paſſer - 1 du ſecond membre dans le premier pour <lb/>avoir y + 2 = z: </s> <s xml:id="echoid-s5379" xml:space="preserve">enſuite je ſubſtitue dans la ſeconde équa-<lb/>tion à la place de z ſa valeur (art. </s> <s xml:id="echoid-s5380" xml:space="preserve">298), & </s> <s xml:id="echoid-s5381" xml:space="preserve">il vient y + 3 <lb/>= 2y - 2, où z ne ſe trouve plus; </s> <s xml:id="echoid-s5382" xml:space="preserve">& </s> <s xml:id="echoid-s5383" xml:space="preserve">faiſant paſſer - 2 du <lb/>ſecond membre dans le premier, il vient y + 5 = 2y, & </s> <s xml:id="echoid-s5384" xml:space="preserve">effa-<lb/>çant y de part & </s> <s xml:id="echoid-s5385" xml:space="preserve">d’autre, j’aurai cette équation 5 = y, qui me <lb/>donne la valeur de y, ſubſtituant cette valeur de y dans l’é-<lb/>quation, où z eſt dégagée, l’on aura 7 = z: </s> <s xml:id="echoid-s5386" xml:space="preserve">par conſéquent <lb/>le premier Grenadier avoit cinq grenades, & </s> <s xml:id="echoid-s5387" xml:space="preserve">le ſecond ſept; <lb/></s> <s xml:id="echoid-s5388" xml:space="preserve">ce qui eſt bien évident, puiſque ces deux nombres rempliſſent <lb/>les conditions du problême.</s> <s xml:id="echoid-s5389" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div289" type="section" level="1" n="262"> <head xml:id="echoid-head304" xml:space="preserve"><emph style="sc">Sixieme question</emph>.</head> <p> <s xml:id="echoid-s5390" xml:space="preserve">Trois Bombardiers ont jetté une certaine quantité de bom-<lb/>bes dans une Ville aſſiégée: </s> <s xml:id="echoid-s5391" xml:space="preserve">le premier & </s> <s xml:id="echoid-s5392" xml:space="preserve">le ſecond en ont <lb/>jetté enſemble 20 plus que le troiſieme; </s> <s xml:id="echoid-s5393" xml:space="preserve">le ſecond & </s> <s xml:id="echoid-s5394" xml:space="preserve">le troi-<lb/>ſieme 32 plus que le premier; </s> <s xml:id="echoid-s5395" xml:space="preserve">& </s> <s xml:id="echoid-s5396" xml:space="preserve">le premier & </s> <s xml:id="echoid-s5397" xml:space="preserve">le troiſieme 28 <lb/>plus que le ſecond: </s> <s xml:id="echoid-s5398" xml:space="preserve">on demande combien chaque Bombardier <lb/>a jetté de bombes?</s> <s xml:id="echoid-s5399" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5400" xml:space="preserve">Comme les quantités connues dans cette queſtion ſont ex-<lb/>primées par des nombres, nous ſubſtituerons à leurs places les <lb/>premieres lettres de l’alphabet: </s> <s xml:id="echoid-s5401" xml:space="preserve">ainſi au lieu des nombres 20, <lb/>32, 28, nous prendrons a, b, c, ſuppoſant que 20 = a, 32 <lb/>= b, 28 = c, pour rendre la réſolution de ce problême plus <lb/>générale, & </s> <s xml:id="echoid-s5402" xml:space="preserve">nous nommerons x la quantité de bombes que le <lb/>premier Bombardier a jetté, y la quantité du ſecond, & </s> <s xml:id="echoid-s5403" xml:space="preserve">z la <lb/>quantité du troiſieme. </s> <s xml:id="echoid-s5404" xml:space="preserve">Cela poſé, je dis ſi de x + y, qui ex-<lb/>prime la quantité de bombes qu’ont jetté le premier & </s> <s xml:id="echoid-s5405" xml:space="preserve">le ſe-<lb/>cond Bombardier, je ſouſtrais a, qui eſt le nombre de bom-<lb/>bes que le premier & </s> <s xml:id="echoid-s5406" xml:space="preserve">le ſecond ont tiré plus que le troiſieme, <lb/>j’aurai x + y - a = z pour la premiere équation; </s> <s xml:id="echoid-s5407" xml:space="preserve">y + z - b <lb/>= x pour la ſeconde, & </s> <s xml:id="echoid-s5408" xml:space="preserve">x + z - c = y pour la troiſieme. <lb/></s> <s xml:id="echoid-s5409" xml:space="preserve">Conſidérant que j’ai trois équations, qui renferment chacune <pb o="154" file="0192" n="192" rhead="NOUVEAU COURS"/> trois inconnues, je cherche la valeur d’une de ces inconnues, <lb/>pour la ſubſtituer dans les autres équations aux endroits où <lb/>cette inconnue ſe trouvera (art. </s> <s xml:id="echoid-s5410" xml:space="preserve">298). </s> <s xml:id="echoid-s5411" xml:space="preserve">Et comme la premiere <lb/>équation x + y - a = z, me donne la valeur de z, qui eſt la <lb/>quantité x + y - a elle-même, je la mets dans la ſeconde & </s> <s xml:id="echoid-s5412" xml:space="preserve"><lb/>troiſieme équation à la place de z; </s> <s xml:id="echoid-s5413" xml:space="preserve">ce qui les changera en <lb/>celles-ci, y + x + y - a - b = x, & </s> <s xml:id="echoid-s5414" xml:space="preserve">x + y - a + x - c = y, <lb/>dont les termes étant rendus poſitifs, & </s> <s xml:id="echoid-s5415" xml:space="preserve">réduits à leur plus <lb/>ſimple expreſſion, donnent 2y = a + b, & </s> <s xml:id="echoid-s5416" xml:space="preserve">2x = a + c, qui <lb/>étant diviſés par 2, donnent enfin y = {a+b/2}, & </s> <s xml:id="echoid-s5417" xml:space="preserve">x = {a+c/2}. </s> <s xml:id="echoid-s5418" xml:space="preserve">Or <lb/>comme il n’y a plus d’inconnues dans ces deux équations, il <lb/>faut revenir à la premiere, qui eſt x + y - a = z, afin de <lb/>ſubſtituer à la place de x & </s> <s xml:id="echoid-s5419" xml:space="preserve">de y leurs valeurs {a+b/2} & </s> <s xml:id="echoid-s5420" xml:space="preserve">{a+c/2} pour <lb/>avoir {1/2}a + {1/2}b + {1/2}a + {1/2}c - a = z, ou bien {b+c/2}, parce queles <lb/>deux termes + {1/2}a + {1/2}a qui valent a, détruiſent - a: </s> <s xml:id="echoid-s5421" xml:space="preserve">on a <lb/>donc la valeur de z, qui eſt la derniere quantité qui reſtoit à <lb/>connoître.</s> <s xml:id="echoid-s5422" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5423" xml:space="preserve">Préſentement que je ſçais que x = {a+c/2}, que y = {a+b/2}, & </s> <s xml:id="echoid-s5424" xml:space="preserve"><lb/>que z = {b + c/2}, je prends à la place de {a + c/2} la moitié <lb/>des nombres repréſentés par a & </s> <s xml:id="echoid-s5425" xml:space="preserve">b, c’eſt-à-dire la moitié de <lb/>20 + 28, qui eſt 24, qui ſera la valeur de x; </s> <s xml:id="echoid-s5426" xml:space="preserve">à la place de <lb/>{a+b/2}, je prends la moitié de 20 + 32 pour avoir 26, qui eſt la <lb/>valeur de y; </s> <s xml:id="echoid-s5427" xml:space="preserve">& </s> <s xml:id="echoid-s5428" xml:space="preserve">enfin à la place de {b + c/2}, je prends la moitié des <lb/>nombres 28 & </s> <s xml:id="echoid-s5429" xml:space="preserve">32 pour avoir 30, qui ſera la valeur de z; </s> <s xml:id="echoid-s5430" xml:space="preserve">d’où <lb/>je conclus que le premier Bombardier a jetté 24 bombes, le <lb/>ſecond 26, & </s> <s xml:id="echoid-s5431" xml:space="preserve">le troiſieme 30, puiſque ces trois nombres ſa-<lb/>tisfont pleinement aux conditions du problême.</s> <s xml:id="echoid-s5432" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div290" type="section" level="1" n="263"> <head xml:id="echoid-head305" xml:space="preserve"><emph style="sc">Septieme question</emph>.</head> <p> <s xml:id="echoid-s5433" xml:space="preserve">L’on aſſiege une Place, dont la garniſon étoit compoſée <lb/>de Troupes Allemandes, Angloiſes, Hollandoiſes & </s> <s xml:id="echoid-s5434" xml:space="preserve">Eſpa-<lb/>gnoles. </s> <s xml:id="echoid-s5435" xml:space="preserve">La Place priſe, on a trouvé qu’il y avoit eu enſemble <lb/>autant d’Allemands, d’ Anglois & </s> <s xml:id="echoid-s5436" xml:space="preserve">de Hollandois de tués que <lb/>d’Eſpagnols, moins 620 hommes; </s> <s xml:id="echoid-s5437" xml:space="preserve">autant d’Allemands, d’An-<lb/>glois & </s> <s xml:id="echoid-s5438" xml:space="preserve">d’Eſpagnols enſemble que de Hollandois, moins 460 <lb/>hommes; </s> <s xml:id="echoid-s5439" xml:space="preserve">autant d’Allemands, de Hollandois & </s> <s xml:id="echoid-s5440" xml:space="preserve">d’Eſpagnols <pb o="155" file="0193" n="193" rhead="DE MATHÉMATIQUE. Liv. II."/> enſemble que d’Anglois, moins 380; </s> <s xml:id="echoid-s5441" xml:space="preserve">enfin autant d’Anglois, <lb/>de Hollandois & </s> <s xml:id="echoid-s5442" xml:space="preserve">d’Eſpagnols, moins 500 hommes que d’Alle-<lb/>mands: </s> <s xml:id="echoid-s5443" xml:space="preserve">on demande combien il y a eu d’Allemands de tués, <lb/>combien d’Anglois, de Hollandois & </s> <s xml:id="echoid-s5444" xml:space="preserve">d’Eſpagnols?</s> <s xml:id="echoid-s5445" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5446" xml:space="preserve">Ayant nommé u le nombre d’Allemands, x celui des An-<lb/>glois, y celui des Hollandois, & </s> <s xml:id="echoid-s5447" xml:space="preserve">z celui des Eſpagnols, nous <lb/>ſuppoſerons que 620 = a, que 460 = b, que 380 = c, & </s> <s xml:id="echoid-s5448" xml:space="preserve">que <lb/>500 = d, afin de rendre la ſolution du problême plus géné-<lb/>rale. </s> <s xml:id="echoid-s5449" xml:space="preserve">Cela poſé, comme les conditions du problême me donnent <lb/>quatre équations, j’ai pour la premiere u + x + y = z + a, <lb/>pour la ſeconde u + x + z = y + b, pour la troiſieme u + y <lb/>+ z = x + c; </s> <s xml:id="echoid-s5450" xml:space="preserve">& </s> <s xml:id="echoid-s5451" xml:space="preserve">enfin pour la quatrieme x + y + z = u + d. <lb/></s> <s xml:id="echoid-s5452" xml:space="preserve">Après cela, je dégage une inconnue dans la premiere équa-<lb/>tion qui ſera, par exemple z, pour avoir u + x + y - a = z, <lb/>qui me donne la valeur de z, que je ſubſtitue dans les trois <lb/>autres équations; </s> <s xml:id="echoid-s5453" xml:space="preserve">ce qui les change en celles-ci, u + x + u <lb/>+ x + y - a = y + b, u + y + u + x + y - a = x + c, <lb/>& </s> <s xml:id="echoid-s5454" xml:space="preserve">x + y + u + x + y - a = u + d, qui deviennent, en <lb/>les réduiſant à leur plus ſimple expreſſion, 2u = a + b - 2x, <lb/>2y = a + c - 2u, & </s> <s xml:id="echoid-s5455" xml:space="preserve">2x = a + d - 2y, en dégageant 2u, <lb/>2x, & </s> <s xml:id="echoid-s5456" xml:space="preserve">2y. </s> <s xml:id="echoid-s5457" xml:space="preserve">Après cela je ſubſtitue la valeur de 2u dans l’équa-<lb/>tion 2y = a + c - 2u, il vient 2y = a + c - a - b + 2x, <lb/>dans laquelle u ne ſe trouve plus; </s> <s xml:id="echoid-s5458" xml:space="preserve">& </s> <s xml:id="echoid-s5459" xml:space="preserve">ſi à la place de 2y je <lb/>mets ſa valeur priſe dans l’égalité 2x = a + d - 2y, il <lb/>viendra cette derniere équation, 2x = a + d - a - c + a <lb/>+ b - 2x, ou bien x = {a + b + d - c/4}, où il n’y a plus d’in-<lb/>connue. </s> <s xml:id="echoid-s5460" xml:space="preserve">Si à la place de 2x dans l’équation 2u = a + b - 2x, <lb/>l’on met la moitié de la valeur de 4x, qui eſt {1/2}a + {1/2}b + {1/2}d <lb/>-{1/2}c, l’on aura 2u = a + b - {1/2}a - {1/2}b - {1/2}d + {1/2}c, ou <lb/>2u = {a + b + c - d/2}, ou bien u = {a + b + c - d/4}, qui donne la <lb/>valeur de u; </s> <s xml:id="echoid-s5461" xml:space="preserve">& </s> <s xml:id="echoid-s5462" xml:space="preserve">ſi l’on met dans l’équation 2y = a + c - 2u <lb/>la moitié de la valeur de 4u, qui eſt {1/1}a + {1/2}b + {1/2}c - {1/2}d, <lb/>l’on aura 2y = a + c - {1/2}a - {1/2}b - {1/2}c + {1/2}d, ou y = <lb/>{a + c + d - b/4}, qui donne la valeur de y; </s> <s xml:id="echoid-s5463" xml:space="preserve">enfin ſi l’on met <lb/>dans l’équation u + x + y - a = z les valeurs de u, de x & </s> <s xml:id="echoid-s5464" xml:space="preserve"><lb/>de y, l’on aura, après les réductions néceſſaires, z = {b + c + d - a/4}</s> </p> <pb o="156" file="0194" n="194" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s5465" xml:space="preserve">Comme l’on vient de trouver u = {a+b+c-d/4}, x= {a+b+d-c/4}, <lb/>y = {a + e + d -b/4}, & </s> <s xml:id="echoid-s5466" xml:space="preserve">z = {b + c + d - a/4}, il s’enſuit que le <lb/>problême eſt réſolu, puiſque ſi l’on diviſe 1460 - 500 par 4, <lb/>qui eſt égal à {a + c + b - d/4}, l’on trouvera 240 pour la valeur <lb/>de u, faiſant de même pour les autres, l’on trouvera 300 pour <lb/>la valeur de x, 260 pour celle de y, & </s> <s xml:id="echoid-s5467" xml:space="preserve">180 pour celle de z. <lb/></s> <s xml:id="echoid-s5468" xml:space="preserve">Ainſi il y a eu 240 Allemands de tués, 300 Anglois, 260 <lb/>Hollandois, & </s> <s xml:id="echoid-s5469" xml:space="preserve">180 Eſpagnols; </s> <s xml:id="echoid-s5470" xml:space="preserve">ce qui eſt bien évident, puiſ-<lb/>que ces nombres répondent aux conditions du problême.</s> <s xml:id="echoid-s5471" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div291" type="section" level="1" n="264"> <head xml:id="echoid-head306" xml:space="preserve"><emph style="sc">Huitieme question</emph>.</head> <p> <s xml:id="echoid-s5472" xml:space="preserve">Un Sergent de Sapeurs s’eſt trouvé à 32 ſieges, & </s> <s xml:id="echoid-s5473" xml:space="preserve">à plu-<lb/>ſieurs batailles, où il a reçu pluſieurs bleſſures: </s> <s xml:id="echoid-s5474" xml:space="preserve">le Roi lui pro-<lb/>met de lui accorder la gratification qu’il lui demandera pour <lb/>ſes ſervices. </s> <s xml:id="echoid-s5475" xml:space="preserve">Le Sergent demande au Roi de lui donner en ar-<lb/>gent la ſomme des gratifications qu’il auroit eu, en ſuppoſant <lb/>qu’on lui eût donné une livre pour la premiere bleſſure, 2 liv. <lb/></s> <s xml:id="echoid-s5476" xml:space="preserve">pour la ſeconde, 4 livres pour la troiſieme, & </s> <s xml:id="echoid-s5477" xml:space="preserve">ainſi de ſuite en <lb/>doublant toujours. </s> <s xml:id="echoid-s5478" xml:space="preserve">Le Roi lui accorde ſa demande, & </s> <s xml:id="echoid-s5479" xml:space="preserve">il re-<lb/>çoit 65535 livres: </s> <s xml:id="echoid-s5480" xml:space="preserve">on demande combien il a reçu de bleſſures.</s> <s xml:id="echoid-s5481" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5482" xml:space="preserve">Pour réſoudre cette queſtion, je la dépouille de tout ce qui <lb/>lui eſt étranger, & </s> <s xml:id="echoid-s5483" xml:space="preserve">je la réduis à ce qu’elle a de plus ſimple; <lb/></s> <s xml:id="echoid-s5484" xml:space="preserve">je vois que le nombre 65535 eſt la ſomme des termes d’une <lb/>progreſſion géométrique, dont le premier terme eſt 1, le ſe-<lb/>cond 2, & </s> <s xml:id="echoid-s5485" xml:space="preserve">dont la raiſon eſt auſſi 2, ou, ce qui eſt la même <lb/>choſe, que ce même nombre eſt la ſomme de pluſieurs puiſ-<lb/>ſances ſucceſſives de 2, dont la derniere, augmentée de l’u-<lb/>nité, marque le nombre des termes de la progreſſion. </s> <s xml:id="echoid-s5486" xml:space="preserve">Je fais <lb/>attention enſuite, que ſi j’avois le dernier terme de cette pro-<lb/>greſſion, il me ſeroit aiſé d’en connoître le nombre, puiſque ce <lb/>dernier terme eſt égal au premier, multiplié par la puiſſance de <lb/>2, exprimée par le nombre des termes qui précédent (art. </s> <s xml:id="echoid-s5487" xml:space="preserve">248). </s> <s xml:id="echoid-s5488" xml:space="preserve"><lb/>J’appelle x ce dernier terme, & </s> <s xml:id="echoid-s5489" xml:space="preserve">je fais encore attention que la <lb/>ſomme des antécédens eſt celle de tous les termes, excepté ce <lb/>dernier, & </s> <s xml:id="echoid-s5490" xml:space="preserve">que la ſomme des conſéquens eſt la même ſomme <lb/>de tous les termes, excepté le premier, qui eſt 1. </s> <s xml:id="echoid-s5491" xml:space="preserve">Or (art. </s> <s xml:id="echoid-s5492" xml:space="preserve">250) <lb/>la ſomme des antécédens eſt à la ſomme des conſéquens, <pb o="157" file="0195" n="195" rhead="DE MATHEMATIQUE. Liv. II."/> comme un ſeul antécédent eſt à ſon conſéquent. </s> <s xml:id="echoid-s5493" xml:space="preserve">Ainſi en <lb/>exprimant cela analitiquement, & </s> <s xml:id="echoid-s5494" xml:space="preserve">appellant s le nombre <lb/>65535, qui eſt la ſomme des termes de la progreſſion, j’aurai <lb/>s - x. </s> <s xml:id="echoid-s5495" xml:space="preserve">s - 1 : </s> <s xml:id="echoid-s5496" xml:space="preserve">: </s> <s xml:id="echoid-s5497" xml:space="preserve">1.</s> <s xml:id="echoid-s5498" xml:space="preserve">2, d’où l’on tire, en faiſant le produit des <lb/>extrêmes & </s> <s xml:id="echoid-s5499" xml:space="preserve">des moyens, 2s - 2x = s - 1, & </s> <s xml:id="echoid-s5500" xml:space="preserve">dégageant <lb/>x, il vient x = {s + 1/2} = {65536/2} = 32768, qui montre que le <lb/>dernier terme de la progreſſion eſt 32768, qui eſt certaine-<lb/>ment une puiſſance de 2. </s> <s xml:id="echoid-s5501" xml:space="preserve">Pour ſçavoir à quelle puiſſance de 2 <lb/>ce nombre eſt égal, j’éleve 2 à ſes puiſſances ſucceſſives, & </s> <s xml:id="echoid-s5502" xml:space="preserve">je <lb/>trouve qu’il eſt égal à la 15<emph style="sub">e</emph> puiſſance de 2: </s> <s xml:id="echoid-s5503" xml:space="preserve">donc ce terme eſt <lb/>le 16<emph style="sub">e</emph>, puiſque le nombre 15 qui marque la puiſſance de 2 à <lb/>laquelle ce terme eſt égal, marque auſſi le nombre des termes <lb/>qui le précédent: </s> <s xml:id="echoid-s5504" xml:space="preserve">ainſi ce Sergent avoit reçu 16 bleſſures.</s> <s xml:id="echoid-s5505" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div292" type="section" level="1" n="265"> <head xml:id="echoid-head307" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s5506" xml:space="preserve">La même proportion, qui nous a ſervi à réſoudre cette <lb/>queſtion, peut auſſi ſervir à la ſolution de toutes les queſtions <lb/>que l’on propoſe ſur les progreſſions géométriques, & </s> <s xml:id="echoid-s5507" xml:space="preserve">parti-<lb/>culiérement dans la ſommation des mêmes ſuites: </s> <s xml:id="echoid-s5508" xml:space="preserve">pour en <lb/>faire ſentir encore mieux l’utilité, nous allons l’appliquer à <lb/>la ſolution du problême ſuivant.</s> <s xml:id="echoid-s5509" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div293" type="section" level="1" n="266"> <head xml:id="echoid-head308" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s5510" xml:space="preserve">305. </s> <s xml:id="echoid-s5511" xml:space="preserve">Trouver la ſomme des termes d’une progreſſion géomé-<lb/>trique décroiſſante à l’infini, dont le premier terme eſt a, & </s> <s xml:id="echoid-s5512" xml:space="preserve">le <lb/>ſecond b.</s> <s xml:id="echoid-s5513" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div294" type="section" level="1" n="267"> <head xml:id="echoid-head309" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s5514" xml:space="preserve">Puiſque le nombre des termes eſt infini, & </s> <s xml:id="echoid-s5515" xml:space="preserve">que d’ailleurs <lb/>la progreſſion eſt ſuppoſée décroiſſante, le dernier terme pourra <lb/>enfin être regardé comme zero: </s> <s xml:id="echoid-s5516" xml:space="preserve">ainſi la ſomme des antécé-<lb/>dens ſera la ſomme de tous les termes, moins zero; </s> <s xml:id="echoid-s5517" xml:space="preserve">la ſomme <lb/>des conſéquens ſera la ſomme de tous les termes, moins le <lb/>premier: </s> <s xml:id="echoid-s5518" xml:space="preserve">donc appellant s cette ſomme, on aura (art. </s> <s xml:id="echoid-s5519" xml:space="preserve">250.) <lb/></s> <s xml:id="echoid-s5520" xml:space="preserve">la ſomme des antécédens eſt à la ſomme des conſéquens, <lb/>comme le premier terme au ſecond, ou analitiquement s - 0. </s> <s xml:id="echoid-s5521" xml:space="preserve"><lb/>s - a : </s> <s xml:id="echoid-s5522" xml:space="preserve">: </s> <s xml:id="echoid-s5523" xml:space="preserve">a. </s> <s xml:id="echoid-s5524" xml:space="preserve">b, d’où l’on tire as - a<emph style="sub">2</emph> = bs, ou as - bs = a<emph style="sub">2</emph>, <lb/>& </s> <s xml:id="echoid-s5525" xml:space="preserve">dégageant s, il vient s = {a<emph style="sub">2</emph>/a-b}; </s> <s xml:id="echoid-s5526" xml:space="preserve">ce qui ſignifie qu’en <lb/>général la ſomme des termes d’une progreſſion géométrique <pb o="158" file="0196" n="196" rhead="NOUVEAU COURS"/> décroiſſante à l’infini, eſt égale au quarré du premier terme, <lb/>diviſé par la différence du premier au ſecond. </s> <s xml:id="echoid-s5527" xml:space="preserve">Par exemple, <lb/>ſi l’on veut ſommer tous les termes de cette progreſſion <lb/>{.</s> <s xml:id="echoid-s5528" xml:space="preserve">./.</s> <s xml:id="echoid-s5529" xml:space="preserve">.} 2. </s> <s xml:id="echoid-s5530" xml:space="preserve">1. </s> <s xml:id="echoid-s5531" xml:space="preserve">{1/2}. </s> <s xml:id="echoid-s5532" xml:space="preserve">{1/4}. </s> <s xml:id="echoid-s5533" xml:space="preserve">{1/8}. </s> <s xml:id="echoid-s5534" xml:space="preserve">{1/16}. </s> <s xml:id="echoid-s5535" xml:space="preserve">{1/32}, &</s> <s xml:id="echoid-s5536" xml:space="preserve">c; </s> <s xml:id="echoid-s5537" xml:space="preserve">j’éleve 2 à ſon quarré, qui eſt 4, que <lb/>je diviſe par 2 --- 1, qui eſt 1: </s> <s xml:id="echoid-s5538" xml:space="preserve">ainſi la ſomme des termes de <lb/>cette progreſſion eſt 4. </s> <s xml:id="echoid-s5539" xml:space="preserve">D’où il ſuit, que toutes les fractions <lb/>{1/2}, {1/4}, {1/8}, {1/16} ne valent qu’un, en les pouſſant juſqu’à l’infini. </s> <s xml:id="echoid-s5540" xml:space="preserve">De <lb/>même ſi l’on a {.</s> <s xml:id="echoid-s5541" xml:space="preserve">./.</s> <s xml:id="echoid-s5542" xml:space="preserve">.} 3. </s> <s xml:id="echoid-s5543" xml:space="preserve">1. </s> <s xml:id="echoid-s5544" xml:space="preserve">{1/3}. </s> <s xml:id="echoid-s5545" xml:space="preserve">{1/9}. </s> <s xml:id="echoid-s5546" xml:space="preserve">{1/27}, &</s> <s xml:id="echoid-s5547" xml:space="preserve">c, je cherche le quarré de 3, qui eſt <lb/>9, que je diviſe par 3 --- 1 ou 2, & </s> <s xml:id="echoid-s5548" xml:space="preserve">j’ai la ſomme des termes de <lb/>la progreſſion s = {9/2} = 4 {1/2}: </s> <s xml:id="echoid-s5549" xml:space="preserve">d’où il ſuit que tous les termes <lb/>{1/3}, {1/9}, {1/27}, {1/81}, &</s> <s xml:id="echoid-s5550" xml:space="preserve">c. </s> <s xml:id="echoid-s5551" xml:space="preserve">ne valent que {1/2}, puiſque les deux premiers <lb/>termes font 4. </s> <s xml:id="echoid-s5552" xml:space="preserve">Il en eſt ainſi des autres progreſſions, ſur leſ-<lb/>quelles il eſt aiſé de faire l’application de la formule générale.</s> <s xml:id="echoid-s5553" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div295" type="section" level="1" n="268"> <head xml:id="echoid-head310" xml:space="preserve">De la réſolution des Equations du ſecond degré. <lb/><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s5554" xml:space="preserve">306. </s> <s xml:id="echoid-s5555" xml:space="preserve">Les équations que nous venons de réſoudre, ſont ap-<lb/>pellées équations du premier degré, ainſi que les problêmes, dont <lb/>elles expriment les conditions, parce que les inconnues n’y <lb/>ſont point multipliées par elles-mêmes, ni les unes par les <lb/>autres: </s> <s xml:id="echoid-s5556" xml:space="preserve">mais ſi cela arrivoit, l’équation qui ſeroit dans ce cas, <lb/>ſeroit plus compliquée que les précédentes, & </s> <s xml:id="echoid-s5557" xml:space="preserve">ſeroit appellée <lb/>du ſecond, troiſieme, quatrieme degré, ſelon que l’inconnue <lb/>y ſeroit élevée à la ſeconde, à la troiſieme ou quatrieme puiſ-<lb/>ſance. </s> <s xml:id="echoid-s5558" xml:space="preserve">Par exemple, xx - 2ax = 30, eſt une équation du <lb/>ſecond degré, x<emph style="sub">3</emph> - 5x<emph style="sub">2</emph> + 7x + 12 = 15, eſt une équation <lb/>du troiſieme degré. </s> <s xml:id="echoid-s5559" xml:space="preserve">Nous ne parlerons ici que des équations <lb/>du ſecond degré, & </s> <s xml:id="echoid-s5560" xml:space="preserve">après les avoir réſolues ſur quel ques exem-<lb/>ples dans des cas particuliers, nous les réſolverons en général <lb/>dans les formules qui comprennent tous les cas poſſibles de <lb/>ces ſortes d’équations.</s> <s xml:id="echoid-s5561" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div296" type="section" level="1" n="269"> <head xml:id="echoid-head311" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s5562" xml:space="preserve">307. </s> <s xml:id="echoid-s5563" xml:space="preserve">Les regles que l’on doit ſuivre pour mettre un pro-<lb/>blême du ſecond degré en équation, ſont préciſément les <lb/>mêmes que celles que nous avons donné pour les autres pro-<lb/>blêmes: </s> <s xml:id="echoid-s5564" xml:space="preserve">le tout conſiſte à bien exprimer analitiquement les <lb/>conditions énoncées ou renfermées dans la queſtion; </s> <s xml:id="echoid-s5565" xml:space="preserve">ce qui <lb/>dépend plutôt de la ſagacité de celui qui réſout le problême, <lb/>que d’aucune regle générale que l’on puiſſe établir.</s> <s xml:id="echoid-s5566" xml:space="preserve"/> </p> <pb o="159" file="0197" n="197" rhead="DE MATHÉMATIQUE. Liv. II."/> <p> <s xml:id="echoid-s5567" xml:space="preserve">308. </s> <s xml:id="echoid-s5568" xml:space="preserve">On remarquera encore avant toutes choſes, que le <lb/>quarré d’une grandeur quelconque peut avoir le ſigne + ou -<lb/>à ſa racine, c’eſt-à-dire que ce quarré aa, peut réſulter de + a <lb/>multiplié par + a, ou de - a x - a, puiſque l’un & </s> <s xml:id="echoid-s5569" xml:space="preserve">l’autre <lb/>donne également a<emph style="sub">2</emph> au produit: </s> <s xml:id="echoid-s5570" xml:space="preserve">d’où il ſuit qu’en général <lb/>une équation du ſecond degré doit avoir deux racines, l’une <lb/>que l’on appelle négative, parce qu’elle eſt précédée du ſigne <lb/>-, & </s> <s xml:id="echoid-s5571" xml:space="preserve">l’autre qu’on appelle poſitive, parce qu’elle eſt précédée <lb/>du ſigne +. </s> <s xml:id="echoid-s5572" xml:space="preserve">L’état de la queſtion détermine ordinairement <lb/>celle que l’on doit prendre; </s> <s xml:id="echoid-s5573" xml:space="preserve">mais on ne doit point, ſurtout <lb/>dans les commencemens, rejetter les valeurs négatives, ſans <lb/>avoir auparavant examiné ce qu’elles peuvent ſignifier, parce <lb/>qu’elles ne réſolvent pas moins le problême, que celles que <lb/>l’on appelle poſitives, quoiqu’elles ne le réſolvent pas dans le <lb/>ſens qu’on s’étoit propoſé d’abord; </s> <s xml:id="echoid-s5574" xml:space="preserve">& </s> <s xml:id="echoid-s5575" xml:space="preserve">parce que d’ailleurs ces <lb/>ſolutions nous découvrent toujours des vérités auxquelles on <lb/>n’auroit peut-être jamais penſé, ſi l’on n’y eût été conduit par <lb/>l’analyſe. </s> <s xml:id="echoid-s5576" xml:space="preserve">On verra dans la ſuite des exemples ſenſibles de ce <lb/>que nous diſons, dans les problêmes que nous allons réſoudre.</s> <s xml:id="echoid-s5577" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div297" type="section" level="1" n="270"> <head xml:id="echoid-head312" xml:space="preserve"><emph style="sc">Premiere question</emph>.</head> <p> <s xml:id="echoid-s5578" xml:space="preserve">309. </s> <s xml:id="echoid-s5579" xml:space="preserve">Un Soldat va rejoindre ſon Régiment, dont il eſt <lb/>éloigné de 64 lieues, il fait une lieue le premier jour, trois le <lb/>ſecond, cinq le troiſieme, & </s> <s xml:id="echoid-s5580" xml:space="preserve">ainſi de ſuite en augmentant <lb/>toujours de deux lieues: </s> <s xml:id="echoid-s5581" xml:space="preserve">on demande combien il ſera de jours <lb/>à rejoindre ſon Régiment?</s> <s xml:id="echoid-s5582" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5583" xml:space="preserve">Pour réſoudre cette queſtion, je la dépouille encore de tout <lb/>ce qui lui eſt étranger (car c’eſt ainſi que l’on accoutume ſon <lb/>eſprit aux idées générales; </s> <s xml:id="echoid-s5584" xml:space="preserve">& </s> <s xml:id="echoid-s5585" xml:space="preserve">d’ailleurs cette regle eſt de la <lb/>derniere importance pour trouver les équations des problêmes <lb/>avec facilité). </s> <s xml:id="echoid-s5586" xml:space="preserve">Je remarque que la queſtion ſe réduit à trouver <lb/>le nombre des termes d’une progreſſion arithmétique, dont le <lb/>premier eſt 1, le ſecond 3, & </s> <s xml:id="echoid-s5587" xml:space="preserve">la ſomme eſt 64. </s> <s xml:id="echoid-s5588" xml:space="preserve">Et pour géné-<lb/>raliſer encore davantage le problême, je ſuppoſe que le pre-<lb/>mier terme de la progreſſion eſt a, le ſecond b, & </s> <s xml:id="echoid-s5589" xml:space="preserve">la ſomme s. <lb/></s> <s xml:id="echoid-s5590" xml:space="preserve">J’appelle x le nombre des termes, & </s> <s xml:id="echoid-s5591" xml:space="preserve">d l’excès de b ſur a. </s> <s xml:id="echoid-s5592" xml:space="preserve">Je <lb/>ſçais que la ſomme des termes d’une progreſſion arithmétique <lb/>eſt égale au produit de la ſomme des extrêmes, multipliée par <lb/>la moitié du nombre des termes (art. </s> <s xml:id="echoid-s5593" xml:space="preserve">238). </s> <s xml:id="echoid-s5594" xml:space="preserve">Je connois le pre- <pb o="160" file="0198" n="198" rhead="NOUVEAU COURS"/> mier extrême, qui eſt a, mais je ne connois pas le dernier; <lb/></s> <s xml:id="echoid-s5595" xml:space="preserve">cependant je ſçais qu’en général ce dernier terme eſt égal au <lb/>premier terme, plus au produit de la différence du ſecond au <lb/>premier, multipliée par le nombre des termes qui le précédent <lb/>(art. </s> <s xml:id="echoid-s5596" xml:space="preserve">240); </s> <s xml:id="echoid-s5597" xml:space="preserve">& </s> <s xml:id="echoid-s5598" xml:space="preserve">comme x eſt le nombre des termes, x- 1 ſera celui <lb/>termes qui précédent le dernier: </s> <s xml:id="echoid-s5599" xml:space="preserve">donc ce dernier ſera a + d X <lb/>√x-1\x{0020}, ou a + dx - d, auquel ajoutant le premier, il vient <lb/>pour la ſomme des extrêmes a + a + dx- d, ou 2a + dx-d, <lb/>que je multiplie par la moitié du nombre des termes {x/2} pour <lb/>former l’équation {2ax + dxx - dx/2} = s; </s> <s xml:id="echoid-s5600" xml:space="preserve">faiſant évanouir le di-<lb/>viſeur 2, il vient 2ax + dxx - dx = 2s, qui eſt l’équation <lb/>qu’il faut réſoudre pour avoir la ſolution du problême.</s> <s xml:id="echoid-s5601" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5602" xml:space="preserve">Pour réſoudre cette équation, je commence par dégager de <lb/>tout coefficient le terme qui contient la plus haute puiſſance <lb/>de l’inconnue, qui eſt xx, en diviſant chaque terme de l’équa-<lb/>tion par d; </s> <s xml:id="echoid-s5603" xml:space="preserve">ce qui me donne xx + {2ax/d} - {dx/d} = {2s/d}, ou xx + <lb/>{2ax/d} - x = {2s/d}, ou xx+x X √{2a/d} - 1\x{0020} = {2s/d}. </s> <s xml:id="echoid-s5604" xml:space="preserve">Pour faciliter encore <lb/>le calcul, je ſuppoſe que le coefficient du ſecond terme, qui eſt <lb/>{2a/d} - 1, eſt égal à une ſeule lettre c, & </s> <s xml:id="echoid-s5605" xml:space="preserve">au lieu de xx + x x <lb/>√{2a/d}-1\x{0020}, j’ai xx + cx = {2s/d}, & </s> <s xml:id="echoid-s5606" xml:space="preserve">c’eſt là la forme la plus ſimple <lb/>que puiſſe avoir une équation du ſecond degré à deux termes. <lb/></s> <s xml:id="echoid-s5607" xml:space="preserve">Préſentement pour rappeller cette équation à celles du premier <lb/>degré, il n’y a qu’à faire enſorte que le premier membre ſoit <lb/>un quarré parfait, dont on puiſſe extraire la racine; </s> <s xml:id="echoid-s5608" xml:space="preserve">& </s> <s xml:id="echoid-s5609" xml:space="preserve">voici <lb/>comment cela ſe pratique. </s> <s xml:id="echoid-s5610" xml:space="preserve">On ajoute à chaque membre de l’é-<lb/>quation le quarré de la moitié du coefficient de x au ſecond <lb/>terme: </s> <s xml:id="echoid-s5611" xml:space="preserve">ainſi je prends la moitié du coefficient de x, qui eſt <lb/>{c/2}, dont le quarré eſt {c c/4} que j’ajoute à chaque membre; </s> <s xml:id="echoid-s5612" xml:space="preserve">ce qui <lb/>me donne la nouvelle équation xx + cx + {c c/4} = {c c/4} + {2s/d}, <lb/>dans laquelle le premier membre eſt un quarré parfait, ſçavoir <lb/>celui de x + {1/2}c, puiſqu’il contient le quarré xx du premier <lb/>terme, le double produit cx, du premier par le ſecond, & </s> <s xml:id="echoid-s5613" xml:space="preserve">le <lb/>quarré du ſecond. </s> <s xml:id="echoid-s5614" xml:space="preserve">Ainſi extrayant les racines de part & </s> <s xml:id="echoid-s5615" xml:space="preserve">d’au-<lb/>tres, il vient x + {1/2} c = ±√{1/4} cc + {2s/d}\x{0020}, & </s> <s xml:id="echoid-s5616" xml:space="preserve">tranſpoſant {1/2} c, <pb o="161" file="0199" n="199" rhead="DE MATHÉMATIQUE. Liv. II."/> x = - {1/2} c ± √{c c/4} + {2s/d}\x{0020}. </s> <s xml:id="echoid-s5617" xml:space="preserve">Pour appliquer cette expreſſion ou <lb/>formule générale à notre problême, je fais a = 1, puiſque 1 <lb/>eſt le premier terme de la progreſſion arithmétique; </s> <s xml:id="echoid-s5618" xml:space="preserve">b = 3, <lb/>puiſque le ſecond jour il fait trois lieues; </s> <s xml:id="echoid-s5619" xml:space="preserve">b - a, ou d = 3 - 1 <lb/>= 2, qui eſt la différence du ſecond au premier terme, & </s> <s xml:id="echoid-s5620" xml:space="preserve"><lb/>s = 64, qui eſt la ſomme de tous les termes. </s> <s xml:id="echoid-s5621" xml:space="preserve">Je cherche par le <lb/>moyen de ces valeurs celle de c, que j’ai fait égal à {2a/d} - 1, que <lb/>je trouve être {2 x 1/2} - 1, ou {2/2} - 1, ou 1 - 1 = 0; </s> <s xml:id="echoid-s5622" xml:space="preserve">ainſi c eſt <lb/>zero, ou rien dans notre queſtion: </s> <s xml:id="echoid-s5623" xml:space="preserve">par conſéquent en l’effa-<lb/>çant partout où il ſe trouve dans l’expreſſion ou formule gé-<lb/>nérale x = - {1/2} c ± √{c c/4} + {2s/d}\x{0020}, elle ſe réduit à ceci, x = ± <lb/>√{2s/d}\x{0020} = ± √{2 x 64/2}\x{0020} = ± √64\x{0020} = ± 8; </s> <s xml:id="echoid-s5624" xml:space="preserve">c’eſt - à - dire que le <lb/>Soldat, dont il eſt queſtion, a été huit jours en chemin: </s> <s xml:id="echoid-s5625" xml:space="preserve">ce <lb/>qui m’apprend en même-tems que le nombre 64, qui eſt la <lb/>ſomme des termes de la progreſſion, eſt auſſi le quarré du nom-<lb/>bre des termes de la même progreſſion: </s> <s xml:id="echoid-s5626" xml:space="preserve">enſorte que les huit <lb/>premiers termes de la progreſſion des nombres impairs · 1. <lb/></s> <s xml:id="echoid-s5627" xml:space="preserve">3. </s> <s xml:id="echoid-s5628" xml:space="preserve">5. </s> <s xml:id="echoid-s5629" xml:space="preserve">7. </s> <s xml:id="echoid-s5630" xml:space="preserve">9. </s> <s xml:id="echoid-s5631" xml:space="preserve">11. </s> <s xml:id="echoid-s5632" xml:space="preserve">13. </s> <s xml:id="echoid-s5633" xml:space="preserve">15 font enſemble 64, & </s> <s xml:id="echoid-s5634" xml:space="preserve">c’eſt une propriété <lb/>commune à tant de termes que l’on voudra de cette progreſ-<lb/>ſion, pourvu que l’on prenne toujours depuis l’unité. </s> <s xml:id="echoid-s5635" xml:space="preserve">Cette <lb/>propriété mérite beaucoup d’attention, comme on le verra <lb/>par la ſuite dans le Traité du jet des bombes.</s> <s xml:id="echoid-s5636" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div298" type="section" level="1" n="271"> <head xml:id="echoid-head313" xml:space="preserve"><emph style="sc">Seconde question</emph>.</head> <p> <s xml:id="echoid-s5637" xml:space="preserve">310. </s> <s xml:id="echoid-s5638" xml:space="preserve">La ſomme de deux nombres eſt 6, la ſomme de leurs <lb/>quarrés eſt 20: </s> <s xml:id="echoid-s5639" xml:space="preserve">on demande chacun de ces deux nombres?</s> <s xml:id="echoid-s5640" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div299" type="section" level="1" n="272"> <head xml:id="echoid-head314" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s5641" xml:space="preserve">Soit x l’un de ces nombres, l’autre ſera 6 - x, puiſque leur <lb/>ſomme eſt 6. </s> <s xml:id="echoid-s5642" xml:space="preserve">Les quarrés de ces nombres ſont xx & </s> <s xml:id="echoid-s5643" xml:space="preserve">36 - 12x <lb/>+ xx, dont la ſomme doit être égale à 20, par la ſeconde <lb/>condition du problême, ce qui donne 2xx - 12x + 36 = 20. <lb/></s> <s xml:id="echoid-s5644" xml:space="preserve">Je fais paſſer d’abord 36 de l’autre côté, ce qui me donne 2xx <lb/>- 12x = 20 - 36, ou en diviſant chaque membre de l’équa-<lb/>tion par 2; </s> <s xml:id="echoid-s5645" xml:space="preserve">xx - 6x = 10 - 18 = - 8. </s> <s xml:id="echoid-s5646" xml:space="preserve">Selon la regle <lb/>générale, pour rendre le premier membre de cette équa- <pb o="162" file="0200" n="200" rhead="NOUVEAU COURS"/> tion un quarré parfait, j’ajoute de part & </s> <s xml:id="echoid-s5647" xml:space="preserve">d’autre le quarré 9 <lb/>de la moitié 3 de 6, coefficient de x au ſecond terme: </s> <s xml:id="echoid-s5648" xml:space="preserve">pour <lb/>avoir xx - 6x + 9 = 9 - 8 = 1, j’extrais les racines de part <lb/>& </s> <s xml:id="echoid-s5649" xml:space="preserve">d’autre, & </s> <s xml:id="echoid-s5650" xml:space="preserve">je trouve x - 3 = ± √1\x{0020} = ± 1, & </s> <s xml:id="echoid-s5651" xml:space="preserve">laiſſant <lb/>x tout ſeul dans un membre, il vient x = 3 ± 1 = 4 ou 2. </s> <s xml:id="echoid-s5652" xml:space="preserve">Si <lb/>je prends 4 pour x, le ſecond membre ſera 2; </s> <s xml:id="echoid-s5653" xml:space="preserve">ſi au contraire <lb/>je prends 2, le ſecond nombre ſera 4, puiſque ces deux nom-<lb/>bres donnent également 6 pour ſomme, & </s> <s xml:id="echoid-s5654" xml:space="preserve">20 pour la ſomme <lb/>de leurs quarrés, où l’on remarquera encore que la racine né-<lb/>gative réſout le problême dans le ſens qu’on s’étoit propoſé <lb/>auſſi-bien que la poſitive.</s> <s xml:id="echoid-s5655" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div300" type="section" level="1" n="273"> <head xml:id="echoid-head315" xml:space="preserve"><emph style="sc">Troisieme question</emph>.</head> <p> <s xml:id="echoid-s5656" xml:space="preserve">311. </s> <s xml:id="echoid-s5657" xml:space="preserve">On propoſe de trouver un nombre qui ſoit tel qu’en <lb/>lui ajoutant la racine quarrée de ſon produit par 10, la ſomme <lb/>ſoit 20.</s> <s xml:id="echoid-s5658" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5659" xml:space="preserve">Soit x le nombre cherché, & </s> <s xml:id="echoid-s5660" xml:space="preserve">ſuppoſons 10 = a, & </s> <s xml:id="echoid-s5661" xml:space="preserve">20 = 2a, <lb/>on aura par les conditions du problême x + √ax\x{0020} = 2a. </s> <s xml:id="echoid-s5662" xml:space="preserve">Je <lb/>laiſſe le radical ſeul dans un membre, & </s> <s xml:id="echoid-s5663" xml:space="preserve">j’ai √ax\x{0020} = 2a - x; <lb/></s> <s xml:id="echoid-s5664" xml:space="preserve">pour faire diſparoître le radical, j’éleve chaque membre au <lb/>quarré, ce qui me donne ax = 4a<emph style="sub">2</emph> - 4ax + xx, & </s> <s xml:id="echoid-s5665" xml:space="preserve">rédui-<lb/>ſant xx - 5ax = - 4a<emph style="sub">2</emph>. </s> <s xml:id="echoid-s5666" xml:space="preserve">Pour completter le quarré, j’ajoute <lb/>de part & </s> <s xml:id="echoid-s5667" xml:space="preserve">d’autre le quarré de la moitié du coefficient, qui eſt <lb/>{25/4}a<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s5668" xml:space="preserve">j’ai xx - 5ax + {25/4}a<emph style="sub">2</emph> = {25/4}a<emph style="sub">2</emph> - {16/4}a<emph style="sub">2</emph> = {9/4}a<emph style="sub">2</emph>, tirant <lb/>la racine de part & </s> <s xml:id="echoid-s5669" xml:space="preserve">d’autre, il vient x - {5/2}a = ± √{9/4}\x{0020}a<emph style="sub">2</emph>= <lb/>± {3/2}a, & </s> <s xml:id="echoid-s5670" xml:space="preserve">laiſſant x tout ſeul, il vient x = {5/2}a ± {3/2}a, ou x = 4a, <lb/>& </s> <s xml:id="echoid-s5671" xml:space="preserve">x = a, c’eſt-à-dire que l’un des nombres eſt 10 & </s> <s xml:id="echoid-s5672" xml:space="preserve">l’autre 40.</s> <s xml:id="echoid-s5673" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5674" xml:space="preserve">Il eſt évident que le nombre 10 eſt tel que la racine quarréc<unsure/> <lb/>de ſon produit par 10, qui eſt 100, & </s> <s xml:id="echoid-s5675" xml:space="preserve">dont la racine eſt 10, <lb/>fait effectivement 20: </s> <s xml:id="echoid-s5676" xml:space="preserve">mais on ne voit pas de même comment <lb/>la racine quarrée de 40 multiplié par 10, ſatisfait auſſi aux con-<lb/>ditions du problême. </s> <s xml:id="echoid-s5677" xml:space="preserve">Pour cela, je remarque que 400 peut <lb/>avoir à ſa racine - 20 ou + 20, puiſque - 20 X - 20 = 400, <lb/>& </s> <s xml:id="echoid-s5678" xml:space="preserve">que + 20 X + 20 = 400: </s> <s xml:id="echoid-s5679" xml:space="preserve">donc en ajoutant cette racine <lb/>de 400, qui eſt - 20 au nombre 40, j’ai 40 - 20 = 20.</s> <s xml:id="echoid-s5680" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div301" type="section" level="1" n="274"> <head xml:id="echoid-head316" xml:space="preserve"><emph style="sc">Quatrieme question</emph>.</head> <p> <s xml:id="echoid-s5681" xml:space="preserve">312. </s> <s xml:id="echoid-s5682" xml:space="preserve">On demande les trois termes d’une progreſſion géomé- <pb o="163" file="0201" n="201" rhead="DE MATHÉMATIQUE. Liv. II."/> trique, dont le premier terme eſt 4, & </s> <s xml:id="echoid-s5683" xml:space="preserve">dont la différence du <lb/>ſecond au troiſieme ſoit 3.</s> <s xml:id="echoid-s5684" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div302" type="section" level="1" n="275"> <head xml:id="echoid-head317" xml:space="preserve"><emph style="sc">Solution.</emph></head> <p> <s xml:id="echoid-s5685" xml:space="preserve">Soit x le ſecond terme, le troiſieme ſera x + 3 par une des <lb/>conditions du problême, & </s> <s xml:id="echoid-s5686" xml:space="preserve">par l’autre on aura 4. </s> <s xml:id="echoid-s5687" xml:space="preserve">x : </s> <s xml:id="echoid-s5688" xml:space="preserve">: </s> <s xml:id="echoid-s5689" xml:space="preserve">x. </s> <s xml:id="echoid-s5690" xml:space="preserve">x+3, <lb/>d’où l’on tire xx = 4x + 12, ou xx - 4x = 12; </s> <s xml:id="echoid-s5691" xml:space="preserve">j’ajoute à <lb/>chaque membre le quarré de la moitié du coefficient, qui eſt <lb/>4, & </s> <s xml:id="echoid-s5692" xml:space="preserve">j’ai xx - 4x + 4 = 16, d’où l’on déduit en prenant les <lb/>racines de chaque membre, x - 2 = ± 4, c’eſt-à-dire que <lb/>l’une des valeurs de x eſt 6, & </s> <s xml:id="echoid-s5693" xml:space="preserve">l’autre eſt 2 - 4 ou - 2, & </s> <s xml:id="echoid-s5694" xml:space="preserve">ces <lb/>valeurs ſont telles, qu’il n’y en a réellement qu’une qui ré-<lb/>ſolve le problême dans le ſens qu’on s’étoit propoſé, en don-<lb/>nant cette progreſſion 4. </s> <s xml:id="echoid-s5695" xml:space="preserve">6 : </s> <s xml:id="echoid-s5696" xml:space="preserve">: </s> <s xml:id="echoid-s5697" xml:space="preserve">6. </s> <s xml:id="echoid-s5698" xml:space="preserve">9; </s> <s xml:id="echoid-s5699" xml:space="preserve">mais on peut dire auſſi que <lb/>l’autre ne réſout pas moins le problême que la premiere, en <lb/>donnant cette autre progreſſion géométrique, 4. </s> <s xml:id="echoid-s5700" xml:space="preserve">- 2 :</s> <s xml:id="echoid-s5701" xml:space="preserve">: - 2. </s> <s xml:id="echoid-s5702" xml:space="preserve">1; <lb/></s> <s xml:id="echoid-s5703" xml:space="preserve">car il eſt évident que ces trois grandeurs ſont en progreſſion <lb/>géométrique, puiſque le produit des extrêmes eſt égal au <lb/>quarré du moyen, & </s> <s xml:id="echoid-s5704" xml:space="preserve">que ſelon la ſeconde condition, la diffé-<lb/>rence du ſecond terme au 3<emph style="sub">e</emph> eſt 3: </s> <s xml:id="echoid-s5705" xml:space="preserve">car il eſt évident que la diffé-<lb/>rence de - 2 à 1 eſt 3, comme on peut voir en ôtant - 2 de 1.</s> <s xml:id="echoid-s5706" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div303" type="section" level="1" n="276"> <head xml:id="echoid-head318" xml:space="preserve"><emph style="sc">Cinquieme question</emph>.</head> <p> <s xml:id="echoid-s5707" xml:space="preserve">313. </s> <s xml:id="echoid-s5708" xml:space="preserve">Deux Commerçans ont placé dans le commerce unc <lb/>ſomme de 1300 liv. </s> <s xml:id="echoid-s5709" xml:space="preserve">ſur laquelle ils gagnent 900; </s> <s xml:id="echoid-s5710" xml:space="preserve">le premier, <lb/>tant pour ſa miſe que pour l’intérêt de ſon argent, qui a été <lb/>trois mois dans le commerce, a retiré 870<emph style="sub">1.</emph>; </s> <s xml:id="echoid-s5711" xml:space="preserve">& </s> <s xml:id="echoid-s5712" xml:space="preserve">le ſecond pa-<lb/>reillement, tant pour ſa miſe que pour l’intérêt de ſon argent, <lb/>qui a été ſix mois dans le commerce, reçoit 1330 livres: </s> <s xml:id="echoid-s5713" xml:space="preserve">on <lb/>demande la miſe de chacun en particulier.</s> <s xml:id="echoid-s5714" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div304" type="section" level="1" n="277"> <head xml:id="echoid-head319" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s5715" xml:space="preserve">Soit x la miſe du premier, celle du ſecond ſera 1300 - x, <lb/>puiſqu’ils ont mis à eux deux 1300 dans le commerce. </s> <s xml:id="echoid-s5716" xml:space="preserve">Le gain <lb/>du premier ſera 870 - x, & </s> <s xml:id="echoid-s5717" xml:space="preserve">celui du ſecond ſera 1330 -<lb/>1300 + x, ou en réduiſant 30 + x: </s> <s xml:id="echoid-s5718" xml:space="preserve">car il eſt clair que pour <lb/>avoir le gain que fait l’un & </s> <s xml:id="echoid-s5719" xml:space="preserve">l’autre, il faut ôter ſa miſe du <lb/>nombre qui contient par hypotheſe la miſe & </s> <s xml:id="echoid-s5720" xml:space="preserve">le gain de cha-<lb/>cun. </s> <s xml:id="echoid-s5721" xml:space="preserve">Or par les conditions du problême, la miſe & </s> <s xml:id="echoid-s5722" xml:space="preserve">le gain du <lb/>premier ſont renfermés dans ſa part 870, & </s> <s xml:id="echoid-s5723" xml:space="preserve">de même la miſe & </s> <s xml:id="echoid-s5724" xml:space="preserve"><lb/>le gain du ſecond ſont contenus dans ſa part, qui eſt 1330.</s> <s xml:id="echoid-s5725" xml:space="preserve"/> </p> <pb o="164" file="0202" n="202" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s5726" xml:space="preserve">On ſçait de plus que les gains ſont dans la raiſon compoſée <lb/>des miſes & </s> <s xml:id="echoid-s5727" xml:space="preserve">des tems, c’eſt - à - dire comme les produits des <lb/>miſes par les tems: </s> <s xml:id="echoid-s5728" xml:space="preserve">car il eſt évident que ſi un homme a placé <lb/>dans le commerce trois fois plus qu’un autre dans le même <lb/>tems, il doit gagner trois fois davantage, & </s> <s xml:id="echoid-s5729" xml:space="preserve">s’il a mis ſon ar-<lb/>gent pendant un tems quadruple, il doit encore par-là gagner <lb/>quatre fois plus que l’autre, c’eſt-à-dire que ſon gain ſera 4 fois <lb/>3 fois plus grand que celui du ſecond, ou qu’il ſera à celui du <lb/>ſecond, comme 12 à 1, qui ſont les produits des miſes par <lb/>les tems; </s> <s xml:id="echoid-s5730" xml:space="preserve">multipliant donc la miſe du premier, qui eſt x, par <lb/>ſon tems 3, & </s> <s xml:id="echoid-s5731" xml:space="preserve">celle du ſecond par ſon tems 6; </s> <s xml:id="echoid-s5732" xml:space="preserve">puis faiſant une <lb/>proportion avec les produits & </s> <s xml:id="echoid-s5733" xml:space="preserve">les gains particuliers, on aura <lb/>3x. </s> <s xml:id="echoid-s5734" xml:space="preserve">√1300 - x\x{0020} x 6 : </s> <s xml:id="echoid-s5735" xml:space="preserve">: </s> <s xml:id="echoid-s5736" xml:space="preserve">870 - x. </s> <s xml:id="echoid-s5737" xml:space="preserve">30 + x, & </s> <s xml:id="echoid-s5738" xml:space="preserve">diviſant chaque <lb/>terme de la premiere raiſon par 3, x. </s> <s xml:id="echoid-s5739" xml:space="preserve">√1300 - x\x{0020} x 2 :</s> <s xml:id="echoid-s5740" xml:space="preserve">: 870 - x. <lb/></s> <s xml:id="echoid-s5741" xml:space="preserve">30 + x: </s> <s xml:id="echoid-s5742" xml:space="preserve">prenant enſuite le produit des extrêmes & </s> <s xml:id="echoid-s5743" xml:space="preserve">des <lb/>moyens, on aura cette égalité 30x + xx = 2262000 -<lb/>4340x + 2xx, qui renferme toutes les conditions du problême. </s> <s xml:id="echoid-s5744" xml:space="preserve"><lb/>Otant xx de chaque membre, & </s> <s xml:id="echoid-s5745" xml:space="preserve">faiſant paſſer 30x de l’autre <lb/>côté, & </s> <s xml:id="echoid-s5746" xml:space="preserve">2262000 dans le premier membre, il vient - 2262000 <lb/>= xx - 4370x, ou xx - 4370x = - 2262000. </s> <s xml:id="echoid-s5747" xml:space="preserve">Ajoutant <lb/>à chaque membre le quarré de 2185, moitié du coefficient, <lb/>pour compléter le quarré, on aura xx - 4370x + 4774225 <lb/>= 4774225 - 2262000 = 2512225; </s> <s xml:id="echoid-s5748" xml:space="preserve">& </s> <s xml:id="echoid-s5749" xml:space="preserve">tirant enſuite la ra-<unsure/> <lb/>cine de chaque membre, il vient x - 2185 = ± √2512225\x{0020} <lb/>= ± 1585, ou enfin x = 2185 ± 1585, qui donne pour une <lb/>des valeurs de x, 3770, & </s> <s xml:id="echoid-s5750" xml:space="preserve">pour l’autre 600 livres, que l’on <lb/>regarde comme celle qui réſout le problême dans le ſens que <lb/>I’on s’étoit propoſé, comme il eſt aiſé de le voir, en détermi-<lb/>nant la part de gain total pour 600, par une Regle de Trois, <lb/>dont le premier terme ſera la ſomme des miſes, multipliées <lb/>par leurs tems, le ſecond terme le gain total, le troiſieme la <lb/>miſe 600 livres du premier, multipliée par ſon tems, & </s> <s xml:id="echoid-s5751" xml:space="preserve">le qua-<lb/>trieme le gain du même premier.</s> <s xml:id="echoid-s5752" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div305" type="section" level="1" n="278"> <head xml:id="echoid-head320" style="it" xml:space="preserve">Remarque générale & importante ſur la ſolution de ce <lb/>Problême.</head> <p> <s xml:id="echoid-s5753" xml:space="preserve">314. </s> <s xml:id="echoid-s5754" xml:space="preserve">On remarquera 1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s5755" xml:space="preserve">que la valeur de l’inconnue qui <lb/>ſatisfait aux conditions du problême, eſt celle qui eſt déter-<lb/>minée par la racine négative du quarré, qui étoit ſous le ſigne <pb o="165" file="0203" n="203" rhead="DE MATHEMATIQUE. Liv. II."/> radical; </s> <s xml:id="echoid-s5756" xml:space="preserve">d’où il ſuit que l’on ne doit pas établir pour regle gé-<lb/>nérale que les quantités déterminées par les racines négatives <lb/>ſont étrangeres à la queſtion, puiſque dans ce cas la négative <lb/>donne la ſolution du problême dans le ſens qu’on s’étoit pro-<lb/>poſé. </s> <s xml:id="echoid-s5757" xml:space="preserve">Pour voir préſentement ce que ſigniſie l’autre racine <lb/>3770, je fais attention que puiſque la ſomme des miſes eſt <lb/>égale à 1300, en ôtant l’une de ce nombre, je dois avoir l’au-<lb/>tre. </s> <s xml:id="echoid-s5758" xml:space="preserve">J’ôte donc 3770 de 1300, & </s> <s xml:id="echoid-s5759" xml:space="preserve">quoique cela ne ſoit pas poſ-<lb/>ſible dans un ſens, cependant de l’autre il eſt vrai de dire qu’en <lb/>ôtant 3770 de 1300, le reſte eſt - 2470, puiſqu’en ajoutant <lb/>ce reſte à la quantité retranchée, il vient 1300, ce qui m’ap-<lb/>prend d’abord que l’un des Commerçans, au lieu d’avoir mis <lb/>dans le commerce, en a réellement ôté 2470 livres; </s> <s xml:id="echoid-s5760" xml:space="preserve">je multi-<lb/>plie enſuite les miſes quelles qu’elles ſoient par leurs temps, <lb/>multipliant 3770 par 3, il vient 11310, & </s> <s xml:id="echoid-s5761" xml:space="preserve">multipliant de <lb/>même la miſe du ſecond - 2470 par ſon tems 6, il vient au <lb/>produit - 14820; </s> <s xml:id="echoid-s5762" xml:space="preserve">la ſomme de ces deux produits, qui eſt cenſée <lb/>la cauſe du gain total eſt - 3510. </s> <s xml:id="echoid-s5763" xml:space="preserve">Je fais après cela une Regle <lb/>de Trois, dont le premier terme ſoit - 3510, le ſecond, la <lb/>miſe du premier multipliée par ſon tems 3, le troiſieme, le gain <lb/>total, que l’on ſuppoſe de 900, & </s> <s xml:id="echoid-s5764" xml:space="preserve">appellant x le quatrieme <lb/>terme, qui ſera le gain du premier, j’ai cette proportion <lb/>- 3510. </s> <s xml:id="echoid-s5765" xml:space="preserve">11310 :</s> <s xml:id="echoid-s5766" xml:space="preserve">: 900. </s> <s xml:id="echoid-s5767" xml:space="preserve">x = {11310 x 900/- 3510} = - 2900, dont le <lb/>quatrieme terme fait voir que le premier, au lieu d’avoir ga-<lb/>gné a réellement perdu 2900, & </s> <s xml:id="echoid-s5768" xml:space="preserve">cette perte eſt telle que la <lb/>ſomme de la perte - 2900, & </s> <s xml:id="echoid-s5769" xml:space="preserve">de la miſe 3770 fait préciſé-<lb/>ment 870. </s> <s xml:id="echoid-s5770" xml:space="preserve">Puiſque le premier perd, il faut néceſſairement <lb/>que le ſecond qui a ôté ſon argent du commerce gagne, puiſ-<lb/>qu’il manque de perdre, & </s> <s xml:id="echoid-s5771" xml:space="preserve">cela d’autant plus qu’il a ôté plus <lb/>d’argent, & </s> <s xml:id="echoid-s5772" xml:space="preserve">qu’il y a plus de tems qu’il a ôté ſon argent, c’eſt-<lb/>à-dire que le gain qu’il fait eſt dans la raiſon compoſée de l’ar-<lb/>gent qu’il a ôté du commerce, multiplié par le tems, ou comme <lb/>le produit de cet argent par le tems qui s’eſt paſſé depuis qu’ll <lb/>l’a retiré. </s> <s xml:id="echoid-s5773" xml:space="preserve">Je fais encore une proportion pour déterminer ſon <lb/>gain, dont le premier terme ſoit la ſomme des produits des <lb/>miſes par leurs tems, le ſecond le produit de la miſe de ce <lb/>Commerçant par ſon tems; </s> <s xml:id="echoid-s5774" xml:space="preserve">le troiſieme le gain total, & </s> <s xml:id="echoid-s5775" xml:space="preserve">le <lb/>quatrieme le gain de ce Commercant, ce qui me donne -<lb/>3510. </s> <s xml:id="echoid-s5776" xml:space="preserve">- 14820 :</s> <s xml:id="echoid-s5777" xml:space="preserve">: 900. </s> <s xml:id="echoid-s5778" xml:space="preserve">x = {- 14820 x 900/- 3510}, ou = {- 13338000/- 3510} = <pb o="166" file="0204" n="204" rhead="NOUVEAU COURS"/> + 3800 livres, puiſque - diviſé par - doit donner +; </s> <s xml:id="echoid-s5779" xml:space="preserve">& </s> <s xml:id="echoid-s5780" xml:space="preserve"><lb/>ce gain eſt encore tel qu’en l’ajoutant avec la miſe négative <lb/>- 2470, il vient pour la ſomme 1330, qui eſt le nombre ex-<lb/>primé par les conditions du problême.</s> <s xml:id="echoid-s5781" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5782" xml:space="preserve">On voit par-là que quoique les valeurs algébriques paroiſ-<lb/>ſent quelquefois ne rien ſignifier, parce qu’elles ſont extrê-<lb/>mement éloignées de ce que nous aurions imaginé, elles n’en <lb/>ſont pas pour cela moins vraies ni moins bien raiſonnées; </s> <s xml:id="echoid-s5783" xml:space="preserve">& </s> <s xml:id="echoid-s5784" xml:space="preserve"><lb/>quoique l’on ne doive pas s’appliquer dans tous les cas à les re-<lb/>connoître, parce que cela deviendroit inutile, il eſt auſſi ridi-<lb/>cule de ne les pas rechercher dans quelques-uns, pour s’accoutu-<lb/>mer aux expreſſions algébriques, & </s> <s xml:id="echoid-s5785" xml:space="preserve">pour être en état d’inter-<lb/>prêter au beſoin les oracles que nous donne l’analyſe.</s> <s xml:id="echoid-s5786" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5787" xml:space="preserve">315. </s> <s xml:id="echoid-s5788" xml:space="preserve">Ces exemples ſuffiſent pour connoître l’uſage que l’on <lb/>doit faire des racines négatives. </s> <s xml:id="echoid-s5789" xml:space="preserve">Nous allons préſentement ré-<lb/>ſoudre en peu de mots les équations du ſecond degré dans leurs <lb/>formules générales, parce que la méthode eſt toujours la même. <lb/></s> <s xml:id="echoid-s5790" xml:space="preserve">Si l’on a une équation du ſecond degré, comme celle-ci, xx <lb/>- 4x = 12, on fait paſſer ordinairement le terme 12 de l’au-<lb/>tre côté du ſigne d’égalité, & </s> <s xml:id="echoid-s5791" xml:space="preserve">alors on dit que l’équation eſt <lb/>égale à zero, & </s> <s xml:id="echoid-s5792" xml:space="preserve">elle ſe marque ainſi : </s> <s xml:id="echoid-s5793" xml:space="preserve">x x - 4x + 12 = 0. </s> <s xml:id="echoid-s5794" xml:space="preserve"><lb/>Cela poſé, toute équation du ſecond degré peut ſe rappeller <lb/>à l’une des ſix formules ſuivantes.</s> <s xml:id="echoid-s5795" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">xx # + # px # + # q # = # 0 <lb/>xx # - # px # - # q # = # 0 <lb/>xx # - # px # + # q # = # 0 <lb/>xx # + # px # - # q # = # 0 <lb/># # xx # - # q # = # 0 <lb/># # xx # + # q # = # 0 <lb/></note> <p> <s xml:id="echoid-s5796" xml:space="preserve">316. </s> <s xml:id="echoid-s5797" xml:space="preserve">Ces équations ſe réſolvent comme les précédentes. </s> <s xml:id="echoid-s5798" xml:space="preserve">Le <lb/>terme q repréſente toutes les quantités connues: </s> <s xml:id="echoid-s5799" xml:space="preserve">la lettre p dé-<lb/>ſigne tous les coefficients qui multiplient l’inconnue au ſecond <lb/>terme. </s> <s xml:id="echoid-s5800" xml:space="preserve">On tranſporte après cela le terme q dans l’autre <lb/>membre, & </s> <s xml:id="echoid-s5801" xml:space="preserve">l’on ajoute à chacun, le quarré de la moitié du <lb/>coefficient p, & </s> <s xml:id="echoid-s5802" xml:space="preserve">l’on prend la racine du premier membre, qui <lb/>devient un quarré parfait, & </s> <s xml:id="echoid-s5803" xml:space="preserve">l’on met les quantités qui ſont <lb/>dans l’autre membre ſous le ſigne radical, pour marquer que <lb/>l’on en prend la racine; </s> <s xml:id="echoid-s5804" xml:space="preserve">ce qui donne les ſix formules ſuivantes <lb/>correſpondantes aux équations précédentes.</s> <s xml:id="echoid-s5805" xml:space="preserve"/> </p> <pb o="167" file="0205" n="205" rhead="DE MATHÉMATIQUE. Liv. II."/> <p> <s xml:id="echoid-s5806" xml:space="preserve">Premiere x = - {1/2}p ± √{1/4}pp - q}\x{0020}</s> </p> <p> <s xml:id="echoid-s5807" xml:space="preserve">Seconde x = {1/2}p ± √{1/4}pp + q}\x{0020}</s> </p> <p> <s xml:id="echoid-s5808" xml:space="preserve">Troiſieme x = {1/2}p ± √{1/4}pp - q\x{0020}</s> </p> <p> <s xml:id="echoid-s5809" xml:space="preserve">Quatrieme x = - {1/2}p ± √{1/4}pp + q\x{0020}</s> </p> <p> <s xml:id="echoid-s5810" xml:space="preserve">Cinquieme x = ± √q\x{0020}</s> </p> <p> <s xml:id="echoid-s5811" xml:space="preserve">Sixieme x = ± √-q\x{0020}.</s> <s xml:id="echoid-s5812" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5813" xml:space="preserve">Voici ce que l’on peut remarquer ſur ces formules. </s> <s xml:id="echoid-s5814" xml:space="preserve">Dans la <lb/>premiere & </s> <s xml:id="echoid-s5815" xml:space="preserve">la troiſieme, le problême ſera toujours poſſible, <lb/>tant que {1/4}pp ſera plus grand que q, ou au moins égal; </s> <s xml:id="echoid-s5816" xml:space="preserve">mais <lb/>s’il étoit moindre, le problême ſeroit impoſſible, puiſque dans <lb/>ce cas √{1/4}pp - q\x{0020} ſeroit une quantité imaginaire. </s> <s xml:id="echoid-s5817" xml:space="preserve">On appelle <lb/>imaginaire une quantité négative, ſoumiſe à un radical, parce <lb/>qu’il n’y a point de quantité qui donne - au quarré. </s> <s xml:id="echoid-s5818" xml:space="preserve">Tous <lb/>les problêmes qui ſe rapportent à la ſeconde & </s> <s xml:id="echoid-s5819" xml:space="preserve">à la troiſieme <lb/>formule, ſeront toujours poſſibles, puiſque jamais la quantité <lb/>√{1/4}pp + q\x{0020} ne pourra être imaginaire.</s> <s xml:id="echoid-s5820" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5821" xml:space="preserve">Enfin la cinquieme formule aura toujours deux valeurs <lb/>égales, l’une poſitive, qui eſt + √q\x{0020}, & </s> <s xml:id="echoid-s5822" xml:space="preserve">l’autre négative, qui <lb/>eſt - √q\x{0020}; </s> <s xml:id="echoid-s5823" xml:space="preserve">& </s> <s xml:id="echoid-s5824" xml:space="preserve">la ſixieme renfermera toujours quelque abſur-<lb/>dité, puiſque ± √-q\x{0020} ſera toujours une quantité imagi-<lb/>naire.</s> <s xml:id="echoid-s5825" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5826" xml:space="preserve">317. </s> <s xml:id="echoid-s5827" xml:space="preserve">Il y a certaines équations du quatrieme degré qui ſe <lb/>réſolvent de même que celles du ſecond, comme on va voit <lb/>dans l’exemple ſuivant.</s> <s xml:id="echoid-s5828" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div306" type="section" level="1" n="279"> <head xml:id="echoid-head321" xml:space="preserve"><emph style="sc">Sixieme question</emph>.</head> <p> <s xml:id="echoid-s5829" xml:space="preserve">On demande deux nombres, dont le produit ſoit 12, & </s> <s xml:id="echoid-s5830" xml:space="preserve">la <lb/>différence des quarrés 7.</s> <s xml:id="echoid-s5831" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div307" type="section" level="1" n="280"> <head xml:id="echoid-head322" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s5832" xml:space="preserve">Soient x & </s> <s xml:id="echoid-s5833" xml:space="preserve">y ces deux nombres, la premiere condition du <lb/>problême donne xy = 12, d’où l’on tire y = {12/x}, & </s> <s xml:id="echoid-s5834" xml:space="preserve">la ſe-<lb/>conde donne xx - yy = 7; </s> <s xml:id="echoid-s5835" xml:space="preserve">& </s> <s xml:id="echoid-s5836" xml:space="preserve">ſubſtituant à la place de yy ſa <lb/>valeur {144/xx}, on aura xx - {144/xx} = 7, multipliant par xx pour <lb/>faire évanouir la fraction {144/xx}, il vient x<emph style="sub">4</emph> - 144 = 7xx, ou <pb o="168" file="0206" n="206" rhead="NOUVEAU COURS"/> x<emph style="sub">4</emph> - 7xx = 144. </s> <s xml:id="echoid-s5837" xml:space="preserve">J’ajoute à chaque membre le quarré de la <lb/>moitié du coefficient de x, qui eſt celui de 3 {1/2}, il vient x<emph style="sub">4</emph> <lb/>- 7xx + 12 {1/4} = 12 + {1/4} + 144, dont le premier membre <lb/>eſt un quarré parfait, & </s> <s xml:id="echoid-s5838" xml:space="preserve">tirant les racines de part & </s> <s xml:id="echoid-s5839" xml:space="preserve">d’autre, <lb/>après avoir réduit le ſecond membre, on aura xx - 3 {1/2} = ± <lb/>√156 {1/4}\x{0020}; </s> <s xml:id="echoid-s5840" xml:space="preserve">la racine de 156 {1/4} eſt 12 {1/2}: </s> <s xml:id="echoid-s5841" xml:space="preserve">ainſi xx - 3 {1/2} = ± 12 {1/2}. <lb/></s> <s xml:id="echoid-s5842" xml:space="preserve">Dégageant xx, on a xx = ± 12 {1/2} + 3 {1/2} = 16 ou - 9, & </s> <s xml:id="echoid-s5843" xml:space="preserve"><lb/>tirant encore les racines pour avoir x au premier degré, on <lb/>aura x = ±√16\x{0020}, & </s> <s xml:id="echoid-s5844" xml:space="preserve">x = ± √- 9\x{0020}, dont les deux premieres <lb/>ſont ±4, & </s> <s xml:id="echoid-s5845" xml:space="preserve">les deux autres ſont imaginaires, c’eſt-à-dire que <lb/>l’une des valeurs de x eſt 4. </s> <s xml:id="echoid-s5846" xml:space="preserve">Je diviſe 12 par 4 pour avoir y = <lb/>{12/x}, & </s> <s xml:id="echoid-s5847" xml:space="preserve">le quotient eſt 3: </s> <s xml:id="echoid-s5848" xml:space="preserve">donc les nombres demandés ſont 3 <lb/>& </s> <s xml:id="echoid-s5849" xml:space="preserve">4, puiſque leur produit eſt 12, & </s> <s xml:id="echoid-s5850" xml:space="preserve">que la différence de leurs <lb/>quarrés 16 & </s> <s xml:id="echoid-s5851" xml:space="preserve">9 eſt 7. </s> <s xml:id="echoid-s5852" xml:space="preserve">On auroit pu réſoudre ce problême, <lb/>en ſe ſervant de la ſeconde formule, & </s> <s xml:id="echoid-s5853" xml:space="preserve">faiſant - 7 = - p, <lb/>& </s> <s xml:id="echoid-s5854" xml:space="preserve">- 144 = - q; </s> <s xml:id="echoid-s5855" xml:space="preserve">ce qui auroit donné la même ſolution.</s> <s xml:id="echoid-s5856" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s5857" xml:space="preserve">Du calcul des radicaux, des opérations qui leur ſont particu-<lb/># lieres, & </s> <s xml:id="echoid-s5858" xml:space="preserve">de la maniere de les réduire, de les ajouter, ſouſtraire, <lb/># multiplier ou diviſer.</s> <s xml:id="echoid-s5859" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5860" xml:space="preserve">318. </s> <s xml:id="echoid-s5861" xml:space="preserve">On appelle radicale une quantité, dont on ne peut pas <lb/>extraire la racine exactement. </s> <s xml:id="echoid-s5862" xml:space="preserve">Pour peu que l’on veuille ré-<lb/>ſoudre quelques problêmes du ſecond degré, on trouve né-<lb/>ceſſairement de ces ſortes d’expreſſions, que l’on appelle radicales <lb/>ou incommenſurables; </s> <s xml:id="echoid-s5863" xml:space="preserve">mais quoiqu’elles ne puiſſent pas avoir <lb/>de racines exactes, il y a cependant bien des cas où on peut <lb/>ſimplifier leurs expreſſions, d’autres dans leſquels on eſt obligé <lb/>d’opérer ſur ces grandeurs par Addition, Multiplication ou <lb/>Diviſion, ce qui arrive principalement dans les équations du <lb/>quatrieme degré réductibles au ſecond; </s> <s xml:id="echoid-s5864" xml:space="preserve">c’eſt pourquoi il eſt à <lb/>propos d’enſeigner de quelle maniere on doit pratiquer toutes <lb/>ces opérations, & </s> <s xml:id="echoid-s5865" xml:space="preserve">c’eſt en cela que conſiſte le calcul des radi-<lb/>caux ou incommenſurables que nous allons expliquer en peu <lb/>de mots. </s> <s xml:id="echoid-s5866" xml:space="preserve">Il y a autant de radicaux qu’il y a de puiſſances diffé-<lb/>rentes; </s> <s xml:id="echoid-s5867" xml:space="preserve">mais pour ne point entrer dans un trop grand détail, <lb/>nous ne parlerons que des radicaux du ſecond degré, auxquels <lb/>on ajoutera quelques exemples de radicaux du troiſieme. </s> <s xml:id="echoid-s5868" xml:space="preserve">Les <pb o="169" file="0207" n="207" rhead="DE MATHÉMATIQUE. Liv. II."/> regles étant générales, on pourra de ſoi-même les appliquer à <lb/>des radicaux plus compliqués.</s> <s xml:id="echoid-s5869" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div308" type="section" level="1" n="281"> <head xml:id="echoid-head323" style="it" xml:space="preserve">Réduire les quantités irrationnelles ou incommenſurables à leur <lb/>plus ſimple expreſſion.</head> <p> <s xml:id="echoid-s5870" xml:space="preserve">319. </s> <s xml:id="echoid-s5871" xml:space="preserve">On examinera ſi la quantité ſoumiſe au radical n’a pas <lb/>parmi ſes facteurs quelque puiſſance de même nom que le ra-<lb/>dical, ſoit que cette puiſſance ſoit une quantité complexe, ſoit <lb/>qu’elle ne ſoit qu’un monome: </s> <s xml:id="echoid-s5872" xml:space="preserve">pour reconnoître ſes facteurs, <lb/>il faut ſçavoir décompoſer une quantité, c’eſt-à-dire trouver <lb/>les autres quantités, de la multiplication deſquelles réſulte <lb/>la grandeur donnée. </s> <s xml:id="echoid-s5873" xml:space="preserve">Cela poſé, lorſqu’on aura trouvé un ou <lb/>pluſieurs facteurs de même puiſſance que la racine, on en ex-<lb/>traira la racine, & </s> <s xml:id="echoid-s5874" xml:space="preserve">l’on mettra le reſte ſous le radical.</s> <s xml:id="echoid-s5875" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5876" xml:space="preserve">Par exemple, √a<emph style="sub">3</emph>b\x{0020} = a√ab\x{0020}: </s> <s xml:id="echoid-s5877" xml:space="preserve">car il eſt évident que a<emph style="sub">3</emph>b = a<emph style="sub">2</emph> <lb/>x ab: </s> <s xml:id="echoid-s5878" xml:space="preserve">donc en prenant la racine du quarré complet a<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s5879" xml:space="preserve">laiſ-<lb/>ſant le reſte ſous le radical, on aura a√ab\x{0020}; </s> <s xml:id="echoid-s5880" xml:space="preserve">tout de même <lb/>√16a<emph style="sub">2</emph>b - 32a<emph style="sub">3</emph>\x{0020} = √16a<emph style="sub">2</emph> x √b - 2a.</s> <s xml:id="echoid-s5881" xml:space="preserve">\x{0020}\x{0020} Or il eſt viſible que <lb/>16a<emph style="sub">2</emph> eſt un quarré parfait, celui de 4a: </s> <s xml:id="echoid-s5882" xml:space="preserve">donc on extraira cette <lb/>racine, & </s> <s xml:id="echoid-s5883" xml:space="preserve">l’on aura pour la plus ſimple expreſſion de ce radical <lb/>4a√b - 2a\x{0020}. </s> <s xml:id="echoid-s5884" xml:space="preserve">Si l’on avoit <emph style="sub">3</emph>√a<emph style="sub">3</emph>c<emph style="sub">2</emph> - a<emph style="sub">@</emph>bd\x{0020}, on voit que a<emph style="sub">3</emph>, qui <lb/>eſt commun aux deux termes, eſt un cube parfait, dont on <lb/>peut prendre la racine cubique; </s> <s xml:id="echoid-s5885" xml:space="preserve">ainſi l’on écrira a <emph style="sub">3</emph>√c<emph style="sub">2</emph> - bd\x{0020}. <lb/></s> <s xml:id="echoid-s5886" xml:space="preserve">De même ſi l’on avoit √50ffgg - 25ffmm + 75bdff\x{0020}, il <lb/>eſt aiſé d’appercevoir qu’il y a dans cette quantité un quarré <lb/>parfait, commun à tous les termes, que l’on peut mettre hors <lb/>du radical, c’eſt 25ff; </s> <s xml:id="echoid-s5887" xml:space="preserve">car on auroit pu écrire cette quantité <lb/>comme il ſuit, √25ff x √2gg - mm + 3bd\x{0020}\x{0020}, & </s> <s xml:id="echoid-s5888" xml:space="preserve">prenant la ra-<lb/>cine, on auroit eu 5f√2gg - mm + 3bd\x{0020}. </s> <s xml:id="echoid-s5889" xml:space="preserve">Il en ſeroit de même <lb/>des autres quantités. </s> <s xml:id="echoid-s5890" xml:space="preserve">Par exemple, √3a<emph style="sub">2</emph>b<emph style="sub">2</emph>fg + 6a<emph style="sub">2</emph>bcfg + 3a<emph style="sub">2</emph>c<emph style="sub">2</emph>fg\x{0020} <lb/>auroit pu s’écrire ainſi: </s> <s xml:id="echoid-s5891" xml:space="preserve">√a<emph style="sub">2</emph> x √b<emph style="sub">2</emph> + 2bc + c<emph style="sub">2</emph>\x{0020} x 3fg\x{0020}, & </s> <s xml:id="echoid-s5892" xml:space="preserve">pre-<lb/>nant la racine des deux facteurs, qui ſont des quarrés parfaits, <lb/>on aura a x √b + c\x{0020} x √3fg\x{0020}. </s> <s xml:id="echoid-s5893" xml:space="preserve">Si l’on avoit à réduire cette autre ex-<lb/>preſſion √27a<emph style="sub">2</emph>b<emph style="sub">2</emph> - 36a<emph style="sub">2</emph>fg + 9a<emph style="sub">3</emph>c\x{0020}, je remarque que cette <lb/>quantité eſt le produit de 9a<emph style="sub">2</emph> par 3b<emph style="sub">2</emph> - 4fg + ac: </s> <s xml:id="echoid-s5894" xml:space="preserve">ainſi j’écrirai <pb o="170" file="0208" n="208" rhead="NOUVEAU COURS"/> en prenant la racine 3a√3b<emph style="sub">2</emph> - 4fg + ac\x{0020}; </s> <s xml:id="echoid-s5895" xml:space="preserve">ſi l’on avoit <lb/>√64m<emph style="sub">2</emph>g<emph style="sub">2</emph> - 36ffgg + 48abgg\x{0020}, on auroit en ſimplifiant ce <lb/>radical, 2g√16mm - 9ff + 12ab\x{0020}, & </s> <s xml:id="echoid-s5896" xml:space="preserve">ainſi de tous les autres.</s> <s xml:id="echoid-s5897" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5898" xml:space="preserve">320. </s> <s xml:id="echoid-s5899" xml:space="preserve">Il eſt quelquefois à propos de compliquer un radical, <lb/>pour faciliter certaines opérations, & </s> <s xml:id="echoid-s5900" xml:space="preserve">de faire préciſément l’in-<lb/>verſe de ce que nous venons d’enſeigner, c’eſt-à-dire de faire <lb/>paſſer ſous le radical une quantité qui eſt hors du même ſigne: <lb/></s> <s xml:id="echoid-s5901" xml:space="preserve">voici comme cela ſe pratique. </s> <s xml:id="echoid-s5902" xml:space="preserve">On éleve la quantité qui eſt hors <lb/>du ſigne, à la puiſſance marquée par l’expoſant du radical, & </s> <s xml:id="echoid-s5903" xml:space="preserve"><lb/>on multiplie cette puiſſance par les quantités ſoumiſes au mê-<lb/>me ſigne. </s> <s xml:id="echoid-s5904" xml:space="preserve">Il eſt aiſé de voir que cette nouvelle expreſſion n’eſt <lb/>différente de la premiere qu’en apparence, & </s> <s xml:id="echoid-s5905" xml:space="preserve">non en valeur; </s> <s xml:id="echoid-s5906" xml:space="preserve"><lb/>car la quantité élevée à la puiſſance du radical & </s> <s xml:id="echoid-s5907" xml:space="preserve">ſoumiſe au <lb/>même radical, ne vaut que la racine de cette même quantité: </s> <s xml:id="echoid-s5908" xml:space="preserve"><lb/>ainſi a√ab\x{0020} = √a<emph style="sub">2</emph> x ab\x{0020}, a + b√fg\x{0020} = √a<emph style="sub">2</emph> + 2ab + b<emph style="sub">2</emph> x fg\x{0020} <lb/>= √a<emph style="sub">2</emph>fg + 2abgf + b<emph style="sub">2</emph>fg\x{0020}.</s> <s xml:id="echoid-s5909" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5910" xml:space="preserve">321. </s> <s xml:id="echoid-s5911" xml:space="preserve">On peut multiplier ou diviſer l’expoſant d’un radical <lb/>ſans en changer la valeur: </s> <s xml:id="echoid-s5912" xml:space="preserve">pour cela, il faut élever la quantité <lb/>qui eſt ſous ce ſigne à la puiſſance marquée par le nombre qui <lb/>multiplie l’expoſant du radical, ou tirer de la quantité qui eſt <lb/>ſoumiſe au même radical, la racine marquée par le diviſeur; <lb/></s> <s xml:id="echoid-s5913" xml:space="preserve">ce qui ſe peut faire en deux manieres, ou bien en indiquant <lb/>cette racine par de nouveaux ſignes radicaux, ou bien en di-<lb/>viſant les expoſans des quantités qui ſont ſous le ſigne, par le <lb/>nombre qui doit diviſer l’expoſant du radical: </s> <s xml:id="echoid-s5914" xml:space="preserve">car on a vu <lb/>qu’en diviſant ainſi les expoſans par des nombres, c’eſt prendre <lb/>la racine marquée par ce même nombre (art. </s> <s xml:id="echoid-s5915" xml:space="preserve">142). </s> <s xml:id="echoid-s5916" xml:space="preserve">D’ailleurs <lb/>ſi l’on multiplie ou ſi l’on diviſe, il eſt évident que la quan-<lb/>tité propoſée reçoit autant par l’élévation de la quantité ſou-<lb/>miſe au radical, à la puiſſance marquée par le multiplicateur <lb/>de l’expoſant du radical; </s> <s xml:id="echoid-s5917" xml:space="preserve">que la racine que l’on prend enſuite <lb/>diminue par la multiplication du même expoſant, & </s> <s xml:id="echoid-s5918" xml:space="preserve">récipro-<lb/>quement lorſque l’on diviſe les expoſans des quantités qui ſont <lb/>ſous le ſigne radical, on diminue ces grandeurs de la quan-<lb/>tité dont elles ont été augmentées par la diviſion de l’expoſant <lb/>du radical. </s> <s xml:id="echoid-s5919" xml:space="preserve">Des exemples éclairciront tout ceci. </s> <s xml:id="echoid-s5920" xml:space="preserve">Si l’on a √ab\x{0020}, <lb/>je dis que l’on peut faire ces égalités, √ab\x{0020} = <emph style="sub">6</emph>√a<emph style="sub">3</emph>b<emph style="sub">3</emph>\x{0020} = <pb o="171" file="0209" n="209" rhead="DE MATHÉMATIQUE. Liv. II."/> <emph style="sub">2m</emph>√a<emph style="sub">m</emph>b<emph style="sub">m</emph>\x{0020}; </s> <s xml:id="echoid-s5921" xml:space="preserve">car <emph style="sub">m</emph>√a<emph style="sub">m</emph>b<emph style="sub">m</emph>\x{0020} = ab, en prenant les racines de chaque <lb/>lettre: </s> <s xml:id="echoid-s5922" xml:space="preserve">donc <emph style="sub">2m</emph>√a<emph style="sub">m</emph>b<emph style="sub">m</emph>\x{0020} = √ab\x{0020}, & </s> <s xml:id="echoid-s5923" xml:space="preserve">ainſi des autres. </s> <s xml:id="echoid-s5924" xml:space="preserve">De même <lb/><emph style="sub">5</emph>√a<emph style="sub">3</emph>b<emph style="sub">2</emph>\x{0020} = <emph style="sub">{5/3}</emph>√a{5/3}b{2/3}\x{0020}, ou en général <emph style="sub">m</emph>√a<emph style="sub">n</emph>b<emph style="sub">p</emph>\x{0020} = <emph style="sub">{m/r}</emph>√a{n/r}b{p/r}\x{0020} = <lb/><emph style="sub">{m/r}</emph>√<emph style="sub">r</emph>√a<emph style="sub">n</emph>b<emph style="sub">p</emph>\x{0020}\x{0020}: </s> <s xml:id="echoid-s5925" xml:space="preserve">car <emph style="sub">{1/r}</emph>√a{n/r}b{p/r}\x{0020} = a<emph style="sub">n</emph>b<emph style="sub">p</emph>: </s> <s xml:id="echoid-s5926" xml:space="preserve">donc <emph style="sub">{m/r}</emph>√a<emph style="sub">{n/r}</emph>b<emph style="sub">{p/r}</emph>\x{0020} = <emph style="sub">m</emph>√a<emph style="sub">n</emph>b<emph style="sub">p</emph>\x{0020}, & </s> <s xml:id="echoid-s5927" xml:space="preserve"><lb/>ainſi des autres: </s> <s xml:id="echoid-s5928" xml:space="preserve">car il eſt évident que lorſque l’expoſant du <lb/>radical eſt égal à l’expoſant des grandeurs ſoumiſes au même <lb/>ſigne, on peut ſupprimer le radical, & </s> <s xml:id="echoid-s5929" xml:space="preserve">écrire les quantités <lb/>toutes ſimples, comme ſi l’on a <emph style="sub">3</emph>√a<emph style="sub">3</emph>\x{0020}, on met a, & </s> <s xml:id="echoid-s5930" xml:space="preserve">pour <emph style="sub">5</emph>√a<emph style="sub">5</emph>b<emph style="sub">10</emph>\x{0020}, <lb/>on met ab<emph style="sub">2</emph>; </s> <s xml:id="echoid-s5931" xml:space="preserve">c’eſt ce qui arrive ici, car l’expoſant {m/r} peut s’é-<lb/>crire ainſi, m x {1/r}, & </s> <s xml:id="echoid-s5932" xml:space="preserve">de même les expoſans {n/r}, {p/r} peuvent ſe <lb/>marquer ainſi, n x {1/r}, p x {1/r}: </s> <s xml:id="echoid-s5933" xml:space="preserve">donc notre quantité deviendroit <lb/><emph style="sub">m x {1/r}</emph>√a<emph style="sub">n x</emph> {1/r}b<emph style="sub">p x</emph> {1/r}\x{0020}, où il eſt viſible que l’on ne fait que multiplier <lb/>les expoſans du radical & </s> <s xml:id="echoid-s5934" xml:space="preserve">des quantités qui lui ſont ſoumiſes <lb/>par la même grandeur {1/r}; </s> <s xml:id="echoid-s5935" xml:space="preserve">ce qui rentre dans le premier cas.</s> <s xml:id="echoid-s5936" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5937" xml:space="preserve">322. </s> <s xml:id="echoid-s5938" xml:space="preserve">On tire delà la méthode de réduire pluſieurs radicaux <lb/>à la même dénomination ſans changer leurs valeurs, c’eſt-à-dire <lb/>de donner à deux radicaux différens un même ſigne. </s> <s xml:id="echoid-s5939" xml:space="preserve">Par exem-<lb/>ple, ſi l’on me donne ces deux incommenſurables √a<emph style="sub">3</emph>\x{0020} & </s> <s xml:id="echoid-s5940" xml:space="preserve"><emph style="sub">3</emph>√a<emph style="sub">2</emph>b<emph style="sub">3</emph>\x{0020}, <lb/>j’éleve le premier a<emph style="sub">3</emph> à ſon cube, & </s> <s xml:id="echoid-s5941" xml:space="preserve">je multiplie l’expoſant 2 <lb/>du radical par 3, ce qui me donne <emph style="sub">6</emph>√a<emph style="sub">9</emph>\x{0020} = √a<emph style="sub">3</emph>\x{0020}: </s> <s xml:id="echoid-s5942" xml:space="preserve">de même j’é-<lb/>leve a<emph style="sub">2</emph>b+ à ſon quarré pour avoir a<emph style="sub">4</emph>b<emph style="sub">8</emph>, & </s> <s xml:id="echoid-s5943" xml:space="preserve">je multiplie l’expo-<lb/>ſant du ſigne radical qui lui eſt joint par l’expoſant 2 du pre-<lb/>mier, ce qui me donne <emph style="sub">6</emph>√a<emph style="sub">4</emph>b<emph style="sub">8</emph>\x{0020} = <emph style="sub">3</emph>√a<emph style="sub">2</emph>b<emph style="sub">4</emph>\x{0020}. </s> <s xml:id="echoid-s5944" xml:space="preserve">De cette maniere il <lb/>eſt viſible que les deux quantités irrationnelles propoſées ont <lb/>changé de forme ou d’expreſſion, ſans avoir changé de va-<lb/>leur, & </s> <s xml:id="echoid-s5945" xml:space="preserve">de plus qu’elles ont le même ſigne radical <emph style="sub">6</emph>√\x{0020}, & </s> <s xml:id="echoid-s5946" xml:space="preserve">ainſi <lb/>des autres. </s> <s xml:id="echoid-s5947" xml:space="preserve">En général pour réduire deux radicaux quelcon-<lb/>ques a <emph style="sub">m</emph>√b<emph style="sub">P</emph>\x{0020}, c <emph style="sub">n</emph>√d<emph style="sub">r</emph>\x{0020}, on écrira a<emph style="sub">mn</emph>√b<emph style="sub">pn</emph>\x{0020}, c <emph style="sub">mn</emph>√d<emph style="sub">mr</emph>\x{0020}. </s> <s xml:id="echoid-s5948" xml:space="preserve">Les opérations <pb o="172" file="0210" n="210" rhead="NOUVEAU COURS"/> que nous venons de voir, ſont particulieres aux quantités irra-<lb/>tionnelles: </s> <s xml:id="echoid-s5949" xml:space="preserve">nous allons préſentement expliquer celles qui leur <lb/>ſont communes avec les autres quantités.</s> <s xml:id="echoid-s5950" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div309" type="section" level="1" n="282"> <head xml:id="echoid-head324" style="it" xml:space="preserve">De l’Addition des Radicaux.</head> <p> <s xml:id="echoid-s5951" xml:space="preserve">323. </s> <s xml:id="echoid-s5952" xml:space="preserve">On ajoutera les radicaux, en les joignant avec leurs <lb/>ſignes tels qu’ils ſont, & </s> <s xml:id="echoid-s5953" xml:space="preserve">obſervant de les réduire avant de <lb/>faire l’addition. </s> <s xml:id="echoid-s5954" xml:space="preserve">De plus, ſi les radicaux ſont les mêmes de part <lb/>& </s> <s xml:id="echoid-s5955" xml:space="preserve">d’autre, il ſuffira d’ajouter les quantités qui précédent le <lb/>ſigne radical, & </s> <s xml:id="echoid-s5956" xml:space="preserve">d’en multiplier la ſomme par le même radical: <lb/></s> <s xml:id="echoid-s5957" xml:space="preserve">ſuivant cette regle, la ſomme de a√b\x{0020} & </s> <s xml:id="echoid-s5958" xml:space="preserve">de c√d\x{0020} eſt a√b\x{0020} + c <lb/>√d\x{0020}; </s> <s xml:id="echoid-s5959" xml:space="preserve">celle de ff<emph style="sub">3</emph>√g<emph style="sub">2</emph>\x{0020}, & </s> <s xml:id="echoid-s5960" xml:space="preserve">de mn√dc\x{0020} eſt ff<emph style="sub">3</emph>√g<emph style="sub">2</emph>\x{0020} + mn√dc\x{0020}; </s> <s xml:id="echoid-s5961" xml:space="preserve"><lb/>celle de af√mn\x{0020} & </s> <s xml:id="echoid-s5962" xml:space="preserve">de bg√mn\x{0020} eſt √af + bg\x{0020}√mn\x{0020}. </s> <s xml:id="echoid-s5963" xml:space="preserve">De même <lb/>en nombres, 3√5\x{0020} & </s> <s xml:id="echoid-s5964" xml:space="preserve">4√7\x{0020} donnent pour ſomme 3√5\x{0020} + 4√7\x{0020}, <lb/>4√8\x{0020} & </s> <s xml:id="echoid-s5965" xml:space="preserve">6√8\x{0020} donnent 10√8\x{0020}, &</s> <s xml:id="echoid-s5966" xml:space="preserve">c.</s> <s xml:id="echoid-s5967" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div310" type="section" level="1" n="283"> <head xml:id="echoid-head325" style="it" xml:space="preserve">De la Souſtraction des Radicaux.</head> <p> <s xml:id="echoid-s5968" xml:space="preserve">324. </s> <s xml:id="echoid-s5969" xml:space="preserve">La Souſtraction des radicaux ſe fait de même que celle <lb/>des autres quantités algébriques, en changeant le ſigne + en <lb/>-, & </s> <s xml:id="echoid-s5970" xml:space="preserve">le ſigne - en + de la quantité que l’on veut ſouſtraire, <lb/>obſervant de ſimplifier auparavant les radicaux propoſés, & </s> <s xml:id="echoid-s5971" xml:space="preserve"><lb/>de multiplier la différence par le même radical, en cas qu’il <lb/>ſoit commun aux deux radicaux. </s> <s xml:id="echoid-s5972" xml:space="preserve">Par exemple, la différence <lb/>de a√c\x{0020}à b√c\x{0020} eſt √a - b\x{0020}√c\x{0020}; </s> <s xml:id="echoid-s5973" xml:space="preserve">celle de 10<emph style="sub">3</emph>√9\x{0020} à 4<emph style="sub">3</emph>√9\x{0020} eſt √10 - 4\x{0020}<emph style="sub">3</emph>√9\x{0020}, <lb/>ou 6<emph style="sub">3</emph>√9\x{0020}, &</s> <s xml:id="echoid-s5974" xml:space="preserve">c.</s> <s xml:id="echoid-s5975" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div311" type="section" level="1" n="284"> <head xml:id="echoid-head326" style="it" xml:space="preserve">De la Multiplication des Radicaux.</head> <p> <s xml:id="echoid-s5976" xml:space="preserve">325. </s> <s xml:id="echoid-s5977" xml:space="preserve">On peut multiplier un radical par un entier, par une <lb/>fraction, ou par un autre radical; </s> <s xml:id="echoid-s5978" xml:space="preserve">ce qui fait trois cas parti-<lb/>culiers, qui n’ont aucune difficulté.</s> <s xml:id="echoid-s5979" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5980" xml:space="preserve">326. </s> <s xml:id="echoid-s5981" xml:space="preserve">Pour multiplier un radical par un entier, s’il a déja <lb/>quelque grandeur qui le précéde, on multipliera cette quan-<lb/>tité qui eſt hors du radical par l’entier propoſé. </s> <s xml:id="echoid-s5982" xml:space="preserve">Par exemple, <lb/>le produit de a√b\x{0020} par 3c eſt 3ac√b\x{0020}; </s> <s xml:id="echoid-s5983" xml:space="preserve">le produit de <emph style="sub">3</emph>√c<emph style="sub">2</emph>\x{0020} par <lb/>a + 2b eſt √a + 2b\x{0020}<emph style="sub">3</emph>√c<emph style="sub">2</emph>\x{0020}, ou a<emph style="sub">3</emph>√c<emph style="sub">2</emph>\x{0020} + 2b<emph style="sub">3</emph>√c<emph style="sub">2</emph>\x{0020}, & </s> <s xml:id="echoid-s5984" xml:space="preserve">ainſi de ſuite. <lb/></s> <s xml:id="echoid-s5985" xml:space="preserve">Si l’on ne vouloit pas que le multiplicateur fût devant le radi- <pb o="173" file="0211" n="211" rhead="DE MATHÉMATIQUE. Liv. II."/> cal, il faudroit l’élever à la puiſſance marquée par l’expoſant <lb/>du radical. </s> <s xml:id="echoid-s5986" xml:space="preserve">Ainſi pour multiplier <emph style="sub">3</emph>√bc\x{0020} par af, j’éleve af à ſon <lb/>cube, & </s> <s xml:id="echoid-s5987" xml:space="preserve">je multiplie ce qui eſt ſous le radical par a<emph style="sub">3</emph>f<emph style="sub">3</emph>, & </s> <s xml:id="echoid-s5988" xml:space="preserve">j’ai <lb/><emph style="sub">3</emph>√a bcf<emph style="sub">3</emph>\x{0020}. </s> <s xml:id="echoid-s5989" xml:space="preserve">Il en ſeroit ainſi des autres en nombres ou en lettres, <lb/>quelque ſoit le multiplicateur incomplexe ou polynome.</s> <s xml:id="echoid-s5990" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s5991" xml:space="preserve">327. </s> <s xml:id="echoid-s5992" xml:space="preserve">Pour multiplier un radical par une ſraction, on mul-<lb/>tipliera la quantité qui eſt hors du ſigne par la fraction pro-<lb/>poſée, & </s> <s xml:id="echoid-s5993" xml:space="preserve">la multiplication ſera faite. </s> <s xml:id="echoid-s5994" xml:space="preserve">Si le radical n’avoit <lb/>d’autre coefficient que l’unité, & </s> <s xml:id="echoid-s5995" xml:space="preserve">qu’on jugeât à propos de ne <lb/>point lui en donner, il faudroit élever la fraction à la puiſ-<lb/>ſance marquée par l’expoſant du radical, & </s> <s xml:id="echoid-s5996" xml:space="preserve">multiplier le nu-<lb/>mérateur de la nouvelle fraction par la quantité ſoumiſe au <lb/>radical. </s> <s xml:id="echoid-s5997" xml:space="preserve">Ainſi pour multiplier le radical f √ab\x{0020} par {c/d}, j’écris <lb/>{cf/d}√ab\x{0020}; </s> <s xml:id="echoid-s5998" xml:space="preserve">de même 3 √c\x{0020} par {6/5}={18/5}√c\x{0020}; </s> <s xml:id="echoid-s5999" xml:space="preserve">de même <emph style="sub">3</emph>√cf\x{0020}, multi-<lb/>plié par {2a/b}=<emph style="sub">3</emph>√{8a<emph style="sub">3</emph>cf/b<emph style="sub">3</emph>}\x{0020}, par la ſeconde partie de cette regle.</s> <s xml:id="echoid-s6000" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6001" xml:space="preserve">328. </s> <s xml:id="echoid-s6002" xml:space="preserve">Si le multiplicateur eſt auſſi un radical de même ex-<lb/>poſant que celui du multiplicande, on multipliera les quan-<lb/>tités ſoumiſes au même radical les unes par les autres, ſuivant <lb/>les regles ordinaires, & </s> <s xml:id="echoid-s6003" xml:space="preserve">on donnera au produit le ſigne du <lb/>multiplicande ou du multiplicateur, obſervant de multiplier <lb/>les quantités qui précédent les radicaux les unes par les autres, <lb/>& </s> <s xml:id="echoid-s6004" xml:space="preserve">de tirer hors du nouveau radical les puiſſances de même <lb/>nom, que la multiplication auroit pu produire. </s> <s xml:id="echoid-s6005" xml:space="preserve">Par exemple, <lb/>a√cb\x{0020}, multiplié par f√cd\x{0020}=af√c<emph style="sub">2</emph>db\x{0020}=a c f√bd\x{0020}; </s> <s xml:id="echoid-s6006" xml:space="preserve">de même <lb/>f<emph style="sub">3</emph>√a<emph style="sub">2</emph>bc\x{0020} x g<emph style="sub">3</emph>√ac<emph style="sub">2</emph>d\x{0020}=fg<emph style="sub">3</emph>√a<emph style="sub">3</emph>bc<emph style="sub">3</emph> d\x{0020} = a c f g<emph style="sub">3</emph>√bd\x{0020}, & </s> <s xml:id="echoid-s6007" xml:space="preserve">ainſi des <lb/>autres.</s> <s xml:id="echoid-s6008" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6009" xml:space="preserve">329. </s> <s xml:id="echoid-s6010" xml:space="preserve">Si le radical n’a pas le même expoſant, on commen-<lb/>cera par les y réduire (art. </s> <s xml:id="echoid-s6011" xml:space="preserve">321), & </s> <s xml:id="echoid-s6012" xml:space="preserve">l’on fera la multiplication <lb/>comme dans le cas précédent. </s> <s xml:id="echoid-s6013" xml:space="preserve">Par exemple, pour multiplier <lb/>a√bc\x{0020} par d<emph style="sub">3</emph>√fg\x{0020}, je réduis d’abord a√bc\x{0020} en a<emph style="sub">6</emph>√b<emph style="sub">3</emph>c<emph style="sub">3</emph>\x{0020}, & </s> <s xml:id="echoid-s6014" xml:space="preserve"><lb/>d<emph style="sub">3</emph>√fg\x{0020} en d<emph style="sub">6</emph>√f<emph style="sub">2</emph>g<emph style="sub">2</emph>\x{0020}, & </s> <s xml:id="echoid-s6015" xml:space="preserve">multipliant enſuite j’ai ad<emph style="sub">6</emph>√b<emph style="sub">3</emph>c<emph style="sub">3</emph>f<emph style="sub">2</emph>g<emph style="sub">2</emph>\x{0020}. <lb/></s> <s xml:id="echoid-s6016" xml:space="preserve">Il en ſeroit de même des radicaux plus compliqués. </s> <s xml:id="echoid-s6017" xml:space="preserve">Il faut <lb/>bien remarquer que ſi le radical du multiplicateur eſt le même <lb/>que celui du multiplicande, la multiplication ſe fait en ſup-<lb/>primant le radical, & </s> <s xml:id="echoid-s6018" xml:space="preserve">multipliant par cette quantité le pro-<lb/>duit des quantités qui précédent. </s> <s xml:id="echoid-s6019" xml:space="preserve">Ainſi a√bc\x{0020} x d√bc\x{0020}=adbc <pb o="174" file="0212" n="212" rhead="NOUVEAU COURS"/> 3√fg\x{0020} x 4√fg\x{0020} = 12fg; </s> <s xml:id="echoid-s6020" xml:space="preserve">ce qui eſt évident, puiſque toute raci-<lb/>ne multipliée par elle-même doit néceſſairement ſe reproduire.</s> <s xml:id="echoid-s6021" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6022" xml:space="preserve">Si l’on avoit des radicaux complexes à multiplier par des <lb/>radicaux monomes ou complexes, la multiplication s’en fe-<lb/>roit, en ſuivant les mêmes regles, & </s> <s xml:id="echoid-s6023" xml:space="preserve">celles de la multiplication <lb/>des polynomes.</s> <s xml:id="echoid-s6024" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div312" type="section" level="1" n="285"> <head xml:id="echoid-head327" style="it" xml:space="preserve">De la Diviſion des Radicaux.</head> <p> <s xml:id="echoid-s6025" xml:space="preserve">330. </s> <s xml:id="echoid-s6026" xml:space="preserve">On peut diviſer un radical par un entier ou par une <lb/>fraction, ou par un autre radical: </s> <s xml:id="echoid-s6027" xml:space="preserve">toutes ces opérations ſont <lb/>les inverſes des précédentes; </s> <s xml:id="echoid-s6028" xml:space="preserve">c’eſt pourquoi nous ne nous y <lb/>arrêterons pas long-tems.</s> <s xml:id="echoid-s6029" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6030" xml:space="preserve">331. </s> <s xml:id="echoid-s6031" xml:space="preserve">Pour diviſer un radical par un entier, on diviſera le <lb/>coefficient par l’entier propoſé: </s> <s xml:id="echoid-s6032" xml:space="preserve">ainſi pour diviſer a√b\x{0020} par c, <lb/>j’écris {a/c}√b\x{0020}; </s> <s xml:id="echoid-s6033" xml:space="preserve">de même 3 √5\x{0020} diviſé par 4 = {3/4} √5\x{0020}, & </s> <s xml:id="echoid-s6034" xml:space="preserve">de même <lb/>des autres.</s> <s xml:id="echoid-s6035" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6036" xml:space="preserve">332. </s> <s xml:id="echoid-s6037" xml:space="preserve">Pour diviſer un radical par une fraction, on multipliera <lb/>le coefficient du radical par la fraction inverſe, à moins que <lb/>l’on ne voulût faire paſſer le diviſeur ſous le ſigne radical; <lb/></s> <s xml:id="echoid-s6038" xml:space="preserve">auquel cas il faudroit multiplier ce qui eſt ſous le radical par <lb/>le quarré de la fraction inverſe. </s> <s xml:id="echoid-s6039" xml:space="preserve">Suivant ces regles, le quo-<lb/>tient de a √bc\x{0020} diviſé par {d/f} = {af/d} √bc\x{0020}, le quotient de 3 √bd\x{0020}, <lb/>diviſé par {4/5} = {15/4}√bd\x{0020}; </s> <s xml:id="echoid-s6040" xml:space="preserve">& </s> <s xml:id="echoid-s6041" xml:space="preserve">celui de <emph style="sub">3</emph>√fg\x{0020} par {a/d} = {d/a}<emph style="sub">3</emph>√fg\x{0020}, ou en <lb/>mettant la fraction ſous le radical <emph style="sub">3</emph>√{d<emph style="sub">3</emph>fg/a<emph style="sub">3</emph>}\x{0020}.</s> <s xml:id="echoid-s6042" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6043" xml:space="preserve">Pour diviſer un radical par un autre, on diviſera les coeffi-<lb/>cients & </s> <s xml:id="echoid-s6044" xml:space="preserve">les radicaux l’un par l’autre, en obſervant d’effacer <lb/>le radical, lorſqu’il eſt commun au diviſeur & </s> <s xml:id="echoid-s6045" xml:space="preserve">au dividende. <lb/></s> <s xml:id="echoid-s6046" xml:space="preserve">Ainſi a√b\x{0020} diviſé par c√d\x{0020} = {a/c} √{b/d}\x{0020}, a√cd\x{0020} diviſé par b√cd\x{0020} <lb/>= {a/b}, & </s> <s xml:id="echoid-s6047" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s6048" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div313" type="section" level="1" n="286"> <head xml:id="echoid-head328" style="it" xml:space="preserve">Formation des Puiſſances des Radicaux.</head> <p> <s xml:id="echoid-s6049" xml:space="preserve">333. </s> <s xml:id="echoid-s6050" xml:space="preserve">Pour élever un radical à une puiſſance propoſée, il <lb/>faut élever à cette puiſſance les quantités qui précédent le ra-<lb/>dical, & </s> <s xml:id="echoid-s6051" xml:space="preserve">celles qui lui ſont ſoumiſes, ou bien diviſer l’expo-<lb/>ſant du radical par l’expoſant de la puiſſance à laquelle on veut <lb/>élever ce radical: </s> <s xml:id="echoid-s6052" xml:space="preserve">ainſi le cube de a√bc\x{0020} eſt a<emph style="sub">3</emph> √b<emph style="sub">3</emph>c<emph style="sub">3</emph>\x{0020}, ou <pb o="175" file="0213" n="213" rhead="DE MATHEMATIQUE. Liv. II."/> a<emph style="sub">3</emph>bc√bc\x{0020}, en ſimplifiant la derniere expreſſion: </s> <s xml:id="echoid-s6053" xml:space="preserve">on peut dire <lb/>auſſi que le cube de cette même quantité eſt a<emph style="sub">3</emph><emph style="sub">{2/3}</emph>√bc\x{0020}: </s> <s xml:id="echoid-s6054" xml:space="preserve">car ſi l’on <lb/>ſe ſouvient de ce que nous avons déja dit ſur les radicaux & </s> <s xml:id="echoid-s6055" xml:space="preserve">les <lb/>expoſans (art. </s> <s xml:id="echoid-s6056" xml:space="preserve">142.) </s> <s xml:id="echoid-s6057" xml:space="preserve">a<emph style="sub">3</emph> <emph style="sub">{2/3}</emph>√bc\x{0020} = a<emph style="sub">3</emph>b<emph style="sub">{1/2/3}</emph>c<emph style="sub">{1/2/3}</emph> = a<emph style="sub">3</emph>b<emph style="sub">{3/2}</emph>c<emph style="sub">{3/2}</emph> = a<emph style="sub">3</emph>√b<emph style="sub">3</emph>c<emph style="sub">3</emph>\x{0020}, par <lb/>le même article. </s> <s xml:id="echoid-s6058" xml:space="preserve">Toutes les fois que l’expoſant du radical ſera <lb/>diviſible par celui de la puiſſance à laquelle on veut l’élever, <lb/>il faudra faire la diviſion préférablement à toute autre méthode.</s> <s xml:id="echoid-s6059" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div314" type="section" level="1" n="287"> <head xml:id="echoid-head329" style="it" xml:space="preserve">Extraction des racines des radicaux.</head> <p> <s xml:id="echoid-s6060" xml:space="preserve">334. </s> <s xml:id="echoid-s6061" xml:space="preserve">Pour tirer la racine d’un radical, il n’y aura qu’à tirer <lb/>la racine de ce qui précéde ce radical, & </s> <s xml:id="echoid-s6062" xml:space="preserve">multiplier l’expoſant <lb/>du ſigne radical par l’expoſant de la racine propoſée; </s> <s xml:id="echoid-s6063" xml:space="preserve">car puiſ-<lb/>que nous venons de voir que la formation des puiſſances de <lb/>ces quantités ſe fait par la diviſion des expoſans, par celui de <lb/>la puiſſance; </s> <s xml:id="echoid-s6064" xml:space="preserve">dans l’extraction des racines, il faut faire le con-<lb/>traire: </s> <s xml:id="echoid-s6065" xml:space="preserve">ainſi la racine cubique de a<emph style="sub">3</emph> √b<emph style="sub">2</emph>c\x{0020} eſt a<emph style="sub">6</emph>√b<emph style="sub">2</emph>c\x{0020}, celle de <lb/>a<emph style="sub">4</emph> <emph style="sub">5</emph>√b<emph style="sub">2</emph>c<emph style="sub">3</emph>\x{0020} eſt a <emph style="sub">3</emph>√a\x{0020}<emph style="sub">15</emph>√b<emph style="sub">2</emph>c<emph style="sub">3</emph>\x{0020}. </s> <s xml:id="echoid-s6066" xml:space="preserve">Si l’on vouloit on pourroit encore <lb/>faire la même choſe, après avoir fait paſſer tout ce qui pré-<lb/>céde le ſigne ſous le même ſigne: </s> <s xml:id="echoid-s6067" xml:space="preserve">ainſi la racine cubique de <lb/>a<emph style="sub">4</emph> <emph style="sub">5</emph>√b<emph style="sub">2</emph>c<emph style="sub">3</emph>\x{0020}, ou celle de <emph style="sub">5</emph>√a<emph style="sub">20</emph>b<emph style="sub">2</emph>c<emph style="sub">3</emph>\x{0020} eſt <emph style="sub">15</emph>√a<emph style="sub">20</emph>b<emph style="sub">2</emph>c<emph style="sub">3</emph>\x{0020}.</s> <s xml:id="echoid-s6068" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6069" xml:space="preserve">335. </s> <s xml:id="echoid-s6070" xml:space="preserve">Il faut bien remarquer que toutes les opérations que <lb/>l’on fait ſur les radicaux peuvent ſe faire d’une autre maniere, <lb/>en cherchant la quantité exponentielle égale au radical pro-<lb/>poſé: </s> <s xml:id="echoid-s6071" xml:space="preserve">car nous avons démontré (art. </s> <s xml:id="echoid-s6072" xml:space="preserve">141 & </s> <s xml:id="echoid-s6073" xml:space="preserve">ſuivans) qu’il n’y <lb/>a point de radical qu’on ne puiſſe convertir en quantité expo-<lb/>nentielle & </s> <s xml:id="echoid-s6074" xml:space="preserve">réciproquement.</s> <s xml:id="echoid-s6075" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6076" xml:space="preserve">Les Commençans confondent quelquefois les racines ima-<lb/>ginaires avecles grandeurs incommenſurables; </s> <s xml:id="echoid-s6077" xml:space="preserve">il y a une diffé-<lb/>rence totale entre les unes & </s> <s xml:id="echoid-s6078" xml:space="preserve">les autres. </s> <s xml:id="echoid-s6079" xml:space="preserve">On peut déterminer <lb/>par la Géométrie la grandeur abſolue des quantités incom-<lb/>menſurables, quoiqu’on ne puiſſe pas déterminer en nombres <lb/>leurs rapports avec l’unité, au lieu que l’on ne peut connoître <lb/>ce que ſignifient les imaginaires; </s> <s xml:id="echoid-s6080" xml:space="preserve">car on ne connoît point de <lb/>racine qui puiſſe donner un quarré négatif: </s> <s xml:id="echoid-s6081" xml:space="preserve">c’eſt ce qui a fait <lb/>regarder ces quantités comme abſolument impoſſibles, & </s> <s xml:id="echoid-s6082" xml:space="preserve">com-<lb/>me abſurdes les équations ou problêmes qui ne donnent que <lb/>de pareilles ſolutions. </s> <s xml:id="echoid-s6083" xml:space="preserve">Mais on a reconnu que l’on ne doit <pb o="176" file="0214" n="214" rhead="NOUVEAU COURS DE MATHEM. Liv. II."/> point établir cette propoſition comme un principe général; <lb/></s> <s xml:id="echoid-s6084" xml:space="preserve">& </s> <s xml:id="echoid-s6085" xml:space="preserve">d’ailleurs ſi l’on conſidere les racines d’une équation dans <lb/>leur nature & </s> <s xml:id="echoid-s6086" xml:space="preserve">leur eſſence, qui eſt d’être des diviſeurs exacts <lb/>de cette même équation, on verra que les imaginaires ne ſont <lb/>pas moins racines d’une équation, que celles que l’on appelle <lb/>vraies ou réelles, puiſque comme celles-ci, elles concourent <lb/>par leur multiplication à former l’équation qui les a données, <lb/>& </s> <s xml:id="echoid-s6087" xml:space="preserve">qu’elles en ſont par conſéquent des diviſeurs exacts, comme <lb/>il eſt aiſé de s’en convaincre par l’exemple ſuivant.</s> <s xml:id="echoid-s6088" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6089" xml:space="preserve">Soit propoſé de réſoudre cette équation du ſecond degré, <lb/>xx - 4x + 12 = 0. </s> <s xml:id="echoid-s6090" xml:space="preserve">On trouvera, en ſuivant les regles ordinaires, <lb/>x = 2 ± √-8\x{0020}, ou, ce qui eſt la même choſe, en égalant les <lb/>deux valeurs de x à zero, les deux équations x - 2 + √-8\x{0020} = 0, <lb/>& </s> <s xml:id="echoid-s6091" xml:space="preserve">x - 2 - √-8\x{0020} = 0, que l’on peut regarder comme des <lb/>racines de la propoſée, parce qu’en les multipliant l’une par <lb/>l’autre, on retrouve au produit, après la réduction & </s> <s xml:id="echoid-s6092" xml:space="preserve">l’évanouiſ-<lb/>ſement des radicaux l’équation propoſée xx - 4x + 12 = 0.</s> <s xml:id="echoid-s6093" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6094" xml:space="preserve">Il faut encore remarquer que dans une équation quelconque, <lb/>délivrée de tout ſigne radical, les racines imaginaires ne peu-<lb/>vent être qu’en nombre pair. </s> <s xml:id="echoid-s6095" xml:space="preserve">Ainſi dans une équation du ſe-<lb/>cond degré, les racines ſont toujours toutes les deux vraies, <lb/>ou toutes deux imaginaires.</s> <s xml:id="echoid-s6096" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6097" xml:space="preserve">Je me borne à ces exemples ſur la maniere de réſoudre les <lb/>équations du ſecond degré, afin d’en faciliter l’uſage qui eſt <lb/>fort fréquent dans les queſtions Mathématiques. </s> <s xml:id="echoid-s6098" xml:space="preserve">L’on trouvera <lb/>vers la fin de ce volume ce qui appartient à celles du troiſieme <lb/>& </s> <s xml:id="echoid-s6099" xml:space="preserve">du quatrieme degré, quoiqu’elle ne ſoient pas auſſi abſolu-<lb/>ment néceſſaires que celles-ci.</s> <s xml:id="echoid-s6100" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div315" type="section" level="1" n="288"> <head xml:id="echoid-head330" style="it" xml:space="preserve">Fin des équations du ſecond degré, & du ſecond Livre.</head> <figure> <image file="0214-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0214-01"/> </figure> <pb o="177" file="0215" n="215"/> <figure> <image file="0215-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0215-01"/> </figure> </div> <div xml:id="echoid-div316" type="section" level="1" n="289"> <head xml:id="echoid-head331" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head332" style="it" xml:space="preserve">LIVRE TROISIEME, <lb/>Où l’on conſidere les différentes poſitions des Lignes droites <lb/>les unes à l’égard des autres. <lb/><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head333" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s6101" xml:space="preserve">336. </s> <s xml:id="echoid-s6102" xml:space="preserve"><emph style="sc">Les</emph> lignes paralleles ſont celles qui, étant prolongées <lb/> <anchor type="note" xlink:label="note-0215-01a" xlink:href="note-0215-01"/> autant que l’on voudra, ſont toujours également éloignées <lb/> <anchor type="note" xlink:label="note-0215-02a" xlink:href="note-0215-02"/> entr’elles, & </s> <s xml:id="echoid-s6103" xml:space="preserve">dont les extrêmités ne peuvent jamais ſe ren-<lb/>contrer, comme les lignes A B & </s> <s xml:id="echoid-s6104" xml:space="preserve">C D.</s> <s xml:id="echoid-s6105" xml:space="preserve"/> </p> <div xml:id="echoid-div316" type="float" level="2" n="1"> <note position="right" xlink:label="note-0215-01" xlink:href="note-0215-01a" xml:space="preserve">Planche I.</note> <note position="right" xlink:label="note-0215-02" xlink:href="note-0215-02a" xml:space="preserve">Figure 7.</note> </div> </div> <div xml:id="echoid-div318" type="section" level="1" n="290"> <head xml:id="echoid-head334" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s6106" xml:space="preserve">337. </s> <s xml:id="echoid-s6107" xml:space="preserve">L’angle eſt l’inclinaiſon d’une ligne ſur une autre: </s> <s xml:id="echoid-s6108" xml:space="preserve">on <lb/>l’appelle angle rectiligne, lorſque les deux lignes qui le forment <lb/>ſont droites, comme l’angle A B C; </s> <s xml:id="echoid-s6109" xml:space="preserve">il eſt appellé curviligne; <lb/></s> <s xml:id="echoid-s6110" xml:space="preserve"> <anchor type="note" xlink:label="note-0215-03a" xlink:href="note-0215-03"/> lorſque les lignes qui le forment ſont des lignes courbes, com-<lb/>me l’angle D E F, & </s> <s xml:id="echoid-s6111" xml:space="preserve">mixtiligne, lorſqu’une des lignes eſt droite <lb/> <anchor type="note" xlink:label="note-0215-04a" xlink:href="note-0215-04"/> & </s> <s xml:id="echoid-s6112" xml:space="preserve">l’autre courbe, comme G H I.</s> <s xml:id="echoid-s6113" xml:space="preserve"/> </p> <div xml:id="echoid-div318" type="float" level="2" n="1"> <note position="right" xlink:label="note-0215-03" xlink:href="note-0215-03a" xml:space="preserve">Figure 8.</note> <note position="right" xlink:label="note-0215-04" xlink:href="note-0215-04a" xml:space="preserve">Figure 9.</note> </div> <note position="right" xml:space="preserve">Figure 10.</note> </div> <div xml:id="echoid-div320" type="section" level="1" n="291"> <head xml:id="echoid-head335" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s6114" xml:space="preserve">338. </s> <s xml:id="echoid-s6115" xml:space="preserve">Les lignes droites ou courbes, dont l’inclinaiſon reſ-<lb/>pective fait un angle quelconque, ſont appellées côtés de l’an-<lb/>gle. </s> <s xml:id="echoid-s6116" xml:space="preserve">Le point où ces deux lignes ſe rencontrent mutuellement, <pb o="178" file="0216" n="216" rhead="NOUVEAU COURS"/> eſt appellé le ſommet de l’angle. </s> <s xml:id="echoid-s6117" xml:space="preserve">Il ſuit delà que la grandeur <lb/>d’un angle ne dépend pas de la longueur de ſes côtés, mais <lb/>ſeulement de l’inclinaiſon de ces lignes l’une ſur l’autre, qui <lb/>ſeule conſtitue la nature de l’angle. </s> <s xml:id="echoid-s6118" xml:space="preserve">Il ſuit encore delà qu’un <lb/>angle ne renferme aucun eſpace fini ou déterminé. </s> <s xml:id="echoid-s6119" xml:space="preserve">Pour mar-<lb/>quer un angle, on ſe ſert ordinairement de trois lettres, & </s> <s xml:id="echoid-s6120" xml:space="preserve"><lb/>celle qui ſe trouve au milieu, déſigne le ſommet de l’angle.</s> <s xml:id="echoid-s6121" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div321" type="section" level="1" n="292"> <head xml:id="echoid-head336" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s6122" xml:space="preserve">339. </s> <s xml:id="echoid-s6123" xml:space="preserve">L’angle droit eſt celui qui eſt formé par la rencontre de <lb/>deux lignes perpendiculaires l’une à l’autre, comme les an-<lb/>gles A B C ou A B D. <lb/></s> </p> </div> <div xml:id="echoid-div322" type="section" level="1" n="293"> <head xml:id="echoid-head337" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s6124" xml:space="preserve">340. </s> <s xml:id="echoid-s6125" xml:space="preserve">L’angle oblique eſt celui qui ſe fait par la rencontre de <lb/>deux lignes qui ne ſont pas perpendiculaires l’une à l’autre, & </s> <s xml:id="echoid-s6126" xml:space="preserve"><lb/>que l’on appelle pour cette raiſon des lignes obliques, comme <lb/>ſont les lignes I H & </s> <s xml:id="echoid-s6127" xml:space="preserve">L K. </s> <s xml:id="echoid-s6128" xml:space="preserve">Il y a deux ſortes d’angles obliques, <lb/> <anchor type="note" xlink:label="note-0216-02a" xlink:href="note-0216-02"/> ſçavoir l’angle aigu & </s> <s xml:id="echoid-s6129" xml:space="preserve">l’angle obtus.</s> <s xml:id="echoid-s6130" xml:space="preserve"/> </p> <div xml:id="echoid-div322" type="float" level="2" n="1"> <note position="left" xlink:label="note-0216-02" xlink:href="note-0216-02a" xml:space="preserve">Figure 12.</note> </div> </div> <div xml:id="echoid-div324" type="section" level="1" n="294"> <head xml:id="echoid-head338" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s6131" xml:space="preserve">341. </s> <s xml:id="echoid-s6132" xml:space="preserve">L’angle aigu eſt celui qui eſt plus petit, ou moins ou-<lb/>vert qu’un droit, comme l’angle H I K; </s> <s xml:id="echoid-s6133" xml:space="preserve">& </s> <s xml:id="echoid-s6134" xml:space="preserve">l’angle obtus eſt <lb/> <anchor type="note" xlink:label="note-0216-03a" xlink:href="note-0216-03"/> celui qui eſt plus grand ou plus ouvert qu’un droit, comme L H I. <lb/></s> <s xml:id="echoid-s6135" xml:space="preserve">Il eſt viſible qu’une ligne H I tombant ſur une autre, forme <lb/>avec elle deux angles inégaux, qui pris enſemble, valent deux <lb/>droits: </s> <s xml:id="echoid-s6136" xml:space="preserve">car ſi l’on imagine la droite I F perpendiculaire à la <lb/>ligne L K au point I, l’angle aigu H I L = F I L - F I H, & </s> <s xml:id="echoid-s6137" xml:space="preserve"><lb/>l’angle obtus H I K = F I K + F I H. </s> <s xml:id="echoid-s6138" xml:space="preserve">Ainſi en ajoutant les <lb/>membres de ces deux équations, on aura H I L + H I K = <lb/>F I L + F I L = 2F I L, puiſque tous les angles droits ſont <lb/>égaux.</s> <s xml:id="echoid-s6139" xml:space="preserve"/> </p> <div xml:id="echoid-div324" type="float" level="2" n="1"> <note position="left" xlink:label="note-0216-03" xlink:href="note-0216-03a" xml:space="preserve">Figure 12.</note> </div> </div> <div xml:id="echoid-div326" type="section" level="1" n="295"> <head xml:id="echoid-head339" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s6140" xml:space="preserve">342. </s> <s xml:id="echoid-s6141" xml:space="preserve">Le cercle eſt une ſurface plane, terminée par une ſeule <lb/> <anchor type="note" xlink:label="note-0216-04a" xlink:href="note-0216-04"/> ligne courbe, qu’on appelle circonférence de cercle, dont tous <lb/>les points ſont également éloignés d’un point A, que l’on ap-<lb/>pelle centre du cercle; </s> <s xml:id="echoid-s6142" xml:space="preserve">les lignes A B, A C, A D menées du <lb/>centre A à la circonférence, ſont appellées rayons du cercle, <lb/>& </s> <s xml:id="echoid-s6143" xml:space="preserve">ſont toutes égales entr’elles, puiſqu’elles meſurent la diſ-<lb/>tance du centre à chaque point de la circonférence, & </s> <s xml:id="echoid-s6144" xml:space="preserve">que <pb o="179" file="0217" n="217" rhead="DE MATHÉMATIQUE. Liv. III."/> cette diſtance eſt partout la même, ſelon la définition du <lb/>cercle.</s> <s xml:id="echoid-s6145" xml:space="preserve"/> </p> <div xml:id="echoid-div326" type="float" level="2" n="1"> <note position="left" xlink:label="note-0216-04" xlink:href="note-0216-04a" xml:space="preserve">Figure 13.</note> </div> </div> <div xml:id="echoid-div328" type="section" level="1" n="296"> <head xml:id="echoid-head340" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s6146" xml:space="preserve">343. </s> <s xml:id="echoid-s6147" xml:space="preserve">Le diametre d’un cercle eſt une ligne droite qui paſſe <lb/> <anchor type="note" xlink:label="note-0217-01a" xlink:href="note-0217-01"/> par le centre, & </s> <s xml:id="echoid-s6148" xml:space="preserve">dont les extrêmités vont aboutir à la circon-<lb/>férence, comme E D: </s> <s xml:id="echoid-s6149" xml:space="preserve">cette ligne diviſe le cercle & </s> <s xml:id="echoid-s6150" xml:space="preserve">ſa circon-<lb/>férence en deux parties égales, que l’on appelle indifféremment <lb/>demi-cercle, & </s> <s xml:id="echoid-s6151" xml:space="preserve">dont la moitié par conſéquent ſe nomme quart <lb/>de cercle.</s> <s xml:id="echoid-s6152" xml:space="preserve"/> </p> <div xml:id="echoid-div328" type="float" level="2" n="1"> <note position="right" xlink:label="note-0217-01" xlink:href="note-0217-01a" xml:space="preserve">Figure 14.</note> </div> </div> <div xml:id="echoid-div330" type="section" level="1" n="297"> <head xml:id="echoid-head341" xml:space="preserve">IX.</head> <p> <s xml:id="echoid-s6153" xml:space="preserve">344. </s> <s xml:id="echoid-s6154" xml:space="preserve">On appelle arc de cercle une partie de la circonférence <lb/>plus petite ou plus grande que la demi-circonférence.</s> <s xml:id="echoid-s6155" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div331" type="section" level="1" n="298"> <head xml:id="echoid-head342" xml:space="preserve">X.</head> <p> <s xml:id="echoid-s6156" xml:space="preserve">345. </s> <s xml:id="echoid-s6157" xml:space="preserve">Les Mathématiciens ont diviſé la circonférence du <lb/>cercle en 360 parties égales, qu’ils ont appellées degrés, & </s> <s xml:id="echoid-s6158" xml:space="preserve">cha-<lb/>que degré en 60 autres parties égales, qu’ils ont appellées mi-<lb/>nutes, dont chacune a été encore diviſée en 60 autres parties <lb/>égales, nommées ſecondes. </s> <s xml:id="echoid-s6159" xml:space="preserve">Ces diviſions ont été imaginées par-<lb/>ticuliérement pour meſurer les angles, & </s> <s xml:id="echoid-s6160" xml:space="preserve">déterminer plus exac-<lb/>tement les rapports qu’ils ont entr’eux. </s> <s xml:id="echoid-s6161" xml:space="preserve">Il ne faut pas s’ima-<lb/>giner que degré ſoit une grandeur fixe & </s> <s xml:id="echoid-s6162" xml:space="preserve">abſolue, mais au <lb/>contraire c’eſt une quantité variable, ſelon les différens cer-<lb/>cles, quoique conſtamment la même, par rapport à chacun en <lb/>particulier, dont chaque degré eſt la 360<emph style="sub">e</emph> partie: </s> <s xml:id="echoid-s6163" xml:space="preserve">d’où il eſt <lb/>aiſé de conclure qu’un grand cercle a des degrés plus grands <lb/>que ceux d’un petit: </s> <s xml:id="echoid-s6164" xml:space="preserve">il en eſt de même des minutes, des ſe-<lb/>condes & </s> <s xml:id="echoid-s6165" xml:space="preserve">des tierces, &</s> <s xml:id="echoid-s6166" xml:space="preserve">c.</s> <s xml:id="echoid-s6167" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div332" type="section" level="1" n="299"> <head xml:id="echoid-head343" xml:space="preserve">XI.</head> <p> <s xml:id="echoid-s6168" xml:space="preserve">346. </s> <s xml:id="echoid-s6169" xml:space="preserve">La meſure d’un angle eſt un arc de cercle décrit à vo-<lb/>lonté de ſa pointe, & </s> <s xml:id="echoid-s6170" xml:space="preserve">terminé par ſes côtés: </s> <s xml:id="echoid-s6171" xml:space="preserve">ainſi l’on con-<lb/>noît que la meſure de l’angle A B C eſt l’arc A C; </s> <s xml:id="echoid-s6172" xml:space="preserve">de ſorte <lb/> <anchor type="note" xlink:label="note-0217-02a" xlink:href="note-0217-02"/> qu’autant l’arc A C contiendra de degrés de minutes, &</s> <s xml:id="echoid-s6173" xml:space="preserve">c, au-<lb/>tant l’angle A B C vaudra de degrés de minutes, &</s> <s xml:id="echoid-s6174" xml:space="preserve">c. </s> <s xml:id="echoid-s6175" xml:space="preserve">Pour <lb/>concevoir comment les arcs de cercles ſont la meſure des an-<lb/>gles, & </s> <s xml:id="echoid-s6176" xml:space="preserve">peuvent ſervir à déterminer leur grandeur, on peut <lb/>imaginer que l’angle C B A a été formé par le mouvement de la <lb/>ligne B C, autour du point B comme d’une charniere, laquelle <lb/>étoit d’abord appliquée ſur la ligne B A: </s> <s xml:id="echoid-s6177" xml:space="preserve">car il eſt évident <pb o="180" file="0218" n="218" rhead="NOUVEAU COURS"/> qu’en prenant ſur cette ligne un point A, & </s> <s xml:id="echoid-s6178" xml:space="preserve">ſur la ligne B C <lb/>un point C, également diſtant du point B, que le point A, <lb/>l’arc A C exprimera la quantité de chemin qu’a parcouru le <lb/>point A pour s’éloigner de la ligne A B. </s> <s xml:id="echoid-s6179" xml:space="preserve">Si cette ligne ſe fût <lb/>éloignée deux fois davantage, l’angle eût été deux fois plus <lb/>grand, ainſi que l’arc qui marque l’eſpace parcouru par le point <lb/>C pour s’éloigner du point A. </s> <s xml:id="echoid-s6180" xml:space="preserve">On peut remarquer que la me-<lb/>ſure d’un angle droit eſt toujours le quart de la circonférence <lb/>d’un cercle, c’eſt-à-dire de 90 degrés: </s> <s xml:id="echoid-s6181" xml:space="preserve">car ſi l’on conſidere les <lb/>deux diametres A B, C D qui ſe coupent à angles droits, on <lb/> <anchor type="note" xlink:label="note-0218-01a" xlink:href="note-0218-01"/> verra qu’ils diviſent la circonférence du cercle en quatre par-<lb/>ties égales, & </s> <s xml:id="echoid-s6182" xml:space="preserve">que chacune eſt la meſure de l’angle droit qui <lb/>lui correſpond: </s> <s xml:id="echoid-s6183" xml:space="preserve">par conſéquent on peut dire encore qu’un <lb/>demi-cercle eſt la meſure de deux angles droits.</s> <s xml:id="echoid-s6184" xml:space="preserve"/> </p> <div xml:id="echoid-div332" type="float" level="2" n="1"> <note position="right" xlink:label="note-0217-02" xlink:href="note-0217-02a" xml:space="preserve">Figure 16.</note> <note position="left" xlink:label="note-0218-01" xlink:href="note-0218-01a" xml:space="preserve">Figure 15.</note> </div> </div> <div xml:id="echoid-div334" type="section" level="1" n="300"> <head xml:id="echoid-head344" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s6185" xml:space="preserve">347. </s> <s xml:id="echoid-s6186" xml:space="preserve">D’un point A donné hors d’une ligne B C ſur le même <lb/> <anchor type="note" xlink:label="note-0218-02a" xlink:href="note-0218-02"/> plan, mener une perpendiculaire A D à cette ligne.</s> <s xml:id="echoid-s6187" xml:space="preserve"/> </p> <div xml:id="echoid-div334" type="float" level="2" n="1"> <note position="left" xlink:label="note-0218-02" xlink:href="note-0218-02a" xml:space="preserve">Figure 17.</note> </div> <p> <s xml:id="echoid-s6188" xml:space="preserve">Pour tirer du point donné A une perpendiculaire ſur la <lb/>ligne B C, décrivez du point A, comme centre, un arc de <lb/>cercle qui vienne couper la ligne donnée dans les points B & </s> <s xml:id="echoid-s6189" xml:space="preserve"><lb/>C; </s> <s xml:id="echoid-s6190" xml:space="preserve">enſuite de ces points & </s> <s xml:id="echoid-s6191" xml:space="preserve">d’une même ouverture de compas, <lb/>moindre que A B, décrivez deux arcs de cercle qui ſe coupe-<lb/>ront en un point E, par lequel & </s> <s xml:id="echoid-s6192" xml:space="preserve">par le point A, faiſant paſſer <lb/>une droite A E D, cette ligne ſera la perpendiculaire demandée. <lb/></s> <s xml:id="echoid-s6193" xml:space="preserve">Pour le prouver, conſidérez que par la conſtruction, les li-<lb/>gnes A B & </s> <s xml:id="echoid-s6194" xml:space="preserve">A C ſont égales, étant rayons d’un même cercle, <lb/>& </s> <s xml:id="echoid-s6195" xml:space="preserve">que les lignes E B & </s> <s xml:id="echoid-s6196" xml:space="preserve">E C le ſont auſſi, par la même raiſon; </s> <s xml:id="echoid-s6197" xml:space="preserve"><lb/>ce qui fait voir que la ligne A D eſt perpendiculaire ſur la <lb/>ligne B C, puiſqu’elle n’eſt pas plus inclinée d’un côté que de <lb/>l’autre.</s> <s xml:id="echoid-s6198" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div336" type="section" level="1" n="301"> <head xml:id="echoid-head345" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s6199" xml:space="preserve">348. </s> <s xml:id="echoid-s6200" xml:space="preserve">D’un point A donné ſur une ligne B C, élever une droite <lb/> <anchor type="note" xlink:label="note-0218-03a" xlink:href="note-0218-03"/> A D perpendiculaire à cette ligne.</s> <s xml:id="echoid-s6201" xml:space="preserve"/> </p> <div xml:id="echoid-div336" type="float" level="2" n="1"> <note position="left" xlink:label="note-0218-03" xlink:href="note-0218-03a" xml:space="preserve">Figure 18.</note> </div> <p> <s xml:id="echoid-s6202" xml:space="preserve">Pour élever une perpendiculaire ſur la ligne B C au point <lb/>donné A, prenez deux points B & </s> <s xml:id="echoid-s6203" xml:space="preserve">C également éloignés de A;</s> <s xml:id="echoid-s6204" xml:space="preserve"> <pb o="181" file="0219" n="219" rhead="DE MATHÉMATIQUE. Liv. III."/> & </s> <s xml:id="echoid-s6205" xml:space="preserve">de ces points comme centre, décrivez avec la même ou-<lb/>verture de compas deux arcs de cercle qui ſe coupent en un <lb/>point comme D; </s> <s xml:id="echoid-s6206" xml:space="preserve">puis tirez du point D au point A la ligne <lb/>D A, elle ſera perpendiculaire ſur B C. </s> <s xml:id="echoid-s6207" xml:space="preserve">Il eſt aiſé d’apperce-<lb/>voir que la ligne A D eſt perpendiculaire ſur B C; </s> <s xml:id="echoid-s6208" xml:space="preserve">car elle a <lb/>par conſtruction deux points A & </s> <s xml:id="echoid-s6209" xml:space="preserve">D, également éloignés de <lb/>deux points B, C, de la ligne B C: </s> <s xml:id="echoid-s6210" xml:space="preserve">donc elle ne penche pas plus <lb/>d’un côté que de l’autre; </s> <s xml:id="echoid-s6211" xml:space="preserve">& </s> <s xml:id="echoid-s6212" xml:space="preserve">par conſéquent elle eſt perpendi-<lb/>culaire ſur B C.</s> <s xml:id="echoid-s6213" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div338" type="section" level="1" n="302"> <head xml:id="echoid-head346" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s6214" xml:space="preserve">349. </s> <s xml:id="echoid-s6215" xml:space="preserve">Diviſer une ligne donnée en deux parties égales.</s> <s xml:id="echoid-s6216" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 19.</note> <p> <s xml:id="echoid-s6217" xml:space="preserve">Pour diviſer une ligne, telle que A B, en deux parties égales, <lb/>décrivez des extrêmités A & </s> <s xml:id="echoid-s6218" xml:space="preserve">B comme centres, avec une <lb/>même ouverture de compas, deux arcs de cercle qui ſe cou-<lb/>pent aux points C & </s> <s xml:id="echoid-s6219" xml:space="preserve">D; </s> <s xml:id="echoid-s6220" xml:space="preserve">tirez par ces deux points la ligne <lb/>C D, qui la coupera en deux également au point E.</s> <s xml:id="echoid-s6221" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6222" xml:space="preserve">Puiſque la ligne C D a deux points C, D, également éloi-<lb/>gnés des extrêmités de la ligne A B, tous ſes points ſeront éga-<lb/>lement éloignés des mêmes extrêmités A & </s> <s xml:id="echoid-s6223" xml:space="preserve">B: </s> <s xml:id="echoid-s6224" xml:space="preserve">donc le point <lb/>E, qui eſt un des points de la ligne C D & </s> <s xml:id="echoid-s6225" xml:space="preserve">de la ligne A B, eſt <lb/>auſſi à égale diſtance de A & </s> <s xml:id="echoid-s6226" xml:space="preserve">de B: </s> <s xml:id="echoid-s6227" xml:space="preserve">donc il eſt le milieu de <lb/>cette ligne. </s> <s xml:id="echoid-s6228" xml:space="preserve">C. </s> <s xml:id="echoid-s6229" xml:space="preserve">Q. </s> <s xml:id="echoid-s6230" xml:space="preserve">F. </s> <s xml:id="echoid-s6231" xml:space="preserve">T.</s> <s xml:id="echoid-s6232" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div339" type="section" level="1" n="303"> <head xml:id="echoid-head347" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6233" xml:space="preserve">350. </s> <s xml:id="echoid-s6234" xml:space="preserve">D’un même point ſur une ligne donnée, on ne peut élever <lb/> <anchor type="note" xlink:label="note-0219-02a" xlink:href="note-0219-02"/> qu’une ſeule perpendiculaire.</s> <s xml:id="echoid-s6235" xml:space="preserve"/> </p> <div xml:id="echoid-div339" type="float" level="2" n="1"> <note position="right" xlink:label="note-0219-02" xlink:href="note-0219-02a" xml:space="preserve">Figure 20.</note> </div> </div> <div xml:id="echoid-div341" type="section" level="1" n="304"> <head xml:id="echoid-head348" xml:space="preserve"><emph style="sc">DÉMONSTRATION</emph>.</head> <p> <s xml:id="echoid-s6236" xml:space="preserve">Si du point C de la ligne A B, on a élevé la ligne C E per-<lb/>pendiculaire à cette ligne, il eſt viſible que ſi on vouloit en <lb/>élever une autre, telle que C D, qui paſſât par le même point <lb/>C, on ne le pourroit faire, ſans que cette ligne ne ſoit plus in-<lb/>clinée d’un côté que d’un autre, comme ici plus vers A que <lb/>vers B; </s> <s xml:id="echoid-s6237" xml:space="preserve">& </s> <s xml:id="echoid-s6238" xml:space="preserve">comme ce ſeroit agir contre la définition des lignes <lb/>perpendiculaires, il s’enſuit qu’on n’en peut élever qu’une d’un <lb/>même point ſur une même ligne. </s> <s xml:id="echoid-s6239" xml:space="preserve">D’ailleurs ſi cette ligne, outre <pb o="182" file="0220" n="220" rhead="NOUVEAU COURS"/> ce point C, a encore un autre point commun avec la perpen-<lb/>diculaire C E, elle ſe confond avec elle, puiſque deux points <lb/>déterminent la poſition d’une ligne droite (art. </s> <s xml:id="echoid-s6240" xml:space="preserve">13): </s> <s xml:id="echoid-s6241" xml:space="preserve">donc <lb/>par un point donné ſur une ligne, on ne peut élever qu’une <lb/>perpendiculaire. </s> <s xml:id="echoid-s6242" xml:space="preserve">C. </s> <s xml:id="echoid-s6243" xml:space="preserve">Q. </s> <s xml:id="echoid-s6244" xml:space="preserve">F. </s> <s xml:id="echoid-s6245" xml:space="preserve">D.</s> <s xml:id="echoid-s6246" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div342" type="section" level="1" n="305"> <head xml:id="echoid-head349" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6247" xml:space="preserve">351. </s> <s xml:id="echoid-s6248" xml:space="preserve">D’un point A donné hors d’une ligne D E, on ne peut <lb/> <anchor type="note" xlink:label="note-0220-01a" xlink:href="note-0220-01"/> abaiſſer qu’une ſeule perpendiculaire A B.</s> <s xml:id="echoid-s6249" xml:space="preserve"/> </p> <div xml:id="echoid-div342" type="float" level="2" n="1"> <note position="left" xlink:label="note-0220-01" xlink:href="note-0220-01a" xml:space="preserve">Figure 21.</note> </div> </div> <div xml:id="echoid-div344" type="section" level="1" n="306"> <head xml:id="echoid-head350" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6250" xml:space="preserve">Si du point A l’on a mené à la ligne D E la perpendicu-<lb/>laire A B, & </s> <s xml:id="echoid-s6251" xml:space="preserve">que les points D, E ſoient également éloignés du <lb/>point A, il eſt certain que le point B, où la perpendiculaire A B <lb/>rencontre la ligne D E, ſera auſſi également éloigné des ex-<lb/>trêmités D, E de la même droite. </s> <s xml:id="echoid-s6252" xml:space="preserve">Mais comme on ne peut tirer <lb/>du point A à la ligne D E aucune ligne, telle que A C, diffé-<lb/>rente de A B, ſans que le point C ne ſoit à droite ou à gauche <lb/>du milieu B, il s’enſuit que les points D, E ne ſeront pas éga-<lb/>lement éloignés du point C; </s> <s xml:id="echoid-s6253" xml:space="preserve">& </s> <s xml:id="echoid-s6254" xml:space="preserve">par conſéquent que la ligne <lb/>A C ne ſera point perpendiculaire ſur D E. </s> <s xml:id="echoid-s6255" xml:space="preserve">C. </s> <s xml:id="echoid-s6256" xml:space="preserve">Q. </s> <s xml:id="echoid-s6257" xml:space="preserve">F. </s> <s xml:id="echoid-s6258" xml:space="preserve">D.</s> <s xml:id="echoid-s6259" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div345" type="section" level="1" n="307"> <head xml:id="echoid-head351" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6260" xml:space="preserve">352. </s> <s xml:id="echoid-s6261" xml:space="preserve">Une ligne perpendiculaire eſt la plus courte de toutes les <lb/> <anchor type="note" xlink:label="note-0220-02a" xlink:href="note-0220-02"/> lignes qu’on peut mener d’un point à une ligne.</s> <s xml:id="echoid-s6262" xml:space="preserve"/> </p> <div xml:id="echoid-div345" type="float" level="2" n="1"> <note position="left" xlink:label="note-0220-02" xlink:href="note-0220-02a" xml:space="preserve">Figure 22.</note> </div> </div> <div xml:id="echoid-div347" type="section" level="1" n="308"> <head xml:id="echoid-head352" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6263" xml:space="preserve">Si l’on a mené du point D la ligne D C perpendiculaire à la <lb/>ligne A B, je dis que cette ligne eſt la plus courte de toutes <lb/>celles que l’on peut mener du point D à la même ligne A B, <lb/>comme la ligne D F.</s> <s xml:id="echoid-s6264" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6265" xml:space="preserve">Pour le prouver, ſoit prolongée la perpendiculaire D C juſ-<lb/>qu’en E, au delà de la ligne A B, par rapport au point D, en-<lb/>ſorte que C E = C D, & </s> <s xml:id="echoid-s6266" xml:space="preserve">ſoit tirée la ligne E F, la ligne D E <lb/>ſera certainement plus courte que la ligne D F E: </s> <s xml:id="echoid-s6267" xml:space="preserve">car, ſelon la <lb/>définition de la ligne droite, elle eſt la plus courte de toutes <lb/>celles que l’on peut mener du point D au point E. </s> <s xml:id="echoid-s6268" xml:space="preserve">D’ailleurs, <pb o="183" file="0221" n="221" rhead="DE MATHÉMATIQUE. Liv. III."/> puiſque la ligne D E eſt perpendiculaire ſur A B, réciproque-<lb/>ment la ligne A B eſt perpendiculaire ſur D E, & </s> <s xml:id="echoid-s6269" xml:space="preserve">par conſtruction <lb/>la coupe en deux également: </s> <s xml:id="echoid-s6270" xml:space="preserve">donc le point F de cette ligne <lb/>eſt également éloigné des extrêmités de la ligne D, E; </s> <s xml:id="echoid-s6271" xml:space="preserve">& </s> <s xml:id="echoid-s6272" xml:space="preserve">par <lb/>conſéquent F D = F E: </s> <s xml:id="echoid-s6273" xml:space="preserve">ainſi prenant les moitié des lignes <lb/>D E, C F E, la droite D C ſera plus courte que la droite D F. <lb/></s> <s xml:id="echoid-s6274" xml:space="preserve">On démontrera la même choſe de toute autre ligne différente <lb/>de D F, priſe à droite ou à gauche de la ligne D C: </s> <s xml:id="echoid-s6275" xml:space="preserve">donc cette <lb/>ligne eſt la plus courte de toutes celles que l’on peut mener du <lb/>point D à la ligne A B.</s> <s xml:id="echoid-s6276" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6277" xml:space="preserve">On pourroit préſentement regarder ce théorême comme <lb/>une définition de la ligne perpendiculaire à une autre, puiſ-<lb/>que cette propriété eſt une des plus importantes, & </s> <s xml:id="echoid-s6278" xml:space="preserve">de laquelle <lb/>on peut déduire les autres.</s> <s xml:id="echoid-s6279" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div348" type="section" level="1" n="309"> <head xml:id="echoid-head353" xml:space="preserve">PROPOSITION VII. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6280" xml:space="preserve">353. </s> <s xml:id="echoid-s6281" xml:space="preserve">Lorſque deux lignes droites ſe coupent, elles forment les <lb/> <anchor type="note" xlink:label="note-0221-01a" xlink:href="note-0221-01"/> angles oppoſés au ſommet qui ſont égaux.</s> <s xml:id="echoid-s6282" xml:space="preserve"/> </p> <div xml:id="echoid-div348" type="float" level="2" n="1"> <note position="right" xlink:label="note-0221-01" xlink:href="note-0221-01a" xml:space="preserve">Figure 24.</note> </div> </div> <div xml:id="echoid-div350" type="section" level="1" n="310"> <head xml:id="echoid-head354" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6283" xml:space="preserve">Soient deux lignes droites quelconques A B, C D, qui ſe <lb/>coupent dans un point E, & </s> <s xml:id="echoid-s6284" xml:space="preserve">forment par leur rencontre ou <lb/>interſection mutuelle, les angles B E D, A E C, que l’on ap-<lb/>pelle oppoſés au ſommet, parce qu’ils ont effectivement leur <lb/>ſommet au même point E, l’un d’un côté, l’autre de l’autre, <lb/>je dis que ces angles ſont égaux. </s> <s xml:id="echoid-s6285" xml:space="preserve">Pour le prouver, du point E <lb/>comme centre, avec un rayon quelconque E B, je décris une <lb/>portion de circonférence qui coupe les lignes A B, C D aux <lb/>points A, C, D, B. </s> <s xml:id="echoid-s6286" xml:space="preserve">Cela poſé, puiſque le centre du cercle <lb/>eſt au point d’interſection des deux lignes, il eſt dans l’une & </s> <s xml:id="echoid-s6287" xml:space="preserve"><lb/>dans l’autre: </s> <s xml:id="echoid-s6288" xml:space="preserve">donc chaque ligne A B, C D eſt à un diametre <lb/>du cercle, & </s> <s xml:id="echoid-s6289" xml:space="preserve">les arcs A D B, D A C ſeront chacuns égaux à <lb/>la demi-circonférence; </s> <s xml:id="echoid-s6290" xml:space="preserve">ce qui donne A D B = D A C, & </s> <s xml:id="echoid-s6291" xml:space="preserve"><lb/>ôtant de part & </s> <s xml:id="echoid-s6292" xml:space="preserve">d’autre l’arc A D commun, on aura l’arc <lb/>D B = A C; </s> <s xml:id="echoid-s6293" xml:space="preserve">mais ces arcs ſont la meſure des angles A F C, <lb/>D E B: </s> <s xml:id="echoid-s6294" xml:space="preserve">donc auſſi les angles oppoſés au ſommet, formés par <lb/>les droites A B, C D, ſont égaux. </s> <s xml:id="echoid-s6295" xml:space="preserve">C. </s> <s xml:id="echoid-s6296" xml:space="preserve">Q. </s> <s xml:id="echoid-s6297" xml:space="preserve">F. </s> <s xml:id="echoid-s6298" xml:space="preserve">D.</s> <s xml:id="echoid-s6299" xml:space="preserve"/> </p> <pb o="184" file="0222" n="222" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div351" type="section" level="1" n="311"> <head xml:id="echoid-head355" xml:space="preserve">PROPOSITION VIII. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6300" xml:space="preserve">354. </s> <s xml:id="echoid-s6301" xml:space="preserve">Lorſque deux lignes droites A B, C D, paralleles en-<lb/> <anchor type="note" xlink:label="note-0222-01a" xlink:href="note-0222-01"/> tr’elles viennent aboutir ſur une troiſieme ligne E F, elles forment <lb/>des angles égaux d’un même côté.</s> <s xml:id="echoid-s6302" xml:space="preserve"/> </p> <div xml:id="echoid-div351" type="float" level="2" n="1"> <note position="left" xlink:label="note-0222-01" xlink:href="note-0222-01a" xml:space="preserve">Figure 25.</note> </div> </div> <div xml:id="echoid-div353" type="section" level="1" n="312"> <head xml:id="echoid-head356" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6303" xml:space="preserve">Pour démontrer que les deux paralleles A B, C D qui vien-<lb/>nent tomber ſur la ligne E F, forment ſur cette ligne d’un <lb/>même côté les angles égaux A B F, C D F, conſidérez que <lb/>l’angle n’étant autre choſe que l’inclinaiſon d’une ligne ſur <lb/>une autre (art. </s> <s xml:id="echoid-s6304" xml:space="preserve">337), l’égalité de ces inclinaiſons fera l’égalité <lb/>des angles, & </s> <s xml:id="echoid-s6305" xml:space="preserve">que les lignes AB, CD ne peuvent être paralleles <lb/>comme on le ſuppoſe, qu’elles ne ſoient également inclinées <lb/>ſur la ligne E F; </s> <s xml:id="echoid-s6306" xml:space="preserve">autrement elles concourroient en quelque <lb/>point: </s> <s xml:id="echoid-s6307" xml:space="preserve">donc l’angle A B F eſt égal à l’angle C D E, puiſque <lb/>la ligne A B eſt autant inclinée ſur E F que la ligne C D. <lb/></s> <s xml:id="echoid-s6308" xml:space="preserve">C. </s> <s xml:id="echoid-s6309" xml:space="preserve">Q. </s> <s xml:id="echoid-s6310" xml:space="preserve">F. </s> <s xml:id="echoid-s6311" xml:space="preserve">D.</s> <s xml:id="echoid-s6312" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div354" type="section" level="1" n="313"> <head xml:id="echoid-head357" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s6313" xml:space="preserve">355. </s> <s xml:id="echoid-s6314" xml:space="preserve">Lorſqu’une droite E F coupe deux paralleles A B, C D, <lb/> <anchor type="note" xlink:label="note-0222-02a" xlink:href="note-0222-02"/> elle forme avec elle des angles auxquels on a donné différens <lb/>noms, ſelon leurs poſitions par rapport à ces mêmes lignes.</s> <s xml:id="echoid-s6315" xml:space="preserve"/> </p> <div xml:id="echoid-div354" type="float" level="2" n="1"> <note position="left" xlink:label="note-0222-02" xlink:href="note-0222-02a" xml:space="preserve">Figure 26.</note> </div> </div> <div xml:id="echoid-div356" type="section" level="1" n="314"> <head xml:id="echoid-head358" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s6316" xml:space="preserve">356. </s> <s xml:id="echoid-s6317" xml:space="preserve">Les angles, tels que B G H, D H G, A G H, C H G, <lb/>ſont appellés angles internes ou intérieurs du même côté.</s> <s xml:id="echoid-s6318" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div357" type="section" level="1" n="315"> <head xml:id="echoid-head359" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s6319" xml:space="preserve">357. </s> <s xml:id="echoid-s6320" xml:space="preserve">Les angles B G E, D H F, ou A G E, C H F ſont ap-<lb/>pellés angles externes ou extérieurs du même côté.</s> <s xml:id="echoid-s6321" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6322" xml:space="preserve">358. </s> <s xml:id="echoid-s6323" xml:space="preserve">Les angles, tels que A G E, D H F, pris, l’un à droite, <lb/>& </s> <s xml:id="echoid-s6324" xml:space="preserve">l’autre à gauche, au dehors des paralleles A B, C D, ſont <lb/>nommés alternes externes, de même que les angles E G B, C H F.</s> <s xml:id="echoid-s6325" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6326" xml:space="preserve">359. </s> <s xml:id="echoid-s6327" xml:space="preserve">Les angles intérieurs, comme A G H, D H G, pris, <lb/>l’un à droite & </s> <s xml:id="echoid-s6328" xml:space="preserve">l’autre à gauche, de la ſécante E F, ſont appellés <lb/>angles alternes internes, ainſi que les angles B G H, C H G.</s> <s xml:id="echoid-s6329" xml:space="preserve"/> </p> <pb o="185" file="0223" n="223" rhead="DE MATHÉMATIQUE. Liv. III."/> </div> <div xml:id="echoid-div358" type="section" level="1" n="316"> <head xml:id="echoid-head360" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6330" xml:space="preserve">360. </s> <s xml:id="echoid-s6331" xml:space="preserve">Si deux lignes droites A B, C D paralleles entr’elles, ſont <lb/> <anchor type="note" xlink:label="note-0223-01a" xlink:href="note-0223-01"/> coupés par une même ligne E F, je dis, 1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6332" xml:space="preserve">que les angles alternes <lb/>internes ou alternes externes ſont égaux; </s> <s xml:id="echoid-s6333" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6334" xml:space="preserve">que les angles internes <lb/>ou externes pris d’un même côté de la ſécante, ſont égaux à deux <lb/>droits.</s> <s xml:id="echoid-s6335" xml:space="preserve"/> </p> <div xml:id="echoid-div358" type="float" level="2" n="1"> <note position="right" xlink:label="note-0223-01" xlink:href="note-0223-01a" xml:space="preserve">Figure 26.</note> </div> </div> <div xml:id="echoid-div360" type="section" level="1" n="317"> <head xml:id="echoid-head361" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6336" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6337" xml:space="preserve">Il faut démontrer que l’angle externe E G B eſt égal à <lb/>ſon alterne C H F. </s> <s xml:id="echoid-s6338" xml:space="preserve">Puiſque les droites A B, C D ſont paral-<lb/>leles, elles ſont également inclinées d’un même côté ſur la ſé-<lb/>cante E F (art. </s> <s xml:id="echoid-s6339" xml:space="preserve">354); </s> <s xml:id="echoid-s6340" xml:space="preserve">ainſi l’on aura l’angle E G B égal à l’an-<lb/>gle G H D, mais G H D eſt égal à l’angle C H F, qui lui eſt <lb/>oppoſé au ſommet (art. </s> <s xml:id="echoid-s6341" xml:space="preserve">353): </s> <s xml:id="echoid-s6342" xml:space="preserve">donc E G B = C H F. </s> <s xml:id="echoid-s6343" xml:space="preserve">On dé-<lb/>montrera de même que l’angle A G E eſt égal à ſon alterne <lb/>D H F; </s> <s xml:id="echoid-s6344" xml:space="preserve">que l’angle interne A G H eſt égal à ſon alterne G H D, <lb/>& </s> <s xml:id="echoid-s6345" xml:space="preserve">que l’angle interne B G H eſt égal à ſon alterne C H E. <lb/></s> <s xml:id="echoid-s6346" xml:space="preserve">C. </s> <s xml:id="echoid-s6347" xml:space="preserve">Q. </s> <s xml:id="echoid-s6348" xml:space="preserve">F. </s> <s xml:id="echoid-s6349" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6350" xml:space="preserve">D.</s> <s xml:id="echoid-s6351" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6352" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6353" xml:space="preserve">Les angles internes B G H, D H G pris d’un même côté <lb/>de la ſécante E F, ou les externes B G E, D H F pris d’un mê-<lb/>me côté, ſont enſemble égaux à deux droits. </s> <s xml:id="echoid-s6354" xml:space="preserve">Puiſque les droites <lb/>A B, C D ſont paralleles, les angles B G E, D H G qu’elles for-<lb/>ment d’un même côté avec la ſécante E F ſont égaux entr’eux, <lb/>ainſi que les angles B G H, D H F; </s> <s xml:id="echoid-s6355" xml:space="preserve">mais (art. </s> <s xml:id="echoid-s6356" xml:space="preserve">341.) </s> <s xml:id="echoid-s6357" xml:space="preserve">B G E <lb/>+ B G H eſt égal à deux droits: </s> <s xml:id="echoid-s6358" xml:space="preserve">donc auſſi D H G + B G H <lb/>eſt égal à deux droits.</s> <s xml:id="echoid-s6359" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6360" xml:space="preserve">On démontrera de même que les angles externes B G E + <lb/>D H F pris enſemble valent deux droits, ou que les angles in-<lb/>ternes A G H + C H G, & </s> <s xml:id="echoid-s6361" xml:space="preserve">les externes du même côté A G E, <lb/>C H F ſont enſemble égaux à deux droits. </s> <s xml:id="echoid-s6362" xml:space="preserve">C. </s> <s xml:id="echoid-s6363" xml:space="preserve">Q. </s> <s xml:id="echoid-s6364" xml:space="preserve">F. </s> <s xml:id="echoid-s6365" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6366" xml:space="preserve">D.</s> <s xml:id="echoid-s6367" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div361" type="section" level="1" n="318"> <head xml:id="echoid-head362" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6368" xml:space="preserve">361. </s> <s xml:id="echoid-s6369" xml:space="preserve">Suppoſant toujours une droite E F qui coupe deux autres <lb/>lignes droites A B, C D, je dis que ces lignes ſeront paralleles, ſi <lb/>les angles alternes internes, ou alternes externes ſont égaux, ou <lb/>bien, ſi les angles internes ou externes d’un même côté valent en-<lb/>ſemble deux droits.</s> <s xml:id="echoid-s6370" xml:space="preserve"/> </p> <pb o="186" file="0224" n="224" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div362" type="section" level="1" n="319"> <head xml:id="echoid-head363" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6371" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6372" xml:space="preserve">Par hypotheſe, l’angle interne D H G eſt égal à ſon al-<lb/>terne A G H, & </s> <s xml:id="echoid-s6373" xml:space="preserve">(art. </s> <s xml:id="echoid-s6374" xml:space="preserve">353.) </s> <s xml:id="echoid-s6375" xml:space="preserve">A G H = B G E qui lui eſt op-<lb/>poſé au ſommet: </s> <s xml:id="echoid-s6376" xml:space="preserve">donc on aura l’angle D H G égal à l’angle <lb/>B G E; </s> <s xml:id="echoid-s6377" xml:space="preserve">ainſi les droites A B, C D ſont parelleles, puiſqu’elles <lb/>forment des angles égaux d’un même côté avec la ſécante E F.</s> <s xml:id="echoid-s6378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6379" xml:space="preserve">On démontrera de même que ces droites ſont paralleles, <lb/>en ſe ſervant des angles alternes internes égaux B G H, C H G, <lb/>ou des angles alternes externes égaux E G B, C H F; </s> <s xml:id="echoid-s6380" xml:space="preserve">A G E, <lb/>D H F. </s> <s xml:id="echoid-s6381" xml:space="preserve">C. </s> <s xml:id="echoid-s6382" xml:space="preserve">Q. </s> <s xml:id="echoid-s6383" xml:space="preserve">F. </s> <s xml:id="echoid-s6384" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6385" xml:space="preserve">D.</s> <s xml:id="echoid-s6386" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6387" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6388" xml:space="preserve">Par hypotheſe, les angles internes D H G, B G H pris <lb/>du même côté de la ſécante E F valent enſemble deux droits, <lb/>& </s> <s xml:id="echoid-s6389" xml:space="preserve">(art. </s> <s xml:id="echoid-s6390" xml:space="preserve">341.) </s> <s xml:id="echoid-s6391" xml:space="preserve">les angles B G H & </s> <s xml:id="echoid-s6392" xml:space="preserve">B G E de ſuite, pris enſem-<lb/>ble, valent auſſi deux droits: </s> <s xml:id="echoid-s6393" xml:space="preserve">donc on aura D H G + B G H <lb/>= B G H + B G E, & </s> <s xml:id="echoid-s6394" xml:space="preserve">ôtant de chaque membre B G H, on <lb/>aura D H G = B G E; </s> <s xml:id="echoid-s6395" xml:space="preserve">ce qui montre que les lignes A B, C D <lb/>font des angles égaux d’un même côté ſur la ſécante E F: </s> <s xml:id="echoid-s6396" xml:space="preserve">donc <lb/>ces mêmes lignes ſont paralleles. </s> <s xml:id="echoid-s6397" xml:space="preserve">C. </s> <s xml:id="echoid-s6398" xml:space="preserve">Q. </s> <s xml:id="echoid-s6399" xml:space="preserve">F. </s> <s xml:id="echoid-s6400" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6401" xml:space="preserve">D.</s> <s xml:id="echoid-s6402" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div363" type="section" level="1" n="320"> <head xml:id="echoid-head364" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s6403" xml:space="preserve">362. </s> <s xml:id="echoid-s6404" xml:space="preserve">Une ligne A B & </s> <s xml:id="echoid-s6405" xml:space="preserve">un point H ſur le même plan étant donnés, <lb/>on propoſe de mener par ce point H une ligne parallele à la ligne A B.</s> <s xml:id="echoid-s6406" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div364" type="section" level="1" n="321"> <head xml:id="echoid-head365" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s6407" xml:space="preserve">Par le point Hon menera une droite quelconque H G, qui <lb/>coupe<unsure/> la droite A B donnée dans un point G; </s> <s xml:id="echoid-s6408" xml:space="preserve">on prendra la <lb/>meſure de l’angle K G H, en décrivant une portion de cercle <lb/>du rayon G H; </s> <s xml:id="echoid-s6409" xml:space="preserve">enſuite du point H comme centre avec le mê-<lb/>me rayon, on décrira un arc de cercle indéfini, ſur lequel on <lb/>prendra l’arc G M égal à l’arc H K, & </s> <s xml:id="echoid-s6410" xml:space="preserve">la ligne H M ſera la <lb/>parallele demandée; </s> <s xml:id="echoid-s6411" xml:space="preserve">car puiſque les arcs de cercles ſont égaux, <lb/>les angles, dont ils ſont la meſure, ſont auſſi égaux, l’angle <lb/>A G H ſera donc égal à ſon alterne G H M: </s> <s xml:id="echoid-s6412" xml:space="preserve">donc par la pro-<lb/>poſition précédente les lignes A B, M H ſont paralleles. <lb/></s> <s xml:id="echoid-s6413" xml:space="preserve">C. </s> <s xml:id="echoid-s6414" xml:space="preserve">Q. </s> <s xml:id="echoid-s6415" xml:space="preserve">F. </s> <s xml:id="echoid-s6416" xml:space="preserve">T. </s> <s xml:id="echoid-s6417" xml:space="preserve">& </s> <s xml:id="echoid-s6418" xml:space="preserve">D.</s> <s xml:id="echoid-s6419" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6420" xml:space="preserve">Il faut remarquer que l’on pourra toujours de la même ma-<lb/>niere faire avec une ligne donnée, un angle égal à un autre angle <lb/>donné.</s> <s xml:id="echoid-s6421" xml:space="preserve"/> </p> <pb o="187" file="0225" n="225" rhead="DE MATHÉMATIQUE. Liv. III."/> </div> <div xml:id="echoid-div365" type="section" level="1" n="322"> <head xml:id="echoid-head366" xml:space="preserve">PROPOSITION XII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s6422" xml:space="preserve">363. </s> <s xml:id="echoid-s6423" xml:space="preserve">Trois points A, D, B étant donnés ſur le même plan, <lb/> <anchor type="note" xlink:label="note-0225-01a" xlink:href="note-0225-01"/> trouver le rayon du cercle qui paſſe par ces trois points.</s> <s xml:id="echoid-s6424" xml:space="preserve"/> </p> <div xml:id="echoid-div365" type="float" level="2" n="1"> <note position="right" xlink:label="note-0225-01" xlink:href="note-0225-01a" xml:space="preserve">Figure 27.</note> </div> </div> <div xml:id="echoid-div367" type="section" level="1" n="323"> <head xml:id="echoid-head367" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s6425" xml:space="preserve">On menera par ces points les droites A B, D B, ſur le mi-<lb/>lieu de la droite A B, on élevera la perpendiculaire indéfinie <lb/>E C; </s> <s xml:id="echoid-s6426" xml:space="preserve">ſur le milieu de B D, on élevera pareillement la droite <lb/>F C perpendiculaire à B D, qui coupera la premiere au point <lb/>C; </s> <s xml:id="echoid-s6427" xml:space="preserve">je dis que ce point ſera le centre du cercle qui paſſe par les <lb/>points A, B, D.</s> <s xml:id="echoid-s6428" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div368" type="section" level="1" n="324"> <head xml:id="echoid-head368" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6429" xml:space="preserve">Le point C, en tant qu’il appartient à la ligne E C perpendi-<lb/>culaire à A B, eſt également éloigné des extrêmités A & </s> <s xml:id="echoid-s6430" xml:space="preserve">B, puiſ-<lb/>que cette ligne diviſe A B en deux également, par conſtruction; <lb/></s> <s xml:id="echoid-s6431" xml:space="preserve">de même en tant qu’il appartient à la droite E F perpendiculaire <lb/>à B D, il eſt auſſi également éloigné des extrêmités B, D de la <lb/>droite B D, par la même raiſon: </s> <s xml:id="echoid-s6432" xml:space="preserve">donc il eſt également éloi-<lb/>gné des trois points A, B, D: </s> <s xml:id="echoid-s6433" xml:space="preserve">donc il eſt le centre du cercle <lb/>qui paſſe par les mêmes points. </s> <s xml:id="echoid-s6434" xml:space="preserve">C. </s> <s xml:id="echoid-s6435" xml:space="preserve">Q. </s> <s xml:id="echoid-s6436" xml:space="preserve">F. </s> <s xml:id="echoid-s6437" xml:space="preserve">T. </s> <s xml:id="echoid-s6438" xml:space="preserve">& </s> <s xml:id="echoid-s6439" xml:space="preserve">D.</s> <s xml:id="echoid-s6440" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div369" type="section" level="1" n="325"> <head xml:id="echoid-head369" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s6441" xml:space="preserve">364. </s> <s xml:id="echoid-s6442" xml:space="preserve">Si les points A, B, D étoient diſpoſés de maniere que <lb/>les perpendiculaires F C, E C ſe trouvaſſent paralleles, le rayon <lb/>du cercle ſeroit inſini; </s> <s xml:id="echoid-s6443" xml:space="preserve">ainſi l’on peut conclure delà qu’un cer-<lb/>cle ne peut pas avoir trois points ſur une ligne droite, à moins <lb/>que la ligne droite ſur laquelle ſe trouvent les trois points ne <lb/>ſoit infiniment petite par rapport au rayon, comme il arrive <lb/>ici, auquel cas cette ligne devient un des côtés du cercle, que <lb/>l’on peut regarder comme un polygone d’une infinité de côtés. <lb/></s> <s xml:id="echoid-s6444" xml:space="preserve">Je dis, que dans notre ſuppoſition les trois points ſont ſur une <lb/>même ligne’droite; </s> <s xml:id="echoid-s6445" xml:space="preserve">car il eſt viſible que les perpendiculaires <lb/>E C, F C ne peuvent être paralleles qu’autant que les droites <lb/>A B, B D formeront une même ligne droite.</s> <s xml:id="echoid-s6446" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div370" type="section" level="1" n="326"> <head xml:id="echoid-head370" style="it" xml:space="preserve">Fin du troiſieme Livre.</head> <pb o="188" file="0226" n="226"/> <figure> <image file="0226-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0226-01"/> </figure> </div> <div xml:id="echoid-div371" type="section" level="1" n="327"> <head xml:id="echoid-head371" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head372" xml:space="preserve">LIVRE QUATRIEME,</head> <head xml:id="echoid-head373" style="it" xml:space="preserve">Qui traite des propriétés des Triangles & des Parallelo-<lb/>grammes.</head> <head xml:id="echoid-head374" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s6447" xml:space="preserve">365. </s> <s xml:id="echoid-s6448" xml:space="preserve"><emph style="sc">Figure</emph> rectiligne eſt une ſurface plane, terminée par <lb/>des lignes droites, appellées côtés; </s> <s xml:id="echoid-s6449" xml:space="preserve">il y a pluſieurs ſortes de <lb/>figures, parmi leſquelles il y en a quelques-unes auxquelles on <lb/>a donné des noms particuliers, ſelon le nombre de leurs côtés, <lb/>& </s> <s xml:id="echoid-s6450" xml:space="preserve">leurs difpoſitions reſpectives les uns à l’égard des autres. <lb/></s> <s xml:id="echoid-s6451" xml:space="preserve">La plus ſimple de toutes les figures eſt celle qui eſt renfermée <lb/>ſous trois côtés, & </s> <s xml:id="echoid-s6452" xml:space="preserve">on l’appelle triangle: </s> <s xml:id="echoid-s6453" xml:space="preserve">on nomme quadri-<lb/>lateres toutes les figures compriſes ſous quatre côtés, & </s> <s xml:id="echoid-s6454" xml:space="preserve">polygones <lb/>en général toutes les figures qui ont plus de quatre côtés.</s> <s xml:id="echoid-s6455" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6456" xml:space="preserve">366. </s> <s xml:id="echoid-s6457" xml:space="preserve">On conſidere le triangle par rapport à ſes côtés, ou par <lb/>rapport à ſes angles. </s> <s xml:id="echoid-s6458" xml:space="preserve">Si le triangle a ſes trois côtés égaux, on <lb/>l’appelle équilatéral, s’il n’a que deux côtés égaux, il eſt appellé <lb/>iſoſcele, & </s> <s xml:id="echoid-s6459" xml:space="preserve">ſcalene, s’il a les trois côtés inégaux; </s> <s xml:id="echoid-s6460" xml:space="preserve">ce qui fait <lb/>trois ſortes de triangles.</s> <s xml:id="echoid-s6461" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6462" xml:space="preserve">Le triangle conſidéré par rapport à ſes angles, eſt encore de <lb/>trois ſortes: </s> <s xml:id="echoid-s6463" xml:space="preserve">on l’appelle rectangle s’il a un angle droit, obtus-<lb/>angle, ou amblygone s’il a un angle obtus, acutangle ou oxygone <lb/>s’il a ſes trois angles aigus ou moindres qu’un droit; </s> <s xml:id="echoid-s6464" xml:space="preserve">d’où il <lb/>ſuit qu’il y a ſix ſortes de triangles en tout.</s> <s xml:id="echoid-s6465" xml:space="preserve"/> </p> <pb o="189" file="0227" n="227" rhead="NOUVEAU COURS DE MATHEM. Liv. IV."/> <p> <s xml:id="echoid-s6466" xml:space="preserve">367. </s> <s xml:id="echoid-s6467" xml:space="preserve">La baſe d’un triangle eſt le côté de ce triangle, ſur <lb/>lequel on a abaiſſé une perpendiculaire de l’angle oppoſé. </s> <s xml:id="echoid-s6468" xml:space="preserve">On <lb/>appelle cette perpendiculaire la hauteur du triangle: </s> <s xml:id="echoid-s6469" xml:space="preserve">ainſi l’on <lb/>voit aiſément, ſuivant ces définitions, que la baſe du trian-<lb/>gle A C B eſt la ligne A B, & </s> <s xml:id="echoid-s6470" xml:space="preserve">que ſa hauteur eſt E D. </s> <s xml:id="echoid-s6471" xml:space="preserve">Si les <lb/> <anchor type="note" xlink:label="note-0227-01a" xlink:href="note-0227-01"/> deux angles fur la baſe ſont aigus, la perpendiculaire tombera <lb/>ſur le côté A B; </s> <s xml:id="echoid-s6472" xml:space="preserve">ſi l’un des angles ſur la même baſe étoit obtus, <lb/>la perpendiculaire ou hauteur du triangle tomberoit ſur le pro-<lb/>longement de la baſe. </s> <s xml:id="echoid-s6473" xml:space="preserve">Comme on peut prendre à volonté dans <lb/>un triangle donné telle ligne que l’on voudra pour baſe de ce <lb/>triangle, il eſt toujours poſſible de faire tomber la perpendi-<lb/>culaire ſur ce côté, que l’on regarde comme baſe; </s> <s xml:id="echoid-s6474" xml:space="preserve">au dedans <lb/>du triangle, les parties dans leſquelles la perpendiculaire C D <lb/>diviſe la baſe A B, ſont appellées ſegmens de cette même baſe. <lb/></s> <s xml:id="echoid-s6475" xml:space="preserve">Dans un triangle rectangle, le côté oppoſé à l’angle droit eſt <lb/>ordinairement regardé comme la baſe de ce triangle, & </s> <s xml:id="echoid-s6476" xml:space="preserve">on <lb/>lui a donnéle nom d’hypothenuſe.</s> <s xml:id="echoid-s6477" xml:space="preserve"/> </p> <div xml:id="echoid-div371" type="float" level="2" n="1"> <note position="right" xlink:label="note-0227-01" xlink:href="note-0227-01a" xml:space="preserve">Figure 28.</note> </div> <p> <s xml:id="echoid-s6478" xml:space="preserve">368. </s> <s xml:id="echoid-s6479" xml:space="preserve">On appelle trapeze un quadrilatere qui n’a aucun de ſes <lb/>côtés paralleles, comme G.</s> <s xml:id="echoid-s6480" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 29.</note> <p> <s xml:id="echoid-s6481" xml:space="preserve">369. </s> <s xml:id="echoid-s6482" xml:space="preserve">Trapezoïde eſt un quadrilatere qui a deux de ſes côtés <lb/>oppoſés paralleles, comme H.</s> <s xml:id="echoid-s6483" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 30.</note> <p> <s xml:id="echoid-s6484" xml:space="preserve">370. </s> <s xml:id="echoid-s6485" xml:space="preserve">Parallelogramme eſt une figure quadrilatere, dont les <lb/>côtés oppoſés ſont égaux & </s> <s xml:id="echoid-s6486" xml:space="preserve">paralleles, comme E F.</s> <s xml:id="echoid-s6487" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 31.</note> <p> <s xml:id="echoid-s6488" xml:space="preserve">371. </s> <s xml:id="echoid-s6489" xml:space="preserve">Diagonale eſt une ligne droite, comme C D, tirée <lb/>dans un parallelogramme ou un rectangle d’un angle quel-<lb/>conque C à celui D qui lui eſt oppoſé.</s> <s xml:id="echoid-s6490" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6491" xml:space="preserve">372. </s> <s xml:id="echoid-s6492" xml:space="preserve">Si par un point quelconque A de la diagonale C D, <lb/>on mene une ligne B A G parallele à E D, & </s> <s xml:id="echoid-s6493" xml:space="preserve">une autre H I <lb/>parallele à D F, l’on aura deux parallelogrammes A E, A F, <lb/>que l’on appellera complémens du parallelogramme E F.</s> <s xml:id="echoid-s6494" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div373" type="section" level="1" n="328"> <head xml:id="echoid-head375" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head376" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6495" xml:space="preserve">373. </s> <s xml:id="echoid-s6496" xml:space="preserve">L’angle extérieur B D C d’un triangle A B D eſt égal aux <lb/> <anchor type="note" xlink:label="note-0227-05a" xlink:href="note-0227-05"/> deux intérieurs oppoſés, & </s> <s xml:id="echoid-s6497" xml:space="preserve">les trois angles du même triangle pris <lb/>enſemble, valent deux droits.</s> <s xml:id="echoid-s6498" xml:space="preserve"/> </p> <div xml:id="echoid-div373" type="float" level="2" n="1"> <note position="right" xlink:label="note-0227-05" xlink:href="note-0227-05a" xml:space="preserve">Figure 33.</note> </div> </div> <div xml:id="echoid-div375" type="section" level="1" n="329"> <head xml:id="echoid-head377" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6499" xml:space="preserve">Pour prouver que l’angle extérieur B D C eſt égal aux deux <pb o="190" file="0228" n="228" rhead="NOUVEAU COURS"/> intérieurs oppoſés, en A & </s> <s xml:id="echoid-s6500" xml:space="preserve">en B: </s> <s xml:id="echoid-s6501" xml:space="preserve">par le point D, ſoit menée <lb/>la droite D E parallele au côté A B du triangle A B D. </s> <s xml:id="echoid-s6502" xml:space="preserve">Cela <lb/>poſé (art. </s> <s xml:id="echoid-s6503" xml:space="preserve">360.) </s> <s xml:id="echoid-s6504" xml:space="preserve">l’angle B D E eſt égal à ſon alterne A B D, <lb/>l’angle E D C eſt égal à l’angle B A D, puiſque les lignes A B, <lb/>D E ſont paralleles entr’elles: </s> <s xml:id="echoid-s6505" xml:space="preserve">donc la ſomme des angles B D E <lb/>& </s> <s xml:id="echoid-s6506" xml:space="preserve">E D C, ou l’angle extérieur B D C eſt égal à la ſomme des <lb/>angles interieurs oppoſés A B D, B D A. </s> <s xml:id="echoid-s6507" xml:space="preserve">C. </s> <s xml:id="echoid-s6508" xml:space="preserve">Q. </s> <s xml:id="echoid-s6509" xml:space="preserve">F. </s> <s xml:id="echoid-s6510" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6511" xml:space="preserve">D.</s> <s xml:id="echoid-s6512" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6513" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6514" xml:space="preserve">Je dis que les trois angles du triangle A B D, pris enſem-<lb/>ble, valent deux droits: </s> <s xml:id="echoid-s6515" xml:space="preserve">car la ligne B D tombant obliquement <lb/>ſur la droite A C, forme deux angles de ſuite B D A, B D C, <lb/>qui pris enſemble, valent deux droits. </s> <s xml:id="echoid-s6516" xml:space="preserve">Mais nous venons de <lb/>voir que l’angle extérieur B D C eſt égal à la ſomme des inté-<lb/>rieurs B A D + A B D; </s> <s xml:id="echoid-s6517" xml:space="preserve">on aura donc en leur ajoutant l’angle <lb/>B D A, B A D + A B D + B D A = B D C + B D A = deux <lb/>droits. </s> <s xml:id="echoid-s6518" xml:space="preserve">C. </s> <s xml:id="echoid-s6519" xml:space="preserve">Q. </s> <s xml:id="echoid-s6520" xml:space="preserve">F. </s> <s xml:id="echoid-s6521" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s6522" xml:space="preserve">D.</s> <s xml:id="echoid-s6523" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div376" type="section" level="1" n="330"> <head xml:id="echoid-head378" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s6524" xml:space="preserve">374. </s> <s xml:id="echoid-s6525" xml:space="preserve">Il ſuit delà que la ſomme des angles d’un polygone <lb/>quelconque vaut toujours autant de fois deux angles droits <lb/>moins quatre, que le polygone a de côtés. </s> <s xml:id="echoid-s6526" xml:space="preserve">Soit le quadrilatere <lb/> <anchor type="note" xlink:label="note-0228-01a" xlink:href="note-0228-01"/> A B C D d’un point G pris au dedans de ce quadrilatere, com-<lb/>me on voudra, ſoient menées les lignes G A, G B, G C, G D <lb/>aux angles A, B, C, D, qui partageront cette figure en quatre <lb/>triangles, il eſt évident que les angles autour du point G, & </s> <s xml:id="echoid-s6527" xml:space="preserve"><lb/>les angles du quadrilatere forment tous les angles des triangles <lb/>dont il eſt compoſé. </s> <s xml:id="echoid-s6528" xml:space="preserve">On aura donc huit angles droits, puiſ-<lb/>que chaque triangle vaut deux droits, mais la ſomme des an-<lb/>gles autour du point G vaut quatre droits: </s> <s xml:id="echoid-s6529" xml:space="preserve">donc les angles du <lb/>polygone valent auſſi quatre droits ou 8 - 4, c’eſt-à-dire au-<lb/>tant de fois deux droits moins quatre que ce polygone a de <lb/>côtés.</s> <s xml:id="echoid-s6530" xml:space="preserve"/> </p> <div xml:id="echoid-div376" type="float" level="2" n="1"> <note position="left" xlink:label="note-0228-01" xlink:href="note-0228-01a" xml:space="preserve">Figure 29.</note> </div> </div> <div xml:id="echoid-div378" type="section" level="1" n="331"> <head xml:id="echoid-head379" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s6531" xml:space="preserve">375. </s> <s xml:id="echoid-s6532" xml:space="preserve">Donc la ſomme des angles extérieurs d’un polygone <lb/>quelconque ne vaut que quatre droits: </s> <s xml:id="echoid-s6533" xml:space="preserve">car tous les angles ex-<lb/>térieurs ſont ſupplémens des angles intérieurs; </s> <s xml:id="echoid-s6534" xml:space="preserve">ainſi la ſomme <lb/>des uns & </s> <s xml:id="echoid-s6535" xml:space="preserve">des autres vaut deux fois autant deux angles droits <lb/>que le polygone a de côtés, & </s> <s xml:id="echoid-s6536" xml:space="preserve">les mêmes angles intérieurs avec <lb/>les angles autour du point G font la même ſomme: </s> <s xml:id="echoid-s6537" xml:space="preserve">donc les <lb/>angles extérieurs ſont égaux à la ſomme des angles autour du <pb o="191" file="0229" n="229" rhead="DE MATHEMATIQUE. Liv. IV."/> point G, c’eſt-à-dire à quatre droits; </s> <s xml:id="echoid-s6538" xml:space="preserve">ce ſeroit la même dé-<lb/>monſtration pour tout autre polygone.</s> <s xml:id="echoid-s6539" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div379" type="section" level="1" n="332"> <head xml:id="echoid-head380" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s6540" xml:space="preserve">376. </s> <s xml:id="echoid-s6541" xml:space="preserve">Il ſuit de cette propoſition, que connoiſſant deux an-<lb/>gles dans un triangle, on pourra connoître le troiſieme, en <lb/>ſouſtrayant la ſomme des deux angles connus de la valeur de <lb/>deux angles droits, & </s> <s xml:id="echoid-s6542" xml:space="preserve">la différence ſera la valeur de l’angle <lb/>inconnu. </s> <s xml:id="echoid-s6543" xml:space="preserve">Ainſi connoiſſant dans le triangle E D F l’angle E <lb/> <anchor type="note" xlink:label="note-0229-01a" xlink:href="note-0229-01"/> de 50 degrés, & </s> <s xml:id="echoid-s6544" xml:space="preserve">l’angle D de 70; </s> <s xml:id="echoid-s6545" xml:space="preserve">pour avoir la valeur de l’an-<lb/>gle F, on ajoutera enſemble 50 & </s> <s xml:id="echoid-s6546" xml:space="preserve">70, qui font 120, qu’il <lb/>faut ſouſtraire de 180 degrés: </s> <s xml:id="echoid-s6547" xml:space="preserve">la différence 60 ſera la valeur <lb/>de l’angle E que l’on cherchoit.</s> <s xml:id="echoid-s6548" xml:space="preserve"/> </p> <div xml:id="echoid-div379" type="float" level="2" n="1"> <note position="right" xlink:label="note-0229-01" xlink:href="note-0229-01a" xml:space="preserve">Figure 32.</note> </div> </div> <div xml:id="echoid-div381" type="section" level="1" n="333"> <head xml:id="echoid-head381" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s6549" xml:space="preserve">377. </s> <s xml:id="echoid-s6550" xml:space="preserve">Il ſuit encore delà, que ſi deux triangles ont deux an-<lb/>gles égaux chacun à chacun, le troiſieme du premier triangle <lb/>ſera égal au troiſieme du ſecond: </s> <s xml:id="echoid-s6551" xml:space="preserve">car ſi l’angle A eſt égal à <lb/>l’angle D, l’angle C à l’angle F, il eſt certain qu’il manquera <lb/>autant de degrés à la ſomme des deux angles A & </s> <s xml:id="echoid-s6552" xml:space="preserve">C pour va-<lb/>loir deux droits, qu’à la ſomme des deux angles D & </s> <s xml:id="echoid-s6553" xml:space="preserve">F pour <lb/>valoir auſſi deux droits, & </s> <s xml:id="echoid-s6554" xml:space="preserve">ces différences égales ne ſont autre <lb/>choſe chacune, que la valeur du troiſieme angle; </s> <s xml:id="echoid-s6555" xml:space="preserve">d’où il ſuit <lb/>que l’angle B ſera égal à l’angle E.</s> <s xml:id="echoid-s6556" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div382" type="section" level="1" n="334"> <head xml:id="echoid-head382" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s6557" xml:space="preserve">378. </s> <s xml:id="echoid-s6558" xml:space="preserve">Deux triangles ſont dits être parfaitement égaux, Iorſ-<lb/>qu’ils ont les trois angles & </s> <s xml:id="echoid-s6559" xml:space="preserve">les trois côtés égaux chacun à <lb/>chacun; </s> <s xml:id="echoid-s6560" xml:space="preserve">& </s> <s xml:id="echoid-s6561" xml:space="preserve">ſimplement égaux, lorſqu’ils ont une égale ſuper-<lb/>ficie compriſe ſous des côtés in égaux.</s> <s xml:id="echoid-s6562" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div383" type="section" level="1" n="335"> <head xml:id="echoid-head383" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head384" xml:space="preserve"><emph style="sc">Theoreme.</emph></head> <p style="it"> <s xml:id="echoid-s6563" xml:space="preserve">379. </s> <s xml:id="echoid-s6564" xml:space="preserve">Deux triangles ſont parfaitement égaux, lorſque les trois <lb/>côtés du premier ſont égaux aux trois côtés du ſecond.</s> <s xml:id="echoid-s6565" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div384" type="section" level="1" n="336"> <head xml:id="echoid-head385" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6566" xml:space="preserve">Pour démontrer que le triangle G, dont on ſuppoſe les côtés <lb/> <anchor type="note" xlink:label="note-0229-02a" xlink:href="note-0229-02"/> A B, B C, A C, égaux aux côtés D E, E F, D F du triangle H, <lb/>eſt entiérement égal à ce dernier triangle, il n’y a qu’à faire voir <pb o="192" file="0230" n="230" rhead="NOUVEAU COURS"/> que l’égalité des côtés emporte néceſſairement l’égalité des <lb/>angles oppoſés aux côtés égaux. </s> <s xml:id="echoid-s6567" xml:space="preserve">Si l’angle D n’eſt pas égal à <lb/>ſon correſpondant A, il ne peut être que plus petit ou plus <lb/>grand: </s> <s xml:id="echoid-s6568" xml:space="preserve">or cela ne peut arriver ſans impliquer contradiction. <lb/></s> <s xml:id="echoid-s6569" xml:space="preserve">Que l’angle D, s’il eſt poſſible, ſoit plus petit que ſon correſ-<lb/>pondant A; </s> <s xml:id="echoid-s6570" xml:space="preserve">ſoit fait l’angle L A C égal à l’angle D, & </s> <s xml:id="echoid-s6571" xml:space="preserve">ſur le <lb/>côté indéfini A L du nouvel angle, ſoit priſe la partie A L = A B <lb/>ou D E, il eſt clair que le côté C L du triangle L A C ſera dans <lb/>ce cas plus petit que le côté C B: </s> <s xml:id="echoid-s6572" xml:space="preserve">car puiſque l’angle eſt plus <lb/>petit, les points C, L, pris à égale diſtance du ſommet A, que <lb/>les points C, B, doivent être plus près l’un de l’autre, que dans <lb/>une plus grande ouverture d’angle, telle que C A B: </s> <s xml:id="echoid-s6573" xml:space="preserve">donc au <lb/>triangle C A L le côté C L ſera plus petit que le côté C B. </s> <s xml:id="echoid-s6574" xml:space="preserve"><lb/>On ne peut donc pas ſuppoſer dans le triangle D E F l’an-<lb/>gle D plus petit que l’angle en A, ſans ſuppoſer en même-<lb/>tems le côté E F plus petit que le côté A B; </s> <s xml:id="echoid-s6575" xml:space="preserve">ce qui eſt contre <lb/>l’hypotheſe: </s> <s xml:id="echoid-s6576" xml:space="preserve">de même on ne pourroit pas ſuppoſer l’angle D <lb/>plus grand que l’angle A ſans une pareille contradiction. </s> <s xml:id="echoid-s6577" xml:space="preserve">L’an-<lb/>gle D eſt donc égal à l’angle A. </s> <s xml:id="echoid-s6578" xml:space="preserve">On fera voir de même que <lb/>l’angle F eſt égal à l’angle C, & </s> <s xml:id="echoid-s6579" xml:space="preserve">l’angle E égal à l’angle B: </s> <s xml:id="echoid-s6580" xml:space="preserve"><lb/>donc ces triangles ſont parfaitement égaux, puiſqu’ils ont, <lb/>outre les côtés égaux, les angles compris entre ces côtés auſſi <lb/>égaux chacun à chacun. </s> <s xml:id="echoid-s6581" xml:space="preserve">C. </s> <s xml:id="echoid-s6582" xml:space="preserve">Q. </s> <s xml:id="echoid-s6583" xml:space="preserve">F. </s> <s xml:id="echoid-s6584" xml:space="preserve">D.</s> <s xml:id="echoid-s6585" xml:space="preserve"/> </p> <div xml:id="echoid-div384" type="float" level="2" n="1"> <note position="right" xlink:label="note-0229-02" xlink:href="note-0229-02a" xml:space="preserve">Figure 34.</note> </div> <p> <s xml:id="echoid-s6586" xml:space="preserve">380. </s> <s xml:id="echoid-s6587" xml:space="preserve">On verra par la ſuite que les trois angles d’un triangle <lb/>peuvent être égaux chacun à chacun aux trois angles d’un au-<lb/>tre triangle, ſans qu’il y ait aucune égalité entre ces deux <lb/>triangles: </s> <s xml:id="echoid-s6588" xml:space="preserve">ainſi de ce que l’égalité des côtés emporte avec <lb/>elle l’égalité des angles, il ne faut pas conclure que l’égalité <lb/>des angles emporte celle des côtés. </s> <s xml:id="echoid-s6589" xml:space="preserve">De plus, il eſt bon d’avertir <lb/>que le triangle eſt le ſeul de toutes les figures qui ait cette pro-<lb/>priété. </s> <s xml:id="echoid-s6590" xml:space="preserve">Par exemple, deux quadrilateres peuvent avoir les côtés <lb/>égaux chacun à chacun, ſans avoir leurs angles égaux ou leurs <lb/>ſuperficies; </s> <s xml:id="echoid-s6591" xml:space="preserve">& </s> <s xml:id="echoid-s6592" xml:space="preserve">par conſéquent ſans être parfaitement égaux.</s> <s xml:id="echoid-s6593" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div386" type="section" level="1" n="337"> <head xml:id="echoid-head386" xml:space="preserve">PROPOSITION III,</head> <head xml:id="echoid-head387" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6594" xml:space="preserve">381. </s> <s xml:id="echoid-s6595" xml:space="preserve">Deux triangles G, H ſont égaux en tout, lorſqu’ils ont <lb/> <anchor type="note" xlink:label="note-0230-01a" xlink:href="note-0230-01"/> un angle égal B, E compris entre deux côtés égaux chacun à <lb/>chacun.</s> <s xml:id="echoid-s6596" xml:space="preserve"/> </p> <div xml:id="echoid-div386" type="float" level="2" n="1"> <note position="right" xlink:label="note-0230-01" xlink:href="note-0230-01a" xml:space="preserve">Figure 34.</note> </div> <pb o="193" file="0231" n="231" rhead="DE MATHÉMATIQUE. Liv. IV."/> </div> <div xml:id="echoid-div388" type="section" level="1" n="338"> <head xml:id="echoid-head388" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6597" xml:space="preserve">Pour démontrer que le triangle G eſt égal au triangle H, <lb/>ſi le côté B A eſt égal au côté D E, le côté B C égal au côté <lb/>E F, & </s> <s xml:id="echoid-s6598" xml:space="preserve">l’angle B égal à l’angle E, imaginons que le côté D E <lb/>eſt appliqué ſur le côté A B: </s> <s xml:id="echoid-s6599" xml:space="preserve">comme ces deux côtés ſont égaux, <lb/>par hypotheſe, en mettant le point E ſur le point B, le point <lb/>D tombera ſur le point A; </s> <s xml:id="echoid-s6600" xml:space="preserve">& </s> <s xml:id="echoid-s6601" xml:space="preserve">parce que l’angle E eſt égal à <lb/>l’angle B, le côté E F tombera ſur le côté B C, & </s> <s xml:id="echoid-s6602" xml:space="preserve">le point F <lb/>ſur le point C, puiſque B C = E F: </s> <s xml:id="echoid-s6603" xml:space="preserve">doncle côté D F tombera <lb/>ſur le côté A C; </s> <s xml:id="echoid-s6604" xml:space="preserve">ce qui montre que les deux triangles con-<lb/>viennent parfaitement: </s> <s xml:id="echoid-s6605" xml:space="preserve">donc ils ſont parfaitement égaux. <lb/></s> <s xml:id="echoid-s6606" xml:space="preserve">C. </s> <s xml:id="echoid-s6607" xml:space="preserve">Q. </s> <s xml:id="echoid-s6608" xml:space="preserve">F. </s> <s xml:id="echoid-s6609" xml:space="preserve">D.</s> <s xml:id="echoid-s6610" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div389" type="section" level="1" n="339"> <head xml:id="echoid-head389" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Theoreme</emph>.</head> <p> <s xml:id="echoid-s6611" xml:space="preserve">382. </s> <s xml:id="echoid-s6612" xml:space="preserve">Deux triangles A B C, D E F ſont parfaitement égaux, <lb/>lorſqu’ils ont un côté A C égal au côté D F, avec les angles en <lb/>A & </s> <s xml:id="echoid-s6613" xml:space="preserve">en C égaux aux angles en D & </s> <s xml:id="echoid-s6614" xml:space="preserve">en F chacun à chacun.</s> <s xml:id="echoid-s6615" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div390" type="section" level="1" n="340"> <head xml:id="echoid-head390" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6616" xml:space="preserve">Si le côté A C du triangle G eſt égal au côté D F du trian-<lb/> <anchor type="note" xlink:label="note-0231-01a" xlink:href="note-0231-01"/> gle H, & </s> <s xml:id="echoid-s6617" xml:space="preserve">que l’angle A ſoit égal à l’angle D, l’angle C à l’an-<lb/>gle F, il eſt aiſé de voir que ces deux triangles ſont parfaite-<lb/>ment égaux: </s> <s xml:id="echoid-s6618" xml:space="preserve">car ſi l’on imagine le côté A C, poſé ſur le côté <lb/>D F, comme ces côtés ſont égaux, par hypotheſe, en mettant <lb/>le point A ſur le point D, le point C tombera ſur le point F; <lb/></s> <s xml:id="echoid-s6619" xml:space="preserve">d’ailleurs à cauſe de l’égalité des angles en A & </s> <s xml:id="echoid-s6620" xml:space="preserve">en C, à ceux <lb/>en D & </s> <s xml:id="echoid-s6621" xml:space="preserve">en F, le côté A B tombera ſur le côté D E, & </s> <s xml:id="echoid-s6622" xml:space="preserve">le côté <lb/>C B ſur le côté F E: </s> <s xml:id="echoid-s6623" xml:space="preserve">donc ces lignes ſe couperont au même <lb/>point E: </s> <s xml:id="echoid-s6624" xml:space="preserve">ainſi les triangles G, H conviendront en tout, & </s> <s xml:id="echoid-s6625" xml:space="preserve">ſe-<lb/>ront parfaitement égaux. </s> <s xml:id="echoid-s6626" xml:space="preserve">C. </s> <s xml:id="echoid-s6627" xml:space="preserve">Q. </s> <s xml:id="echoid-s6628" xml:space="preserve">F. </s> <s xml:id="echoid-s6629" xml:space="preserve">D.</s> <s xml:id="echoid-s6630" xml:space="preserve"/> </p> <div xml:id="echoid-div390" type="float" level="2" n="1"> <note position="right" xlink:label="note-0231-01" xlink:href="note-0231-01a" xml:space="preserve">Figure 34.</note> </div> </div> <div xml:id="echoid-div392" type="section" level="1" n="341"> <head xml:id="echoid-head391" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6631" xml:space="preserve">383. </s> <s xml:id="echoid-s6632" xml:space="preserve">Deux parallelogrammes A B D C, E B D F ſont égaux, <lb/> <anchor type="note" xlink:label="note-0231-02a" xlink:href="note-0231-02"/> lorſqu’ils ont une baſe commune, & </s> <s xml:id="echoid-s6633" xml:space="preserve">ſont compris entre les mêmes <lb/>paralleles.</s> <s xml:id="echoid-s6634" xml:space="preserve"/> </p> <div xml:id="echoid-div392" type="float" level="2" n="1"> <note position="right" xlink:label="note-0231-02" xlink:href="note-0231-02a" xml:space="preserve">Figure 35.</note> </div> </div> <div xml:id="echoid-div394" type="section" level="1" n="342"> <head xml:id="echoid-head392" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6635" xml:space="preserve">Il eſt aiſé de voir que les triangles A B E, C D F ſont égaux <pb o="194" file="0232" n="232" rhead="NOUVEAU COURS"/> en tout: </s> <s xml:id="echoid-s6636" xml:space="preserve">car puiſque A B D C eſt un parallelogramme, le côté <lb/>A B du premier eſt égal au côté C D du ſecond; </s> <s xml:id="echoid-s6637" xml:space="preserve">par la même <lb/>raiſon, puiſque E B D F eſt auſſi un parallelogramme, le côté <lb/>B E du premier triangle eſt égal au côté D F du ſecond: </s> <s xml:id="echoid-s6638" xml:space="preserve">enfin <lb/>le troiſieme côté A E eſt égal au troiſieme côté C F; </s> <s xml:id="echoid-s6639" xml:space="preserve">car A C = <lb/>B D, & </s> <s xml:id="echoid-s6640" xml:space="preserve">B D = E F, puiſque ceſont des côtés oppoſés des paralle-<lb/>logrammes A D, B F: </s> <s xml:id="echoid-s6641" xml:space="preserve">donc A C = E F, & </s> <s xml:id="echoid-s6642" xml:space="preserve">ajoutant à chacun <lb/>la ligne C E, on a A E = C F; </s> <s xml:id="echoid-s6643" xml:space="preserve">d’où il ſuit que ces triangles <lb/>ſont parfaitement égaux (art. </s> <s xml:id="echoid-s6644" xml:space="preserve">378): </s> <s xml:id="echoid-s6645" xml:space="preserve">donc en leur ôtant la <lb/>partie commune C G E, on aura le trapeze A B G C égal au <lb/>trapeze E G D F; </s> <s xml:id="echoid-s6646" xml:space="preserve">& </s> <s xml:id="echoid-s6647" xml:space="preserve">en leur ajoutant à chacun le triangle <lb/>B D G, on aura le parallelogramme A B D C égal au paralle-<lb/>logramme E B D F, compris entre les mêmes paralleles. <lb/></s> <s xml:id="echoid-s6648" xml:space="preserve">C. </s> <s xml:id="echoid-s6649" xml:space="preserve">Q. </s> <s xml:id="echoid-s6650" xml:space="preserve">F. </s> <s xml:id="echoid-s6651" xml:space="preserve">D.</s> <s xml:id="echoid-s6652" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div395" type="section" level="1" n="343"> <head xml:id="echoid-head393" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s6653" xml:space="preserve">384. </s> <s xml:id="echoid-s6654" xml:space="preserve">Il ſuit de la propoſition précédente, que les parallelo-<lb/>grammes qui ont des baſes égales, & </s> <s xml:id="echoid-s6655" xml:space="preserve">qui ſont renfermés entre <lb/>les mêmes paralleles, ſont égaux: </s> <s xml:id="echoid-s6656" xml:space="preserve">car pour prouver que le pa-<lb/>rallelogramme A D eſt égal au parallelogramme G F; </s> <s xml:id="echoid-s6657" xml:space="preserve">ſi les <lb/> <anchor type="note" xlink:label="note-0232-01a" xlink:href="note-0232-01"/> baſes C D & </s> <s xml:id="echoid-s6658" xml:space="preserve">E F ſont égales, il n’y a qu’à tirer les lignes C G <lb/>& </s> <s xml:id="echoid-s6659" xml:space="preserve">D H, qui formeront le parallelogramme C H, & </s> <s xml:id="echoid-s6660" xml:space="preserve">conſi-<lb/>dérer que ce parallelogramme eſt égal au parallelogramme <lb/>A D, parce qu’ils ont la même baſe C D, & </s> <s xml:id="echoid-s6661" xml:space="preserve">qu’il eſt auſſi égal <lb/>au parallelogramme G F, parce qu’ils ont la même baſe G H; <lb/></s> <s xml:id="echoid-s6662" xml:space="preserve">d’ou il ſuit évidemment que les parallelogrammes A D, G F <lb/>ſont égaux, puiſque chacun d’eux eſt égal à un même troi-<lb/>ſieme.</s> <s xml:id="echoid-s6663" xml:space="preserve"/> </p> <div xml:id="echoid-div395" type="float" level="2" n="1"> <note position="left" xlink:label="note-0232-01" xlink:href="note-0232-01a" xml:space="preserve">Figure 36.</note> </div> </div> <div xml:id="echoid-div397" type="section" level="1" n="344"> <head xml:id="echoid-head394" xml:space="preserve">PROPOSITION VI <lb/><emph style="sc">Theoreme</emph>.</head> <p> <s xml:id="echoid-s6664" xml:space="preserve">385. </s> <s xml:id="echoid-s6665" xml:space="preserve">Deux triangles B C D, B F D ſont égaux, lorſqu’ayant <lb/> <anchor type="note" xlink:label="note-0232-02a" xlink:href="note-0232-02"/> une baſe commune B D ils ſont compris entre les mêmes paralleles <lb/>B D, C F.</s> <s xml:id="echoid-s6666" xml:space="preserve"/> </p> <div xml:id="echoid-div397" type="float" level="2" n="1"> <note position="left" xlink:label="note-0232-02" xlink:href="note-0232-02a" xml:space="preserve">Figure 37.</note> </div> </div> <div xml:id="echoid-div399" type="section" level="1" n="345"> <head xml:id="echoid-head395" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6667" xml:space="preserve">Par le point D, ſoit menée la ligne D A parallele au côté <lb/>C B, & </s> <s xml:id="echoid-s6668" xml:space="preserve">la ligne D E parallele au côté B F, on aura deux pa-<lb/>rallelogrammes A B, B E, qui ſeront égaux entr’eux, puiſ-<lb/>qu’ils ont même baſe, & </s> <s xml:id="echoid-s6669" xml:space="preserve">qu’ils ſont compris entre paralleles; <lb/></s> <s xml:id="echoid-s6670" xml:space="preserve">d’ailleurs ces parallelogrammes ſont doubles des triangles <pb o="195" file="0233" n="233" rhead="DE MATHÉMATIQUE. Liv. IV."/> B C D, B F D, puiſque les triangles C A D, D E F ont les côtés <lb/>égaux chacun à chacun à ceux des triangles C B D, D B F: <lb/></s> <s xml:id="echoid-s6671" xml:space="preserve">donc les triangles B C D, B F D ou les moitiés des parallelo-<lb/>grammes A B, B E ſont égaux entr’eux. </s> <s xml:id="echoid-s6672" xml:space="preserve">C. </s> <s xml:id="echoid-s6673" xml:space="preserve">Q. </s> <s xml:id="echoid-s6674" xml:space="preserve">F. </s> <s xml:id="echoid-s6675" xml:space="preserve">D.</s> <s xml:id="echoid-s6676" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div400" type="section" level="1" n="346"> <head xml:id="echoid-head396" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s6677" xml:space="preserve">386. </s> <s xml:id="echoid-s6678" xml:space="preserve">Il ſuit de cette propoſition, que ſi un parallelogramme <lb/> <anchor type="note" xlink:label="note-0233-01a" xlink:href="note-0233-01"/> A D, & </s> <s xml:id="echoid-s6679" xml:space="preserve">un triangle A E C, renfermés entre les mêmes pa-<lb/>ralleles, ont la même baſe A C, le triangle eſt la moitié du pa-<lb/>rallelogramme, parce que le triangle B A C qui lui eſt égal, <lb/>eſt auſſi la moitié du même parallelogramme.</s> <s xml:id="echoid-s6680" xml:space="preserve"/> </p> <div xml:id="echoid-div400" type="float" level="2" n="1"> <note position="right" xlink:label="note-0233-01" xlink:href="note-0233-01a" xml:space="preserve">Figure 38.</note> </div> </div> <div xml:id="echoid-div402" type="section" level="1" n="347"> <head xml:id="echoid-head397" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s6681" xml:space="preserve">387. </s> <s xml:id="echoid-s6682" xml:space="preserve">Comme le triangle B A C eſt égal au triangle A E C, <lb/>il eſt conſtant qu’ayant la même baſe, ils doivent avoir la <lb/>même hauteur; </s> <s xml:id="echoid-s6683" xml:space="preserve">& </s> <s xml:id="echoid-s6684" xml:space="preserve">comme la hauteur du premier eſt la per-<lb/>pendiculaire B A, la hauteur du ſecond ſera auſſi la même per-<lb/>pendiculaire B A, ou ſa parallele E F, abaiſſée de l’angle E ſur <lb/>la baſe A C prolongée; </s> <s xml:id="echoid-s6685" xml:space="preserve">ce qui fait voir que la hauteur d’un <lb/>triangle incliné eſt la perpendiculaire abaiſſée de ſon ſommet <lb/>ſur le prolongement de ſa baſe. </s> <s xml:id="echoid-s6686" xml:space="preserve">Ce ſera la même choſe pour <lb/>les parallelogrammes inclinés.</s> <s xml:id="echoid-s6687" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div403" type="section" level="1" n="348"> <head xml:id="echoid-head398" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s6688" xml:space="preserve">388. </s> <s xml:id="echoid-s6689" xml:space="preserve">Un triangle A B C étant la moitié d’un parallelo-<lb/> <anchor type="note" xlink:label="note-0233-02a" xlink:href="note-0233-02"/> gramme A G, il ſera égal au parallelogramme A D E C, dont <lb/>la hauteur H F eſt ſuppoſée la moitié de la perpendiculaire <lb/>B F, qui ſert de hauteur commune au triangle & </s> <s xml:id="echoid-s6690" xml:space="preserve">au parallelo-<lb/>gramme. </s> <s xml:id="echoid-s6691" xml:space="preserve">Or, comme pour trouver la ſuperficie du paralle-<lb/>logramme A D E C, il faut multiplier la baſe A C par ſa hau-<lb/>teur H F, moitié de la perpendiculaire B F; </s> <s xml:id="echoid-s6692" xml:space="preserve">il s’enſuit qu’en <lb/>multipliant la baſe d’un triangle par la moitié de la perpendicu-<lb/>laire, qui en meſure la hauteur, ou, ce qui revient au même, la <lb/>hauteur entiere par la moitié de la baſe, le produit donnera la ſu-<lb/>perficie du triangle.</s> <s xml:id="echoid-s6693" xml:space="preserve"/> </p> <div xml:id="echoid-div403" type="float" level="2" n="1"> <note position="right" xlink:label="note-0233-02" xlink:href="note-0233-02a" xml:space="preserve">Figure 39.</note> </div> </div> <div xml:id="echoid-div405" type="section" level="1" n="349"> <head xml:id="echoid-head399" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s6694" xml:space="preserve">389. </s> <s xml:id="echoid-s6695" xml:space="preserve">Si l’on conſidere qu’un triangle A B C eſt compoſé <lb/>d’une infinité de lignes paralleles, qui en ſont les élémens, <lb/>& </s> <s xml:id="echoid-s6696" xml:space="preserve">que toutes ces lignes étant également éloignées ſe ſurpaſ-<lb/>ſent de la même quantité, on verra qu’elles compoſent une <pb o="196" file="0234" n="234" rhead="NOUVEAU COURS"/> progreſſion arithmétique d’une quantité infinie de termes, <lb/>dont le premier eſt 0, & </s> <s xml:id="echoid-s6697" xml:space="preserve">dont la ſomme eſt exprimée par la <lb/>perpendiculaire B D. </s> <s xml:id="echoid-s6698" xml:space="preserve">Or comme on trouvera la valeur du <lb/>triangle, ou autrement la ſomme de toutes ces paralleles, en <lb/>multipliant la plus grande, qui eſt la baſe, par la moitié de la <lb/>grandeur qui exprime le nombre des termes, il s’enſuit que l’on <lb/>peut tirer de ce raiſonnement le principe ſuivant: </s> <s xml:id="echoid-s6699" xml:space="preserve">Qui eſt que <lb/>la ſomme des termes des quantités infinies en progreſſion arithmé-<lb/>tique, à commencer par 0, eſt égale au produit du plus grand <lb/>terme, par la moitié de la grandeur qui exprime la quantité de <lb/>ces termes. </s> <s xml:id="echoid-s6700" xml:space="preserve">C’eſt ce que nous avons déja démontré directe-<lb/>ment (art. </s> <s xml:id="echoid-s6701" xml:space="preserve">238).</s> <s xml:id="echoid-s6702" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6703" xml:space="preserve">Il faut s’attacher à bien comprendre ce corollaire, parce que <lb/>nous en ſervirons utilement dans la ſuite.</s> <s xml:id="echoid-s6704" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div406" type="section" level="1" n="350"> <head xml:id="echoid-head400" xml:space="preserve">PROPOSITION VII. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6705" xml:space="preserve">390. </s> <s xml:id="echoid-s6706" xml:space="preserve">Les complémens A E, A F d’un parallelogramme E F ſont <lb/> <anchor type="note" xlink:label="note-0234-01a" xlink:href="note-0234-01"/> égaux entr’eux.</s> <s xml:id="echoid-s6707" xml:space="preserve"/> </p> <div xml:id="echoid-div406" type="float" level="2" n="1"> <note position="left" xlink:label="note-0234-01" xlink:href="note-0234-01a" xml:space="preserve">Figure 31.</note> </div> </div> <div xml:id="echoid-div408" type="section" level="1" n="351"> <head xml:id="echoid-head401" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6708" xml:space="preserve">Pour prouver que les complémens A E & </s> <s xml:id="echoid-s6709" xml:space="preserve">A F du parallelo-<lb/>gramme E F ſont égaux, conſidérez que le parallelogramme <lb/>E F eſt diviſé en deux triangles égaux D E C, D F C, de <lb/>même que les parallelogrammes B I, G H, formés ſur les par-<lb/>ties A D, A C de la diagonale C D: </s> <s xml:id="echoid-s6710" xml:space="preserve">donc ſi l’on retranche du <lb/>triangle D E C les triangles A D H, A B C, & </s> <s xml:id="echoid-s6711" xml:space="preserve">de ſon égal D C F <lb/>les triangles égaux correſpondans A D G, A I C, il reſtera <lb/>d’une part le complément A E égal au complément A F. <lb/></s> <s xml:id="echoid-s6712" xml:space="preserve">C. </s> <s xml:id="echoid-s6713" xml:space="preserve">Q. </s> <s xml:id="echoid-s6714" xml:space="preserve">F. </s> <s xml:id="echoid-s6715" xml:space="preserve">D.</s> <s xml:id="echoid-s6716" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div409" type="section" level="1" n="352"> <head xml:id="echoid-head402" xml:space="preserve">PROPOSITION VIII. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6717" xml:space="preserve">391. </s> <s xml:id="echoid-s6718" xml:space="preserve">Les parallelogrammes, qui ont même hauteur, ſont en-<lb/>tr’eux comme leurs baſes.</s> <s xml:id="echoid-s6719" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div410" type="section" level="1" n="353"> <head xml:id="echoid-head403" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6720" xml:space="preserve">Je dis que ſi les parallelogrammes E F ont même hauteur, <lb/> <anchor type="note" xlink:label="note-0234-02a" xlink:href="note-0234-02"/> ou, ce qui revient au même, ſont compris entre paralleles, <lb/>ils ſeront entr’eux dans la raiſon de leurs baſes. </s> <s xml:id="echoid-s6721" xml:space="preserve">Pour le <pb o="197" file="0235" n="235" rhead="DE MATHÉMATIQUE. Liv. IV."/> prouver, ſoit a la baſe du premier, b celle du ſecond, & </s> <s xml:id="echoid-s6722" xml:space="preserve">c la <lb/>hauteur commune, la ſurface du premier ſera repréſentée par <lb/>a c, & </s> <s xml:id="echoid-s6723" xml:space="preserve">celle du ſecond par b c: </s> <s xml:id="echoid-s6724" xml:space="preserve">or il eſt évident que l’on a a c: <lb/></s> <s xml:id="echoid-s6725" xml:space="preserve">b c :</s> <s xml:id="echoid-s6726" xml:space="preserve">: a : </s> <s xml:id="echoid-s6727" xml:space="preserve">b, puiſque a b c = a b c. </s> <s xml:id="echoid-s6728" xml:space="preserve">C. </s> <s xml:id="echoid-s6729" xml:space="preserve">Q. </s> <s xml:id="echoid-s6730" xml:space="preserve">F. </s> <s xml:id="echoid-s6731" xml:space="preserve">D.</s> <s xml:id="echoid-s6732" xml:space="preserve"/> </p> <div xml:id="echoid-div410" type="float" level="2" n="1"> <note position="left" xlink:label="note-0234-02" xlink:href="note-0234-02a" xml:space="preserve">Figure 41.</note> </div> </div> <div xml:id="echoid-div412" type="section" level="1" n="354"> <head xml:id="echoid-head404" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s6733" xml:space="preserve">392. </s> <s xml:id="echoid-s6734" xml:space="preserve">Il ſuit de cette propoſition, que ſi deux triangles A B C, <lb/> <anchor type="note" xlink:label="note-0235-01a" xlink:href="note-0235-01"/> C D B, ont même hauteur, ou bien leur ſommet au même <lb/>point, ils ſeront entr’eux dans la raiſon de leurs baſes A C <lb/>C D: </s> <s xml:id="echoid-s6735" xml:space="preserve">car ces triangles étant moitié des parallelogrammes cor-<lb/>reſpondans de même baſe & </s> <s xml:id="echoid-s6736" xml:space="preserve">de même hauteur, il en ſera des <lb/>moitiés comme des tous.</s> <s xml:id="echoid-s6737" xml:space="preserve"/> </p> <div xml:id="echoid-div412" type="float" level="2" n="1"> <note position="right" xlink:label="note-0235-01" xlink:href="note-0235-01a" xml:space="preserve">Figure 42.</note> </div> </div> <div xml:id="echoid-div414" type="section" level="1" n="355"> <head xml:id="echoid-head405" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Théoreme</emph>.</head> <p> <s xml:id="echoid-s6738" xml:space="preserve">393. </s> <s xml:id="echoid-s6739" xml:space="preserve">Si l’on coupe les deux côtés A B, A C d’un triangle B A C <lb/> <anchor type="note" xlink:label="note-0235-02a" xlink:href="note-0235-02"/> par une ligne D E, parallele à la baſe B C de ce triangle, je dis <lb/>que les côtés A B, A C ſeront coupés proportionnellement, ou, ce <lb/>ce qui eſt la même choſe, que l’on aura cette proportion A D : </s> <s xml:id="echoid-s6740" xml:space="preserve">D B :</s> <s xml:id="echoid-s6741" xml:space="preserve">: <lb/>A E : </s> <s xml:id="echoid-s6742" xml:space="preserve">E C.</s> <s xml:id="echoid-s6743" xml:space="preserve"/> </p> <div xml:id="echoid-div414" type="float" level="2" n="1"> <note position="right" xlink:label="note-0235-02" xlink:href="note-0235-02a" xml:space="preserve">Figure 43.</note> </div> </div> <div xml:id="echoid-div416" type="section" level="1" n="356"> <head xml:id="echoid-head406" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6744" xml:space="preserve">Pour démontrer cette propoſition, ſoient tirées les lignes <lb/>B E, D C. </s> <s xml:id="echoid-s6745" xml:space="preserve">Cela poſé, il eſt évident que les triangles D B E, <lb/>D C E ſont égaux, puiſqu’ils ont même baſe D E, & </s> <s xml:id="echoid-s6746" xml:space="preserve">qu’ils <lb/>ſont compris entre paralleles. </s> <s xml:id="echoid-s6747" xml:space="preserve">Mais les triangles A D E & </s> <s xml:id="echoid-s6748" xml:space="preserve">D E B <lb/>ayant même ſommet, ſont entr’eux comme leurs baſes (art. </s> <s xml:id="echoid-s6749" xml:space="preserve">392); <lb/></s> <s xml:id="echoid-s6750" xml:space="preserve">ainſi que le même triangle A D E, & </s> <s xml:id="echoid-s6751" xml:space="preserve">le triangle C D E qui ont <lb/>auſſi même ſommet en D.</s> <s xml:id="echoid-s6752" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6753" xml:space="preserve">On aura donc A D : </s> <s xml:id="echoid-s6754" xml:space="preserve">D B :</s> <s xml:id="echoid-s6755" xml:space="preserve">: A D E : </s> <s xml:id="echoid-s6756" xml:space="preserve">D E B; </s> <s xml:id="echoid-s6757" xml:space="preserve">& </s> <s xml:id="echoid-s6758" xml:space="preserve">parce que <lb/>D E B = D C E . </s> <s xml:id="echoid-s6759" xml:space="preserve">. </s> <s xml:id="echoid-s6760" xml:space="preserve">. </s> <s xml:id="echoid-s6761" xml:space="preserve">A D E : </s> <s xml:id="echoid-s6762" xml:space="preserve">D E B :</s> <s xml:id="echoid-s6763" xml:space="preserve">: A D E : </s> <s xml:id="echoid-s6764" xml:space="preserve">D C E :</s> <s xml:id="echoid-s6765" xml:space="preserve">: A E : </s> <s xml:id="echoid-s6766" xml:space="preserve">E C; <lb/></s> <s xml:id="echoid-s6767" xml:space="preserve">& </s> <s xml:id="echoid-s6768" xml:space="preserve">comme la ſuite des rapports égaux n’eſt pas interrompue, on <lb/>en concluera que A D : </s> <s xml:id="echoid-s6769" xml:space="preserve">D B :</s> <s xml:id="echoid-s6770" xml:space="preserve">: A E : </s> <s xml:id="echoid-s6771" xml:space="preserve">E C, c’eſt-à-dire que les côtés <lb/>A B, A C ſont coupés proportionnellement. </s> <s xml:id="echoid-s6772" xml:space="preserve">C. </s> <s xml:id="echoid-s6773" xml:space="preserve">Q. </s> <s xml:id="echoid-s6774" xml:space="preserve">F. </s> <s xml:id="echoid-s6775" xml:space="preserve">D.</s> <s xml:id="echoid-s6776" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div417" type="section" level="1" n="357"> <head xml:id="echoid-head407" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s6777" xml:space="preserve">394. </s> <s xml:id="echoid-s6778" xml:space="preserve">Puiſque A D : </s> <s xml:id="echoid-s6779" xml:space="preserve">D B :</s> <s xml:id="echoid-s6780" xml:space="preserve">: A E : </s> <s xml:id="echoid-s6781" xml:space="preserve">E C, on aura componendo <lb/>A D : </s> <s xml:id="echoid-s6782" xml:space="preserve">A D + D B :</s> <s xml:id="echoid-s6783" xml:space="preserve">: A E : </s> <s xml:id="echoid-s6784" xml:space="preserve">A E + E C, ou en réduiſant A D: <lb/></s> <s xml:id="echoid-s6785" xml:space="preserve">A B :</s> <s xml:id="echoid-s6786" xml:space="preserve">: A E : </s> <s xml:id="echoid-s6787" xml:space="preserve">A C, c’eſt-à-dire que les côtés A B, A C ſont pro-<lb/>portionnels à leurs parties A D, A E.</s> <s xml:id="echoid-s6788" xml:space="preserve"/> </p> <pb o="198" file="0236" n="236" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div418" type="section" level="1" n="358"> <head xml:id="echoid-head408" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s6789" xml:space="preserve">395. </s> <s xml:id="echoid-s6790" xml:space="preserve">Il ſuit delà que deux triangles ſont égaux, lorſqu’ils <lb/>ont un angle égal compris entre côtés réciproques, c’eſt-à-<lb/>dire que les côtés de l’un ſont les extrêmes d’une proportion, <lb/>dont les côtés de l’autre ſont les moyens : </s> <s xml:id="echoid-s6791" xml:space="preserve">car ſi aux triangles <lb/>égaux D B E, D C E, on ajoute le même triangle A D E, on <lb/>aura deux nouveaux triangles égaux en ſuperficie A D C, <lb/>A E B, qui ont un angle en A commun, & </s> <s xml:id="echoid-s6792" xml:space="preserve">par conſéquent <lb/>égal; </s> <s xml:id="echoid-s6793" xml:space="preserve">d’ailleurs, par le corollaire précédent, on a A D : </s> <s xml:id="echoid-s6794" xml:space="preserve">A B :</s> <s xml:id="echoid-s6795" xml:space="preserve">: <lb/>A E : </s> <s xml:id="echoid-s6796" xml:space="preserve">A C, où l’on voit que les côtés A D, A C du triangle A D C <lb/>ſont les extrêmes, tandis que les côtés A B, A E du triangle B A E <lb/>ſont les moyens. </s> <s xml:id="echoid-s6797" xml:space="preserve">Comme les parallelogrammes ſont doubles des <lb/>triangles, il ſuit encore des deux articles précédens, que deux <lb/>parallelogrammes ſont égaux, lorſqu’ils ont un angle égal <lb/>compris entre côtés réciproques.</s> <s xml:id="echoid-s6798" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div419" type="section" level="1" n="359"> <head xml:id="echoid-head409" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s6799" xml:space="preserve">396. </s> <s xml:id="echoid-s6800" xml:space="preserve">Si par le point E on mene la ligne E F parallele au côté <lb/>A B, les côtés A C, C B ſeront auſſi coupés en parties propor-<lb/>tionnelles, & </s> <s xml:id="echoid-s6801" xml:space="preserve">l’on aura A C : </s> <s xml:id="echoid-s6802" xml:space="preserve">C E :</s> <s xml:id="echoid-s6803" xml:space="preserve">: B C : </s> <s xml:id="echoid-s6804" xml:space="preserve">C F, & </s> <s xml:id="echoid-s6805" xml:space="preserve">A E : </s> <s xml:id="echoid-s6806" xml:space="preserve">A C :</s> <s xml:id="echoid-s6807" xml:space="preserve">: <lb/>B F : </s> <s xml:id="echoid-s6808" xml:space="preserve">B C; </s> <s xml:id="echoid-s6809" xml:space="preserve">mais à cauſe des paralleles B D, E F; </s> <s xml:id="echoid-s6810" xml:space="preserve">B F eſt égale <lb/>à D E : </s> <s xml:id="echoid-s6811" xml:space="preserve">on aura donc A E : </s> <s xml:id="echoid-s6812" xml:space="preserve">A C :</s> <s xml:id="echoid-s6813" xml:space="preserve">: D E : </s> <s xml:id="echoid-s6814" xml:space="preserve">B C, c’eſt-à-dire que <lb/>les parties A C, A E ſont proportionnelles au côté B C, & </s> <s xml:id="echoid-s6815" xml:space="preserve">à la <lb/>ſécante D E.</s> <s xml:id="echoid-s6816" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div420" type="section" level="1" n="360"> <head xml:id="echoid-head410" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s6817" xml:space="preserve">397. </s> <s xml:id="echoid-s6818" xml:space="preserve">Deux triangles, ou en général deux figures quelcon-<lb/>ques, ſont dites être ſemblables, lorſque tous les angles de l’une <lb/>ſont égaux aux angles de l’autre, & </s> <s xml:id="echoid-s6819" xml:space="preserve">que les côtés oppoſés aux <lb/>angles égaux ſont proportionnels. </s> <s xml:id="echoid-s6820" xml:space="preserve">Par exemple, les deux trian-<lb/>M N ſeront ſemblables, ſi l’on a l’angle A égal à l’angle D, <lb/> <anchor type="note" xlink:label="note-0236-01a" xlink:href="note-0236-01"/> l’angle C égal à l’angle F, l’angle B égal à l’angle E; </s> <s xml:id="echoid-s6821" xml:space="preserve">& </s> <s xml:id="echoid-s6822" xml:space="preserve">les côtés <lb/>A B, B C, A C proportionnels aux côtés D E, E F, D F.</s> <s xml:id="echoid-s6823" xml:space="preserve"/> </p> <div xml:id="echoid-div420" type="float" level="2" n="1"> <note position="left" xlink:label="note-0236-01" xlink:href="note-0236-01a" xml:space="preserve">Figure 44.</note> </div> </div> <div xml:id="echoid-div422" type="section" level="1" n="361"> <head xml:id="echoid-head411" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s6824" xml:space="preserve">398. </s> <s xml:id="echoid-s6825" xml:space="preserve">Il faut bien remarquer que le triangle eſt le ſeul de <lb/>toutes les figures qui puiſſe être ſemblable à un autre, ayant <lb/>ſes trois angles égaux chacun à chacun, ou ſes côtés propor-<lb/>tionnels; </s> <s xml:id="echoid-s6826" xml:space="preserve">enſorte que l’une de ces conditions emporte l’autre, <lb/>au lieu que dans une figure, tous les côtés peuvent être pro- <pb o="199" file="0237" n="237" rhead="DE MATHÉMATIQUE. Liv. IV."/> portionnels à ceux d’une autre, ſans que les angles oppoſés à <lb/>ces côtés ſoient égaux, comme on le verra par la ſuite.</s> <s xml:id="echoid-s6827" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div423" type="section" level="1" n="362"> <head xml:id="echoid-head412" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6828" xml:space="preserve">399. </s> <s xml:id="echoid-s6829" xml:space="preserve">Deux triangles A B C, D E F ſont ſemblables, lorſque <lb/>les trois côtés A B, B C, A C du premier ſont proportionnels aux <lb/>trois côtés D E, E F, D F du ſecond.</s> <s xml:id="echoid-s6830" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div424" type="section" level="1" n="363"> <head xml:id="echoid-head413" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6831" xml:space="preserve">Pour démontrer cette propoſition, il n’y a qu’à faire voir <lb/>que les angles A, B, C du premier triangles ſont égaux aux an-<lb/>gles D, E, F du ſecond, oppoſés aux côtés proportionnels à <lb/>ceux du triangle A B C : </s> <s xml:id="echoid-s6832" xml:space="preserve">pour cela, ſur le côté A B propor-<lb/>tionnel au côté D E du triangle D E F, ſoit priſe la ligne B G <lb/>égale à D E, & </s> <s xml:id="echoid-s6833" xml:space="preserve">ſoit menée par ce point la parallele G K au côté <lb/>A C, on aura (art. </s> <s xml:id="echoid-s6834" xml:space="preserve">393.) </s> <s xml:id="echoid-s6835" xml:space="preserve">A B: </s> <s xml:id="echoid-s6836" xml:space="preserve">B G :</s> <s xml:id="echoid-s6837" xml:space="preserve">: B C : </s> <s xml:id="echoid-s6838" xml:space="preserve">B K = {BG x BC/AB}= <lb/>{D E x B C/A B}, puiſque par conſtruction D E = B G: </s> <s xml:id="echoid-s6839" xml:space="preserve">mais par hy-<lb/>potheſe, puiſque les trois côtés du premier triangle ſont pro-<lb/>portionnels aux trois côtés du ſecond, A B : </s> <s xml:id="echoid-s6840" xml:space="preserve">D E :</s> <s xml:id="echoid-s6841" xml:space="preserve">: B C : </s> <s xml:id="echoid-s6842" xml:space="preserve">E F <lb/>= {D E x B C/A B}; </s> <s xml:id="echoid-s6843" xml:space="preserve">d’où il ſuit que le triangle B G K a le côté B K <lb/>égal au côté E F du triangle D E F: </s> <s xml:id="echoid-s6844" xml:space="preserve">on démontrera de même, <lb/>que ce même triangle B G K a auſſi le côté G K égal au côté <lb/>D F du triangle D E F: </s> <s xml:id="echoid-s6845" xml:space="preserve">donc ces triangles ſont parfaitement <lb/>égaux, puiſqu’ils ont les trois côtés égaux chacun à chacun <lb/>(art. </s> <s xml:id="echoid-s6846" xml:space="preserve">378) : </s> <s xml:id="echoid-s6847" xml:space="preserve">donc les angles en D & </s> <s xml:id="echoid-s6848" xml:space="preserve">en F ſont égaux aux an-<lb/>gles en G & </s> <s xml:id="echoid-s6849" xml:space="preserve">en K, ou aux angles en A & </s> <s xml:id="echoid-s6850" xml:space="preserve">en C, à cauſe des <lb/>paralleles : </s> <s xml:id="echoid-s6851" xml:space="preserve">donc le triangle D E F eſt ſemblable au triangle <lb/>A B C. </s> <s xml:id="echoid-s6852" xml:space="preserve">C. </s> <s xml:id="echoid-s6853" xml:space="preserve">Q. </s> <s xml:id="echoid-s6854" xml:space="preserve">F. </s> <s xml:id="echoid-s6855" xml:space="preserve">D.</s> <s xml:id="echoid-s6856" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div425" type="section" level="1" n="364"> <head xml:id="echoid-head414" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s6857" xml:space="preserve">400. </s> <s xml:id="echoid-s6858" xml:space="preserve">Réciproquement ſi deux triangles ſont ſemblables, ils <lb/>auront les côtés proportionnels; </s> <s xml:id="echoid-s6859" xml:space="preserve">car s’ils étoient ſemblables <lb/>ſans avoir les côtés proportionnels, la propoſition que nous <lb/>venons de démontrer ſeroit fauſſe; </s> <s xml:id="echoid-s6860" xml:space="preserve">ce qui ne peut arriver.</s> <s xml:id="echoid-s6861" xml:space="preserve"/> </p> <pb o="200" file="0238" n="238" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div426" type="section" level="1" n="365"> <head xml:id="echoid-head415" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6862" xml:space="preserve">401. </s> <s xml:id="echoid-s6863" xml:space="preserve">Deux triangles A B C, D E F ſont ſemblables, lorſqu’ils <lb/>ont un angle égal compris entre côtés proportionnels.</s> <s xml:id="echoid-s6864" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div427" type="section" level="1" n="366"> <head xml:id="echoid-head416" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6865" xml:space="preserve">Suppoſons que l’angle E du triangle D E F eſt égal à l’angle <lb/>B du triangle A B C, & </s> <s xml:id="echoid-s6866" xml:space="preserve">que l’on a A B : </s> <s xml:id="echoid-s6867" xml:space="preserve">B C :</s> <s xml:id="echoid-s6868" xml:space="preserve">: D E : </s> <s xml:id="echoid-s6869" xml:space="preserve">E F, il <lb/>faut démontrer que les angles en A & </s> <s xml:id="echoid-s6870" xml:space="preserve">en C ſeront égaux aux <lb/>angles en D & </s> <s xml:id="echoid-s6871" xml:space="preserve">en F, & </s> <s xml:id="echoid-s6872" xml:space="preserve">que l’on aura A B : </s> <s xml:id="echoid-s6873" xml:space="preserve">A C :</s> <s xml:id="echoid-s6874" xml:space="preserve">: D E : </s> <s xml:id="echoid-s6875" xml:space="preserve">D F. <lb/></s> <s xml:id="echoid-s6876" xml:space="preserve">Soit pris ſur le côté A B la ligne B G égale à D E, & </s> <s xml:id="echoid-s6877" xml:space="preserve">la ligne <lb/>B K égale à E F, à cauſe de l’angle en B, ſuppoſé égal à l’angle <lb/>en E, le triangle B G K ſera parfaitement égal au triangle <lb/>E D F (art. </s> <s xml:id="echoid-s6878" xml:space="preserve">381): </s> <s xml:id="echoid-s6879" xml:space="preserve">donc G K eſt égal à D F, & </s> <s xml:id="echoid-s6880" xml:space="preserve">l’angle D <lb/>eſt égal à l’angle G, de même que l’angle K à l’angle F. </s> <s xml:id="echoid-s6881" xml:space="preserve">De <lb/>plus les côtés B A, B C ſont coupés proportionnellement par <lb/>la ligne G K : </s> <s xml:id="echoid-s6882" xml:space="preserve">donc la ligne G K eſt parallele à la baſe A C, <lb/>& </s> <s xml:id="echoid-s6883" xml:space="preserve">le triangle B G K eſt ſemblable au triangle B A C : </s> <s xml:id="echoid-s6884" xml:space="preserve">donc on <lb/>aura A B : </s> <s xml:id="echoid-s6885" xml:space="preserve">B G :</s> <s xml:id="echoid-s6886" xml:space="preserve">: A C : </s> <s xml:id="echoid-s6887" xml:space="preserve">G K, ou A B : </s> <s xml:id="echoid-s6888" xml:space="preserve">D E :</s> <s xml:id="echoid-s6889" xml:space="preserve">: A C : </s> <s xml:id="echoid-s6890" xml:space="preserve">D F, ou <lb/>alternando A B : </s> <s xml:id="echoid-s6891" xml:space="preserve">A C :</s> <s xml:id="echoid-s6892" xml:space="preserve">: D E : </s> <s xml:id="echoid-s6893" xml:space="preserve">D F. </s> <s xml:id="echoid-s6894" xml:space="preserve">D’où il ſuit que les angles <lb/>du triangle D E F ſont égaux aux angles du triangle A B C; </s> <s xml:id="echoid-s6895" xml:space="preserve"><lb/>d’ailleurs les côtés oppoſés à ces angles ſont proportionnels à <lb/>ceux qui ſont oppoſés aux mêmes angles dans le triangle A B C: </s> <s xml:id="echoid-s6896" xml:space="preserve"><lb/>doncle triangle D E F eſt ſemblable au triangle A B C. </s> <s xml:id="echoid-s6897" xml:space="preserve">C. </s> <s xml:id="echoid-s6898" xml:space="preserve">Q. </s> <s xml:id="echoid-s6899" xml:space="preserve">F. </s> <s xml:id="echoid-s6900" xml:space="preserve">D.</s> <s xml:id="echoid-s6901" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div428" type="section" level="1" n="367"> <head xml:id="echoid-head417" xml:space="preserve">PROPOSITION XII. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6902" xml:space="preserve">402. </s> <s xml:id="echoid-s6903" xml:space="preserve">Deux triangles A B C, D E F ſont ſemblables, lorſque <lb/>deux angles de l’un ſont égaux aux deux angles de l’autre.</s> <s xml:id="echoid-s6904" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div429" type="section" level="1" n="368"> <head xml:id="echoid-head418" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6905" xml:space="preserve">Suppoſons que l’angle A eſt égal à l’angle D, & </s> <s xml:id="echoid-s6906" xml:space="preserve">que l’angle <lb/> <anchor type="note" xlink:label="note-0238-01a" xlink:href="note-0238-01"/> C eſt égal à l’angle F. </s> <s xml:id="echoid-s6907" xml:space="preserve">Sur le côté A C prolongé, on prendra <lb/>une partie C D = D F, & </s> <s xml:id="echoid-s6908" xml:space="preserve">par le point C, on menera la droite <lb/>C E parallele au côté A B, & </s> <s xml:id="echoid-s6909" xml:space="preserve">par le point D, la droite D E <lb/>parallele au côté C B. </s> <s xml:id="echoid-s6910" xml:space="preserve">Le triangle C E D ſera entiérement égal <lb/>au triangle D E F (art. </s> <s xml:id="echoid-s6911" xml:space="preserve">352), puiſque ces triangles ont deux <lb/>angles égaux chacun à chacun ſur un même côté: </s> <s xml:id="echoid-s6912" xml:space="preserve">reſte à faire <pb o="201" file="0239" n="239" rhead="DE MATHEMATIQUE. Liv. IV."/> voir que le triangle C E D eſt ſemblable au triangle A B C. <lb/></s> <s xml:id="echoid-s6913" xml:space="preserve">Pour cela ſoient prolongées les lignes A B, D E, juſqu’à ce <lb/>qu’elles ſe rencontrent en F; </s> <s xml:id="echoid-s6914" xml:space="preserve">les côtés A D, A F ſeront coupés <lb/>proportionnellement par la ligne B C, & </s> <s xml:id="echoid-s6915" xml:space="preserve">l’on aura A B : </s> <s xml:id="echoid-s6916" xml:space="preserve">A C :</s> <s xml:id="echoid-s6917" xml:space="preserve">: <lb/>B F : </s> <s xml:id="echoid-s6918" xml:space="preserve">C D, ou en mettant à la place de B F, C E = B F, à cauſe <lb/>du parallelogramme C F, A B : </s> <s xml:id="echoid-s6919" xml:space="preserve">A C :</s> <s xml:id="echoid-s6920" xml:space="preserve">: C E : </s> <s xml:id="echoid-s6921" xml:space="preserve">C D : </s> <s xml:id="echoid-s6922" xml:space="preserve">donc le trian-<lb/>gle C D E ou ſon égal D E F a les côtés proportionnels à ceux <lb/>du triangle ABC, & </s> <s xml:id="echoid-s6923" xml:space="preserve">lui eſt par conſéquent ſemblable. </s> <s xml:id="echoid-s6924" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s6925" xml:space="preserve">Q. </s> <s xml:id="echoid-s6926" xml:space="preserve">F. </s> <s xml:id="echoid-s6927" xml:space="preserve">D.</s> <s xml:id="echoid-s6928" xml:space="preserve"/> </p> <div xml:id="echoid-div429" type="float" level="2" n="1"> <note position="left" xlink:label="note-0238-01" xlink:href="note-0238-01a" xml:space="preserve">Figure 44 <lb/>& 45.</note> </div> </div> <div xml:id="echoid-div431" type="section" level="1" n="369"> <head xml:id="echoid-head419" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s6929" xml:space="preserve">403. </s> <s xml:id="echoid-s6930" xml:space="preserve">Il ſuit de tout ce que nous venons de voir, que lorſ-<lb/>qu’on aura des triangles ſemblables, on pourra toujours faire <lb/>une proportion par la comparaiſon des côtés du premier aux <lb/>côtés du ſecond. </s> <s xml:id="echoid-s6931" xml:space="preserve">Par exemple, ſi les triangles, M, N ſont <lb/> <anchor type="note" xlink:label="note-0239-01a" xlink:href="note-0239-01"/> ſemblables, & </s> <s xml:id="echoid-s6932" xml:space="preserve">que l’on repréſente les côtés AB, AC du pre-<lb/>mier par a & </s> <s xml:id="echoid-s6933" xml:space="preserve">par b, & </s> <s xml:id="echoid-s6934" xml:space="preserve">les côtés correſpondans du triangle N, <lb/>DE, DF par c & </s> <s xml:id="echoid-s6935" xml:space="preserve">d, on aura a : </s> <s xml:id="echoid-s6936" xml:space="preserve">b :</s> <s xml:id="echoid-s6937" xml:space="preserve">: c : </s> <s xml:id="echoid-s6938" xml:space="preserve">d; </s> <s xml:id="echoid-s6939" xml:space="preserve">donc ad = bc : </s> <s xml:id="echoid-s6940" xml:space="preserve">ce <lb/>qui montre qu’avec deux côtés, pris dans deux triangles ſem-<lb/>blables, & </s> <s xml:id="echoid-s6941" xml:space="preserve">deux autres pris dans les mêmes triangles, on peut <lb/>toujours faire des rectangles égaux, pourvu que les côtés ſoient <lb/>oppoſés à des angles égaux.</s> <s xml:id="echoid-s6942" xml:space="preserve"/> </p> <div xml:id="echoid-div431" type="float" level="2" n="1"> <note position="right" xlink:label="note-0239-01" xlink:href="note-0239-01a" xml:space="preserve">Figure 44.</note> </div> </div> <div xml:id="echoid-div433" type="section" level="1" n="370"> <head xml:id="echoid-head420" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s6943" xml:space="preserve">404. </s> <s xml:id="echoid-s6944" xml:space="preserve">Il ſuit encore que ſi l’on a deux triangles ſemblables, <lb/>dont on connoît deux côtés dans l’un, & </s> <s xml:id="echoid-s6945" xml:space="preserve">un côté dans l’autre, <lb/>qu’on pourra trouver ce ſecond côté : </s> <s xml:id="echoid-s6946" xml:space="preserve">car ſuppoſant, par exem-<lb/>ple, que dans les triangles M, N les côtés a, b ſoient de 12 <lb/>pieds, & </s> <s xml:id="echoid-s6947" xml:space="preserve">8 pieds, & </s> <s xml:id="echoid-s6948" xml:space="preserve">le côté c de 9 pieds, & </s> <s xml:id="echoid-s6949" xml:space="preserve">que l’on veuille <lb/>connoître le côté d, il n’y aura qu’à faire une Regle de Trois, <lb/>& </s> <s xml:id="echoid-s6950" xml:space="preserve">dire 12 : </s> <s xml:id="echoid-s6951" xml:space="preserve">8 :</s> <s xml:id="echoid-s6952" xml:space="preserve">: 9 : </s> <s xml:id="echoid-s6953" xml:space="preserve">x = {9 x 8/12} = 6, qui ſera la valeur du côté <lb/>d, & </s> <s xml:id="echoid-s6954" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s6955" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div434" type="section" level="1" n="371"> <head xml:id="echoid-head421" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s6956" xml:space="preserve">405. </s> <s xml:id="echoid-s6957" xml:space="preserve">On appelle dans des triangles ſemblables, & </s> <s xml:id="echoid-s6958" xml:space="preserve">dans toutes <lb/>les autres figures, côtés homologues ou correſpondans, ceux qui <lb/>ſont oppoſés à des angles égaux dans l’un & </s> <s xml:id="echoid-s6959" xml:space="preserve">dans l’autre trian-<lb/>gles; </s> <s xml:id="echoid-s6960" xml:space="preserve">& </s> <s xml:id="echoid-s6961" xml:space="preserve">l’on ne peut former de proportion qu’avec des côtés <lb/>homologues, ſoit dans les triangles, ſoit dans les autres figures.</s> <s xml:id="echoid-s6962" xml:space="preserve"/> </p> <pb o="202" file="0240" n="240" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div435" type="section" level="1" n="372"> <head xml:id="echoid-head422" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s6963" xml:space="preserve">Les propoſitions précédentes ſont les plus importantes de la <lb/>Géométrie, dont elles font la baſe & </s> <s xml:id="echoid-s6964" xml:space="preserve">le fondement; </s> <s xml:id="echoid-s6965" xml:space="preserve">c’eſt pour-<lb/>quoi il faut s’appliquer à les bien comprendre, ſi l’on veut en-<lb/>tendre les ſuivantes, & </s> <s xml:id="echoid-s6966" xml:space="preserve">faire quelque progrès dans toutes les <lb/>parties des Mathématiques qui ne peuvent ſe paſſer de ces pro-<lb/>poſitions.</s> <s xml:id="echoid-s6967" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div436" type="section" level="1" n="373"> <head xml:id="echoid-head423" xml:space="preserve">PROPOSITION XIII. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6968" xml:space="preserve">406. </s> <s xml:id="echoid-s6969" xml:space="preserve">Si de l’angle droit B d’un triangle rectangle A B C, on <lb/> <anchor type="note" xlink:label="note-0240-01a" xlink:href="note-0240-01"/> abaiſſe une perpendiculaire B D ſur l’hypoténuſe A C, elle divi-<lb/>ſera le même triangle en deux autres triangles A B D, B D C, qui <lb/>lui ſeront ſemblables, & </s> <s xml:id="echoid-s6970" xml:space="preserve">par conſéquent ſemblables entr’eux.</s> <s xml:id="echoid-s6971" xml:space="preserve"/> </p> <div xml:id="echoid-div436" type="float" level="2" n="1"> <note position="left" xlink:label="note-0240-01" xlink:href="note-0240-01a" xml:space="preserve">Figure 46.</note> </div> </div> <div xml:id="echoid-div438" type="section" level="1" n="374"> <head xml:id="echoid-head424" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s6972" xml:space="preserve">Pour démontrer que la perpendiculaire B D diviſe le triangle <lb/>A B C en deux autres ſemblables A B D, B D C; </s> <s xml:id="echoid-s6973" xml:space="preserve">conſidérez que <lb/>chacun de ces triangles a un angle communavec le grand trian-<lb/>gle & </s> <s xml:id="echoid-s6974" xml:space="preserve">un angle droit. </s> <s xml:id="echoid-s6975" xml:space="preserve">L’angle A pour le triangle A B D & </s> <s xml:id="echoid-s6976" xml:space="preserve">le <lb/>triangle A B C, l’angle C au triangle B D C & </s> <s xml:id="echoid-s6977" xml:space="preserve">au triangle <lb/>A B C : </s> <s xml:id="echoid-s6978" xml:space="preserve">donc ils ſont chacun ſemblables au grand triang le, & </s> <s xml:id="echoid-s6979" xml:space="preserve"><lb/>ſemblables entr’eux. </s> <s xml:id="echoid-s6980" xml:space="preserve">C. </s> <s xml:id="echoid-s6981" xml:space="preserve">Q. </s> <s xml:id="echoid-s6982" xml:space="preserve">F. </s> <s xml:id="echoid-s6983" xml:space="preserve">D.</s> <s xml:id="echoid-s6984" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div439" type="section" level="1" n="375"> <head xml:id="echoid-head425" xml:space="preserve">PROPOSITION XIV. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s6985" xml:space="preserve">407. </s> <s xml:id="echoid-s6986" xml:space="preserve">Dans un triangle rectangle A B C, le quarré de l’hypo-<lb/> <anchor type="note" xlink:label="note-0240-02a" xlink:href="note-0240-02"/> ténuſe A C eſt égal à la ſomme des quarrés des deux autres côtés.</s> <s xml:id="echoid-s6987" xml:space="preserve"/> </p> <div xml:id="echoid-div439" type="float" level="2" n="1"> <note position="left" xlink:label="note-0240-02" xlink:href="note-0240-02a" xml:space="preserve">Figure 47.</note> </div> </div> <div xml:id="echoid-div441" type="section" level="1" n="376"> <head xml:id="echoid-head426" xml:space="preserve">DÉMONSTRATION.</head> <p> <s xml:id="echoid-s6988" xml:space="preserve">Soit abaiſſée de l’angle droit la perpendiculaire B D ſur la <lb/>baſe A C, ſoit nommé A C, a, B A, b, B B, c, A D, x; </s> <s xml:id="echoid-s6989" xml:space="preserve">D C ſera <lb/>a - x. </s> <s xml:id="echoid-s6990" xml:space="preserve">Cela poſé, nous ferons voir aiſément que A C<emph style="sub">2</emph> (aa) <lb/>= A B<emph style="sub">2</emph> + B C<emph style="sub">2</emph> (bb + cc).</s> <s xml:id="echoid-s6991" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s6992" xml:space="preserve">Comme la perpendiculaire B D diviſe le triangle rectangle <lb/>en deux autres qui lui ſont ſemblables, A D B, B D C, les <lb/>côtés homologues de ces triangles ſeront proportionnels à <lb/>ceux du grand triangle A B C, & </s> <s xml:id="echoid-s6993" xml:space="preserve">donneront A C (a) : </s> <s xml:id="echoid-s6994" xml:space="preserve">A B (b) :</s> <s xml:id="echoid-s6995" xml:space="preserve">: <lb/>A B (b) : </s> <s xml:id="echoid-s6996" xml:space="preserve">A D (x), & </s> <s xml:id="echoid-s6997" xml:space="preserve">A C (a) : </s> <s xml:id="echoid-s6998" xml:space="preserve">C B (c) :</s> <s xml:id="echoid-s6999" xml:space="preserve">: C B (c) : </s> <s xml:id="echoid-s7000" xml:space="preserve">D C (a-x);</s> <s xml:id="echoid-s7001" xml:space="preserve"> <pb o="203" file="0241" n="241" rhead="DE MATHÉMATIQUE. Liv. IV."/> d’où l’on tire ces équations a x = b b, & </s> <s xml:id="echoid-s7002" xml:space="preserve">c c = a a - a x, en <lb/>prenant les produits des extrêmes & </s> <s xml:id="echoid-s7003" xml:space="preserve">des moyens. </s> <s xml:id="echoid-s7004" xml:space="preserve">En ajoutant <lb/>enſemble ces deux équations, on aura a x + a a - a x = b b <lb/>+ c c, ou en réduiſant a a = b b + c c, ou enfin A C<emph style="sub">2</emph> = A B<emph style="sub">2</emph> <lb/>+ B C<emph style="sub">2</emph>. </s> <s xml:id="echoid-s7005" xml:space="preserve">C. </s> <s xml:id="echoid-s7006" xml:space="preserve">Q. </s> <s xml:id="echoid-s7007" xml:space="preserve">F. </s> <s xml:id="echoid-s7008" xml:space="preserve">D.</s> <s xml:id="echoid-s7009" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7010" xml:space="preserve">Si aſſuré que l’on ſoit d’une propoſition, l’eſprit, ou <lb/>plutôt la raiſon qui veut toujours être éclairée, a encore <lb/>quelque choſe à déſirer, lorſqu’elle ne joint pas la derniere <lb/>évidence à la certitude entiere, & </s> <s xml:id="echoid-s7011" xml:space="preserve">cette évidence eſt d’autant <lb/>plus à déſirer, que les propoſitions ſont plus importantes.</s> <s xml:id="echoid-s7012" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7013" xml:space="preserve">Comme celle-ci eſt une des plus belles propoſitions qu’il y <lb/>ait, tous les grands Géometres ſe ſont appliqués à en donner <lb/>des démonſtrations palpables, parmi leſquelles je regarde la <lb/>ſuivante comme une des plus belles & </s> <s xml:id="echoid-s7014" xml:space="preserve">des plus claires que l’on <lb/>puiſſe donner, attendu qu’elle ne ſuppoſe pas d’autre principe <lb/>que celui-ci, que deux triangles ſont égaux en tout, lorſqu’ils <lb/>ont les trois côtés égaux chacun à chacun.</s> <s xml:id="echoid-s7015" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div442" type="section" level="1" n="377"> <head xml:id="echoid-head427" xml:space="preserve"><emph style="sc">Seconde demonstration</emph>.</head> <p> <s xml:id="echoid-s7016" xml:space="preserve">Soit prolongé le côté A B en K, enſorte que l’on ait B K <lb/>= B C; </s> <s xml:id="echoid-s7017" xml:space="preserve">ſoit de même prolongé le côté B C, enſorte que B L <lb/>= A B. </s> <s xml:id="echoid-s7018" xml:space="preserve">Soient achevés les quarrés ſur les côtés B C, A B, dont <lb/>les côtés I K, H L, prolongés autant qu’il le faut, ſe rencon-<lb/>trent en G: </s> <s xml:id="echoid-s7019" xml:space="preserve">enfin ſoit menée la droite G B, & </s> <s xml:id="echoid-s7020" xml:space="preserve">la perpendicu-<lb/>laire à la baſe B D, & </s> <s xml:id="echoid-s7021" xml:space="preserve">conſtruit le quarré A C E F ſur l’hypo-<lb/>ténuſe A C.</s> <s xml:id="echoid-s7022" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7023" xml:space="preserve">Il eſt aiſé de voir que la droite B G eſt parallele à la droite <lb/>C E: </s> <s xml:id="echoid-s7024" xml:space="preserve">car le triangle G B K eſt égal au triangle A B C, puiſque <lb/>G K = B L = A B, que B K = B C, & </s> <s xml:id="echoid-s7025" xml:space="preserve">quel’angle en K eſt droit: <lb/></s> <s xml:id="echoid-s7026" xml:space="preserve">donc on aura G B = A C = C E: </s> <s xml:id="echoid-s7027" xml:space="preserve">donc l’angle G B K eſt égal <lb/>à l’angle B C A, ou à l’angle A B D du triangle A B D ſem-<lb/>blable au grand triangle, c’eſt-à-dire que l’angle G B K eſt <lb/>égal à ſon oppoſé ou ſommet: </s> <s xml:id="echoid-s7028" xml:space="preserve">donc les lignes G B, B D ne <lb/>font qu’une ſeule ligne droite, & </s> <s xml:id="echoid-s7029" xml:space="preserve">cette ligne G B D eſt pa-<lb/>rallele à C E, puiſque chacune eſt perpendiculaire ſur le côté <lb/>A C. </s> <s xml:id="echoid-s7030" xml:space="preserve">G B C E ſera donc un parallélogramme, ainſi que A B G F, <lb/>puiſque les lignes B C, G I ſont paralleles auſſi-bien que les li-<lb/>gnes B K, G F, & </s> <s xml:id="echoid-s7031" xml:space="preserve">les droites A F, G D, C E. </s> <s xml:id="echoid-s7032" xml:space="preserve">De plus ces pa-<lb/>rallélogrammes ont même baſe que les quarrés B I, B H, & </s> <s xml:id="echoid-s7033" xml:space="preserve"><lb/>ſont compris entre les mêmes paralleles: </s> <s xml:id="echoid-s7034" xml:space="preserve">donc ils leur ſont <pb o="204" file="0242" n="242" rhead="NOUVEAU COURS"/> égaux (art. </s> <s xml:id="echoid-s7035" xml:space="preserve">383): </s> <s xml:id="echoid-s7036" xml:space="preserve">reſte à faire voir que la ſomme de ces pa-<lb/>rallélogrammes ou la figure A B C E G F eſt égale au quarré <lb/>C F fait ſur A E; </s> <s xml:id="echoid-s7037" xml:space="preserve">ce qu’il eſt aiſé de reconnoître: </s> <s xml:id="echoid-s7038" xml:space="preserve">car le côté <lb/>F E = A C, le côté G E = B C, & </s> <s xml:id="echoid-s7039" xml:space="preserve">le côté G F = A B: </s> <s xml:id="echoid-s7040" xml:space="preserve">donc en <lb/>ôtant le triangle FGE de la figure A B C E G F, & </s> <s xml:id="echoid-s7041" xml:space="preserve">mettant à ſa <lb/>place le triangle A B C, ſon égal, on aura la ſomme des paral-<lb/>lélogrammes C G, B F, ou des quarrés B I, B H égale au quarré <lb/>de l’hypoténuſe A C. </s> <s xml:id="echoid-s7042" xml:space="preserve">C. </s> <s xml:id="echoid-s7043" xml:space="preserve">Q. </s> <s xml:id="echoid-s7044" xml:space="preserve">F. </s> <s xml:id="echoid-s7045" xml:space="preserve">D.</s> <s xml:id="echoid-s7046" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div443" type="section" level="1" n="378"> <head xml:id="echoid-head428" xml:space="preserve"><emph style="sc">Troisieme démonstration</emph>.</head> <p> <s xml:id="echoid-s7047" xml:space="preserve">Soit prolongée la perpendiculaire BD, juſqu’à ce qu’elle ren-<lb/>contre en O le côté N M du quarré fait ſur l’hypoténuſe, <lb/>qu’elle diviſera en deux rectangles D M, D N; </s> <s xml:id="echoid-s7048" xml:space="preserve">du point B, <lb/>ſommet de l’angle droit, ſoient menées aux points M, N les <lb/>droites B M, B N, & </s> <s xml:id="echoid-s7049" xml:space="preserve">par les points A, C aux points I, H les <lb/>lignes A I, C H, on aura quatre triangles A C I, BCM; </s> <s xml:id="echoid-s7050" xml:space="preserve">C A H, <lb/>B A N, qui ſeront parfaitement égaux deux à deux: </s> <s xml:id="echoid-s7051" xml:space="preserve">car l’an-<lb/>gle A C I du premier eſt égal à l’angle B C M du ſecond, puiſ-<lb/>que chacun eſt la ſomme d’un angle droit & </s> <s xml:id="echoid-s7052" xml:space="preserve">de l’angle com-<lb/>mun B C D. </s> <s xml:id="echoid-s7053" xml:space="preserve">De plus, le côté C I du premier eſt égal au côté <lb/>B C du ſecond, puiſque ce ſont les côtés d’un même quarré, <lb/>& </s> <s xml:id="echoid-s7054" xml:space="preserve">le côté A C du triangle A C I eſt, par la même raiſon, égal <lb/>au côté C M: </s> <s xml:id="echoid-s7055" xml:space="preserve">donc ces triangles ſont parfaitement égaux, puiſ-<lb/>qu’ils ont un angle égal, compris entre côtés égaux chacun à <lb/>chacun (art. </s> <s xml:id="echoid-s7056" xml:space="preserve">381): </s> <s xml:id="echoid-s7057" xml:space="preserve">donc les rectangles A D N O, D C M O, <lb/>dont ces triangles ſont les moitiés, ſeront auſſi égaux. </s> <s xml:id="echoid-s7058" xml:space="preserve">Or il <lb/>eſt viſible que le triangle A C I eſt moitié du quarré fait ſur <lb/>B C, puiſqu’ils ont même baſe C I, & </s> <s xml:id="echoid-s7059" xml:space="preserve">qu’ils ſont compris en-<lb/>tre paralleles A K, C I. </s> <s xml:id="echoid-s7060" xml:space="preserve">Il eſt encore évident que le triangle <lb/>B C M eſt moitié du rectangle D M, puiſqu’ils ont même baſe <lb/>B M, & </s> <s xml:id="echoid-s7061" xml:space="preserve">ſont compris entre les mêmes paralleles B M, B D O: <lb/></s> <s xml:id="echoid-s7062" xml:space="preserve">donc le quarré fait ſur B C eſt égal au rectangle D M. </s> <s xml:id="echoid-s7063" xml:space="preserve">On dé-<lb/>montrera préciſément de la même maniere que le quarré fait <lb/>ſur A B eſt égal au rectangle D N; </s> <s xml:id="echoid-s7064" xml:space="preserve">mais la ſomme des rectan-<lb/>gles D M, D N eſt égal au quarré conſtruit ſur l’hypoténuſe: </s> <s xml:id="echoid-s7065" xml:space="preserve"><lb/>donc la ſomme des quarrés faits ſur les deux côtés A B, B C eſt <lb/>égale au quarré de l’hypoténuſe A C. </s> <s xml:id="echoid-s7066" xml:space="preserve">C. </s> <s xml:id="echoid-s7067" xml:space="preserve">Q. </s> <s xml:id="echoid-s7068" xml:space="preserve">F. </s> <s xml:id="echoid-s7069" xml:space="preserve">D.</s> <s xml:id="echoid-s7070" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div444" type="section" level="1" n="379"> <head xml:id="echoid-head429" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s7071" xml:space="preserve">408. </s> <s xml:id="echoid-s7072" xml:space="preserve">Cette propoſition eſt la fameuſe 47<emph style="sub">e</emph> du premier Livre <pb o="205" file="0243" n="243" rhead="DE MATHÉMATIQUE. Liv. IV."/> d’Euclide, pour la découverte de laquelle Pythagore offrit aux <lb/>Muſes un ſacrifice de cent bœufs, en reconnoiſſance de la fa-<lb/>veur qu’il croyoit avoir reçu d’elles. </s> <s xml:id="echoid-s7073" xml:space="preserve">Pour être prévenu de <lb/>l’uſage que nous en ferons dans la ſuite, il faut remarquer que <lb/>connoiſſant les quarrés de deux côtés d’un triangle rectangle, <lb/>on pourra toujours connoître celui du troiſieme: </s> <s xml:id="echoid-s7074" xml:space="preserve">car ſi l’on <lb/>connoît A C<emph style="sub">2</emph> (aa), & </s> <s xml:id="echoid-s7075" xml:space="preserve">A B<emph style="sub">2</emph>(bb), on voit qu’on aura toujours <lb/>A C<emph style="sub">2</emph> - A B<emph style="sub">2</emph> (a a - b b) = B C<emph style="sub">2</emph>(cc), qui donne la valeur <lb/>du côté B C: </s> <s xml:id="echoid-s7076" xml:space="preserve">on voit de plus, que connoiſſant les deux côtés <lb/>qui comprennent l’angle droit d’un triangle rectangle, on <lb/>pourra connoître l’hypoténuſe, en quarrant ces deux côtés, <lb/>& </s> <s xml:id="echoid-s7077" xml:space="preserve">extrayant la racine des deux membres de l’équation aa=bb <lb/>+ cc, on aura a = √bb + cc\x{0020}; </s> <s xml:id="echoid-s7078" xml:space="preserve">& </s> <s xml:id="echoid-s7079" xml:space="preserve">ſi connoiſſant l’hypoté-<lb/>nuſe avec un autre côté, on vouloit trouver le troiſieme, on <lb/>n’auroit qu’à ſouſtraire du quarré de l’hypoténuſe le quarré <lb/>du ſecond côté que l’on connoît; </s> <s xml:id="echoid-s7080" xml:space="preserve">& </s> <s xml:id="echoid-s7081" xml:space="preserve">la racine quarrée de la <lb/>différence, donnera la valeur du côté qu’on cherche. </s> <s xml:id="echoid-s7082" xml:space="preserve">Ainſi <lb/>connoiſſant les deux côtés B C & </s> <s xml:id="echoid-s7083" xml:space="preserve">A C, on voit que A B = <lb/>√aa - cc\x{0020}.</s> <s xml:id="echoid-s7084" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div445" type="section" level="1" n="380"> <head xml:id="echoid-head430" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s7085" xml:space="preserve">409. </s> <s xml:id="echoid-s7086" xml:space="preserve">Il ſuit de cette propoſition, que la perpendiculaire tirée <lb/>de l’angle droit d’un triangle rectangle ſur l’hypoténuſe, eſt <lb/>moyenne proportionnelle entre les parties de l’hypothenuſe: <lb/></s> <s xml:id="echoid-s7087" xml:space="preserve">car comme cette perpendiculaire diviſe le triangle A B C en <lb/>deux autres triangles ſemblables A D B, B D C, en les compa-<lb/>rant enſemble, on aura A D : </s> <s xml:id="echoid-s7088" xml:space="preserve">D B :</s> <s xml:id="echoid-s7089" xml:space="preserve">: D B : </s> <s xml:id="echoid-s7090" xml:space="preserve">D C. </s> <s xml:id="echoid-s7091" xml:space="preserve">Ainſi connoiſ-<lb/>ſant la baſe d’un triangle rectangle A B C, & </s> <s xml:id="echoid-s7092" xml:space="preserve">les deux ſegmens <lb/>A D, D C de cette baſe, on pourra connoître les autres côtés <lb/>de ce triangle; </s> <s xml:id="echoid-s7093" xml:space="preserve">il n’y aura qu’à chercher une moyenne propor-<lb/>tionnelle entre les ſegmens donnés, ajouter le quarré de cette <lb/>ligne au quarré de chaque ſegment, & </s> <s xml:id="echoid-s7094" xml:space="preserve">extraire les racines des <lb/>deux ſommes qui ſeront les deux côtés demandés.</s> <s xml:id="echoid-s7095" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div446" type="section" level="1" n="381"> <head xml:id="echoid-head431" xml:space="preserve">PROPOSITION XV.</head> <head xml:id="echoid-head432" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7096" xml:space="preserve">410. </s> <s xml:id="echoid-s7097" xml:space="preserve">Dans un triangle obtus-angle A B C, le quarré du côté <lb/> <anchor type="note" xlink:label="note-0243-01a" xlink:href="note-0243-01"/> A C, oppoſé à l’angle obtus, eſt égal au quarré des deux auires <lb/>côtés, plus à deux rectangles compris ſous le côté B C prolongé juſ- <pb o="206" file="0244" n="244" rhead="NOUVEAU COURS"/> qu’à la rencontre de la perpendiculaire abaiſſée de l’angle A, & </s> <s xml:id="echoid-s7098" xml:space="preserve">la <lb/>partie B D ou le prolongement du même côté B D; </s> <s xml:id="echoid-s7099" xml:space="preserve">c’eſt-à-dire que <lb/>l’on aura A C<emph style="sub">2</emph> = A B<emph style="sub">2</emph> + B C<emph style="sub">2</emph> + <emph style="super">2</emph>B C x B D.</s> <s xml:id="echoid-s7100" xml:space="preserve"/> </p> <div xml:id="echoid-div446" type="float" level="2" n="1"> <note position="right" xlink:label="note-0243-01" xlink:href="note-0243-01a" xml:space="preserve">Figure 48.</note> </div> </div> <div xml:id="echoid-div448" type="section" level="1" n="382"> <head xml:id="echoid-head433" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s7101" xml:space="preserve">Soit fait A C = a, A B = c, B C = b, B D = x, A D = d; <lb/></s> <s xml:id="echoid-s7102" xml:space="preserve">il faut démontrer que aa = cc+bb + 2bx, ou que A C<emph style="sub">2</emph> = A B<emph style="sub">2</emph> <lb/>+ B C<emph style="sub">2</emph> + <emph style="super">2</emph>B C x B D. </s> <s xml:id="echoid-s7103" xml:space="preserve">Le triangle rectangle A D C donne <lb/>A C<emph style="sub">2</emph> = A D<emph style="sub">2</emph> + D C<emph style="sub">2</emph>, ou aa = dd + bb + 2bx + xx: </s> <s xml:id="echoid-s7104" xml:space="preserve">car <lb/>D C = D B + B C = b + x; </s> <s xml:id="echoid-s7105" xml:space="preserve">& </s> <s xml:id="echoid-s7106" xml:space="preserve">le triangle rectangle A D B <lb/>donne A B<emph style="sub">2</emph> = A D<emph style="sub">2</emph> + D B<emph style="sub">2</emph>, ou cc = dd + xx; </s> <s xml:id="echoid-s7107" xml:space="preserve">ſi l’on retran-<lb/>che les deux membres de cette équation des deux membres de <lb/>la premiere, on aura aa - cc = dd + bb + 2bx + xx - dd <lb/>- xx, & </s> <s xml:id="echoid-s7108" xml:space="preserve">réduiſant le dernier membre aa - cc = bb + 2bx, <lb/>& </s> <s xml:id="echoid-s7109" xml:space="preserve">faiſant paſſer cc de l’autre côté du ſigne d’égalité, aa = bb <lb/>+ cc + 2bx, ou A C<emph style="sub">2</emph> = A B<emph style="sub">2</emph> + B C<emph style="sub">2</emph> + <emph style="super">2</emph>B C x B D. </s> <s xml:id="echoid-s7110" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s7111" xml:space="preserve">Q. </s> <s xml:id="echoid-s7112" xml:space="preserve">F. </s> <s xml:id="echoid-s7113" xml:space="preserve">D.</s> <s xml:id="echoid-s7114" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div449" type="section" level="1" n="383"> <head xml:id="echoid-head434" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7115" xml:space="preserve">411. </s> <s xml:id="echoid-s7116" xml:space="preserve">Si l’on avoit un triangle A B C, dont on connût les <lb/>trois côtés, on pourroit par cette propoſition trouver la per-<lb/>pendiculaire A D, qui détermine la hauteur du triangle: </s> <s xml:id="echoid-s7117" xml:space="preserve">car <lb/>comme on a aa = cc + bb + 2bx, ſi l’on fait paſſer cc + bb <lb/>du ſecond membre dans le premier, il viendra aa - cc - bb <lb/>= 2bx, qui étant diviſé par 2b, donne la valeur de x, ou <lb/>{aa - cc - bb/2b} = x, qui fait voir qu’on trouvera la valeur de la <lb/>ligne D B, en ſouſtrayant du quarré du côté A C oppoſé à l’an-<lb/>gle obtus; </s> <s xml:id="echoid-s7118" xml:space="preserve">les quarrés des côtés A B & </s> <s xml:id="echoid-s7119" xml:space="preserve">B C, & </s> <s xml:id="echoid-s7120" xml:space="preserve">en diviſant le <lb/>reſte par le double du côté B C. </s> <s xml:id="echoid-s7121" xml:space="preserve">Mais dans le triangle rectan-<lb/>gle A D B, on connoît le côté A B par l’hypotheſe, on connoît <lb/>le côté B D par le préſent corollaire: </s> <s xml:id="echoid-s7122" xml:space="preserve">donc on pourra con-<lb/>noître l’autre côté A D, ou la perpendiculaire qui meſure la <lb/>hauteur du triangle, & </s> <s xml:id="echoid-s7123" xml:space="preserve">l’on aura A D = √A B<emph style="sub">2</emph> - B D<emph style="sub">2</emph>\x{0020}, ou <lb/>d = √cc - xx\x{0020}. </s> <s xml:id="echoid-s7124" xml:space="preserve">Si le triangle donné étoit rectangle, la per-<lb/>pendiculaire A D ſe confondroit avec le côté A B, & </s> <s xml:id="echoid-s7125" xml:space="preserve">l’on auroit <lb/>{aa - cc - bb/2b} = o<unsure/> = B D.</s> <s xml:id="echoid-s7126" xml:space="preserve"/> </p> <pb o="207" file="0245" n="245" rhead="DE MATHÉMATIQUE. Liv. IV."/> </div> <div xml:id="echoid-div450" type="section" level="1" n="384"> <head xml:id="echoid-head435" xml:space="preserve">PROPOSITION XVI. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7127" xml:space="preserve">412. </s> <s xml:id="echoid-s7128" xml:space="preserve">Dans tout triangle A B C, le quarré d’un côté A B oppoſé <lb/> <anchor type="note" xlink:label="note-0245-01a" xlink:href="note-0245-01"/> à un angle aigu, eſt égal à la ſomme des quarrés des deux autres <lb/>côtés, moins deux rectangles égaux, compris ſous le côté A C, op-<lb/>poſé au plus grand angle, ſur lequel on a abaiſſé une perpendicu-<lb/>laire B D; </s> <s xml:id="echoid-s7129" xml:space="preserve">& </s> <s xml:id="echoid-s7130" xml:space="preserve">la partie C D du même côté A C, compriſe entre l’an-<lb/>gle C, auquel ce côté A B eſt oppoſé, & </s> <s xml:id="echoid-s7131" xml:space="preserve">la perpendiculaire B D; <lb/></s> <s xml:id="echoid-s7132" xml:space="preserve">c’eſt-à-dire que l’on aura A B<emph style="sub">2</emph> = A C<emph style="sub">2</emph> + B C<emph style="sub">2</emph> - 2A C x D C.</s> <s xml:id="echoid-s7133" xml:space="preserve"/> </p> <div xml:id="echoid-div450" type="float" level="2" n="1"> <note position="right" xlink:label="note-0245-01" xlink:href="note-0245-01a" xml:space="preserve">Figure 49.</note> </div> </div> <div xml:id="echoid-div452" type="section" level="1" n="385"> <head xml:id="echoid-head436" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7134" xml:space="preserve">Soit fait A B = a, B C = b, A C = c, B D = d, D C = x, <lb/>A D ſera c - x. </s> <s xml:id="echoid-s7135" xml:space="preserve">Cela poſé, le triangle rectangle B A D donne <lb/>A B<emph style="sub">2</emph> = B D<emph style="sub">2</emph> + A D<emph style="sub">2</emph>, ou analytiquement aa = dd + cc - 2cx <lb/>+ xx; </s> <s xml:id="echoid-s7136" xml:space="preserve">& </s> <s xml:id="echoid-s7137" xml:space="preserve">par la même raiſon, le triangle rectangle B D C <lb/>donne B C<emph style="sub">2</emph> = B D<emph style="sub">2</emph> + D C<emph style="sub">2</emph>, ou en termes analytiques, <lb/>bb = dd + xx. </s> <s xml:id="echoid-s7138" xml:space="preserve">Si l’on retranche les termes de cette derniere <lb/>égalité des termes de la précédente, on aura aa - bb = dd <lb/>+ cc - 2cx + xx - dd - xx = cc - 2cx; </s> <s xml:id="echoid-s7139" xml:space="preserve">en effaçant ce <lb/>qui ſe détruit, & </s> <s xml:id="echoid-s7140" xml:space="preserve">faiſant paſſer dans l’autre membre le terme <lb/>- bb, on aura aa = bb + cc - 2cx, ou A B<emph style="sub">2</emph> = A C<emph style="sub">2</emph> + B C<emph style="sub">2</emph> <lb/>- <emph style="super">2</emph>A C x D C. </s> <s xml:id="echoid-s7141" xml:space="preserve">C. </s> <s xml:id="echoid-s7142" xml:space="preserve">Q. </s> <s xml:id="echoid-s7143" xml:space="preserve">F. </s> <s xml:id="echoid-s7144" xml:space="preserve">D.</s> <s xml:id="echoid-s7145" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7146" xml:space="preserve">On démontreroit de la même maniere que l’on auroit <lb/>B C<emph style="sub">2</emph> = A B<emph style="sub">2</emph> + A C<emph style="sub">2</emph> - <emph style="super">2</emph>A C x A D.</s> <s xml:id="echoid-s7147" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div453" type="section" level="1" n="386"> <head xml:id="echoid-head437" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7148" xml:space="preserve">413. </s> <s xml:id="echoid-s7149" xml:space="preserve">Puiſque l’on a aa = bb + cc - 2cx, on aura, en <lb/>faiſant paſſer - 2cx dans le premier membre, & </s> <s xml:id="echoid-s7150" xml:space="preserve">aa dans le <lb/>ſecond, 2cx = bb + cc - aa, d’où l’on tire x = {bb + cc - aa/2c}. <lb/></s> <s xml:id="echoid-s7151" xml:space="preserve">Ce qui fait voir que pour avoir la valeur du ſegment D C, il <lb/>faut de la ſomme des quarrés des côtés A C, B C, ôter le quarré <lb/>du côté A B oppoſé à l’angle C, & </s> <s xml:id="echoid-s7152" xml:space="preserve">diviſer le reſte par 2c, ou <lb/>deux fois le côté ſur lequel on a abaiſſé la perpendiculaire B D. </s> <s xml:id="echoid-s7153" xml:space="preserve"><lb/>D’où il ſuit que par la connoiſſance des trois côtés d’un trian-<lb/>gle quelconque, on peut toujours trouver la ſurface; </s> <s xml:id="echoid-s7154" xml:space="preserve">car la <pb o="208" file="0246" n="246" rhead="NOUVEAU COURS DE MATH. Liv. IV."/> ſurface d’un triangle eſt égal au produit de ſa baſe par la per-<lb/>pendiculaire, & </s> <s xml:id="echoid-s7155" xml:space="preserve">nous voyons par le préſent corollaire, que <lb/>l’on peut toujours avoir la perpendiculaire B D. </s> <s xml:id="echoid-s7156" xml:space="preserve">Pour cela, <lb/>il n’y a qu’à ôter le quarré du ſegment D C du quarré de B C, <lb/>& </s> <s xml:id="echoid-s7157" xml:space="preserve">prendre la racine quarrée de la différence, que l’on multi-<lb/>pliera par le côté A C, pour avoir la ſurface du triangle A B C.</s> <s xml:id="echoid-s7158" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div454" type="section" level="1" n="387"> <head xml:id="echoid-head438" xml:space="preserve">Fin du quatrieme Livre.</head> <figure> <image file="0246-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0246-01"/> </figure> <pb o="209" file="0247" n="247"/> <figure> <image file="0247-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0247-01"/> </figure> </div> <div xml:id="echoid-div455" type="section" level="1" n="388"> <head xml:id="echoid-head439" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE. <lb/>LIVRE CINQUIEME, <lb/>Où l’on traite des propriétés du Cercle. <lb/><emph style="sc">Définitions</emph>. <lb/>I.</head> <p> <s xml:id="echoid-s7159" xml:space="preserve">414. </s> <s xml:id="echoid-s7160" xml:space="preserve">L’<emph style="sc">On</emph> nomme cercles concentriques, ceux qui ayant été <lb/> <anchor type="note" xlink:label="note-0247-01a" xlink:href="note-0247-01"/> décrits du même centre, ont leurs circonférences paralleles: <lb/></s> <s xml:id="echoid-s7161" xml:space="preserve"> <anchor type="note" xlink:label="note-0247-02a" xlink:href="note-0247-02"/> tels ſont les deux cercles qui ont pour centre commun le <lb/>point A.</s> <s xml:id="echoid-s7162" xml:space="preserve"/> </p> <div xml:id="echoid-div455" type="float" level="2" n="1"> <note position="right" xlink:label="note-0247-01" xlink:href="note-0247-01a" xml:space="preserve">Planche III.</note> <note position="right" xlink:label="note-0247-02" xlink:href="note-0247-02a" xml:space="preserve">Figure 50.</note> </div> </div> <div xml:id="echoid-div457" type="section" level="1" n="389"> <head xml:id="echoid-head440" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s7163" xml:space="preserve">415. </s> <s xml:id="echoid-s7164" xml:space="preserve">Les cercles excentriques, ſont ceux qui ayant été décrits <lb/> <anchor type="note" xlink:label="note-0247-03a" xlink:href="note-0247-03"/> par des centres différens, n’ont pas leurs circonférences pa-<lb/>ralleles, comme B & </s> <s xml:id="echoid-s7165" xml:space="preserve">C.</s> <s xml:id="echoid-s7166" xml:space="preserve"/> </p> <div xml:id="echoid-div457" type="float" level="2" n="1"> <note position="right" xlink:label="note-0247-03" xlink:href="note-0247-03a" xml:space="preserve">Figure 51.</note> </div> </div> <div xml:id="echoid-div459" type="section" level="1" n="390"> <head xml:id="echoid-head441" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s7167" xml:space="preserve">416. </s> <s xml:id="echoid-s7168" xml:space="preserve">L’on nomme couronne, l’eſpace renfermé entre les <lb/> <anchor type="note" xlink:label="note-0247-04a" xlink:href="note-0247-04"/> circonférences de deux cercles concentriques, comme eſt l’eſ-<lb/>pace B B, terminé par les circonférences E & </s> <s xml:id="echoid-s7169" xml:space="preserve">F.</s> <s xml:id="echoid-s7170" xml:space="preserve"/> </p> <div xml:id="echoid-div459" type="float" level="2" n="1"> <note position="right" xlink:label="note-0247-04" xlink:href="note-0247-04a" xml:space="preserve">Figure 50.</note> </div> </div> <div xml:id="echoid-div461" type="section" level="1" n="391"> <head xml:id="echoid-head442" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s7171" xml:space="preserve">417. </s> <s xml:id="echoid-s7172" xml:space="preserve">Le ſegment de cercle eſt la partie de la ſurface d’un cer-<lb/> <anchor type="note" xlink:label="note-0247-05a" xlink:href="note-0247-05"/> cle, terminée par une ligne droite, & </s> <s xml:id="echoid-s7173" xml:space="preserve">par une partie de ſa <lb/>circonférence, comme A B C. </s> <s xml:id="echoid-s7174" xml:space="preserve">Si la ligne droite A C ne <pb o="210" file="0248" n="248" rhead="NOUVEAU COURS"/> paſſe pas par le centre, le cercle ſera diviſé en deux ſegmens <lb/>inégaux.</s> <s xml:id="echoid-s7175" xml:space="preserve"/> </p> <div xml:id="echoid-div461" type="float" level="2" n="1"> <note position="right" xlink:label="note-0247-05" xlink:href="note-0247-05a" xml:space="preserve">Figure 52.</note> </div> </div> <div xml:id="echoid-div463" type="section" level="1" n="392"> <head xml:id="echoid-head443" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s7176" xml:space="preserve">418. </s> <s xml:id="echoid-s7177" xml:space="preserve">Le ſecteur de cercle eſt une partie de ſa ſurface, termi-<lb/> <anchor type="note" xlink:label="note-0248-01a" xlink:href="note-0248-01"/> née par deux rayons D C, D E, & </s> <s xml:id="echoid-s7178" xml:space="preserve">par une partie de ſa circon-<lb/>férence, comme C D E.</s> <s xml:id="echoid-s7179" xml:space="preserve"/> </p> <div xml:id="echoid-div463" type="float" level="2" n="1"> <note position="left" xlink:label="note-0248-01" xlink:href="note-0248-01a" xml:space="preserve">Figure 53.</note> </div> </div> <div xml:id="echoid-div465" type="section" level="1" n="393"> <head xml:id="echoid-head444" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s7180" xml:space="preserve">419. </s> <s xml:id="echoid-s7181" xml:space="preserve">L’arc de cercle eſt une partie de la circonférence, plus <lb/>grande ou plus petite que la demi-circonférence.</s> <s xml:id="echoid-s7182" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div466" type="section" level="1" n="394"> <head xml:id="echoid-head445" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s7183" xml:space="preserve">420. </s> <s xml:id="echoid-s7184" xml:space="preserve">On nomme cordes toutes les lignes droites, comme <lb/> <anchor type="note" xlink:label="note-0248-02a" xlink:href="note-0248-02"/> A C, terminées par la circonférence d’un cercle.</s> <s xml:id="echoid-s7185" xml:space="preserve"/> </p> <div xml:id="echoid-div466" type="float" level="2" n="1"> <note position="left" xlink:label="note-0248-02" xlink:href="note-0248-02a" xml:space="preserve">Figure 52.</note> </div> </div> <div xml:id="echoid-div468" type="section" level="1" n="395"> <head xml:id="echoid-head446" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s7186" xml:space="preserve">421. </s> <s xml:id="echoid-s7187" xml:space="preserve">Quand une ligne touche la circonférence d’un cercle <lb/> <anchor type="note" xlink:label="note-0248-03a" xlink:href="note-0248-03"/> ſans le couper, cette ligne eſt nommée tangente: </s> <s xml:id="echoid-s7188" xml:space="preserve">ainſi la ligne <lb/>A B, qui ne touche la circonférence du cercle D qu’au point <lb/>d, eſt dite tangente à ce cercle au point d.</s> <s xml:id="echoid-s7189" xml:space="preserve"/> </p> <div xml:id="echoid-div468" type="float" level="2" n="1"> <note position="left" xlink:label="note-0248-03" xlink:href="note-0248-03a" xml:space="preserve">Figure 54.</note> </div> </div> <div xml:id="echoid-div470" type="section" level="1" n="396"> <head xml:id="echoid-head447" xml:space="preserve">IX.</head> <p> <s xml:id="echoid-s7190" xml:space="preserve">422. </s> <s xml:id="echoid-s7191" xml:space="preserve">Si une ligne rencontre la circonférence d’un cercle, <lb/>de maniere qu’elle paſſe au dedans, cette ligne eſt appellée ſé-<lb/>cante, comme eſt, par exemple, la ligne B E.</s> <s xml:id="echoid-s7192" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div471" type="section" level="1" n="397"> <head xml:id="echoid-head448" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7193" xml:space="preserve">423. </s> <s xml:id="echoid-s7194" xml:space="preserve">Si du centre d’un cercle on abaiſſe une perpendiculaire <lb/> <anchor type="note" xlink:label="note-0248-04a" xlink:href="note-0248-04"/> B D E ſur une corde A C, elle la diviſera en deux parties égales <lb/>auſſi-bien que l’arc A E C ſoutenu par cette corde.</s> <s xml:id="echoid-s7195" xml:space="preserve"/> </p> <div xml:id="echoid-div471" type="float" level="2" n="1"> <note position="left" xlink:label="note-0248-04" xlink:href="note-0248-04a" xml:space="preserve">Figure 55.</note> </div> </div> <div xml:id="echoid-div473" type="section" level="1" n="398"> <head xml:id="echoid-head449" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7196" xml:space="preserve">Soient menés aux extrêmités A, C de la corde A C les rayons <lb/>A B, B C; </s> <s xml:id="echoid-s7197" xml:space="preserve">il eſt aiſé de voir que les triangles rectangles A B D, <lb/>C B D ſont égaux en tout; </s> <s xml:id="echoid-s7198" xml:space="preserve">car ils ont, outre l’angle droit, deux <lb/>côtés A B, B C égaux, puiſque ce ſont les rayons d’un même <lb/>cercle; </s> <s xml:id="echoid-s7199" xml:space="preserve">& </s> <s xml:id="echoid-s7200" xml:space="preserve">de plus, le côté B D eſt commun à l’un & </s> <s xml:id="echoid-s7201" xml:space="preserve">à l’autre : <lb/></s> <s xml:id="echoid-s7202" xml:space="preserve">donc la ligne A D eſt égale à la ligne D C. </s> <s xml:id="echoid-s7203" xml:space="preserve">On peut encore <lb/>démontrer cette propoſition par la propriété des triangles rec- <pb o="211" file="0249" n="249" rhead="DE MATHÉMATIQUE. Liv. V."/> tangles: </s> <s xml:id="echoid-s7204" xml:space="preserve">car puiſque par hypotheſe B D eſt perpendiculaire ſur <lb/>A C, on aura A D<emph style="sub">2</emph> = A B<emph style="sub">2</emph> - B D<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s7205" xml:space="preserve">D C<emph style="sub">2</emph> = B C<emph style="sub">2</emph> - B D<emph style="sub">2</emph> <lb/>= A B<emph style="sub">2</emph> - B D<emph style="sub">2</emph>: </s> <s xml:id="echoid-s7206" xml:space="preserve">donc A D<emph style="sub">2</emph> = D C<emph style="sub">2</emph>, ou A D = A C.</s> <s xml:id="echoid-s7207" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7208" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s7209" xml:space="preserve">Puiſque les triangles A B D, C B D ſont égaux en tout, <lb/>l’angle A B D ſera égal à l’angle C B D; </s> <s xml:id="echoid-s7210" xml:space="preserve">& </s> <s xml:id="echoid-s7211" xml:space="preserve">prolongeant le côté <lb/>B D juſqu à la circonférence du cercle en E, les arcs A E, E C <lb/>qui meſurent les angles A B E, C B E ſont égaux; </s> <s xml:id="echoid-s7212" xml:space="preserve">& </s> <s xml:id="echoid-s7213" xml:space="preserve">par con-<lb/>ſéquent l’arc A C eſt auſſi diviſé en deux parties égales au <lb/>point E.</s> <s xml:id="echoid-s7214" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div474" type="section" level="1" n="399"> <head xml:id="echoid-head450" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7215" xml:space="preserve">424. </s> <s xml:id="echoid-s7216" xml:space="preserve">Si une droite B D paſſe par le centre, & </s> <s xml:id="echoid-s7217" xml:space="preserve">diviſe la corde ou <lb/>ſon arc A C en deux parties égales; </s> <s xml:id="echoid-s7218" xml:space="preserve">elle ſera perpendiculaire à cette <lb/>corde.</s> <s xml:id="echoid-s7219" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div475" type="section" level="1" n="400"> <head xml:id="echoid-head451" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7220" xml:space="preserve">Soient tirés les rayons A B, B C aux extrêmités de la corde <lb/>A C. </s> <s xml:id="echoid-s7221" xml:space="preserve">Cela poſé, puiſque la droite B D diviſe la corde A C en <lb/>deux parties égales, le point D de cette droite ſera également <lb/>éloigné des extrêmités A, C de la droite A C; </s> <s xml:id="echoid-s7222" xml:space="preserve">& </s> <s xml:id="echoid-s7223" xml:space="preserve">parce que, <lb/>par hypotheſe, la même droite B D paſſe par le centre B, ſon <lb/>point B ſera encore également éloigné des mêmes extrêmités <lb/>A, C : </s> <s xml:id="echoid-s7224" xml:space="preserve">donc elle ſera perpendiculaire à cette corde.</s> <s xml:id="echoid-s7225" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7226" xml:space="preserve">Si l’on ſuppoſe que l’arc A C eſt coupé en deux également <lb/>par la droite B D, prolongée en E, il eſt viſible que les an-<lb/>gles A B E, C B E, meſurés par ces arcs, ſeront égaux; </s> <s xml:id="echoid-s7227" xml:space="preserve">& </s> <s xml:id="echoid-s7228" xml:space="preserve">parce <lb/>que le point B eſt le centre du cercle, les rayons B C, A B <lb/>ſeront auſſi égaux: </s> <s xml:id="echoid-s7229" xml:space="preserve">donc les triangles A B D, C B D auront un <lb/>angle égal compris entre côtés égaux; </s> <s xml:id="echoid-s7230" xml:space="preserve">ainſi ils ſeront parfai-<lb/>tement égaux (art. </s> <s xml:id="echoid-s7231" xml:space="preserve">381). </s> <s xml:id="echoid-s7232" xml:space="preserve">Donc l’angle B D C eſt égal à l’an-<lb/>gle B D A: </s> <s xml:id="echoid-s7233" xml:space="preserve">donc la ligne B D ne penche pas plus d’un côté <lb/>que de l’autre ſur la ligne A C, & </s> <s xml:id="echoid-s7234" xml:space="preserve">par conſéquent lui eſt per-<lb/>pendiculaire. </s> <s xml:id="echoid-s7235" xml:space="preserve">C. </s> <s xml:id="echoid-s7236" xml:space="preserve">Q. </s> <s xml:id="echoid-s7237" xml:space="preserve">F. </s> <s xml:id="echoid-s7238" xml:space="preserve">D.</s> <s xml:id="echoid-s7239" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div476" type="section" level="1" n="401"> <head xml:id="echoid-head452" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7240" xml:space="preserve">425. </s> <s xml:id="echoid-s7241" xml:space="preserve">Si une ligne droite E D B perpendiculaire à une corde <lb/>A C, diviſe cette corde ou ſon arc en deux parties égales, je dis <lb/>que cette ligne paſſe néceſſairement par le centre.</s> <s xml:id="echoid-s7242" xml:space="preserve"/> </p> <pb o="212" file="0250" n="250" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div477" type="section" level="1" n="402"> <head xml:id="echoid-head453" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7243" xml:space="preserve">Puiſque la ligne E B eſt perpendiculaire ſur le milieu de la <lb/>corde A C, elle paſſe néceſſairement par tous les points égale-<lb/>ment éloignés de A & </s> <s xml:id="echoid-s7244" xml:space="preserve">de C; </s> <s xml:id="echoid-s7245" xml:space="preserve">mais le centre B eſt également <lb/>éloigné des points A & </s> <s xml:id="echoid-s7246" xml:space="preserve">C, qui ſont à la circonférence, par la <lb/>définition du cercle & </s> <s xml:id="echoid-s7247" xml:space="preserve">de ſon centre : </s> <s xml:id="echoid-s7248" xml:space="preserve">donc la ligne E D B <lb/>paſſe néceſſairement par le centre B. </s> <s xml:id="echoid-s7249" xml:space="preserve">C. </s> <s xml:id="echoid-s7250" xml:space="preserve">Q. </s> <s xml:id="echoid-s7251" xml:space="preserve">F. </s> <s xml:id="echoid-s7252" xml:space="preserve">D.</s> <s xml:id="echoid-s7253" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div478" type="section" level="1" n="403"> <head xml:id="echoid-head454" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7254" xml:space="preserve">426. </s> <s xml:id="echoid-s7255" xml:space="preserve">Il ſuit des trois propoſitions précédentes, que de ces <lb/>trois conditions, paſſer par le centre, être perpendiculaire à la <lb/>corde, & </s> <s xml:id="echoid-s7256" xml:space="preserve">la couper en deux parties égales, deux, comme l’on <lb/>voudra, étant poſées, la troiſieme s’enſuit néceſſairement.</s> <s xml:id="echoid-s7257" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div479" type="section" level="1" n="404"> <head xml:id="echoid-head455" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7258" xml:space="preserve">427. </s> <s xml:id="echoid-s7259" xml:space="preserve">Si du centre D d’un cercle on mene une ligne DC au point <lb/> <anchor type="note" xlink:label="note-0250-01a" xlink:href="note-0250-01"/> C, où une tangente A B touche le cercle, je dis que cette ligne ſera <lb/>perpendiculaire à la tangente.</s> <s xml:id="echoid-s7260" xml:space="preserve"/> </p> <div xml:id="echoid-div479" type="float" level="2" n="1"> <note position="left" xlink:label="note-0250-01" xlink:href="note-0250-01a" xml:space="preserve">Figure 56.</note> </div> </div> <div xml:id="echoid-div481" type="section" level="1" n="405"> <head xml:id="echoid-head456" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7261" xml:space="preserve">Puiſque la ligne A B eſt ſuppoſée tangente en C, tout autre <lb/>point de cette ligne, comme F, ſera au dehors du cercle, & </s> <s xml:id="echoid-s7262" xml:space="preserve"><lb/>partant la ligne DF, menée du centre D à ce point, ſera plus <lb/>grande que le rayon D C: </s> <s xml:id="echoid-s7263" xml:space="preserve">donc le rayon D C eſt la plus courte <lb/>de toutes les lignes qu’on puiſſe mener du point D à la tan-<lb/>gente A B: </s> <s xml:id="echoid-s7264" xml:space="preserve">donc ce rayon D C eſt perpendiculaire à la même <lb/>tangente. </s> <s xml:id="echoid-s7265" xml:space="preserve">C. </s> <s xml:id="echoid-s7266" xml:space="preserve">Q. </s> <s xml:id="echoid-s7267" xml:space="preserve">F. </s> <s xml:id="echoid-s7268" xml:space="preserve">D.</s> <s xml:id="echoid-s7269" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div482" type="section" level="1" n="406"> <head xml:id="echoid-head457" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7270" xml:space="preserve">428. </s> <s xml:id="echoid-s7271" xml:space="preserve">Réciproquement ſi une ligne C B eſt perpendiculaire <lb/>à l’extrêmité d’un rayon D C, elle ſera tangente en C; </s> <s xml:id="echoid-s7272" xml:space="preserve">car <lb/>toute autre ligne, comme D F, étant plus longue que le rayon <lb/>D C, aura ſon extrêmité F ſur la ligne A B hors du cercle; </s> <s xml:id="echoid-s7273" xml:space="preserve">& </s> <s xml:id="echoid-s7274" xml:space="preserve"><lb/>par conſéquent la ligne A B perpendiculaire à l’extrêmité du <lb/>rayon, ſera tangente au cercle en ce point. </s> <s xml:id="echoid-s7275" xml:space="preserve">C. </s> <s xml:id="echoid-s7276" xml:space="preserve">Q. </s> <s xml:id="echoid-s7277" xml:space="preserve">F. </s> <s xml:id="echoid-s7278" xml:space="preserve">D.</s> <s xml:id="echoid-s7279" xml:space="preserve"/> </p> <pb o="213" file="0251" n="251" rhead="DE MATHÉMATIQUE. Liv. V."/> </div> <div xml:id="echoid-div483" type="section" level="1" n="407"> <head xml:id="echoid-head458" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head459" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7280" xml:space="preserve">429. </s> <s xml:id="echoid-s7281" xml:space="preserve">L’angle A B C, qui a ſon ſommet à la circonférence d’un <lb/> <anchor type="note" xlink:label="note-0251-01a" xlink:href="note-0251-01"/> cercle, a pour meſure la moitié de l’arc compris entre ſes côtés.</s> <s xml:id="echoid-s7282" xml:space="preserve"/> </p> <div xml:id="echoid-div483" type="float" level="2" n="1"> <note position="right" xlink:label="note-0251-01" xlink:href="note-0251-01a" xml:space="preserve">Figure 57</note> </div> </div> <div xml:id="echoid-div485" type="section" level="1" n="408"> <head xml:id="echoid-head460" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7283" xml:space="preserve">Par le ſommet B de l’angle A B C, & </s> <s xml:id="echoid-s7284" xml:space="preserve">le centre D, ſoit me-<lb/>née la ligne B D E, & </s> <s xml:id="echoid-s7285" xml:space="preserve">les rayons D A, D C; </s> <s xml:id="echoid-s7286" xml:space="preserve">il eſt évident <lb/>que l’angle total A B C eſt égal à la ſomme des angles A B E, <lb/>C B E, & </s> <s xml:id="echoid-s7287" xml:space="preserve">que l’angle au centre A D C eſt égal à la ſomme des <lb/>angles A D E, C D E. </s> <s xml:id="echoid-s7288" xml:space="preserve">Cela poſé, l’angle C D E extérieur au <lb/>triangle iſoſcele C D B eſt égal aux deux angles intérieurs en B <lb/>& </s> <s xml:id="echoid-s7289" xml:space="preserve">en C, ou double de l’angle en B; </s> <s xml:id="echoid-s7290" xml:space="preserve">& </s> <s xml:id="echoid-s7291" xml:space="preserve">de même l’angle A D E <lb/>étant extérieur au triangle iſoſcele A D B, eſt égal à la ſomme <lb/>des intérieurs oppoſés en B & </s> <s xml:id="echoid-s7292" xml:space="preserve">en A, ou double de l’angle <lb/>A B D: </s> <s xml:id="echoid-s7293" xml:space="preserve">donc l’angle total A D C eſt double de l’angle total <lb/>A B C: </s> <s xml:id="echoid-s7294" xml:space="preserve">donc l’angle à la circonférence n’eſt que la moitié de <lb/>l’angle au centre. </s> <s xml:id="echoid-s7295" xml:space="preserve">C.</s> <s xml:id="echoid-s7296" xml:space="preserve">Q.</s> <s xml:id="echoid-s7297" xml:space="preserve">F.</s> <s xml:id="echoid-s7298" xml:space="preserve">D.</s> <s xml:id="echoid-s7299" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7300" xml:space="preserve">430. </s> <s xml:id="echoid-s7301" xml:space="preserve">On déduit de cette propoſition pluſieurs conſéquences, <lb/>qui ſont d’un très-grand uſage. </s> <s xml:id="echoid-s7302" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s7303" xml:space="preserve">Qu’un angle, tel que A B C, <lb/> <anchor type="note" xlink:label="note-0251-02a" xlink:href="note-0251-02"/> eſt droit, lorſqu’il eſt appuyé ſur le diametre, ou ſur une demi-<lb/>circonférence, puiſqu’il a pour meſure la moitié de l’arc A O C, <lb/>qui eſt de 90 degrés, ou un quart de cercle.</s> <s xml:id="echoid-s7304" xml:space="preserve"/> </p> <div xml:id="echoid-div485" type="float" level="2" n="1"> <note position="right" xlink:label="note-0251-02" xlink:href="note-0251-02a" xml:space="preserve">Figure 59.</note> </div> <p> <s xml:id="echoid-s7305" xml:space="preserve">431. </s> <s xml:id="echoid-s7306" xml:space="preserve">2<emph style="sub">0</emph>, Qu’un angle, comme D E F, qui eſt renfermé dans <lb/> <anchor type="note" xlink:label="note-0251-03a" xlink:href="note-0251-03"/> un ſegment plus petit qu’un demi-cercle eſt obtus, puiſqu’il a <lb/>pour meſure un arc plus grand qu’un quart de cercle, étant <lb/>appuyé ſur l’arc D O F, plus grand que la demi-circonférence.</s> <s xml:id="echoid-s7307" xml:space="preserve"/> </p> <div xml:id="echoid-div486" type="float" level="2" n="2"> <note position="right" xlink:label="note-0251-03" xlink:href="note-0251-03a" xml:space="preserve">Figure 60</note> </div> <p> <s xml:id="echoid-s7308" xml:space="preserve">432. </s> <s xml:id="echoid-s7309" xml:space="preserve">3<emph style="sub">0</emph>. </s> <s xml:id="echoid-s7310" xml:space="preserve">Qu’un angle, comme G H I, qui eſt renfermé dans <lb/> <anchor type="note" xlink:label="note-0251-04a" xlink:href="note-0251-04"/> un ſegment plus petit qu’un demi-cercle eſt obtus, puiſqu’il <lb/>a pour meſure la moitié de l’arc G O I, qui eſt plus petite qu’un <lb/>quart de cercle.</s> <s xml:id="echoid-s7311" xml:space="preserve"/> </p> <div xml:id="echoid-div487" type="float" level="2" n="3"> <note position="right" xlink:label="note-0251-04" xlink:href="note-0251-04a" xml:space="preserve">Figure 60</note> </div> <p> <s xml:id="echoid-s7312" xml:space="preserve">433. </s> <s xml:id="echoid-s7313" xml:space="preserve">4<emph style="sub">0</emph>. </s> <s xml:id="echoid-s7314" xml:space="preserve">Que les angles, comme A B C & </s> <s xml:id="echoid-s7315" xml:space="preserve">A D C, qui ſont <lb/> <anchor type="note" xlink:label="note-0251-05a" xlink:href="note-0251-05"/> renfermés dans le même ſegment ſont égaux, puiſqu’ils ont <lb/>chacun pour meſure la moitié de l’arc A O C.</s> <s xml:id="echoid-s7316" xml:space="preserve"/> </p> <div xml:id="echoid-div488" type="float" level="2" n="4"> <note position="right" xlink:label="note-0251-05" xlink:href="note-0251-05a" xml:space="preserve">Figure 62.</note> </div> <p> <s xml:id="echoid-s7317" xml:space="preserve">434. </s> <s xml:id="echoid-s7318" xml:space="preserve">Que deux angles qui ſont appuyés ſur une même corde <lb/> <anchor type="note" xlink:label="note-0251-06a" xlink:href="note-0251-06"/> D F, l’un d’un côté, l’autre de l’autre, ſont ſupplémens l’un <lb/>de l’autre, puiſqu’ils ont enſemble pour meſure la moitié de <lb/>la circonférence, tels ſont les angles D E F, D O F.</s> <s xml:id="echoid-s7319" xml:space="preserve"/> </p> <div xml:id="echoid-div489" type="float" level="2" n="5"> <note position="right" xlink:label="note-0251-06" xlink:href="note-0251-06a" xml:space="preserve">Figure 60.</note> </div> <pb o="214" file="0252" n="252" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div491" type="section" level="1" n="409"> <head xml:id="echoid-head461" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head462" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7320" xml:space="preserve">435. </s> <s xml:id="echoid-s7321" xml:space="preserve">Si l’on a un angle B A D, formé par une tangente & </s> <s xml:id="echoid-s7322" xml:space="preserve">par <lb/> <anchor type="note" xlink:label="note-0252-01a" xlink:href="note-0252-01"/> une corde A D, cet angle aura pour meſure la moitié de l’arc <lb/>A F D, compris entre la corde & </s> <s xml:id="echoid-s7323" xml:space="preserve">la tangente.</s> <s xml:id="echoid-s7324" xml:space="preserve"/> </p> <div xml:id="echoid-div491" type="float" level="2" n="1"> <note position="left" xlink:label="note-0252-01" xlink:href="note-0252-01a" xml:space="preserve">Figure 58.</note> </div> </div> <div xml:id="echoid-div493" type="section" level="1" n="410"> <head xml:id="echoid-head463" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7325" xml:space="preserve">Tirez du centre E le rayon E A au point d’attouchement A, <lb/>qui ſera perpendiculaire ſur la tangente A B (art. </s> <s xml:id="echoid-s7326" xml:space="preserve">427), & </s> <s xml:id="echoid-s7327" xml:space="preserve"><lb/>tirez du centre E la droite E G F perpendiculaire ſur A D, <lb/>qui la diviſera en deux également, auſſi-bien que l’arc A F D <lb/>(art. </s> <s xml:id="echoid-s7328" xml:space="preserve">423). </s> <s xml:id="echoid-s7329" xml:space="preserve">Cela poſé, à cauſe du triangle rectangle A G E, <lb/>l’angle G A E, joint à l’angle A E G vaut un droit, & </s> <s xml:id="echoid-s7330" xml:space="preserve">le même <lb/>angle G A E, joint à G A B vaut auſſi un droit: </s> <s xml:id="echoid-s7331" xml:space="preserve">donc l’angle <lb/>G A B eſt égal à l’angle A E G; </s> <s xml:id="echoid-s7332" xml:space="preserve">mais l’angle A E G étant au <lb/>centre du cercle, a pour meſure l’arc A F compris entre ſes <lb/>côtés, & </s> <s xml:id="echoid-s7333" xml:space="preserve">moitié de l’arc A F D ſoutenu par la corde A D: </s> <s xml:id="echoid-s7334" xml:space="preserve">donc <lb/>l’angle B A D formé par une tangente & </s> <s xml:id="echoid-s7335" xml:space="preserve">par une corde, a pour <lb/>meſure la moitié de l’arc compris entre la corde & </s> <s xml:id="echoid-s7336" xml:space="preserve">cette tan-<lb/>gente. </s> <s xml:id="echoid-s7337" xml:space="preserve">C.</s> <s xml:id="echoid-s7338" xml:space="preserve">Q.</s> <s xml:id="echoid-s7339" xml:space="preserve">F.</s> <s xml:id="echoid-s7340" xml:space="preserve">D.</s> <s xml:id="echoid-s7341" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div494" type="section" level="1" n="411"> <head xml:id="echoid-head464" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head465" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7342" xml:space="preserve">436. </s> <s xml:id="echoid-s7343" xml:space="preserve">Un angle A E C qui a ſon ſommet placé au dedans du <lb/> <anchor type="note" xlink:label="note-0252-02a" xlink:href="note-0252-02"/> cercle dans un point quelconque E, différent du centre & </s> <s xml:id="echoid-s7344" xml:space="preserve">des points <lb/>de la circonférence, a pour meſure la moitié de l’arc A C, ſur <lb/>lequel il eſt appuyé; </s> <s xml:id="echoid-s7345" xml:space="preserve">plus la moitié de l’arc B D compris entre le <lb/>prolongement de ſes côtés A E, EC.</s> <s xml:id="echoid-s7346" xml:space="preserve"/> </p> <div xml:id="echoid-div494" type="float" level="2" n="1"> <note position="left" xlink:label="note-0252-02" xlink:href="note-0252-02a" xml:space="preserve">Figure 63.</note> </div> </div> <div xml:id="echoid-div496" type="section" level="1" n="412"> <head xml:id="echoid-head466" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7347" xml:space="preserve">Soit tirée la droite B C du point B au point C. </s> <s xml:id="echoid-s7348" xml:space="preserve">L’angle A E C <lb/>étant extérieur au triangle B E C, eſt égal à la ſomme des an-<lb/>gles intérieurs B C E, C B E: </s> <s xml:id="echoid-s7349" xml:space="preserve">mais ces mêmes angles ayant leur <lb/>ſommet à la circonférence, ont pour meſure la moitié de l’arc <lb/>compris entre leurs côtés; </s> <s xml:id="echoid-s7350" xml:space="preserve">ſçavoir, l’angle C B E ou C B A, la <lb/>moitié de l’arc A C, & </s> <s xml:id="echoid-s7351" xml:space="preserve">l’angle B C E ou B C D ſon égal, la <lb/>moitié de l’arc B D: </s> <s xml:id="echoid-s7352" xml:space="preserve">donc l’angle A E C, qui eſt égal à leur <pb o="215" file="0253" n="253" rhead="DE MATHÉMATIQUE. Liv. V."/> ſomme, a pour meſure la ſomme de la moitié des mêmes arcs, <lb/>c’eſt-à-dire la moitié de l’arc A C compris entre ſes côtés, plus <lb/>la moitié de l’arc B D, compris entre le prolongement des <lb/>mêmes côtés. </s> <s xml:id="echoid-s7353" xml:space="preserve">C.</s> <s xml:id="echoid-s7354" xml:space="preserve">Q.</s> <s xml:id="echoid-s7355" xml:space="preserve">F.</s> <s xml:id="echoid-s7356" xml:space="preserve">D.</s> <s xml:id="echoid-s7357" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div497" type="section" level="1" n="413"> <head xml:id="echoid-head467" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head468" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7358" xml:space="preserve">437. </s> <s xml:id="echoid-s7359" xml:space="preserve">L’angle B A C, dont le ſommet eſt au dehors de la cir-<lb/> <anchor type="note" xlink:label="note-0253-01a" xlink:href="note-0253-01"/> conférence d’un cercle, & </s> <s xml:id="echoid-s7360" xml:space="preserve">dont les côtés ſe terminent à la partie <lb/>concave de la même circonférence en B & </s> <s xml:id="echoid-s7361" xml:space="preserve">en C, a pour meſure la <lb/>moitié de l’arc concave B C, moins la moitié de l’arc conyexe <lb/>D E compris entre ſes côtés.</s> <s xml:id="echoid-s7362" xml:space="preserve"/> </p> <div xml:id="echoid-div497" type="float" level="2" n="1"> <note position="right" xlink:label="note-0253-01" xlink:href="note-0253-01a" xml:space="preserve">Figure 64.</note> </div> </div> <div xml:id="echoid-div499" type="section" level="1" n="414"> <head xml:id="echoid-head469" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7363" xml:space="preserve">Soient menées les lignes B E, C D, qui donneront les trian-<lb/>gles B A E, D A C. </s> <s xml:id="echoid-s7364" xml:space="preserve">L’angle B D C étant extérieur au triangle <lb/>D A C eſt égal à l’angle D A C, plus à l’angle A C D: </s> <s xml:id="echoid-s7365" xml:space="preserve">donc <lb/>l’angle D A C, ou ſon égal B A C, eſt égal à l’angle B D C, <lb/>moins l’angle D C E: </s> <s xml:id="echoid-s7366" xml:space="preserve">mais chacun de ces angles étant à la <lb/>circonférence, a pour meſure la moitié de l’arc compris entre <lb/>ſes côtés; </s> <s xml:id="echoid-s7367" xml:space="preserve">ſçavoir l’angle B D C, la moitié de l’arc B C, & </s> <s xml:id="echoid-s7368" xml:space="preserve"><lb/>l’angle A C D, la moitié de l’arc D E: </s> <s xml:id="echoid-s7369" xml:space="preserve">donc l’angle B A C a <lb/>pour meſure la moitié de la différence des mêmes arcs, c’eſt-<lb/>à-dire la moitié de l’arc concave B C ſur lequel il eſt appuyé, <lb/>moins la moitié de l’arc convexe D E. </s> <s xml:id="echoid-s7370" xml:space="preserve">C.</s> <s xml:id="echoid-s7371" xml:space="preserve">Q.</s> <s xml:id="echoid-s7372" xml:space="preserve">F.</s> <s xml:id="echoid-s7373" xml:space="preserve">D.</s> <s xml:id="echoid-s7374" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div500" type="section" level="1" n="415"> <head xml:id="echoid-head470" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7375" xml:space="preserve">438. </s> <s xml:id="echoid-s7376" xml:space="preserve">Il ſuit de tout ce que nous venons de dire, que, ſi l’on <lb/>a un angle à la circonférence, tel que A D C, formé par une <lb/> <anchor type="note" xlink:label="note-0253-02a" xlink:href="note-0253-02"/> corde D C & </s> <s xml:id="echoid-s7377" xml:space="preserve">une droite A D, dont le prolongement coupe <lb/>le cercle, cet angle aura pour meſure la moitié de l’arc <lb/>compris entre la corde D C, plus la moitié de l’arc ſoutenu <lb/>par le côté A D, prolongé juſqu’à la circonférence du cercle <lb/>en B: </s> <s xml:id="echoid-s7378" xml:space="preserve">car puiſque la ligne A D B eſt une ligne droite, ainſi que <lb/>la ligne D C, les angles B D C, A D C ſont enſemble égaux à <lb/>deux droits, & </s> <s xml:id="echoid-s7379" xml:space="preserve">par conſéquent doivent avoir pour meſure la <lb/>moitié de la circonférence; </s> <s xml:id="echoid-s7380" xml:space="preserve">mais l’angle B D C ayant ſon ſom-<lb/>met à la circonférence, a pour meſure la moitié de l’arc B C: <lb/></s> <s xml:id="echoid-s7381" xml:space="preserve">donc l’angle A D C doit avoir pour meſure la moitié de l’arc <lb/>D C, plus la moitié de l’arc B D, qui font enſemble la moitié <lb/>du reſte de la circonférence.</s> <s xml:id="echoid-s7382" xml:space="preserve"/> </p> <div xml:id="echoid-div500" type="float" level="2" n="1"> <note position="right" xlink:label="note-0253-02" xlink:href="note-0253-02a" xml:space="preserve">Figure 64.</note> </div> <pb o="216" file="0254" n="254" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div502" type="section" level="1" n="416"> <head xml:id="echoid-head471" xml:space="preserve">PROPOSITION IX.</head> <head xml:id="echoid-head472" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7383" xml:space="preserve">439. </s> <s xml:id="echoid-s7384" xml:space="preserve">Si l’on a deux droites quelconques A B, C D, qui ſe cou-<lb/> <anchor type="note" xlink:label="note-0254-01a" xlink:href="note-0254-01"/> pent au dedans d’un cercle dans un point E, je dis que le rectangle <lb/>compris ſous les parties A E & </s> <s xml:id="echoid-s7385" xml:space="preserve">E B de l’une, eſt égal au rectangle <lb/>compris ſous les parties D E & </s> <s xml:id="echoid-s7386" xml:space="preserve">E C de l’autre.</s> <s xml:id="echoid-s7387" xml:space="preserve"/> </p> <div xml:id="echoid-div502" type="float" level="2" n="1"> <note position="left" xlink:label="note-0254-01" xlink:href="note-0254-01a" xml:space="preserve">Figure 63.</note> </div> </div> <div xml:id="echoid-div504" type="section" level="1" n="417"> <head xml:id="echoid-head473" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7388" xml:space="preserve">Soient menées les cordes A C & </s> <s xml:id="echoid-s7389" xml:space="preserve">D B; </s> <s xml:id="echoid-s7390" xml:space="preserve">conſidérez que les <lb/>triangles A C E & </s> <s xml:id="echoid-s7391" xml:space="preserve">D B E ſont ſemblables, ayant les angles <lb/>égaux en E, puiſqu’ils ſont oppoſés au ſommet, & </s> <s xml:id="echoid-s7392" xml:space="preserve">que de plus <lb/>l’angle en C eſt égal à l’angle en B, puiſque chacun d’eux eſt <lb/>appuyé ſur le même arc: </s> <s xml:id="echoid-s7393" xml:space="preserve">donc les côtés oppoſés aux angles <lb/>égaux ſeront proportionnels, & </s> <s xml:id="echoid-s7394" xml:space="preserve">donneront A E: </s> <s xml:id="echoid-s7395" xml:space="preserve">E D:</s> <s xml:id="echoid-s7396" xml:space="preserve">: E C: </s> <s xml:id="echoid-s7397" xml:space="preserve">EB <lb/>(art. </s> <s xml:id="echoid-s7398" xml:space="preserve">402): </s> <s xml:id="echoid-s7399" xml:space="preserve">donc en prenant le produit des extrêmes & </s> <s xml:id="echoid-s7400" xml:space="preserve">des <lb/>moyens, on aura A E x E B = E D x E C. </s> <s xml:id="echoid-s7401" xml:space="preserve">C.</s> <s xml:id="echoid-s7402" xml:space="preserve">Q.</s> <s xml:id="echoid-s7403" xml:space="preserve">F.</s> <s xml:id="echoid-s7404" xml:space="preserve">D.</s> <s xml:id="echoid-s7405" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div505" type="section" level="1" n="418"> <head xml:id="echoid-head474" xml:space="preserve">PROPOSITION X.</head> <head xml:id="echoid-head475" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7406" xml:space="preserve">440. </s> <s xml:id="echoid-s7407" xml:space="preserve">Si du point A, pris au dehors d’un cercle ſur le même <lb/> <anchor type="note" xlink:label="note-0254-02a" xlink:href="note-0254-02"/> plan, on mene deux lignes droites A B, A C qui aillent ſe terminer <lb/>à la partie concave de la circonférence; </s> <s xml:id="echoid-s7408" xml:space="preserve">je dis que le rectangle com-<lb/>pris ſous une ſécante entiere A B, & </s> <s xml:id="echoid-s7409" xml:space="preserve">ſa partie A D extérieure au <lb/>cercle, eſt égal au rectangle compris ſous l’autre ſecante entiere A C, <lb/>& </s> <s xml:id="echoid-s7410" xml:space="preserve">ſa partie extérieure A E.</s> <s xml:id="echoid-s7411" xml:space="preserve"/> </p> <div xml:id="echoid-div505" type="float" level="2" n="1"> <note position="left" xlink:label="note-0254-02" xlink:href="note-0254-02a" xml:space="preserve">Figure 64.</note> </div> </div> <div xml:id="echoid-div507" type="section" level="1" n="419"> <head xml:id="echoid-head476" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7412" xml:space="preserve">Si l’on tire les lignes B E & </s> <s xml:id="echoid-s7413" xml:space="preserve">C D, on aura deux triangles <lb/>ſemblables A B E & </s> <s xml:id="echoid-s7414" xml:space="preserve">A C D: </s> <s xml:id="echoid-s7415" xml:space="preserve">car l’angle A leur eſt commun, <lb/>& </s> <s xml:id="echoid-s7416" xml:space="preserve">les angles B & </s> <s xml:id="echoid-s7417" xml:space="preserve">C ont chacun pour meſure la moitié de l’arc <lb/>D E (art. </s> <s xml:id="echoid-s7418" xml:space="preserve">429): </s> <s xml:id="echoid-s7419" xml:space="preserve">donc les côtés oppoſés aux angles égaux ſeront <lb/>proportionnels (art. </s> <s xml:id="echoid-s7420" xml:space="preserve">403), & </s> <s xml:id="echoid-s7421" xml:space="preserve">donneront A B: </s> <s xml:id="echoid-s7422" xml:space="preserve">A C:</s> <s xml:id="echoid-s7423" xml:space="preserve">: A E: </s> <s xml:id="echoid-s7424" xml:space="preserve">A D: <lb/></s> <s xml:id="echoid-s7425" xml:space="preserve">par conſéquent en prenant le produit des extrêmes & </s> <s xml:id="echoid-s7426" xml:space="preserve">des <lb/>moyens, on aura A B x A D = A C x A E. </s> <s xml:id="echoid-s7427" xml:space="preserve">C.</s> <s xml:id="echoid-s7428" xml:space="preserve">Q.</s> <s xml:id="echoid-s7429" xml:space="preserve">F.</s> <s xml:id="echoid-s7430" xml:space="preserve">D.</s> <s xml:id="echoid-s7431" xml:space="preserve"/> </p> <pb o="217" file="0255" n="255" rhead="DE MATHÉMATIQUE. Liv. V."/> </div> <div xml:id="echoid-div508" type="section" level="1" n="420"> <head xml:id="echoid-head477" xml:space="preserve">PROPOSITION XI.</head> <head xml:id="echoid-head478" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7432" xml:space="preserve">441. </s> <s xml:id="echoid-s7433" xml:space="preserve">Si d’un point B quelconque de la circonférence A B C, on <lb/> <anchor type="note" xlink:label="note-0255-01a" xlink:href="note-0255-01"/> abaiſſe une perpendiculaire B D ſur le diametre A C; </s> <s xml:id="echoid-s7434" xml:space="preserve">je dis que le <lb/>quarré de cette perpendiculaire ſera égal au rectangle des parties <lb/>A D, D C du diametre.</s> <s xml:id="echoid-s7435" xml:space="preserve"/> </p> <div xml:id="echoid-div508" type="float" level="2" n="1"> <note position="right" xlink:label="note-0255-01" xlink:href="note-0255-01a" xml:space="preserve">Figure 65.</note> </div> </div> <div xml:id="echoid-div510" type="section" level="1" n="421"> <head xml:id="echoid-head479" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7436" xml:space="preserve">Soient tirées les droites A B, B C du point B aux extrê-<lb/>mités du diametre A C, le triangle A B C ſera rectangle en B, <lb/>puiſque l’angle A B C eſt appuyé ſur la demi-circonférence <lb/>(art. </s> <s xml:id="echoid-s7437" xml:space="preserve">430), & </s> <s xml:id="echoid-s7438" xml:space="preserve">ſera partagé en deux autres triangles A B D, <lb/>B D C auſſi rectangles, & </s> <s xml:id="echoid-s7439" xml:space="preserve">qui lui ſeront ſemblables (art. </s> <s xml:id="echoid-s7440" xml:space="preserve">406). <lb/></s> <s xml:id="echoid-s7441" xml:space="preserve">Comparant ces deux triangles ſemblables, & </s> <s xml:id="echoid-s7442" xml:space="preserve">prenant les côtés <lb/>homologues, on aura A D: </s> <s xml:id="echoid-s7443" xml:space="preserve">B D:</s> <s xml:id="echoid-s7444" xml:space="preserve">: B D: </s> <s xml:id="echoid-s7445" xml:space="preserve">D C: </s> <s xml:id="echoid-s7446" xml:space="preserve">donc en pre-<lb/>nant le produit des extrêmes & </s> <s xml:id="echoid-s7447" xml:space="preserve">celui des moyens, A D x D C <lb/>= B D<emph style="sub">2</emph>. </s> <s xml:id="echoid-s7448" xml:space="preserve">C.</s> <s xml:id="echoid-s7449" xml:space="preserve">Q.</s> <s xml:id="echoid-s7450" xml:space="preserve">F.</s> <s xml:id="echoid-s7451" xml:space="preserve">D.</s> <s xml:id="echoid-s7452" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div511" type="section" level="1" n="422"> <head xml:id="echoid-head480" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s7453" xml:space="preserve">442. </s> <s xml:id="echoid-s7454" xml:space="preserve">Il ſuit de cette propoſition, qu’à quelque point du dia-<lb/>metre qu’on éleve une perpendiculaire, elle eſt toujours <lb/>moyenne proportionnelle entre les deux parties du même dia-<lb/>metre; </s> <s xml:id="echoid-s7455" xml:space="preserve">& </s> <s xml:id="echoid-s7456" xml:space="preserve">c’eſt ce que nous appellerons dans laſuite, la pro-<lb/>priété principale du cercle, de laquelle on déduit ſon équation.</s> <s xml:id="echoid-s7457" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div512" type="section" level="1" n="423"> <head xml:id="echoid-head481" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s7458" xml:space="preserve">443. </s> <s xml:id="echoid-s7459" xml:space="preserve">Il ſuit auſſi de la démonſtration précédente, qu’une <lb/>corde quelconque A B eſt moyenne proportionnelle entre le <lb/>diametre entier A C, & </s> <s xml:id="echoid-s7460" xml:space="preserve">la partie compriſe entre l’origine de <lb/>cette corde & </s> <s xml:id="echoid-s7461" xml:space="preserve">la perpendiculaire B D, abaiſſée de ſon extrê-<lb/>mité: </s> <s xml:id="echoid-s7462" xml:space="preserve">car le triangle rectangle B D A eſt ſemblable au grand <lb/>triangle C B A, puiſqu’ils ont un angle commun en A, outre <lb/>l’angle droit: </s> <s xml:id="echoid-s7463" xml:space="preserve">donc en comparant les côtés homologues, on <lb/>aura A C: </s> <s xml:id="echoid-s7464" xml:space="preserve">A B:</s> <s xml:id="echoid-s7465" xml:space="preserve">: A B: </s> <s xml:id="echoid-s7466" xml:space="preserve">A D: </s> <s xml:id="echoid-s7467" xml:space="preserve">donc A D x A C = A B<emph style="sub">2</emph>. </s> <s xml:id="echoid-s7468" xml:space="preserve">On <lb/>démontreroit de même que B C eſt moyenne proportionnelle <lb/>entre A C & </s> <s xml:id="echoid-s7469" xml:space="preserve">C D.</s> <s xml:id="echoid-s7470" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div513" type="section" level="1" n="424"> <head xml:id="echoid-head482" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s7471" xml:space="preserve">444. </s> <s xml:id="echoid-s7472" xml:space="preserve">On auroit pu déduire cette derniere propoſition de la <pb o="218" file="0256" n="256" rhead="NOUVEAU COURS"/> propoſition neuvieme; </s> <s xml:id="echoid-s7473" xml:space="preserve">car puiſque deux droites quelconques, <lb/>qui ſe coupent dans le cercle, s’y coupent de maniere que les <lb/>produits de leurs parties ſont égaux; </s> <s xml:id="echoid-s7474" xml:space="preserve">lorſque l’une des ſécantes <lb/>ſera coupée en deux également par une droite AC, comme la <lb/>ligne B F, le produit B D x D F deviendra le quarré B D; </s> <s xml:id="echoid-s7475" xml:space="preserve">& </s> <s xml:id="echoid-s7476" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0256-01a" xlink:href="note-0256-01"/> ſi l’on ſuppoſe de plus que l’autre ſécante AC paſſe par le cen-<lb/>tre, ou qu’elle eſt perpendiculaire au milieu de la ſécante B F; <lb/></s> <s xml:id="echoid-s7477" xml:space="preserve">cette ſuppoſition nous donnera préciſément l’énoncé du der-<lb/>nier théorême.</s> <s xml:id="echoid-s7478" xml:space="preserve"/> </p> <div xml:id="echoid-div513" type="float" level="2" n="1"> <note position="left" xlink:label="note-0256-01" xlink:href="note-0256-01a" xml:space="preserve">Figure 65.</note> </div> </div> <div xml:id="echoid-div515" type="section" level="1" n="425"> <head xml:id="echoid-head483" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s7479" xml:space="preserve">445. </s> <s xml:id="echoid-s7480" xml:space="preserve">La perpendiculaire BD, menée d’un point B de la circon-<lb/>férence du cercle ſur le diametre AC, eſt appellée ordonnée à ce <lb/>diametre, & </s> <s xml:id="echoid-s7481" xml:space="preserve">les parties du diametre déterminées ou coupées du <lb/>en D, comme A D, D C ſont appellées abſciſſes ou coupées du <lb/>même diametre. </s> <s xml:id="echoid-s7482" xml:space="preserve">On exprime généralement le théorême pré-<lb/>cédent, en diſant que dans un cercle, les quarrés des ordonnées <lb/>ſont égaux aux produits de leurs abſciſſes.</s> <s xml:id="echoid-s7483" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div516" type="section" level="1" n="426"> <head xml:id="echoid-head484" xml:space="preserve">PROPOSITION XII.</head> <head xml:id="echoid-head485" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7484" xml:space="preserve">446. </s> <s xml:id="echoid-s7485" xml:space="preserve">Un cercle B E étant donné avec un point D ſur le même <lb/> <anchor type="note" xlink:label="note-0256-02a" xlink:href="note-0256-02"/> plan, mener une droite DB qui aille toucher le cercle en un point B.</s> <s xml:id="echoid-s7486" xml:space="preserve"/> </p> <div xml:id="echoid-div516" type="float" level="2" n="1"> <note position="left" xlink:label="note-0256-02" xlink:href="note-0256-02a" xml:space="preserve">Figure 66.</note> </div> </div> <div xml:id="echoid-div518" type="section" level="1" n="427"> <head xml:id="echoid-head486" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s7487" xml:space="preserve">Par le centre C & </s> <s xml:id="echoid-s7488" xml:space="preserve">le point donné D, menez une ligne C D; <lb/></s> <s xml:id="echoid-s7489" xml:space="preserve">ſur cette ligne, comme diametre, décrivez un demi-cercle <lb/>C B D qui coupe le cercle donné dans un point B; </s> <s xml:id="echoid-s7490" xml:space="preserve">menez la <lb/>ligne B D, qui ſera la tangenre demandée, & </s> <s xml:id="echoid-s7491" xml:space="preserve">qui ne rencontre <lb/>le cercle qu’au ſeul point B.</s> <s xml:id="echoid-s7492" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div519" type="section" level="1" n="428"> <head xml:id="echoid-head487" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7493" xml:space="preserve">Pour concevoir la raiſon de cette opération, tirez encore au <lb/>centre C du point B la ligne BC. </s> <s xml:id="echoid-s7494" xml:space="preserve">Il eſt viſible que l’angle C B D <lb/>eſt droit (art. </s> <s xml:id="echoid-s7495" xml:space="preserve">430), étant appuyé ſur le diametre; </s> <s xml:id="echoid-s7496" xml:space="preserve">d’ailleurs, <lb/>la ligne B D eſt perpendiculaire à l’extrêmité du rayon C B, <lb/>& </s> <s xml:id="echoid-s7497" xml:space="preserve">paſſe par le point D: </s> <s xml:id="echoid-s7498" xml:space="preserve">donc elle eſt la tangente demandée. <lb/></s> <s xml:id="echoid-s7499" xml:space="preserve">C. </s> <s xml:id="echoid-s7500" xml:space="preserve">Q. </s> <s xml:id="echoid-s7501" xml:space="preserve">F. </s> <s xml:id="echoid-s7502" xml:space="preserve">T. </s> <s xml:id="echoid-s7503" xml:space="preserve">& </s> <s xml:id="echoid-s7504" xml:space="preserve">D.</s> <s xml:id="echoid-s7505" xml:space="preserve"/> </p> <pb o="219" file="0257" n="257" rhead="DE MATHÉMATIQUE. Liv. V."/> </div> <div xml:id="echoid-div520" type="section" level="1" n="429"> <head xml:id="echoid-head488" xml:space="preserve">PROPOSITION XIII.</head> <head xml:id="echoid-head489" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7506" xml:space="preserve">447. </s> <s xml:id="echoid-s7507" xml:space="preserve">Si d’un point B hors d’un cercle, on mene une tangente BA, <lb/> <anchor type="note" xlink:label="note-0257-01a" xlink:href="note-0257-01"/> & </s> <s xml:id="echoid-s7508" xml:space="preserve">une ſécante B C, je dis que le quarré de la tangente A B eſt égal <lb/>au rectangle, compris ſous la ſécante entiere BC, & </s> <s xml:id="echoid-s7509" xml:space="preserve">ſa partie ex-<lb/>térieure B D.</s> <s xml:id="echoid-s7510" xml:space="preserve"/> </p> <div xml:id="echoid-div520" type="float" level="2" n="1"> <note position="right" xlink:label="note-0257-01" xlink:href="note-0257-01a" xml:space="preserve">Figure 67.</note> </div> </div> <div xml:id="echoid-div522" type="section" level="1" n="430"> <head xml:id="echoid-head490" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7511" xml:space="preserve">Soient menées les cordes A C & </s> <s xml:id="echoid-s7512" xml:space="preserve">A D du point de contin-<lb/>gence A, aux points C, D, où la ſécante BC rencontre le cer-<lb/>cle. </s> <s xml:id="echoid-s7513" xml:space="preserve">Les triangles C A B, A D B ſeront ſemblables, car ils ont <lb/>un angle commun en B; </s> <s xml:id="echoid-s7514" xml:space="preserve">& </s> <s xml:id="echoid-s7515" xml:space="preserve">de plus, l’angle A C B, formé par <lb/>la corde A C & </s> <s xml:id="echoid-s7516" xml:space="preserve">la ſécante C B, eſt égal à l’angle B A D, formé <lb/>par la tangente A B & </s> <s xml:id="echoid-s7517" xml:space="preserve">la corde A D, puiſqu’ils ont chacun <lb/>pour meſure la moitié de l’arc A D, compris entre leurs côtés <lb/>(art. </s> <s xml:id="echoid-s7518" xml:space="preserve">429 & </s> <s xml:id="echoid-s7519" xml:space="preserve">435): </s> <s xml:id="echoid-s7520" xml:space="preserve">donc ces triangles ſont ſemblables (art. </s> <s xml:id="echoid-s7521" xml:space="preserve">402); <lb/></s> <s xml:id="echoid-s7522" xml:space="preserve">& </s> <s xml:id="echoid-s7523" xml:space="preserve">par conſéquent les côtés homologues ſont proportionnels, <lb/>& </s> <s xml:id="echoid-s7524" xml:space="preserve">donnent B C : </s> <s xml:id="echoid-s7525" xml:space="preserve">A B :</s> <s xml:id="echoid-s7526" xml:space="preserve">: A B : </s> <s xml:id="echoid-s7527" xml:space="preserve">B D : </s> <s xml:id="echoid-s7528" xml:space="preserve">donc A B<emph style="sub">2</emph> = B C x B D. </s> <s xml:id="echoid-s7529" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s7530" xml:space="preserve">Q. </s> <s xml:id="echoid-s7531" xml:space="preserve">F. </s> <s xml:id="echoid-s7532" xml:space="preserve">D.</s> <s xml:id="echoid-s7533" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div523" type="section" level="1" n="431"> <head xml:id="echoid-head491" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7534" xml:space="preserve">448. </s> <s xml:id="echoid-s7535" xml:space="preserve">Il ſuit delà, que ſi deux tangentes A B, B F ſe rencon-<lb/>trent dans un point A, les parties A B, B F de ces tangentes, <lb/>priſes depuis le point de rencontre juſqu’aux points de con-<lb/>tact, ſont égales entr’elles: </s> <s xml:id="echoid-s7536" xml:space="preserve">car on démontrera de même que <lb/>pour la tangente A B, que l’on auroit B F<emph style="sub">2</emph> = B D x B C: <lb/></s> <s xml:id="echoid-s7537" xml:space="preserve">donc puiſque A B<emph style="sub">2</emph> = B C x B D, on aura A B<emph style="sub">2</emph> = B F<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s7538" xml:space="preserve"><lb/>par conſéquent A B = B F.</s> <s xml:id="echoid-s7539" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7540" xml:space="preserve">Il eſt à remarquer, que l’on auroit pu déduire cette propo-<lb/>ſition, immédiatement de la dixieme: </s> <s xml:id="echoid-s7541" xml:space="preserve">car ſi l’on imagine <lb/>que la ſécante A B tourne au tour du point A comme d’une char-<lb/> <anchor type="note" xlink:label="note-0257-02a" xlink:href="note-0257-02"/> niere, on verra que les points B, D s’approchant continuelle-<lb/>ment l’un de l’autre, ſe confondront enfin, lorſque la ligne <lb/>A B ſera devenue tangente dans un ſeul point, & </s> <s xml:id="echoid-s7542" xml:space="preserve">alors le rec-<lb/>tangle A B x A D deviendra le quarré de la même tangente, <lb/>qui ſera égal au produit de la ſécante entiere A C par ſa partie <lb/>extérieure A E.</s> <s xml:id="echoid-s7543" xml:space="preserve"/> </p> <div xml:id="echoid-div523" type="float" level="2" n="1"> <note position="right" xlink:label="note-0257-02" xlink:href="note-0257-02a" xml:space="preserve">Figure 64.</note> </div> <pb o="220" file="0258" n="258" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div525" type="section" level="1" n="432"> <head xml:id="echoid-head492" xml:space="preserve">PROPOSITION XIV.</head> <head xml:id="echoid-head493" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7544" xml:space="preserve">449. </s> <s xml:id="echoid-s7545" xml:space="preserve">Si l’on a une tangente C D perpendiculaire à l’extrêmité <lb/> <anchor type="note" xlink:label="note-0258-01a" xlink:href="note-0258-01"/> d’un diametre A B, je dis que ſi l’on tire autant de lignes qu’on <lb/>voudra du point A à la tangente, telles que A C, A D, le quarré <lb/>du diamettre A B ſera égal au produit de cette ligne A C par la <lb/>partie intérieure A E.</s> <s xml:id="echoid-s7546" xml:space="preserve"/> </p> <div xml:id="echoid-div525" type="float" level="2" n="1"> <note position="left" xlink:label="note-0258-01" xlink:href="note-0258-01a" xml:space="preserve">Figure 68.</note> </div> </div> <div xml:id="echoid-div527" type="section" level="1" n="433"> <head xml:id="echoid-head494" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7547" xml:space="preserve">Soit menée la droite B E de l’extrêmité inférieure du dia-<lb/>metre au point E, où la droite A C coupe le cercle: </s> <s xml:id="echoid-s7548" xml:space="preserve">on aura <lb/>deux triangles rectangles ſemblables A B C, A E B: </s> <s xml:id="echoid-s7549" xml:space="preserve">car le pre-<lb/>mier A B C eſt rectangle en B, à cauſe de la tangente A D, qui <lb/>eſt perpendiculaire au diametre A B, le ſecond A E B eſt rec-<lb/>tangle en E, puiſque cet angle eſt appuyé ſur le diametre; </s> <s xml:id="echoid-s7550" xml:space="preserve">de <lb/>plus, ces triangles ont un angle commun en A: </s> <s xml:id="echoid-s7551" xml:space="preserve">donc ils ſont <lb/>ſemblables (art. </s> <s xml:id="echoid-s7552" xml:space="preserve">402), & </s> <s xml:id="echoid-s7553" xml:space="preserve">les côtés homologues nous donnent <lb/>A C : </s> <s xml:id="echoid-s7554" xml:space="preserve">A B :</s> <s xml:id="echoid-s7555" xml:space="preserve">: A B : </s> <s xml:id="echoid-s7556" xml:space="preserve">A E; </s> <s xml:id="echoid-s7557" xml:space="preserve">donc A B<emph style="sub">2</emph> = A C x A E. </s> <s xml:id="echoid-s7558" xml:space="preserve">C. </s> <s xml:id="echoid-s7559" xml:space="preserve">Q. </s> <s xml:id="echoid-s7560" xml:space="preserve">F. </s> <s xml:id="echoid-s7561" xml:space="preserve">D.</s> <s xml:id="echoid-s7562" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div528" type="section" level="1" n="434"> <head xml:id="echoid-head495" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s7563" xml:space="preserve">450. </s> <s xml:id="echoid-s7564" xml:space="preserve">L’on dit qu’une ligne eſt diviſée en moyenne & </s> <s xml:id="echoid-s7565" xml:space="preserve">ex-<lb/>trême raiſon, lorſque la ligne entiere eſt à la plus grande par-<lb/>tie; </s> <s xml:id="echoid-s7566" xml:space="preserve">comme la même plus grande partie eſt à la plus petite: <lb/></s> <s xml:id="echoid-s7567" xml:space="preserve">& </s> <s xml:id="echoid-s7568" xml:space="preserve">la plus grande partie eſt appellée médiane.</s> <s xml:id="echoid-s7569" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div529" type="section" level="1" n="435"> <head xml:id="echoid-head496" xml:space="preserve">PROPOSITION XV.</head> <head xml:id="echoid-head497" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7570" xml:space="preserve">451. </s> <s xml:id="echoid-s7571" xml:space="preserve">Diviſer une ligne donnée A B en moyenne & </s> <s xml:id="echoid-s7572" xml:space="preserve">extrême rai-<lb/> <anchor type="note" xlink:label="note-0258-02a" xlink:href="note-0258-02"/> ſon, c’eſt-à-dire de maniere que l’on ait A B : </s> <s xml:id="echoid-s7573" xml:space="preserve">A F :</s> <s xml:id="echoid-s7574" xml:space="preserve">: A F : </s> <s xml:id="echoid-s7575" xml:space="preserve">F B.</s> <s xml:id="echoid-s7576" xml:space="preserve"/> </p> <div xml:id="echoid-div529" type="float" level="2" n="1"> <note position="left" xlink:label="note-0258-02" xlink:href="note-0258-02a" xml:space="preserve">Figure 69.</note> </div> </div> <div xml:id="echoid-div531" type="section" level="1" n="436"> <head xml:id="echoid-head498" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s7577" xml:space="preserve">A l’extrêmité B de la ligne donnée A B, ſoit élevée la per-<lb/>pendiculaire B D, égale à la moitié de la même ligne A B: </s> <s xml:id="echoid-s7578" xml:space="preserve">du <lb/>point D, & </s> <s xml:id="echoid-s7579" xml:space="preserve">de l’intervale ou rayon B D, ſoit décrit un cercle <lb/>E B C, enſuite par le point A & </s> <s xml:id="echoid-s7580" xml:space="preserve">le centre D, ſoit menée la ſé-<lb/>cante A C: </s> <s xml:id="echoid-s7581" xml:space="preserve">enfin ſoit priſe A F égale à la partie extérieure A E <lb/>de la ſécante A C; </s> <s xml:id="echoid-s7582" xml:space="preserve">je dis que le point F diviſe la ligne A B en <lb/>moyenne & </s> <s xml:id="echoid-s7583" xml:space="preserve">extrême raiſon, ou, ce qui revient au même, que <lb/>l’on a A B : </s> <s xml:id="echoid-s7584" xml:space="preserve">A F :</s> <s xml:id="echoid-s7585" xml:space="preserve">: A F : </s> <s xml:id="echoid-s7586" xml:space="preserve">F B.</s> <s xml:id="echoid-s7587" xml:space="preserve"/> </p> <pb o="221" file="0259" n="259" rhead="DE MATHEMATIQUE. Liv. V."/> </div> <div xml:id="echoid-div532" type="section" level="1" n="437"> <head xml:id="echoid-head499" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7588" xml:space="preserve">Soit nommé A F ou A E x, A B a, C E ſera auſſi a, A C <lb/>ſera a + x, & </s> <s xml:id="echoid-s7589" xml:space="preserve">F B ſera a - x. </s> <s xml:id="echoid-s7590" xml:space="preserve">Cela poſé, par la propoſi-<lb/>tion 13, on a A C : </s> <s xml:id="echoid-s7591" xml:space="preserve">A B :</s> <s xml:id="echoid-s7592" xml:space="preserve">: A B : </s> <s xml:id="echoid-s7593" xml:space="preserve">A E, ou A F; </s> <s xml:id="echoid-s7594" xml:space="preserve">& </s> <s xml:id="echoid-s7595" xml:space="preserve">en termes <lb/>analytiques, a + x : </s> <s xml:id="echoid-s7596" xml:space="preserve">a :</s> <s xml:id="echoid-s7597" xml:space="preserve">: a: </s> <s xml:id="echoid-s7598" xml:space="preserve">x, & </s> <s xml:id="echoid-s7599" xml:space="preserve">faiſant le produit des extrê-<lb/>mes & </s> <s xml:id="echoid-s7600" xml:space="preserve">des moyens, il vient aa = ax + xx, faiſant paſſer en-<lb/>ſuite ax du ſecond membre dans le premier, il vient aa - ax <lb/>= xx, ou √a - x\x{0020} x a = xx, d’où l’on déduit cette proportion <lb/>a : </s> <s xml:id="echoid-s7601" xml:space="preserve">x :</s> <s xml:id="echoid-s7602" xml:space="preserve">: x : </s> <s xml:id="echoid-s7603" xml:space="preserve">a - x, ou A B : </s> <s xml:id="echoid-s7604" xml:space="preserve">A F :</s> <s xml:id="echoid-s7605" xml:space="preserve">: A F : </s> <s xml:id="echoid-s7606" xml:space="preserve">F B. </s> <s xml:id="echoid-s7607" xml:space="preserve">C. </s> <s xml:id="echoid-s7608" xml:space="preserve">Q. </s> <s xml:id="echoid-s7609" xml:space="preserve">F. </s> <s xml:id="echoid-s7610" xml:space="preserve">T. </s> <s xml:id="echoid-s7611" xml:space="preserve">& </s> <s xml:id="echoid-s7612" xml:space="preserve">D.</s> <s xml:id="echoid-s7613" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div533" type="section" level="1" n="438"> <head xml:id="echoid-head500" style="it" xml:space="preserve">Fin du cinquieme Livre.</head> <figure> <image file="0259-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0259-01"/> </figure> <pb o="222" file="0260" n="260"/> <figure> <image file="0260-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0260-01"/> </figure> </div> <div xml:id="echoid-div534" type="section" level="1" n="439"> <head xml:id="echoid-head501" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head502" xml:space="preserve">LIVRE SIXIEME,</head> <head xml:id="echoid-head503" style="it" xml:space="preserve">Qui traite des Polygones réguliers, inſcrits & circonſcrits <lb/>au cercle.</head> <head xml:id="echoid-head504" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head505" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s7614" xml:space="preserve">452. </s> <s xml:id="echoid-s7615" xml:space="preserve">ON dit qu’un polygone régulier ou irrégulier eſt inſcrit <lb/>au cercle, lorſque tous les ſommets de ſes angles ſont à la cir-<lb/>conférence du cercle.</s> <s xml:id="echoid-s7616" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div535" type="section" level="1" n="440"> <head xml:id="echoid-head506" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s7617" xml:space="preserve">453. </s> <s xml:id="echoid-s7618" xml:space="preserve">On dit qu’une figure rectiligne, réguliere ou irréguliere <lb/>eſt circonſcrite au cercle, quand chacun de ſes côtés touche la <lb/>circonférence du cercle, ou autrement, quand chaque côté eſt <lb/>une tangente au cercle.</s> <s xml:id="echoid-s7619" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div536" type="section" level="1" n="441"> <head xml:id="echoid-head507" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s7620" xml:space="preserve">454. </s> <s xml:id="echoid-s7621" xml:space="preserve">On appelle polygone régulier, une figure dont tous les <lb/>angles & </s> <s xml:id="echoid-s7622" xml:space="preserve">les côtés ſont égaux entr’eux, & </s> <s xml:id="echoid-s7623" xml:space="preserve">polygones ſymétriques, <lb/>ceux dont les côtés oppoſés ſont égaux, & </s> <s xml:id="echoid-s7624" xml:space="preserve">paralleles deux à <lb/>deux.</s> <s xml:id="echoid-s7625" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div537" type="section" level="1" n="442"> <head xml:id="echoid-head508" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s7626" xml:space="preserve">455. </s> <s xml:id="echoid-s7627" xml:space="preserve">Un polygone régulier ſe nomme pentagone, lorſqu’il a <lb/>cinq côtés; </s> <s xml:id="echoid-s7628" xml:space="preserve">exagone, quand il a ſix côtés, eptagone, quand il <pb o="223" file="0261" n="261" rhead="NOUVEAU COURS DE MATHEM. Liv. VI."/> en a ſept; </s> <s xml:id="echoid-s7629" xml:space="preserve">octogone, quand il en a huit; </s> <s xml:id="echoid-s7630" xml:space="preserve">ennéagone, quand il <lb/>en a neuf; </s> <s xml:id="echoid-s7631" xml:space="preserve">décagone, quand il en a dix; </s> <s xml:id="echoid-s7632" xml:space="preserve">& </s> <s xml:id="echoid-s7633" xml:space="preserve">enfin ondécagone <lb/>ou dodécagone, quand il en a onze ou douze.</s> <s xml:id="echoid-s7634" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div538" type="section" level="1" n="443"> <head xml:id="echoid-head509" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s7635" xml:space="preserve">456. </s> <s xml:id="echoid-s7636" xml:space="preserve">Comme tout polygone régulier peut être inſcrit dans <lb/>un cercle, on diſtingue dans tout polygone régulier deux ſortes <lb/>d’angles, les angles du centre, & </s> <s xml:id="echoid-s7637" xml:space="preserve">les angles du polygone ou de <lb/>la circonférence.</s> <s xml:id="echoid-s7638" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div539" type="section" level="1" n="444"> <head xml:id="echoid-head510" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s7639" xml:space="preserve">457. </s> <s xml:id="echoid-s7640" xml:space="preserve">L’angle au centre eſt un angle, comme B A C, formé <lb/> <anchor type="note" xlink:label="note-0261-01a" xlink:href="note-0261-01"/> par deux rayons A B & </s> <s xml:id="echoid-s7641" xml:space="preserve">A C, tirés du centre aux extrêmités d’un <lb/> <anchor type="note" xlink:label="note-0261-02a" xlink:href="note-0261-02"/> des côtés du polygone.</s> <s xml:id="echoid-s7642" xml:space="preserve"/> </p> <div xml:id="echoid-div539" type="float" level="2" n="1"> <note position="right" xlink:label="note-0261-01" xlink:href="note-0261-01a" xml:space="preserve">Planche IV.</note> <note position="right" xlink:label="note-0261-02" xlink:href="note-0261-02a" xml:space="preserve">Figure 70.</note> </div> </div> <div xml:id="echoid-div541" type="section" level="1" n="445"> <head xml:id="echoid-head511" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s7643" xml:space="preserve">458. </s> <s xml:id="echoid-s7644" xml:space="preserve">L’angle du polygone, eſt un angle comme B C D, formé <lb/>par la rencontre des deux côtés B C & </s> <s xml:id="echoid-s7645" xml:space="preserve">C D du même polygone.</s> <s xml:id="echoid-s7646" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div542" type="section" level="1" n="446"> <head xml:id="echoid-head512" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s7647" xml:space="preserve">459. </s> <s xml:id="echoid-s7648" xml:space="preserve">Comme l’angle du centre du polygone a pour meſure <lb/>l’arc, dont un des côtés du polygone eſt la corde, l’on trou-<lb/>vera toujours la valeur de cet angle, en diviſant 360, ou les <lb/>degrés de la circonférence entiere, par le nombre des côtés <lb/>du polygone. </s> <s xml:id="echoid-s7649" xml:space="preserve">Ainſi pour trouver l’angle au centre d’un exa-<lb/>gone, je diviſe 360 par 6, & </s> <s xml:id="echoid-s7650" xml:space="preserve">le quotient 60, eſt la meſure de <lb/>l’angle que je cherche. </s> <s xml:id="echoid-s7651" xml:space="preserve">Or comme l’angle B C D du polygone <lb/>eſt double de l’angle A B C, & </s> <s xml:id="echoid-s7652" xml:space="preserve">que par conſéquent il eſt égal <lb/>aux deux angles de la baſe du triangle iſoſcele A B C, il s’enſuit <lb/>qu’il eſt égal à la différence de l’angle du centre à deux droits: <lb/></s> <s xml:id="echoid-s7653" xml:space="preserve">ainſi on trouvera la valeur de l’angle du polygone, en retran-<lb/>chant l’angle du centre de 180 degrés.</s> <s xml:id="echoid-s7654" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div543" type="section" level="1" n="447"> <head xml:id="echoid-head513" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head514" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7655" xml:space="preserve">460. </s> <s xml:id="echoid-s7656" xml:space="preserve">Inſcrire un exagone dans un cercle.</s> <s xml:id="echoid-s7657" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div544" type="section" level="1" n="448"> <head xml:id="echoid-head515" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s7658" xml:space="preserve">Pour inſcrire un exagone dans un cercle, il faut prendre le <lb/> <anchor type="note" xlink:label="note-0261-03a" xlink:href="note-0261-03"/> rayon du cercle avec le compas, & </s> <s xml:id="echoid-s7659" xml:space="preserve">le porter ſix fois ſur la cir-<lb/>conférence; </s> <s xml:id="echoid-s7660" xml:space="preserve">cette opération détermine les points qui ſervent à <lb/>tracer l’exagone.</s> <s xml:id="echoid-s7661" xml:space="preserve"/> </p> <div xml:id="echoid-div544" type="float" level="2" n="1"> <note position="right" xlink:label="note-0261-03" xlink:href="note-0261-03a" xml:space="preserve">Figure 70.</note> </div> <pb o="224" file="0262" n="262" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div546" type="section" level="1" n="449"> <head xml:id="echoid-head516" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7662" xml:space="preserve">Conſidérez que le côté B C de l’exagone eſt égal au rayon <lb/>A B; </s> <s xml:id="echoid-s7663" xml:space="preserve">car comme l’angle du centre B A C de l’exagone eſt de <lb/>60 degrés, la ſomme des deux angles de la baſe du triangle <lb/>iſoſcele B A C ſera de 120 degrés, double de l’angle au centre; <lb/></s> <s xml:id="echoid-s7664" xml:space="preserve">chacun d’eux ſera donc de 60 degrés: </s> <s xml:id="echoid-s7665" xml:space="preserve">donc le triangle <lb/>B A C eſt équilatéral, & </s> <s xml:id="echoid-s7666" xml:space="preserve">le côté B C eſt égal au rayon A C. </s> <s xml:id="echoid-s7667" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s7668" xml:space="preserve">Q. </s> <s xml:id="echoid-s7669" xml:space="preserve">F. </s> <s xml:id="echoid-s7670" xml:space="preserve">D.</s> <s xml:id="echoid-s7671" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div547" type="section" level="1" n="450"> <head xml:id="echoid-head517" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head518" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7672" xml:space="preserve">461. </s> <s xml:id="echoid-s7673" xml:space="preserve">Décrire un dodécagone dans un cercle, ou, ce qui eſt la <lb/> <anchor type="note" xlink:label="note-0262-01a" xlink:href="note-0262-01"/> même choſe, une figure de douze côtés.</s> <s xml:id="echoid-s7674" xml:space="preserve"/> </p> <div xml:id="echoid-div547" type="float" level="2" n="1"> <note position="left" xlink:label="note-0262-01" xlink:href="note-0262-01a" xml:space="preserve">Figure 71.</note> </div> </div> <div xml:id="echoid-div549" type="section" level="1" n="451"> <head xml:id="echoid-head519" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s7675" xml:space="preserve">Pour déctire un dodécagone dans un cercle, il faut porter <lb/>le rayon A C ſur la circonférence, afin d’avoir l’arc C D de 60 <lb/>degrés, ou autrement égal à la ſixieme partie de la même cir-<lb/>conférence, & </s> <s xml:id="echoid-s7676" xml:space="preserve">diviſer enſuite cet arc en deux également en E, <lb/>la corde D E ſera le côté du dodécagone, puiſqu’elle eſt la <lb/>corde d’un angle de 30 degrés, qui font la douzieme partie de <lb/>la circonférence. </s> <s xml:id="echoid-s7677" xml:space="preserve">C. </s> <s xml:id="echoid-s7678" xml:space="preserve">Q. </s> <s xml:id="echoid-s7679" xml:space="preserve">F. </s> <s xml:id="echoid-s7680" xml:space="preserve">D.</s> <s xml:id="echoid-s7681" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div550" type="section" level="1" n="452"> <head xml:id="echoid-head520" xml:space="preserve"><emph style="sc">Lemme</emph>.</head> <p style="it"> <s xml:id="echoid-s7682" xml:space="preserve">462. </s> <s xml:id="echoid-s7683" xml:space="preserve">Si l’on a un triangle iſoſcele A B C, dont chaque angle de <lb/> <anchor type="note" xlink:label="note-0262-02a" xlink:href="note-0262-02"/> la baſe ſoit double de celui du ſommet; </s> <s xml:id="echoid-s7684" xml:space="preserve">je dis que ſi l’on diviſe l’un <lb/>des angles de la baſe, comme B A C en deux également par une <lb/>ligne A D, qui va rencontrer le côté oppoſé en D, cette ligne divi-<lb/>ſera ce même côté A C en moyenne & </s> <s xml:id="echoid-s7685" xml:space="preserve">extrême raiſon au point D, <lb/>enſorte que l’on aura B C : </s> <s xml:id="echoid-s7686" xml:space="preserve">B D : </s> <s xml:id="echoid-s7687" xml:space="preserve">: </s> <s xml:id="echoid-s7688" xml:space="preserve">B D : </s> <s xml:id="echoid-s7689" xml:space="preserve">D C.</s> <s xml:id="echoid-s7690" xml:space="preserve"/> </p> <div xml:id="echoid-div550" type="float" level="2" n="1"> <note position="left" xlink:label="note-0262-02" xlink:href="note-0262-02a" xml:space="preserve">Figure 72.</note> </div> </div> <div xml:id="echoid-div552" type="section" level="1" n="453"> <head xml:id="echoid-head521" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7691" xml:space="preserve">Conſidérez que les triangles A B C & </s> <s xml:id="echoid-s7692" xml:space="preserve">D A C ſont ſembla-<lb/>bles, puiſqu’ils ont un angle commun en C, & </s> <s xml:id="echoid-s7693" xml:space="preserve">que l’angle <lb/>D A C eſt égal à l’angle B, puiſque l’angle B eſt par ſuppoſi-<lb/>tion moitié de l’angle B A C, dont celui-ci eſt auſſi la moitié. <lb/></s> <s xml:id="echoid-s7694" xml:space="preserve">On aura de plus le triangle B D A, qui ſera iſoſcele, puiſque <lb/>l’angle D B A eſt égal à l’angle B A D: </s> <s xml:id="echoid-s7695" xml:space="preserve">donc les côtés A D, B D <lb/>ſeront égaux. </s> <s xml:id="echoid-s7696" xml:space="preserve">Cela poſé, les triangles ſemblables A B C, D A C <pb o="225" file="0263" n="263" rhead="DE MATHÉMATIQUE. Liv. VI."/> nous donnent par la comparaiſon des côtés homologues <lb/>B C : </s> <s xml:id="echoid-s7697" xml:space="preserve">A C : </s> <s xml:id="echoid-s7698" xml:space="preserve">: </s> <s xml:id="echoid-s7699" xml:space="preserve">A C : </s> <s xml:id="echoid-s7700" xml:space="preserve">D C, & </s> <s xml:id="echoid-s7701" xml:space="preserve">mettant B D à la place de A C, au-<lb/>quel il eſt égal, on aura B C : </s> <s xml:id="echoid-s7702" xml:space="preserve">B D : </s> <s xml:id="echoid-s7703" xml:space="preserve">: </s> <s xml:id="echoid-s7704" xml:space="preserve">B D : </s> <s xml:id="echoid-s7705" xml:space="preserve">D C. </s> <s xml:id="echoid-s7706" xml:space="preserve">C. </s> <s xml:id="echoid-s7707" xml:space="preserve">Q. </s> <s xml:id="echoid-s7708" xml:space="preserve">F. </s> <s xml:id="echoid-s7709" xml:space="preserve">D.</s> <s xml:id="echoid-s7710" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div553" type="section" level="1" n="454"> <head xml:id="echoid-head522" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s7711" xml:space="preserve">463. </s> <s xml:id="echoid-s7712" xml:space="preserve">Cette propoſition donne un moyen de faire un trian-<lb/>gle iſoſcele, dont les angles de la baſe ſoient chacun doubles <lb/>de celui du ſommet; </s> <s xml:id="echoid-s7713" xml:space="preserve">car pour faire, par exemple, un triangle <lb/>comme A B C, l’on n’aura qu’à diviſer le côté B C en moyen ne <lb/>& </s> <s xml:id="echoid-s7714" xml:space="preserve">extrême raiſon (art. </s> <s xml:id="echoid-s7715" xml:space="preserve">451), & </s> <s xml:id="echoid-s7716" xml:space="preserve">ſur la plus petite partie D C <lb/>comme baſe, faire un triangle iſoſcele par le moyen de deux <lb/>ſections, avec une ouverture de compas de la grandeur de la <lb/>médiane B D, & </s> <s xml:id="echoid-s7717" xml:space="preserve">l’on aura le point A, qui ſervira à former le <lb/>triangle A B C. </s> <s xml:id="echoid-s7718" xml:space="preserve">Comme il n’y a qu’une maniere de diviſer une <lb/>ligne en moyenne & </s> <s xml:id="echoid-s7719" xml:space="preserve">extrême raiſon, il n’y a auſſi qu’un trian-<lb/>gle qui ait la propriété que nous venons de voir.</s> <s xml:id="echoid-s7720" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div554" type="section" level="1" n="455"> <head xml:id="echoid-head523" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s7721" xml:space="preserve">464. </s> <s xml:id="echoid-s7722" xml:space="preserve">Il ſuit encore delà que ſi du point B, comme centre, <lb/>l’on décrit un cercle, dont le rayon ſoit B A ou B C, la baſe A C <lb/>du triangle iſoſcele A B C ſera le côté du décagone inſcrit dans <lb/>ce cercle: </s> <s xml:id="echoid-s7723" xml:space="preserve">car puiſque, par conſtruction, les deux angles de <lb/>la baſe ſont chacun doubles de l’angle au ſommet, les trois an-<lb/>gles du même triangle, pris enſemble, vaudront cinq fois <lb/>l’angle du ſommet; </s> <s xml:id="echoid-s7724" xml:space="preserve">& </s> <s xml:id="echoid-s7725" xml:space="preserve">comme la valeur des trois angles d’un <lb/>triangle quelconque eſt de deux angles droits, on aura la va-<lb/>leur de l’angle au ſommet, en diviſant deux droits ou 180 <lb/>degrés par 5, & </s> <s xml:id="echoid-s7726" xml:space="preserve">ce qui donnera 36 pour le nombre des degrés de <lb/>l’angle au centre B, lequel nombre eſt préciſément la dixieme <lb/>partie de la circonférence, ou de 360 degrés.</s> <s xml:id="echoid-s7727" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div555" type="section" level="1" n="456"> <head xml:id="echoid-head524" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head525" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7728" xml:space="preserve">465. </s> <s xml:id="echoid-s7729" xml:space="preserve">Inſcrire un décagone dans un cercle.</s> <s xml:id="echoid-s7730" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7731" xml:space="preserve">Pour inſcrire un décagone dans un cercle, il faut diviſer le <lb/>rayon de ce cercle en moyenne & </s> <s xml:id="echoid-s7732" xml:space="preserve">extrême raiſon, la médiane <lb/>ſera le côté du décagone; </s> <s xml:id="echoid-s7733" xml:space="preserve">ainſi l’on n’aura qu’à porter dix fois <lb/>cette ligne ſur la circonférence, & </s> <s xml:id="echoid-s7734" xml:space="preserve">l’on aura les points qui ſer-<lb/>viront à tracer le décagone; </s> <s xml:id="echoid-s7735" xml:space="preserve">ce qui eſt évident, puiſque par le <pb o="226" file="0264" n="264" rhead="NOUVEAU COURS"/> corollaire précédent, la médiane d’une ligne quelconque, di-<lb/>viſée en moyenne & </s> <s xml:id="echoid-s7736" xml:space="preserve">extrême raiſon, eſt le côté du décagone <lb/>inſcrit au cercle qui auroit cette ligne pour rayon.</s> <s xml:id="echoid-s7737" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div556" type="section" level="1" n="457"> <head xml:id="echoid-head526" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head527" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7738" xml:space="preserve">466. </s> <s xml:id="echoid-s7739" xml:space="preserve">Si l’on a une ligne droite B D égale à la ſomme des côtés <lb/> <anchor type="note" xlink:label="note-0264-01a" xlink:href="note-0264-01"/> de l’exagone & </s> <s xml:id="echoid-s7740" xml:space="preserve">du décagone inſcrit au même cercle, elle ſera diviſée <lb/>en moyenne & </s> <s xml:id="echoid-s7741" xml:space="preserve">extrême raiſon au point de jonction de ces deux côtés.</s> <s xml:id="echoid-s7742" xml:space="preserve"/> </p> <div xml:id="echoid-div556" type="float" level="2" n="1"> <note position="left" xlink:label="note-0264-01" xlink:href="note-0264-01a" xml:space="preserve">Figure 74.</note> </div> </div> <div xml:id="echoid-div558" type="section" level="1" n="458"> <head xml:id="echoid-head528" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7743" xml:space="preserve">Soit la ligne B C égale au côté du décagone inſcrit au cer-<lb/>cle, qui a pour rayon B A ou A C. </s> <s xml:id="echoid-s7744" xml:space="preserve">Soit prolongée cette ligne <lb/>en D, de maniere que l’on ait D C = A C, puiſque le rayon <lb/>eſt le côté de l’exagone; </s> <s xml:id="echoid-s7745" xml:space="preserve">il faut faire voir que l’on aura B D: <lb/></s> <s xml:id="echoid-s7746" xml:space="preserve">D C : </s> <s xml:id="echoid-s7747" xml:space="preserve">: </s> <s xml:id="echoid-s7748" xml:space="preserve">D C : </s> <s xml:id="echoid-s7749" xml:space="preserve">B C. </s> <s xml:id="echoid-s7750" xml:space="preserve">Pour cela, ſoit tirée la ligne A D qui nous <lb/>donnera le triangle iſoſcele D C A, & </s> <s xml:id="echoid-s7751" xml:space="preserve">le nouveau triangle <lb/>B D A ſemblable au triangle B A C, puiſque ces triangles ont <lb/>l’angle B commun, & </s> <s xml:id="echoid-s7752" xml:space="preserve">que l’angle B D A eſt égal à l’angle <lb/>C A B; </s> <s xml:id="echoid-s7753" xml:space="preserve">car à cauſe du triangle iſoſcele D C A, l’angle A C B <lb/>qui lui eſt extérieur, eſt double de l’angle C A D, ou C D A; </s> <s xml:id="echoid-s7754" xml:space="preserve"><lb/>mais par la nature du côté du décagone, le même angle eſt <lb/>double de l’angle B A C au centre A: </s> <s xml:id="echoid-s7755" xml:space="preserve">donc l’angle B D A eſt <lb/>égal à l’angle C A B: </s> <s xml:id="echoid-s7756" xml:space="preserve">donc les triangles B D A, B A C ſont <lb/>ſemblables, & </s> <s xml:id="echoid-s7757" xml:space="preserve">les côtés homologues ſeront proportionnels; </s> <s xml:id="echoid-s7758" xml:space="preserve"><lb/>ainſi l’on aura B D : </s> <s xml:id="echoid-s7759" xml:space="preserve">B A : </s> <s xml:id="echoid-s7760" xml:space="preserve">: </s> <s xml:id="echoid-s7761" xml:space="preserve">B A : </s> <s xml:id="echoid-s7762" xml:space="preserve">B C, ou en mettant D C au lieu <lb/>de B A qui lui eſt égal, B D : </s> <s xml:id="echoid-s7763" xml:space="preserve">D C : </s> <s xml:id="echoid-s7764" xml:space="preserve">: </s> <s xml:id="echoid-s7765" xml:space="preserve">D C : </s> <s xml:id="echoid-s7766" xml:space="preserve">B C. </s> <s xml:id="echoid-s7767" xml:space="preserve">C. </s> <s xml:id="echoid-s7768" xml:space="preserve">Q. </s> <s xml:id="echoid-s7769" xml:space="preserve">F. </s> <s xml:id="echoid-s7770" xml:space="preserve">D.</s> <s xml:id="echoid-s7771" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div559" type="section" level="1" n="459"> <head xml:id="echoid-head529" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head530" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s7772" xml:space="preserve">467. </s> <s xml:id="echoid-s7773" xml:space="preserve">Le quarré du côté du pentagone inſcrit dans un cercle eſt <lb/> <anchor type="note" xlink:label="note-0264-02a" xlink:href="note-0264-02"/> égal à la ſomme des quarrés de l’exagone & </s> <s xml:id="echoid-s7774" xml:space="preserve">du décagone inſcrits au <lb/>même cercle.</s> <s xml:id="echoid-s7775" xml:space="preserve"/> </p> <div xml:id="echoid-div559" type="float" level="2" n="1"> <note position="left" xlink:label="note-0264-02" xlink:href="note-0264-02a" xml:space="preserve">Figure 75.</note> </div> </div> <div xml:id="echoid-div561" type="section" level="1" n="460"> <head xml:id="echoid-head531" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7776" xml:space="preserve">Si l’on a dans un cercle le côté A B du pentagone, & </s> <s xml:id="echoid-s7777" xml:space="preserve">qu’on <lb/>diviſe en deux également au point C l’are A C B, la corde A C <lb/>ou C B ſera le côté du décagone, & </s> <s xml:id="echoid-s7778" xml:space="preserve">le rayon A D ou B D le <lb/>côté de l’exagone. </s> <s xml:id="echoid-s7779" xml:space="preserve">Il faut démontrer que l’on aura A B<emph style="sub">2</emph> = B D<emph style="sub">2</emph> <pb o="227" file="0265" n="265" rhead="DE MATHEMATIQUE. Liv. VI."/> + A C. </s> <s xml:id="echoid-s7780" xml:space="preserve">Pour cela, ſoit encore diviſé l’arc A C en deux éga-<lb/>lement en F, ſoit mené le rayon F D & </s> <s xml:id="echoid-s7781" xml:space="preserve">du point E, où il coupe <lb/>le côté A B du pentagone, ſoit tirée la droite E C. </s> <s xml:id="echoid-s7782" xml:space="preserve">Le triangle <lb/>A E C ſera iſoſcele & </s> <s xml:id="echoid-s7783" xml:space="preserve">ſemblable au triangle A C D; </s> <s xml:id="echoid-s7784" xml:space="preserve">car puiſ-<lb/>que la droite F D coupe l’arc A C en deux parties égales, & </s> <s xml:id="echoid-s7785" xml:space="preserve"><lb/>paſſe par le centre; </s> <s xml:id="echoid-s7786" xml:space="preserve">elle coupe auſſi la corde en deux parties <lb/>égales, & </s> <s xml:id="echoid-s7787" xml:space="preserve">lui eſt perpendiculaire: </s> <s xml:id="echoid-s7788" xml:space="preserve">donc tous les points de cette <lb/>droite F D ſont également éloignés des extrêmités A C, ainſi <lb/>l’on aura A E = E C. </s> <s xml:id="echoid-s7789" xml:space="preserve">De plus, ce triangle a un angle com-<lb/>mun avec le triangle iſoſcele A C B: </s> <s xml:id="echoid-s7790" xml:space="preserve">donc ils ſont ſemblables; <lb/></s> <s xml:id="echoid-s7791" xml:space="preserve">& </s> <s xml:id="echoid-s7792" xml:space="preserve">comparant les côtés homologues on aura A B : </s> <s xml:id="echoid-s7793" xml:space="preserve">A C : </s> <s xml:id="echoid-s7794" xml:space="preserve">: </s> <s xml:id="echoid-s7795" xml:space="preserve">A C : </s> <s xml:id="echoid-s7796" xml:space="preserve">A E; </s> <s xml:id="echoid-s7797" xml:space="preserve"><lb/>donc A C<emph style="sub">2</emph> = A B x A E. </s> <s xml:id="echoid-s7798" xml:space="preserve">De même le triangle A D B eſt ſem-<lb/>blable au triangle D E B, car ces triangles ont un angle com-<lb/>mun en B, qui vaut 54 degrés; </s> <s xml:id="echoid-s7799" xml:space="preserve">mais l’angle B D F eſt auſſi de <lb/>54 degrés, ayant pour meſure l’arc F B, qui vaut C B de 36 <lb/>degrés, plus F C de 18 degrés, puiſque F C eſt moitié de l’arc <lb/>A C; </s> <s xml:id="echoid-s7800" xml:space="preserve">ce triangle D E B ſera donc iſoſcele, ainſi que le trian-<lb/>gle A D B, & </s> <s xml:id="echoid-s7801" xml:space="preserve">comparant les côtés homologues, on aura <lb/>A B : </s> <s xml:id="echoid-s7802" xml:space="preserve">B D : </s> <s xml:id="echoid-s7803" xml:space="preserve">: </s> <s xml:id="echoid-s7804" xml:space="preserve">B D : </s> <s xml:id="echoid-s7805" xml:space="preserve">B E; </s> <s xml:id="echoid-s7806" xml:space="preserve">donc B D<emph style="sub">2</emph> = A B x B E. </s> <s xml:id="echoid-s7807" xml:space="preserve">Et ajoutant <lb/>aux membres de cette équation ceux de l’équation précédente, <lb/>on aura B D<emph style="sub">2</emph> + A C<emph style="sub">2</emph> = A B x A E + A B x B E. </s> <s xml:id="echoid-s7808" xml:space="preserve">Mais A B <lb/>x A E + A B x B E = A B x (A E + B E) = A B x A B = <lb/>A B<emph style="sub">2</emph>: </s> <s xml:id="echoid-s7809" xml:space="preserve">donc B D<emph style="sub">2</emph> + A C<emph style="sub">2</emph> = A B<emph style="sub">2</emph>. </s> <s xml:id="echoid-s7810" xml:space="preserve">C. </s> <s xml:id="echoid-s7811" xml:space="preserve">Q. </s> <s xml:id="echoid-s7812" xml:space="preserve">F. </s> <s xml:id="echoid-s7813" xml:space="preserve">D.</s> <s xml:id="echoid-s7814" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div562" type="section" level="1" n="461"> <head xml:id="echoid-head532" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head533" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7815" xml:space="preserve">468. </s> <s xml:id="echoid-s7816" xml:space="preserve">Inſcrire un Pentagone dans un cercle.</s> <s xml:id="echoid-s7817" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div563" type="section" level="1" n="462"> <head xml:id="echoid-head534" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s7818" xml:space="preserve">Pour inſcrire un pentagone dans un cercle, tirez le rayon <lb/> <anchor type="note" xlink:label="note-0265-01a" xlink:href="note-0265-01"/> C F, perpendiculaire ſur le diametre A B, & </s> <s xml:id="echoid-s7819" xml:space="preserve">diviſez le demi-<lb/>diametre C B en deux également au point E; </s> <s xml:id="echoid-s7820" xml:space="preserve">de ce point <lb/>comme centre, & </s> <s xml:id="echoid-s7821" xml:space="preserve">de l’intervalle E F, décrivez l’arc F D, & </s> <s xml:id="echoid-s7822" xml:space="preserve"><lb/>la ligne F D ſera le côté du pentagone inſcrit au cercle A F D.</s> <s xml:id="echoid-s7823" xml:space="preserve"/> </p> <div xml:id="echoid-div563" type="float" level="2" n="1"> <note position="right" xlink:label="note-0265-01" xlink:href="note-0265-01a" xml:space="preserve">Figure 76.</note> </div> </div> <div xml:id="echoid-div565" type="section" level="1" n="463"> <head xml:id="echoid-head535" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s7824" xml:space="preserve">Pour le prouver, conſidérez que le triangle D F C eſt rec-<lb/>tangle, par conſtruction, & </s> <s xml:id="echoid-s7825" xml:space="preserve">que le côté C F étant celui de <lb/>l’exagone, il ſuffira de faire voir que le côté D C eſt celui du <lb/>décagone: </s> <s xml:id="echoid-s7826" xml:space="preserve">car pour que la ligne F D ſoit le côté du pentagone, <pb o="228" file="0266" n="266" rhead="NOUVEAU COURS"/> on ſçait, par le théorême précédent, qu’il faut que le quarré <lb/>de ce côté ſoit égal aux quarrés des côtés de l’exagone & </s> <s xml:id="echoid-s7827" xml:space="preserve">du <lb/>décagone. </s> <s xml:id="echoid-s7828" xml:space="preserve">Pour cela, nous nommerons C F ou C B, a; </s> <s xml:id="echoid-s7829" xml:space="preserve">par <lb/>conſéquent C E ſera {1/2} a, l’inconnue D C ſera nommée x, ainſi <lb/>B D ſera a+x. </s> <s xml:id="echoid-s7830" xml:space="preserve">Cela poſé, comme E F eſt égal à E D, on <lb/>aura, à cauſe du triangle rectangle E C F, C E<emph style="sub">2</emph> + C F<emph style="sub">2</emph> = E F<emph style="sub">2</emph>, <lb/>ou en termes analytiques aa + {1/4} aa = xx + ax + {1/4} aa, ou <lb/>aa = xx + ax, en effaçant {1/4} aa de part & </s> <s xml:id="echoid-s7831" xml:space="preserve">d’autre; </s> <s xml:id="echoid-s7832" xml:space="preserve">d’où l’on <lb/>tire a + x: </s> <s xml:id="echoid-s7833" xml:space="preserve">a:</s> <s xml:id="echoid-s7834" xml:space="preserve">: a: </s> <s xml:id="echoid-s7835" xml:space="preserve">x, ou D B: </s> <s xml:id="echoid-s7836" xml:space="preserve">C B:</s> <s xml:id="echoid-s7837" xml:space="preserve">: C B: </s> <s xml:id="echoid-s7838" xml:space="preserve">D C, qui montre <lb/>que la ligne D B eſt diviſée en moyenne & </s> <s xml:id="echoid-s7839" xml:space="preserve">extrême raiſon au <lb/>point C; </s> <s xml:id="echoid-s7840" xml:space="preserve">& </s> <s xml:id="echoid-s7841" xml:space="preserve">par conſéquent (art. </s> <s xml:id="echoid-s7842" xml:space="preserve">466) la ligne D C eſt le côté <lb/>du décagone, puiſque B C eſt celui de l’exagone. </s> <s xml:id="echoid-s7843" xml:space="preserve">C. </s> <s xml:id="echoid-s7844" xml:space="preserve">Q. </s> <s xml:id="echoid-s7845" xml:space="preserve">F. </s> <s xml:id="echoid-s7846" xml:space="preserve">D.</s> <s xml:id="echoid-s7847" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div566" type="section" level="1" n="464"> <head xml:id="echoid-head536" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head537" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7848" xml:space="preserve">469. </s> <s xml:id="echoid-s7849" xml:space="preserve">Inſcrire un quarré dans un cercle.</s> <s xml:id="echoid-s7850" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7851" xml:space="preserve">Pour inſcrire un quarré dans un cercle, tirez le diametre <lb/> <anchor type="note" xlink:label="note-0266-01a" xlink:href="note-0266-01"/> A B, ſur le milieu de ce diametre, élevez un ſecond diametre <lb/>C E D perpendiculaire au premier: </s> <s xml:id="echoid-s7852" xml:space="preserve">ces deux diametres coupe-<lb/>ront la circonférence en quatre parties égales dans les points <lb/>A, C, B, D, par leſquels vous menerez les droites A C, C B, <lb/>B D & </s> <s xml:id="echoid-s7853" xml:space="preserve">D A, qui formeront un quarré; </s> <s xml:id="echoid-s7854" xml:space="preserve">car toutes ces lignes <lb/>ſont égales, puiſqu’elles ſont des cordes d’arcs égaux; </s> <s xml:id="echoid-s7855" xml:space="preserve">& </s> <s xml:id="echoid-s7856" xml:space="preserve">de <lb/>plus, chacun des angles de cette figure eſt droit, puiſqu’il eft <lb/>appuyé ſur le diametre. </s> <s xml:id="echoid-s7857" xml:space="preserve">C. </s> <s xml:id="echoid-s7858" xml:space="preserve">Q. </s> <s xml:id="echoid-s7859" xml:space="preserve">F. </s> <s xml:id="echoid-s7860" xml:space="preserve">T. </s> <s xml:id="echoid-s7861" xml:space="preserve">& </s> <s xml:id="echoid-s7862" xml:space="preserve">D.</s> <s xml:id="echoid-s7863" xml:space="preserve"/> </p> <div xml:id="echoid-div566" type="float" level="2" n="1"> <note position="left" xlink:label="note-0266-01" xlink:href="note-0266-01a" xml:space="preserve">Figure 77.</note> </div> </div> <div xml:id="echoid-div568" type="section" level="1" n="465"> <head xml:id="echoid-head538" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head539" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7864" xml:space="preserve">470. </s> <s xml:id="echoid-s7865" xml:space="preserve">Inſcrire un octogone dans un cercle.</s> <s xml:id="echoid-s7866" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 77.</note> <p> <s xml:id="echoid-s7867" xml:space="preserve">Pour inſcrire un octogone dans un cercle, il faut d’abord <lb/>diviſer ſa circonférence en quatre parties égales, comme ſi l’on <lb/>vouloit y inſcrire un quarré, & </s> <s xml:id="echoid-s7868" xml:space="preserve">diviſer en deux également <lb/>chaque quart de cercle, tel que C B; </s> <s xml:id="echoid-s7869" xml:space="preserve">la corde C F ou F B ſera <lb/>le côté de l’octogone.</s> <s xml:id="echoid-s7870" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div569" type="section" level="1" n="466"> <head xml:id="echoid-head540" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s7871" xml:space="preserve">Nous n’avons point parlé de la maniere d’inſcrire dans un <lb/>cercle l’eptagone, l’ennéagone, ni l’ondécagone, parce que l’on <pb o="229" file="0267" n="267" rhead="DE MATHÉMATIQUE. Liv. VI."/> n’a pas encore trouvé le moyen de tracer géométriquement <lb/>ces trois polygones, ſimplement avec la regle & </s> <s xml:id="echoid-s7872" xml:space="preserve">le compas, <lb/>étant obligé d’avoir recours à la Géométrie compoſée, c’eſt-<lb/>à-dire à la Géométrie des courbes. </s> <s xml:id="echoid-s7873" xml:space="preserve">Il s’en faut beaucoup que <lb/>que les ſolutions des problêmes, par le moyen des courbes, <lb/>ſoient auſſi ſimples que celles que l’on trouve par la regle & </s> <s xml:id="echoid-s7874" xml:space="preserve">le <lb/>compas, c’eſt ce qui a fait regarder juſqu’ici ces ſortes de pro-<lb/>blêmes comme très-difficiles, ainſi que celui de la triſection <lb/>de l’angle, où il s’agit de diviſer un angle donné en trois par-<lb/>ties égales, & </s> <s xml:id="echoid-s7875" xml:space="preserve">dont l’équation monte au troiſieme degré. <lb/></s> <s xml:id="echoid-s7876" xml:space="preserve">Comme nous ne parlons pas de ces ſortes d’équations dans <lb/>ce Traité, nous allons donner le moyen de tracer une courbe, <lb/>que l’on a nommé quadratrice de Dinoſtrate, du nom de ſon <lb/>inventeur, par le moyen de laquelle on pourra diviſer les an-<lb/>gles & </s> <s xml:id="echoid-s7877" xml:space="preserve">les arcs de cercles, en autant des parties égales que l’on <lb/>voudra; </s> <s xml:id="echoid-s7878" xml:space="preserve">mais auparavant il faut être prévenu des deux pro-<lb/>blêmes ſuivans.</s> <s xml:id="echoid-s7879" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div570" type="section" level="1" n="467"> <head xml:id="echoid-head541" xml:space="preserve"><emph style="sc">Probleme</emph> I.</head> <p style="it"> <s xml:id="echoid-s7880" xml:space="preserve">471. </s> <s xml:id="echoid-s7881" xml:space="preserve">Diviſer une ligne droite en autant de parties égales que <lb/> <anchor type="note" xlink:label="note-0267-01a" xlink:href="note-0267-01"/> l’on voudra.</s> <s xml:id="echoid-s7882" xml:space="preserve"/> </p> <div xml:id="echoid-div570" type="float" level="2" n="1"> <note position="right" xlink:label="note-0267-01" xlink:href="note-0267-01a" xml:space="preserve">Figure 80.</note> </div> <p> <s xml:id="echoid-s7883" xml:space="preserve">Pour diviſer une ligne A B, par exemple, en neuſ parties <lb/>égales, tirez la ligne A C, qui faſſe avec A B un angle à <lb/>volonté; </s> <s xml:id="echoid-s7884" xml:space="preserve">du point A comme centre, & </s> <s xml:id="echoid-s7885" xml:space="preserve">du rayon A B, <lb/>décrivez l’arc B C, qui ſera la meſure de l’angle C A B; </s> <s xml:id="echoid-s7886" xml:space="preserve">en-<lb/>ſuite avec la même ouverture de compas, & </s> <s xml:id="echoid-s7887" xml:space="preserve">du point B com-<lb/>me centre, décrivez l’arc A D égal à B C, & </s> <s xml:id="echoid-s7888" xml:space="preserve">tirez la ligne <lb/>B D, qui donnera l’angle A B D égal à l’angle C A B. </s> <s xml:id="echoid-s7889" xml:space="preserve">Cela <lb/>poſé, marquez ſur le côté A C avec une ouverture de compas <lb/>à volonté, un nombre de parties égales, tel que celui dans le-<lb/>quel on veut diviſer la ligne A B, c’eſt-à-dire qu’en commen-<lb/>cant du point A, il faut marquer neuf parties égales ſur la <lb/>ligne A C; </s> <s xml:id="echoid-s7890" xml:space="preserve">aprés quoi il en faudra faire autant ſur la ligne <lb/>B D, en commençant du point B: </s> <s xml:id="echoid-s7891" xml:space="preserve">après cela, ſi l’on tire les <lb/>lignes 9 A, 81, 72, &</s> <s xml:id="echoid-s7892" xml:space="preserve">c. </s> <s xml:id="echoid-s7893" xml:space="preserve">elles diviſeront la ligne A B en neuf <lb/>parties égales; </s> <s xml:id="echoid-s7894" xml:space="preserve">ce qui eſt bien évident: </s> <s xml:id="echoid-s7895" xml:space="preserve">car comme les lignes <lb/>que l’on a tirées ſont paralleles entr’elles, elles donneront les <lb/>triangles ſemblables A1E, A9B, qui font voir que puiſque <lb/>A1 eſt la neuvieme partie de A9, A E ſera la neuvieme partie <lb/>de A B, ainſi des autres.</s> <s xml:id="echoid-s7896" xml:space="preserve"/> </p> <pb o="230" file="0268" n="268" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div572" type="section" level="1" n="468"> <head xml:id="echoid-head542" xml:space="preserve"><emph style="sc">Probleme</emph> II.</head> <p style="it"> <s xml:id="echoid-s7897" xml:space="preserve">472. </s> <s xml:id="echoid-s7898" xml:space="preserve">Diviſer un arc de cercle en un nombre de parties égales, <lb/>pairement paires, c’eſt-à-dire qui ſoit diviſible par les nombres <lb/>deux, & </s> <s xml:id="echoid-s7899" xml:space="preserve">ſes puiſſances 4, 8, 16, 32.</s> <s xml:id="echoid-s7900" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div573" type="section" level="1" n="469"> <head xml:id="echoid-head543" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s7901" xml:space="preserve">Si l’on veut diviſer, par exemple, le quart de cercle A B C <lb/> <anchor type="note" xlink:label="note-0268-01a" xlink:href="note-0268-01"/> en ſeize parties égales, il faut des points A & </s> <s xml:id="echoid-s7902" xml:space="preserve">C décrire avec la <lb/>même ouverture de compas la ſection D, & </s> <s xml:id="echoid-s7903" xml:space="preserve">tirer la ligne B D, <lb/>qui diviſera l’arc A C en deux également au point E; </s> <s xml:id="echoid-s7904" xml:space="preserve">diviſer <lb/>de la même maniere l’arc E C en deux également au point F, <lb/>l’arc F C encore en deux également au point G, & </s> <s xml:id="echoid-s7905" xml:space="preserve">l’arc G C <lb/>en deux également au point H, pour avoir l’arc C H, quiſera <lb/>la ſeizieme partie de A C, & </s> <s xml:id="echoid-s7906" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s7907" xml:space="preserve"/> </p> <div xml:id="echoid-div573" type="float" level="2" n="1"> <note position="left" xlink:label="note-0268-01" xlink:href="note-0268-01a" xml:space="preserve">Figure 81.</note> </div> <p> <s xml:id="echoid-s7908" xml:space="preserve">C’eſt ainſi qu’on pourra diviſer géométriquement un arc de <lb/>cercle en un nombre infini de parties égales, pourvu que l’on <lb/>diviſe le tout, & </s> <s xml:id="echoid-s7909" xml:space="preserve">ſes parties toujours de deux en deux.</s> <s xml:id="echoid-s7910" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div575" type="section" level="1" n="470"> <head xml:id="echoid-head544" style="it" xml:space="preserve">Maniere de décrire la Quadratrice.</head> <p> <s xml:id="echoid-s7911" xml:space="preserve">473. </s> <s xml:id="echoid-s7912" xml:space="preserve">Pour décrire cette courbe, il faut diviſer le rayon A B <lb/>en un grand nombres de parties égales; </s> <s xml:id="echoid-s7913" xml:space="preserve">de maniere que le <lb/>quart de cercle puiſſe être diviſé dans le même nombre de <lb/>parties égales. </s> <s xml:id="echoid-s7914" xml:space="preserve">Nous ſuppoſerons donc que l’on a diviſé le quart <lb/>de cercle en ſeize parties égales, ainſi que le rayon A B. </s> <s xml:id="echoid-s7915" xml:space="preserve">Cela <lb/>poſé, après avoir tiré du centre B à l’extrêmité de chaque par-<lb/>rie égale du quart de cercle, les droites BC, BD, BE, BF, &</s> <s xml:id="echoid-s7916" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s7917" xml:space="preserve">l’on tirera par les points G,H,I,K des parties égales du rayon, <lb/>parallélement au diametre B F, les droites G L, H M, I N, K G; </s> <s xml:id="echoid-s7918" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s7919" xml:space="preserve">les rencontres de ces droites, avec les rayons qui diviſent le <lb/>quart de cercle, donneront les points L, M, N, O, &</s> <s xml:id="echoid-s7920" xml:space="preserve">c. </s> <s xml:id="echoid-s7921" xml:space="preserve">avec <lb/>leſquels on tracera la courbe A S, que l’on pourra faire beau-<lb/>coup plus juſte, en diviſant le quart de cercle & </s> <s xml:id="echoid-s7922" xml:space="preserve">le rayon B A <lb/>en un plus grand nombre de parties égales que l’on n’a fait <lb/>ici, afin d’avoir les points L, M, N, O beaucoup plus près les <lb/>uns des autres, & </s> <s xml:id="echoid-s7923" xml:space="preserve">que le point R, formé par la rencontre du <lb/>dernier rayon B P, & </s> <s xml:id="echoid-s7924" xml:space="preserve">la parallele Q R approche le plus près <lb/>qu’il eſt poſſible du demi-diametre B T, pour rendre inſen-<lb/>ſible l’erreur que l’on pourroit faire, en continuant méchani- <pb o="231" file="0269" n="269" rhead="DE MATHÉMATIQUE. Liv. VI."/> quement la courbe AR, juſqu’à la rencontre du demi-diametre.</s> <s xml:id="echoid-s7925" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7926" xml:space="preserve">Il faut bien remarquer que par la génération de cette courbe, <lb/>ſi l’on mene des paralleles H M & </s> <s xml:id="echoid-s7927" xml:space="preserve">K O, qui aillent rencontrer <lb/>la courbe aux points M & </s> <s xml:id="echoid-s7928" xml:space="preserve">O, & </s> <s xml:id="echoid-s7929" xml:space="preserve">que l’on tire par ces points <lb/>des rayons B D & </s> <s xml:id="echoid-s7930" xml:space="preserve">B F, qu’il y aura même raiſon de l’arc A D <lb/>à l’arc D F, que de la ligne A H à la ligne H K; </s> <s xml:id="echoid-s7931" xml:space="preserve">& </s> <s xml:id="echoid-s7932" xml:space="preserve">c’eſt dans <lb/>cette proportion que conſiſte la nature de cette courbe.</s> <s xml:id="echoid-s7933" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div576" type="section" level="1" n="471"> <head xml:id="echoid-head545" xml:space="preserve">PROPOSITION IX.</head> <head xml:id="echoid-head546" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7934" xml:space="preserve">474. </s> <s xml:id="echoid-s7935" xml:space="preserve">Diviſer un angle en trois parties égales.</s> <s xml:id="echoid-s7936" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7937" xml:space="preserve">Suppoſant que l’on ait tracé ſur un morceau de corne ou de <lb/> <anchor type="note" xlink:label="note-0269-01a" xlink:href="note-0269-01"/> carton bien uni la courbe A D, de la façon qu’on vient de l’en-<lb/>ſeigner, on propoſe de diviſer l’angle O P Q en trois parties <lb/>égales.</s> <s xml:id="echoid-s7938" xml:space="preserve"/> </p> <div xml:id="echoid-div576" type="float" level="2" n="1"> <note position="right" xlink:label="note-0269-01" xlink:href="note-0269-01a" xml:space="preserve">Figure 83 <lb/>& 85.</note> </div> <p> <s xml:id="echoid-s7939" xml:space="preserve">Pour réſoudre ce problême, ſuppoſant que la courbe ſoit <lb/>accompagnée de ſon quart de cercle A C, je fais l’angle A B E <lb/>égal à l’angle donné, & </s> <s xml:id="echoid-s7940" xml:space="preserve">au point F, où le rayon B E coupe la <lb/>courbe A D, j’abaiſſe la perpendiculaire F G ſur le demi-dia-<lb/>metre A B, qui me donne la partie A G, que je diviſe en au-<lb/>tant de parties égales qu’on veut que l’angle donné ſoit diviſé: <lb/></s> <s xml:id="echoid-s7941" xml:space="preserve">ainſi je la partage en trois parties égales, aux points H & </s> <s xml:id="echoid-s7942" xml:space="preserve">K, <lb/>deſquels je mene les paralleles K L & </s> <s xml:id="echoid-s7943" xml:space="preserve">H I, qui me coupent la <lb/>courbe aux points L & </s> <s xml:id="echoid-s7944" xml:space="preserve">I, par leſquels je mene les rayons B M <lb/>& </s> <s xml:id="echoid-s7945" xml:space="preserve">B N, qui diviſent l’arc A E en trois parties égales, aux points <lb/>M & </s> <s xml:id="echoid-s7946" xml:space="preserve">N; </s> <s xml:id="echoid-s7947" xml:space="preserve">puiſque par la propriété de la courbe, il y a même <lb/>raiſon de A K à A G, que de A M à A E; </s> <s xml:id="echoid-s7948" xml:space="preserve">& </s> <s xml:id="echoid-s7949" xml:space="preserve">comme A K eſt la <lb/>troiſieme partie de A G, l’arc A M ſera auſſi la troiſieme partie <lb/>de l’arc A E.</s> <s xml:id="echoid-s7950" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7951" xml:space="preserve">Mais ſi l’on propoſoit de diviſer un angle obtus, comme <lb/> <anchor type="note" xlink:label="note-0269-02a" xlink:href="note-0269-02"/> R S T en trois parties égales, il ſemble que cela ſouffriroit <lb/>quelque difficulté, parce que l’arc R T ne peut pas être con-<lb/>tenu dans l’arc A C, puiſqu’il eſt ſuppoſé plus grand que lui: <lb/></s> <s xml:id="echoid-s7952" xml:space="preserve">en ce cas, il faut diviſer en deux également l’angle obtus <lb/>donné, pour avoir l’angle aigu R S V, que nous ſuppoſerons <lb/>être le même que l’angle A B E: </s> <s xml:id="echoid-s7953" xml:space="preserve">ainſi diviſant l’angle aigu en <lb/>trois parties égales, aux points M & </s> <s xml:id="echoid-s7954" xml:space="preserve">N, l’on n’aura qu’à pren-<lb/>dre l’arc A N, qui étant double de la ſixieme partie de l’arc <lb/>R T, ſera par conſéquent le tiers du même arc R T. </s> <s xml:id="echoid-s7955" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s7956" xml:space="preserve">Q. </s> <s xml:id="echoid-s7957" xml:space="preserve">F. </s> <s xml:id="echoid-s7958" xml:space="preserve">T. </s> <s xml:id="echoid-s7959" xml:space="preserve">& </s> <s xml:id="echoid-s7960" xml:space="preserve">D.</s> <s xml:id="echoid-s7961" xml:space="preserve"/> </p> <div xml:id="echoid-div577" type="float" level="2" n="2"> <note position="right" xlink:label="note-0269-02" xlink:href="note-0269-02a" xml:space="preserve">Figure 84.</note> </div> <pb o="232" file="0270" n="270" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div579" type="section" level="1" n="472"> <head xml:id="echoid-head547" xml:space="preserve">PROPOSITION X.</head> <head xml:id="echoid-head548" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7962" xml:space="preserve">475. </s> <s xml:id="echoid-s7963" xml:space="preserve">Décrire un ennéagone dans un cercle.</s> <s xml:id="echoid-s7964" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 78.</note> <p> <s xml:id="echoid-s7965" xml:space="preserve">Pour décrire un ennéagone dans un cercle, il faut porter <lb/>le rayon du cercle ſix fois ſur la circonférence, pour avoir les <lb/>points B, C, D, E, F, G, qui la diviſeront en ſix parties égales; <lb/></s> <s xml:id="echoid-s7966" xml:space="preserve">& </s> <s xml:id="echoid-s7967" xml:space="preserve">tirant des lignes du premier point au troiſieme, du troi-<lb/>ſieme au cinquieme, & </s> <s xml:id="echoid-s7968" xml:space="preserve">du cinquieme au premier, on aura un <lb/>triangle équilatéral B D E, qui diviſera la circonférence en <lb/>trois parties égales; </s> <s xml:id="echoid-s7969" xml:space="preserve">ſi on diviſe après cela un de ces arcs, <lb/>comme B C D, en trois parties égales, par le problême précé-<lb/>dent, l’on aura la neuvieme partie de la circonférence du cer-<lb/>cle, dont la corde ſera le côté de l’ennéagone. </s> <s xml:id="echoid-s7970" xml:space="preserve">C. </s> <s xml:id="echoid-s7971" xml:space="preserve">Q. </s> <s xml:id="echoid-s7972" xml:space="preserve">F. </s> <s xml:id="echoid-s7973" xml:space="preserve">T. </s> <s xml:id="echoid-s7974" xml:space="preserve">& </s> <s xml:id="echoid-s7975" xml:space="preserve">D.</s> <s xml:id="echoid-s7976" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div580" type="section" level="1" n="473"> <head xml:id="echoid-head549" xml:space="preserve">PROPOSITION XI.</head> <head xml:id="echoid-head550" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7977" xml:space="preserve">476. </s> <s xml:id="echoid-s7978" xml:space="preserve">Décrire un eptagone dans un cercle.</s> <s xml:id="echoid-s7979" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7980" xml:space="preserve">Pour décrire un eptagone dans un cercle, il faut diviſer le <lb/>quart de la circonférence du cercle en ſept parties égales: </s> <s xml:id="echoid-s7981" xml:space="preserve">ainſi <lb/>chacune de ces parties ſera la 28<emph style="sub">e</emph> partie de toute la circonfé-<lb/>rence. </s> <s xml:id="echoid-s7982" xml:space="preserve">Or prenant un arc égal au quatre ſeptiemes du quart de <lb/>cercle, il ſera égal à la ſeptieme partie de la circonférence du <lb/>même cercle, & </s> <s xml:id="echoid-s7983" xml:space="preserve">par conſéquent la corde qui le ſoutient ſera <lb/>le côté de l’eptagone demandé. </s> <s xml:id="echoid-s7984" xml:space="preserve">C. </s> <s xml:id="echoid-s7985" xml:space="preserve">Q. </s> <s xml:id="echoid-s7986" xml:space="preserve">F. </s> <s xml:id="echoid-s7987" xml:space="preserve">T. </s> <s xml:id="echoid-s7988" xml:space="preserve">& </s> <s xml:id="echoid-s7989" xml:space="preserve">D.</s> <s xml:id="echoid-s7990" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div581" type="section" level="1" n="474"> <head xml:id="echoid-head551" xml:space="preserve">PROPOSITION XII.</head> <head xml:id="echoid-head552" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s7991" xml:space="preserve">477. </s> <s xml:id="echoid-s7992" xml:space="preserve">Décrire un ondécagone dans un cercle.</s> <s xml:id="echoid-s7993" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s7994" xml:space="preserve">Pour décrire un polygone de onze côtés qui ſoit inſcrit au <lb/>cercle, il faut diviſer le quart de cercle en onze parties égales, <lb/>& </s> <s xml:id="echoid-s7995" xml:space="preserve">ſi l’on prend la corde d’un angle, qui ſeroit les quatre on-<lb/>ziemes du quart de cercle, elle ſera le côté de l’ondécagone <lb/>demandé. </s> <s xml:id="echoid-s7996" xml:space="preserve">C. </s> <s xml:id="echoid-s7997" xml:space="preserve">Q. </s> <s xml:id="echoid-s7998" xml:space="preserve">F. </s> <s xml:id="echoid-s7999" xml:space="preserve">T. </s> <s xml:id="echoid-s8000" xml:space="preserve">& </s> <s xml:id="echoid-s8001" xml:space="preserve">D.</s> <s xml:id="echoid-s8002" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div582" type="section" level="1" n="475"> <head xml:id="echoid-head553" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s8003" xml:space="preserve">L’on a nommé quadratrice, la courbe que nous venons d’exa- <pb o="233" file="0271" n="271" rhead="DE MATHEMATIQUE. Liv. VI."/> miner, parce qu’elle contribue à la quadrature méchanique <lb/>du cercle: </s> <s xml:id="echoid-s8004" xml:space="preserve">car ſuppoſons qu’on ait trouvé le point D, où cette <lb/>courbe rencontre le rayon B C, il eſt démontré dans Pappus, <lb/>dans Clavius, & </s> <s xml:id="echoid-s8005" xml:space="preserve">dans pluſieurs autres Auteurs, que le demi-<lb/>diametre B C eſt moyen proportionnel entre la baſe B D de la <lb/>quadratrice & </s> <s xml:id="echoid-s8006" xml:space="preserve">la circonférence A E C du quart de cercle; </s> <s xml:id="echoid-s8007" xml:space="preserve">en-<lb/>ſorte que l’on a cette proportion B D : </s> <s xml:id="echoid-s8008" xml:space="preserve">B C :</s> <s xml:id="echoid-s8009" xml:space="preserve">: B C : </s> <s xml:id="echoid-s8010" xml:space="preserve">A E C. <lb/></s> <s xml:id="echoid-s8011" xml:space="preserve">D’où il ſuit qu’en connoiſſant cette baſe, on pourroit déter-<lb/>miner une ligne droite égale à la circonférence du quart de <lb/>cercle.</s> <s xml:id="echoid-s8012" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div583" type="section" level="1" n="476"> <head xml:id="echoid-head554" xml:space="preserve">PROPOSITION XIII.</head> <head xml:id="echoid-head555" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8013" xml:space="preserve">478. </s> <s xml:id="echoid-s8014" xml:space="preserve">Circonſcrire un polygone quelconque autour d’un cercle <lb/>donné.</s> <s xml:id="echoid-s8015" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8016" xml:space="preserve">Quand on veut circonſcrire un polygone autour d’un cer-<lb/>cle, il faut commencer par en inſcrire un ſemblable dans le <lb/>même cercle: </s> <s xml:id="echoid-s8017" xml:space="preserve">ainſi voulant, par exemple, circonſcrire un <lb/>exagone autour du cercle BEC, il faut commencer par en <lb/> <anchor type="note" xlink:label="note-0271-01a" xlink:href="note-0271-01"/> tracer un dans le cercle, & </s> <s xml:id="echoid-s8018" xml:space="preserve">diviſer un de ſes côtés, tels que <lb/>B C, en deux également, par un rayon A E, & </s> <s xml:id="echoid-s8019" xml:space="preserve">à l’extrêmité <lb/>E, mener la tangente F G, qu’il faut terminer par les rayons <lb/>A B, A C prolongés, juſqu’à la rencontre de la tangente, & </s> <s xml:id="echoid-s8020" xml:space="preserve"><lb/>l’on aura le côté F G de l’exagone circonſcrit: </s> <s xml:id="echoid-s8021" xml:space="preserve">ainſi on trou-<lb/>vera tous les autres, en faiſant la même opération. </s> <s xml:id="echoid-s8022" xml:space="preserve">Mais pour <lb/>avoir plutôt fait, après avoir trouvé les points F, G, il vaut <lb/>mieux décrire un cercle du centre A avec le rayon A G, ſur la <lb/>circonférence duquel on pourra marquer les points, qui ſer-<lb/>viront à tracer le polygone, en y portant avec le compas la <lb/>longueur du côté F G.</s> <s xml:id="echoid-s8023" xml:space="preserve"/> </p> <div xml:id="echoid-div583" type="float" level="2" n="1"> <note position="right" xlink:label="note-0271-01" xlink:href="note-0271-01a" xml:space="preserve">Figure 79.</note> </div> </div> <div xml:id="echoid-div585" type="section" level="1" n="477"> <head xml:id="echoid-head556" style="it" xml:space="preserve">Fin du ſixieme Livre.</head> <figure> <image file="0271-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0271-01"/> </figure> <pb o="234" file="0272" n="272"/> <figure> <image file="0272-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0272-01"/> </figure> </div> <div xml:id="echoid-div586" type="section" level="1" n="478"> <head xml:id="echoid-head557" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head558" xml:space="preserve">LIVRE SEPTIEME, <lb/>Où l’on conſidere les rapports qu’ont entr’eux les circuits des <lb/>figures ſemblables, & la proportion de leurs ſuperficies.</head> <head xml:id="echoid-head559" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s8024" xml:space="preserve">479. </s> <s xml:id="echoid-s8025" xml:space="preserve">ON dit que deux quadrilateres ont leurs baſes & </s> <s xml:id="echoid-s8026" xml:space="preserve">leurs <lb/>hauteurs réciproques, lorſque la baſe du premier eſt à la baſe <lb/>du ſecond, comme la hauteur du même ſecond eſt à celle du <lb/>premier. </s> <s xml:id="echoid-s8027" xml:space="preserve">En général on dit que deux grandeurs quelconques <lb/>ſont réciproques à deux autres, lorſque les deux premieres <lb/>ſont les extrêmes ou les moyens d’une proportion, dont les <lb/>deux autres ſont les moyens ou les extrêmes. </s> <s xml:id="echoid-s8028" xml:space="preserve">Par exemple, <lb/>a & </s> <s xml:id="echoid-s8029" xml:space="preserve">b ſont réciproques aux grandeurs c & </s> <s xml:id="echoid-s8030" xml:space="preserve">d, ſi l’on a a : </s> <s xml:id="echoid-s8031" xml:space="preserve">c :</s> <s xml:id="echoid-s8032" xml:space="preserve">: d : </s> <s xml:id="echoid-s8033" xml:space="preserve">b, <lb/>ou c : </s> <s xml:id="echoid-s8034" xml:space="preserve">a :</s> <s xml:id="echoid-s8035" xml:space="preserve">: b : </s> <s xml:id="echoid-s8036" xml:space="preserve">d.</s> <s xml:id="echoid-s8037" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div587" type="section" level="1" n="479"> <head xml:id="echoid-head560" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head561" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8038" xml:space="preserve">480. </s> <s xml:id="echoid-s8039" xml:space="preserve">Si l’on a deux polygones réguliers ſemblables, A & </s> <s xml:id="echoid-s8040" xml:space="preserve">B, <lb/> <anchor type="note" xlink:label="note-0272-01a" xlink:href="note-0272-01"/> je dis que le circuit ou le contour du polygone A eſt au contour du <lb/> <anchor type="note" xlink:label="note-0272-02a" xlink:href="note-0272-02"/> polygone B, comme le rayon A C eſt au rayon B F.</s> <s xml:id="echoid-s8041" xml:space="preserve"/> </p> <div xml:id="echoid-div587" type="float" level="2" n="1"> <note position="left" xlink:label="note-0272-01" xlink:href="note-0272-01a" xml:space="preserve">Planche V.</note> <note position="left" xlink:label="note-0272-02" xlink:href="note-0272-02a" xml:space="preserve">Figure 86 <lb/>& 87.</note> </div> </div> <div xml:id="echoid-div589" type="section" level="1" n="480"> <head xml:id="echoid-head562" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8042" xml:space="preserve">Nous nommerons C D, a, F G, b, A C, c, & </s> <s xml:id="echoid-s8043" xml:space="preserve">B F, d. </s> <s xml:id="echoid-s8044" xml:space="preserve">Cela <pb o="235" file="0273" n="273" rhead="NOUVEAU COURS DE MATH. Liv. VII."/> poſé, ſi chaque polygone eſt un exagone, le circuit du poly-<lb/>gone A ſera 6a, & </s> <s xml:id="echoid-s8045" xml:space="preserve">le circuit du polygone b ſera 6b: </s> <s xml:id="echoid-s8046" xml:space="preserve">ainſi il <lb/>faut prouver que l’on aura 6a : </s> <s xml:id="echoid-s8047" xml:space="preserve">6b :</s> <s xml:id="echoid-s8048" xml:space="preserve">: c: </s> <s xml:id="echoid-s8049" xml:space="preserve">d. </s> <s xml:id="echoid-s8050" xml:space="preserve">Les triangles D A C, <lb/>G B F ſont ſemblables; </s> <s xml:id="echoid-s8051" xml:space="preserve">car puiſque les polygones ſont ſem-<lb/>blables, les angles de chacun des triangles qui les compoſent <lb/>ſont égaux chacun à chacun, & </s> <s xml:id="echoid-s8052" xml:space="preserve">les côtés oppoſés aux angles <lb/>égaux ſont proportionnels (art. </s> <s xml:id="echoid-s8053" xml:space="preserve">405): </s> <s xml:id="echoid-s8054" xml:space="preserve">on aura donc a : </s> <s xml:id="echoid-s8055" xml:space="preserve">b :</s> <s xml:id="echoid-s8056" xml:space="preserve">: c : </s> <s xml:id="echoid-s8057" xml:space="preserve">d, <lb/>& </s> <s xml:id="echoid-s8058" xml:space="preserve">multipliant les deux termes a & </s> <s xml:id="echoid-s8059" xml:space="preserve">b par 6, on aura 6a : </s> <s xml:id="echoid-s8060" xml:space="preserve">6b :</s> <s xml:id="echoid-s8061" xml:space="preserve">: c : </s> <s xml:id="echoid-s8062" xml:space="preserve">d. <lb/></s> <s xml:id="echoid-s8063" xml:space="preserve">C. </s> <s xml:id="echoid-s8064" xml:space="preserve">Q. </s> <s xml:id="echoid-s8065" xml:space="preserve">F. </s> <s xml:id="echoid-s8066" xml:space="preserve">D.</s> <s xml:id="echoid-s8067" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div590" type="section" level="1" n="481"> <head xml:id="echoid-head563" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s8068" xml:space="preserve">Cette propoſition ſe doit entendre de toutes les figures ſem-<lb/>blables, régulieres ou irrégulieres, à commencer par les trian-<lb/>gles: </s> <s xml:id="echoid-s8069" xml:space="preserve">car quoique des figures irrégulieres ne ſoient pas inſcrip-<lb/>tibles au cercle, on peut dire cependant que les contours de <lb/>ces polygones, ſuppoſés ſemblables, ſont entr’eux comme les <lb/>rayons de deux cercles qui paſſeront par les ſommets de trois <lb/>angles égaux, pris comme l’on voudra dans l’une & </s> <s xml:id="echoid-s8070" xml:space="preserve">dans <lb/>l’autre figure, pourvu que ces cercles paſſent par les angles de <lb/>deux triangles ſemblables, & </s> <s xml:id="echoid-s8071" xml:space="preserve">ſemblablement placés dans l’une <lb/>& </s> <s xml:id="echoid-s8072" xml:space="preserve">dans l’autre figure.</s> <s xml:id="echoid-s8073" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div591" type="section" level="1" n="482"> <head xml:id="echoid-head564" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s8074" xml:space="preserve">481. </s> <s xml:id="echoid-s8075" xml:space="preserve">Il ſuit de cette propoſition, que les circonférences des <lb/>cercles ſont entr’elles comme les rayons de ces cercles: </s> <s xml:id="echoid-s8076" xml:space="preserve">car ſi <lb/>l’on conſidere les cercles X & </s> <s xml:id="echoid-s8077" xml:space="preserve">Y, comme étant des polygones <lb/> <anchor type="note" xlink:label="note-0273-01a" xlink:href="note-0273-01"/> ſemblables d’une infinité de côtés, nommant a la circonfé-<lb/>rence du premier, c le rayon, b la circonférence du ſecond, & </s> <s xml:id="echoid-s8078" xml:space="preserve"><lb/>d ſon rayon, on aura encore a : </s> <s xml:id="echoid-s8079" xml:space="preserve">b :</s> <s xml:id="echoid-s8080" xml:space="preserve">: c : </s> <s xml:id="echoid-s8081" xml:space="preserve">d.</s> <s xml:id="echoid-s8082" xml:space="preserve"/> </p> <div xml:id="echoid-div591" type="float" level="2" n="1"> <note position="right" xlink:label="note-0273-01" xlink:href="note-0273-01a" xml:space="preserve">Figure 88 <lb/>& 89.</note> </div> </div> <div xml:id="echoid-div593" type="section" level="1" n="483"> <head xml:id="echoid-head565" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head566" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8083" xml:space="preserve">482. </s> <s xml:id="echoid-s8084" xml:space="preserve">Si du centre A d’un polygone régulier, on abaiſſe une <lb/> <anchor type="note" xlink:label="note-0273-02a" xlink:href="note-0273-02"/> perpendiculaire A E ſur l’un de ſes côtés, je dis que la ſuperficie de <lb/>ce polygone ſera égale à un triangle rectangle I K L, qui auroit pour <lb/>hauteur la ligne I K égale à la perpendiculaire A E, & </s> <s xml:id="echoid-s8085" xml:space="preserve">pour baſe <lb/>une ligne K L égale au circuit du polygone.</s> <s xml:id="echoid-s8086" xml:space="preserve"/> </p> <div xml:id="echoid-div593" type="float" level="2" n="1"> <note position="right" xlink:label="note-0273-02" xlink:href="note-0273-02a" xml:space="preserve">Figure 86 <lb/>& 90.</note> </div> </div> <div xml:id="echoid-div595" type="section" level="1" n="484"> <head xml:id="echoid-head567" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8087" xml:space="preserve">Si le polygone régulier eſt un exagone, & </s> <s xml:id="echoid-s8088" xml:space="preserve">que l’on tire du <pb o="236" file="0274" n="274" rhead="NOUVEAU COURS"/> centre des rayons à tous les angles, l’on aura autant de trian-<lb/>gles égaux, que le polygone a de côtés: </s> <s xml:id="echoid-s8089" xml:space="preserve">ainſi le polygone A <lb/>ſera compoſé de ſix triangles, tels que C A D; </s> <s xml:id="echoid-s8090" xml:space="preserve">mais comme <lb/>les triangles C A D & </s> <s xml:id="echoid-s8091" xml:space="preserve">K I L ont la même hauteur, ils ſeront <lb/>dans la même raiſon que leurs baſes (art. </s> <s xml:id="echoid-s8092" xml:space="preserve">392); </s> <s xml:id="echoid-s8093" xml:space="preserve">& </s> <s xml:id="echoid-s8094" xml:space="preserve">comme la <lb/>baſe K L eſt ſextuple de la baſe C D, par conſtruction, le trian-<lb/>gle K I L ſera auſſi ſextuple du triangle C A D, & </s> <s xml:id="echoid-s8095" xml:space="preserve">par conſé-<lb/>quent égal au polygone. </s> <s xml:id="echoid-s8096" xml:space="preserve">C. </s> <s xml:id="echoid-s8097" xml:space="preserve">Q. </s> <s xml:id="echoid-s8098" xml:space="preserve">F. </s> <s xml:id="echoid-s8099" xml:space="preserve">D.</s> <s xml:id="echoid-s8100" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div596" type="section" level="1" n="485"> <head xml:id="echoid-head568" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s8101" xml:space="preserve">483. </s> <s xml:id="echoid-s8102" xml:space="preserve">Il ſuit de cette propoſition, que pour trouver la ſu-<lb/>perficie d’un polygone régulier, il faut multiplier la moitié de <lb/>ſon circuit par la perpendiculaire abaiſſée du centre de ce po-<lb/>lygone ſur ſon côté, puiſque pour trouver la ſurface du trian-<lb/>gle I K L, qui eſt égal à ce polygone, il faut multiplier la <lb/>moitié de ſa baſe K L par la perpendiculaire I K.</s> <s xml:id="echoid-s8103" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div597" type="section" level="1" n="486"> <head xml:id="echoid-head569" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head570" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8104" xml:space="preserve">484. </s> <s xml:id="echoid-s8105" xml:space="preserve">La ſuperficie d’un cercle eſt égale à un triangle, qui au-<lb/>roit pour hauteur le rayon du cercle, & </s> <s xml:id="echoid-s8106" xml:space="preserve">pour baſe la circonférence <lb/>du même cercle.</s> <s xml:id="echoid-s8107" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div598" type="section" level="1" n="487"> <head xml:id="echoid-head571" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8108" xml:space="preserve">Comme un cercle eſt un polygone d’une infinité de côtés, <lb/>on peut prendre la circonférence du cercle pour la ſomme de <lb/>ces côtés, & </s> <s xml:id="echoid-s8109" xml:space="preserve">ce rayon pour la perpendiculaire du polygone; <lb/></s> <s xml:id="echoid-s8110" xml:space="preserve">d’ou il ſuit qu’il ſera égal à un triangle qui auroit pour hau-<lb/>teur le rayon M N, & </s> <s xml:id="echoid-s8111" xml:space="preserve">pour baſe une ligne N O égale à la cir-<lb/>conférence. </s> <s xml:id="echoid-s8112" xml:space="preserve">C. </s> <s xml:id="echoid-s8113" xml:space="preserve">Q. </s> <s xml:id="echoid-s8114" xml:space="preserve">F. </s> <s xml:id="echoid-s8115" xml:space="preserve">D.</s> <s xml:id="echoid-s8116" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div599" type="section" level="1" n="488"> <head xml:id="echoid-head572" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s8117" xml:space="preserve">485. </s> <s xml:id="echoid-s8118" xml:space="preserve">Puiſque le triangle M N O eſt égal au cercle, & </s> <s xml:id="echoid-s8119" xml:space="preserve">qu’il <lb/>eſt auſſi égal à un rectangle qui auroit pour baſe la moitié de <lb/>la baſe N O, & </s> <s xml:id="echoid-s8120" xml:space="preserve">pour hauteur la ligne M N, il s’enſuit qu’un <lb/>cercle eſt égal à un rectangle qui auroit pour baſe la moitié <lb/>de la circonférence, & </s> <s xml:id="echoid-s8121" xml:space="preserve">pour hauteur le rayon; </s> <s xml:id="echoid-s8122" xml:space="preserve">& </s> <s xml:id="echoid-s8123" xml:space="preserve">que pour <lb/>en trouver la ſuperficie, il faut multiplier la moitié du dia-<lb/>metre, par la moitié de la circonférence.</s> <s xml:id="echoid-s8124" xml:space="preserve"/> </p> <pb o="237" file="0275" n="275" rhead="DE MATHÉMATIQUE. Liv. VII."/> </div> <div xml:id="echoid-div600" type="section" level="1" n="489"> <head xml:id="echoid-head573" xml:space="preserve"><emph style="sc">Remarque</emph> I.</head> <p> <s xml:id="echoid-s8125" xml:space="preserve">486. </s> <s xml:id="echoid-s8126" xml:space="preserve">Si l’on conſidere la ſurface du cercle, comme étant <lb/>compoſée d’une infinité de circonférences concentriques, dont <lb/>les rayons ſe ſurpaſſent également, toutes ces circonférences <lb/>compoſeront une progreſſion infinie arithmétique, dont le cen-<lb/>tre ſera le plus petit terme, & </s> <s xml:id="echoid-s8127" xml:space="preserve">la circonférence le plus grand. <lb/></s> <s xml:id="echoid-s8128" xml:space="preserve">Or comme le demi-diametre A B exprime le nombre des ter-<lb/>mes de la progreſſion, il s’enſuit qu’on en trouvera la ſomme <lb/>en multipliant le plus grand terme, qui eſt la circonférence, <lb/>par la moitié du rayon A B.</s> <s xml:id="echoid-s8129" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div601" type="section" level="1" n="490"> <head xml:id="echoid-head574" xml:space="preserve"><emph style="sc">Remarque</emph> II.</head> <p> <s xml:id="echoid-s8130" xml:space="preserve">487. </s> <s xml:id="echoid-s8131" xml:space="preserve">Il ſemble d’abord que la propoſition précédente donne <lb/>la quadrature du cercle, parce qu’elle prouve qu’un cercle eſt <lb/>égal à un triangle, qui auroit pour baſe la circonférence du <lb/>cercle, & </s> <s xml:id="echoid-s8132" xml:space="preserve">pour hauteur le rayon; </s> <s xml:id="echoid-s8133" xml:space="preserve">mais comme on n’a pas en-<lb/>core trouvé géométriquement une ligne droite, parfaitement <lb/>égale à la circonférence d’un cercle, l’on n’a pu par conſéquent <lb/>trouver un triangle parfaitement égal au cercle. </s> <s xml:id="echoid-s8134" xml:space="preserve">Quand je dis <lb/>un triangle, on peut auſſi entendre un quarré égal au cercle, <lb/>parce que l’on peut faire géométriquement un quarré égal à un <lb/>triangle, comme on le verra ailleurs. </s> <s xml:id="echoid-s8135" xml:space="preserve">Mais pour qu’il n’y ait <lb/>point d’équivoque ſur le mot de quadrature du cercle, il eſt <lb/>bon que les Commençans ſçachent que la quadrature du cer-<lb/>cle conſiſte à trouver une propoſition qui donne le moyen de <lb/>faire un quarré égal en ſurface à un cercle donné, & </s> <s xml:id="echoid-s8136" xml:space="preserve">qui dé-<lb/>qu’on le fait réellement.</s> <s xml:id="echoid-s8137" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8138" xml:space="preserve">Quoique les Géometres n’aient pas encore trouvé une ligne <lb/>droite parfaitement égale à la circonférence d’un cercle, cela <lb/>n’empêche pas que dans la pratique on ne ſuppoſe que cela ſe <lb/>puiſſe faire, en ſe ſervant de quelques regles qui ſont des <lb/>approximations de la quadrature du cercle, comme on le va <lb/>voir.</s> <s xml:id="echoid-s8139" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8140" xml:space="preserve">488. </s> <s xml:id="echoid-s8141" xml:space="preserve">Archimede a trouvé que le rapport du diametre à la <lb/>circonférence, eſt à peu près celui de 7 à 22, c’eſt-à-dire que <lb/>ſi le diametre contient ſept parties égales, la circonférence <lb/>en contiendra à peu près 22, ou, ce qui revient au même, que <lb/>la circonférence vaut trois fois le diametre & </s> <s xml:id="echoid-s8142" xml:space="preserve">un ſeptieme. <lb/></s> <s xml:id="echoid-s8143" xml:space="preserve">Or comme les diametres des cercles ſont dans la raiſon de <pb o="238" file="0276" n="276" rhead="NOUVEAU COURS"/> leurs circonférences (art. </s> <s xml:id="echoid-s8144" xml:space="preserve">481), ſi l’on avoit un cercle, dont <lb/>le diametre fût, par exemple, de 28 pieds, pour en trouver la <lb/>circonférence, on diroit: </s> <s xml:id="echoid-s8145" xml:space="preserve">Si 7, diametre d’un cercle, donne <lb/>22 pour la circonférence du même cercle, combien donneront <lb/>28, diametre d’un autre cercle pour ſa circonférence, que l’on <lb/>trouvera de 88 pieds?</s> <s xml:id="echoid-s8146" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8147" xml:space="preserve">Mais ſi l’on avoit un cercle, dont on connût ſeulement la <lb/>circonférence, que nous ſuppoſerons de 66 pieds, pour en <lb/>trouver le diametre, il faudroit encore faire une Regle de <lb/>Trois, en diſant: </s> <s xml:id="echoid-s8148" xml:space="preserve">Si la circonférence d’un cercle qui auroit <lb/>22 pieds, donne 7 pour ſon diametre, combien donnera la cir-<lb/>conférence d’un autre cercle, qui ſeroit de 66 pieds, pour le <lb/>diametre du même cercle? </s> <s xml:id="echoid-s8149" xml:space="preserve">L’on trouvera 21 pieds pour le dia-<lb/>metre qu’on cherche. </s> <s xml:id="echoid-s8150" xml:space="preserve">Outre le rapport de 7 à 22, dont on peut <lb/>toujours ſe ſervir, lorſqu’on ne veut pas arriver à la derniere <lb/>préciſion, on peut encore faire uſage de celui de 113 à 355, <lb/>trouvé par Métius, & </s> <s xml:id="echoid-s8151" xml:space="preserve">plus exact que le précédent; </s> <s xml:id="echoid-s8152" xml:space="preserve">c’eſt pour-<lb/>quoi il ſera à propos de ſe ſervir de ce dernier dans les opéra-<lb/>tions où il faudra déterminer la circonférence du cercle avec <lb/>plus de juſteſſe que dans les opérations ordinaires.</s> <s xml:id="echoid-s8153" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div602" type="section" level="1" n="491"> <head xml:id="echoid-head575" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s8154" xml:space="preserve">489. </s> <s xml:id="echoid-s8155" xml:space="preserve">Il ſuit encore delà, qu’un cercle eſt égal à un rectan-<lb/>gle N S R Q, qui auroit pour baſe le quart de la circonférence, <lb/>& </s> <s xml:id="echoid-s8156" xml:space="preserve">pour hauteur le diametre, puiſque ce rectangle eſt égal au <lb/>rectangle N T, qui a pour hauteur le rayon, & </s> <s xml:id="echoid-s8157" xml:space="preserve">pour baſe la moi-<lb/>tié de la circonférence: </s> <s xml:id="echoid-s8158" xml:space="preserve">par conſéquent ſi l’on avoit un quarré <lb/>V X Y Z fait ſur le diametre du cercle, le quarré & </s> <s xml:id="echoid-s8159" xml:space="preserve">le rectangle <lb/>N R égal à la ſurface du cercle, ayant même hauteur, ſeront <lb/>entr’eux comme leurs baſes V Z & </s> <s xml:id="echoid-s8160" xml:space="preserve">N Q. </s> <s xml:id="echoid-s8161" xml:space="preserve">On peut donc dire <lb/>que le diametre d’un cercle eſt au quart de la circonférence, <lb/>comme le quarré de ce même diamette eſt à la ſuperficie du <lb/>même cercle.</s> <s xml:id="echoid-s8162" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div603" type="section" level="1" n="492"> <head xml:id="echoid-head576" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s8163" xml:space="preserve">490. </s> <s xml:id="echoid-s8164" xml:space="preserve">Si l’on ſuppoſe que le diametre d’un cercle ſoit diviſé <lb/>en ſept parties égales, & </s> <s xml:id="echoid-s8165" xml:space="preserve">que ſa circonférence en contienne <lb/>exactement vingt-deux (ce qui ne peut faire une erreur ſenſi-<lb/>ble dans la pratique), en doublant les mêmes nombres, on <lb/>aura 14 & </s> <s xml:id="echoid-s8166" xml:space="preserve">44 pour le diametre & </s> <s xml:id="echoid-s8167" xml:space="preserve">la circonférence, ſur quoi <lb/>l’on remarquera que le dernier étant diviſible par 4, & </s> <s xml:id="echoid-s8168" xml:space="preserve">don- <pb o="239" file="0277" n="277" rhead="DE MATHÉMATIQUE. Liv. VII."/> nant 11 au quotient, on pourra prendre ce même quotient <lb/>pour le quart de la circonférence; </s> <s xml:id="echoid-s8169" xml:space="preserve">d’où il ſuit, par le corollaire <lb/>précédent, que le rapport de 14 à 11 eſt le même que celui <lb/>du quarré du diametre à la ſurface du cercle: </s> <s xml:id="echoid-s8170" xml:space="preserve">ainſi pour avoir <lb/>la ſuperficie d’un cercle, dont on connoît le diametre, que <lb/>je ſuppoſe = a, on n’aura qu’à faire cette Regle de Trois, <lb/>14 : </s> <s xml:id="echoid-s8171" xml:space="preserve">11 :</s> <s xml:id="echoid-s8172" xml:space="preserve">: aa : </s> <s xml:id="echoid-s8173" xml:space="preserve">{11aa/14}, ou, ce qui revient au même, pour avoir <lb/>l’aire d’un cercle quelconque, il ſuffira de prendre les onze <lb/>quatorziemes du quarré du diametre de ce cercle.</s> <s xml:id="echoid-s8174" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div604" type="section" level="1" n="493"> <head xml:id="echoid-head577" xml:space="preserve"><emph style="sc">Scholie</emph>.</head> <p> <s xml:id="echoid-s8175" xml:space="preserve">491. </s> <s xml:id="echoid-s8176" xml:space="preserve">Les Commençans ne ſeront peut-être pas fâchés de <lb/>connoître la route qu’ Archimede a ſuivie pour découvrir le <lb/>rapport dont nous venons de parler. </s> <s xml:id="echoid-s8177" xml:space="preserve">La connoiſſance des pre-<lb/>miers axiomes de géométrie ſuffit pour nous faire concevoir <lb/>clairement que la circonférence d’un cercle eſt plus grande <lb/>que le contour d’un polygone quelconque inſcrit à ce cercle, <lb/>& </s> <s xml:id="echoid-s8178" xml:space="preserve">plus petite que le contour d’un polygone quelconque cir-<lb/>conſcrit au même cercle. </s> <s xml:id="echoid-s8179" xml:space="preserve">Il faut entendre la même choſe pour <lb/>la ſuperficie du cercle & </s> <s xml:id="echoid-s8180" xml:space="preserve">celle des polygones inſcrits & </s> <s xml:id="echoid-s8181" xml:space="preserve">cir-<lb/>conſcrits. </s> <s xml:id="echoid-s8182" xml:space="preserve">Cela poſé, voici ce que fit Archimede pour décou-<lb/>vrir le rapport approché du diametre à la circonférence. </s> <s xml:id="echoid-s8183" xml:space="preserve">Il <lb/>inſcrivit à un cercle un polygone de 96 côtés, & </s> <s xml:id="echoid-s8184" xml:space="preserve">circonſcrivit <lb/>au même cercle un polygone ſemblable d’un pareil nombre de <lb/>côtés; </s> <s xml:id="echoid-s8185" xml:space="preserve">il calcula enſuite par les propriétés des lignes ou des <lb/>cordes de cercle, la longueur d’un des côtés de chaque poly-<lb/>gone, dont il trouva par conſéquent le contour, en multipliant <lb/>le nombre trouvé par 96. </s> <s xml:id="echoid-s8186" xml:space="preserve">Ayant donc ſuppoſé que le diametre <lb/>du cercle étoit l’unité, il trouva que le périmetre de polygone <lb/>inſcrit étoit plus grand que 3 {10/71} du diametre, & </s> <s xml:id="echoid-s8187" xml:space="preserve">que celui du <lb/>polygone circonſcrit étoit moindre que 3 {10/70}, ou 3 & </s> <s xml:id="echoid-s8188" xml:space="preserve">{1/7}; </s> <s xml:id="echoid-s8189" xml:space="preserve">d’où <lb/>il faut conclure que la circonférence, qui eſt néceſſairement <lb/>entre ces deux contours, eſt auſſi à plus forte raiſon plus grande <lb/>que 3 {10/71}, & </s> <s xml:id="echoid-s8190" xml:space="preserve">moindre que 3 & </s> <s xml:id="echoid-s8191" xml:space="preserve">{10/70}: </s> <s xml:id="echoid-s8192" xml:space="preserve">ainſi le diametre du cercle <lb/>étant 7, il faut néceſſairement que la circonférence ſoit plus <lb/>grande que 21, & </s> <s xml:id="echoid-s8193" xml:space="preserve">moindre que 22, qui vaut trois fois le dia-<lb/>metre & </s> <s xml:id="echoid-s8194" xml:space="preserve">{1/7}, de maniere que cette même circonférence eſt beau-<lb/>coup plus proche de 22, qu’elle ne l’eſt de 21. </s> <s xml:id="echoid-s8195" xml:space="preserve">Il eſt aiſé de <lb/>voir qu’ Archimede partagea d’abord ſon cercle en quatre parties <pb o="240" file="0278" n="278" rhead="NOUVEAU COURS"/> égales, ou, ce qui eſt la même choſe, qu’il chercha la valeur <lb/>d’une corde de 90 degrés; </s> <s xml:id="echoid-s8196" xml:space="preserve">enſuite il chercha la corde d’un arc <lb/>de 45 degrés pour avoir le côté de l’octogone; </s> <s xml:id="echoid-s8197" xml:space="preserve">il chercha en-<lb/>ſuite le côté d’un polygone de 16 côtés, & </s> <s xml:id="echoid-s8198" xml:space="preserve">enfin celui d’un <lb/>polygone de 32 côtés; </s> <s xml:id="echoid-s8199" xml:space="preserve">après quoi il chercha la corde d’un arc, <lb/>qui n’eſt plus que le tiers du dernier polygone de 32 côtés, & </s> <s xml:id="echoid-s8200" xml:space="preserve"><lb/>cette corde eſt le côté de ſon polygone de 96 côtés; </s> <s xml:id="echoid-s8201" xml:space="preserve">car il eſt <lb/>évident que 32 x 3 = 96.</s> <s xml:id="echoid-s8202" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div605" type="section" level="1" n="494"> <head xml:id="echoid-head578" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head579" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8203" xml:space="preserve">492. </s> <s xml:id="echoid-s8204" xml:space="preserve">Si l’on a deux polygones ſemblables A & </s> <s xml:id="echoid-s8205" xml:space="preserve">B, la ſurface <lb/> <anchor type="note" xlink:label="note-0278-01a" xlink:href="note-0278-01"/> du premier ſera à celle du ſecond, comme le quarré de la perpendi-<lb/>culaire A E au quarré de la perpendiculaire B H, ou comme le <lb/>quarré du rayon A C au quarré du rayon B F.</s> <s xml:id="echoid-s8206" xml:space="preserve"/> </p> <div xml:id="echoid-div605" type="float" level="2" n="1"> <note position="left" xlink:label="note-0278-01" xlink:href="note-0278-01a" xml:space="preserve">Figure 86 <lb/>& 87.</note> </div> <p> <s xml:id="echoid-s8207" xml:space="preserve">Soit nommé le côté C D du 1<emph style="sub">er</emph> polygone, a, la perpendicu-<lb/>laire A E, b, le côté F G de l’autre polygone, c, la perpendiculaire <lb/>B H, d: </s> <s xml:id="echoid-s8208" xml:space="preserve">le circuit du premier polygone ſera 6a, & </s> <s xml:id="echoid-s8209" xml:space="preserve">celui du ſe-<lb/>cond ſera 6c: </s> <s xml:id="echoid-s8210" xml:space="preserve">multipliant les moitiés de ces circuits par leurs <lb/>perpendiculaires, les produits donneront les ſurfaces des po-<lb/>lygones, & </s> <s xml:id="echoid-s8211" xml:space="preserve">l’on aura 3ab pour le premier A, & </s> <s xml:id="echoid-s8212" xml:space="preserve">3cd pour le <lb/>ſecond B: </s> <s xml:id="echoid-s8213" xml:space="preserve">ainſi il faut démontrer que 3ab : </s> <s xml:id="echoid-s8214" xml:space="preserve">3cd :</s> <s xml:id="echoid-s8215" xml:space="preserve">: bb : </s> <s xml:id="echoid-s8216" xml:space="preserve">dd.</s> <s xml:id="echoid-s8217" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div607" type="section" level="1" n="495"> <head xml:id="echoid-head580" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8218" xml:space="preserve">Pour prouver que 3ab : </s> <s xml:id="echoid-s8219" xml:space="preserve">3cd :</s> <s xml:id="echoid-s8220" xml:space="preserve">: bb : </s> <s xml:id="echoid-s8221" xml:space="preserve">dd, nous ferons voir que <lb/>dans cette proportion le produit des extrêmes eſt égal au pro-<lb/>duit des moyens, & </s> <s xml:id="echoid-s8222" xml:space="preserve">que l’on a 3abdd = 3cbbd. </s> <s xml:id="echoid-s8223" xml:space="preserve">Pour cela, <lb/>conſidérez qu’à cauſe des triangles ſemblables, A C D & </s> <s xml:id="echoid-s8224" xml:space="preserve">B F G, <lb/>a : </s> <s xml:id="echoid-s8225" xml:space="preserve">c :</s> <s xml:id="echoid-s8226" xml:space="preserve">: b : </s> <s xml:id="echoid-s8227" xml:space="preserve">d, d’où l’on tire ad = bc: </s> <s xml:id="echoid-s8228" xml:space="preserve">ainſi mettant ad dans le <lb/>ſecond membre de la premiere équation à la place de bc, <lb/>auquel il eſt égal, il viendra 3abdd = 3abdd, C. </s> <s xml:id="echoid-s8229" xml:space="preserve">Q. </s> <s xml:id="echoid-s8230" xml:space="preserve">F. </s> <s xml:id="echoid-s8231" xml:space="preserve">D.</s> <s xml:id="echoid-s8232" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div608" type="section" level="1" n="496"> <head xml:id="echoid-head581" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s8233" xml:space="preserve">493. </s> <s xml:id="echoid-s8234" xml:space="preserve">Puiſque les figures A & </s> <s xml:id="echoid-s8235" xml:space="preserve">B ſont ſemblables, les trian-<lb/>gles dont elles ſont compoſées le ſeront auſſi; </s> <s xml:id="echoid-s8236" xml:space="preserve">ainſi le triangle <lb/>A C E ſera ſemblable au triangle B F H, puiſqu’ils ont deux <lb/>angles égaux chacun à chacun: </s> <s xml:id="echoid-s8237" xml:space="preserve">donc on aura AE : </s> <s xml:id="echoid-s8238" xml:space="preserve">BH :</s> <s xml:id="echoid-s8239" xml:space="preserve">: AC : </s> <s xml:id="echoid-s8240" xml:space="preserve">BF, <lb/>& </s> <s xml:id="echoid-s8241" xml:space="preserve">A E<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8242" xml:space="preserve">B H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s8243" xml:space="preserve">: A C<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8244" xml:space="preserve">B F<emph style="sub">2</emph>. </s> <s xml:id="echoid-s8245" xml:space="preserve">Mais les polygones ſont entr’eux <lb/>comme les quarrés des perpendiculaires A E, B H, par la pré- <pb o="241" file="0279" n="279" rhead="DE MATHÉMATIQUE. Liv. VII."/> ſente propoſition: </s> <s xml:id="echoid-s8246" xml:space="preserve">donc ils ſont auſſi entr’eux comme les <lb/>quarrés des rayons A C, B F, ou des côtés C D, F G, puiſque <lb/>ces quarrés ſont en même raiſon que les quarrés des perpendi-<lb/>culaires.</s> <s xml:id="echoid-s8247" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div609" type="section" level="1" n="497"> <head xml:id="echoid-head582" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s8248" xml:space="preserve">494. </s> <s xml:id="echoid-s8249" xml:space="preserve">Cette propoſition ſe doit entendre, non ſeulement de <lb/>tous les polygones réguliers ſemblables inſcriptibles à un cer-<lb/>cle, mais encore de tous les autres autres polygones irréguliers <lb/>ſemblables, qui ſont entr’eux comme les quarrés des perpen-<lb/>diculaires abaiſſées d’un point ſemblablement placé dans l’une <lb/>& </s> <s xml:id="echoid-s8250" xml:space="preserve">dans l’autre figure, ſur des côtés homologues. </s> <s xml:id="echoid-s8251" xml:space="preserve">En un mot, les <lb/>ſuperficies de deux polygones ſemblables quelconques, ſont en-<lb/>tr’elles comme les quarrés des côtés homologues, des lignes <lb/>tirées dans les figures par des angles égaux, des perpendicu-<lb/>laires abaiſſées ſur deux côtés correſpondans, ou en général <lb/>des lignes ſemblablement placées.</s> <s xml:id="echoid-s8252" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div610" type="section" level="1" n="498"> <head xml:id="echoid-head583" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head584" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8253" xml:space="preserve">495. </s> <s xml:id="echoid-s8254" xml:space="preserve">Les ſurfaces de deux cercles ſont entr’elles comme les quar-<lb/>rés des rayons.</s> <s xml:id="echoid-s8255" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8256" xml:space="preserve">Si l’on a deux cercles X & </s> <s xml:id="echoid-s8257" xml:space="preserve">Y, & </s> <s xml:id="echoid-s8258" xml:space="preserve">que l’on nomme a la cir-<lb/> <anchor type="note" xlink:label="note-0279-01a" xlink:href="note-0279-01"/> conférence du cercle X, c ſon rayon, b la circonférence du cer-<lb/>cle Y, & </s> <s xml:id="echoid-s8259" xml:space="preserve">d ſon rayon, la ſurface du premier ſera {ac/2}, & </s> <s xml:id="echoid-s8260" xml:space="preserve">la ſur-<lb/>face du ſecond ſera {bd/2}. </s> <s xml:id="echoid-s8261" xml:space="preserve">Cela poſé, il faut prouver que {ac/2} : </s> <s xml:id="echoid-s8262" xml:space="preserve">{bd/2} :</s> <s xml:id="echoid-s8263" xml:space="preserve">: <lb/>cc : </s> <s xml:id="echoid-s8264" xml:space="preserve">dd.</s> <s xml:id="echoid-s8265" xml:space="preserve"/> </p> <div xml:id="echoid-div610" type="float" level="2" n="1"> <note position="right" xlink:label="note-0279-01" xlink:href="note-0279-01a" xml:space="preserve">Figure 89.</note> </div> </div> <div xml:id="echoid-div612" type="section" level="1" n="499"> <head xml:id="echoid-head585" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8266" xml:space="preserve">Pour prouver que {ac/2} : </s> <s xml:id="echoid-s8267" xml:space="preserve">{bd/2} :</s> <s xml:id="echoid-s8268" xml:space="preserve">: cc : </s> <s xml:id="echoid-s8269" xml:space="preserve">dd, nous ferons voir que le <lb/>produit des extrêmes de ces quatre quantités, eſt égal au pro-<lb/>duit des moyens, ou que {acdd/2} = {bdcc/2}. </s> <s xml:id="echoid-s8270" xml:space="preserve">Pour cela, faites atten-<lb/>tion que les circonférences des cercles étant entr’elles comme <lb/>les rayons (art. </s> <s xml:id="echoid-s8271" xml:space="preserve">481), on aura a : </s> <s xml:id="echoid-s8272" xml:space="preserve">b :</s> <s xml:id="echoid-s8273" xml:space="preserve">: c : </s> <s xml:id="echoid-s8274" xml:space="preserve">d, d’où l’on tire ad = bc. <lb/></s> <s xml:id="echoid-s8275" xml:space="preserve">Si donc on met dans le ſecond membre de l’équation précé-<lb/>dente, a d à la place de b c, on aura {acdd/2} = {acdd/2}. </s> <s xml:id="echoid-s8276" xml:space="preserve">C. </s> <s xml:id="echoid-s8277" xml:space="preserve">Q. </s> <s xml:id="echoid-s8278" xml:space="preserve">F. </s> <s xml:id="echoid-s8279" xml:space="preserve">D.</s> <s xml:id="echoid-s8280" xml:space="preserve"/> </p> <pb o="242" file="0280" n="280" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div613" type="section" level="1" n="500"> <head xml:id="echoid-head586" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s8281" xml:space="preserve">496. </s> <s xml:id="echoid-s8282" xml:space="preserve">Puiſque les rayons des cercles ſont entr’eux comme les <lb/>cordes des arcs d’un même nombre de degrés, comme les dia-<lb/>metres ou les côtésdes polygones ſemblables inſcrits dans ces mê-<lb/>mes cercles: </s> <s xml:id="echoid-s8283" xml:space="preserve">donc les ſurfaces des cercles, qui ſont comme les <lb/>quarrés des rayons, ſont auſſi entr’elles comme les quarrés des <lb/>diametres, des cordes d’un même nombre de degrés, comme <lb/>les quarrés des côtés de polygones ſemblables inſcrits ou cir-<lb/>conſcrits à ces cercles.</s> <s xml:id="echoid-s8284" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div614" type="section" level="1" n="501"> <head xml:id="echoid-head587" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s8285" xml:space="preserve">Cette propoſition, ainſi que la précédente, ſont d’un grand <lb/>uſage dans la Géométrie, & </s> <s xml:id="echoid-s8286" xml:space="preserve">l’on ne peut trop s’appliquer à les <lb/>concevoir dans toute leur étendue. </s> <s xml:id="echoid-s8287" xml:space="preserve">Quoique l’on puiſſe dé-<lb/>duire la propoſition ſuivante de la précédente, nous allons la <lb/>démontrer en particulier d’une maniere différente, en avertiſ-<lb/>ſant que l’on pourroit auſſi déduire de cette même propoſition <lb/>ſuivante, toutes les propriétés des figures ſemblables, puiſque <lb/>par la définition des figures ſemblables, tous les polygones <lb/>ſemblables ſont compoſés de triangles ſemblables.</s> <s xml:id="echoid-s8288" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div615" type="section" level="1" n="502"> <head xml:id="echoid-head588" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head589" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8289" xml:space="preserve">497. </s> <s xml:id="echoid-s8290" xml:space="preserve">Deux triangles ſemblables A B C, D E G ſont entr’eux <lb/> <anchor type="note" xlink:label="note-0280-01a" xlink:href="note-0280-01"/> comme les quarrés de leurs côtés homologues, c’eſt-à-dire que l’on <lb/>aura A B C : </s> <s xml:id="echoid-s8291" xml:space="preserve">D E G :</s> <s xml:id="echoid-s8292" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8293" xml:space="preserve">D E<emph style="sub">2</emph>, ou :</s> <s xml:id="echoid-s8294" xml:space="preserve">: A C<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8295" xml:space="preserve">D G<emph style="sub">2</emph>.</s> <s xml:id="echoid-s8296" xml:space="preserve"/> </p> <div xml:id="echoid-div615" type="float" level="2" n="1"> <note position="left" xlink:label="note-0280-01" xlink:href="note-0280-01a" xml:space="preserve">Figure 92 <lb/>& 93.</note> </div> </div> <div xml:id="echoid-div617" type="section" level="1" n="503"> <head xml:id="echoid-head590" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s8297" xml:space="preserve">Soient abaiſſées des angles égaux C, G les perpendiculaires <lb/>C H, G F: </s> <s xml:id="echoid-s8298" xml:space="preserve">le triangle A C B eſt égal au produit de ſa baſe <lb/>A B par la moitié de la perpendiculaire C H; </s> <s xml:id="echoid-s8299" xml:space="preserve">& </s> <s xml:id="echoid-s8300" xml:space="preserve">de même le <lb/>triangle D G E eſt égal au produit de ſa baſe D E par la moi-<lb/>tié de la perpendiculaire G F; </s> <s xml:id="echoid-s8301" xml:space="preserve">on aura donc A C B = A B <lb/>x {C H/2}, & </s> <s xml:id="echoid-s8302" xml:space="preserve">D G E = D E x {G F/2}, & </s> <s xml:id="echoid-s8303" xml:space="preserve">faiſant une proportion avec <lb/>les termes de ces équations, on aura A C B : </s> <s xml:id="echoid-s8304" xml:space="preserve">D G E :</s> <s xml:id="echoid-s8305" xml:space="preserve">: A B x {CH/2}: <lb/></s> <s xml:id="echoid-s8306" xml:space="preserve">D E x {G F/2}. </s> <s xml:id="echoid-s8307" xml:space="preserve">Mais puiſque les triangles ſont ſuppoſés ſembla-<lb/>bles, les triangles rectangles A C H, D G F le ſeront auſſi, <pb o="243" file="0281" n="281" rhead="DE MATHÉMATIQUE. Liv. VII."/> ayant un angle égal, outre l’angle droit, l’angle A de l’un <lb/>égal à l’angle D de l’autre: </s> <s xml:id="echoid-s8308" xml:space="preserve">donc on aura C H: </s> <s xml:id="echoid-s8309" xml:space="preserve">G F:</s> <s xml:id="echoid-s8310" xml:space="preserve">: C A: </s> <s xml:id="echoid-s8311" xml:space="preserve">G D, <lb/>& </s> <s xml:id="echoid-s8312" xml:space="preserve">l’on a pour les premiers triangles C A : </s> <s xml:id="echoid-s8313" xml:space="preserve">G D :</s> <s xml:id="echoid-s8314" xml:space="preserve">: A B: </s> <s xml:id="echoid-s8315" xml:space="preserve">D E; <lb/></s> <s xml:id="echoid-s8316" xml:space="preserve">donc A B : </s> <s xml:id="echoid-s8317" xml:space="preserve">D E :</s> <s xml:id="echoid-s8318" xml:space="preserve">: C H: </s> <s xml:id="echoid-s8319" xml:space="preserve">G E; </s> <s xml:id="echoid-s8320" xml:space="preserve">on aura auſſi A B: </s> <s xml:id="echoid-s8321" xml:space="preserve">D F:</s> <s xml:id="echoid-s8322" xml:space="preserve">: {A B/2}: </s> <s xml:id="echoid-s8323" xml:space="preserve">{D E/2}; </s> <s xml:id="echoid-s8324" xml:space="preserve"><lb/>donc en multipliant par ordre les deux dernieres proportions, <lb/>il viendra A B<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8325" xml:space="preserve">D E<emph style="sub">2</emph> :</s> <s xml:id="echoid-s8326" xml:space="preserve">: C H x {A B/2}: </s> <s xml:id="echoid-s8327" xml:space="preserve">G F x {D E/2}; </s> <s xml:id="echoid-s8328" xml:space="preserve">donc puiſque la <lb/>derniere raiſon de cette proportion eſt la même que la derniere <lb/>de notre premiere proportion, on aura A C B : </s> <s xml:id="echoid-s8329" xml:space="preserve">C G E :</s> <s xml:id="echoid-s8330" xml:space="preserve">: A B<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8331" xml:space="preserve">D E<emph style="sub">2</emph>. </s> <s xml:id="echoid-s8332" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s8333" xml:space="preserve">Q. </s> <s xml:id="echoid-s8334" xml:space="preserve">F. </s> <s xml:id="echoid-s8335" xml:space="preserve">D.</s> <s xml:id="echoid-s8336" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div618" type="section" level="1" n="504"> <head xml:id="echoid-head591" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s8337" xml:space="preserve">498. </s> <s xml:id="echoid-s8338" xml:space="preserve">On peut encore ſe ſervir de cette propoſition, pour <lb/> <anchor type="note" xlink:label="note-0281-01a" xlink:href="note-0281-01"/> démontrer que le quarré de l’hypoténuſe eſt égal au quarré des <lb/>deux autres côtés dans un triangle rectangle quelconque, <lb/>comme A B C: </s> <s xml:id="echoid-s8339" xml:space="preserve">car abaiſſant de l’angle droit la perpendicu-<lb/>laire B D, on aura trois triangles ſemblables A B C, A D B, <lb/>B D C; </s> <s xml:id="echoid-s8340" xml:space="preserve">& </s> <s xml:id="echoid-s8341" xml:space="preserve">prenant pour côtés homologues de ces triangles rec-<lb/>tangles les hypoténuſes A C, A B, B C, on aura A B C: </s> <s xml:id="echoid-s8342" xml:space="preserve">A D B: <lb/></s> <s xml:id="echoid-s8343" xml:space="preserve">B D C :</s> <s xml:id="echoid-s8344" xml:space="preserve">: A C<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8345" xml:space="preserve">A B<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8346" xml:space="preserve">B C<emph style="sub">2</emph>; </s> <s xml:id="echoid-s8347" xml:space="preserve">mais le triangles A B C eſt égal à <lb/>la ſomme des triangles A D B, B D C: </s> <s xml:id="echoid-s8348" xml:space="preserve">donc auſſi le quarré <lb/>A C<emph style="sub">2</emph> de l’hypoténuſe A C ſera égal aux quarrés des autres hy-<lb/>poténuſes A B, B C, qui ſont les côtés du même triangle A B C.</s> <s xml:id="echoid-s8349" xml:space="preserve"/> </p> <div xml:id="echoid-div618" type="float" level="2" n="1"> <note position="right" xlink:label="note-0281-01" xlink:href="note-0281-01a" xml:space="preserve">Figure 94.</note> </div> </div> <div xml:id="echoid-div620" type="section" level="1" n="505"> <head xml:id="echoid-head592" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head593" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8350" xml:space="preserve">499. </s> <s xml:id="echoid-s8351" xml:space="preserve">Les parallélogrammes ſont dans la raiſon compoſée des <lb/> <anchor type="note" xlink:label="note-0281-02a" xlink:href="note-0281-02"/> baſes & </s> <s xml:id="echoid-s8352" xml:space="preserve">des hauteurs, c’eſt-à-dire comme les produits de leurs baſes <lb/>par leurs hauteurs.</s> <s xml:id="echoid-s8353" xml:space="preserve"/> </p> <div xml:id="echoid-div620" type="float" level="2" n="1"> <note position="right" xlink:label="note-0281-02" xlink:href="note-0281-02a" xml:space="preserve">Figure 97 <lb/>& 98.</note> </div> </div> <div xml:id="echoid-div622" type="section" level="1" n="506"> <head xml:id="echoid-head594" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8354" xml:space="preserve">Ayant les parallélogrammes G & </s> <s xml:id="echoid-s8355" xml:space="preserve">H, ſi l’on nomme a la baſe <lb/>du premier, & </s> <s xml:id="echoid-s8356" xml:space="preserve">b ſa hauteur, c la baſe du ſecond, & </s> <s xml:id="echoid-s8357" xml:space="preserve">d ſa hau-<lb/>teur, le premier G ſera égal au produit ab, & </s> <s xml:id="echoid-s8358" xml:space="preserve">le ſecond H <lb/>ſera égal au produit c d de ſa baſe par ſa hauteur: </s> <s xml:id="echoid-s8359" xml:space="preserve">ainſi on <lb/>aura G: </s> <s xml:id="echoid-s8360" xml:space="preserve">H :</s> <s xml:id="echoid-s8361" xml:space="preserve">: a b : </s> <s xml:id="echoid-s8362" xml:space="preserve">c d. </s> <s xml:id="echoid-s8363" xml:space="preserve">C. </s> <s xml:id="echoid-s8364" xml:space="preserve">Q. </s> <s xml:id="echoid-s8365" xml:space="preserve">F. </s> <s xml:id="echoid-s8366" xml:space="preserve">D.</s> <s xml:id="echoid-s8367" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div623" type="section" level="1" n="507"> <head xml:id="echoid-head595" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s8368" xml:space="preserve">500. </s> <s xml:id="echoid-s8369" xml:space="preserve">Commelestriangles ſont moitiés des parallélogrammes <pb o="244" file="0282" n="282" rhead="NOUVEAU COURS"/> de même baſe & </s> <s xml:id="echoid-s8370" xml:space="preserve">de même hauteur, ils ſeront auſſi entr’eux <lb/>comme les produits de leurs baſes par leurs hauteurs.</s> <s xml:id="echoid-s8371" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div624" type="section" level="1" n="508"> <head xml:id="echoid-head596" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s8372" xml:space="preserve">501. </s> <s xml:id="echoid-s8373" xml:space="preserve">Si les produits a b, c d des baſes par les hauteurs ſont <lb/>égaux, les parallélogrammes G, H, qui ſont comme ces pro-<lb/>duits, ſeront auſſi égaux; </s> <s xml:id="echoid-s8374" xml:space="preserve">auſſi-bien que les triangles, qui ſont <lb/>la moitié des mêmes parallélogrammes; </s> <s xml:id="echoid-s8375" xml:space="preserve">d’où l’on déduit cette <lb/>propoſition générale: </s> <s xml:id="echoid-s8376" xml:space="preserve">deux parallélogrammes ou deux trian-<lb/>gles ſont égaux, lorſqu’ils ont des baſes réciproques à leurs <lb/>hauteurs; </s> <s xml:id="echoid-s8377" xml:space="preserve">& </s> <s xml:id="echoid-s8378" xml:space="preserve">réciproquement, ſi deux triangles ou deux paral-<lb/>lélogrammes ſont égaux, ils ont des baſes réciproques à leurs <lb/>hauteurs; </s> <s xml:id="echoid-s8379" xml:space="preserve">car puiſque a b = c d, on aura a: </s> <s xml:id="echoid-s8380" xml:space="preserve">c :</s> <s xml:id="echoid-s8381" xml:space="preserve">: d: </s> <s xml:id="echoid-s8382" xml:space="preserve">b.</s> <s xml:id="echoid-s8383" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div625" type="section" level="1" n="509"> <head xml:id="echoid-head597" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s8384" xml:space="preserve">502. </s> <s xml:id="echoid-s8385" xml:space="preserve">Il ſuit encore de cette propoſition, que ſi deux trian-<lb/>gles ou deux parallélogrammes ſont ſemblables, ils ſeront en-<lb/>tr’eux comme les quarrés de leurs baſes ou de leurs hauteurs: <lb/></s> <s xml:id="echoid-s8386" xml:space="preserve">car puiſque ces triangles ſont ſuppoſés ſemblables, les baſes & </s> <s xml:id="echoid-s8387" xml:space="preserve"><lb/>les hauteurs ſeront proportionnelles: </s> <s xml:id="echoid-s8388" xml:space="preserve">ainſi on aura a: </s> <s xml:id="echoid-s8389" xml:space="preserve">c :</s> <s xml:id="echoid-s8390" xml:space="preserve">: b: </s> <s xml:id="echoid-s8391" xml:space="preserve">d, <lb/>& </s> <s xml:id="echoid-s8392" xml:space="preserve">a: </s> <s xml:id="echoid-s8393" xml:space="preserve">c :</s> <s xml:id="echoid-s8394" xml:space="preserve">: a: </s> <s xml:id="echoid-s8395" xml:space="preserve">c, multipliant ces deux proportions par ordre, il <lb/>viendra a a: </s> <s xml:id="echoid-s8396" xml:space="preserve">cc :</s> <s xml:id="echoid-s8397" xml:space="preserve">: a b: </s> <s xml:id="echoid-s8398" xml:space="preserve">c d; </s> <s xml:id="echoid-s8399" xml:space="preserve">donc puiſque la raiſon de a<emph style="sub">2</emph> à c<emph style="sub">2</emph> eſt <lb/>égale à celle de a b à c d, on aura G: </s> <s xml:id="echoid-s8400" xml:space="preserve">H :</s> <s xml:id="echoid-s8401" xml:space="preserve">: a<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8402" xml:space="preserve">c<emph style="sub">2</emph>, c’eſt-à-dire <lb/>que les parallélogrammes ſemblables, ou les triangles qui en <lb/>ſont les moitiés, ſont entr’eux comme les quarrés de leurs <lb/>baſes, ou comme s’expriment les Géometres en raiſon dou-<lb/>blée de leurs baſes.</s> <s xml:id="echoid-s8403" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div626" type="section" level="1" n="510"> <head xml:id="echoid-head598" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head599" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8404" xml:space="preserve">503. </s> <s xml:id="echoid-s8405" xml:space="preserve">Si l’on a trois lignes en proportion continue, je dis que le <lb/>quarré fait ſur la premiere, eſt au quarré fait ſur la ſeconde, comme <lb/>la premiere ligne eſt à la troiſieme, c’eſt-à-dire, en repréſentant ces <lb/>lignes par les lettres a, b, c, que ſi l’on a, a: </s> <s xml:id="echoid-s8406" xml:space="preserve">b :</s> <s xml:id="echoid-s8407" xml:space="preserve">: b : </s> <s xml:id="echoid-s8408" xml:space="preserve">c, on aura <lb/>a<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8409" xml:space="preserve">b<emph style="sub">2</emph> :</s> <s xml:id="echoid-s8410" xml:space="preserve">: a : </s> <s xml:id="echoid-s8411" xml:space="preserve">c.</s> <s xml:id="echoid-s8412" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div627" type="section" level="1" n="511"> <head xml:id="echoid-head600" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8413" xml:space="preserve">Pour prouver que aa: </s> <s xml:id="echoid-s8414" xml:space="preserve">bb :</s> <s xml:id="echoid-s8415" xml:space="preserve">: a: </s> <s xml:id="echoid-s8416" xml:space="preserve">c, nous ferons voir que le pro-<lb/>duit des extrêmes eſt égal à celui des moyens, ou que abb = aac. <lb/></s> <s xml:id="echoid-s8417" xml:space="preserve">Pour cela, faites attention que puiſque par hypotheſe les trois <pb o="245" file="0283" n="283" rhead="DE MATHÉMATIQUE. Liv. VII."/> lignes ſont en proportion continue, on aura a: </s> <s xml:id="echoid-s8418" xml:space="preserve">b :</s> <s xml:id="echoid-s8419" xml:space="preserve">: b: </s> <s xml:id="echoid-s8420" xml:space="preserve">c, d’où <lb/>l’on tire a c = b<emph style="sub">2</emph>. </s> <s xml:id="echoid-s8421" xml:space="preserve">Si donc on multiplie chaque membre de <lb/>cette équation par a, on aura a<emph style="sub">2</emph>c = ab<emph style="sub">2</emph>, qui eſt préciſément <lb/>le produit des extrêmes & </s> <s xml:id="echoid-s8422" xml:space="preserve">celui des moyens de la proportion <lb/>qu’il s’agiſſoit de prouver. </s> <s xml:id="echoid-s8423" xml:space="preserve">C. </s> <s xml:id="echoid-s8424" xml:space="preserve">Q. </s> <s xml:id="echoid-s8425" xml:space="preserve">F. </s> <s xml:id="echoid-s8426" xml:space="preserve">D.</s> <s xml:id="echoid-s8427" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div628" type="section" level="1" n="512"> <head xml:id="echoid-head601" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s8428" xml:space="preserve">504. </s> <s xml:id="echoid-s8429" xml:space="preserve">Il ſuit de cette propoſition, que ſi l’on a trois lignes <lb/>proportionnelles, non ſeulement le quarré fait ſur la premiere <lb/>eſt au quarré fait ſur la ſeconde, comme la premiere eſt à la <lb/>troiſieme; </s> <s xml:id="echoid-s8430" xml:space="preserve">mais que tous polygones ſemblables, réguliers ou <lb/>irréguliers, faits ſur ces deux lignes, ſeront entr’eux comme <lb/>la premiere eſt à la troiſieme: </s> <s xml:id="echoid-s8431" xml:space="preserve">car comme les polygones ſem-<lb/>blables ſont entr’eux comme les quarrés de leurs rayons ou des <lb/>côtés homologues, & </s> <s xml:id="echoid-s8432" xml:space="preserve">que par hypotheſe, nos deux premieres <lb/>lignes ſont des côtés homologues de ces polygones ſemblables, <lb/>le premier polygone ſera au ſecond, comme le quarré de la <lb/>premiere ligne au quarré de la ſeconde, ou comme la premiere <lb/>ligne à la troiſieme. </s> <s xml:id="echoid-s8433" xml:space="preserve">D’où il ſuit, qu’ayant les ſurfaces de deux <lb/>polygones ſemblables, on peut toujours aſſigner deux lignes <lb/>qui ſoient entr’elles, comme ces ſurfaces.</s> <s xml:id="echoid-s8434" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div629" type="section" level="1" n="513"> <head xml:id="echoid-head602" xml:space="preserve">PROPOSITION IX.</head> <head xml:id="echoid-head603" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8435" xml:space="preserve">505. </s> <s xml:id="echoid-s8436" xml:space="preserve">Si l’on a deux lignes droites, que nous nommerons a & </s> <s xml:id="echoid-s8437" xml:space="preserve">b, <lb/>je dis que le rectangle compris ſous ces deux lignes, eſt moyen pro-<lb/>portionnel entre les quarrés des mêmes lignes, c’eſt-à-dire que l’on <lb/>aura aa: </s> <s xml:id="echoid-s8438" xml:space="preserve">ab :</s> <s xml:id="echoid-s8439" xml:space="preserve">: ab: </s> <s xml:id="echoid-s8440" xml:space="preserve">bb.</s> <s xml:id="echoid-s8441" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div630" type="section" level="1" n="514"> <head xml:id="echoid-head604" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8442" xml:space="preserve">Il eſt certain que la proportion aa: </s> <s xml:id="echoid-s8443" xml:space="preserve">ab :</s> <s xml:id="echoid-s8444" xml:space="preserve">: ab: </s> <s xml:id="echoid-s8445" xml:space="preserve">bb, doit avoir <lb/>lieu, puiſque le produit des extrêmes & </s> <s xml:id="echoid-s8446" xml:space="preserve">celui des moyens don-<lb/>nent a a b b = a a b b. </s> <s xml:id="echoid-s8447" xml:space="preserve">C. </s> <s xml:id="echoid-s8448" xml:space="preserve">Q. </s> <s xml:id="echoid-s8449" xml:space="preserve">F. </s> <s xml:id="echoid-s8450" xml:space="preserve">D.</s> <s xml:id="echoid-s8451" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div631" type="section" level="1" n="515"> <head xml:id="echoid-head605" xml:space="preserve">PROPOSITION X.</head> <head xml:id="echoid-head606" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8452" xml:space="preserve">506. </s> <s xml:id="echoid-s8453" xml:space="preserve">Trouver une moyenne proportionnelle entre deux lignes <lb/> <anchor type="note" xlink:label="note-0283-01a" xlink:href="note-0283-01"/> données.</s> <s xml:id="echoid-s8454" xml:space="preserve"/> </p> <div xml:id="echoid-div631" type="float" level="2" n="1"> <note position="right" xlink:label="note-0283-01" xlink:href="note-0283-01a" xml:space="preserve">Figure 99.</note> </div> <pb o="246" file="0284" n="284" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s8455" xml:space="preserve">Pour trouver une moyenne proportionnelle entre les deux <lb/>lignes A & </s> <s xml:id="echoid-s8456" xml:space="preserve">B, il faut joindre ces deux lignes, enſorte qu’elles <lb/>n’en faſſent qu’une ſeule C D, obſervant de marquer le point <lb/>E où elles ſe joignent; </s> <s xml:id="echoid-s8457" xml:space="preserve">il faut enſuite diviſer la ligne entiere <lb/>en deux également au point F, & </s> <s xml:id="echoid-s8458" xml:space="preserve">de cepoint, comme centre, <lb/>décrire un demi-cercle. </s> <s xml:id="echoid-s8459" xml:space="preserve">Préſentement ſi au point E, où les deux <lb/>lignes ſe joignent, on éleve une perpendiculaire E H, qui aille <lb/>ſe terminer à la circonférence, elle ſera la moyenne que l’on <lb/>cherche; </s> <s xml:id="echoid-s8460" xml:space="preserve">ce qui eſt bien évident, puiſque par la propriété du <lb/>cercle (art. </s> <s xml:id="echoid-s8461" xml:space="preserve">444), toute perpendiculaire, comme H E, eſt <lb/>moyenne proportionnelle entre les parties C E & </s> <s xml:id="echoid-s8462" xml:space="preserve">E D du dia-<lb/>metre. </s> <s xml:id="echoid-s8463" xml:space="preserve">Ainſi ſuppoſant que la ligne K ſoit égale à H E, l’on <lb/>aura les trois lignes proportionnelles A, K, B.</s> <s xml:id="echoid-s8464" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8465" xml:space="preserve">507. </s> <s xml:id="echoid-s8466" xml:space="preserve">Si l’on vouloit avoir une moyenne proportionnelle <lb/>entre deux nombres donnés, comme 4 & </s> <s xml:id="echoid-s8467" xml:space="preserve">9, il faudroit mul-<lb/>tiplier ces deux nombres l’un par l’autre, & </s> <s xml:id="echoid-s8468" xml:space="preserve">extraire la racine <lb/>du produit 36, que l’on regardera comme le quarré de la <lb/>moyenne, qui eſt 6, puiſque le quarré de cette moyenne eſt <lb/>égal au produit des extrêmes 4 & </s> <s xml:id="echoid-s8469" xml:space="preserve">9; </s> <s xml:id="echoid-s8470" xml:space="preserve">ce qui donne 4: </s> <s xml:id="echoid-s8471" xml:space="preserve">6 :</s> <s xml:id="echoid-s8472" xml:space="preserve">: 6: </s> <s xml:id="echoid-s8473" xml:space="preserve">9.</s> <s xml:id="echoid-s8474" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8475" xml:space="preserve">Si le produit des deux nombres donnés n’eſt pas un quarré, <lb/>ce qui arrivera toutes les fois que l’un des nombres, ou tous les <lb/>deux, ne ſeront point des quarrés, on ne pourra avoir la <lb/>moyenne que l’on demande que par approximation, en ſe ſer-<lb/>vant des décimales pour extraire la racine du produit. </s> <s xml:id="echoid-s8476" xml:space="preserve">Il eſt <lb/>encore à remarquer que la Géométrie nous donne exacte-<lb/>ment ces lignes, quoiqu’elles ſoient ce qu’on appelle incom-<lb/>menſurables, c’eſt-à-dire qu’elles n’aient aucune meſure com-<lb/>mune, ſi petite qu’elle ſoit, avec les lignes propoſées. </s> <s xml:id="echoid-s8477" xml:space="preserve">Par <lb/>exemple, quoiqu’il puiſſe arriver que le nombre des parties de <lb/>la ligne A ne ſoit pas un nombre quarré, ainſi que ceux des <lb/>parties de la ligne B, on trouve cependant la longueur exacte <lb/>de la moyenne K, que l’on ne pourroit pas déterminer en <lb/>nombres dans cette ſuppoſition.</s> <s xml:id="echoid-s8478" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div633" type="section" level="1" n="516"> <head xml:id="echoid-head607" xml:space="preserve">PROPOSITION XI.</head> <head xml:id="echoid-head608" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8479" xml:space="preserve">508. </s> <s xml:id="echoid-s8480" xml:space="preserve">Trouver une troiſieme proportionnelle à deux lignes don-<lb/>nées.</s> <s xml:id="echoid-s8481" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8482" xml:space="preserve">Si l’on veut trouver une troiſieme proportionnelle à deux <pb o="247" file="0285" n="285" rhead="DE MATHÉMATIQUE. Liv. VII."/> lignes données M & </s> <s xml:id="echoid-s8483" xml:space="preserve">N, enſorte que la premiere ligne M ſoit <lb/>à la ſeconde N, comme la même ſeconde N à celle que l’on <lb/>cherche; </s> <s xml:id="echoid-s8484" xml:space="preserve">il faut faire à volonté un angle A B C, prendre ſur <lb/> <anchor type="note" xlink:label="note-0285-01a" xlink:href="note-0285-01"/> le côté B C la partie B D égale à la premiere M, & </s> <s xml:id="echoid-s8485" xml:space="preserve">la partie <lb/>D F égale à la ſeconde N, & </s> <s xml:id="echoid-s8486" xml:space="preserve">ſur le côté B A la partie B E égale <lb/>à la même ſeconde N, & </s> <s xml:id="echoid-s8487" xml:space="preserve">tirer la ligne E D; </s> <s xml:id="echoid-s8488" xml:space="preserve">ſi du point F <lb/>on mene la ligne F G parallele à la ligne E D, je dis que la ligne <lb/>E G ſera la troiſieme proportionnelle demandée.</s> <s xml:id="echoid-s8489" xml:space="preserve"/> </p> <div xml:id="echoid-div633" type="float" level="2" n="1"> <note position="right" xlink:label="note-0285-01" xlink:href="note-0285-01a" xml:space="preserve">Figure 100.</note> </div> </div> <div xml:id="echoid-div635" type="section" level="1" n="517"> <head xml:id="echoid-head609" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s8490" xml:space="preserve">Conſidérez que le triangle B G F a ſes deux côtés B G, B F <lb/>coupés proportionnellement par la ligne D E parallele à ſa baſe <lb/>F G, par conſtruction, & </s> <s xml:id="echoid-s8491" xml:space="preserve">que par conſéquent (art. </s> <s xml:id="echoid-s8492" xml:space="preserve">397) on a <lb/>B D : </s> <s xml:id="echoid-s8493" xml:space="preserve">D F :</s> <s xml:id="echoid-s8494" xml:space="preserve">: B E: </s> <s xml:id="echoid-s8495" xml:space="preserve">E G, mais B E étant égal à D F, par con-<lb/>ſtruction, on aura B D: </s> <s xml:id="echoid-s8496" xml:space="preserve">D F :</s> <s xml:id="echoid-s8497" xml:space="preserve">: D F: </s> <s xml:id="echoid-s8498" xml:space="preserve">E G. </s> <s xml:id="echoid-s8499" xml:space="preserve">Ainſi faiſant la ligne <lb/>O égale à E G, on aura les trois lignes continuement propor-<lb/>tionnelles M, N, O.</s> <s xml:id="echoid-s8500" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8501" xml:space="preserve">509. </s> <s xml:id="echoid-s8502" xml:space="preserve">Si l’on vouloit trouver une troiſieme proportionnelle <lb/>à deux nombres, il faut quarrer le ſecond, & </s> <s xml:id="echoid-s8503" xml:space="preserve">diviſer ce quarré <lb/>par le premier; </s> <s xml:id="echoid-s8504" xml:space="preserve">le quotient ſera la troiſieme proportionnelle <lb/>demandée. </s> <s xml:id="echoid-s8505" xml:space="preserve">Si le ſecond nombre n’eſt pas diviſible par le pre-<lb/>mier, ſon quarré ne ſera pas non plus diviſible par ce même <lb/>premier nombre: </s> <s xml:id="echoid-s8506" xml:space="preserve">ainſi l’on ne pourra trouver la troiſieme pro-<lb/>portionnelle que par approximation, en ſe ſervant des fractions <lb/>décimales. </s> <s xml:id="echoid-s8507" xml:space="preserve">Surquoi l’on remarquera encore la différence de la <lb/>Géométrie à l’Arithmétique dans la détermination des quan-<lb/>tités, en ce que la premiere donne exactement la longueur <lb/>des lignes que l’on cherche, ſans déterminer le nombre de <lb/>leurs parties, & </s> <s xml:id="echoid-s8508" xml:space="preserve">la ſeconde donne leur valeur exacte dans cer-<lb/>tains cas, en fixant le nombre de lcurs parties; </s> <s xml:id="echoid-s8509" xml:space="preserve">& </s> <s xml:id="echoid-s8510" xml:space="preserve">dans d’au-<lb/>tres, ne peut la donner que par une approximation, que l’on <lb/>pouſſeroit juſqu’à l’infini, ſans jamais arriver à la juſte valeur.</s> <s xml:id="echoid-s8511" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8512" xml:space="preserve">On pourroit encore réſoudre le dernier problême d’une au-<lb/>tre maniere, en ſe ſervant du cercle. </s> <s xml:id="echoid-s8513" xml:space="preserve">Qu’il faille, par exemple, <lb/>trouver une troiſieme proportionnelle aux lignes B, K, on pren-<lb/>dra la ligne C E égale à la ligne B; </s> <s xml:id="echoid-s8514" xml:space="preserve">ſur cette ligne on élevera <lb/>la perpendiculaire E H égale à la ligne K; </s> <s xml:id="echoid-s8515" xml:space="preserve">on menera la ligne <lb/>C H, ſur laquelle on élevera la droite H D perpendiculaire, <lb/>qui ira rencontrer le prolongement de la ligne C E en D, & </s> <s xml:id="echoid-s8516" xml:space="preserve"><lb/>déterminera la ligne E D, qui ſera la troiſieme proportionnelle <pb o="248" file="0286" n="286" rhead="NOUVEAU COURS"/> demandée: </s> <s xml:id="echoid-s8517" xml:space="preserve">car il eſt viſible que l’angle C H D étant droit, la <lb/>droite E H ſera moyenne entre les ſegmens de la baſe, ou, ce <lb/>qui revient au même, la droite E D ſera troiſieme proportion-<lb/>nelle aux lignes C E, E H, ou à leurs égales B, K. </s> <s xml:id="echoid-s8518" xml:space="preserve">C. </s> <s xml:id="echoid-s8519" xml:space="preserve">Q. </s> <s xml:id="echoid-s8520" xml:space="preserve">F. <lb/></s> <s xml:id="echoid-s8521" xml:space="preserve">T. </s> <s xml:id="echoid-s8522" xml:space="preserve">& </s> <s xml:id="echoid-s8523" xml:space="preserve">D.</s> <s xml:id="echoid-s8524" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div636" type="section" level="1" n="518"> <head xml:id="echoid-head610" xml:space="preserve">PROPOSITION XII.</head> <head xml:id="echoid-head611" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8525" xml:space="preserve">510. </s> <s xml:id="echoid-s8526" xml:space="preserve">Trouver une quatrieme proportionnelle à trois lignes don-<lb/> <anchor type="note" xlink:label="note-0286-01a" xlink:href="note-0286-01"/> nées.</s> <s xml:id="echoid-s8527" xml:space="preserve"/> </p> <div xml:id="echoid-div636" type="float" level="2" n="1"> <note position="left" xlink:label="note-0286-01" xlink:href="note-0286-01a" xml:space="preserve">Figure 101 <lb/>& 102.</note> </div> <p> <s xml:id="echoid-s8528" xml:space="preserve">Pour trouver une quatrieme proportionnelle aux trois lignes <lb/>P, Q, R, il faut, comme dans la propoſition précédente, faire <lb/>un angle à volonté C S X; </s> <s xml:id="echoid-s8529" xml:space="preserve">prendre ſur le côté C S la partie <lb/>S V égale à la ligne P, & </s> <s xml:id="echoid-s8530" xml:space="preserve">la partie V Z ſur le même côté égale <lb/>à la ligne Q, & </s> <s xml:id="echoid-s8531" xml:space="preserve">ſur l’autre côté S X, la partie S T égale à la <lb/>ligne R; </s> <s xml:id="echoid-s8532" xml:space="preserve">après quoi tirer la ligne T V, à laquelle on menera <lb/>du point Z la parallele Z X, qui donnera la ligne T X égale à <lb/>la quatrieme proportionnelle que l’on cherche.</s> <s xml:id="echoid-s8533" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div638" type="section" level="1" n="519"> <head xml:id="echoid-head612" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8534" xml:space="preserve">Les côtés du triangle Z S X étant coupés par la ligne T V, <lb/>parallele à la baſe Z X, l’on aura (art. </s> <s xml:id="echoid-s8535" xml:space="preserve">393) S V : </s> <s xml:id="echoid-s8536" xml:space="preserve">V Z :</s> <s xml:id="echoid-s8537" xml:space="preserve">: ST : </s> <s xml:id="echoid-s8538" xml:space="preserve">TX. <lb/></s> <s xml:id="echoid-s8539" xml:space="preserve">Ainſi faiſant la ligne Y égale à T X, l’on aura les quatre lignes <lb/>proportionnelles, P, Q, R, Y. </s> <s xml:id="echoid-s8540" xml:space="preserve">C. </s> <s xml:id="echoid-s8541" xml:space="preserve">Q. </s> <s xml:id="echoid-s8542" xml:space="preserve">F. </s> <s xml:id="echoid-s8543" xml:space="preserve">T. </s> <s xml:id="echoid-s8544" xml:space="preserve">& </s> <s xml:id="echoid-s8545" xml:space="preserve">D.</s> <s xml:id="echoid-s8546" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8547" xml:space="preserve">511. </s> <s xml:id="echoid-s8548" xml:space="preserve">Pour trouver une quatrieme proportionnelle à trois <lb/>nombres donnés, il n’y a qu’à faire la Regle de Trois ordi-<lb/>naire, puiſque cette Regle n’eſt autre choſe que l’art de trouver <lb/>une grandeur quatrieme proportionnelle à trois autres don-<lb/>nées. </s> <s xml:id="echoid-s8549" xml:space="preserve">On va voir dans les problêmes ſuivans, l’uſage qu’on <lb/>peut faire des précédens, & </s> <s xml:id="echoid-s8550" xml:space="preserve">les propriétés des lignes propor-<lb/>tionnelles.</s> <s xml:id="echoid-s8551" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div639" type="section" level="1" n="520"> <head xml:id="echoid-head613" xml:space="preserve">PROPOSITION XIII.</head> <head xml:id="echoid-head614" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8552" xml:space="preserve">512. </s> <s xml:id="echoid-s8553" xml:space="preserve">Faire un quarré égal à un rectangle.</s> <s xml:id="echoid-s8554" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 97 <lb/>& 98.</note> <p> <s xml:id="echoid-s8555" xml:space="preserve">Pour faire un quarré égal à un rectangle A C, il faut cher-<lb/>cher une moyenne proportionnelle entre les côtés inégaux A B <lb/>& </s> <s xml:id="echoid-s8556" xml:space="preserve">B C du rectangle donné, & </s> <s xml:id="echoid-s8557" xml:space="preserve">le quarré de cette moyenne ſera <lb/>égal au rectangle donné. </s> <s xml:id="echoid-s8558" xml:space="preserve">Puiſque la ligne D E eſt moyenne <pb o="249" file="0287" n="287" rhead="DE MATHÉMATIQUE. Liv. VII."/> proportionnelle entre les côtés A B & </s> <s xml:id="echoid-s8559" xml:space="preserve">B C du rectangle A C, <lb/>il eſt certain que ſon quarré D F ſera égal au rectangle A C, <lb/>puiſque ce rectangle eſt égal au produit des extrêmes A B, B C.</s> <s xml:id="echoid-s8560" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div640" type="section" level="1" n="521"> <head xml:id="echoid-head615" xml:space="preserve"><emph style="sc">Corollaire.</emph></head> <p> <s xml:id="echoid-s8561" xml:space="preserve">513. </s> <s xml:id="echoid-s8562" xml:space="preserve">Comme nous avons prouvé qu’un cercle eſt égal à <lb/>un rectangle compris ſous la moitié de la circonférence, & </s> <s xml:id="echoid-s8563" xml:space="preserve">la <lb/>moitié du diametre (art. </s> <s xml:id="echoid-s8564" xml:space="preserve">485), il s’enſuit que le quarré d’une <lb/>ligne qui ſeroit moyenne proportionnelle entre le demi-dia-<lb/>metre & </s> <s xml:id="echoid-s8565" xml:space="preserve">la demi-circonférence, ſeroit égal au cercle.</s> <s xml:id="echoid-s8566" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div641" type="section" level="1" n="522"> <head xml:id="echoid-head616" xml:space="preserve">PROPOSITION XIV.</head> <head xml:id="echoid-head617" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8567" xml:space="preserve">514. </s> <s xml:id="echoid-s8568" xml:space="preserve">Trouver un quarré qui ſoit à un autre dans une raiſon <lb/> <anchor type="note" xlink:label="note-0287-01a" xlink:href="note-0287-01"/> donnée.</s> <s xml:id="echoid-s8569" xml:space="preserve"/> </p> <div xml:id="echoid-div641" type="float" level="2" n="1"> <note position="right" xlink:label="note-0287-01" xlink:href="note-0287-01a" xml:space="preserve">Figure 105 <lb/>& 106.</note> </div> <p> <s xml:id="echoid-s8570" xml:space="preserve">Pour trouver un quarré qui ſoit au quarré C B dans une rai-<lb/>ſon donnée, par exemple, de 3 à 5, je fais une ligne G H, <lb/>égale aux trois cinquiemes du côté A B; </s> <s xml:id="echoid-s8571" xml:space="preserve">enſuite entre les li-<lb/>gnes A B & </s> <s xml:id="echoid-s8572" xml:space="preserve">G H, je cherche une moyenne proportionnelle <lb/>E F, ſur laquelle je fais le quarré I F, qui ſera les trois cin-<lb/>quiemes du quarré C B: </s> <s xml:id="echoid-s8573" xml:space="preserve">car comme les trois lignes A B, E F, <lb/>G H ſont en proportion continue, on aura A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8574" xml:space="preserve">E F<emph style="sub">2</emph> :</s> <s xml:id="echoid-s8575" xml:space="preserve">: A B : </s> <s xml:id="echoid-s8576" xml:space="preserve">G H; <lb/></s> <s xml:id="echoid-s8577" xml:space="preserve">mais G H eſt, par conſtruction, les trois cinquiemes de A B: </s> <s xml:id="echoid-s8578" xml:space="preserve"><lb/>donc auſſi E F<emph style="sub">2</emph> ſera les trois cinquiemes du quarré A B<emph style="sub">2</emph>.</s> <s xml:id="echoid-s8579" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8580" xml:space="preserve">515. </s> <s xml:id="echoid-s8581" xml:space="preserve">Cette propoſition doit s’entendre, non ſeulement des <lb/>quarrés, mais encore de toutes les figures. </s> <s xml:id="echoid-s8582" xml:space="preserve">Par exemple, ſi l’on <lb/>vouloit faire un pentagone irrégulier quelconque ſemblable à <lb/>un autre pentagone irrégulier, & </s> <s xml:id="echoid-s8583" xml:space="preserve">qui eût avec lui une raiſon <lb/>donnée, on chercheroit une moyenne proportionnelle entre un <lb/>côté quelconque du pentagone propoſé, & </s> <s xml:id="echoid-s8584" xml:space="preserve">une ligne qui au-<lb/>roit avec ce côté, la raiſon donnée: </s> <s xml:id="echoid-s8585" xml:space="preserve">ſur cette moyenne ainſi <lb/>déterminée, comme côté homologue, on décriroit le penta-<lb/>gone demandé, & </s> <s xml:id="echoid-s8586" xml:space="preserve">l’on trouveroit les autres côtés par une ſim-<lb/>ple Regle de Trois, en ſe ſervant des triangles ſemblables, <lb/>comme on a vu (art. </s> <s xml:id="echoid-s8587" xml:space="preserve">510). </s> <s xml:id="echoid-s8588" xml:space="preserve">Cette propoſition fournit un moyen <lb/>pour réduire des figures quelconques de grand en petit, ou de <lb/>petit en grand, dans un rapport quelconque.</s> <s xml:id="echoid-s8589" xml:space="preserve"/> </p> <pb o="250" file="0288" n="288" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div643" type="section" level="1" n="523"> <head xml:id="echoid-head618" xml:space="preserve">PROPOSITION XV.</head> <head xml:id="echoid-head619" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8590" xml:space="preserve">516. </s> <s xml:id="echoid-s8591" xml:space="preserve">Trouver le rapport de deux figures ſemblables.</s> <s xml:id="echoid-s8592" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8593" xml:space="preserve">Pour trouver le rapport de deux figures ſemblables A & </s> <s xml:id="echoid-s8594" xml:space="preserve">B, <lb/> <anchor type="note" xlink:label="note-0288-01a" xlink:href="note-0288-01"/> il faut chercher une troiſieme proportionnelle, telle que G H <lb/>à leurs côtés homologues, C D & </s> <s xml:id="echoid-s8595" xml:space="preserve">E F; </s> <s xml:id="echoid-s8596" xml:space="preserve">le rapport de la ligne <lb/>C D à la ligne G H, ſera le même que celui du polygone A au <lb/>polygone B.</s> <s xml:id="echoid-s8597" xml:space="preserve"/> </p> <div xml:id="echoid-div643" type="float" level="2" n="1"> <note position="left" xlink:label="note-0288-01" xlink:href="note-0288-01a" xml:space="preserve">Figure 107 <lb/>& 108.</note> </div> <p> <s xml:id="echoid-s8598" xml:space="preserve">Pour le prouver, conſidérez que puiſque les polygones A <lb/>& </s> <s xml:id="echoid-s8599" xml:space="preserve">B ſont ſemblables, on a A : </s> <s xml:id="echoid-s8600" xml:space="preserve">B :</s> <s xml:id="echoid-s8601" xml:space="preserve">: C D<emph style="sub">2</emph>: </s> <s xml:id="echoid-s8602" xml:space="preserve">E F<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s8603" xml:space="preserve">que puiſ-<lb/>que les trois lignes C D, E F, G H ſont en proportion conti-<lb/>nue, on a C D<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8604" xml:space="preserve">E F<emph style="sub">2</emph> :</s> <s xml:id="echoid-s8605" xml:space="preserve">: C D : </s> <s xml:id="echoid-s8606" xml:space="preserve">GH, d’où l’on tire A : </s> <s xml:id="echoid-s8607" xml:space="preserve">B :</s> <s xml:id="echoid-s8608" xml:space="preserve">: CD : </s> <s xml:id="echoid-s8609" xml:space="preserve">GH. <lb/></s> <s xml:id="echoid-s8610" xml:space="preserve">C. </s> <s xml:id="echoid-s8611" xml:space="preserve">Q. </s> <s xml:id="echoid-s8612" xml:space="preserve">F. </s> <s xml:id="echoid-s8613" xml:space="preserve">T. </s> <s xml:id="echoid-s8614" xml:space="preserve">& </s> <s xml:id="echoid-s8615" xml:space="preserve">D.</s> <s xml:id="echoid-s8616" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div645" type="section" level="1" n="524"> <head xml:id="echoid-head620" xml:space="preserve">PROPOSITION XVI.</head> <head xml:id="echoid-head621" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8617" xml:space="preserve">517. </s> <s xml:id="echoid-s8618" xml:space="preserve">Faire un rectangle égal à un autre qui ait un côté dé-<lb/> <anchor type="note" xlink:label="note-0288-02a" xlink:href="note-0288-02"/> terminé.</s> <s xml:id="echoid-s8619" xml:space="preserve"/> </p> <div xml:id="echoid-div645" type="float" level="2" n="1"> <note position="left" xlink:label="note-0288-02" xlink:href="note-0288-02a" xml:space="preserve">Figure 109 <lb/>& 110.</note> </div> <p> <s xml:id="echoid-s8620" xml:space="preserve">L’on demande de faire un rectangle égal au rectangle B C, <lb/>enſorte qu’il ait un de ſes côtés égal à la ligne donnée D E.</s> <s xml:id="echoid-s8621" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8622" xml:space="preserve">Pour cela, il faut chercher une ligne qui ſoit quatrieme pro-<lb/>portionnelle à la ligne donnée D E (art. </s> <s xml:id="echoid-s8623" xml:space="preserve">510), & </s> <s xml:id="echoid-s8624" xml:space="preserve">aux deux <lb/>côtés A C & </s> <s xml:id="echoid-s8625" xml:space="preserve">A B du rectangle; </s> <s xml:id="echoid-s8626" xml:space="preserve">enſuite ſi l’on fait un rectan-<lb/>gle ſous la ligne donnée D E, & </s> <s xml:id="echoid-s8627" xml:space="preserve">ſous la quatrieme que l’on <lb/>aura trouvée, ce rectangle ſera égal au rectangle B C.</s> <s xml:id="echoid-s8628" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8629" xml:space="preserve">Pour le prouver, conſidérez que ſi l’on a fait le rectangle <lb/>G H, dont le côté F G ſoit égal à la proportionnelle trouvée, <lb/>& </s> <s xml:id="echoid-s8630" xml:space="preserve">le côté F H égal à D E, on aura F G: </s> <s xml:id="echoid-s8631" xml:space="preserve">A B :</s> <s xml:id="echoid-s8632" xml:space="preserve">: A C: </s> <s xml:id="echoid-s8633" xml:space="preserve">F H; <lb/></s> <s xml:id="echoid-s8634" xml:space="preserve">donc F G x F H = A B x A C. </s> <s xml:id="echoid-s8635" xml:space="preserve">C. </s> <s xml:id="echoid-s8636" xml:space="preserve">Q. </s> <s xml:id="echoid-s8637" xml:space="preserve">F. </s> <s xml:id="echoid-s8638" xml:space="preserve">D.</s> <s xml:id="echoid-s8639" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div647" type="section" level="1" n="525"> <head xml:id="echoid-head622" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s8640" xml:space="preserve">518. </s> <s xml:id="echoid-s8641" xml:space="preserve">Il ſuit de cette propoſition, que ſi l’on a pluſieurs rectan-<lb/>gles, dont les baſes & </s> <s xml:id="echoid-s8642" xml:space="preserve">les hauteurs ſoient inégales, on pourra <lb/>les réduire tous à la même hauteur; </s> <s xml:id="echoid-s8643" xml:space="preserve">& </s> <s xml:id="echoid-s8644" xml:space="preserve">après cela, ſi l’on <lb/>veut, n’en faire qu’un ſeul, égal à tous les autres pris enſem-<lb/>ble, en lui donnant pour baſe une ligne égale à la ſomme de <lb/>toutes les baſes, & </s> <s xml:id="echoid-s8645" xml:space="preserve">pour hauteur, la hauteur commune.</s> <s xml:id="echoid-s8646" xml:space="preserve"/> </p> <pb o="251" file="0289" n="289" rhead="DE MATHÉMATIQUE. Liv. VII."/> </div> <div xml:id="echoid-div648" type="section" level="1" n="526"> <head xml:id="echoid-head623" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s8647" xml:space="preserve">519. </s> <s xml:id="echoid-s8648" xml:space="preserve">Comme on peut réduire toutes les figures rectiligne des <lb/>triangles, & </s> <s xml:id="echoid-s8649" xml:space="preserve">que de chaque triangle on peut faire un rectan-<lb/>gle, il ſuit encore, que ſi l’on donne la même hauteur aux rec-<lb/>tangles provenus des triangles, on pourra, en les réduiſant <lb/>tous dans un ſeul, faire un quarré égal à une figure rectiligne, <lb/>compoſée d’un grand nombre de côtés, & </s> <s xml:id="echoid-s8650" xml:space="preserve">même à la ſomme <lb/>de pluſieurs figures rectilignes, puiſqu’on n’aura qu’à chercher <lb/>une moyenne proportionnelle entre les côtés du rectangle égal <lb/>à la figure rectiligne propoſée, ou à la ſomme des figures <lb/>données.</s> <s xml:id="echoid-s8651" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div649" type="section" level="1" n="527"> <head xml:id="echoid-head624" xml:space="preserve"><emph style="sc">Scholie</emph>.</head> <p> <s xml:id="echoid-s8652" xml:space="preserve">520. </s> <s xml:id="echoid-s8653" xml:space="preserve">Toutes la théorie des rapports des figures ſemblables <lb/>ou non ſemblables, eſt fondée ſur les propoſitions que nous <lb/>venons de démontrer. </s> <s xml:id="echoid-s8654" xml:space="preserve">Mais comme toutes les figures géomé-<lb/>triques droites ou courbes ſont compoſées de triangles, pour <lb/>rendre cette partie encore plus complette, nous allons ajouter <lb/>deux Théorêmes ſur les propriétés des triangles conſidérés <lb/>par rapport à leurs ſuperficies, & </s> <s xml:id="echoid-s8655" xml:space="preserve">dont la connoiſſance ne peut <lb/>être que très-utile dans la Géométrie pratique.</s> <s xml:id="echoid-s8656" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8657" xml:space="preserve">Le premier que j’ai tiré d’un Livre de M. </s> <s xml:id="echoid-s8658" xml:space="preserve">Scooten, Com-<lb/>mentateur de la Géométrie de M. </s> <s xml:id="echoid-s8659" xml:space="preserve">Deſcartes, & </s> <s xml:id="echoid-s8660" xml:space="preserve">qu’on ne trouve <lb/>dans aucun Livre d’Elément, peut-être mis au rang des pro-<lb/>poſitions les plus géérales que l’on puiſſe donner ſur les rap-<lb/>ports des triangles. </s> <s xml:id="echoid-s8661" xml:space="preserve">J’aurois même pu commencer par cette <lb/>propoſition le Traité des raiſons des figures géométriques, & </s> <s xml:id="echoid-s8662" xml:space="preserve"><lb/>en déduire toutes les propoſitions que nous venons de voir, ſi <lb/>cela ne m’eût engagé dans des changemens trop conſidérables, <lb/>aimant mieux le faire ici en peu de mots; </s> <s xml:id="echoid-s8663" xml:space="preserve">ce qui ne peut qu’af-<lb/>fermir les Commençans dans cette partie, qui eſt abſolument <lb/>néceſſaire pour entendre la ſuite. </s> <s xml:id="echoid-s8664" xml:space="preserve">On peut encore faire un <lb/>grand uſage de cette propoſition dans la Géodéſie ou diviſion <lb/>des champs. </s> <s xml:id="echoid-s8665" xml:space="preserve">Rien de plus curieux que la ſimplicité avec la-<lb/>quelle M. </s> <s xml:id="echoid-s8666" xml:space="preserve">Scooten réſout pluſieurs problêmes, qui ſans le ſe-<lb/>cours de cette propoſition, paroîtroient très - compliqués. </s> <s xml:id="echoid-s8667" xml:space="preserve">Le <lb/>ſecond théorême donne la maniere de trouver la ſurface d’un <lb/>triangle quelconque, dont on connoît les trois côtés. </s> <s xml:id="echoid-s8668" xml:space="preserve">Nous <lb/>avons déja vu que cette connoiſſance ſuffit pour en avoir la <pb o="252" file="0290" n="290" rhead="NOUVEAU COURS"/> ſurface, puiſque les trois côtés déterminent la perpendiculaire <lb/>qu’il faut multiplier par la moitié de la baſe pour avoir l’aire <lb/>du triangle (art. </s> <s xml:id="echoid-s8669" xml:space="preserve">411).</s> <s xml:id="echoid-s8670" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div650" type="section" level="1" n="528"> <head xml:id="echoid-head625" xml:space="preserve">PROPOSITION XVII.</head> <head xml:id="echoid-head626" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8671" xml:space="preserve">521. </s> <s xml:id="echoid-s8672" xml:space="preserve">Deux triangles quelconques B A C, E D F qui ont un <lb/> <anchor type="note" xlink:label="note-0290-01a" xlink:href="note-0290-01"/> angle égal, l’un en A & </s> <s xml:id="echoid-s8673" xml:space="preserve">l’autre en D, compris entre deux côtés <lb/>quelconques, ſont entr’eux comme les produits des côtés qui con-<lb/>tiennent l’angle égal.</s> <s xml:id="echoid-s8674" xml:space="preserve"/> </p> <div xml:id="echoid-div650" type="float" level="2" n="1"> <note position="left" xlink:label="note-0290-01" xlink:href="note-0290-01a" xml:space="preserve">Figure 103 <lb/>& 104.</note> </div> </div> <div xml:id="echoid-div652" type="section" level="1" n="529"> <head xml:id="echoid-head627" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8675" xml:space="preserve">Sur le côté A C du triangle B A C, ſoit priſe la partie A H <lb/>= D F, & </s> <s xml:id="echoid-s8676" xml:space="preserve">ſur A B la ligne A L = D E, & </s> <s xml:id="echoid-s8677" xml:space="preserve">ſoient menées les <lb/>lignes L H, B H. </s> <s xml:id="echoid-s8678" xml:space="preserve">Les triangles L A H, E D F ayant, par hypo-<lb/>theſe, un angle égal compris entre côtés égaux, par conſtruc-<lb/>tion, ſeront égaux en tout. </s> <s xml:id="echoid-s8679" xml:space="preserve">Cela poſé, à cauſe des triangles <lb/>A H L, A H B, qui ont même ſommet en H, & </s> <s xml:id="echoid-s8680" xml:space="preserve">des triangles <lb/>A B H, A B C, qui ont même ſommet en B, & </s> <s xml:id="echoid-s8681" xml:space="preserve">qui ſont en-<lb/>tr’eux dans la raiſon de leurs baſes, on aura les proportions <lb/>ſuivantes. </s> <s xml:id="echoid-s8682" xml:space="preserve">A L H : </s> <s xml:id="echoid-s8683" xml:space="preserve">A B H :</s> <s xml:id="echoid-s8684" xml:space="preserve">: A L : </s> <s xml:id="echoid-s8685" xml:space="preserve">A B, & </s> <s xml:id="echoid-s8686" xml:space="preserve">A B H : </s> <s xml:id="echoid-s8687" xml:space="preserve">A B C :</s> <s xml:id="echoid-s8688" xml:space="preserve">: A H : </s> <s xml:id="echoid-s8689" xml:space="preserve">A C; <lb/></s> <s xml:id="echoid-s8690" xml:space="preserve">donc en multipliant par ordre A L H x A B H : </s> <s xml:id="echoid-s8691" xml:space="preserve">A B C x A B H :</s> <s xml:id="echoid-s8692" xml:space="preserve">: <lb/>A L x A H, ou E D x D F: </s> <s xml:id="echoid-s8693" xml:space="preserve">A B x A C, ou en diviſant les deux <lb/>premiers termes par la même grandeur A B H, & </s> <s xml:id="echoid-s8694" xml:space="preserve">mettant à <lb/>la place du triangle A L H ſon égal D E F, on aura E D F : </s> <s xml:id="echoid-s8695" xml:space="preserve"><lb/>A B C :</s> <s xml:id="echoid-s8696" xml:space="preserve">: E D x D F : </s> <s xml:id="echoid-s8697" xml:space="preserve">A B x A C. </s> <s xml:id="echoid-s8698" xml:space="preserve">C. </s> <s xml:id="echoid-s8699" xml:space="preserve">Q. </s> <s xml:id="echoid-s8700" xml:space="preserve">F. </s> <s xml:id="echoid-s8701" xml:space="preserve">D.</s> <s xml:id="echoid-s8702" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div653" type="section" level="1" n="530"> <head xml:id="echoid-head628" xml:space="preserve"><emph style="sc">Autre démonstration</emph>.</head> <p> <s xml:id="echoid-s8703" xml:space="preserve">Des ſommets B, E de chaque triangle, ſoient abaiſſées ſur <lb/>les baſes A C, D F les perpendiculaires B K, E M: </s> <s xml:id="echoid-s8704" xml:space="preserve">les ſurfaces <lb/>des triangles étant égales aux produits des hauteurs par les <lb/>moitiés des baſes, ſeront proportionnelles aux produits des <lb/>baſes par les hauteurs, & </s> <s xml:id="echoid-s8705" xml:space="preserve">donneront A B C: </s> <s xml:id="echoid-s8706" xml:space="preserve">D E F :</s> <s xml:id="echoid-s8707" xml:space="preserve">: A C x <lb/>B K : </s> <s xml:id="echoid-s8708" xml:space="preserve">D F x E M; </s> <s xml:id="echoid-s8709" xml:space="preserve">mais les triangles A B K, D E M ſont ſem-<lb/>blables, ayant, outre l’angle droit, un angle égal de part & </s> <s xml:id="echoid-s8710" xml:space="preserve"><lb/>d’autre, l’angle A du premier égal à l’angle D du ſecond: <lb/></s> <s xml:id="echoid-s8711" xml:space="preserve">donc A B : </s> <s xml:id="echoid-s8712" xml:space="preserve">D E :</s> <s xml:id="echoid-s8713" xml:space="preserve">: B K : </s> <s xml:id="echoid-s8714" xml:space="preserve">E M, ou en multipliant les deux anté-<lb/>cédens par A C, & </s> <s xml:id="echoid-s8715" xml:space="preserve">les deux conſéquens par D F, A B x A C : </s> <s xml:id="echoid-s8716" xml:space="preserve"><lb/>D E x D F :</s> <s xml:id="echoid-s8717" xml:space="preserve">: B K x A C : </s> <s xml:id="echoid-s8718" xml:space="preserve">E M x D F; </s> <s xml:id="echoid-s8719" xml:space="preserve">mais nous venons de voir que <lb/>A B C : </s> <s xml:id="echoid-s8720" xml:space="preserve">D E F :</s> <s xml:id="echoid-s8721" xml:space="preserve">: B K x A C : </s> <s xml:id="echoid-s8722" xml:space="preserve">E M x D F; </s> <s xml:id="echoid-s8723" xml:space="preserve">donc A B C : </s> <s xml:id="echoid-s8724" xml:space="preserve">D E F :</s> <s xml:id="echoid-s8725" xml:space="preserve">: A B x <lb/>A C : </s> <s xml:id="echoid-s8726" xml:space="preserve">D E x D F. </s> <s xml:id="echoid-s8727" xml:space="preserve">C. </s> <s xml:id="echoid-s8728" xml:space="preserve">Q. </s> <s xml:id="echoid-s8729" xml:space="preserve">F. </s> <s xml:id="echoid-s8730" xml:space="preserve">D.</s> <s xml:id="echoid-s8731" xml:space="preserve"/> </p> <pb o="253" file="0291" n="291" rhead="DE MATHÉMATIQUE. Liv. VII."/> </div> <div xml:id="echoid-div654" type="section" level="1" n="531"> <head xml:id="echoid-head629" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s8732" xml:space="preserve">522. </s> <s xml:id="echoid-s8733" xml:space="preserve">Il ſuit des deux démonſtrations précédentes, que la <lb/>propoſition eſt encore vraie dans le cas où les angles des deux <lb/>triangles ſeroient ſeulement ſupplément l’un de l’autre. </s> <s xml:id="echoid-s8734" xml:space="preserve">Pour <lb/>le prouver, ſoit prolongée la ligne F D en G, de maniere que <lb/>G D = F D, & </s> <s xml:id="echoid-s8735" xml:space="preserve">ſoit tirée E D: </s> <s xml:id="echoid-s8736" xml:space="preserve">les triangles G E D, D E F, <lb/>ayant des baſes égales, & </s> <s xml:id="echoid-s8737" xml:space="preserve">leur ſommet au même point ſeront <lb/>égaux en ſuperficie: </s> <s xml:id="echoid-s8738" xml:space="preserve">donc puiſque A B C: </s> <s xml:id="echoid-s8739" xml:space="preserve">D E F : </s> <s xml:id="echoid-s8740" xml:space="preserve">: </s> <s xml:id="echoid-s8741" xml:space="preserve">A B x A C: <lb/></s> <s xml:id="echoid-s8742" xml:space="preserve">D E x D F, on aura auſſi, en mettant à la place du triangle <lb/>D E F ſon égal G D E, & </s> <s xml:id="echoid-s8743" xml:space="preserve">à la place du rectangle D E x D F ſon <lb/>égal D E x D G, A B C: </s> <s xml:id="echoid-s8744" xml:space="preserve">G D E :</s> <s xml:id="echoid-s8745" xml:space="preserve">: A B x A C : </s> <s xml:id="echoid-s8746" xml:space="preserve">G D x D E.</s> <s xml:id="echoid-s8747" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div655" type="section" level="1" n="532"> <head xml:id="echoid-head630" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s8748" xml:space="preserve">523. </s> <s xml:id="echoid-s8749" xml:space="preserve">Comme les parallélogrammes ſont doubles des trian-<lb/>gles de même baſe & </s> <s xml:id="echoid-s8750" xml:space="preserve">de même hauteur, il s’enſuit que deux <lb/>parallélogrammes quelconques, qui ont un angle égal ou ſup-<lb/>plément l’un de l’autre, ſont entr’eux comme les produits des <lb/>côtés qui comprennent cet angle.</s> <s xml:id="echoid-s8751" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div656" type="section" level="1" n="533"> <head xml:id="echoid-head631" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s8752" xml:space="preserve">524. </s> <s xml:id="echoid-s8753" xml:space="preserve">Si les côtés qui comprennent l’angle égal ſont réci-<lb/>proques, c’eſt-à-dire ſi l’on a cette analogie AB : </s> <s xml:id="echoid-s8754" xml:space="preserve">DE :</s> <s xml:id="echoid-s8755" xml:space="preserve">: DF : </s> <s xml:id="echoid-s8756" xml:space="preserve">AC, <lb/>les rectangles A B x A C, D E x D F ſeront égaux: </s> <s xml:id="echoid-s8757" xml:space="preserve">donc les <lb/>triangles ou les parallélogrammes qui ſont dans la raiſon de <lb/>ces rectangles ſeront auſſi égaux. </s> <s xml:id="echoid-s8758" xml:space="preserve">On voit par-là que les ar-<lb/>ticles 390 & </s> <s xml:id="echoid-s8759" xml:space="preserve">395 deviennent des corollaires trés-ſimples de <lb/>cette propoſition. </s> <s xml:id="echoid-s8760" xml:space="preserve">On peut donc établir généralement, que <lb/>deux triangles ou deux parallélogrammes ſont égaux, lorſqu’ils <lb/>ont un angle égal ou des angles ſupplémens l’un de l’autre, compris <lb/>entre des côtés réciproques.</s> <s xml:id="echoid-s8761" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div657" type="section" level="1" n="534"> <head xml:id="echoid-head632" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s8762" xml:space="preserve">525. </s> <s xml:id="echoid-s8763" xml:space="preserve">On pourroit auſſi déduire de cette propoſition la pro-<lb/>priété commune à toutes les figures ſemblables, d’être en-<lb/>tr’elles comme les quarrés des côtés homologues: </s> <s xml:id="echoid-s8764" xml:space="preserve">car les figures <lb/>ſemblables étant toutes compoſées de triangles ſemblables, <lb/>& </s> <s xml:id="echoid-s8765" xml:space="preserve">les triangles ſemblables ayant les côtés homologues propor-<lb/>tionnels, ceux qui contiendront des angles égaux, ſeront des <lb/>côtés homologues: </s> <s xml:id="echoid-s8766" xml:space="preserve">donc puiſque ces triangles ſont entr’eux <pb o="254" file="0292" n="292" rhead="NOUVEAU COURS"/> comme les produits de ces côtés, ils ſeront auſſi dans la raiſon <lb/>des quarrés des mêmes côtés: </s> <s xml:id="echoid-s8767" xml:space="preserve">car il eſt évident que ſi l’on a <lb/>A B : </s> <s xml:id="echoid-s8768" xml:space="preserve">A C :</s> <s xml:id="echoid-s8769" xml:space="preserve">: D E : </s> <s xml:id="echoid-s8770" xml:space="preserve">D F, on a auſſi A B : </s> <s xml:id="echoid-s8771" xml:space="preserve">A B :</s> <s xml:id="echoid-s8772" xml:space="preserve">: D E : </s> <s xml:id="echoid-s8773" xml:space="preserve">D E : </s> <s xml:id="echoid-s8774" xml:space="preserve">donc <lb/>en multipliant par ordre A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8775" xml:space="preserve">A B x A C :</s> <s xml:id="echoid-s8776" xml:space="preserve">: D E<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8777" xml:space="preserve">D E x D F, <lb/>& </s> <s xml:id="echoid-s8778" xml:space="preserve">alternando A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8779" xml:space="preserve">DE<emph style="sub">2</emph> :</s> <s xml:id="echoid-s8780" xml:space="preserve">: A B x A C : </s> <s xml:id="echoid-s8781" xml:space="preserve">D E x D F; </s> <s xml:id="echoid-s8782" xml:space="preserve">mais par la <lb/>préſente propoſition, ABC : </s> <s xml:id="echoid-s8783" xml:space="preserve">DEF :</s> <s xml:id="echoid-s8784" xml:space="preserve">: AB x AC : </s> <s xml:id="echoid-s8785" xml:space="preserve">DE x DF : </s> <s xml:id="echoid-s8786" xml:space="preserve">donc <lb/>dans le cas des triangles ſemblables, ABC : </s> <s xml:id="echoid-s8787" xml:space="preserve">DEF :</s> <s xml:id="echoid-s8788" xml:space="preserve">: AB<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8789" xml:space="preserve">DE<emph style="sub">2</emph>.</s> <s xml:id="echoid-s8790" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div658" type="section" level="1" n="535"> <head xml:id="echoid-head633" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s8791" xml:space="preserve">526. </s> <s xml:id="echoid-s8792" xml:space="preserve">On peut encore faire uſage de cette propoſition pour <lb/> <anchor type="note" xlink:label="note-0292-01a" xlink:href="note-0292-01"/> trouver un triangle A L H, qui ait un côté déterminé A L ſur <lb/>le côté A B du triangle A B C, & </s> <s xml:id="echoid-s8793" xml:space="preserve">qui ait avec ce triangle une <lb/>raiſon donnée. </s> <s xml:id="echoid-s8794" xml:space="preserve">Par exemple, ſi je veux que le triangle A L H <lb/>ſoit le tiers du triangle B A C, après avoir fait A B = a, <lb/>A C = b, A L = c, & </s> <s xml:id="echoid-s8795" xml:space="preserve">A H = x; </s> <s xml:id="echoid-s8796" xml:space="preserve">j’ai par la propoſition pré-<lb/>ſente, ab : </s> <s xml:id="echoid-s8797" xml:space="preserve">cx :</s> <s xml:id="echoid-s8798" xml:space="preserve">: 3 : </s> <s xml:id="echoid-s8799" xml:space="preserve">1; </s> <s xml:id="echoid-s8800" xml:space="preserve">donc 3cx = ab, & </s> <s xml:id="echoid-s8801" xml:space="preserve">dégageant l’incon-<lb/>nue, x = {ab/3c}; </s> <s xml:id="echoid-s8802" xml:space="preserve">d’où il ſuit que pour avoir x, il faut chercher une <lb/>quatrieme proportionnelle aux lignes 3A L, A B & </s> <s xml:id="echoid-s8803" xml:space="preserve">A C : </s> <s xml:id="echoid-s8804" xml:space="preserve">car <lb/>de l’équation 3cx = ab, on tire cette proportion, 3c : </s> <s xml:id="echoid-s8805" xml:space="preserve">a :</s> <s xml:id="echoid-s8806" xml:space="preserve">: b : </s> <s xml:id="echoid-s8807" xml:space="preserve">x.</s> <s xml:id="echoid-s8808" xml:space="preserve"/> </p> <div xml:id="echoid-div658" type="float" level="2" n="1"> <note position="left" xlink:label="note-0292-01" xlink:href="note-0292-01a" xml:space="preserve">Figure 103.</note> </div> </div> <div xml:id="echoid-div660" type="section" level="1" n="536"> <head xml:id="echoid-head634" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s8809" xml:space="preserve">Pour faciliter l’intelligence de la propoſition ſuivante, qui <lb/>ſeroit un peu compliquée pour des Commençans, nous allons <lb/>expliquer dans les deux Lemmes ſuivans tout ce qu’il eſt né-<lb/>ceſſaire de ſçavoir pour la comprendre aiſément.</s> <s xml:id="echoid-s8810" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div661" type="section" level="1" n="537"> <head xml:id="echoid-head635" xml:space="preserve">LEMME PREMIER.</head> <head xml:id="echoid-head636" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s8811" xml:space="preserve">527. </s> <s xml:id="echoid-s8812" xml:space="preserve">Un triangle B A C étant donné, lui inſcrire un cercle <lb/> <anchor type="note" xlink:label="note-0292-02a" xlink:href="note-0292-02"/> E D F.</s> <s xml:id="echoid-s8813" xml:space="preserve"/> </p> <div xml:id="echoid-div661" type="float" level="2" n="1"> <note position="left" xlink:label="note-0292-02" xlink:href="note-0292-02a" xml:space="preserve">Figure 111.</note> </div> </div> <div xml:id="echoid-div663" type="section" level="1" n="538"> <head xml:id="echoid-head637" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s8814" xml:space="preserve">Il eſt aiſé de voir que tout ſe réduit à trouver un point G <lb/>au dedans du triangle, qui ſoit tel qu’en abaiſſant ſur chaque <lb/>côté les perpendiculaires G D, G E, G F, ces trois lignes ſoient <lb/>égales entr’elles: </s> <s xml:id="echoid-s8815" xml:space="preserve">car puiſque le cercle doit être inſcrit au trian-<lb/>gle, chaque côté ſera une tangente de ce cercle, & </s> <s xml:id="echoid-s8816" xml:space="preserve">par conſé-<lb/>quent perpendiculaire à l’extrêmité des rayons G D, G E, G F. <lb/></s> <s xml:id="echoid-s8817" xml:space="preserve">Suppoſons pour un moment que le point G eſt celui qu’on de-<lb/>mande, & </s> <s xml:id="echoid-s8818" xml:space="preserve">qu’on ait menées les perpendiculaires GD, GE, GF <pb o="255" file="0293" n="293" rhead="DE MATHEMATIQUE. Liv. VII."/> aux côtés A C, A B, B C; </s> <s xml:id="echoid-s8819" xml:space="preserve">nous avons déja vu (art. </s> <s xml:id="echoid-s8820" xml:space="preserve">448) que <lb/>les parties A E, A D des tangentes, compriſes entre le point A <lb/>de rencontre, & </s> <s xml:id="echoid-s8821" xml:space="preserve">les points E, D de contact ſont égales en-<lb/>tr’elles, mais les droites E G, D G le ſontauſſi; </s> <s xml:id="echoid-s8822" xml:space="preserve">donc les trian-<lb/>gles rectangles A G D, A G E ſont égaux en tout, puiſque les <lb/>trois côtés de l’un ſont égaux aux trois côtés de l’autre: </s> <s xml:id="echoid-s8823" xml:space="preserve">donc <lb/>les angles E A G, D A G ſont égaux; </s> <s xml:id="echoid-s8824" xml:space="preserve">& </s> <s xml:id="echoid-s8825" xml:space="preserve">par conſéquent le <lb/>centre du cercle ſe trouvera quelque part ſur la ligne A G qui <lb/>diviſe l’angle B A C en deux également. </s> <s xml:id="echoid-s8826" xml:space="preserve">On fera voir de la <lb/>même maniere, que les triangles rectangles B E G, B F G ſont <lb/>égaux, & </s> <s xml:id="echoid-s8827" xml:space="preserve">que le centre du cercle ſe trouvera dans la ligne B G <lb/>qui diviſe l’angle A B C en deux également: </s> <s xml:id="echoid-s8828" xml:space="preserve">donc il ſera au <lb/>point d’interſection des lignes A G, B G. </s> <s xml:id="echoid-s8829" xml:space="preserve">Ainſi pour avoir le <lb/>centre G, on n’aura qu’à diviſer deux angles quelconques A <lb/>& </s> <s xml:id="echoid-s8830" xml:space="preserve">C, ou bien A & </s> <s xml:id="echoid-s8831" xml:space="preserve">B, chacun en deux angles égaux, & </s> <s xml:id="echoid-s8832" xml:space="preserve">le point <lb/>G, où les lignes de diviſion ſe couperont, ſera le point de-<lb/>mandé. </s> <s xml:id="echoid-s8833" xml:space="preserve">Abaiſſant enſuite de ce point la perpendiculaire G D <lb/>ſur le côté A C, on aura le rayon avec lequel on pourra dé-<lb/>crire le cercle demandé.</s> <s xml:id="echoid-s8834" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div664" type="section" level="1" n="539"> <head xml:id="echoid-head638" xml:space="preserve"><emph style="sc">Lemme</emph> II.</head> <p style="it"> <s xml:id="echoid-s8835" xml:space="preserve">528. </s> <s xml:id="echoid-s8836" xml:space="preserve">Suppoſant toutes choſes, comme dans le problême précé-<lb/>dent, ſi l’on prolonge le côté A B d’une quantité B K = F C, je <lb/>dis 1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s8837" xml:space="preserve">que la ligne A K ſera égale à la demi-ſomme des trois côtés: <lb/></s> <s xml:id="echoid-s8838" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s8839" xml:space="preserve">Quelle ſera la ſomme des trois différences de la demi-ſomme des <lb/>trois côtés à chacun des mêmes côtés?</s> <s xml:id="echoid-s8840" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div665" type="section" level="1" n="540"> <head xml:id="echoid-head639" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8841" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s8842" xml:space="preserve">Puiſque l’on a AE = AD, BE = BF, DC = CF, la ſom <lb/>me des trois côtés ſera 2A E + 2B E + 2C F, ou 2A E + 2B E <lb/>+ 2B K, puiſque B K = C F (conſtruction) : </s> <s xml:id="echoid-s8843" xml:space="preserve">donc la demi-<lb/>ſomme des trois côtés ſera A E + E B + B K = A K. <lb/></s> <s xml:id="echoid-s8844" xml:space="preserve">C. </s> <s xml:id="echoid-s8845" xml:space="preserve">Q. </s> <s xml:id="echoid-s8846" xml:space="preserve">F. </s> <s xml:id="echoid-s8847" xml:space="preserve">1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s8848" xml:space="preserve">D.</s> <s xml:id="echoid-s8849" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8850" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s8851" xml:space="preserve">Puiſque A K eſt égal à la demi-ſomme des trois côtés, <lb/>il eſt évident que B K eſt l’excès de la même demi-ſomme ſur <lb/>le côté A B; </s> <s xml:id="echoid-s8852" xml:space="preserve">de même A E eſt l’excès de la demi-ſomme ſur <lb/>B E + B K, ou ſur ſon égal B F + F C, c’eſt-à-dire ſur le côté <lb/>B C; </s> <s xml:id="echoid-s8853" xml:space="preserve">&</s> <s xml:id="echoid-s8854" xml:space="preserve">enfin B E eſt l’excès de la demi-ſomme ſur B K + A E, <lb/>ou ſur leurs égales D C + A D, c’eſt-à-dire ſur le troiſieme <lb/>côté A C: </s> <s xml:id="echoid-s8855" xml:space="preserve">donc A K eſt la ſomme des trois différences de <pb o="256" file="0294" n="294" rhead="NOUVEAU COURS"/> chacun des trois côtés à la demi - ſomme des mêmes côtés. <lb/></s> <s xml:id="echoid-s8856" xml:space="preserve">C. </s> <s xml:id="echoid-s8857" xml:space="preserve">Q. </s> <s xml:id="echoid-s8858" xml:space="preserve">F. </s> <s xml:id="echoid-s8859" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s8860" xml:space="preserve">D.</s> <s xml:id="echoid-s8861" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8862" xml:space="preserve">529. </s> <s xml:id="echoid-s8863" xml:space="preserve">On remarquera encore que le triangle B A C eſt par-<lb/>tagé par les lignes G B, G C, G A en trois triangles A G C, <lb/>A G B, B G C, qui ont tous pour hauteur le rayon du même <lb/>cercle: </s> <s xml:id="echoid-s8864" xml:space="preserve">donc la ſurface de ce triangle ſera égale à la ſomme <lb/>de celles des trois triangles, c’eſt-à-dire que l’on aura cette <lb/>égalité B A C = {AB/2} x G E + {AC/2} x G E + {BC/2} x G E = <lb/>{A B + A C + B C/2} x G E = A K x G E. </s> <s xml:id="echoid-s8865" xml:space="preserve">Cette remarque eſt en-<lb/>core abſolument néceſſaire pour l’intelligence du théorême <lb/>ſuivant.</s> <s xml:id="echoid-s8866" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div666" type="section" level="1" n="541"> <head xml:id="echoid-head640" xml:space="preserve">PROPOSITION XVIII.</head> <head xml:id="echoid-head641" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s8867" xml:space="preserve">530. </s> <s xml:id="echoid-s8868" xml:space="preserve">La ſurface d’un triangle quelconque B A C eſt égale à la <lb/>racine quarrée d’un produit de quatre dimenſions, fait de la demi-<lb/>ſomme des trois côtés, multipliée par les différences de chacun des <lb/>côtés à la même demi-ſomme.</s> <s xml:id="echoid-s8869" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div667" type="section" level="1" n="542"> <head xml:id="echoid-head642" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s8870" xml:space="preserve">Sur le côté B C ſoit priſe la ligne B M = F C, qui donnera <lb/>C M = F B, en ôtant des lignes égales la partie commune <lb/>F M; </s> <s xml:id="echoid-s8871" xml:space="preserve">ſoit prolongé le côté A C d’une quantité C H = B F ou <lb/>C M: </s> <s xml:id="echoid-s8872" xml:space="preserve">on aura A H = A K, puiſque les parties qui compoſent <lb/>ces deux lignes ſont égales. </s> <s xml:id="echoid-s8873" xml:space="preserve">Aux points K, M, H, ſoient éle-<lb/>vées ſur chacune des lignes correſpondantes B K, B C, C H <lb/>les perpendiculaires K I, M I, H I qui ſe rencontreront toutes <lb/>en un ſeul & </s> <s xml:id="echoid-s8874" xml:space="preserve">même point I, & </s> <s xml:id="echoid-s8875" xml:space="preserve">ſeront toutes égales entr’elles; <lb/></s> <s xml:id="echoid-s8876" xml:space="preserve">car puiſque B M = B K, en tirant B I, les triangles rectangles <lb/>B M I, B K I auront, outre l’angle droit, deux côtés égaux cha-<lb/>cun à chacun B M = B K, & </s> <s xml:id="echoid-s8877" xml:space="preserve">le côté B I qui leur eſt com-<lb/>mun: </s> <s xml:id="echoid-s8878" xml:space="preserve">donc K I = M I; </s> <s xml:id="echoid-s8879" xml:space="preserve">on feroit voir de même que M I = H I, <lb/>puiſque les lignes C M & </s> <s xml:id="echoid-s8880" xml:space="preserve">C H ſont égales: </s> <s xml:id="echoid-s8881" xml:space="preserve">on prolongera en-<lb/>ſuite la ligne A G, qui paſſera auſſi par le point I, comme il eſt <lb/>aiſé de le voir, à cauſe des quadrilateres A E G D, A K I H, <lb/>qui ſont évidemment ſemblables, puiſque les lignes G D, G E <lb/>ſont égales entr’elles, & </s> <s xml:id="echoid-s8882" xml:space="preserve">paralleles aux lignes I H, I K auſſi <lb/>égales entr’elles; </s> <s xml:id="echoid-s8883" xml:space="preserve">& </s> <s xml:id="echoid-s8884" xml:space="preserve">que les lignes A D, A E ſont auſſi égales <lb/>entr’elles, ainſi que les lignes A H, A K.</s> <s xml:id="echoid-s8885" xml:space="preserve"/> </p> <pb o="257" file="0295" n="295" rhead="DE MATHÉMATIQUE. Liv. VII."/> <p> <s xml:id="echoid-s8886" xml:space="preserve">Cette conſtruction ſuppoſée, il eſt aiſé de voir que les qua-<lb/>drilateres E G B F, M B K I ſont ſemblables, ayant chacun deux <lb/>angles droits, les côtés E G, G F égaux entr’eux, de même <lb/>que les côtés B K, B M, l’angle E B F du premier égal à l’angle <lb/>en I du ſecond, puiſqu’ils ſont chacun ſupplément du même <lb/>angle M B K, & </s> <s xml:id="echoid-s8887" xml:space="preserve">que dans tout quadrilatere, les quatre angles <lb/>valent quatre droits: </s> <s xml:id="echoid-s8888" xml:space="preserve">donc les triangles G E B, B K I, qui ſont <lb/>les moitiés de ces quadrilateres, ſeront ſemblables, & </s> <s xml:id="echoid-s8889" xml:space="preserve">donne-<lb/>ront I K: </s> <s xml:id="echoid-s8890" xml:space="preserve">B K : </s> <s xml:id="echoid-s8891" xml:space="preserve">: </s> <s xml:id="echoid-s8892" xml:space="preserve">BE : </s> <s xml:id="echoid-s8893" xml:space="preserve">GE, d’où l’on tire IK x GE = BK x BE; <lb/></s> <s xml:id="echoid-s8894" xml:space="preserve">mais G E<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8895" xml:space="preserve">IK x GE:</s> <s xml:id="echoid-s8896" xml:space="preserve">:GE: </s> <s xml:id="echoid-s8897" xml:space="preserve">I K, & </s> <s xml:id="echoid-s8898" xml:space="preserve">à cauſe des triangles ſemblables <lb/>A E G, A K I; </s> <s xml:id="echoid-s8899" xml:space="preserve">GE : </s> <s xml:id="echoid-s8900" xml:space="preserve">I K :</s> <s xml:id="echoid-s8901" xml:space="preserve">: AE : </s> <s xml:id="echoid-s8902" xml:space="preserve">AK; </s> <s xml:id="echoid-s8903" xml:space="preserve">donc GE<emph style="sub">2</emph> : </s> <s xml:id="echoid-s8904" xml:space="preserve">IK x GE <lb/>ou B K x B E :</s> <s xml:id="echoid-s8905" xml:space="preserve">: A E : </s> <s xml:id="echoid-s8906" xml:space="preserve">A K; </s> <s xml:id="echoid-s8907" xml:space="preserve">& </s> <s xml:id="echoid-s8908" xml:space="preserve">prenant le produit des extrêmes <lb/>& </s> <s xml:id="echoid-s8909" xml:space="preserve">des moyens G E<emph style="sub">2</emph> x A K = B K x B E x A E : </s> <s xml:id="echoid-s8910" xml:space="preserve">& </s> <s xml:id="echoid-s8911" xml:space="preserve">multipliant <lb/>encore chaque membre par A K, G E<emph style="sub">2</emph> x A K<emph style="sub">2</emph> = B K x B E <lb/>x A E x A K, d’où l’on déduit, en prenant les racines de part <lb/>& </s> <s xml:id="echoid-s8912" xml:space="preserve">d’autre, G E x A K, ou (art. </s> <s xml:id="echoid-s8913" xml:space="preserve">529) la ſurface du triangle <lb/>B A C = √B K x B E x A E x A K\x{0020}. </s> <s xml:id="echoid-s8914" xml:space="preserve">Or il eſt viſible que les <lb/>facteurs ſoumis au radical ſont les trois différences de la demi-<lb/>ſomme des trois côtés, à chacun de ces côtés, multipliées par <lb/>la même demi-ſomme A K : </s> <s xml:id="echoid-s8915" xml:space="preserve">donc la ſurface du triangle B A C <lb/>eſt égale à la racine quarrée d’un produit de quatre dimen-<lb/>ſions, fait de la demi-ſomme des trois côtés, multipliée par <lb/>la différence de la même demi-ſomme à chacundes trois côtés. </s> <s xml:id="echoid-s8916" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s8917" xml:space="preserve">Q. </s> <s xml:id="echoid-s8918" xml:space="preserve">F. </s> <s xml:id="echoid-s8919" xml:space="preserve">D.</s> <s xml:id="echoid-s8920" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div668" type="section" level="1" n="543"> <head xml:id="echoid-head643" style="it" xml:space="preserve">Fin du ſeptieme Livre.</head> <figure> <image file="0295-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0295-01"/> </figure> <pb o="258" file="0296" n="296"/> <figure> <image file="0296-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0296-01"/> </figure> </div> <div xml:id="echoid-div669" type="section" level="1" n="544"> <head xml:id="echoid-head644" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head645" xml:space="preserve">LIVRE HUITIEME, <lb/>Qui traite des propriétés des corps, de leurs ſurfaces, & <lb/>de leurs ſolidités.</head> <head xml:id="echoid-head646" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head647" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s8921" xml:space="preserve">531. </s> <s xml:id="echoid-s8922" xml:space="preserve">ON appelle priſme, un ſolide terminé par deux poly-<lb/>gones ſemblables & </s> <s xml:id="echoid-s8923" xml:space="preserve">égaux, paralleles entr’eux, & </s> <s xml:id="echoid-s8924" xml:space="preserve">par autant <lb/>de parallélogrammes que le polygone qui lui ſert de baſe a de <lb/>côtés: </s> <s xml:id="echoid-s8925" xml:space="preserve">tel eſt le ſolide cotté A. </s> <s xml:id="echoid-s8926" xml:space="preserve">On appelle axe du priſme une <lb/> <anchor type="note" xlink:label="note-0296-01a" xlink:href="note-0296-01"/> droite, telle que C B, tirée du centre C du polygone qui ſert <lb/> <anchor type="note" xlink:label="note-0296-02a" xlink:href="note-0296-02"/> de baſe au centre B du polygone ſupérieur. </s> <s xml:id="echoid-s8927" xml:space="preserve">Si cette ligne eſt <lb/>perpendiculaire à la baſe du priſme, le priſme eſt appellé droit, <lb/>& </s> <s xml:id="echoid-s8928" xml:space="preserve">on l’appelle priſme oblique ou incliné, lorſque cette ligne eſt <lb/>inclinée ſur le plan de la baſe.</s> <s xml:id="echoid-s8929" xml:space="preserve"/> </p> <div xml:id="echoid-div669" type="float" level="2" n="1"> <note position="left" xlink:label="note-0296-01" xlink:href="note-0296-01a" xml:space="preserve">Planche VI.</note> <note position="left" xlink:label="note-0296-02" xlink:href="note-0296-02a" xml:space="preserve">Figure 112.</note> </div> </div> <div xml:id="echoid-div671" type="section" level="1" n="545"> <head xml:id="echoid-head648" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s8930" xml:space="preserve">532. </s> <s xml:id="echoid-s8931" xml:space="preserve">On appelle cylindre, un ſolide engendré par le mou-<lb/>vement d’un cercle qui ſe meut parallélement à lui même le <lb/>long d’une ligne A B. </s> <s xml:id="echoid-s8932" xml:space="preserve">Le cercle inférieur de ce ſolide eſt ap-<lb/> <anchor type="note" xlink:label="note-0296-03a" xlink:href="note-0296-03"/> pellé baſe du cylindre ou cercle générateur: </s> <s xml:id="echoid-s8933" xml:space="preserve">une ligne menée du <lb/>centre du cercle inférieur au centre du cercle ſupérieur eſt ap-<lb/>pellée l’axe du cylindre: </s> <s xml:id="echoid-s8934" xml:space="preserve">Si cette ligne eſt perpendiculaire ſur <lb/>le cercle inférieur, le cylindre eſt appellé droit; </s> <s xml:id="echoid-s8935" xml:space="preserve">& </s> <s xml:id="echoid-s8936" xml:space="preserve">ſi cette <lb/>ligne eſt inclinée à la même baſe, on l’appelle cylindre oblique.</s> <s xml:id="echoid-s8937" xml:space="preserve"> <pb o="259" file="0297" n="297" rhead="NOUVEAU COURS DE MATHEM. Liv. VIII."/> Il ſuit de cette génération du cylindre, que ſi l’on coupe un <lb/>cylindre par un plan parallele à la baſe de ce cylindre, la coupe <lb/>repréſentera un cercle, puiſque le cercle générateur a néceſ-<lb/>ſairement paſſé par ce plan pour engendrer le ſolide.</s> <s xml:id="echoid-s8938" xml:space="preserve"/> </p> <div xml:id="echoid-div671" type="float" level="2" n="1"> <note position="left" xlink:label="note-0296-03" xlink:href="note-0296-03a" xml:space="preserve">Figure 113.</note> </div> </div> <div xml:id="echoid-div673" type="section" level="1" n="546"> <head xml:id="echoid-head649" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s8939" xml:space="preserve">533. </s> <s xml:id="echoid-s8940" xml:space="preserve">Si d’un point quelconque A, pris au dehors d’un poly-<lb/> <anchor type="note" xlink:label="note-0297-01a" xlink:href="note-0297-01"/> gone quelconque, on mene des droites A B, A C, A D, A E à <lb/>tous les angles d’un polygone, il en réſultera un ſolide, que <lb/>l’on appelle pyramide, dont la baſe ſera le polygone donné, <lb/>& </s> <s xml:id="echoid-s8941" xml:space="preserve">qui ſera terminée par autant de triangles que le polygone a <lb/>de côtés. </s> <s xml:id="echoid-s8942" xml:space="preserve">Les ſolides, repréſentés par les figures 114 & </s> <s xml:id="echoid-s8943" xml:space="preserve">115, <lb/>ſont des pyramides. </s> <s xml:id="echoid-s8944" xml:space="preserve">Le point A, d’où l’on mene les lignes aux <lb/>angles de la baſe, eſt appellé le ſommet de la pyramide. </s> <s xml:id="echoid-s8945" xml:space="preserve">Si la <lb/>baſe de la pyramide eſt un polygone régulier, la ligne A H, <lb/>menée du centre H de cette baſe au ſommet de la pyramide, <lb/>eſt appellée l’axe de la pyramide. </s> <s xml:id="echoid-s8946" xml:space="preserve">Lorſque cet axe eſt perpen-<lb/>diculaire à la baſe, la pyramide eſt droite; </s> <s xml:id="echoid-s8947" xml:space="preserve">autrement elle eſt <lb/>inclinée.</s> <s xml:id="echoid-s8948" xml:space="preserve"/> </p> <div xml:id="echoid-div673" type="float" level="2" n="1"> <note position="right" xlink:label="note-0297-01" xlink:href="note-0297-01a" xml:space="preserve">Figure 115.</note> </div> </div> <div xml:id="echoid-div675" type="section" level="1" n="547"> <head xml:id="echoid-head650" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s8949" xml:space="preserve">534. </s> <s xml:id="echoid-s8950" xml:space="preserve">Si le polygone qui ſert de baſe à la pyramide eſt un <lb/>cercle, alors on lui-donne le nom de cône. </s> <s xml:id="echoid-s8951" xml:space="preserve">On peut donc ima-<lb/>giner qu’un cône eſt formé par la révolution d’une droite C A, <lb/> <anchor type="note" xlink:label="note-0297-02a" xlink:href="note-0297-02"/> qui eſt attachée fixement en C, & </s> <s xml:id="echoid-s8952" xml:space="preserve">dont l’extrêmité inférieure <lb/>tourne autour d’un cercle A D B A, au dehors duquel eſt placé <lb/>le point C. </s> <s xml:id="echoid-s8953" xml:space="preserve">Le cercle A D B A eſt appellé la baſe du cône; </s> <s xml:id="echoid-s8954" xml:space="preserve">le <lb/>point C eſt appellé le ſommet du cône. </s> <s xml:id="echoid-s8955" xml:space="preserve">Une ligne menée du <lb/>centre de la baſe du cône au ſommet, eſt appellée axe du cône. <lb/></s> <s xml:id="echoid-s8956" xml:space="preserve">Si l’axe eſt perpendiculaire à la baſe du cône, le cône eſt droit. </s> <s xml:id="echoid-s8957" xml:space="preserve"><lb/>Si l’axe eſt incliné à la même baſe, le cône eſt oblique. </s> <s xml:id="echoid-s8958" xml:space="preserve">Les <lb/>figures 116 & </s> <s xml:id="echoid-s8959" xml:space="preserve">117 repréſentent des cônes.</s> <s xml:id="echoid-s8960" xml:space="preserve"/> </p> <div xml:id="echoid-div675" type="float" level="2" n="1"> <note position="right" xlink:label="note-0297-02" xlink:href="note-0297-02a" xml:space="preserve">Figure 116.</note> </div> <p> <s xml:id="echoid-s8961" xml:space="preserve">On peut encore imaginer que le cône droit eſt formé par <lb/>la révolution d’un triangle rectangle A D C, autour d’un des <lb/>côtés de l’angle droit C D; </s> <s xml:id="echoid-s8962" xml:space="preserve">mais on ne peut pas ſuppoſer que le <lb/>cône oblique ſoit formé par la révolution d’un triangle obli-<lb/>qu’angle, autour de quelqu’un de ſes côtés; </s> <s xml:id="echoid-s8963" xml:space="preserve">ainſi la premiere <lb/>définition étant plus générale, eſt auſſi la meilleure.</s> <s xml:id="echoid-s8964" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div677" type="section" level="1" n="548"> <head xml:id="echoid-head651" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s8965" xml:space="preserve">535. </s> <s xml:id="echoid-s8966" xml:space="preserve">On appelle cône tronqué droit, un ſolide formé par la <pb o="260" file="0298" n="298" rhead="NOUVEAU COURS"/> révolution d’un trapezoïde rectangle, tel que F G H I, autour <lb/> <anchor type="note" xlink:label="note-0298-01a" xlink:href="note-0298-01"/> d’un de ſes côtés G F, qui ſoutient les deux angles droits. </s> <s xml:id="echoid-s8967" xml:space="preserve">On <lb/>peut encore dire qu’un cône tronqué eſt ce qui reſte d’un cône <lb/>A B C, après en avoir ôté le petit cône D B E, qui a été coupé <lb/> <anchor type="note" xlink:label="note-0298-02a" xlink:href="note-0298-02"/> par un plan parallele à la baſe du cône.</s> <s xml:id="echoid-s8968" xml:space="preserve"/> </p> <div xml:id="echoid-div677" type="float" level="2" n="1"> <note position="left" xlink:label="note-0298-01" xlink:href="note-0298-01a" xml:space="preserve">Figure 118.</note> <note position="left" xlink:label="note-0298-02" xlink:href="note-0298-02a" xml:space="preserve">Figure 117.</note> </div> </div> <div xml:id="echoid-div679" type="section" level="1" n="549"> <head xml:id="echoid-head652" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s8969" xml:space="preserve">536. </s> <s xml:id="echoid-s8970" xml:space="preserve">La ſphere eſt un ſolide terminée par une ſeule ſurface <lb/> <anchor type="note" xlink:label="note-0298-03a" xlink:href="note-0298-03"/> courbe, qu’on appelle ſurface ſphérique, comme A D C B, au-<lb/>dedans de laquelle il y a un point qu’on appelle centre de la <lb/>ſphere, duquel toutes les lignes droites menées à la ſurface ſont <lb/>égales entr’elles. </s> <s xml:id="echoid-s8971" xml:space="preserve">On peut imaginer que la ſphere a été en-<lb/>gendrée par la révolution d’un demi-cercle autour d’un dia-<lb/>metre. </s> <s xml:id="echoid-s8972" xml:space="preserve">Le demi-cercle engendre la ſolidité de la ſphere, & </s> <s xml:id="echoid-s8973" xml:space="preserve">la <lb/>demi-circonférence engendre la ſurface de la même ſphere.</s> <s xml:id="echoid-s8974" xml:space="preserve"/> </p> <div xml:id="echoid-div679" type="float" level="2" n="1"> <note position="left" xlink:label="note-0298-03" xlink:href="note-0298-03a" xml:space="preserve">Figure 119.</note> </div> </div> <div xml:id="echoid-div681" type="section" level="1" n="550"> <head xml:id="echoid-head653" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s8975" xml:space="preserve">537. </s> <s xml:id="echoid-s8976" xml:space="preserve">Segment ſphérique ou portion de ſphere, eſt un ſolide <lb/>compris ſous une partie de la ſurface de la ſphere & </s> <s xml:id="echoid-s8977" xml:space="preserve">la ſurface <lb/>d’un cercle; </s> <s xml:id="echoid-s8978" xml:space="preserve">où l’une des deux parties inégales A B C & </s> <s xml:id="echoid-s8979" xml:space="preserve">A D C <lb/> <anchor type="note" xlink:label="note-0298-04a" xlink:href="note-0298-04"/> d’une ſphere coupée par un plan qui ne paſſe pas par ſon centre. <lb/></s> <s xml:id="echoid-s8980" xml:space="preserve">Si le plan de ſection paſſe par le centre de la ſphere, il la diviſe <lb/>en deux ſegmens égaux, que l’on appelle hemiſpheres. </s> <s xml:id="echoid-s8981" xml:space="preserve">On peut <lb/>imaginer que le ſegment ſphérique eſt formé par la révolution <lb/>d’un ſegment de cercle autour d’une ligne, qui diviſe la corde <lb/>de ce ſegment en deux parties égales, & </s> <s xml:id="echoid-s8982" xml:space="preserve">qui lui eſt perpendi-<lb/>culaire.</s> <s xml:id="echoid-s8983" xml:space="preserve"/> </p> <div xml:id="echoid-div681" type="float" level="2" n="1"> <note position="left" xlink:label="note-0298-04" xlink:href="note-0298-04a" xml:space="preserve">Figure 119.</note> </div> </div> <div xml:id="echoid-div683" type="section" level="1" n="551"> <head xml:id="echoid-head654" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s8984" xml:space="preserve">538. </s> <s xml:id="echoid-s8985" xml:space="preserve">On appelle zone une partie A B C D de la ſurface d’une <lb/> <anchor type="note" xlink:label="note-0298-05a" xlink:href="note-0298-05"/> ſphere, terminée par deux cercles B C & </s> <s xml:id="echoid-s8986" xml:space="preserve">A D, de la même <lb/>ſphere paralleles entr’eux.</s> <s xml:id="echoid-s8987" xml:space="preserve"/> </p> <div xml:id="echoid-div683" type="float" level="2" n="1"> <note position="left" xlink:label="note-0298-05" xlink:href="note-0298-05a" xml:space="preserve">Figure 120.</note> </div> </div> <div xml:id="echoid-div685" type="section" level="1" n="552"> <head xml:id="echoid-head655" xml:space="preserve">IX.</head> <p> <s xml:id="echoid-s8988" xml:space="preserve">539. </s> <s xml:id="echoid-s8989" xml:space="preserve">Le ſecteur de ſphere eſt un ſolide terminé en pointe au <lb/>centre de la ſphere, qui a pour baſe une partie de la ſurface de <lb/>la ſphere, comme C O H. </s> <s xml:id="echoid-s8990" xml:space="preserve">On peut imaginer que le ſecteur <lb/> <anchor type="note" xlink:label="note-0298-06a" xlink:href="note-0298-06"/> ſphérique a été produit par la révolution d’un ſecteur de cercle <lb/>autour d’une ligne qui paſſe par le centre, & </s> <s xml:id="echoid-s8991" xml:space="preserve">qui diviſe ſa corde <lb/>en deux parties égales.</s> <s xml:id="echoid-s8992" xml:space="preserve"/> </p> <div xml:id="echoid-div685" type="float" level="2" n="1"> <note position="left" xlink:label="note-0298-06" xlink:href="note-0298-06a" xml:space="preserve">Figure 121.</note> </div> </div> <div xml:id="echoid-div687" type="section" level="1" n="553"> <head xml:id="echoid-head656" xml:space="preserve">X.</head> <p> <s xml:id="echoid-s8993" xml:space="preserve">540. </s> <s xml:id="echoid-s8994" xml:space="preserve">Orbe eſt un corps ſphérique, qui eſt terminé par deux <pb o="261" file="0299" n="299" rhead="DE MATHéMATIQUE. Liv. VIII."/> ſuperficies ſphériques & </s> <s xml:id="echoid-s8995" xml:space="preserve">concentriques, l’une concave, & </s> <s xml:id="echoid-s8996" xml:space="preserve">l’au-<lb/>tre convexe, comme le corps qui eſt borné par les deux ſuper-<lb/> <anchor type="note" xlink:label="note-0299-01a" xlink:href="note-0299-01"/> ficies ſphériques, l’une B C D E, qui eſt convexe, & </s> <s xml:id="echoid-s8997" xml:space="preserve">l’autre <lb/>F G H I, qui eſt concave: </s> <s xml:id="echoid-s8998" xml:space="preserve">ainſi vous voyez que l’orbe eſt ce <lb/>qui reſte, lorſque d’une grande ſphere, comme B C D E on <lb/>en a ôté une plus petite concentrique à la plus grande, comme <lb/>F G H I. </s> <s xml:id="echoid-s8999" xml:space="preserve">On peut concevoir un orbe comme formé, par la ré-<lb/>volution d’une couronne autour d’un diametre.</s> <s xml:id="echoid-s9000" xml:space="preserve"/> </p> <div xml:id="echoid-div687" type="float" level="2" n="1"> <note position="right" xlink:label="note-0299-01" xlink:href="note-0299-01a" xml:space="preserve">Figure 112.</note> </div> <p> <s xml:id="echoid-s9001" xml:space="preserve">541. </s> <s xml:id="echoid-s9002" xml:space="preserve">Comme on peut concevoir un orbe d’une épaiſſeur in-<lb/>finiment petite, il s’enſuit qu’une ſphere peut être conſidérée <lb/>comme compoſée d’une infinité d’orbes, dont le plus grand <lb/>eſt la ſurface de la ſphere, & </s> <s xml:id="echoid-s9003" xml:space="preserve">le plus petit eſt celui qui va ſe <lb/>terminer à zero, au centre de la ſphere.</s> <s xml:id="echoid-s9004" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div689" type="section" level="1" n="554"> <head xml:id="echoid-head657" xml:space="preserve">XI.</head> <p> <s xml:id="echoid-s9005" xml:space="preserve">542. </s> <s xml:id="echoid-s9006" xml:space="preserve">On appelle angle ſolide celui qui eſt formé par la ren-<lb/>contre de pluſieurs plans qui ſe terminent à un même point, <lb/>tel eſt, par exemple, l’angle E qui eſt compoſé des plans <lb/> <anchor type="note" xlink:label="note-0299-02a" xlink:href="note-0299-02"/> B E A, A E D, D E C & </s> <s xml:id="echoid-s9007" xml:space="preserve">B E C: </s> <s xml:id="echoid-s9008" xml:space="preserve">pour mieux comprendre cette <lb/>définition, il faut conſidérer le ſommet des pyramides, les <lb/>coins des cubes & </s> <s xml:id="echoid-s9009" xml:space="preserve">des parallelepipedes, qui ſont des angles <lb/>ſolides. </s> <s xml:id="echoid-s9010" xml:space="preserve">Il faut au moins trois plans pour former un angle ſo-<lb/>lide, de même qu’il faut deux lignes pour former un angle <lb/>plan.</s> <s xml:id="echoid-s9011" xml:space="preserve"/> </p> <div xml:id="echoid-div689" type="float" level="2" n="1"> <note position="right" xlink:label="note-0299-02" xlink:href="note-0299-02a" xml:space="preserve">Figure 127.</note> </div> </div> <div xml:id="echoid-div691" type="section" level="1" n="555"> <head xml:id="echoid-head658" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head659" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9012" xml:space="preserve">543. </s> <s xml:id="echoid-s9013" xml:space="preserve">La ſurface de tout priſme droit, ſans y comprendre les <lb/> <anchor type="note" xlink:label="note-0299-03a" xlink:href="note-0299-03"/> baſes, eſt égale à celle d’un rectangle, qui auroit pour baſe une li-<lb/>gne F G égale à la ſomme des côtés de la baſe du priſme, & </s> <s xml:id="echoid-s9014" xml:space="preserve">pour <lb/>hauteur une ligne G H égale à la hauteur A E du priſme.</s> <s xml:id="echoid-s9015" xml:space="preserve"/> </p> <div xml:id="echoid-div691" type="float" level="2" n="1"> <note position="right" xlink:label="note-0299-03" xlink:href="note-0299-03a" xml:space="preserve">Figure 123 <lb/>& 124.</note> </div> </div> <div xml:id="echoid-div693" type="section" level="1" n="556"> <head xml:id="echoid-head660" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9016" xml:space="preserve">Si le priſme droit a pour baſe un exagone régulier, il ſera <lb/>renfermé par ſix rectangles, tels que D E: </s> <s xml:id="echoid-s9017" xml:space="preserve">donc ſi la ligne F G <lb/>eſt égale à la ſomme des côtés du polygone, pris enſemble, <lb/>elle ſera ſextuple du côté A D; </s> <s xml:id="echoid-s9018" xml:space="preserve">& </s> <s xml:id="echoid-s9019" xml:space="preserve">comme les rectangles E D, <lb/>F H ont la même hauteur, le rectangle F H ſera ſextuple du <lb/>rectangle E D, & </s> <s xml:id="echoid-s9020" xml:space="preserve">par conſéquent égal à la ſurface du priſme. <lb/></s> <s xml:id="echoid-s9021" xml:space="preserve">C. </s> <s xml:id="echoid-s9022" xml:space="preserve">Q. </s> <s xml:id="echoid-s9023" xml:space="preserve">F. </s> <s xml:id="echoid-s9024" xml:space="preserve">D.</s> <s xml:id="echoid-s9025" xml:space="preserve"/> </p> <pb o="262" file="0300" n="300" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div694" type="section" level="1" n="557"> <head xml:id="echoid-head661" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9026" xml:space="preserve">544. </s> <s xml:id="echoid-s9027" xml:space="preserve">Le cylindre ayant pour baſe un cercle, que l’on peut <lb/>regarder comme un polygone d’une infinité de côtés, il s’en-<lb/>ſuit que le rectangle qui aura pour baſe une ligne droite égale <lb/>à la circonférence du cercle qui ſert de baſe au cylindre, & </s> <s xml:id="echoid-s9028" xml:space="preserve"><lb/>pour hauteur celle du cylindre, que l’on ſuppoſe droit, ſera <lb/>égal à la ſurface du même cylindre.</s> <s xml:id="echoid-s9029" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9030" xml:space="preserve">On démontreroit de même que la ſurface d’un priſme droit <lb/>quelconque, dont la baſe ſeroit un polygone irrégulier, comme <lb/>on voudra, eſt égale à celle d’un rectangle qui auroit même <lb/>hauteur que le priſme, & </s> <s xml:id="echoid-s9031" xml:space="preserve">une baſe égale à la ſomme des côtés <lb/>du polygone.</s> <s xml:id="echoid-s9032" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div695" type="section" level="1" n="558"> <head xml:id="echoid-head662" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head663" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9033" xml:space="preserve">545. </s> <s xml:id="echoid-s9034" xml:space="preserve">La ſurface d’une pyramide droite quelconque, comme ABC, <lb/> <anchor type="note" xlink:label="note-0300-01a" xlink:href="note-0300-01"/> eſt égale à celle d’un triangle, qui auroit pour baſe une ligne G I <lb/>égale à la ſomme des côtés du polygone régulier qui lui ſert de <lb/>baſe, & </s> <s xml:id="echoid-s9035" xml:space="preserve">pour hauteur une ligne G H égale à une perpendiculaire <lb/>B F abaiſſée du ſommet de la pyramide ſur un des côtés D E.</s> <s xml:id="echoid-s9036" xml:space="preserve"/> </p> <div xml:id="echoid-div695" type="float" level="2" n="1"> <note position="left" xlink:label="note-0300-01" xlink:href="note-0300-01a" xml:space="preserve">Figure 125. <lb/>& 126.</note> </div> </div> <div xml:id="echoid-div697" type="section" level="1" n="559"> <head xml:id="echoid-head664" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9037" xml:space="preserve">Imaginons que la pyramide A B C D E a pour baſe un exa-<lb/>gone régulier; </s> <s xml:id="echoid-s9038" xml:space="preserve">comme elle eſt ſuppoſée droite, elle ſera renfer-<lb/>mée par ſix triangles égaux au triangle D B E: </s> <s xml:id="echoid-s9039" xml:space="preserve">donc ſi l’on a <lb/>un triangle G H I, dont la baſe H I ſoit ſextuple de la baſe <lb/>D E du triangle D B E, & </s> <s xml:id="echoid-s9040" xml:space="preserve">dont la hauteur ſoit égale à celle du <lb/>même triangle, la ſurface de ce dernier triangle G H I ſera <lb/>ſextuple de celle du triangle D B E: </s> <s xml:id="echoid-s9041" xml:space="preserve">donc elle ſera égale à la <lb/>ſurface dela pyramide, ſans y comprendre la baſe. </s> <s xml:id="echoid-s9042" xml:space="preserve">C.</s> <s xml:id="echoid-s9043" xml:space="preserve">Q.</s> <s xml:id="echoid-s9044" xml:space="preserve">F.</s> <s xml:id="echoid-s9045" xml:space="preserve">D.</s> <s xml:id="echoid-s9046" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9047" xml:space="preserve">546. </s> <s xml:id="echoid-s9048" xml:space="preserve">Si la pyramide n’avoit pas pour baſe un polygone régu-<lb/>lier, la perpendiculaire menée du ſommet de la pyramide ſur <lb/>chaque côté ne ſeroit pas la même pour tous les triangles, quoi-<lb/>que la pyramide fût droite, & </s> <s xml:id="echoid-s9049" xml:space="preserve">cela arriveroit encore dans le <lb/>cas où la pyramide ayant pour baſe un polygone régulier, ne <lb/>ſeroit pas droite. </s> <s xml:id="echoid-s9050" xml:space="preserve">Dans ces deux cas, il faut chercher la ſurface <lb/>de chacun des triangles en particulier, & </s> <s xml:id="echoid-s9051" xml:space="preserve">la ſomme de ces ſur-<lb/>faces ſera la ſurface de la pyramide.</s> <s xml:id="echoid-s9052" xml:space="preserve"/> </p> <pb o="263" file="0301" n="301" rhead="DE MATHEMATIQUE. Liv. VIII."/> </div> <div xml:id="echoid-div698" type="section" level="1" n="560"> <head xml:id="echoid-head665" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9053" xml:space="preserve">547. </s> <s xml:id="echoid-s9054" xml:space="preserve">Un cône droit pouvant être regardé comme une py-<lb/>ramide droite d’une infinité de côtés, il s’enſuit que ſa ſurface <lb/>ſera égale à celle d’un triangle, qui auroit pour baſe une ligne <lb/>égale à la circonférence du cercle qui lui ſert de baſe, & </s> <s xml:id="echoid-s9055" xml:space="preserve">pour <lb/>hauteur une ligne égale au côté du cône.</s> <s xml:id="echoid-s9056" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div699" type="section" level="1" n="561"> <head xml:id="echoid-head666" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9057" xml:space="preserve">548. </s> <s xml:id="echoid-s9058" xml:space="preserve">Les parallelepipedes & </s> <s xml:id="echoid-s9059" xml:space="preserve">les priſmes droits ſont dans la rai-<lb/>ſon compoſée des raiſons de leurs trois dimenſions, ou comme les <lb/>produits de leurs trois dimenſions.</s> <s xml:id="echoid-s9060" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div700" type="section" level="1" n="562"> <head xml:id="echoid-head667" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9061" xml:space="preserve">Nous avons vu (art. </s> <s xml:id="echoid-s9062" xml:space="preserve">26), que pour trouver la ſolidité des <lb/>parallelepipedes, il falloit multiplier le produit des deux di-<lb/>menſions de leurs baſes par leurs hauteurs. </s> <s xml:id="echoid-s9063" xml:space="preserve">Si donc on a deux <lb/>priſmes, dont l’un ſoit A & </s> <s xml:id="echoid-s9064" xml:space="preserve">l’autre B, dont les dimenſions du <lb/>premier ſoient a, b, c; </s> <s xml:id="echoid-s9065" xml:space="preserve">& </s> <s xml:id="echoid-s9066" xml:space="preserve">les dimenſions du ſecond d, e, f; <lb/></s> <s xml:id="echoid-s9067" xml:space="preserve">le ſolide du premier priſme, ou ce priſme lui-même, ſera égal <lb/>à abc, & </s> <s xml:id="echoid-s9068" xml:space="preserve">le ſolide du ſecond priſme, ou ce prime lui-même, <lb/>ſera d e f: </s> <s xml:id="echoid-s9069" xml:space="preserve">donc on aura A: </s> <s xml:id="echoid-s9070" xml:space="preserve">B:</s> <s xml:id="echoid-s9071" xml:space="preserve">: a b c: </s> <s xml:id="echoid-s9072" xml:space="preserve">d e f; </s> <s xml:id="echoid-s9073" xml:space="preserve">mais la raiſon de <lb/>a b c à d e f eſt compoſée des trois raiſons de a à d, de b à e, <lb/>de c à f: </s> <s xml:id="echoid-s9074" xml:space="preserve">donc les priſmes ſont en raiſon compoſée de leurs <lb/>trois dimenſions, ou comme les produits de leurs dimenſions. </s> <s xml:id="echoid-s9075" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s9076" xml:space="preserve">Q. </s> <s xml:id="echoid-s9077" xml:space="preserve">F. </s> <s xml:id="echoid-s9078" xml:space="preserve">D.</s> <s xml:id="echoid-s9079" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div701" type="section" level="1" n="563"> <head xml:id="echoid-head668" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9080" xml:space="preserve">549. </s> <s xml:id="echoid-s9081" xml:space="preserve">Les priſmes & </s> <s xml:id="echoid-s9082" xml:space="preserve">les cylindres étant compoſés d’un nom-<lb/>bre infini de plans égaux, & </s> <s xml:id="echoid-s9083" xml:space="preserve">ſemblables à ceux de leurs baſes, <lb/>on peut dire que puiſque le nombre de ces plans eſt exprimé <lb/>par la hauteur de ces ſolides, il faudra, pour en trouver la va-<lb/>leur, multiplier la baſe par la hauteur: </s> <s xml:id="echoid-s9084" xml:space="preserve">donc puiſque la ſolidité <lb/>des priſmes & </s> <s xml:id="echoid-s9085" xml:space="preserve">des cylindres dépend du produit de leur trois <lb/>dimenſions, il s’enſuit qu’ils ſeront entr’eux dans la raiſon <lb/>compoſée de celles des mêmes dimenſions.</s> <s xml:id="echoid-s9086" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div702" type="section" level="1" n="564"> <head xml:id="echoid-head669" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9087" xml:space="preserve">550. </s> <s xml:id="echoid-s9088" xml:space="preserve">Il ſuit encore delà que l’on trouvera toujours le rap- <pb o="264" file="0302" n="302" rhead="NOUVEAU COURS"/> port des ſolides de même eſpece, en multipliant leurs baſes <lb/>par leurs hauteurs: </s> <s xml:id="echoid-s9089" xml:space="preserve">quand je dis de même eſpece, j’entends, <lb/>par exemple, les pyramides, les cônes, &</s> <s xml:id="echoid-s9090" xml:space="preserve">c. </s> <s xml:id="echoid-s9091" xml:space="preserve">Car quoique nous <lb/>n’ayons pas encore donné la maniere de trouver la ſolidité des <lb/>pyramides & </s> <s xml:id="echoid-s9092" xml:space="preserve">des cônes, cela n’empêche pas qu’on ne ſoit con-<lb/>vaincu qu’elles dépendent des produits de leur trois dimen-<lb/>ſions: </s> <s xml:id="echoid-s9093" xml:space="preserve">car ſi pour trouver le ſolide d’une pyramide, il faut mul-<lb/>tiplier la baſe par le tiers ou lamoitié de ſa hauteur, il eſt certain <lb/>que pour trouver la ſolidité d’une autre pyramide, il faudra auſſi <lb/>multiplier ſa baſe par le tiers ou la moitié de ſa hauteur: </s> <s xml:id="echoid-s9094" xml:space="preserve">ainſi <lb/>en multipliant de la même maniere les trois dimenſions d’une <lb/>pyramide, & </s> <s xml:id="echoid-s9095" xml:space="preserve">les trois dimenſions d’une autre; </s> <s xml:id="echoid-s9096" xml:space="preserve">ſi ces produits <lb/>ne donnent pas les ſolidités, ils donneront au moins le rap-<lb/>port que ces pyramides ont entr’elles.</s> <s xml:id="echoid-s9097" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div703" type="section" level="1" n="565"> <head xml:id="echoid-head670" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Theoreme.</emph></head> <p style="it"> <s xml:id="echoid-s9098" xml:space="preserve">551. </s> <s xml:id="echoid-s9099" xml:space="preserve">Toute pyramide, comme A B C D E, eſt le tiers d’un priſme <lb/> <anchor type="note" xlink:label="note-0302-01a" xlink:href="note-0302-01"/> de même baſe & </s> <s xml:id="echoid-s9100" xml:space="preserve">de même hauteur.</s> <s xml:id="echoid-s9101" xml:space="preserve"/> </p> <div xml:id="echoid-div703" type="float" level="2" n="1"> <note position="left" xlink:label="note-0302-01" xlink:href="note-0302-01a" xml:space="preserve">Figure 128.</note> </div> <p> <s xml:id="echoid-s9102" xml:space="preserve">Suppoſant que la baſe A C ſoit un quarré, nous nommerons <lb/>A D ou D C a, A H ou E F b, & </s> <s xml:id="echoid-s9103" xml:space="preserve">la perpendiculaire E G {1/2} a, <lb/>puiſqu’elle eſt moitié de I K ou de A D.</s> <s xml:id="echoid-s9104" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div705" type="section" level="1" n="566"> <head xml:id="echoid-head671" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9105" xml:space="preserve">Conſidérez que ſi du priſme A K on retranche la pyramide <lb/>A B C D E, il reſtera quatre autres pyramides telles que A H I E B, <lb/>qui ſont toutes égales entr’elles, ayant chacune pour baſe un <lb/>des rectangles A H I B de la ſurface du priſme, & </s> <s xml:id="echoid-s9106" xml:space="preserve">pour hauteur <lb/>une perpendiculaire égale à E G. </s> <s xml:id="echoid-s9107" xml:space="preserve">Or ſi l’on multiplie a a, qui <lb/>eſt la baſe A C, de la pyramide A E C par ſa hauteur E F, qui <lb/>eſt b, on aura a a b pour le produit de ſes trois dimenſions; </s> <s xml:id="echoid-s9108" xml:space="preserve">& </s> <s xml:id="echoid-s9109" xml:space="preserve"><lb/>multipliant auſſi a b, qui eſt la baſe de la pyramide A H I E B, <lb/>par ſa hauteur E G, qui eſt {1/2} a, on aura {aab/2} pour le produit de <lb/>ſes trois dimenſions. </s> <s xml:id="echoid-s9110" xml:space="preserve">Ainſi la pyramide A B C D E eſt à la <lb/>pyramide A H I E B, comme a a b eſt à {aab/2}; </s> <s xml:id="echoid-s9111" xml:space="preserve">donc la premiere <lb/>eſt double de la ſeconde (art. </s> <s xml:id="echoid-s9112" xml:space="preserve">550), puiſque ces pyramides ſont <lb/>entr’elles comme les produits de leurs trois dimenſions. </s> <s xml:id="echoid-s9113" xml:space="preserve">Mais <pb o="265" file="0303" n="303" rhead="DE MATHEMATIQUE. Liv. VIII."/> <gap/>me il y a quatre pyramides égales à {aab/2}, leur ſomme ſera <lb/>{<gap/>/2} ou 2aab, & </s> <s xml:id="echoid-s9114" xml:space="preserve">ſi l’on joint encore à cette pyramide la py-<lb/>ramide A B C D E = a a b, on aura le ſolide entier, égal à <lb/>3aab: </s> <s xml:id="echoid-s9115" xml:space="preserve">donc la pyramide A E C ſera le tiers du ſolide ou priſme <lb/>droit A K. </s> <s xml:id="echoid-s9116" xml:space="preserve">C. </s> <s xml:id="echoid-s9117" xml:space="preserve">Q. </s> <s xml:id="echoid-s9118" xml:space="preserve">F. </s> <s xml:id="echoid-s9119" xml:space="preserve">D.</s> <s xml:id="echoid-s9120" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div706" type="section" level="1" n="567"> <head xml:id="echoid-head672" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9121" xml:space="preserve">552. </s> <s xml:id="echoid-s9122" xml:space="preserve">Puiſque la pyramide A B C D E eſt le tiers du priſme <lb/>A K, ſi l’on coupe cette pyramide par un plan B E D, qui paſſe <lb/>par le ſommet E & </s> <s xml:id="echoid-s9123" xml:space="preserve">les angles oppoſés de la baſe, ce plan di-<lb/>viſera la pyramide totale en deux autres pyramides égales, <lb/>& </s> <s xml:id="echoid-s9124" xml:space="preserve">le priſme quarré en deux autres priſmes, pareillement égaux <lb/>entr’eux, puiſque chacun a même baſe & </s> <s xml:id="echoid-s9125" xml:space="preserve">même hauteur: <lb/></s> <s xml:id="echoid-s9126" xml:space="preserve">donc puiſque la pyramide totale eſt le tiers du priſme total, la <lb/>pyramide triangulaire ſera auſſi le tiers du priſme triangulaire. </s> <s xml:id="echoid-s9127" xml:space="preserve"><lb/>D’où il ſuit qu’une pyramide quelconque eſt toujours le tiers <lb/>d’un priſme de même baſe & </s> <s xml:id="echoid-s9128" xml:space="preserve">de même hauteur, parce que l’on <lb/>peut concevoir un priſme pentagonal, par exemple, comme <lb/>compoſé de cinq priſmes triangulaires, & </s> <s xml:id="echoid-s9129" xml:space="preserve">une pyramide pen-<lb/>tagonale, comme auſſi compoſée de cinq pyramides triangu-<lb/>laires, & </s> <s xml:id="echoid-s9130" xml:space="preserve">comme chacune ſera le tiers du priſme correſpon-<lb/>dant, la pyramide totale ſera auſſi le tiers du priſme total.</s> <s xml:id="echoid-s9131" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div707" type="section" level="1" n="568"> <head xml:id="echoid-head673" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9132" xml:space="preserve">553. </s> <s xml:id="echoid-s9133" xml:space="preserve">Il ſuit de cette propoſition, que pour trouver la ſolidité <lb/>d’une pyramide telle que A B C D E, qui a pour baſe un quarré, <lb/>il faut multiplier la baſe, c’eſt-à-dire le quarré A D, par le tiers <lb/>de la hauteur de la pyramide, qui eſt la perpendiculaire C H, <lb/>ou bien multiplier la baſe par toute la hauteur, & </s> <s xml:id="echoid-s9134" xml:space="preserve">prendre le <lb/>tiers du produit.</s> <s xml:id="echoid-s9135" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div708" type="section" level="1" n="569"> <head xml:id="echoid-head674" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s9136" xml:space="preserve">554. </s> <s xml:id="echoid-s9137" xml:space="preserve">Si l’on coupe la pyramide droite A C D par un plan F C G, <lb/> <anchor type="note" xlink:label="note-0303-01a" xlink:href="note-0303-01"/> qui paſſant par l’axe, ſoit parallele à un des côtés de la baſe, <lb/>la ſection donnera un triangle iſoſcele F C G, dont tous les <lb/>élémens, tels que I K, ſont en progreſſion arithmétique; </s> <s xml:id="echoid-s9138" xml:space="preserve">mais <lb/>comme tous ces élémens ſont autant de lignes égales aux côtés <lb/>des quarrés qui compoſent la pyramide, il s’enſuit que la py-<lb/>ramide eſt compoſée d’un nombre infini de quarrés, dont <pb o="266" file="0304" n="304" rhead="NOUVEAU COURS"/> tous les côtés ſont en progreſſion arithmétique; </s> <s xml:id="echoid-s9139" xml:space="preserve">& </s> <s xml:id="echoid-s9140" xml:space="preserve">comme <lb/>pour trouver la ſomme de tous ces quarrés, il faut multiplier <lb/>le quarré A D par le tiers de la perpendiculaire C H, l’on pourra <lb/>tirer de ce raiſonnement un principe général, qui eſt que ſi l’on <lb/>a une progreſſion arithmétique infinie, compoſée de lignes, dont <lb/>la plus petite va ſe terminer à o, l’on trouvera la ſomme des quarrés <lb/>de toutes ces lignes, en multipliant le quarré de la plus grande li-<lb/>gne par le tiers de la grandeur qui exprime la quantité des lignes <lb/>ou des quarrés. </s> <s xml:id="echoid-s9141" xml:space="preserve">Comme la ſuite des nombres naturels eſt une <lb/>ſuite de grandeurs qui croiſſent en progreſſion arithmétique, <lb/>on peut par cette propoſition, prouver que la ſomme des quarrés <lb/>de tous les nombres poſſibles, depuis zero juſqu’à l’infini, eſt <lb/>égale au tiers du cube du dernier nombre que l’on puiſſe ima-<lb/>giner, ou bien au tiers du cube de l’infini.</s> <s xml:id="echoid-s9142" xml:space="preserve"/> </p> <div xml:id="echoid-div708" type="float" level="2" n="1"> <note position="right" xlink:label="note-0303-01" xlink:href="note-0303-01a" xml:space="preserve">Figure 129.</note> </div> <p> <s xml:id="echoid-s9143" xml:space="preserve">Il eſt bien important de comprendre ce corollaire, parce <lb/>que nous nous en ſervirons dans les démonſtrations ſuivantes.</s> <s xml:id="echoid-s9144" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div710" type="section" level="1" n="570"> <head xml:id="echoid-head675" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s9145" xml:space="preserve">555. </s> <s xml:id="echoid-s9146" xml:space="preserve">Il ſuit encore delà, que pour trouver la ſolidité d’une <lb/>pyramide droite A B C, qui a pour baſe un polygone quelcon-<lb/> <anchor type="note" xlink:label="note-0304-01a" xlink:href="note-0304-01"/> que A C, il faut multiplier la baſe par le tiers de l’axe B D; <lb/></s> <s xml:id="echoid-s9147" xml:space="preserve">car comme cette pyramide eſt compoſée d’une infinité de po-<lb/>lygones ſemblables à la baſe, & </s> <s xml:id="echoid-s9148" xml:space="preserve">tous ces polygones ſemblables <lb/>étant dans la raiſon des quarrés de leurs côtés homologues <lb/>(art. </s> <s xml:id="echoid-s9149" xml:space="preserve">493), ou de leurs rayons, tels que E F & </s> <s xml:id="echoid-s9150" xml:space="preserve">A D, leſquels <lb/>ſont les mêmes que les élémens du triangle A B D, on peut <lb/>dire que ces polygones ſont dans la raiſon des quarrés des li-<lb/>gnes d’une progreſſion infinie arithmétique, & </s> <s xml:id="echoid-s9151" xml:space="preserve">que par conſé-<lb/>quent pour en trouver la valeur, il faudra multiplier le plus <lb/>grand polygone A C par le tiers de la perpendiculaire B D.</s> <s xml:id="echoid-s9152" xml:space="preserve"/> </p> <div xml:id="echoid-div710" type="float" level="2" n="1"> <note position="left" xlink:label="note-0304-01" xlink:href="note-0304-01a" xml:space="preserve">Figure 130.</note> </div> </div> <div xml:id="echoid-div712" type="section" level="1" n="571"> <head xml:id="echoid-head676" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s9153" xml:space="preserve">556. </s> <s xml:id="echoid-s9154" xml:space="preserve">Comme le cône A B C eſt compoſé d’une infinité de <lb/> <anchor type="note" xlink:label="note-0304-02a" xlink:href="note-0304-02"/> cercles, qui ont pour rayons les élémens, tels que E F & </s> <s xml:id="echoid-s9155" xml:space="preserve">A D <lb/>du triangle A B D, il s’enſuit que les cercles étant dans la <lb/>même raiſon que les quarrés de leurs rayons, il faudra, pour <lb/>trouver la valeur de tous les cercles dont le cône eſt compoſé, <lb/>multiplier le plus grand cercle A C par le tiers de la perpendi-<lb/>culaire B D qui en exprime la quantité.</s> <s xml:id="echoid-s9156" xml:space="preserve"/> </p> <div xml:id="echoid-div712" type="float" level="2" n="1"> <note position="left" xlink:label="note-0304-02" xlink:href="note-0304-02a" xml:space="preserve">Figure 132.</note> </div> <pb o="267" file="0305" n="305" rhead="DE MATHÉMATIQUE. Liv. VIII."/> </div> <div xml:id="echoid-div714" type="section" level="1" n="572"> <head xml:id="echoid-head677" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9157" xml:space="preserve">557. </s> <s xml:id="echoid-s9158" xml:space="preserve">Si l’on a deux pyramides, A B C & </s> <s xml:id="echoid-s9159" xml:space="preserve">H L K, dont la hau-<lb/> <anchor type="note" xlink:label="note-0305-01a" xlink:href="note-0305-01"/> teur B D de la premiere ſoit égale à la hauteur L O de la ſeconde, <lb/>je dis qu’elles ſeront entr’elles dans la raiſon de la baſe A C à la <lb/>baſe H K.</s> <s xml:id="echoid-s9160" xml:space="preserve"/> </p> <div xml:id="echoid-div714" type="float" level="2" n="1"> <note position="right" xlink:label="note-0305-01" xlink:href="note-0305-01a" xml:space="preserve">Figure 130 <lb/>& 131.</note> </div> <p> <s xml:id="echoid-s9161" xml:space="preserve">Suppoſant que la baſe A C ſoit un exagone régulier, & </s> <s xml:id="echoid-s9162" xml:space="preserve">la <lb/>baſe H K un quarré, nous nommerons le côté M N, a; </s> <s xml:id="echoid-s9163" xml:space="preserve">la <lb/>perpendiculaire D G, b; </s> <s xml:id="echoid-s9164" xml:space="preserve">le côté H I ou I K, c; </s> <s xml:id="echoid-s9165" xml:space="preserve">& </s> <s xml:id="echoid-s9166" xml:space="preserve">la hauteur <lb/>B D ou L O, d. </s> <s xml:id="echoid-s9167" xml:space="preserve">Cela poſé, la baſe A C ſera {6ab/2} ou 3ab, & </s> <s xml:id="echoid-s9168" xml:space="preserve">la <lb/>baſe H K ſera c c, & </s> <s xml:id="echoid-s9169" xml:space="preserve">multipliant les deux baſes par le tiers de <lb/>la hauteur commune (art. </s> <s xml:id="echoid-s9170" xml:space="preserve">553), c’eſt-à-dire par {d/3}, l’on aura <lb/>{3abd/3} pour la valeur de la premiere pyramide A B C, & </s> <s xml:id="echoid-s9171" xml:space="preserve">{ccd/3} pour <lb/>la valeur de la pyramide H K L: </s> <s xml:id="echoid-s9172" xml:space="preserve">ainſi il faut démontrer que <lb/>abd: </s> <s xml:id="echoid-s9173" xml:space="preserve">{ccd/3}:</s> <s xml:id="echoid-s9174" xml:space="preserve">: 3ab: </s> <s xml:id="echoid-s9175" xml:space="preserve">c c.</s> <s xml:id="echoid-s9176" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div716" type="section" level="1" n="573"> <head xml:id="echoid-head678" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9177" xml:space="preserve">Cette proportion eſt évidente, puiſque le produit des ex-<lb/>trêmes eſt égal à celui des moyens: </s> <s xml:id="echoid-s9178" xml:space="preserve">car a b d c c = {3a b d c c/3} = <lb/>a b d c c. </s> <s xml:id="echoid-s9179" xml:space="preserve">C. </s> <s xml:id="echoid-s9180" xml:space="preserve">Q. </s> <s xml:id="echoid-s9181" xml:space="preserve">F. </s> <s xml:id="echoid-s9182" xml:space="preserve">D.</s> <s xml:id="echoid-s9183" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div717" type="section" level="1" n="574"> <head xml:id="echoid-head679" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9184" xml:space="preserve">558. </s> <s xml:id="echoid-s9185" xml:space="preserve">Les cônes étant des pyramides d’une infinité de côtés, <lb/>il s’enſuit que lorſqu’ils auront la même hauteur, ils ſeront <lb/>dans la raiſon de leurs baſes. </s> <s xml:id="echoid-s9186" xml:space="preserve">Il en ſera de même pour les <lb/>priſmes & </s> <s xml:id="echoid-s9187" xml:space="preserve">les cylindres qui ſont triples des pyramides ou des <lb/>cônes de même baſe & </s> <s xml:id="echoid-s9188" xml:space="preserve">de même hauteur: </s> <s xml:id="echoid-s9189" xml:space="preserve">car ſi les parties <lb/>ſont entr’elles comme les tous, réciproquement les tous ſont <lb/>entr’eux comme leurs parties de même nom.</s> <s xml:id="echoid-s9190" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div718" type="section" level="1" n="575"> <head xml:id="echoid-head680" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9191" xml:space="preserve">559. </s> <s xml:id="echoid-s9192" xml:space="preserve">Si l’on a deux priſmes X & </s> <s xml:id="echoid-s9193" xml:space="preserve">Y, dont les baſes & </s> <s xml:id="echoid-s9194" xml:space="preserve">les hau-<lb/> <anchor type="note" xlink:label="note-0305-02a" xlink:href="note-0305-02"/> teurs ſoient réciproques, je dis qu’ils ſont égaux.</s> <s xml:id="echoid-s9195" xml:space="preserve"/> </p> <div xml:id="echoid-div718" type="float" level="2" n="1"> <note position="right" xlink:label="note-0305-02" xlink:href="note-0305-02a" xml:space="preserve">Pl. VII.</note> </div> <note position="right" xml:space="preserve">Figure 133 <lb/>& 134.</note> <pb o="268" file="0306" n="306" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div720" type="section" level="1" n="576"> <head xml:id="echoid-head681" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9196" xml:space="preserve">Pour le prouver, nous ſuppoſerons que a b eſt la baſe du <lb/>priſme X, & </s> <s xml:id="echoid-s9197" xml:space="preserve">c d celle du priſme Y, e la hauteur du priſme Y, <lb/>& </s> <s xml:id="echoid-s9198" xml:space="preserve">f celle du prime X; </s> <s xml:id="echoid-s9199" xml:space="preserve">cela étant, par hypotheſe, on a a b: <lb/></s> <s xml:id="echoid-s9200" xml:space="preserve">c d:</s> <s xml:id="echoid-s9201" xml:space="preserve">:e:</s> <s xml:id="echoid-s9202" xml:space="preserve">f; </s> <s xml:id="echoid-s9203" xml:space="preserve">donc a b f=c d e: </s> <s xml:id="echoid-s9204" xml:space="preserve">or comme le premier membre <lb/>de cette équation eſt le produit des trois dimenſions du priſme <lb/>X, & </s> <s xml:id="echoid-s9205" xml:space="preserve">le ſecond le produit des trois dimenſions du priſme Y, il <lb/>s’enſuit évidemment que ces priſmes ſont égaux. </s> <s xml:id="echoid-s9206" xml:space="preserve">C. </s> <s xml:id="echoid-s9207" xml:space="preserve">Q. </s> <s xml:id="echoid-s9208" xml:space="preserve">F. </s> <s xml:id="echoid-s9209" xml:space="preserve">D.</s> <s xml:id="echoid-s9210" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div721" type="section" level="1" n="577"> <head xml:id="echoid-head682" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9211" xml:space="preserve">560. </s> <s xml:id="echoid-s9212" xml:space="preserve">Il ſuit de cette propoſition, que les cylindres, les pyra-<lb/>mides & </s> <s xml:id="echoid-s9213" xml:space="preserve">les cônes qui ont leurs baſes & </s> <s xml:id="echoid-s9214" xml:space="preserve">leurs hauteurs réci-<lb/>proques, ſont égaux chacun à chacun. </s> <s xml:id="echoid-s9215" xml:space="preserve">La démonſtration eſt <lb/>la même que la précédente.</s> <s xml:id="echoid-s9216" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div722" type="section" level="1" n="578"> <head xml:id="echoid-head683" xml:space="preserve">PROPOSITION VII <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9217" xml:space="preserve">561. </s> <s xml:id="echoid-s9218" xml:space="preserve">Une pyramide tronquée, comme A B E D, eſt égale à une <lb/> <anchor type="note" xlink:label="note-0306-01a" xlink:href="note-0306-01"/> pyramide qui auroit pour baſe un plan égal aux deux quarrés B E <lb/>& </s> <s xml:id="echoid-s9219" xml:space="preserve">A H, pris enſemble; </s> <s xml:id="echoid-s9220" xml:space="preserve">plus un plan qui ſeroit moyen géométrique <lb/>entre ces deux quarrés, & </s> <s xml:id="echoid-s9221" xml:space="preserve">pour hauteur l’axe F G.</s> <s xml:id="echoid-s9222" xml:space="preserve"/> </p> <div xml:id="echoid-div722" type="float" level="2" n="1"> <note position="left" xlink:label="note-0306-01" xlink:href="note-0306-01a" xml:space="preserve">Figure 135. <lb/>& 136.</note> </div> <p> <s xml:id="echoid-s9223" xml:space="preserve">Conſidérant la figure H K L I, comme étant la coupe de la <lb/>pyramide tronquée, coupée par un plan perpendiculaire à ſa <lb/>baſe, & </s> <s xml:id="echoid-s9224" xml:space="preserve">qui paſſeroit par ſon ſommet, & </s> <s xml:id="echoid-s9225" xml:space="preserve">le triangle H M I, <lb/>comme la coupe de la pyramide entiere, nous nommerons le <lb/>côté A D, a; </s> <s xml:id="echoid-s9226" xml:space="preserve">K L ou B C, b; </s> <s xml:id="echoid-s9227" xml:space="preserve">l’axe M G, c; </s> <s xml:id="echoid-s9228" xml:space="preserve">le petit axe M F de <lb/>la pyramide K M L, d: </s> <s xml:id="echoid-s9229" xml:space="preserve">ainſi l’axe F G de la pyramide tronquée <lb/>ſera c-d, & </s> <s xml:id="echoid-s9230" xml:space="preserve">l’on aura aa+bb+ab pour la baſe de la pyramide <lb/>égale à la pyramide tronquée; </s> <s xml:id="echoid-s9231" xml:space="preserve">car a b eſt moyen proportionnel <lb/>entre a a & </s> <s xml:id="echoid-s9232" xml:space="preserve">b b (art. </s> <s xml:id="echoid-s9233" xml:space="preserve">505). </s> <s xml:id="echoid-s9234" xml:space="preserve">Ainſi il faut prouver que le produit <lb/>de aa + bb + ab par {c-d/3}, qui eſt {aac+bbc+abc-aad-bbd-abd/3}, <lb/>eſt égal au ſolide de la pyramide tronquée.</s> <s xml:id="echoid-s9235" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div724" type="section" level="1" n="579"> <head xml:id="echoid-head684" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9236" xml:space="preserve">Faites attention que la pyramide tronquée eſt égale à la diſ-<lb/>férence de la pyramide entiere & </s> <s xml:id="echoid-s9237" xml:space="preserve">de la pyramide emportée; <lb/></s> <s xml:id="echoid-s9238" xml:space="preserve">que la pyramide entiere H M I eſt {aac/3}, & </s> <s xml:id="echoid-s9239" xml:space="preserve">que la petite pyra- <pb o="269" file="0307" n="307" rhead="DE MATHÉMATIQUE. Liv. VIII."/> mide K M L eſt {bbd/3}, & </s> <s xml:id="echoid-s9240" xml:space="preserve">que ſi l’on ôte la petite de la grande, <lb/>la différence ſera la valeur de la pyramide tronquée, qui eſt <lb/>{aac-bbd/3}, & </s> <s xml:id="echoid-s9241" xml:space="preserve">qui doit être égale au produit <lb/>{aac+bbc+abc-aad-bbd-abd/3}; </s> <s xml:id="echoid-s9242" xml:space="preserve">ce qui fournit cette équation, <lb/>{aac-bbd/3}={aac+bbc+abc-aad-bbd-abd/3}.</s> <s xml:id="echoid-s9243" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9244" xml:space="preserve">Pour prouver cette équation, on fera attention qu’à cauſe <lb/>des triangles ſemblables H M I, K M L, on a HI:</s> <s xml:id="echoid-s9245" xml:space="preserve">KL:</s> <s xml:id="echoid-s9246" xml:space="preserve">:MG:</s> <s xml:id="echoid-s9247" xml:space="preserve">MF, <lb/>ou a:</s> <s xml:id="echoid-s9248" xml:space="preserve">b:</s> <s xml:id="echoid-s9249" xml:space="preserve">:c:</s> <s xml:id="echoid-s9250" xml:space="preserve">d; </s> <s xml:id="echoid-s9251" xml:space="preserve">ce qui donne ad=bc: </s> <s xml:id="echoid-s9252" xml:space="preserve">en mettant donc b c <lb/>à la place de a d dans le quatrieme & </s> <s xml:id="echoid-s9253" xml:space="preserve">ſixieme terme du ſecond <lb/>membre de cette équation, on aura celle-ci {aac-bbd/3}= <lb/>{aac+bbc+abc-abc-bbd-bbc/3}, dans laquelle, effaçant ce qui <lb/>ſe détruit, on aura {aac-bbd/3}={aac-bbd/3}. </s> <s xml:id="echoid-s9254" xml:space="preserve">C. </s> <s xml:id="echoid-s9255" xml:space="preserve">Q. </s> <s xml:id="echoid-s9256" xml:space="preserve">F. </s> <s xml:id="echoid-s9257" xml:space="preserve">D.</s> <s xml:id="echoid-s9258" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div725" type="section" level="1" n="580"> <head xml:id="echoid-head685" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9259" xml:space="preserve">562. </s> <s xml:id="echoid-s9260" xml:space="preserve">Il ſuit de cette propoſition, que pour trouver la valeur <lb/>d’une pyramide quarrée tronquée, il faut multiplier le côté de <lb/>la baſe inférieure de cette pyramide par le côté de la baſe ſupé-<lb/>rieure, pour avoir le plan a b, moyen entre les deux, & </s> <s xml:id="echoid-s9261" xml:space="preserve">ajouter <lb/>ce plan à la ſomme des deux baſes inférieure & </s> <s xml:id="echoid-s9262" xml:space="preserve">ſupérieure, puis <lb/>multiplier le tout par le tiers de la perpendiculaire F G.</s> <s xml:id="echoid-s9263" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div726" type="section" level="1" n="581"> <head xml:id="echoid-head686" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s9264" xml:space="preserve">563. </s> <s xml:id="echoid-s9265" xml:space="preserve">Si la baſe de la pyramide n’étoit pas un quarré, pour <lb/>avoir le plan moyen, il faudroit multiplier les deux plans l’un <lb/>par l’autre, & </s> <s xml:id="echoid-s9266" xml:space="preserve">en extraire la racine: </s> <s xml:id="echoid-s9267" xml:space="preserve">mais on peut trouver ce <lb/>plan d’une maniere plus ſimple, comme on le va voir.</s> <s xml:id="echoid-s9268" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9269" xml:space="preserve">Suppoſons que la baſe de la pyramide eſt un pentagone ré-<lb/>gulier, la baſe ſupérieure de la pyramide ſera auſſi un penta-<lb/>gone régulier, & </s> <s xml:id="echoid-s9270" xml:space="preserve">ſemblable à celui de la baſe inférieure, parce <lb/>que l’on ſuppoſe la pyramide coupée par un plan parallele à <lb/>cette baſe. </s> <s xml:id="echoid-s9271" xml:space="preserve">Soit 2a le contour du premier polygone, & </s> <s xml:id="echoid-s9272" xml:space="preserve">b la per-<lb/>pendiculaire qui meſure la hauteur d’un triangle: </s> <s xml:id="echoid-s9273" xml:space="preserve">ſoit pareil-<lb/>lement 2c le contour du polygone, qui eſt la baſe de la pyra-<lb/>mide emportée, & </s> <s xml:id="echoid-s9274" xml:space="preserve">d la perpendiculaire qui meſure la hauteur <lb/>d’un triangle: </s> <s xml:id="echoid-s9275" xml:space="preserve">on aura la ſurface du premier polygone, en <lb/>multipliant la hauteur d’un triangle par la moitié du contour:</s> <s xml:id="echoid-s9276" xml:space="preserve"> <pb o="270" file="0308" n="308" rhead="NOUVEAU COURS"/> on aura de même la ſurface du ſecond polygone, ou de la baſe <lb/>ſupérieure, en multipliant ſa perpendiculaire par la moitié du <lb/>contour (art. </s> <s xml:id="echoid-s9277" xml:space="preserve">483). </s> <s xml:id="echoid-s9278" xml:space="preserve">La baſe inférieure ſera donc a b, & </s> <s xml:id="echoid-s9279" xml:space="preserve">la <lb/>baſe ſupérieure c d: </s> <s xml:id="echoid-s9280" xml:space="preserve">multipliant ces deux ſurfaces l’une par <lb/>l’autre, le produit ſera a b c d, dont la racine donneroit le <lb/>moyen cherché entre les deux baſes: </s> <s xml:id="echoid-s9281" xml:space="preserve">mais je fais attention que <lb/>puiſque ces polygones ſont ſemblables, leurs contours, ou les <lb/>moitiés de ces contours ſeront entr’elles comme les perpen-<lb/>diculaires: </s> <s xml:id="echoid-s9282" xml:space="preserve">on aura donc a:</s> <s xml:id="echoid-s9283" xml:space="preserve">b:</s> <s xml:id="echoid-s9284" xml:space="preserve">:c:</s> <s xml:id="echoid-s9285" xml:space="preserve">d, d’où l’on tire ad=bc. <lb/></s> <s xml:id="echoid-s9286" xml:space="preserve">Si donc dans le produit a b c d, on met à la place de bc le pro-<lb/>duit a d, qui lui eſt égal, on aura a a d d pour le quarré du plan <lb/>moyen géométrique entre les deux baſes, dont la racine a d, <lb/>que l’on peut prendre ſur le champ, donne ce même plan <lb/>moyen. </s> <s xml:id="echoid-s9287" xml:space="preserve">D’où il ſuit, que pour trouver un polygone quel-<lb/>conque ſemblable à deux autres polygones ſemblables en-<lb/>tr’eux, & </s> <s xml:id="echoid-s9288" xml:space="preserve">qui ſoit moyen géométrique entre ces deux poly-<lb/>gones, il faut multiplier la moitié du contour du plus grand <lb/>par la perpendiculaire de l’autre, ou le demi-contour du plus <lb/>petit par la perpendiculaire du plus grand. </s> <s xml:id="echoid-s9289" xml:space="preserve">J’ai inſiſté ſur cette <lb/>remarque, parce qu’elle donne une méthode fort commode <lb/>de trouver une ſurface moyenne géométrique entre deux au-<lb/>tres ſurfaces ſemblables, & </s> <s xml:id="echoid-s9290" xml:space="preserve">que d’ailleurs on ne le trouve pas <lb/>dans les autres élémens. </s> <s xml:id="echoid-s9291" xml:space="preserve">Par exemple, pour trouver un cercle <lb/>moyen géométrique entre deux cercles donnés, dont les rayons <lb/>ſont a & </s> <s xml:id="echoid-s9292" xml:space="preserve">b, les circonférences 2c & </s> <s xml:id="echoid-s9293" xml:space="preserve">2d, le cercle moyen ſera <lb/>également a d ou b c, que l’on trouve ſur le champ, ſans être <lb/>obligé d’extraire de racines.</s> <s xml:id="echoid-s9294" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div727" type="section" level="1" n="582"> <head xml:id="echoid-head687" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9295" xml:space="preserve">564. </s> <s xml:id="echoid-s9296" xml:space="preserve">Comme un cône tronqué eſt compoſé d’une infinité <lb/>de cercles, qui ſont tous dans la raiſon des quarrés qui com-<lb/>poſent une pyramide tronquée, il s’enſuit que pour en trouver <lb/>la ſolidité, il faut chercher un cercle moyen entre les deux <lb/>cercles oppoſés, ajouter cette ſomme avec les deux qui ſervent <lb/>de baſe, & </s> <s xml:id="echoid-s9297" xml:space="preserve">multiplier le tout par le tiers de l’axe compris entre <lb/>les deux cercles; </s> <s xml:id="echoid-s9298" xml:space="preserve">il faut auſſi entendre la même choſe de toute <lb/>autre pyramide tronquée, ſoit que ſa baſe ſoit réguliere, ſoit <lb/>qu’elle ſoit irréguliere.</s> <s xml:id="echoid-s9299" xml:space="preserve"/> </p> <pb o="271" file="0309" n="309" rhead="DE MATHÉMATIQUE. Liv. VIII."/> </div> <div xml:id="echoid-div728" type="section" level="1" n="583"> <head xml:id="echoid-head688" xml:space="preserve"><emph style="sc">Lemme</emph>.</head> <p style="it"> <s xml:id="echoid-s9300" xml:space="preserve">565. </s> <s xml:id="echoid-s9301" xml:space="preserve">Une ligne moyenne proportionnelle entre les parties E G <lb/> <anchor type="note" xlink:label="note-0309-01a" xlink:href="note-0309-01"/> & </s> <s xml:id="echoid-s9302" xml:space="preserve">G F du diametre E F d’un cercle, ſera le rayon d’un cercle égal <lb/>à la couronne X.</s> <s xml:id="echoid-s9303" xml:space="preserve"/> </p> <div xml:id="echoid-div728" type="float" level="2" n="1"> <note position="right" xlink:label="note-0309-01" xlink:href="note-0309-01a" xml:space="preserve">Figure 137.</note> </div> </div> <div xml:id="echoid-div730" type="section" level="1" n="584"> <head xml:id="echoid-head689" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9304" xml:space="preserve">Conſidérez que par la nature du cercle, la ligne G H eſt <lb/>moyenne proportionnelle entre les parties E G & </s> <s xml:id="echoid-s9305" xml:space="preserve">G F du dia-<lb/>metre; </s> <s xml:id="echoid-s9306" xml:space="preserve">& </s> <s xml:id="echoid-s9307" xml:space="preserve">à cauſe du triangle rectangle D G H, on a GH<emph style="sub">2</emph> = <lb/>DH<emph style="sub">2</emph>-DG<emph style="sub">2</emph>: </s> <s xml:id="echoid-s9308" xml:space="preserve">& </s> <s xml:id="echoid-s9309" xml:space="preserve">commeles cercles ſont en mêmeraiſon que les <lb/>quarrés de leurs rayons, on aura le cercle de G H égal au cer-<lb/>cle de D H moins le cercle de D G; </s> <s xml:id="echoid-s9310" xml:space="preserve">mais la couronne eſt auſſi <lb/>égale à la différence des cercles décrits du rayon DH & </s> <s xml:id="echoid-s9311" xml:space="preserve">du <lb/>rayon D G: </s> <s xml:id="echoid-s9312" xml:space="preserve">donc la couronne eſt égale au cercle du rayon <lb/>G H, ou d’une ligne moyenne entre les parties du diametre. <lb/></s> <s xml:id="echoid-s9313" xml:space="preserve">C. </s> <s xml:id="echoid-s9314" xml:space="preserve">Q. </s> <s xml:id="echoid-s9315" xml:space="preserve">F. </s> <s xml:id="echoid-s9316" xml:space="preserve">D.</s> <s xml:id="echoid-s9317" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div731" type="section" level="1" n="585"> <head xml:id="echoid-head690" xml:space="preserve">PROPOSITION VIII. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9318" xml:space="preserve">566. </s> <s xml:id="echoid-s9319" xml:space="preserve">Si l’on a une demi-ſphere A E D inſcrite dans un cylindre <lb/> <anchor type="note" xlink:label="note-0309-02a" xlink:href="note-0309-02"/> A B C D, je dis que la demi - ſphere eſt égale aux deux tiers du <lb/>cylindre.</s> <s xml:id="echoid-s9320" xml:space="preserve"/> </p> <div xml:id="echoid-div731" type="float" level="2" n="1"> <note position="right" xlink:label="note-0309-02" xlink:href="note-0309-02a" xml:space="preserve">Figure 138.</note> </div> <p> <s xml:id="echoid-s9321" xml:space="preserve">Prolongez le diametre B C juſqu’en F, enſorte que B F ſoit <lb/>égale à B A, & </s> <s xml:id="echoid-s9322" xml:space="preserve">tirez la ligne F A, qui donnera letriangle iſoſ-<lb/>cele A B F.</s> <s xml:id="echoid-s9323" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div733" type="section" level="1" n="586"> <head xml:id="echoid-head691" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9324" xml:space="preserve">Si l’on ſuppoſe que la demi-ſphere & </s> <s xml:id="echoid-s9325" xml:space="preserve">le cylindre ſont coupés <lb/>par un plan GL parallele à la baſe A D, cette ſection formera <lb/>la couronne G H, & </s> <s xml:id="echoid-s9326" xml:space="preserve">ſi l’on abaiſſe du point H la perpendi-<lb/>culaire H I ſur le diametre A D, elle ſera, par le lemme précé-<lb/>dent, le rayon du cercle égal à la couronne G H, puiſqu’elle <lb/>eſt moyenne proportionnelle entre les parties A I & </s> <s xml:id="echoid-s9327" xml:space="preserve">I D, ou <lb/>G H & </s> <s xml:id="echoid-s9328" xml:space="preserve">H L qui leur ſont égales. </s> <s xml:id="echoid-s9329" xml:space="preserve">Or comme les lignes H I, <lb/>G A, G K ſont égales, par conſtruction, il s’enſuit que la cou-<lb/>ronne G H ſera égale au cercle, qui auroit pour rayon la ligne <lb/>correſpondante G K, qui eſt un des élémens du triangle A B E; <lb/></s> <s xml:id="echoid-s9330" xml:space="preserve">& </s> <s xml:id="echoid-s9331" xml:space="preserve">comme le triangle eſt compoſé d’autant d’élémens qu’il y a <lb/>de couronnes dans l’eſpace qui eſt entre la demi-ſphere & </s> <s xml:id="echoid-s9332" xml:space="preserve">le <pb o="272" file="0310" n="310" rhead="NOUVEAU COURS"/> cylindre. </s> <s xml:id="echoid-s9333" xml:space="preserve">La ſomme des élémens & </s> <s xml:id="echoid-s9334" xml:space="preserve">des couronnes étant ex-<lb/>primée par la ligne B A, il s’enſuit que tous les cercles qui au-<lb/>ront pour rayons les élémens du triangle, vaudront, pris en-<lb/>ſemble, toutes les couronnes; </s> <s xml:id="echoid-s9335" xml:space="preserve">& </s> <s xml:id="echoid-s9336" xml:space="preserve">comme pour trouver la va-<lb/>leur de tous ces cercles, il faut multiplier le cercle du plus <lb/>grand élément F B par le tiers de la ligne B A (art. </s> <s xml:id="echoid-s9337" xml:space="preserve">554), il <lb/>faudra donc pour trouver la ſomme de toutes les couronnes, <lb/>multiplier la plus grande couronne B C, qui eſt le cercle qui <lb/>ſert de baſe au cylindre, par le tiers de la ligne A B, hauteur <lb/>du cylindre; </s> <s xml:id="echoid-s9338" xml:space="preserve">ce qui fait voir que toutes les couronnes, priſes <lb/>enſemble, ſont égales au tiers du cylindre, & </s> <s xml:id="echoid-s9339" xml:space="preserve">que par conſé-<lb/>quent la demi-ſphere en eſt les deux tiers. </s> <s xml:id="echoid-s9340" xml:space="preserve">C. </s> <s xml:id="echoid-s9341" xml:space="preserve">Q. </s> <s xml:id="echoid-s9342" xml:space="preserve">F. </s> <s xml:id="echoid-s9343" xml:space="preserve">D.</s> <s xml:id="echoid-s9344" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div734" type="section" level="1" n="587"> <head xml:id="echoid-head692" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9345" xml:space="preserve">567. </s> <s xml:id="echoid-s9346" xml:space="preserve">Puiſqu’une demi-ſphere eſt les deux tiers du cylindre <lb/>où elle ſeroit inſcrite, c’eſt-à-dire de même baſe & </s> <s xml:id="echoid-s9347" xml:space="preserve">de même <lb/>hauteur, il s’enſuit que pour en trouver la ſolidité, il faut mul-<lb/>tiplier ſon plus grand cercle A D par les deux tiers du rayon <lb/>M E.</s> <s xml:id="echoid-s9348" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div735" type="section" level="1" n="588"> <head xml:id="echoid-head693" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9349" xml:space="preserve">568. </s> <s xml:id="echoid-s9350" xml:space="preserve">Une demi-ſphere étant les deux tiers d’un cylindre de <lb/>même baſe & </s> <s xml:id="echoid-s9351" xml:space="preserve">de même hauteur, une ſphere ſera par conſé-<lb/>quent les deux tiers du cylindre, qui auroit pour baſe le grand <lb/>cercle de la ſphere, & </s> <s xml:id="echoid-s9352" xml:space="preserve">pour hauteur le diametre: </s> <s xml:id="echoid-s9353" xml:space="preserve">ainſi il faut, <lb/>pour trouver la ſolidité d’une ſphere, multiplier ſon grand cercle <lb/>par les deux tiers du diametre, ou bien multiplier le grand cer-<lb/>cle par le diametre, & </s> <s xml:id="echoid-s9354" xml:space="preserve">prendre les deux tiers du produit.</s> <s xml:id="echoid-s9355" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div736" type="section" level="1" n="589"> <head xml:id="echoid-head694" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s9356" xml:space="preserve">569. </s> <s xml:id="echoid-s9357" xml:space="preserve">Si l’on conſidere qu’un quart de cercle eſt compoſé <lb/>d’un nombre infini d’élémens, tels que D E, on verra que ſi <lb/> <anchor type="note" xlink:label="note-0310-01a" xlink:href="note-0310-01"/> le quart de cercle fait une révolution autour du rayon A B, il <lb/>décrira une demi-ſphere telle que X, qui ſera compoſée d’une <lb/> <anchor type="note" xlink:label="note-0310-02a" xlink:href="note-0310-02"/> infinité de cercles, dont tous les élémens du quart de cercle <lb/>ſeront les rayons. </s> <s xml:id="echoid-s9358" xml:space="preserve">Or comme les cercles ſont dans la même <lb/>raiſon que les quarrés de leurs rayons, & </s> <s xml:id="echoid-s9359" xml:space="preserve">que pour trouver la <lb/>valeur de tous les cercles, qui ont pour rayon les élémens du <lb/>quart de cercle, il faut multiplier le cercle du plus grand rayon <lb/>B C par les deux tiers du demi-diametre A B, il ſuit delà, que <pb o="273" file="0311" n="311" rhead="DE MATHÉMATIQUE. Liv. VIII."/> pour trouver tous les quarrés des élémens du quart de cercle <lb/>A C, il faut multiplier le quarré du plus grand élément par <lb/>les deux tiers de la ligne A B, & </s> <s xml:id="echoid-s9360" xml:space="preserve">l’on peut tirer de ce raiſon-<lb/>nement le principe général ſuivant, qui eſt que, dans une ſuite <lb/>qui ſeroit compoſee des élémens infinis du quart de cercle, la ſomme <lb/>de tous les élémens ſeroit égale au produit du quarré du plus grand <lb/>élément, c’eſt-à-dire du rayon par les deux tiers du même rayon.</s> <s xml:id="echoid-s9361" xml:space="preserve"/> </p> <div xml:id="echoid-div736" type="float" level="2" n="1"> <note position="left" xlink:label="note-0310-01" xlink:href="note-0310-01a" xml:space="preserve">Figure 139.</note> <note position="left" xlink:label="note-0310-02" xlink:href="note-0310-02a" xml:space="preserve">Figure 142.</note> </div> </div> <div xml:id="echoid-div738" type="section" level="1" n="590"> <head xml:id="echoid-head695" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9362" xml:space="preserve">570. </s> <s xml:id="echoid-s9363" xml:space="preserve">Les ſolidités des ſpheres ſont dans la même raiſon que les <lb/> <anchor type="note" xlink:label="note-0311-01a" xlink:href="note-0311-01"/> cubes de leurs diametres.</s> <s xml:id="echoid-s9364" xml:space="preserve"/> </p> <div xml:id="echoid-div738" type="float" level="2" n="1"> <note position="right" xlink:label="note-0311-01" xlink:href="note-0311-01a" xml:space="preserve">Figure 143.</note> </div> <p> <s xml:id="echoid-s9365" xml:space="preserve">Si l’on nomme le diametre A B, a, ſa circonférence, b, <lb/>le diametre C D, c, & </s> <s xml:id="echoid-s9366" xml:space="preserve">ſa circonférence, d, la ſuperficie du <lb/>grand cercle de la premiere ſphere ſera {ab/4}, puiſqu’il faut mul-<lb/>tiplier la demi-circonférence par le rayon pour avoir la ſur-<lb/>face d’un cercle; </s> <s xml:id="echoid-s9367" xml:space="preserve">de même la ſuperficie du grand cercle de la <lb/>ſeconde ſphere ſera {cd/4} multipliant enſuite l’un & </s> <s xml:id="echoid-s9368" xml:space="preserve">l’autre, cha-<lb/>cun par les deux tiers de ſon diametre, l’on aura {2a<emph style="sub">2</emph>b/12} ou {a<emph style="sub">2</emph>b/6} <lb/>pour la ſolidité de la premiere ſphere (art. </s> <s xml:id="echoid-s9369" xml:space="preserve">568), & </s> <s xml:id="echoid-s9370" xml:space="preserve">par la <lb/>même raiſon {cd/6} pour la ſolidité de la ſeconde ſphere: </s> <s xml:id="echoid-s9371" xml:space="preserve">il faut <lb/>donc démontrer que {aab/6}: </s> <s xml:id="echoid-s9372" xml:space="preserve">{ccd/6}:</s> <s xml:id="echoid-s9373" xml:space="preserve">: a<emph style="sub">3</emph>: </s> <s xml:id="echoid-s9374" xml:space="preserve">c<emph style="sub">3</emph>.</s> <s xml:id="echoid-s9375" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div740" type="section" level="1" n="591"> <head xml:id="echoid-head696" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9376" xml:space="preserve">Pour prouver que {aab/6}: </s> <s xml:id="echoid-s9377" xml:space="preserve">{ccd/6}:</s> <s xml:id="echoid-s9378" xml:space="preserve">: a<emph style="sub">3</emph>: </s> <s xml:id="echoid-s9379" xml:space="preserve">c<emph style="sub">3</emph>, nous ferons voir que <lb/>dans ces quatre termes le produit des extrêmes eſt égal à celui <lb/>des moyens, c’eſt-à-dire que {aabc<emph style="sub">3</emph>/6} = {a<emph style="sub">3</emph>dcc/6}. </s> <s xml:id="echoid-s9380" xml:space="preserve">Pour cela, con-<lb/>ſidérez que les diametres des cercles étant en même raiſon <lb/>que leurs circonférences (art. </s> <s xml:id="echoid-s9381" xml:space="preserve">481), on aura a: </s> <s xml:id="echoid-s9382" xml:space="preserve">b:</s> <s xml:id="echoid-s9383" xml:space="preserve">: c: </s> <s xml:id="echoid-s9384" xml:space="preserve">d, d’où <lb/>l’on tire a d = b c, & </s> <s xml:id="echoid-s9385" xml:space="preserve">que ſi l’on met a d à la place de b c dans <lb/>le premier membre de l’équation précédente, elle deviendra, en <lb/>multipliant chaque membre par 6, aaadcc = aaadcc. </s> <s xml:id="echoid-s9386" xml:space="preserve">C. </s> <s xml:id="echoid-s9387" xml:space="preserve">Q. </s> <s xml:id="echoid-s9388" xml:space="preserve">F. </s> <s xml:id="echoid-s9389" xml:space="preserve">D.</s> <s xml:id="echoid-s9390" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div741" type="section" level="1" n="592"> <head xml:id="echoid-head697" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s9391" xml:space="preserve">571. </s> <s xml:id="echoid-s9392" xml:space="preserve">On appelle corps ou ſolides ſemblables ceux dont <pb o="274" file="0312" n="312" rhead="NOUVEAU COURS"/> toutes les dimenſions ſont proportionnelles, par exemple, <lb/>deux pyramides ſont ſemblables, lorſqu’elles ont chacune <lb/>pour baſes des polygones ſemblables, & </s> <s xml:id="echoid-s9393" xml:space="preserve">que leurs axes ſont <lb/>diſpoſés de la même maniere par rapport au plan de leur baſe, <lb/>& </s> <s xml:id="echoid-s9394" xml:space="preserve">ſont proportionnels aux côtés homologues, ou aux rayons <lb/>de ces polygones: </s> <s xml:id="echoid-s9395" xml:space="preserve">car il faut bien faire attention que les axes <lb/>de deux pyramides, ou même leurs hauteurs, peuvent être pro-<lb/>portionnelles à leurs rayons, ou aux côtés homologues des baſes <lb/>ſemblables, ſans que ces pyramides ſoient des corps ſembla-<lb/>bles; </s> <s xml:id="echoid-s9396" xml:space="preserve">ce qui arriveroit ſi l’une des pyramides étoit droite & </s> <s xml:id="echoid-s9397" xml:space="preserve">l’au-<lb/>tre oblique.</s> <s xml:id="echoid-s9398" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div742" type="section" level="1" n="593"> <head xml:id="echoid-head698" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9399" xml:space="preserve">572. </s> <s xml:id="echoid-s9400" xml:space="preserve">Il ſuit de la définition précédente & </s> <s xml:id="echoid-s9401" xml:space="preserve">de la derniere <lb/>propoſition, que toutes les pyramides, priſmes, cylindres, ou <lb/>cônes ſemblables, ſeront entr’eux comme les cubes des dimen-<lb/>ſions homologues; </s> <s xml:id="echoid-s9402" xml:space="preserve">de leurs axes, par exemple, de leurs hauteurs, <lb/>ou, comme s’expriment les Géometres, dans la raiſon triplée <lb/>de leurs dimenſions homologues.</s> <s xml:id="echoid-s9403" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div743" type="section" level="1" n="594"> <head xml:id="echoid-head699" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s9404" xml:space="preserve">Il pourroit arriver, comme nous l’avons déja inſinué, que <lb/>deux corps qui ont des baſes ſemblables, fuſſent entr’eux com-<lb/>me les cubes de leurs hauteurs, ſans qu’on en puiſſe conclure <lb/>qu’ils ſont ſemblables. </s> <s xml:id="echoid-s9405" xml:space="preserve">Imaginons deux priſmes, qui ont cha-<lb/>cun pour baſe des pentagones ſemblables, & </s> <s xml:id="echoid-s9406" xml:space="preserve">des hauteurs <lb/>proportionnelles aux côtés homologues de ces pentagones, mais <lb/>le premier droit, & </s> <s xml:id="echoid-s9407" xml:space="preserve">le ſecond oblique. </s> <s xml:id="echoid-s9408" xml:space="preserve">Soit 2a le contour de la <lb/>baſe du premier; </s> <s xml:id="echoid-s9409" xml:space="preserve">b, la perpendiculaire qui meſure la hauteur <lb/>d’un des triangles de la baſe, & </s> <s xml:id="echoid-s9410" xml:space="preserve">c ſa hauteur: </s> <s xml:id="echoid-s9411" xml:space="preserve">ſoit de même <lb/>2d le contour du polygone qui ſert de baſe au ſecond priſme, f <lb/>la hauteur d’un triangle, & </s> <s xml:id="echoid-s9412" xml:space="preserve">g la hauteur de ce priſme. </s> <s xml:id="echoid-s9413" xml:space="preserve">La ſoli-<lb/>dité du premier ſera a b c, & </s> <s xml:id="echoid-s9414" xml:space="preserve">celle du ſecond ſera d f g, puiſ-<lb/>qu’il faut multiplier la baſe de chacun par ſa hauteur, & </s> <s xml:id="echoid-s9415" xml:space="preserve">l’on <lb/>auroit dans ce cas a b c: </s> <s xml:id="echoid-s9416" xml:space="preserve">d f g:</s> <s xml:id="echoid-s9417" xml:space="preserve">: a<emph style="sub">3</emph>: </s> <s xml:id="echoid-s9418" xml:space="preserve">d; </s> <s xml:id="echoid-s9419" xml:space="preserve">ce qu’il eſt aiſé de prou-<lb/>ver, en faiſant voir que le produit des extrêmes eſt égal à ce-<lb/>lui des moyens, ou que a b c d = d f g a<emph style="sub">3</emph>: </s> <s xml:id="echoid-s9420" xml:space="preserve">car puiſque les po-<lb/>lygones qui ſervent de baſes ſont ſemblables, leurs contours <lb/>ou les moitiés de ces contours ſont proportionnels aux per-<lb/>pendiculaires qui meſurent les hauteurs des triangles: </s> <s xml:id="echoid-s9421" xml:space="preserve">donc <pb o="275" file="0313" n="313" rhead="DE MATHÉMATIQUE. Liv. VIII."/> a: </s> <s xml:id="echoid-s9422" xml:space="preserve">b :</s> <s xml:id="echoid-s9423" xml:space="preserve">: d: </s> <s xml:id="echoid-s9424" xml:space="preserve">f; </s> <s xml:id="echoid-s9425" xml:space="preserve">donc a f = b d, & </s> <s xml:id="echoid-s9426" xml:space="preserve">puiſque, par hypotheſe, les <lb/>hauteurs de ces priſmes ſont proportionnelles aux circuits des <lb/>baſes, on aura a: </s> <s xml:id="echoid-s9427" xml:space="preserve">c :</s> <s xml:id="echoid-s9428" xml:space="preserve">: d: </s> <s xml:id="echoid-s9429" xml:space="preserve">g; </s> <s xml:id="echoid-s9430" xml:space="preserve">donc a g=c d. </s> <s xml:id="echoid-s9431" xml:space="preserve">Si dans le premier <lb/>membre de l’équation, qu’il faut prouver, on met a f à la place <lb/>de bd, & </s> <s xml:id="echoid-s9432" xml:space="preserve">ag à la place de cd, il viendra celle-ci, a<emph style="sub">3</emph>d f g = a<emph style="sub">3</emph>d f g, <lb/>qui fait voir que ces priſmes ſont entr’eux comme les cubes <lb/>des côtés de leurs baſes ou de leurs rayons, quoiqu’ils ne <lb/>ſoient pas ſemblables. </s> <s xml:id="echoid-s9433" xml:space="preserve">Il eſt donc vrai de dire que lorſque deux <lb/>ſolides ſont ſemblables, ils ſont entr’eux comme les cubes des <lb/>côtés homologues de leurs baſes, ou comme les cubes de leurs <lb/>hauteurs; </s> <s xml:id="echoid-s9434" xml:space="preserve">mais de ce que deux ſolides ſeroient entr’eux com-<lb/>me les cubes de leurs côtés homologues ou de leurs hauteurs, <lb/>il ne s’enſuit pas qu’ils ſoient ſemblables.</s> <s xml:id="echoid-s9435" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9436" xml:space="preserve">On a ſuppoſé dans cette remarque & </s> <s xml:id="echoid-s9437" xml:space="preserve">dans ce qui pré-<lb/>cede, qu’un priſme oblique eſt égal au produit de ſa baſe par <lb/>ſa hauteur; </s> <s xml:id="echoid-s9438" xml:space="preserve">ou, ce qui revient au même, que deux priſmes <lb/>ſont égaux, lorſqu’ils ſont compris entre deux plans paral-<lb/>leles: </s> <s xml:id="echoid-s9439" xml:space="preserve">ſi l’on veut ſe convaincre de cette vérité, il n’y a qu’à <lb/>faire attention qu’un priſme peut être engendré par le mouve-<lb/>ment d’un parallélogramme qui ſe meut parallélement à lui-<lb/>même, & </s> <s xml:id="echoid-s9440" xml:space="preserve">comme les parallélogrammes inclinés ſont égaux <lb/>au rectangle de même baſe, & </s> <s xml:id="echoid-s9441" xml:space="preserve">compris entre les mêmes pa-<lb/>ralleles, il s’enſuit que les priſmes droits & </s> <s xml:id="echoid-s9442" xml:space="preserve">obliques, engen-<lb/>drés par les mouvemens de ces ſurfaces, ſeront auſſi égaux, <lb/>puiſque les ſurfaces génératrices ſont égales, & </s> <s xml:id="echoid-s9443" xml:space="preserve">parcourent le <lb/>même eſpace parallélement à elles-mêmes.</s> <s xml:id="echoid-s9444" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div744" type="section" level="1" n="595"> <head xml:id="echoid-head700" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9445" xml:space="preserve">573. </s> <s xml:id="echoid-s9446" xml:space="preserve">La ſurface d’une demi-ſphere A E D eſt égale à celle du <lb/> <anchor type="note" xlink:label="note-0313-01a" xlink:href="note-0313-01"/> cylindre A B C D, dans lequel elle eſt inſcrite.</s> <s xml:id="echoid-s9447" xml:space="preserve"/> </p> <div xml:id="echoid-div744" type="float" level="2" n="1"> <note position="right" xlink:label="note-0313-01" xlink:href="note-0313-01a" xml:space="preserve">Figure 140 <lb/>& 141.</note> </div> <p> <s xml:id="echoid-s9448" xml:space="preserve">Suppoſant que le cylindre A C & </s> <s xml:id="echoid-s9449" xml:space="preserve">le cône G H I ont la même <lb/>baſe & </s> <s xml:id="echoid-s9450" xml:space="preserve">la même hauteur, nous nommerons a les lignes égales <lb/>F E, F D, K H, K I, & </s> <s xml:id="echoid-s9451" xml:space="preserve">b les circonférences A D & </s> <s xml:id="echoid-s9452" xml:space="preserve">G I. </s> <s xml:id="echoid-s9453" xml:space="preserve">Cela <lb/>poſé, on aura {a b/2} pour la valeur du cercle A D ou G I, qui <lb/>étant multiplié par les deux tiers de F E ({2a/3}) donnera {2aab/6} <lb/>= {aab/3} pour la valeur de la demi-ſphere (art. </s> <s xml:id="echoid-s9454" xml:space="preserve">567 & </s> <s xml:id="echoid-s9455" xml:space="preserve">568), &</s> <s xml:id="echoid-s9456" xml:space="preserve"> <pb o="276" file="0314" n="314" rhead="NOUVEAU COURS"/> multipliant {ab/2} par le tiers de H K ({a/3}), il viendra {aab/6} pour la <lb/>ſolidité du cône G H I</s> </p> </div> <div xml:id="echoid-div746" type="section" level="1" n="596"> <head xml:id="echoid-head701" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9457" xml:space="preserve">Si l’on imagine la demi-ſphere, comme étant compoſée <lb/>d’une infinité de petits cônes, qui ont leurs baſes égales, ré-<lb/>pandues ſur la ſurface de la ſphere, & </s> <s xml:id="echoid-s9458" xml:space="preserve">dont tous les fommets <lb/>venant aboutir au centre F, ont pour hauteur commune le <lb/>rayon, on pourra dire que tous ces petits cônes ſont égaux, <lb/>pris enſemble, à un ſeul qui auroit pour baſe la ſurface de la <lb/>ſphere, & </s> <s xml:id="echoid-s9459" xml:space="preserve">pour hauteur le rayon. </s> <s xml:id="echoid-s9460" xml:space="preserve">Or comme la valeur de <lb/>ce cône, égal à la demi-ſphere, eſt {aab/3}, & </s> <s xml:id="echoid-s9461" xml:space="preserve">que celle du cône <lb/>G H I eſt {aab/6}, ces deux cônes ayant la même hauteur, il s’en-<lb/>ſuit qu’ils ſeront dans la raiſon des baſes, c’eſt-à-dire comme <lb/>le cercle G I eſt à la ſurface de la ſphere, que l’on trouvera, <lb/>en diſant: </s> <s xml:id="echoid-s9462" xml:space="preserve">Comme {aab/6}, valeur du cône G H I, eſt à {aab/3}, valeur <lb/>du cône égal à la ſphere, ainſi {ab/2}, baſe du cône G H I, eſt à <lb/>la baſe du ſecond cône, ou autrement à la ſurface de la demi-<lb/>ſphere, que l’on trouvera {6a<emph style="sub">3</emph>b<emph style="sub">2</emph>/6a<emph style="sub">2</emph>b} = a b, qui eſt un rectangle égal <lb/>à la ſurface du cylindre, puiſqu’il eſt compris ſous la hauteur a <lb/>& </s> <s xml:id="echoid-s9463" xml:space="preserve">la circonférence b. </s> <s xml:id="echoid-s9464" xml:space="preserve">C. </s> <s xml:id="echoid-s9465" xml:space="preserve">Q. </s> <s xml:id="echoid-s9466" xml:space="preserve">F. </s> <s xml:id="echoid-s9467" xml:space="preserve">D.</s> <s xml:id="echoid-s9468" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div747" type="section" level="1" n="597"> <head xml:id="echoid-head702" xml:space="preserve"><emph style="sc">Autre demonstration</emph>.</head> <p> <s xml:id="echoid-s9469" xml:space="preserve">Conſidérez que ſi du cylindre A C l’on retranche le cône <lb/>B F C, qui en eſt le tiers, le ſolide A B F C D qui reſtera, que <lb/> <anchor type="note" xlink:label="note-0314-01a" xlink:href="note-0314-01"/> nous nommerons entonnoir, en ſera les deux tiers; </s> <s xml:id="echoid-s9470" xml:space="preserve">& </s> <s xml:id="echoid-s9471" xml:space="preserve">comme <lb/>la demi-ſphere inſcrite eſt auſſi les deux tiers du cylindre, elle <lb/>ſera par conſéquent égale à l’entonnoir. </s> <s xml:id="echoid-s9472" xml:space="preserve">Mais ſi l’on imagine <lb/>l’entonnoir compoſé d’une infinité de petites pyramides, dont <lb/>toutes les baſes ſont à la ſurface du cylindre, & </s> <s xml:id="echoid-s9473" xml:space="preserve">dont la hau-<lb/>teur commune eſt le rayon F D, il s’enſuit que toutes les pyra-<lb/>mides de la demi-ſphere étant égales à toutes celles de l’en-<lb/>tonnoir, toutes les baſes des unes, priſes enſemble, ſeront <lb/>égales à toutes les baſes des autres, auſſi priſes enſemble, puiſ-<lb/>que ces pyramides ont la même hauteur; </s> <s xml:id="echoid-s9474" xml:space="preserve">mais toutes les baſes <lb/>des unes valent la ſurface de la ſphere, & </s> <s xml:id="echoid-s9475" xml:space="preserve">toutes les baſes des <pb o="277" file="0315" n="315" rhead="DE MATHÉMATIQUE. Liv. VIII."/> autres valent la ſurface du cylindre: </s> <s xml:id="echoid-s9476" xml:space="preserve">donc la ſurface de la <lb/>ſphere eſt égale à la ſurface du cylindre qui lui eſt circonſcrit. <lb/></s> <s xml:id="echoid-s9477" xml:space="preserve">C. </s> <s xml:id="echoid-s9478" xml:space="preserve">Q. </s> <s xml:id="echoid-s9479" xml:space="preserve">F. </s> <s xml:id="echoid-s9480" xml:space="preserve">D.</s> <s xml:id="echoid-s9481" xml:space="preserve"/> </p> <div xml:id="echoid-div747" type="float" level="2" n="1"> <note position="left" xlink:label="note-0314-01" xlink:href="note-0314-01a" xml:space="preserve">Figure 140.</note> </div> </div> <div xml:id="echoid-div749" type="section" level="1" n="598"> <head xml:id="echoid-head703" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9482" xml:space="preserve">574. </s> <s xml:id="echoid-s9483" xml:space="preserve">La ſurface du cylindre A C ayant pour baſe la circon-<lb/>férence du grand cercle de la ſphere, & </s> <s xml:id="echoid-s9484" xml:space="preserve">pour hauteur le rayon, <lb/>il s’enſuit que la ſurface d’une demi-ſphere eſt égale au rectan-<lb/>gle compris ſous une ligne droite égale à la circonférence de <lb/>ſon grand cercle, & </s> <s xml:id="echoid-s9485" xml:space="preserve">ſous le rayon; </s> <s xml:id="echoid-s9486" xml:space="preserve">& </s> <s xml:id="echoid-s9487" xml:space="preserve">que par conſéquent la <lb/>ſurface d’une ſphere eſt égale au rectangle compris ſous une <lb/>ligne égale à la circonférence de ſon grand cercle & </s> <s xml:id="echoid-s9488" xml:space="preserve">ſous ſon <lb/>axe: </s> <s xml:id="echoid-s9489" xml:space="preserve">ainſi pour trouver la ſurface d’une ſphere, il faut mul-<lb/>tiplier le diametre de ſon grand cercle par ſa circonférence.</s> <s xml:id="echoid-s9490" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div750" type="section" level="1" n="599"> <head xml:id="echoid-head704" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9491" xml:space="preserve">575. </s> <s xml:id="echoid-s9492" xml:space="preserve">Le grand cercle d’une demi-ſphere étant la moitié du <lb/>rectangle compris ſous la circonférence & </s> <s xml:id="echoid-s9493" xml:space="preserve">ſous le rayon, il <lb/>s’enſuit que la ſurface d’une demi-ſphere eſt double de ſon <lb/>grand cercle; </s> <s xml:id="echoid-s9494" xml:space="preserve">& </s> <s xml:id="echoid-s9495" xml:space="preserve">par conſéquent la ſurface de la ſphere entiere <lb/>eſt quadruple de celle du même grand cercle.</s> <s xml:id="echoid-s9496" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div751" type="section" level="1" n="600"> <head xml:id="echoid-head705" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s9497" xml:space="preserve">576. </s> <s xml:id="echoid-s9498" xml:space="preserve">Comme les cercles ſont dans la même raiſon que les <lb/>quarrés de leurs rayons (art. </s> <s xml:id="echoid-s9499" xml:space="preserve">495), il s’enſuit qu’un cercle qui <lb/>aura un rayon double d’un autre, aura une ſurface quadruple: <lb/></s> <s xml:id="echoid-s9500" xml:space="preserve">par conſéquent la ſurface d’une ſphere eſt égale à celle d’un <lb/>cercle, qui auroit pour rayon l’axe de la même ſphere.</s> <s xml:id="echoid-s9501" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div752" type="section" level="1" n="601"> <head xml:id="echoid-head706" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s9502" xml:space="preserve">577. </s> <s xml:id="echoid-s9503" xml:space="preserve">Comme les ſurfaces de ſpheres ſont égales à des cer-<lb/>cles qui auroient pour rayons les diametres des ſpheres, & </s> <s xml:id="echoid-s9504" xml:space="preserve"><lb/>ces cercles étant comme les quarrés de leurs rayons, qui ſont <lb/>ici les diametres des ſpheres, il s’enſuit que les ſurfaces des <lb/>ſpheres ſont entr’elles comme les quarrés de leurs diametres.</s> <s xml:id="echoid-s9505" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div753" type="section" level="1" n="602"> <head xml:id="echoid-head707" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9506" xml:space="preserve">578. </s> <s xml:id="echoid-s9507" xml:space="preserve">La ſolidité d’une zone A B C D eſt égale aux deux tiers <lb/> <anchor type="note" xlink:label="note-0315-01a" xlink:href="note-0315-01"/> du cylindre A E F D du grand cercle A D, plus au tiers du cylin-<lb/>dre G B C H du plus petit cercle B C.</s> <s xml:id="echoid-s9508" xml:space="preserve"/> </p> <div xml:id="echoid-div753" type="float" level="2" n="1"> <note position="right" xlink:label="note-0315-01" xlink:href="note-0315-01a" xml:space="preserve">Figure 144.</note> </div> <pb o="278" file="0316" n="316" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div755" type="section" level="1" n="603"> <head xml:id="echoid-head708" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s9509" xml:space="preserve">Comme l’on trouve la valeur de toutes les couronnes qui <lb/>ſont entre la zone & </s> <s xml:id="echoid-s9510" xml:space="preserve">le cylindre A E F D, en multipliant la <lb/>plus grande couronne E B par le tiers de la ligne E A ou O I <lb/>(art. </s> <s xml:id="echoid-s9511" xml:space="preserve">566), il s’enſuit que ce produit eſt égal au tiers de l’eſ-<lb/>pace E G ou F H qui regne entre les deux cylindres A E F D <lb/>G B C H; </s> <s xml:id="echoid-s9512" xml:space="preserve">& </s> <s xml:id="echoid-s9513" xml:space="preserve">que par conſéquent la partie A B G de la zone <lb/>qui regne autour du cylindre en eſt les deux tiers. </s> <s xml:id="echoid-s9514" xml:space="preserve">Or ſi l’on <lb/>retranche de ce cylindre le cône B I C, qui en eſt le tiers, il <lb/>reſtera l’entonnoir G B I C H, qui en ſera les deux tiers, ainſi <lb/>la partie A B I C D de la zone vaudra les deux tiers du cylin-<lb/>dre A E F D; </s> <s xml:id="echoid-s9515" xml:space="preserve">mais comme le cône B I C, qui fait auſſi partie <lb/>de la zone, eſt le tiers du cylindre G B C H, il faut ajouter <lb/>ce cône aux deux tiers du cylindre A E F D pour avoir la ſo-<lb/>lidité de la zone: </s> <s xml:id="echoid-s9516" xml:space="preserve">ainſi cette ſolidité eſt égale aux deux tiers <lb/>du cylindre A E F D, plus au tiers du cylindre G B C H. <lb/></s> <s xml:id="echoid-s9517" xml:space="preserve">C. </s> <s xml:id="echoid-s9518" xml:space="preserve">Q. </s> <s xml:id="echoid-s9519" xml:space="preserve">F. </s> <s xml:id="echoid-s9520" xml:space="preserve">D.</s> <s xml:id="echoid-s9521" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div756" type="section" level="1" n="604"> <head xml:id="echoid-head709" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9522" xml:space="preserve">579. </s> <s xml:id="echoid-s9523" xml:space="preserve">Il ſuit de cette propoſition, que ſi l’on coupe une demi-<lb/> <anchor type="note" xlink:label="note-0316-01a" xlink:href="note-0316-01"/> ſphere inſcrite dans un cylindre, par un plan F G, parallele <lb/>à la baſe A E, la partie A B C D E (qui eſt la différence de <lb/>la demi-ſphere au ſecteur ſphérique C B H D) eſt égale à <lb/>l’entonnoir A F C G E du cylindre correſpondant A G, puiſ-<lb/>que l’une & </s> <s xml:id="echoid-s9524" xml:space="preserve">l’autre ſont les deux tiers du même cylindre A G.</s> <s xml:id="echoid-s9525" xml:space="preserve"/> </p> <div xml:id="echoid-div756" type="float" level="2" n="1"> <note position="left" xlink:label="note-0316-01" xlink:href="note-0316-01a" xml:space="preserve">Figure 145.</note> </div> </div> <div xml:id="echoid-div758" type="section" level="1" n="605"> <head xml:id="echoid-head710" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9526" xml:space="preserve">580. </s> <s xml:id="echoid-s9527" xml:space="preserve">Il ſuit encore delà que la ſolidité d’un ſecteur ſphé-<lb/>rique tel que C I B P, eſt égale aux deux tiers du cylindre <lb/> <anchor type="note" xlink:label="note-0316-02a" xlink:href="note-0316-02"/> E F L K, qui a pour baſe le grand cercle de la ſphere, & </s> <s xml:id="echoid-s9528" xml:space="preserve">pour <lb/>hauteur la fleche P O du ſegment ſphérique B P C, plus au <lb/>tiers du cylindre G B C H: </s> <s xml:id="echoid-s9529" xml:space="preserve">car puiſque la demi-ſphere eſt les <lb/>deux tiers du cylindre qui lui eſt circonſcrit, & </s> <s xml:id="echoid-s9530" xml:space="preserve">que la zone <lb/>A B C D eſt les deux tiers du cylindre A E F D, plus le tiers <lb/>du cylindre G B C H, il faut que le ſecteur C I A P ſoit les deux <lb/>tiers du cylindre E K L F, plus le tiers du cylindre G B C H.</s> <s xml:id="echoid-s9531" xml:space="preserve"/> </p> <div xml:id="echoid-div758" type="float" level="2" n="1"> <note position="left" xlink:label="note-0316-02" xlink:href="note-0316-02a" xml:space="preserve">Figure 144.</note> </div> </div> <div xml:id="echoid-div760" type="section" level="1" n="606"> <head xml:id="echoid-head711" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s9532" xml:space="preserve">581. </s> <s xml:id="echoid-s9533" xml:space="preserve">Il ſuit encore de cette propoſition, que le ſegment ſphé- <pb o="279" file="0317" n="317" rhead="DE MATHÉMATIQUE. Liv. VIII."/> rique B P C eſt égal aux deux tiers du cylindre E K L F, moinsle <lb/>tiers du cylindre G B C H: </s> <s xml:id="echoid-s9534" xml:space="preserve">car la demi-ſphere entiere étant les <lb/>deux tiers du cylindre A K L D, ſera auſſi les deux tiers des cy-<lb/>lindres A E F D & </s> <s xml:id="echoid-s9535" xml:space="preserve">E K L F, dont la ſomme eſt égale au cylin-<lb/>dre circonſcrit; </s> <s xml:id="echoid-s9536" xml:space="preserve">mais la zone eſt égale aux deux tiers du cy-<lb/>lindre A E F D, plus au tiers du cylindre G B C H: </s> <s xml:id="echoid-s9537" xml:space="preserve">donc en <lb/>ôtant la zone de la demi-ſphere, on aura pour le ſolide de la <lb/>calotte deux tiers du cylindre E K L F, moins le tiers du cy-<lb/>lindre G B C H; </s> <s xml:id="echoid-s9538" xml:space="preserve">d’où il ſuit que le ſolide d’une calotte ſphé-<lb/>rique eſt les deux tiers d’un cylindre qui auroit pour baſe le <lb/>grand cercle de la ſphere, & </s> <s xml:id="echoid-s9539" xml:space="preserve">pour hauteur, la fleche P O de <lb/>la calotte, moins un cône, qui auroit pour baſe le cercle ou la <lb/>baſe de la calotte, & </s> <s xml:id="echoid-s9540" xml:space="preserve">pour hauteur le rayon I P, moins la <lb/>fleche P O.</s> <s xml:id="echoid-s9541" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div761" type="section" level="1" n="607"> <head xml:id="echoid-head712" xml:space="preserve">PROPOSITION XII.</head> <head xml:id="echoid-head713" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9542" xml:space="preserve">582. </s> <s xml:id="echoid-s9543" xml:space="preserve">Si l’on coupe une demi-ſphere inſcrite dans un cylindre <lb/> <anchor type="note" xlink:label="note-0317-01a" xlink:href="note-0317-01"/> par un plan F G parallele à la baſe A E, je dis que la ſurface de <lb/>la zone A B D E eſt égale à celle du cylindre correſpondant A G.</s> <s xml:id="echoid-s9544" xml:space="preserve"/> </p> <div xml:id="echoid-div761" type="float" level="2" n="1"> <note position="right" xlink:label="note-0317-01" xlink:href="note-0317-01a" xml:space="preserve">Figure 145.</note> </div> </div> <div xml:id="echoid-div763" type="section" level="1" n="608"> <head xml:id="echoid-head714" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9545" xml:space="preserve">L’entonnoir A F C G E étant égal à la partie A B C D E <lb/> <anchor type="note" xlink:label="note-0317-02a" xlink:href="note-0317-02"/> de la zone (art. </s> <s xml:id="echoid-s9546" xml:space="preserve">579), ſi l’on imagine l’entonnoir compoſé <lb/>d’une infinité de petites pyramides qui ont toutes leurs baſes <lb/>dans la ſurface du cylindre A G, & </s> <s xml:id="echoid-s9547" xml:space="preserve">pour hauteur le rayon <lb/>C E; </s> <s xml:id="echoid-s9548" xml:space="preserve">& </s> <s xml:id="echoid-s9549" xml:space="preserve">la partie A B C D E de la demi-ſphere, comme étant <lb/>auſſi compoſée de petites pyramides, dont les baſes ſont dans <lb/>la ſurface de la zone, & </s> <s xml:id="echoid-s9550" xml:space="preserve">qui ont pour hauteur commune le <lb/>rayon C E, il s’enſuivra (toutes les pyramides d’une part étant <lb/>égales à toutes celles de l’autre, & </s> <s xml:id="echoid-s9551" xml:space="preserve">ayant toutes la même hau-<lb/>teur) que néceſſairement toutes les baſes d’une part ſeront <lb/>égales à toutes les baſes de l’autre, & </s> <s xml:id="echoid-s9552" xml:space="preserve">qu’ainſi la ſurface de la <lb/>zone A B D E ſera égale à celle du cylindre A F G E. </s> <s xml:id="echoid-s9553" xml:space="preserve">C. </s> <s xml:id="echoid-s9554" xml:space="preserve">Q. </s> <s xml:id="echoid-s9555" xml:space="preserve">F. </s> <s xml:id="echoid-s9556" xml:space="preserve">D.</s> <s xml:id="echoid-s9557" xml:space="preserve"/> </p> <div xml:id="echoid-div763" type="float" level="2" n="1"> <note position="right" xlink:label="note-0317-02" xlink:href="note-0317-02a" xml:space="preserve">Figure 145.</note> </div> </div> <div xml:id="echoid-div765" type="section" level="1" n="609"> <head xml:id="echoid-head715" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9558" xml:space="preserve">583. </s> <s xml:id="echoid-s9559" xml:space="preserve">Comme la ſurface de la demi-ſphere A H E eſt égale <lb/>à celle du cylindre A I, & </s> <s xml:id="echoid-s9560" xml:space="preserve">que la ſurface de la zone A B D E <lb/>eſt égale à celle du cylindre A G, il s’enſuit que la ſurface du <lb/>ſegment B H D de la ſphere, eſt égale à celle du cylindre cor- <pb o="280" file="0318" n="318" rhead="NOUVEAU COURS"/> reſpondant F I, ou bien au rectangle compris ſous une ligne <lb/>égale à la circonférence du grand cercle de la ſphere, & </s> <s xml:id="echoid-s9561" xml:space="preserve">ſous <lb/>la partie H K.</s> <s xml:id="echoid-s9562" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div766" type="section" level="1" n="610"> <head xml:id="echoid-head716" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9563" xml:space="preserve">584. </s> <s xml:id="echoid-s9564" xml:space="preserve">Il ſuit encore de cette propoſition, que ſi l’on coupe <lb/>une demi-ſphere inſcrite dans un cylindre par un plan parallele <lb/>à la baſe, les parties de la ſurface de la demi-ſphere ſeront <lb/>égales aux zones correſpondantes du cylindre.</s> <s xml:id="echoid-s9565" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div767" type="section" level="1" n="611"> <head xml:id="echoid-head717" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s9566" xml:space="preserve">585. </s> <s xml:id="echoid-s9567" xml:space="preserve">Les ſurfaces des cylindres F I & </s> <s xml:id="echoid-s9568" xml:space="preserve">A G ayant des baſes <lb/>égales, ſeront dans la même raiſon que leurs hauteurs H K <lb/>& </s> <s xml:id="echoid-s9569" xml:space="preserve">K C; </s> <s xml:id="echoid-s9570" xml:space="preserve">& </s> <s xml:id="echoid-s9571" xml:space="preserve">comme le premier cylindre eſt égal à la partie de <lb/>la ſurface B H D de la demi-ſphere, & </s> <s xml:id="echoid-s9572" xml:space="preserve">le ſecond à la partie <lb/>A B D E, il s’enſuit que les parties de la ſurface de la demi-<lb/>ſphere ſont dans la même raiſon que les parties H K & </s> <s xml:id="echoid-s9573" xml:space="preserve">K C <lb/>du demi-diametre, la demi-ſphere étant coupée par un plan <lb/>B D parallele à ſon grand cercle.</s> <s xml:id="echoid-s9574" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9575" xml:space="preserve">586. </s> <s xml:id="echoid-s9576" xml:space="preserve">L’on peut dire encore que ſi l’on coupe une ſphere par <lb/>un plan perpendiculaire à l’axe, les parties de la ſurface ſphé-<lb/>rique ſeront dans la même raiſon que les parties de l’axe.</s> <s xml:id="echoid-s9577" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div768" type="section" level="1" n="612"> <head xml:id="echoid-head718" xml:space="preserve">PROPOSITION XIII.</head> <head xml:id="echoid-head719" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9578" xml:space="preserve">587. </s> <s xml:id="echoid-s9579" xml:space="preserve">Lorſque trois lignes a, b, c ſont en proportion continue, <lb/>le parallelepipede fait ſur ces trois lignes, eſt égal au cube fait <lb/>ſur la moyenne: </s> <s xml:id="echoid-s9580" xml:space="preserve">ainſi il faut prouver que ſi l’on a, a : </s> <s xml:id="echoid-s9581" xml:space="preserve">b :</s> <s xml:id="echoid-s9582" xml:space="preserve">: b : </s> <s xml:id="echoid-s9583" xml:space="preserve">c, <lb/>on aura a b c = b b b.</s> <s xml:id="echoid-s9584" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div769" type="section" level="1" n="613"> <head xml:id="echoid-head720" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9585" xml:space="preserve">Puiſque par hypotheſe a : </s> <s xml:id="echoid-s9586" xml:space="preserve">b :</s> <s xml:id="echoid-s9587" xml:space="preserve">: b : </s> <s xml:id="echoid-s9588" xml:space="preserve">c, on aura a c = b b: </s> <s xml:id="echoid-s9589" xml:space="preserve">ainſi <lb/>en mettant dans l’équation a b c = b b b, a c à la place de b b, <lb/>on aura a b c = a b c. </s> <s xml:id="echoid-s9590" xml:space="preserve">C. </s> <s xml:id="echoid-s9591" xml:space="preserve">Q. </s> <s xml:id="echoid-s9592" xml:space="preserve">F. </s> <s xml:id="echoid-s9593" xml:space="preserve">D.</s> <s xml:id="echoid-s9594" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div770" type="section" level="1" n="614"> <head xml:id="echoid-head721" xml:space="preserve">PROPOSITION XIV.</head> <head xml:id="echoid-head722" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9595" xml:space="preserve">588. </s> <s xml:id="echoid-s9596" xml:space="preserve">Lorſque quatre lignes ſont en progreſſion géométrique, le <lb/>cube fait ſur la premiere, eſt au cube fait ſur la ſeconde, comme <pb o="281" file="0319" n="319" rhead="DE MATHÉMATIQUE. Liv. VIII."/> la premiere ligne eſt à la quatrieme, c’eſt-à-dire que ſi l’on a <lb/>{.</s> <s xml:id="echoid-s9597" xml:space="preserve">./.</s> <s xml:id="echoid-s9598" xml:space="preserve">.} a. </s> <s xml:id="echoid-s9599" xml:space="preserve">b. </s> <s xml:id="echoid-s9600" xml:space="preserve">c. </s> <s xml:id="echoid-s9601" xml:space="preserve">d, on aura auſſi a a a : </s> <s xml:id="echoid-s9602" xml:space="preserve">b b b :</s> <s xml:id="echoid-s9603" xml:space="preserve">: a : </s> <s xml:id="echoid-s9604" xml:space="preserve">d.</s> <s xml:id="echoid-s9605" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div771" type="section" level="1" n="615"> <head xml:id="echoid-head723" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9606" xml:space="preserve">Conſidérez que dans la progreſſion {.</s> <s xml:id="echoid-s9607" xml:space="preserve">./.</s> <s xml:id="echoid-s9608" xml:space="preserve">.} a. </s> <s xml:id="echoid-s9609" xml:space="preserve">b. </s> <s xml:id="echoid-s9610" xml:space="preserve">c. </s> <s xml:id="echoid-s9611" xml:space="preserve">d, les trois pre-<lb/>miers termes donnent a c = b b, puiſque l’on a a : </s> <s xml:id="echoid-s9612" xml:space="preserve">b :</s> <s xml:id="echoid-s9613" xml:space="preserve">: b : </s> <s xml:id="echoid-s9614" xml:space="preserve">c, & </s> <s xml:id="echoid-s9615" xml:space="preserve"><lb/>que l’on aura auſſi a d = b c, puiſque a : </s> <s xml:id="echoid-s9616" xml:space="preserve">b :</s> <s xml:id="echoid-s9617" xml:space="preserve">: c : </s> <s xml:id="echoid-s9618" xml:space="preserve">d. </s> <s xml:id="echoid-s9619" xml:space="preserve">Ainſi pour <lb/>prouver que a : </s> <s xml:id="echoid-s9620" xml:space="preserve">b :</s> <s xml:id="echoid-s9621" xml:space="preserve">: a : </s> <s xml:id="echoid-s9622" xml:space="preserve">d, il ſuffit de faire voir que le produit <lb/>des extrêmes & </s> <s xml:id="echoid-s9623" xml:space="preserve">celui des moyens donnent a d = ab. </s> <s xml:id="echoid-s9624" xml:space="preserve">Pour <lb/>cela, il n’y a qu’à mettre a c à la place de b b dans le ſecond <lb/>membre de l’équation, & </s> <s xml:id="echoid-s9625" xml:space="preserve">b c à la place de a d dans le premier, <lb/>& </s> <s xml:id="echoid-s9626" xml:space="preserve">l’on aura a a b c = a a b c. </s> <s xml:id="echoid-s9627" xml:space="preserve">C. </s> <s xml:id="echoid-s9628" xml:space="preserve">Q. </s> <s xml:id="echoid-s9629" xml:space="preserve">F. </s> <s xml:id="echoid-s9630" xml:space="preserve">D.</s> <s xml:id="echoid-s9631" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div772" type="section" level="1" n="616"> <head xml:id="echoid-head724" xml:space="preserve">PROPOSITION XV.</head> <head xml:id="echoid-head725" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s9632" xml:space="preserve">589. </s> <s xml:id="echoid-s9633" xml:space="preserve">Trouver deux moyennes proportionnelles entre deux lignes <lb/> <anchor type="note" xlink:label="note-0319-01a" xlink:href="note-0319-01"/> données.</s> <s xml:id="echoid-s9634" xml:space="preserve"/> </p> <div xml:id="echoid-div772" type="float" level="2" n="1"> <note position="right" xlink:label="note-0319-01" xlink:href="note-0319-01a" xml:space="preserve">Figure 136.</note> </div> </div> <div xml:id="echoid-div774" type="section" level="1" n="617"> <head xml:id="echoid-head726" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s9635" xml:space="preserve">Pour trouver deux moyennes proportionnelles entre deux <lb/>lignes données A B & </s> <s xml:id="echoid-s9636" xml:space="preserve">C D, il faut faire un rectangle <lb/>ſous les deux lignes, tel que E F ſoit égale à C D, & </s> <s xml:id="echoid-s9637" xml:space="preserve">E G <lb/>égal à A B; </s> <s xml:id="echoid-s9638" xml:space="preserve">enſuite prolonger indéfiniment les côtés E F, <lb/>E G, & </s> <s xml:id="echoid-s9639" xml:space="preserve">du centre I du rectangle, décrire un cercle de ma-<lb/>niere que la circonférence venant couper les lignes prolongées <lb/>G K & </s> <s xml:id="echoid-s9640" xml:space="preserve">F L, on puiſſe mener du point K au point L une ligne <lb/>K L, qui ne faſſe que toucher l’angle H, & </s> <s xml:id="echoid-s9641" xml:space="preserve">l’on aura les lignes <lb/>G K & </s> <s xml:id="echoid-s9642" xml:space="preserve">F L, qui ſeront moyennes proportionnelles entre G E <lb/>& </s> <s xml:id="echoid-s9643" xml:space="preserve">E F, c’eſt-à-dire entre les données A B & </s> <s xml:id="echoid-s9644" xml:space="preserve">C D.</s> <s xml:id="echoid-s9645" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div775" type="section" level="1" n="618"> <head xml:id="echoid-head727" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9646" xml:space="preserve">Conſidérez que ſi l’on abaiſſe les perpendiculaires I M & </s> <s xml:id="echoid-s9647" xml:space="preserve"><lb/>I N, la corde O L ſera diviſée en deux également au point M <lb/>(art. </s> <s xml:id="echoid-s9648" xml:space="preserve">423) auſſi-bien que la ligne E F, & </s> <s xml:id="echoid-s9649" xml:space="preserve">que par conſéquent <lb/>O E eſt égale à F L, & </s> <s xml:id="echoid-s9650" xml:space="preserve">que K P étant diviſée en deux égale-<lb/>ment au point N, auſſi-bien que G E, G K ſera égale à E P. <lb/></s> <s xml:id="echoid-s9651" xml:space="preserve">Cela poſé, comme les triangles O E P, H F L, K G H ſont <lb/>ſemblables, on aura H F : </s> <s xml:id="echoid-s9652" xml:space="preserve">F L :</s> <s xml:id="echoid-s9653" xml:space="preserve">: E O : </s> <s xml:id="echoid-s9654" xml:space="preserve">E P; </s> <s xml:id="echoid-s9655" xml:space="preserve">mais puiſque O E <lb/>eſt égal à F L, on aura H F : </s> <s xml:id="echoid-s9656" xml:space="preserve">F L : </s> <s xml:id="echoid-s9657" xml:space="preserve">: </s> <s xml:id="echoid-s9658" xml:space="preserve">F L : </s> <s xml:id="echoid-s9659" xml:space="preserve">E P; </s> <s xml:id="echoid-s9660" xml:space="preserve">& </s> <s xml:id="echoid-s9661" xml:space="preserve">comme les <lb/>deux triangles ſemblables E O P, G K H donnent encore <pb o="282" file="0320" n="320" rhead="NOUVEAU COURS"/> O E : </s> <s xml:id="echoid-s9662" xml:space="preserve">E P :</s> <s xml:id="echoid-s9663" xml:space="preserve">: G K : </s> <s xml:id="echoid-s9664" xml:space="preserve">G H, ſi à la place de E P on met G K, qui <lb/>lui eſt égal, on aura O E : </s> <s xml:id="echoid-s9665" xml:space="preserve">G K :</s> <s xml:id="echoid-s9666" xml:space="preserve">: G K : </s> <s xml:id="echoid-s9667" xml:space="preserve">G H; </s> <s xml:id="echoid-s9668" xml:space="preserve">ce qui prouve <lb/>qu’il y a même raiſon de H F à F L, que de F L à G K, & </s> <s xml:id="echoid-s9669" xml:space="preserve">que <lb/>de G K à G H, & </s> <s xml:id="echoid-s9670" xml:space="preserve">que par conſéquent les lignes F L & </s> <s xml:id="echoid-s9671" xml:space="preserve">G K <lb/>ſontmoyennes proportionnelles entre G E & </s> <s xml:id="echoid-s9672" xml:space="preserve">E F. </s> <s xml:id="echoid-s9673" xml:space="preserve">C. </s> <s xml:id="echoid-s9674" xml:space="preserve">Q. </s> <s xml:id="echoid-s9675" xml:space="preserve">F. </s> <s xml:id="echoid-s9676" xml:space="preserve">D.</s> <s xml:id="echoid-s9677" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div776" type="section" level="1" n="619"> <head xml:id="echoid-head728" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s9678" xml:space="preserve">590. </s> <s xml:id="echoid-s9679" xml:space="preserve">Le problême précédent eſt celui qu’on appelle com-<lb/>munément la duplication du cube, parce qu’il ſert à faire un <lb/>cube double d’un autre, ou qui ait avec lui une raiſon don-<lb/>née; </s> <s xml:id="echoid-s9680" xml:space="preserve">il ſeroit à ſouhaiter qu’on pût le réſoudre géométrique-<lb/>ment ſans tâtonner: </s> <s xml:id="echoid-s9681" xml:space="preserve">car on peut aiſément reconnoître dans <lb/>la conſtruction précédente, qu’il faut décrire pluſieurs cer-<lb/>cles avant d’en trouver un, dont la circonférence venant à <lb/>couper aux points K, L les lignes prolongées, l’on puiſſe tirer <lb/>la ligne K L, qui ne faſſe que toucher l’angle H; </s> <s xml:id="echoid-s9682" xml:space="preserve">il eſt vrai <lb/>qu’on peut encore le réſoudre d’une autre façon, comme on <lb/>le verra à la ſuite des ſections coniques. </s> <s xml:id="echoid-s9683" xml:space="preserve">Mais quoique la mé-<lb/>thode que nous donnerons ſoit plus géométrique que celle-ci, <lb/>elle ne laiſſe pas d’avoir ſes difficultés; </s> <s xml:id="echoid-s9684" xml:space="preserve">cependant comme on <lb/>ſe ſert plus volontiers des nombres que des lignes dans la pra-<lb/>tique, l’on va voir dans le problême ſuivant la maniere dont <lb/>on peut trouver en nombres deux grandeurs moyennes géo-<lb/>métriques entre deux nombres donnés.</s> <s xml:id="echoid-s9685" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div777" type="section" level="1" n="620"> <head xml:id="echoid-head729" xml:space="preserve">PROPOSITION XVI.</head> <head xml:id="echoid-head730" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s9686" xml:space="preserve">591. </s> <s xml:id="echoid-s9687" xml:space="preserve">Trouver entre deux nombres donnés deux moyennes pro-<lb/>portionnelles.</s> <s xml:id="echoid-s9688" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9689" xml:space="preserve">Pour trouver entre deux nombres deux moyennes propor-<lb/>tionnelles, il faut cuber le premier nombre, & </s> <s xml:id="echoid-s9690" xml:space="preserve">faire une Regle <lb/>de Trois, dont les deux premiers termes ſoient le premier & </s> <s xml:id="echoid-s9691" xml:space="preserve"><lb/>le ſecond nombre donnés, le troiſieme le cube du premier <lb/>nombre donné, & </s> <s xml:id="echoid-s9692" xml:space="preserve">le quatrieme terme étant trouvé, ſera le <lb/>cube de la premiere moyenne proportionnelle: </s> <s xml:id="echoid-s9693" xml:space="preserve">ainſi pour trou-<lb/>ver cette premiere moyenne, il faudra extraire la racine cube <lb/>du quatrieme terme. </s> <s xml:id="echoid-s9694" xml:space="preserve">Pour trouver enſuite la ſeconde moyenne, <lb/>il faudra chercher une moyenne entre cette premiere trouvée <lb/>& </s> <s xml:id="echoid-s9695" xml:space="preserve">le dernier nombre donné.</s> <s xml:id="echoid-s9696" xml:space="preserve"/> </p> <pb o="283" file="0321" n="321" rhead="DE MATHÉMATIQUE. Liv. VIII."/> <p> <s xml:id="echoid-s9697" xml:space="preserve">Ainſi pour trouver deux moyennes proportionnelles entre <lb/>2 & </s> <s xml:id="echoid-s9698" xml:space="preserve">16, je cube le premier nombre 2, qui donne 8, & </s> <s xml:id="echoid-s9699" xml:space="preserve">je fais <lb/>la proportion 2 : </s> <s xml:id="echoid-s9700" xml:space="preserve">16 : </s> <s xml:id="echoid-s9701" xml:space="preserve">: </s> <s xml:id="echoid-s9702" xml:space="preserve">8 : </s> <s xml:id="echoid-s9703" xml:space="preserve">{8 x 16/2} = 4 x 16 = 64, dont la racine <lb/>cube eſt 4, que je regarde comme la premiere de mes deux <lb/>moyennes proportionnelles; </s> <s xml:id="echoid-s9704" xml:space="preserve">pour avoir la ſeconde, je cherche <lb/>un moyen géométrique entre cette premiere 4, & </s> <s xml:id="echoid-s9705" xml:space="preserve">le ſecond <lb/>nombre donné 16, en faiſant 4 : </s> <s xml:id="echoid-s9706" xml:space="preserve">x : </s> <s xml:id="echoid-s9707" xml:space="preserve">: </s> <s xml:id="echoid-s9708" xml:space="preserve">x : </s> <s xml:id="echoid-s9709" xml:space="preserve">16, d’où je tire <lb/>xx = 64, & </s> <s xml:id="echoid-s9710" xml:space="preserve">x = 8 en prenant la racine, mes deux moyennes <lb/>ſeront donc 4 & </s> <s xml:id="echoid-s9711" xml:space="preserve">8: </s> <s xml:id="echoid-s9712" xml:space="preserve">en effet, l’on a la progreſſion {.</s> <s xml:id="echoid-s9713" xml:space="preserve">./.</s> <s xml:id="echoid-s9714" xml:space="preserve">.} 2 : </s> <s xml:id="echoid-s9715" xml:space="preserve">4 : </s> <s xml:id="echoid-s9716" xml:space="preserve">: </s> <s xml:id="echoid-s9717" xml:space="preserve">8 : </s> <s xml:id="echoid-s9718" xml:space="preserve">16.</s> <s xml:id="echoid-s9719" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9720" xml:space="preserve">Si les nombres donnés étoient tels qu’on ne pût pas dans les <lb/>opérations extraire les racines cubes & </s> <s xml:id="echoid-s9721" xml:space="preserve">quarrées avec exacti-<lb/>tude, il faudroit en ce cas ſe ſervir des décimales, ſuivant les <lb/>méthodes expliquées (art. </s> <s xml:id="echoid-s9722" xml:space="preserve">158 & </s> <s xml:id="echoid-s9723" xml:space="preserve">159), afin d’approcher le <lb/>plus près qu’il eſt poſſible des racines, & </s> <s xml:id="echoid-s9724" xml:space="preserve">d’avoir le plus exacte-<lb/>ment qu’on pourra les moyennes demandées. </s> <s xml:id="echoid-s9725" xml:space="preserve">Comme les <lb/>Commençans pourroient ne pas entendre d’eux-mêmes la rai-<lb/>ſon des opérations que nous venons d’enſeigner pour trouver <lb/>deux moyennes proportionnelles entre deux nombres donnés, <lb/>en voici la démonſtration.</s> <s xml:id="echoid-s9726" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9727" xml:space="preserve">L’on a vu (art. </s> <s xml:id="echoid-s9728" xml:space="preserve">588), que lorſque quatre lignes ſont en <lb/>progreſſion géométrique, le cube fait ſur la premiere eſt au <lb/>cube fait ſur la ſeconde, comme la premiere ligne à la qua-<lb/>trieme. </s> <s xml:id="echoid-s9729" xml:space="preserve">On peut donc dire invertendo, la premiere eſt à la ſe-<lb/>conde, comme le cube de la premiere eſt au cube de la ſeconde: <lb/></s> <s xml:id="echoid-s9730" xml:space="preserve">ainſi connoiſſant la premiere ligne & </s> <s xml:id="echoid-s9731" xml:space="preserve">la quatrieme, avec le cube <lb/>de la premiere, on a les trois premiers termes de cette Regle <lb/>de Trois: </s> <s xml:id="echoid-s9732" xml:space="preserve">donc on pourra trouver le cube de la ſeconde, dont <lb/>la racine cube ſera la même ſeconde. </s> <s xml:id="echoid-s9733" xml:space="preserve">Mais quand on a une <lb/>fois la ſeconde, on voit qu’il n’y a plus qu’à chercher une <lb/>moyenne proportionnelle entre cette ſeconde & </s> <s xml:id="echoid-s9734" xml:space="preserve">la quatrieme <lb/>(qui n’eſt autre choſe que le ſecond nombre donné), & </s> <s xml:id="echoid-s9735" xml:space="preserve">l’on <lb/>aura la troiſieme des quatre proportionnelles, qui ſera en <lb/>même-tems la ſeconde des deux inconnues que l’on cherche. </s> <s xml:id="echoid-s9736" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s9737" xml:space="preserve">Q. </s> <s xml:id="echoid-s9738" xml:space="preserve">F. </s> <s xml:id="echoid-s9739" xml:space="preserve">D.</s> <s xml:id="echoid-s9740" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div778" type="section" level="1" n="621"> <head xml:id="echoid-head731" xml:space="preserve">PROPOSITION XVII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s9741" xml:space="preserve">592. </s> <s xml:id="echoid-s9742" xml:space="preserve">Faire un cube qui ſoit à un autre dans une raiſon donnée.</s> <s xml:id="echoid-s9743" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 147. <lb/>& 148.</note> <p> <s xml:id="echoid-s9744" xml:space="preserve">Pour faire un cube qui ſoit au cube C, dans une raiſon <pb o="284" file="0322" n="322" rhead="NOUVEAU COURS"/> donnée de 2 à 3, par exemple, c’eſt-à-dire un cube qui ſoit les <lb/>deux tiers du cube C, il faut diviſer le côté A B du cube C en <lb/>trois parties égales, & </s> <s xml:id="echoid-s9745" xml:space="preserve">faire une ligne D E égale à deux de ces <lb/>parties, enſuite chercher entre A B & </s> <s xml:id="echoid-s9746" xml:space="preserve">D E deux moyennes <lb/>proportionnelles telles que F G & </s> <s xml:id="echoid-s9747" xml:space="preserve">H I, & </s> <s xml:id="echoid-s9748" xml:space="preserve">le cube qui aura pour <lb/>côté la premiere F G de ces deux moyennes proportionnelles, <lb/>ſera le cube demandé; </s> <s xml:id="echoid-s9749" xml:space="preserve">car nous allons prouver qu’il eſt les <lb/>deux tiers du cube C.</s> <s xml:id="echoid-s9750" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div779" type="section" level="1" n="622"> <head xml:id="echoid-head732" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9751" xml:space="preserve">Les quatre lignes A B, F G, H I, D E étant en proportion <lb/>continue, on aura le cube de la premiere au cube de la ſeconde, <lb/>comme la premiere à la quatrieme; </s> <s xml:id="echoid-s9752" xml:space="preserve">mais par conſtruction, la <lb/>quatrieme eſt les deux tiers de la premiere: </s> <s xml:id="echoid-s9753" xml:space="preserve">donc le cube de la <lb/>ſeconde F G eſt les deux tiers du cube C fait ſur la premiere. <lb/></s> <s xml:id="echoid-s9754" xml:space="preserve">C. </s> <s xml:id="echoid-s9755" xml:space="preserve">Q. </s> <s xml:id="echoid-s9756" xml:space="preserve">F. </s> <s xml:id="echoid-s9757" xml:space="preserve">D.</s> <s xml:id="echoid-s9758" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9759" xml:space="preserve">Si le côté du cube étoit exprimé en nombres, il faudroit de <lb/>même en prendre les deux tiers, & </s> <s xml:id="echoid-s9760" xml:space="preserve">chercher entre le tout & </s> <s xml:id="echoid-s9761" xml:space="preserve"><lb/>les deux tiers, deux moyennes proportionnelles; </s> <s xml:id="echoid-s9762" xml:space="preserve">le cube fait <lb/>ſur la premiere ſera celui que l’on demande.</s> <s xml:id="echoid-s9763" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div780" type="section" level="1" n="623"> <head xml:id="echoid-head733" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9764" xml:space="preserve">593. </s> <s xml:id="echoid-s9765" xml:space="preserve">Comme les ſpheres ſont dans la raiſon des cubes de <lb/>leurs diametres ou de leurs rayons (art. </s> <s xml:id="echoid-s9766" xml:space="preserve">570), de même que <lb/>les cylindres, les priſmes, les pyramides & </s> <s xml:id="echoid-s9767" xml:space="preserve">les cônes ſembla-<lb/>bles; </s> <s xml:id="echoid-s9768" xml:space="preserve">il s’enſuit que pour trouver quelqu’un de ces ſolides qui <lb/>ſoit à ſon ſemblable dans une raiſon donnée, il faut agir à <lb/>l’égard de leurs dimenſions homologues, des axes, par exem-<lb/>ple, comme on vient de faire à l’égard des côtés des cubes; <lb/></s> <s xml:id="echoid-s9769" xml:space="preserve">& </s> <s xml:id="echoid-s9770" xml:space="preserve">après avoir trouvé la dimenſion homologue, qui eſt ici l’axe, <lb/>l’on n’aura qu’à en faire l’axe d’un ſolide ſemblable au ſolide <lb/>propoſé, en cherchant les autres dimenſions qui ſoient toutes <lb/>proportionnelles aux dimenſions correſpondantes, & </s> <s xml:id="echoid-s9771" xml:space="preserve">dans la <lb/>raiſon de l’axe du premier à l’axe du ſecond.</s> <s xml:id="echoid-s9772" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div781" type="section" level="1" n="624"> <head xml:id="echoid-head734" xml:space="preserve">PROPOSITION XVIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s9773" xml:space="preserve">594. </s> <s xml:id="echoid-s9774" xml:space="preserve">Faire un cube égal à un parallelepipede.</s> <s xml:id="echoid-s9775" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 149 <lb/>& 150.</note> <p> <s xml:id="echoid-s9776" xml:space="preserve">Pour faire un cube qui ſoit égal au parallelepipede A E, il <pb o="285" file="0323" n="323" rhead="DE MATHÉMATIQUE. Liv. VIII."/> faut, ſi les trois dimenſions du parallelepipede ſont inégales, <lb/>comme on le ſuppoſe ici, chercher une moyenne proportion-<lb/>nelle entre les deux plus petites, A B, B C (art. </s> <s xml:id="echoid-s9777" xml:space="preserve">506), qui ſera, <lb/>par exemple F G, & </s> <s xml:id="echoid-s9778" xml:space="preserve">faire ſur cette ligne un quarré F H, qui <lb/>doit ſervir de baſe à un parallelepipede F I, qui doit avoir la <lb/>même hauteur que le parallelepipede A E, puiſque le rectangle <lb/>A C, qui lui ſert de baſe, eſt égal au quarré F H, qui ſert de <lb/>baſe au ſecond. </s> <s xml:id="echoid-s9779" xml:space="preserve">Cela poſé, il faut chercher deux moyennes <lb/>proportionnelles entre F G & </s> <s xml:id="echoid-s9780" xml:space="preserve">G K (art. </s> <s xml:id="echoid-s9781" xml:space="preserve">589), qui ſeront, par <lb/>exemple, N O & </s> <s xml:id="echoid-s9782" xml:space="preserve">P Q, & </s> <s xml:id="echoid-s9783" xml:space="preserve">je dis que le cube fait ſur la premiere <lb/>N O ſera égal au parallelepipede F I ou A E.</s> <s xml:id="echoid-s9784" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9785" xml:space="preserve">Pour le prouver, nous prendrons G D égal à F G, pour <lb/>avoir le cube G O, nous nommerons F G ou G H, ou G D, a; <lb/></s> <s xml:id="echoid-s9786" xml:space="preserve">G K, b; </s> <s xml:id="echoid-s9787" xml:space="preserve">& </s> <s xml:id="echoid-s9788" xml:space="preserve">N O, c: </s> <s xml:id="echoid-s9789" xml:space="preserve">ainſi le parallelepipede F I ſera a a b, le cube <lb/>F M ſera a a a, le cube de N O ſera c c c: </s> <s xml:id="echoid-s9790" xml:space="preserve">il faut donc prouver <lb/>que a a b = c c c.</s> <s xml:id="echoid-s9791" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div782" type="section" level="1" n="625"> <head xml:id="echoid-head735" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s9792" xml:space="preserve">Le cube F M & </s> <s xml:id="echoid-s9793" xml:space="preserve">le parallelepipede F I ayant la même baſe <lb/>F H, ſeront dans la raiſon de leurs hauteurs G D & </s> <s xml:id="echoid-s9794" xml:space="preserve">G K, d’où <lb/>l’on tire a a a: </s> <s xml:id="echoid-s9795" xml:space="preserve">a a b : </s> <s xml:id="echoid-s9796" xml:space="preserve">: </s> <s xml:id="echoid-s9797" xml:space="preserve">a: </s> <s xml:id="echoid-s9798" xml:space="preserve">b; </s> <s xml:id="echoid-s9799" xml:space="preserve">& </s> <s xml:id="echoid-s9800" xml:space="preserve">à cauſe des quatre proportion-<lb/>nelles, on verra que le cube fait ſur la premiere, eſt au cube <lb/>fait ſur la ſeconde, comme la premiere à la quatrieme, ce qui <lb/>donne a a a : </s> <s xml:id="echoid-s9801" xml:space="preserve">c c c : </s> <s xml:id="echoid-s9802" xml:space="preserve">: </s> <s xml:id="echoid-s9803" xml:space="preserve">a : </s> <s xml:id="echoid-s9804" xml:space="preserve">b; </s> <s xml:id="echoid-s9805" xml:space="preserve">donc puiſque ces deux proportions <lb/>ont la même derniere raiſon, on aura a a a : </s> <s xml:id="echoid-s9806" xml:space="preserve">a a b : </s> <s xml:id="echoid-s9807" xml:space="preserve">: </s> <s xml:id="echoid-s9808" xml:space="preserve">a a a : </s> <s xml:id="echoid-s9809" xml:space="preserve">c c c; <lb/></s> <s xml:id="echoid-s9810" xml:space="preserve">mais a<emph style="sub">3</emph> = a<emph style="sub">3</emph> : </s> <s xml:id="echoid-s9811" xml:space="preserve">donc a a b = c c c. </s> <s xml:id="echoid-s9812" xml:space="preserve">C. </s> <s xml:id="echoid-s9813" xml:space="preserve">Q. </s> <s xml:id="echoid-s9814" xml:space="preserve">F. </s> <s xml:id="echoid-s9815" xml:space="preserve">D.</s> <s xml:id="echoid-s9816" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9817" xml:space="preserve">Si les dimenſions du parallelepipede donné étoient expri-<lb/>mées en nombres, on n’auroit (pour trouver un cube égal au <lb/>parallelepipede) qu’à multiplier les trois dimenſions l’une par <lb/>l’autre pour avoir le ſolide du parallelepipede, & </s> <s xml:id="echoid-s9818" xml:space="preserve">extraire la <lb/>racine cube du produit, qui ſera le côté du cube demandé.</s> <s xml:id="echoid-s9819" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div783" type="section" level="1" n="626"> <head xml:id="echoid-head736" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s9820" xml:space="preserve">595. </s> <s xml:id="echoid-s9821" xml:space="preserve">L’on voit par cette propoſition, qu’il n’y a point de <lb/>ſolide qu’on ne puiſſe réduire en cube; </s> <s xml:id="echoid-s9822" xml:space="preserve">car les cônes & </s> <s xml:id="echoid-s9823" xml:space="preserve">les <lb/>ſpheres pouvant ſe réduire en cylindres, & </s> <s xml:id="echoid-s9824" xml:space="preserve">les pyramides en <lb/>priſmes, ſi on change la baſe des cylindres & </s> <s xml:id="echoid-s9825" xml:space="preserve">des priſmes en <lb/>quarrés qui leur ſoient égaux, on aura des parallelepipedes, <lb/>que l’on réduira aiſément en cube par le problême que nous <lb/>venons de réſoudre.</s> <s xml:id="echoid-s9826" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div784" type="section" level="1" n="627"> <head xml:id="echoid-head737" style="it" xml:space="preserve">Fin du huitieme Livre.</head> <pb o="286" file="0324" n="324"/> <figure> <image file="0324-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0324-01"/> </figure> </div> <div xml:id="echoid-div785" type="section" level="1" n="628"> <head xml:id="echoid-head738" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head739" xml:space="preserve">LIVRE NEUVIEME. <lb/>DES SECTIONS CONIQUES.</head> <p style="it"> <s xml:id="echoid-s9827" xml:space="preserve">COmme tous les Livres qui traitent des Elémens de Géométrie <lb/>ne parlent point des Sections Coniques, la plûpart de ceux qui <lb/>étudient ces Elémens s’en tiennent là, ſans s’embarraſſer de les <lb/>chercher ailleurs, dans la penſée que cette étude eſt plus curieuſe <lb/>que néceſſaire, & </s> <s xml:id="echoid-s9828" xml:space="preserve">ne convient qu’aux perſonnes qui veulent ſe <lb/>donner toutes entieres aux Mathématiques: </s> <s xml:id="echoid-s9829" xml:space="preserve">cependant il eſt ſi <lb/>utile de les ſçavoir, que ſi on les ignore, il n’eſt pas poſſible de <lb/>réſoudre les Problêmes les plus communs de la Géométrie pratique, <lb/>particuliérement de cette Géométrie pratique qui convient à l’In-<lb/>génieur & </s> <s xml:id="echoid-s9830" xml:space="preserve">à l’Officier d’Artillerie: </s> <s xml:id="echoid-s9831" xml:space="preserve">car ſi le premier veut toiſer <lb/>des voûtes ſurbaiſſées, il faut qu’il ſçache comme on trouve la <lb/>ſuperficie d’une ellipſe, que l’on appelle communément ovale, & </s> <s xml:id="echoid-s9832" xml:space="preserve"><lb/>qui eſt une des Sections coniques. </s> <s xml:id="echoid-s9833" xml:space="preserve">Si le ſecond veut ſçavoir l’art <lb/>de jetter les bombes, il ne le peut encore ſans connoître les pro-<lb/>priétés de la Parabole, qui eſt auſſi une des Sections coniques. <lb/></s> <s xml:id="echoid-s9834" xml:space="preserve">Et pour être bien convaincu de la néceſſité de ſçavoir au moins les <lb/>principales propriétés des Sections coniques, il ne faut que lire <lb/>l’Application de la Géométrie à la pratique, l’on verra que les <lb/>plus belles opérations en dépendent abſolument. </s> <s xml:id="echoid-s9835" xml:space="preserve">Cependant malgré <lb/>cela, les Sections coniques ſeroient bien peu de choſe, ſi elles n’a-<lb/>voient d’autres uſages que ceux que l’on trouvera ici; </s> <s xml:id="echoid-s9836" xml:space="preserve">elles ſont <lb/>ſi néceſſaires à un homme, qui ſans vouloir devenir grand Géo- <pb o="287" file="0325" n="325" rhead="NOUVEAU COURS DE MATH. Liv. IX."/> metre, veut ſeulement ſçavoir cette ſcience paſſablement, qu’il <lb/>ne peut pas les perdre de vue d’un moment: </s> <s xml:id="echoid-s9837" xml:space="preserve">car s’il veut réſoudre <lb/>un problême un peu compoſé, il trouvera des équations qui lui <lb/>indiqueront les courbes, dont il faudra qu’il ſe ſerve pour conſtruire <lb/>les égalités, c’eſt-à-dire pour conſtruire une figure qui donne la <lb/>ſolution du Problême.</s> <s xml:id="echoid-s9838" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s9839" xml:space="preserve">Je ne parle point de ceci dans cet Ouvrage, parce que je ne <lb/>donne que les principales propriétés des Sections coniques, ayant <lb/>eu ſeulement pour objet de les faire connoître à ceux qui ont du <lb/>goût pour la Géométrie, afin de leur inſpirer l’envie d’aller plus <lb/>loin, & </s> <s xml:id="echoid-s9840" xml:space="preserve">d’ailleurs pour m’en ſervir dans les endroits où je ne <lb/>pourrois m’en paſſer. </s> <s xml:id="echoid-s9841" xml:space="preserve">Mais s’il ſe trouvoit de ces perſonnes dont <lb/>je viens de parler, qui ne ſe bornent point à voir un Livre de <lb/>Géométrie, je leur conſeille d’étudier l’excellent Traité des Sec-<lb/>tions Coniques de M. </s> <s xml:id="echoid-s9842" xml:space="preserve">le Marquis de l’Hôpital, qui eſt ce que <lb/>nous avons de meilleur dans ce genre. </s> <s xml:id="echoid-s9843" xml:space="preserve">Et comme je me ſuis ſervi <lb/>dans ce que je donne ici d’une façon de démontrer fort approchante <lb/>de la ſienne, je ne doute pas qu’on n’ait une grande facilité à <lb/>comprendre cet Auteur, ſi l’on entend bien ce qui ſuit, qui en eſt <lb/>en quelque ſorte l’introduction.</s> <s xml:id="echoid-s9844" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div786" type="section" level="1" n="629"> <head xml:id="echoid-head740" xml:space="preserve">CHAPITRE PREMIER.</head> <head xml:id="echoid-head741" style="it" xml:space="preserve">Qui traite des propriétés de la Parabole.</head> <head xml:id="echoid-head742" xml:space="preserve"><emph style="sc">Définitions</emph>. <lb/>I.</head> <p> <s xml:id="echoid-s9845" xml:space="preserve">596. </s> <s xml:id="echoid-s9846" xml:space="preserve">SI l’on a une ligne droite A B perpendiculaire ſur la <lb/> <anchor type="note" xlink:label="note-0325-01a" xlink:href="note-0325-01"/> ligne O P, ſur laquelle on aura pris les parties A C & </s> <s xml:id="echoid-s9847" xml:space="preserve">C D <lb/>égales entr’elles; </s> <s xml:id="echoid-s9848" xml:space="preserve">& </s> <s xml:id="echoid-s9849" xml:space="preserve">que de C, en venant vers B, l’on mene <lb/>ſur la ligne A B une quantité de paralleles, comme E F, G H <lb/>à la ligne O P, & </s> <s xml:id="echoid-s9850" xml:space="preserve">qu’on faſſe D E ou D F égale à A K, & </s> <s xml:id="echoid-s9851" xml:space="preserve">de <lb/>même D G ou D H égal à A I, & </s> <s xml:id="echoid-s9852" xml:space="preserve">que l’on continue à trouver <lb/>une quantité de points, tels que E, G, M, en faiſant tou-<lb/>jours D M égal à A L; </s> <s xml:id="echoid-s9853" xml:space="preserve">la ligne que l’on ſera paſſer par tous <lb/>ces points ſera une courbe nommée parabole.</s> <s xml:id="echoid-s9854" xml:space="preserve"/> </p> <div xml:id="echoid-div786" type="float" level="2" n="1"> <note position="right" xlink:label="note-0325-01" xlink:href="note-0325-01a" xml:space="preserve">Figure 151.</note> </div> </div> <div xml:id="echoid-div788" type="section" level="1" n="630"> <head xml:id="echoid-head743" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s9855" xml:space="preserve">597. </s> <s xml:id="echoid-s9856" xml:space="preserve">La ligne A C B eſt nommée l’axe de la parabole.</s> <s xml:id="echoid-s9857" xml:space="preserve"/> </p> <pb o="288" file="0326" n="326" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div789" type="section" level="1" n="631"> <head xml:id="echoid-head744" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s9858" xml:space="preserve">598. </s> <s xml:id="echoid-s9859" xml:space="preserve">Le point A eſt appellé le point générateur, la ligne <lb/>O P directrice, & </s> <s xml:id="echoid-s9860" xml:space="preserve">le point D le foyer.</s> <s xml:id="echoid-s9861" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div790" type="section" level="1" n="632"> <head xml:id="echoid-head745" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s9862" xml:space="preserve">599. </s> <s xml:id="echoid-s9863" xml:space="preserve">Le point C eſt appellé origine de l’axe ou ſommet de la <lb/>parabole, parce que c’eſt de ce point que l’on ſuppoſe avoir <lb/>commencé les lignes paralleles qui forment la parabole.</s> <s xml:id="echoid-s9864" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div791" type="section" level="1" n="633"> <head xml:id="echoid-head746" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s9865" xml:space="preserve">600. </s> <s xml:id="echoid-s9866" xml:space="preserve">Chaque perpendiculaire, comme K E ou I G, ou M L, <lb/>eſt appellée ordonnée à l’axe A B.</s> <s xml:id="echoid-s9867" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div792" type="section" level="1" n="634"> <head xml:id="echoid-head747" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s9868" xml:space="preserve">601. </s> <s xml:id="echoid-s9869" xml:space="preserve">Les parties C K, C I, C L de l’axe, compriſes entre le <lb/>ſommet & </s> <s xml:id="echoid-s9870" xml:space="preserve">la rencontre d’une ordonnée, ſont appellées abſciſſes <lb/>ou coupées de l’axe C B.</s> <s xml:id="echoid-s9871" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div793" type="section" level="1" n="635"> <head xml:id="echoid-head748" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s9872" xml:space="preserve">602. </s> <s xml:id="echoid-s9873" xml:space="preserve">Si au ſommet de la courbe on éleve une perpendicu-<lb/>laire C N à l’axe C B, quadruple de A C, elle ſera appellée <lb/>parametre de la parabole.</s> <s xml:id="echoid-s9874" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div794" type="section" level="1" n="636"> <head xml:id="echoid-head749" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s9875" xml:space="preserve">603. </s> <s xml:id="echoid-s9876" xml:space="preserve">Une ligne droite qui ne rencontre la parabole qu’en <lb/>un ſeul point, & </s> <s xml:id="echoid-s9877" xml:space="preserve">qui étant prolongée à droite ou à gauche, <lb/>ne peut pas la couper, mais tombe toujours au dehors, eſt ap-<lb/>pellée tangente.</s> <s xml:id="echoid-s9878" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div795" type="section" level="1" n="637"> <head xml:id="echoid-head750" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9879" xml:space="preserve">604. </s> <s xml:id="echoid-s9880" xml:space="preserve">Dans la parabole, le rectangle compris ſous l’abſciſſe C I <lb/> <anchor type="note" xlink:label="note-0326-01a" xlink:href="note-0326-01"/> & </s> <s xml:id="echoid-s9881" xml:space="preserve">le parametre C N, eſt égal au quarré de l’ordonnée G I.</s> <s xml:id="echoid-s9882" xml:space="preserve"/> </p> <div xml:id="echoid-div795" type="float" level="2" n="1"> <note position="left" xlink:label="note-0326-01" xlink:href="note-0326-01a" xml:space="preserve">Figure 151.</note> </div> <p> <s xml:id="echoid-s9883" xml:space="preserve">Ayant nommé les données A C ou C D, a; </s> <s xml:id="echoid-s9884" xml:space="preserve">les indétermi-<lb/>nées ou lignes variables C I, x, & </s> <s xml:id="echoid-s9885" xml:space="preserve">G I, y; </s> <s xml:id="echoid-s9886" xml:space="preserve">A I ou D G qui lui <lb/>eſt égal, par la définition de la courbe, ſera x + a; </s> <s xml:id="echoid-s9887" xml:space="preserve">& </s> <s xml:id="echoid-s9888" xml:space="preserve">D I ou <lb/>C I - C D, ſera x - a, le parametre C N, par ſa définition, <lb/>ſera 4a: </s> <s xml:id="echoid-s9889" xml:space="preserve">il faut donc prouver que C I x C N = G I<emph style="sub">2</emph>, ou que <lb/>4ax = yy.</s> <s xml:id="echoid-s9890" xml:space="preserve"/> </p> <pb file="0327" n="327"/> <pb file="0327a" n="328"/> <figure> <image file="0327a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0327a-01"/> </figure> <pb file="0328" n="329"/> <pb file="0329" n="330"/> <pb file="0329a" n="331"/> <figure> <image file="0329a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0329a-01"/> </figure> <pb file="0330" n="332"/> <pb file="0331" n="333"/> <pb file="0331a" n="334"/> <figure> <image file="0331a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0331a-01"/> </figure> <pb file="0332" n="335"/> <pb file="0333" n="336"/> <pb file="0333a" n="337"/> <figure> <image file="0333a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0333a-01"/> </figure> <pb file="0334" n="338"/> <pb file="0335" n="339"/> <pb file="0335a" n="340"/> <figure> <image file="0335a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0335a-01"/> </figure> <pb file="0336" n="341"/> <pb file="0337" n="342"/> <pb file="0337a" n="343"/> <figure> <image file="0337a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0337a-01"/> </figure> <pb file="0338" n="344"/> <pb file="0339" n="345"/> <pb file="0339a" n="346"/> <figure> <image file="0339a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0339a-01"/> </figure> <pb file="0340" n="347"/> <pb file="0341" n="348"/> <pb file="0341a" n="349"/> <figure> <image file="0341a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0341a-01"/> </figure> <pb file="0342" n="350"/> <pb o="289" file="0343" n="351" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div797" type="section" level="1" n="638"> <head xml:id="echoid-head751" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9891" xml:space="preserve">Conſidérez qu’à cauſe du triangle rectangle G I D, on a <lb/>G D<emph style="sub">2</emph> = G I<emph style="sub">2</emph> + D I<emph style="sub">2</emph>, d’où l’on tire G I<emph style="sub">2</emph> = G D<emph style="sub">2</emph> - D I<emph style="sub">2</emph>; <lb/></s> <s xml:id="echoid-s9892" xml:space="preserve">mais G D = A I = x + a, ainſi G D<emph style="sub">2</emph> ſera x<emph style="sub">2</emph> + 2ax + aa, <lb/>& </s> <s xml:id="echoid-s9893" xml:space="preserve">D I = x - a: </s> <s xml:id="echoid-s9894" xml:space="preserve">donc D I<emph style="sub">2</emph> ſera xx - 2ax + aa, & </s> <s xml:id="echoid-s9895" xml:space="preserve">GI<emph style="sub">2</emph> = yy: </s> <s xml:id="echoid-s9896" xml:space="preserve"><lb/>on aura donc cette équation, yy = xx + 2ax + aa - xx <lb/>+ 2ax - aa = 4ax, en effaçant ce qui ſe détruit. </s> <s xml:id="echoid-s9897" xml:space="preserve">C. </s> <s xml:id="echoid-s9898" xml:space="preserve">Q. </s> <s xml:id="echoid-s9899" xml:space="preserve">F. </s> <s xml:id="echoid-s9900" xml:space="preserve">D.</s> <s xml:id="echoid-s9901" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div798" type="section" level="1" n="639"> <head xml:id="echoid-head752" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head753" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9902" xml:space="preserve">605. </s> <s xml:id="echoid-s9903" xml:space="preserve">Dans la parabole, je dis que les quarrés des ordonnées <lb/>E K, G I ſont entr’ eux comme leurs abſciſſes C K, C I; </s> <s xml:id="echoid-s9904" xml:space="preserve">ou, ce qui <lb/>eſt la même choſe, que les quarrés de deux ordonnées quelconques & </s> <s xml:id="echoid-s9905" xml:space="preserve"><lb/>de leurs abſciſſes, donneront cette proportion EK<emph style="sub">2</emph> : </s> <s xml:id="echoid-s9906" xml:space="preserve">GI<emph style="sub">2</emph> :</s> <s xml:id="echoid-s9907" xml:space="preserve">: CK : </s> <s xml:id="echoid-s9908" xml:space="preserve">CI.</s> <s xml:id="echoid-s9909" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div799" type="section" level="1" n="640"> <head xml:id="echoid-head754" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9910" xml:space="preserve">Les quarrés des ordonnées étant égaux aux rectangles com-<lb/>pris ſous leurs abſciſſes & </s> <s xml:id="echoid-s9911" xml:space="preserve">le parametre, ces quarrés ſont en-<lb/>tr’eux comme les rectangles auxquels ils ſont égaux; </s> <s xml:id="echoid-s9912" xml:space="preserve">mais <lb/>comme tous ces rectangles ont une hauteur commune, qui eſt <lb/>le parametre, ils ſeront dans la raiſon de leurs baſes (art. </s> <s xml:id="echoid-s9913" xml:space="preserve">391): <lb/></s> <s xml:id="echoid-s9914" xml:space="preserve">donc on aura E K<emph style="sub">2</emph> : </s> <s xml:id="echoid-s9915" xml:space="preserve">G I<emph style="sub">2</emph> :</s> <s xml:id="echoid-s9916" xml:space="preserve">: CK : </s> <s xml:id="echoid-s9917" xml:space="preserve">CI. </s> <s xml:id="echoid-s9918" xml:space="preserve">C. </s> <s xml:id="echoid-s9919" xml:space="preserve">Q. </s> <s xml:id="echoid-s9920" xml:space="preserve">F. </s> <s xml:id="echoid-s9921" xml:space="preserve">D.</s> <s xml:id="echoid-s9922" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div800" type="section" level="1" n="641"> <head xml:id="echoid-head755" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9923" xml:space="preserve">606. </s> <s xml:id="echoid-s9924" xml:space="preserve">Si à l’origine de l’axe C B on mene une perpendiculaire <lb/>C S, & </s> <s xml:id="echoid-s9925" xml:space="preserve">que des points E, G, M de la courbe, on mene les per-<lb/>pendiculaires ſur la ligne C S, il s’enſuit qu’il y aura même rai-<lb/>ſon du quarré C Q<emph style="sub">2</emph> au quarré C R<emph style="sub">2</emph>, que de la ligne Q E à la <lb/>ligne R G, puiſque les lignes C Q & </s> <s xml:id="echoid-s9926" xml:space="preserve">C R ſont égales aux or-<lb/>données K E & </s> <s xml:id="echoid-s9927" xml:space="preserve">I G, & </s> <s xml:id="echoid-s9928" xml:space="preserve">que les lignes Q E & </s> <s xml:id="echoid-s9929" xml:space="preserve">R G ſont égales <lb/>aux abſciſſes C K & </s> <s xml:id="echoid-s9930" xml:space="preserve">C I.</s> <s xml:id="echoid-s9931" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s9932" xml:space="preserve">Nous nous ſervirons de ce corollaire dans la ſuite, pour faire <lb/>voir que les boulets & </s> <s xml:id="echoid-s9933" xml:space="preserve">les bombes décrivent des paraboles dans <lb/>l’eſpace qu’ils parcourent, depuis le lieu d’où ils ſont pouſſés, <lb/>juſqu’à l’endroit où ils vont tomber.</s> <s xml:id="echoid-s9934" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div801" type="section" level="1" n="642"> <head xml:id="echoid-head756" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9935" xml:space="preserve">607. </s> <s xml:id="echoid-s9936" xml:space="preserve">Comme les quarrés des ordonnées qui ſont à droite & </s> <s xml:id="echoid-s9937" xml:space="preserve"><lb/>à gauche de l’axe ſur une même ligne ſont égaux au rectangle <pb o="290" file="0344" n="352" rhead="NOUVEAU COURS"/> de la même abſciſſe par le même parametre, il s’enſuit qu’ils <lb/>ſont égaux entr’eux; </s> <s xml:id="echoid-s9938" xml:space="preserve">ainſi les ordonnées ſont égales entr’elles: <lb/></s> <s xml:id="echoid-s9939" xml:space="preserve">donc l’axe diviſe l’eſpace indéfini, terminé par la courbe, en <lb/>deux parties égales, puiſqu’il diviſe en deux également toutes <lb/>les ordonnées qui lui ſont perpendiculaires, & </s> <s xml:id="echoid-s9940" xml:space="preserve">que l’on peut <lb/>regarder comme les élémens de cette ſurface.</s> <s xml:id="echoid-s9941" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div802" type="section" level="1" n="643"> <head xml:id="echoid-head757" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s9942" xml:space="preserve">608. </s> <s xml:id="echoid-s9943" xml:space="preserve">Comme l’on peut prendre des lignes C L ſi grandes <lb/>que l’on voudra, & </s> <s xml:id="echoid-s9944" xml:space="preserve">terminer le point M toujours de la même <lb/>maniere, en faiſant D M = C L, il s’enſuit que la courbe peut <lb/>s’étendre à l’infini, & </s> <s xml:id="echoid-s9945" xml:space="preserve">que ſes deux branches s’éloignent con-<lb/>tinuellement de l’axe.</s> <s xml:id="echoid-s9946" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div803" type="section" level="1" n="644"> <head xml:id="echoid-head758" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head759" xml:space="preserve"><emph style="sc">Probleme</emph></head> <p style="it"> <s xml:id="echoid-s9947" xml:space="preserve">609. </s> <s xml:id="echoid-s9948" xml:space="preserve">Mener une tangente à une parabole par un point donné.</s> <s xml:id="echoid-s9949" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 152.</note> <p> <s xml:id="echoid-s9950" xml:space="preserve">Pour mener une tangente à une parabole par un point donné <lb/>E, tirez de ce point au foyer C la ligne E C, & </s> <s xml:id="echoid-s9951" xml:space="preserve">du mêmepoint <lb/>la parallele E D à l’axe, qui ſera perpendiculaire à la directrice <lb/>A H, qu’elle rencontrera dans un point D; </s> <s xml:id="echoid-s9952" xml:space="preserve">joignez la ligne <lb/>D C, & </s> <s xml:id="echoid-s9953" xml:space="preserve">ſi vous menez la ligne E G qui paſſe par le milieu I <lb/>de la ligne D C, & </s> <s xml:id="echoid-s9954" xml:space="preserve">par le point E donné; </s> <s xml:id="echoid-s9955" xml:space="preserve">je dis qu’elle ſera <lb/>tangente à la parabole, ou, ce qui revient au même, qu’elle ne <lb/>la touchera qu’au ſeul point E; </s> <s xml:id="echoid-s9956" xml:space="preserve">tirez les lignes F D & </s> <s xml:id="echoid-s9957" xml:space="preserve">F C par <lb/>deux points quelconques de la ligne E I, & </s> <s xml:id="echoid-s9958" xml:space="preserve">les paralleles F H, <lb/>F H à l’axe A K, & </s> <s xml:id="echoid-s9959" xml:space="preserve">la ligne E K perpendiculaire au même axe.</s> <s xml:id="echoid-s9960" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div804" type="section" level="1" n="645"> <head xml:id="echoid-head760" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s9961" xml:space="preserve">Puiſque le point E eſt à la parabole, la ligne E C menée de <lb/>ce point au foyer C eſt égale à la ligne A K, par la définition <lb/>de la parabole, ou à la ligne E D qui lui eſt égale, à cauſe du <lb/>rectangle E D A K. </s> <s xml:id="echoid-s9962" xml:space="preserve">De plus, par conſtruction, la ligne E G <lb/>diviſe la ligne D C en deux également au point I: </s> <s xml:id="echoid-s9963" xml:space="preserve">donc cette <lb/>ligne eſt perpendiculaire ſur D C, puiſqu’elle a deux points <lb/>E, I, également éloignés de ſes extrêmités; </s> <s xml:id="echoid-s9964" xml:space="preserve">donc cette ligne <lb/>paſſera par tous les points également éloignés des mêmes ex-<lb/>trêmités, tels que ſont les points F, F; </s> <s xml:id="echoid-s9965" xml:space="preserve">mais dans les triangles <lb/>rectangles D H F, l’hypoténuſe D F = F C, eſt plus grande <pb o="291" file="0345" n="353" rhead="DE MATHÉMATIQUE. Liv. IX."/> qu’un des côtés F H: </s> <s xml:id="echoid-s9966" xml:space="preserve">donc F C eſt plus grande que F H ou que <lb/>A C, ainſi le point F n’eſt pas à la parabole. </s> <s xml:id="echoid-s9967" xml:space="preserve">On démon-<lb/>trera la même choſe de tout autre point: </s> <s xml:id="echoid-s9968" xml:space="preserve">donc la ligne E G <lb/>touche la parabole au ſeul point E, & </s> <s xml:id="echoid-s9969" xml:space="preserve">par conſéquent elle eſt <lb/>tangente à la courbe. </s> <s xml:id="echoid-s9970" xml:space="preserve">C. </s> <s xml:id="echoid-s9971" xml:space="preserve">Q. </s> <s xml:id="echoid-s9972" xml:space="preserve">F. </s> <s xml:id="echoid-s9973" xml:space="preserve">D.</s> <s xml:id="echoid-s9974" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div805" type="section" level="1" n="646"> <head xml:id="echoid-head761" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s9975" xml:space="preserve">610. </s> <s xml:id="echoid-s9976" xml:space="preserve">Il ſuit de cette conſtruction que l’angle D E C eſt <lb/> <anchor type="note" xlink:label="note-0345-01a" xlink:href="note-0345-01"/> coupé en deux également par la tangente E G, puiſque cette <lb/>ligne diviſe la ligne D C en deux parties égales. </s> <s xml:id="echoid-s9977" xml:space="preserve">D’où il ſuit <lb/>encore que l’angle R E L formé par la tangente E G, & </s> <s xml:id="echoid-s9978" xml:space="preserve">le dia-<lb/>metre D E R mené par le point de contact, eſt égal à l’angle <lb/>C E I formé par la même tangente, & </s> <s xml:id="echoid-s9979" xml:space="preserve">la ligne menée du point <lb/>de contingence au foyer C; </s> <s xml:id="echoid-s9980" xml:space="preserve">car comme on vient de voir l’an-<lb/>gle C E I = D E I, mais D E I = L E R qui lui eſt oppoſé au <lb/>ſommet: </s> <s xml:id="echoid-s9981" xml:space="preserve">donc C E I = L E R.</s> <s xml:id="echoid-s9982" xml:space="preserve"/> </p> <div xml:id="echoid-div805" type="float" level="2" n="1"> <note position="right" xlink:label="note-0345-01" xlink:href="note-0345-01a" xml:space="preserve">Figure 153.</note> </div> </div> <div xml:id="echoid-div807" type="section" level="1" n="647"> <head xml:id="echoid-head762" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s9983" xml:space="preserve">611. </s> <s xml:id="echoid-s9984" xml:space="preserve">Il ſuit du dernier corollaire, que ſi l’on place un point <lb/>lumineux au foyer C, tous les rayons qui partiront de ce point, <lb/>ſe réflechiront à la rencontre de la parabole, ſuivant des lignes <lb/>paralleles à l’axe; </s> <s xml:id="echoid-s9985" xml:space="preserve">car c’eſt un principe dans la catoptrique, <lb/>que tout rayon réflechi fait avec le plan de réflexion, l’angle <lb/>de réflexion égal à celui d’incidence. </s> <s xml:id="echoid-s9986" xml:space="preserve">Or il eſt viſible que la <lb/>tangente au point E peut repréſenter le plan de réflexion; </s> <s xml:id="echoid-s9987" xml:space="preserve">& </s> <s xml:id="echoid-s9988" xml:space="preserve"><lb/>par conſéquent le rayon parti du foyer C, ſuivant la ligne C E, <lb/>ſe réflechira ſuivant la ligne E R. </s> <s xml:id="echoid-s9989" xml:space="preserve">Réciproquement tous les <lb/>rayons paralleles à l’axe d’une parabole, interceptés par le péri-<lb/>metre de cette courbe, ſe réfléchiront au foyer F. </s> <s xml:id="echoid-s9990" xml:space="preserve">Il faut en-<lb/>tendre la même choſe de tout corps à reſſort différent de la lu-<lb/>miere. </s> <s xml:id="echoid-s9991" xml:space="preserve">Ainſi une petite bille d’yvoire que l’on pouſſeroit, ſui-<lb/>vant R E, ſe détourneroit à la rencontre de la courbe pour <lb/>ſuivre la ligne E C.</s> <s xml:id="echoid-s9992" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div808" type="section" level="1" n="648"> <head xml:id="echoid-head763" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s9993" xml:space="preserve">612. </s> <s xml:id="echoid-s9994" xml:space="preserve">Si du point d’attouchement E l’on mene l’ordonnée <lb/>E K à l’axe de la parabole, la ligne G K ſera nommée ſoutan-<lb/>gente.</s> <s xml:id="echoid-s9995" xml:space="preserve"/> </p> <pb o="292" file="0346" n="354" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div809" type="section" level="1" n="649"> <head xml:id="echoid-head764" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head765" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s9996" xml:space="preserve">613. </s> <s xml:id="echoid-s9997" xml:space="preserve">Si on éleve une perpendiculaire E M au point de contin-<lb/> <anchor type="note" xlink:label="note-0346-01a" xlink:href="note-0346-01"/> gence E, & </s> <s xml:id="echoid-s9998" xml:space="preserve">que de ce même point l’on tire une ordonnée E K à l’axe <lb/>B M, je dis que la partie K M de l’axe ſera toujours égale à la <lb/>moitié du parametre de cette parabole, c’eſt-à-dire à 2a.</s> <s xml:id="echoid-s9999" xml:space="preserve"/> </p> <div xml:id="echoid-div809" type="float" level="2" n="1"> <note position="left" xlink:label="note-0346-01" xlink:href="note-0346-01a" xml:space="preserve">Figure 153.</note> </div> </div> <div xml:id="echoid-div811" type="section" level="1" n="650"> <head xml:id="echoid-head766" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10000" xml:space="preserve">Comme les lignes D C & </s> <s xml:id="echoid-s10001" xml:space="preserve">E M ſont paralleles, étant toutes <lb/>deux, par conſtruction, perpendiculaires ſur L G, ainſi que les <lb/>lignes E K & </s> <s xml:id="echoid-s10002" xml:space="preserve">A D, qui ſont toutes deux perpendiculaires à <lb/>l’axe, il s’enſuit que les triangles rectangles D A C, E K M ſont <lb/>égaux en tout: </s> <s xml:id="echoid-s10003" xml:space="preserve">donc A C = K M, ou la moitié du parametre <lb/>qui eſt 2a. </s> <s xml:id="echoid-s10004" xml:space="preserve">C. </s> <s xml:id="echoid-s10005" xml:space="preserve">Q. </s> <s xml:id="echoid-s10006" xml:space="preserve">F. </s> <s xml:id="echoid-s10007" xml:space="preserve">D.</s> <s xml:id="echoid-s10008" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div812" type="section" level="1" n="651"> <head xml:id="echoid-head767" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head768" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10009" xml:space="preserve">614. </s> <s xml:id="echoid-s10010" xml:space="preserve">Nous ſervant de la même figure, je dis que la ſoutan-<lb/> <anchor type="note" xlink:label="note-0346-02a" xlink:href="note-0346-02"/> gente G K eſt double de l’abſciſſe B K.</s> <s xml:id="echoid-s10011" xml:space="preserve"/> </p> <div xml:id="echoid-div812" type="float" level="2" n="1"> <note position="left" xlink:label="note-0346-02" xlink:href="note-0346-02a" xml:space="preserve">Figure 153.</note> </div> </div> <div xml:id="echoid-div814" type="section" level="1" n="652"> <head xml:id="echoid-head769" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10012" xml:space="preserve">Le parametre de cette parabole étant 4a (art. </s> <s xml:id="echoid-s10013" xml:space="preserve">604), K M <lb/>ſera 2a, par la derniere propoſition; </s> <s xml:id="echoid-s10014" xml:space="preserve">& </s> <s xml:id="echoid-s10015" xml:space="preserve">à cauſe des triangles <lb/>rectangles ſemblables G K E, E K M (art. </s> <s xml:id="echoid-s10016" xml:space="preserve">406), l’on aura cette <lb/>proportion K M (2a) : </s> <s xml:id="echoid-s10017" xml:space="preserve">K E (y) :</s> <s xml:id="echoid-s10018" xml:space="preserve">: K E (y): </s> <s xml:id="echoid-s10019" xml:space="preserve">{K E<emph style="sub">2</emph>/K M} ({y y/2a})=K G, <lb/>& </s> <s xml:id="echoid-s10020" xml:space="preserve">ſi dans l’équation K G = {yy/2a}, on met 4ax à la place de yy, <lb/>auquel il eſt égal (art, 605), on aura K G = {4ax/2a} = 2x. <lb/></s> <s xml:id="echoid-s10021" xml:space="preserve">C. </s> <s xml:id="echoid-s10022" xml:space="preserve">Q. </s> <s xml:id="echoid-s10023" xml:space="preserve">F. </s> <s xml:id="echoid-s10024" xml:space="preserve">D.</s> <s xml:id="echoid-s10025" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div815" type="section" level="1" n="653"> <head xml:id="echoid-head770" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10026" xml:space="preserve">615. </s> <s xml:id="echoid-s10027" xml:space="preserve">L’on tire de cette propoſition un moyen fort aiſé de <lb/>mener une tangente à une parabole: </s> <s xml:id="echoid-s10028" xml:space="preserve">car, par exemple, pour <lb/>mener la ligne L G qui ſoit tangente à la parabole au point E, <lb/>il n’y a qu’à abaiſſer du point E la perpendiculaire E K ſur l’axe <lb/>B M, & </s> <s xml:id="echoid-s10029" xml:space="preserve">faire la ligne B G égale à l’abſciſſe B K; </s> <s xml:id="echoid-s10030" xml:space="preserve">& </s> <s xml:id="echoid-s10031" xml:space="preserve">par les <lb/>points G, E, mener la ligne G E L.</s> <s xml:id="echoid-s10032" xml:space="preserve"/> </p> <pb o="293" file="0347" n="355" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div816" type="section" level="1" n="654"> <head xml:id="echoid-head771" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s10033" xml:space="preserve">616. </s> <s xml:id="echoid-s10034" xml:space="preserve">Si du point A, où une droite A B touche la parabole, <lb/> <anchor type="note" xlink:label="note-0347-01a" xlink:href="note-0347-01"/> on mene une ligne A O parallele à l’axe M N, cette ligne ſera <lb/>nommée un diametre de la parabole.</s> <s xml:id="echoid-s10035" xml:space="preserve"/> </p> <div xml:id="echoid-div816" type="float" level="2" n="1"> <note position="right" xlink:label="note-0347-01" xlink:href="note-0347-01a" xml:space="preserve">Figure 154.</note> </div> </div> <div xml:id="echoid-div818" type="section" level="1" n="655"> <head xml:id="echoid-head772" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head773" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10036" xml:space="preserve">617. </s> <s xml:id="echoid-s10037" xml:space="preserve">Si l’on tire une ligne C D parallele à la tangente N B, <lb/> <anchor type="note" xlink:label="note-0347-02a" xlink:href="note-0347-02"/> je dis qu’elle ſera diviſée en deux également au point E par le dia-<lb/>metre A O.</s> <s xml:id="echoid-s10038" xml:space="preserve"/> </p> <div xml:id="echoid-div818" type="float" level="2" n="1"> <note position="right" xlink:label="note-0347-02" xlink:href="note-0347-02a" xml:space="preserve">Figure 154.</note> </div> <p> <s xml:id="echoid-s10039" xml:space="preserve">Du point A menez l’ordonnée A G, & </s> <s xml:id="echoid-s10040" xml:space="preserve">des points C, E, D <lb/>les lignes H C I, E F, D L paralleles à l’ordonnée A G; </s> <s xml:id="echoid-s10041" xml:space="preserve">pro-<lb/>longez le diametre O A juſqu’à la rencontre de la ligne H C. <lb/></s> <s xml:id="echoid-s10042" xml:space="preserve">Cela poſé, nous nommerons M F, m; </s> <s xml:id="echoid-s10043" xml:space="preserve">I F ou H E, t; </s> <s xml:id="echoid-s10044" xml:space="preserve">F L ou <lb/>E K, u: </s> <s xml:id="echoid-s10045" xml:space="preserve">ainſi F I ſera m - t; </s> <s xml:id="echoid-s10046" xml:space="preserve">& </s> <s xml:id="echoid-s10047" xml:space="preserve">M L m + u: </s> <s xml:id="echoid-s10048" xml:space="preserve">nous nommerons <lb/>de même M G, x; </s> <s xml:id="echoid-s10049" xml:space="preserve">A G, y; </s> <s xml:id="echoid-s10050" xml:space="preserve">G F ſera m - x. </s> <s xml:id="echoid-s10051" xml:space="preserve">Ainſi il faut <lb/>prouver que E C eſt égal à E D, ou que H E (t) = E K (u); </s> <s xml:id="echoid-s10052" xml:space="preserve"><lb/>ce qui eſt la même choſe: </s> <s xml:id="echoid-s10053" xml:space="preserve">car ſi H K eſt diviſé en deux égale-<lb/>ment au point E, la droite C D le ſera auſſi au même point.</s> <s xml:id="echoid-s10054" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div820" type="section" level="1" n="656"> <head xml:id="echoid-head774" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10055" xml:space="preserve">Les triangles B G A & </s> <s xml:id="echoid-s10056" xml:space="preserve">E H C, E K D ſont ſemblables, parce <lb/>qu’ils ont les côtés paralleles chacun à chacun, & </s> <s xml:id="echoid-s10057" xml:space="preserve">donnent les <lb/>deux proportions ſuivantes B G (2x) : </s> <s xml:id="echoid-s10058" xml:space="preserve">A G (y) :</s> <s xml:id="echoid-s10059" xml:space="preserve">: E K (u): <lb/></s> <s xml:id="echoid-s10060" xml:space="preserve">D K ({uy/2x}), & </s> <s xml:id="echoid-s10061" xml:space="preserve">B G (2x) : </s> <s xml:id="echoid-s10062" xml:space="preserve">A G (y) :</s> <s xml:id="echoid-s10063" xml:space="preserve">: E H (t): </s> <s xml:id="echoid-s10064" xml:space="preserve">C H ({ty/2x}). </s> <s xml:id="echoid-s10065" xml:space="preserve"><lb/>Ayant ainſi déterminé les valeurs des lignes D K, C H, on a <lb/>celles des ordonnées C I, D L: </s> <s xml:id="echoid-s10066" xml:space="preserve">car C I = I H - C H, ou <lb/>A G - C H = y - {ty/2x}; </s> <s xml:id="echoid-s10067" xml:space="preserve">& </s> <s xml:id="echoid-s10068" xml:space="preserve">de même D L = K L + D K = <lb/>A G + D K = y + {uy/2x}. </s> <s xml:id="echoid-s10069" xml:space="preserve">Mais par la propriété de la parabole, <lb/>les quarrés des ordonnées C I, A G, D L ſont entr’eux comme <lb/>leurs abſciſſes; </s> <s xml:id="echoid-s10070" xml:space="preserve">ce qui donne les deux proportions ſuivantes: </s> <s xml:id="echoid-s10071" xml:space="preserve"><lb/>A G<emph style="sub">2</emph> (yy) : </s> <s xml:id="echoid-s10072" xml:space="preserve">C I<emph style="sub">2</emph> (yy - {ty<emph style="sub">2</emph>/x} + {ttyy/4xx}) :</s> <s xml:id="echoid-s10073" xml:space="preserve">: MG (x) : </s> <s xml:id="echoid-s10074" xml:space="preserve">MI (m-t). </s> <s xml:id="echoid-s10075" xml:space="preserve">Et <lb/>A G<emph style="sub">2</emph> (yy) : </s> <s xml:id="echoid-s10076" xml:space="preserve">D L<emph style="sub">2</emph> (yy + {uy<emph style="sub">2</emph>/x} + {uuyy/4xx}) :</s> <s xml:id="echoid-s10077" xml:space="preserve">: MG (x) : </s> <s xml:id="echoid-s10078" xml:space="preserve">ML (m+u), <lb/>d’où l’on tire les deux équations ſuivantes, myy - tyy = xyy <lb/>- ty<emph style="sub">2</emph> + {ttyy/4x}, & </s> <s xml:id="echoid-s10079" xml:space="preserve">myy + uyy = xyy + uy<emph style="sub">2</emph> + {uuyy/4x). </s> <s xml:id="echoid-s10080" xml:space="preserve">Préſen- <pb o="294" file="0348" n="356" rhead="NOUVEAU COURS"/> tement ſi l’on retranche la premiere équation de la ſeconde, <lb/>c’eſt-à-dire le premier membre de la premiere du premier <lb/>membre de la ſeconde, & </s> <s xml:id="echoid-s10081" xml:space="preserve">le ſecond membre de la premiere <lb/>du ſecond membre de la ſeconde, on aura myy + uyy - myy <lb/>+ tyy = xyy + uyy + {uuyy/4x} - xyy + tyy - {ttyy/4x}, ou en ré-<lb/>duiſant le premier & </s> <s xml:id="echoid-s10082" xml:space="preserve">le ſecond membre, & </s> <s xml:id="echoid-s10083" xml:space="preserve">ôtant de chaque <lb/>membre les quantités égales uyy + tyy; </s> <s xml:id="echoid-s10084" xml:space="preserve">0 = {uuyy/4x} - {ttyy/4x}, & </s> <s xml:id="echoid-s10085" xml:space="preserve"><lb/>tranſpoſant {uuyy/4x} = {ttyy/4x}, d’où l’on tire uu = tt, ou u = t, en <lb/>tirant les racines, & </s> <s xml:id="echoid-s10086" xml:space="preserve">diviſant chaque membre par la fraction <lb/>{yy/4x}. </s> <s xml:id="echoid-s10087" xml:space="preserve">C. </s> <s xml:id="echoid-s10088" xml:space="preserve">Q. </s> <s xml:id="echoid-s10089" xml:space="preserve">F. </s> <s xml:id="echoid-s10090" xml:space="preserve">D.</s> <s xml:id="echoid-s10091" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div821" type="section" level="1" n="657"> <head xml:id="echoid-head775" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head776" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s10092" xml:space="preserve">618. </s> <s xml:id="echoid-s10093" xml:space="preserve">Toute ligne, comme E C ou E D, menée paralléle-<lb/>ment à la tangente A B, eſt nommée ordonnée au diametre <lb/>A O.</s> <s xml:id="echoid-s10094" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div822" type="section" level="1" n="658"> <head xml:id="echoid-head777" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s10095" xml:space="preserve">619. </s> <s xml:id="echoid-s10096" xml:space="preserve">Si l’on cherche une troiſieme proportionnelle à la ligne <lb/>B M & </s> <s xml:id="echoid-s10097" xml:space="preserve">à la tangente A B, cette ligne ſera appellée le para-<lb/>metre du diametre A O.</s> <s xml:id="echoid-s10098" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div823" type="section" level="1" n="659"> <head xml:id="echoid-head778" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10099" xml:space="preserve">620. </s> <s xml:id="echoid-s10100" xml:space="preserve">Il ſuit de la définition précédente, que ſi l’on tire une <lb/>ligne du foyer P au point d’attouchement A, une ligne qua-<lb/>druple A P ſera égale au parametre du diametre A O.</s> <s xml:id="echoid-s10101" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10102" xml:space="preserve">Pour le prouver, nous ſuppoſerons que le point S eſt le point <lb/>générateur; </s> <s xml:id="echoid-s10103" xml:space="preserve">ce qui donnera G S = P A (art. </s> <s xml:id="echoid-s10104" xml:space="preserve">596). </s> <s xml:id="echoid-s10105" xml:space="preserve">Et ſi l’on <lb/>nomme S M ou M P, a; </s> <s xml:id="echoid-s10106" xml:space="preserve">M G, x; </s> <s xml:id="echoid-s10107" xml:space="preserve">A G, y; </s> <s xml:id="echoid-s10108" xml:space="preserve">nous aurons G S ou <lb/>A P = x + a, & </s> <s xml:id="echoid-s10109" xml:space="preserve">par la premiere propoſition 4ax = yy. </s> <s xml:id="echoid-s10110" xml:space="preserve">Cela <lb/>poſé, ſi on nomme p le parametre du diametre A O, on aura <lb/>par la définition précédente (art. </s> <s xml:id="echoid-s10111" xml:space="preserve">619) M B (x) : </s> <s xml:id="echoid-s10112" xml:space="preserve">AB :</s> <s xml:id="echoid-s10113" xml:space="preserve">: AB : </s> <s xml:id="echoid-s10114" xml:space="preserve">p; <lb/></s> <s xml:id="echoid-s10115" xml:space="preserve">donc p x = A B<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10116" xml:space="preserve">mais à cauſe du triangle rectangle A B G, <lb/>A B<emph style="sub">2</emph> = A G<emph style="sub">2</emph> + G B<emph style="sub">2</emph> = 4ax + 4xx: </s> <s xml:id="echoid-s10117" xml:space="preserve">donc px = 4ax + 4xx, <lb/>ou en diviſant tout par x, p = 4a + 4x = 4A P. </s> <s xml:id="echoid-s10118" xml:space="preserve">C. </s> <s xml:id="echoid-s10119" xml:space="preserve">Q. </s> <s xml:id="echoid-s10120" xml:space="preserve">F. </s> <s xml:id="echoid-s10121" xml:space="preserve">D.</s> <s xml:id="echoid-s10122" xml:space="preserve"/> </p> <pb o="295" file="0349" n="357" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div824" type="section" level="1" n="660"> <head xml:id="echoid-head779" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head780" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10123" xml:space="preserve">621. </s> <s xml:id="echoid-s10124" xml:space="preserve">Le quarré d’une ordonnée quelconque E C à un diametre <lb/>A O eſt égal au rectangle compris ſous l’abſciſſe A E, & </s> <s xml:id="echoid-s10125" xml:space="preserve">ſous le <lb/>parametre du diametre A O (ou, ce qui eſt la même choſe, ſous <lb/>une ligne quadruple de A P). </s> <s xml:id="echoid-s10126" xml:space="preserve">Les choſes demeurant les mêmes que <lb/>dans la propoſition précédente; </s> <s xml:id="echoid-s10127" xml:space="preserve">les lignes ſeront nommées avec les <lb/>mêmes lettres, excepté la ligne A E, que nous nommerons z, qui <lb/>étant égale à F G, ſera m - x.</s> <s xml:id="echoid-s10128" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div825" type="section" level="1" n="661"> <head xml:id="echoid-head781" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10129" xml:space="preserve">Il faut d’abord ajouter les deux équations que nous avons <lb/>trouvées dans le théorême précédent, après avoir mis t à la <lb/>place de u qui lui eſt égal; </s> <s xml:id="echoid-s10130" xml:space="preserve">ce qui donnera myy + tyy + myy <lb/>- tyy = xyy + tyy + {ttyy/4x} + xyy - tyy + {ttyy/4x}; </s> <s xml:id="echoid-s10131" xml:space="preserve">d’où l’on <lb/>tire, en faiſant la réduction, 2myy = 2xyy + {ttyy/2x}, ou en fai-<lb/>ſant évanouir la fraction, 4mxyy = 4xxyy + ttyy, qui étant <lb/>diviſée par yy, donne 4mx = 4xx + tt; </s> <s xml:id="echoid-s10132" xml:space="preserve">& </s> <s xml:id="echoid-s10133" xml:space="preserve">faiſant paſſer 4xx <lb/>du ſecond membre dans le 1<emph style="sub">er</emph> 4mx-4xx ou √m-x\x{0020} x 4x = tt, <lb/>& </s> <s xml:id="echoid-s10134" xml:space="preserve">comme m - x = z, on aura 4zx = tt; </s> <s xml:id="echoid-s10135" xml:space="preserve">mais à cauſe <lb/>du triangle rectangle E H C, l’on aura E C<emph style="sub">2</emph> = E H<emph style="sub">2</emph> + C H<emph style="sub">2</emph> <lb/>= tt + {ttyy/4xx}, & </s> <s xml:id="echoid-s10136" xml:space="preserve">mettant 4xz à la place de tt, & </s> <s xml:id="echoid-s10137" xml:space="preserve">4ax à la place <lb/>de yy, il viendra E C<emph style="sub">2</emph> = 4xz + {4xz x 4ax/4xx}, ou 4xz + 4az = z <lb/>x √4x+4a\x{0020}, ou E C<emph style="sub">2</emph> = 4A P x A E. </s> <s xml:id="echoid-s10138" xml:space="preserve">C. </s> <s xml:id="echoid-s10139" xml:space="preserve">Q. </s> <s xml:id="echoid-s10140" xml:space="preserve">F. </s> <s xml:id="echoid-s10141" xml:space="preserve">D.</s> <s xml:id="echoid-s10142" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div826" type="section" level="1" n="662"> <head xml:id="echoid-head782" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s10143" xml:space="preserve">622. </s> <s xml:id="echoid-s10144" xml:space="preserve">On voit par ce théorême que la propoſition premiere <lb/>devient générale, puiſque non ſeulement le quarré d’une or-<lb/>donnée à l’axe eſt égal au rectangle compris ſous le parametre <lb/>de l’axe & </s> <s xml:id="echoid-s10145" xml:space="preserve">ſous l’abſciſſe, mais que le quarré de toute ordon-<lb/>née à un diametre quelconque, eſt auſſi égal au rectangle com-<lb/>pris ſous l’abſciſſe correſpondante & </s> <s xml:id="echoid-s10146" xml:space="preserve">le parametre de ce dia-<lb/>metre. </s> <s xml:id="echoid-s10147" xml:space="preserve">Mais pour mieux faire entendre ceci, conſidérez que <lb/>ſi la ligne R T eſt tangente au point M, extrêmité de l’axe, <lb/>toutes les ordonnées à l’axe ſeront paralleles à cette tangente, <pb o="296" file="0350" n="358" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s10148" xml:space="preserve">par la propoſition premiere, le quarré de chacune de ces <lb/>ordonnées ſera égal au rectangle compris ſous l’abſciſſe cor-<lb/>reſpondante, & </s> <s xml:id="echoid-s10149" xml:space="preserve">ſous une ligne quadruple de P M, qui eſt la <lb/>diſtance du foyer au point d’attouchement. </s> <s xml:id="echoid-s10150" xml:space="preserve">Si donc l’on ima-<lb/>gine que l’axe M L ſe ſoit mu parallélement à lui-même juſ-<lb/>qu’au point A, où il devient le diametre A O, & </s> <s xml:id="echoid-s10151" xml:space="preserve">que la tan-<lb/>gente R T ait gliſſée ſur la parabole, ne la touchant toujours <lb/>qu’en un ſeul point, juſqu’à ce que le point M devienne le <lb/>point A; </s> <s xml:id="echoid-s10152" xml:space="preserve">pour lors la tangente R T deviendra la tangente N B, <lb/>& </s> <s xml:id="echoid-s10153" xml:space="preserve">la ligne P M deviendra la ligne P A; </s> <s xml:id="echoid-s10154" xml:space="preserve">& </s> <s xml:id="echoid-s10155" xml:space="preserve">par conſéquent elle <lb/>ſera encore la quatrieme partie du parametre de l’axe, devenue <lb/>le diametre A O, & </s> <s xml:id="echoid-s10156" xml:space="preserve">les ordonnées que l’on auroit menées pa-<lb/>rallélement à la tangente R T, telles que V X, ſeront toujours <lb/>paralleles à la tangente, ſi elles ont accompagné l’axe, & </s> <s xml:id="echoid-s10157" xml:space="preserve">ſi <lb/>l’abſciſſe M V eſt égale à l’abſciſſe A E, l’ordonnée V X de-<lb/>viendra l’ordonnée E C, & </s> <s xml:id="echoid-s10158" xml:space="preserve">l’on aura toujours le quarré de <lb/>E C égal au rectangle compris ſous l’abſciſſe A E, & </s> <s xml:id="echoid-s10159" xml:space="preserve">ſous une <lb/>ligne quadruple de la diſtance du point d’attouchement A au <lb/>foyer P, comme on l’a démontré dans la propoſition précé-<lb/>dente.</s> <s xml:id="echoid-s10160" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10161" xml:space="preserve">On pourra remarquer que ſi le point A approchoit plus du <lb/>point M, il pourroit arriver que le point C tomberoit au-delà <lb/>de l’axe M L, & </s> <s xml:id="echoid-s10162" xml:space="preserve">qu’il y tombât encore dans le cas où l’on <lb/>prendroit une abſciſſe A E plus grande ſur le diametre, ſuppoſé <lb/>toujours au même point A; </s> <s xml:id="echoid-s10163" xml:space="preserve">mais cela n’empêcheroit pas que <lb/>tout ce que nous avons démontré ne ſubſiſtât de même, de <lb/>quelque façon que la ligne D C puiſſe ſe trouver dans la para-<lb/>bole, puiſqu’elle ſera toujours diviſée en deux également par <lb/>le diametre, lorſqu’elle ſera parallele à la tangente.</s> <s xml:id="echoid-s10164" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div827" type="section" level="1" n="663"> <head xml:id="echoid-head783" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s10165" xml:space="preserve">623. </s> <s xml:id="echoid-s10166" xml:space="preserve">Il ſuit auſſi de ce que nous avons vu, & </s> <s xml:id="echoid-s10167" xml:space="preserve">de la remarque <lb/>précédente, 1<emph style="sub">0</emph>. </s> <s xml:id="echoid-s10168" xml:space="preserve">que le parametre de l’axe eſt le plus petit de <lb/>tous les parametres: </s> <s xml:id="echoid-s10169" xml:space="preserve">2<emph style="sub">0</emph>. </s> <s xml:id="echoid-s10170" xml:space="preserve">Que ſi l’on prend ſur l’axe & </s> <s xml:id="echoid-s10171" xml:space="preserve">ſur un <lb/>diametre quelconque des abſciſſes égales, les ordonnées au <lb/>diametre ſeront plus grandes que celles de l’axe, puiſque leurs <lb/>quarrés ſont égaux aux rectangles d’une même abſciſſe par <lb/>des parametres différens, & </s> <s xml:id="echoid-s10172" xml:space="preserve">que d’ailleurs le parametre d’un <lb/>diametre quelconque eſt plus grand que celui de l’axe.</s> <s xml:id="echoid-s10173" xml:space="preserve"/> </p> <pb o="297" file="0351" n="359" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div828" type="section" level="1" n="664"> <head xml:id="echoid-head784" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s10174" xml:space="preserve">624. </s> <s xml:id="echoid-s10175" xml:space="preserve">Puiſque le quarré d’une ordonnée à un diametre quel-<lb/>conque eſt égal au produit de l’abſciſſe par le parametre, qui <lb/>eſt une grandeur conſtante pour chaque diametre, & </s> <s xml:id="echoid-s10176" xml:space="preserve">variable <lb/>ſuivant les différens diametres, il ſuit qu’en déſignant par p le <lb/>parametre d’un diametre quelconque, par x, l’abſciſſe priſe <lb/>ſur le même diametre, à commencer de l’origine du diametre, <lb/>& </s> <s xml:id="echoid-s10177" xml:space="preserve">par y, l’ordonnée correſpondante à cette abſciſſe, on aura <lb/>toujours y y = p x pour l’équation qui renferme les propriétés <lb/>de la parabole, ſoit par rapport aux diametres, ſoit par rap-<lb/>port à l’axe. </s> <s xml:id="echoid-s10178" xml:space="preserve">Si l’on ſuppoſe que l’abſciſſe ſoit priſe ſur l’axe, <lb/>& </s> <s xml:id="echoid-s10179" xml:space="preserve">qu’elle ſoit égale au quart du parametre, cette équation <lb/>deviendra y y = {1/4}pp, d’où l’on tire y = {1/2}p, & </s> <s xml:id="echoid-s10180" xml:space="preserve">en doublant <lb/>2y = p; </s> <s xml:id="echoid-s10181" xml:space="preserve">ce qui montre que la double ordonnée qui paſſe par <lb/>le foyer eſt égale au parametre; </s> <s xml:id="echoid-s10182" xml:space="preserve">ce qui eſt encore vrai par rap-<lb/>port à un diametre quelconque, comme on peut aiſément le <lb/>reconnoître, ſi l’on conçoit bien ce que nous avons expliqué <lb/>(art. </s> <s xml:id="echoid-s10183" xml:space="preserve">622).</s> <s xml:id="echoid-s10184" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div829" type="section" level="1" n="665"> <head xml:id="echoid-head785" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head786" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10185" xml:space="preserve">625. </s> <s xml:id="echoid-s10186" xml:space="preserve">Si l’on coupe un cône par un plan parallele à un de ſes <lb/> <anchor type="note" xlink:label="note-0351-01a" xlink:href="note-0351-01"/> côtés, la ſection ſera une parabole.</s> <s xml:id="echoid-s10187" xml:space="preserve"/> </p> <div xml:id="echoid-div829" type="float" level="2" n="1"> <note position="left" xlink:label="note-0351-01" xlink:href="note-0351-01a" xml:space="preserve">Figure 155.</note> </div> <p> <s xml:id="echoid-s10188" xml:space="preserve">Si l’on a coupé le cône A B C par un plan parallele à un de <lb/>ſes côtés B C, je dis que la ſection qui ſera, par exemple <lb/>D E I, aura formé ſur la ſurface du cône une courbe DHEKI <lb/>qui ſera une parabole. </s> <s xml:id="echoid-s10189" xml:space="preserve">Suppoſons encore que le cône a été <lb/>coupé par un plan L M parallele à ſa baſe, la ſection ſera un <lb/>cercle, dont les lignes F K & </s> <s xml:id="echoid-s10190" xml:space="preserve">F H ſeront des perpendiculaires <lb/>au diametre LM, & </s> <s xml:id="echoid-s10191" xml:space="preserve">en même-tems des ordonnées de la courbe, <lb/>parce que l’on ſuppoſe que le plan coupant E D I eſt perpen-<lb/>diculaire au plan du triangle A B C, que l’on appelle le triangle <lb/>par l’axe. </s> <s xml:id="echoid-s10192" xml:space="preserve">Cela poſé, prenez ſur le côté B C la partie B O <lb/>égale à F M, & </s> <s xml:id="echoid-s10193" xml:space="preserve">du point O, menez à F M la parallele O N, <lb/>qui ſera le parametre de la parabole; </s> <s xml:id="echoid-s10194" xml:space="preserve">car nous démontrerons <lb/>que le rectangle compris ſous N O, & </s> <s xml:id="echoid-s10195" xml:space="preserve">l’abſciſſe E F, eſt égal <lb/>au quarré de l’ordonnée F K; </s> <s xml:id="echoid-s10196" xml:space="preserve">après avoir nommé les lignes <lb/>B O ou F M, a; </s> <s xml:id="echoid-s10197" xml:space="preserve">N O, p; </s> <s xml:id="echoid-s10198" xml:space="preserve">E F, x, & </s> <s xml:id="echoid-s10199" xml:space="preserve">F K, y.</s> <s xml:id="echoid-s10200" xml:space="preserve"/> </p> <pb o="298" file="0352" n="360" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div831" type="section" level="1" n="666"> <head xml:id="echoid-head787" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10201" xml:space="preserve">Les triangles BNO, EFL ayant les côtés paralleles chacun <lb/>à chacun, ſeront ſemblables, & </s> <s xml:id="echoid-s10202" xml:space="preserve">donneront BO (a) : </s> <s xml:id="echoid-s10203" xml:space="preserve">ON (p) :</s> <s xml:id="echoid-s10204" xml:space="preserve">: <lb/>EF (x) : </s> <s xml:id="echoid-s10205" xml:space="preserve">F L ({px/a}), d’où l’on tire B O x F L, ou F M x F L <lb/>= O N x EF, & </s> <s xml:id="echoid-s10206" xml:space="preserve">analytiquement p x = {apx/a}; </s> <s xml:id="echoid-s10207" xml:space="preserve">mais par la pro-<lb/>priété du cercle F M x F L = F K: </s> <s xml:id="echoid-s10208" xml:space="preserve">donc on aura p x = y y. <lb/></s> <s xml:id="echoid-s10209" xml:space="preserve">C. </s> <s xml:id="echoid-s10210" xml:space="preserve">Q. </s> <s xml:id="echoid-s10211" xml:space="preserve">F. </s> <s xml:id="echoid-s10212" xml:space="preserve">D.</s> <s xml:id="echoid-s10213" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div832" type="section" level="1" n="667"> <head xml:id="echoid-head788" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10214" xml:space="preserve">626. </s> <s xml:id="echoid-s10215" xml:space="preserve">Si le triangle par l’axe eſt équilatéral, la ligne F M <lb/>compriſe entre l’axe de la parabole & </s> <s xml:id="echoid-s10216" xml:space="preserve">le côté B C du cône, <lb/>ſera égale au parametre de la parabole; </s> <s xml:id="echoid-s10217" xml:space="preserve">car il eſt évident que <lb/>l’abſciſſe LF ſera dans ce cas égale à l’abſciſſe E F.</s> <s xml:id="echoid-s10218" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div833" type="section" level="1" n="668"> <head xml:id="echoid-head789" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s10219" xml:space="preserve">627. </s> <s xml:id="echoid-s10220" xml:space="preserve">Décrire une parabole, le parametre étant donné. <lb/></s> </p> <p> <s xml:id="echoid-s10221" xml:space="preserve">Pour décrire une parabole, dont la ligne A B ſoit le pa-<lb/>rametre, prenez dans une ligne telle que E K, les parties <lb/>C E & </s> <s xml:id="echoid-s10222" xml:space="preserve">C F, chacune égale au quart du parametre A B; </s> <s xml:id="echoid-s10223" xml:space="preserve">en-<lb/>ſuite tirez ſur la ligne E K un nombre in déterminé de perpen-<lb/>diculaires telles que G H, & </s> <s xml:id="echoid-s10224" xml:space="preserve">faites les lignes F G, F H chacune <lb/>égale à la ligne E I, ou, ce qui eſt la même choſe, du point F <lb/>comme centre avec le rayon E I, décrivez un arc de cercle qui <lb/>coupe la ligne G H aux points déterminés H & </s> <s xml:id="echoid-s10225" xml:space="preserve">G. </s> <s xml:id="echoid-s10226" xml:space="preserve">La courbe <lb/>qui paſſera par ces points ſera une parabole. </s> <s xml:id="echoid-s10227" xml:space="preserve">La démonſtration <lb/>eſt la même que celle de la premiere propoſition.</s> <s xml:id="echoid-s10228" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div834" type="section" level="1" n="669"> <head xml:id="echoid-head790" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s10229" xml:space="preserve">628. </s> <s xml:id="echoid-s10230" xml:space="preserve">Trouver l’axe d’une parabole donnée.</s> <s xml:id="echoid-s10231" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 157.</note> <p> <s xml:id="echoid-s10232" xml:space="preserve">Pour trouver l’axe d’une parabole donnée CLI, on n’a <lb/>qu’à tirer par tels points que l’on voudra de la parabole deux <lb/>lignes A B & </s> <s xml:id="echoid-s10233" xml:space="preserve">C D paralleles entr’elles, diviſer chacune de ces <lb/>lignes en deux également aux points E, F, & </s> <s xml:id="echoid-s10234" xml:space="preserve">tirer par ces <lb/>points la ligne G F H qui ſera un diametre, puiſqu’elle diviſe <pb o="299" file="0353" n="361" rhead="DE MATHÉMATIQUES. Liv. IX."/> deux lignes paralleles en deux également; </s> <s xml:id="echoid-s10235" xml:space="preserve">enſuite du point C <lb/>tirer la ligne C I perpendiculaire ſur G H, diviſer cette ligne <lb/>en deux également au point K; </s> <s xml:id="echoid-s10236" xml:space="preserve">& </s> <s xml:id="echoid-s10237" xml:space="preserve">ſi à ce point vous élevez <lb/>la perpendiculaire K L, elle ſera l’axe de la parabole.</s> <s xml:id="echoid-s10238" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div835" type="section" level="1" n="670"> <head xml:id="echoid-head791" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10239" xml:space="preserve">Les lignes A B & </s> <s xml:id="echoid-s10240" xml:space="preserve">C D étant des ordonnées au diametre G H, <lb/>la ligne C I perpendiculaire à ce diametre, ſera auſſi perpen-<lb/>diculaire à l’axe, puiſque l’axe eſt parallele au diametre, & </s> <s xml:id="echoid-s10241" xml:space="preserve">cette <lb/>même ligne ſera une double ordonnée à l’axe: </s> <s xml:id="echoid-s10242" xml:space="preserve">donc la ligne <lb/>K L qui paſſe par ſon milieu eſt l’axe demandé, puiſque l’axe <lb/>diviſe ſes doubles ordonnées en deux également.</s> <s xml:id="echoid-s10243" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div836" type="section" level="1" n="671"> <head xml:id="echoid-head792" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s10244" xml:space="preserve">629. </s> <s xml:id="echoid-s10245" xml:space="preserve">Trouver le parametre d’une parabole donnée.</s> <s xml:id="echoid-s10246" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 157.</note> <p> <s xml:id="echoid-s10247" xml:space="preserve">Pour trouver le parametre d’une parabole donnée, il ne faut <lb/>que chercher à une abſciſſe quelconque L M, & </s> <s xml:id="echoid-s10248" xml:space="preserve">à l’ordonnée <lb/>correſpondante M N, une troiſieme proportionnelle (art. </s> <s xml:id="echoid-s10249" xml:space="preserve">602) <lb/>qui ſera, par exemple O P, & </s> <s xml:id="echoid-s10250" xml:space="preserve">cette ligne O P ſera le para-<lb/>metre que l’on demande, puiſque le rectangle compris ſous <lb/>L M & </s> <s xml:id="echoid-s10251" xml:space="preserve">O P ſera égal au quarré de l’ordonnée M N. </s> <s xml:id="echoid-s10252" xml:space="preserve">(art. </s> <s xml:id="echoid-s10253" xml:space="preserve">604).</s> <s xml:id="echoid-s10254" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div837" type="section" level="1" n="672"> <head xml:id="echoid-head793" xml:space="preserve">PROPOSITION XII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s10255" xml:space="preserve">630. </s> <s xml:id="echoid-s10256" xml:space="preserve">Trouver le foyer d’une parabole dont on connoît le para-<lb/> <anchor type="note" xlink:label="note-0353-02a" xlink:href="note-0353-02"/> metre.</s> <s xml:id="echoid-s10257" xml:space="preserve"/> </p> <div xml:id="echoid-div837" type="float" level="2" n="1"> <note position="right" xlink:label="note-0353-02" xlink:href="note-0353-02a" xml:space="preserve">Figure 157.</note> </div> <p> <s xml:id="echoid-s10258" xml:space="preserve">Pour trouver le foyer d’une parabole, il faut prendre dans <lb/>l’axe L K une partie L Q, égale au quart du parametre O P, <lb/>& </s> <s xml:id="echoid-s10259" xml:space="preserve">le point Q ſera le foyer qu’on demande; </s> <s xml:id="echoid-s10260" xml:space="preserve">ce qui eſt bien <lb/>évident, puiſque par la génération de la parabole, le parametre <lb/>eſt quadruple de la diſtance du foyer Q au ſommet L de la para-<lb/>bole (art. </s> <s xml:id="echoid-s10261" xml:space="preserve">620).</s> <s xml:id="echoid-s10262" xml:space="preserve"/> </p> <figure> <image file="0353-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0353-01"/> </figure> <pb o="300" file="0354" n="362" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div839" type="section" level="1" n="673"> <head xml:id="echoid-head794" xml:space="preserve">CHAPITRE II.</head> <head xml:id="echoid-head795" xml:space="preserve">Qui traite de l’Ellipſe. <lb/><emph style="sc">Definitions</emph>.</head> <p> <s xml:id="echoid-s10263" xml:space="preserve">631. </s> <s xml:id="echoid-s10264" xml:space="preserve">A Yant tiré ſur un plan deux lignes droites & </s> <s xml:id="echoid-s10265" xml:space="preserve">inégales <lb/> <anchor type="note" xlink:label="note-0354-01a" xlink:href="note-0354-01"/> A B & </s> <s xml:id="echoid-s10266" xml:space="preserve">C D, qui ſe coupent par le milieu à angles droits au <lb/> <anchor type="note" xlink:label="note-0354-02a" xlink:href="note-0354-02"/> point E; </s> <s xml:id="echoid-s10267" xml:space="preserve">ſi l’on décrit un demi-cercle, dont le diametre ſoit <lb/>la plus grande A B, & </s> <s xml:id="echoid-s10268" xml:space="preserve">que l’on éleve ſur ce diametre quantité <lb/>de perpendiculaires, comme F G & </s> <s xml:id="echoid-s10269" xml:space="preserve">I K, &</s> <s xml:id="echoid-s10270" xml:space="preserve">c. </s> <s xml:id="echoid-s10271" xml:space="preserve">& </s> <s xml:id="echoid-s10272" xml:space="preserve">qu’enſuite <lb/>on faſſe F H quatrieme proportionnelle aux lignes A B, C D, <lb/>F G, & </s> <s xml:id="echoid-s10273" xml:space="preserve">de même I L, quatrieme proportionnelle à A B, C D <lb/>& </s> <s xml:id="echoid-s10274" xml:space="preserve">I K, & </s> <s xml:id="echoid-s10275" xml:space="preserve">que l’on continue à trouver de la même maniere <lb/>une quantité de points, tels que H & </s> <s xml:id="echoid-s10276" xml:space="preserve">L, la courbe qu’on fera <lb/>paſſer par tous ces points ſera nommée ellipſe.</s> <s xml:id="echoid-s10277" xml:space="preserve"/> </p> <div xml:id="echoid-div839" type="float" level="2" n="1"> <note position="left" xlink:label="note-0354-01" xlink:href="note-0354-01a" xml:space="preserve">Planche IX.</note> <note position="left" xlink:label="note-0354-02" xlink:href="note-0354-02a" xml:space="preserve">Figure 158.</note> </div> <p> <s xml:id="echoid-s10278" xml:space="preserve">632. </s> <s xml:id="echoid-s10279" xml:space="preserve">La ligne A B eſt nommée grand axe de l’ellipſe, & </s> <s xml:id="echoid-s10280" xml:space="preserve">la <lb/>ligne C D, qu’on ſuppoſe perpendiculaire ſur le milieu de A B, <lb/>eſt appellée petit axe. </s> <s xml:id="echoid-s10281" xml:space="preserve">On dit auſſi que la ligne C D eſt l’axe <lb/>conjugué à l’axe A B, & </s> <s xml:id="echoid-s10282" xml:space="preserve">réciproquement que l’axe A B eſt con-<lb/>jugué à l’axe C D.</s> <s xml:id="echoid-s10283" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10284" xml:space="preserve">633. </s> <s xml:id="echoid-s10285" xml:space="preserve">Les lignes telles que F H, I L perpendiculaires à l’axe <lb/>A B ſont appellées ordonnées au même axe; </s> <s xml:id="echoid-s10286" xml:space="preserve">les lignes I K, F G <lb/>ſont appellées ordonnées du cercle, & </s> <s xml:id="echoid-s10287" xml:space="preserve">en les comparant aux or-<lb/>données de l’ellipſe, qui en font partie, on les appelle toutes <lb/>ordonnées correſpondantes. </s> <s xml:id="echoid-s10288" xml:space="preserve">D’où il ſuit que l’ellipſe eſt une courbe <lb/>dont les ordonnées ſont toujours aux ordonnées d’un cercle décrit <lb/>ſur ſon grand axe dans un rapport conſtant, qui eſt celui du grand <lb/>axe A B à ſon conjugué C D; </s> <s xml:id="echoid-s10289" xml:space="preserve">ce qui donne cette analogie pour <lb/>une ordonnée quelconque F H; </s> <s xml:id="echoid-s10290" xml:space="preserve">A B : </s> <s xml:id="echoid-s10291" xml:space="preserve">C D :</s> <s xml:id="echoid-s10292" xml:space="preserve">: F G : </s> <s xml:id="echoid-s10293" xml:space="preserve">F H.</s> <s xml:id="echoid-s10294" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10295" xml:space="preserve">634. </s> <s xml:id="echoid-s10296" xml:space="preserve">Si l’on cherche une troiſieme proportionnelle aux axes <lb/>A B & </s> <s xml:id="echoid-s10297" xml:space="preserve">C D, telle que M N; </s> <s xml:id="echoid-s10298" xml:space="preserve">cette ligne eſt nommée parametre <lb/>de l’axe qui occupe le premier terme de la proportion con-<lb/>tinue.</s> <s xml:id="echoid-s10299" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10300" xml:space="preserve">635. </s> <s xml:id="echoid-s10301" xml:space="preserve">Le point E, où les axes ſe coupent à angles droits, eſt <lb/>appellé centre de l’ellipſe.</s> <s xml:id="echoid-s10302" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10303" xml:space="preserve">636. </s> <s xml:id="echoid-s10304" xml:space="preserve">Si dans le grand axe A B d’une ellipſe on prend les <lb/> <anchor type="note" xlink:label="note-0354-03a" xlink:href="note-0354-03"/> points K, K, chacun éloigné des extrêmités du petit axe de <lb/>la quantité K D = A E, c’eſt-à-dire de la diſtance du grand <pb o="301" file="0355" n="363" rhead="DE MATHÉMATIQUE. Liv. IX."/> demi-grand axe, ces points ſeront nommés foyers de l’ellipſe.</s> <s xml:id="echoid-s10305" xml:space="preserve"/> </p> <div xml:id="echoid-div840" type="float" level="2" n="2"> <note position="left" xlink:label="note-0354-03" xlink:href="note-0354-03a" xml:space="preserve">Figure 159.</note> </div> <p> <s xml:id="echoid-s10306" xml:space="preserve">637. </s> <s xml:id="echoid-s10307" xml:space="preserve">Les parties A F, F B d’un axe faites par la rencontre <lb/>d’une ordonnée F G à cet axe, ſont appellées abſciſſes ou cou-<lb/>pées de cet axe, par rapport à l’ordonnée F G: </s> <s xml:id="echoid-s10308" xml:space="preserve">on appelle auſſi <lb/>quelquefois abſciſſes les parties compriſes entre le centre & </s> <s xml:id="echoid-s10309" xml:space="preserve"><lb/>la rencontre d’une ordonnée, comme E F; </s> <s xml:id="echoid-s10310" xml:space="preserve">alors on dit que <lb/>les abſciſſes ont leur origine au centre.</s> <s xml:id="echoid-s10311" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div842" type="section" level="1" n="674"> <head xml:id="echoid-head796" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10312" xml:space="preserve">638. </s> <s xml:id="echoid-s10313" xml:space="preserve">Dans l’ellipſe ſi l’on mene une ordonnée F H au premier <lb/> <anchor type="note" xlink:label="note-0355-01a" xlink:href="note-0355-01"/> axe, je dis que le rectangle des abſciſſes A F, F B de cet axe eſt au <lb/>quarré de l’ordonnée F H, comme le quarré du premier axe A B eſt <lb/>au quarré du ſecond axe C D; </s> <s xml:id="echoid-s10314" xml:space="preserve">ou, ce qui eſt la même choſe, comme <lb/>le quarré de A E eſt au quarré de D E.</s> <s xml:id="echoid-s10315" xml:space="preserve"/> </p> <div xml:id="echoid-div842" type="float" level="2" n="1"> <note position="right" xlink:label="note-0355-01" xlink:href="note-0355-01a" xml:space="preserve">Figure 159.</note> </div> <p> <s xml:id="echoid-s10316" xml:space="preserve">Ayant nommé les données A E ou E B, a; </s> <s xml:id="echoid-s10317" xml:space="preserve">C E ou E D, b; <lb/></s> <s xml:id="echoid-s10318" xml:space="preserve">& </s> <s xml:id="echoid-s10319" xml:space="preserve">les indéterminées E F, x; </s> <s xml:id="echoid-s10320" xml:space="preserve">F H, y; </s> <s xml:id="echoid-s10321" xml:space="preserve">F G, s; </s> <s xml:id="echoid-s10322" xml:space="preserve">A F ſera a - x; </s> <s xml:id="echoid-s10323" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s10324" xml:space="preserve">F B a + x. </s> <s xml:id="echoid-s10325" xml:space="preserve">Cela poſé, il faut démontrer que l’on aura <lb/>A F x F B : </s> <s xml:id="echoid-s10326" xml:space="preserve">FH<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10327" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10328" xml:space="preserve">CD<emph style="sub">2</emph>, ou :</s> <s xml:id="echoid-s10329" xml:space="preserve">: AE<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10330" xml:space="preserve">DE<emph style="sub">2</emph>, ou que <lb/>aa - x x : </s> <s xml:id="echoid-s10331" xml:space="preserve">y y :</s> <s xml:id="echoid-s10332" xml:space="preserve">: a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10333" xml:space="preserve">b<emph style="sub">2</emph>.</s> <s xml:id="echoid-s10334" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div844" type="section" level="1" n="675"> <head xml:id="echoid-head797" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s10335" xml:space="preserve">Par la définition de l’ellipſe, chaque ordonnée étant qua-<lb/>trieme proportionnelle au grand axe A B, au petit axe C D, <lb/>& </s> <s xml:id="echoid-s10336" xml:space="preserve">à l’ordonnée F G, on a A B : </s> <s xml:id="echoid-s10337" xml:space="preserve">C D :</s> <s xml:id="echoid-s10338" xml:space="preserve">: F G : </s> <s xml:id="echoid-s10339" xml:space="preserve">F H, ou <lb/>2a : </s> <s xml:id="echoid-s10340" xml:space="preserve">2b :</s> <s xml:id="echoid-s10341" xml:space="preserve">: s : </s> <s xml:id="echoid-s10342" xml:space="preserve">y : </s> <s xml:id="echoid-s10343" xml:space="preserve">donc AB<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10344" xml:space="preserve">CD<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10345" xml:space="preserve">: FG<emph style="sub">2</emph>:</s> <s xml:id="echoid-s10346" xml:space="preserve">FH<emph style="sub">2</emph>, ou 4a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10347" xml:space="preserve">4b<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10348" xml:space="preserve">: ss : </s> <s xml:id="echoid-s10349" xml:space="preserve">yy. <lb/></s> <s xml:id="echoid-s10350" xml:space="preserve">Mais par la propriété du cercle, le quarré de l’ordonnée F G <lb/>eſt égal au produit de ſes abſciſſes, ou A F x F B = F G<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s10351" xml:space="preserve"><lb/>analytiquements ss = a a - x x : </s> <s xml:id="echoid-s10352" xml:space="preserve">donc en mettant cette expreſ-<lb/>ſion au lieu de ss dans la proportion précédente, on aura <lb/>4a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10353" xml:space="preserve">4b<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10354" xml:space="preserve">: aa - xx : </s> <s xml:id="echoid-s10355" xml:space="preserve">yy, ou bien invertendo aa - x x : </s> <s xml:id="echoid-s10356" xml:space="preserve">y y :</s> <s xml:id="echoid-s10357" xml:space="preserve">: <lb/>4a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10358" xml:space="preserve">4b<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10359" xml:space="preserve">: a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10360" xml:space="preserve">b<emph style="sub">2</emph>, en diviſant les termes de la ſeconde raiſon <lb/>par 4.</s> <s xml:id="echoid-s10361" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div845" type="section" level="1" n="676"> <head xml:id="echoid-head798" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s10362" xml:space="preserve">639. </s> <s xml:id="echoid-s10363" xml:space="preserve">Si l’on a deux ordonnées F H & </s> <s xml:id="echoid-s10364" xml:space="preserve">I L, l’on aura par la <lb/> <anchor type="note" xlink:label="note-0355-02a" xlink:href="note-0355-02"/> propoſition précédente, A F x F B : </s> <s xml:id="echoid-s10365" xml:space="preserve">F H<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10366" xml:space="preserve">: </s> <s xml:id="echoid-s10367" xml:space="preserve">A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10368" xml:space="preserve">C D<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s10369" xml:space="preserve"><lb/>AI x IB : </s> <s xml:id="echoid-s10370" xml:space="preserve">IL<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10371" xml:space="preserve">: AB<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10372" xml:space="preserve">CD<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10373" xml:space="preserve">donc AF x FB : </s> <s xml:id="echoid-s10374" xml:space="preserve">FH<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10375" xml:space="preserve">: AI x IB : </s> <s xml:id="echoid-s10376" xml:space="preserve">IL<emph style="sub">2</emph>, <lb/>ou alternando, A F x F B : </s> <s xml:id="echoid-s10377" xml:space="preserve">A I x I B :</s> <s xml:id="echoid-s10378" xml:space="preserve">: F H<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10379" xml:space="preserve">I L<emph style="sub">2</emph>, c’eſt-à-dire <pb o="302" file="0356" n="364" rhead="NOUVEAU COURS"/> que les quarrés des ordonnées F H, I L ſont entr’eux comme <lb/>les produits de leurs abſciſſes.</s> <s xml:id="echoid-s10380" xml:space="preserve"/> </p> <div xml:id="echoid-div845" type="float" level="2" n="1"> <note position="right" xlink:label="note-0355-02" xlink:href="note-0355-02a" xml:space="preserve">Figure 158.</note> </div> </div> <div xml:id="echoid-div847" type="section" level="1" n="677"> <head xml:id="echoid-head799" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s10381" xml:space="preserve">640. </s> <s xml:id="echoid-s10382" xml:space="preserve">Il ſuit encore delà, que ſi du point H l’on mene l’or-<lb/>donnée H I au ſecond axe C D, le rectangle compris ſous les <lb/> <anchor type="note" xlink:label="note-0356-01a" xlink:href="note-0356-01"/> les parties I C, I D eſt au quarré de l’ordonnée correſpondante <lb/>I H, comme le quarré du même axe C D eſt au quarré de ſon <lb/>conjugué A B.</s> <s xml:id="echoid-s10383" xml:space="preserve"/> </p> <div xml:id="echoid-div847" type="float" level="2" n="1"> <note position="left" xlink:label="note-0356-01" xlink:href="note-0356-01a" xml:space="preserve">Figure 159.</note> </div> <p> <s xml:id="echoid-s10384" xml:space="preserve">Pour le prouver, conſidérez que F H étant égale à E I, on <lb/>aura E I = y, & </s> <s xml:id="echoid-s10385" xml:space="preserve">que F E étant égale à H I, on aura encore <lb/>H I = x; </s> <s xml:id="echoid-s10386" xml:space="preserve">ainſi I D ſera b - y, & </s> <s xml:id="echoid-s10387" xml:space="preserve">C I ſera b + y. </s> <s xml:id="echoid-s10388" xml:space="preserve">Cela poſé, <lb/>puiſque par la propoſition préſente, on a aa - xx : </s> <s xml:id="echoid-s10389" xml:space="preserve">yy :</s> <s xml:id="echoid-s10390" xml:space="preserve">: aa : </s> <s xml:id="echoid-s10391" xml:space="preserve">bb, <lb/>en prenant le produit des extrêmes & </s> <s xml:id="echoid-s10392" xml:space="preserve">des moyens, on aura <lb/>a a y y = a a b b - b b x x. </s> <s xml:id="echoid-s10393" xml:space="preserve">Si l’on fait paſſer - bbxx du ſecond <lb/>membre dans le premier, & </s> <s xml:id="echoid-s10394" xml:space="preserve">aayy du premier dans le ſecond, <lb/>il viendra b b x x = a a b b - a a y y, d’où l’on tire cette pro-<lb/>portion b b - y y : </s> <s xml:id="echoid-s10395" xml:space="preserve">x x :</s> <s xml:id="echoid-s10396" xml:space="preserve">: b b : </s> <s xml:id="echoid-s10397" xml:space="preserve">aa, c’eſt-à-dire que I D x D C: <lb/></s> <s xml:id="echoid-s10398" xml:space="preserve">I H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10399" xml:space="preserve">: D E<emph style="sub">2</emph>: </s> <s xml:id="echoid-s10400" xml:space="preserve">A E<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10401" xml:space="preserve">Ainſi l’on voit que les propriétés des or-<lb/>données au petit axe ſont préciſément les mêmes que celles du <lb/>grand axe; </s> <s xml:id="echoid-s10402" xml:space="preserve">d’où l’on peut conclure que les ordonnées H I au <lb/>petit axe de l’ellipſe, ſont troiſiemes proportionnelles au demi <lb/>petit axe, au demi-grand axe, & </s> <s xml:id="echoid-s10403" xml:space="preserve">à l’ordonnée I N d’un cercle <lb/>décrit ſur le petit axe; </s> <s xml:id="echoid-s10404" xml:space="preserve">c’eſt ce qu’il eſt aiſé de voir, ſi l’on fait <lb/>attention que dans la proportion I D x D C : </s> <s xml:id="echoid-s10405" xml:space="preserve">I H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10406" xml:space="preserve">: A E<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10407" xml:space="preserve">D E<emph style="sub">2</emph>, <lb/>on peut mettre au lieu du rectangle I D x D C le quarré de l’or-<lb/>donnée IN, qui lui eſt égal; </s> <s xml:id="echoid-s10408" xml:space="preserve">d’où l’on déduit, en prenant les <lb/>racines, & </s> <s xml:id="echoid-s10409" xml:space="preserve">faiſant un invertendo D E : </s> <s xml:id="echoid-s10410" xml:space="preserve">A E :</s> <s xml:id="echoid-s10411" xml:space="preserve">: I N : </s> <s xml:id="echoid-s10412" xml:space="preserve">I H. </s> <s xml:id="echoid-s10413" xml:space="preserve">On <lb/>peut donc définir l’ellipſe d’une maniere plus générale, en di-<lb/>ſant que c’eſt une courbe, dont toutes les ordonnées ont été <lb/>alongées ou raccourcies proportionnellement; </s> <s xml:id="echoid-s10414" xml:space="preserve">alongées, lorſ-<lb/>que le cercle eſt décrit ſur le petit axe, & </s> <s xml:id="echoid-s10415" xml:space="preserve">raccourcies, lorſ-<lb/>qu’il eſt décrit ſur le grand axe.</s> <s xml:id="echoid-s10416" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div849" type="section" level="1" n="678"> <head xml:id="echoid-head800" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s10417" xml:space="preserve">641. </s> <s xml:id="echoid-s10418" xml:space="preserve">Si l’on nomme a le premier axe d’une ellipſe, & </s> <s xml:id="echoid-s10419" xml:space="preserve">b le <lb/>ſecond, p le parametre du premier axe, on aura (art. </s> <s xml:id="echoid-s10420" xml:space="preserve">634) <lb/>a : </s> <s xml:id="echoid-s10421" xml:space="preserve">b :</s> <s xml:id="echoid-s10422" xml:space="preserve">: b : </s> <s xml:id="echoid-s10423" xml:space="preserve">p, & </s> <s xml:id="echoid-s10424" xml:space="preserve">(art. </s> <s xml:id="echoid-s10425" xml:space="preserve">503) a a : </s> <s xml:id="echoid-s10426" xml:space="preserve">b b :</s> <s xml:id="echoid-s10427" xml:space="preserve">: a : </s> <s xml:id="echoid-s10428" xml:space="preserve">p. </s> <s xml:id="echoid-s10429" xml:space="preserve">Mais par la propriété <lb/>de l’ellipſe, on a a a - x x : </s> <s xml:id="echoid-s10430" xml:space="preserve">y y :</s> <s xml:id="echoid-s10431" xml:space="preserve">: a a : </s> <s xml:id="echoid-s10432" xml:space="preserve">b b; </s> <s xml:id="echoid-s10433" xml:space="preserve">donc on aura auſſi <lb/>aa - xx : </s> <s xml:id="echoid-s10434" xml:space="preserve">y y :</s> <s xml:id="echoid-s10435" xml:space="preserve">: a : </s> <s xml:id="echoid-s10436" xml:space="preserve">p; </s> <s xml:id="echoid-s10437" xml:space="preserve">d’où l’on tire y y = aa - xx x {p/a}, c’eſt- <pb o="303" file="0357" n="365" rhead="DE MATHEMATIQUE. Liv. IX."/> à-dire que le quarré d’une ordonnée quelconque eſt égal au <lb/>produit de ſes abſciſſes, multiplié par le rapport du parame-<lb/>tre à l’axe: </s> <s xml:id="echoid-s10438" xml:space="preserve">ainſi, ſi l’on ſçait que le parametre eſt les deux tiers <lb/>de l’axe, le quarré de chaque ordonnée ſera égal aux deux <lb/>tiers du rectangle des abſciſſes correſpondantes.</s> <s xml:id="echoid-s10439" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div850" type="section" level="1" n="679"> <head xml:id="echoid-head801" xml:space="preserve"><emph style="sc">Remarque</emph> I.</head> <p> <s xml:id="echoid-s10440" xml:space="preserve">642. </s> <s xml:id="echoid-s10441" xml:space="preserve">Il eſt à remarquer que puiſque l’on a A F x F B : </s> <s xml:id="echoid-s10442" xml:space="preserve">F H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10443" xml:space="preserve">: <lb/> <anchor type="note" xlink:label="note-0357-01a" xlink:href="note-0357-01"/> A I x I B : </s> <s xml:id="echoid-s10444" xml:space="preserve">I L<emph style="sub">2</emph>, ſi l’on met à la place des rectangles A F x F B, <lb/>A I x I B, les quarrés des ordonnées F G, I K, qui leur ſont <lb/>égaux par la propriété du cercle, on aura F G<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10445" xml:space="preserve">F H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10446" xml:space="preserve">: I K<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10447" xml:space="preserve">IL<emph style="sub">2</emph>; <lb/></s> <s xml:id="echoid-s10448" xml:space="preserve">& </s> <s xml:id="echoid-s10449" xml:space="preserve">en tirant les racines de chaque terme, F G : </s> <s xml:id="echoid-s10450" xml:space="preserve">F H :</s> <s xml:id="echoid-s10451" xml:space="preserve">: I K : </s> <s xml:id="echoid-s10452" xml:space="preserve">I L, <lb/>& </s> <s xml:id="echoid-s10453" xml:space="preserve">alternando, F G : </s> <s xml:id="echoid-s10454" xml:space="preserve">I K :</s> <s xml:id="echoid-s10455" xml:space="preserve">: F H : </s> <s xml:id="echoid-s10456" xml:space="preserve">I L, qui fait voir que ſi l’on <lb/>prend les lignes F H, I L pour les élémens de la ſuperficie du <lb/>quart d’ellipſe E A D, & </s> <s xml:id="echoid-s10457" xml:space="preserve">les lignes F G, I K pour les élémens <lb/>du quart de cercle E A M; </s> <s xml:id="echoid-s10458" xml:space="preserve">les élémens du quart d’ellipſe ſont <lb/>dans la même raiſon que les élémens correſpondans du quart <lb/>de cercle.</s> <s xml:id="echoid-s10459" xml:space="preserve"/> </p> <div xml:id="echoid-div850" type="float" level="2" n="1"> <note position="right" xlink:label="note-0357-01" xlink:href="note-0357-01a" xml:space="preserve">Figure 158.</note> </div> </div> <div xml:id="echoid-div852" type="section" level="1" n="680"> <head xml:id="echoid-head802" xml:space="preserve"><emph style="sc">Remarque</emph> II.</head> <p> <s xml:id="echoid-s10460" xml:space="preserve">643. </s> <s xml:id="echoid-s10461" xml:space="preserve">On a vu (art. </s> <s xml:id="echoid-s10462" xml:space="preserve">569) que dans une progreſſion qui ſe-<lb/>roit compoſée des élémens infinis tels que F G & </s> <s xml:id="echoid-s10463" xml:space="preserve">I K d’un quart <lb/>de cercle, la ſomme des quarrés de tous ces élémens ſeroit <lb/>égale au produit du quarré du plus grand élément E M, par <lb/>les deux tiers de la ligne A E, qui en exprime le nombre: <lb/></s> <s xml:id="echoid-s10464" xml:space="preserve">or comme les élémens de l’ellipſe ont tous un rapport conſtant <lb/>avec les élémens correſpondans du quart de cercle, il s’enſuit <lb/>qu’ils auront la même propriété que ceux du cercle; </s> <s xml:id="echoid-s10465" xml:space="preserve">& </s> <s xml:id="echoid-s10466" xml:space="preserve">que <lb/>par conſéquent ſi l’on a une progreſſion compoſée de termes <lb/>infinis des élémens d’un quart d’ellipſe E A D, la ſomme des <lb/>quarrés de tous les élémens, tels que F H & </s> <s xml:id="echoid-s10467" xml:space="preserve">I L, eſt égale au <lb/>produit du quarré du plus grand élément E D, par les deux <lb/>tiers de la grandeur qui en exprime le nombre, c’eſt-à-dire <lb/>par les deux tiers de la ligne A E.</s> <s xml:id="echoid-s10468" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10469" xml:space="preserve">Comme ces deux remarques nous ſervent beaucoup dans <lb/>la Géométrie pratique, il faut s’attacher à les bien com-<lb/>prendre.</s> <s xml:id="echoid-s10470" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div853" type="section" level="1" n="681"> <head xml:id="echoid-head803" xml:space="preserve"><emph style="sc">Definitions</emph>.</head> <head xml:id="echoid-head804" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s10471" xml:space="preserve">644. </s> <s xml:id="echoid-s10472" xml:space="preserve">L’on nomme diametres d’une ellipſe, deux lignes, <lb/> <anchor type="note" xlink:label="note-0357-02a" xlink:href="note-0357-02"/> <pb o="304" file="0358" n="366" rhead="NOUVEAU COURS"/> comme C D, E F, qui paſſent par le centre de l’ellipſe, & </s> <s xml:id="echoid-s10473" xml:space="preserve">qui <lb/>ſont terminées à cette courbe.</s> <s xml:id="echoid-s10474" xml:space="preserve"/> </p> <div xml:id="echoid-div853" type="float" level="2" n="1"> <note position="right" xlink:label="note-0357-02" xlink:href="note-0357-02a" xml:space="preserve">Figure 160.</note> </div> </div> <div xml:id="echoid-div855" type="section" level="1" n="682"> <head xml:id="echoid-head805" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s10475" xml:space="preserve">645. </s> <s xml:id="echoid-s10476" xml:space="preserve">Ayant mené d’un point quelconque C de l’ellipſe un <lb/> <anchor type="note" xlink:label="note-0358-01a" xlink:href="note-0358-01"/> diametre C D, & </s> <s xml:id="echoid-s10477" xml:space="preserve">une ordonnée C K à l’axe A B, ſi l’on fait <lb/>G O troiſieme proportionnelle à G K & </s> <s xml:id="echoid-s10478" xml:space="preserve">G A, le diametre E F, <lb/>que l’on aura mené parallele à la ligne C O, eſt appellé dia-<lb/>metre conjugué au diametre C D; </s> <s xml:id="echoid-s10479" xml:space="preserve">& </s> <s xml:id="echoid-s10480" xml:space="preserve">réciproquement le dia-<lb/>metre C D eſt dit conjugué au diametre E F.</s> <s xml:id="echoid-s10481" xml:space="preserve"/> </p> <div xml:id="echoid-div855" type="float" level="2" n="1"> <note position="left" xlink:label="note-0358-01" xlink:href="note-0358-01a" xml:space="preserve">Figure 160.</note> </div> </div> <div xml:id="echoid-div857" type="section" level="1" n="683"> <head xml:id="echoid-head806" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s10482" xml:space="preserve">646. </s> <s xml:id="echoid-s10483" xml:space="preserve">Toute ligne, comme H I, menée d’un point quelcon-<lb/>que H, pris dans le diametre C D, parallélement à ſon con-<lb/>jugué E F, eſt appellée ordonnée au diametre C D.</s> <s xml:id="echoid-s10484" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div858" type="section" level="1" n="684"> <head xml:id="echoid-head807" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s10485" xml:space="preserve">647. </s> <s xml:id="echoid-s10486" xml:space="preserve">Si l’on cherche une troiſieme proportionnelle aux dia-<lb/>metres conjugués C D, E F, elle ſera nommée parametre du <lb/>diametre, qui occupe le premier terme de la proportion.</s> <s xml:id="echoid-s10487" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div859" type="section" level="1" n="685"> <head xml:id="echoid-head808" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10488" xml:space="preserve">648. </s> <s xml:id="echoid-s10489" xml:space="preserve">Puiſque l’on a fait (art. </s> <s xml:id="echoid-s10490" xml:space="preserve">645) G K : </s> <s xml:id="echoid-s10491" xml:space="preserve">G A :</s> <s xml:id="echoid-s10492" xml:space="preserve">: G A : </s> <s xml:id="echoid-s10493" xml:space="preserve">G O, <lb/>il s’enſuit que ſi l’on nomme G K, x; </s> <s xml:id="echoid-s10494" xml:space="preserve">G A, a; </s> <s xml:id="echoid-s10495" xml:space="preserve">K O, z, l’on <lb/>aura G K (x) : </s> <s xml:id="echoid-s10496" xml:space="preserve">G A (a) :</s> <s xml:id="echoid-s10497" xml:space="preserve">: G A (a) : </s> <s xml:id="echoid-s10498" xml:space="preserve">G O (x + z), d’où l’on <lb/>tire x x + zx = aa; </s> <s xml:id="echoid-s10499" xml:space="preserve">& </s> <s xml:id="echoid-s10500" xml:space="preserve">en faiſant paſſer xx du premier mem-<lb/>bre dans le ſecond, z x = aa - xx, ou bien O K x K G = <lb/>A K x K B. </s> <s xml:id="echoid-s10501" xml:space="preserve">Comme ce corollaire nous ſervira beaucoup dans <lb/>les propoſitions ſuivantes, il eſt à propos de le bien retenir.</s> <s xml:id="echoid-s10502" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div860" type="section" level="1" n="686"> <head xml:id="echoid-head809" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head810" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10503" xml:space="preserve">649. </s> <s xml:id="echoid-s10504" xml:space="preserve">Si des extrêmités C & </s> <s xml:id="echoid-s10505" xml:space="preserve">E de deux diametres conjugués <lb/> <anchor type="note" xlink:label="note-0358-02a" xlink:href="note-0358-02"/> C D, E F on mene à l’axe A B les ordonnées C K, E P, je dis <lb/>que le quarré de la partie G P ſera égal au rectangle de A K par K B.</s> <s xml:id="echoid-s10506" xml:space="preserve"/> </p> <div xml:id="echoid-div860" type="float" level="2" n="1"> <note position="left" xlink:label="note-0358-02" xlink:href="note-0358-02a" xml:space="preserve">Figure 160.</note> </div> <p> <s xml:id="echoid-s10507" xml:space="preserve">Ayant fait A G = a, G P = f, G K = x, K O = z, G O <lb/>ſera x + z. </s> <s xml:id="echoid-s10508" xml:space="preserve">Cela poſé, nous ferons voir que A K x K B (aa-xx) <lb/>ou bien x z (art. </s> <s xml:id="echoid-s10509" xml:space="preserve">648) = f f.</s> <s xml:id="echoid-s10510" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div862" type="section" level="1" n="687"> <head xml:id="echoid-head811" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s10511" xml:space="preserve">Conſidérez que l’on a par la propriété de l’ellipſe (art. </s> <s xml:id="echoid-s10512" xml:space="preserve">639) <pb o="305" file="0359" n="367" rhead="DE MATHÉMATIQUE. Liv. IX."/> A K x K B (x z) : </s> <s xml:id="echoid-s10513" xml:space="preserve">A P x P B (aa - f f) :</s> <s xml:id="echoid-s10514" xml:space="preserve">: K C<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10515" xml:space="preserve">P E<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10516" xml:space="preserve">& </s> <s xml:id="echoid-s10517" xml:space="preserve">que <lb/>ſi au lieu de aa dans le ſecond terme de cette proportion on <lb/>met x x + x z, qui lui eſt égal (art. </s> <s xml:id="echoid-s10518" xml:space="preserve">648), & </s> <s xml:id="echoid-s10519" xml:space="preserve">au lieu de K C<emph style="sub">2</emph> <lb/>& </s> <s xml:id="echoid-s10520" xml:space="preserve">P E<emph style="sub">2</emph>, on met K O<emph style="sub">2</emph> (z z) & </s> <s xml:id="echoid-s10521" xml:space="preserve">P G<emph style="sub">2</emph> (f f) qui ſont dans la <lb/>même raiſon, à cauſe des triangles ſemblables C O K, E G P, <lb/>qui donnent C K : </s> <s xml:id="echoid-s10522" xml:space="preserve">P E :</s> <s xml:id="echoid-s10523" xml:space="preserve">: K O : </s> <s xml:id="echoid-s10524" xml:space="preserve">P G, on aura A K x K B : </s> <s xml:id="echoid-s10525" xml:space="preserve">A P <lb/>x P B :</s> <s xml:id="echoid-s10526" xml:space="preserve">: K C<emph style="sub">2</emph>, : </s> <s xml:id="echoid-s10527" xml:space="preserve">P E<emph style="sub">2</emph>, ou x z : </s> <s xml:id="echoid-s10528" xml:space="preserve">xx + xz - f f :</s> <s xml:id="echoid-s10529" xml:space="preserve">: z z : </s> <s xml:id="echoid-s10530" xml:space="preserve">f f, dont le <lb/>produit des extrêmes & </s> <s xml:id="echoid-s10531" xml:space="preserve">des moyens donnent cette équation <lb/>x x z z + x z<emph style="sub">3</emph> - f f z z = f f x z; </s> <s xml:id="echoid-s10532" xml:space="preserve">d’où tranſpoſant f f z z du pre-<lb/>mier membre dans le ſecond, vient xxzz+xz<emph style="sub">3</emph>=ffzz+ffxz, <lb/>& </s> <s xml:id="echoid-s10533" xml:space="preserve">diviſant chaque membre de l’équation par z, il vient x x z <lb/>+ z<emph style="sub">2</emph>x = f f z + f f x, & </s> <s xml:id="echoid-s10534" xml:space="preserve">diviſant encore chaque membre par <lb/>z + x, il vient x z = f f, ou A K x K B = G P<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10535" xml:space="preserve">C. </s> <s xml:id="echoid-s10536" xml:space="preserve">Q. </s> <s xml:id="echoid-s10537" xml:space="preserve">F. </s> <s xml:id="echoid-s10538" xml:space="preserve">D.</s> <s xml:id="echoid-s10539" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div863" type="section" level="1" n="688"> <head xml:id="echoid-head812" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10540" xml:space="preserve">650. </s> <s xml:id="echoid-s10541" xml:space="preserve">Comme on a xx + xz = aa (art. </s> <s xml:id="echoid-s10542" xml:space="preserve">648), il ſuit de cette <lb/>propoſition, que ſi l’on met ff à la place de xz qui lui eſt égal, <lb/>on aura x x + f f = a a; </s> <s xml:id="echoid-s10543" xml:space="preserve">& </s> <s xml:id="echoid-s10544" xml:space="preserve">faiſant paſſer f f du premier mem-<lb/>bre dans le ſecond, on aura G K<emph style="sub">2</emph> (xx) = A P x P B (aa-ff).</s> <s xml:id="echoid-s10545" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div864" type="section" level="1" n="689"> <head xml:id="echoid-head813" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head814" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10546" xml:space="preserve">651. </s> <s xml:id="echoid-s10547" xml:space="preserve">Le rectangle fait des parties C H, H D du diametre C D, <lb/>eſt au quarré d’une ordonnée H I, comme le quarré de ce même dia-<lb/>metre eſt à celui de ſon conjugué E F.</s> <s xml:id="echoid-s10548" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10549" xml:space="preserve">Après avoir tiré les lignes I N, H L paralleles à C K, & </s> <s xml:id="echoid-s10550" xml:space="preserve">la <lb/>ligne H M parallele à A B, nous nommerons G K, x; </s> <s xml:id="echoid-s10551" xml:space="preserve">C K, y; <lb/></s> <s xml:id="echoid-s10552" xml:space="preserve">A G, a; </s> <s xml:id="echoid-s10553" xml:space="preserve">K O, z; </s> <s xml:id="echoid-s10554" xml:space="preserve">M H, ou L N, c; </s> <s xml:id="echoid-s10555" xml:space="preserve">G L, g; </s> <s xml:id="echoid-s10556" xml:space="preserve">G C, s.</s> <s xml:id="echoid-s10557" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div865" type="section" level="1" n="690"> <head xml:id="echoid-head815" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10558" xml:space="preserve">Les triangles G K C, G L H ſont évidemment ſemblables, <lb/>& </s> <s xml:id="echoid-s10559" xml:space="preserve">donnent G K (x) : </s> <s xml:id="echoid-s10560" xml:space="preserve">K C (y) :</s> <s xml:id="echoid-s10561" xml:space="preserve">: G L (g) : </s> <s xml:id="echoid-s10562" xml:space="preserve">L H {g y/x}; </s> <s xml:id="echoid-s10563" xml:space="preserve">& </s> <s xml:id="echoid-s10564" xml:space="preserve">les trian-<lb/>gles C O K, I H M, qui ſont auſſi ſemblables, puiſqu’ils ont les <lb/>côtés paralleles chacun à chacun, nous donnent O K (z) : </s> <s xml:id="echoid-s10565" xml:space="preserve">K C (y) <lb/>:</s> <s xml:id="echoid-s10566" xml:space="preserve">: H M (c) : </s> <s xml:id="echoid-s10567" xml:space="preserve">I M ({cy/z};) </s> <s xml:id="echoid-s10568" xml:space="preserve">d’où l’on tire IM+HL = IM+MN <lb/>ou I N = {g y/x} + {c y/z}, dont le quarré eſt {yygg/xx} + {2cgy<emph style="sub">2</emph>/xz} + {ccyy/zz}. </s> <s xml:id="echoid-s10569" xml:space="preserve">De <lb/>plus, conſidérez que L N - L G = G N = c - g, dont le <pb o="306" file="0360" n="368" rhead="NOUVEAU COURS"/> quarré eſt c c · 2cg + gg. </s> <s xml:id="echoid-s10570" xml:space="preserve">Cela poſé, il faut encore chercher <lb/>une ſeconde valeur de I N<emph style="sub">2</emph>, que l’on trouvera par la propriété <lb/>de l’ellipſe (art. </s> <s xml:id="echoid-s10571" xml:space="preserve">639) : </s> <s xml:id="echoid-s10572" xml:space="preserve">car A K x K B : </s> <s xml:id="echoid-s10573" xml:space="preserve">A N x N B, ou G B<emph style="sub">2</emph> <lb/>- G N<emph style="sub">2</emph> (art. </s> <s xml:id="echoid-s10574" xml:space="preserve">62) :</s> <s xml:id="echoid-s10575" xml:space="preserve">: C K<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10576" xml:space="preserve">I N<emph style="sub">2</emph>, ou analytiquement aa - xx : <lb/></s> <s xml:id="echoid-s10577" xml:space="preserve">aa - cc + 2cg - gg :</s> <s xml:id="echoid-s10578" xml:space="preserve">: yy : </s> <s xml:id="echoid-s10579" xml:space="preserve">yy x √{aa - cc + 2cg - gg/aa - xx}\x{0020} = IN<emph style="sub">2</emph>, ou <lb/>en faiſant la multiplication {aayy - ccyy + 2cgyy - ggyy/aa - xx}. </s> <s xml:id="echoid-s10580" xml:space="preserve">Préſen-<lb/>tement ſi l’on forme une égalité avec ces deux valeurs, on <lb/>aura {ggyy/xx} + {2cgy<emph style="sub">2</emph>/xz} + {ccyy/zz} = {aayy - ccyy + 2cgyy - ggyy/aa - xx}. </s> <s xml:id="echoid-s10581" xml:space="preserve">Mais com-<lb/>me on ſçait que aa - xx = xz, on aura {ccyy/zz} + {2cgyy/zx} + {ggyy/xx} <lb/>= {aayy - ccyy + 2cgyy - ggyy/zx}, ou en effaçant dans chaque mem-<lb/>bre le terme égal {2cgyy/zx}, & </s> <s xml:id="echoid-s10582" xml:space="preserve">diviſant enſuite tout par yy, {c c/zz} + <lb/>{g g/xx} = {aa - cc - gg/aa - xx}. </s> <s xml:id="echoid-s10583" xml:space="preserve">Préſentement il faut multiplier tout par xx, <lb/>afin de n’avoir plus gg en fraction; </s> <s xml:id="echoid-s10584" xml:space="preserve">ce qui donnera {ccxx/zz} + gg <lb/>= {aaxx - ccxx - ggxx/aa - xx}: </s> <s xml:id="echoid-s10585" xml:space="preserve">on fera paſſer gg du premier membre <lb/>dans le ſecond, & </s> <s xml:id="echoid-s10586" xml:space="preserve">on le réduira en fraction, dont le dénomina-<lb/>teur ſoit aa - xx; </s> <s xml:id="echoid-s10587" xml:space="preserve">ce qui donnera cette nouvelle équation <lb/>{ccxx/zz} ou {ccx<emph style="sub">4</emph>/zzxx} = {aaxx - ccxx - aagg + ggxx - ggxx/aa - xx}, faiſant attention <lb/>que le premier membre {ccxx/zz} eſt la même choſe que {ccx<emph style="sub">4</emph>/zzxx}, puiſ-<lb/>que l’on n’a fait que multiplier les deux termes de chaque frac-<lb/>tion par la même grandeur x x. </s> <s xml:id="echoid-s10588" xml:space="preserve">Mais le premier membre de <lb/>cette équation eſt diviſé par le quarré de x z ou de a a - x x, <lb/>qui diviſe le ſecond membre. </s> <s xml:id="echoid-s10589" xml:space="preserve">D’où il ſuit que l’on fera éva-<lb/>nouir toute fraction, en multipliant le numérateur du ſecond <lb/>membre par aa - xx: </s> <s xml:id="echoid-s10590" xml:space="preserve">on aura donc ccx<emph style="sub">4</emph> = √aaxx-ccxx-aagg\x{0020} <lb/>x √aa - xx\x{0020} = a<emph style="sub">4</emph>x<emph style="sub">2</emph> - a<emph style="sub">2</emph>c<emph style="sub">2</emph>x<emph style="sub">2</emph> - a<emph style="sub">4</emph>g<emph style="sub">2</emph> - a<emph style="sub">2</emph>x<emph style="sub">4</emph> + c<emph style="sub">2</emph>x<emph style="sub">4</emph> + a<emph style="sub">2</emph>g<emph style="sub">2</emph>x<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10591" xml:space="preserve"><lb/>d’où l’on tire en effaçant de part & </s> <s xml:id="echoid-s10592" xml:space="preserve">d’autre c<emph style="sub">2</emph>x<emph style="sub">4</emph>, & </s> <s xml:id="echoid-s10593" xml:space="preserve">tranſpo-<lb/>ſant a<emph style="sub">2</emph>c<emph style="sub">2</emph>x<emph style="sub">2</emph> du ſecond membre dans le premier, a<emph style="sub">2</emph>c<emph style="sub">2</emph>x<emph style="sub">2</emph> = a x<emph style="sub">2</emph> <lb/>- a<emph style="sub">4</emph>g<emph style="sub">2</emph> - a<emph style="sub">2</emph>x<emph style="sub">4</emph> + a<emph style="sub">2</emph>g<emph style="sub">2</emph>x<emph style="sub">2</emph>, qu’il faut diviſer par a<emph style="sub">2</emph>x<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10594" xml:space="preserve">ce qui <lb/>donne cc = a<emph style="sub">2</emph> - x<emph style="sub">2</emph> + g<emph style="sub">2</emph> - {a<emph style="sub">2</emph>g<emph style="sub">2</emph>/x<emph style="sub">2</emph>} = LN<emph style="sub">2</emph> = HM<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10595" xml:space="preserve">Cela poſé, <lb/>conſidérez que les triangles ſemblables G K C, G L H donnent <pb o="307" file="0361" n="369" rhead="DE MATHÉMATIQUE. Liv. IX."/> G K (x) : </s> <s xml:id="echoid-s10596" xml:space="preserve">G C (s) : </s> <s xml:id="echoid-s10597" xml:space="preserve">: </s> <s xml:id="echoid-s10598" xml:space="preserve">G L (g) : </s> <s xml:id="echoid-s10599" xml:space="preserve">G H ({gs/x}); </s> <s xml:id="echoid-s10600" xml:space="preserve">par conſéquent <lb/>GC<emph style="sub">2</emph> - GH<emph style="sub">2</emph>, ou C H x H D (art. </s> <s xml:id="echoid-s10601" xml:space="preserve">62)=ss - {g<emph style="sub">2</emph>s<emph style="sub">2</emph>/xx} = {s<emph style="sub">2</emph>xx-g<emph style="sub">2</emph>s<emph style="sub">2</emph>/xx}. <lb/></s> <s xml:id="echoid-s10602" xml:space="preserve">Pour voir préſentement ſi la proportion énoncée au théorême <lb/>eſt vraie, je fais attention que les quatre grandeurs ſuivantes <lb/>C H x H D, H M<emph style="sub">2</emph>, C G<emph style="sub">2</emph>, G P<emph style="sub">2</emph> ſont en proportion, puiſque <lb/>l’on trouve, en diſpoſant leurs expreſſions analytiques, ſelon le <lb/>même ordre, que le produit des extrêmes eſt égal au produit des <lb/>moyens, ou, ce qui eſt la même choſe, que CH x HD ({ssxx-ggss/xx}) <lb/>: </s> <s xml:id="echoid-s10603" xml:space="preserve">H M<emph style="sub">2</emph> (a<emph style="sub">2</emph> - x<emph style="sub">2</emph> + g<emph style="sub">2</emph> - {a<emph style="sub">2</emph>g<emph style="sub">2</emph>/x<emph style="sub">2</emph>}) :</s> <s xml:id="echoid-s10604" xml:space="preserve">: C G<emph style="sub">2</emph> (ss) : </s> <s xml:id="echoid-s10605" xml:space="preserve">G P<emph style="sub">2</emph> (aa-xx) : </s> <s xml:id="echoid-s10606" xml:space="preserve"><lb/>donc en ſubſtituant à la place des conſéquens des quantités <lb/>qui leur ſoient proportionnelles, ſçavoir HI<emph style="sub">2</emph> & </s> <s xml:id="echoid-s10607" xml:space="preserve">GE<emph style="sub">2</emph>, comme <lb/>il eſt évident, à cauſe des triangles ſemblables, MIH, PEG, <lb/>on aura C H x H D : </s> <s xml:id="echoid-s10608" xml:space="preserve">H I<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10609" xml:space="preserve">: C G<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10610" xml:space="preserve">G E<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10611" xml:space="preserve">C. </s> <s xml:id="echoid-s10612" xml:space="preserve">Q. </s> <s xml:id="echoid-s10613" xml:space="preserve">F. </s> <s xml:id="echoid-s10614" xml:space="preserve">D.</s> <s xml:id="echoid-s10615" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div866" type="section" level="1" n="691"> <head xml:id="echoid-head816" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s10616" xml:space="preserve">652. </s> <s xml:id="echoid-s10617" xml:space="preserve">L’on voit que ce qui a été démontré dans la propoſi-<lb/>tion premiere par rapport aux deux axes, s’étend par le moyen <lb/>de celle-ci à deux diametres quelconques : </s> <s xml:id="echoid-s10618" xml:space="preserve">car ſi l’on fait le <lb/>même raiſonnement pour l’ellipſe que pour la parabole (art. </s> <s xml:id="echoid-s10619" xml:space="preserve">622), <lb/> <anchor type="note" xlink:label="note-0361-01a" xlink:href="note-0361-01"/> l’on verra que la tangente H I, à l’extrêmité A de l’axe A B, <lb/>ayant gliſſé le long de la courbe pour prendre la ſituation <lb/>Q R, & </s> <s xml:id="echoid-s10620" xml:space="preserve">l’axe A B ayant tourné pour prendre la ſituation F G, <lb/>l’ordonnée K L qui l’aura accompagnée toujours parallélement <lb/>à la tangente H I, deviendra l’ordonnée O P; </s> <s xml:id="echoid-s10621" xml:space="preserve">& </s> <s xml:id="echoid-s10622" xml:space="preserve">comme l’axe <lb/>conjugué C D aura auſſi tourné parallélement à la tangente <lb/>H I, il deviendra le diametre conjugué M N; </s> <s xml:id="echoid-s10623" xml:space="preserve">& </s> <s xml:id="echoid-s10624" xml:space="preserve">par conſé-<lb/>quent toutes ces lignes demeurant dans des rapports conſtans <lb/>les unes avec les autres, il s’enſuit que le rectangle compris <lb/>ſous les abſciſſes O F, O G eſt quarré de l’ordonnée O P, comme <lb/>le quarré du diametre F G eſt au quarré de ſon conjugué M N.</s> <s xml:id="echoid-s10625" xml:space="preserve"/> </p> <div xml:id="echoid-div866" type="float" level="2" n="1"> <note position="right" xlink:label="note-0361-01" xlink:href="note-0361-01a" xml:space="preserve">Figure 161.</note> </div> </div> <div xml:id="echoid-div868" type="section" level="1" n="692"> <head xml:id="echoid-head817" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s10626" xml:space="preserve">653. </s> <s xml:id="echoid-s10627" xml:space="preserve">Il ſuit encore delà, que pour mener par un point F <lb/>une tangente Q R à l’ellipſe, il faut de ce point abaiſſer une <lb/>perpendiculaire F S à l’axe A B, & </s> <s xml:id="echoid-s10628" xml:space="preserve">faire E Q troiſieme pro-<lb/>portionnelle aux droites ES, EA (art. </s> <s xml:id="echoid-s10629" xml:space="preserve">645) pour avoir le point <lb/>Q, duquel on n’aura qu’à mener la tangente par le point <lb/>donné.</s> <s xml:id="echoid-s10630" xml:space="preserve"/> </p> <pb o="308" file="0362" n="370" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div869" type="section" level="1" n="693"> <head xml:id="echoid-head818" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s10631" xml:space="preserve">654. </s> <s xml:id="echoid-s10632" xml:space="preserve">Il ſuit encore de cette propoſition, que toute ligne <lb/>comme T P parallele à la tangente R Q eſt diviſée en deux <lb/>également par le diametre F G; </s> <s xml:id="echoid-s10633" xml:space="preserve">car le rectangle de F O par <lb/>O G eſt au quarré de O P, comme le quarré de F G au quarré <lb/>de N M, & </s> <s xml:id="echoid-s10634" xml:space="preserve">le même rectangle de F O par O G eſt encore au <lb/>quarré de O T, comme le quarré de F G eſt au quarré de N M, <lb/>il s’enſuit donc que le quarré de O P eſt égal au quarré de O T, <lb/>& </s> <s xml:id="echoid-s10635" xml:space="preserve">que par conſéquent O T = O P.</s> <s xml:id="echoid-s10636" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div870" type="section" level="1" n="694"> <head xml:id="echoid-head819" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s10637" xml:space="preserve">655. </s> <s xml:id="echoid-s10638" xml:space="preserve">Il ſuit encore delà que les quarrés des ordonnées à un <lb/>même diametre ſont entr’eux comme les rectangles faits ſur <lb/>les abſciſſes correſpondantes; </s> <s xml:id="echoid-s10639" xml:space="preserve">d’où l’on voit que ſi l’on ap-<lb/>pelle un diametre quelconque 2a, ſon conjugué 2b, le para-<lb/>metre du premier p, x & </s> <s xml:id="echoid-s10640" xml:space="preserve">y l’abſciſſe & </s> <s xml:id="echoid-s10641" xml:space="preserve">l’ordonnée correſ-<lb/>pondante, on aura comme pour les axes yy : </s> <s xml:id="echoid-s10642" xml:space="preserve">aa-xx :</s> <s xml:id="echoid-s10643" xml:space="preserve">: 4aa <lb/>: </s> <s xml:id="echoid-s10644" xml:space="preserve">4bb :</s> <s xml:id="echoid-s10645" xml:space="preserve">: 2a : </s> <s xml:id="echoid-s10646" xml:space="preserve">p, d’où l’on tire yy = {aa - xx x p/2a}, c’eſt-à-dire que <lb/>le quarré d’une ordonnée à un diametre quelconque eſt égal <lb/>au rectangle des abſciſſes, multiplié par le rapport du parame-<lb/>tre au diametre. </s> <s xml:id="echoid-s10647" xml:space="preserve">Si le diametre eſt plus grand que ſon para-<lb/>metre, le quarré d’une ordonnée quelconque ſera plus grand <lb/>que le rectangle des abſciſſes. </s> <s xml:id="echoid-s10648" xml:space="preserve">Si les deux diametres ſont égaux, <lb/>le parametre ſera égal au diametre, & </s> <s xml:id="echoid-s10649" xml:space="preserve">par conſéquent le rec-<lb/>tangle des abſciſſes ſera égal au quarré de chaque ordonnée, <lb/>& </s> <s xml:id="echoid-s10650" xml:space="preserve">alors les ordonnées ſeroient égales à celle d’un cercle décrit <lb/>ſur un des diametres, mais obliques à ce diametre, parce que <lb/>dans cette courbe il n’y a que les ordonnées aux axes qui puiſ-<lb/>ſent être à angles droits, comme il eſt aiſé de le remarquer, ſi <lb/>l’on fait attention que les ordonnées étant toujours paralleles <lb/>aux tangentes, il faut néceſſairement qu’elles faſſent avec leurs <lb/>diametres les mêmes angles que ces tangentes.</s> <s xml:id="echoid-s10651" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div871" type="section" level="1" n="695"> <head xml:id="echoid-head820" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10652" xml:space="preserve">656. </s> <s xml:id="echoid-s10653" xml:space="preserve">La ſomme des quarrés de deux diametres conjugués C D, <lb/> <anchor type="note" xlink:label="note-0362-01a" xlink:href="note-0362-01"/> E F eſt égale à celle des quarrés des deux axes A B, Q R.</s> <s xml:id="echoid-s10654" xml:space="preserve"/> </p> <div xml:id="echoid-div871" type="float" level="2" n="1"> <note position="left" xlink:label="note-0362-01" xlink:href="note-0362-01a" xml:space="preserve">Figure 160.</note> </div> <pb o="309" file="0363" n="371" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div873" type="section" level="1" n="696"> <head xml:id="echoid-head821" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10655" xml:space="preserve">Les choſes étant toujours les mêmes que ci-devant, nous <lb/>aurons (art. </s> <s xml:id="echoid-s10656" xml:space="preserve">649) G P<emph style="sub">2</emph> = aa-xx, & </s> <s xml:id="echoid-s10657" xml:space="preserve">(art. </s> <s xml:id="echoid-s10658" xml:space="preserve">650) G A<emph style="sub">2</emph>-G P<emph style="sub">2</emph> <lb/>= ou A P x P B = G K<emph style="sub">2</emph> = xx. </s> <s xml:id="echoid-s10659" xml:space="preserve">Or par la propriété de l’ellipſe, <lb/>l’on aura G A<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10660" xml:space="preserve">G R<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10661" xml:space="preserve">: A P x P B : </s> <s xml:id="echoid-s10662" xml:space="preserve">P E<emph style="sub">2</emph>, ou analytiquement <lb/>a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10663" xml:space="preserve">b<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10664" xml:space="preserve">: xx : </s> <s xml:id="echoid-s10665" xml:space="preserve">{bbxx/aa}=P E<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s10666" xml:space="preserve">d’une autre part G A<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10667" xml:space="preserve">G R<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10668" xml:space="preserve">: <lb/>A K X K B : </s> <s xml:id="echoid-s10669" xml:space="preserve">C K<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s10670" xml:space="preserve">en lettres a<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10671" xml:space="preserve">b<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10672" xml:space="preserve">: aa-xx : </s> <s xml:id="echoid-s10673" xml:space="preserve">{aabb-bbxx/aa}. <lb/></s> <s xml:id="echoid-s10674" xml:space="preserve">Or lestriangles rectangles G P E, G K C donnent E G<emph style="sub">2</emph>=E P<emph style="sub">2</emph> <lb/>+ PG<emph style="sub">2</emph>=aa - xx+{bbxx/aa}, ou E G<emph style="sub">2</emph>={a<emph style="sub">4</emph> - aaxx + bbxx/aa}, & </s> <s xml:id="echoid-s10675" xml:space="preserve"><lb/>encore CG<emph style="sub">2</emph>=CK<emph style="sub">2</emph>+GK<emph style="sub">2</emph>={aabb-bbxx/aa}+xx={aabb-bbxx+aaxx/aa}: </s> <s xml:id="echoid-s10676" xml:space="preserve"><lb/>donc E G<emph style="sub">2</emph>+C G<emph style="sub">2</emph>={a<emph style="sub">4</emph>-a<emph style="sub">2</emph>x<emph style="sub">2</emph>+b<emph style="sub">2</emph>x<emph style="sub">2</emph>+a<emph style="sub">2</emph>b<emph style="sub">2</emph>-bbxx+a<emph style="sub">2</emph>x<emph style="sub">2</emph>/a<emph style="sub">2</emph>}={a<emph style="sub">4</emph>+a<emph style="sub">2</emph>b<emph style="sub">2</emph>/a<emph style="sub">2</emph>}, <lb/>& </s> <s xml:id="echoid-s10677" xml:space="preserve">diviſant par a<emph style="sub">2</emph>, aa+bb=E G<emph style="sub">2</emph>+C G<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s10678" xml:space="preserve">en quadruplant <lb/>les termes de chaque membre A B<emph style="sub">2</emph>+Q R<emph style="sub">2</emph>=C D<emph style="sub">2</emph>+E F<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10679" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s10680" xml:space="preserve">Q. </s> <s xml:id="echoid-s10681" xml:space="preserve">F. </s> <s xml:id="echoid-s10682" xml:space="preserve">D.</s> <s xml:id="echoid-s10683" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div874" type="section" level="1" n="697"> <head xml:id="echoid-head822" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10684" xml:space="preserve">657. </s> <s xml:id="echoid-s10685" xml:space="preserve">Il ſuit de cette propoſition, qu’il ne peut y avoir dans <lb/>une ellipſe que deux diametres conjugués qui ſoient égaux: <lb/></s> <s xml:id="echoid-s10686" xml:space="preserve">car puiſque la ſomme des quarrés de deux demi-diametres <lb/>conjugués eſt égale à celle des quarrés des deux demi-axes, ſi <lb/>l’on prend l’expreſſion générale de l’un de ces diametres pour <lb/>le quarré d’un des deux diametres conjugués égaux, par exem-<lb/>ple, celle de C G<emph style="sub">2</emph>, on aura cette équation {2aabb-2bbxx+2aaxx/aa} <lb/>= aa + bb, & </s> <s xml:id="echoid-s10687" xml:space="preserve">multipliant tout par aa, 2aabb - 2bbxx+ <lb/>2aaxx=a<emph style="sub">4</emph>+aabb, d’où l’on déduit, en effaçant aabb dans <lb/>chaque membre aabb-2bbxx+2aaxx=a<emph style="sub">4</emph>, ou en tranſ-<lb/>poſant aabb-a<emph style="sub">4</emph>=2bbxx-2aax, & </s> <s xml:id="echoid-s10688" xml:space="preserve">diviſant tout par <lb/>bb-aa, il vient a<emph style="sub">2</emph>=2xx, ou x<emph style="sub">2</emph>={aa/2}, d’où l’on déduit <lb/>cette propoſition {1/2} a : </s> <s xml:id="echoid-s10689" xml:space="preserve">x :</s> <s xml:id="echoid-s10690" xml:space="preserve">: x : </s> <s xml:id="echoid-s10691" xml:space="preserve">a, qui fait voir que l’abſciſſe qui <lb/>détermine les deux diametres conjugués égaux, eſt moyenne <lb/>proportionnelle entre le quart & </s> <s xml:id="echoid-s10692" xml:space="preserve">la moitié du grand axe. </s> <s xml:id="echoid-s10693" xml:space="preserve">Et <lb/>comme il n’y a qu’une moyenne proportionnelle entre ces <lb/>deux grandeurs, il s’enſuit qu’il n’y a auſſi dans une ellipſe que <lb/>deux diametres conjugués égaux entr’eux. </s> <s xml:id="echoid-s10694" xml:space="preserve">C. </s> <s xml:id="echoid-s10695" xml:space="preserve">Q. </s> <s xml:id="echoid-s10696" xml:space="preserve">F. </s> <s xml:id="echoid-s10697" xml:space="preserve">D.</s> <s xml:id="echoid-s10698" xml:space="preserve"/> </p> <pb o="310" file="0364" n="372" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div875" type="section" level="1" n="698"> <head xml:id="echoid-head823" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head824" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10699" xml:space="preserve">658. </s> <s xml:id="echoid-s10700" xml:space="preserve">Si par l’extrêmité A de l’axe A B l’on mene une tan-<lb/> <anchor type="note" xlink:label="note-0364-01a" xlink:href="note-0364-01"/> gente qui aille rencontrer aux points N & </s> <s xml:id="echoid-s10701" xml:space="preserve">F, les deux diametres <lb/>conjugués M G, I H prolongés autant qu’il eſt néceſſaire, je dis <lb/>que le rectangle des parties A N, A F eſt égal au quarré de la <lb/>moitié de l’axe C D. </s> <s xml:id="echoid-s10702" xml:space="preserve">Ainſi il faut prouver A N x A F = C E<emph style="sub">2</emph>.</s> <s xml:id="echoid-s10703" xml:space="preserve"/> </p> <div xml:id="echoid-div875" type="float" level="2" n="1"> <note position="left" xlink:label="note-0364-01" xlink:href="note-0364-01a" xml:space="preserve">Figure 162.</note> </div> </div> <div xml:id="echoid-div877" type="section" level="1" n="699"> <head xml:id="echoid-head825" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10704" xml:space="preserve">Conſidérez que l’on a A L x L B égal au quarré de E K, <lb/>qui eſt xx (art. </s> <s xml:id="echoid-s10705" xml:space="preserve">650), & </s> <s xml:id="echoid-s10706" xml:space="preserve">que par conſéquent AE<emph style="sub">2</emph> (aa) : </s> <s xml:id="echoid-s10707" xml:space="preserve">EC<emph style="sub">2</emph> (bb) <lb/>:</s> <s xml:id="echoid-s10708" xml:space="preserve">: A L x L B (xx) : </s> <s xml:id="echoid-s10709" xml:space="preserve">L M<emph style="sub">2</emph> ({bbxx/aa}); </s> <s xml:id="echoid-s10710" xml:space="preserve">& </s> <s xml:id="echoid-s10711" xml:space="preserve">comme ce dernier terme <lb/>eſt un quarré parfait en extrayant laracine, on aura L M = {bx/a}. <lb/></s> <s xml:id="echoid-s10712" xml:space="preserve">Mais comme on a auſſi (art. </s> <s xml:id="echoid-s10713" xml:space="preserve">649) A K x K B = L E<emph style="sub">2</emph>, on aura <lb/>encore C E<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10714" xml:space="preserve">A E<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10715" xml:space="preserve">: I K<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10716" xml:space="preserve">A K x K B ou E L<emph style="sub">2</emph>, & </s> <s xml:id="echoid-s10717" xml:space="preserve">analytique-<lb/>ment bb : </s> <s xml:id="echoid-s10718" xml:space="preserve">aa :</s> <s xml:id="echoid-s10719" xml:space="preserve">: yy : </s> <s xml:id="echoid-s10720" xml:space="preserve">{aayy/bb} = L E<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10721" xml:space="preserve">& </s> <s xml:id="echoid-s10722" xml:space="preserve">comme cette quantité <lb/>eſt auſſi un quarré, ſi on en extrait la racine, on aura EL={ay/b}. </s> <s xml:id="echoid-s10723" xml:space="preserve"><lb/>Cela poſé, à cauſe des triangles ſemblables E A F, E L M, on <lb/>pourra former cette proportion E L : </s> <s xml:id="echoid-s10724" xml:space="preserve">L M :</s> <s xml:id="echoid-s10725" xml:space="preserve">: E A : </s> <s xml:id="echoid-s10726" xml:space="preserve">A F; </s> <s xml:id="echoid-s10727" xml:space="preserve">& </s> <s xml:id="echoid-s10728" xml:space="preserve"><lb/>mettant les valeurs analytiques trouvées précédemment, <lb/>{ay/b} : </s> <s xml:id="echoid-s10729" xml:space="preserve">{bx/a} :</s> <s xml:id="echoid-s10730" xml:space="preserve">: a : </s> <s xml:id="echoid-s10731" xml:space="preserve">{abxb/aay} = {bbx/ay} = A F. </s> <s xml:id="echoid-s10732" xml:space="preserve">Et de même à cauſe des trian-<lb/>gles ſemblables E A N, E K I, on aura E K : </s> <s xml:id="echoid-s10733" xml:space="preserve">E A :</s> <s xml:id="echoid-s10734" xml:space="preserve">: I K : </s> <s xml:id="echoid-s10735" xml:space="preserve">A N, <lb/>ou x : </s> <s xml:id="echoid-s10736" xml:space="preserve">a :</s> <s xml:id="echoid-s10737" xml:space="preserve">: y : </s> <s xml:id="echoid-s10738" xml:space="preserve">{ay/x} = AN: </s> <s xml:id="echoid-s10739" xml:space="preserve">donc AN x A F={bbx/ay}x{ay/x}=bb=CE<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10740" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s10741" xml:space="preserve">Q. </s> <s xml:id="echoid-s10742" xml:space="preserve">F. </s> <s xml:id="echoid-s10743" xml:space="preserve">D.</s> <s xml:id="echoid-s10744" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div878" type="section" level="1" n="700"> <head xml:id="echoid-head826" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s10745" xml:space="preserve">659. </s> <s xml:id="echoid-s10746" xml:space="preserve">On peut aiſément, par le moyen de cette propoſition, <lb/>déterminer dans l’ellipſe les diametres conjugués égaux: </s> <s xml:id="echoid-s10747" xml:space="preserve">car <lb/>pour cela, il n’y a qu’à prendre ſur la perpendiculaire A N à l’ori-<lb/>gine de l’axe, une partie AR égale à CE, moitié du petit axe, & </s> <s xml:id="echoid-s10748" xml:space="preserve"><lb/>par le centre E & </s> <s xml:id="echoid-s10749" xml:space="preserve">lepoint R mener la ligne E R, dont la partie <lb/>compriſe entre le centre & </s> <s xml:id="echoid-s10750" xml:space="preserve">la courbe, ſera l’un des demi-dia-<lb/>metres conjugués égaux: </s> <s xml:id="echoid-s10751" xml:space="preserve">car puiſque l’on a toujours A N <lb/>x A F=C E<emph style="sub">2</emph>, lorſque les diametres conjugués ſont égaux, <lb/>les parties A N, A F ſont égales; </s> <s xml:id="echoid-s10752" xml:space="preserve">& </s> <s xml:id="echoid-s10753" xml:space="preserve">par conſéquent A R doit <lb/>être égale à C E.</s> <s xml:id="echoid-s10754" xml:space="preserve"/> </p> <pb o="311" file="0365" n="373" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div879" type="section" level="1" n="701"> <head xml:id="echoid-head827" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10755" xml:space="preserve">660. </s> <s xml:id="echoid-s10756" xml:space="preserve">Si l’on coupe un cône par un plan oblique à la baſe, de <lb/> <anchor type="note" xlink:label="note-0365-01a" xlink:href="note-0365-01"/> maniere que les deux côtés du cône ſoient coupés entre le ſommet <lb/>& </s> <s xml:id="echoid-s10757" xml:space="preserve">la baſe, la ſection ſera une ellipſe.</s> <s xml:id="echoid-s10758" xml:space="preserve"/> </p> <div xml:id="echoid-div879" type="float" level="2" n="1"> <note position="right" xlink:label="note-0365-01" xlink:href="note-0365-01a" xml:space="preserve">Figure 164.</note> </div> <p> <s xml:id="echoid-s10759" xml:space="preserve">Si l’on coupe le cône X par un plan A B, oblique à ſa baſe, <lb/>& </s> <s xml:id="echoid-s10760" xml:space="preserve">perpendiculaire au plan du triangle N O X qui paſſe par <lb/>l’axe de ce cône, la ſection B E A F ſera une ellipſe. </s> <s xml:id="echoid-s10761" xml:space="preserve">Nous ſup-<lb/>poſerons que le cône eſt auſſi coupé parallélement à ſa baſe <lb/>par un plan C M, qui paſſe par le milieu de la ligne A B, qui <lb/>eſt l’interſection des plans NOX, AEBF, & </s> <s xml:id="echoid-s10762" xml:space="preserve">l’axe de la courbe; <lb/></s> <s xml:id="echoid-s10763" xml:space="preserve">& </s> <s xml:id="echoid-s10764" xml:space="preserve">encore par un plan L D, auſſi parallele à la baſe, & </s> <s xml:id="echoid-s10765" xml:space="preserve">qui <lb/>paſſera par un point quelconque I de l’axe A B. </s> <s xml:id="echoid-s10766" xml:space="preserve">Comme ces <lb/>deux ſections formeront des cercles, nous tirerons les lignes <lb/>E F & </s> <s xml:id="echoid-s10767" xml:space="preserve">H K, qui couperont les diametres L D, C M à angles <lb/>droits aux points I, G, & </s> <s xml:id="echoid-s10768" xml:space="preserve">la ligne E F deviendra le petit axe <lb/>de l’ellipſe, & </s> <s xml:id="echoid-s10769" xml:space="preserve">les lignes I K & </s> <s xml:id="echoid-s10770" xml:space="preserve">I H en ſeront des ordonnées. </s> <s xml:id="echoid-s10771" xml:space="preserve"><lb/>Cela poſé, nous ferons A G ou G B=a, E G ou G F=b, <lb/>G M=c, C G=d, G I = x, I K = y, ainſi I B ſera a+x, <lb/>& </s> <s xml:id="echoid-s10772" xml:space="preserve">AI ſera a-x. </s> <s xml:id="echoid-s10773" xml:space="preserve">Nous ferons voir que A I x I B (aa-xx): </s> <s xml:id="echoid-s10774" xml:space="preserve"><lb/>I K<emph style="sub">2</emph> (yy) : </s> <s xml:id="echoid-s10775" xml:space="preserve">: </s> <s xml:id="echoid-s10776" xml:space="preserve">A G<emph style="sub">2</emph> (aa) : </s> <s xml:id="echoid-s10777" xml:space="preserve">G F<emph style="sub">2</emph>(bb).</s> <s xml:id="echoid-s10778" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div881" type="section" level="1" n="702"> <head xml:id="echoid-head828" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s10779" xml:space="preserve">Les triangles ſemblables B G M, B I D nous donnent B G : <lb/></s> <s xml:id="echoid-s10780" xml:space="preserve">B I :</s> <s xml:id="echoid-s10781" xml:space="preserve">: G M : </s> <s xml:id="echoid-s10782" xml:space="preserve">I D, ou en lettres a : </s> <s xml:id="echoid-s10783" xml:space="preserve">a+x :</s> <s xml:id="echoid-s10784" xml:space="preserve">: c : </s> <s xml:id="echoid-s10785" xml:space="preserve">{ac+cx/a}; </s> <s xml:id="echoid-s10786" xml:space="preserve">& </s> <s xml:id="echoid-s10787" xml:space="preserve">de <lb/>même les triangles ſemblables A L I, A C G nous donnent <lb/>A G : </s> <s xml:id="echoid-s10788" xml:space="preserve">A I :</s> <s xml:id="echoid-s10789" xml:space="preserve">: C G : </s> <s xml:id="echoid-s10790" xml:space="preserve">L I, & </s> <s xml:id="echoid-s10791" xml:space="preserve">en lettres a : </s> <s xml:id="echoid-s10792" xml:space="preserve">a-x :</s> <s xml:id="echoid-s10793" xml:space="preserve">: d : </s> <s xml:id="echoid-s10794" xml:space="preserve">{ad-dx/a}: </s> <s xml:id="echoid-s10795" xml:space="preserve">donc <lb/>en multipliant ces deux proportions termes par termes, on <lb/>aura aa : </s> <s xml:id="echoid-s10796" xml:space="preserve">aa-xx :</s> <s xml:id="echoid-s10797" xml:space="preserve">: cd : </s> <s xml:id="echoid-s10798" xml:space="preserve">{ac+cx/a}x{ad-ax/a}, ou A G<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10799" xml:space="preserve">A I x I B <lb/>: </s> <s xml:id="echoid-s10800" xml:space="preserve">: </s> <s xml:id="echoid-s10801" xml:space="preserve">C G x G M : </s> <s xml:id="echoid-s10802" xml:space="preserve">L I x I D. </s> <s xml:id="echoid-s10803" xml:space="preserve">Mais à cauſe des cercles G E M, <lb/>K D H L, on a C G x G M = G E<emph style="sub">2</emph>, ou G F<emph style="sub">2</emph> = bb, & </s> <s xml:id="echoid-s10804" xml:space="preserve">ID <lb/>x IL = IH<emph style="sub">2</emph> ou IK<emph style="sub">2</emph>=yy; </s> <s xml:id="echoid-s10805" xml:space="preserve">on aura donc AG : </s> <s xml:id="echoid-s10806" xml:space="preserve">AI x IB :</s> <s xml:id="echoid-s10807" xml:space="preserve">: IH<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10808" xml:space="preserve">EF<emph style="sub">2</emph>, <lb/>ou invertendo & </s> <s xml:id="echoid-s10809" xml:space="preserve">alternando A I x I B : </s> <s xml:id="echoid-s10810" xml:space="preserve">I H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10811" xml:space="preserve">: A G<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10812" xml:space="preserve">E F<emph style="sub">2</emph>, ou <lb/>aa-xx : </s> <s xml:id="echoid-s10813" xml:space="preserve">yy :</s> <s xml:id="echoid-s10814" xml:space="preserve">: aa : </s> <s xml:id="echoid-s10815" xml:space="preserve">bb.</s> <s xml:id="echoid-s10816" xml:space="preserve"/> </p> <pb o="312" file="0366" n="374" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div882" type="section" level="1" n="703"> <head xml:id="echoid-head829" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head830" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10817" xml:space="preserve">661. </s> <s xml:id="echoid-s10818" xml:space="preserve">Si l’on coupe un cylindre par un plan oblique à la baſe, <lb/> <anchor type="note" xlink:label="note-0366-01a" xlink:href="note-0366-01"/> je dis que la ſection ſera une ellipſe.</s> <s xml:id="echoid-s10819" xml:space="preserve"/> </p> <div xml:id="echoid-div882" type="float" level="2" n="1"> <note position="left" xlink:label="note-0366-01" xlink:href="note-0366-01a" xml:space="preserve">Figure 165.</note> </div> <p> <s xml:id="echoid-s10820" xml:space="preserve">Pour être convaincu que la ſection B E A F du cylindre Y <lb/>eſt une ellipſe, il ne faut que lire la démonſtration du théo-<lb/>rême précédent, & </s> <s xml:id="echoid-s10821" xml:space="preserve">partout où il y aura le nom de cône ſub-<lb/>ſtituer celui de cylindre, la démonſtration étant la même.</s> <s xml:id="echoid-s10822" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div884" type="section" level="1" n="704"> <head xml:id="echoid-head831" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head832" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10823" xml:space="preserve">662. </s> <s xml:id="echoid-s10824" xml:space="preserve">Si du point quelconque G de l’ellipſe on mene des droites <lb/> <anchor type="note" xlink:label="note-0366-02a" xlink:href="note-0366-02"/> G F, G E aux foyers E, F, je dis que la ſomme de ces deux lignes <lb/>priſes où l’on voudra, ſera toujours égale au grand axe A B.</s> <s xml:id="echoid-s10825" xml:space="preserve"/> </p> <div xml:id="echoid-div884" type="float" level="2" n="1"> <note position="left" xlink:label="note-0366-02" xlink:href="note-0366-02a" xml:space="preserve">Figure 166.</note> </div> </div> <div xml:id="echoid-div886" type="section" level="1" n="705"> <head xml:id="echoid-head833" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10826" xml:space="preserve">Il faut ſe reſſouvenir que l’on détermine les foyers E, F en <lb/>décrivant du point D, extrêmité du petit axe comme centre, <lb/>un arc de cercle avec le rayon D F égal à la moitié du grand <lb/>axe, qui coupe cet axe dans les points E, F; </s> <s xml:id="echoid-s10827" xml:space="preserve">d’où il ſuit évidem-<lb/>ment que le point D eſt tel que E D + D F = A B. </s> <s xml:id="echoid-s10828" xml:space="preserve">Pour <lb/>démontrer cette propoſition par rapport à un point quelconque <lb/>G différent du point D, nous ferons A I = a, I D = b, <lb/>E I = c, I K = x, G K ordonnée à l’axe y. </s> <s xml:id="echoid-s10829" xml:space="preserve">Cela poſé, à cauſe <lb/>du triangle rectangle E K G, on a E G<emph style="sub">2</emph> = E K<emph style="sub">2</emph> + G K<emph style="sub">2</emph>; </s> <s xml:id="echoid-s10830" xml:space="preserve">mais <lb/>E K eſt c+x, dont le quarré eſt cc + 2cx + xx, & </s> <s xml:id="echoid-s10831" xml:space="preserve">G K étant <lb/>ordonnée à l’axe, on aura G K<emph style="sub">2</emph> = bb - {bbxx/aa}: </s> <s xml:id="echoid-s10832" xml:space="preserve">donc E G<emph style="sub">2</emph> = <lb/>cc + 2cx + xx + bb - {bbxx/aa}, & </s> <s xml:id="echoid-s10833" xml:space="preserve">tirant les racines de chaque <lb/>membre E G = √cc + 2cx + xx + bb - {bbxx/aa}\x{0020}. </s> <s xml:id="echoid-s10834" xml:space="preserve">De même à <lb/>cauſe du triangle rectangle F K G, on a F G<emph style="sub">2</emph> = F K<emph style="sub">2</emph> + G K<emph style="sub">2</emph>; <lb/></s> <s xml:id="echoid-s10835" xml:space="preserve">mais F K = c - x: </s> <s xml:id="echoid-s10836" xml:space="preserve">donc F K<emph style="sub">2</emph> = cc - 2cx + xx; </s> <s xml:id="echoid-s10837" xml:space="preserve">& </s> <s xml:id="echoid-s10838" xml:space="preserve">partant <lb/>F G<emph style="sub">2</emph> = cc - 2cx + xx + bb - {bbxx/aa}; </s> <s xml:id="echoid-s10839" xml:space="preserve">& </s> <s xml:id="echoid-s10840" xml:space="preserve">tirant les racines de part <lb/>& </s> <s xml:id="echoid-s10841" xml:space="preserve">d’autre, on aura F G = √cc - 2cx + xx + bb - {bbxx/aa}\x{0020}. </s> <s xml:id="echoid-s10842" xml:space="preserve">Pré-<lb/>ſentement ſi la propoſition eſt vraie, il faut qu’en égalant la <lb/>ſomme de ces deux lignes au grand axe 2a, on arrive à quel- <pb o="313" file="0367" n="375" rhead="DE MATHÉMATIQUE. Liv. IX."/> que principe qui nous démontre que nous avons ſuppoſé vrai, <lb/>ou qui nous faſſe voir que nous avons mal ſuppoſé, en nous <lb/>conduiſant à quelque abſurdité. </s> <s xml:id="echoid-s10843" xml:space="preserve">Je fais donc cette équation <lb/>2a = √cc + 2cx + xx + bb - {bbxx/aa}\x{0020} + √cc - 2cx + xx + bb - {bbxx/aa}\x{0020}, <lb/>d’où je tire, en tranſpoſant, 2a - √cc - 2cx + xx + bb - {bbxx/aa}\x{0020} <lb/>= √cc + 2cx + xx + bb - {bbxx/aa}\x{0020}, & </s> <s xml:id="echoid-s10844" xml:space="preserve">en quarrant chaque mem-<lb/>bre 4a<emph style="sub">2</emph> - 4a x √cc - 2cx + xx + bb - {bbxx/aa}\x{0020} + cc -2cx + xx <lb/>+ bb - {bbxx/aa} = cc + 2cx + xx + bb - {bbxx/aa}; </s> <s xml:id="echoid-s10845" xml:space="preserve">& </s> <s xml:id="echoid-s10846" xml:space="preserve">en effaçant <lb/>de part & </s> <s xml:id="echoid-s10847" xml:space="preserve">d’autre les quantités égales, & </s> <s xml:id="echoid-s10848" xml:space="preserve">tranſpoſant la quan-<lb/>tité - 2cx du premier membre dans le ſecond, on aura <lb/>4a<emph style="sub">2</emph> - 4a √cc - 2cx + xx + bb - {bbxx/aa}\x{0020} = 4cx; </s> <s xml:id="echoid-s10849" xml:space="preserve">d’où l’on tire <lb/>en ſaiſant paſſer 4cx dans le premier membre, & </s> <s xml:id="echoid-s10850" xml:space="preserve">le terme ra-<lb/>dical dans le ſecond, après avoir diviſé par 4, a<emph style="sub">2</emph> - cx = <lb/>a √cc - 2cx + xx + bb - {bbxx/aa}\x{0020}. </s> <s xml:id="echoid-s10851" xml:space="preserve">Si l’on quarre chaque membre <lb/>de l’équation, on aura celle - ci, a<emph style="sub">4</emph> - 2a<emph style="sub">2</emph>cx + ccxx = a<emph style="sub">2</emph>c<emph style="sub">2</emph> <lb/>- 2a<emph style="sub">2</emph>cx + a<emph style="sub">2</emph>x<emph style="sub">2</emph> + a<emph style="sub">2</emph>b<emph style="sub">2</emph> - bbx<emph style="sub">2</emph>, dans laquelle effaçant de <lb/>chaque terme les quantités égales - 2a<emph style="sub">2</emph>cx, on aura a<emph style="sub">4</emph> + c<emph style="sub">2</emph>x<emph style="sub">2</emph> <lb/>= a<emph style="sub">2</emph>c<emph style="sub">2</emph> + a<emph style="sub">2</emph>x<emph style="sub">2</emph> + a<emph style="sub">2</emph>b<emph style="sub">2</emph> - bbxx. </s> <s xml:id="echoid-s10852" xml:space="preserve">Enfin ſi dans cette derniere <lb/>équation on met à la place de c<emph style="sub">2</emph> ſa valeur, qui eſt aa - bb, <lb/>comme il eſt viſible dans la figure, à cauſe du triangle rectan-<lb/>gle E I B, il viendra a<emph style="sub">4</emph> + a x<emph style="sub">2</emph> - b<emph style="sub">2</emph>x<emph style="sub">2</emph> = a<emph style="sub">4</emph> - a b<emph style="sub">2</emph> + a<emph style="sub">2</emph>x<emph style="sub">2</emph> <lb/>+ a<emph style="sub">2</emph>b<emph style="sub">2</emph> - bbxx; </s> <s xml:id="echoid-s10853" xml:space="preserve">d’où l’on déduit, en effaçant toutes les quan-<lb/>tités égales de part & </s> <s xml:id="echoid-s10854" xml:space="preserve">d’autre, & </s> <s xml:id="echoid-s10855" xml:space="preserve">réduiſant le ſecond membre, <lb/>0 = 0; </s> <s xml:id="echoid-s10856" xml:space="preserve">d’où il ſuit que la propoſition eſt vraie.</s> <s xml:id="echoid-s10857" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div887" type="section" level="1" n="706"> <head xml:id="echoid-head834" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s10858" xml:space="preserve">663. </s> <s xml:id="echoid-s10859" xml:space="preserve">Un réſultat ſemblable au dernier 0 = 0 doit paroître <lb/>d’abord bien ſingulier, & </s> <s xml:id="echoid-s10860" xml:space="preserve">les Commençans pourroient être <lb/>embarraſſés à concevoir comment ſur cette équation on peut <lb/>établir la vérité d’un théorême, ou de toute autre propoſition. <lb/></s> <s xml:id="echoid-s10861" xml:space="preserve">Pour comprendre ce qu’il ſignifie, il faut faire attention que <lb/>toutes les démonſtrations étant fondées ſur des axiomes, il <lb/>ſuffit de faire voir la liaiſon d’une propoſition avec quel qu’un <lb/>de ces axiomes, pour en établir la certitude. </s> <s xml:id="echoid-s10862" xml:space="preserve">Préſentement ſi <pb o="314" file="0368" n="376" rhead="NOUVEAU COURS"/> l’on réfléchit à toutes les opérations que nous avons faites, on <lb/>verra que notre ſuppoſition nous a conduit à cet axiome, que <lb/>le rien eſt égal au rien, que l’on pourroit mettre au rang des <lb/>premiers axiomes, puiſque cette vérité ne peut pas être con-<lb/>çue autrement que par ſon énoncé: </s> <s xml:id="echoid-s10863" xml:space="preserve">donc notre propoſition <lb/>eſt vraie, puiſqu’elle a une liaiſon néceſſaire avec ce dernier <lb/>axiome Ceux qui liront les Auteurs qui ont beaucoup écrit <lb/>ſur les Mathématiques, verront combien ce principe eſt d’u-<lb/>ſage pour la démonſtration d’un grand nombre de théorêmes, <lb/>& </s> <s xml:id="echoid-s10864" xml:space="preserve">l’on peut dire que c’eſt, à proprement parler, la méthode la <lb/>plus convenable de démontrer les propoſitions, & </s> <s xml:id="echoid-s10865" xml:space="preserve">de décou-<lb/>vrir les vérités par Algebre: </s> <s xml:id="echoid-s10866" xml:space="preserve">car il n’y a qu’à ſuppoſer que la <lb/>choſe ſoit; </s> <s xml:id="echoid-s10867" xml:space="preserve">ſi cette ſuppoſition vous conduit à quelqu’abſur-<lb/>dité, vous en concluez qu’elle eſt fauſſe, & </s> <s xml:id="echoid-s10868" xml:space="preserve">qu’elle eſt vraie, <lb/>ſi vous pouvez arriver, en partant delà, à quelqu’axiome ou à <lb/>quelqu’autre vérité connue par elle-même ou déja démontrée.</s> <s xml:id="echoid-s10869" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div888" type="section" level="1" n="707"> <head xml:id="echoid-head835" xml:space="preserve">PROPOSITION IX.</head> <head xml:id="echoid-head836" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s10870" xml:space="preserve">664. </s> <s xml:id="echoid-s10871" xml:space="preserve">Les deux axes conjugués A B & </s> <s xml:id="echoid-s10872" xml:space="preserve">C E d’une ellipſe étant <lb/> <anchor type="note" xlink:label="note-0368-01a" xlink:href="note-0368-01"/> donnés, la décrire par un mouvement continu.</s> <s xml:id="echoid-s10873" xml:space="preserve"/> </p> <div xml:id="echoid-div888" type="float" level="2" n="1"> <note position="left" xlink:label="note-0368-01" xlink:href="note-0368-01a" xml:space="preserve">Figure 166.</note> </div> </div> <div xml:id="echoid-div890" type="section" level="1" n="708"> <head xml:id="echoid-head837" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s10874" xml:space="preserve">Il faut du point D comme centre, & </s> <s xml:id="echoid-s10875" xml:space="preserve">d’un intervalle égal à <lb/>la moitié A I du grand axc décrire un arc de cercle qui coupe <lb/>ce grand axe dans les points E, F qui ſeront les foyers de l’el-<lb/>lipſe. </s> <s xml:id="echoid-s10876" xml:space="preserve">Il faut enſuite avoir un fil de la longueur du même axe <lb/>A B, dont on attachera les extrêmités aux points E, F, en ſe <lb/>ſervant d’un ſtyle G pour tenir le fil tendu; </s> <s xml:id="echoid-s10877" xml:space="preserve">l’on ira du point <lb/>A au point D, du point D au point B, & </s> <s xml:id="echoid-s10878" xml:space="preserve">l’on décrira avec le <lb/>bout du ſtyle la demi-ellipſe A D B. </s> <s xml:id="echoid-s10879" xml:space="preserve">Si l’on fait paſſer le ſtyle <lb/>de l’autre côté de l’axe A B, on décrira de la même maniere <lb/>avec le ſtyle G l’autre moitié de l’ellipſe A C B.</s> <s xml:id="echoid-s10880" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 166.</note> <p> <s xml:id="echoid-s10881" xml:space="preserve">La démonſtration de cette pratique ſe tire de ce que l’on <lb/>a démontré dans la propoſition précédente, que la ſomme <lb/>des lignes menées d’un des points de l’ellipſe à chaque foyer, <lb/>eſt égal au grand axe, & </s> <s xml:id="echoid-s10882" xml:space="preserve">l’on auroit pu définir l’ellipſe en <lb/>partant de cette propriété de laquelle on auroit déduit toutes <lb/>les autres.</s> <s xml:id="echoid-s10883" xml:space="preserve"/> </p> <pb o="315" file="0369" n="377" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div891" type="section" level="1" n="709"> <head xml:id="echoid-head838" xml:space="preserve">PROPOSITION X.</head> <head xml:id="echoid-head839" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s10884" xml:space="preserve">665. </s> <s xml:id="echoid-s10885" xml:space="preserve">Trouver le centre & </s> <s xml:id="echoid-s10886" xml:space="preserve">les deux axes conjugués d’une ellipſe <lb/> <anchor type="note" xlink:label="note-0369-01a" xlink:href="note-0369-01"/> donnée.</s> <s xml:id="echoid-s10887" xml:space="preserve"/> </p> <div xml:id="echoid-div891" type="float" level="2" n="1"> <note position="right" xlink:label="note-0369-01" xlink:href="note-0369-01a" xml:space="preserve">Figure 163.</note> </div> </div> <div xml:id="echoid-div893" type="section" level="1" n="710"> <head xml:id="echoid-head840" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s10888" xml:space="preserve">Par deux points quelconques A, C, tirez les lignes A B & </s> <s xml:id="echoid-s10889" xml:space="preserve"><lb/>C D paralleles, que vous diviſerez chacune en deux également <lb/>aux points E, F; </s> <s xml:id="echoid-s10890" xml:space="preserve">pour avoir les ordonnées au diametre G H <lb/>(art. </s> <s xml:id="echoid-s10891" xml:space="preserve">654), qui paſſant par les points E & </s> <s xml:id="echoid-s10892" xml:space="preserve">F, paſſera auſſi par <lb/>le centre de l’ellipſe. </s> <s xml:id="echoid-s10893" xml:space="preserve">Ainſi en diviſant en deux également <lb/>cette ligne G H au point I, ce point ſera le centre de l’ellipſe, <lb/>duquel décrivant l’arc G L, on aura deux points ſur la courbe <lb/>également éloignés du centre, qui ſerviront à tracer la ſec-<lb/>tion M, par laquelle & </s> <s xml:id="echoid-s10894" xml:space="preserve">par le point I faiſant paſſer une ligne <lb/>M I, la partie N O de cette ligne renfermée dans la courbe <lb/>ſera le grand axe. </s> <s xml:id="echoid-s10895" xml:space="preserve">Si l’on veut trouver le petit axe, il n’y a <lb/>qu’à élever au point I une perpendiculaire à la ligne N O. <lb/></s> <s xml:id="echoid-s10896" xml:space="preserve">Cette propoſition eſt ſuffiſamment démontrée par tout ce que <lb/>nous avons vu précédemment.</s> <s xml:id="echoid-s10897" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div894" type="section" level="1" n="711"> <head xml:id="echoid-head841" xml:space="preserve">CHAPITRE III,</head> <head xml:id="echoid-head842" style="it" xml:space="preserve">Qui traite de l’Hyperbole.</head> <head xml:id="echoid-head843" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s10898" xml:space="preserve">666. </s> <s xml:id="echoid-s10899" xml:space="preserve">AYant tiré ſur un plan deux lignes droites A B, D E qui <lb/>ſe coupent à angles droits au point C, on élevera la perpendi-<lb/>culaire B S à l’extrêmité B; </s> <s xml:id="echoid-s10900" xml:space="preserve">& </s> <s xml:id="echoid-s10901" xml:space="preserve">après avoir prolongé A B in-<lb/>définiment vers O & </s> <s xml:id="echoid-s10902" xml:space="preserve">vers P, on prendra ſur la ligne B O un <lb/>nombre de parties égales telles que B G, G L, pour décrire <lb/>enſuite du point C comme centre les demi-cercles G Q I, <lb/>L R K, &</s> <s xml:id="echoid-s10903" xml:space="preserve">c. </s> <s xml:id="echoid-s10904" xml:space="preserve">qui couperont la perpendiculaire B S aux points <lb/>F, N; </s> <s xml:id="echoid-s10905" xml:space="preserve">enſuite on cherchera aux lignes A B, D E, B F une qua-<lb/>trieme proportionnelle G H qu’on élevera perpendiculaire au <lb/>point G; </s> <s xml:id="echoid-s10906" xml:space="preserve">on cherchera de même une quatrieme proportion-<lb/>nelle L M aux droites A B, D E, B N, qu’on élevera perpendi-<lb/>culaire au point L, & </s> <s xml:id="echoid-s10907" xml:space="preserve">continuant à trouver de même un nom- <pb o="316" file="0370" n="378" rhead="NOUVEAU COURS"/> bre de points, tels que H, M, la courbe que l’on fera paſſer <lb/>par tous ces points, ſera nommée hyperbole.</s> <s xml:id="echoid-s10908" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10909" xml:space="preserve">667. </s> <s xml:id="echoid-s10910" xml:space="preserve">Si dans le même tems on décrit deux hyperboles, l’une <lb/>à l’extrêmité A, l’autre à l’extrêmité B, elles ſeront nommées <lb/>enſemble hyperboles oppoſées.</s> <s xml:id="echoid-s10911" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10912" xml:space="preserve">668. </s> <s xml:id="echoid-s10913" xml:space="preserve">La ligne A B eſt nommée premier axe, & </s> <s xml:id="echoid-s10914" xml:space="preserve">la ligne D E <lb/>ſecond axe de chacune des hyperboles oppoſées.</s> <s xml:id="echoid-s10915" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10916" xml:space="preserve">669. </s> <s xml:id="echoid-s10917" xml:space="preserve">Les deux axes A B & </s> <s xml:id="echoid-s10918" xml:space="preserve">D E ſont appellés enſemble <lb/>conjugués, de ſorte que le premier A B eſt dit conjugué au ſe-<lb/>cond D E, & </s> <s xml:id="echoid-s10919" xml:space="preserve">réciproquement le ſecond D E conjugué au pre-<lb/>mier A B.</s> <s xml:id="echoid-s10920" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10921" xml:space="preserve">670. </s> <s xml:id="echoid-s10922" xml:space="preserve">Le point C où ſe coupent les deux axes à angles droits, <lb/>eſt nommé centre de l’hyperbole ou des hyperboles oppoſées.</s> <s xml:id="echoid-s10923" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10924" xml:space="preserve">671. </s> <s xml:id="echoid-s10925" xml:space="preserve">Toutes lignes comme G H ou L M perpendiculaires <lb/>au prolongement de l’axe A B, ſont appellées ordonnées au <lb/>premier axe A B; </s> <s xml:id="echoid-s10926" xml:space="preserve">& </s> <s xml:id="echoid-s10927" xml:space="preserve">toute ligne comme TV, menée perpen-<lb/>diculairement au ſecond axe D E, eſt appellée ordonnée au même <lb/>ſecond axe.</s> <s xml:id="echoid-s10928" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10929" xml:space="preserve">672. </s> <s xml:id="echoid-s10930" xml:space="preserve">Les parties A G, B G de l’axe & </s> <s xml:id="echoid-s10931" xml:space="preserve">de ſon prolongement <lb/>ſont appellées abſciſſes de l’ordonnée correſpondante G H, de <lb/>même A L, B L ſont les abſciſſes de l’ordonnée M L.</s> <s xml:id="echoid-s10932" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10933" xml:space="preserve">673. </s> <s xml:id="echoid-s10934" xml:space="preserve">Une ligne troiſieme proportionnelle aux deux axes, eſt <lb/>appellée le parametre de celui qui occupe le premier terme de la <lb/>proportion.</s> <s xml:id="echoid-s10935" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div895" type="section" level="1" n="712"> <head xml:id="echoid-head844" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head845" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s10936" xml:space="preserve">674. </s> <s xml:id="echoid-s10937" xml:space="preserve">Dans l’hyperbole, le rectangle des abſciſſes A G, B G de <lb/>l’axe A B, eſt au quarré de l’ordonnée G H correſpondante, comme <lb/>le quarré du grand axe A B au quarré de ſon conjugué D E.</s> <s xml:id="echoid-s10938" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10939" xml:space="preserve">Ayant nommé C A ou C B, a; </s> <s xml:id="echoid-s10940" xml:space="preserve">C D ou C E, b; </s> <s xml:id="echoid-s10941" xml:space="preserve">B F, c; </s> <s xml:id="echoid-s10942" xml:space="preserve">les <lb/>indéterminées C G ou C I, x; </s> <s xml:id="echoid-s10943" xml:space="preserve">G H, y; </s> <s xml:id="echoid-s10944" xml:space="preserve">A G ſera x + a, & </s> <s xml:id="echoid-s10945" xml:space="preserve">BG <lb/>x - a.</s> <s xml:id="echoid-s10946" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div896" type="section" level="1" n="713"> <head xml:id="echoid-head846" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s10947" xml:space="preserve">Par la définition de l’hyperbole, on a A B : </s> <s xml:id="echoid-s10948" xml:space="preserve">D E :</s> <s xml:id="echoid-s10949" xml:space="preserve">: B F : </s> <s xml:id="echoid-s10950" xml:space="preserve">G H, <lb/>ou 2a : </s> <s xml:id="echoid-s10951" xml:space="preserve">2b :</s> <s xml:id="echoid-s10952" xml:space="preserve">: c : </s> <s xml:id="echoid-s10953" xml:space="preserve">y : </s> <s xml:id="echoid-s10954" xml:space="preserve">donc en quarrant les termes de cette pro-<lb/>portion, 4aa : </s> <s xml:id="echoid-s10955" xml:space="preserve">4bb :</s> <s xml:id="echoid-s10956" xml:space="preserve">: cc : </s> <s xml:id="echoid-s10957" xml:space="preserve">yy; </s> <s xml:id="echoid-s10958" xml:space="preserve">mais par la nature du cercle, <lb/>B F ou cc = I G x B G, ou A G x B G = (a + x) x (x - a) <lb/>= xx - aa, & </s> <s xml:id="echoid-s10959" xml:space="preserve">mettant cette expreſſion à la place de cc, on <pb o="317" file="0371" n="379" rhead="DE MATHÉMATIQUE. Liv. IX."/> aura 4aa: </s> <s xml:id="echoid-s10960" xml:space="preserve">4bb:</s> <s xml:id="echoid-s10961" xml:space="preserve">: xx--aa : </s> <s xml:id="echoid-s10962" xml:space="preserve">yy, ou bien xx--aa : </s> <s xml:id="echoid-s10963" xml:space="preserve">yy :</s> <s xml:id="echoid-s10964" xml:space="preserve">: 4aa : </s> <s xml:id="echoid-s10965" xml:space="preserve">4bb, <lb/>c’eſt-à-dire que A G x B G : </s> <s xml:id="echoid-s10966" xml:space="preserve">G H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10967" xml:space="preserve">: A B<emph style="sub">2</emph>: </s> <s xml:id="echoid-s10968" xml:space="preserve">D E<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10969" xml:space="preserve">C. </s> <s xml:id="echoid-s10970" xml:space="preserve">Q. </s> <s xml:id="echoid-s10971" xml:space="preserve">F. </s> <s xml:id="echoid-s10972" xml:space="preserve">D.</s> <s xml:id="echoid-s10973" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div897" type="section" level="1" n="714"> <head xml:id="echoid-head847" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s10974" xml:space="preserve">675. </s> <s xml:id="echoid-s10975" xml:space="preserve">Il ſuit de cette propoſition, que les quarrés des ordon-<lb/>nées ſont entr’eux comme les rectangles de leurs abſciſſes: <lb/></s> <s xml:id="echoid-s10976" xml:space="preserve">car puiſque l’on a A G x B G: </s> <s xml:id="echoid-s10977" xml:space="preserve">G H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10978" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10979" xml:space="preserve">D E<emph style="sub">2</emph>, on aura <lb/>par la même raiſon, A L x B L : </s> <s xml:id="echoid-s10980" xml:space="preserve">L M<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10981" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10982" xml:space="preserve">D E<emph style="sub">2</emph> : </s> <s xml:id="echoid-s10983" xml:space="preserve">donc <lb/>puiſque les deux dernieres raiſons ſont égales, on aura A G <lb/>x B G : </s> <s xml:id="echoid-s10984" xml:space="preserve">G H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s10985" xml:space="preserve">: A L x B L : </s> <s xml:id="echoid-s10986" xml:space="preserve">L M<emph style="sub">2</emph>, ou alternando A G x B G: </s> <s xml:id="echoid-s10987" xml:space="preserve"><lb/>A L x B L :</s> <s xml:id="echoid-s10988" xml:space="preserve">: G H<emph style="sub">2</emph>: </s> <s xml:id="echoid-s10989" xml:space="preserve">L M<emph style="sub">2</emph>. </s> <s xml:id="echoid-s10990" xml:space="preserve">Donc, &</s> <s xml:id="echoid-s10991" xml:space="preserve">c.</s> <s xml:id="echoid-s10992" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div898" type="section" level="1" n="715"> <head xml:id="echoid-head848" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s10993" xml:space="preserve">676. </s> <s xml:id="echoid-s10994" xml:space="preserve">Il ſuit de cette propoſition, que ſi l’on mene une or-<lb/>donnée T V au ſecond axe D E, le quarré de cette ordonnée <lb/>eſt au quarré de T C, plus celui de D C, moitié du ſecond <lb/>axe, comme le quarré de ſon conjugué A B eſt au quarré du <lb/>même axe D E. </s> <s xml:id="echoid-s10995" xml:space="preserve">Pour le prouver, conſidérez que T V = G C <lb/>= x, & </s> <s xml:id="echoid-s10996" xml:space="preserve">que T C = V G = y. </s> <s xml:id="echoid-s10997" xml:space="preserve">Or comme la propoſition pré-<lb/>cédente donne x x - aa : </s> <s xml:id="echoid-s10998" xml:space="preserve">yy :</s> <s xml:id="echoid-s10999" xml:space="preserve">: 4aa : </s> <s xml:id="echoid-s11000" xml:space="preserve">4bb, on peut en tirer <lb/>cette équation, 4a<emph style="sub">2</emph>y<emph style="sub">2</emph> = 4bbxx - 4aabb, & </s> <s xml:id="echoid-s11001" xml:space="preserve">faiſant paſſer <lb/>- 4aabb du ſecond membre dans le premier, on aura 4a<emph style="sub">2</emph>y<emph style="sub">2</emph> <lb/>+ 4a b<emph style="sub">2</emph> = 4b<emph style="sub">2</emph>x<emph style="sub">2</emph>, d’où l’on tire xx : </s> <s xml:id="echoid-s11002" xml:space="preserve">yy + bb :</s> <s xml:id="echoid-s11003" xml:space="preserve">: 4aa:</s> <s xml:id="echoid-s11004" xml:space="preserve">4bb, <lb/>ou T V<emph style="sub">2</emph> : </s> <s xml:id="echoid-s11005" xml:space="preserve">C T<emph style="sub">2</emph> + C D<emph style="sub">2</emph> :</s> <s xml:id="echoid-s11006" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s11007" xml:space="preserve">D E<emph style="sub">2</emph>.</s> <s xml:id="echoid-s11008" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div899" type="section" level="1" n="716"> <head xml:id="echoid-head849" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s11009" xml:space="preserve">677. </s> <s xml:id="echoid-s11010" xml:space="preserve">Comme on a trouvé dans le corollaire précédent <lb/>cette équation, 4aayy = 4bbxx - 4aabb, il eſt viſible qu’en <lb/>diviſant par 4aa chaque membre de l’équation, on aura yy <lb/>= {bbxx/aa} - bb, qui eſt une équation dont nous aurons beſoin <lb/>par la ſuite.</s> <s xml:id="echoid-s11011" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div900" type="section" level="1" n="717"> <head xml:id="echoid-head850" xml:space="preserve"><emph style="sc">Définition.</emph></head> <p> <s xml:id="echoid-s11012" xml:space="preserve">678. </s> <s xml:id="echoid-s11013" xml:space="preserve">Si par l’extrêmité B de l’axe A B on mene une ligne <lb/> <anchor type="note" xlink:label="note-0371-01a" xlink:href="note-0371-01"/> droite F G parallele au ſecond axe D E, enſorte que B F ou <lb/>B G ſoient chacune égale à la moitié du même axe, & </s> <s xml:id="echoid-s11014" xml:space="preserve">que <lb/>du centre C on tire par les extrêmités F, G les lignes CF, CG, <lb/>prolongées indéfiniment; </s> <s xml:id="echoid-s11015" xml:space="preserve">ces lignes ſeront nommées les <lb/>aſymptotes de l’hyperbole L B M; </s> <s xml:id="echoid-s11016" xml:space="preserve">& </s> <s xml:id="echoid-s11017" xml:space="preserve">ſi on les prolonge auſſi <lb/>indéfiniment de l’autre côté du centre, elles deviendront <lb/>aſymptotes de l’autre hyperbole oppoſée.</s> <s xml:id="echoid-s11018" xml:space="preserve"/> </p> <div xml:id="echoid-div900" type="float" level="2" n="1"> <note position="right" xlink:label="note-0371-01" xlink:href="note-0371-01a" xml:space="preserve">Figure 168.</note> </div> <pb o="318" file="0372" n="380" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div902" type="section" level="1" n="718"> <head xml:id="echoid-head851" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head852" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s11019" xml:space="preserve">679. </s> <s xml:id="echoid-s11020" xml:space="preserve">Si l’on mene une droite H I parallele au ſecond axe D E, <lb/>enſorte qu’elle coupe une des hyperboles, & </s> <s xml:id="echoid-s11021" xml:space="preserve">qu’elle ſoit terminée aux <lb/>aſymptotes, je dis que le rectangle de H K par K I ſera égal au <lb/>quarré de D C ou F B, moitié du ſecond axe D E.</s> <s xml:id="echoid-s11022" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11023" xml:space="preserve">Ayant nommé C B, a; </s> <s xml:id="echoid-s11024" xml:space="preserve">C D ou B F, b; </s> <s xml:id="echoid-s11025" xml:space="preserve">les indéterminées <lb/>C P, x; </s> <s xml:id="echoid-s11026" xml:space="preserve">P K, y, il faut prouver que D C<emph style="sub">2</emph> ou F B<emph style="sub">2</emph> = K H x K I.</s> <s xml:id="echoid-s11027" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div903" type="section" level="1" n="719"> <head xml:id="echoid-head853" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11028" xml:space="preserve">Conſidérez que les triangles ſemblables C B F & </s> <s xml:id="echoid-s11029" xml:space="preserve">C P H don-<lb/>nent C B : </s> <s xml:id="echoid-s11030" xml:space="preserve">B F :</s> <s xml:id="echoid-s11031" xml:space="preserve">: C P : </s> <s xml:id="echoid-s11032" xml:space="preserve">P H, ou en lettres a : </s> <s xml:id="echoid-s11033" xml:space="preserve">b :</s> <s xml:id="echoid-s11034" xml:space="preserve">: x : </s> <s xml:id="echoid-s11035" xml:space="preserve">{bx/a} = PH; <lb/></s> <s xml:id="echoid-s11036" xml:space="preserve">ainſi l’on aura H P - P K = {bx/a} - y, & </s> <s xml:id="echoid-s11037" xml:space="preserve">P I + P K = {bx/a} + y: </s> <s xml:id="echoid-s11038" xml:space="preserve"><lb/>donc (H P-P K) x (H P+P K) ou K H x K I = √{bx/a} - y\x{0020} x <lb/>√{bx/a} + y\x{0020}, ou en faiſant la multiplication {bbxx/aa} - yy=KHxKI, <lb/>mais (art. </s> <s xml:id="echoid-s11039" xml:space="preserve">677) yy = {bbxx/aa} -bb: </s> <s xml:id="echoid-s11040" xml:space="preserve">on aura donc, en ſubſtituant <lb/>cette valeur {bbxx/aa} - {bbxx/aa} + bb = H K x K I, ou bb = F B<emph style="sub">2</emph> <lb/>= H K x K I. </s> <s xml:id="echoid-s11041" xml:space="preserve">C. </s> <s xml:id="echoid-s11042" xml:space="preserve">Q. </s> <s xml:id="echoid-s11043" xml:space="preserve">F. </s> <s xml:id="echoid-s11044" xml:space="preserve">D.</s> <s xml:id="echoid-s11045" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div904" type="section" level="1" n="720"> <head xml:id="echoid-head854" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s11046" xml:space="preserve">680. </s> <s xml:id="echoid-s11047" xml:space="preserve">Il ſuit delà que ſi l’on mene des lignes T S & </s> <s xml:id="echoid-s11048" xml:space="preserve">Q R <lb/>paralleles au ſecond axe D E, & </s> <s xml:id="echoid-s11049" xml:space="preserve">terminées aux aſymptotes, <lb/>les rectangles T O x O S, H K x K I, & </s> <s xml:id="echoid-s11050" xml:space="preserve">Q L x Q R ſont <lb/>égaux entr’eux, puiſque chacun eſt égal au quarré de F B; <lb/></s> <s xml:id="echoid-s11051" xml:space="preserve">d’où l’on peut tirer ces proportions, O S : </s> <s xml:id="echoid-s11052" xml:space="preserve">H K :</s> <s xml:id="echoid-s11053" xml:space="preserve">: K I : </s> <s xml:id="echoid-s11054" xml:space="preserve">O T, & </s> <s xml:id="echoid-s11055" xml:space="preserve"><lb/>H K : </s> <s xml:id="echoid-s11056" xml:space="preserve">Q L :</s> <s xml:id="echoid-s11057" xml:space="preserve">: L R : </s> <s xml:id="echoid-s11058" xml:space="preserve">K I.</s> <s xml:id="echoid-s11059" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div905" type="section" level="1" n="721"> <head xml:id="echoid-head855" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s11060" xml:space="preserve">681. </s> <s xml:id="echoid-s11061" xml:space="preserve">Il ſuit encore delà que les parties M R, Q L compriſes <lb/>entre la courbe & </s> <s xml:id="echoid-s11062" xml:space="preserve">les aſymptotes ſont égales entr’elles: </s> <s xml:id="echoid-s11063" xml:space="preserve">car <lb/>on démontreroit de même que M R x M Q = F B<emph style="sub">2</emph>; </s> <s xml:id="echoid-s11064" xml:space="preserve">& </s> <s xml:id="echoid-s11065" xml:space="preserve">comme <lb/>les ordonnées ſont égales, il faut que les lignes M R, Q L le <lb/>ſoient auſſi.</s> <s xml:id="echoid-s11066" xml:space="preserve"/> </p> <pb o="319" file="0373" n="381" rhead="DE MATHÉMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div906" type="section" level="1" n="722"> <head xml:id="echoid-head856" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s11067" xml:space="preserve">682. </s> <s xml:id="echoid-s11068" xml:space="preserve">Il ſuit encore delà que ſi loin que l’on prolonge la <lb/>courbe & </s> <s xml:id="echoid-s11069" xml:space="preserve">ſes aſymptotes, jamais ces deux lignes ne ſe rencon-<lb/>treront, puiſque l’on aura toujours QL x LR = FB<emph style="sub">2</emph>; </s> <s xml:id="echoid-s11070" xml:space="preserve">ce qui ne <lb/>pourroit arriver ſi ces lignes ſe rencontroient, puiſque dans <lb/>ce cas Q L ſeroit égal à zero; </s> <s xml:id="echoid-s11071" xml:space="preserve">& </s> <s xml:id="echoid-s11072" xml:space="preserve">c’eſt par cette raiſon que les <lb/>lignes C Q, C R ont été nommées aſymptotes, c’eſt-à-dire <lb/>qui ne peuvent rencontrer (l’hyperbole).</s> <s xml:id="echoid-s11073" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div907" type="section" level="1" n="723"> <head xml:id="echoid-head857" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head858" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s11074" xml:space="preserve">683. </s> <s xml:id="echoid-s11075" xml:space="preserve">Si l’on mene par deux points quelconques K, O de deux <lb/> <anchor type="note" xlink:label="note-0373-01a" xlink:href="note-0373-01"/> hyperboles oppoſées deux lignes droites V X & </s> <s xml:id="echoid-s11076" xml:space="preserve">Y Z paralleles en-<lb/>tr’elles, & </s> <s xml:id="echoid-s11077" xml:space="preserve">terminées par les aſymptotes, je dis que le rectangle de <lb/>V O par O X eſt égal à celui de Y K par K Z.</s> <s xml:id="echoid-s11078" xml:space="preserve"/> </p> <div xml:id="echoid-div907" type="float" level="2" n="1"> <note position="right" xlink:label="note-0373-01" xlink:href="note-0373-01a" xml:space="preserve">Figure 168.</note> </div> </div> <div xml:id="echoid-div909" type="section" level="1" n="724"> <head xml:id="echoid-head859" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11079" xml:space="preserve">Pour démontrer cette propoſition, tirez par les points O, K les <lb/>lignes T O S, H K I paralleles entr’elles & </s> <s xml:id="echoid-s11080" xml:space="preserve">au ſecond axe D E, <lb/>pour avoir les triangles ſemblables OSX, YHK, OTV, KIZ, <lb/>qui donnent les proportions ſuivantes, OS : </s> <s xml:id="echoid-s11081" xml:space="preserve">KH :</s> <s xml:id="echoid-s11082" xml:space="preserve">: OX : </s> <s xml:id="echoid-s11083" xml:space="preserve">KY, <lb/>& </s> <s xml:id="echoid-s11084" xml:space="preserve">O T : </s> <s xml:id="echoid-s11085" xml:space="preserve">K I :</s> <s xml:id="echoid-s11086" xml:space="preserve">: O V : </s> <s xml:id="echoid-s11087" xml:space="preserve">KZ : </s> <s xml:id="echoid-s11088" xml:space="preserve">donc en multipliant ces deux pro-<lb/>portions, termes par termes, on aura O S x O T : </s> <s xml:id="echoid-s11089" xml:space="preserve">K H x K I <lb/>:</s> <s xml:id="echoid-s11090" xml:space="preserve">: O X x O V : </s> <s xml:id="echoid-s11091" xml:space="preserve">K Y x K Z, & </s> <s xml:id="echoid-s11092" xml:space="preserve">(art. </s> <s xml:id="echoid-s11093" xml:space="preserve">680) O S x O T = KH x KI: <lb/></s> <s xml:id="echoid-s11094" xml:space="preserve">donc O X x O V = K Y x K Z. </s> <s xml:id="echoid-s11095" xml:space="preserve">C. </s> <s xml:id="echoid-s11096" xml:space="preserve">Q. </s> <s xml:id="echoid-s11097" xml:space="preserve">F. </s> <s xml:id="echoid-s11098" xml:space="preserve">D.</s> <s xml:id="echoid-s11099" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div910" type="section" level="1" n="725"> <head xml:id="echoid-head860" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head861" xml:space="preserve"><emph style="sc">Theoreme.</emph></head> <p style="it"> <s xml:id="echoid-s11100" xml:space="preserve">684. </s> <s xml:id="echoid-s11101" xml:space="preserve">Si l’on mene par deux points quelconques A & </s> <s xml:id="echoid-s11102" xml:space="preserve">C d’une <lb/> <anchor type="note" xlink:label="note-0373-02a" xlink:href="note-0373-02"/> hyperbole, ou des hyperboles oppoſées deux lignes droites A B & </s> <s xml:id="echoid-s11103" xml:space="preserve"><lb/>C D paralleles entr’elles, & </s> <s xml:id="echoid-s11104" xml:space="preserve">deux autres A E, C F auſſi paralleles <lb/>entr’elles, & </s> <s xml:id="echoid-s11105" xml:space="preserve">terminées aux aſymptotes, je dis que le rectangle <lb/>A E x A B ſera égal à celui de F C par C D.</s> <s xml:id="echoid-s11106" xml:space="preserve"/> </p> <div xml:id="echoid-div910" type="float" level="2" n="1"> <note position="right" xlink:label="note-0373-02" xlink:href="note-0373-02a" xml:space="preserve">Figure 169.</note> </div> </div> <div xml:id="echoid-div912" type="section" level="1" n="726"> <head xml:id="echoid-head862" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11107" xml:space="preserve">Soient tirées par les points A, C les lignes G A H, I C K pa-<lb/>ralleles entr’elles; </s> <s xml:id="echoid-s11108" xml:space="preserve">& </s> <s xml:id="echoid-s11109" xml:space="preserve">conſidérez que les triangles ſemblables <lb/>EAG, FCI, BAH, DCK, nous donneront AG : </s> <s xml:id="echoid-s11110" xml:space="preserve">AE :</s> <s xml:id="echoid-s11111" xml:space="preserve">: CI : </s> <s xml:id="echoid-s11112" xml:space="preserve">CE, <pb o="320" file="0374" n="382" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s11113" xml:space="preserve">A H : </s> <s xml:id="echoid-s11114" xml:space="preserve">A B :</s> <s xml:id="echoid-s11115" xml:space="preserve">: C K : </s> <s xml:id="echoid-s11116" xml:space="preserve">C D : </s> <s xml:id="echoid-s11117" xml:space="preserve">donc en multipliant par ordre les <lb/>termes de ces proportions, on aura A G x A H : </s> <s xml:id="echoid-s11118" xml:space="preserve">A E x A B :</s> <s xml:id="echoid-s11119" xml:space="preserve">: <lb/>C K x C I : </s> <s xml:id="echoid-s11120" xml:space="preserve">C F x C D. </s> <s xml:id="echoid-s11121" xml:space="preserve">Mais (art. </s> <s xml:id="echoid-s11122" xml:space="preserve">680) A G x A H = IC x CK : <lb/></s> <s xml:id="echoid-s11123" xml:space="preserve">donc auſſi A E x A B = C F x C D. </s> <s xml:id="echoid-s11124" xml:space="preserve">C. </s> <s xml:id="echoid-s11125" xml:space="preserve">Q. </s> <s xml:id="echoid-s11126" xml:space="preserve">F. </s> <s xml:id="echoid-s11127" xml:space="preserve">D.</s> <s xml:id="echoid-s11128" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div913" type="section" level="1" n="727"> <head xml:id="echoid-head863" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s11129" xml:space="preserve">685. </s> <s xml:id="echoid-s11130" xml:space="preserve">Il ſuit de cette propoſition, que ſi l’on mene par des <lb/>points quelconques A & </s> <s xml:id="echoid-s11131" xml:space="preserve">C, pris ſur une hyperbole ou les hy-<lb/>perboles oppoſées, des lignes A P, C O, A E, C F paralleles <lb/>aux aſymptotes, les rectangles A E x A P, C F x C O ſeront <lb/>égaux entr’eux: </s> <s xml:id="echoid-s11132" xml:space="preserve">car les lignes étant paralleles aux aſymptotes, <lb/>ſont paralleles entr’elles, & </s> <s xml:id="echoid-s11133" xml:space="preserve">ſont par conſéquent dans le cas des <lb/>lignes A B, C D.</s> <s xml:id="echoid-s11134" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div914" type="section" level="1" n="728"> <head xml:id="echoid-head864" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s11135" xml:space="preserve">686. </s> <s xml:id="echoid-s11136" xml:space="preserve">Comme le point L, extrêmité de l’axe eſt un des points <lb/>de l’hyperbole, il s’enſuit qu’en menant les lignes L M & </s> <s xml:id="echoid-s11137" xml:space="preserve">L N <lb/>paralleles aux aſymptotes, on aura encore L M x L N = A E <lb/>x AP, ou L M x L N = C F x C O. </s> <s xml:id="echoid-s11138" xml:space="preserve">Mais comme L M x L N <lb/>n’eſt autre choſe que le quarré de M L, l’on voit qu’en nom-<lb/>mant L M, a; </s> <s xml:id="echoid-s11139" xml:space="preserve">A P, x; </s> <s xml:id="echoid-s11140" xml:space="preserve">A E, y, on aura toujours A P x A E, <lb/>ou C F x C O (xy) = L M<emph style="sub">2</emph> (aa), qui eſt une équation qui <lb/>exprime parfaitement la propriété de l’hyperbole, & </s> <s xml:id="echoid-s11141" xml:space="preserve">par le <lb/>moyen de laquelle on peut déterminer tous ſes points.</s> <s xml:id="echoid-s11142" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div915" type="section" level="1" n="729"> <head xml:id="echoid-head865" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head866" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11143" xml:space="preserve">687. </s> <s xml:id="echoid-s11144" xml:space="preserve">Par un point donné, mener une tangente à une hyperbole, <lb/> <anchor type="note" xlink:label="note-0374-01a" xlink:href="note-0374-01"/> dont les aſymptotes ſont données.</s> <s xml:id="echoid-s11145" xml:space="preserve"/> </p> <div xml:id="echoid-div915" type="float" level="2" n="1"> <note position="left" xlink:label="note-0374-01" xlink:href="note-0374-01a" xml:space="preserve">Figure 171.</note> </div> <p> <s xml:id="echoid-s11146" xml:space="preserve">Pour mener une tangente à une hyperbole, par un point <lb/>donné A, il faut de ce point mener la ligne A B parallele à <lb/>l’aſymptote oppoſée E F, faire la partie B D égale à B E, & </s> <s xml:id="echoid-s11147" xml:space="preserve"><lb/>tirer la ligne D A C, qui ſera tangente au ſeul point A: </s> <s xml:id="echoid-s11148" xml:space="preserve">car à <lb/>cauſe des triangles ſemblables, D C E, D A B, on voit que <lb/>A C eſt égal à A D. </s> <s xml:id="echoid-s11149" xml:space="preserve">Et ſi on vouloit que l’hyperbole rencon-<lb/>trât encore cette ligne dans un point H, il faudroit qu’on eût <lb/>A C = H D, ce qui eſt impoſſible, à moins que le point H <lb/>ne tombe ſur le point A: </s> <s xml:id="echoid-s11150" xml:space="preserve">donc cette ligne eſt tangente au ſeul <lb/>point A. </s> <s xml:id="echoid-s11151" xml:space="preserve">C. </s> <s xml:id="echoid-s11152" xml:space="preserve">Q. </s> <s xml:id="echoid-s11153" xml:space="preserve">F. </s> <s xml:id="echoid-s11154" xml:space="preserve">D.</s> <s xml:id="echoid-s11155" xml:space="preserve"/> </p> <pb o="321" file="0375" n="383" rhead="DE MATHEMATIQUE. Liv. IX."/> </div> <div xml:id="echoid-div917" type="section" level="1" n="730"> <head xml:id="echoid-head867" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s11156" xml:space="preserve">688. </s> <s xml:id="echoid-s11157" xml:space="preserve">Comme il n’y a que la ſeule ligne C D, qui étant ter-<lb/>minée aux aſymptotes, ſoit coupée en deux également au point <lb/>A, il s’enſuit que ſi une ligne droite C D, terminée par les <lb/>aſymptotes d’une hyperbole, eſt tangente au point A, où elle <lb/>ſeroit coupée par une ligne I K, elle y ſera diviſée par cette <lb/>ligne en deux parties égales A C & </s> <s xml:id="echoid-s11158" xml:space="preserve">A D.</s> <s xml:id="echoid-s11159" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div918" type="section" level="1" n="731"> <head xml:id="echoid-head868" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s11160" xml:space="preserve">689. </s> <s xml:id="echoid-s11161" xml:space="preserve">Si l’on a deux lignes A B, C D qui s’entrecoupent au <lb/> <anchor type="note" xlink:label="note-0375-01a" xlink:href="note-0375-01"/> centre de l’hyperbole, ou des hyperboles oppoſées, dont l’une <lb/>A B ſoit menée par le point touchant B, milieu d’une tan-<lb/>gente F G, & </s> <s xml:id="echoid-s11162" xml:space="preserve">l’autre C D parallele, & </s> <s xml:id="echoid-s11163" xml:space="preserve">égale à la même tan-<lb/>gente; </s> <s xml:id="echoid-s11164" xml:space="preserve">ces deux lignes ſeront nommées diametres des hyper-<lb/>boles, & </s> <s xml:id="echoid-s11165" xml:space="preserve">enſemble diametres conjugués l’un à l’autre.</s> <s xml:id="echoid-s11166" xml:space="preserve"/> </p> <div xml:id="echoid-div918" type="float" level="2" n="1"> <note position="right" xlink:label="note-0375-01" xlink:href="note-0375-01a" xml:space="preserve">Figure 170.</note> </div> <p> <s xml:id="echoid-s11167" xml:space="preserve">690. </s> <s xml:id="echoid-s11168" xml:space="preserve">Si par un point H quelconque de l’hyperbole, on mene <lb/>une ligne H K I, terminée de part & </s> <s xml:id="echoid-s11169" xml:space="preserve">d’autre à la courbe, & </s> <s xml:id="echoid-s11170" xml:space="preserve"><lb/>parallele à la tangente F G; </s> <s xml:id="echoid-s11171" xml:space="preserve">cette ligne ſera nommée une <lb/>double ordonnée au diametre E B, dont la ligne H K ſera l’or-<lb/>donnée. </s> <s xml:id="echoid-s11172" xml:space="preserve">Les parties E K, B K du diametre ſeront nommées les <lb/>abſciſſes de l’ordonnée H K.</s> <s xml:id="echoid-s11173" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div920" type="section" level="1" n="732"> <head xml:id="echoid-head869" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head870" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s11174" xml:space="preserve">691. </s> <s xml:id="echoid-s11175" xml:space="preserve">Le quarré d’une ordonnée quelconque H K parallele à une <lb/> <anchor type="note" xlink:label="note-0375-02a" xlink:href="note-0375-02"/> tangente F G, eſt au rectangle A K x K B de ſes abſciſſes, comme <lb/>le quarré du diametre C D eſt au quarré du diametre A B.</s> <s xml:id="echoid-s11176" xml:space="preserve"/> </p> <div xml:id="echoid-div920" type="float" level="2" n="1"> <note position="right" xlink:label="note-0375-02" xlink:href="note-0375-02a" xml:space="preserve">Figure 170.</note> </div> <p> <s xml:id="echoid-s11177" xml:space="preserve">Par l’une des extrêmités B du diametre A B, ſoient me-<lb/>nées les lignes B C, B D, & </s> <s xml:id="echoid-s11178" xml:space="preserve">la tangente F G parallele au dia-<lb/>metre C D; </s> <s xml:id="echoid-s11179" xml:space="preserve">& </s> <s xml:id="echoid-s11180" xml:space="preserve">par conſéquent par le corollaire précédent, <lb/>diviſée en deux également en B; </s> <s xml:id="echoid-s11181" xml:space="preserve">ſoit prolongé la ligne H I <lb/>juſqu’aux aſymptotes; </s> <s xml:id="echoid-s11182" xml:space="preserve">ce qui donnera les parties égales K M, <lb/>K L, & </s> <s xml:id="echoid-s11183" xml:space="preserve">ſoit fait A E ou E B = a, C E ou D E = b, E K = x, <lb/>K H ou K I = y; </s> <s xml:id="echoid-s11184" xml:space="preserve">d’où l’on tire B K = x - a, & </s> <s xml:id="echoid-s11185" xml:space="preserve">A K =x + a.</s> <s xml:id="echoid-s11186" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div922" type="section" level="1" n="733"> <head xml:id="echoid-head871" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11187" xml:space="preserve">Il eſt viſible que les triangles E B F, E B D ſont égaux, ainſi <lb/>que les triangles E B G, C B E; </s> <s xml:id="echoid-s11188" xml:space="preserve">car ces triangles ont les côtés <pb o="322" file="0376" n="384" rhead="NOUVEAU COURS"/> paralleles chacun à chacun, & </s> <s xml:id="echoid-s11189" xml:space="preserve">un côté commun E B: </s> <s xml:id="echoid-s11190" xml:space="preserve">donc <lb/>F B = C E, ou E D = a. </s> <s xml:id="echoid-s11191" xml:space="preserve">Cela poſé, les triangles ſemblables <lb/>EBF, E K L nous donnent E B : </s> <s xml:id="echoid-s11192" xml:space="preserve">B F :</s> <s xml:id="echoid-s11193" xml:space="preserve">: E K : </s> <s xml:id="echoid-s11194" xml:space="preserve">K L, ou a : </s> <s xml:id="echoid-s11195" xml:space="preserve">b :</s> <s xml:id="echoid-s11196" xml:space="preserve">: x : <lb/></s> <s xml:id="echoid-s11197" xml:space="preserve">{bx/a} = K L : </s> <s xml:id="echoid-s11198" xml:space="preserve">donc L H = K L - K H = {bx/a}-y, & </s> <s xml:id="echoid-s11199" xml:space="preserve">HM=KM, <lb/>ou KL + KH = {bx/a} + y; </s> <s xml:id="echoid-s11200" xml:space="preserve">mais par la propriété des aſymptotes, <lb/>H M x H L = F B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s11201" xml:space="preserve">donc √{bx/a}+y\x{0020} x √{bx/a}-y\x{0020}=bb, ou {bbxx/aa} <lb/>-yy=bb, d’où l’on tire yy = {bbxx/aa}-bb={bbxx/aa} - {aabb/aa}, <lb/>ou aayy=bbxx-aabb; </s> <s xml:id="echoid-s11202" xml:space="preserve">ce qui donne cette proportion <lb/>xx - aa : </s> <s xml:id="echoid-s11203" xml:space="preserve">yy :</s> <s xml:id="echoid-s11204" xml:space="preserve">: aa : </s> <s xml:id="echoid-s11205" xml:space="preserve">bb, ou A K x K B : </s> <s xml:id="echoid-s11206" xml:space="preserve">K H<emph style="sub">2</emph> :</s> <s xml:id="echoid-s11207" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s11208" xml:space="preserve">C D<emph style="sub">2</emph>. </s> <s xml:id="echoid-s11209" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s11210" xml:space="preserve">Q. </s> <s xml:id="echoid-s11211" xml:space="preserve">F. </s> <s xml:id="echoid-s11212" xml:space="preserve">D.</s> <s xml:id="echoid-s11213" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div923" type="section" level="1" n="734"> <head xml:id="echoid-head872" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s11214" xml:space="preserve">692. </s> <s xml:id="echoid-s11215" xml:space="preserve">Il ſuit delà que ce que l’on a démontré dans la pre-<lb/>miere propoſition à l’égard des deux axes d’une hyperbole, <lb/>s’étend par celle-ci à deux diametres conjugués quelconques <lb/>A B & </s> <s xml:id="echoid-s11216" xml:space="preserve">C D, auſſi bien que toutes les autres propriétés que l’on <lb/>a démontrées d’une hyperbole avec ſes aſymptotes: </s> <s xml:id="echoid-s11217" xml:space="preserve">car pour <lb/>s’en convaincre, il ne faut que relire les articles précédens, <lb/>& </s> <s xml:id="echoid-s11218" xml:space="preserve">mettre diametre partout où il y aura le mot d’axe; </s> <s xml:id="echoid-s11219" xml:space="preserve">car tout <lb/>ſubſiſtera également, ſoit que l’angle E B F ſoit droit ou non.</s> <s xml:id="echoid-s11220" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div924" type="section" level="1" n="735"> <head xml:id="echoid-head873" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head874" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s11221" xml:space="preserve">693. </s> <s xml:id="echoid-s11222" xml:space="preserve">Si l’on coupe un cône droit A B C par un plan parallele à <lb/> <anchor type="note" xlink:label="note-0376-01a" xlink:href="note-0376-01"/> l’axe B Q, je dis que la courbe F H D K G ſera une hyperbole.</s> <s xml:id="echoid-s11223" xml:space="preserve"/> </p> <div xml:id="echoid-div924" type="float" level="2" n="1"> <note position="left" xlink:label="note-0376-01" xlink:href="note-0376-01a" xml:space="preserve">Figure 172.</note> </div> <p> <s xml:id="echoid-s11224" xml:space="preserve">Ayant prolongé le côté C B du cône juſqu’en P, enſorte <lb/>que B P ſoit égal à B D, la ligne P D ſera le premier axe de <lb/>l’hyperbole, & </s> <s xml:id="echoid-s11225" xml:space="preserve">la ligne B N tirée du point B perpendiculaire <lb/>au milieu de la ligne P D, ſera la moitié du ſecond axe; </s> <s xml:id="echoid-s11226" xml:space="preserve">en-<lb/>ſorte que ſi l’on fait N O = B N, O B ſera le ſecond axe. <lb/></s> <s xml:id="echoid-s11227" xml:space="preserve">Ayant nommé les données N P ou N D, a; </s> <s xml:id="echoid-s11228" xml:space="preserve">N O ou N B, b; </s> <s xml:id="echoid-s11229" xml:space="preserve"><lb/>les indéterminées N I, x; </s> <s xml:id="echoid-s11230" xml:space="preserve">I K ou I H, y, D I ſera x - a, & </s> <s xml:id="echoid-s11231" xml:space="preserve">P I <lb/>ſera x + a; </s> <s xml:id="echoid-s11232" xml:space="preserve">& </s> <s xml:id="echoid-s11233" xml:space="preserve">nous allons ſaire voir que l’on a xx - aa : </s> <s xml:id="echoid-s11234" xml:space="preserve">yy <lb/>:</s> <s xml:id="echoid-s11235" xml:space="preserve">: 4aa : </s> <s xml:id="echoid-s11236" xml:space="preserve">4bb, ou P I x I D : </s> <s xml:id="echoid-s11237" xml:space="preserve">IK<emph style="sub">2</emph> :</s> <s xml:id="echoid-s11238" xml:space="preserve">: P D<emph style="sub">2</emph> : </s> <s xml:id="echoid-s11239" xml:space="preserve">B O<emph style="sub">2</emph>.</s> <s xml:id="echoid-s11240" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div926" type="section" level="1" n="736"> <head xml:id="echoid-head875" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11241" xml:space="preserve">Les triangles ſemblables PNB, PIM donnent PN:</s> <s xml:id="echoid-s11242" xml:space="preserve">NB:</s> <s xml:id="echoid-s11243" xml:space="preserve">:PI:</s> <s xml:id="echoid-s11244" xml:space="preserve">IM, <pb o="323" file="0377" n="385" rhead="DE MATHÉMATIQUE. Liv. IX."/> ou a : </s> <s xml:id="echoid-s11245" xml:space="preserve">b :</s> <s xml:id="echoid-s11246" xml:space="preserve">: a + x : </s> <s xml:id="echoid-s11247" xml:space="preserve">√{a+x x b/a}\x{0020}; </s> <s xml:id="echoid-s11248" xml:space="preserve">de même les triangles ſemblables <lb/>D N B, D I L donnent D N : </s> <s xml:id="echoid-s11249" xml:space="preserve">N B :</s> <s xml:id="echoid-s11250" xml:space="preserve">: D I : </s> <s xml:id="echoid-s11251" xml:space="preserve">I L, ou bien a : </s> <s xml:id="echoid-s11252" xml:space="preserve">b :</s> <s xml:id="echoid-s11253" xml:space="preserve">: <lb/>x - a : </s> <s xml:id="echoid-s11254" xml:space="preserve">√{x-a x b/a}\x{0020} = I L: </s> <s xml:id="echoid-s11255" xml:space="preserve">on aura donc, en multipliant les termes <lb/>de ces deux proportions les uns par les autres, aa : </s> <s xml:id="echoid-s11256" xml:space="preserve">bb :</s> <s xml:id="echoid-s11257" xml:space="preserve">: xx - aa: <lb/></s> <s xml:id="echoid-s11258" xml:space="preserve">I M x I L; </s> <s xml:id="echoid-s11259" xml:space="preserve">mais par la propriété du cercle, I M x I L = I K<emph style="sub">2</emph> <lb/>ou I H<emph style="sub">2</emph>, ou yy : </s> <s xml:id="echoid-s11260" xml:space="preserve">donc on aura aa : </s> <s xml:id="echoid-s11261" xml:space="preserve">bb :</s> <s xml:id="echoid-s11262" xml:space="preserve">: xx - aa : </s> <s xml:id="echoid-s11263" xml:space="preserve">yy, ou <lb/>4aa : </s> <s xml:id="echoid-s11264" xml:space="preserve">4bb :</s> <s xml:id="echoid-s11265" xml:space="preserve">: xx - aa : </s> <s xml:id="echoid-s11266" xml:space="preserve">yy, c’eſt-à-dire qu’en faiſant invertendo <lb/>P I x I D : </s> <s xml:id="echoid-s11267" xml:space="preserve">I K<emph style="sub">2</emph> :</s> <s xml:id="echoid-s11268" xml:space="preserve">: P D<emph style="sub">2</emph> : </s> <s xml:id="echoid-s11269" xml:space="preserve">B O<emph style="sub">2</emph>. </s> <s xml:id="echoid-s11270" xml:space="preserve">C. </s> <s xml:id="echoid-s11271" xml:space="preserve">Q. </s> <s xml:id="echoid-s11272" xml:space="preserve">F. </s> <s xml:id="echoid-s11273" xml:space="preserve">D.</s> <s xml:id="echoid-s11274" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11275" xml:space="preserve">Nous ne parlerons point des différentes manieres de tracer <lb/>l’hyperbole, parce que cette courbe n’a guere lieu dans la <lb/>Géométrie pratique; </s> <s xml:id="echoid-s11276" xml:space="preserve">c’eſt pourquoi l’on pourra paſſer légére-<lb/>ment ce chapitre, pour s’attacher à ce qui va ſuivre, qui eſt <lb/>de la derniere importance dans tout ce qui s’appelle Géométrie <lb/>pratique, & </s> <s xml:id="echoid-s11277" xml:space="preserve">ſurtout dans la Géométrie qui regarde particulié-<lb/>rement l’Ingénieur.</s> <s xml:id="echoid-s11278" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div927" type="section" level="1" n="737"> <head xml:id="echoid-head876" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s11279" xml:space="preserve">Quand on eſt né avec le goût des Mathématiques, l’on <lb/>ne s’en tient guere à la lecture des ſimples Elémens; </s> <s xml:id="echoid-s11280" xml:space="preserve">il ſuffit <lb/>qu’ils nous aient montré qu’on peut aller beaucoup plus loin <lb/>pour deſirer des Livres qui nous apprennent des choſes nou-<lb/>velles; </s> <s xml:id="echoid-s11281" xml:space="preserve">car ceux qui ont l’eſprit Géometre, cherchent à ſe le <lb/>nourrir des vérités d’une ſcience qu’il eſt difficile de connoître <lb/>ſans l’aimer. </s> <s xml:id="echoid-s11282" xml:space="preserve">L’on cherche, l’on s’informe quels ſont les bons <lb/>Livres de Mathématiques qu’on n’a pas vus; </s> <s xml:id="echoid-s11283" xml:space="preserve">mais ſouvent à <lb/>qui s’en informer? </s> <s xml:id="echoid-s11284" xml:space="preserve">Je ferai donc plaiſir de rapporter ici une <lb/>liſte des meilleurs Ouvrages de Mathématique qu’ils pour-<lb/>ront étudier. </s> <s xml:id="echoid-s11285" xml:space="preserve">Je ne prétends parler que des principaux Livres <lb/>qui ont été imprimés à Paris; </s> <s xml:id="echoid-s11286" xml:space="preserve">s’il falloit citer tous les bons <lb/>qu’on a faits chez les Etrangers, & </s> <s xml:id="echoid-s11287" xml:space="preserve">particuliérement en An-<lb/>gleterre, il faudroit un volume entier pour en faire le dé-<lb/>nombrement.</s> <s xml:id="echoid-s11288" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11289" xml:space="preserve">Indépendamment de ce que j’ai donné d’Algebre dans mes <lb/>Elémens de Géométrie pour en ſçavoir parfaitement toutes <lb/>les opérations, l’on pourra avoir recours au Livre de la Science <lb/>du Calcul du R. </s> <s xml:id="echoid-s11290" xml:space="preserve">P. </s> <s xml:id="echoid-s11291" xml:space="preserve">Reyneau. </s> <s xml:id="echoid-s11292" xml:space="preserve">Cet Ouvrage ſert d’intro-<lb/>duction à un autre du même Auteur, intitulé l’Analyſe dé-<lb/>montrée, qui eſt ce que nous avons de meilleur ſur l’Algebre;</s> <s xml:id="echoid-s11293" xml:space="preserve"> <pb o="324" file="0378" n="386" rhead="NOUVEAU COURS DE MATHEM. Liv. IX."/> ce Livre eſt en deux vol. </s> <s xml:id="echoid-s11294" xml:space="preserve">in-4. </s> <s xml:id="echoid-s11295" xml:space="preserve">Dans le premier on enſeigne <lb/>la réſolution des problêmes qui ſe réduiſent à des équations <lb/>ſimples & </s> <s xml:id="echoid-s11296" xml:space="preserve">compoſées; </s> <s xml:id="echoid-s11297" xml:space="preserve">ce qui eſt uniquement l’objet de l’ana-<lb/>lyſe; </s> <s xml:id="echoid-s11298" xml:space="preserve">& </s> <s xml:id="echoid-s11299" xml:space="preserve">dans le ſecond, l’on trouve les nouveaux calculs, c’eſt-<lb/>à-dire le calcul différentiel, & </s> <s xml:id="echoid-s11300" xml:space="preserve">le calcul intégral, qui eſt une <lb/>autre ſorte d’Algebre; </s> <s xml:id="echoid-s11301" xml:space="preserve">& </s> <s xml:id="echoid-s11302" xml:space="preserve">ces calculs font enſuite appliqués à <lb/>la réſolution d’un grand nombre de problêmes Phyſico-Mathé-<lb/>matiques, qui font voir la beauté de ces calculs, & </s> <s xml:id="echoid-s11303" xml:space="preserve">une partie <lb/>des belles découvertes qu’on a faites dans ces derniers tems.</s> <s xml:id="echoid-s11304" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11305" xml:space="preserve">L’on peut voir après cela l’excellent Livre des Infiniment <lb/>petits de M. </s> <s xml:id="echoid-s11306" xml:space="preserve">le Marquis de l’Hôpital, qui traite uniquement <lb/>du calcul différentiel appliqué à la Géométrie des Courbes. </s> <s xml:id="echoid-s11307" xml:space="preserve">Cet <lb/>Ouvrage eſt le plus beau morceau que nous ayons en France <lb/>ſur les Mathématiques; </s> <s xml:id="echoid-s11308" xml:space="preserve">& </s> <s xml:id="echoid-s11309" xml:space="preserve">comme il eſt un peu abſtrait, on <lb/>pourra recourir au Commentaire qu’en a donné M. </s> <s xml:id="echoid-s11310" xml:space="preserve">de Crouſas, <lb/>qui ſervira beaucoup à ſoulager les Commençans.</s> <s xml:id="echoid-s11311" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11312" xml:space="preserve">Quoique j’aie déja parlé du Traité des Sections coniques de <lb/>M. </s> <s xml:id="echoid-s11313" xml:space="preserve">de l’Hôpital, je crois devoir recommander encore une <lb/>fois aux Commençans d’étudier ſérieuſcment cet Ouvrage, <lb/>s’ils ont envie de faire du progrès, & </s> <s xml:id="echoid-s11314" xml:space="preserve">de le lire même immédia-<lb/>tement après qu’ils auront étudié le premier tome de l’Analyſe <lb/>démontrée, parce qu’ils s’y fortifieront, & </s> <s xml:id="echoid-s11315" xml:space="preserve">auront l’eſprit <lb/>plus diſpoſé à voir enſuite le ſecond tome de l’Analyſe.</s> <s xml:id="echoid-s11316" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11317" xml:space="preserve">Il y a auſſi un Livre de M. </s> <s xml:id="echoid-s11318" xml:space="preserve">Carré ſur le calcul intégral, qui <lb/>eſt une application de ce calcul à la meſure des ſurfaces, des <lb/>ſolides, & </s> <s xml:id="echoid-s11319" xml:space="preserve">à la maniere de trouver leur centre de gravité, &</s> <s xml:id="echoid-s11320" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s11321" xml:space="preserve">qu’il eſt bon auſſi de ſçavoir, pour connoître l’uſage de ce <lb/>calcul.</s> <s xml:id="echoid-s11322" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div928" type="section" level="1" n="738"> <head xml:id="echoid-head877" style="it" xml:space="preserve">Fin du neuvieme Livre.</head> <figure> <image file="0378-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0378-01"/> </figure> <pb o="325" file="0379" n="387"/> </div> <div xml:id="echoid-div929" type="section" level="1" n="739"> <head xml:id="echoid-head878" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head879" xml:space="preserve">LIVRE DIXIEME, <lb/>Qui traite de la Trigonométrie rectiligne, & du <lb/>Nivellement.</head> <p style="it"> <s xml:id="echoid-s11323" xml:space="preserve">DE toutes les parties des Mathématiques, il n’y en a point que <lb/>les Commençans étudient plus volontiers que la Trigonométrie, <lb/>parce qu’elle préſente à l’eſprit des problêmes fort curieux, dont la <lb/>ſolution eſt aiſée, n’ayant beſoin que du ſimple calcul de l’Arith-<lb/>métique. </s> <s xml:id="echoid-s11324" xml:space="preserve">Cependant il faut ſe rendre bien familieres les analogies <lb/>de ce calcul, afin d’en placer les termes à propos; </s> <s xml:id="echoid-s11325" xml:space="preserve">car la Trigono-<lb/>métrie eſt d’un ſi grand uſage dans le métier de la guerre, qu’un <lb/>homme qui eſt chargé des moindres choſes dans le Génie, ou dans <lb/>l’Artillerie, ne peut abſolument l’ignorer; </s> <s xml:id="echoid-s11326" xml:space="preserve">puiſque ſi l’on veut <lb/>conduire quelque galerie de mines, jetter des bombes avec regles, <lb/>calculer les parties d’une fortification réguliere pour la tracer ſur le <lb/>terrein, lever un Camp, une Carte, le plan d’une tranchée, orien-<lb/>ter des batteries, il faut néceſſairement avoir recours à la Trigono-<lb/>métrie.</s> <s xml:id="echoid-s11327" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s11328" xml:space="preserve">Et pour dire un mot du Traité que j’en donne ici, l’on ſçaura <lb/>que je ne parle que des triangles rectilignes, parce que ceux qu’on <lb/>nomme Sphériques, à cauſe qu’ils ſont formés par des cercles de <lb/>la Sphere, ne ſont d’aucune utilité à un homme de guerre, au-<lb/>quel il ne faut apprendre que les choſes néceſſaires, crainte de le <lb/>rebuter, en voulant lui fatiguer la mémoire par celles qui ſont pu- <pb o="326" file="0380" n="388" rhead="NOUVEAU COURS"/> rement curieuſes, ou dont l’uſage ne ſe rencontre point dans les <lb/>choſes de ſon miniſtere. </s> <s xml:id="echoid-s11329" xml:space="preserve">J’ai fait enſorte d’éviter ce défaut, parti-<lb/>culiérement dans ce petit Traité, que j’ai tâché de rendre le plus <lb/>clair & </s> <s xml:id="echoid-s11330" xml:space="preserve">le plus intéreſſant qu’il m’a été poſſible, en appliquant la <lb/>Trigonométrie à quantité d’opérations, qui feront plaiſir à ceux <lb/>qui n’aiment point à s’appliquer, ſans voir dans le moment l’uſage <lb/>des propoſitions qu’ils apprennent. </s> <s xml:id="echoid-s11331" xml:space="preserve">Outre les problêmes généraux <lb/>de la Trigonométrie, on a joint ici pluſieurs problêmes particu-<lb/>liers très-intéreſſans pour un Ingénieur militaire. </s> <s xml:id="echoid-s11332" xml:space="preserve">Comme il y a <lb/>toujours du danger de meſurer des baſes dans un terrein expoſe au <lb/>feu de l’ennemi: </s> <s xml:id="echoid-s11333" xml:space="preserve">je donne la maniere de connoître la diſtance du <lb/>lieu où l’on eſt à celui que l’on veut attaquer par une ſeule opéra-<lb/>tion ſans ſortir de l’endroit où je ſuppoſe l’Ingénieur. </s> <s xml:id="echoid-s11334" xml:space="preserve">Cette opéra-<lb/>tion ſera toujours praticable, pourvu que d’un même lieu on puiſſe <lb/>appercevoir trois objets au dedans, ou au dehors de la ville, <lb/>& </s> <s xml:id="echoid-s11335" xml:space="preserve">dont la poſition a été déterminée géométriquement avec toute <lb/>la préciſion poſſible dans des endroits où l’on pouvoit faire toutes <lb/>les opérations néceſſaires ſans aucun danger. </s> <s xml:id="echoid-s11336" xml:space="preserve">Je donne des ſolu-<lb/>tions numériques & </s> <s xml:id="echoid-s11337" xml:space="preserve">géométriques du même problême, afin que l’on <lb/>puiſſe ſe ſervir de l’une dans le cas où l’on a beſoin de toute la pré-<lb/>ciſion poſſible, & </s> <s xml:id="echoid-s11338" xml:space="preserve">de l’autre, lorſqu’on peut ſe contenter d’un à <lb/>peu près qui peut être ſuffiſant dans un grand nombre d’occaſions; <lb/></s> <s xml:id="echoid-s11339" xml:space="preserve">c’eſt à l’Ingénieur à ſçavoir de laquelle des deux méthodes il doit <lb/>faire uſage, & </s> <s xml:id="echoid-s11340" xml:space="preserve">je lui laiſſe faire l’application de ce problême dans <lb/>toutes les circonſtances où il peut s’en ſervir avec avantage.</s> <s xml:id="echoid-s11341" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s11342" xml:space="preserve">Comme en meſurant la diſtance d’un lieu à un autre, il arrive <lb/>quelquefois qu’on eſt obligé d’en connoître auſſi les différentes hau-<lb/>teurs par rapport au centre de la terre, il ſemble que le nivelle-<lb/>ment eſt une partie des Mathématiques qui doit ſuivre immédia-<lb/>tement la Trigonométrie: </s> <s xml:id="echoid-s11343" xml:space="preserve">auſſi ai-je obſervé cet ordre, puiſqu’a-<lb/>près la Trigonométrie l’on trouvera un Traité du Nivellement, où <lb/>l’on fait voir l’uſage du niveau d’eau, & </s> <s xml:id="echoid-s11344" xml:space="preserve">celui d’un autre niveau, <lb/>pour niveler des grandes diſtances. </s> <s xml:id="echoid-s11345" xml:space="preserve">Ces inſtrumens ſont d’un ſi <lb/>grand uſage dans la pratique, qu’on ne ſçauroit trop engager ceux <lb/>qui peuvent ſe trouver dans le cas de s’en ſervir, de s’appliquer à <lb/>ce que l’on verra dans la ſuite ſur ce ſujet. </s> <s xml:id="echoid-s11346" xml:space="preserve">Tout le monde ſçait que <lb/>quand on veut faire un canal de navigation, joindre une riviere <lb/>avec une autre, conduire des eaux aux endroits où il en man-<lb/>que, les projets de ces ſortes de choſes ne peuvent avoir lieu, ſans <lb/>avoir fait auparavant des nivellemens fort exacts; </s> <s xml:id="echoid-s11347" xml:space="preserve">& </s> <s xml:id="echoid-s11348" xml:space="preserve">c’eſt-là par- <pb o="327" file="0381" n="389" rhead="DE MATHÉMATIQUE. Liv. X."/> ticuliérement où la théorie & </s> <s xml:id="echoid-s11349" xml:space="preserve">la pratique doivent travailler de con-<lb/>cert. </s> <s xml:id="echoid-s11350" xml:space="preserve">Combien de grands ouvrages n’a-t’on pas exécutés, depuis <lb/>qu’on a ſçu réduire à des principes l’art du nivellement! Auroit-<lb/>on oſé tenter autrefois un travail auſſi admirable que celui de la <lb/>jonction des deux Mers? </s> <s xml:id="echoid-s11351" xml:space="preserve">Toute la magnificence des Anciens a-<lb/>t’elle jamais été juſqu’ à faire naître des jets d’eau dans des lieux <lb/>fort éloignés de tous réſervoirs? </s> <s xml:id="echoid-s11352" xml:space="preserve">Et ſi cela s’eſt fait, étoit-on sûr <lb/>de la réuſſite avant l’exécution? </s> <s xml:id="echoid-s11353" xml:space="preserve">Combien eſt-il arrivé de fois <lb/>qu’après avoir commencé un grand projet, on s’eſt apperçu trop <lb/>tard, & </s> <s xml:id="echoid-s11354" xml:space="preserve">après de grandes dépenſes, de l’impoſſibilité du deſſein! <lb/>au lieu qu’à préſent on trouve avec toute l’exactitude poſſible la <lb/>différence du niveau de pluſieurs endroits, lorſqu’on entend bien <lb/>le nivellement, & </s> <s xml:id="echoid-s11355" xml:space="preserve">l’on ſçait ſi le projet qu’on a en vue, eſt poſſi-<lb/>ble, ou non; </s> <s xml:id="echoid-s11356" xml:space="preserve">s’il faut des écluſes, à quelle diſtance il faut les <lb/>conſtruire; </s> <s xml:id="echoid-s11357" xml:space="preserve">enfin on eſt en état de ne rien craindre du ſuccès d’une <lb/>grande entrepriſe, ſi après en avoir fait le nivellement, l’on a re-<lb/>connu le projet poſſible.</s> <s xml:id="echoid-s11358" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div930" type="section" level="1" n="740"> <head xml:id="echoid-head880" xml:space="preserve"><emph style="sc">De la</emph> <emph style="sc">Trigonometrie rectiligne</emph>. <lb/><emph style="sc">Definitions</emph>.</head> <head xml:id="echoid-head881" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s11359" xml:space="preserve">694. </s> <s xml:id="echoid-s11360" xml:space="preserve">La Trigonométrie eſt une partie de la Géométrie, par <lb/>le moyen de laquelle trois choſes étant données ou connues <lb/>dans un triangle, l’on vient à la connoiſſance du reſte.</s> <s xml:id="echoid-s11361" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div931" type="section" level="1" n="741"> <head xml:id="echoid-head882" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s11362" xml:space="preserve">695. </s> <s xml:id="echoid-s11363" xml:space="preserve">Comme l’on ne parvient à trouver ce que l’on cher-<lb/>che dans la Trigonométrie que par le calcul ordinaire de l’A-<lb/>rithmétique, l’on ſe ſert de certaines Tables dreſſées pour ce <lb/>ſujet, qu’on appelle Tables des Sinus, Tangentes, Sécantes, <lb/>dont je donnerai l’uſage ſeulement, ſans en enſeigner la con-<lb/>ſtruction, que l’on trouvera dans pluſieurs Livres, ne voulant <lb/>parler que des choſes qu’il faut abſolument ſçavoir.</s> <s xml:id="echoid-s11364" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div932" type="section" level="1" n="742"> <head xml:id="echoid-head883" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s11365" xml:space="preserve">696. </s> <s xml:id="echoid-s11366" xml:space="preserve">Nous avons ſix choſes à conſidérer dans un triangle; <lb/></s> <s xml:id="echoid-s11367" xml:space="preserve">ſçavoir, les trois côtés & </s> <s xml:id="echoid-s11368" xml:space="preserve">les trois angles, ſans s’embarraſſer <lb/>de la ſuperficie: </s> <s xml:id="echoid-s11369" xml:space="preserve">& </s> <s xml:id="echoid-s11370" xml:space="preserve">comme il y a trois de ces ſix termes, qui <lb/>peuvent être donnés, pour arriver à la connoiſſance des au-<lb/>tres, il faut toujours que ce ſoit deux angles & </s> <s xml:id="echoid-s11371" xml:space="preserve">un côté, ou <lb/>un angle & </s> <s xml:id="echoid-s11372" xml:space="preserve">deux côtés, ou bien enfin les trois côtés; </s> <s xml:id="echoid-s11373" xml:space="preserve">car les <pb o="328" file="0382" n="390" rhead="NOUVEAU COURS"/> trois angles ne ſuffiſent pas pour connoître la valeur des trois <lb/>côtés, parce qu’on peut former deux triangles, tels que les <lb/>angles de l’un ſoient égaux aux angles de l’autre, chacun à ſon <lb/>correſpondant, ſans que pour cela les côtés du premier ſoient <lb/>égaux à ceux du ſecond. </s> <s xml:id="echoid-s11374" xml:space="preserve">Il eſt bien vrai qu’on peut trouver <lb/>la proportion de ces côtés, mais non pas leur juſte valeur.</s> <s xml:id="echoid-s11375" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div933" type="section" level="1" n="743"> <head xml:id="echoid-head884" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s11376" xml:space="preserve">697. </s> <s xml:id="echoid-s11377" xml:space="preserve">Nous avons déja dit que la meſure d’un angle n’étoit <lb/>autre choſe que la quantité de degrés, ou de degrés & </s> <s xml:id="echoid-s11378" xml:space="preserve">de mi-<lb/>nutes, que l’arc terminé par les lignes qui forment cet angle <lb/>peut contenir. </s> <s xml:id="echoid-s11379" xml:space="preserve">Mais comme cette meſure eſt relative dans la <lb/>Trigonométrie à certaines lignes, qui en font le principal <lb/>objet, voici leurs noms.</s> <s xml:id="echoid-s11380" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div934" type="section" level="1" n="744"> <head xml:id="echoid-head885" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s11381" xml:space="preserve">698. </s> <s xml:id="echoid-s11382" xml:space="preserve">Sinus droit d’un arc, ou d’un angle dont cet arc eſt la <lb/> <anchor type="note" xlink:label="note-0382-01a" xlink:href="note-0382-01"/> meſure, eſt une ligne droite, abaiſſée de l’extrêmité F de <lb/>cet arc perpendiculairement au côté qui paſſe par l’autre ex-<lb/>trêmité B du même arc F B. </s> <s xml:id="echoid-s11383" xml:space="preserve">Ainſi la ligne F H tirée de l’ex-<lb/>trêmité F de l’arc F B perpendiculaire ſur le côté B C, eſt le <lb/>ſinus de l’angle F C B.</s> <s xml:id="echoid-s11384" xml:space="preserve"/> </p> <div xml:id="echoid-div934" type="float" level="2" n="1"> <note position="left" xlink:label="note-0382-01" xlink:href="note-0382-01a" xml:space="preserve">Figure 174.</note> </div> </div> <div xml:id="echoid-div936" type="section" level="1" n="745"> <head xml:id="echoid-head886" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s11385" xml:space="preserve">699. </s> <s xml:id="echoid-s11386" xml:space="preserve">Si l’on prolonge la ligne F H juſqu’en G, le rayon <lb/>C B étant perpendiculaire ſur la ligne F G, la diviſera en deux <lb/>également au point H (art. </s> <s xml:id="echoid-s11387" xml:space="preserve">423), auſſi-bien que l’arc F B G; <lb/></s> <s xml:id="echoid-s11388" xml:space="preserve">& </s> <s xml:id="echoid-s11389" xml:space="preserve">comme la ligne F G eſt la corde de cet arc, & </s> <s xml:id="echoid-s11390" xml:space="preserve">que la ligne <lb/>F H eſt le ſinus de l’arc F B, il s’enſuit que le ſinus d’un arc eſt <lb/>la moitié de la corde d’un arc double.</s> <s xml:id="echoid-s11391" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div937" type="section" level="1" n="746"> <head xml:id="echoid-head887" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s11392" xml:space="preserve">700. </s> <s xml:id="echoid-s11393" xml:space="preserve">Comme le ſinus F H augmentera d’autant plus que <lb/>l’angle F C B ſera grand, il s’enſuit que lorſque le rayon C F <lb/>ſera perpendiculaire ſur A B, comme eſt le côté C I, le ſinus <lb/>F H, & </s> <s xml:id="echoid-s11394" xml:space="preserve">le côté C F ſe joindront pour ne faire qu’une ſeule <lb/>ligne C I, & </s> <s xml:id="echoid-s11395" xml:space="preserve">que dans ce cas le ſinus de l’angle droit I C H <lb/>ſera le rayon même du cercle; </s> <s xml:id="echoid-s11396" xml:space="preserve">ce qui fait voir que l’angle <lb/>droit a le plus grand de tous les ſinus, que l’on nomme à <lb/>cauſe de cela, Sinus total.</s> <s xml:id="echoid-s11397" xml:space="preserve"/> </p> <pb o="329" file="0383" n="391" rhead="DE MATHEMATIQUES. Liv. X."/> </div> <div xml:id="echoid-div938" type="section" level="1" n="747"> <head xml:id="echoid-head888" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s11398" xml:space="preserve">701. </s> <s xml:id="echoid-s11399" xml:space="preserve">Le ſinus de l’angle droit n’étant autre choſe que le <lb/>rayon du cercle, dont l’angle tire ſa meſure, nous nomme-<lb/>rons dans la ſuite le rayon C B ſinus total. </s> <s xml:id="echoid-s11400" xml:space="preserve">On voit par ce qui <lb/>précede, que les ſinus des angles moindres qu’un droit, croiſ-<lb/>ſent depuis zero juſqu’à la grandeur du rayon. </s> <s xml:id="echoid-s11401" xml:space="preserve">Il ſuit auſſi de cette <lb/>définition, que le ſinus d’un angle plus grand qu’un droit, eſt <lb/>égal au ſinus de ſon ſupplément. </s> <s xml:id="echoid-s11402" xml:space="preserve">Ainſi le ſinus de 120 degrés <lb/>eſt le même que celui de 60 degrés, & </s> <s xml:id="echoid-s11403" xml:space="preserve">plus les angles ſeront <lb/>obtus, plus leurs ſinus ſeront petits, puiſqu’ils auront pour <lb/>ſinus ceux de leurs ſupplémens.</s> <s xml:id="echoid-s11404" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div939" type="section" level="1" n="748"> <head xml:id="echoid-head889" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s11405" xml:space="preserve">702. </s> <s xml:id="echoid-s11406" xml:space="preserve">Sinus verſe d’un arc ou de l’angle dont cet arc eſt la <lb/> <anchor type="note" xlink:label="note-0383-01a" xlink:href="note-0383-01"/> meſure, eſt la partie du rayon compriſe entre le ſinus droit & </s> <s xml:id="echoid-s11407" xml:space="preserve"><lb/>l’extrêmité de cet arc: </s> <s xml:id="echoid-s11408" xml:space="preserve">ainſi la ligne droite, ou la partie B H <lb/>du rayon C B, eſt le ſinus verſe de l’arc F B ou de l’angle F C B, <lb/>dont cet arc eſt la meſure.</s> <s xml:id="echoid-s11409" xml:space="preserve"/> </p> <div xml:id="echoid-div939" type="float" level="2" n="1"> <note position="right" xlink:label="note-0383-01" xlink:href="note-0383-01a" xml:space="preserve">Figure 174.</note> </div> </div> <div xml:id="echoid-div941" type="section" level="1" n="749"> <head xml:id="echoid-head890" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s11410" xml:space="preserve">703. </s> <s xml:id="echoid-s11411" xml:space="preserve">Tangente d’un arc ou d’un angle dont cet arc eſt la <lb/>meſure, eſt une ligne perpendiculaire ſur l’extrêmité d’un des <lb/>côtés de l’angle, & </s> <s xml:id="echoid-s11412" xml:space="preserve">terminée par l’autre côté prolongé: </s> <s xml:id="echoid-s11413" xml:space="preserve">ainſi <lb/>la ligne B E perpendiculaire à l’extrêmité B du côté C B, & </s> <s xml:id="echoid-s11414" xml:space="preserve"><lb/>terminée par la rencontre du côté C F prolongé juſqu’en E, <lb/>eſt la tangente de l’angle F C B. </s> <s xml:id="echoid-s11415" xml:space="preserve">On voit auſſi par cette défini-<lb/>tion, que la tangente d’un angle obtus eſt la même que celle <lb/>d’un angle aigu, qui eſt ſon ſupplément: </s> <s xml:id="echoid-s11416" xml:space="preserve">car la ligne A B eſt <lb/>le côté de l’angle obtus A C F, & </s> <s xml:id="echoid-s11417" xml:space="preserve">cette ligne rencontre le pro-<lb/>longement de l’autre côté en F; </s> <s xml:id="echoid-s11418" xml:space="preserve">ainſi plus l’angle ſera obtus, <lb/>plus ſa tangente ſera petite.</s> <s xml:id="echoid-s11419" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div942" type="section" level="1" n="750"> <head xml:id="echoid-head891" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s11420" xml:space="preserve">704. </s> <s xml:id="echoid-s11421" xml:space="preserve">On appelle coſinus d’un angle ou d’un arc le ſinus de <lb/>ſon complément. </s> <s xml:id="echoid-s11422" xml:space="preserve">L F eſt le coſinus de l’angle BCF, ou de l’arc <lb/>B F. </s> <s xml:id="echoid-s11423" xml:space="preserve">On voit par-là que le coſinus d’un arc ou d’un angle eſt <lb/>la partie du rayon compriſe entre le centre & </s> <s xml:id="echoid-s11424" xml:space="preserve">la rencontre de <lb/>ſon ſinus: </s> <s xml:id="echoid-s11425" xml:space="preserve">car il eſt clair que L F = C H. </s> <s xml:id="echoid-s11426" xml:space="preserve">Une ligne comme <lb/>I K, tangente de l’arc I F complément de l’arc B F, eſt appellée <lb/>cotangente ou tangente de complément de l’angle B C F.</s> <s xml:id="echoid-s11427" xml:space="preserve"/> </p> <pb o="330" file="0384" n="392" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div943" type="section" level="1" n="751"> <head xml:id="echoid-head892" xml:space="preserve">IX.</head> <p> <s xml:id="echoid-s11428" xml:space="preserve">705. </s> <s xml:id="echoid-s11429" xml:space="preserve">Sécante d’un arc ou d’un angle, dont cet arc eſt la <lb/>meſure, n’eſt autre choſe que le côté de l’angle prolongé, & </s> <s xml:id="echoid-s11430" xml:space="preserve"><lb/>terminé à la tangente: </s> <s xml:id="echoid-s11431" xml:space="preserve">ainſi la ligne C E eſt ſécante de l’angle <lb/>F C B. </s> <s xml:id="echoid-s11432" xml:space="preserve">La ligne C K eſt appellée la co-ſécante de l’arc B F, parce <lb/>qu’elle eſt la ſécante de ſon complément. </s> <s xml:id="echoid-s11433" xml:space="preserve">On peut auſſi remar-<lb/>quer que la ſécante d’un angle obtus eſt égale à la ſécante d’un <lb/>angle aigu, qui eſt ſon ſupplément. </s> <s xml:id="echoid-s11434" xml:space="preserve">La ſécante d’un angle <lb/>droit eſt infinie: </s> <s xml:id="echoid-s11435" xml:space="preserve">car étant alors parallele à la tangente, qui <lb/>paſſe par l’extrêmité de l’autre côté de l’angle droit, elle ne <lb/>peut la rencontrer qu’à l’infini; </s> <s xml:id="echoid-s11436" xml:space="preserve">ainſi les ſécantes croiſſent <lb/>depuis zero juſqu’à l’infini.</s> <s xml:id="echoid-s11437" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11438" xml:space="preserve">706. </s> <s xml:id="echoid-s11439" xml:space="preserve">Quand on a conſtruit les Tables des Sinus, l’on a <lb/>ſuppoſé le rayon C B, ou autrement le ſinus total diviſé en <lb/>10000000 parties, & </s> <s xml:id="echoid-s11440" xml:space="preserve">l’on a cherché combien le ſinus de cha-<lb/>que angle, depuis une minute juſqu’à 90 degrés, pouvoit <lb/>contenir de parties du ſinus total, afin de connoître les ſinus <lb/>en nombre; </s> <s xml:id="echoid-s11441" xml:space="preserve">& </s> <s xml:id="echoid-s11442" xml:space="preserve">c’eſt ainſi que l’on a trouvé que le ſinus d’un <lb/>angle de 20 degrés, par exemple, contenoit 3420202 de ces <lb/>parties, que le ſinus de 55 degrés 10 minutes en contenoit <lb/>8208170, ainſi des autres qui en contiennent plus ou moins, <lb/>ſelon que l’angle approche plus ou moins de la valeur d’un <lb/>droit; </s> <s xml:id="echoid-s11443" xml:space="preserve">& </s> <s xml:id="echoid-s11444" xml:space="preserve">ce ſont tous ces différens ſinus que l’on trouve dans <lb/>la ſeconde colonne des Tables ſur chacun des feuillets.</s> <s xml:id="echoid-s11445" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11446" xml:space="preserve">707. </s> <s xml:id="echoid-s11447" xml:space="preserve">Comme une tangente telle que B E augmente ou di-<lb/>minue, ſelon que l’angle E C B approche ou s’éloigne plus <lb/>ou moins de l’angle droit, l’on a cherché auſſi la valeur des <lb/>tangentes de tous les angles, depuis celle d’une minute juſ-<lb/>qu’à celle de 90 degrés, en conſidérant combien elle conte-<lb/>noit de parties de ſinus total, c’eſt-à-dire de 10000000, & </s> <s xml:id="echoid-s11448" xml:space="preserve"><lb/>l’on en a compoſé la troiſieme colonne des Tables, qui ſuit <lb/>immédiatement celle des ſinus; </s> <s xml:id="echoid-s11449" xml:space="preserve">de ſorte que l’on trouve à <lb/>côté des ſinus de chaque angle la valeur de la tangente du <lb/>même angle. </s> <s xml:id="echoid-s11450" xml:space="preserve">Ainſi l’on verra que la tangente de 20 degrés eſt <lb/>de 3639702, & </s> <s xml:id="echoid-s11451" xml:space="preserve">que la tangente de 55 degrés 10 minutes eſt <lb/>14370267 parties du ſinus total, diviſé en 10000000.</s> <s xml:id="echoid-s11452" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11453" xml:space="preserve">708. </s> <s xml:id="echoid-s11454" xml:space="preserve">Enfin l’on a cherché auſſi la valeur de la ſécante de <lb/>chaque angle, que l’on a trouvée par le moyen du ſinus total <lb/>& </s> <s xml:id="echoid-s11455" xml:space="preserve">de la tangente; </s> <s xml:id="echoid-s11456" xml:space="preserve">car comme une ſécante telle que C E, n’eſt <pb o="331" file="0385" n="393" rhead="DE MATHÉMATIQUE. Liv. X."/> autre choſe que l’hypoténuſe d’un triangle rectangle C B E, <lb/>dont l’angle droit eſt compris par le ſinus total C B, & </s> <s xml:id="echoid-s11457" xml:space="preserve">la <lb/>tangente B E de l’angle F C B; </s> <s xml:id="echoid-s11458" xml:space="preserve">l’on a quarré le ſinus total C B, <lb/>& </s> <s xml:id="echoid-s11459" xml:space="preserve">la tangente B E pour avoir la racine quarrée de la ſomme de <lb/>ces deux produits, qui donne la valeur de la ſécante; </s> <s xml:id="echoid-s11460" xml:space="preserve">& </s> <s xml:id="echoid-s11461" xml:space="preserve">c’eſt <lb/>ainſi que l’on a trouvé les ſécantes de tous les angles, depuis <lb/>une minute juſqu’à 90 degrés, dont on a compoſé la troiſieme <lb/>colonne qui ſe trouve dans les Tables.</s> <s xml:id="echoid-s11462" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11463" xml:space="preserve">709. </s> <s xml:id="echoid-s11464" xml:space="preserve">Si donc on veut ſçavoir quel eſt le ſinus, la tangente, <lb/>& </s> <s xml:id="echoid-s11465" xml:space="preserve">la ſécante d’un angle, il faut conſidérer d’abord combien <lb/>la meſure de l’angle contient de degrés, ou de degrés & </s> <s xml:id="echoid-s11466" xml:space="preserve">de <lb/>minutes, & </s> <s xml:id="echoid-s11467" xml:space="preserve">chercher dans la Table le feuillet, où il y a mar-<lb/>qué en haut le nombre de degrés de cet angle: </s> <s xml:id="echoid-s11468" xml:space="preserve">par exemple, <lb/>ſi l’angle eſt de 15 degrés, je cherche la page où eſt le nom-<lb/>bre 15 en haut, & </s> <s xml:id="echoid-s11469" xml:space="preserve">je trouve dans la premiere ligne que le <lb/>ſinus de 15 degrés eſt 2588190, que ſa tangente eſt 2679492, <lb/>& </s> <s xml:id="echoid-s11470" xml:space="preserve">que la ſécante eſt 10352762.</s> <s xml:id="echoid-s11471" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11472" xml:space="preserve">710. </s> <s xml:id="echoid-s11473" xml:space="preserve">Mais comme les degrés de chaque page ſont accom-<lb/>pagnés d’un nombre de minutes, qui ſont en progreſſion Arith-<lb/>métique, depuis 1 juſqu’à 60, qui ſe trouvent dans une petite <lb/>colonne, où il y a au commencement ce mot minute, ſi l’on <lb/>vouloit ſçavoir le ſinus de 15 degrés 24 minutes, je cherche <lb/>d’abord, comme ci-devant, la page où il y a 15 degrés en <lb/>haut, & </s> <s xml:id="echoid-s11474" xml:space="preserve">je deſcends juſqu’à l’endroit de la colonne des mi-<lb/>nutes, où 24 ſe trouve marqué, & </s> <s xml:id="echoid-s11475" xml:space="preserve">je prends le ſinus qui lui <lb/>correſpond, qui eſt de 2655561.</s> <s xml:id="echoid-s11476" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11477" xml:space="preserve">711. </s> <s xml:id="echoid-s11478" xml:space="preserve">Comme le ſinus total, ou autrement le côté C B, de-<lb/>vient le côté commun de tous les angles, puiſqu’il n’y a que <lb/>l’autre côté C F qui varie pour faire l’angle plus ou moins <lb/>ouvert: </s> <s xml:id="echoid-s11479" xml:space="preserve">il eſt à remarquer que le ſinus total, la tangente & </s> <s xml:id="echoid-s11480" xml:space="preserve"><lb/>la ſécante d’un angle peuvent toujours former les côtés d’un <lb/>triangle rectangle, dont la grandeur eſt indéterminée, parce <lb/>qu’il n’eſt queſtion que de la proportion de ces côtés avec <lb/>ceux d’un autre triangle qui lui ſeroit ſemblable; </s> <s xml:id="echoid-s11481" xml:space="preserve">& </s> <s xml:id="echoid-s11482" xml:space="preserve">pour faire <lb/>voir ceci plus clairement, conſidérez le triangle rectangle <lb/>C E F, ſi du point C l’on décrit l’arc B D, qui ſera, par exem-<lb/> <anchor type="note" xlink:label="note-0385-01a" xlink:href="note-0385-01"/> ple, de 35 degrés, & </s> <s xml:id="echoid-s11483" xml:space="preserve">qu’on éleve au point B la perpendicu-<lb/>laire B A, l’on aura le triangle rectangle C B A, dont le côté <lb/>C B pourra être pris pour le ſinus total, le côté A B pour la <lb/>tangente de l’angle C, & </s> <s xml:id="echoid-s11484" xml:space="preserve">le côté C A pour la ſécante du <pb o="332" file="0386" n="394" rhead="NOUVEAU COURS"/> même angle; </s> <s xml:id="echoid-s11485" xml:space="preserve">mais tous les côtés de ce triangle ſont connus: <lb/></s> <s xml:id="echoid-s11486" xml:space="preserve">car le côté C B étant le ſinus total, ſera de 10000000, le côté <lb/>B A étant la tangente d’un angle de 35 degrés, ſera de 7002075, <lb/>& </s> <s xml:id="echoid-s11487" xml:space="preserve">le côté C A étant la fécante du même angle, ſera par conſé-<lb/>quent de 12207746, & </s> <s xml:id="echoid-s11488" xml:space="preserve">c’eſt par le moyen de ces triangles <lb/>qu’on va réſoudre les problêmes ſuivans.</s> <s xml:id="echoid-s11489" xml:space="preserve"/> </p> <div xml:id="echoid-div943" type="float" level="2" n="1"> <note position="right" xlink:label="note-0385-01" xlink:href="note-0385-01a" xml:space="preserve">Figure 175.</note> </div> <p> <s xml:id="echoid-s11490" xml:space="preserve">712. </s> <s xml:id="echoid-s11491" xml:space="preserve">Pour conſtruire les tables, l’on a diviſé le ſinus total <lb/>en un grand nombre de parties, afin que dans les diviſions <lb/>que les opérations demandent, l’on puiſſe négliger les reſtes, <lb/>quand ils ſont compoſés de ces petites parties; </s> <s xml:id="echoid-s11492" xml:space="preserve">mais comme <lb/>dans la pratique ordinaire de la Géométrie l’on peut ſe diſ-<lb/>penſer d’entrer dans une ſi grande cxactitude, l’on pourra, <lb/>au lieu de ſuppoſer que le ſinus total eſt diviſé en 10000000, <lb/>le ſuppoſer ſeulement en 100000; </s> <s xml:id="echoid-s11493" xml:space="preserve">& </s> <s xml:id="echoid-s11494" xml:space="preserve">pour lors il faudra, au <lb/>lieu de prendre toutes les figures qui ſont dans les colonnes des <lb/>ſinus, des tangentes & </s> <s xml:id="echoid-s11495" xml:space="preserve">ſécantes, prendre ſeulement les pre-<lb/>mieres, & </s> <s xml:id="echoid-s11496" xml:space="preserve">négliger les deux dernieres, que l’on voit ſéparées <lb/>à droite par un petit point, c’eſt-à-dire, que pour la tangente <lb/>de 30 degrés, au lieu de prendre 57735:</s> <s xml:id="echoid-s11497" xml:space="preserve">03, on ne prendra <lb/>que 57735; </s> <s xml:id="echoid-s11498" xml:space="preserve">& </s> <s xml:id="echoid-s11499" xml:space="preserve">c’eſt de cette façon que ſeront faits tous les cal-<lb/>culs que l’on verra dans la ſuite.</s> <s xml:id="echoid-s11500" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div945" type="section" level="1" n="752"> <head xml:id="echoid-head893" style="it" xml:space="preserve"><emph style="sc">Calcul des</emph> <emph style="sc">Triangles rectangles</emph>.</head> <head xml:id="echoid-head894" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head895" xml:space="preserve"><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s11501" xml:space="preserve">713. </s> <s xml:id="echoid-s11502" xml:space="preserve">Dans un triangle rectangle A D E, dont on connoît un <lb/>angle aigu A, & </s> <s xml:id="echoid-s11503" xml:space="preserve">le côté A D, trouver le côté D E oppoſé à l’angle <lb/>aigu.</s> <s xml:id="echoid-s11504" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11505" xml:space="preserve">Suppoſant que l’angle A ſoit de 30 degrés, & </s> <s xml:id="echoid-s11506" xml:space="preserve">le côté A D <lb/>de 20 toiſes, il faut chercher dans la table la tangente de 30 <lb/>degrés, que l’on trouvera de 57735, & </s> <s xml:id="echoid-s11507" xml:space="preserve">conſidérer que les <lb/>triangles A B C & </s> <s xml:id="echoid-s11508" xml:space="preserve">A D E étant ſemblables, l’on a A B: </s> <s xml:id="echoid-s11509" xml:space="preserve">B C :</s> <s xml:id="echoid-s11510" xml:space="preserve">: <lb/>A D: </s> <s xml:id="echoid-s11511" xml:space="preserve">D E, qui nous fournit cette regle, ſi A B, qui eſt le ſinus <lb/>total de 1000000, donne la tangente B C de 57735, que don-<lb/>nera le côté A D de 20 toiſes pour le côté D E, que l’on trou-<lb/>vera de 11 toiſes 3 pieds & </s> <s xml:id="echoid-s11512" xml:space="preserve">quelques pouces.</s> <s xml:id="echoid-s11513" xml:space="preserve"/> </p> <pb o="333" file="0387" n="395" rhead="DE MATHÉMATIQUE. Liv. X."/> </div> <div xml:id="echoid-div946" type="section" level="1" n="753"> <head xml:id="echoid-head896" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head897" xml:space="preserve"><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s11514" xml:space="preserve">714. </s> <s xml:id="echoid-s11515" xml:space="preserve">Connoiſſant dans un triangle rectangle A D E, un angle <lb/> <anchor type="note" xlink:label="note-0387-01a" xlink:href="note-0387-01"/> aigu A de 30 degrés, & </s> <s xml:id="echoid-s11516" xml:space="preserve">le côté A D de 20 toiſes, trouver l’hy-<lb/>poténuſe A E.</s> <s xml:id="echoid-s11517" xml:space="preserve"/> </p> <div xml:id="echoid-div946" type="float" level="2" n="1"> <note position="right" xlink:label="note-0387-01" xlink:href="note-0387-01a" xml:space="preserve">Figure 176.</note> </div> <p> <s xml:id="echoid-s11518" xml:space="preserve">Il faut chercher la ſécante de 30 degrés, qui eſt 115470, <lb/>& </s> <s xml:id="echoid-s11519" xml:space="preserve">conſidérer que le triangle A B C étant ſemblable au triangle <lb/>A D E, A B: </s> <s xml:id="echoid-s11520" xml:space="preserve">A C:</s> <s xml:id="echoid-s11521" xml:space="preserve">: A D: </s> <s xml:id="echoid-s11522" xml:space="preserve">A E. </s> <s xml:id="echoid-s11523" xml:space="preserve">d’où l’on tire cette regle, ſi <lb/>A B, qui eſt le ſinus total de 100000, m’a donnné 115470 <lb/>pour la ſécante A C, que me donnera le côté A D de 20 <lb/>toiſes pour le côté A E, que l’on trouvera de 23 toiſes & </s> <s xml:id="echoid-s11524" xml:space="preserve"><lb/>quelques pouces.</s> <s xml:id="echoid-s11525" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div948" type="section" level="1" n="754"> <head xml:id="echoid-head898" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head899" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11526" xml:space="preserve">715. </s> <s xml:id="echoid-s11527" xml:space="preserve">Dans un triangle rectangle A B C, dont on connoît un <lb/> <anchor type="note" xlink:label="note-0387-02a" xlink:href="note-0387-02"/> angle aigu A, & </s> <s xml:id="echoid-s11528" xml:space="preserve">le côté B C oppoſé à cet angle, trouver le côté <lb/>A B oppoſé à l’autre angle aigu C.</s> <s xml:id="echoid-s11529" xml:space="preserve"/> </p> <div xml:id="echoid-div948" type="float" level="2" n="1"> <note position="right" xlink:label="note-0387-02" xlink:href="note-0387-02a" xml:space="preserve">Figure 177.</note> </div> <p> <s xml:id="echoid-s11530" xml:space="preserve">Si l’angle aigu A eſt de 40 degrés, & </s> <s xml:id="echoid-s11531" xml:space="preserve">le côté C B de 25 toi-<lb/>ſes, il faut chercher la tangente de 40 degrés, qui eſt 83909, <lb/>& </s> <s xml:id="echoid-s11532" xml:space="preserve">conſidérer que les triangles A E D & </s> <s xml:id="echoid-s11533" xml:space="preserve">A B C étant ſembla-<lb/>bles, l’on a D E: </s> <s xml:id="echoid-s11534" xml:space="preserve">E A:</s> <s xml:id="echoid-s11535" xml:space="preserve">: C B: </s> <s xml:id="echoid-s11536" xml:space="preserve">B A, d’où l’on tire cette regle, <lb/>comme la tangente D E de 83909 eſt au côté E A, ſinus total <lb/>de 100000; </s> <s xml:id="echoid-s11537" xml:space="preserve">ainſi le côté C B de 25 toiſes eſt au côté B A, que <lb/>l’on trouvera de 29 toiſes & </s> <s xml:id="echoid-s11538" xml:space="preserve">quelque choſe.</s> <s xml:id="echoid-s11539" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11540" xml:space="preserve">716. </s> <s xml:id="echoid-s11541" xml:space="preserve">Autrement, comme l’angle A eſt de 40 degrés, ſi l’on <lb/> <anchor type="note" xlink:label="note-0387-03a" xlink:href="note-0387-03"/> retranche ce nombre de 90, l’on aura 50 degrés pour l’angle <lb/>C; </s> <s xml:id="echoid-s11542" xml:space="preserve">& </s> <s xml:id="echoid-s11543" xml:space="preserve">comme les triangles C E D & </s> <s xml:id="echoid-s11544" xml:space="preserve">C B A ſont ſemblables, <lb/>l’on pourra, en cherchant la tangente de l’angle C, dire, <lb/>comme le côté C E, qui eſt le ſinus total, eſt au côté E D, <lb/>qui eſt la tangente de 40 degrés, ainſi le côté C B de 25 toiſes, <lb/>eſt au côté B A, que l’on trouvera encore de 29 toiſes & </s> <s xml:id="echoid-s11545" xml:space="preserve">quel-<lb/>que choſe.</s> <s xml:id="echoid-s11546" xml:space="preserve"/> </p> <div xml:id="echoid-div949" type="float" level="2" n="2"> <note position="right" xlink:label="note-0387-03" xlink:href="note-0387-03a" xml:space="preserve">Figure 178.</note> </div> <pb o="334" file="0388" n="396" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div951" type="section" level="1" n="755"> <head xml:id="echoid-head900" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head901" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11547" xml:space="preserve">717. </s> <s xml:id="echoid-s11548" xml:space="preserve">Dans un triangle rectangle A B C, dont on connoît les <lb/> <anchor type="note" xlink:label="note-0388-01a" xlink:href="note-0388-01"/> deux côtés A B & </s> <s xml:id="echoid-s11549" xml:space="preserve">B C, qui comprennent l’angle droit, trouver <lb/>l’angle aigu A.</s> <s xml:id="echoid-s11550" xml:space="preserve"/> </p> <div xml:id="echoid-div951" type="float" level="2" n="1"> <note position="left" xlink:label="note-0388-01" xlink:href="note-0388-01a" xml:space="preserve">Figure 179.</note> </div> <p> <s xml:id="echoid-s11551" xml:space="preserve">Suppoſant que le côté A B ſoit de 16 toiſes, & </s> <s xml:id="echoid-s11552" xml:space="preserve">le côté B C <lb/>de 14, remarquez que les triangles A D E & </s> <s xml:id="echoid-s11553" xml:space="preserve">A B C étant ſem-<lb/>blables, A B: </s> <s xml:id="echoid-s11554" xml:space="preserve">B C:</s> <s xml:id="echoid-s11555" xml:space="preserve">: A D: </s> <s xml:id="echoid-s11556" xml:space="preserve">D E, d’où l’on tire cette regle, ſi <lb/>le côté A B de 16 toiſes, donne le côté B C de 14, que don-<lb/>nera 100000, qui eſt le côté A D pour le côté D E, qui eſt <lb/>la tangente de l’angle A, que l’on trouvera de 875000; </s> <s xml:id="echoid-s11557" xml:space="preserve">& </s> <s xml:id="echoid-s11558" xml:space="preserve"><lb/>cherchant le nombre le plus approchant de celui-là dans la <lb/>colonne des tangentes, l’on trouvera qu’il correſpond à 41 <lb/>degrés & </s> <s xml:id="echoid-s11559" xml:space="preserve">12 minutes, qui eſt la valeur de l’angle A.</s> <s xml:id="echoid-s11560" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div953" type="section" level="1" n="756"> <head xml:id="echoid-head902" xml:space="preserve">PROPOSITION V.</head> <head xml:id="echoid-head903" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11561" xml:space="preserve">718. </s> <s xml:id="echoid-s11562" xml:space="preserve">Dans un triangle rectangle A B C, où l’on connoît deux <lb/> <anchor type="note" xlink:label="note-0388-02a" xlink:href="note-0388-02"/> côtés A B & </s> <s xml:id="echoid-s11563" xml:space="preserve">A C, qui comprennent un angle aigu A, trouver la <lb/>valeur de cet angle.</s> <s xml:id="echoid-s11564" xml:space="preserve"/> </p> <div xml:id="echoid-div953" type="float" level="2" n="1"> <note position="left" xlink:label="note-0388-02" xlink:href="note-0388-02a" xml:space="preserve">Figure 180.</note> </div> <p> <s xml:id="echoid-s11565" xml:space="preserve">Suppoſant le côté A B de 35 toiſes, & </s> <s xml:id="echoid-s11566" xml:space="preserve">le côté A C de 40, <lb/>l’on aura, à cauſe des triangles ſemblables A D E & </s> <s xml:id="echoid-s11567" xml:space="preserve">A B C, <lb/>A B:</s> <s xml:id="echoid-s11568" xml:space="preserve">A C:</s> <s xml:id="echoid-s11569" xml:space="preserve">: A D: </s> <s xml:id="echoid-s11570" xml:space="preserve">A E, d’où l’on tire cette regle, ſi le côté <lb/>A B de 35 toiſes, donne 40 toiſes pour le côté A C, que don-<lb/>nera le ſinus total A D de 100000 pour la ſécante A E de l’an-<lb/>gle A, que l’on trouvera de 114285, & </s> <s xml:id="echoid-s11571" xml:space="preserve">ayant recours à la <lb/>table pour y chercher dans la colonne des ſécantes le nombre <lb/>qui approche le plus de celui-ci, on trouvera qu’il correſpond <lb/>à 28 degrés 57 minutes, qui eſt la valeur de l’angle A.</s> <s xml:id="echoid-s11572" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div955" type="section" level="1" n="757"> <head xml:id="echoid-head904" xml:space="preserve">PROPOSITION VI.</head> <head xml:id="echoid-head905" xml:space="preserve"><emph style="sc">Theoreme.</emph></head> <p style="it"> <s xml:id="echoid-s11573" xml:space="preserve">719. </s> <s xml:id="echoid-s11574" xml:space="preserve">Dans tous triangles les ſinus des angles ſont dans la même <lb/> <anchor type="note" xlink:label="note-0388-03a" xlink:href="note-0388-03"/> raiſon que leurs côtés oppoſés.</s> <s xml:id="echoid-s11575" xml:space="preserve"/> </p> <div xml:id="echoid-div955" type="float" level="2" n="1"> <note position="left" xlink:label="note-0388-03" xlink:href="note-0388-03a" xml:space="preserve">Figure 181.</note> </div> <p> <s xml:id="echoid-s11576" xml:space="preserve">Je dis que dans un triangle A B C, il y a même raiſon du <lb/>ſinus de l’angle A à ſon côté oppoſé B C, que du ſinus de l’an-<lb/>gle B à ſon côté oppoſé A C.</s> <s xml:id="echoid-s11577" xml:space="preserve"/> </p> <pb o="335" file="0389" n="397" rhead="DE MATHÉMATIQUE. Liv. X."/> </div> <div xml:id="echoid-div957" type="section" level="1" n="758"> <head xml:id="echoid-head906" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s11578" xml:space="preserve">Ayant circonſcrit un cercle autour de ce triangle, on voit <lb/>que l’angle A ayant pour meſure la moitié de l’arc B D C, la <lb/>ligne B C ſera la corde d’un arc double de celui qui meſure <lb/>l’angle A: </s> <s xml:id="echoid-s11579" xml:space="preserve">par conſéquent la moitié de la ligne B C ſera le <lb/>ſinus de l’angle A; </s> <s xml:id="echoid-s11580" xml:space="preserve">& </s> <s xml:id="echoid-s11581" xml:space="preserve">par la même raiſon le ſinus de l’angle <lb/>B ſera la moitié de la ligne A C, comme le ſinus de l’angle C <lb/>eſt à la moitié du côté A B; </s> <s xml:id="echoid-s11582" xml:space="preserve">ainſi l’on aura {B C/2}: </s> <s xml:id="echoid-s11583" xml:space="preserve">B C:</s> <s xml:id="echoid-s11584" xml:space="preserve">: {A C/2}: </s> <s xml:id="echoid-s11585" xml:space="preserve">A C, <lb/>ou bien {A C/2} : </s> <s xml:id="echoid-s11586" xml:space="preserve">A C :</s> <s xml:id="echoid-s11587" xml:space="preserve">: {A B/2} : </s> <s xml:id="echoid-s11588" xml:space="preserve">A B. </s> <s xml:id="echoid-s11589" xml:space="preserve">C. </s> <s xml:id="echoid-s11590" xml:space="preserve">Q. </s> <s xml:id="echoid-s11591" xml:space="preserve">F. </s> <s xml:id="echoid-s11592" xml:space="preserve">D.</s> <s xml:id="echoid-s11593" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div958" type="section" level="1" n="759"> <head xml:id="echoid-head907" xml:space="preserve">PROPOSITION VII.</head> <head xml:id="echoid-head908" xml:space="preserve"><emph style="sc">Theoreme.</emph></head> <p style="it"> <s xml:id="echoid-s11594" xml:space="preserve">720. </s> <s xml:id="echoid-s11595" xml:space="preserve">Dans un triangle obtus-angle, le ſinus de l’angle obtus <lb/> <anchor type="note" xlink:label="note-0389-01a" xlink:href="note-0389-01"/> eſt le même que celui de ſon ſupplément.</s> <s xml:id="echoid-s11596" xml:space="preserve"/> </p> <div xml:id="echoid-div958" type="float" level="2" n="1"> <note position="right" xlink:label="note-0389-01" xlink:href="note-0389-01a" xml:space="preserve">Figure 184.</note> </div> <p> <s xml:id="echoid-s11597" xml:space="preserve">Ayant abaiſſé la perpendiculaire C D ſur la baſe prolongée <lb/>B D, & </s> <s xml:id="echoid-s11598" xml:space="preserve">décrit les arcs F E & </s> <s xml:id="echoid-s11599" xml:space="preserve">H G avec une même ouverture <lb/>de compas A F & </s> <s xml:id="echoid-s11600" xml:space="preserve">B H, l’on abaiſſera les perpendiculaires F I <lb/>& </s> <s xml:id="echoid-s11601" xml:space="preserve">H L. </s> <s xml:id="echoid-s11602" xml:space="preserve">Cela poſé, comme A F eſt égal à B H, l’un & </s> <s xml:id="echoid-s11603" xml:space="preserve">l’autre <lb/>ſera nommé a; </s> <s xml:id="echoid-s11604" xml:space="preserve">A C, b; </s> <s xml:id="echoid-s11605" xml:space="preserve">C D, c; </s> <s xml:id="echoid-s11606" xml:space="preserve">F I, d; </s> <s xml:id="echoid-s11607" xml:space="preserve">H L, e; </s> <s xml:id="echoid-s11608" xml:space="preserve">C B, f; </s> <s xml:id="echoid-s11609" xml:space="preserve">& </s> <s xml:id="echoid-s11610" xml:space="preserve"><lb/>nous ferons voir que F I (d) : </s> <s xml:id="echoid-s11611" xml:space="preserve">C B (f):</s> <s xml:id="echoid-s11612" xml:space="preserve">: H L (e): </s> <s xml:id="echoid-s11613" xml:space="preserve">A C (b).</s> <s xml:id="echoid-s11614" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div960" type="section" level="1" n="760"> <head xml:id="echoid-head909" xml:space="preserve"><emph style="sc">Demonstration.</emph></head> <p> <s xml:id="echoid-s11615" xml:space="preserve">Les triangles C A D & </s> <s xml:id="echoid-s11616" xml:space="preserve">F A I étant ſemblables, l’on aura <lb/>C D (c): </s> <s xml:id="echoid-s11617" xml:space="preserve">C A (b):</s> <s xml:id="echoid-s11618" xml:space="preserve">: F I (d): </s> <s xml:id="echoid-s11619" xml:space="preserve">A F (a). </s> <s xml:id="echoid-s11620" xml:space="preserve">Et comme les triangles <lb/>C B D & </s> <s xml:id="echoid-s11621" xml:space="preserve">H B L ſont auſſi ſemblables, l’on aura encore C D (c): <lb/></s> <s xml:id="echoid-s11622" xml:space="preserve">H L (e):</s> <s xml:id="echoid-s11623" xml:space="preserve">: C B (f): </s> <s xml:id="echoid-s11624" xml:space="preserve">H B (a), d’où l’on tire ces deux équations <lb/>a c=b d, & </s> <s xml:id="echoid-s11625" xml:space="preserve">a c=e f, dont les premiers membres étant égaux, <lb/>l’on aura par conſéquent b d=e f, d’où l’on tire F I (d): </s> <s xml:id="echoid-s11626" xml:space="preserve"><lb/>C B (f):</s> <s xml:id="echoid-s11627" xml:space="preserve">: H L (e): </s> <s xml:id="echoid-s11628" xml:space="preserve">A C (b), qui fait voir que le ſinus H L <lb/>du ſupplément de l’angle A B C a même raiſon au côté A C, <lb/>que le ſinus F I au côté B C; </s> <s xml:id="echoid-s11629" xml:space="preserve">& </s> <s xml:id="echoid-s11630" xml:space="preserve">que par conſéquent le ſinus <lb/>d’un angle obtus eſt toujours celui de ſon ſupplément. </s> <s xml:id="echoid-s11631" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s11632" xml:space="preserve">Q. </s> <s xml:id="echoid-s11633" xml:space="preserve">F. </s> <s xml:id="echoid-s11634" xml:space="preserve">D.</s> <s xml:id="echoid-s11635" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11636" xml:space="preserve">Ces deux théorêmes nous fourniſſent le moyen de connoître <lb/>les angles & </s> <s xml:id="echoid-s11637" xml:space="preserve">les côtés de la plûpart des triangles qui ne ſont <lb/>pas rectangles, comme on le va voir dans les problêmes <lb/>ſuivans.</s> <s xml:id="echoid-s11638" xml:space="preserve"/> </p> <pb o="336" file="0390" n="398" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div961" type="section" level="1" n="761"> <head xml:id="echoid-head910" xml:space="preserve">PROPOSITION VIII.</head> <head xml:id="echoid-head911" xml:space="preserve"><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s11639" xml:space="preserve">721. </s> <s xml:id="echoid-s11640" xml:space="preserve">Dans un triangle A B C, dont on connoît deux angles <lb/> <anchor type="note" xlink:label="note-0390-01a" xlink:href="note-0390-01"/> & </s> <s xml:id="echoid-s11641" xml:space="preserve">un côté; </s> <s xml:id="echoid-s11642" xml:space="preserve">on demande de trouver les deux autres côtés.</s> <s xml:id="echoid-s11643" xml:space="preserve"/> </p> <div xml:id="echoid-div961" type="float" level="2" n="1"> <note position="left" xlink:label="note-0390-01" xlink:href="note-0390-01a" xml:space="preserve">Figure 182.</note> </div> <p> <s xml:id="echoid-s11644" xml:space="preserve">Le côté B C étant ſuppoſé de 15 toiſes, l’angle A de 40 de-<lb/>grés, & </s> <s xml:id="echoid-s11645" xml:space="preserve">l’angle B de 60, l’on connoîtra le troiſieme angle, en <lb/>ſouſtrayant de la valeur de deux droits, c’eſt-à-dire de 180 <lb/>degrés, la ſomme des angles A & </s> <s xml:id="echoid-s11646" xml:space="preserve">B, & </s> <s xml:id="echoid-s11647" xml:space="preserve">l’on trouvera 80 degrés <lb/>pour l’angle C. </s> <s xml:id="echoid-s11648" xml:space="preserve">Cela poſé, pour connoître le côté A C, je <lb/>cherche dans les Tables le ſinus de l’angle A, c’eſt-à-dire le <lb/>ſinus de 40 degrés, qui ſera celui de l’angle oppoſé au côté <lb/>que je connois, & </s> <s xml:id="echoid-s11649" xml:space="preserve">je trouve qu’il eſt 64278; </s> <s xml:id="echoid-s11650" xml:space="preserve">& </s> <s xml:id="echoid-s11651" xml:space="preserve">cherchant <lb/>auſſi celui de l’angle B oppoſé au côté que je cherche, je <lb/>trouve qu’il eſt de 86602: </s> <s xml:id="echoid-s11652" xml:space="preserve">préſentement je dis: </s> <s xml:id="echoid-s11653" xml:space="preserve">Si 64278, <lb/>qui eſt le ſinus de l’angle A, donne 15 toiſes pour le côté B C, <lb/>que donnera 86602, qui eſt le ſinus de l’angle B, pour le côté <lb/>A C, que l’on trouvera de 20 toiſes & </s> <s xml:id="echoid-s11654" xml:space="preserve">quelque choſe: </s> <s xml:id="echoid-s11655" xml:space="preserve">pour <lb/>trouver la valeur du côté A B, il faut chercher le ſinus de <lb/>l’angle C, qui eſt de 98480, & </s> <s xml:id="echoid-s11656" xml:space="preserve">dire encore: </s> <s xml:id="echoid-s11657" xml:space="preserve">Si le ſinus de <lb/>l’angle A, qui eſt 64278, donne 15 toiſes pour le côté B C, <lb/>que donnera le ſinus de l’angle C, qui eſt 98480 pour le côté <lb/>A B, que l’on trouvera de 23 toiſes & </s> <s xml:id="echoid-s11658" xml:space="preserve">quelque choſe.</s> <s xml:id="echoid-s11659" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div963" type="section" level="1" n="762"> <head xml:id="echoid-head912" xml:space="preserve"><emph style="sc">Lemme</emph>.</head> <p style="it"> <s xml:id="echoid-s11660" xml:space="preserve">722. </s> <s xml:id="echoid-s11661" xml:space="preserve">Si l’on a deux grandeurs x & </s> <s xml:id="echoid-s11662" xml:space="preserve">y, dont la ſomme eſt a, <lb/>& </s> <s xml:id="echoid-s11663" xml:space="preserve">la différence d, la plus grande eſt égale à la moitié de la ſomme, <lb/>plus la moitié de la différence, & </s> <s xml:id="echoid-s11664" xml:space="preserve">la plus petite eſt égale à la <lb/>moitié de la ſomme, moins la moitié de la différence.</s> <s xml:id="echoid-s11665" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11666" xml:space="preserve">Suppoſant que x ſoit la plus grande, & </s> <s xml:id="echoid-s11667" xml:space="preserve">y la plus petite, il <lb/>faut démontrer que x = {a+d/2}, & </s> <s xml:id="echoid-s11668" xml:space="preserve">que y={a-d/2}.</s> <s xml:id="echoid-s11669" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div964" type="section" level="1" n="763"> <head xml:id="echoid-head913" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11670" xml:space="preserve">Puiſque la ſomme des deux grandeurs eſt a, on aura x+y <lb/>=a, & </s> <s xml:id="echoid-s11671" xml:space="preserve">puiſque leur différence eſt d, on aura x-y=d. </s> <s xml:id="echoid-s11672" xml:space="preserve">De <lb/>la premiere équation, on tire y=a-x: </s> <s xml:id="echoid-s11673" xml:space="preserve">donc en mettant <lb/>cette valeur de y dans la ſeconde équation, on aura x-a <lb/>+x=d, ou 2x=a+d: </s> <s xml:id="echoid-s11674" xml:space="preserve">donc x={a+d/2}. </s> <s xml:id="echoid-s11675" xml:space="preserve">Si l’on met cette <pb o="337" file="0391" n="399" rhead="DE MATHÉMATIQUE. Liv. X."/> valeur de x dans la premiere équation, on aura {a+d/2}+y=a <lb/>ou a + d + 2y = 2a: </s> <s xml:id="echoid-s11676" xml:space="preserve">donc 2y = 2a - a - d = a - d, & </s> <s xml:id="echoid-s11677" xml:space="preserve"><lb/>y = {a-d/2}. </s> <s xml:id="echoid-s11678" xml:space="preserve">C. </s> <s xml:id="echoid-s11679" xml:space="preserve">Q. </s> <s xml:id="echoid-s11680" xml:space="preserve">F. </s> <s xml:id="echoid-s11681" xml:space="preserve">D.</s> <s xml:id="echoid-s11682" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div965" type="section" level="1" n="764"> <head xml:id="echoid-head914" xml:space="preserve">PROPOSITION IX.</head> <head xml:id="echoid-head915" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11683" xml:space="preserve">723. </s> <s xml:id="echoid-s11684" xml:space="preserve">Dans un triangle A B C, dont on connoît deux côtés A C <lb/> <anchor type="note" xlink:label="note-0391-01a" xlink:href="note-0391-01"/> & </s> <s xml:id="echoid-s11685" xml:space="preserve">B C avec un angle A, oppoſé à l’un des côtés connus, trouver <lb/>les deux autres angles.</s> <s xml:id="echoid-s11686" xml:space="preserve"/> </p> <div xml:id="echoid-div965" type="float" level="2" n="1"> <note position="right" xlink:label="note-0391-01" xlink:href="note-0391-01a" xml:space="preserve">Figure 183.</note> </div> <p> <s xml:id="echoid-s11687" xml:space="preserve">Pour trouver d’abord l’angle B, ſuppoſant que le côté A C <lb/>ſoit de 26 toiſes, le côté B C de 20, & </s> <s xml:id="echoid-s11688" xml:space="preserve">l’angle A de 50 de-<lb/>grés, il faut chercher le ſinus de cet angle, qui eſt de 76604, <lb/>& </s> <s xml:id="echoid-s11689" xml:space="preserve">dire: </s> <s xml:id="echoid-s11690" xml:space="preserve">Si le côté B C de 20 toiſes donne 76604 pour ſinus de <lb/>l’angle A, que donnera le côté A C de 26 toiſes pour le ſinus <lb/>de l’angle B, que l’on trouvera de 99585; </s> <s xml:id="echoid-s11691" xml:space="preserve">& </s> <s xml:id="echoid-s11692" xml:space="preserve">cherchant dans la <lb/>colonne des ſinus le nombre qui approche le plus de celui-ci, <lb/>l’on verra qu’il correſpond à 84 degrés 45 minutes, qui eſt la <lb/>valeur de l’angle B.</s> <s xml:id="echoid-s11693" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11694" xml:space="preserve">Comme l’on connoît les angles A & </s> <s xml:id="echoid-s11695" xml:space="preserve">B, l’on n’aura qu’à <lb/>ſouſtraire la ſomme de 180, le reſte ſera la différence 45 de-<lb/>grés 15 minutes pour l’angle C.</s> <s xml:id="echoid-s11696" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11697" xml:space="preserve">724. </s> <s xml:id="echoid-s11698" xml:space="preserve">Mais ſi l’angle donné étoit plus ouvert qu’un droit, <lb/> <anchor type="note" xlink:label="note-0391-02a" xlink:href="note-0391-02"/> comme dans le triangle A B C, où l’angle B eſt de 120 degrés, <lb/>le côté A C de 18 toiſes, & </s> <s xml:id="echoid-s11699" xml:space="preserve">le côté B C de 12, il faudra, pour <lb/>connoître l’angle A, chercher le ſinus du ſupplément de l’an-<lb/>gle obtus, c’eſt-à-dire le ſinus de 60 degrés, qui eſt 86602, & </s> <s xml:id="echoid-s11700" xml:space="preserve"><lb/>dire: </s> <s xml:id="echoid-s11701" xml:space="preserve">Si le côté A C de 18 toiſes donne 86602 pour le ſinus <lb/>du ſupplément de l’angle obtus, que donnera le côté B C de <lb/>12 toiſes pour le ſinus de l’angle A, que l’on trouvera de <lb/>57734, qui correſpond à 35 degrés 16 minutes?</s> <s xml:id="echoid-s11702" xml:space="preserve"/> </p> <div xml:id="echoid-div966" type="float" level="2" n="2"> <note position="right" xlink:label="note-0391-02" xlink:href="note-0391-02a" xml:space="preserve">Figure 185.</note> </div> </div> <div xml:id="echoid-div968" type="section" level="1" n="765"> <head xml:id="echoid-head916" xml:space="preserve">PROPOSITION X.</head> <head xml:id="echoid-head917" xml:space="preserve"><emph style="sc">Theoreme</emph></head> <p style="it"> <s xml:id="echoid-s11703" xml:space="preserve">725. </s> <s xml:id="echoid-s11704" xml:space="preserve">Dans tous triangles, comme A B C, dont on connoît deux <lb/> <anchor type="note" xlink:label="note-0391-03a" xlink:href="note-0391-03"/> côtés B A & </s> <s xml:id="echoid-s11705" xml:space="preserve">B C avec l’angle compris A B C, la ſomme des deux <lb/>côtés connus eſt à leur différence, comme la tangente de la moitié <lb/>de la ſomme des deux angles inconnus B A C, & </s> <s xml:id="echoid-s11706" xml:space="preserve">B C A eſt la tan-<lb/>gente de la moitié de leur différence.</s> <s xml:id="echoid-s11707" xml:space="preserve"/> </p> <div xml:id="echoid-div968" type="float" level="2" n="1"> <note position="right" xlink:label="note-0391-03" xlink:href="note-0391-03a" xml:space="preserve">Figure 186.</note> </div> <pb o="338" file="0392" n="400" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div970" type="section" level="1" n="766"> <head xml:id="echoid-head918" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11708" xml:space="preserve">Si du point angulaire B l’on décrit un cercle, dont le rayon <lb/>ſoit le côté B C, & </s> <s xml:id="echoid-s11709" xml:space="preserve">que l’on prolonge le côté A B juſqu’à la <lb/>circonférence en D & </s> <s xml:id="echoid-s11710" xml:space="preserve">E, la ligne A D ſera la ſomme des deux <lb/>côtés connus, puiſque B D eſt égal à B C, & </s> <s xml:id="echoid-s11711" xml:space="preserve">la ligne A E ſera <lb/>la différence de ces deux côtés, puiſque B A eſt plus petit que <lb/>B D de toute la ligne A E. </s> <s xml:id="echoid-s11712" xml:space="preserve">Cela poſé, comme l’angle D B C <lb/>eſt extérieur au triangle A B C, il ſera égal aux deux inté-<lb/>rieurs B A C & </s> <s xml:id="echoid-s11713" xml:space="preserve">B C A: </s> <s xml:id="echoid-s11714" xml:space="preserve">ainſi il vaudra la ſomme des deux an-<lb/>gles inconnus; </s> <s xml:id="echoid-s11715" xml:space="preserve">& </s> <s xml:id="echoid-s11716" xml:space="preserve">ſi l’on tire la ligne E C, l’angle D E C, qui <lb/>eſt à la circonférence, ſera moitié de celui du centre D B C: <lb/></s> <s xml:id="echoid-s11717" xml:space="preserve">ainſi il vaudra la moitié de la ſomme des deux angles incon-<lb/>nus; </s> <s xml:id="echoid-s11718" xml:space="preserve">& </s> <s xml:id="echoid-s11719" xml:space="preserve">ſi l’on tire la ligne D C, qui ſe trouve perpendiculaire <lb/>ſur E C, à cauſe que l’angle E C D eſt renfermé dans un demi-<lb/>cercle, cette ligne ſera la tangente de l’angle D E C, c’eſt-à-<lb/>dire de la moitié de la ſomme des deux angles inconnus. </s> <s xml:id="echoid-s11720" xml:space="preserve">Pré-<lb/>ſentement conſidérez que le triangle E B C eſt iſoſcele, & </s> <s xml:id="echoid-s11721" xml:space="preserve"><lb/>que les angles B E C & </s> <s xml:id="echoid-s11722" xml:space="preserve">B C E de la baſe ſont égaux; </s> <s xml:id="echoid-s11723" xml:space="preserve">par con-<lb/>ſéquent l’angle B E C ſera plus grand que l’angle B C A de <lb/>tout l’angle F C E; </s> <s xml:id="echoid-s11724" xml:space="preserve">& </s> <s xml:id="echoid-s11725" xml:space="preserve">comme l’angle extérieur B A C du trian-<lb/>gle B A C eſt plus grand que l’angle B E C de tout l’angle A C E, <lb/>il s’enſuit donc que l’angle B A C eſt plus grand que B C A de <lb/>deux fois l’angle A C E; </s> <s xml:id="echoid-s11726" xml:space="preserve">ce qui fait voir que l’angle A C E eſt <lb/>la moitié de la différence des deux angles inconnus B A C & </s> <s xml:id="echoid-s11727" xml:space="preserve"><lb/>B C A. </s> <s xml:id="echoid-s11728" xml:space="preserve">Or ſi la ligne E F eſt perpendiculaire ſur E C, elle ſera <lb/>la tangente de la moitié de la différence des deux angles in-<lb/>connus, étant tangente de l’angle F C E; </s> <s xml:id="echoid-s11729" xml:space="preserve">mais les lignes D C <lb/>& </s> <s xml:id="echoid-s11730" xml:space="preserve">F E ſont paralleles, puiſqu’elles ſont perpendiculaires ſur <lb/>E C; </s> <s xml:id="echoid-s11731" xml:space="preserve">par conſéquent l’angle F E A ſera égal à ſon alterne <lb/>E D C. </s> <s xml:id="echoid-s11732" xml:space="preserve">Et comme les angles F A E & </s> <s xml:id="echoid-s11733" xml:space="preserve">D A C ſont auſſi égaux, <lb/>il s’enſuit que les triangles A F E & </s> <s xml:id="echoid-s11734" xml:space="preserve">A D C ſont ſemblables, <lb/>d’où l’on tire A D: </s> <s xml:id="echoid-s11735" xml:space="preserve">A E : </s> <s xml:id="echoid-s11736" xml:space="preserve">: </s> <s xml:id="echoid-s11737" xml:space="preserve">D C: </s> <s xml:id="echoid-s11738" xml:space="preserve">F E, qui fait voir que la ſomme <lb/>des deux côtés A D eſt à leur différence A E, comme la ligne <lb/>D C, tangente de la moitié de la ſomme des deux angles in-<lb/>connus, eſt à la ligne F E tangente de la moitié de leur diffé-<lb/>rence. </s> <s xml:id="echoid-s11739" xml:space="preserve">C. </s> <s xml:id="echoid-s11740" xml:space="preserve">Q. </s> <s xml:id="echoid-s11741" xml:space="preserve">F. </s> <s xml:id="echoid-s11742" xml:space="preserve">D.</s> <s xml:id="echoid-s11743" xml:space="preserve"/> </p> <pb o="339" file="0393" n="401" rhead="DE MATHÉMATIQUE. Liv. X."/> </div> <div xml:id="echoid-div971" type="section" level="1" n="767"> <head xml:id="echoid-head919" xml:space="preserve">PROPOSITION XI.</head> <head xml:id="echoid-head920" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11744" xml:space="preserve">726. </s> <s xml:id="echoid-s11745" xml:space="preserve">Dans un triangle A B C, dont on connoît deux côtés A C <lb/> <anchor type="note" xlink:label="note-0393-01a" xlink:href="note-0393-01"/> & </s> <s xml:id="echoid-s11746" xml:space="preserve">B C avec l’angle compris C, trouver les angles A & </s> <s xml:id="echoid-s11747" xml:space="preserve">B.</s> <s xml:id="echoid-s11748" xml:space="preserve"/> </p> <div xml:id="echoid-div971" type="float" level="2" n="1"> <note position="right" xlink:label="note-0393-01" xlink:href="note-0393-01a" xml:space="preserve">Figure 185.</note> </div> <p> <s xml:id="echoid-s11749" xml:space="preserve">Comme ce Problême eſt une application du théorême pré-<lb/>cédent, il faut, pour le réſoudre, ajouter les deux côtés C B & </s> <s xml:id="echoid-s11750" xml:space="preserve"><lb/>C A enſemble, c’eſt-à-dire 25, & </s> <s xml:id="echoid-s11751" xml:space="preserve">20 pour avoir la ſomme des <lb/>deux côtés connus, & </s> <s xml:id="echoid-s11752" xml:space="preserve">ſouſtraire le plus petit côté du grand <lb/>pour en avoir la différence, qui ſera 5; </s> <s xml:id="echoid-s11753" xml:space="preserve">& </s> <s xml:id="echoid-s11754" xml:space="preserve">comme l’angle C <lb/>eſt ſuppoſé de 40 degrés, l’on cherchera ſa différence avec <lb/>deux droits, que l’on trouvera de 140, dont la moitié 70 ſera <lb/>la moitié de la ſomme des deux angles inconnus A & </s> <s xml:id="echoid-s11755" xml:space="preserve">B. </s> <s xml:id="echoid-s11756" xml:space="preserve">Or <lb/>cherchant la tangente de cet angle, qui eſt 274747, l’on dira: <lb/></s> <s xml:id="echoid-s11757" xml:space="preserve">Si 45, ſomme des deux côtés connus, donne 5 pour leur dif-<lb/>férence, que donnera 274747, tangente de la moitié de la <lb/>ſomme des deux angles inconnus pour la tangente de la moitié <lb/>de la différence des deux angles inconnus, que l’on trouvera <lb/>30527.</s> <s xml:id="echoid-s11758" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11759" xml:space="preserve">Préſentement ſi l’on cherche dans la colonne des tangentes <lb/>le nombre le plus approchant de celui-ci, l’on verra qu’il cor-<lb/>reſpond à 16 degrés & </s> <s xml:id="echoid-s11760" xml:space="preserve">59 minutes: </s> <s xml:id="echoid-s11761" xml:space="preserve">& </s> <s xml:id="echoid-s11762" xml:space="preserve">comme cette quantité <lb/>n’eſt que la moitié de la différence, il faut la doubler pour <lb/>avoir la différence entiere, qui ſera 33 degrés 58 minutes, <lb/>qu’il faut ſouſtraire de la ſomme des deux angles inconnus, <lb/>c’eſt-à-dire de 140 degrés, & </s> <s xml:id="echoid-s11763" xml:space="preserve">l’on trouvera pour la différence <lb/>106 degrés 2 minutes, dont on n’a plus qu’à prendre la moitié <lb/>pour avoir la valeur de l’angle oppoſé au pluspetit côté, c’eſt-<lb/>à-dire de l’angle B, qui ſera de 53 degrés une minute: </s> <s xml:id="echoid-s11764" xml:space="preserve">car par <lb/>le lemme de l’art. </s> <s xml:id="echoid-s11765" xml:space="preserve">722, le plus petit angle doit être égal à la <lb/>moitié de la ſomme, moins la moitié de la différence, & </s> <s xml:id="echoid-s11766" xml:space="preserve">c’eſt <lb/>ce que l’on trouve en ôtant la différence de la ſomme, & </s> <s xml:id="echoid-s11767" xml:space="preserve">pre-<lb/>nant la moitié du reſte.</s> <s xml:id="echoid-s11768" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11769" xml:space="preserve">Pour avoir l’angle A, on n’a qu’à ajouter la différence 33 <lb/>degrés 58 minutes à la valeur de l’angle B, & </s> <s xml:id="echoid-s11770" xml:space="preserve">l’on trouvera <lb/>qu’il eſt de 86 degrés 59 minutes.</s> <s xml:id="echoid-s11771" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11772" xml:space="preserve">Si l’on veut connoître le côté A B, il ſera facile de le trouver <lb/>par la ſeptieme propoſition.</s> <s xml:id="echoid-s11773" xml:space="preserve"/> </p> <pb o="340" file="0394" n="402" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div973" type="section" level="1" n="768"> <head xml:id="echoid-head921" xml:space="preserve">PROPOSITION XII.</head> <head xml:id="echoid-head922" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s11774" xml:space="preserve">727. </s> <s xml:id="echoid-s11775" xml:space="preserve">Dans tous triangles comme A B C, dont on connoît les <lb/> <anchor type="note" xlink:label="note-0394-01a" xlink:href="note-0394-01"/> trois côtés, le plus grand côté A C eſt à la ſomme des deux autres <lb/>côtés A B & </s> <s xml:id="echoid-s11776" xml:space="preserve">B C, comme la différence de ces deux mêmes côtés eſt <lb/>à la différence des ſegmens A G & </s> <s xml:id="echoid-s11777" xml:space="preserve">G C de la baſe.</s> <s xml:id="echoid-s11778" xml:space="preserve"/> </p> <div xml:id="echoid-div973" type="float" level="2" n="1"> <note position="left" xlink:label="note-0394-01" xlink:href="note-0394-01a" xml:space="preserve">Figure 188.</note> </div> </div> <div xml:id="echoid-div975" type="section" level="1" n="769"> <head xml:id="echoid-head923" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s11779" xml:space="preserve">Si du point B l’on décrit un cercle, dont le rayon ſoit le <lb/>côté B C plus grand que B A, & </s> <s xml:id="echoid-s11780" xml:space="preserve">que l’on prolonge le côté A B <lb/>juſqu’à la circonférence, B D étant égal à B C, A D ſera la <lb/>ſomme des deux côtés A B & </s> <s xml:id="echoid-s11781" xml:space="preserve">B C, & </s> <s xml:id="echoid-s11782" xml:space="preserve">A F en ſera la différence: <lb/></s> <s xml:id="echoid-s11783" xml:space="preserve">& </s> <s xml:id="echoid-s11784" xml:space="preserve">comme la ligne E C eſt diviſée en deux également par la <lb/>perpendiculaire B G, E A ſera la différence des deux ſegmens <lb/>A G & </s> <s xml:id="echoid-s11785" xml:space="preserve">G C. </s> <s xml:id="echoid-s11786" xml:space="preserve">Si l’on tire les lignes D C & </s> <s xml:id="echoid-s11787" xml:space="preserve">E F, l’on aura les <lb/>deux triangles ſemblables A E F & </s> <s xml:id="echoid-s11788" xml:space="preserve">A D C: </s> <s xml:id="echoid-s11789" xml:space="preserve">car ils ont un an-<lb/>gle oppoſé au ſommet en A, & </s> <s xml:id="echoid-s11790" xml:space="preserve">de plus l’angle en E eſt égal <lb/>à l’angle en D, puiſqu’ils ſont appuyés ſur le même arc F C. </s> <s xml:id="echoid-s11791" xml:space="preserve"><lb/>On aura donc cette proportion, A C qui eſt la baſe, eſt à A D <lb/>qui eſt la ſomme des deux côtés, comme A F, qui eſt la <lb/>différence de ces deux côtés eſt à A E, qui eſt la différence des <lb/>ſegmens de la baſe. </s> <s xml:id="echoid-s11792" xml:space="preserve">C. </s> <s xml:id="echoid-s11793" xml:space="preserve">Q. </s> <s xml:id="echoid-s11794" xml:space="preserve">F. </s> <s xml:id="echoid-s11795" xml:space="preserve">D.</s> <s xml:id="echoid-s11796" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11797" xml:space="preserve">Ce théorême nous donne un moyen de connoître les trois <lb/>angles d’un triangle dont on connoît les trois côtés, comme <lb/>on le va voir dans le problême ſuivant, qui en eſt une appli-<lb/>cation.</s> <s xml:id="echoid-s11798" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div976" type="section" level="1" n="770"> <head xml:id="echoid-head924" xml:space="preserve">PROPOSITION XIII.</head> <head xml:id="echoid-head925" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11799" xml:space="preserve">728. </s> <s xml:id="echoid-s11800" xml:space="preserve">Connoiſſant les trois côtés d’un triangle A B C, l’on de-<lb/> <anchor type="note" xlink:label="note-0394-02a" xlink:href="note-0394-02"/> mande de trouver la valeur d’un des ſegmens de la baſe.</s> <s xml:id="echoid-s11801" xml:space="preserve"/> </p> <div xml:id="echoid-div976" type="float" level="2" n="1"> <note position="left" xlink:label="note-0394-02" xlink:href="note-0394-02a" xml:space="preserve">Figure 189.</note> </div> <p> <s xml:id="echoid-s11802" xml:space="preserve">Suppoſant que la baſe A C ſoit de 15 toiſes, le côté A B de <lb/>8, & </s> <s xml:id="echoid-s11803" xml:space="preserve">le côté B C de 12, il faut dire: </s> <s xml:id="echoid-s11804" xml:space="preserve">Comme la baſe A C <lb/>de 15 eſt à la ſomme des deux autres côtés, qui eſt 20: </s> <s xml:id="echoid-s11805" xml:space="preserve">ainſi <lb/>la différence de ces deux côtés, qui eſt 4, eſt à la différence <lb/>des deux ſegmens, que l’on trouvera de 5 toiſes 2 pieds. </s> <s xml:id="echoid-s11806" xml:space="preserve">Pré-<lb/>ſentement ſi l’on ajoute cette quantité à la valeur de la baſe <lb/>A C, l’on aura 20 toiſes 2 pieds, qui ſera la valeur d’une ligne <pb o="341" file="0395" n="403" rhead="DE MATHÉMATIQUE. Liv. X."/> telle que E C; </s> <s xml:id="echoid-s11807" xml:space="preserve">par conſéquent ſi on en prend la moitié, on <lb/>connoîtra le plus grand ſegment D C, qui eſt de 10 toiſes <lb/>un pied: </s> <s xml:id="echoid-s11808" xml:space="preserve">mais comme l’on connoît dans le triangle rectangle <lb/>D B C les côtés B C & </s> <s xml:id="echoid-s11809" xml:space="preserve">D C, l’on pourra donc connoître auſſi <lb/>l’angle C, & </s> <s xml:id="echoid-s11810" xml:space="preserve">enſuite les angles A & </s> <s xml:id="echoid-s11811" xml:space="preserve">B. </s> <s xml:id="echoid-s11812" xml:space="preserve">Pour cela, on fera <lb/>cette proportion, comme le côté B C eſt au ſinus total, ainſi <lb/>le ſegment D C eſt au ſinus de l’angle D B C. </s> <s xml:id="echoid-s11813" xml:space="preserve">Connoiſſant cet <lb/>angle, on n’aura qu’à ôter ſa valeur de 90 degrés, & </s> <s xml:id="echoid-s11814" xml:space="preserve">l’on aura <lb/>la valeur de l’angle C. </s> <s xml:id="echoid-s11815" xml:space="preserve">On trouveroit de même l’angle A B D <lb/>& </s> <s xml:id="echoid-s11816" xml:space="preserve">l’angle A.</s> <s xml:id="echoid-s11817" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div978" type="section" level="1" n="771"> <head xml:id="echoid-head926" style="it" xml:space="preserve">Uſages des Logarithmes pour le calcul des Triangles.</head> <p> <s xml:id="echoid-s11818" xml:space="preserve">729. </s> <s xml:id="echoid-s11819" xml:space="preserve">On a pu voir dans les Tables qu’il y a trois colonnes <lb/>ſur la droite de celles dont nous nous ſommes ſervis, au haut <lb/>deſquelles on trouve ces mots, Logarithmes des ſinus, Loga-<lb/>rithmes des tangentes, Logarithmes des ſécantes. </s> <s xml:id="echoid-s11820" xml:space="preserve">Pour concevoir <lb/>comment on peut faire uſage des logarithmes dans le calcul <lb/>des triangles, il faut ſe rappeller ce que nous avons démontré <lb/>ſur les propriétés des logarithmes, par le moyen deſquels toute <lb/>multiplication eſt réduite à l’addition des logarithmes du mul-<lb/>tiplicande & </s> <s xml:id="echoid-s11821" xml:space="preserve">du multiplicateur, & </s> <s xml:id="echoid-s11822" xml:space="preserve">toute diviſion à une ſouſ-<lb/>traction du logarithme du diviſeur de celui du dividende. </s> <s xml:id="echoid-s11823" xml:space="preserve">Il <lb/>faut encore ſe rappeller que toute Regle de Trois ſe réduit à <lb/>l’addition des logarithmes des deux moyens, & </s> <s xml:id="echoid-s11824" xml:space="preserve">à la ſouſtrac-<lb/>tion du logarithme du premier extrême de la ſomme de ceux <lb/>des moyens. </s> <s xml:id="echoid-s11825" xml:space="preserve">Cela poſé, il eſt évident que ſi l’on connoît les <lb/>logarithmes des ſinus, tangentes & </s> <s xml:id="echoid-s11826" xml:space="preserve">ſécantes, comme on a ceux <lb/>des nombres naturels qui expriment les côtés des triangles que <lb/>l’on veut calculer, les proportions qu’il faut faire ſe réduiront <lb/>à l’addition de deux logarithmes, & </s> <s xml:id="echoid-s11827" xml:space="preserve">à la ſouſtraction du lo-<lb/>garithme du premier terme de la ſomme des logarithmes des <lb/>moyens. </s> <s xml:id="echoid-s11828" xml:space="preserve">Ainſi en cherchant les ſinus, il faudra prendre le <lb/>logarithme du ſinus; </s> <s xml:id="echoid-s11829" xml:space="preserve">en cherchant une tangente, il faudra <lb/>prendre le logarithme de cette tangente, & </s> <s xml:id="echoid-s11830" xml:space="preserve">en cherchant la <lb/>ſécante, il faudra prendre le logarithme de cette ſécante au <lb/>lieu des ſinus des tangentes & </s> <s xml:id="echoid-s11831" xml:space="preserve">des ſécantes. </s> <s xml:id="echoid-s11832" xml:space="preserve">Enſuite au lieu de <lb/>mettre les nombres naturels qui expriment le nombre de toiſes <lb/>ou de pieds contenus dans les côtés connus, il faudra prendre <lb/>les logarithmes de ces nombres, que l’on cherchera dans les <pb o="342" file="0396" n="404" rhead="NOUVEAU COURS"/> Tables des Logarithmes calculées depuis l’unité juſqu’à 200000, <lb/>que l’on trouve dans le même Livre que les Tables des ſinus <lb/>tangentes & </s> <s xml:id="echoid-s11833" xml:space="preserve">ſécantes. </s> <s xml:id="echoid-s11834" xml:space="preserve">On en va voir des exemples dans les <lb/>articles ſuivans.</s> <s xml:id="echoid-s11835" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div979" type="section" level="1" n="772"> <head xml:id="echoid-head927" xml:space="preserve"><emph style="sc">Exemple</emph> I.</head> <p> <s xml:id="echoid-s11836" xml:space="preserve">730. </s> <s xml:id="echoid-s11837" xml:space="preserve">Ayant un triangle rectangle A D E, dont on connoît <lb/> <anchor type="note" xlink:label="note-0396-01a" xlink:href="note-0396-01"/> l’angle A de 30 degrés, & </s> <s xml:id="echoid-s11838" xml:space="preserve">le côté A D de 20 toiſes; </s> <s xml:id="echoid-s11839" xml:space="preserve">l’on de-<lb/>mande de trouver le côté D E, en ſe ſervant des logarithmes.</s> <s xml:id="echoid-s11840" xml:space="preserve"/> </p> <div xml:id="echoid-div979" type="float" level="2" n="1"> <note position="left" xlink:label="note-0396-01" xlink:href="note-0396-01a" xml:space="preserve">Figure 180.</note> </div> <p> <s xml:id="echoid-s11841" xml:space="preserve">Pour le trouver, je cherche dans la Table la page, au ſom-<lb/>met de laquelle il y a 30 degrés; </s> <s xml:id="echoid-s11842" xml:space="preserve">& </s> <s xml:id="echoid-s11843" xml:space="preserve">au lieu de prendre la tan-<lb/>gente de la troiſieme colonne, je prends ſon logarithme, qui <lb/>eſt 97614394. </s> <s xml:id="echoid-s11844" xml:space="preserve">Et comme j’ai auſſi beſoin du ſinus total, au <lb/>lieu de prendre celui qui eſt diviſé en 100000 parties, je <lb/>prends ſon logarithme, qui eſt diviſé en 100000000 parties; <lb/></s> <s xml:id="echoid-s11845" xml:space="preserve">& </s> <s xml:id="echoid-s11846" xml:space="preserve">comme il faut faire une Regle pour trouver le côté D E, <lb/>dont le premier terme doit être le ſinus total dont je viens de <lb/>parler, le ſecond la tangente que nous venons de trouver, & </s> <s xml:id="echoid-s11847" xml:space="preserve"><lb/>le troiſieme la valeur du côté A D. </s> <s xml:id="echoid-s11848" xml:space="preserve">Il faut auſſi, au lieu de <lb/>mettre ſimplement 20 toiſes au troiſieme terme, mettre à ſa <lb/>place le logarithme de ce nombre, que l’on trouvera dans le <lb/>premier feuillet de la Table des Logarithmes des nombres na-<lb/>turels à côté du nombre 20, dont le logarithme eſt 13010300. </s> <s xml:id="echoid-s11849" xml:space="preserve"><lb/>Préſentement il faut faire cette proportion arithmétique: </s> <s xml:id="echoid-s11850" xml:space="preserve">Si <lb/>le ſinus total 100000000 donne 97614394 pour le logarithme <lb/>de la tangente de 30 degrés, combien donneront 13010300, <lb/>logarithme de 20 toiſes, pour le logarithme du nombre que je <lb/>cherche; </s> <s xml:id="echoid-s11851" xml:space="preserve">& </s> <s xml:id="echoid-s11852" xml:space="preserve">pour le trouver, j’additionne le ſecond & </s> <s xml:id="echoid-s11853" xml:space="preserve">le troi-<lb/>ſieme terme, & </s> <s xml:id="echoid-s11854" xml:space="preserve">de la ſomme j’en ſouſtrais le premier pour <lb/>avoir 10624694, qui eſt le logarithme du nombre que je <lb/>cherche: </s> <s xml:id="echoid-s11855" xml:space="preserve">& </s> <s xml:id="echoid-s11856" xml:space="preserve">pour ſçavoir quel eſt ce nombre, j’ai recours à la <lb/>Table des Logarithmes des nombres naturels pour chercher <lb/>un logarithme qui approche le plus de celui-ci, & </s> <s xml:id="echoid-s11857" xml:space="preserve">j’en trouve <lb/>un qui eſt un peu trop petit, qui correſpond au nombre 11, <lb/>& </s> <s xml:id="echoid-s11858" xml:space="preserve">un autre qui eſt un peu trop grand, qui correſpond au nom-<lb/>bre 12; </s> <s xml:id="echoid-s11859" xml:space="preserve">c’eſt pourquoi j’en cherche un qui ſoit à peu près <lb/>moyen entre ces deux-là, comme eſt, par exemple, 11 {1/2}; </s> <s xml:id="echoid-s11860" xml:space="preserve">ce <lb/>qui fait voir que le côté D E eſt à peu près de 11 toiſes <lb/>3 pieds.</s> <s xml:id="echoid-s11861" xml:space="preserve"/> </p> <pb o="343" file="0397" n="405" rhead="DE MATHÉMATIQUE. Liv. X."/> </div> <div xml:id="echoid-div981" type="section" level="1" n="773"> <head xml:id="echoid-head928" xml:space="preserve"><emph style="sc">Exemple</emph> II.</head> <p> <s xml:id="echoid-s11862" xml:space="preserve">731. </s> <s xml:id="echoid-s11863" xml:space="preserve">Si l’on a un triangle rectangle A B C, dont on con-<lb/> <anchor type="note" xlink:label="note-0397-01a" xlink:href="note-0397-01"/> noît le côté A B de 16 toiſes, & </s> <s xml:id="echoid-s11864" xml:space="preserve">le côté B C de 14, pour con-<lb/>noître l’angle A, il faut chercher dans la ſeconde Table le <lb/>logarithme de 16, qui eſt 12041200, & </s> <s xml:id="echoid-s11865" xml:space="preserve">le logarithme de 14, <lb/>qui éſt 11461280; </s> <s xml:id="echoid-s11866" xml:space="preserve">& </s> <s xml:id="echoid-s11867" xml:space="preserve">à cauſe des triangles ſemblables A B C <lb/>& </s> <s xml:id="echoid-s11868" xml:space="preserve">A D E, l’on dira: </s> <s xml:id="echoid-s11869" xml:space="preserve">Si 12041200, logarithme du côté A B, <lb/>donne 11461280 pour le logarithme du côté B C, que donnera <lb/>le logarithme du côté A D, qui eſt 100000000 pour le loga-<lb/>rithme de la tangente D E, l’on trouvera (après avoir ajouté <lb/>le ſecond & </s> <s xml:id="echoid-s11870" xml:space="preserve">le troiſieme terme, & </s> <s xml:id="echoid-s11871" xml:space="preserve">ſouſtrait de leur ſomme le <lb/>premier) que la différence eſt 99420080 pour le logarithme <lb/>de la tangente, lequel correſpond dans les Tables à 41 degrés <lb/>12 minutes, qui eſt la valeur de l’angle A.</s> <s xml:id="echoid-s11872" xml:space="preserve"/> </p> <div xml:id="echoid-div981" type="float" level="2" n="1"> <note position="right" xlink:label="note-0397-01" xlink:href="note-0397-01a" xml:space="preserve">Figure 179.</note> </div> </div> <div xml:id="echoid-div983" type="section" level="1" n="774"> <head xml:id="echoid-head929" xml:space="preserve"><emph style="sc">Exemple</emph> III.</head> <p> <s xml:id="echoid-s11873" xml:space="preserve">732. </s> <s xml:id="echoid-s11874" xml:space="preserve">Ayant un triangle A B C, dont on connoît l’angle A <lb/> <anchor type="note" xlink:label="note-0397-02a" xlink:href="note-0397-02"/> de 40 degrés, & </s> <s xml:id="echoid-s11875" xml:space="preserve">l’angle B de 60, & </s> <s xml:id="echoid-s11876" xml:space="preserve">le côté B C de 15 toiſes, <lb/>l’on demande la valeur du côté A C.</s> <s xml:id="echoid-s11877" xml:space="preserve"/> </p> <div xml:id="echoid-div983" type="float" level="2" n="1"> <note position="right" xlink:label="note-0397-02" xlink:href="note-0397-02a" xml:space="preserve">Figure 182.</note> </div> <p> <s xml:id="echoid-s11878" xml:space="preserve">Je cherche le logarithme du ſinus de 40 degrés, qui eſt <lb/>98080675, & </s> <s xml:id="echoid-s11879" xml:space="preserve">le logarithme de 60 degrés, qui eſt 99375306; <lb/></s> <s xml:id="echoid-s11880" xml:space="preserve">& </s> <s xml:id="echoid-s11881" xml:space="preserve">enfin dans la ſeconde Table le logarithme du nombre 15, <lb/>qui eſt 11760913; </s> <s xml:id="echoid-s11882" xml:space="preserve">& </s> <s xml:id="echoid-s11883" xml:space="preserve">faiſant l’analogie ordinaire, je dis: </s> <s xml:id="echoid-s11884" xml:space="preserve">Si le <lb/>logarithme du ſinus de l’angle A, qui eſt 98080675, donne <lb/>11760913 pour le logarithme du côté B C, que donnera le lo-<lb/>garithme du ſinus de l’angle B, qui eſt 99375306 pour le lo-<lb/>garithme du côté A C, que je trouve de 13055544; </s> <s xml:id="echoid-s11885" xml:space="preserve">& </s> <s xml:id="echoid-s11886" xml:space="preserve">cher-<lb/>chant dans la ſeconde Table le logarithme qui approche le plus <lb/>de celui-ci, je trouve qu’il correſpond au nombre 20; </s> <s xml:id="echoid-s11887" xml:space="preserve">ce qui <lb/>fait voir que le côté A C eſt de 20 toiſes.</s> <s xml:id="echoid-s11888" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div985" type="section" level="1" n="775"> <head xml:id="echoid-head930" xml:space="preserve"><emph style="sc">Application de la</emph> <emph style="sc">Trigonometrie a la pratique</emph>. <lb/>PROPOSITION XIV.</head> <head xml:id="echoid-head931" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p> <s xml:id="echoid-s11889" xml:space="preserve">733. </s> <s xml:id="echoid-s11890" xml:space="preserve">Trouver une diſtance inacceſſible. <lb/></s> </p> <p> <s xml:id="echoid-s11891" xml:space="preserve">Un objet quelconque tel que C étant donnée, duquel <lb/> <anchor type="note" xlink:label="note-0397-04a" xlink:href="note-0397-04"/> on ſuppoſe qu’on ne peut pas approcher, on demande la <lb/>quantité de toiſes qu’il peut y avoir de cet objet à l’endroit D.</s> <s xml:id="echoid-s11892" xml:space="preserve"> <pb o="344" file="0398" n="406" rhead="NOUVEAU COURS"/> Pour la trouver, il faut envoyer une perſonne avec un jalon à <lb/>l’endroit A, éloigné d’une diſtance proportionnée à l’inter-<lb/>valle qu’il peut y avoir du point D au point C. </s> <s xml:id="echoid-s11893" xml:space="preserve">Cette diſtance <lb/>ſera, par exemple, ici de 20 toiſes, qui eſt une quantité qui <lb/>doit ſervir de baſe pour faire l’opération. </s> <s xml:id="echoid-s11894" xml:space="preserve">Après cela vous <lb/>prendrez l’ouverture de l’angle formé par la baſe D A, & </s> <s xml:id="echoid-s11895" xml:space="preserve">le <lb/>rayon viſuel D C; </s> <s xml:id="echoid-s11896" xml:space="preserve">& </s> <s xml:id="echoid-s11897" xml:space="preserve">pour bien prendre cet angle, il faut com-<lb/>mencer par mettre les deux pinulles du graphometre, qui ſont <lb/>immobiles d’alignement avec les points D & </s> <s xml:id="echoid-s11898" xml:space="preserve">A: </s> <s xml:id="echoid-s11899" xml:space="preserve">après quoi <lb/>vous faites tourner l’alidale de maniere que vous puiſſiez ap-<lb/>percevoir par les fentes des pinulles (qui ſont à ſes extrêmités) <lb/>l’objet C. </s> <s xml:id="echoid-s11900" xml:space="preserve">Après quoi vous comptez la quantité de degrés que <lb/>contient l’angle marqué ſur le graphometre, c’eſt-à-dire l’angle <lb/>compris par le côté du graphometre, qui eſt d’alignement avec <lb/>les points D & </s> <s xml:id="echoid-s11901" xml:space="preserve">A, & </s> <s xml:id="echoid-s11902" xml:space="preserve">le rayon viſuel qui apperçoit l’objet C; <lb/></s> <s xml:id="echoid-s11903" xml:space="preserve">& </s> <s xml:id="echoid-s11904" xml:space="preserve">je ſuppoſe que c’eſt ici de 70 degrés. </s> <s xml:id="echoid-s11905" xml:space="preserve">Cela étant fait, il faut <lb/>poſer un autre jalon à l’endroit où étoit poſé le pied du gra-<lb/>phometre, c’eſt-à-dire au point D, & </s> <s xml:id="echoid-s11906" xml:space="preserve">puis venir à l’endroit A <lb/>pour y prendre la valeur de l’angle D A C, j’entends l’angle <lb/>formé par la baſe, & </s> <s xml:id="echoid-s11907" xml:space="preserve">par un ſecond rayon viſuel, qui doit <lb/>obſerver l’objet C, & </s> <s xml:id="echoid-s11908" xml:space="preserve">je ſuppoſe que cet angle eſt de 80 de-<lb/>grés. </s> <s xml:id="echoid-s11909" xml:space="preserve">Cela poſé, il ne s’agit plus que de connoître l’angle C, <lb/>que l’on trouvera aiſément en ſouſtrayant la ſomme des deux <lb/>angles A & </s> <s xml:id="echoid-s11910" xml:space="preserve">D de la valeur de deux droits, & </s> <s xml:id="echoid-s11911" xml:space="preserve">vous trouverez <lb/>que cet angle eſt de 30 degrés. </s> <s xml:id="echoid-s11912" xml:space="preserve">Or pour connoître le côté C D, <lb/>il n’y a qu’à dire: </s> <s xml:id="echoid-s11913" xml:space="preserve">Si le ſinus de 30 degrés m’a donné 20 toiſes <lb/>pour le côté A D, que me donnera le ſinus de l’angle A de <lb/>80 degrés pour la valeur du côté C D? </s> <s xml:id="echoid-s11914" xml:space="preserve">L’on trouvera 39 toiſes <lb/>deux pieds pour la diſtance que l’on cherche.</s> <s xml:id="echoid-s11915" xml:space="preserve"/> </p> <div xml:id="echoid-div985" type="float" level="2" n="1"> <note position="right" xlink:label="note-0397-04" xlink:href="note-0397-04a" xml:space="preserve">Figure 190.</note> </div> </div> <div xml:id="echoid-div987" type="section" level="1" n="776"> <head xml:id="echoid-head932" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s11916" xml:space="preserve">734. </s> <s xml:id="echoid-s11917" xml:space="preserve">Il arrive quelquefois que l’on eſt embarraſſé de trouver <lb/>une diſtance inacceſſible, lorſqu’elle eſt extrêmement éloignée, <lb/>comme ſi elle avoit deux ou trois lieues La difficulté pour lors <lb/>eſt d’avoir une baſe aſſez grande, qu’il faut dans ce cas-là au <lb/>moins de 1000 toiſes. </s> <s xml:id="echoid-s11918" xml:space="preserve">Comme il ſeroit fort pénible de meſurer <lb/>une ſi longue diſtance, jointe à l’inégalité du terrein, & </s> <s xml:id="echoid-s11919" xml:space="preserve">aux <lb/>obſtacles qu’on peut rencontrer, le parti qu’il faut prendre, <lb/>c’eſt de ſe donner d’abord une petite baſe, par le moyen de <lb/>laquelle vous pouvez en avoir une trois ou quatre fois plus <pb o="345" file="0399" n="407" rhead="DE MATHÉMATIQUE. Liv. X."/> grande; </s> <s xml:id="echoid-s11920" xml:space="preserve">& </s> <s xml:id="echoid-s11921" xml:space="preserve">avec cette ſeconde, une troiſieme plus grande eſt <lb/>ſuffiſante pour faire votre opération.</s> <s xml:id="echoid-s11922" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11923" xml:space="preserve">Les opérations précédentes ſont très-utiles pour lever des <lb/>Cartes, afin de ſe donner des points capitaux pour y rapporter <lb/>tous les lieux qui y ont rapport; </s> <s xml:id="echoid-s11924" xml:space="preserve">ou bien ſi l’on veut lever la <lb/>campagne qu’occupe une armée, pour y marquer les quartiers, <lb/>les lignes de circonvallation, les poſtes de conſéquence; </s> <s xml:id="echoid-s11925" xml:space="preserve">enfin <lb/>tout ce qui peut devenir intéreſſant en pareil cas.</s> <s xml:id="echoid-s11926" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11927" xml:space="preserve">Si on aſſiége une Place, & </s> <s xml:id="echoid-s11928" xml:space="preserve">que l’on ſoit obligé de faire <lb/>quelques galeries pour établir des fourneaux ſous les angles du <lb/>chemin couvert, ou ſous quelque ouvrage avancé, il faut ab-<lb/>ſolument avoir recours à cette opération, afin qu’étant pré-<lb/>venu de la diſtance de l’entrée de la galerie à l’objet vers lequel <lb/>on chemine, on ſçache donner à cette galerie la longueur <lb/>néceſſaire pour être poſitivement ſous l’objet qu’on veut faire <lb/>ſauter.</s> <s xml:id="echoid-s11929" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div988" type="section" level="1" n="777"> <head xml:id="echoid-head933" xml:space="preserve">PROPOSITION XV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11930" xml:space="preserve">735. </s> <s xml:id="echoid-s11931" xml:space="preserve">Trouver la diſtance inacceſſible d’un lieu à un autre, <lb/> <anchor type="note" xlink:label="note-0399-01a" xlink:href="note-0399-01"/> comme de l’endroit D à l’endroit C.</s> <s xml:id="echoid-s11932" xml:space="preserve"/> </p> <div xml:id="echoid-div988" type="float" level="2" n="1"> <note position="right" xlink:label="note-0399-01" xlink:href="note-0399-01a" xml:space="preserve">Figure 191.</note> </div> <p> <s xml:id="echoid-s11933" xml:space="preserve">Pour faire cette opération, il faut commencer par ſe don-<lb/>ner une baſe telle que A B, que je ſuppoſe ici de 100 toiſes, <lb/>& </s> <s xml:id="echoid-s11934" xml:space="preserve">de l’extrêmité B prendre avec l’inſtrument l’ouverture de <lb/>l’angle A B C, formé par la baſe A B, & </s> <s xml:id="echoid-s11935" xml:space="preserve">le rayon viſuel B C; <lb/></s> <s xml:id="echoid-s11936" xml:space="preserve">on ſuppoſe cet angle de 92 degrés: </s> <s xml:id="echoid-s11937" xml:space="preserve">du même endroit B il <lb/>faut prendre auſſi l’ouverture de l’angle A B D, qui ſera, par <lb/>exemple, de 45 degrés; </s> <s xml:id="echoid-s11938" xml:space="preserve">& </s> <s xml:id="echoid-s11939" xml:space="preserve">cette opération étant faite, il faut <lb/>venir à l’autre extrêmité A de la baſe A B, pour y prendre <lb/>l’ouverture de l’angle D A B, que je ſuppoſe ici de 98 degrés; </s> <s xml:id="echoid-s11940" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s11941" xml:space="preserve">du même endroit prendre encore l’ouverture de l’angle <lb/>D A C, qui ſera, par exemple, de 50 degrés. </s> <s xml:id="echoid-s11942" xml:space="preserve">Les angles étant <lb/>connus, auſſi-bien que la baſe A B, l’on n’aura aucune diffi-<lb/>culté de trouver la diſtance D C, non plus que celle de D en <lb/>A, & </s> <s xml:id="echoid-s11943" xml:space="preserve">celle de B en C: </s> <s xml:id="echoid-s11944" xml:space="preserve">car conſidérez qu’il eſt facile de trouver <lb/>la valeur des côtés A C & </s> <s xml:id="echoid-s11945" xml:space="preserve">B C du triangle C A B, parce que <lb/>l’on connoît le côté A B de 100 toiſes, l’angle B de 92 de-<lb/>grés, & </s> <s xml:id="echoid-s11946" xml:space="preserve">l’angle C A B de 48, & </s> <s xml:id="echoid-s11947" xml:space="preserve">par conſéquent l’angle A C B <lb/>de 40 degrés. </s> <s xml:id="echoid-s11948" xml:space="preserve">Cela poſé, pour trouver la valeur du côté C B, <pb o="346" file="0400" n="408" rhead="NOUVEAU COURS"/> il n’y a qu’à dire: </s> <s xml:id="echoid-s11949" xml:space="preserve">Si le ſinus de l’angle A C B m’a donné le côté <lb/>A B de 100 toiſes, que me donnera le ſinus de l’angle C A B <lb/>pour la valeur du côté C B que je cherche? </s> <s xml:id="echoid-s11950" xml:space="preserve">& </s> <s xml:id="echoid-s11951" xml:space="preserve">pour trouver le <lb/>côté A C, il faut dire encore: </s> <s xml:id="echoid-s11952" xml:space="preserve">Si le ſinus de l’angle A C B <lb/>m’a donné la valeur du côté A B, que me donnera le ſinus <lb/>de l’angle du complément de 92 degrés, qui ſera celui de 88 <lb/>degrés pour la valeur du côté A C, parce que l’angle A B C <lb/>eſt obtus?</s> <s xml:id="echoid-s11953" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11954" xml:space="preserve">Comme on ne peut pas connoître la valeur du côté D C <lb/>ſans celle du côté D A, pour le trouver il faut dire: </s> <s xml:id="echoid-s11955" xml:space="preserve">Si le ſinus <lb/>de l’angle A D B de 37 degrés m’a donné la valeur du côté A B <lb/>de 100 toiſes, que me donnera le ſinus de 45 degrés pour la <lb/>valeur du côté D A, lequel étant connu, auſſi-bien que le <lb/>côté A C, & </s> <s xml:id="echoid-s11956" xml:space="preserve">l’angle D A C, nous aurons deux côtés connus, <lb/>& </s> <s xml:id="echoid-s11957" xml:space="preserve">l’angle compris dans un triangle, qui pourra nous donner <lb/>les deux angles inconnus; </s> <s xml:id="echoid-s11958" xml:space="preserve">& </s> <s xml:id="echoid-s11959" xml:space="preserve">en ſuivant ce qui eſt dit dans la <lb/>propoſition 10<emph style="sub">e</emph>, art. </s> <s xml:id="echoid-s11960" xml:space="preserve">725, il faudra d’abord chercher les angles <lb/>en D & </s> <s xml:id="echoid-s11961" xml:space="preserve">en C: </s> <s xml:id="echoid-s11962" xml:space="preserve">par cette proportion, la ſomme des deux côtés <lb/>A C, A D (que l’on vient de trouver), eſt à leur différence, <lb/>comme la tangente de la moitié de la ſomme des angles en C <lb/>& </s> <s xml:id="echoid-s11963" xml:space="preserve">en D eſt à la tangente de la moitié de la différence. </s> <s xml:id="echoid-s11964" xml:space="preserve">A yant <lb/>l’angle C, que je ſuppoſerai le plus grand, pour avoir le côté <lb/>C D, on fera cette autre proportion: </s> <s xml:id="echoid-s11965" xml:space="preserve">Le ſinus de l’angle C eſt <lb/>au côté A D connu, comme le ſinus de l’angle A eſt au côté <lb/>D C que je cherche; </s> <s xml:id="echoid-s11966" xml:space="preserve">& </s> <s xml:id="echoid-s11967" xml:space="preserve">l’on aura ainſi le côté D C, qui eſt la <lb/>diſtance que l’on demande.</s> <s xml:id="echoid-s11968" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s11969" xml:space="preserve">Comme il arrive preſque toujours que la campagne n’eſt pas <lb/>marquée ſur le plan des Villes que l’on aſſiege, & </s> <s xml:id="echoid-s11970" xml:space="preserve">que ſi elle y <lb/>eſt figurée, l’on ne peut pas, ſans faire de grandes erreurs, ſe fier <lb/>à la préciſion de ceux qui les ont levés ou copiés, l’opération pré-<lb/>cédente nous donne un excellent moyen pour orienter ſur le plan <lb/>par rapport à la place, la queue de la tranchée de chaque attaque, <lb/>afin de pouvoir enſuite projetter les travaux que l’on a envie de <lb/>faire d’une nuit à l’autre, ou ſeulement les y marquer à meſure <lb/>qu’on les avance, parce qu’ayant une fois un bout de parallele, <lb/>l’on peut de dedans la tranchée meſurer les boyaux, & </s> <s xml:id="echoid-s11971" xml:space="preserve">prendre <lb/>l’ouverture des angles qui font les retours; </s> <s xml:id="echoid-s11972" xml:space="preserve">marquer la poſition <lb/>des batteries; </s> <s xml:id="echoid-s11973" xml:space="preserve">enfin lever le plan de la tranchée avec autant d’exac-<lb/>titude que s’il n’y avoit aucun obſtacle.</s> <s xml:id="echoid-s11974" xml:space="preserve"/> </p> <pb o="347" file="0401" n="409" rhead="DE MATHÈMATIQUE. Liv. X."/> </div> <div xml:id="echoid-div990" type="section" level="1" n="778"> <head xml:id="echoid-head934" xml:space="preserve"><emph style="sc">Remarque generale</emph>.</head> <p> <s xml:id="echoid-s11975" xml:space="preserve">736. </s> <s xml:id="echoid-s11976" xml:space="preserve">Il faut bien remarquer que lorſque l’on cherche un <lb/>côté, on doit toujours commencer la proportion par un ſinus; <lb/></s> <s xml:id="echoid-s11977" xml:space="preserve">& </s> <s xml:id="echoid-s11978" xml:space="preserve">ſi c’eſt un angle que l’on veut avoir, il faut commencer la <lb/>proportion par un côté: </s> <s xml:id="echoid-s11979" xml:space="preserve">de cette maniere la grandeur que <lb/>l’on cherche ſera toujours le quatrieme terme d’une proportion <lb/>géométrique, dont les trois premiers termes ſont connus, en <lb/>cas que l’on ſe ſerve des ſinus & </s> <s xml:id="echoid-s11980" xml:space="preserve">des nombres naturels, ou ce <lb/>quatrieme terme ſera le logarithme de ce que l’on cherche, en <lb/>cas que l’on prenne les logarithmes des ſinus & </s> <s xml:id="echoid-s11981" xml:space="preserve">ceux des nom-<lb/>bres naturels.</s> <s xml:id="echoid-s11982" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div991" type="section" level="1" n="779"> <head xml:id="echoid-head935" xml:space="preserve">PROPOSITION XVI.</head> <head xml:id="echoid-head936" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s11983" xml:space="preserve">737. </s> <s xml:id="echoid-s11984" xml:space="preserve">Tirer une ligne parallele à une autre inacceſſible.</s> <s xml:id="echoid-s11985" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 192.</note> <p> <s xml:id="echoid-s11986" xml:space="preserve">On demande de tirer par le point C une parallele à une <lb/>ligne inacceſſible A B.</s> <s xml:id="echoid-s11987" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11988" xml:space="preserve">Pour réſoudre ce problême, il faut commencer par ſe donner <lb/>une baſe telle que C D, qui doit être, comme nous l’avons <lb/>dit ailleurs, proportionnée à la diſtance de l’objet, afin que <lb/>l’opération en ſoit plus juſte, & </s> <s xml:id="echoid-s11989" xml:space="preserve">nous ſuppoſons que 150 toiſes <lb/>eſt la longueur qui lui convient.</s> <s xml:id="echoid-s11990" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s11991" xml:space="preserve">Nous ſçavons que deux lignes paralleles étant coupées <lb/>par une troiſieme, forment les angles alternes égaux, & </s> <s xml:id="echoid-s11992" xml:space="preserve">que <lb/>par conſéquent lorſque les angles alternes ſeront égaux, les <lb/>lignes ſeront paralleles; </s> <s xml:id="echoid-s11993" xml:space="preserve">d’où il ſuit que ſi l’on connoît l’angle <lb/>A B C, formé par la parallele A B, & </s> <s xml:id="echoid-s11994" xml:space="preserve">le rayon viſuel C B, on <lb/>n’aura qu’à faire l’angle B C E égal au précédent, pour que <lb/>la ligne C E ſoit parallele à la ligne A B: </s> <s xml:id="echoid-s11995" xml:space="preserve">ainſi toute la queſ-<lb/>tion eſt réduite à trouver la valeur de l’angle A B C. </s> <s xml:id="echoid-s11996" xml:space="preserve">Afin de <lb/>la connoître, je commence du point C par prendre l’ouver-<lb/>ture de l’angle A C B, que je trouve de 40 degrés: </s> <s xml:id="echoid-s11997" xml:space="preserve">enſuite je <lb/>viens au point D pour y prendre l’ouverture de l’angle C D B, <lb/>qui eſt de 86 degrés; </s> <s xml:id="echoid-s11998" xml:space="preserve">& </s> <s xml:id="echoid-s11999" xml:space="preserve">je prends auſſi l’ouverture de l’angle <lb/>A D B, qui ſera, par exemple, de 60 degrés. </s> <s xml:id="echoid-s12000" xml:space="preserve">Ces choſes étant <lb/>connues, je fais enſorte de trouver par leur moyen la valeur <lb/>des lignes C A & </s> <s xml:id="echoid-s12001" xml:space="preserve">C B. </s> <s xml:id="echoid-s12002" xml:space="preserve">Pour cela, je cherche dans le triangle <lb/>C D B la valeur du côté C B. </s> <s xml:id="echoid-s12003" xml:space="preserve">Pour le trouver, je conſidere <pb o="348" file="0402" n="410" rhead="NOUVEAU COURS"/> que l’angle B C D eſt de 80 degrés, & </s> <s xml:id="echoid-s12004" xml:space="preserve">que l’angle C D B eſt <lb/>de 86; </s> <s xml:id="echoid-s12005" xml:space="preserve">d’où il ſuit que l’angle C B D eſt de 14 degrés. </s> <s xml:id="echoid-s12006" xml:space="preserve">Cela <lb/>poſé, il faut dire: </s> <s xml:id="echoid-s12007" xml:space="preserve">Si le ſinus de l’angle de 14 degrés m’a <lb/>donné 150, que me donnera le ſinus de 86 pour la valeur du <lb/>côté oppoſé C B?</s> <s xml:id="echoid-s12008" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12009" xml:space="preserve">Pour trouver le côté C A, je fais attention que l’angle C D A <lb/>eſt de 26 degrés, & </s> <s xml:id="echoid-s12010" xml:space="preserve">que l’angle A C D étant de 120 degrés, <lb/>l’angle C A D doit être de 34 degrés. </s> <s xml:id="echoid-s12011" xml:space="preserve">Cela étant, je dis encore: <lb/></s> <s xml:id="echoid-s12012" xml:space="preserve">Si le ſinus de l’angle C A D de 34 degrés, m’a donné 150 toiſes <lb/>pour le côté C D, que me donnera le ſinus de l’angle C D A <lb/>de 26 degrés pour la valeur du côté C A? </s> <s xml:id="echoid-s12013" xml:space="preserve">Or comme nous <lb/>avons dans le triangle A C B les deux côtés A C & </s> <s xml:id="echoid-s12014" xml:space="preserve">C B de <lb/>connus avec l’angle compris A C B, il s’enſuit que l’on trou-<lb/>vera aiſément par la propoſition 10<emph style="sub">e</emph> la valeur de l’angle A B C, <lb/>dont la connoiſſance eſt la ſolution du problême.</s> <s xml:id="echoid-s12015" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s12016" xml:space="preserve">L’on eſt ſouvent obligé de mener une parallele à une ligne inac-<lb/>ceſſible dans une infinité d’occaſions, ſoit qu’on veuille percer des <lb/>routes dans un bois avec certaines précautions, ou ſoit dans les <lb/>ſiéges, quand on veut dreſſer une batterie qui ſoit parallele à la face <lb/>de l’ouvrage que l’on veut battre, ou quand on en veut faire une <lb/>autre en écharpe, dont les feux aillent ſe diriger ſelon un angle <lb/>donné avec la face.</s> <s xml:id="echoid-s12017" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div992" type="section" level="1" n="780"> <head xml:id="echoid-head937" xml:space="preserve">PROPOSITION XVII.</head> <head xml:id="echoid-head938" xml:space="preserve"><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s12018" xml:space="preserve">738. </s> <s xml:id="echoid-s12019" xml:space="preserve">Meſurer une hauteur acceſſible ou inacceſſible.</s> <s xml:id="echoid-s12020" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 193.</note> <p> <s xml:id="echoid-s12021" xml:space="preserve">Pour meſurer la hauteur A B d’une Tour, il faut ſe donner <lb/>une baſe telle que E B, qu’il faut meſurer exactement depuis <lb/>le point du milieu B de la Tour juſqu’à l’endroit E, qui eſt le <lb/>lieu où l’on aura planté le graphometre; </s> <s xml:id="echoid-s12022" xml:space="preserve">& </s> <s xml:id="echoid-s12023" xml:space="preserve">ſuppoſant que cette <lb/>baſe ſoit de 25 toiſes, l’on prendra l’ouverture de l’angle A C D <lb/>formé par deux rayons viſuels, dont le premier C D doit être <lb/>parallele à l’horizon, & </s> <s xml:id="echoid-s12024" xml:space="preserve">le ſecond C A doit aboutir au ſommet <lb/>de la Tour; </s> <s xml:id="echoid-s12025" xml:space="preserve">& </s> <s xml:id="echoid-s12026" xml:space="preserve">ſuppoſant que l’angle ſoit de 35 degrés, l’on <lb/>cherchera dans le triangle A C D le côté A D, en diſant: <lb/></s> <s xml:id="echoid-s12027" xml:space="preserve">Comme le ſinus total eſt à la tangente de l’angle C, ainſi le <lb/>côté C D de 25 toiſes eſt au côté D A, que l’on trouvera de <lb/>17 toiſes 3 pieds; </s> <s xml:id="echoid-s12028" xml:space="preserve">à quoi ajoutant la hauteur D B ou C E du <lb/>pied de l’inſtrument, qui eſt ordinairement de 4 pieds, on <pb o="349" file="0403" n="411" rhead="DE MATHÉMATIQUE. Liv. X."/> trouvera que la hauteur A B de la Tour eſt de 18 toiſes un pied.</s> <s xml:id="echoid-s12029" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12030" xml:space="preserve">Mais ſi l’on avoit à prendre la hauteur d’une Tour ou d’une <lb/> <anchor type="note" xlink:label="note-0403-01a" xlink:href="note-0403-01"/> éminence qui fût inacceſſible, comme on le voit dans la figure <lb/>194, il faudroit de l’endroit F prendre l’ouverture de l’angle <lb/>A D G, formé par deux rayons; </s> <s xml:id="echoid-s12031" xml:space="preserve">& </s> <s xml:id="echoid-s12032" xml:space="preserve">ſuppoſant qu’on a trouvé <lb/>cet angle de 50 degrés, il faudra ſe reculer ſur l’alignement <lb/>des points D & </s> <s xml:id="echoid-s12033" xml:space="preserve">G juſqu’à l’endroit C, afin d’avoir une baſe <lb/>E F d’une longueur ſuffiſante pour que l’angle C A D ne ſoit <lb/>pas trop aigu; </s> <s xml:id="echoid-s12034" xml:space="preserve">& </s> <s xml:id="echoid-s12035" xml:space="preserve">cette baſe ayant été trouvée de 40 toiſes, <lb/>l’on prendra encore l’ouverture de l’angle A C G, qui ſera, <lb/>par exemple, de 30 degrés. </s> <s xml:id="echoid-s12036" xml:space="preserve">Or comme l’angle A D G eſt égal <lb/>aux deux autres intérieurs oppoſés du triangle C A D, la dif-<lb/>férence de cet angle, qui eſt de 50 degrés à l’angle A C D, <lb/>qui eſt de 30 degrés, ſera la valeur de l’angle C A D, que l’on <lb/>trouvera de 20 degrés. </s> <s xml:id="echoid-s12037" xml:space="preserve">Or comme dans le triangle rectangle <lb/>A D G nous avons beſoin de connoître le côté D A pour con-<lb/>noître le côté A G, l’on dira: </s> <s xml:id="echoid-s12038" xml:space="preserve">Si le ſinus de l’angle C A D de <lb/>20 degrés m’a donné 40 toiſes pour le côté C D, que donnera <lb/>le ſinus de l’angle A C D de 30 degrés pour le côté A D, que <lb/>l’on trouvera de 63 toiſes 2 pieds?</s> <s xml:id="echoid-s12039" xml:space="preserve"/> </p> <div xml:id="echoid-div992" type="float" level="2" n="1"> <note position="right" xlink:label="note-0403-01" xlink:href="note-0403-01a" xml:space="preserve">Figure 194.</note> </div> <p> <s xml:id="echoid-s12040" xml:space="preserve">Pour donc trouver le côté A G, je dis: </s> <s xml:id="echoid-s12041" xml:space="preserve">Comme la ſécante <lb/>de l’angle A D G eſt à ſa tangente, ainſi le côté D A de 63 <lb/>toiſes 2 pieds, eſt au côté A G, que l’on trouvera de 48 toiſes <lb/>3 pieds: </s> <s xml:id="echoid-s12042" xml:space="preserve">à quoi il ne faut plus qu’ajouter la hauteur du pied <lb/>de l’inſtrument pour avoir la ligne A B.</s> <s xml:id="echoid-s12043" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s12044" xml:space="preserve">Maniere de lever une Carte par le moyen de la trigonométrie.</s> <s xml:id="echoid-s12045" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12046" xml:space="preserve">739. </s> <s xml:id="echoid-s12047" xml:space="preserve">L’on doit diſtinguer deux ſortes de cartes: </s> <s xml:id="echoid-s12048" xml:space="preserve">les unes <lb/> <anchor type="note" xlink:label="note-0403-02a" xlink:href="note-0403-02"/> ſont des cartes générales, & </s> <s xml:id="echoid-s12049" xml:space="preserve">les autres des cartes particulieres: <lb/></s> <s xml:id="echoid-s12050" xml:space="preserve">les dernieres ſont celles que l’on leve avec beaucoup d’atten-<lb/>tion, n’oubliant rien de tout ce qui peut avoir lieu dans la <lb/>carte, tel que la grandeur & </s> <s xml:id="echoid-s12051" xml:space="preserve">la figure des Villages, des <lb/>Bourgs & </s> <s xml:id="echoid-s12052" xml:space="preserve">des Villes, les Bois, les Ponts, les Rivieres, les <lb/>Chemins, les Fontaines, les Croix, Chapelles, Juſtices, &</s> <s xml:id="echoid-s12053" xml:space="preserve">c.</s> <s xml:id="echoid-s12054" xml:space="preserve"/> </p> <div xml:id="echoid-div993" type="float" level="2" n="2"> <note position="right" xlink:label="note-0403-02" xlink:href="note-0403-02a" xml:space="preserve">Figure 195.</note> </div> <p> <s xml:id="echoid-s12055" xml:space="preserve">Pour les cartes générales, l’on ne prend que la poſition <lb/>des lieux les plus conſidérables, & </s> <s xml:id="echoid-s12056" xml:space="preserve">la figure des grands che-<lb/>mins, omettant quantité de choſes qui ne pourroient ſe placer <lb/>ſur ces ſortes de cartes, parce qu’elles ſont ordinairement <lb/>dreſſées ſur de petites échelles. </s> <s xml:id="echoid-s12057" xml:space="preserve">Telles ſont les Cartes des <pb o="350" file="0404" n="412" rhead="NOUVEAU COURS"/> Royaumes & </s> <s xml:id="echoid-s12058" xml:space="preserve">des grandes Provinces. </s> <s xml:id="echoid-s12059" xml:space="preserve">Cependant l’on peut dire <lb/>que l’on s’y prend de la même façon pour lever les cartes par-<lb/>ticulieres & </s> <s xml:id="echoid-s12060" xml:space="preserve">générales, parce que pour les unes & </s> <s xml:id="echoid-s12061" xml:space="preserve">les autres <lb/>l’on commence par faire un canevas, qui n’eſt autre choſe <lb/>que la grandeur de la carte déterminée avec les principales <lb/>poſitions, après quoi l’on entre dans le détail de chaque choſe, <lb/>comme nous le ferons voir après avoir enſeignè la maniere de <lb/>prendre les poſitions qui doivent faire les principaux points de <lb/>la carte.</s> <s xml:id="echoid-s12062" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12063" xml:space="preserve">Si l’on vouloit, par exemple, lever la carte des lieux mar-<lb/>qués par les lettres de cette figure, l’on voit que l’objet qu’on <lb/>ſe propoſe n’eſt autre choſe que de placer ſur le papier les dif-<lb/>férens endroits qui ſont ici, enſorte que la diſtance qu’il y a <lb/>d’un lieu à un autre ait le même rapport ſur la carte que ſur le <lb/>terrein; </s> <s xml:id="echoid-s12064" xml:space="preserve">ce qui eſt proprement faire une réduction de grand en <lb/>petit. </s> <s xml:id="echoid-s12065" xml:space="preserve">Comme ces réductions ne peuvent ſe faire que par les <lb/>triangles ſemblables, il s’enſuit qu’en levant la carte d’un pays <lb/>par le moyen de la Trigonométrie, il ne s’agit que de trouver <lb/>la valeur des angles & </s> <s xml:id="echoid-s12066" xml:space="preserve">des côtés qui ſont formés par la diſ-<lb/>tance des lieux. </s> <s xml:id="echoid-s12067" xml:space="preserve">Cela poſé, je commence par établir une baſe <lb/>la plus grande qu’il eſt poſſible, aſin que les lieux qui doivent <lb/>s’y rapporter ſoient plus exactement levés. </s> <s xml:id="echoid-s12068" xml:space="preserve">Pour cela il faut <lb/>éviter, autant qu’il eſt poſſible, d’avoir des angles trop obtus <lb/>& </s> <s xml:id="echoid-s12069" xml:space="preserve">trop aigus. </s> <s xml:id="echoid-s12070" xml:space="preserve">Ayant donc choiſi les points de ſtation A & </s> <s xml:id="echoid-s12071" xml:space="preserve">B, <lb/>je commence par en chercher la diſtance de la maniere que <lb/>nous l’avons enſeigné dans la ſeconde propoſition: </s> <s xml:id="echoid-s12072" xml:space="preserve">l’ayant <lb/>trouvée, je viens à l’endroit B, pour y prendre l’ouverture <lb/>des angles formés par la baſe A B, & </s> <s xml:id="echoid-s12073" xml:space="preserve">les différens endroits <lb/>que je me propoſe de lever. </s> <s xml:id="echoid-s12074" xml:space="preserve">Pour cela, je prends l’ouverture <lb/>de l’angle A B C, de l’angle A B D, de l’angle A B E, je paſſe <lb/>le point F, parce que l’angle qu’il formeroit avec la baſe ſe-<lb/>roit trop obtus, & </s> <s xml:id="echoid-s12075" xml:space="preserve">qu’on auroit trop de peine à couper le <lb/>rayon qui ſeroit tiré de B en F: </s> <s xml:id="echoid-s12076" xml:space="preserve">je continue à prendre l’ouver-<lb/>ture des angles A B G, A B H, A B I, & </s> <s xml:id="echoid-s12077" xml:space="preserve">A B K: </s> <s xml:id="echoid-s12078" xml:space="preserve">je paſſe auſſi <lb/>le point L, parce que l’angle formé par la baſe A B, & </s> <s xml:id="echoid-s12079" xml:space="preserve">le <lb/>rayon de B en L ſeroit trop aigu.</s> <s xml:id="echoid-s12080" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12081" xml:space="preserve">Préſentement il ne s’agit plus, pour avoir la poſition des <lb/>endroits qu’on voit marqués ci - deſſus, que de couper les <lb/>rayons qu’on vient de tirer. </s> <s xml:id="echoid-s12082" xml:space="preserve">Pour cela, je viens au point A, <lb/>pour y prendre l’ouverture de l’angle B A E, qui me donnera <pb o="351" file="0405" n="413" rhead="DE MATHÉMATIQUE. Liv. X."/> le point E, parce que dans le triangle A B E, je connois le côté <lb/>A B, & </s> <s xml:id="echoid-s12083" xml:space="preserve">la valeur des angles E A B & </s> <s xml:id="echoid-s12084" xml:space="preserve">A B E, par le moyen deſ-<lb/>quels je trouverai les diſtances A E & </s> <s xml:id="echoid-s12085" xml:space="preserve">B E. </s> <s xml:id="echoid-s12086" xml:space="preserve">Pour les autres <lb/>points, je continue à couper les rayons que j’ai tirés dans la <lb/>premiere opération, en prenant l’ouverture des angles B A D, <lb/>B A C, B A G, B A H, B A I, B A K. </s> <s xml:id="echoid-s12087" xml:space="preserve">Comme tous les trian-<lb/>gles formés par ces rayons ont la baſe A B pour côté commun, <lb/>il s’enſuit qu’on pourra en trouver la longueur, puiſqu’il n’y <lb/>a point de triangle dans lequel on ne connoiſſe deux angles & </s> <s xml:id="echoid-s12088" xml:space="preserve"><lb/>un côté. </s> <s xml:id="echoid-s12089" xml:space="preserve">Comme nous avons paſſé deux endroits pour les <lb/>raiſons que nous avons dites, il faut faire voir comment on <lb/>en peut trouver la poſition, ſans ſe ſervir de la baſe A B: <lb/></s> <s xml:id="echoid-s12090" xml:space="preserve">pour donc trouver le point F, je prends la diſtance B E ou B G <lb/>pour baſe, ou toute autre qui pourroit mieux convenir; </s> <s xml:id="echoid-s12091" xml:space="preserve">mais <lb/>je choiſis ici le côté B E, & </s> <s xml:id="echoid-s12092" xml:space="preserve">du point B je prends l’ouverture de <lb/>l’angle E B F, & </s> <s xml:id="echoid-s12093" xml:space="preserve">du point E l’ouverture de l’angle B E F, qui <lb/>me donne le point F. </s> <s xml:id="echoid-s12094" xml:space="preserve">Je fais la même choſe pour trouver le <lb/>point L, & </s> <s xml:id="echoid-s12095" xml:space="preserve">même le point M, que je ſuppoſe n’avoir pu pren-<lb/>dre dans les opérations précédentes, c’eſt-à-dire, je choiſis la <lb/>baſe A C, & </s> <s xml:id="echoid-s12096" xml:space="preserve">du point A je prends les ouvertures des angles <lb/>C A M & </s> <s xml:id="echoid-s12097" xml:space="preserve">C A L, & </s> <s xml:id="echoid-s12098" xml:space="preserve">du point C je prends encore l’ouverture <lb/>des angles A C L & </s> <s xml:id="echoid-s12099" xml:space="preserve">A C M.</s> <s xml:id="echoid-s12100" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12101" xml:space="preserve">Après avoir trouvé la valeur de tous les côtés des triangles <lb/>qui ſont ici, il faut les rapporter ſur le papier, en donnant à <lb/>chaque ligne la valeur qu’elle doit avoir; </s> <s xml:id="echoid-s12102" xml:space="preserve">ce qui ſe fera ſans <lb/>difficulté par le moyen d’une échelle: </s> <s xml:id="echoid-s12103" xml:space="preserve">& </s> <s xml:id="echoid-s12104" xml:space="preserve">après que toutes ces <lb/>poſitions ſeront rapportées bien exactement, l’on pourra, en <lb/>ſuivant la même méthode, continuer à lever les lieux qu’on <lb/>aura pu découvrir dans les premieres opérations; </s> <s xml:id="echoid-s12105" xml:space="preserve">ce qui ſera <lb/>bien aiſé, puiſqu’on aura de toutes parts des baſes, dont la <lb/>valeur ſera connue. </s> <s xml:id="echoid-s12106" xml:space="preserve">Par exemple, pour lever les objets au-<lb/>delà des points C & </s> <s xml:id="echoid-s12107" xml:space="preserve">D, on pourra prendre la diſtance C D <lb/>pour baſe; </s> <s xml:id="echoid-s12108" xml:space="preserve">d’un autre côté on pourra prendre la ligne I H; <lb/></s> <s xml:id="echoid-s12109" xml:space="preserve">enfin ſur la gauche la diſtance L K, ſur la droite toute autre <lb/>ligne que l’on choiſira de même.</s> <s xml:id="echoid-s12110" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s12111" xml:space="preserve">Des attentions qu’il faut avoir pour lever une Carte particuliere.</s> <s xml:id="echoid-s12112" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12113" xml:space="preserve">740. </s> <s xml:id="echoid-s12114" xml:space="preserve">Quand on veut lever une carte d’une façon à ne rien <lb/>omettre de toutes les particularités qui entrent dans le détail <pb o="352" file="0406" n="414" rhead="NOUVEAU COURS"/> d’une carte, ceux qui conduiſent le travail doivent envoyer <lb/>des perſonnes entendues dans les Villages pour lever leurs <lb/>ſituations, leurs figures, la forme des rues, la poſition des fon-<lb/>taines, s’il s’y en trouve, des carrieres, des montagnes, col-<lb/>lines & </s> <s xml:id="echoid-s12115" xml:space="preserve">vallons, qui peuvent ſe rencontrer dans les environs. <lb/></s> <s xml:id="echoid-s12116" xml:space="preserve">On réduit chaque village ſur l’échelle de la carte; </s> <s xml:id="echoid-s12117" xml:space="preserve">& </s> <s xml:id="echoid-s12118" xml:space="preserve">pour les <lb/>rapporter, on a ſoin que l’Egliſe ſoit poſitivement au point qui <lb/>eſt marqué ſur le canevas, parce que ces points ſont ordinai-<lb/>rement des clochers & </s> <s xml:id="echoid-s12119" xml:space="preserve">des tours. </s> <s xml:id="echoid-s12120" xml:space="preserve">Pour les Villes, on fait en-<lb/>ſorte d’en avoir les plans, qu’on réduit à l’échelle de la carte. </s> <s xml:id="echoid-s12121" xml:space="preserve"><lb/>Quand il ſe rencontre des bois ou des forêts, l’on commence <lb/>par lever exactement les villages & </s> <s xml:id="echoid-s12122" xml:space="preserve">les hameaux qui ſont les <lb/>plus proches, pour avoir des baſes, qui ne ſont autre choſe <lb/>que la diſtance d’un lieu à un autre, deſquels on forme un eſ-<lb/>pece de polygone qui entoure le bois: </s> <s xml:id="echoid-s12123" xml:space="preserve">après quoi il eſt aiſé de <lb/>rapporter à ce polygone un nombre de points, qui marquent <lb/>les limites du bois, pour en tracer enſuite à la vue la figure <lb/>extérieure, quand il ne s’agira que de quelque ſinuoſité peu <lb/>conſidérable, Après cela, il faut entrer dans le bois pour y <lb/>conſiderer les principaux chemins, les ruiſſeaux, les fontaines, <lb/>les maiſons & </s> <s xml:id="echoid-s12124" xml:space="preserve">les châteaux qui pourroient s’y rencontrer. </s> <s xml:id="echoid-s12125" xml:space="preserve">Toutes <lb/>ces choſes doivent être levées avec le plus de préciſion qu’il eſt <lb/>poſſible. </s> <s xml:id="echoid-s12126" xml:space="preserve">Pour cela l’on ſe donne des points de poſition que <lb/>l’on prend dans les bois, par des opérations que l’on fait ſur <lb/>quelque éminence hors du bois. </s> <s xml:id="echoid-s12127" xml:space="preserve">Ces points de poſition ſont <lb/>ordinairement des clochers, des châteaux, ou bien quelques <lb/>grands arbres qui ſe font diſtinguer au deſſus des autres: </s> <s xml:id="echoid-s12128" xml:space="preserve">& </s> <s xml:id="echoid-s12129" xml:space="preserve"><lb/>lorſqu’on eſt une fois parvenu à la connoiſſance de quelqu’un <lb/>de ces points, l’on peut, ſans aucune difficulté, orienter les <lb/>différens endroits qui ſe trouvent dans le bois, à l’aide des <lb/>poſitions connues.</s> <s xml:id="echoid-s12130" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div995" type="section" level="1" n="781"> <head xml:id="echoid-head939" xml:space="preserve">Application de la Trigonométrie à la Fortification.</head> <p> <s xml:id="echoid-s12131" xml:space="preserve">741. </s> <s xml:id="echoid-s12132" xml:space="preserve">Quand on veut tracer une fortification ſur le terrein, <lb/> <anchor type="note" xlink:label="note-0406-01a" xlink:href="note-0406-01"/> il eſt abſolument néceſſaire de connoître toutes les lignes & </s> <s xml:id="echoid-s12133" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0406-02a" xlink:href="note-0406-02"/> les angles qui en compoſent le projet; </s> <s xml:id="echoid-s12134" xml:space="preserve">& </s> <s xml:id="echoid-s12135" xml:space="preserve">comme cette con-<lb/>noiſſance doit être la plus exacte qu’il eſt poſſible, il ne con-<lb/>viendroit pas que l’on ſe ſervît du compas pour trouver avec <lb/>l’échelle les lignes que l’on ne connoît pas, non plus que du <pb o="353" file="0407" n="415" rhead="DE MATHEMATIQUES. Liv. X."/> rapporteur pour trouver la valeur des angles, puiſque l’on peut <lb/>faire des erreurs inſenſibles ſur le papier, qui deviendroient <lb/>de conſéquence ſur le terrein; </s> <s xml:id="echoid-s12136" xml:space="preserve">c’eſt pourquoi il eſt à propos <lb/>d’avoir recours à la Trigonométrie, pour déterminer par le <lb/>moyen des lignes que l’on connoît, celles que l’on ne connoît <lb/>pas: </s> <s xml:id="echoid-s12137" xml:space="preserve">& </s> <s xml:id="echoid-s12138" xml:space="preserve">comme dans la fortification, ſelon la méthode de <lb/>M. </s> <s xml:id="echoid-s12139" xml:space="preserve">de Vauban, l’on connoît la baſe de 180 toiſes, la perpen-<lb/>diculaire C F de 30, & </s> <s xml:id="echoid-s12140" xml:space="preserve">la face A D de 50, voici de quelle <lb/>maniere on pourra connoître l’angle de l’épaule, l’angle flan-<lb/>quant, le flanc & </s> <s xml:id="echoid-s12141" xml:space="preserve">la courtine; </s> <s xml:id="echoid-s12142" xml:space="preserve">ſuppoſant qu’on eſt prévenu <lb/>que la ligne D H eſt égale à la ligne D E.</s> <s xml:id="echoid-s12143" xml:space="preserve"/> </p> <div xml:id="echoid-div995" type="float" level="2" n="1"> <note position="left" xlink:label="note-0406-01" xlink:href="note-0406-01a" xml:space="preserve">Pl. XIII.</note> <note position="left" xlink:label="note-0406-02" xlink:href="note-0406-02a" xml:space="preserve">Figure 196.</note> </div> <p> <s xml:id="echoid-s12144" xml:space="preserve">Il faut avant toutes choſes chercher la valeur de l’angle <lb/>F A C, en diſant: </s> <s xml:id="echoid-s12145" xml:space="preserve">Comme le côté A C de 90 toiſes eſt au côté <lb/>C F de 30, ainſi le ſinus total A I eſt à la tangente I D, qui <lb/>étant trouvée, l’on verra qu’elle correſpond à un angle de 18 <lb/>degrés 26 minutes, qui eſt la valeur de l’angle F A C: </s> <s xml:id="echoid-s12146" xml:space="preserve">par <lb/>conſéquent celle de l’angle H D E, à cauſe des paralleles A B <lb/>& </s> <s xml:id="echoid-s12147" xml:space="preserve">D E qui aboutiſſent ſur A H.</s> <s xml:id="echoid-s12148" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12149" xml:space="preserve">Or comme nous avons beſoin dans le triangle D A I du <lb/>côté A I, on n’aura qu’à dire (pour le connoître): </s> <s xml:id="echoid-s12150" xml:space="preserve">comme la <lb/>ſécante de l’angle D A I eſt au ſinus total, ainſi le côté A D <lb/>de 50 toiſes eſt au côté A I; </s> <s xml:id="echoid-s12151" xml:space="preserve">que l’on trouvera de 47 toiſes <lb/>2 pieds, qu’on n’aura qu’à retrancher de la ligne A C de 90 <lb/>toiſes pour avoir la ligne I C de 42 toiſes 4 pieds; </s> <s xml:id="echoid-s12152" xml:space="preserve">& </s> <s xml:id="echoid-s12153" xml:space="preserve">comme <lb/>cette ligne eſt moitié du côté D E, on verra que ce même côté <lb/>eſt de 85 toiſes 2 pieds.</s> <s xml:id="echoid-s12154" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12155" xml:space="preserve">Comme le triangle H D E eſt iſoſcele, & </s> <s xml:id="echoid-s12156" xml:space="preserve">que l’on connoît <lb/>l’angle du ſommet avec les deux côtés qui le comprennent, <lb/>parce que la ligne D H eſt le prolongement de la ligne A D; <lb/></s> <s xml:id="echoid-s12157" xml:space="preserve">& </s> <s xml:id="echoid-s12158" xml:space="preserve">que la ligne D E eſt parallele à la ligne A B, par conſtruction, <lb/>on aura l’angle en H ou l’angle en E, en retranchant l’an-<lb/>gle D de 180 degrés, & </s> <s xml:id="echoid-s12159" xml:space="preserve">prenant la moitié pour cet angle. </s> <s xml:id="echoid-s12160" xml:space="preserve"><lb/>Ainſi l’on dira (pour avoir le flanc H E): </s> <s xml:id="echoid-s12161" xml:space="preserve">Si le ſinus de l’an-<lb/>gle D H E m’a donné le côté D E, que me donnera le ſinus <lb/>de l’angle H D E pour le flanc ou côté H E, que l’on trouvera <lb/>de 27 toiſes 2 pieds?</s> <s xml:id="echoid-s12162" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12163" xml:space="preserve">Comme les angles de la baſe dutriangle iſoſcele ſont chacun <lb/>de 80 degrés & </s> <s xml:id="echoid-s12164" xml:space="preserve">47 minutes, puiſque l’angle du ſommet eſt de <lb/>18 degrés 26 minutes; </s> <s xml:id="echoid-s12165" xml:space="preserve">il s’enſuit, à cauſe des angles alternes <lb/>formés par les lignes paralleles G H & </s> <s xml:id="echoid-s12166" xml:space="preserve">D E, que ſi de l’angle <pb o="354" file="0408" n="416" rhead="NOUVEAU COURS"/> H E D on retranche l’angle G E D de 18 degrés 26 minutes, <lb/>il reſtera 62 degrés 21 minutes pour l’angle G E H, dont le <lb/>ſupplément à 180, qui eſt l’angle de l’épaule H E B, eſt de 117 <lb/>degrés 39 minutes: </s> <s xml:id="echoid-s12167" xml:space="preserve">& </s> <s xml:id="echoid-s12168" xml:space="preserve">ſi l’on ajoute au contraire à D H E l’angle <lb/>G H D, qui eſt auſſi de 18 degrés 26 minutes, l’on trouvera <lb/>que l’angle flanquant G H E eſt de 99 degrés 13 minutes.</s> <s xml:id="echoid-s12169" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12170" xml:space="preserve">Or comme du triangle G H E l’on connoît les angles & </s> <s xml:id="echoid-s12171" xml:space="preserve">le <lb/>côté H E, l’on n’aura (pour connoître la courtine) qu’à dire: <lb/></s> <s xml:id="echoid-s12172" xml:space="preserve">Comme le ſinus de l’angle H G E eſt au côté H E, ainſi le <lb/>ſinus de l’angle G E H eſt au côté G H, que l’on trouvera de <lb/>76 toiſes 3 pieds.</s> <s xml:id="echoid-s12173" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12174" xml:space="preserve">Pour connoître I’angle flanqué, conſidérez qu’il eſt plus <lb/>petit que l’angle de la circonférence de deux fois l’angle D A I, <lb/>qui eſt de 18 degrés 26 minutes: </s> <s xml:id="echoid-s12175" xml:space="preserve">& </s> <s xml:id="echoid-s12176" xml:space="preserve">comme l’on ſuppoſe qu’il <lb/>s’agit ici d’un exagone, dont l’angle de la circonférence eſt <lb/>de 120 degrés, l’on n’aura qu’à retrancher 36 degrés 52 mi-<lb/>nutes de 120 degrés pour avoir l’angle flanqué, qui ſera de <lb/>83 degrés 8 minutes.</s> <s xml:id="echoid-s12177" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12178" xml:space="preserve">L’on pourra calculer de même tous les autres fronts de for-<lb/>tification, dont le côté extérieur auroit plus ou moins de 180 <lb/>toiſes, parce que les proportions ſe trouveront toujours. </s> <s xml:id="echoid-s12179" xml:space="preserve">Ainſi <lb/>quand il s’agira de calculer les lignes & </s> <s xml:id="echoid-s12180" xml:space="preserve">les angles dont un <lb/>ouvrage à corne, ou un ouvrage à couronne eſt compoſé, <lb/>il ſuffira de connoître le côté extérieur, la perpendiculaire, & </s> <s xml:id="echoid-s12181" xml:space="preserve"><lb/>la face d’un baſtion pour connoître le reſte: </s> <s xml:id="echoid-s12182" xml:space="preserve">c’eſt pourquoi <lb/>cette pratique peut avoir également lieu dans la fortification <lb/>irréguliere comme dans la réguliere; </s> <s xml:id="echoid-s12183" xml:space="preserve">car ſoit que l’on faſſe les <lb/>flancs perpendiculaires ſur la ligne de défenſe, ou ſur la cour-<lb/>tine, ſelon les cas où l’on ſeroit obligé de ſuivre une méthode <lb/>plutôt qu’une autre, l’on trouvera le calcul également aiſé, <lb/>pourvu que l’on ait ſeulement quelques grandeurs connues, <lb/>par le moyen deſquelles on puiſſe opérer.</s> <s xml:id="echoid-s12184" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12185" xml:space="preserve">742. </s> <s xml:id="echoid-s12186" xml:space="preserve">De tout ce qui regarde le calcul d’une fortification, je <lb/>n’ai point trouvé de partie plus difficile à calculer que la valeur <lb/>de la face de la demi-lune; </s> <s xml:id="echoid-s12187" xml:space="preserve">& </s> <s xml:id="echoid-s12188" xml:space="preserve">l’on peut même regarder ce <lb/>cas-là comme un petit problême de fortification: </s> <s xml:id="echoid-s12189" xml:space="preserve">c’eſt pour-<lb/>quoi je crois qu’on ſera bien aiſe d’en voir la ſolution; </s> <s xml:id="echoid-s12190" xml:space="preserve">car <lb/>quoiqu’elle paroiſſe peu de choſe, elle ne laiſſeroit pas que <lb/>d’embarraſſer un Commençant: </s> <s xml:id="echoid-s12191" xml:space="preserve">ainſi pour bien ſçavoir de quoi <lb/>il eſt queſtion, voici comme on ſuppoſe que la demi-lune a été <lb/>tracée.</s> <s xml:id="echoid-s12192" xml:space="preserve"/> </p> <pb o="355" file="0409" n="417" rhead="DE MATHÉMATIQUE. Liv. X."/> <p> <s xml:id="echoid-s12193" xml:space="preserve">Après avoir pris le point E ſur la face d’un baſtion à 5 toiſes <lb/> <anchor type="note" xlink:label="note-0409-01a" xlink:href="note-0409-01"/> au deſſus de l’angle de l’épaule, l’on a du point C comme cen-<lb/>tre, & </s> <s xml:id="echoid-s12194" xml:space="preserve">de l’intervalle C E, décrit un arc, qui venant rencon-<lb/>trer la capitale, a donné le point F pour la pointe de la demi-<lb/>lune; </s> <s xml:id="echoid-s12195" xml:space="preserve">enſuite l’on a pris le point D à trois toiſes au deſſus de <lb/>l’angle de l’épaule, & </s> <s xml:id="echoid-s12196" xml:space="preserve">l’on a tiré la ligne F D: </s> <s xml:id="echoid-s12197" xml:space="preserve">après quoi l’on <lb/>a fait le foſſé de 20 toiſes ſur le prolongement de la face à <lb/>l’endroit A H, & </s> <s xml:id="echoid-s12198" xml:space="preserve">l’on a tiré la ligne H K, qui détermine la <lb/>longueur I F de la face de la demi-lune, dont il s’agit de <lb/>trouver la valeur.</s> <s xml:id="echoid-s12199" xml:space="preserve"/> </p> <div xml:id="echoid-div996" type="float" level="2" n="2"> <note position="right" xlink:label="note-0409-01" xlink:href="note-0409-01a" xml:space="preserve">Figure 197.</note> </div> <p> <s xml:id="echoid-s12200" xml:space="preserve">Comme il ſeroit facile de trouver la longueur I F, ſi l’on <lb/>connoiſſoit la valeur des lignes D I & </s> <s xml:id="echoid-s12201" xml:space="preserve">D F, nous allons voir <lb/>comment on peut y parvenir, en tirant les lignes D H, D K, <lb/>C F, & </s> <s xml:id="echoid-s12202" xml:space="preserve">en connoiſſant les parties du corps de la Place que <lb/>nous venons de trouver. </s> <s xml:id="echoid-s12203" xml:space="preserve">Pour y arriver, je cherche dans le <lb/>triangle rectangle C L F la valeur de l’angle L C F, par le moyen <lb/>des deux côtés L C & </s> <s xml:id="echoid-s12204" xml:space="preserve">C F, qui me ſont connus (puiſque l’un <lb/>vaut la moitié de la courtine, & </s> <s xml:id="echoid-s12205" xml:space="preserve">que l’autre eſt égal à la ligne <lb/>C E), en diſant: </s> <s xml:id="echoid-s12206" xml:space="preserve">Comme le côté L C eſt au côté C F; </s> <s xml:id="echoid-s12207" xml:space="preserve">ainſi le <lb/>ſinus total eſt à la ſécante, qui donnera 65 degrés pour l’an-<lb/>gle L C F, duquel ayant retranché l’angle M C D de 18 degrés <lb/>26 minutes, reſtera 46 degrés 34 minutes pour l’angle D C F.</s> <s xml:id="echoid-s12208" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12209" xml:space="preserve">Or comme le côté D C eſt de 88 toiſes 2 pieds, & </s> <s xml:id="echoid-s12210" xml:space="preserve">le côté <lb/>C F de 90 toiſes 2 pieds, & </s> <s xml:id="echoid-s12211" xml:space="preserve">que l’on connoît l’angle qu’ils <lb/>comprennent, on trouvera par l’analogie ordinaire que le côté <lb/>D F eſt de 70 toiſes 2 pieds, & </s> <s xml:id="echoid-s12212" xml:space="preserve">que l’angle C D F eſt de 68 de-<lb/>grés 15 minutes.</s> <s xml:id="echoid-s12213" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12214" xml:space="preserve">Comme nous avons beſoin de connoître l’angle C D K, auſſi-<lb/>bien que le côté D K, conſidérez que dans le triangle C D K, <lb/>l’on connoît les deux côtés D C & </s> <s xml:id="echoid-s12215" xml:space="preserve">C K avec l’angle qu’ils com-<lb/>prennent, & </s> <s xml:id="echoid-s12216" xml:space="preserve">que par conſéquent il ſera facile de trouver ce <lb/>que l’on cherche. </s> <s xml:id="echoid-s12217" xml:space="preserve">Auſſi verra-t’on que C D K eſt de 17 degrés <lb/>49 minutes, & </s> <s xml:id="echoid-s12218" xml:space="preserve">le côté D K de 88 toiſes.</s> <s xml:id="echoid-s12219" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12220" xml:space="preserve">Or comme il faut dans le triangle H D K connoître, outre <lb/>le côté D K, le côté H D avec l’angle qu’ils comprennent pour <lb/>parvenir à la ſolution du problême, conſidérez que dans le <lb/>triangle A H D l’on connoît le côté A D de 47 toiſes, & </s> <s xml:id="echoid-s12221" xml:space="preserve">le <lb/>côté A H de 20, & </s> <s xml:id="echoid-s12222" xml:space="preserve">qu’on connoîtra l’angle H A D, quand on <lb/>ſçaura la valeur de l’angle flanqué, puiſqu’il en eſt la différence <lb/>avec deux droits; </s> <s xml:id="echoid-s12223" xml:space="preserve">& </s> <s xml:id="echoid-s12224" xml:space="preserve">comme l’on ſuppoſe que c’eſt ici un exa- <pb o="356" file="0410" n="418" rhead="NOUVEAU COURS"/> gone, l’angle flanqué ſera par conſéquent de 83 degrés 8 mi-<lb/>nutes: </s> <s xml:id="echoid-s12225" xml:space="preserve">ainſi l’angle D A H ſera de 96 degrés 52 minutes; </s> <s xml:id="echoid-s12226" xml:space="preserve">& </s> <s xml:id="echoid-s12227" xml:space="preserve">en <lb/>faiſant la regle ordinaire, l’on trouvera (art. </s> <s xml:id="echoid-s12228" xml:space="preserve">725) que le côté <lb/>H D eſt de 53 toiſes un pied, & </s> <s xml:id="echoid-s12229" xml:space="preserve">que l’angle A D H eſt de 21 <lb/>degrés 59 minutes.</s> <s xml:id="echoid-s12230" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12231" xml:space="preserve">Préſentement ſi l’on retranche de 180 degrés la ſomme des <lb/>deux angles C D K & </s> <s xml:id="echoid-s12232" xml:space="preserve">A D H, il reſtera 140 degrés 12 minutes <lb/>pour la valeur de l’angle H D K.</s> <s xml:id="echoid-s12233" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12234" xml:space="preserve">743. </s> <s xml:id="echoid-s12235" xml:space="preserve">Or comme l’on connoît dans le triangle H D K deux <lb/>côtés & </s> <s xml:id="echoid-s12236" xml:space="preserve">l’angle compris, on trouvera par conſéquent (art. </s> <s xml:id="echoid-s12237" xml:space="preserve">725) <lb/>les deux autres angles, particuliérement l’angle D K I, dont <lb/>nous avons beſoin, qui eſt de 14 degrés 4 minutes; </s> <s xml:id="echoid-s12238" xml:space="preserve">& </s> <s xml:id="echoid-s12239" xml:space="preserve">comme <lb/>il nous faut auſſi l’angle F D K, on trouvera qu’il eſt de 50 de-<lb/>grés 26 minutes, ſi l’on retranche de l’angle F D C l’angle <lb/>K D C: </s> <s xml:id="echoid-s12240" xml:space="preserve">mais comme ceci nous donne la valeur de l’angle <lb/>D I K, qui eſt de 115 degrés 30 minutes, l’on pourra donc <lb/>dire pour trouver le côté D I: </s> <s xml:id="echoid-s12241" xml:space="preserve">Si le ſinus du ſupplément de <lb/>l’angle D I K a donné le côté D K, que donnera le ſinus de <lb/>l’angle D K I pour la valeur du côté D I, que l’on trouvera de <lb/>23 toiſes 4 pieds, qu’on n’aura qu’à retrancher de la ligne D F, <lb/>qui vaut, comme nous l’avons vu, 70 toiſes 2 pieds, l’on trou-<lb/>vera que la face I F de la demi-lune eſt de 46 toiſes 4 pieds?</s> <s xml:id="echoid-s12242" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12243" xml:space="preserve">744. </s> <s xml:id="echoid-s12244" xml:space="preserve">Pour trouver la demi-gorge I N de la demi-lune, faites <lb/>attention que dans le triangle O D F, l’on connoît les deux <lb/>angles F O D & </s> <s xml:id="echoid-s12245" xml:space="preserve">O D F, & </s> <s xml:id="echoid-s12246" xml:space="preserve">que par conſéquent on connoîtra <lb/>l’angle O F D, qui ſe trouve de 40 degrés 11 minutes; </s> <s xml:id="echoid-s12247" xml:space="preserve">& </s> <s xml:id="echoid-s12248" xml:space="preserve">comme <lb/>cet angle ſe trouve auſſi dans le triangle I N F, dont on connoît <lb/>l’angle N I F, puiſqu’il eſt ſupplément de l’angle D I K, il s’en-<lb/>ſuit qu’ayant deux angles dans le triangle I F N, l’on connoîtra <lb/>le troiſieme I N F; </s> <s xml:id="echoid-s12249" xml:space="preserve">par conſéquent l’on pourra dire: </s> <s xml:id="echoid-s12250" xml:space="preserve">Si le ſinus <lb/>de l’angle I N F de 75 degrés 19 minutes a donné le côté I F, <lb/>que donnera le ſinus de l’angle I F N pour le côté I N, que <lb/>l’on trouvera de .</s> <s xml:id="echoid-s12251" xml:space="preserve">..</s> <s xml:id="echoid-s12252" xml:space="preserve">.?</s> <s xml:id="echoid-s12253" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12254" xml:space="preserve">Enfin ſi pour tracer la demi - lune l’on avoit beſoin de la <lb/>diſtance du milieu L de la courtine au point F, il ſeroit facile <lb/>de la trouver, en diſant: </s> <s xml:id="echoid-s12255" xml:space="preserve">Comme le ſinus total eſt à la tangente <lb/>de l’angle L C F, ainſi le côté C L eſt au côté L F, que l’on <lb/>trouvera de 82. </s> <s xml:id="echoid-s12256" xml:space="preserve">0. </s> <s xml:id="echoid-s12257" xml:space="preserve">9 pouces.</s> <s xml:id="echoid-s12258" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12259" xml:space="preserve">Je ne parle point de la maniere de calculer les lignes, tant <lb/>droites que courbes, qui forment la contreſcarpe, parce que <pb o="357" file="0411" n="419" rhead="DE MATHÉMATIQUE. Liv. X."/> c’eſt une choſe qui m’a paru fort aiſée, & </s> <s xml:id="echoid-s12260" xml:space="preserve">que les Commen-<lb/>çans pourront faire d’eux-mêmes. </s> <s xml:id="echoid-s12261" xml:space="preserve">Je ne dis rien non plus de <lb/>la maniere de calculer une fortification, dont les baſtions ſe-<lb/>roient à orillons, pour leur laiſſer le plaiſir de faire quelque <lb/>choſe par eux-mêmes, ayant mieux aimé leur donner, au lieu <lb/>de cela, une idée de la façon de tracer une fortification ſur le <lb/>terrein.</s> <s xml:id="echoid-s12262" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div998" type="section" level="1" n="782"> <head xml:id="echoid-head940" style="it" xml:space="preserve">Maniere de tracer les Fortifications ſur le terrein.</head> <p> <s xml:id="echoid-s12263" xml:space="preserve">745. </s> <s xml:id="echoid-s12264" xml:space="preserve">Après que l’on a fait le calcul des lignes & </s> <s xml:id="echoid-s12265" xml:space="preserve">des angles <lb/>qui compoſent la fortification, on commence, pour la tracer <lb/>ſur le terrein, par planter des piquets à tous les angles qui doi-<lb/>vent former le polygone; </s> <s xml:id="echoid-s12266" xml:space="preserve">enſuite l’on s’attache à tracer la for-<lb/>tification de chaque front, juſqu’à ce que tout ſoit achevé.</s> <s xml:id="echoid-s12267" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12268" xml:space="preserve">Si l’on ſuppoſe que les points A & </s> <s xml:id="echoid-s12269" xml:space="preserve">B repréſentent deux en-<lb/>droits auxquels l’on a planté des piquets, qui déterminent la <lb/>longueur A B d’un des côtés du polygone, qui ſera, par exem-<lb/>ple de 180 toiſes, voici comment il faut s’y prendre pour <lb/>tracer le front qui correſpond à ſes côtés.</s> <s xml:id="echoid-s12270" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12271" xml:space="preserve">Ayant marqué ſur un plan le projet de la fortification avec <lb/> <anchor type="note" xlink:label="note-0411-01a" xlink:href="note-0411-01"/> la valeur des lignes & </s> <s xml:id="echoid-s12272" xml:space="preserve">des angles, comme on le voit dans la <lb/>figure 198, l’on commencera par poſer le pied du graphometre <lb/>à l’endroit du piquet A: </s> <s xml:id="echoid-s12273" xml:space="preserve">l’on fera avec la baſe A B, & </s> <s xml:id="echoid-s12274" xml:space="preserve">les pin-<lb/>nules immobiles, un angle E A B de 18 degrés 26 minutes; <lb/></s> <s xml:id="echoid-s12275" xml:space="preserve">& </s> <s xml:id="echoid-s12276" xml:space="preserve">ayant fait porter un piquet ſur l’alignement du rayon viſuel <lb/>A E, on déterminera, en toiſant fort juſte, une longueur <lb/>comme A C de 50 toiſes, qui donnera une des faces du pre-<lb/>mier baſtion. </s> <s xml:id="echoid-s12277" xml:space="preserve">Après quoi l’on portera l’inſtrument à l’extrê-<lb/>mité C, & </s> <s xml:id="echoid-s12278" xml:space="preserve">l’on fera avec la ligne C A un angle A C D de <lb/>117 degrés 39 minutes, qui ſera l’angle de l’épaule, & </s> <s xml:id="echoid-s12279" xml:space="preserve">l’on <lb/>prendra dans la longueur C D une quantité de 27 toiſes 2 pieds, <lb/>en commençant du point C pour avoir le flanc C D.</s> <s xml:id="echoid-s12280" xml:space="preserve"/> </p> <div xml:id="echoid-div998" type="float" level="2" n="1"> <note position="right" xlink:label="note-0411-01" xlink:href="note-0411-01a" xml:space="preserve">Figure 199.</note> </div> <p> <s xml:id="echoid-s12281" xml:space="preserve">L’on fera la même opération au piquet B, comme on vient <lb/>de faire à l’autre; </s> <s xml:id="echoid-s12282" xml:space="preserve">& </s> <s xml:id="echoid-s12283" xml:space="preserve">aprés avoir tracé, ou ſeulement planté <lb/>des piquets aux points F & </s> <s xml:id="echoid-s12284" xml:space="preserve">E, l’on ſe portera au point E pour <lb/>voir s’il ſe trouve de même alignement que les deux C & </s> <s xml:id="echoid-s12285" xml:space="preserve">A, <lb/>afin de remarquer ſi la face A C ſe termine préciſément dans <lb/>l’angle flanquant; </s> <s xml:id="echoid-s12286" xml:space="preserve">& </s> <s xml:id="echoid-s12287" xml:space="preserve">l’on fera la même choſe pour être aſſuré <lb/>de la juſteſſe de la face B F; </s> <s xml:id="echoid-s12288" xml:space="preserve">enſuite l’on n’aura plus qu’à tracer <lb/>avec un cordeau la courtine D E, auſſi-bien que les faces &</s> <s xml:id="echoid-s12289" xml:space="preserve"> <pb o="358" file="0412" n="420" rhead="NOUVEAU COURS"/> les flancs des baſtions: </s> <s xml:id="echoid-s12290" xml:space="preserve">& </s> <s xml:id="echoid-s12291" xml:space="preserve">pour voir ſi on ne s’eſt pas trompé <lb/>en traçant les faces & </s> <s xml:id="echoid-s12292" xml:space="preserve">les flancs, on meſurera la courtine, afin <lb/>de la vérifier avec le calcul.</s> <s xml:id="echoid-s12293" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12294" xml:space="preserve">Pour faire ſentir encore davantage l’utilité de la Trigono-<lb/>métrie dans ce qui concerne les fortifications, nous allons <lb/>ajouter quelques problêmes, dont la ſolution dépend des prin-<lb/>cipes précédens, & </s> <s xml:id="echoid-s12295" xml:space="preserve">qui peuvent être d’un grand uſage dans <lb/>l’attaque des places, & </s> <s xml:id="echoid-s12296" xml:space="preserve">dans la conduite des ouvrages, pour <lb/>connoître par une ſeule obſervation la diſtance où l’on eſt de <lb/>certains endroits remarquables que l’on a intérêt d’attaquer.</s> <s xml:id="echoid-s12297" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1000" type="section" level="1" n="783"> <head xml:id="echoid-head941" style="it" xml:space="preserve">Problêmes de Trigonométrie applicables à la Fortification.</head> <head xml:id="echoid-head942" xml:space="preserve"><emph style="sc">Probleme</emph> I.</head> <p style="it"> <s xml:id="echoid-s12298" xml:space="preserve">746. </s> <s xml:id="echoid-s12299" xml:space="preserve">Connoiſſant une ligne A B, dont on ne peut approcher, <lb/> <anchor type="note" xlink:label="note-0412-01a" xlink:href="note-0412-01"/> avec les angles A D C, A D B; </s> <s xml:id="echoid-s12300" xml:space="preserve">& </s> <s xml:id="echoid-s12301" xml:space="preserve">les angles B C D, B C A ob-<lb/>ſervés aux points de ſtation C & </s> <s xml:id="echoid-s12302" xml:space="preserve">D, connoître tous les angles & </s> <s xml:id="echoid-s12303" xml:space="preserve"><lb/>les lignes de cette figure.</s> <s xml:id="echoid-s12304" xml:space="preserve"/> </p> <div xml:id="echoid-div1000" type="float" level="2" n="1"> <note position="left" xlink:label="note-0412-01" xlink:href="note-0412-01a" xml:space="preserve">Figure 173.</note> </div> </div> <div xml:id="echoid-div1002" type="section" level="1" n="784"> <head xml:id="echoid-head943" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s12305" xml:space="preserve">Puiſque l’on connoît l’angle A C D & </s> <s xml:id="echoid-s12306" xml:space="preserve">l’angle A D C, on <lb/>connoît auſſi l’angle en A, en ôtant les deux premiers de 180 <lb/>degrés; </s> <s xml:id="echoid-s12307" xml:space="preserve">de même dans le triangle B C D on connoît l’angle <lb/>C B D, puiſque, par hypotheſe, les angles B C D, B D C ſont <lb/>connus. </s> <s xml:id="echoid-s12308" xml:space="preserve">Quoique je ne connoiſſe point les côtés A C, A D, <lb/>D C, B C, B D de ces triangles, je ſçais cependant que ces <lb/>côtés ſont entr’eux comme les ſinus des angles qui leur ſont <lb/>oppoſés; </s> <s xml:id="echoid-s12309" xml:space="preserve">& </s> <s xml:id="echoid-s12310" xml:space="preserve">comme ces angles ſont connus, les rapports des <lb/>côtés le ſeront auſſi. </s> <s xml:id="echoid-s12311" xml:space="preserve">Cela poſé, dans le triangle C A D, on aura <lb/>cette proportion, S. </s> <s xml:id="echoid-s12312" xml:space="preserve">CAD: </s> <s xml:id="echoid-s12313" xml:space="preserve">S. </s> <s xml:id="echoid-s12314" xml:space="preserve">ADC:</s> <s xml:id="echoid-s12315" xml:space="preserve">: DC: </s> <s xml:id="echoid-s12316" xml:space="preserve">AC, & </s> <s xml:id="echoid-s12317" xml:space="preserve">dans le trian-<lb/>gle C B D, on aura cette autre, S. </s> <s xml:id="echoid-s12318" xml:space="preserve">BDC: </s> <s xml:id="echoid-s12319" xml:space="preserve">S. </s> <s xml:id="echoid-s12320" xml:space="preserve">CBD:</s> <s xml:id="echoid-s12321" xml:space="preserve">: BC: </s> <s xml:id="echoid-s12322" xml:space="preserve">DC: <lb/></s> <s xml:id="echoid-s12323" xml:space="preserve">donc en multipliant terme par terme ces deux proportions, on <lb/>aura S. </s> <s xml:id="echoid-s12324" xml:space="preserve">C A D x S. </s> <s xml:id="echoid-s12325" xml:space="preserve">B D C: </s> <s xml:id="echoid-s12326" xml:space="preserve">S. </s> <s xml:id="echoid-s12327" xml:space="preserve">A D C x S. </s> <s xml:id="echoid-s12328" xml:space="preserve">C B D:</s> <s xml:id="echoid-s12329" xml:space="preserve">: B C x D C: </s> <s xml:id="echoid-s12330" xml:space="preserve"><lb/>A C x D C:</s> <s xml:id="echoid-s12331" xml:space="preserve">: B C: </s> <s xml:id="echoid-s12332" xml:space="preserve">A C. </s> <s xml:id="echoid-s12333" xml:space="preserve">D’où il ſuit que dans le triangle B C A, <lb/>on a le rapport exact des côtés A C, C B qui comprennent <lb/>l’angle connu A C B; </s> <s xml:id="echoid-s12334" xml:space="preserve">ainſi on ſuppoſera pour un inſtant que <lb/>ces côtés ſont effectivement égaux aux produits des ſinus des <lb/>angles C A D, B D C, A D C, C B D; </s> <s xml:id="echoid-s12335" xml:space="preserve">& </s> <s xml:id="echoid-s12336" xml:space="preserve">pour avoir les angles <lb/>en A & </s> <s xml:id="echoid-s12337" xml:space="preserve">en B du triangle A B C, on fera cette proportion: </s> <s xml:id="echoid-s12338" xml:space="preserve">La <lb/>ſomme des deux côtés A C + B C eſt à leur différence, comme <pb o="359" file="0413" n="421" rhead="DE MATHÉMATIQUE. Liv. X."/> la tangente de la moitié de la ſomme des angles oppoſés à ces <lb/>côtés, eſt à la tangente de la moitié de la différence des mêmes <lb/>angles (art. </s> <s xml:id="echoid-s12339" xml:space="preserve">725). </s> <s xml:id="echoid-s12340" xml:space="preserve">Ayant ainſi déterminé les angles en A & </s> <s xml:id="echoid-s12341" xml:space="preserve"><lb/>en B, on calculera de nouveau le triangle A B C, pour avoir <lb/>la véritable expreſſion des côtés A C, B C que l’on trouvera en <lb/>faiſant cette proportion: </s> <s xml:id="echoid-s12342" xml:space="preserve">Le ſinus de l’angle A C B eſt au côté <lb/>comme A B, comme le ſinus de l’angle A B C, que l’on vient <lb/>de trouver, eſt au côté A C. </s> <s xml:id="echoid-s12343" xml:space="preserve">On calculera par le ſecours de la <lb/>même analogie tous les autres côtés de la figure: </s> <s xml:id="echoid-s12344" xml:space="preserve">ainſi le pro-<lb/>blême eſt réſolu.</s> <s xml:id="echoid-s12345" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1003" type="section" level="1" n="785"> <head xml:id="echoid-head944" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s12346" xml:space="preserve">747. </s> <s xml:id="echoid-s12347" xml:space="preserve">On a dû remarquer que dans le triangle A C B l’on <lb/>ne connoiſſoit d’abord que le côté A B & </s> <s xml:id="echoid-s12348" xml:space="preserve">l’angle oppoſé C, <lb/>& </s> <s xml:id="echoid-s12349" xml:space="preserve">rien de plus. </s> <s xml:id="echoid-s12350" xml:space="preserve">L’on pourroit donc être tenté de croire que <lb/>la connoiſſance d’un angle & </s> <s xml:id="echoid-s12351" xml:space="preserve">du côté oppoſé ſuffit pour con-<lb/>noître toutes les parties d’un triangle, mais il eſt aiſé d’ap-<lb/>percevoir la fauſſeté d’une pareille induction; </s> <s xml:id="echoid-s12352" xml:space="preserve">il eſt vrai que <lb/>l’on ne connoît qu’un angle & </s> <s xml:id="echoid-s12353" xml:space="preserve">le côté oppoſé, mais les ob-<lb/>ſervations des angles en D ſuppléent à ce qui nous manque, en <lb/>donnant le rapport des côtés A C & </s> <s xml:id="echoid-s12354" xml:space="preserve">C B, par le moyen deſ-<lb/>quels on a calculé les angles en A & </s> <s xml:id="echoid-s12355" xml:space="preserve">en B.</s> <s xml:id="echoid-s12356" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12357" xml:space="preserve">748. </s> <s xml:id="echoid-s12358" xml:space="preserve">Si l’on appelle le ſinus de l’angle A D C, a; </s> <s xml:id="echoid-s12359" xml:space="preserve">celui de <lb/>l’angle C A D, b; </s> <s xml:id="echoid-s12360" xml:space="preserve">celui de l’angle C B D, c; </s> <s xml:id="echoid-s12361" xml:space="preserve">& </s> <s xml:id="echoid-s12362" xml:space="preserve">enfin celui de <lb/>l’angle B D C, d; </s> <s xml:id="echoid-s12363" xml:space="preserve">on aura au lieu de la proportion S. </s> <s xml:id="echoid-s12364" xml:space="preserve">C A D <lb/>x S. </s> <s xml:id="echoid-s12365" xml:space="preserve">B D C: </s> <s xml:id="echoid-s12366" xml:space="preserve">S. </s> <s xml:id="echoid-s12367" xml:space="preserve">A D C x S. </s> <s xml:id="echoid-s12368" xml:space="preserve">C B D :</s> <s xml:id="echoid-s12369" xml:space="preserve">: B C: </s> <s xml:id="echoid-s12370" xml:space="preserve">A C, celle-ci bd: </s> <s xml:id="echoid-s12371" xml:space="preserve">ac <lb/>:</s> <s xml:id="echoid-s12372" xml:space="preserve">: B C: </s> <s xml:id="echoid-s12373" xml:space="preserve">A C: </s> <s xml:id="echoid-s12374" xml:space="preserve">donc en diviſant les deux premiers termes par c, <lb/>on auroit {bd/c}: </s> <s xml:id="echoid-s12375" xml:space="preserve">a :</s> <s xml:id="echoid-s12376" xml:space="preserve">: B C: </s> <s xml:id="echoid-s12377" xml:space="preserve">A C.</s> <s xml:id="echoid-s12378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12379" xml:space="preserve">Si l’on vouloit ſe ſervir des logarithmes pour faire cette <lb/>opération, voici comment on pourroit s’y prendre. </s> <s xml:id="echoid-s12380" xml:space="preserve">On cher-<lb/>chera d’abord les ſinus des angles a, b, c, d, que l’on regar-<lb/>dera comme des nombres naturels; </s> <s xml:id="echoid-s12381" xml:space="preserve">on cherchera enſuite dans <lb/>la Table des Logarithmes des nombres naturels, les logarithmes <lb/>de ces ſinus conſidérés comme tels; </s> <s xml:id="echoid-s12382" xml:space="preserve">on ajoutera enſemble les <lb/>logarithmes des ſinus s, b, d, & </s> <s xml:id="echoid-s12383" xml:space="preserve">de la ſomme on retranchera le <lb/>logarithme de c: </s> <s xml:id="echoid-s12384" xml:space="preserve">ce qui viendra ſera le logarithme de la frac-<lb/>tion {bd/c}; </s> <s xml:id="echoid-s12385" xml:space="preserve">on cherchera ce logarithme dans la Table des Loga-<lb/>rithmes pour voir le nombre qui lui répond. </s> <s xml:id="echoid-s12386" xml:space="preserve">Après cela on <lb/>prendra la ſomme de ce nombre & </s> <s xml:id="echoid-s12387" xml:space="preserve">du ſinus a, & </s> <s xml:id="echoid-s12388" xml:space="preserve">la diffé- <pb o="360" file="0414" n="422" rhead="NOUVEAU COURS"/> rence des mêmes nombres; </s> <s xml:id="echoid-s12389" xml:space="preserve">on cherchera encore les logarithmes <lb/>de ces nouvelles quantités, & </s> <s xml:id="echoid-s12390" xml:space="preserve">dans les Tables des Sinus le <lb/>logarithme de la tangente de la moitié de la ſomme des angles <lb/>oppoſés aux côtés A C & </s> <s xml:id="echoid-s12391" xml:space="preserve">B C. </s> <s xml:id="echoid-s12392" xml:space="preserve">Ajoutant les logarithmes de <lb/>cette tangente, & </s> <s xml:id="echoid-s12393" xml:space="preserve">de la différence des nombres a & </s> <s xml:id="echoid-s12394" xml:space="preserve">{bd/c}, on <lb/>aura, aprés en avoir retranché le logarithme de la ſomme des <lb/>mêmes nombres, celui de la tangente de la moitié de la diffé-<lb/>rence des angles que l’on cherche, & </s> <s xml:id="echoid-s12395" xml:space="preserve">le problême ſera réſolu.</s> <s xml:id="echoid-s12396" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1004" type="section" level="1" n="786"> <head xml:id="echoid-head945" xml:space="preserve"><emph style="sc">Probleme</emph> II.</head> <p style="it"> <s xml:id="echoid-s12397" xml:space="preserve">749. </s> <s xml:id="echoid-s12398" xml:space="preserve">La ligne A C & </s> <s xml:id="echoid-s12399" xml:space="preserve">ſes parties A B, B C étant connues avec <lb/> <anchor type="note" xlink:label="note-0414-01a" xlink:href="note-0414-01"/> les angles A F B, B F C, obſervés dans un point F, trouver les <lb/>diſtances du point F aux points A, B, C.</s> <s xml:id="echoid-s12400" xml:space="preserve"/> </p> <div xml:id="echoid-div1004" type="float" level="2" n="1"> <note position="left" xlink:label="note-0414-01" xlink:href="note-0414-01a" xml:space="preserve">Figure 200.</note> </div> <p> <s xml:id="echoid-s12401" xml:space="preserve">Ce problême peut être réſolu géométriquement, & </s> <s xml:id="echoid-s12402" xml:space="preserve">par le <lb/>calcul trigonométrique: </s> <s xml:id="echoid-s12403" xml:space="preserve">nous allons donner la ſolution qui dé-<lb/>pend du calcul, & </s> <s xml:id="echoid-s12404" xml:space="preserve">nous donnerons enſuite la ſolution géomé-<lb/>trique.</s> <s xml:id="echoid-s12405" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1006" type="section" level="1" n="787"> <head xml:id="echoid-head946" xml:space="preserve"><emph style="sc">Solution</emph> I.</head> <p> <s xml:id="echoid-s12406" xml:space="preserve">Imaginons les lignes A F, B F, C F, tirées des extrêmités <lb/> <anchor type="note" xlink:label="note-0414-02a" xlink:href="note-0414-02"/> A, B, C des parties de la ligne A C, au point d’obſervation F; <lb/></s> <s xml:id="echoid-s12407" xml:space="preserve">imaginons encore, que par chacun des deux triangles A B F, <lb/>B C F, on ait fait paſſer un cercle F B C, A B F. </s> <s xml:id="echoid-s12408" xml:space="preserve">Comme les <lb/>angles en F ſeront à la circonférence, ils ne ſeront que la <lb/>moitié des angles B E C, B D A, appuyés ſur les mêmes baſes <lb/>B C, A B, & </s> <s xml:id="echoid-s12409" xml:space="preserve">dont le ſommet eſt au centre E ou D. </s> <s xml:id="echoid-s12410" xml:space="preserve">Cela poſé, <lb/>puiſque les angles B F C, A FB ſont connus, les angles au <lb/>centre doubles des angles obſervés le ſeront auſſi, & </s> <s xml:id="echoid-s12411" xml:space="preserve">dans les <lb/>triangles iſoſceles B E C, B D A, on connoîtra les trois angles <lb/>& </s> <s xml:id="echoid-s12412" xml:space="preserve">un côté; </s> <s xml:id="echoid-s12413" xml:space="preserve">ainſi on connoîtra les côtés ou les rayons B E, <lb/>B D, puiſque l’on connoît les angles C B E, A B D; </s> <s xml:id="echoid-s12414" xml:space="preserve">on con-<lb/>noîtra auſſi l’angle D B E, qui joint avec ces angles, fait la va-<lb/>leur de deux droits; </s> <s xml:id="echoid-s12415" xml:space="preserve">de plus on vient de trouver les côtés <lb/>B E, B D: </s> <s xml:id="echoid-s12416" xml:space="preserve">donc on connoîtra toutes les parties de ce triangle <lb/>dans lequel on a les côtés B D, B E, & </s> <s xml:id="echoid-s12417" xml:space="preserve">l’angle compris entre <lb/>ces côtés: </s> <s xml:id="echoid-s12418" xml:space="preserve">ainſi on connoîtra l’angle en E & </s> <s xml:id="echoid-s12419" xml:space="preserve">l’angle en D. </s> <s xml:id="echoid-s12420" xml:space="preserve"><lb/>Puiſque les cercles décrits ſur les triangles C B F, A B F ſe cou-<lb/>pent en deux points B, F, le centre F ſera également éloigné <lb/>des points B, F, & </s> <s xml:id="echoid-s12421" xml:space="preserve">le point D de la même ligne D E ſera auſſi <lb/>également éloigné des mêmes points B, F; </s> <s xml:id="echoid-s12422" xml:space="preserve">ainſi la ligne D E <pb o="361" file="0415" n="423" rhead="DE MATHÉMATIQUE. Liv. X."/> ſera perpendiculaire à la ligne B F, & </s> <s xml:id="echoid-s12423" xml:space="preserve">partout dans le triangle <lb/>rectangle B G E, dans lequel on connoît déja l’angle en E, <lb/>comme on vient de voir, on connoîtra auſſi l’angle G B E; <lb/></s> <s xml:id="echoid-s12424" xml:space="preserve">ajoutant cet angle à l’angle connu C B E du triangle iſoſcele <lb/>B E C, on aura l’angle total C B F; </s> <s xml:id="echoid-s12425" xml:space="preserve">ainſi dans le triangle CBF <lb/>on connoît deux angles & </s> <s xml:id="echoid-s12426" xml:space="preserve">un côté: </s> <s xml:id="echoid-s12427" xml:space="preserve">donc on peut connoître <lb/>toutes les autres parties.</s> <s xml:id="echoid-s12428" xml:space="preserve"/> </p> <div xml:id="echoid-div1006" type="float" level="2" n="1"> <note position="left" xlink:label="note-0414-02" xlink:href="note-0414-02a" xml:space="preserve">Figure 200.</note> </div> </div> <div xml:id="echoid-div1008" type="section" level="1" n="788"> <head xml:id="echoid-head947" xml:space="preserve"><emph style="sc">Solution geométrique</emph>.</head> <p> <s xml:id="echoid-s12429" xml:space="preserve">750. </s> <s xml:id="echoid-s12430" xml:space="preserve">Puiſque les parties de la ligne A C ſont connues, ainſi <lb/>que les angles A F B, B F C, je prends une ligne A B qui con-<lb/>tienne autant de parties égales que la ligne A B, que l’on ſup-<lb/>poſe ſur le terrein, contient de toiſes: </s> <s xml:id="echoid-s12431" xml:space="preserve">je prends de même ſur <lb/>la ligne A B prolongée une partie B C qui contienne autant <lb/>de parties égales, que la ligne B C obſervée ſur le terrein con-<lb/>tient de toiſes. </s> <s xml:id="echoid-s12432" xml:space="preserve">Je double l’angle A F B, j’ôte cet angle de 180 <lb/>degrés, & </s> <s xml:id="echoid-s12433" xml:space="preserve">je diviſe le reſte en deux parties égales. </s> <s xml:id="echoid-s12434" xml:space="preserve">Au point <lb/>A & </s> <s xml:id="echoid-s12435" xml:space="preserve">au point B, je fais les angles B A D, A B D égaux cha-<lb/>cun à la moitié de cette différence; </s> <s xml:id="echoid-s12436" xml:space="preserve">ce qui me détermine le <lb/>point D. </s> <s xml:id="echoid-s12437" xml:space="preserve">Je double de même l’angle obſervé B F C, & </s> <s xml:id="echoid-s12438" xml:space="preserve">ôtant <lb/>ce double de 180 degrés, je fais en B & </s> <s xml:id="echoid-s12439" xml:space="preserve">en C les angles C B E, <lb/>B C E égaux chacun à la moitié de la différence du double de <lb/>l’angle obſervé; </s> <s xml:id="echoid-s12440" xml:space="preserve">ce qui me donne le point E: </s> <s xml:id="echoid-s12441" xml:space="preserve">je mene la ligne <lb/>E D; </s> <s xml:id="echoid-s12442" xml:space="preserve">du point B j’abaiſſe ſur cette ligne E D la perpendicu-<lb/>laire B G F, & </s> <s xml:id="echoid-s12443" xml:space="preserve">je prends G F = B G; </s> <s xml:id="echoid-s12444" xml:space="preserve">le point F eſt le point <lb/>qui me donne tout ce dont j’ai beſoin: </s> <s xml:id="echoid-s12445" xml:space="preserve">ainſi je n’ai qu’à voir <lb/>combien les lignes BF, CF, AF contiennent de parties égales, <lb/>& </s> <s xml:id="echoid-s12446" xml:space="preserve">j’aurai les diſtances du point F aux points donnés A, B, C.</s> <s xml:id="echoid-s12447" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12448" xml:space="preserve">751. </s> <s xml:id="echoid-s12449" xml:space="preserve">On pourroit encore réſoudre le problême géométri-<lb/>quement d’une autre maniere: </s> <s xml:id="echoid-s12450" xml:space="preserve">il n’y auroit qu’à décrire ſur <lb/>les lignes A B & </s> <s xml:id="echoid-s12451" xml:space="preserve">B C des ſegmens capables des angles donnés <lb/>A F B, B F C, & </s> <s xml:id="echoid-s12452" xml:space="preserve">le point où ces cercles s’entrecouperoient au <lb/>dehors de la ligne A C, ſeroit celui qui donneroit les diſtances <lb/>demandées.</s> <s xml:id="echoid-s12453" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1009" type="section" level="1" n="789"> <head xml:id="echoid-head948" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s12454" xml:space="preserve">752. </s> <s xml:id="echoid-s12455" xml:space="preserve">On pourroit encore réſoudre le problême par les mé-<lb/>thodes que nous venons de propoſer dans le cas où les parties <lb/>A B & </s> <s xml:id="echoid-s12456" xml:space="preserve">B C ne ſeroient pas en lignes droites, comme dans les <lb/>figures 202, 203, pourvu que l’on connût l’angle A B C qu’elles <pb o="362" file="0416" n="424" rhead="NOUVEAU COURS"/> forment entr’elles, ou bien les trois côtés du triangle B A C. <lb/></s> <s xml:id="echoid-s12457" xml:space="preserve">Pour s’en convaincre, il n’y a qu’à relire les deux ſolutions <lb/>précédentes, en les appliquant ſur les figures ſuivantes, & </s> <s xml:id="echoid-s12458" xml:space="preserve">ob-<lb/>ſervant que dans la figure 201, l’angle D B E eſt égal à la dif-<lb/>férence de l’angle A B C aux angles A B D, C B E, & </s> <s xml:id="echoid-s12459" xml:space="preserve">que dans <lb/>la figure 202 on trouve l’angle D B E, en prenant la différence <lb/>des trois angles A B C, A B D, C B E à quatre droits. </s> <s xml:id="echoid-s12460" xml:space="preserve">Enfin <lb/>l’on remarquera que ſi le point F d’obſervation eſt placé au <lb/>dedans du triangle A B C, & </s> <s xml:id="echoid-s12461" xml:space="preserve">que l’un des angles obſervés ſoit <lb/>obtus, on fera de l’autre côté de la ligne A B un triangle iſoſ-<lb/>cele A D B, dont l’angle D ſoit double du ſupplément de l’an-<lb/>gle A F B; </s> <s xml:id="echoid-s12462" xml:space="preserve">& </s> <s xml:id="echoid-s12463" xml:space="preserve">dans ce cas l’angle D B E eſt égal à la ſomme <lb/>des angles B B A, A B C, moins l’angle E B C, que l’on con-<lb/>noîtra, puiſque l’on connoît, par conſtruction ou par hypotheſe, <lb/>les angles qui le déterminent.</s> <s xml:id="echoid-s12464" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1010" type="section" level="1" n="790"> <head xml:id="echoid-head949" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s12465" xml:space="preserve">753. </s> <s xml:id="echoid-s12466" xml:space="preserve">Il ſuit delà que ſi l’on connoît la poſition de trois <lb/>objets placés au dedans d’une Ville aſſiégée par le moyen d’un <lb/>plan, ou bien parce qu’on l’aura déterminée géométrique-<lb/>ment; </s> <s xml:id="echoid-s12467" xml:space="preserve">on pourra toujours par une ſeule opération déterminer <lb/>la diſtance de l’endroit où l’on eſt aux mêmes objets que l’on <lb/>a intérêt d’attaquer; </s> <s xml:id="echoid-s12468" xml:space="preserve">& </s> <s xml:id="echoid-s12469" xml:space="preserve">par conſéquent on ſera le maître d’y <lb/>conduire des galeries de mines, ou d’y jetter des bombes, <lb/>ou enfin de diriger ſes batteries de la maniere qui paroîtra la <lb/>plus avantageuſe. </s> <s xml:id="echoid-s12470" xml:space="preserve">Il faut dans le cas où l’on auroit beſoin d’une <lb/>grande préciſion ſe ſervir des ſolutions numériques préféra-<lb/>blement aux ſolutions géométriques, parce que le calcul donne <lb/>toujours les diſtances avec la derniere exactitude.</s> <s xml:id="echoid-s12471" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1011" type="section" level="1" n="791"> <head xml:id="echoid-head950" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s12472" xml:space="preserve">754. </s> <s xml:id="echoid-s12473" xml:space="preserve">Il ſuit encore delà que l’on peut, par le moyen des <lb/> <anchor type="note" xlink:label="note-0416-01a" xlink:href="note-0416-01"/> mêmes objets, que nous ſuppoſons toujours viſibles, lever <lb/> <anchor type="note" xlink:label="note-0416-02a" xlink:href="note-0416-02"/> très-promptement les dehors d’une place par deux obſerva-<lb/>tions, ſans être obligé de meſurer des baſes dans un terrein <lb/>expoſé au feu de l’ennemi. </s> <s xml:id="echoid-s12474" xml:space="preserve">Suppoſons, par exemple, qu’on <lb/>veuille avoir la poſition des baſtions F, G, H d’une place que <lb/>l’on aſſiége, par deux obſervations faites aux points D, E. </s> <s xml:id="echoid-s12475" xml:space="preserve">On <lb/>prendra par Trigonométrie la poſition des trois objets qui <lb/>@voiſinent la place, tels que A, B, C, ce que l’on pourra faire <pb o="363" file="0417" n="425" rhead="DE MATHÉMATIQUE. Liv. X."/> ſans aucun danger, en meſurant une baſe dans un terrein qui <lb/>ſoit à l’abri du canon; </s> <s xml:id="echoid-s12476" xml:space="preserve">enſuite par le moyen de ces trois ob-<lb/>jets, deux Ingénieurs placés l’un en E, l’autre en D, obſer-<lb/>veront les angles formés par les rayons viſuels, dirigés des <lb/>points de ſtations aux points A, B, C, & </s> <s xml:id="echoid-s12477" xml:space="preserve">aux angles du flanc <lb/>des baſtions F, G, H; </s> <s xml:id="echoid-s12478" xml:space="preserve">ſçavoir, celui qui eſt placé en E, les <lb/>angles C E H, C E G, C E F, & </s> <s xml:id="echoid-s12479" xml:space="preserve">celui qui eſt placé en D les angles <lb/>B D H, B D G, B D F; </s> <s xml:id="echoid-s12480" xml:space="preserve">de cette maniere on aura tout d’un <lb/>coup la poſition reſpective des baſtions les uns à l’égard des <lb/>autres, & </s> <s xml:id="echoid-s12481" xml:space="preserve">leurs diſtances aux points d’obſervations: </s> <s xml:id="echoid-s12482" xml:space="preserve">car il eſt <lb/>évident que les ſtations D, E ſont déterminées par rapport <lb/>aux points A, B, C; </s> <s xml:id="echoid-s12483" xml:space="preserve">ce qui ſuffit pour déterminer tout le reſte.</s> <s xml:id="echoid-s12484" xml:space="preserve"/> </p> <div xml:id="echoid-div1011" type="float" level="2" n="1"> <note position="left" xlink:label="note-0416-01" xlink:href="note-0416-01a" xml:space="preserve">Pl. XIV.</note> <note position="left" xlink:label="note-0416-02" xlink:href="note-0416-02a" xml:space="preserve">Figure 210.</note> </div> <p> <s xml:id="echoid-s12485" xml:space="preserve">Nota. </s> <s xml:id="echoid-s12486" xml:space="preserve">Le problême propoſé (art. </s> <s xml:id="echoid-s12487" xml:space="preserve">746) pourroit auſſi ſervir <lb/>aux mêmes uſages, & </s> <s xml:id="echoid-s12488" xml:space="preserve">l’on peut en faire l’application à bien <lb/>d’autres opérations qu’il ſeroit inutile de détailler ici. </s> <s xml:id="echoid-s12489" xml:space="preserve">L’occa-<lb/>ſion fournit des reſſources & </s> <s xml:id="echoid-s12490" xml:space="preserve">des expédiens lorſque l’on eſt <lb/>d’abord muni d’une bonne théorie; </s> <s xml:id="echoid-s12491" xml:space="preserve">ainſi chacun pourra s’exer-<lb/>cer à mettre en œuvre les propoſitions que nous venons de <lb/>démontrer.</s> <s xml:id="echoid-s12492" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1013" type="section" level="1" n="792"> <head xml:id="echoid-head951" xml:space="preserve"><emph style="sc">Theorie et pratique du</emph> <emph style="sc">Nivellement</emph>.</head> <head xml:id="echoid-head952" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head953" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s12493" xml:space="preserve">755. </s> <s xml:id="echoid-s12494" xml:space="preserve">L’on dit que deux points ſont de niveau, lorſqu’ils <lb/> <anchor type="note" xlink:label="note-0417-01a" xlink:href="note-0417-01"/> ſont également éloignés du centre de la terre:</s> <s xml:id="echoid-s12495" xml:space="preserve"/> </p> <div xml:id="echoid-div1013" type="float" level="2" n="1"> <note position="right" xlink:label="note-0417-01" xlink:href="note-0417-01a" xml:space="preserve">Pl. XIV.</note> </div> <note position="right" xml:space="preserve">Figure 203.</note> <p> <s xml:id="echoid-s12496" xml:space="preserve">756. </s> <s xml:id="echoid-s12497" xml:space="preserve">De ſorte qu’une ligne qui a tous ſes points également <lb/>éloignés du centre de la terre, eſt appellée ligne du vrai niveau, <lb/>qui ne peut être qu’une ligne courbe.</s> <s xml:id="echoid-s12498" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12499" xml:space="preserve">757. </s> <s xml:id="echoid-s12500" xml:space="preserve">L’on peut donc dire que la ſuperficie des lacs, des <lb/>étangs, & </s> <s xml:id="echoid-s12501" xml:space="preserve">de toutes les eaux qui ne ſont guere agitées, ren-<lb/>ferment une infinité de points de niveau, puiſqu’ils ſont tous <lb/>également éloignés du centre de la terre.</s> <s xml:id="echoid-s12502" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1015" type="section" level="1" n="793"> <head xml:id="echoid-head954" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s12503" xml:space="preserve">758. </s> <s xml:id="echoid-s12504" xml:space="preserve">Ligne de niveau apparent, eſt une ligne telle que B D, <lb/>tangente au cercle de la terre, & </s> <s xml:id="echoid-s12505" xml:space="preserve">par conſéquent perpendicu-<lb/>laire au demi-diametre A B; </s> <s xml:id="echoid-s12506" xml:space="preserve">cette ligne eſt nommée ligne de <lb/>niveau apparent, parce que ſes extrêmités B & </s> <s xml:id="echoid-s12507" xml:space="preserve">D ne ſont pas <pb o="364" file="0418" n="426" rhead="NOUVEAU COURS"/> également éloignées du centre de la terre: </s> <s xml:id="echoid-s12508" xml:space="preserve">ainſi toute ligne <lb/>parallele à l’horizon, & </s> <s xml:id="echoid-s12509" xml:space="preserve">qui étant prolongée par une de ſes <lb/>extrêmités, s’écarte de la ſuperficie de la terre, comme une <lb/>tangente s’écarte de la circonférence d’un cercle, eſt une ligne <lb/>de niveau apparent.</s> <s xml:id="echoid-s12510" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12511" xml:space="preserve">Comme le point B eſt de niveau avec le point C, puiſqu’ils <lb/>ſont également éloignés du centre A de la terre, l’on voit <lb/>qu’il s’en faut toute la ligne C D, que le point B ne ſoit de <lb/>niveau avec le point D; </s> <s xml:id="echoid-s12512" xml:space="preserve">l’on peut donc dire que la ligne C D <lb/>eſt la différence du niveau apparent au deſſus du vrai.</s> <s xml:id="echoid-s12513" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12514" xml:space="preserve">759 Quand une ligne de niveau apparent n’a pas plus de <lb/>100 ou 150 toiſes, il s’en faut ſi peu que ſes extrêmités ne <lb/>ſoient également éloignées du centre de la terre, qu’on peut <lb/>la regarder comme étant parfaitement de niveau; </s> <s xml:id="echoid-s12515" xml:space="preserve">mais ſi elle <lb/>ſurpaſſe cette longueur, il faut avoir égard à la différence du <lb/>niveau apparent au deſſus du vrai, comme nous le ferons voir <lb/>en ſon lieu.</s> <s xml:id="echoid-s12516" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1016" type="section" level="1" n="794"> <head xml:id="echoid-head955" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s12517" xml:space="preserve">Quand on veut niveler deux endroits pour ſçavoir de com-<lb/>bien l’un eſt plus élevé que l’autre, ces deux endroits ſont <lb/>nommés termes, & </s> <s xml:id="echoid-s12518" xml:space="preserve">pour lors l’endroit par où l’on commence <lb/>le nivellement, eſt nommé premier terme, & </s> <s xml:id="echoid-s12519" xml:space="preserve">celui où ſe va ter-<lb/>miner la ligne de niveau apparent, eſt nommé le ſecond terme.</s> <s xml:id="echoid-s12520" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1017" type="section" level="1" n="795"> <head xml:id="echoid-head956" xml:space="preserve">CHAPITRE PREMIER,</head> <head xml:id="echoid-head957" xml:space="preserve">Où l’on donne l’uſage du Niveau d’eau.</head> <p> <s xml:id="echoid-s12521" xml:space="preserve">760. </s> <s xml:id="echoid-s12522" xml:space="preserve">LA principale piece du niveau d’eau eſt un tuyau A B de <lb/> <anchor type="note" xlink:label="note-0418-01a" xlink:href="note-0418-01"/> 5 ou 6 pieds de long, recourbé par ſes extrêmités C & </s> <s xml:id="echoid-s12523" xml:space="preserve">D; </s> <s xml:id="echoid-s12524" xml:space="preserve">ce <lb/>tuyau peut avoir un pouce de diametre: </s> <s xml:id="echoid-s12525" xml:space="preserve">aux extrêmités ſont <lb/>deux bouteilles F C & </s> <s xml:id="echoid-s12526" xml:space="preserve">G D, qui font le principal du niveau: <lb/></s> <s xml:id="echoid-s12527" xml:space="preserve">ces bouteilles, pour être d’un bon uſage, doivent être d’un <lb/>verre fort blanc, bien clair & </s> <s xml:id="echoid-s12528" xml:space="preserve">tranſparent, faites exprès pour <lb/>être plus commodes; </s> <s xml:id="echoid-s12529" xml:space="preserve">car les deux cercles F & </s> <s xml:id="echoid-s12530" xml:space="preserve">G, qui ont <lb/>environ trois pouces de diametre, ſont proprement les culs <lb/>de ces bouteilles, dans le milieu deſquels il y a un trou cir-<lb/>culaire d’environ un pouce: </s> <s xml:id="echoid-s12531" xml:space="preserve">ces bouteilles, qui ont 5 pouces <lb/>de hauteur, ont un petit goulot, dont la groſſeur eſt plus <pb o="365" file="0419" n="427" rhead="DE MATHÉMATIQUE. Liv. X."/> petite que celle du tuyau, parce qu’elles doivent être maſti-<lb/>quées dedans aux extrêmités C & </s> <s xml:id="echoid-s12532" xml:space="preserve">D: </s> <s xml:id="echoid-s12533" xml:space="preserve">dans le milieu du tuyau <lb/>A B eſt une virole avec un genou, qui répond à un bâton <lb/>M N de 4 pieds, de ſorte que le niveau étant poſé à un en-<lb/>droit, on le peut faire tourner en tous ſens, comme ſur un <lb/>pivot ſans bouger le pied.</s> <s xml:id="echoid-s12534" xml:space="preserve"/> </p> <div xml:id="echoid-div1017" type="float" level="2" n="1"> <note position="left" xlink:label="note-0418-01" xlink:href="note-0418-01a" xml:space="preserve">Figure 204.</note> </div> <p> <s xml:id="echoid-s12535" xml:space="preserve">Pour ſe ſervir de cet inſtrument, l’on verſe de l’eau dans <lb/>une des bouteilles, qui va auſſi-tôt ſe communiquer dans l’au-<lb/>tre, à cauſe du tuyau qui eſt ouvert par les deux bouts; </s> <s xml:id="echoid-s12536" xml:space="preserve">& </s> <s xml:id="echoid-s12537" xml:space="preserve"><lb/>quand les bouteilles ont de l’eau environ juſqu’aux deux tiers, <lb/>l’eau donne deux ſurfaces H & </s> <s xml:id="echoid-s12538" xml:space="preserve">I, qui ſont parfaitement de <lb/>niveau. </s> <s xml:id="echoid-s12539" xml:space="preserve">Cela poſé, ſi l’on veut ſçavoir de combien le terme <lb/>Q eſt plus élevé que le terme P, celui qui fait l’opération en-<lb/>voie un aide au ſecond terme Q, où il poſe une toiſe, ou une <lb/>double toiſe, le plus perpendiculairement qu’il eſt poſſible, <lb/>qu’il doit tenir de la main gauche, parce que dans la droite <lb/>il doit avoir un carton blanc de la grandeur d’un cul de cha-<lb/>peau, & </s> <s xml:id="echoid-s12540" xml:space="preserve">dans le milieu duquel on fait un petit rond noir d’un <lb/>pouce de diametre; </s> <s xml:id="echoid-s12541" xml:space="preserve">& </s> <s xml:id="echoid-s12542" xml:space="preserve">ſuppoſant que cet aide ſoit bien inſtruit <lb/>des mouvemens qu’il doit faire, ſoit pour aller ſur la droite ou <lb/>ſur la gauche, ou pour faire monter ou deſcendre le carton le <lb/>long de la toiſe, aux différens ſignes qu’on lui fera, celui <lb/>qui fait l’opération viſe horizontalement aux ſurfaces de l’eau, <lb/>l’endroit de la toiſe qui ſe rencontre dans le rayon de mire <lb/>K L; </s> <s xml:id="echoid-s12543" xml:space="preserve">& </s> <s xml:id="echoid-s12544" xml:space="preserve">ayant fait ſigne à l’aide de faire gliſſer le carton le <lb/>long de la toiſe, pour que le bord ſupérieur du rond noir ſe <lb/>rencontre au point L, on lui fera enſuite un autre ſigne, pour <lb/>lui faire entendre qu’il s’eſt rencontré juſte, & </s> <s xml:id="echoid-s12545" xml:space="preserve">pour lors un <lb/>autre aide, qui eſt avec celui-ci, meſure exactement la hau-<lb/>teur Q L, que je ſuppoſe de 2 pieds 9 pouces, & </s> <s xml:id="echoid-s12546" xml:space="preserve">pendant ce <lb/>tems-là un autre aide, qui ne quitte point celui qui fait l’o-<lb/>pération, meſure la hauteur K P, qui ſera, par exemple, de <lb/>4 pieds 6 pouces: </s> <s xml:id="echoid-s12547" xml:space="preserve">après avoir mis en écrit de part & </s> <s xml:id="echoid-s12548" xml:space="preserve">d’autre <lb/>les hauteurs que l’on aura trouvées, & </s> <s xml:id="echoid-s12549" xml:space="preserve">les deux aides que l’on <lb/>a détachés, étant venus joindre celui qui fait l’opération, l’on <lb/>cherche quelle eſt la différence de la ligne K P à la ligne L Q, <lb/>en les ſouſtrayant l’une de l’autre, & </s> <s xml:id="echoid-s12550" xml:space="preserve">l’on trouve un pied <lb/>9 pouces, qui eſt la hauteur du ſecond terme au deſſus du pre-<lb/>mier: </s> <s xml:id="echoid-s12551" xml:space="preserve">ainſi l’on voit que tout l’objet du nivellement eſt de <lb/>connoître de combien un lieu eſt plus élevé qu’un autre.</s> <s xml:id="echoid-s12552" xml:space="preserve"/> </p> <pb o="366" file="0420" n="428" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s12553" xml:space="preserve">761. </s> <s xml:id="echoid-s12554" xml:space="preserve">Comme les coups de niveau, qui ſe donnent avec cet <lb/>inſtrument, ne vont guere au-delà de 100 à 120 toiſes, l’on <lb/>n’a point égard au niveau apparent dans les petites opérations <lb/>comme celle-ci, parce que le niveau apparent peut être pris <lb/>pour le vrai.</s> <s xml:id="echoid-s12555" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12556" xml:space="preserve">A cauſe de la petite portée des coups de niveau, on eſt <lb/> <anchor type="note" xlink:label="note-0420-01a" xlink:href="note-0420-01"/> obligé d’en donner pluſieurs de diſtance en diſtance, quand <lb/>les objets que l’on veut niveler ſont beaucoup plus éloignés <lb/>l’un de l’autre que l’on ne l’a ſuppoſé ici: </s> <s xml:id="echoid-s12557" xml:space="preserve">cependant quand <lb/>cette diſtance eſt environ double de la portée du coup de ni-<lb/>veau, on peut par une ſeule ſtation trouver la différence des <lb/>hauteurs du niveau de ces deux endroits, pourvu que l’on <lb/>puiſſe les appercevoir tous les deux, quand on ſe ſera placé à <lb/>peu près dans le milieu de leur diſtance.</s> <s xml:id="echoid-s12558" xml:space="preserve"/> </p> <div xml:id="echoid-div1018" type="float" level="2" n="2"> <note position="left" xlink:label="note-0420-01" xlink:href="note-0420-01a" xml:space="preserve">Figure 205.</note> </div> <p> <s xml:id="echoid-s12559" xml:space="preserve">Par exemple, ſuppoſant que la diſtance de A en B ſoit de <lb/> <anchor type="note" xlink:label="note-0420-02a" xlink:href="note-0420-02"/> 220 toiſes, & </s> <s xml:id="echoid-s12560" xml:space="preserve">qu’on veuille ſçavoir de combien le terme A eſt <lb/>plus bas que le terme B, il faut poſer le niveau en C, qui <lb/>ſera à peu près le milieu de la diſtance A B, enſuite viſer de <lb/>D en E, le rond noir du carton que l’aide aura poſé au point <lb/>G, que je ſuppoſe élevé de 2 pieds 4 pouces. </s> <s xml:id="echoid-s12561" xml:space="preserve">Cela poſé, celui <lb/>qui fait l’opération, quitte la bouteille D, & </s> <s xml:id="echoid-s12562" xml:space="preserve">vient à la bou-<lb/>teille E, pour viſer de E en F, parce qu’il doit y avoir à l’en-<lb/>droit A un autre aide pour tenir la toiſe & </s> <s xml:id="echoid-s12563" xml:space="preserve">le carton: </s> <s xml:id="echoid-s12564" xml:space="preserve">& </s> <s xml:id="echoid-s12565" xml:space="preserve">comme <lb/>il peut arriver que la ligne A F ſoit élevé au deſſus de l’endroit <lb/>A de plus de 6 pieds, en ce cas on a une autre toiſe, au bout <lb/>de laquelle eſt un carton, comme celui dont nous avons déja <lb/>parlé, & </s> <s xml:id="echoid-s12566" xml:space="preserve">l’aide fait gliſſer cette toiſe le long de l’autre, la <lb/>faiſant monter & </s> <s xml:id="echoid-s12567" xml:space="preserve">deſcendre tant que le rond noir du carton <lb/>ſe rencontre dans le rayon de mire E F; </s> <s xml:id="echoid-s12568" xml:space="preserve">après quoi un autre <lb/>aide meſure exactement la hauteur F A. </s> <s xml:id="echoid-s12569" xml:space="preserve">Or ſuppoſant qu’ayant <lb/>meſuré avec autant de préciſion qu’il eſt poſſible, l’on ait trouvé <lb/>9 pieds 6 pouces pour la hauteur A F, on ſouſtraira de cette <lb/>quantité 2 pieds 4 pouces, qui eſt l’élevation du point G, & </s> <s xml:id="echoid-s12570" xml:space="preserve">la <lb/>différence ſera 7 pieds 2 pouces, qui fait voir que l’endroit A <lb/>eſt plus bas que B de 7 pieds 2 pouces.</s> <s xml:id="echoid-s12571" xml:space="preserve"/> </p> <div xml:id="echoid-div1019" type="float" level="2" n="3"> <note position="left" xlink:label="note-0420-02" xlink:href="note-0420-02a" xml:space="preserve">Figure 205.</note> </div> <p> <s xml:id="echoid-s12572" xml:space="preserve">Cette maniere de niveler eſt la meilleure de toutes, parce <lb/>qu’elle eſt moins ſujette à erreur, ſoit de la part du niveau ap-<lb/>parent, ou des réfractions; </s> <s xml:id="echoid-s12573" xml:space="preserve">car tant que le point C ſera dans <lb/>le milieu de deux termes, les points F & </s> <s xml:id="echoid-s12574" xml:space="preserve">G ſeront parfaite-<lb/>ment de niveau, puiſqu’ils ſont également éloignés du centre <pb o="367" file="0421" n="429" rhead="DE MATHÉMATIQUE. Liv. X."/> de la terre: </s> <s xml:id="echoid-s12575" xml:space="preserve">d’ailleurs par cette pratique, on fait beaucoup <lb/>moins de ſtations que ſi l’on alloit par pluſieurs coups de ni-<lb/>veau d’un terme à l’autre.</s> <s xml:id="echoid-s12576" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1021" type="section" level="1" n="796"> <head xml:id="echoid-head958" xml:space="preserve">CHAPITRE II,</head> <head xml:id="echoid-head959" style="it" xml:space="preserve">Où l’on donne la maniere de faire le Nivellement compoſé.</head> <p> <s xml:id="echoid-s12577" xml:space="preserve">762. </s> <s xml:id="echoid-s12578" xml:space="preserve">QUand les deux termes que l’on veut niveler ſont beau-<lb/> <anchor type="note" xlink:label="note-0421-01a" xlink:href="note-0421-01"/> coup plus éloignés l’un de l’autre qu’on l’a ſuppoſé dans l’opé-<lb/>ration précédente, on eſt obligé de faire pluſieurs ſtations; <lb/></s> <s xml:id="echoid-s12579" xml:space="preserve">& </s> <s xml:id="echoid-s12580" xml:space="preserve">en ce cas l’on dit que le nivellement eſt compoſé: </s> <s xml:id="echoid-s12581" xml:space="preserve">car en <lb/>effet il eſt compoſé de pluſieurs coups de niveau, que l’on fait <lb/>enſorte d’abréger, comme on le va voir dans l’opération ſui-<lb/>vante.</s> <s xml:id="echoid-s12582" xml:space="preserve"/> </p> <div xml:id="echoid-div1021" type="float" level="2" n="1"> <note position="right" xlink:label="note-0421-01" xlink:href="note-0421-01a" xml:space="preserve">Figure 206.</note> </div> <p> <s xml:id="echoid-s12583" xml:space="preserve">Pour niveler deux objets A & </s> <s xml:id="echoid-s12584" xml:space="preserve">B, éloignés l’un de l’autre de <lb/>680 toiſes, il faut diviſer ce nombre par 200 ou 220 toiſes, <lb/>pour voir combien l’on ſera obligé de faire de ſtations: </s> <s xml:id="echoid-s12585" xml:space="preserve">car <lb/>dans l’opération précédente on a nivelé par une ſeule ſtation <lb/>une diſtance de 220 toiſes: </s> <s xml:id="echoid-s12586" xml:space="preserve">ainſi comme 680, diviſé par 220, <lb/>donne 3 au quotient, je vois qu’en trois ſtations on peut ni-<lb/>veler les deux termes A & </s> <s xml:id="echoid-s12587" xml:space="preserve">B. </s> <s xml:id="echoid-s12588" xml:space="preserve">Pour cela, je commence par <lb/>chercher dans la diſtance A B les trois endroits qui ſont les <lb/>plus commodes pour faire les ſtations: </s> <s xml:id="echoid-s12589" xml:space="preserve">je choiſis d’abord le <lb/>point C à peu près dans le milieu de A B, où je fais planter un <lb/>piquet, & </s> <s xml:id="echoid-s12590" xml:space="preserve">à une diſtance de 100 ou 110 toiſes du point A j’en <lb/>fais planter un autre en D, & </s> <s xml:id="echoid-s12591" xml:space="preserve">à la même diſtance du point B <lb/>j’en fais placer un troiſieme E, & </s> <s xml:id="echoid-s12592" xml:space="preserve">autant qu’il ſe peut, il faut <lb/>que ces trois piquets ſoient d’alignement avec les deux termes <lb/>A & </s> <s xml:id="echoid-s12593" xml:space="preserve">B. </s> <s xml:id="echoid-s12594" xml:space="preserve">Ayant donc déterminé les trois ſtations D, C, E, il <lb/>faut envoyer deux aides au premier terme A, dont le premier <lb/>porte une ou deux toiſes, & </s> <s xml:id="echoid-s12595" xml:space="preserve">le ſecond ſoit chargé d’écrire les <lb/>hauteurs; </s> <s xml:id="echoid-s12596" xml:space="preserve">on en envoie un troiſieme à peu près dans le milieu <lb/>de la diſtance D C, lequel ne doit point bouger de ſa place, <lb/>qu’on n’ait achevé les opérations de la premiere & </s> <s xml:id="echoid-s12597" xml:space="preserve">de la ſe-<lb/>conde ſtation, parce que la toiſe qu’il tiendra en main doit <lb/>ſervir de terme commun pour les deux premieres ſtations.</s> <s xml:id="echoid-s12598" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12599" xml:space="preserve">Ayant donc fait porter le niveau au point D, il faut viſer <lb/>de T en S, pour que le rayon de mire T M aille rencontrer <pb o="368" file="0422" n="430" rhead="NOUVEAU COURS"/> le bord ſupérieur du rond noir, qui ſera au point M, & </s> <s xml:id="echoid-s12600" xml:space="preserve">le <lb/>ſecond Aide meſure la hauteur M A, que je ſuppoſe de <lb/>8 pieds 2 pouces, qu’il a ſoin d’écrire ſur des tablettes: </s> <s xml:id="echoid-s12601" xml:space="preserve">enſuite <lb/>on viſe de S en T, pour découvrir le rond noir au point K; <lb/></s> <s xml:id="echoid-s12602" xml:space="preserve">& </s> <s xml:id="echoid-s12603" xml:space="preserve">comme il n’eſt pas néceſſaire de connoître la hauteur K F, <lb/>qui ſeroit plus embarraſſante qu’utile, l’Aide qui tient la toiſe <lb/>ſe contente de marquer un trait de crayon à l’endroit de la <lb/>toiſe où le rayon de mire S K s’eſt terminé: </s> <s xml:id="echoid-s12604" xml:space="preserve">delà on vient à <lb/>la ſeconde ſtation C, & </s> <s xml:id="echoid-s12605" xml:space="preserve">on envoie à peu prés dans le milieu <lb/>de la diſtance C E un Aide à l’endroit G, qui ne doit pas <lb/>bouger de ſa place, que les opérations de la ſeconde & </s> <s xml:id="echoid-s12606" xml:space="preserve">de la <lb/>troiſieme ſtation ne ſoient finies. </s> <s xml:id="echoid-s12607" xml:space="preserve">Préſentement il faut donner <lb/>un coup de niveau de Q en R, pour découvrir le point L du <lb/>rond noir; </s> <s xml:id="echoid-s12608" xml:space="preserve">& </s> <s xml:id="echoid-s12609" xml:space="preserve">quand on l’aura rencontré, on meſurera la <lb/>hauteur K L, qui eſt la diſtance du trait de crayon que l’on a <lb/>marqué ſur la toiſe au point L, & </s> <s xml:id="echoid-s12610" xml:space="preserve">celui qui tenoit les tablettes <lb/>à l’endroit A, a eu ſoin de ſe rendre à la ſeconde ſtation, pour <lb/>y écrire la hauteur K L, qui ſera, par exemple, de 3 pieds <lb/>6 pouces: </s> <s xml:id="echoid-s12611" xml:space="preserve">après cela il faut viſer de R en Q, pour que celui <lb/>qui eſt en G puiſſe marquer ſur la toiſe le point H par un <lb/>trait de crayon, ſans s’embarraſſer de ſon élévation, qu’il eſt <lb/>inutile de connoître, comme nous l’avons déja dit. </s> <s xml:id="echoid-s12612" xml:space="preserve">Enfin, <lb/>l’on fait porter le niveau à la troiſieme ſtation E, pour donner <lb/>un coup de niveau de P en O, qui déterminera enſuite le point I; </s> <s xml:id="echoid-s12613" xml:space="preserve"><lb/>on meſurera la ligne H I, que je ſuppoſe de 4 pieds 3 pouces, <lb/>qu’on aura ſoin d’écrire ſur les tablettes; </s> <s xml:id="echoid-s12614" xml:space="preserve">après quoi on donnera <lb/>le dernier coup de niveau O N, & </s> <s xml:id="echoid-s12615" xml:space="preserve">l’Aide qui eſt en B, me-<lb/>ſurera la hauteur B N, que je ſuppoſe d’un pied 6 pouces, <lb/>qu’il faudra écrire à part, parce que cette hauteur n’a rien de <lb/>commun avec ce que l’on a marqué ſur les tablettes.</s> <s xml:id="echoid-s12616" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12617" xml:space="preserve">Le nivellement étant achevé, l’on ajoutera enſemble les <lb/>hauteurs que l’on a écrites ſur les tablettes, c’eſt-à-dire 8 pieds <lb/>2 pouces, 3 pieds 6 pouces, & </s> <s xml:id="echoid-s12618" xml:space="preserve">4 pieds 3 pouces, qui font <lb/>15 pieds 11 pouces, d’où il faudra ſouſtraire la hauteur B N <lb/>d’un pied 6 pouces, & </s> <s xml:id="echoid-s12619" xml:space="preserve">la différence ſera 14 pieds 5 pouces, <lb/>qui eſt l’élévation de l’endroit B au deſſus de l’endroit A.</s> <s xml:id="echoid-s12620" xml:space="preserve"/> </p> <pb file="0423" n="431"/> <pb file="0423a" n="432"/> <figure> <image file="0423a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0423a-01"/> </figure> <pb file="0424" n="433"/> <pb file="0425" n="434"/> <pb file="0425a" n="435"/> <figure> <image file="0425a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0425a-01"/> </figure> <pb file="0426" n="436"/> <pb file="0427" n="437"/> <pb file="0427a" n="438"/> <figure> <image file="0427a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0427a-01"/> </figure> <pb file="0428" n="439"/> <pb file="0429" n="440"/> <pb file="0429a" n="441"/> <figure> <image file="0429a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0429a-01"/> </figure> <pb file="0430" n="442"/> <pb file="0431" n="443"/> <pb file="0431a" n="444"/> <figure> <image file="0431a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0431a-01"/> </figure> <pb file="0432" n="445"/> <pb file="0433" n="446"/> <pb file="0433a" n="447"/> <figure> <image file="0433a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0433a-01"/> </figure> <pb file="0434" n="448"/> <pb o="369" file="0435" n="449" rhead="DE MATHÉMATIQUE. Liv. X."/> </div> <div xml:id="echoid-div1023" type="section" level="1" n="797"> <head xml:id="echoid-head960" xml:space="preserve">CHAPITRE III,</head> <head xml:id="echoid-head961" style="it" xml:space="preserve">Où l’on donne la maniere de niveler deux termes, entre leſquels il <lb/>ſe trouve des hauteurs & des fonds.</head> <p> <s xml:id="echoid-s12621" xml:space="preserve">763. </s> <s xml:id="echoid-s12622" xml:space="preserve">QUand on veut niveler deux objets fort éloignés l’un <lb/> <anchor type="note" xlink:label="note-0435-01a" xlink:href="note-0435-01"/> de l’autre, il eſt aſſez rare qu’on ne rencontre en chemin des <lb/>hauteurs & </s> <s xml:id="echoid-s12623" xml:space="preserve">des fonds, qui obligent de niveler tantôt en mon-<lb/>tant, tantôt en deſcendant: </s> <s xml:id="echoid-s12624" xml:space="preserve">en ce cas, il faut obſerver cer-<lb/>taines choſes dont nous n’avons pas encore parlé, qui ſont, <lb/>d’écrire ſur les tablettes dans une colonne toutes les hauteurs <lb/>que l’on trouvera en montant, & </s> <s xml:id="echoid-s12625" xml:space="preserve">dans une autre colonne <lb/>toutes celles que l’on trouvera en deſcendant; </s> <s xml:id="echoid-s12626" xml:space="preserve">& </s> <s xml:id="echoid-s12627" xml:space="preserve">pour les diſ-<lb/>tinguer à l’avenir, nous nommerons premiere colonne celle <lb/>dans laquelle il faudra écrire les hauteurs que l’on trouvera en <lb/>montant, & </s> <s xml:id="echoid-s12628" xml:space="preserve">ſeconde colonne celle dans laquelle on écrira <lb/>toutes les hauteurs que l’on trouvera en deſcendant. </s> <s xml:id="echoid-s12629" xml:space="preserve">L’on va <lb/>voir ceci dans l’opération ſuivante.</s> <s xml:id="echoid-s12630" xml:space="preserve"/> </p> <div xml:id="echoid-div1023" type="float" level="2" n="1"> <note position="right" xlink:label="note-0435-01" xlink:href="note-0435-01a" xml:space="preserve">Figure 207.</note> </div> <p> <s xml:id="echoid-s12631" xml:space="preserve">Pour niveler deux lieux A & </s> <s xml:id="echoid-s12632" xml:space="preserve">B, il faut commencer par poſer <lb/>le niveau au point D, éloigné d’environ 100 toiſes des endroits <lb/>A & </s> <s xml:id="echoid-s12633" xml:space="preserve">3, où l’on aura envoyé des Aides avec des toiſes; </s> <s xml:id="echoid-s12634" xml:space="preserve">enſuite <lb/>il faut donner les coups de niveau D C & </s> <s xml:id="echoid-s12635" xml:space="preserve">D E, & </s> <s xml:id="echoid-s12636" xml:space="preserve">écrire la <lb/>hauteur A C de 9 pieds 4 pouces dans la premiere colonne, & </s> <s xml:id="echoid-s12637" xml:space="preserve"><lb/>marquer un trait de crayon à l’endroit E: </s> <s xml:id="echoid-s12638" xml:space="preserve">delà il faut faire <lb/>porter le niveau au point 4, qui n’eſt pas dans le milieu de la <lb/>ligne F H, à cauſe que la rampe de trois en 5 ne le permet <lb/>point, mais cela n’empêche pas que les coups de niveau G F <lb/>& </s> <s xml:id="echoid-s12639" xml:space="preserve">G H ne ſoient juſtes, parce qu’ils ne ſont pas d’une grande <lb/>portée. </s> <s xml:id="echoid-s12640" xml:space="preserve">Ayant donc déterminé les points F & </s> <s xml:id="echoid-s12641" xml:space="preserve">H, il faut me-<lb/>ſurer la hauteur F E, qui ſera, par exemple, de 9 pieds 6 pouces, <lb/>qu’il faut écrire dans la premiere colonne, & </s> <s xml:id="echoid-s12642" xml:space="preserve">ne pas oublier de <lb/>marquer un trait de crayon au point H de la toiſe 5: </s> <s xml:id="echoid-s12643" xml:space="preserve">delà il <lb/>faut venir à la ſtation 6, & </s> <s xml:id="echoid-s12644" xml:space="preserve">donner les coups de niveau K I & </s> <s xml:id="echoid-s12645" xml:space="preserve"><lb/>K L; </s> <s xml:id="echoid-s12646" xml:space="preserve">l’on marquera, comme à l’ordinaire, un trait de crayon <lb/>au point L, & </s> <s xml:id="echoid-s12647" xml:space="preserve">l’on écrira dans la premiere colonne la hauteur <lb/>I H, que je ſuppoſe de 7 pieds: </s> <s xml:id="echoid-s12648" xml:space="preserve">delà on viendra à la ſtation 8, <lb/>de laquelle je ſuppoſe qu’on ne peut donner que le coup de <lb/>niveau N M, à cauſe que la rampe eſt trop grande pour pouvoir <pb o="370" file="0436" n="450" rhead="NOUVEAU COURS"/> en donner un ſecond de l’autre côté, l’on meſurera la hau-<lb/>teur L M depuis le point L, que l’on a marqué ſur la toiſe <lb/>juſqu’au point M du rayon de mire, qui ſe trouve de 8 pieds <lb/>2 pouces; </s> <s xml:id="echoid-s12649" xml:space="preserve">l’on aura ſoin de l’écrire dans la ſeconde colonne, <lb/>parce que c’eſt une hauteur que l’on a trouvée en deſcendant:</s> <s xml:id="echoid-s12650" xml:space="preserve"><unsure/> <lb/>mais comme la hauteur N O du niveau fait voir de combien le <lb/>point O eſt plus bas que le point M, il faudra meſurer cette <lb/>hauteur, que je ſuppoſe de 4 pieds & </s> <s xml:id="echoid-s12651" xml:space="preserve">demi, pour l’écrire auſſi <lb/>dans la ſeconde colonne; </s> <s xml:id="echoid-s12652" xml:space="preserve">enſuite il faudra faire planter un <lb/>piquet à l’endroit O, & </s> <s xml:id="echoid-s12653" xml:space="preserve">deſcendre le niveau au point 9, qu’il <lb/>faudra trouver; </s> <s xml:id="echoid-s12654" xml:space="preserve">de ſorte que le rayon de mire P O aille ren-<lb/>contrer le pied du piquet: </s> <s xml:id="echoid-s12655" xml:space="preserve">après quoi l’on donnera le coup de <lb/>niveau P Q, & </s> <s xml:id="echoid-s12656" xml:space="preserve">l’Aide qui tient la toiſe aura ſoin de marquer <lb/>un trait de crayon au point Q. </s> <s xml:id="echoid-s12657" xml:space="preserve">Delà on ira à la ſtation 11, <lb/>pour y donner les coups de niveau S R & </s> <s xml:id="echoid-s12658" xml:space="preserve">S T, afin d’avoir la <lb/>hauteur R Q, qui ſera, par exemple, de 3 pieds, qu’il faudra <lb/>écrire dans la premiere colonne, parce que c’eſt une hauteur <lb/>que l’on a trouvée en montant; </s> <s xml:id="echoid-s12659" xml:space="preserve">il faut aller après cela au <lb/>point 13, pour y donner les coups de niveau X V, X Y, & </s> <s xml:id="echoid-s12660" xml:space="preserve">l’on <lb/>écrira dans la premiere colonne la hauteur V T, qu’on ſuppoſe <lb/>de 5 pieds 5 pouces: </s> <s xml:id="echoid-s12661" xml:space="preserve">& </s> <s xml:id="echoid-s12662" xml:space="preserve">comme il arrive que le rayon X Y va <lb/>ſe terminer à un point Y de la hauteur, il n’y aura pas de trait <lb/>de crayon à marquer ſur la toiſe à cet endroit-là, on y laiſſera <lb/>ſeulement un Aide pour ſervir à l’opération 15, laquelle ayant <lb/>déterminé les points Z & </s> <s xml:id="echoid-s12663" xml:space="preserve">B, des coups de niveau A Z & </s> <s xml:id="echoid-s12664" xml:space="preserve">A B, <lb/>l’on meſurera la hauteur Z Y, que je ſuppoſe de 7 pieds 4 pouces, <lb/>qu’il faudra encore écrire dans la premiere colonne: </s> <s xml:id="echoid-s12665" xml:space="preserve">delà il <lb/>faut venir à la ſtation 17, pour y donner les coups de niveau <lb/>D C & </s> <s xml:id="echoid-s12666" xml:space="preserve">D E, marquer un trait de crayon au point E, & </s> <s xml:id="echoid-s12667" xml:space="preserve">conſi-<lb/>dérer que la hauteur B C, qu’on ſuppoſe de 6 pieds 6 pouces; <lb/></s> <s xml:id="echoid-s12668" xml:space="preserve">a été trouvée en deſcendant, & </s> <s xml:id="echoid-s12669" xml:space="preserve">que par conſéquent il faut <lb/>l’écrire dans la ſeconde colonne. </s> <s xml:id="echoid-s12670" xml:space="preserve">Enfin l’on portera le niveau <lb/>à la derniere ſtation B, pour déterminer par le rayon G F la <lb/>hauteur E F, qui ſera, par exemple, de 8 pieds 5 pouces, qu’il <lb/>faudra écrire dans la ſeconde colonne, auſſi-bien que la hau-<lb/>teur G B du niveau, qui eſt ordinairement de 4 pieds 6 pouces.</s> <s xml:id="echoid-s12671" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12672" xml:space="preserve">Préſentement ſi l’on additionne les nombres de la premiere <lb/>colonne, l’on trouvera 41 pieds 7 pouces; </s> <s xml:id="echoid-s12673" xml:space="preserve">& </s> <s xml:id="echoid-s12674" xml:space="preserve">faiſant la même <lb/>choſe pour la ſeconde, l’on aura 32 pieds un pouce. </s> <s xml:id="echoid-s12675" xml:space="preserve">Or ſi l’on <lb/>retranche la plus petite ſomme de la plus grande, c’eſt-à-dire <pb o="371" file="0437" n="451" rhead="DE MATHÉMATIQUE. Liv. X."/> 32 pieds un pouce, de 41 pieds 7 pouces, la différence ſera <lb/>9 pieds 6 pouces, qui fait voir que le terme A eſt plus bas que <lb/>le terme B de 9 pieds 9 pouces.</s> <s xml:id="echoid-s12676" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12677" xml:space="preserve">Il eſt bon de remarquer que lorſque l’on a un nivellement <lb/>à faire en montant, & </s> <s xml:id="echoid-s12678" xml:space="preserve">qu’on s’apperçoit que les coups de ni-<lb/>veau ſont trop courts, de ſorte qu’on eſt obligé d’en donner <lb/>trop ſouvent, il vaut mieux monter au ſommet de la hauteur, <lb/>& </s> <s xml:id="echoid-s12679" xml:space="preserve">faire le nivellement en deſcendant, obſervant d’écrire dans <lb/>la premiere colonne les hauteurs que l’on trouvera en allant <lb/>vers un terme, & </s> <s xml:id="echoid-s12680" xml:space="preserve">dans la ſeconde colonne, celles que l’on <lb/>trouvera en allant vers l’autre.</s> <s xml:id="echoid-s12681" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12682" xml:space="preserve">Par exemple, ſi l’on veut connoître la différence des hau-<lb/>teurs de deux termes A & </s> <s xml:id="echoid-s12683" xml:space="preserve">B, & </s> <s xml:id="echoid-s12684" xml:space="preserve">qu’on s’apperçoive qu’il fau-<lb/>dra trop de tems & </s> <s xml:id="echoid-s12685" xml:space="preserve">trop d’opérations pour niveler de A en B <lb/>par une ſuite de coups de niveau, on fera porter le niveau à <lb/>l’endroit 6, que je ſuppoſe être le ſommet de la hauteur, & </s> <s xml:id="echoid-s12686" xml:space="preserve"><lb/>l’on nivellera de 6 en A, en obſervant d’écrire dans la pre-<lb/>miere colonne les hauteurs que l’on trouvera; </s> <s xml:id="echoid-s12687" xml:space="preserve">après cela l’on <lb/>viendra à l’endroit 6 pour niveler de 6 en 10, & </s> <s xml:id="echoid-s12688" xml:space="preserve">les hauteurs <lb/>que l’on trouvera, on les écrira dans la ſeconde colonne. </s> <s xml:id="echoid-s12689" xml:space="preserve">Enfin <lb/>on viendra au ſommet 15 de la ſeconde éminence, pour ni-<lb/>veler de 15 en 10, mettant toutes les hauteurs que l’on trou-<lb/>vera dans la premiere colonne; </s> <s xml:id="echoid-s12690" xml:space="preserve">après quoi l’on nivellera de <lb/>15 en B, & </s> <s xml:id="echoid-s12691" xml:space="preserve">on écrira les hauteurs de cette derniere opération <lb/>dans la ſeconde colonne, & </s> <s xml:id="echoid-s12692" xml:space="preserve">le reſte ſera comme dans l’opé-<lb/>ration précédente.</s> <s xml:id="echoid-s12693" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12694" xml:space="preserve">L’on peut faire beaucoup d’ouvrage en peu de tems par cette <lb/>maniere de niveler, parce que tandis qu’une perſonne enten-<lb/>due fait le nivellement de 6 en A, une autre peut faire celui <lb/>de 6 en 10; </s> <s xml:id="echoid-s12695" xml:space="preserve">& </s> <s xml:id="echoid-s12696" xml:space="preserve">de la même façon celui de 15 en 10, & </s> <s xml:id="echoid-s12697" xml:space="preserve">de 15 <lb/>en B.</s> <s xml:id="echoid-s12698" xml:space="preserve"/> </p> <pb o="372" file="0438" n="452" rhead="NOUVEAU COURS"/> <note position="right" xml:space="preserve"># # # <emph style="sc">Premiere colonne</emph>. <lb/>Pieds. # Pouces. # Lignes. <lb/>9<emph style="sub">//--//</emph> # 4<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>9<emph style="sub">//--//</emph> # 6<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>7<emph style="sub">//--//</emph> # 0<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>3<emph style="sub">//--//</emph> # 0<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>5<emph style="sub">//--//</emph> # 5<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>7<emph style="sub">//--//</emph> # 4<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>41 pieds<emph style="sub">//</emph> # 7 pou<emph style="sub">//</emph> # 0 lig.<emph style="sub">//</emph> <lb/></note> <note position="right" xml:space="preserve"># # # <emph style="sc">Seconde colonne</emph>. <lb/>Pieds. # Pouces. # Lignes. <lb/>8<emph style="sub">//--//</emph> # 2<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>4<emph style="sub">//--//</emph> # 6<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>6<emph style="sub">//--//</emph> # 6<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>8<emph style="sub">//--//</emph> # 5<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>4<emph style="sub">//--//</emph> # 6<emph style="sub">//--//</emph> # 0<emph style="sub">//</emph> <lb/>32 pieds<emph style="sub">//</emph> # 1 pouce<emph style="sub">//</emph> # 0 lig.<emph style="sub">//</emph> <lb/></note> <note position="right" xml:space="preserve">pieds. # pouces. <lb/>41<emph style="sub">//--//</emph> # 7<emph style="sub">//--</emph>0<emph style="sub">//</emph> <lb/>32<emph style="sub">//--//</emph> # 1<emph style="sub">//--</emph>0<emph style="sub">//</emph> <lb/>Différence # 9 pieds # 6 pouces. <lb/></note> </div> <div xml:id="echoid-div1025" type="section" level="1" n="798"> <head xml:id="echoid-head962" xml:space="preserve">CHAPITRE IV,</head> <head xml:id="echoid-head963" style="it" xml:space="preserve">Qù l’on fait voir la maniere de connoître de combien le Niveau <lb/>apparent eſt élevé au deſſus du vrai, pour une ligne de telle <lb/>longueur que l’on voudra.</head> <p> <s xml:id="echoid-s12699" xml:space="preserve">764. </s> <s xml:id="echoid-s12700" xml:space="preserve">L’On n’a pas eu égard à la différence du niveau appa-<lb/>rent au deſſus du vrai dans les nivellemens que nous venons <lb/>d’enſeigner, parce que les coups de niveau étoient fortpetits; <lb/></s> <s xml:id="echoid-s12701" xml:space="preserve">d’ailleurs les opérations ont été faites d’une maniere à ne pas <lb/>donner lieu à cette différence: </s> <s xml:id="echoid-s12702" xml:space="preserve">mais comme le niveau d’eau <lb/>ne peut ſervir que pour des petits nivellemens, & </s> <s xml:id="echoid-s12703" xml:space="preserve">qu’il de-<lb/>mande une grande exactitude, pour ne point faire d’erreur, <lb/>quand le nivellement eſt fort compoſé, on a inventé une autre <lb/>eſpece de niveau, avec lequel, par le moyen d’une lunette, <lb/>l’on peut donner des coups de niveau extrêmement grands; </s> <s xml:id="echoid-s12704" xml:space="preserve"><lb/>c’eſt l’uſage de ce niveau que nous allons enſeigner, après <lb/>avoir donné dans ce chapitre la maniere de calculer la hauteur <lb/>du niveau apparent au deſſus du vrai, dont la connoiſſance <lb/>eſt abſolument néceſſaire, quand on fait de grands nivel-<lb/>lemens.</s> <s xml:id="echoid-s12705" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12706" xml:space="preserve">765. </s> <s xml:id="echoid-s12707" xml:space="preserve">Nous avons vu dans la Géométrie que le quarré de <pb o="373" file="0439" n="453" rhead="DE MATHÉMATIQUE. Liv. X."/> la tangente B D étoit égal au rectangle compris ſous la ſécante <lb/> <anchor type="note" xlink:label="note-0439-01a" xlink:href="note-0439-01"/> G D, & </s> <s xml:id="echoid-s12708" xml:space="preserve">ſous la partie C D: </s> <s xml:id="echoid-s12709" xml:space="preserve">ainſi diviſant le quarré de la <lb/>ligne B D par la valeur de la ligne G D, on trouvera la ligne <lb/>C D. </s> <s xml:id="echoid-s12710" xml:space="preserve">Mais comme la ligne G C, qui eſt le diametre de la terre, <lb/>a été trouvée de 6538594 toiſes, elle ne differe de la ligne G D <lb/>que d’une quantité infiniment petite, il s’enſuit que l’on pourra <lb/>prendre la ligne G C pour la ligne G D, & </s> <s xml:id="echoid-s12711" xml:space="preserve">que diviſant le <lb/>quarré de la ligne B D par le diametre G C de la terre, c’eſt-<lb/>à-dire par 6538594, l’on aura la valeur de la ligne C D, qui <lb/>eſt la différence du niveau apparent avec le vrai. </s> <s xml:id="echoid-s12712" xml:space="preserve">Or ſuppoſant <lb/>que la ligne de niveau apparent B D ſoit de 800 toiſes, il fau-<lb/>dra les réduire en lignes, & </s> <s xml:id="echoid-s12713" xml:space="preserve">l’on aura 691200 lignes, qu’il faut <lb/>enſuite quarrer pour avoir 477754440000, qui eſt le quarré <lb/>de la ligne B D. </s> <s xml:id="echoid-s12714" xml:space="preserve">Préſentement ſi l’on réduit le diametre de <lb/>la terre, qui eſt de 6538594 toiſes en lignes, on aura 5649345216 <lb/>lignes; </s> <s xml:id="echoid-s12715" xml:space="preserve">& </s> <s xml:id="echoid-s12716" xml:space="preserve">diviſant le quarré de la ligne B D par le nombre pré-<lb/>cédent, l’on aura environ 85 lignes, qui font 7 pouces une <lb/>ligne, pour la différence C D du niveau apparent au deſſus du <lb/>vrai.</s> <s xml:id="echoid-s12717" xml:space="preserve"/> </p> <div xml:id="echoid-div1025" type="float" level="2" n="1"> <note position="right" xlink:label="note-0439-01" xlink:href="note-0439-01a" xml:space="preserve">Figure 203.</note> </div> <p> <s xml:id="echoid-s12718" xml:space="preserve">766. </s> <s xml:id="echoid-s12719" xml:space="preserve">L’on peut encore d’une maniere plus géométrique <lb/>que la précédente, trouver la valeur C D du niveau apparent <lb/>au deſſus du vrai: </s> <s xml:id="echoid-s12720" xml:space="preserve">car à cauſe du triangle rectangle A B D, les <lb/>quarrés A B & </s> <s xml:id="echoid-s12721" xml:space="preserve">B D, pris enſemble, valent le quarré de l’hy-<lb/>poténuſe A D. </s> <s xml:id="echoid-s12722" xml:space="preserve">Ainſi il n’y a qu’à quarrer la valeur du demi-<lb/>diametre de la terre, & </s> <s xml:id="echoid-s12723" xml:space="preserve">la valeur de B D de la ligne de niveau <lb/>apparent, & </s> <s xml:id="echoid-s12724" xml:space="preserve">additionner ces deux quarrés, dont la racine ſera <lb/>la ligne A D, de laquelle il faudra retrancher la valeur du <lb/>demi-diametre A B ou A C de la terre, & </s> <s xml:id="echoid-s12725" xml:space="preserve">la différence ſera <lb/>la valeur de la ligne C D.</s> <s xml:id="echoid-s12726" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12727" xml:space="preserve">767. </s> <s xml:id="echoid-s12728" xml:space="preserve">L’on peut remarquer que les hauteurs de deux points <lb/>de niveau apparent au deſſus du vrai, ſont dans la même raiſon <lb/>que les quarrés des lignes des niveaux apparens; </s> <s xml:id="echoid-s12729" xml:space="preserve">car prenant <lb/>le diametre G C pour la ligne G D, & </s> <s xml:id="echoid-s12730" xml:space="preserve">le diametre H K pour <lb/>la ligne H I, le quarré de la ligne B I étant auſſi égal au rec-<lb/>tangle compris ſous H K & </s> <s xml:id="echoid-s12731" xml:space="preserve">K I, les quarrés des lignes B D & </s> <s xml:id="echoid-s12732" xml:space="preserve"><lb/>B I ſeront dans la même raiſon que les rectangles qui leur ſont <lb/>égaux: </s> <s xml:id="echoid-s12733" xml:space="preserve">mais ces rectangles ayant chacun pour baſe le dia-<lb/>metre G C ou H K de la terre, ſeront comme leurs hauteurs <lb/>C D & </s> <s xml:id="echoid-s12734" xml:space="preserve">K I: </s> <s xml:id="echoid-s12735" xml:space="preserve">ainſi les quarrés B D & </s> <s xml:id="echoid-s12736" xml:space="preserve">B I ſeront donc dans la <lb/>raiſon des lignes C D & </s> <s xml:id="echoid-s12737" xml:space="preserve">K I.</s> <s xml:id="echoid-s12738" xml:space="preserve"/> </p> <pb o="374" file="0440" n="454" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s12739" xml:space="preserve">768. </s> <s xml:id="echoid-s12740" xml:space="preserve">L’on peut tirer de cette conſéquence une regle gé-<lb/>nérale pour trouver la hauteur du niveau apparent au deſſus <lb/>du vrai, d’une façon bien plus courte, que par les deux mé-<lb/>thodes précédentes: </s> <s xml:id="echoid-s12741" xml:space="preserve">car ſi on connoît une fois la hauteur du <lb/>niveau apparent au deſſus du vrai pour une ligne d’une certaine <lb/>longueur, l’on pourra trouver la même choſe pour toutes les <lb/>autres.</s> <s xml:id="echoid-s12742" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12743" xml:space="preserve">Par exemple, étant prévenu que pour une diſtance de 600 <lb/>toiſes, le niveau apparent eſt élevé au deſſus du vrai de 4 pouces, <lb/>pour ſçavoir combien il eſt élevé pour une diſtance de 1000 <lb/>toiſes, je fais une Regle de Trois, en diſant: </s> <s xml:id="echoid-s12744" xml:space="preserve">Si le quarré de <lb/>600, qui eſt 360000, donne 4 pouces, combien donnera le <lb/>quarré de 1000, qui eſt 1000000? </s> <s xml:id="echoid-s12745" xml:space="preserve">La Regle étant faite, on <lb/>trouvera 11 pouces une ligne 4 points pour la hauteur du ni-<lb/>veau apparent au deſſus du vrai, d’un coup de niveau de <lb/>1000 toiſes.</s> <s xml:id="echoid-s12746" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1027" type="section" level="1" n="799"> <head xml:id="echoid-head964" xml:space="preserve">CHAPITRE V,</head> <head xml:id="echoid-head965" style="it" xml:space="preserve">Où l’on fait la deſcription du Niveau de M. Huyghens.</head> <p> <s xml:id="echoid-s12747" xml:space="preserve">769. </s> <s xml:id="echoid-s12748" xml:space="preserve">NOus n’avons parlé juſqu’à préſent que du niveau d’eau, <lb/>parce que c’eſt celui qui eſt le plus en uſage dans les nivelle-<lb/>mens qui ne ſont pas d’une grande étendue. </s> <s xml:id="echoid-s12749" xml:space="preserve">Cependant comme <lb/>les niveaux qui ont des lunettes ſont bien plus commodes, <lb/>parce que l’on peut en deux ou trois coups de niveau, ou quel-<lb/>quefois même en un ſeul, niveler deux objets, dont on ne <lb/>pourroit connoître la différence des hauteurs avec le niveau <lb/>d’eau, ſans faire beaucoup plus d’opérations, voici celui qui <lb/>a été inventé par M. </s> <s xml:id="echoid-s12750" xml:space="preserve">Huyghens, qui peut paſſer pour le plus <lb/>commode & </s> <s xml:id="echoid-s12751" xml:space="preserve">le plus juſte de tous ceux qui ont été faits dans ce <lb/>goût-là.</s> <s xml:id="echoid-s12752" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12753" xml:space="preserve">Une des principales parties de cet inſtrument eſt la virole <lb/>D, qui a deux branches plates, C & </s> <s xml:id="echoid-s12754" xml:space="preserve">E qui ſont ſemblables, <lb/>chacune d’environ un demi-pied de long; </s> <s xml:id="echoid-s12755" xml:space="preserve">de ſorte que le tout <lb/>fait une eſpece de croix. </s> <s xml:id="echoid-s12756" xml:space="preserve">Cette virole D porte la lunette A B <lb/>longue de deux pieds: </s> <s xml:id="echoid-s12757" xml:space="preserve">ſi elle n’a que deux verres convexes, <lb/>elle repréſentera les objets renverſés, mais avec beaucoup plus <lb/>de clarté que ſi elle en a quatre, qui les remettroient dans leur <pb o="375" file="0441" n="455" rhead="DE MATHÉMATIQUE. Liv. X."/> ſituation naturelle. </s> <s xml:id="echoid-s12758" xml:space="preserve">Le tuyau de cette lunette doit être de <lb/>cuivre, ou de quelqu’autre matiere forte, & </s> <s xml:id="echoid-s12759" xml:space="preserve">à l’épreuve des <lb/>injures de l’air.</s> <s xml:id="echoid-s12760" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12761" xml:space="preserve">Au bout des branches de la virole D ſont attachés deux <lb/>filets doubles paſſés dans des petits anneaux, & </s> <s xml:id="echoid-s12762" xml:space="preserve">ſerrés entre <lb/>des pinces à deux dents, dont l’une eſt fixée au bout de ſa <lb/>branche, & </s> <s xml:id="echoid-s12763" xml:space="preserve">l’autre y eſt attachée de telle maniere qu’elle ſe <lb/>puiſſe ouvrir.</s> <s xml:id="echoid-s12764" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12765" xml:space="preserve">Comme la lunette eſt ſuſpendue par la virole D au cro-<lb/>chet F, elle eſt tendue horizontalement par le poids qui eſt <lb/>enfermé dans la boîte G, dont il ne ſort que ſon crochet. </s> <s xml:id="echoid-s12766" xml:space="preserve">La <lb/>peſanteur de ce poids ne doit être qu’environ la peſanteur de <lb/>la croix, & </s> <s xml:id="echoid-s12767" xml:space="preserve">le vuide qui reſte dans cette boîte eſt rempli <lb/>d’huile de noix ou de lin, ou de quelqu’autre liqueur qui ne <lb/>ſe glace ni ne ſe fige point; </s> <s xml:id="echoid-s12768" xml:space="preserve">& </s> <s xml:id="echoid-s12769" xml:space="preserve">c’eſt par cette liqueur que ſont <lb/>arrêtés les balancemens du poids & </s> <s xml:id="echoid-s12770" xml:space="preserve">de la lunette. </s> <s xml:id="echoid-s12771" xml:space="preserve">Il doit y <lb/>avoir au dedans de la lunette un fil de ſoie tendu horizontale-<lb/>ment au foyer du verre objectif; </s> <s xml:id="echoid-s12772" xml:space="preserve">& </s> <s xml:id="echoid-s12773" xml:space="preserve">c’eſt par une vis que l’on <lb/>tourne au travers du trou H, percé dans le tuyau de la lunette, <lb/>que l’on abaiſſe ou éleve ce fil ſelon le beſoin. </s> <s xml:id="echoid-s12774" xml:space="preserve">Il faut mettre <lb/>au tuyau de la lunette une petite virole, qui doit être fort <lb/>legere, & </s> <s xml:id="echoid-s12775" xml:space="preserve">ne pas peſer plus d’une 80<emph style="sub">e</emph> partie de la croix: </s> <s xml:id="echoid-s12776" xml:space="preserve">elle <lb/>n’eſt point attachée au tuyau de la lunette, parce qu’il faut la <lb/>pouſſer vers le bout, ou l’en reculer autant qu’il eſt néceſſaire <lb/>pour trouver l’équilibre de la lunette, & </s> <s xml:id="echoid-s12777" xml:space="preserve">la mettre parallele à <lb/>l’horizon.</s> <s xml:id="echoid-s12778" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12779" xml:space="preserve">Cette machine eſt ſuſpendue au haut d’une eſpece de croix <lb/>de bois plate, où il y a pour cela le crochet F, qui peut ſe <lb/>hauſſer ou baiſſer par le moyen de la vis qui tient à l’anneau <lb/>qui ſuſpend la machine: </s> <s xml:id="echoid-s12780" xml:space="preserve">cette même croix tient la boîte qui <lb/>contient le plomb & </s> <s xml:id="echoid-s12781" xml:space="preserve">l’huile; </s> <s xml:id="echoid-s12782" xml:space="preserve">& </s> <s xml:id="echoid-s12783" xml:space="preserve">cette boîte eſt enfermée par <lb/>les côtés & </s> <s xml:id="echoid-s12784" xml:space="preserve">par le fond.</s> <s xml:id="echoid-s12785" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12786" xml:space="preserve">On couvre le niveau par une autre eſpece de croix, qui eſt <lb/>creuſe, que l’on applique contre la croix de bois plate, avec <lb/>pluſieurs crochets, afin de couvrir le niveau contre les injures <lb/>du tems; </s> <s xml:id="echoid-s12787" xml:space="preserve">de ſorte que le tout fait une boîte.</s> <s xml:id="echoid-s12788" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12789" xml:space="preserve">Pour rectifier ce niveau, on le ſuſpendra par l’anneau d’une <lb/>de ſes branches, ſans attacher de poids par en bas, & </s> <s xml:id="echoid-s12790" xml:space="preserve">l’on <lb/>viſera par la lunette à quelque objet éloigné, remarquant l’en-<lb/>droit où le point de l’objet eſt coupé par le fil de la lunette, <pb o="376" file="0442" n="456" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s12791" xml:space="preserve">enſuite on mettra le poids, en l’accrochant dans l’anneau <lb/>d’en bas: </s> <s xml:id="echoid-s12792" xml:space="preserve">& </s> <s xml:id="echoid-s12793" xml:space="preserve">ſi alors le fil de la lunette répond à la même <lb/>marque de l’objet, c’eſt une preuve certaine que le centre de <lb/>gravité, ou les deux points de la ſuſpenſion de la croix répon-<lb/>dent au centre du tuyau de la lunette, ou au centre de la <lb/>terre; </s> <s xml:id="echoid-s12794" xml:space="preserve">mais ſi cela ne ſe trouve pas préciſément au même point, <lb/>on la vérifiera par le moyen de la virole I, en la faiſant couler <lb/>de part ou d’autre, pour réparer le défaut, & </s> <s xml:id="echoid-s12795" xml:space="preserve">mettre la lu-<lb/>nette en équilibre; </s> <s xml:id="echoid-s12796" xml:space="preserve">& </s> <s xml:id="echoid-s12797" xml:space="preserve">la lunette étant miſe horizontalement <lb/>par la virole ſans poids & </s> <s xml:id="echoid-s12798" xml:space="preserve">avec poids, on la tournera ſans <lb/>deſſus deſſous, mettant en haut la branche d’en bas, & </s> <s xml:id="echoid-s12799" xml:space="preserve">atta-<lb/>chant le poids à la branche que l’on a abaiſſée.</s> <s xml:id="echoid-s12800" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12801" xml:space="preserve">Si après cette rectification, le fil qui eſt dans la lunette ſe <lb/>trouve à la même hauteur de l’objet que devant; </s> <s xml:id="echoid-s12802" xml:space="preserve">c’eſt une <lb/>marque que le fil du foyer de la lunette eſt directement au <lb/>milieu de ce foyer: </s> <s xml:id="echoid-s12803" xml:space="preserve">mais ſi le fil ne viſe pas au même point, & </s> <s xml:id="echoid-s12804" xml:space="preserve"><lb/>qu’il coupe l’objet au deſſus ou au deſſous, on hauſſera ou baiſ-<lb/>ſera moyennant la vis qui eſt pour cela, juſqu’à ce que le fil <lb/>coupe le point moyen, qui eſt entre les deux points remar-<lb/>qués, & </s> <s xml:id="echoid-s12805" xml:space="preserve">après cela le niveau ſera bien rectifié.</s> <s xml:id="echoid-s12806" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12807" xml:space="preserve">Le pied qui doit porter la machine eſt une eſpece de table <lb/>de fer ou de cuivre, qui eſt ronde & </s> <s xml:id="echoid-s12808" xml:space="preserve">un peu concave, afin <lb/>que la machine ſoit plus ſolidement établie dans la concavité: <lb/></s> <s xml:id="echoid-s12809" xml:space="preserve">elle eſt élevée ſur trois pieds, qui y ſont attachés en char-<lb/>niere, & </s> <s xml:id="echoid-s12810" xml:space="preserve">dont la hauteur eſt de trois ou quatre pieds.</s> <s xml:id="echoid-s12811" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12812" xml:space="preserve">La figure N repréſente en grand le tuyau qui porte en de-<lb/>dans de la lunette le fil horizontal, qui eſt attaché à la four-<lb/>chette K avec de la cire.</s> <s xml:id="echoid-s12813" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12814" xml:space="preserve">Il faut ſi peu de choſe pour faire de grandes erreurs en ni-<lb/>velant, que l’on ne ſçauroit prendre trop de précautions à ſe <lb/>bien ſervir des inſtrumens: </s> <s xml:id="echoid-s12815" xml:space="preserve">pour cela, il faut les connoître <lb/>parfaitement; </s> <s xml:id="echoid-s12816" xml:space="preserve">quand je dis les connoître, j’entends que l’on <lb/>doit ſi bien les examiner, que l’on puiſſe en ſçavoir juſqu’au <lb/>moindre défaut, entre leſquels il n’y en a point de plus con-<lb/>ſidérable que de baiſſer ou hauſſer la mire. </s> <s xml:id="echoid-s12817" xml:space="preserve">Il eſt vrai que pour <lb/>le niveau de M. </s> <s xml:id="echoid-s12818" xml:space="preserve">Huyghens, quand même il n’auroit pas été fait <lb/>avec aſſez de précaution pour avoir cet inconvénient, il ne <lb/>faut pas beaucoup s’en embarraſſer; </s> <s xml:id="echoid-s12819" xml:space="preserve">car s’il baiſſe la mire dans <lb/>un ſens, il la hauſſera d’autant dans un autre; </s> <s xml:id="echoid-s12820" xml:space="preserve">& </s> <s xml:id="echoid-s12821" xml:space="preserve">prenant <lb/>le point milieu des deux objets, l’on aura toujours le vrai <pb o="377" file="0443" n="457" rhead="DE MATHÉMATIQUE. Liv. X."/> niveau apparent, qui eſt un avantage particulier de ce niveau, <lb/>de pouvoir être renverſé de bas en haut, & </s> <s xml:id="echoid-s12822" xml:space="preserve">de haut en bas; <lb/></s> <s xml:id="echoid-s12823" xml:space="preserve">mais comme on peut ſe ſervir de tout autre inſtrument qui <lb/>n’aura pas cet avantage, voici le moyen de corriger un rayon <lb/>de mire faux.</s> <s xml:id="echoid-s12824" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12825" xml:space="preserve">Ayant poſé un inſtrument à l’endroit A, pour pointer vers <lb/> <anchor type="note" xlink:label="note-0443-01a" xlink:href="note-0443-01"/> D G, je ſuppoſe que l’on a reconnu que la lunette, au lieu de <lb/>donner le point C du niveau apparent B C, donne le point D, <lb/>qui eſt plus élevé que le point C, parce que l’inſtrument hauſſe <lb/>la mire; </s> <s xml:id="echoid-s12826" xml:space="preserve">& </s> <s xml:id="echoid-s12827" xml:space="preserve">ayant remarqué que ſur une diſtance B C de 200 <lb/>toiſes, le point D eſt élevé de deux pouces au deſſus du point <lb/>C. </s> <s xml:id="echoid-s12828" xml:space="preserve">Après en être bien aſſuré, ſi je vois que cette faute ne ſe <lb/>puiſſe pas réparer, parce que l’on ſuppoſe que l’inſtrument a <lb/>été mal fait, j’ai égard, dans toutes les opérations que je fais, <lb/>à la correction de l’inſtrument; </s> <s xml:id="echoid-s12829" xml:space="preserve">de ſorte qu’ayant donné un <lb/>autre coup de niveau B E de 600 toiſes, je cherche à quel point <lb/>de la hauteur E H doit être le niveau apparent, parce que je <lb/>ſuis prévenu que ce n’eſt pas le point E, mais que ce doit être <lb/>un autre point au deſſus de celui-ci. </s> <s xml:id="echoid-s12830" xml:space="preserve">Pour le trouver, je dis: <lb/></s> <s xml:id="echoid-s12831" xml:space="preserve">Si 200 toiſes donnent 2 poucespour le hauſſement du rayon de <lb/>mire, combien donneront 600 toiſes? </s> <s xml:id="echoid-s12832" xml:space="preserve">La Regle étant faite, <lb/>je trouve 6 pouces; </s> <s xml:id="echoid-s12833" xml:space="preserve">ainſi je prends le point F, ſix pouces au <lb/>deſſous du point E, & </s> <s xml:id="echoid-s12834" xml:space="preserve">pour lors la ligne B F eſt celle du ni-<lb/>veau apparent: </s> <s xml:id="echoid-s12835" xml:space="preserve">mais ſi l’inſtrument baiſſe la mire, au lieu de <lb/>la hauſſer, on trouvera toujours le point du vrai niveau appa-<lb/>rent en ſuivant la même regle, qui eſt fondée ſur ce que les <lb/>triangles B C D & </s> <s xml:id="echoid-s12836" xml:space="preserve">B F E ſont ſemblables.</s> <s xml:id="echoid-s12837" xml:space="preserve"/> </p> <div xml:id="echoid-div1027" type="float" level="2" n="1"> <note position="right" xlink:label="note-0443-01" xlink:href="note-0443-01a" xml:space="preserve">Figure 209.</note> </div> </div> <div xml:id="echoid-div1029" type="section" level="1" n="800"> <head xml:id="echoid-head966" xml:space="preserve">CHAPITRE VI,</head> <head xml:id="echoid-head967" style="it" xml:space="preserve">Où l’on donne la maniere de ſe ſervir du Niveau de M. Huyghens.</head> <p> <s xml:id="echoid-s12838" xml:space="preserve">770. </s> <s xml:id="echoid-s12839" xml:space="preserve">LE niveau ayant été poſé au lieu deſtiné pour l’opéra-<lb/>tion, on envoyera, comme à l’ordinaire, un Aide à une diſ-<lb/>tance convenable, & </s> <s xml:id="echoid-s12840" xml:space="preserve">on regardera exactement par la lunette <lb/>l’endroit de la perche où le fil répondra; </s> <s xml:id="echoid-s12841" xml:space="preserve">& </s> <s xml:id="echoid-s12842" xml:space="preserve">l’Aide qui tient la <lb/>carte l’ayant hauſſée & </s> <s xml:id="echoid-s12843" xml:space="preserve">baiſſée tant que le petit rond noir ré-<lb/>ponde au rayon de mire, il a ſoin de marquer un trait de crayon <pb o="378" file="0444" n="458" rhead="NOUVEAU COURS"/> ſur la perche à l’endroit où le rayon de mire a répondu, & </s> <s xml:id="echoid-s12844" xml:space="preserve">il <lb/>ne bouge point de ſa place juſqu’à ce qu’il ſoit averti; </s> <s xml:id="echoid-s12845" xml:space="preserve">& </s> <s xml:id="echoid-s12846" xml:space="preserve">alors <lb/>celui qui eſt à l’inſtrument, le change de diſpoſition, mettant <lb/>le deſſus au deſſous, c’eſt-à-dire qu’il faut accrocher la croix <lb/>par l’anneau d’en bas; </s> <s xml:id="echoid-s12847" xml:space="preserve">après quoi on viſe de rechef avec la <lb/>lunette, & </s> <s xml:id="echoid-s12848" xml:space="preserve">celui qui eſt à la perche hauſſe & </s> <s xml:id="echoid-s12849" xml:space="preserve">baiſſe encore le <lb/>carton, pour marquer à quelle hauteur porte le rayon de mire, <lb/>qui doit répondre au même endroit que l’on a marqué. </s> <s xml:id="echoid-s12850" xml:space="preserve">Or <lb/>ſuppoſant qu’il donne au deſſous de la marque, il faut mar-<lb/>quer exactement à quel endroit; </s> <s xml:id="echoid-s12851" xml:space="preserve">enſuite diviſer en deux égale-<lb/>ment l’intervalle des deux coups de niveau différens, & </s> <s xml:id="echoid-s12852" xml:space="preserve">l’on <lb/>aura au juſte la hauteur du niveau apparent, de laquelle il fau-<lb/>dra retrancher la hauteur du niveau apparent au deſſus du vrai, <lb/>que l’on trouvera, ſelon qu’il a été enſeigné au quatrieme <lb/>chapitre, & </s> <s xml:id="echoid-s12853" xml:space="preserve">la différence ſera la hauteur du vrai niveau, la-<lb/>quelle on pourroit encore trouver ſans faire de calcul, comme <lb/>on le va voir.</s> <s xml:id="echoid-s12854" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12855" xml:space="preserve">Ayant deux perches C A & </s> <s xml:id="echoid-s12856" xml:space="preserve">B E, éloignées l’une de l’autre, <lb/> <anchor type="note" xlink:label="note-0444-01a" xlink:href="note-0444-01"/> je ſuppoſe d’une diſtance de 600 toiſes, l’on demande quelle <lb/>ſeroit la hauteur du niveau apparent au deſſus du vrai.</s> <s xml:id="echoid-s12857" xml:space="preserve"/> </p> <div xml:id="echoid-div1029" type="float" level="2" n="1"> <note position="left" xlink:label="note-0444-01" xlink:href="note-0444-01a" xml:space="preserve">Figure 210.</note> </div> <p> <s xml:id="echoid-s12858" xml:space="preserve">Pour la trouver, poſez le niveau à l’endroit A, & </s> <s xml:id="echoid-s12859" xml:space="preserve">pointez <lb/>avec la lunette l’endroit de la perche B E, où le rayon de mire <lb/>ira la rencontrer, ſuppoſant que ce ſoit au point B, il faut y <lb/>faire une marque, & </s> <s xml:id="echoid-s12860" xml:space="preserve">vérifier ce coup de niveau, en renverſant <lb/>l’inſtrument, pour voir ſi dans cette ſituation le rayon de <lb/>mire ſe termine au point B. </s> <s xml:id="echoid-s12861" xml:space="preserve">Cela poſé, faites porter l’inſtru-<lb/>ment à l’endroit E, & </s> <s xml:id="echoid-s12862" xml:space="preserve">diſpoſez-le de maniere que le foyer du <lb/>verre de la lunette ſoit préciſément à la hauteur B. </s> <s xml:id="echoid-s12863" xml:space="preserve">Après cela <lb/>donnez un autre coup de niveau B C, qui aille rencontrer la <lb/>perche A C au point C, qu’il faudra marquer ſur la perche, <lb/>après l’avoir vérifié comme ci-devant; </s> <s xml:id="echoid-s12864" xml:space="preserve">& </s> <s xml:id="echoid-s12865" xml:space="preserve">ſi l’on meſure exac-<lb/>tement la diſtance C A, je dis qu’elle ſera double de la hauteur <lb/>du niveau apparent au deſſus du vrai; </s> <s xml:id="echoid-s12866" xml:space="preserve">de ſorte que C A doit ſe <lb/>trouver ici de 8 pouces: </s> <s xml:id="echoid-s12867" xml:space="preserve">car en diviſant C A en deux également <lb/>au point D, l’on aura la ligne C D de 4 pouces, qui ſera la <lb/>différence du niveau apparent au deſſus du vrai, pour une diſ-<lb/>tance de 600 toiſes, comme on le peut voir par le calcul: <lb/></s> <s xml:id="echoid-s12868" xml:space="preserve">ainſi les points B & </s> <s xml:id="echoid-s12869" xml:space="preserve">D ſont de niveau, étant également éloi-<lb/>gnés du centre de la terre, comme vous l’allez voir.</s> <s xml:id="echoid-s12870" xml:space="preserve"/> </p> <pb o="379" file="0445" n="459" rhead="DE MATHÉMATIQUE. Liv. X."/> <p> <s xml:id="echoid-s12871" xml:space="preserve">Si l’on prend le point A pour l’extrêmité d’un des rayons <lb/>de la terre, le point B ſera plus éloigné du centre de la terre <lb/>que le point A de 4 pouces: </s> <s xml:id="echoid-s12872" xml:space="preserve">mais le point C étant plus éloigné <lb/>du centre de la terre que le point B auſſi de 4 pouces, le point <lb/>C ſera donc plus éloigné que le point A du centre de la terre <lb/>de 8 pouces: </s> <s xml:id="echoid-s12873" xml:space="preserve">donc les points D & </s> <s xml:id="echoid-s12874" xml:space="preserve">B étant chacun plus éloignés <lb/>du centre de la terre que le point A de 4 pouces, il s’enſuit <lb/>qu’ils ſeront de niveau, & </s> <s xml:id="echoid-s12875" xml:space="preserve">que la moitié de la ligne C A ſera <lb/>la hauteur du niveau apparent au deſſus du vrai.</s> <s xml:id="echoid-s12876" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12877" xml:space="preserve">L’on voit que par le nivellement réciproque l’on peut d’une <lb/>maniere fort ſimple déterminer deux points parfaitement de <lb/>niveau, ſans s’embarraſſer de leur diſtance. </s> <s xml:id="echoid-s12878" xml:space="preserve">Il eſt vrai que l’on <lb/>peut encore trouver deux points de niveau, ſans même faire <lb/>de nivellement réciproque, en poſant l’inſtrument dans le <lb/>milieu de la diſtance de deux objets que l’on a à niveler; </s> <s xml:id="echoid-s12879" xml:space="preserve">ce <lb/>qui ſe fait à peu près de la maniere qu’on a expliqué dans <lb/>l’uſage du niveau d’eau.</s> <s xml:id="echoid-s12880" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1031" type="section" level="1" n="801"> <head xml:id="echoid-head968" xml:space="preserve">CHAPITRE VII,</head> <head xml:id="echoid-head969" style="it" xml:space="preserve">Où l’on donne la maniere de faire le Nivellement compoſé, avec <lb/>le niveau de M. Huyghens.</head> <p> <s xml:id="echoid-s12881" xml:space="preserve">771. </s> <s xml:id="echoid-s12882" xml:space="preserve">NOus avons dit que pour faire un nivellement com-<lb/> <anchor type="note" xlink:label="note-0445-01a" xlink:href="note-0445-01"/> poſé, il falloit ajouter toutes les hauteurs que l’on trouveroit <lb/>en montant, & </s> <s xml:id="echoid-s12883" xml:space="preserve">que l’on auroit miſes dans la premiere co-<lb/>lonne, & </s> <s xml:id="echoid-s12884" xml:space="preserve">ajouter auſſi enſemble toutes celles que l’on aura <lb/>trouvées en deſcendant, qui ſont dans la ſeconde colonne, <lb/>afin de ſouſtraire la ſomme des unes de la ſomme des autres, <lb/>pour avoir la différence, qui fait voir de combien l’un des en-<lb/>droits eſt plus élevé que l’autre: </s> <s xml:id="echoid-s12885" xml:space="preserve">mais comme dans cette pra-<lb/>tique nous nous ſommes ſervis du niveau d’eau, dont les coups <lb/>de niveau ne ſont pas conſidérables, & </s> <s xml:id="echoid-s12886" xml:space="preserve">que d’ailleurs l’inſtru-<lb/>ment pour chaque ſtation a été placé dans le milieu des deux <lb/>termes, on n’a pas eu égard à la différence du niveau appa-<lb/>rent au deſſus du vrai, ni en deſcendant, ni en montant, <lb/>parce que, ſelon cette pratique, la différence du niveau ap-<lb/>parent n’a pas lieu: </s> <s xml:id="echoid-s12887" xml:space="preserve">mais il n’en eſt pas de même, lorſqu’on <pb o="380" file="0446" n="460" rhead="NOUVEAU COURS"/> ſe ſert d’un inſtrument à pouvoir donner des grands coups de <lb/>niveau, ou il faut avoir égard à la différence du niveau appa-<lb/>rent au deſſus du vrai, en montant comme en deſcendant, <lb/>ſurtout quand l’inſtrument eſt placé au premier terme, pour <lb/>niveler d’un terme à l’autre: </s> <s xml:id="echoid-s12888" xml:space="preserve">car dans cette occaſion, il faut <lb/>non ſeulement mettre dans la premiere colonne les hauteurs <lb/>que l’on a trouvées en montant, & </s> <s xml:id="echoid-s12889" xml:space="preserve">dans la ſeconde celles que <lb/>l’on a trouvées en deſcendant; </s> <s xml:id="echoid-s12890" xml:space="preserve">mais encore écrire à côté de <lb/>chaque colonne la différence du niveau apparent au deſſus du <lb/>vrai, pour chaque diſtance qui ſont dans les colonnes, tant <lb/>en montant qu’en deſcendant: </s> <s xml:id="echoid-s12891" xml:space="preserve">& </s> <s xml:id="echoid-s12892" xml:space="preserve">ce qu’il y a de particulier <lb/>en ceci, c’eſt qu’après avoir mis dans une ſomme les hauteurs <lb/>du niveau apparent au deſſus du vrai, que l’on aura trouvées <lb/>en montant, il faut l’ajouter à la ſomme des hauteurs de la <lb/>premiere colonne, pour ne faire qu’une ſomme des hauteurs <lb/>de la premiere colonne, & </s> <s xml:id="echoid-s12893" xml:space="preserve">des différences de leur niveau ap-<lb/>parent au deſſus du vrai.</s> <s xml:id="echoid-s12894" xml:space="preserve"/> </p> <div xml:id="echoid-div1031" type="float" level="2" n="1"> <note position="right" xlink:label="note-0445-01" xlink:href="note-0445-01a" xml:space="preserve">Figure 212.</note> </div> <p> <s xml:id="echoid-s12895" xml:space="preserve">L’on écrira de même à côté de la ſeconde colonne, la dif-<lb/>férence du niveau apparent au deſſus du vrai, pour chaque <lb/>hauteur que l’on aura trouvée en deſcendant; </s> <s xml:id="echoid-s12896" xml:space="preserve">& </s> <s xml:id="echoid-s12897" xml:space="preserve">l’on fera une <lb/>ſomme de toutes ces différences, qu’il faudra enſuite ſouſtraire <lb/>de celles des hauteurs, tellement qu’il faut regarder comme <lb/>une regle générale, qu’en montant il faut ajouter la différence <lb/>du niveau apparent au deſſus du vrai, aux hauteurs que l’on <lb/>trouvera en deſcendant, & </s> <s xml:id="echoid-s12898" xml:space="preserve">qu’en deſcendant il les faut ſouſ-<lb/>traire; </s> <s xml:id="echoid-s12899" xml:space="preserve">& </s> <s xml:id="echoid-s12900" xml:space="preserve">en voici la raiſon.</s> <s xml:id="echoid-s12901" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12902" xml:space="preserve">Suppoſons qu’en montant l’on ait donné des coups de ni-<lb/> <anchor type="note" xlink:label="note-0446-01a" xlink:href="note-0446-01"/> veau B C & </s> <s xml:id="echoid-s12903" xml:space="preserve">F G, & </s> <s xml:id="echoid-s12904" xml:space="preserve">en deſcendant les coups de niveau K N <lb/>& </s> <s xml:id="echoid-s12905" xml:space="preserve">Q R. </s> <s xml:id="echoid-s12906" xml:space="preserve">Cela poſé, conſidérez qu’ayant mené à la ligne B C <lb/>la parallele A D, cette parallele ſera une tangente à la terre, <lb/>& </s> <s xml:id="echoid-s12907" xml:space="preserve">la ligne D E marquera la hauteur du niveau apparent au <lb/>deſſus du vrai. </s> <s xml:id="echoid-s12908" xml:space="preserve">Or comme les lignes B A & </s> <s xml:id="echoid-s12909" xml:space="preserve">C D ſont égales, <lb/>le point C ſera plus éloigné du centre de la terre que le point <lb/>B de toute la ligne D E: </s> <s xml:id="echoid-s12910" xml:space="preserve">ainſi pour que le point B ſoit de ni-<lb/>veau avec le point C, il faudra ajouter à la hauteur B A la <lb/>ligne D E, c’eſt-à-dire la ligne de la différence du niveau ap-<lb/>parent au deſſus du vrai. </s> <s xml:id="echoid-s12911" xml:space="preserve">De même ſi à la ligne de niveau ap-<lb/>parent F G l’on mene la parallele E H, la ligne H I ſera en-<lb/>core la différence du niveau apparent au deſſus du vrai. </s> <s xml:id="echoid-s12912" xml:space="preserve">Or <pb o="381" file="0447" n="461" rhead="DE MATHÉMATIQUE. Liv. X."/> les lignes F E & </s> <s xml:id="echoid-s12913" xml:space="preserve">G H étant égales, le point G ſera plus éloigné <lb/>du centre de la terre que le point F de toute la ligne H I: </s> <s xml:id="echoid-s12914" xml:space="preserve">il <lb/>faut donc, pour que le point F ſoit de niveau avec le point G, <lb/>ajouter à la hauteur F C la ligne H I.</s> <s xml:id="echoid-s12915" xml:space="preserve"/> </p> <div xml:id="echoid-div1032" type="float" level="2" n="2"> <note position="left" xlink:label="note-0446-01" xlink:href="note-0446-01a" xml:space="preserve">Figure 211.</note> </div> <p> <s xml:id="echoid-s12916" xml:space="preserve">A l’égard des coups de niveau K N & </s> <s xml:id="echoid-s12917" xml:space="preserve">Q R, que l’on a donnés <lb/>en deſcendant, l’on voit que leur ayant mené les paralleles <lb/>L O & </s> <s xml:id="echoid-s12918" xml:space="preserve">P S, qui ſont des tangentes à la terre, le point N eſt <lb/>plus éloigné du centre de la terre que le point K de toute la <lb/>ligne O P; </s> <s xml:id="echoid-s12919" xml:space="preserve">& </s> <s xml:id="echoid-s12920" xml:space="preserve">que pour trouver un point de niveau avec le <lb/>point K, il faut ôter de la hauteur N Q la ligne O P, qui eſt <lb/>la différence du niveau apparent au deſſus du vrai pour la lon-<lb/>gueur K N. </s> <s xml:id="echoid-s12921" xml:space="preserve">Enfin comme le point R n’eſt pas de niveau avec <lb/>le point Q, parce que le premier eſt plus éloigné du centre de <lb/>la terre que le ſecond de toute la ligne S T, il faudra donc <lb/>encore ôter la ligne S T de la hauteur R T, pour mettre le <lb/>point R de niveau avec le point Q. </s> <s xml:id="echoid-s12922" xml:space="preserve">Il en ſera de même des <lb/>autres.</s> <s xml:id="echoid-s12923" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12924" xml:space="preserve">L’on a ſuppoſé que les lignes B A & </s> <s xml:id="echoid-s12925" xml:space="preserve">C D, F E & </s> <s xml:id="echoid-s12926" xml:space="preserve">G H, &</s> <s xml:id="echoid-s12927" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s12928" xml:space="preserve"> <anchor type="note" xlink:label="note-0447-01a" xlink:href="note-0447-01"/> étoient paralleles, quoiqu’elles ſoient des demi-diametres de <lb/>la terre prolongés; </s> <s xml:id="echoid-s12929" xml:space="preserve">mais à cauſe de la grande diſtance au cen-<lb/>tre, on les peut regarder comme telles, ſans que cela puiſſe <lb/>faire une erreur ſenſible.</s> <s xml:id="echoid-s12930" xml:space="preserve"/> </p> <div xml:id="echoid-div1033" type="float" level="2" n="3"> <note position="right" xlink:label="note-0447-01" xlink:href="note-0447-01a" xml:space="preserve">Figure 212.</note> </div> <p> <s xml:id="echoid-s12931" xml:space="preserve">Pour appliquer à un exemple ce que nous venons d’enſei-<lb/>gner, ſoient les lieux A & </s> <s xml:id="echoid-s12932" xml:space="preserve">F, dont on veut connoître la dif-<lb/>férence de niveau.</s> <s xml:id="echoid-s12933" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12934" xml:space="preserve">Pour cela je me ſers d’un niveau à lunettes, que je poſe <lb/>au premier terme A, pour donner le coup de niveau G B, qui <lb/>ſe termine à un point B de la hauteur, auquel j’envoie un Aide <lb/>pour y planter un piquet, & </s> <s xml:id="echoid-s12935" xml:space="preserve">je conſidere que la différence du <lb/>niveau apparent eſt de 4 pieds & </s> <s xml:id="echoid-s12936" xml:space="preserve">demi, qui eſt la hauteur G Q <lb/>du niveau, que j’écris dans la premiere colonne; </s> <s xml:id="echoid-s12937" xml:space="preserve">enſuite je <lb/>fais meſurer la longueur G B, que je ſuppoſe de 600 toiſes, <lb/>& </s> <s xml:id="echoid-s12938" xml:space="preserve">je cherche quelle eſt la hauteur du niveau apparent au deſſus <lb/>du vrai, que je trouve de 4 pouces: </s> <s xml:id="echoid-s12939" xml:space="preserve">j’écris cette hauteur à côté <lb/>de la premiere colonne, vis-à-vis de 4 pieds & </s> <s xml:id="echoid-s12940" xml:space="preserve">demi. </s> <s xml:id="echoid-s12941" xml:space="preserve">Après <lb/>cela je fais porter le niveau au point B, & </s> <s xml:id="echoid-s12942" xml:space="preserve">j’envoie un Aide <lb/>à l’endroit C, qui eſt une diſtance que l’on aura jugé conve-<lb/>nable; </s> <s xml:id="echoid-s12943" xml:space="preserve">& </s> <s xml:id="echoid-s12944" xml:space="preserve">après avoir donné le coup de niveau H I, je ſuppoſe <lb/>que l’on a trouvé I C de 2 pieds, que je ſouſtrais de 4 pieds &</s> <s xml:id="echoid-s12945" xml:space="preserve"> <pb o="382" file="0448" n="462" rhead="NOUVEAU COURS"/> demi, & </s> <s xml:id="echoid-s12946" xml:space="preserve">il reſte 2 pieds & </s> <s xml:id="echoid-s12947" xml:space="preserve">demi pour la hauteur du point C <lb/>au deſſus du point B. </s> <s xml:id="echoid-s12948" xml:space="preserve">Ayant donc écrit cette quantité dans <lb/>la premiere colonne, je fais meſurer la longueur H I, que je <lb/>trouve de 380 toiſes, qui donnent un pouce 7 lignes pour la <lb/>différence du niveau apparent au deſſus du vrai, que j’écris à <lb/>côté de la premiere colonne, vis-à-vis 2 pieds 6 pouces.</s> <s xml:id="echoid-s12949" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12950" xml:space="preserve">Delà je viens au point C, & </s> <s xml:id="echoid-s12951" xml:space="preserve">j’envoie un Aide au point D <lb/>avec une perche; </s> <s xml:id="echoid-s12952" xml:space="preserve">enſuite je donne le coup de niveau K L, & </s> <s xml:id="echoid-s12953" xml:space="preserve"><lb/>l’Aide qui eſt en L, marque un trait de crayon à l’endroit de <lb/>la perche où a répondu le rayon de mire, & </s> <s xml:id="echoid-s12954" xml:space="preserve">on meſure la hau-<lb/>teur L D, qui ſera, par exemple, de 9 pieds; </s> <s xml:id="echoid-s12955" xml:space="preserve">d’où ayant ſouſ-<lb/>trait la hauteur du niveau, il vient 4 pieds & </s> <s xml:id="echoid-s12956" xml:space="preserve">demi, qui fait <lb/>voir la différence de niveau apparent des points C & </s> <s xml:id="echoid-s12957" xml:space="preserve">D. </s> <s xml:id="echoid-s12958" xml:space="preserve">Mais <lb/>comme 4 pieds & </s> <s xml:id="echoid-s12959" xml:space="preserve">demi eſt une hauteur que l’on a trouvée <lb/>en deſcendant, je l’écris dans la ſeconde colonne, à côté de <lb/>laquelle j’écris auſſi 2 pouces 4 lignes, qui eſt la différence du <lb/>niveau apparent au deſſus du vrai pour la longueur K L. </s> <s xml:id="echoid-s12960" xml:space="preserve">Après <lb/>cela je fais porter le niveau au point D, & </s> <s xml:id="echoid-s12961" xml:space="preserve">j’envoie un Aide <lb/>en E, pour marquer le point M ſur la perche, après que j’aurai <lb/>donné le coup de niveau M N: </s> <s xml:id="echoid-s12962" xml:space="preserve">ayant trouvé 10 pieds & </s> <s xml:id="echoid-s12963" xml:space="preserve">demi <lb/>pour la hauteur E N, j’en ſouſtrais celle du niveau, qui eſt de <lb/>4 pieds & </s> <s xml:id="echoid-s12964" xml:space="preserve">demi, & </s> <s xml:id="echoid-s12965" xml:space="preserve">la différence eſt 6 pieds, que j’écris dans <lb/>la ſeconde colonne: </s> <s xml:id="echoid-s12966" xml:space="preserve">& </s> <s xml:id="echoid-s12967" xml:space="preserve">ſuppoſant que la diſtance M N ſoit de <lb/>650 toiſes, je cherche la hauteur du niveau apparent au deſſus <lb/>du vrai pour une pareille diſtance, & </s> <s xml:id="echoid-s12968" xml:space="preserve">je trouve qu’elle eſt de <lb/>4 pouces 8 lignes, que j’écris à côté de la ſeconde colonne, <lb/>vis-à-vis le dernier nombre que j’y ai marqué, c’eſt-à-dire <lb/>vis-à-vis 6 pieds. </s> <s xml:id="echoid-s12969" xml:space="preserve">Enfin je fais porter le niveau en E, pour <lb/>faire la derniere opération O P, qui donne 8 pieds pour la <lb/>hauteur P F; </s> <s xml:id="echoid-s12970" xml:space="preserve">d’où ayant retranché celle du niveau, la diffé-<lb/>rence eſt 3 pieds & </s> <s xml:id="echoid-s12971" xml:space="preserve">demi, que j’écris dans la ſeconde colonne, <lb/>à côté de laquelle je mets 5 pouces 4 lignes, qui eſt la diffé-<lb/>rence du niveau apparent au deſſus du vrai pour la diſtance <lb/>O P, que nous ſuppoſons de 700 toiſes.</s> <s xml:id="echoid-s12972" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12973" xml:space="preserve">Après que l’on a fait l’opération, il faut faire l’addition <lb/>des hauteurs de la premiere colonne, & </s> <s xml:id="echoid-s12974" xml:space="preserve">l’on aura 6 pieds, <lb/>& </s> <s xml:id="echoid-s12975" xml:space="preserve">ajouter auſſi enſemble les hauteurs des niveaux apparens <lb/>au deſſus du vrai, pour avoir 5 pouces 7 lignes, qu’il faut <lb/>ajouter avec la premiere colonne, & </s> <s xml:id="echoid-s12976" xml:space="preserve">le tout ſera 6 pieds <lb/>5 pouces 7 lignes.</s> <s xml:id="echoid-s12977" xml:space="preserve"/> </p> <pb o="383" file="0449" n="463" rhead="DE MATHÉMATIQUE. Liv. X."/> <p> <s xml:id="echoid-s12978" xml:space="preserve">Enſuite il faut ajouter les hauteurs de la ſeconde co-<lb/>lonne, qui font 14 pieds; </s> <s xml:id="echoid-s12979" xml:space="preserve">mettre auſſi dans une ſomme les <lb/>hauteurs du niveau apparent au deſſus du vrai, qui ſont à <lb/>côté, pour avoir un pied 4 lignes, qu’il faut ſouſtraire de <lb/>la ſomme des hauteurs de la ſeconde colonne, c’eſt-à-dire, <lb/>de 14 pieds, & </s> <s xml:id="echoid-s12980" xml:space="preserve">la différence ſera 12 pieds 11 pouces 8 lignes. <lb/></s> <s xml:id="echoid-s12981" xml:space="preserve">Enfin il faut ſouſtraire 6 pieds 5 pouces 7 lignes de cette quan-<lb/>tité, & </s> <s xml:id="echoid-s12982" xml:space="preserve">le reſte ſera 6 pieds 6 pouces une ligne, qui fait voir <lb/>que le lieu A eſt plus élevé que le lieu F de 6 pieds 6 pouces <lb/>une ligne.</s> <s xml:id="echoid-s12983" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12984" xml:space="preserve">772. </s> <s xml:id="echoid-s12985" xml:space="preserve">Quand le terrein le permet, il vaut beaucoup mieux <lb/>faire le nivellement entre deux termes, que de ſuivre ce qui <lb/>vient d’être dit, parce que l’on n’a point d’égard à la diffé-<lb/>rence du niveau apparent au deſſus du vrai, non plus que <lb/>dans les pratiques que nous avons données au ſujet du ni-<lb/>veau d’eau: </s> <s xml:id="echoid-s12986" xml:space="preserve">mais pour cela il ſeroit à propos que le niveau <lb/>eût deux lunettes, l’une pour pointer de la droite à la gau-<lb/>che, & </s> <s xml:id="echoid-s12987" xml:space="preserve">l’autre pour pointer de la gauche à la droite. </s> <s xml:id="echoid-s12988" xml:space="preserve">Les <lb/>corrections des coups de niveau ſe feront toujours de la même <lb/>façon qu’il a été enſeigné.</s> <s xml:id="echoid-s12989" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12990" xml:space="preserve">Par exemple, voulant connoître la différence des hau-<lb/> <anchor type="note" xlink:label="note-0449-01a" xlink:href="note-0449-01"/> teurs de deux endroits I & </s> <s xml:id="echoid-s12991" xml:space="preserve">E, je partage la diſtance de ces <lb/>deux termes, pour faire des ſtations aux endroits les plus <lb/>convenables; </s> <s xml:id="echoid-s12992" xml:space="preserve">& </s> <s xml:id="echoid-s12993" xml:space="preserve">ayant fait planter des piquets aux endroits <lb/>F, G, H, je fais ma premiere ſtation au point A, à peu <lb/>près dans le milieu de E F, la ſeconde au point B, auſſi dans <lb/>le milieu de F G, la troiſieme au point C, & </s> <s xml:id="echoid-s12994" xml:space="preserve">la quatrieme <lb/>au point D; </s> <s xml:id="echoid-s12995" xml:space="preserve">obſervant toujours d’écrire dans la premiere co-<lb/>lonne les hauteurs que l’on trouvera en montant, & </s> <s xml:id="echoid-s12996" xml:space="preserve">dans <lb/>la ſeconde celles que l’on trouvera en deſcendant, ſans ſe <lb/>mettre en peine des hauteurs du niveau apparent au deſſus <lb/>du vrai. </s> <s xml:id="echoid-s12997" xml:space="preserve">Je crois avoir aſſez dit pour ne rien laiſſer à déſirer <lb/>ſur tout ce qui regarde le nivellement; </s> <s xml:id="echoid-s12998" xml:space="preserve">& </s> <s xml:id="echoid-s12999" xml:space="preserve">pour peu qu’on <lb/>s’attache à le bien entendre, il ne faudra qu’un peu de pra-<lb/>tique pour être en état de faire toutes les opérations qui ſe <lb/>pourront préſenter.</s> <s xml:id="echoid-s13000" xml:space="preserve"/> </p> <div xml:id="echoid-div1034" type="float" level="2" n="4"> <note position="right" xlink:label="note-0449-01" xlink:href="note-0449-01a" xml:space="preserve">Figure 213.</note> </div> <pb o="384" file="0450" n="464" rhead="NOUVEAU COURS DE MATHEM. Liv. X."/> </div> <div xml:id="echoid-div1036" type="section" level="1" n="802"> <head xml:id="echoid-head970" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s13001" xml:space="preserve">M’étant apperçu qu’une grande partie de ceux qui ſe ſer-<lb/>vent tous les jours du toiſé, n’en ont que la routine, & </s> <s xml:id="echoid-s13002" xml:space="preserve">que <lb/>les perſonnes qui en ont écrit ne ſe ſont attachées qu’à don-<lb/>ner la pratique de ce calcul, ſans rien dire des raiſons ſur <lb/>leſquelles il eſt établi; </s> <s xml:id="echoid-s13003" xml:space="preserve">j’ai cru devoir en donner un petit <lb/>Traité avant de parler de la meſure des corps, afin que ceux <lb/>qui commencent puiſſent les calculer, & </s> <s xml:id="echoid-s13004" xml:space="preserve">trouvent dans cet <lb/>Ouvrage tout ce qu’il faut qu’ils ſçachent, pour être en état <lb/>de ſe ſervir utilement de ce qui a été enſeigné dans la pre-<lb/>miere Partie.</s> <s xml:id="echoid-s13005" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1037" type="section" level="1" n="803"> <head xml:id="echoid-head971" style="it" xml:space="preserve">Fin du dixieme Livre.</head> <figure> <image file="0450-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0450-01"/> </figure> <pb o="385" file="0451" n="465"/> <figure> <image file="0451-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0451-01"/> </figure> </div> <div xml:id="echoid-div1038" type="section" level="1" n="804"> <head xml:id="echoid-head972" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head973" xml:space="preserve">LIVRE ONZIEME.</head> <head xml:id="echoid-head974" style="it" xml:space="preserve">Du Toiſé en général, où l’on enſeigne la maniere de faire le <lb/>calcul du toiſé des plans, des ſolides, & de la charpente.</head> <p> <s xml:id="echoid-s13006" xml:space="preserve">773. </s> <s xml:id="echoid-s13007" xml:space="preserve">L’<emph style="sc">On</emph> entend ordinairement par le toiſé, la maniere de <lb/>calculer les dimenſions de tous les ouvrages qui font partie de <lb/>la fortification d’une place, & </s> <s xml:id="echoid-s13008" xml:space="preserve">même de tous les édifices civils. <lb/></s> <s xml:id="echoid-s13009" xml:space="preserve">Quoique chaque pays ait ſa meſure particuliere, & </s> <s xml:id="echoid-s13010" xml:space="preserve">que le pied <lb/>ne ſoit pas le même partout, cela n’empêche pas que pour les <lb/>ouvrages du Roi, l’on ne ſe ſerve toujours de la toiſe, qui eſt <lb/>(comme nous l’avons dit ailleurs) compoſée de ſix pieds. </s> <s xml:id="echoid-s13011" xml:space="preserve">Mais <lb/>comme le pied eſt dans un endroit de dix pouces, dans un <lb/>autre de onze pouces, on a nommé celui dont on ſe ſert en <lb/>France pour les fortifications, pied de Roi, lequel eſt compoſé <lb/>de 12 pouces; </s> <s xml:id="echoid-s13012" xml:space="preserve">ainſi la toiſe vaut 72 pouces. </s> <s xml:id="echoid-s13013" xml:space="preserve">L’on a auſſi diviſé <lb/>le pouce en douze parties, que l’on nomme lignes, & </s> <s xml:id="echoid-s13014" xml:space="preserve">la ligne <lb/>en douze autres parties, que l’on nomme points.</s> <s xml:id="echoid-s13015" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13016" xml:space="preserve">Cependant on diſtingue trois ſortes de toiſes; </s> <s xml:id="echoid-s13017" xml:space="preserve">la toiſe cou-<lb/>rante, la toiſe quarrée, & </s> <s xml:id="echoid-s13018" xml:space="preserve">la toiſe cube. </s> <s xml:id="echoid-s13019" xml:space="preserve">La toiſe courante eſt <lb/>celle qui a 6 pieds de longueur, ſans largeur ni profondeur; <lb/></s> <s xml:id="echoid-s13020" xml:space="preserve">la toiſe quarrée eſt celle qui a 6 pieds de longueur ſur 6 pieds <lb/>de largeur, ſans hauteur ou profondeur; </s> <s xml:id="echoid-s13021" xml:space="preserve">& </s> <s xml:id="echoid-s13022" xml:space="preserve">la toiſe cube eſt <lb/>celle qui a 6 pieds de longueur, 6 pieds de largeur, ſur 6 pieds <lb/>de hauteur, & </s> <s xml:id="echoid-s13023" xml:space="preserve">qui a par conſéquent les trois dimenſions égales:</s> <s xml:id="echoid-s13024" xml:space="preserve"> <pb o="386" file="0452" n="466" rhead="NOUVEAU COURS"/> auſſi cette toiſe ſert-elle à meſurer les ſolides, au lieu que la <lb/>toiſe quarrée ne ſert qu’à meſurer les ſuperficies, & </s> <s xml:id="echoid-s13025" xml:space="preserve">la toiſe <lb/>courante les longueurs, & </s> <s xml:id="echoid-s13026" xml:space="preserve">à déterminer les dimenſions des <lb/>plans & </s> <s xml:id="echoid-s13027" xml:space="preserve">des ſolides.</s> <s xml:id="echoid-s13028" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13029" xml:space="preserve">Ainſi ce que nous venons d’expliquer à l’égard de la toiſe, <lb/>eſt la même choſe que ce que l’on a dit à l’égard du pied au <lb/>commencement du premier Livre.</s> <s xml:id="echoid-s13030" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13031" xml:space="preserve">La toiſe quarrée ayant 6 pieds de longueur ſur 6 pieds de <lb/>largeur, l’on peut dire que ſa ſuperficie eſt compoſée de 36 <lb/>pieds quarrés, puiſque multipliant les deux dimenſions de <lb/>cette toiſe l’une par l’autre, c’eſt-à-dire 6 pieds par 6 pieds, <lb/>l’on aura 36 pieds quarrés: </s> <s xml:id="echoid-s13032" xml:space="preserve">à l’égard de la toiſe cube, comme <lb/>ſes trois dimenſions ſont chacune compoſées de 6 pieds, on <lb/>voit qu’elle doit être compoſée de 216 pieds cubes; </s> <s xml:id="echoid-s13033" xml:space="preserve">car multi-<lb/>pliant la toiſe quarrée, qui vaut 36 pieds quarrés par 6 pieds, <lb/>qui eſt la hauteur de la toiſe cube, l’on aura 216 pieds cubes.</s> <s xml:id="echoid-s13034" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13035" xml:space="preserve">774. </s> <s xml:id="echoid-s13036" xml:space="preserve">Il eſt bon de remarquer ici que dans le toiſé des plans <lb/>& </s> <s xml:id="echoid-s13037" xml:space="preserve">des ſolides, tel que nous l’allons expliquer, on ne conſidere <lb/>point combien il faut de pieds quarrés pour compoſer une toiſe <lb/>quarrée, ni combien il faut de pieds cubes pour compoſer une <lb/>toiſe cube; </s> <s xml:id="echoid-s13038" xml:space="preserve">parce que pour rendre le calcul plus court, l’on a <lb/>pris pour le pied de la toiſe quarrée, la ſixieme partie de la <lb/>même toiſe, & </s> <s xml:id="echoid-s13039" xml:space="preserve">pour le pied de la toiſe cube, la ſixieme partie <lb/>de cette toiſe; </s> <s xml:id="echoid-s13040" xml:space="preserve">tellement que ſi l’on conſidere le quarré A B <lb/> <anchor type="note" xlink:label="note-0452-01a" xlink:href="note-0452-01"/> comme une toiſe quarrée, dont le côté A C eſt diviſé en ſix <lb/>parties égales, le rectangle D E étant la ſixieme partie du quarré <lb/>A B, il ſera par conſéquent un pied de toiſe quarrée, de même <lb/>que le rectangle D F renferme 3 pieds de toiſe quarrée, puiſ-<lb/>qu’il eſt la moitié du quarré A B. </s> <s xml:id="echoid-s13041" xml:space="preserve">Mais comme la toiſe quarrée <lb/>vaut 36 pieds quarrés, & </s> <s xml:id="echoid-s13042" xml:space="preserve">que le rectangle D E eſt la ſixieme <lb/>partie de la toiſe, il s’enſuit qu’un pied de toiſe quarrée vaut <lb/>6 pieds quarrés, & </s> <s xml:id="echoid-s13043" xml:space="preserve">que le rectangle D F, qui eſt la moitié de <lb/>la toiſe, en vaut 18.</s> <s xml:id="echoid-s13044" xml:space="preserve"/> </p> <div xml:id="echoid-div1038" type="float" level="2" n="1"> <note position="left" xlink:label="note-0452-01" xlink:href="note-0452-01a" xml:space="preserve">Figure 214.</note> </div> <p> <s xml:id="echoid-s13045" xml:space="preserve">L’on pourroit dire la même choſe des pouces, des lignes, <lb/>des points de toiſe quarrée; </s> <s xml:id="echoid-s13046" xml:space="preserve">car un pouce tel que celui-ci eſt <lb/>un rectangle, qui a un pouce de baſe ſur une toiſe de hau-<lb/>teur; </s> <s xml:id="echoid-s13047" xml:space="preserve">de même une ligne eſt un rectangle, qui a une ligne de <lb/>baſe ſur une toiſe de hauteur. </s> <s xml:id="echoid-s13048" xml:space="preserve">Enfin un point eſt encore un <lb/>rectangle, qui a pour baſe la douzieme partie d’une ligne, & </s> <s xml:id="echoid-s13049" xml:space="preserve"><lb/>pour hauteur une toiſe: </s> <s xml:id="echoid-s13050" xml:space="preserve">ainſi l’on voit que 12 points de toiſe <pb o="387" file="0453" n="467" rhead="DE MATHÉMATIQUE. Liv. XI."/> quarrée font une ligne de la même toiſe, que 12 lignes font <lb/>un pouce, que 12 pouces font un pied, & </s> <s xml:id="echoid-s13051" xml:space="preserve">que 6 pieds font <lb/>une toiſe quarrée, puiſque toutes ces quantités ont la même <lb/>hauteur. </s> <s xml:id="echoid-s13052" xml:space="preserve">Nous ferons voir la même choſe à l’égard des pieds, <lb/>des pouces, des lignes & </s> <s xml:id="echoid-s13053" xml:space="preserve">des points de la toiſe cube, après <lb/>que nous aurons ſuffiſamment expliqué la maniere de multi-<lb/>plier deux dimenſions exprimées par des toiſes & </s> <s xml:id="echoid-s13054" xml:space="preserve">des parties <lb/>de toiſes courantes.</s> <s xml:id="echoid-s13055" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1040" type="section" level="1" n="805"> <head xml:id="echoid-head975" xml:space="preserve">CHAPITRE PREMIER,</head> <head xml:id="echoid-head976" style="it" xml:space="preserve">Où l’on fait voir comment on multiplie deux dimenſions, dont <lb/>la premiere eſt compoſée de toiſes & de parties de toiſes, & <lb/>la ſeconde de toiſes ſeulement.</head> <p> <s xml:id="echoid-s13056" xml:space="preserve">775. </s> <s xml:id="echoid-s13057" xml:space="preserve">A Yant une longueur A B de 6 toiſes, à laquelle on a <lb/>ajouté une petite longueur C B de 2 pieds, & </s> <s xml:id="echoid-s13058" xml:space="preserve">une autre C D <lb/>de 6 pouces, toute la ligne A D vaudra 6 toiſes 2 pieds 6 pouces; <lb/></s> <s xml:id="echoid-s13059" xml:space="preserve">laquelle étant multipliée par la ligne A E d’une toiſe, le pro-<lb/>duit donnera le rectangle E A D H, dont on aura la valeur, <lb/>en multipliant 6 toiſes 2 pieds 6 pouces par une toiſe, pour en <lb/>faire le calcul.</s> <s xml:id="echoid-s13060" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13061" xml:space="preserve">Je poſe les deux dimenſions comme on les <lb/> <anchor type="note" xlink:label="note-0453-01a" xlink:href="note-0453-01"/> voit ici; </s> <s xml:id="echoid-s13062" xml:space="preserve">enſuite je multiplie les plus petites <lb/>parties, en commençant par la droite, & </s> <s xml:id="echoid-s13063" xml:space="preserve">finiſ-<lb/>ſant par la gauche, en diſant: </s> <s xml:id="echoid-s13064" xml:space="preserve">une fois 6 eſt 6, <lb/>que je poſe à la colonne des pouces, parce que <lb/>ce ſont 6 pouces de toiſe quarrée, & </s> <s xml:id="echoid-s13065" xml:space="preserve">puis une fois 2 eſt 2, que <lb/>je poſe au rang des pieds, parce que ce ſont des pieds de toiſe <lb/>quarrée: </s> <s xml:id="echoid-s13066" xml:space="preserve">enfin une fois 6 eſt 6, que je poſe au rang des toiſes, <lb/>parce que ce ſont autant de toiſes quarrées: </s> <s xml:id="echoid-s13067" xml:space="preserve">ainſi le produit <lb/>6 toiſes 2 pieds 6 pouces, eſt la valeur du rectangle A H, le-<lb/>quel eſt compoſé du rectangle A F, qui vaut 6 toiſes du rec-<lb/>tangle B G, qui vaut 2 pieds, & </s> <s xml:id="echoid-s13068" xml:space="preserve">du rectangle C H, qui vaut <lb/>6 pouces.</s> <s xml:id="echoid-s13069" xml:space="preserve"/> </p> <div xml:id="echoid-div1040" type="float" level="2" n="1"> <note position="right" xlink:label="note-0453-01" xlink:href="note-0453-01a" xml:space="preserve">toiſes. # pieds. # pou. <lb/>6. # 2. # 6. <lb/>1. # 0. # 0. <lb/>6. # 2. # 6. <lb/></note> </div> <p> <s xml:id="echoid-s13070" xml:space="preserve">Pour multiplier 10 toiſes 4 pieds 8 pouces <lb/> <anchor type="note" xlink:label="note-0453-02a" xlink:href="note-0453-02"/> par 5 toiſes, je diſpoſe ce nombre comme on <lb/>le voit ici, & </s> <s xml:id="echoid-s13071" xml:space="preserve">je dis, 5 fois 8 font 40, faiſant <lb/>attention que ce font 40 unités, qui valent <pb o="388" file="0454" n="468" rhead="NOUVEAU COURS"/> chacune un petit rectangle, qui a pour baſe un pouce ſur une <lb/>toiſe de hauteur; </s> <s xml:id="echoid-s13072" xml:space="preserve">& </s> <s xml:id="echoid-s13073" xml:space="preserve">comme ce ſont autant de pouces de toiſe <lb/>quarrée, je conſidere en 40 combien il y a de fois 12, parce <lb/>que 12 pouces de toiſe quarrée valent un pied de la même <lb/>toiſe: </s> <s xml:id="echoid-s13074" xml:space="preserve">& </s> <s xml:id="echoid-s13075" xml:space="preserve">comme je trouve qu’en 40 il y a trois fois 12, & </s> <s xml:id="echoid-s13076" xml:space="preserve">4 de <lb/>reſte, je poſe 4 au rang des pouces, & </s> <s xml:id="echoid-s13077" xml:space="preserve">je retiens 3 pieds: </s> <s xml:id="echoid-s13078" xml:space="preserve">en-<lb/>ſuite je dis, 5 fois 4 font 20, & </s> <s xml:id="echoid-s13079" xml:space="preserve">3 de retenu, font 23, dont <lb/>chaque unité vaut un pied de toiſe quarrée; </s> <s xml:id="echoid-s13080" xml:space="preserve">& </s> <s xml:id="echoid-s13081" xml:space="preserve">comme il faut <lb/>6 de ces pieds pour faire une toiſe, je conſidere combien 6 ſe <lb/>trouve de fois dans 23; </s> <s xml:id="echoid-s13082" xml:space="preserve">& </s> <s xml:id="echoid-s13083" xml:space="preserve">comme il y eſt 3, & </s> <s xml:id="echoid-s13084" xml:space="preserve">qu’il reſte 5, <lb/>je poſe 5 au rang des pieds, & </s> <s xml:id="echoid-s13085" xml:space="preserve">je retiens 3, qui ſont autant de <lb/>toiſes quarrées, que j’ajoute avec le produit de 10 par 5, pour <lb/>avoir 53: </s> <s xml:id="echoid-s13086" xml:space="preserve">ainſi l’opération étant faite, on trouvera 53 toiſes <lb/>5 pieds 4 pouces.</s> <s xml:id="echoid-s13087" xml:space="preserve"/> </p> <div xml:id="echoid-div1041" type="float" level="2" n="2"> <note position="right" xlink:label="note-0453-02" xlink:href="note-0453-02a" xml:space="preserve">toiſes. # pieds. # pou. <lb/>10. # 4. # 8. <lb/>5. # 0. # 0. <lb/>53. # 5. # 4. <lb/></note> </div> <p> <s xml:id="echoid-s13088" xml:space="preserve">Pour multiplier 60 toiſ. </s> <s xml:id="echoid-s13089" xml:space="preserve">3 pieds 9 pouces <lb/> <anchor type="note" xlink:label="note-0454-01a" xlink:href="note-0454-01"/> par 84 toiſes, je remarque que le nombre <lb/>84 étant conſidérable, la mémoire ſeroit <lb/>fatiguée en multipliant les pieds & </s> <s xml:id="echoid-s13090" xml:space="preserve">les <lb/>pouces, comme on le voit dans cette opé-<lb/>ration: </s> <s xml:id="echoid-s13091" xml:space="preserve">car d’aller dire 84 fois 9, on n’ap-<lb/>perçoit pas d’abord combien ce produit <lb/>doit donner de pouces; </s> <s xml:id="echoid-s13092" xml:space="preserve">& </s> <s xml:id="echoid-s13093" xml:space="preserve">ſuppoſé qu’on <lb/>le ſçache à l’inſtant, l’on trouveroit en-<lb/>core un autre embarras, en cherchant combien ce produit <lb/>contient de pieds, à moins qu’on ne faſſe une diviſion par <lb/>12; </s> <s xml:id="echoid-s13094" xml:space="preserve">& </s> <s xml:id="echoid-s13095" xml:space="preserve">ceci ſe rencontrera, non ſeulement à l’égard des pouces, <lb/>mais encore pour les pieds, les lignes, & </s> <s xml:id="echoid-s13096" xml:space="preserve">les points. </s> <s xml:id="echoid-s13097" xml:space="preserve">Or pour <lb/>éviter les difficultés que pourroit donner un pareil calcul, on <lb/>agit d’une façon fort ſimple pour multiplier les pieds, les <lb/>pouces, les lignes & </s> <s xml:id="echoid-s13098" xml:space="preserve">les points de la premiere dimenſion, quand <lb/>le nombre de toiſes de la ſeconde eſt compoſé de plus d’une <lb/>figure. </s> <s xml:id="echoid-s13099" xml:space="preserve">Pour cela, il faut commencer par multiplier les entiers <lb/>par les entiers: </s> <s xml:id="echoid-s13100" xml:space="preserve">ainſi je multiplie 60 par 84, & </s> <s xml:id="echoid-s13101" xml:space="preserve">j’écris le pro-<lb/>duit comme à l’ordinaire; </s> <s xml:id="echoid-s13102" xml:space="preserve">enſuite je remarque que ſi au lieu de <lb/>3 pieds j’avois une toiſe à multiplier par 84, le produit ſeroit <lb/>84 toiſes; </s> <s xml:id="echoid-s13103" xml:space="preserve">mais comme 3 pieds ne valent que la moitié d’une <lb/>toiſe, la moitié de 84 ſera donc le produit de 3 pieds; </s> <s xml:id="echoid-s13104" xml:space="preserve">ainſi je <lb/>dis: </s> <s xml:id="echoid-s13105" xml:space="preserve">La moitié de 8 eſt 4, & </s> <s xml:id="echoid-s13106" xml:space="preserve">la moitié de 4. </s> <s xml:id="echoid-s13107" xml:space="preserve">eſt 2, ce qui donne <lb/>42 pour le produit; </s> <s xml:id="echoid-s13108" xml:space="preserve">mais il faut remarquer que dans le tems <lb/>que je prends la moitié de 84 pour le produit de 3 pieds, j’agis <pb o="389" file="0455" n="469" rhead="DE MATHÉMATIQUE. Liv. XI."/> comme ſi 84 contenoit des toiſes quarrées: </s> <s xml:id="echoid-s13109" xml:space="preserve">car pour que 42 <lb/>toiſes ſoient le produit de deux dimenſions, ou autrement <lb/>ſoient des toiſes quarrées, il faut que 84 ſoient regardées comme <lb/>des toiſes quarrées.</s> <s xml:id="echoid-s13110" xml:space="preserve"/> </p> <div xml:id="echoid-div1042" type="float" level="2" n="3"> <note position="right" xlink:label="note-0454-01" xlink:href="note-0454-01a" xml:space="preserve">toiſes. # pieds. # pou. <lb/>60. # 3. # 9. <lb/>84. # 0. # 0. <lb/>240. <lb/>480. <lb/>42. # 0. # 0. <lb/>10. # 3. # 0. <lb/>5092. # 3. # 0. <lb/></note> </div> <p> <s xml:id="echoid-s13111" xml:space="preserve">Mais comme il y a encore 9 pouces qui n’ont pas été mul-<lb/>tipliés, je conſidere quel eſt le rapport de 9 pouces avec 3 pieds, <lb/>de même que j’ai conſidéré celui de 3 pieds avec la toiſe. </s> <s xml:id="echoid-s13112" xml:space="preserve">Or <lb/>comme 3 pieds valent 36 pouces, je vois que le rapport de 9 <lb/>à 36 eſt un quart, & </s> <s xml:id="echoid-s13113" xml:space="preserve">que ſi le produit de 84 par 3 pieds a <lb/>donné 42 toiſes, le produit de 9 pouces par 84 ne doit donner <lb/>que le quart de 42: </s> <s xml:id="echoid-s13114" xml:space="preserve">je dis donc, le quart de 4 eſt 1, que je <lb/>poſe ſous le 4, & </s> <s xml:id="echoid-s13115" xml:space="preserve">le quart de 2 eſt 0; </s> <s xml:id="echoid-s13116" xml:space="preserve">mais comme 2 toiſes <lb/>valent 12 pieds, n’ayant pu prendre le quart de 2 toiſes en <lb/>nombres entiers, je les réduis en pieds pour en prendre le <lb/>quart, qui eſt 3; </s> <s xml:id="echoid-s13117" xml:space="preserve">après quoi je fais l’addition de tous ces pro-<lb/>duits, afin d’avoir le produit total, qui eſt 5092 toiſes & </s> <s xml:id="echoid-s13118" xml:space="preserve">3 pieds.</s> <s xml:id="echoid-s13119" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13120" xml:space="preserve">Pour rendre ce calcul plus familier aux Commençans, voici <lb/>encore pluſieurs exemples des mêmes Regles.</s> <s xml:id="echoid-s13121" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13122" xml:space="preserve">Pour multiplier 18 toiſes 2 pieds 8 pouces <lb/> <anchor type="note" xlink:label="note-0455-01a" xlink:href="note-0455-01"/> par 24 toiſes, l’on commence par multiplier <lb/>les toiſes par les toiſes, comme à l’ordinaire: <lb/></s> <s xml:id="echoid-s13123" xml:space="preserve">après cela il faut conſidérer le rapport de <lb/>2 pieds avec la toiſe; </s> <s xml:id="echoid-s13124" xml:space="preserve">& </s> <s xml:id="echoid-s13125" xml:space="preserve">comme 2 pieds en <lb/>eſt le tiers, je prends le tiers de 24, qui eſt <lb/>8; </s> <s xml:id="echoid-s13126" xml:space="preserve">& </s> <s xml:id="echoid-s13127" xml:space="preserve">comme ce ſont autant de toiſes, je les <lb/>place au rang des toiſes.</s> <s xml:id="echoid-s13128" xml:space="preserve"/> </p> <div xml:id="echoid-div1043" type="float" level="2" n="4"> <note position="right" xlink:label="note-0455-01" xlink:href="note-0455-01a" xml:space="preserve">toiſes. # pieds. # pou. <lb/>18. # 2. # 8. <lb/>24. # 0. # 0. <lb/>72. <lb/>36. <lb/>8. # 0. # 0. <lb/>2. # 4. # 0. <lb/>442. # 4. # 0. <lb/></note> </div> <p> <s xml:id="echoid-s13129" xml:space="preserve">Pour être convaincu que 24 multipliés par 2 pieds, donne <lb/>8 toiſes, faiſons-en la multiplication comme à l’ordinaire, <lb/>l’on verra que le produit eſt 48 pieds, c’eſt-à-dire 48 petits rec-<lb/>tangles, dont chacun a un pied pour baſe, & </s> <s xml:id="echoid-s13130" xml:space="preserve">une toiſe pour <lb/>hauteur: </s> <s xml:id="echoid-s13131" xml:space="preserve">& </s> <s xml:id="echoid-s13132" xml:space="preserve">comme il en faut 6 pour faire une toiſe quarrée, <lb/>l’on voit que diviſant 48 par 6, le quotient ſera 8, qui eſt le <lb/>même nombre que nous avons trouvé de l’autre façon.</s> <s xml:id="echoid-s13133" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13134" xml:space="preserve">Mais il nous reſte encore à multiplier 24 toiſes par 8 pouces; <lb/></s> <s xml:id="echoid-s13135" xml:space="preserve">& </s> <s xml:id="echoid-s13136" xml:space="preserve">comme cela ſe peut faire par le moyen du produit de 2 pieds, <lb/>je conſidere le rapport que 2 pieds ont avec 8 pouces, parce que <lb/>le rapport du produit de 8 pouces avec celui de 2 pieds ſera le <lb/>même que 8 pouces avec 2 pieds. </s> <s xml:id="echoid-s13137" xml:space="preserve">Or comme 2 pieds valent <lb/>24 pouces, & </s> <s xml:id="echoid-s13138" xml:space="preserve">que 8 en eſt le tiers, je prends le tiers du produit <lb/>de 2 pieds, c’eſt-à-dire le tiers de 8 toiſes, en diſant: </s> <s xml:id="echoid-s13139" xml:space="preserve">Le tiers <pb o="390" file="0456" n="470" rhead="NOUVEAU COURS"/> de 8 eſt 2, il reſte 2 toiſes, qui valent 12 pieds, dont le tiers <lb/>eſt 4 pieds, que je poſe au rang des pieds; </s> <s xml:id="echoid-s13140" xml:space="preserve">aprés quoi je fais <lb/>l’addition de tous les produits pour avoir le total, qui eſt 442 <lb/>toiſes 4 pieds.</s> <s xml:id="echoid-s13141" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13142" xml:space="preserve">Pour multiplier 36 toiſes 5 pieds <lb/> <anchor type="note" xlink:label="note-0456-01a" xlink:href="note-0456-01"/> 6 pouces 9 lignes par 28 toiſes, je <lb/>commence, comme à l’ordinaire, à <lb/>multiplier les toiſes par lestoiſes; </s> <s xml:id="echoid-s13143" xml:space="preserve">en-<lb/>ſuite je compare le rapport de 5 pieds <lb/>avec la toiſe, & </s> <s xml:id="echoid-s13144" xml:space="preserve">je vois que c’eſt les <lb/>{5/6}, & </s> <s xml:id="echoid-s13145" xml:space="preserve">par conſéquent il faut pour mul-<lb/>tiplier 28 toiſes par 5 pieds, prendre <lb/>les {5/6} de 28 toiſes; </s> <s xml:id="echoid-s13146" xml:space="preserve">& </s> <s xml:id="echoid-s13147" xml:space="preserve">comme il n’eſt <lb/>pas aiſé de prendre cela tout d’un <lb/>coup, je cherche des parties aliquotes pour rendre le calcul <lb/>plus aiſé; </s> <s xml:id="echoid-s13148" xml:space="preserve">& </s> <s xml:id="echoid-s13149" xml:space="preserve">comme 5 eſt compoſé de 3 & </s> <s xml:id="echoid-s13150" xml:space="preserve">de 2, dont 3 eſt la <lb/>moitié de la toiſe, & </s> <s xml:id="echoid-s13151" xml:space="preserve">2 le tiers, je prends d’abord pour 3 la <lb/>moitié de 28, qui eſt 14, enſuite pour 2 pieds le tiers, en di-<lb/>ſant: </s> <s xml:id="echoid-s13152" xml:space="preserve">Le tiers de 28 eſt 9; </s> <s xml:id="echoid-s13153" xml:space="preserve">& </s> <s xml:id="echoid-s13154" xml:space="preserve">comme il reſte une toiſe, j’en <lb/>prends encore le tiers, qui eſt 2 pieds.</s> <s xml:id="echoid-s13155" xml:space="preserve"/> </p> <div xml:id="echoid-div1044" type="float" level="2" n="5"> <note position="right" xlink:label="note-0456-01" xlink:href="note-0456-01a" xml:space="preserve">toiſes. # pieds. # pouces. # lig. <lb/>36. # 5. # 6. # 9. <lb/>28. # 0. # 0. # 0. <lb/>288. <lb/>72. <lb/>14. # 0. # 0. # 0. <lb/>9. # 2. # 0. # 0. <lb/>2. # 2. # 0. # 0. <lb/>0. # 1. # 9. <lb/>1033. # 5. # 9. # 0. <lb/></note> </div> <p> <s xml:id="echoid-s13156" xml:space="preserve">Pour multiplier les 6 pouces, j’ai recours au produit de 2 <lb/>pieds, qui paroît le plus commode, parce que 6 pouces eſt le <lb/>quart de 2 pieds, puiſque 2 pieds valent 24 pouces; </s> <s xml:id="echoid-s13157" xml:space="preserve">ainſi le <lb/>produit de 6 pouces ſera le quart de celui de 2 pieds; </s> <s xml:id="echoid-s13158" xml:space="preserve">& </s> <s xml:id="echoid-s13159" xml:space="preserve">comme <lb/>ce produit eſt 9 toiſes 2 pieds, je dis: </s> <s xml:id="echoid-s13160" xml:space="preserve">Le quart de 9 eſt 2, il <lb/>reſte une toiſe, qui vaut 6 pieds, leſquels étant ajoutés avec <lb/>les 2 pieds qui reſtent, font 8 pieds, dont le quart eſt 2: </s> <s xml:id="echoid-s13161" xml:space="preserve">ainſi <lb/>le produit de 6 pouces eſt 2 toiſes 2 pieds.</s> <s xml:id="echoid-s13162" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13163" xml:space="preserve">Comme il reſte encore 9 lignes, qui n’ont pas été multi-<lb/>pliées, je cherche le rapport de 9 lignes avec 6 pouces. </s> <s xml:id="echoid-s13164" xml:space="preserve">Or <lb/>comme 6 pouces valent 72 lignes, & </s> <s xml:id="echoid-s13165" xml:space="preserve">que 9 lignes en font la <lb/>huitieme partie, le produit de 9 lignes ſera donc la huitieme <lb/>partie de celui de 6 pouces, je dis donc: </s> <s xml:id="echoid-s13166" xml:space="preserve">La huitieme partie <lb/>de 2 eſt o; </s> <s xml:id="echoid-s13167" xml:space="preserve">mais ce ſont 2 toiſes qui valent 12 pieds, auxquels <lb/>ajoutant 2 pieds qui reſtent, on aura 14, dont la huitieme <lb/>partie eſt un pied, il reſte 6 pieds, que je réduis en pouces <lb/>pour avoir 72 pouces, dont la huitieme partie eſt 9, que je <lb/>poſe au rang des pouces; </s> <s xml:id="echoid-s13168" xml:space="preserve">après quoi je fais l’addition, qui <lb/>donne 1033 toiſes 5 pieds 9 pouces pour produit total.</s> <s xml:id="echoid-s13169" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13170" xml:space="preserve">Pour multiplier 12 toiſes 9 pouces par 18 toiſes, je fais la <pb o="391" file="0457" n="471" rhead="DE MATHÉMATIQUE. Liv. XI."/> multiplication des toiſes comme à l’ordi-<lb/> <anchor type="note" xlink:label="note-0457-01a" xlink:href="note-0457-01"/> naire; </s> <s xml:id="echoid-s13171" xml:space="preserve">enſuite pour multiplier 18 toiſes <lb/>par 9 pouces, je cherche le rapport de 9 <lb/>pouces avec la toiſe, & </s> <s xml:id="echoid-s13172" xml:space="preserve">je trouve qu’ils <lb/>en ſont la huitieme partie, puiſqu’une toiſe <lb/>vaut 72 pouces; </s> <s xml:id="echoid-s13173" xml:space="preserve">mais comme il ſe peut <lb/>rencontrer une quantité de nombres, <lb/>7, 11, 10, où ce rapport ne ſe fera pas ap-<lb/>percevoir aiſément, il vaut mieux faire <lb/>une fauſſe poſition, c’eſt-à-dire ſuppoſer <lb/>le produit d’un pied. </s> <s xml:id="echoid-s13174" xml:space="preserve">Faiſant donc comme s’il y avoit un pied <lb/>à la place du zero, je multiplie ce pied ſuppoſé par 18 toiſes; <lb/></s> <s xml:id="echoid-s13175" xml:space="preserve">& </s> <s xml:id="echoid-s13176" xml:space="preserve">comme un pied eſt la ſixieme partie de la toiſe, je prends <lb/>la ſixieme partie de 18, qui eſt 3 toiſes, que je poſe au rang des <lb/>toiſes, ayant ſoin de couper le 3 par un trait de plume, pour <lb/>faire voir qu’il ne doit point être compris dans l’addition. </s> <s xml:id="echoid-s13177" xml:space="preserve"><lb/>Cela poſé, je cherche le rapport de 9 pouces avec un pied, <lb/>qui eſt les {3/4}: </s> <s xml:id="echoid-s13178" xml:space="preserve">je prends donc d’abord pour 6 pouces, qui eſt la <lb/>moitié; </s> <s xml:id="echoid-s13179" xml:space="preserve">ainſi je dis: </s> <s xml:id="echoid-s13180" xml:space="preserve">la moitié de 3 eſt 1, il reſte une toiſe, <lb/>qui vaut 6 pieds, dont la moitié eſt 3; </s> <s xml:id="echoid-s13181" xml:space="preserve">enſuite je prends la moitié <lb/>de ce produit pour 3 pouces, en diſant: </s> <s xml:id="echoid-s13182" xml:space="preserve">la moitié d’un n’eſt <lb/>rien, mais c’eſt une toiſe qui vaut 6 pieds, leſquels étant <lb/>joints avec les 3 pieds qui reſtent, font 9 pieds, dont la moitié <lb/>eſt 4 pieds 6 pouces, que j’additionne avec les autres produits, <lb/>& </s> <s xml:id="echoid-s13183" xml:space="preserve">il vient 218 toiſes un pied 6 pouces pour le produit total.</s> <s xml:id="echoid-s13184" xml:space="preserve"/> </p> <div xml:id="echoid-div1045" type="float" level="2" n="6"> <note position="right" xlink:label="note-0457-01" xlink:href="note-0457-01a" xml:space="preserve">toiſes. # pieds. # pou. <lb/>12. # 0. # 9. <lb/>18. # 0. # 0. <lb/>96. <lb/>12. <lb/>1. # 3. # 0. <lb/>0. # 4. # 6. <lb/>218. # 1. # 6. <lb/></note> </div> <p> <s xml:id="echoid-s13185" xml:space="preserve">Pour multiplier 24 toiſes 2 pieds <lb/> <anchor type="note" xlink:label="note-0457-02a" xlink:href="note-0457-02"/> 6 lignes par 52 toiſes, il faut, après <lb/>avoir multiplié les toiſes par les toi-<lb/>ſes, chercher le rapport de 2 pieds <lb/>avec la toiſe; </s> <s xml:id="echoid-s13186" xml:space="preserve">& </s> <s xml:id="echoid-s13187" xml:space="preserve">comme c’eſt le tiers, <lb/>on prendra donc le tiers de 52, qui <lb/>eſt 17 toiſes 2 pieds. </s> <s xml:id="echoid-s13188" xml:space="preserve">Comme il reſte <lb/>6 lignes à multiplier par 52 toiſes, il <lb/>n’eſt pas aiſé de voir le rapport de 6 <lb/>lignes avec 2 pieds; </s> <s xml:id="echoid-s13189" xml:space="preserve">l’on auroit bien <lb/>plus de facilité, ſi l’on avoit le pro-<lb/>duit de quelque pouce: </s> <s xml:id="echoid-s13190" xml:space="preserve">cependant comme il n’y a pas de pouces <lb/>dans la premiere dimenſion, il faut ſe donner un produit ſup-<lb/>poſé d’un pouce; </s> <s xml:id="echoid-s13191" xml:space="preserve">& </s> <s xml:id="echoid-s13192" xml:space="preserve">comme un pouce eſt la 24<emph style="sub">e</emph> partie de 2 pieds, <lb/>je m’apperçois qu’il n’eſt pas encore aiſé de prendre la 24<emph style="sub">e</emph> partie <pb o="392" file="0458" n="472" rhead="NOUVEAU COURS"/> de 17 toiſes 2 pieds; </s> <s xml:id="echoid-s13193" xml:space="preserve">c’eſt pourquoi j’en prends la moitié pour <lb/>avoir le produit d’un pied ſeulement, qui ſera 8 toiſes 4 pieds. <lb/></s> <s xml:id="echoid-s13194" xml:space="preserve">Ayant poſé ces nombres à leurs places ordinaires, je les coupe <lb/>par un trait de plume, pour qu’ils ne ſoient pas compris dans <lb/>l’addition: </s> <s xml:id="echoid-s13195" xml:space="preserve">après cela je conſidere qu’un pouce étant la dou-<lb/>zieme partie d’un pied, ſi je prends la douzieme de 8 toiſes <lb/>4 pieds, j’aurai 4 pieds 4 pouces pour le produit d’un pied: </s> <s xml:id="echoid-s13196" xml:space="preserve"><lb/>après quoi je barre ces deux nombres, parce qu’ils compoſent <lb/>un produit ſuppoſé. </s> <s xml:id="echoid-s13197" xml:space="preserve">Or comme 6 lignes ſont la moitié d’un <lb/>pouce, il n’y a donc qu’à prendre la moitié de 4 pieds 4 pouces, <lb/>qui eſt 2 pieds 2 pouces, pour avoir le produit de 6 lignes: </s> <s xml:id="echoid-s13198" xml:space="preserve">ſi <lb/>l’on fait l’addition de tous les produits, l’on aura 1265 toiſes <lb/>4 pieds 2 pouces pour le produit total.</s> <s xml:id="echoid-s13199" xml:space="preserve"/> </p> <div xml:id="echoid-div1046" type="float" level="2" n="7"> <note position="right" xlink:label="note-0457-02" xlink:href="note-0457-02a" xml:space="preserve">toiſes. # pieds. # pouces. # lig. <lb/>24. # 2. # 0. # 6. <lb/>52. # 0. # 0. # 0. <lb/>48. <lb/>120. <lb/>17. # 2. # 0. # 0. <lb/># # 0. # 0. <lb/>0. # # # 0. <lb/>0. # 2. # 2. # 0. <lb/>1265. # 4. # 2. # 0. <lb/></note> </div> <p> <s xml:id="echoid-s13200" xml:space="preserve">Si l’on avoit eu à multiplier 24 toiſes 6 lignespar 52 toiſes, <lb/>& </s> <s xml:id="echoid-s13201" xml:space="preserve">que dans la premiere dimenſion il n’y eût eu ni pieds ni <lb/>pouces, comme on le ſuppoſe ici, il auroit fallu pour trouver <lb/>le produit de 6 lignes, ſuppoſer celui d’un pied, enſuite celui <lb/>d’un pouce pour avoir celui de 6 lignes, qui ſera la moitié de <lb/>celui d’un pouce.</s> <s xml:id="echoid-s13202" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1048" type="section" level="1" n="806"> <head xml:id="echoid-head977" xml:space="preserve">CHAPITRE II,</head> <head xml:id="echoid-head978" style="it" xml:space="preserve">Où l’on donne la maniere de multiplier deux dimenſions, dont <lb/>chacune eſt compoſée de toiſes, pieds, pouces, &c.</head> <p> <s xml:id="echoid-s13203" xml:space="preserve">776. </s> <s xml:id="echoid-s13204" xml:space="preserve">NOus avons affecté de ne pas mettre des pieds, pouces, <lb/>& </s> <s xml:id="echoid-s13205" xml:space="preserve">des lignes dans la ſeconde dimenſion des multiplications <lb/>que l’on a faites dans le chapitre précédent, afin de rendre les <lb/>opérations plus ſimples: </s> <s xml:id="echoid-s13206" xml:space="preserve">mais comme il arrive preſque toujours <lb/>que s’il y a des pieds, des pouces dans la premiere dimenſion, <lb/>il y en a auſſi dans la ſeconde, voici la maniere de multiplier <lb/>les parties de toiſes qui peuvent ſe rencontrer dans l’une & </s> <s xml:id="echoid-s13207" xml:space="preserve"><lb/>dans l’autre.</s> <s xml:id="echoid-s13208" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13209" xml:space="preserve">Pour multiplier 15 toiſes 4 pieds 8 pouces 7 lignes par 6 toiſes <lb/>3 pieds 6 pouces, je conſidere que le nombre des toiſes de la <lb/>ſeconde dimenſion étant exprimé par un chiffre ſeulement, <lb/>je puis faire la multiplication de toute la premiere dimenſion <lb/>par 6 toiſes, par un calcul de mémoire, comme on l’a fait au <lb/>commencement du chapitre précédent: </s> <s xml:id="echoid-s13210" xml:space="preserve">ainſi faiſant abſtrac- <pb o="393" file="0459" n="473" rhead="DE MATHÉMATIQUE. Liv. XI."/> tion pour un moment des 3 pieds <lb/> <anchor type="note" xlink:label="note-0459-01a" xlink:href="note-0459-01"/> 6 pouces de la ſeconde dimenſion, <lb/>je commence par multiplier les plus <lb/>petites parties de la premiere dimen-<lb/>ſion par 6 toiſes, en diſant: </s> <s xml:id="echoid-s13211" xml:space="preserve">ſix fois <lb/>7 font 42 lignes, qui valent 3 pouces <lb/>6 lignes. </s> <s xml:id="echoid-s13212" xml:space="preserve">Ayant poſé 6 lignes en leur <lb/>place, je retiens 3 pouces; </s> <s xml:id="echoid-s13213" xml:space="preserve">je dis en-<lb/>ſuite: </s> <s xml:id="echoid-s13214" xml:space="preserve">ſix fois 8 font 48, & </s> <s xml:id="echoid-s13215" xml:space="preserve">3 de retenus font 51 pouces, qui <lb/>valent 4 pieds 3 pouces: </s> <s xml:id="echoid-s13216" xml:space="preserve">je poſe 3 pouces, & </s> <s xml:id="echoid-s13217" xml:space="preserve">retiens 4 pieds, <lb/>& </s> <s xml:id="echoid-s13218" xml:space="preserve">je viens à la multiplication des pieds, en diſant: </s> <s xml:id="echoid-s13219" xml:space="preserve">ſix fois <lb/>4 font 24, & </s> <s xml:id="echoid-s13220" xml:space="preserve">4 de retenus font 28 pieds, qui valent 4 toiſes <lb/>4 pieds, je poſe 4 pieds, & </s> <s xml:id="echoid-s13221" xml:space="preserve">retiens 4 toiſes, que j’ajoute au <lb/>produit de 15 toiſes par 6 pour avoir 94: </s> <s xml:id="echoid-s13222" xml:space="preserve">ainſi le produit de <lb/>6 toiſes par la premiere dimenſion eſt 94 toiſes 4 pieds 3 pouces <lb/>6 lignes, qui eſt une quantité qui contient autant de fois la <lb/>premiere dimenſion, qu’il y a d’unités dans le nombre 6.</s> <s xml:id="echoid-s13223" xml:space="preserve"/> </p> <div xml:id="echoid-div1048" type="float" level="2" n="1"> <note position="right" xlink:label="note-0459-01" xlink:href="note-0459-01a" xml:space="preserve">toiſes. # pieds. # pouces. # lig. # poi. <lb/>15. # 4. # 8. # 7. # 0. <lb/>6. # 3. # 6. # 0. # 0. <lb/>94. # 4. # 3. # 6. # 0. <lb/>7. # 5. # 4. # 3. # 6. <lb/>1. # 1. # 10. # 8. # 7. <lb/>103. # 5. # 6. # 6. # 1. <lb/></note> </div> <p> <s xml:id="echoid-s13224" xml:space="preserve">Préſentement je conſidere que puiſque chaque toiſe du nom-<lb/>bre 6 a donné pour ſon produit une quantité ſemblable à celle <lb/>de la premiere dimenſion, ſi j’ai à multiplier cette premiere <lb/>dimenſion par des parties de la toiſe, il faut que le produit ait <lb/>le même rapport avec celui de la toiſe par la premiere dimen-<lb/>ſion, que ſes parties avec la toiſe même. </s> <s xml:id="echoid-s13225" xml:space="preserve">Cela poſé, comme <lb/>la premiere dimenſion doit être multipliée encore par 3 pieds, <lb/>je conſidere que 3 pieds étant la moitié de la toiſe, le pro-<lb/>duit de 3 pieds ſera la moitié de la premiere dimenſion, qui eſt <lb/>ſuppoſée dans ce cas avoir été multipliée par la toiſe; </s> <s xml:id="echoid-s13226" xml:space="preserve">ainſi je <lb/>dis: </s> <s xml:id="echoid-s13227" xml:space="preserve">la moitié de 15 eſt 7, il reſte une toiſe qui vaut 6 pieds, <lb/>qui étant ajoutés avec 4 pieds, font 10 pieds, dont la moitié eſt <lb/>5; </s> <s xml:id="echoid-s13228" xml:space="preserve">je dis enſuite: </s> <s xml:id="echoid-s13229" xml:space="preserve">la moitié de 8 eſt 4, & </s> <s xml:id="echoid-s13230" xml:space="preserve">la moitié de 7 lignes <lb/>eſt 3 lignes 6 points.</s> <s xml:id="echoid-s13231" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13232" xml:space="preserve">Comme il nous reſte encore 6 pouces à multiplier, je con-<lb/>ſidere que 6 pouces étant la ſixieme partie de 3 pieds, le pro-<lb/>duit de 6 pouces ſera la ſixieme partie de celui de 3 pieds; </s> <s xml:id="echoid-s13233" xml:space="preserve">ainſi <lb/>je prends la ſixieme partie de ce produit, qui donne une toiſe <lb/>un pied 10 pouces 8 lignes 7 points, qui étant ajoutés avec le <lb/>reſte, il vient 103 toiſes 5 pieds 6 pouces 6 lignes un point <lb/>pour le produit total.</s> <s xml:id="echoid-s13234" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13235" xml:space="preserve">Pour multiplier 68 toiſes 3 pieds 4 pouces 9 lignes par 9 toiſes <lb/>4 pieds 9 pouces, je commence par multiplier la premiere di- <pb o="394" file="0460" n="474" rhead="NOUVEAU COURS"/> menſion par 9, & </s> <s xml:id="echoid-s13236" xml:space="preserve">le produit donne <lb/> <anchor type="note" xlink:label="note-0460-01a" xlink:href="note-0460-01"/> 617 toiſes 6 pouces 9 lignes; </s> <s xml:id="echoid-s13237" xml:space="preserve">enſuite <lb/>je conſidere que 4 pieds ſont les <lb/>deux tiers de la toiſe: </s> <s xml:id="echoid-s13238" xml:space="preserve">ainſi je prends <lb/>deux fois le tiers pour avoir moins <lb/>d’embarras, c’eſt-à-dire, je prends <lb/>chaque fois pour deux pieds, en <lb/>diſant: </s> <s xml:id="echoid-s13239" xml:space="preserve">le tiers de 6 eſt 2, le tiers <lb/>de 8 eſt encore 2, & </s> <s xml:id="echoid-s13240" xml:space="preserve">il reſte 2 toiſes, <lb/>qui valent 12 pieds, qui étant ajou-<lb/>tés avec les 3 pieds qui ſont ſur la droite, font 15, dont le <lb/>tiers eſt 5. </s> <s xml:id="echoid-s13241" xml:space="preserve">Après cela le tiers de 4 eſt 1, & </s> <s xml:id="echoid-s13242" xml:space="preserve">il reſte un pouce, <lb/>qui vaut 12 lignes, qui étant ajoutées avec 9, font 21 lignes, <lb/>dont le tiers eſt 7: </s> <s xml:id="echoid-s13243" xml:space="preserve">ainſi le produit de 2 pieds étant 22 toiſes <lb/>5 pieds un pouce 7 lignes, j’écris encore une ſeconde fois ce <lb/>produit, afin que les deux faſſent celui de 4 pieds; </s> <s xml:id="echoid-s13244" xml:space="preserve">& </s> <s xml:id="echoid-s13245" xml:space="preserve">comme <lb/>il y a encore 9 pouces à multiplier, je prends ſeulement pour <lb/>6 pouces le quart du produit de 2 pieds, en diſant: </s> <s xml:id="echoid-s13246" xml:space="preserve">le quart <lb/>de 22 eſt 5, il reſte 2, qui valent 12 pieds, & </s> <s xml:id="echoid-s13247" xml:space="preserve">5 font 17, dont <lb/>le quart eſt 4, il reſte un pied, qui vaut 12 pouces, dont le <lb/>quart eſt 3, il reſte encore un pouce, qui vaut 12 lignes, & </s> <s xml:id="echoid-s13248" xml:space="preserve">7 <lb/>font 19, dont le quart eſt 4: </s> <s xml:id="echoid-s13249" xml:space="preserve">enfin il reſte 3 lignes, qui valent <lb/>36 points, dont le quart eſt 9 points; </s> <s xml:id="echoid-s13250" xml:space="preserve">de ſorte que le produit <lb/>de 6 pouces eſt 5 toiſes 4 pieds 3 pouces 4 lignes 9 points. </s> <s xml:id="echoid-s13251" xml:space="preserve">Mais <lb/>comme je dois avoir le produit de 9 pouces, & </s> <s xml:id="echoid-s13252" xml:space="preserve">que je n’ai <lb/>encore que celui de 6, je prends pour le produit de 3 pouces la <lb/>moitié de celui de 6 pouces, qui eſt 2 toiſes 5 pieds un pouce <lb/>8 lignes 4 points & </s> <s xml:id="echoid-s13253" xml:space="preserve">demi: </s> <s xml:id="echoid-s13254" xml:space="preserve">après quoi je fais l’addition de tous <lb/>ces produits, qui font enſemble 671 toiſes 2 pieds 3 pouces un <lb/>point & </s> <s xml:id="echoid-s13255" xml:space="preserve">demi.</s> <s xml:id="echoid-s13256" xml:space="preserve"/> </p> <div xml:id="echoid-div1049" type="float" level="2" n="2"> <note position="right" xlink:label="note-0460-01" xlink:href="note-0460-01a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>68. # 3. # 4. # 9. # 0. <lb/>9. # 4. # 9. # 0. # 0. <lb/>617. # 0. # 6. # 9. # 0. <lb/>22. # 5. # 1. # 7. # 0. <lb/>22. # 5. # 1. # 7. # 0. <lb/>5. # 4. # 3. # 4. # 9. <lb/>2. # 5. # 1. # 8. # 4. {1/2} <lb/>671. # 2. # 3. # 0. # 1. {1/2} <lb/></note> </div> <p> <s xml:id="echoid-s13257" xml:space="preserve">Pour multiplier 12 toiſes 5 pieds <lb/> <anchor type="note" xlink:label="note-0460-02a" xlink:href="note-0460-02"/> 6 pouces 4 lignes par 6 toiſes 4 pouces <lb/>8 lignes, je commence, comme à <lb/>l’ordinaire, par multiplier la premiere <lb/>dimenſion par 6 toiſes; </s> <s xml:id="echoid-s13258" xml:space="preserve">aprés quoi je <lb/>remarque que comme il n’y a point <lb/>de pieds dans la ſeconde dimenſion, <lb/>il n’eſt pas aiſé de trouver le produit <lb/>de 4 pouces, ſans faire une fauſſe poſition: </s> <s xml:id="echoid-s13259" xml:space="preserve">c’eſt pourquoi je <lb/>ſuppoſe le produit d’un pied, en prenant la ſixieme partie de <pb o="395" file="0461" n="475" rhead="DE MATHÉMATIQUE. Liv. XI."/> la premiere dimenſion, qui eſt 2 toiſes 11 pouces 8 points, <lb/>dont j’ai ſoin de barrer les chiffres; </s> <s xml:id="echoid-s13260" xml:space="preserve">& </s> <s xml:id="echoid-s13261" xml:space="preserve">comme 4 pouces eſt le <lb/>riers d’un pied, je prends le tiers du produit d’un pied, qui eſt <lb/>4 pieds 3 pouces 8 lignes 2 points & </s> <s xml:id="echoid-s13262" xml:space="preserve">deux tiers; </s> <s xml:id="echoid-s13263" xml:space="preserve">& </s> <s xml:id="echoid-s13264" xml:space="preserve">comme il y <lb/>a encore 8 lignes à multiplier, je vois que 8 lignes étant la <lb/>ſixieme partie de 4 pouces (puiſque 4 pouces valent 48 lignes) <lb/>le produit de 8 lignes ſera la ſixieme partie de celui de 4 pouces: <lb/></s> <s xml:id="echoid-s13265" xml:space="preserve">après avoir pris cette ſixieme partie, qui eſt 8 pouces 7 lignes <lb/>4 points & </s> <s xml:id="echoid-s13266" xml:space="preserve">4 neuviemes, j’additionne le tout pour avoir le <lb/>produit total, qui eſt 78 toiſes 2 pieds 2 pouces 3 lignes 7 <lb/>points {1/9}.</s> <s xml:id="echoid-s13267" xml:space="preserve"/> </p> <div xml:id="echoid-div1050" type="float" level="2" n="3"> <note position="right" xlink:label="note-0460-02" xlink:href="note-0460-02a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>12. # 5. # 6. # 4. # 0. <lb/>6. # 0. # 4. # 8. # 0. <lb/>77. # 3. # 2. # 0. # 0. <lb/>. # 4. # 3. # 8. # 2. {2/3} <lb/>. # 0. # 8. # 7. # 4. {4/9} <lb/>78. # 2. # 2. # 3. # 7. {1/9} <lb/></note> </div> <p> <s xml:id="echoid-s13268" xml:space="preserve">Pour multiplier 40 toiſ. </s> <s xml:id="echoid-s13269" xml:space="preserve">3 pieds <lb/> <anchor type="note" xlink:label="note-0461-01a" xlink:href="note-0461-01"/> 6 pouces 8 lignes par 24 toiſes 6 <lb/>pieds 8 pouces, je commence par <lb/>multiplier les toiſes par les toiſes, <lb/>au lieu de multiplier d’abord les <lb/>lignes, les pouces, & </s> <s xml:id="echoid-s13270" xml:space="preserve">les pieds de <lb/>la premiere dimenſion, à cauſe <lb/>qu’il y a plus d’une figure dans le <lb/>nombre des toiſes de la ſeconde <lb/>dimenſion; </s> <s xml:id="echoid-s13271" xml:space="preserve">enſuite j’agis comme <lb/>j’ai fait dans le chapitre précédent, <lb/>en prenant pour 3 pieds la moitié <lb/>de 24, qui eſt 12, n’ayant égard qu’aux nombres entiers de la <lb/>ſeconde dimenſion: </s> <s xml:id="echoid-s13272" xml:space="preserve">ainſi je fais abſtraction de 5 pieds & </s> <s xml:id="echoid-s13273" xml:space="preserve">de 8 <lb/>pouces qui s’y trouvent, parce qu’il n’eſt pas encore tems de <lb/>les multiplier. </s> <s xml:id="echoid-s13274" xml:space="preserve">Ayant donc trouvé le produit de 3 pieds, qui <lb/>eſt 12 toiſes, je conſidere que les 6 pouces qui ſont dans la pre-<lb/>miere dimenſion, font la ſixieme partie de 3 pieds, c’eſt-à-<lb/>dire la ſixieme partie de 12, qui eſt 2; </s> <s xml:id="echoid-s13275" xml:space="preserve">& </s> <s xml:id="echoid-s13276" xml:space="preserve">ayant encore 8 lignes <lb/>de la premiere dimenſion à multiplier, je vois que 6 pouces <lb/>valant 72 lignes, les 8 lignes en font la neuvieme partie, & </s> <s xml:id="echoid-s13277" xml:space="preserve">par <lb/>conſéquent le produit de ces 8 lignes ſera la neuvieme partie <lb/>du produit de 6 pouces. </s> <s xml:id="echoid-s13278" xml:space="preserve">Or comme le produit de 6 pouces eſt <lb/>2 toiſes, je dis: </s> <s xml:id="echoid-s13279" xml:space="preserve">la neuvieme partie de 2 n’eſt rien, mais ce ſont <lb/>2 toiſes, qui valent 12 pieds, dont la neuvieme partie eſt un <lb/>pied, & </s> <s xml:id="echoid-s13280" xml:space="preserve">il en reſte 3, qui valent 36 pouces, dont la neuvieme <lb/>partie eſt 4, que je place au rang des pouces.</s> <s xml:id="echoid-s13281" xml:space="preserve"/> </p> <div xml:id="echoid-div1051" type="float" level="2" n="4"> <note position="right" xlink:label="note-0461-01" xlink:href="note-0461-01a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>40. # 3. # 6. # 8. # 0. <lb/>24. # 5. # 8. # 0. # 0. <lb/>160. <lb/>80. <lb/>12. # 0. # 0. # 0. # 0. <lb/>2. # 0. # 0. # 0. # 0. <lb/>0. # 1. # 4. # 0. # 0. <lb/>20. # 1. # 9. # 4. # 0. <lb/>13. # 3. # 2. # 2. # 8. <lb/>4. # 3. # 0. # 8. # 10. {2/3} <lb/>1012. # 3. # 4. # 3. # 6. {2/3} <lb/></note> </div> <p> <s xml:id="echoid-s13282" xml:space="preserve">Juſqu’ici nous n’avons fait que multiplier la premiere di-<lb/>menſion par les 24 toiſes qui ſont dans la ſeconde: </s> <s xml:id="echoid-s13283" xml:space="preserve">mais comme <pb o="396" file="0462" n="476" rhead="NOUVEAU COURS"/> ces 24 toiſes ſont accompagnées de 5 pieds 8 pouces, il faut, <lb/>comme dans les opérations précédentes, chercher le produit <lb/>de ces deux quantités: </s> <s xml:id="echoid-s13284" xml:space="preserve">ainſi je conſidere que 5 pieds valent 3 <lb/>& </s> <s xml:id="echoid-s13285" xml:space="preserve">2, c’eſt-à-dire la moitié & </s> <s xml:id="echoid-s13286" xml:space="preserve">le tiers de la toiſe: </s> <s xml:id="echoid-s13287" xml:space="preserve">je prends <lb/>donc pour 3 pieds la moitié de toutes les quantités qui ſe trou-<lb/>vent dans la premiere dimenſion, & </s> <s xml:id="echoid-s13288" xml:space="preserve">pour 2 pieds le tiers de <lb/>ces mêmes quantités. </s> <s xml:id="echoid-s13289" xml:space="preserve">Or comme ce dernier produit eſt celui <lb/>de 2 pieds, je remarque que 8 pouces étant le tiers de 2 pieds, <lb/>le produit de 8 pouces ſera le tiers de celui de 2 pieds. </s> <s xml:id="echoid-s13290" xml:space="preserve">Ayant <lb/>donc pris le tiers de ce produit, je l’additionne avec les autres, <lb/>pour avoir le produit total, qui eſt 1012 toiſes 3 pieds 4 pouces <lb/>3 lignes 6 points {2/3}.</s> <s xml:id="echoid-s13291" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13292" xml:space="preserve">Pour multiplier 36 toiſes 3 pou-<lb/> <anchor type="note" xlink:label="note-0462-01a" xlink:href="note-0462-01"/> ces 9 lignes par 50 toiſes 8 lignes, <lb/>je multiplie les toiſes par les toiſes, <lb/>comme à l’ordinaire; </s> <s xml:id="echoid-s13293" xml:space="preserve">enſuite pour <lb/>trouver le produit de 3 pouces, je <lb/>vois que j’ai beſoin de ſuppoſer <lb/>celui d’un pied: </s> <s xml:id="echoid-s13294" xml:space="preserve">ainſi je prends la <lb/>ſixieme partie de 50 toiſes, qui eſt <lb/>8 toiſes 2 pieds; </s> <s xml:id="echoid-s13295" xml:space="preserve">& </s> <s xml:id="echoid-s13296" xml:space="preserve">comme 3 pou-<lb/>ces font le quart d’un pied, je <lb/>prends le quart de 8 toiſes 2 pieds, <lb/>qui eſt 2 toiſes 6 pouces: </s> <s xml:id="echoid-s13297" xml:space="preserve">après cela <lb/>je cherche le produit de 9 lignes, en conſidérant que 9 lignes <lb/>étant le quart de 3 pouces, qui valent 36 lignes, le quart du <lb/>produit de 3 pouces ſera par conſéquent celui de 9 lignes; </s> <s xml:id="echoid-s13298" xml:space="preserve">je <lb/>prends donc le quart de 2 toiſes 6 pouces, qui eſt 3 pieds un <lb/>pouce 6 lignes.</s> <s xml:id="echoid-s13299" xml:space="preserve"/> </p> <div xml:id="echoid-div1052" type="float" level="2" n="5"> <note position="right" xlink:label="note-0462-01" xlink:href="note-0462-01a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>36. # 0. # 3. # 9. # 0. <lb/>50. # 0. # 0. # 8. # 0. <lb/>1800. <lb/># # 0. # 0. # 0. <lb/>2. # 0. # 6. # 0. # 0. <lb/>0. # 3. # 1. # 6. # 0. <lb/># 0. # 0. <lb/>0. # # 0. # 0. # {1/2} <lb/>0. # 1. # 0. # 0. # 2. {1/2} <lb/>0. # 1. # 0. # 0. # 2. {1/2} <lb/>1802. # 5. # 7. # 6. # 5. <lb/></note> </div> <p> <s xml:id="echoid-s13300" xml:space="preserve">Après cela je vois que j’ai 8 lignes dans la ſeconde dimen-<lb/>ſion, & </s> <s xml:id="echoid-s13301" xml:space="preserve">que n’ayant ni pieds ni pouces dans cette dimenſion, <lb/>il faut néceſſairement ſuppoſer des faux produits pour trouver <lb/>celui de 8 lignes. </s> <s xml:id="echoid-s13302" xml:space="preserve">Je cherche donc d’abord celui d’un pied, en <lb/>prenant la ſixieme partie des quantités qui compoſent la pre-<lb/>miere dimenſion, & </s> <s xml:id="echoid-s13303" xml:space="preserve">je trouve 6 toiſes 7 lignes & </s> <s xml:id="echoid-s13304" xml:space="preserve">6 points: <lb/></s> <s xml:id="echoid-s13305" xml:space="preserve">mais comme le rapport de 8 lignes à un pied eſt encore trop <lb/>grand, pour ne point fatiguer la mémoire, je prends la dou-<lb/>zieme partie de ce produit, qui eſt 3 pieds 7 points & </s> <s xml:id="echoid-s13306" xml:space="preserve">demi pour <lb/>le produit d’un pouce; </s> <s xml:id="echoid-s13307" xml:space="preserve">& </s> <s xml:id="echoid-s13308" xml:space="preserve">comme 8 lignes ſont les deux tiers <lb/>d’un pouce, je prends pour leur produit les deux tiers de celui <pb o="397" file="0463" n="477" rhead="DE MATHÉMATIQUE. Liv. XI."/> d’un pouce, lequel ayant été additionné, donne pour le pro-<lb/>duit total 1802 toiſes 5 pieds 7 pouces 6 lignes & </s> <s xml:id="echoid-s13309" xml:space="preserve">5 points.</s> <s xml:id="echoid-s13310" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1054" type="section" level="1" n="807"> <head xml:id="echoid-head979" xml:space="preserve">CHAPITRE III,</head> <head xml:id="echoid-head980" style="it" xml:space="preserve">Où l’on donne la maniere de multiplier trois dimenſions exprimées <lb/>en toiſes, pieds, pouces, &c.</head> <p> <s xml:id="echoid-s13311" xml:space="preserve">777. </s> <s xml:id="echoid-s13312" xml:space="preserve">LE calcul que l’on a enſeigné dans les deux chapitres <lb/>précédens, ne convient qu’aux ſuperficies, parce que nous n’y <lb/>avons ſuppoſé que deux dimenſions; </s> <s xml:id="echoid-s13313" xml:space="preserve">il eſt vrai que le calcul <lb/>de trois dimenſions ne differe pas beaucoup de celui-ci, puiſ-<lb/>que pour en avoir le produit, il ne faut que multiplier celui <lb/>des deux premieres dimenſions par la troiſieme: </s> <s xml:id="echoid-s13314" xml:space="preserve">mais comme <lb/>le produit de trois dimenſions donne non ſeulement des toiſes <lb/>cubes, mais auſſi des pieds, des pouces, & </s> <s xml:id="echoid-s13315" xml:space="preserve">des lignes de toiſe <lb/>cube, voici l’idée qu’il faut avoir de ces différentes parties.</s> <s xml:id="echoid-s13316" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13317" xml:space="preserve">Nous avons dit que la toiſe cube étoit compoſée de 216 <lb/>pieds cubes; </s> <s xml:id="echoid-s13318" xml:space="preserve">mais dans le calcul on ne s’embarraſſe point de <lb/>ces ſortes de pieds: </s> <s xml:id="echoid-s13319" xml:space="preserve">car on entend par un pied de toiſe cube <lb/>la ſixieme partie de la même toiſe, qui eſt (ſi l’on veut) de <lb/>36 pieds cubes, qui font un parallelepipede E A F G H I D, <lb/>qui a pour baſe une toiſe quarrée E A H D, & </s> <s xml:id="echoid-s13320" xml:space="preserve">pour hauteur <lb/>la ligne H G d’un pied: </s> <s xml:id="echoid-s13321" xml:space="preserve">de ſorte que ce ſolide eſt la ſixieme <lb/>partie du corps E A B C, qui eſt une toiſe cube. </s> <s xml:id="echoid-s13322" xml:space="preserve">On conſidé-<lb/>rera de même que le pouce de toiſe cube eſt un parallelepi-<lb/>pede, qui a une toiſe quarrée pour baſe ſur un pouce de hau-<lb/>teur, & </s> <s xml:id="echoid-s13323" xml:space="preserve">qu’une ligne de toiſe cube eſt un parallelepipede, qui <lb/>a pour baſe une toiſe quarrée, & </s> <s xml:id="echoid-s13324" xml:space="preserve">une ligne pour hauteur; </s> <s xml:id="echoid-s13325" xml:space="preserve">ainſi <lb/>des autres parties.</s> <s xml:id="echoid-s13326" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13327" xml:space="preserve">778. </s> <s xml:id="echoid-s13328" xml:space="preserve">Il ſuit de cette définition, que 12 lignes de toiſe cube <lb/>font un pouce de la même toiſe; </s> <s xml:id="echoid-s13329" xml:space="preserve">que 12 pouces font un pied, <lb/>& </s> <s xml:id="echoid-s13330" xml:space="preserve">que 6 pieds font une toiſe cube; </s> <s xml:id="echoid-s13331" xml:space="preserve">puiſque tous ces ſolides <lb/>ont pour baſe une toiſe quarrée, & </s> <s xml:id="echoid-s13332" xml:space="preserve">des hauteurs, qui étant <lb/>jointes enſemble, peuvent donner des toiſes cubes, ou des <lb/>parties de toiſes cubes, comme on le va voir dans les opéra-<lb/>tions ſuivantes.</s> <s xml:id="echoid-s13333" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13334" xml:space="preserve">Pour multiplier trois dimenſions, dont la premiere eſt de <lb/>8 toiſes 2 pieds 4 pouces, la ſeconde 6 toiſes 4 pieds 8 pouces, <pb o="398" file="0464" n="478" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s13335" xml:space="preserve">la troiſieme 5 toiſes 3 pieds 6 <lb/> <anchor type="note" xlink:label="note-0464-01a" xlink:href="note-0464-01"/> pouces, il faut commencer par mul-<lb/>tiplier la ſeconde dimenſion par la <lb/>premiere, & </s> <s xml:id="echoid-s13336" xml:space="preserve">le produit ſera 56 toiſ. <lb/></s> <s xml:id="echoid-s13337" xml:space="preserve">5 pieds un pouce 9 lignes 4 points, <lb/>qu’il faut enſuite multiplier par la <lb/>troiſieme dimenſion, agiſſant com-<lb/>me dans les regles des chapitres <lb/>précédens, c’eſt-à-dire qu’il faut <lb/>faire comme ſi le produit des deux <lb/>premieres dimenſions ne faiſoit <lb/>qu’une dimenſion. </s> <s xml:id="echoid-s13338" xml:space="preserve">Je dis donc: </s> <s xml:id="echoid-s13339" xml:space="preserve"><lb/>cinq fois 4 font 20, qui ſont autant de points de toiſe cube, <lb/>c’eſt-à-dire que ce ſont autant de petits parallelepipedes, qui <lb/>ont pour baſe une toiſe quarrée, & </s> <s xml:id="echoid-s13340" xml:space="preserve">pour hauteur un point: </s> <s xml:id="echoid-s13341" xml:space="preserve"><lb/>car ſi l’on fait attention que chaque unité du nombre 4 eſt un <lb/>petit parallélogramme, qui a pour baſe un point, & </s> <s xml:id="echoid-s13342" xml:space="preserve">pour hau-<lb/>teur une toiſe, puiſque ce ſont des points de toiſe quarrée <lb/>(art. </s> <s xml:id="echoid-s13343" xml:space="preserve">774), l’on verra que multipliant ce parallélogramme <lb/>par une ou pluſieurs toiſes, ils ſeront changés en parallele-<lb/>pipedes, qui auront deux dimenſions d’une toiſe, qui font <lb/>enſemble une toiſe quarrée; </s> <s xml:id="echoid-s13344" xml:space="preserve">ce qui répond à la définition. </s> <s xml:id="echoid-s13345" xml:space="preserve"><lb/>De même ſi l’on multiplie 9 lignes de toiſe quarrée par des <lb/>toiſes, l’on aura encore des petits parallelepipedes, qui auront <lb/>pour baſe une toiſe quarrée, & </s> <s xml:id="echoid-s13346" xml:space="preserve">pour hauteur une ligne, puiſ-<lb/>que l’on aura multiplié par des toiſes les rectangles qui ont <lb/>une de leurs dimenſions, qui vaut une toiſe; </s> <s xml:id="echoid-s13347" xml:space="preserve">il en ſera ainſi <lb/>des pouces & </s> <s xml:id="echoid-s13348" xml:space="preserve">des pieds. </s> <s xml:id="echoid-s13349" xml:space="preserve">A l’égard des toiſes, il n’y a point <lb/>de doute que multipliant des toiſes quarrées par des toiſes cou-<lb/>rantes, le produit ne donne des toiſes cubes.</s> <s xml:id="echoid-s13350" xml:space="preserve"/> </p> <div xml:id="echoid-div1054" type="float" level="2" n="1"> <note position="right" xlink:label="note-0464-01" xlink:href="note-0464-01a" xml:space="preserve">toiſes. # pieds. # pouces. # lig. # points. <lb/>8. # 2. # 4. # 0. # 0. <lb/>6. # 4. # 8. # 0. # 0. <lb/>5. # 3. # 6. # 0. # 0. <lb/>8. # 2. # 4. # 0. # 0. <lb/>6. # 4. # 8. # 0. # 0. <lb/>50. # 2. # 0. # 0. # 0. <lb/>2. # 4. # 9. # 4. # 0. <lb/>2. # 4. # 9. # 4. # 0. <lb/># 5. # 7. # 1. # 4. <lb/>56. # 5. # 1. # 9. # 4. <lb/></note> </div> <p> <s xml:id="echoid-s13351" xml:space="preserve">Ainſi multipliant 56 toiſes 5 pieds 1 pouce 9 lignes 4 points <lb/>de toiſe quarrée par 5 toiſes courantes, le produit ſera 284 <lb/>toiſes 1 pied 8 pouces 10 lignes 8 points de toiſe cube.</s> <s xml:id="echoid-s13352" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13353" xml:space="preserve">Or comme 56 toiſes 5 pieds 1 pouce 9 lignes 4 points étant <lb/>multipliés par une toiſe, donneront des toiſes & </s> <s xml:id="echoid-s13354" xml:space="preserve">des parties de <lb/>toiſe cube, qui ſeront toujours exprimées par les mêmes <lb/>nombres qui ſont ici, c’eſt-à-dire par 56 toiſes 5 pieds, &</s> <s xml:id="echoid-s13355" xml:space="preserve">c. </s> <s xml:id="echoid-s13356" xml:space="preserve">ſi <lb/>l’on ſuppoſe que cette multiplication a été faite, la moitié de <lb/>cette quantité ſera donc le produit de 3 pieds: </s> <s xml:id="echoid-s13357" xml:space="preserve">ainſi comme il <lb/>y a 3 pieds dans la ſeconde dimenſion, je prends la moitié de <pb o="399" file="0465" n="479" rhead="DE MATHÉMATIQUE. Liv. XI."/> cette quantité, qui ſera 28 toiſes 2 pieds 6 pouces 10 lignes <lb/>8 points, que je regarde comme des toiſes & </s> <s xml:id="echoid-s13358" xml:space="preserve">des parties de <lb/>toiſe cube, qui compoſent le produit de 3 pieds.</s> <s xml:id="echoid-s13359" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13360" xml:space="preserve">Enfin comme il y a encore 6 pouces dans la troiſieme di-<lb/>menſion, je conſidere que 6 pouces étant la ſixieme partie de <lb/>3 pieds, le produit de 6 pouces ſera la ſixieme partie de celui <lb/>de 3 pieds: </s> <s xml:id="echoid-s13361" xml:space="preserve">ainſi prenant la ſixieme partie de ce produit, l’on <lb/>aura 4 toiſes 4 pieds 5 pouces une ligne 9 points & </s> <s xml:id="echoid-s13362" xml:space="preserve">un tiers pour <lb/>le produit de 6 pouces, qui étant ajoutés avec les autres, don-<lb/>neront le produit total de 317 toiſes 2 pieds 8 pouces 11 lignes <lb/>1 point & </s> <s xml:id="echoid-s13363" xml:space="preserve">un tiers.</s> <s xml:id="echoid-s13364" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13365" xml:space="preserve">Pour multiplier trois dimenſions, <lb/> <anchor type="note" xlink:label="note-0465-01a" xlink:href="note-0465-01"/> dont la premiere eſt 15 toiſes 5 pieds <lb/>3 pouces, la ſeconde 8 toiſes 3 pieds <lb/>9 pouces, & </s> <s xml:id="echoid-s13366" xml:space="preserve">la troiſieme 6 toiſes 2 <lb/>pieds 6 pouces, je multiplie, comme <lb/>ci-devant, les deux premieres di-<lb/>menſions l’une par l’autre pour avoir <lb/>leur produit, qui eſt 136 toiſes 5 <lb/>pieds 6 pouces 4 lignes 6 points; </s> <s xml:id="echoid-s13367" xml:space="preserve">& </s> <s xml:id="echoid-s13368" xml:space="preserve"><lb/>comme ce produit donne des toiſes <lb/>& </s> <s xml:id="echoid-s13369" xml:space="preserve">des parties de toiſes quarrées, je <lb/>multiplie encore le tout par la troi-<lb/>ſieme dimenſion, c’eſt-à-dire par 6 <lb/>toiſes 2 pieds 6 pouces, & </s> <s xml:id="echoid-s13370" xml:space="preserve">le pro-<lb/>duit donne 878 toiſes 3 pieds 5 pou-<lb/>ces 10 lignes 10 points & </s> <s xml:id="echoid-s13371" xml:space="preserve">demi.</s> <s xml:id="echoid-s13372" xml:space="preserve"/> </p> <div xml:id="echoid-div1055" type="float" level="2" n="2"> <note position="right" xlink:label="note-0465-01" xlink:href="note-0465-01a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>15. # 5. # 3. # 0. # 0. <lb/>8. # 3. # 9. # 0. # 0. <lb/>6. # 2. # 6. # 0. # 0. <lb/>15. # 5. # 3. # 0. # 0. <lb/>8. # 3. # 9. # 0. # 0. <lb/>127. # 0. # 0. # 0. # 0. <lb/>7. # 5. # 7. # 6. # 0. <lb/>1. # 5. # 10. # 10. # 6. <lb/>136. # 5. # 6. # 4. # 6. <lb/>6. # 2. # 6. # 0. # 0. <lb/>821. # 3. # 2. # 3. # 0. <lb/>45. # 3. # 10. # 1. # 6. <lb/>11. # 2. # 5. # 6. # 4. {1/2} <lb/>878. # 3. # 5. # 10. # 10. {1/2} <lb/></note> </div> <p> <s xml:id="echoid-s13373" xml:space="preserve">Pour multiplier trois dimenſions, dont la premiere eſt <lb/>4 toiſes 2 pieds 5 pouces, la ſeconde 3 toiſes 1 pied 6 pouces, <lb/>& </s> <s xml:id="echoid-s13374" xml:space="preserve">la troiſieme 5 pieds 4 pouces, je commence par multiplier <lb/>les deux premieres dimenſions, dont le produit eſt 14 toiſes <lb/>1 pied 10 pouces 3 lignes; </s> <s xml:id="echoid-s13375" xml:space="preserve">enſuite je multiplie ce produit par <lb/>5 pieds 4 pouces; </s> <s xml:id="echoid-s13376" xml:space="preserve">& </s> <s xml:id="echoid-s13377" xml:space="preserve">comme il n’y a point de toiſes dans la <lb/>troiſieme dimenſion, je poſe un zero en leur place, & </s> <s xml:id="echoid-s13378" xml:space="preserve">je mul-<lb/>tiplie par 5 pieds 4 pouces, commençant par prendre pour <lb/>5 pieds la moitié de 14 toiſes 1 pied, &</s> <s xml:id="echoid-s13379" xml:space="preserve">c; </s> <s xml:id="echoid-s13380" xml:space="preserve">enſuite je prends <lb/>pour 2 pieds le tiers de la même quantité, & </s> <s xml:id="echoid-s13381" xml:space="preserve">le produit donne <lb/>4 toiſes 4 pieds 7 pouces 5 lignes, dont je prends la ſixieme <lb/>partie pour le produit de 4 pouces, parce que 4 pouces eſt la <lb/>ſixieme partie de 2 pieds: </s> <s xml:id="echoid-s13382" xml:space="preserve">enfin j’additionne ce produit avec <pb o="400" file="0466" n="480" rhead="NOUVEAU COURS"/> les autres pour avoir 12 toiſes 4 pieds 3 pouces 9 lignes 4 points; <lb/></s> <s xml:id="echoid-s13383" xml:space="preserve">ce qui eſt le produit total.</s> <s xml:id="echoid-s13384" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">toiſes. # pieds. # pouces. # lignes. # points. <lb/>4. # 2. # 5. # 0. # 0. <lb/>3. # 1. # 6. # 0. # 0. <lb/>0. # 5. # 4. # 0. # 0. <lb/>4. # 2. # 5. # 0. # 0. <lb/>3. # 1. # 6. # 0. # 0. <lb/>13. # 1. # 3. # 0. # 0. <lb/># 4. # 4. # 10. # 0. <lb/># 2. # 2. # 5. # 0. <lb/>14. # 1. # 10. # 3. # 0. <lb/>0. # 5. # 4. # 0. # 0. <lb/>7. # 0. # 11. # 1. # 6. <lb/>4. # 4. # 7. # 5. # 0. <lb/>0. # 4. # 9. # 2. # 10. <lb/>12. # 4. # 3. # 9. # 4. <lb/></note> <p> <s xml:id="echoid-s13385" xml:space="preserve">Pour multiplier trois dimenſions, dont la premiere eſt 5 pieds <lb/>9 pouces 6 lignes, la ſeconde 3 pieds 6 pouces, & </s> <s xml:id="echoid-s13386" xml:space="preserve">la troiſieme <lb/>4 pieds 8 pouces 6 lignes, je range <lb/> <anchor type="note" xlink:label="note-0466-02a" xlink:href="note-0466-02"/> les deux premieres dimenſions l’une <lb/>ſur l’autre, en mettant des zero à la <lb/>place des toiſes; </s> <s xml:id="echoid-s13387" xml:space="preserve">enſuite comme il ſe <lb/>trouve 3 pieds dans la ſeconde di-<lb/>menſion, je prends la moitié des <lb/>termes de la premiere dimenſion, <lb/>pour avoir le produit de 3 pieds; </s> <s xml:id="echoid-s13388" xml:space="preserve">& </s> <s xml:id="echoid-s13389" xml:space="preserve"><lb/>comme il y a encore 6 pouces, qui <lb/>valent la ſixieme partie de 3 pieds, <lb/>je prends pour le produit de 6 pouces <lb/>la ſixieme partie du produit de 3 <lb/>pieds; </s> <s xml:id="echoid-s13390" xml:space="preserve">& </s> <s xml:id="echoid-s13391" xml:space="preserve">l’addition étant faite, il <lb/>vient 3 pieds 4 pouces, 6 lignes 6 <lb/>points pour le produit des deux pre-<lb/>mieres dimenſions, que je multiplie <lb/>enſuite par la 3<emph style="sub">me</emph>, qui eſt, comme <pb o="401" file="0467" n="481" rhead="DE MATHÉMATIQUE. Liv. XI."/> nous l’avons dit, compoſée de 4 pouces 8 lignes 6 points: <lb/></s> <s xml:id="echoid-s13392" xml:space="preserve">ainſi je commence par prendre deux fois le tiers de ce pro-<lb/>duit pour avoir celui de 4 pieds; </s> <s xml:id="echoid-s13393" xml:space="preserve">& </s> <s xml:id="echoid-s13394" xml:space="preserve">comme celui de 2 pieds <lb/>eſt 1 pied 1 pouce 6 lignes 2 points, je conſidere que 8 pouces <lb/>étant le tiers de 2 pieds, le produit de 8 pouces ſera le tiers de <lb/>celui de 2 pieds, qui donne 4 pouces 6 lignes & </s> <s xml:id="echoid-s13395" xml:space="preserve">{2/3} de points: </s> <s xml:id="echoid-s13396" xml:space="preserve"><lb/>mais nous avons encore 6 lignes dans la troiſieme dimenſion, <lb/>dont le rapport étant un peu éloigné de 8 pouces, je trouve <lb/>qu’il eſt moins embarraſſant de faire un faux produit; </s> <s xml:id="echoid-s13397" xml:space="preserve">& </s> <s xml:id="echoid-s13398" xml:space="preserve"><lb/>comme celui de 2 pouces conviendroit fort, parce qu’on n’au-<lb/>roit qu’à prendre le quart pour avoir celui de 6 lignes, je prends <lb/>donc le quart du produit de 8 pouces pour avoir celui de 2 <lb/>pouces, qui eſt 1 pouce 1 ligne 6 points & </s> <s xml:id="echoid-s13399" xml:space="preserve">{1/6}, dont je coupe les <lb/>figures; </s> <s xml:id="echoid-s13400" xml:space="preserve">& </s> <s xml:id="echoid-s13401" xml:space="preserve">prenant le quart de ce produit, il vient 3 lignes <lb/>4 points & </s> <s xml:id="echoid-s13402" xml:space="preserve">{7/12} pour le produit de 6 lignes: </s> <s xml:id="echoid-s13403" xml:space="preserve">& </s> <s xml:id="echoid-s13404" xml:space="preserve">comme il ne reſte <lb/>plus rien à multiplier, je fais l’addition de tous les produits <lb/>pour avoir le total, qui eſt 2 pieds 7 pouces 9 lignes 9 points <lb/>& </s> <s xml:id="echoid-s13405" xml:space="preserve">{1/4} de points cubes.</s> <s xml:id="echoid-s13406" xml:space="preserve"/> </p> <div xml:id="echoid-div1056" type="float" level="2" n="3"> <note position="right" xlink:label="note-0466-02" xlink:href="note-0466-02a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>0. # 5. # 9. # 6. # 0. <lb/>0. # 3. # 6. # 0. # 0. <lb/>0. # 4. # 8. # 6. # 0. <lb/>0. # 5. # 9. # 6. # 0. <lb/>0. # 3. # 6. # 0. # 0. <lb/>0. # 2. # 10. # 9. # 0. <lb/>0. # 0. # 5. # 9. # 6. <lb/>0. # 3. # 4. # 6. # 6. <lb/>0. # 4. # 8. # 6. # 0. <lb/>0. # 1. # 1. # 6. # 2. <lb/>0. # 1. # 1. # 6. # 2. <lb/>0. # 0. # 4. # 6. # 0.{2/3} <lb/>0. # 0. # # # {1/6} <lb/>0. # 0. # 0. # 3. # 4.{7/12} <lb/>0. # 2. # 7. # 9. # 4.{1/3} <lb/></note> </div> </div> <div xml:id="echoid-div1058" type="section" level="1" n="808"> <head xml:id="echoid-head981" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s13407" xml:space="preserve">779. </s> <s xml:id="echoid-s13408" xml:space="preserve">Comme les preuves de toutes les Regles d’Arithmé-<lb/>tique ſe font par des Regles contraires, il ſemble que la meil-<lb/>leure preuve que l’on puiſſe donner du calcul du toiſé, ſeroit <lb/>qu’aprés avoir multiplié deux dimenſions, l’on divisât le pro-<lb/>duit par la premiere dimenſion pour avoir la ſeconde au quo-<lb/>tient, ou bien diviſer par la ſeconde pour avoir la premiere: <lb/></s> <s xml:id="echoid-s13409" xml:space="preserve">il y en a qui pratiquent cette preuve, mais ils ſont obligés de <lb/>réduire tous les termes du produit en leur moindre eſpece, <lb/>auſſi-bien qu’une des dimenſions, c’eſt-à-dire que ſi l’on a ré-<lb/>duit le produit en lignes, il faut auſſi réduire une des di-<lb/>menſions en lignes: </s> <s xml:id="echoid-s13410" xml:space="preserve">après cela on fait une diviſion, dont on <lb/>réduit le quotient en toiſes, en pieds, &</s> <s xml:id="echoid-s13411" xml:space="preserve">c. </s> <s xml:id="echoid-s13412" xml:space="preserve">pour avoir l’autre <lb/>dimenſion; </s> <s xml:id="echoid-s13413" xml:space="preserve">mais comme cette preuve demande beaucoup d’o-<lb/>pération, en voici une beaucoup plus ſimple.</s> <s xml:id="echoid-s13414" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13415" xml:space="preserve">Après que l’on a trouvé le produit des deux dimenſions, <lb/>pour voir ſi l’opération eſt juſte, l’on prend la moitié de la <lb/>premiere dimenſion, & </s> <s xml:id="echoid-s13416" xml:space="preserve">l’on double la ſeconde; </s> <s xml:id="echoid-s13417" xml:space="preserve">enſuite l’on <lb/>multiplie les deux dimenſions ainſi changées l’une par l’autre, <lb/>& </s> <s xml:id="echoid-s13418" xml:space="preserve">il vient un ſecond produit, qui doit être égal au premier. <lb/></s> <s xml:id="echoid-s13419" xml:space="preserve">Par exemple, pour ſçavoir ſi le produit de 6 toiſes 5 pieds <pb o="402" file="0468" n="482" rhead="NOUVEAU COURS"/> 4 pouces par 4 toiſes 2 pieds 6 pouces, qui eſt 30 toiſes 2 pieds <lb/>6 pouces 8 lignes eſt bon, il faut prendre la moitié de la pre-<lb/>miere dimenſion pour avoir 3 toiſes 2 pieds 8 pouces, & </s> <s xml:id="echoid-s13420" xml:space="preserve">dou-<lb/>bler la ſeconde, qui vaudra 8 toiſes 5 pieds: </s> <s xml:id="echoid-s13421" xml:space="preserve">après cela ſi l’on <lb/>multiplie ces deux quantités l’une par l’autre, l’on trouvera <lb/>que le produit eſt encore 30 toiſes 2 pieds 6 pouces 8 lignes; <lb/></s> <s xml:id="echoid-s13422" xml:space="preserve">ce qui ne peut arriver autrement, ſi l’opération eſt bien faite.</s> <s xml:id="echoid-s13423" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1059" type="section" level="1" n="809"> <head xml:id="echoid-head982" xml:space="preserve">CHAPITRE IV,</head> <head xml:id="echoid-head983" style="it" xml:space="preserve">Où l’on donne la maniere de calculer le Toiſé de la charpente.</head> <p> <s xml:id="echoid-s13424" xml:space="preserve">780. </s> <s xml:id="echoid-s13425" xml:space="preserve">LE toiſé de la charpente eſt fort différent de celui des <lb/>autres ouvrages, parce que ce toiſé a une meſure particuliere, <lb/>que l’on nomme ſolive, qui eſt une quantité qui contient <lb/>3 pieds cubes de bois; </s> <s xml:id="echoid-s13426" xml:space="preserve">de ſorte que ſi l’on a une piece de bois <lb/>D C, dont la longueur A D ſoit de 6 pieds, la largeur A B de <lb/>12 pouces, & </s> <s xml:id="echoid-s13427" xml:space="preserve">l’épaiſſeur B C de 6 pouces, cette piece compo-<lb/>ſera une ſolive, puiſqu’elle vaut 3 pieds cubes. </s> <s xml:id="echoid-s13428" xml:space="preserve">Or comme la <lb/>toiſe cube vaut 216 pieds cubes, & </s> <s xml:id="echoid-s13429" xml:space="preserve">que 216 diviſé par 3 donne <lb/>72, il s’enſuit qu’une ſolive eſt la ſoixante & </s> <s xml:id="echoid-s13430" xml:space="preserve">douzieme partie <lb/>d’une toiſe cube.</s> <s xml:id="echoid-s13431" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13432" xml:space="preserve">La ſolive, ainſi que la toiſe, eſt diviſée en 6 pieds, que <lb/>l’on nomme pieds de ſolive, qui eſt une quantité d’une toiſe <lb/>de longueur ſur un pied de largeur, & </s> <s xml:id="echoid-s13433" xml:space="preserve">un pouce d’épaiſſeur: <lb/></s> <s xml:id="echoid-s13434" xml:space="preserve">de ſorte que ſi la ligne B G eſt la ſixieme partie de la ligne B C, <lb/>la ſolive D A F G B E H ſera un pied de ſolive, puiſqu’il eſt la <lb/>ſixieme partie de D C.</s> <s xml:id="echoid-s13435" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13436" xml:space="preserve">Comme un pied de toiſe cube vaut 36 pieds cubes, la ſolive <lb/>en ſera donc la douzieme partie: </s> <s xml:id="echoid-s13437" xml:space="preserve">& </s> <s xml:id="echoid-s13438" xml:space="preserve">comme un pied de ſolive <lb/>eſt la ſixieme partie de la ſolive, il s’enſuit qu’un pied de ſo-<lb/>live eſt la ſoixante & </s> <s xml:id="echoid-s13439" xml:space="preserve">douzieme partie d’un pied de toiſe cube, <lb/>puiſqu’il faut 6 pieds de ſolive pour faire une ſolive, & </s> <s xml:id="echoid-s13440" xml:space="preserve">12 ſo-<lb/>lives pour faire un pied de toiſe cube. </s> <s xml:id="echoid-s13441" xml:space="preserve">Comme le pouce de <lb/>ſolive eſt la douzieme partie du pied de folive, l’on verra de <lb/>même qu’il eſt la ſoixante & </s> <s xml:id="echoid-s13442" xml:space="preserve">douzieme partie d’un pouce de <lb/>toiſe cube: </s> <s xml:id="echoid-s13443" xml:space="preserve">il en ſera ainſi des lignes & </s> <s xml:id="echoid-s13444" xml:space="preserve">des points.</s> <s xml:id="echoid-s13445" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13446" xml:space="preserve">Il ſuit de ce qu’on vient de dire, que ſi l’on a une piece de <lb/>bois qui contienne un certain nombre de toiſes, de pieds &</s> <s xml:id="echoid-s13447" xml:space="preserve"> <pb o="403" file="0469" n="483" rhead="DE MATHÉMATIQUE. Liv. XI."/> de pouces cubes, pour réduire cette piece en ſolives, il faut <lb/>multiplier ſa valeur par 72, & </s> <s xml:id="echoid-s13448" xml:space="preserve">le produit ſera la quantité de <lb/>ſolives contenues dans la piece.</s> <s xml:id="echoid-s13449" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13450" xml:space="preserve">Par exemple, ſi l’on ſuppoſe que <lb/> <anchor type="note" xlink:label="note-0469-01a" xlink:href="note-0469-01"/> 2 toiſes 3 pieds 6 pouces cubes ſoient <lb/>la valeur d’une piece de bois, je con-<lb/>ſidere que chaque toiſe de cette quan-<lb/>tité vaut 72 ſolives, chaque pied 72 <lb/>pieds de ſolive, & </s> <s xml:id="echoid-s13451" xml:space="preserve">chaque pouce <lb/>72 pouces de ſolive: </s> <s xml:id="echoid-s13452" xml:space="preserve">ainſi ſi l’on mul-<lb/>tiplie 2 toiſes 3 pieds 6 pouces cubes <lb/>par 72, on aura 186 ſolives.</s> <s xml:id="echoid-s13453" xml:space="preserve"/> </p> <div xml:id="echoid-div1059" type="float" level="2" n="1"> <note position="right" xlink:label="note-0469-01" xlink:href="note-0469-01a" xml:space="preserve">toiſes. # pieds. # pou. # cubes. <lb/>2. # 3. # 6. <lb/>72. <lb/>144. <lb/>46. <lb/>6. <lb/>186 ſolives. <lb/></note> </div> <p> <s xml:id="echoid-s13454" xml:space="preserve">Pour meſurer une piece de bois, dont la premiere dimen-<lb/>ſion a 4 toiſes 5 pieds 9 pouces, la ſeconde 1 pied 6 pouces, <lb/>& </s> <s xml:id="echoid-s13455" xml:space="preserve">la troiſieme 1 pied 3 pouces, je <lb/> <anchor type="note" xlink:label="note-0469-02a" xlink:href="note-0469-02"/> multiplie, comme à l’ordinaire, la <lb/>premiere dimenſion par la ſeconde, <lb/>& </s> <s xml:id="echoid-s13456" xml:space="preserve">le produit donne une toiſe 1 pied <lb/>5 pouces 3 lignes, que je multiplie <lb/>par la troiſieme dimenſion pour <lb/>avoir 1 pied 6 pouces 7 lig. </s> <s xml:id="echoid-s13457" xml:space="preserve">1 point <lb/>& </s> <s xml:id="echoid-s13458" xml:space="preserve">demi. </s> <s xml:id="echoid-s13459" xml:space="preserve">Préſentement pour ré-<lb/>duire cette quantité en ſolives, je la <lb/>multiplie par 72. </s> <s xml:id="echoid-s13460" xml:space="preserve">Pour cela je prends <lb/>pour 1 pied la ſixieme partie de 72, <lb/>qui eſt 12, & </s> <s xml:id="echoid-s13461" xml:space="preserve">pour 6 pouces la moi-<lb/>tié du produit d’un pied, qui eſt 6: <lb/></s> <s xml:id="echoid-s13462" xml:space="preserve">& </s> <s xml:id="echoid-s13463" xml:space="preserve">comme il y a 7 lignes, je prends <lb/>d’abord pour 6 la douzieme partie <lb/>du produit de 6 pouces, qui eſt 3 <lb/>pieds; </s> <s xml:id="echoid-s13464" xml:space="preserve">enſuite pour une ligne la <lb/>ſixieme partie du produit précé-<lb/>dent, qui donne 6 pouces, il reſte <lb/>encore un point & </s> <s xml:id="echoid-s13465" xml:space="preserve">demi; </s> <s xml:id="echoid-s13466" xml:space="preserve">je prends premiérement pour un <lb/>point la douzieme partie de 6 pouces, qui eſt 6 lignes; </s> <s xml:id="echoid-s13467" xml:space="preserve">enfin <lb/>pour la moitié d’un point la moitié du dernier produit pour <lb/>avoir 3 lignes; </s> <s xml:id="echoid-s13468" xml:space="preserve">après quoi j’additionne le tout, qui donne <lb/>18 ſolives 3 pieds 6 pouces 9 lignes de ſolive, pour la valeur de <lb/>la piece de bois.</s> <s xml:id="echoid-s13469" xml:space="preserve"/> </p> <div xml:id="echoid-div1060" type="float" level="2" n="2"> <note position="right" xlink:label="note-0469-02" xlink:href="note-0469-02a" xml:space="preserve">toiſes. # pieds. # pouces. # lig. # points. <lb/>4. # 5. # 9. # 0. # 0. <lb/># 1. # 6. # 0. # 0. <lb/>0. # 4. # 11. # 6. # 0. <lb/>0. # 2. # 5. # 9. # 0. <lb/>1. # 1. # 5. # 3. # 0. <lb/>0. # 1. # 3. # 0. # 0. <lb/>0. # 1. # 2. # 10. # 6. <lb/>0. # 0. # 3. # 8. # 7.{1/2} <lb/>0. # 1. # 6. # 7 # 1.{1/2} <lb/>72. <lb/>12. # 0. # 0. # 0. # 0. <lb/>6. # 0. # 0. # 0. # 0. <lb/>0. # 3. # 0. # 0. # 0. <lb/>0. # 0. # 6. # 0. # 0. <lb/>0. # 0. # 0. # 6. # 0. <lb/>0. # 0. # 0. # 3. # 0. <lb/>18. # 3. # 6. # 9. # 0. <lb/></note> </div> <p> <s xml:id="echoid-s13470" xml:space="preserve">Il y a une maniere de calcuer les bois, qui eſt bien plus <pb o="404" file="0470" n="484" rhead="NOUVEAU COURS"/> courte que la précédente; </s> <s xml:id="echoid-s13471" xml:space="preserve">c’eſt de réduire d’abord une des <lb/>deux dimenſions de l’équarriſſage en pouces, enſuite les mettre <lb/>au rang des toiſes, & </s> <s xml:id="echoid-s13472" xml:space="preserve">l’autre à la place qu’elle doit occuper na-<lb/>turellement. </s> <s xml:id="echoid-s13473" xml:space="preserve">L’on multiplie ces deux dimenſions l’une par <lb/>l’autre, comme dans les regles précédentes, regardant celle <lb/>qu’on a miſe au rang des toiſes, comme des toiſes mêmes; <lb/></s> <s xml:id="echoid-s13474" xml:space="preserve">après quoi on multiplie le produit qui en vient par la longueur <lb/>de la piece pour avoir un ſecond produit, qui donne le nombre <lb/>des ſolives, des pieds & </s> <s xml:id="echoid-s13475" xml:space="preserve">des pouces de ſolive, qui ſont conte-<lb/>nues dans la piece.</s> <s xml:id="echoid-s13476" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13477" xml:space="preserve">Par exemple, pour calculer la même <lb/> <anchor type="note" xlink:label="note-0470-01a" xlink:href="note-0470-01"/> piece de bois que ci-devant, qui a 1 pied <lb/>6 pouces ſur 1 pied 3 pouces d’équarriſ-<lb/>ſage, & </s> <s xml:id="echoid-s13478" xml:space="preserve">4 toiſes 5 pieds 9 pouces de lon-<lb/>gueur, je réduis une des dimenſions de <lb/>l’équarriſſage en pouces, qui ſera, par <lb/>exemple, 1 pied 6 pouces pour avoir 18 <lb/>pouces, que je mets au rang des toiſes, & </s> <s xml:id="echoid-s13479" xml:space="preserve"><lb/>1 pied 3 pouces de l’autre dimenſion à leur <lb/>place ordinaire; </s> <s xml:id="echoid-s13480" xml:space="preserve">enſuite je prends pour <lb/>1 pied la ſixieme partie de 18, qui eſt 3; <lb/></s> <s xml:id="echoid-s13481" xml:space="preserve">& </s> <s xml:id="echoid-s13482" xml:space="preserve">comme il y a encore 3 pouces qui ſont <lb/>le quart d’un pied, je prends le quart du <lb/>produit d’un pied, pour avoir celui de 3 pouces, qui eſt <lb/>4 pieds 6 pouces, & </s> <s xml:id="echoid-s13483" xml:space="preserve">j’additionne le tout pour avoir le produit <lb/>de 3 toiſes 4 pieds 6 pouces, qu’il faut multiplier par la lon-<lb/>gueur de la piece, c’eſt-à-dire par 4 toiſes 5 pieds 9 pouces, & </s> <s xml:id="echoid-s13484" xml:space="preserve"><lb/>l’on aura 18 ſolives 3 pieds 6 pouces 9 lignes de ſolive.</s> <s xml:id="echoid-s13485" xml:space="preserve"/> </p> <div xml:id="echoid-div1061" type="float" level="2" n="3"> <note position="right" xlink:label="note-0470-01" xlink:href="note-0470-01a" xml:space="preserve">toiſes. # pieds. # pouces. # lig. <lb/>18. # 0. # 0. # 0. <lb/>0. # 1. # 3. # 0. <lb/>3. # 0. # 0. # 0. <lb/># 4. # 6. # 0. <lb/>3. # 4. # 6. # 0. <lb/>4. # 5. # 9. # 0. <lb/>15. # 0. # 0. # 0. <lb/>1. # 1. # 6. # 0. <lb/>1. # 5. # 3. # 0. <lb/># 2. # 9. # 9. <lb/>18. # 3. # 6. # 6. <lb/></note> </div> <p> <s xml:id="echoid-s13486" xml:space="preserve">Pour entendre ceci, conſidérez que ſi l’on a trois quantités <lb/>a, b, c à multiplier l’une par l’autre, le produit ſera a b c; <lb/></s> <s xml:id="echoid-s13487" xml:space="preserve">& </s> <s xml:id="echoid-s13488" xml:space="preserve">que ſi ce produit doit être multiplié par d, l’on aura a b c d; </s> <s xml:id="echoid-s13489" xml:space="preserve"><lb/>mais ſi au lieu de multiplier le produit a b c par d, l’on multi-<lb/>plioit ſeulement une des dimenſions, comme a par d, l’on <lb/>aura a d, b c, dont le produit donne encore a b c d; </s> <s xml:id="echoid-s13490" xml:space="preserve">ainſi c’eſt <lb/>la même choſe de multiplier le produit de trois dimenſions <lb/>par une quantité, ou de multiplier une des dimenſions par la <lb/>même quantité, & </s> <s xml:id="echoid-s13491" xml:space="preserve">enſuite ce produit par les autres dimenſions, <lb/>puiſqu’à la fin l’on trouvera toujours la même choſe pour le <lb/>produit total.</s> <s xml:id="echoid-s13492" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13493" xml:space="preserve">781. </s> <s xml:id="echoid-s13494" xml:space="preserve">Or ſi l’on fait attention qu’une toiſe vaut 72 pouces, <pb o="405" file="0471" n="485" rhead="DE MATHÉMATIQUE. Liv. XI."/> l’on verra que mettant un pouce au rang des toiſes, c’eſt <lb/>comme ſi on l’avoit multiplié par 72: </s> <s xml:id="echoid-s13495" xml:space="preserve">ainſi quand nous avons <lb/>mis 18 pouces au rang des toiſes, on les a donc multipliés par <lb/>72; </s> <s xml:id="echoid-s13496" xml:space="preserve">& </s> <s xml:id="echoid-s13497" xml:space="preserve">par conſéquent le produit de cette quantité par les deux <lb/>autres dimenſions eſt devenu 72 fois plus grand qu’il n’eût été, <lb/>ſi l’on avoit mis les 18 pouces à leur place ordinaire; </s> <s xml:id="echoid-s13498" xml:space="preserve">ce qui <lb/>fait voir que le produit doit donner des ſolives: </s> <s xml:id="echoid-s13499" xml:space="preserve">car le produit <lb/>total devient 72 fois plus grand qu’il n’eût été, ſi l’on n’avoit <lb/>pas mis les 18 pouces au rang des toiſes, & </s> <s xml:id="echoid-s13500" xml:space="preserve">que l’on eût fait <lb/>l’opération à l’ordinaire. </s> <s xml:id="echoid-s13501" xml:space="preserve">Mais pour donner aux Commençans <lb/>plus de facilité de ſe ſervir de cette méthode, voici encore <lb/>quelques exemples ſur le même ſujet.</s> <s xml:id="echoid-s13502" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13503" xml:space="preserve">Pour ſçavoir combien il y a de ſo-<lb/> <anchor type="note" xlink:label="note-0471-01a" xlink:href="note-0471-01"/> lives dans une piece de bois, qui a 3 <lb/>toiſes 4 pieds 8 pouces de longueur <lb/>ſur 8 à 14 pouces d’équarriſſage, je <lb/>poſe 8 pouces au rang des toiſes, & </s> <s xml:id="echoid-s13504" xml:space="preserve"><lb/>l’autre dimenſion, qui vaut 1 pied 2 <lb/>pouces, au rang qu’elle doit occuper, <lb/>& </s> <s xml:id="echoid-s13505" xml:space="preserve">je dis: </s> <s xml:id="echoid-s13506" xml:space="preserve">la ſixieme partie de 8 eſt 1, il <lb/>reſte 2, qui valent 12, dont la ſixieme <lb/>partie eſt 2; </s> <s xml:id="echoid-s13507" xml:space="preserve">& </s> <s xml:id="echoid-s13508" xml:space="preserve">comme il y a encore <lb/>2 pouces, qui font la ſixieme partie <lb/>d’un pied, je prends pour 2 pouces la <lb/>fixieme partie du produit d’un pied <lb/>pour avoir 1 pied 4 pouces, & </s> <s xml:id="echoid-s13509" xml:space="preserve">le produit total eſt une toiſe <lb/>3 pieds 4 pouces, que je multiplie par la longueur, c’eſt-à-dire <lb/>par 3 toiſes 4 pieds 8 pouces, & </s> <s xml:id="echoid-s13510" xml:space="preserve">le produit donne 5 ſolives 5 <lb/>pieds 3 pouces une ligne 4 points de ſolive pour la valeur de la <lb/>piece.</s> <s xml:id="echoid-s13511" xml:space="preserve"/> </p> <div xml:id="echoid-div1062" type="float" level="2" n="4"> <note position="right" xlink:label="note-0471-01" xlink:href="note-0471-01a" xml:space="preserve">toiſes. # pieds. # pon. # lig. # points <lb/>8. # 0. # 0. # 0. # 0. <lb/>0. # 1. # 2. # 0. # 0. <lb/>1. # 2. # 0. # 0. # 0. <lb/>0. # 1. # 4. # 0. # 0. <lb/>1. # 3. # 4. # 0. # 0. <lb/>3. # 4. # 8. # 0. # 0. <lb/>4. # 4. # 0. # 0. # 0. <lb/>0. # 3. # 1. # 4. # 0. <lb/>0. # 3. # 1. # 4. # 0. <lb/>0. # 1. # 0. # 5. # 4. <lb/>5. # 5. # 3. # 1. # 4. <lb/></note> </div> <p> <s xml:id="echoid-s13512" xml:space="preserve">L’on peut remarquer que ce n’eſt <lb/> <anchor type="note" xlink:label="note-0471-02a" xlink:href="note-0471-02"/> pas une néceſſité abſolue de commen-<lb/>cer par multiplier les deux dimenſions <lb/>de l’équarriſſage l’une par l’autre: </s> <s xml:id="echoid-s13513" xml:space="preserve">car <lb/>ſi l’on veut, il n’y a qu’à multiplier <lb/>la longueur par la dimenſion de l’é-<lb/>quarriſſage, qui doit être miſe au rang <lb/>des toiſes: </s> <s xml:id="echoid-s13514" xml:space="preserve">ainſi pour avoir la valeur <lb/>de la piece de bois précédente, je <lb/>prends pour premiere dimenſion la longueur, qui eſt 3 toiſes <pb o="406" file="0472" n="486" rhead="NOUVEAU COURS"/> 4 pieds 8 pouces; </s> <s xml:id="echoid-s13515" xml:space="preserve">& </s> <s xml:id="echoid-s13516" xml:space="preserve">ſuppoſant que 8 pouces de l’équarriſſage <lb/>valent 8 toiſes, je les poſe pour ſeconde dimenſion, & </s> <s xml:id="echoid-s13517" xml:space="preserve">la mul-<lb/>tiplication étant faite, il vient 30 toiſes 1 pied 4 pouces, qui <lb/>étant multipliés par 1 pied 2 pouces, donnent encore 5 ſolives <lb/>5 pieds 3 pouces une ligne 4 points de ſolive.</s> <s xml:id="echoid-s13518" xml:space="preserve"/> </p> <div xml:id="echoid-div1063" type="float" level="2" n="5"> <note position="right" xlink:label="note-0471-02" xlink:href="note-0471-02a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>3. # 4. # 8. # 0. # 0. <lb/>8. # 0. # 0. # 0. # 0. <lb/>30. # 1. # 4. # 0. # 0. <lb/>0. # 1. # 2. # 0. # 0. <lb/>5. # 0. # 2. # 8. # 0. <lb/>0. # 5. # 0. # 5. # 4. <lb/>5. # 5. # 3. # 1. # 4. <lb/></note> </div> <p> <s xml:id="echoid-s13519" xml:space="preserve">Pour calculer la valeur d’une piece <lb/> <anchor type="note" xlink:label="note-0472-01a" xlink:href="note-0472-01"/> de bois, qui a 3 toiſes 4 pieds de lon-<lb/>gueur ſur 10 à 9 pouces 6 lignes d’é-<lb/>quarriſſage, je prends la plus ſimple <lb/>de deux dimenſions de l’équarriſſage, <lb/>c’eſt-à-dire celle qui eſt compoſée des <lb/>pouces ſeulement, pour la mettre au <lb/>rang des toiſes: </s> <s xml:id="echoid-s13520" xml:space="preserve">ainſi ayant pris 10 <lb/>pour la premiere dimenſion, je la <lb/>multiplie par la longueur de la piece, <lb/>ou par l’autre dimenſion de l’équar-<lb/>riſſage: </s> <s xml:id="echoid-s13521" xml:space="preserve">car il eſt indifférent de mul-<lb/>tiplier d’abord par l’une ou l’autre de <lb/>ces quantités, comme on l’a déja dit: <lb/></s> <s xml:id="echoid-s13522" xml:space="preserve">ainſi je multiplie 10 par 3 toiſes 4 pieds pour avoir le produit, <lb/>qui eſt 36 toiſes 4 pieds, que je multiplie enſuite par 9 pouces <lb/>6 lignes, & </s> <s xml:id="echoid-s13523" xml:space="preserve">il vient 4 ſolives 5 pieds 4 lignes de ſolives pour la <lb/>valeur de la piece de bois.</s> <s xml:id="echoid-s13524" xml:space="preserve"/> </p> <div xml:id="echoid-div1064" type="float" level="2" n="6"> <note position="right" xlink:label="note-0472-01" xlink:href="note-0472-01a" xml:space="preserve">toiſes. # pieds. # pou. # lig. # points. <lb/>10. # 0. # 0. # 0. # 0. <lb/>3. # 4. # 0. # 0. # 0. <lb/>30. <lb/>5. # 0. # 0. # 0. # 0. <lb/>1. # 4. # 0. # 0. # 0. <lb/>36. # 4. # 0. # 0. # 0. <lb/>0. # 0. # 9. # 6. # 0. <lb/># 0. # # 0. <lb/>3. # 0. # 4. # 0. <lb/>1. # 3. # 2. # 0. <lb/># # 1. # 6. # 4. <lb/>4. # 5. # 0. # 4. <lb/></note> </div> <p> <s xml:id="echoid-s13525" xml:space="preserve">782. </s> <s xml:id="echoid-s13526" xml:space="preserve">S’il arrive que dans les deux dimenſions de l’équar-<lb/>riſſage il ſe trouve des pouces & </s> <s xml:id="echoid-s13527" xml:space="preserve">des lignes, il faut pour la <lb/>dimenſion qu’on doit changer de valeur, mettre les pouces <lb/>au rang des toiſes, comme à l’ordinaire, & </s> <s xml:id="echoid-s13528" xml:space="preserve">regarder les lignes <lb/>de cette dimenſion comme des pieds: </s> <s xml:id="echoid-s13529" xml:space="preserve">ainſi on les mettra au <lb/>rang des pieds, avec cette attention, qu’au lieu de mettre au-<lb/>tant de pieds qu’il y a de lignes, il n’en faut mettre que la <lb/>moitié, c’eſt-à-dire que ſi cette dimenſion eſt compoſée de <lb/>6 pouces 8 lignes, l’on mettra 6 pouces au rang des toiſes, & </s> <s xml:id="echoid-s13530" xml:space="preserve"><lb/>la moitié des lignes au rang des pieds, pour avoir 6 toiſes <lb/>4 pieds; </s> <s xml:id="echoid-s13531" xml:space="preserve">& </s> <s xml:id="echoid-s13532" xml:space="preserve">ſi au lieu de 8 on en avoit 7 ou 9, ou tout autre <lb/>nombre impair, on en prendra toujours la moitié, & </s> <s xml:id="echoid-s13533" xml:space="preserve">l’on mar-<lb/>quera 3 pieds 6 pouces, ou bien 4 pieds 6 pouces. </s> <s xml:id="echoid-s13534" xml:space="preserve">L’on va voir <lb/>ceci dans les deux exemples ſuivans.</s> <s xml:id="echoid-s13535" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13536" xml:space="preserve">Pour toiſer une piece de bois qui a 6 toiſes 3 pieds de lon-<lb/>gueur ſur 9 pouces 6 lignes à 10 pouces 8 lignes d’équarriſſage, <lb/>ilfaut, pour changer une des deux dimenſions de l’équarriſſage, <pb o="407" file="0473" n="487" rhead="DE MATHÉMATIQUE. Liv. XI."/> qui ſera, par exemple, 9 pouces 6 lignes, mettre 9 pouces au <lb/>rang des toiſes, & </s> <s xml:id="echoid-s13537" xml:space="preserve">la moitié de 6 lignes au rang des pieds, <lb/>pour avoir 9 toiſes trois pieds, qu’il faut multiplier par l’autre <lb/>dimenſion, c’eſt-à-dire par 10 pouces 8 lignes, pour avoir une <lb/>toiſe 2 pieds 5 pouces 4 lignes au produit, qui étant multiplié <lb/>par la longueur de la piece, l’on verra qu’elle contient 9 ſolives <lb/>10 pouces 8 lignes.</s> <s xml:id="echoid-s13538" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1066" type="section" level="1" n="810"> <head xml:id="echoid-head984" xml:space="preserve"><emph style="sc">Exemple</emph> I</head> <note position="right" xml:space="preserve">toiſ. # pieds. # pou. # lig. # points. <lb/>9. # 3. # 0. # 0. # 0. <lb/>0. # 0. # 10. # 8. # 0. <lb/># # # 0. # 0. <lb/>0. # 4. # 9. # 0. # 0. <lb/>0. # 3. # 2. # 0. # 0. <lb/>0. # 0. # 6. # 4. # 0. <lb/>1. # 2. # 5. # 4. # 0. <lb/>6. # 3. # 0. # 0. # 0. <lb/>8. # 2. # 8. # 0. # 0. <lb/>0. # 4. # 2. # 8. # 0. <lb/>9. # 0. # 10. # 8. # 0. <lb/></note> </div> <div xml:id="echoid-div1067" type="section" level="1" n="811"> <head xml:id="echoid-head985" xml:space="preserve"><emph style="sc">Exemple</emph> II</head> <note position="right" xml:space="preserve">toiſ. # pieds. # pon. # lig. # points. <lb/>8. # 3. # 6. # 0. # 0. <lb/>0. # 0. # 9. # 6. # 0. <lb/># # # 0. # 0. <lb/>0. # 4. # 3. # 6. # 0. <lb/>0. # 2. # 1. # 9. # 0. <lb/>0. # 0. # 4. # 3. # 6. <lb/>1. # 0. # 9. # 6. # 6. <lb/>0. # 5. # 8. # 0. # 0. <lb/>0. # 3. # 4. # 9. # 3. <lb/>0. # 2. # 2. # 2. # 2. <lb/>0. # 0. # 9. # 0. # 8.{2/3} <lb/>1. # 0. # 5. # 0. # 1.{2/3} <lb/></note> <p> <s xml:id="echoid-s13539" xml:space="preserve">Pour trouver la valeur d’une piece de bois, qui a 5 pieds 8 <lb/>pouces de longueur ſur 8 pouces 7 lignes à 9 pouces 4 lignes <lb/>d’équarriſſage, je porte 8 pouces à l’endroit des toiſes; </s> <s xml:id="echoid-s13540" xml:space="preserve">& </s> <s xml:id="echoid-s13541" xml:space="preserve">con-<lb/>ſidérant les 7 lignes de cette dimenſion comme valant des <lb/>pieds, je marque 3 pieds 6 pouces; </s> <s xml:id="echoid-s13542" xml:space="preserve">enſuite je multiplie cette <lb/>dimenſion ainſi changée par 9 pouces 6 lignes, & </s> <s xml:id="echoid-s13543" xml:space="preserve">le produit <lb/>donne une toiſe 9 pouces 6 lignes 6 points, quiétant multipliés <lb/>par 5 pieds 8 pouces, il vient une ſolive 5 pouces 1 point {2/3} pour <lb/>la valeur de la piece.</s> <s xml:id="echoid-s13544" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13545" xml:space="preserve">783. </s> <s xml:id="echoid-s13546" xml:space="preserve">Pour rendre raiſon de ce que nous avons dit qu’il fal-<lb/>loit regarder les lignes comme des pieds, après en avoir pris <lb/>la moitié, conſidérez que nous avons dit qu’il falloit multi-<lb/>plier une des dimenſions par 72, pour que la ſuite de la Regle <lb/>donnât des ſolives: </s> <s xml:id="echoid-s13547" xml:space="preserve">pour cela, ſi la dimenſion eſt 8 pouces <lb/>7 lignes, nous ſçavons que mettant 8 pouces à l’endroit des <lb/>toiſes, la multiplication par 72 ſe fait tout d’un coup; </s> <s xml:id="echoid-s13548" xml:space="preserve">mais à <lb/>l’égard de ces lignes qui reſtent, remarquez que ſi on les met-<lb/>toit au rang des pouces, c’eſt comme ſi on les multiplioit par <lb/>12; </s> <s xml:id="echoid-s13549" xml:space="preserve">& </s> <s xml:id="echoid-s13550" xml:space="preserve">que ſi du rang des pouces on les porte au rang des pieds, <pb o="408" file="0474" n="488" rhead="NOUVEAU COURS DE MATHEM. Liv. XI."/> c’eſt comme ſi on les multiplioit encore par 12: </s> <s xml:id="echoid-s13551" xml:space="preserve">ainſi quand <lb/>on poſe des lignes au rang des pieds, c’eſt proprement les mul-<lb/>tiplier par 144; </s> <s xml:id="echoid-s13552" xml:space="preserve">mais comme, ſelon notre regle, elles ne doi-<lb/>vent être multipliées que par 72, qui eſt la moitié de 144, il <lb/>faut donc, ſi l’on porte les lignes au rang des pieds, n’en pren-<lb/>dre que la moitié, pour n’voir que la moitié de 144.</s> <s xml:id="echoid-s13553" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13554" xml:space="preserve">Pour trouver la quantité de ſolives & </s> <s xml:id="echoid-s13555" xml:space="preserve">de ſes parties conte-<lb/>nues dans un pilot non équarri, dont le diametre ſeroit, par <lb/>exemple, de 14 pouces, pris à la tête ou dans le milieu, ſelon <lb/>qu’on le jugera plus à propos, & </s> <s xml:id="echoid-s13556" xml:space="preserve">dont la longueur ſeroit de <lb/>27 pieds 6 pouces, il faut quarrer le diametre pour avoir 196; <lb/></s> <s xml:id="echoid-s13557" xml:space="preserve">& </s> <s xml:id="echoid-s13558" xml:space="preserve">comme le rapport au quarré du diametre d’un cercle eſt à la <lb/>ſuperficie du même cercle, à peu de choſe près, comme 14 eſt <lb/>à 11, l’on dira comme 14 eſt à 11; </s> <s xml:id="echoid-s13559" xml:space="preserve">ainſi 196, quarré du dia-<lb/>metre du pilot eſt à la ſuperficie de ſon cercle, qu’on trouvera <lb/>de 154 pouces quarrés, qu’il faut diviſer par 72, pour avoir des <lb/>baſes de ſolives; </s> <s xml:id="echoid-s13560" xml:space="preserve">l’on trouvera 2 au quotient qu’il faut poſer au <lb/>rang des ſolives: </s> <s xml:id="echoid-s13561" xml:space="preserve">comme il reſte 10 pouces, qui ne ſuffiſent <lb/>pas pour faire un pied, on mettra zero au rang des pieds, & </s> <s xml:id="echoid-s13562" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s13563" xml:space="preserve">les 10 pouces immédiatement après, pour avoir 2 ſolives <lb/>0 pieds 10 pouces, qu’il faut enſuite multiplier par la longueur <lb/>du pilot, c’eſt-à-dire par 4 toiſes 3 pieds 6 pouces, comme au <lb/>calcul ordinaire du toiſé, & </s> <s xml:id="echoid-s13564" xml:space="preserve">l’on trouvera 9 ſolives 4 pieds <lb/>9 pouces 10 lignes pour la valeur du pilot.</s> <s xml:id="echoid-s13565" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13566" xml:space="preserve">Si l’on avoit pluſieurs pilots de même groſſeur, il faudroit <lb/>trouver, comme l’on vient de faire, la ſuperficie de leurs cer-<lb/>cles communs, la diviſer de même par 72, afin d’avoir des <lb/>baſes de ſolives, & </s> <s xml:id="echoid-s13567" xml:space="preserve">multiplier ce qui viendra par la ſomme de <lb/>to utes les longueurs différentes.</s> <s xml:id="echoid-s13568" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1068" type="section" level="1" n="812"> <head xml:id="echoid-head986" style="it" xml:space="preserve">Fin du onzieme Livre.</head> <figure> <image file="0474-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0474-01"/> </figure> <pb o="409" file="0475" n="489"/> <figure> <image file="0475-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0475-01"/> </figure> </div> <div xml:id="echoid-div1069" type="section" level="1" n="813"> <head xml:id="echoid-head987" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head988" xml:space="preserve">LIVRE DOUZIEME, <lb/>Où l’on applique la Géométrie à la meſure des Superficies <lb/>& des Solides.</head> <head xml:id="echoid-head989" xml:space="preserve">CHAPITRE PREMIER. <lb/>De la meſure des ſuperficies. <lb/>PROPOSITION I. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13569" xml:space="preserve">784. </s> <s xml:id="echoid-s13570" xml:space="preserve"><emph style="sc">Me</emph>ſurer les figures triangulaires.</s> <s xml:id="echoid-s13571" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Pl. XVI.</note> <p> <s xml:id="echoid-s13572" xml:space="preserve">Si l’on a un triangle rectangle ABC, dont la baſe B C ſoit <lb/> <anchor type="note" xlink:label="note-0475-02a" xlink:href="note-0475-02"/> de 8 pieds, & </s> <s xml:id="echoid-s13573" xml:space="preserve">la hauteur A B de 5, il faut, pour en trouver la <lb/>ſuperficie, multiplier la moitié de la baſe par toute la perpen-<lb/>diculaire, ou la moitié de la perpendiculaire par toute la baſe, <lb/>& </s> <s xml:id="echoid-s13574" xml:space="preserve">l’on aura 20 pieds quarrés pour la valeur du triangle (art. </s> <s xml:id="echoid-s13575" xml:space="preserve">388).</s> <s xml:id="echoid-s13576" xml:space="preserve"/> </p> <div xml:id="echoid-div1069" type="float" level="2" n="1"> <note position="right" xlink:label="note-0475-02" xlink:href="note-0475-02a" xml:space="preserve">Figure 216.</note> </div> <p> <s xml:id="echoid-s13577" xml:space="preserve">785. </s> <s xml:id="echoid-s13578" xml:space="preserve">Si le triangle n’étoit pas rectangle, comme D E F, il <lb/>faudroit, en connoiſſant les trois côtés, chercher la valeur de <lb/>la perpendiculaire E G (art. </s> <s xml:id="echoid-s13579" xml:space="preserve">413), & </s> <s xml:id="echoid-s13580" xml:space="preserve">multiplier encore la <lb/>moitié de la baſe par toute la perpendiculaire, ou toute la per-<lb/>pendiculaire par la moitié de la baſe.</s> <s xml:id="echoid-s13581" xml:space="preserve"/> </p> <pb o="410" file="0476" n="490" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s13582" xml:space="preserve">786. </s> <s xml:id="echoid-s13583" xml:space="preserve">Mais comme il peut arriver que la perpendiculaire au <lb/> <anchor type="note" xlink:label="note-0476-01a" xlink:href="note-0476-01"/> lieu de tomber dans le triangle tombe en dehors, comme H L, <lb/>en ce cas il en faut chercher la valeur (art. </s> <s xml:id="echoid-s13584" xml:space="preserve">411), & </s> <s xml:id="echoid-s13585" xml:space="preserve">la multi-<lb/>plier par la moitié de la baſe I K.</s> <s xml:id="echoid-s13586" xml:space="preserve"/> </p> <div xml:id="echoid-div1070" type="float" level="2" n="2"> <note position="left" xlink:label="note-0476-01" xlink:href="note-0476-01a" xml:space="preserve">Figure 217.</note> </div> <p> <s xml:id="echoid-s13587" xml:space="preserve">787. </s> <s xml:id="echoid-s13588" xml:space="preserve">Enfin ſi l’on avoit ſeulement les trois côtés d’un trian-<lb/>gle, l’on pourra également avoir ſa ſuperficie, en ſuivant ce <lb/>qui eſt enſeigné dans l’art. </s> <s xml:id="echoid-s13589" xml:space="preserve">530, c’eſt-à-dire que ſuppoſant le <lb/>côté D E de 10 pieds, le côté E F de 11, & </s> <s xml:id="echoid-s13590" xml:space="preserve">le côté D F de 13, <lb/>il faut les ajouter enſemble pour avoir 34 pieds, dont on pren-<lb/>dra la moitié, qui eſt 17; </s> <s xml:id="echoid-s13591" xml:space="preserve">enſuite la différence des mêmes <lb/>côtés avec cette moitié, qui font 7, 6 & </s> <s xml:id="echoid-s13592" xml:space="preserve">4: </s> <s xml:id="echoid-s13593" xml:space="preserve">après quoi l’on <lb/>multipliera de ſuite les quatre termes, 17, 7, 6 & </s> <s xml:id="echoid-s13594" xml:space="preserve">4 l’un par <lb/>l’autre, j’entends 17 par 7, qui donneront 119; </s> <s xml:id="echoid-s13595" xml:space="preserve">enſuite ce <lb/>produit par ſix pour avoir 714, & </s> <s xml:id="echoid-s13596" xml:space="preserve">ce dernier par 4, qui donne <lb/>2856, dont il faut extraire la racine qu’on trouvera de 52 pieds <lb/>5 pouces & </s> <s xml:id="echoid-s13597" xml:space="preserve">3 lignes de pied quarré pour la ſuperficie du trian-<lb/>gle D E F.</s> <s xml:id="echoid-s13598" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1072" type="section" level="1" n="814"> <head xml:id="echoid-head990" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13599" xml:space="preserve">788. </s> <s xml:id="echoid-s13600" xml:space="preserve">Trouver la ſuperficie des figures quadrilateres. <lb/></s> </p> <p> <s xml:id="echoid-s13601" xml:space="preserve">Pour trouver la ſuperficie du quarré A C, dont le côté ſe-<lb/>roit, par exemple, de 7 pieds, il faut multiplier 7 par lui-<lb/>même, c’eſt-à-dire A B par B C, & </s> <s xml:id="echoid-s13602" xml:space="preserve">le produit ſera 49 pieds, <lb/>qui eſt la valeur du quarré A C.</s> <s xml:id="echoid-s13603" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13604" xml:space="preserve">789. </s> <s xml:id="echoid-s13605" xml:space="preserve">Si au lieu d’un quarré l’on a un rectangle D F, dont <lb/> <anchor type="note" xlink:label="note-0476-03a" xlink:href="note-0476-03"/> la baſe D E eſt ſuppoſée de 5 pieds, & </s> <s xml:id="echoid-s13606" xml:space="preserve">la hauteur E F de 12, <lb/>l’on multipliera 5 par 12 pour avoir au produit 60 pieds, qui <lb/>ſeront la valeur du rectangle.</s> <s xml:id="echoid-s13607" xml:space="preserve"/> </p> <div xml:id="echoid-div1072" type="float" level="2" n="1"> <note position="left" xlink:label="note-0476-03" xlink:href="note-0476-03a" xml:space="preserve">Figure 219.</note> </div> <p> <s xml:id="echoid-s13608" xml:space="preserve">790. </s> <s xml:id="echoid-s13609" xml:space="preserve">Mais ſi au lieu d’un rectangle D F l’on avoit un pa-<lb/>rallélogramme G K, dont on voulut avoir la ſuperficie, il <lb/>faudroit prolonger la baſe G L, & </s> <s xml:id="echoid-s13610" xml:space="preserve">abaiſſer la perpendiculaire <lb/>K I, qui ſera la hauteur du parallélogramme (art. </s> <s xml:id="echoid-s13611" xml:space="preserve">383); </s> <s xml:id="echoid-s13612" xml:space="preserve">& </s> <s xml:id="echoid-s13613" xml:space="preserve"><lb/>ſuppoſant que cette perpendiculaire ſoit de 10 pieds, & </s> <s xml:id="echoid-s13614" xml:space="preserve">la <lb/>baſe G L de 4, l’on multipliera 10 par 4, & </s> <s xml:id="echoid-s13615" xml:space="preserve">le produit ſera <lb/>40 pieds pour la valeur du parallélogramme.</s> <s xml:id="echoid-s13616" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13617" xml:space="preserve">791. </s> <s xml:id="echoid-s13618" xml:space="preserve">Si la figure eſt trapezoïde, comme A B C D, & </s> <s xml:id="echoid-s13619" xml:space="preserve">que <lb/> <anchor type="note" xlink:label="note-0476-04a" xlink:href="note-0476-04"/> le côté B A ſoit perpendiculaire ſur les deux côtés paralleles <lb/>B C & </s> <s xml:id="echoid-s13620" xml:space="preserve">A D, il faut joindre ces deux côtés enſemble pour <pb o="411" file="0477" n="491" rhead="DE MATHÉMATIQUE. Liv. XII."/> avoir la baſe A E du triangle A B E, qui ſera égal au trapezoïde. <lb/></s> <s xml:id="echoid-s13621" xml:space="preserve">Ainſi ſuppoſant que le côté B C ſoit de 4 pieds, le côté A D <lb/>de 10, la hauteur B A de 12, la baſe A E, ou autrement la <lb/>ſomme des deux côtés ſera de 14, qu’il faut multiplier par 6, <lb/>moitié de la perpendiculaire, l’on aura 48 au produit pour la <lb/>ſuperficie du triangle A B E, qui eſt la même que celle du tra-<lb/>pezoïde, parce que les triangles B C F & </s> <s xml:id="echoid-s13622" xml:space="preserve">F D E ſont égaux.</s> <s xml:id="echoid-s13623" xml:space="preserve"/> </p> <div xml:id="echoid-div1073" type="float" level="2" n="2"> <note position="left" xlink:label="note-0476-04" xlink:href="note-0476-04a" xml:space="preserve">Figure 221.</note> </div> <p> <s xml:id="echoid-s13624" xml:space="preserve">792. </s> <s xml:id="echoid-s13625" xml:space="preserve">Si l’on veut encore d’une autre façon trouver la ſuper-<lb/>ficie du trapezoïde, il n’y a qu’à chercher une moyenne arith-<lb/>métique (art. </s> <s xml:id="echoid-s13626" xml:space="preserve">232) G F entre B C & </s> <s xml:id="echoid-s13627" xml:space="preserve">A D, c’eſt-à-dire entre <lb/>4 & </s> <s xml:id="echoid-s13628" xml:space="preserve">10, l’on trouvera qu’elle eſt 7; </s> <s xml:id="echoid-s13629" xml:space="preserve">& </s> <s xml:id="echoid-s13630" xml:space="preserve">ſi l’on multiplie cette <lb/>moyenne par toute la hauteur B A, qui eſt 12, l’on aura 84 <lb/>pour la ſuperficie; </s> <s xml:id="echoid-s13631" xml:space="preserve">ce qui eſt évident, puiſque le rectangle <lb/>A B H I eſt égal au trapezoïde A B C D, à cauſe que le trian-<lb/>gle C H F eſt le même que F I D.</s> <s xml:id="echoid-s13632" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1075" type="section" level="1" n="815"> <head xml:id="echoid-head991" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13633" xml:space="preserve">793. </s> <s xml:id="echoid-s13634" xml:space="preserve">Meſurer la ſuperficie des polygones réguliers & </s> <s xml:id="echoid-s13635" xml:space="preserve">irréguliers. <lb/></s> </p> <p> <s xml:id="echoid-s13636" xml:space="preserve">Si l’on veut ſçavoir la ſuperficie d’un polygone régulier, il <lb/>faut du centre E abaiſſer une perpendiculaire E B ſur un des <lb/>côtés C D, & </s> <s xml:id="echoid-s13637" xml:space="preserve">tirer les rayons E C & </s> <s xml:id="echoid-s13638" xml:space="preserve">E D, qui donneront le <lb/>triangle iſoſcele E C D. </s> <s xml:id="echoid-s13639" xml:space="preserve">Or comme on connoîtra les angles <lb/>de la baſe de ce triangle, puiſque le polygone eſt régulier, & </s> <s xml:id="echoid-s13640" xml:space="preserve"><lb/>que d’ailleurs on connoît le côté C D, on aura le triangle rec-<lb/>tangle E B D, duquel il ſera facile de connoître le côté E B <lb/>(art. </s> <s xml:id="echoid-s13641" xml:space="preserve">713): </s> <s xml:id="echoid-s13642" xml:space="preserve">& </s> <s xml:id="echoid-s13643" xml:space="preserve">ſuppoſant qu’on l’a trouvé de 6 pieds, on ajou-<lb/>tera enſemble tous les côtés du polygone, dont la ſomme <lb/>ſera, par exemple, 48, qu’il faudra multiplier par 3, moitié <lb/>de la perpendiculaire, pour avoir 144 pieds, qui ſera la va-<lb/>leur du polygone.</s> <s xml:id="echoid-s13644" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13645" xml:space="preserve">794. </s> <s xml:id="echoid-s13646" xml:space="preserve">Si le polygone eſt irrégulier, comme A B C D E F, <lb/> <anchor type="note" xlink:label="note-0477-02a" xlink:href="note-0477-02"/> l’on tirera du point E les lignes E C, E B, E A, qui diviſe-<lb/>ront le polygone en quatre triangles, dont le premier aura <lb/>pour hauteur la perpendiculaire F G, le ſecond, la perpendi-<lb/>culaire A H; </s> <s xml:id="echoid-s13647" xml:space="preserve">le troiſieme, la perpendiculaire C I; </s> <s xml:id="echoid-s13648" xml:space="preserve">& </s> <s xml:id="echoid-s13649" xml:space="preserve">le qua-<lb/>trieme, la perpendiculaire D K. </s> <s xml:id="echoid-s13650" xml:space="preserve">Cela poſé, ſi l’on meſure ſur <lb/>le terrein avec la toiſe, ou ſur le papier avec une échelle, la <lb/>valeur des perpendiculaires, auſſi-bien que celles des lignes ſur <pb o="412" file="0478" n="492" rhead="NOUVEAU COURS"/> leſquelles ces perpendiculaires tombent, l’on n’aura qu’à faire <lb/>autant de multiplications qu’il y a de triangles; </s> <s xml:id="echoid-s13651" xml:space="preserve">& </s> <s xml:id="echoid-s13652" xml:space="preserve">ajoutant <lb/>tous les produits enſemble, l’on aura la valeur du polygone.</s> <s xml:id="echoid-s13653" xml:space="preserve"/> </p> <div xml:id="echoid-div1075" type="float" level="2" n="1"> <note position="right" xlink:label="note-0477-02" xlink:href="note-0477-02a" xml:space="preserve">Figure 223.</note> </div> </div> <div xml:id="echoid-div1077" type="section" level="1" n="816"> <head xml:id="echoid-head992" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13654" xml:space="preserve">795. </s> <s xml:id="echoid-s13655" xml:space="preserve">Meſurer la ſuperficie des cercles & </s> <s xml:id="echoid-s13656" xml:space="preserve">de leurs parties. <lb/></s> </p> <p> <s xml:id="echoid-s13657" xml:space="preserve">Pour meſurer la ſuperficie d’un cercle A B, il faut connoître <lb/>la valeur de ſon diametre & </s> <s xml:id="echoid-s13658" xml:space="preserve">de ſa circonférence, comme on <lb/>l’a dit (art. </s> <s xml:id="echoid-s13659" xml:space="preserve">484), & </s> <s xml:id="echoid-s13660" xml:space="preserve">multiplier la moitié de la circonférence <lb/>par la moitié du diametre, & </s> <s xml:id="echoid-s13661" xml:space="preserve">le produit donnera la valeur du <lb/>cercle. </s> <s xml:id="echoid-s13662" xml:space="preserve">Par exemple, pour trouver la ſuperficie d’un cercle, <lb/>dont le diametre eſt 14, je cherche ſa circonférence, qui ſera <lb/>44; </s> <s xml:id="echoid-s13663" xml:space="preserve">& </s> <s xml:id="echoid-s13664" xml:space="preserve">prenant la moitié de 44, qui eſt 22, & </s> <s xml:id="echoid-s13665" xml:space="preserve">la moitié de <lb/>14, qui eſt 7, je multiplie ces deux nombres l’un par l’autre <lb/>pour avoir 154, qui ſera la ſuperficie du cercle.</s> <s xml:id="echoid-s13666" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13667" xml:space="preserve">L’on peut auſſi ſe ſervir du rapport de 14 à 11, qui exprime <lb/>celui du quarré du diametre d’un cercle à la ſuperficie du même <lb/>cercle, ſelon l’art. </s> <s xml:id="echoid-s13668" xml:space="preserve">490. </s> <s xml:id="echoid-s13669" xml:space="preserve">Ainſi ſuppoſant que le diametre ſoit <lb/>de 15 pieds, je quarre ce diametre pour avoir 225; </s> <s xml:id="echoid-s13670" xml:space="preserve">enſuite je <lb/>dis: </s> <s xml:id="echoid-s13671" xml:space="preserve">comme 14 eſt à 11, ainſi 225, quarré du diametre, eſt à <lb/>la ſuperficie du cercle que l’on trouvera de 176 {11/14}.</s> <s xml:id="echoid-s13672" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13673" xml:space="preserve">796. </s> <s xml:id="echoid-s13674" xml:space="preserve">Si l’on veut ſçavoir la ſuperficie d’un ſecteur de cer-<lb/> <anchor type="note" xlink:label="note-0478-02a" xlink:href="note-0478-02"/> cle, il faut connoître l’angle formé par les deux rayons, & </s> <s xml:id="echoid-s13675" xml:space="preserve"><lb/>la valeur du rayon. </s> <s xml:id="echoid-s13676" xml:space="preserve">Ainſi ſuppoſant que l’angle du ſecteur <lb/>A B C eſt de 60 degrés, & </s> <s xml:id="echoid-s13677" xml:space="preserve">le rayon de 7 pieds, je commence <lb/>par trouver la valeur du cercle d’où eſt provenu le ſecteur, la-<lb/>quelle ſe trouve de 154, & </s> <s xml:id="echoid-s13678" xml:space="preserve">puis je fais une Regle de Trois, <lb/>en diſant: </s> <s xml:id="echoid-s13679" xml:space="preserve">Si 360, valeur de toute la circonférence, m’a donné <lb/>154 pour la ſuperficie qu’elle renferme, combien me donne-<lb/>ront 60, valeur de la circonférence du ſecteur, pour la ſuper-<lb/>ficie qu’elle renferme, l’on trouvera 25 pieds 8 pouces.</s> <s xml:id="echoid-s13680" xml:space="preserve"/> </p> <div xml:id="echoid-div1077" type="float" level="2" n="1"> <note position="left" xlink:label="note-0478-02" xlink:href="note-0478-02a" xml:space="preserve">Figure 225.</note> </div> <p> <s xml:id="echoid-s13681" xml:space="preserve">797. </s> <s xml:id="echoid-s13682" xml:space="preserve">Enfin pour trouver la valeur d’un ſegment de cercle, <lb/> <anchor type="note" xlink:label="note-0478-03a" xlink:href="note-0478-03"/> tel que D G F, il faudra commencer par en faire un ſecteur, <lb/>dont on cherchera la ſuperficie, que je ſuppoſe encore être <lb/>25 pieds 8 pouces. </s> <s xml:id="echoid-s13683" xml:space="preserve">Cela poſé, on cherchera la ſuperficie du <lb/>triangle D E F, que l’on trouvera à peu près de 21 pieds; </s> <s xml:id="echoid-s13684" xml:space="preserve">& </s> <s xml:id="echoid-s13685" xml:space="preserve"><lb/>ſouſtrayant cette quantité de 25 pieds 8 pouces, le reſte ſera <lb/>la valeur du ſegment qui ſera environ de 4 pieds 8 pouces.</s> <s xml:id="echoid-s13686" xml:space="preserve"/> </p> <div xml:id="echoid-div1078" type="float" level="2" n="2"> <note position="left" xlink:label="note-0478-03" xlink:href="note-0478-03a" xml:space="preserve">Figure 226.</note> </div> <pb o="413" file="0479" n="493" rhead="DE MATHÉMATIQUE. Liv. XII."/> </div> <div xml:id="echoid-div1080" type="section" level="1" n="817"> <head xml:id="echoid-head993" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13687" xml:space="preserve">798. </s> <s xml:id="echoid-s13688" xml:space="preserve">Meſurer la ſuperficie d’une Ellipſe.</s> <s xml:id="echoid-s13689" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13690" xml:space="preserve">Nous avons vu (art. </s> <s xml:id="echoid-s13691" xml:space="preserve">240) que les élémens F H & </s> <s xml:id="echoid-s13692" xml:space="preserve">E I d’un <lb/> <anchor type="note" xlink:label="note-0479-01a" xlink:href="note-0479-01"/> quart de cercle étoient en même raiſon avec les élémens F G <lb/>& </s> <s xml:id="echoid-s13693" xml:space="preserve">E D d’un quart d’ellipſe; </s> <s xml:id="echoid-s13694" xml:space="preserve">par conſéquent il y aura donc <lb/>même raiſon de la ſomme de tous les antécédens à la ſomme <lb/>de tous les conſéquens, que d’un antécédent à ſon conſé-<lb/>quent (art. </s> <s xml:id="echoid-s13695" xml:space="preserve">633), c’eſt-à-dire que le quart de cercle E A I eſt <lb/>au quart d’ellipſe E A D, comme la ligne E I eſt à la ligne E D, <lb/>ou bien comme la ligne A B eſt à la ligne C D: </s> <s xml:id="echoid-s13696" xml:space="preserve">& </s> <s xml:id="echoid-s13697" xml:space="preserve">ſi au lieu <lb/>du quart de cercle, & </s> <s xml:id="echoid-s13698" xml:space="preserve">du quart d’ellipſe, l’on prend tout le <lb/>cercle & </s> <s xml:id="echoid-s13699" xml:space="preserve">toute l’ellipſe; </s> <s xml:id="echoid-s13700" xml:space="preserve">il y aura encore même raiſon du <lb/>cercle à l’ellipſe, que de la ligne A B à la ligne C D; </s> <s xml:id="echoid-s13701" xml:space="preserve">ce qui <lb/>fait voir que la ſuperficie d’un cercle qui auroit pour diametre <lb/>le grand axe d’une ellipſe, eſt à la ſuperficie de l’ellipſe, comme <lb/>le grand axe eſt au petit. </s> <s xml:id="echoid-s13702" xml:space="preserve">Or ſuppoſant que le grand axe A B <lb/>ſoit de 14 pieds, & </s> <s xml:id="echoid-s13703" xml:space="preserve">le petit C D de 8, il faut pour trouver la <lb/>ſuperficie de l’ellipſe, chercher d’abord celle du cercle de ſon <lb/>grand axe, que l’on trouvera de 154, & </s> <s xml:id="echoid-s13704" xml:space="preserve">puis dire: </s> <s xml:id="echoid-s13705" xml:space="preserve">Si le grand <lb/>axe de 14 m’a donné 8 pouces pour le petit, que me donne-<lb/>ront 154, ſuperficie du cercle pour celle de l’ellipſe, que l’on <lb/>trouvera de 88 pieds.</s> <s xml:id="echoid-s13706" xml:space="preserve"/> </p> <div xml:id="echoid-div1080" type="float" level="2" n="1"> <note position="right" xlink:label="note-0479-01" xlink:href="note-0479-01a" xml:space="preserve">Figure 227.</note> </div> <p> <s xml:id="echoid-s13707" xml:space="preserve">Les ſuperficies des cercles étant dans la raiſon des quarrés <lb/>de leurs diametres, l’on peut dire que celles des ellipſes ſont <lb/>dans la raiſon compoſée de leurs axes, que par conſéquent <lb/>l’on peut prendre à la place de leurs diametres les rectangles <lb/>compris ſous les mêmes axes; </s> <s xml:id="echoid-s13708" xml:space="preserve">& </s> <s xml:id="echoid-s13709" xml:space="preserve">comme il n’y a point de <lb/>quarré qui ne puiſſe être produit par les dimenſions d’un rec-<lb/>tangle qui lui ſeroit égal, l’on peut trouver la ſuperficie de <lb/>l’ellipſe précédente, en multipliant ces deux axes 14 & </s> <s xml:id="echoid-s13710" xml:space="preserve">8 l’un <lb/>par l’autre pour avoir 112, qui tiendra lieu du quarré de ſon <lb/>diametre, enſuite dire, comme 14 eſt à 11, ainſi 112 eſt à la <lb/>ſuperficie de l’ellipſe, que l’on trouvera encore de 88 pieds.</s> <s xml:id="echoid-s13711" xml:space="preserve"/> </p> <pb o="414" file="0480" n="494" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1082" type="section" level="1" n="818"> <head xml:id="echoid-head994" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13712" xml:space="preserve">799. </s> <s xml:id="echoid-s13713" xml:space="preserve">Meſurer l’eſpace renfermé par une parabole. <lb/></s> </p> <p> <s xml:id="echoid-s13714" xml:space="preserve">Si l’on a une parabole A B C, dont l’axe B D ſoit de 9 <lb/>pieds, & </s> <s xml:id="echoid-s13715" xml:space="preserve">la plus grande ordonnée D A de 12, toute la ligne <lb/>A C ſera de 24. </s> <s xml:id="echoid-s13716" xml:space="preserve">Cela étant, je dis que pour trouver l’eſpace <lb/>renfermé par la parabole A B C, il faut multiplier la ligne <lb/>A C par les deux tiers de l’axe B D, c’eſt-à-dire 24 par 6, <lb/>pour avoir 144 au produit, qui ſera l’eſpace que l’on demande.</s> <s xml:id="echoid-s13717" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13718" xml:space="preserve">La raiſon de cette opération eſt que l’eſpace A B C eſt les <lb/>deux tiers du rectangle A E F C; </s> <s xml:id="echoid-s13719" xml:space="preserve">pour le prouver nous fe-<lb/>rons voir que l’eſpace A E B K eſt le tiers du rectangle A E B D.</s> <s xml:id="echoid-s13720" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13721" xml:space="preserve">Ayant diviſé la ligne E B en un nombre de parties égales, <lb/>& </s> <s xml:id="echoid-s13722" xml:space="preserve">tiré par tous les points de diviſion des lignes telles que G H <lb/>& </s> <s xml:id="echoid-s13723" xml:space="preserve">I K, paralleles à A E, l’on verra (art. </s> <s xml:id="echoid-s13724" xml:space="preserve">605) que par la pro-<lb/>priété de la parabole, le quarré B G eſt au quarré B I, comme <lb/>G H eſt à I K; </s> <s xml:id="echoid-s13725" xml:space="preserve">mais les parties de ſuite de la ligne E B étant <lb/>en progreſſion arithmétique, les quarrés des lignes B G & </s> <s xml:id="echoid-s13726" xml:space="preserve">B I <lb/>ſeront ceux des termes d’une progreſſion arithmétique; </s> <s xml:id="echoid-s13727" xml:space="preserve">par <lb/>conſéquent les élémens G H & </s> <s xml:id="echoid-s13728" xml:space="preserve">I K ſont en même raiſon que <lb/>les quarrés des termes d’une progreſſion arithmétique: </s> <s xml:id="echoid-s13729" xml:space="preserve">ainſi <lb/>l’eſpace A E B K contient une quantité infinie d’élémens, qui <lb/>ſont tous dans la même raiſon que les quarrés des termes in-<lb/>finis d’une progreſſion arithmétique: </s> <s xml:id="echoid-s13730" xml:space="preserve">mais comme pour trouver <lb/>la valeur de tous ces quarrés, il faut (art. </s> <s xml:id="echoid-s13731" xml:space="preserve">551) multiplier le <lb/>plus grand quarré par le tiers de la grandeur qui exprime la <lb/>quantité des termes; </s> <s xml:id="echoid-s13732" xml:space="preserve">il faut donc pour trouver la valeur de <lb/>tous les élémens qui compoſent l’eſpace A E B K, multiplier <lb/>le plus grand élément E A par le tiers de la ligne E B, qui en <lb/>exprime la quantité: </s> <s xml:id="echoid-s13733" xml:space="preserve">ce qui fait voir que cet eſpace eſt le tiers <lb/>du rectangle A E B D, & </s> <s xml:id="echoid-s13734" xml:space="preserve">que par conſéquent l’eſpace A K B D <lb/>de la parabole en eſt les deux tiers.</s> <s xml:id="echoid-s13735" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1083" type="section" level="1" n="819"> <head xml:id="echoid-head995" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s13736" xml:space="preserve">Il eſt abſolument néceſſaire pour ceux qui veulent s’atta-<lb/>cher au Génie, de ſçavoir bien meſurer les figures planes, <lb/>parce qu’elles ſe rencontrent continuellement dans le toiſé des <lb/>fortifications & </s> <s xml:id="echoid-s13737" xml:space="preserve">des bâtimens civils: </s> <s xml:id="echoid-s13738" xml:space="preserve">car les couvertures de <pb o="415" file="0481" n="495" rhead="DE MATHÉMATIQUE. Liv. XII."/> tuiles & </s> <s xml:id="echoid-s13739" xml:space="preserve">d’ardoiſes, les planchers, les pavés, le blanchiſſage <lb/>des murs recrepis, les vitres, le gazon avec lequel on revêtit <lb/>les ouvrages de terraſſe, ſe meſurent à la toiſe quarrée, & </s> <s xml:id="echoid-s13740" xml:space="preserve"><lb/>toutes les figures que toutes ces choſes peuvent former, ſe ré-<lb/>duiſent toujours à des rectangles ou à des triangles.</s> <s xml:id="echoid-s13741" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1084" type="section" level="1" n="820"> <head xml:id="echoid-head996" xml:space="preserve">PROPOSITION VII <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s13742" xml:space="preserve">800. </s> <s xml:id="echoid-s13743" xml:space="preserve">Meſurer les ſurfaces des Priſmes & </s> <s xml:id="echoid-s13744" xml:space="preserve">des Cylindres. <lb/></s> </p> <p> <s xml:id="echoid-s13745" xml:space="preserve">Pour meſurer la ſurface d’un priſme A E, il faut multiplier <lb/>la ſomme des côtés du polygone, qui lui ſert de baſe par la <lb/>hauteur du priſme: </s> <s xml:id="echoid-s13746" xml:space="preserve">ainſi ſi le priſme a pour baſe un exa-<lb/>gone, dont chaque côté B C ſoit de 4 pieds, & </s> <s xml:id="echoid-s13747" xml:space="preserve">la hauteur <lb/>B E de 6, la ſomme des côtés ſera 24, qui étant multiplié par <lb/>6, le produit ſera 144 pieds pour la valeur de la ſurface.</s> <s xml:id="echoid-s13748" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13749" xml:space="preserve">801. </s> <s xml:id="echoid-s13750" xml:space="preserve">Pour meſurer la ſurface d’un cylindre, tel que B C, <lb/> <anchor type="note" xlink:label="note-0481-02a" xlink:href="note-0481-02"/> dont le diametre A C eſt de 14 pieds, & </s> <s xml:id="echoid-s13751" xml:space="preserve">la hauteur A B de 8, il <lb/>faut commencer par chercher la circonférence du cercle qui <lb/>lui ſert de baſe, qu’on trouvera de 44 pieds. </s> <s xml:id="echoid-s13752" xml:space="preserve">Après cela, il <lb/>faut multiplier cette circonférence par 8, hauteur du cylindre, <lb/>& </s> <s xml:id="echoid-s13753" xml:space="preserve">l’on trouvera 352 pieds pour la ſurface du cylindre.</s> <s xml:id="echoid-s13754" xml:space="preserve"/> </p> <div xml:id="echoid-div1084" type="float" level="2" n="1"> <note position="right" xlink:label="note-0481-02" xlink:href="note-0481-02a" xml:space="preserve">Figure 230.</note> </div> </div> <div xml:id="echoid-div1086" type="section" level="1" n="821"> <head xml:id="echoid-head997" xml:space="preserve">PROPOSITION VIII. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s13755" xml:space="preserve">802. </s> <s xml:id="echoid-s13756" xml:space="preserve">Meſurer les ſurfaces des Pyramides & </s> <s xml:id="echoid-s13757" xml:space="preserve">des Cônes. <lb/></s> </p> <p> <s xml:id="echoid-s13758" xml:space="preserve">Pour meſurer la ſurface d’une pyramide droite, qui a pour <lb/>baſe un exagone, dont chaque côté, tel que A B, eſt ſuppoſé <lb/>de 6 pieds, & </s> <s xml:id="echoid-s13759" xml:space="preserve">la perpendiculaire tirée du ſommet ſur un de <lb/>ſes côtés de 10 pieds, il faut multiplier la ſomme de la moitié <lb/>de tous ces côtés par toute la perpendiculaire (art. </s> <s xml:id="echoid-s13760" xml:space="preserve">545), c’eſt-<lb/>à-dire 18 par 10: </s> <s xml:id="echoid-s13761" xml:space="preserve">l’on trouvera 180 pour la ſurface de la py-<lb/>ramide.</s> <s xml:id="echoid-s13762" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13763" xml:space="preserve">803. </s> <s xml:id="echoid-s13764" xml:space="preserve">Pour trouver la ſurface d’un cône droit, dont le dia-<lb/> <anchor type="note" xlink:label="note-0481-04a" xlink:href="note-0481-04"/> metre A B du cercle de ſa baſe eſt de 14 pieds, & </s> <s xml:id="echoid-s13765" xml:space="preserve">le côté A D <lb/>de 12, il faut multiplier la circonférence du cercle, que l’on <lb/>trouvera de 44, par la moitié du côté A D (art. </s> <s xml:id="echoid-s13766" xml:space="preserve">547), c’eſt-<lb/>à-dire par 6, & </s> <s xml:id="echoid-s13767" xml:space="preserve">l’on verra que la ſurface du cône eſt de 264;</s> <s xml:id="echoid-s13768" xml:space="preserve"> <pb o="416" file="0482" n="496" rhead="NOUVEAU COURS"/> ou bien multiplier la moitié de la circonférence par tout le <lb/>côté A D, & </s> <s xml:id="echoid-s13769" xml:space="preserve">l’on aura encore la même choſe.</s> <s xml:id="echoid-s13770" xml:space="preserve"/> </p> <div xml:id="echoid-div1086" type="float" level="2" n="1"> <note position="right" xlink:label="note-0481-04" xlink:href="note-0481-04a" xml:space="preserve">Figure 232.</note> </div> </div> <div xml:id="echoid-div1088" type="section" level="1" n="822"> <head xml:id="echoid-head998" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s13771" xml:space="preserve">804. </s> <s xml:id="echoid-s13772" xml:space="preserve">Meſurer les ſurfaces des Spheres, celles de leurs Seg-<lb/> <anchor type="note" xlink:label="note-0482-01a" xlink:href="note-0482-01"/> mens, & </s> <s xml:id="echoid-s13773" xml:space="preserve">celles de leurs Zones.</s> <s xml:id="echoid-s13774" xml:space="preserve"/> </p> <div xml:id="echoid-div1088" type="float" level="2" n="1"> <note position="left" xlink:label="note-0482-01" xlink:href="note-0482-01a" xml:space="preserve">Figure 233.</note> </div> <p> <s xml:id="echoid-s13775" xml:space="preserve">Pour meſurer la ſurface d’une ſphere, dont le diametre <lb/>H G eſt ſuppoſé de 14 pieds, il faut commencer par chercher <lb/>la circonférence de ce diametre, que l’on trouvera de 44; </s> <s xml:id="echoid-s13776" xml:space="preserve">& </s> <s xml:id="echoid-s13777" xml:space="preserve"><lb/>il faut la multiplier par le diametre, c’eſt-à-dire par 14, & </s> <s xml:id="echoid-s13778" xml:space="preserve">le <lb/>produit donnera la valeur de la ſurface de la ſphere (art. </s> <s xml:id="echoid-s13779" xml:space="preserve">575) <lb/>que l’on trouvera de 616.</s> <s xml:id="echoid-s13780" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13781" xml:space="preserve">805. </s> <s xml:id="echoid-s13782" xml:space="preserve">si au lieu de la ſurface de toute une ſphere, on vou-<lb/>loit meſurer ſeulement celle d’un ſegment, tel que A B C, il <lb/>faudroit chercher d’abord la circonférence du grand cercle de <lb/>la ſphere d’où le ſegment a été tiré; </s> <s xml:id="echoid-s13783" xml:space="preserve">& </s> <s xml:id="echoid-s13784" xml:space="preserve">de plus connoître <lb/>exactement la perpendiculaire C D élevée ſur le centre du cer-<lb/>cle A B, & </s> <s xml:id="echoid-s13785" xml:space="preserve">puis multiplier la circonférence du grand cercle <lb/>par la valeur de cette perpendiculaire (582): </s> <s xml:id="echoid-s13786" xml:space="preserve">ainſi ſuppoſant <lb/>que la circonférence du cercle ſoit 44, & </s> <s xml:id="echoid-s13787" xml:space="preserve">la perpendiculaire <lb/>C D de 4, multipliant l’un par l’autre, on aura 176 pieds pour <lb/>la valeur de la ſurface du ſegment.</s> <s xml:id="echoid-s13788" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13789" xml:space="preserve">806. </s> <s xml:id="echoid-s13790" xml:space="preserve">Enfin pour meſurer la ſurface d’une zone, telle que <lb/>E H F G, il faut connoître auſſi la circonférence du grand <lb/>cercle de la ſphere d’où elle a été tirée, & </s> <s xml:id="echoid-s13791" xml:space="preserve">la valeur de la per-<lb/>pendiculaire I K, tirée d’un centre à l’autre des deux cercles <lb/>oppoſés, & </s> <s xml:id="echoid-s13792" xml:space="preserve">multiplier cette perpendiculaire par la circonfé-<lb/>rence du grand cercle (art. </s> <s xml:id="echoid-s13793" xml:space="preserve">582), dont nous venons de parler. <lb/></s> <s xml:id="echoid-s13794" xml:space="preserve">Ainſi ſuppoſant qu’elle ſoit encore de 44 pieds, & </s> <s xml:id="echoid-s13795" xml:space="preserve">la perpen-<lb/>diculaire I K de 5, multipliant l’un par l’autre, l’on trouvera <lb/>220 pieds pour la valeur de la ſurface de la zone.</s> <s xml:id="echoid-s13796" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1090" type="section" level="1" n="823"> <head xml:id="echoid-head999" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s13797" xml:space="preserve">La plûpart de ceux qui étudient la Géométrie ſçavent bien <lb/>que cette ſcience eſt fort utile, & </s> <s xml:id="echoid-s13798" xml:space="preserve">qu’en général toutes les <lb/>propoſitions qu’elle renferme ont leur uſage; </s> <s xml:id="echoid-s13799" xml:space="preserve">cependant comme <lb/>ils n’en connoiſſent point l’application, faute de s’être trouvés <lb/>dans le cas de s’en ſervir, ils en viennent toujours à demander <pb o="417" file="0483" n="497" rhead="DE MATHÉMATIQUE. Liv. XII."/> à quoi tels & </s> <s xml:id="echoid-s13800" xml:space="preserve">tels problêmes peuvent ſervir: </s> <s xml:id="echoid-s13801" xml:space="preserve">c’eſt pourquoi <lb/>ayant deſſein de leur ôter cette inquiétude, je tâcherai, autant <lb/>qu’il me ſera poſſible, de leur faire voir l’application des moin-<lb/>dres choſes: </s> <s xml:id="echoid-s13802" xml:space="preserve">& </s> <s xml:id="echoid-s13803" xml:space="preserve">pour dire un mot des propoſitions précédentes, <lb/>ils feront attention que les cloches étant toujours des pyra-<lb/>mides ou des cônes, que les dômes étant ordinairement des <lb/>figures ſphériques, & </s> <s xml:id="echoid-s13804" xml:space="preserve">les tours des châteaux étant couvertes <lb/>par des toits faits en cône ou en pyramide, il faut, pour en <lb/>toiſer la couverture, ſçavoir meſurer ces différentes ſurfaces.</s> <s xml:id="echoid-s13805" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1091" type="section" level="1" n="824"> <head xml:id="echoid-head1000" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s13806" xml:space="preserve">807 Meſurer la ſolidité des Cubes, des Parallelepipedes, des <lb/> <anchor type="note" xlink:label="note-0483-01a" xlink:href="note-0483-01"/> Priſmes & </s> <s xml:id="echoid-s13807" xml:space="preserve">des cylindres.</s> <s xml:id="echoid-s13808" xml:space="preserve"/> </p> <div xml:id="echoid-div1091" type="float" level="2" n="1"> <note position="right" xlink:label="note-0483-01" xlink:href="note-0483-01a" xml:space="preserve">Figure 234.</note> </div> <p> <s xml:id="echoid-s13809" xml:space="preserve">Pour meſurer la ſolidité d’un cube A D, dont le côté A B <lb/>ſeroit, par exemple, de 6 pieds, il faut quarrer 6 pour avoir <lb/>la ſuperficie de la baſe, qui ſera 36; </s> <s xml:id="echoid-s13810" xml:space="preserve">& </s> <s xml:id="echoid-s13811" xml:space="preserve">multipliant cette baſe <lb/>par la hauteur du cube, c’eſt-à-dire par 6 pieds, l’on aura 216 <lb/>pieds pour la valeur du cube.</s> <s xml:id="echoid-s13812" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13813" xml:space="preserve">808. </s> <s xml:id="echoid-s13814" xml:space="preserve">L’on trouvera de même la valeur d’un parallelepipede, <lb/> <anchor type="note" xlink:label="note-0483-02a" xlink:href="note-0483-02"/> en multipliant la ſuperficie de ſa baſe par la hauteur. </s> <s xml:id="echoid-s13815" xml:space="preserve">Ainſi <lb/>voulant meſurer le parallelepipede E H, ſuppoſant que ſa baſe <lb/>ait 10 pieds de long ſur 4 pieds de large, & </s> <s xml:id="echoid-s13816" xml:space="preserve">que ſa hauteur <lb/>H F ſoit de 5 pieds, il faut multiplier 4 par 10 pour avoir 40, <lb/>qui ſera la ſuperficie de la baſe, qui étant multipliée par la <lb/>hauteur 5, donnera 200 pieds cubes pour le parallelepipede.</s> <s xml:id="echoid-s13817" xml:space="preserve"/> </p> <div xml:id="echoid-div1092" type="float" level="2" n="2"> <note position="right" xlink:label="note-0483-02" xlink:href="note-0483-02a" xml:space="preserve">Figure 235.</note> </div> <p> <s xml:id="echoid-s13818" xml:space="preserve">809. </s> <s xml:id="echoid-s13819" xml:space="preserve">Pour meſurer la ſolidité d’un priſme C E, dont la baſe <lb/> <anchor type="note" xlink:label="note-0483-03a" xlink:href="note-0483-03"/> eſt un exagone, il faut d’abord connoître la ſuperficie de l’exa-<lb/>gone, que l’on trouvera en multipliant la ſomme de ſes côtés <lb/>par la moitié de la perpendiculaire A D: </s> <s xml:id="echoid-s13820" xml:space="preserve">ainſi ce côté B C étant <lb/>de 4 pieds, la perpendiculaire de 3 {1/2}, la ſomme des côtés ſera <lb/>24, qui étant multipliée par 1 {3/4}, on aura 42 pieds quarrés <lb/>pour la valeur de la baſe, qu’il faut enſuite multiplier par la <lb/>hauteur B E, que je ſuppoſe de 6 pieds: </s> <s xml:id="echoid-s13821" xml:space="preserve">la multiplication étant <lb/>faite, l’on trouvera 252 pieds cubes pour la valeur du priſme.</s> <s xml:id="echoid-s13822" xml:space="preserve"/> </p> <div xml:id="echoid-div1093" type="float" level="2" n="3"> <note position="right" xlink:label="note-0483-03" xlink:href="note-0483-03a" xml:space="preserve">Figure 229.</note> </div> <p> <s xml:id="echoid-s13823" xml:space="preserve">810. </s> <s xml:id="echoid-s13824" xml:space="preserve">Pour meſurer la ſolidité d’un cylindre C B, dont le <lb/> <anchor type="note" xlink:label="note-0483-04a" xlink:href="note-0483-04"/> diametre B D du cercle de la baſe eſt de 14 pieds, & </s> <s xml:id="echoid-s13825" xml:space="preserve">la hauteur <lb/>A B de 8 pieds, il faut commencer par avoir la valeur du cercle <lb/>qui ſert de baſe au cylindre: </s> <s xml:id="echoid-s13826" xml:space="preserve">pour cela, il faut chercher la cir- <pb o="418" file="0484" n="498" rhead="NOUVEAU COURS"/> conférence, que l’on trouvera de 44, dont la moitié étant <lb/>multipliée par le rayon du même cercle, donnera 154 pieds <lb/>quarrés pour la valeur de la baſe du cylindre: </s> <s xml:id="echoid-s13827" xml:space="preserve">il faut enſuite la <lb/>multiplier par 8 pour avoir 1232 pieds cubes pour la ſolidité <lb/>du cylindre.</s> <s xml:id="echoid-s13828" xml:space="preserve"/> </p> <div xml:id="echoid-div1094" type="float" level="2" n="4"> <note position="right" xlink:label="note-0483-04" xlink:href="note-0483-04a" xml:space="preserve">Figure 230.</note> </div> <p> <s xml:id="echoid-s13829" xml:space="preserve">Comme la ſolidité des cubes, des parallelepipedes, des priſ-<lb/>mes & </s> <s xml:id="echoid-s13830" xml:space="preserve">des cylindres, eſt compoſée d’une infinité de plans ſem-<lb/>blables à celui qui ſert de baſe à chacun de ces corps, & </s> <s xml:id="echoid-s13831" xml:space="preserve">que <lb/>leur hauteur exprime la quantité de plans dont ils ſont com-<lb/>poſés; </s> <s xml:id="echoid-s13832" xml:space="preserve">il s’enſuit que pour trouver la ſolidité d’un corps tel <lb/>que les précédens, il faut multiplier ſa baſe par toute ſa hau-<lb/>teur.</s> <s xml:id="echoid-s13833" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1096" type="section" level="1" n="825"> <head xml:id="echoid-head1001" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s13834" xml:space="preserve">811. </s> <s xml:id="echoid-s13835" xml:space="preserve">Meſurer la ſolidité des Pyramides & </s> <s xml:id="echoid-s13836" xml:space="preserve">des Cônes. <lb/></s> </p> <p> <s xml:id="echoid-s13837" xml:space="preserve">Pour meſurer la ſolidité d’une pyramide qui a pour baſe <lb/>un exagone, il faut commencer par connoître la ſuperficie de <lb/>la baſe. </s> <s xml:id="echoid-s13838" xml:space="preserve">Ainſi ſuppoſant que le côté A B ſoit de 6 pieds, & </s> <s xml:id="echoid-s13839" xml:space="preserve">la <lb/>perpendiculaire C E de 6 {3/4}, l’on trouvera 121 pieds {1/2} quarrés <lb/>pour la ſuperficie de la baſe, qu’il faut multiplier par le tiers de <lb/>l’axe D C de la pyramide. </s> <s xml:id="echoid-s13840" xml:space="preserve">Comme cet axe eſt ſuppoſé de 10 <lb/>pieds, il faudra multiplier 121 {1/2} par 3 {1/3}, & </s> <s xml:id="echoid-s13841" xml:space="preserve">le produit ſera 405 <lb/>pieds cubes pour la ſolidité de la pyramide.</s> <s xml:id="echoid-s13842" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13843" xml:space="preserve">812. </s> <s xml:id="echoid-s13844" xml:space="preserve">Pour trouver la ſolidité d’un cône, l’on agira comme <lb/> <anchor type="note" xlink:label="note-0484-02a" xlink:href="note-0484-02"/> on vient de faire pour trouver celle de la pyramide: </s> <s xml:id="echoid-s13845" xml:space="preserve">on <lb/>commencera par connoître la ſuperficie du cercle, qui ſert de <lb/>baſe au cône, il faudra la multiplier par le tiers de l’axe du <lb/>cône. </s> <s xml:id="echoid-s13846" xml:space="preserve">Ainſi voulant meſurer la ſolidité d’un cône A D B, dont <lb/>le diametre de ſon cercle eſt de 14 pieds, & </s> <s xml:id="echoid-s13847" xml:space="preserve">la valeur de ſon <lb/>axe de 9 {1/2}, l’on trouvera que la ſuperficie de la baſe eſt de 154 <lb/>pieds quarrés, qui étant multipliés par 3 {1/6}, qui eſt le tiers de <lb/>l’axe, l’on trouvera 456 pieds cubes pour la ſolidité du cône.</s> <s xml:id="echoid-s13848" xml:space="preserve"/> </p> <div xml:id="echoid-div1096" type="float" level="2" n="1"> <note position="left" xlink:label="note-0484-02" xlink:href="note-0484-02a" xml:space="preserve">Figure 232.</note> </div> <p> <s xml:id="echoid-s13849" xml:space="preserve">Si nous avons multiplié la baſe de la pyramide, auſſi-bien <lb/>que celle du cône, par le tiers de la hauteur de l’un & </s> <s xml:id="echoid-s13850" xml:space="preserve">de l’autre, <lb/>c’eſt que nous avons vu (art. </s> <s xml:id="echoid-s13851" xml:space="preserve">551) que la pyramide étoit le <lb/>tiers du priſme de même baſe & </s> <s xml:id="echoid-s13852" xml:space="preserve">de même hauteur, comme <lb/>le cône étoit auſſi le tiers du cylindre de même baſe & </s> <s xml:id="echoid-s13853" xml:space="preserve">de <lb/>même hauteur.</s> <s xml:id="echoid-s13854" xml:space="preserve"/> </p> <pb o="319" file="0485" n="499" rhead="DE MATHÉMATIQUE. Liv. XII."/> <p> <s xml:id="echoid-s13855" xml:space="preserve">813. </s> <s xml:id="echoid-s13856" xml:space="preserve">Si les parallelepipedes, les priſmes, les cylindres, les <lb/>pyramides, les cônes que l’on veut meſurer étoient inclinés, <lb/>il faudroit tirer une perpendiculaire de leur ſommet ſur leurs <lb/>baſes prolongées; </s> <s xml:id="echoid-s13857" xml:space="preserve">enſuite connoître la valeur de cette per-<lb/>pendiculaire, & </s> <s xml:id="echoid-s13858" xml:space="preserve">la regarder comme celle de la hauteur du <lb/>ſolide, qui ſera incliné; </s> <s xml:id="echoid-s13859" xml:space="preserve">& </s> <s xml:id="echoid-s13860" xml:space="preserve">ſi cela arrive à l’égard d’un paralle-<lb/>lepipede, d’un priſme, ou d’un cylindre, on multipliera toute <lb/>la perpendiculaire par la baſe du ſolide auquel elle correſpond: <lb/></s> <s xml:id="echoid-s13861" xml:space="preserve">& </s> <s xml:id="echoid-s13862" xml:space="preserve">ſi cela arrive à l’égard des pyramides, des cônes, on mul-<lb/>tipliera la baſe de l’un ou l’autre de ces ſolides par le tiers de <lb/>la perpendiculaire.</s> <s xml:id="echoid-s13863" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1098" type="section" level="1" n="826"> <head xml:id="echoid-head1002" xml:space="preserve">PROPOSITION XII. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s13864" xml:space="preserve">814. </s> <s xml:id="echoid-s13865" xml:space="preserve">Meſurer la ſolidité des Pyramides & </s> <s xml:id="echoid-s13866" xml:space="preserve">des cônes tronqués. <lb/></s> </p> <p> <s xml:id="echoid-s13867" xml:space="preserve">Si l’on a une pyramide D B, dont les plans oppoſés D F & </s> <s xml:id="echoid-s13868" xml:space="preserve"><lb/>A B ſoient des quarrés, pour en ſçavoir la ſolidité, nous ſup-<lb/>poſerons que le côté D E eſt de 9 pieds, le côté A C de 4, & </s> <s xml:id="echoid-s13869" xml:space="preserve"><lb/>l’axe G H de 12. </s> <s xml:id="echoid-s13870" xml:space="preserve">Cela poſé, il faut chercher la valeur des <lb/>plans A B & </s> <s xml:id="echoid-s13871" xml:space="preserve">D F, qui ſeront de 16 & </s> <s xml:id="echoid-s13872" xml:space="preserve">de 81 pieds, entre leſ-<lb/>quels il faut chercher une moyenne proportionnelle, qui ſera <lb/>36 pour le plan moyen, qu’il faut ajouter avec les deux autres, <lb/>pour avoir 133, qui ſera la ſomme des trois plans, qu’il faut <lb/>multiplier par le tiers de l’axe, c’eſt-à-dire par 4 pour avoir <lb/>532 pieds pour la ſolidité de la pyramide tronquée (art. </s> <s xml:id="echoid-s13873" xml:space="preserve">561).</s> <s xml:id="echoid-s13874" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13875" xml:space="preserve">Si l’on avoit un cône tronqué, l’on en trouveroit de même <lb/>la valeur, en cherchant un cercle moyen entre les deux op-<lb/>poſés, & </s> <s xml:id="echoid-s13876" xml:space="preserve">en multipliant la ſomme de la valeur des trois cer-<lb/>cles par le tiers de l’axe, pour avoir un produit, qui ſera ce <lb/>que l’on demande.</s> <s xml:id="echoid-s13877" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13878" xml:space="preserve">815. </s> <s xml:id="echoid-s13879" xml:space="preserve">Voici encore une autre maniere de trouver la valeur <lb/> <anchor type="note" xlink:label="note-0485-02a" xlink:href="note-0485-02"/> d’une pyramide ou d’un cône tronqué, qui eſt plus d’uſage <lb/>que la précédente: </s> <s xml:id="echoid-s13880" xml:space="preserve">par exemple, pour connoître la ſolidité <lb/>du cône tronqué A D E B, dont l’axe G C eſt de 15 pieds, le <lb/>diametre D E de 7, & </s> <s xml:id="echoid-s13881" xml:space="preserve">le diametre A B de 21: </s> <s xml:id="echoid-s13882" xml:space="preserve">j’abaiſſe la per-<lb/>pendiculaire D H, & </s> <s xml:id="echoid-s13883" xml:space="preserve">j’acheve le cône pour avoir l’axe entier <lb/>C F, dont je cherche la valeur comme il ſuit.</s> <s xml:id="echoid-s13884" xml:space="preserve"/> </p> <div xml:id="echoid-div1098" type="float" level="2" n="1"> <note position="right" xlink:label="note-0485-02" xlink:href="note-0485-02a" xml:space="preserve">Figure 237.</note> </div> <p> <s xml:id="echoid-s13885" xml:space="preserve">Le rayon D G étant de 3 pieds {1/2}, & </s> <s xml:id="echoid-s13886" xml:space="preserve">le rayon A C de 10 {1/2}, <lb/>la ligne A H ſera la différence de D G à A C: </s> <s xml:id="echoid-s13887" xml:space="preserve">par conſéquent <pb o="420" file="0486" n="500" rhead="NOUVEAU COURS"/> de 7 pieds. </s> <s xml:id="echoid-s13888" xml:space="preserve">Or ayant les deux triangles ſemblables A H D & </s> <s xml:id="echoid-s13889" xml:space="preserve"><lb/>A C F, je dis: </s> <s xml:id="echoid-s13890" xml:space="preserve">Si le côté A H de 7 pieds donne 15 pieds pour <lb/>le côté H D, que donnera le côté A C de 10 {1/2} pour le côté <lb/>C F, que l’on trouvera de 22 pieds {1/2}.</s> <s xml:id="echoid-s13891" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13892" xml:space="preserve">Préſentement que l’on a trouvé le grand axe, il faut cher-<lb/>cher la valeur du cône A B F, & </s> <s xml:id="echoid-s13893" xml:space="preserve">celle du petit cône D F E, & </s> <s xml:id="echoid-s13894" xml:space="preserve"><lb/>retran cher celle-ci de l’autre pour avoir la différence, qui ſera <lb/>la valeur du cône tronqué.</s> <s xml:id="echoid-s13895" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13896" xml:space="preserve">816. </s> <s xml:id="echoid-s13897" xml:space="preserve">Ou bien à cauſe que les cônes D F E & </s> <s xml:id="echoid-s13898" xml:space="preserve">A F B ſont <lb/>ſemblables, l’on pourra cuber les diametres A B & </s> <s xml:id="echoid-s13899" xml:space="preserve">D E, & </s> <s xml:id="echoid-s13900" xml:space="preserve"><lb/>dire. </s> <s xml:id="echoid-s13901" xml:space="preserve">Comme le cube du diametre A B eſt au cube du diametre <lb/>D E, ainſi la valeur du cône A F B eſt à celle du cône D F E, <lb/>qui étant trouvée, ſera retranchée de celle du cône A F B <lb/>pour avoir la différence, qui ſera la partie tronquée.</s> <s xml:id="echoid-s13902" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1100" type="section" level="1" n="827"> <head xml:id="echoid-head1003" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s13903" xml:space="preserve">L’on verra dans la ſuite la néceſſité de ſçavoir meſurer les <lb/>priſmes, les cylindres, les pyramides & </s> <s xml:id="echoid-s13904" xml:space="preserve">les cônes, auſſi-bien <lb/>que leurs parties tronquées: </s> <s xml:id="echoid-s13905" xml:space="preserve">car on ne peut faire le toiſé de la <lb/>maçonnerie du revêtement d’une fortification, ſans qu’il ne <lb/>ſe rencontre des parties ſemblables à celles-ci; </s> <s xml:id="echoid-s13906" xml:space="preserve">ce qui arrive <lb/>toujours aux angles rentrans & </s> <s xml:id="echoid-s13907" xml:space="preserve">ſaillans: </s> <s xml:id="echoid-s13908" xml:space="preserve">il ſe rencontre même <lb/>bien des cas où la figure bizarre de ce que l’on veut meſurer, <lb/>demande beaucoup d’uſage de la Géométrie pour en venir à <lb/>bout: </s> <s xml:id="echoid-s13909" xml:space="preserve">& </s> <s xml:id="echoid-s13910" xml:space="preserve">comme bien des Ingénieurs ſe contentent de les <lb/>toiſer par approximation, voici quelques propoſitions qui <lb/>donneront beaucoup d’éclairciſſemens pour réſoudre les dif-<lb/>ficultés que je ferai appercevoir à ce ſujet.</s> <s xml:id="echoid-s13911" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1101" type="section" level="1" n="828"> <head xml:id="echoid-head1004" xml:space="preserve">PROPOSITION XIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13912" xml:space="preserve">817. </s> <s xml:id="echoid-s13913" xml:space="preserve">Meſurer la ſolidité des Secteurs de cylindre & </s> <s xml:id="echoid-s13914" xml:space="preserve">de Cônes <lb/> <anchor type="note" xlink:label="note-0486-01a" xlink:href="note-0486-01"/> tronqués.</s> <s xml:id="echoid-s13915" xml:space="preserve"/> </p> <div xml:id="echoid-div1101" type="float" level="2" n="1"> <note position="left" xlink:label="note-0486-01" xlink:href="note-0486-01a" xml:space="preserve">Figure 238.</note> </div> <p> <s xml:id="echoid-s13916" xml:space="preserve">Pour trouver la ſolidité d’un ſecteur A B C D E F d’un <lb/>cylindre formé par deux plans C A & </s> <s xml:id="echoid-s13917" xml:space="preserve">C E, il faut commencer <lb/>par ſçavoir la valeur du cylindre entier, & </s> <s xml:id="echoid-s13918" xml:space="preserve">connoître l’angle <lb/>B C D du ſecteur. </s> <s xml:id="echoid-s13919" xml:space="preserve">Ainſi ſuppoſant que cet angle ſoit de 50 <lb/>degrés, & </s> <s xml:id="echoid-s13920" xml:space="preserve">que la ſolidité du cylindre ſoit de 425 pieds, il faut <lb/>dire: </s> <s xml:id="echoid-s13921" xml:space="preserve">Si 360 degrés, valeur du cercle qui renferme le cylindre, <pb o="421" file="0487" n="501" rhead="DE MATHÉMATIQUE. Liv. XII."/> m’a donné 425 pieds pour la valeur du cylindre, que me don-<lb/>neront 50 degrés pour la valeur du ſecteur, l’on trouvera qu’il <lb/>eſt de 59 pieds & </s> <s xml:id="echoid-s13922" xml:space="preserve">quelque choſe.</s> <s xml:id="echoid-s13923" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13924" xml:space="preserve">818. </s> <s xml:id="echoid-s13925" xml:space="preserve">Pour meſurer un ſecteur G H K L M N d’un cône <lb/> <anchor type="note" xlink:label="note-0487-01a" xlink:href="note-0487-01"/> tronqué, il faut, comme ci-devant, connoître l’angle H K L <lb/>du ſecteur, & </s> <s xml:id="echoid-s13926" xml:space="preserve">la valeur du cône tronqué: </s> <s xml:id="echoid-s13927" xml:space="preserve">ainſi ſuppoſant que <lb/>l’angle eſt de 60 degrés, & </s> <s xml:id="echoid-s13928" xml:space="preserve">que le cône tronqué eſt de 600 pieds, <lb/>l’on dira encore: </s> <s xml:id="echoid-s13929" xml:space="preserve">Si 360 m’ont donné 600 pour la valeur du <lb/>cône tronqué, que me donneront 60 pour la valeur du ſecteur, <lb/>que l’on trouvera de 100 pieds.</s> <s xml:id="echoid-s13930" xml:space="preserve"/> </p> <div xml:id="echoid-div1102" type="float" level="2" n="2"> <note position="right" xlink:label="note-0487-01" xlink:href="note-0487-01a" xml:space="preserve">Figure 239.</note> </div> <p> <s xml:id="echoid-s13931" xml:space="preserve">819. </s> <s xml:id="echoid-s13932" xml:space="preserve">Mais ſi l’on avoit un cône tronqué A B C D, dans le <lb/> <anchor type="note" xlink:label="note-0487-02a" xlink:href="note-0487-02"/> milieu duquel il y auroit un vuide cylindrique G E F H, & </s> <s xml:id="echoid-s13933" xml:space="preserve"><lb/>qu’on voulût ſçavoir la valeur du fragment L N P Q O M S R <lb/>formé par des parties de couronnes, il faudroit commencer <lb/>par trouver la ſolidité de tout le cône tronqué A B C D, <lb/>comme s’il n’y avoit point de vuide pour avoir la valeur du <lb/>ſecteur L N K O M I, tant plein que vuide, de la façon qu’on <lb/>vient de le pratiquer; </s> <s xml:id="echoid-s13934" xml:space="preserve">enſuite en retrancher le ſecteur du cy-<lb/>lindre R P K Q S I, & </s> <s xml:id="echoid-s13935" xml:space="preserve">la différence ſera la ſolidité du fragment <lb/>L N P Q O M S R que l’on demande.</s> <s xml:id="echoid-s13936" xml:space="preserve"/> </p> <div xml:id="echoid-div1103" type="float" level="2" n="3"> <note position="right" xlink:label="note-0487-02" xlink:href="note-0487-02a" xml:space="preserve">Figure 240.</note> </div> <p> <s xml:id="echoid-s13937" xml:space="preserve">820. </s> <s xml:id="echoid-s13938" xml:space="preserve">Si au contraire on avoit un cylindre A B C D, dans le <lb/> <anchor type="note" xlink:label="note-0487-03a" xlink:href="note-0487-03"/> milieu duquel il y eût un vuide en forme de cône tronqué <lb/>E F G H, & </s> <s xml:id="echoid-s13939" xml:space="preserve">qu’on voulût ſçavoir la valeur de la ſolidité du <lb/>fragment Q O N P R L M S terminé par des plans qui ſoient <lb/>dans les rayons I N & </s> <s xml:id="echoid-s13940" xml:space="preserve">I L, il faudra chercher la valeur du <lb/>ſecteur cylindrique K O N I L M, & </s> <s xml:id="echoid-s13941" xml:space="preserve">celle du ſecteur K Q P I R S <lb/>du cône tronqué pour le retrancher de celle du ſecteur du cylin-<lb/>dre, & </s> <s xml:id="echoid-s13942" xml:space="preserve">la différence ſera la valeur du fragment QONPRLMS <lb/>que l’on demande.</s> <s xml:id="echoid-s13943" xml:space="preserve"/> </p> <div xml:id="echoid-div1104" type="float" level="2" n="4"> <note position="right" xlink:label="note-0487-03" xlink:href="note-0487-03a" xml:space="preserve">Figure 241.</note> </div> <p> <s xml:id="echoid-s13944" xml:space="preserve">Il faut, pour ſe rendre familier ce que l’on vient de voir, <lb/>donner des dimenſions aux lignes qui compoſent ces figures, <lb/>en faire le calcul, & </s> <s xml:id="echoid-s13945" xml:space="preserve">bien entendre les raiſons de chaque opé-<lb/>ration: </s> <s xml:id="echoid-s13946" xml:space="preserve">car, comme je l’ai déja dit, nous ſerons obligés d’avoit <lb/>recours à lui pour donner la ſolution de quelques-uns des pro-<lb/>blêmes les plus difficiles du toiſé de fortification.</s> <s xml:id="echoid-s13947" xml:space="preserve"/> </p> <pb o="422" file="0488" n="502" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1106" type="section" level="1" n="829"> <head xml:id="echoid-head1005" xml:space="preserve">PROPOSITION XIV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s13948" xml:space="preserve">821. </s> <s xml:id="echoid-s13949" xml:space="preserve">Meſurer la ſolidité d’une Sphere. <lb/></s> </p> <p> <s xml:id="echoid-s13950" xml:space="preserve">Pour avoir la ſolidité d’une ſphere, dont le diametre A B <lb/> <anchor type="note" xlink:label="note-0488-02a" xlink:href="note-0488-02"/> eſt de 14 pieds, il faut chercher la circonférence de ce diametre, <lb/>qui ſera 44, & </s> <s xml:id="echoid-s13951" xml:space="preserve">la multiplier par le diametre même pour avoir <lb/>la ſurface de la ſphere (art. </s> <s xml:id="echoid-s13952" xml:space="preserve">576), qui ſera de 616 pieds, qu’il <lb/>faut multiplier par le tiers du rayon (art. </s> <s xml:id="echoid-s13953" xml:space="preserve">575), c’eſt-à-dire <lb/>par le tiers de 7, pour avoir 1437 {1/2} pieds cubes pour la ſolidité <lb/>de la ſphere.</s> <s xml:id="echoid-s13954" xml:space="preserve"/> </p> <div xml:id="echoid-div1106" type="float" level="2" n="1"> <note position="left" xlink:label="note-0488-02" xlink:href="note-0488-02a" xml:space="preserve">Figure 242.</note> </div> <p> <s xml:id="echoid-s13955" xml:space="preserve">L’on trouvera encore la ſolidité de la ſphere d’une autre <lb/>maniere, en multipliant la ſuperficie de ſon grand cercle par <lb/>les deux tiers du diametre (art. </s> <s xml:id="echoid-s13956" xml:space="preserve">568).</s> <s xml:id="echoid-s13957" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13958" xml:space="preserve">L’on peut encore trouver la ſolidité des ſpheres par une ſeule <lb/>Regle de Trois, ayant ſeulement les cubes de leurs axes, avec <lb/>la même facilité que l’on trouve la ſuperficie des cercles à <lb/>l’aide du quarré de leur diametre; </s> <s xml:id="echoid-s13959" xml:space="preserve">car il y a même raiſon du <lb/>cube de l’axe d’une ſphere à la ſolidité de la même ſphere, que <lb/>de ſon diametre à la ſixieme partie de la circonférence du <lb/>même diametre. </s> <s xml:id="echoid-s13960" xml:space="preserve">Pour en être convaincus, nous nommerons a <lb/>le diametre où l’axe de cette ſphere, & </s> <s xml:id="echoid-s13961" xml:space="preserve">b ſa circonférence; </s> <s xml:id="echoid-s13962" xml:space="preserve">la <lb/>ſuperficie de ſon grand cercle ſera par conſéquent {ab/4}, qui <lb/>étant multiplié par les deux tiers du diametre, c’eſt-à-dire <lb/>par {2a/3} donne {2aab/12} = {aab/6} pour la ſolidité de la ſphere: </s> <s xml:id="echoid-s13963" xml:space="preserve">ainſi <lb/>l’on aura a a a: </s> <s xml:id="echoid-s13964" xml:space="preserve">{aab/6}:</s> <s xml:id="echoid-s13965" xml:space="preserve">: a: </s> <s xml:id="echoid-s13966" xml:space="preserve">{b/6}: </s> <s xml:id="echoid-s13967" xml:space="preserve">& </s> <s xml:id="echoid-s13968" xml:space="preserve">ſuppoſant une ſphere de <lb/>21 pieds de diametre, dont la circonférence eſt de 66 pieds, <lb/>en prenant la ſixieme partie, qui eſt 11, on n’aura plus qu’à <lb/>dire, comme 21 eſt à 11: </s> <s xml:id="echoid-s13969" xml:space="preserve">ainſi le cube de 14, qui eſt 2744 <lb/>eſt à la ſolidité de la ſphere que l’on trouvera encore de 1437 <lb/>pieds & </s> <s xml:id="echoid-s13970" xml:space="preserve">{1/7}.</s> <s xml:id="echoid-s13971" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s13972" xml:space="preserve">822. </s> <s xml:id="echoid-s13973" xml:space="preserve">Pour meſurer un ſecteur de ſphere, tel que A B C D, <lb/> <anchor type="note" xlink:label="note-0488-03a" xlink:href="note-0488-03"/> il faut connoître le rayon & </s> <s xml:id="echoid-s13974" xml:space="preserve">la perpendiculaire D E, élevée <lb/>ſur le milieu de la corde A C. </s> <s xml:id="echoid-s13975" xml:space="preserve">Or ſi nous ſuppoſons le rayon <lb/>de 7 pieds, & </s> <s xml:id="echoid-s13976" xml:space="preserve">la perpendiculaire de 3, il faut chercher, par le <lb/>moyen du rayon, la circonférence du grand cercle de la ſphere, <lb/>d’où le ſecteur a été tiré, & </s> <s xml:id="echoid-s13977" xml:space="preserve">on la trouvera de 44 pieds: </s> <s xml:id="echoid-s13978" xml:space="preserve">il <pb o="423" file="0489" n="503" rhead="DE MATHÉMATIQUE. Liv. XII."/> faut enſuite multiplier cette circonférence par la perpendicu-<lb/>laire D E, c’eſt-à-dire 44 par 3; </s> <s xml:id="echoid-s13979" xml:space="preserve">& </s> <s xml:id="echoid-s13980" xml:space="preserve">le produit 132 ſera la ſur-<lb/>face A D C du ſecteur (art. </s> <s xml:id="echoid-s13981" xml:space="preserve">805), qu’il faudra multiplier par <lb/>le tiers du rayon B C, c’eſt-à-dire par 2 {1/3}, pour avoir 308 pieds <lb/>cubes, qui eſt la ſolidité du ſecteur.</s> <s xml:id="echoid-s13982" xml:space="preserve"/> </p> <div xml:id="echoid-div1107" type="float" level="2" n="2"> <note position="left" xlink:label="note-0488-03" xlink:href="note-0488-03a" xml:space="preserve">Figure 243.</note> </div> <p> <s xml:id="echoid-s13983" xml:space="preserve">823. </s> <s xml:id="echoid-s13984" xml:space="preserve">Si au lieu d’un ſecteur l’on avoit un ſegment de ſphere <lb/> <anchor type="note" xlink:label="note-0489-01a" xlink:href="note-0489-01"/> D G F, il faudroit, pour en trouver la ſolidité, le réduire en <lb/>ſecteur, & </s> <s xml:id="echoid-s13985" xml:space="preserve">chercher la ſolidité de ce ſecteur, de laquelle il <lb/>faudroit retrancher le cône D E F, & </s> <s xml:id="echoid-s13986" xml:space="preserve">le reſtant ſeroit la va-<lb/>leur du ſegment.</s> <s xml:id="echoid-s13987" xml:space="preserve"/> </p> <div xml:id="echoid-div1108" type="float" level="2" n="3"> <note position="right" xlink:label="note-0489-01" xlink:href="note-0489-01a" xml:space="preserve">Figure 244.</note> </div> <p> <s xml:id="echoid-s13988" xml:space="preserve">824. </s> <s xml:id="echoid-s13989" xml:space="preserve">Mais ſi la partie de la ſphere que l’on veut meſurer <lb/> <anchor type="note" xlink:label="note-0489-02a" xlink:href="note-0489-02"/> étoit une zone compriſe par le grand cercle de la ſphere, & </s> <s xml:id="echoid-s13990" xml:space="preserve"><lb/>par un autre quelconque, qui lui ſeroit parallelement oppoſé, <lb/>comme eſt la zone A F H E, on en trouveroit la ſolidité en <lb/>prenant les deux tiers du cylindre qui auroit pour baſe le <lb/>grand cercle A E, & </s> <s xml:id="echoid-s13991" xml:space="preserve">pour hauteur la partie de l’axe G C; </s> <s xml:id="echoid-s13992" xml:space="preserve">& </s> <s xml:id="echoid-s13993" xml:space="preserve"><lb/>de plus le tiers du cylindre qui auroit pour baſe le petit cer-<lb/>cle F H, & </s> <s xml:id="echoid-s13994" xml:space="preserve">pour hauteur la même ligne G C (art. </s> <s xml:id="echoid-s13995" xml:space="preserve">578). </s> <s xml:id="echoid-s13996" xml:space="preserve">Or <lb/>pour en faire l’opération, nous ſuppoſerons le rayon C E de <lb/>14 pieds, & </s> <s xml:id="echoid-s13997" xml:space="preserve">la perpendiculaire C G de 8; </s> <s xml:id="echoid-s13998" xml:space="preserve">& </s> <s xml:id="echoid-s13999" xml:space="preserve">comme nous <lb/>avons le triangle rectangle C H K, dont l’hypoténuſe C H eſt <lb/>de 14 pieds, & </s> <s xml:id="echoid-s14000" xml:space="preserve">le côté H K de 8, l’on trouvera par la racine <lb/>quarrée le côté C K de 11 pieds: </s> <s xml:id="echoid-s14001" xml:space="preserve">ainſi l’on aura le rayon du <lb/>cercle F H; </s> <s xml:id="echoid-s14002" xml:space="preserve">& </s> <s xml:id="echoid-s14003" xml:space="preserve">par conſéquent l’on trouvera la ſolidité du cy-<lb/>lindre I H, qui eſt de 3036 pieds cubes, & </s> <s xml:id="echoid-s14004" xml:space="preserve">la ſolidité du <lb/>grand cylindre A D ſe trouvera de 4928 pieds cubes. </s> <s xml:id="echoid-s14005" xml:space="preserve">Or ſi <lb/>l’on prend les deux tiers du plus grand cylindre, l’on aura <lb/>3285 {1/3}, qui étant ajouté avec 1012, qui eſt le tiers du petit <lb/>cylindre, nous donnera 4297 {1/3} pieds cubes pour la ſolidité de <lb/>la zone.</s> <s xml:id="echoid-s14006" xml:space="preserve"/> </p> <div xml:id="echoid-div1109" type="float" level="2" n="4"> <note position="right" xlink:label="note-0489-02" xlink:href="note-0489-02a" xml:space="preserve">Figure 245.</note> </div> </div> <div xml:id="echoid-div1111" type="section" level="1" n="830"> <head xml:id="echoid-head1006" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s14007" xml:space="preserve">825. </s> <s xml:id="echoid-s14008" xml:space="preserve">La génération de la plûpart des ſolides ayant été for-<lb/> <anchor type="note" xlink:label="note-0489-03a" xlink:href="note-0489-03"/> mée par la circonvolution d’un plan ſur ſon axe, l’on peut <lb/>avoir autant de ſolides différens, que l’on peut avoir de plans <lb/>générateurs différens: </s> <s xml:id="echoid-s14009" xml:space="preserve">mais pour ne parler que de ceux qui <lb/>ſont formés par le plan des courbes des ſections coniques, <lb/>l’on ſçaura que ſi une demi-parabole A C B fait une circonvo-<lb/>lution autour de ſon axe A B, elle décrira un corps H I K, <lb/>que l’on nomme parabolique, qui eſt compoſé d’une infinité <pb o="424" file="0490" n="504" rhead="NOUVEAU COURS"/> de cercles qui auront tous pour rayons les ordonnées, telles <lb/>que D E & </s> <s xml:id="echoid-s14010" xml:space="preserve">F G, que l’on regarde ici comme les élémens du <lb/>plan A B C de la parabole.</s> <s xml:id="echoid-s14011" xml:space="preserve"/> </p> <div xml:id="echoid-div1111" type="float" level="2" n="1"> <note position="right" xlink:label="note-0489-03" xlink:href="note-0489-03a" xml:space="preserve">Figure 246. <lb/>& 247.</note> </div> <p> <s xml:id="echoid-s14012" xml:space="preserve">826. </s> <s xml:id="echoid-s14013" xml:space="preserve">Si l’on a une demi-ellipſe H L I qui faſſe une circon-<lb/> <anchor type="note" xlink:label="note-0490-01a" xlink:href="note-0490-01"/> volution autour de ſon axe H I, toutes les ordonnées, comme <lb/>O P & </s> <s xml:id="echoid-s14014" xml:space="preserve">R S, que l’on peut regarder comme les élémens du plan <lb/>de l’ellipſe, décriront une infinité de cercles, qui tous enſem-<lb/>ble formeront le corps A B C D, que l’on nomme ſphéroïde, <lb/>parce qu’ayant pour plan générateur une ellipſe, qui eſt pro-<lb/>prement un cercle alongé, le ſphéroïde eſt regardé comme <lb/>une ſphere alongée.</s> <s xml:id="echoid-s14015" xml:space="preserve"/> </p> <div xml:id="echoid-div1112" type="float" level="2" n="2"> <note position="left" xlink:label="note-0490-01" xlink:href="note-0490-01a" xml:space="preserve">Figure 250. <lb/>& 251.</note> </div> <p> <s xml:id="echoid-s14016" xml:space="preserve">827. </s> <s xml:id="echoid-s14017" xml:space="preserve">Enfin ſi l’on fait faire à une demi - hyperbole A B C <lb/> <anchor type="note" xlink:label="note-0490-02a" xlink:href="note-0490-02"/> une circonvolution ſur ſon axe B C, elle décrira un ſolide, <lb/>que l’on nomme hyperboloïde; </s> <s xml:id="echoid-s14018" xml:space="preserve">& </s> <s xml:id="echoid-s14019" xml:space="preserve">ſi la demi-hyperbole eſt ac-<lb/>compagnée d’une aſymptote E F, & </s> <s xml:id="echoid-s14020" xml:space="preserve">des lignes D B & </s> <s xml:id="echoid-s14021" xml:space="preserve">D G pa-<lb/>ralleles à A C & </s> <s xml:id="echoid-s14022" xml:space="preserve">B C, le triangle E F C décrira un cône, & </s> <s xml:id="echoid-s14023" xml:space="preserve">le <lb/>rectangle G D B C un cylindre.</s> <s xml:id="echoid-s14024" xml:space="preserve"/> </p> <div xml:id="echoid-div1113" type="float" level="2" n="3"> <note position="left" xlink:label="note-0490-02" xlink:href="note-0490-02a" xml:space="preserve">Figure 252.</note> </div> <p> <s xml:id="echoid-s14025" xml:space="preserve">Comme la plûpart de ces ſolides ont lieu dans bien des oc-<lb/>caſions, nous en ferons voir l’application, après que nous <lb/>aurons donné dans les propoſitions ſuivantes la maniere de les <lb/>meſurer.</s> <s xml:id="echoid-s14026" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1115" type="section" level="1" n="831"> <head xml:id="echoid-head1007" xml:space="preserve">PROPOSITION XV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14027" xml:space="preserve">828. </s> <s xml:id="echoid-s14028" xml:space="preserve">Meſurer la ſolidité d’un Paraboloïde.</s> <s xml:id="echoid-s14029" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14030" xml:space="preserve">Pour avoir la ſolidité d’un paraboloïde, dont le rayon L K <lb/> <anchor type="note" xlink:label="note-0490-03a" xlink:href="note-0490-03"/> du cercle de la baſe ſeroit de 7 pieds, l’axe I L de 10, il faut <lb/>chercher la valeur du cercle de la baſe, qui ſera de 154 pieds, <lb/>qu’il faut multiplier par la moitié de l’axe I L, c’eſt-à-dire par <lb/>5 pour avoir 770 au produit, qui ſera ce que l’on demande.</s> <s xml:id="echoid-s14031" xml:space="preserve"/> </p> <div xml:id="echoid-div1115" type="float" level="2" n="1"> <note position="left" xlink:label="note-0490-03" xlink:href="note-0490-03a" xml:space="preserve">Figure 246. <lb/>& 247.</note> </div> <p> <s xml:id="echoid-s14032" xml:space="preserve">Pour ſçavoir la raiſon de cette opération, conſidérez que <lb/>l’axe AB de la parabole eſt compoſé d’une infinité de parties, <lb/>comme A E & </s> <s xml:id="echoid-s14033" xml:space="preserve">A G, qui ſont en progreſſion arithmétique, & </s> <s xml:id="echoid-s14034" xml:space="preserve"><lb/>que les quarrés des ordonnées E D & </s> <s xml:id="echoid-s14035" xml:space="preserve">G F étant dans la <lb/>même raiſon que les parties A E & </s> <s xml:id="echoid-s14036" xml:space="preserve">E G (art. </s> <s xml:id="echoid-s14037" xml:space="preserve">605); </s> <s xml:id="echoid-s14038" xml:space="preserve">ces quarrés <lb/>ſeront auſſi en progreſſion arithmétique. </s> <s xml:id="echoid-s14039" xml:space="preserve">Or comme les cercles <lb/>ſont dans la même raiſon que les quarrés de leurs rayons <lb/>(art. </s> <s xml:id="echoid-s14040" xml:space="preserve">455), il s’enſuit que les cercles qui compoſent le para-<lb/>boloïde H I K ſont en progreſſion arithmétique, puiſqu’ils ſont <pb o="425" file="0491" n="505" rhead="DE MATHEMATIQUE. Liv. XII."/> comme les quarrés des ordonnées de la parabole: </s> <s xml:id="echoid-s14041" xml:space="preserve">mais comme <lb/>pour trouver la valeur des termes infinis d’une progreſſion arith-<lb/>métique (art. </s> <s xml:id="echoid-s14042" xml:space="preserve">389), il faut multiplier le plus grand terme de <lb/>la progreſſion par la moitié de la grandeur qui exprime la <lb/>quantité de ces termes, il faut donc, pour trouver la valeur <lb/>de tous les cercles qui compoſent le paraboloïde, multiplier le <lb/>plus grand cercle H K par la moitié de l’axe I L.</s> <s xml:id="echoid-s14043" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1117" type="section" level="1" n="832"> <head xml:id="echoid-head1008" xml:space="preserve">PROPOSITION XVI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14044" xml:space="preserve">829. </s> <s xml:id="echoid-s14045" xml:space="preserve">Meſurer la ſolidité d’un Sphéroïde. <lb/></s> <s xml:id="echoid-s14046" xml:space="preserve"> <anchor type="note" xlink:label="note-0491-01a" xlink:href="note-0491-01"/> & </s> <s xml:id="echoid-s14047" xml:space="preserve">251. que l’on demande. née N L de l’ellipſe par les deux tiers de l’axe H I. <anchor type="note" xlink:label="note-0491-02a" xlink:href="note-0491-02"/> convolution à l’entour de ſon grand axe A B, en faiſoit une ſur ſon petit axe C D, l’on auroit encore un ſphéroïde A C B D, dont on trouvera la ſolidité, comme ci-devant, en multi- pliant la ſuperficie du cercle du grand axe A B par les deux tiers du petit axe C D: car ſi l’on a un cercle E C F D, qui <pb o="426" file="0492" n="506" rhead="NOUVEAU COURS"/> ait pour diametre le petit axe C D, & que l’on mene les or- données G H & K L, l’on aura par la propriété de l’ellipſe (art. C G x G D : C K x K D :: <emph style="ol">G H</emph><emph style="sub">2</emph>: <emph style="ol">K L</emph><emph style="sub">2</emph>; & ſi à la place des rectangles C G x G D & C K x K D, l’on prend les quarrés G I<emph style="sub">2</emph> & K M<emph style="sub">2</emph>, qui leur ſont égaux par la propriété du cercle, l’on aura <emph style="ol">G I</emph><emph style="sub">2</emph>: <emph style="ol">K M</emph><emph style="sub">2</emph>:: <emph style="ol">G H</emph><emph style="sub">2</emph>: <emph style="ol">K L</emph><emph style="sub">2</emph>. Or ſi à la place des quar- rés de toutes les ordonnées du demi-cercle C F D, l’on prend les cercles dont ces ordonnées ſont les rayons, & qu’on faſſe la même choſe pour la demi-ellipſe C B D, l’on verra que tous les cercles de la ſphere ſont dans la même raiſon que tous les cercles du ſphéroïde, & que la quantité des uns & des autres étant exprimée par la ligne C D, ſi l’on multiplie le cercle E F par les deux tiers de la ligne C D, pour avoir la valeur de tous les cercles qui compoſent la ſphere, il faudra multiplier le cercle de A B par les deux tiers de la ligne C D, pour avoir la valeur de tous les cercles qui compoſent le ſphéroïde.</s></p> <div xml:id="echoid-div1117" type="float" level="2" n="1"> <note position="right" xlink:label="note-0491-02" xlink:href="note-0491-02a" xml:space="preserve">Figure 248. <lb/>& 249.</note> </div> <p> <s xml:id="echoid-s14048" xml:space="preserve">831. </s> <s xml:id="echoid-s14049" xml:space="preserve">L’on peut dire auſſi que ſi l’on n’avoit que la moitié <lb/>d’un ſphéroïde A C B, il faudroit de même, pour en trouver <lb/>la ſolidité, multiplier le cercle A B par les deux tiers de la <lb/>ligne C N.</s> <s xml:id="echoid-s14050" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14051" xml:space="preserve">Quoique l’hyperboloïde n’ait guere lieu dans la Géométrie <lb/>pratique, cela n’empêche pas que je ne diſe un mot ſur la <lb/>maniere de meſurer ce ſolide, pour ſatisfaire la curioſité de <lb/>ceux qui n’aiment pas qu’on leur ſupprime rien.</s> <s xml:id="echoid-s14052" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1119" type="section" level="1" n="833"> <head xml:id="echoid-head1009" xml:space="preserve">PROPOSITION XVII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14053" xml:space="preserve">832. </s> <s xml:id="echoid-s14054" xml:space="preserve">Meſurer la ſolidité d’un Hyperboloïde. <lb/></s> </p> <p> <s xml:id="echoid-s14055" xml:space="preserve">Pour avoir la ſolidité d’un hyperboloïde D E F, il faut ac-<lb/>compagner la courbe D E F de ſes aſymptotes B A & </s> <s xml:id="echoid-s14056" xml:space="preserve">B C, & </s> <s xml:id="echoid-s14057" xml:space="preserve"><lb/>de la ligne G H, qui ſera égale à un de ſes axes. </s> <s xml:id="echoid-s14058" xml:space="preserve">Cela poſé. </s> <s xml:id="echoid-s14059" xml:space="preserve">il <lb/>faut chercher la ſolidité d’un cône tronqué A G H C (art. </s> <s xml:id="echoid-s14060" xml:space="preserve">815), <lb/>& </s> <s xml:id="echoid-s14061" xml:space="preserve">en retrancher le cylindre I G H K pour avoir la différence, <lb/>qui ſera la ſolidité de l’hyperboloïde.</s> <s xml:id="echoid-s14062" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14063" xml:space="preserve">Pour entendre la raiſon de l’opération que nous indiquons <lb/>ici, il faut ſe rappeller que nous avons fait voir dans l’hyper-<lb/>bole (art. </s> <s xml:id="echoid-s14064" xml:space="preserve">679), que ſi l’on menoit une ligne telle que A C, <lb/>parallele à G H, le rectangle compris ſous les parties A D &</s> <s xml:id="echoid-s14065" xml:space="preserve"> <pb o="427" file="0493" n="507" rhead="DE MATHÉMATIQUE. Liv. XII."/> D C, ſeroit égal au quarré de la ligne G E. </s> <s xml:id="echoid-s14066" xml:space="preserve">Or comme le <lb/>rectangle compris ſous A D & </s> <s xml:id="echoid-s14067" xml:space="preserve">D C, eſt égal au quarré de la <lb/>perpendiculaire D M (art. </s> <s xml:id="echoid-s14068" xml:space="preserve">441), à cauſe du demi-cercle A D C, <lb/>il s’enſuit que la ligne D M eſt égale à la ligne G E. </s> <s xml:id="echoid-s14069" xml:space="preserve">Cela <lb/>poſé, l’on ſçait que le cercle, qui auroit pour rayon la ligne <lb/>D M, eſt égal à la couronne formée par les deux circonfé-<lb/>rences (art. </s> <s xml:id="echoid-s14070" xml:space="preserve">565) A N C O & </s> <s xml:id="echoid-s14071" xml:space="preserve">D P F Q. </s> <s xml:id="echoid-s14072" xml:space="preserve">Cela étant, cette cou-<lb/>ronne ſera égale au cercle, qui aura pour rayon la ligne G E, <lb/>& </s> <s xml:id="echoid-s14073" xml:space="preserve">qui ſera un des cercles du cylindre G H I K; </s> <s xml:id="echoid-s14074" xml:space="preserve">& </s> <s xml:id="echoid-s14075" xml:space="preserve">comme il <lb/>arrivera la même choſe pour toutes les lignes telles que A C, <lb/>qu’on tirera parallele à G H par tel point que l’on voudra de la <lb/>ligne G A; </s> <s xml:id="echoid-s14076" xml:space="preserve">il s’enſuit que toutes les couronnes ſeront égales <lb/>entr’elles, puiſque chacune ſera égale à des cercles du cylin-<lb/>dre. </s> <s xml:id="echoid-s14077" xml:space="preserve">Or comme il y a autant de couronnes que de cercles, les <lb/>uns & </s> <s xml:id="echoid-s14078" xml:space="preserve">les autres étant exprimés par la ligne E L, il s’enſuit <lb/>que l’eſpace qui eſt renfermé entre l’hyperboloïde D P F Q E <lb/>& </s> <s xml:id="echoid-s14079" xml:space="preserve">le cône tronqué A N C O G F (qui n’eſt autre choſe que la <lb/>ſomme de toutes les couronnes), eſt égal au cylindre I G H K; <lb/></s> <s xml:id="echoid-s14080" xml:space="preserve">& </s> <s xml:id="echoid-s14081" xml:space="preserve">par conſéquent le cône tronqué eſt plus grand que l’hyper-<lb/>boloïde de tout le même cylindre I G H K.</s> <s xml:id="echoid-s14082" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s14083" xml:space="preserve">Application de la Géométrie au Toiſé des Voûtes.</s> <s xml:id="echoid-s14084" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1120" type="section" level="1" n="834"> <head xml:id="echoid-head1010" xml:space="preserve">PROPOSITION XVIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14085" xml:space="preserve">833. </s> <s xml:id="echoid-s14086" xml:space="preserve">Meſurer la ſolidité de la Maçonnerie de toutes ſortes de <lb/> <anchor type="note" xlink:label="note-0493-01a" xlink:href="note-0493-01"/> voûtes.</s> <s xml:id="echoid-s14087" xml:space="preserve"/> </p> <div xml:id="echoid-div1120" type="float" level="2" n="1"> <note position="right" xlink:label="note-0493-01" xlink:href="note-0493-01a" xml:space="preserve">Pl. XVIII.</note> </div> <p> <s xml:id="echoid-s14088" xml:space="preserve">Il n’y a guere que trois ſortes de voûtes parmi les ouvrages <lb/> <anchor type="note" xlink:label="note-0493-02a" xlink:href="note-0493-02"/> de fortification. </s> <s xml:id="echoid-s14089" xml:space="preserve">Les premieres ſont celles des ſouterreins, les <lb/>ſecondes, celles des magaſins à poudre, & </s> <s xml:id="echoid-s14090" xml:space="preserve">les troiſiemes, <lb/>celles des tours auxquelles il y a des plates-formes: </s> <s xml:id="echoid-s14091" xml:space="preserve">les unes <lb/>& </s> <s xml:id="echoid-s14092" xml:space="preserve">les autres ſont ou à plein ceintre, comme dans la figure 256, <lb/>ou ſurbaiſſées, comme dans la figure 257, ou gothique, que l’on <lb/>nomme auſſi voûte en tiers point, ou voûte en arc de cloître, <lb/>comme dans la figure 258, & </s> <s xml:id="echoid-s14093" xml:space="preserve">ſoit qu’elles ſervent aux ma-<lb/>gaſins ou aux ſouterreins, elles ſont toujours diſpoſées en de-<lb/>hors en dos d’âne, comme un toit, parce qu’on y applique <lb/>deſſus une chape de ciment pour les garantir des eaux de pluies.</s> <s xml:id="echoid-s14094" xml:space="preserve"/> </p> <div xml:id="echoid-div1121" type="float" level="2" n="2"> <note position="right" xlink:label="note-0493-02" xlink:href="note-0493-02a" xml:space="preserve">Figure 256, <lb/>257 & 258.</note> </div> <p> <s xml:id="echoid-s14095" xml:space="preserve">834. </s> <s xml:id="echoid-s14096" xml:space="preserve">Si l’on a donc à toiſer la maçonnerie d’un ſouterrein <pb o="428" file="0494" n="508" rhead="NOUVEAU COURS"/> ou d’un magaſin, dont la figure 250 ſoit le plan, l’on com-<lb/> <anchor type="note" xlink:label="note-0494-01a" xlink:href="note-0494-01"/> mence par toiſer les pignons P R S T & </s> <s xml:id="echoid-s14097" xml:space="preserve">M K O L, ſans aucune <lb/>difficulté, parce que ce ſont des parallelepipedes; </s> <s xml:id="echoid-s14098" xml:space="preserve">enſuite on <lb/>toiſe auſſi les pieds-droits A D F G, depuis la retraite des fon-<lb/>demens juſqu’à la naiſſance A C de la voûte; </s> <s xml:id="echoid-s14099" xml:space="preserve">& </s> <s xml:id="echoid-s14100" xml:space="preserve">pour la voûte, <lb/>l’on toiſe la ſuperficie du triangle A B C, que l’on multiplie <lb/>par la longueur dans œ uvre de la voûte; </s> <s xml:id="echoid-s14101" xml:space="preserve">ce qui s’appelle toiſer <lb/>tant plein que vuide: </s> <s xml:id="echoid-s14102" xml:space="preserve">& </s> <s xml:id="echoid-s14103" xml:space="preserve">comme il faut du produit en déduire <lb/>le vuide D K E, ſi la voûte eſt en plein ceintre, l’on meſure <lb/>la ſuperficie du demi-cercle (art. </s> <s xml:id="echoid-s14104" xml:space="preserve">484) D K E, que l’on mul-<lb/>tiplie par la même longueur qui a ſervi à meſurer le triangle <lb/>A B C; </s> <s xml:id="echoid-s14105" xml:space="preserve">& </s> <s xml:id="echoid-s14106" xml:space="preserve">ſouſtrayant ce produit-ci du précédent, la diffé-<lb/>rence eſt la valeur de la voûte.</s> <s xml:id="echoid-s14107" xml:space="preserve"/> </p> <div xml:id="echoid-div1122" type="float" level="2" n="3"> <note position="left" xlink:label="note-0494-01" xlink:href="note-0494-01a" xml:space="preserve">Figure 256. <lb/>& 259.</note> </div> <p> <s xml:id="echoid-s14108" xml:space="preserve">835. </s> <s xml:id="echoid-s14109" xml:space="preserve">Si la voûte eſt ſurbaiſſée, comme F E G, dont la figure <lb/> <anchor type="note" xlink:label="note-0494-02a" xlink:href="note-0494-02"/> eſt une demi-ellipſe, il faut meſurer le triangle A B C comme <lb/>ci-devant, & </s> <s xml:id="echoid-s14110" xml:space="preserve">le multiplier par la longueur dans œuvre de la <lb/>voûte: </s> <s xml:id="echoid-s14111" xml:space="preserve">après quoi l’on cherchera la ſuperficie de la demi-ellipſe <lb/>F E G (art. </s> <s xml:id="echoid-s14112" xml:space="preserve">798), pour la multiplier auſſi par la même lon-<lb/>gueur; </s> <s xml:id="echoid-s14113" xml:space="preserve">& </s> <s xml:id="echoid-s14114" xml:space="preserve">ſouſtrayant ce produit-ci du précédent, on aura la <lb/>valeur de la voûte.</s> <s xml:id="echoid-s14115" xml:space="preserve"/> </p> <div xml:id="echoid-div1123" type="float" level="2" n="4"> <note position="left" xlink:label="note-0494-02" xlink:href="note-0494-02a" xml:space="preserve">Figure 257.</note> </div> <p> <s xml:id="echoid-s14116" xml:space="preserve">836. </s> <s xml:id="echoid-s14117" xml:space="preserve">Enfin ſi la voûte que l’on veut meſurer eſt en tiers <lb/> <anchor type="note" xlink:label="note-0494-03a" xlink:href="note-0494-03"/> point, comme I L M, on cherchera la ſuperficie du triangle <lb/>I L M, à laquelle on joindra celle des ſegmens (art. </s> <s xml:id="echoid-s14118" xml:space="preserve">797) des <lb/>cercles, dont les lignes L I & </s> <s xml:id="echoid-s14119" xml:space="preserve">L M ſont les cordes; </s> <s xml:id="echoid-s14120" xml:space="preserve">& </s> <s xml:id="echoid-s14121" xml:space="preserve">ayant <lb/>multiplié cette quantité par la longueur de la voûte dans œu-<lb/>vre, on ſouſtraira le produit de celui du triangle H K N, <lb/>multiplié par la même longueur, & </s> <s xml:id="echoid-s14122" xml:space="preserve">l’on aura la ſolidité que <lb/>l’on demande.</s> <s xml:id="echoid-s14123" xml:space="preserve"/> </p> <div xml:id="echoid-div1124" type="float" level="2" n="5"> <note position="left" xlink:label="note-0494-03" xlink:href="note-0494-03a" xml:space="preserve">Figure 258.</note> </div> <p> <s xml:id="echoid-s14124" xml:space="preserve">837. </s> <s xml:id="echoid-s14125" xml:space="preserve">Pour les voûtes au deſſus deſquelles il y a des plates-<lb/>formes, comme, par exemple, celles qui couvrent les ſalles <lb/>de l’Obſervatoire Royal de Paris, le toiſé en eſt un peu plus <lb/>difficile; </s> <s xml:id="echoid-s14126" xml:space="preserve">& </s> <s xml:id="echoid-s14127" xml:space="preserve">je ne ſçache pas même que perſonne ait recherché <lb/>la maniere de le faire géométriquement: </s> <s xml:id="echoid-s14128" xml:space="preserve">comme ces ſortes <lb/>d’endroits ont pour baſe un quarré ou un polygone régulier, <lb/>le vuide & </s> <s xml:id="echoid-s14129" xml:space="preserve">le plein de la voûte font ordinairement un priſme, <lb/>qui eſt facile à meſurer: </s> <s xml:id="echoid-s14130" xml:space="preserve">& </s> <s xml:id="echoid-s14131" xml:space="preserve">comme il n’y a que le vuide qu’il <lb/>faut déduire, qui peut faire quelque difficulté, nous conſidé-<lb/>rerons ici les différentes figures qu’il peut avoir, afin de les <lb/>réduire à des corps réguliers.</s> <s xml:id="echoid-s14132" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14133" xml:space="preserve">Suppoſant donc que les lieux dont il s’agit, ayent pour baſe <pb o="429" file="0495" n="509" rhead="DE MATHÉMATIQUE. Liv. XII."/> un quarré A B, ou un polygone régulier G H, voici comment <lb/>on peut conſidérer la nature de leurs voûtes.</s> <s xml:id="echoid-s14134" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14135" xml:space="preserve">Si la baſe eſt un quarré, les diagonales A B & </s> <s xml:id="echoid-s14136" xml:space="preserve">C D ſervi-<lb/> <anchor type="note" xlink:label="note-0495-01a" xlink:href="note-0495-01"/> ront de diametre à des demi-cercles A E B & </s> <s xml:id="echoid-s14137" xml:space="preserve">C F D, qui par-<lb/>tagent la voûte en quatre, & </s> <s xml:id="echoid-s14138" xml:space="preserve">qui forment des arrêtes dans les <lb/>angles. </s> <s xml:id="echoid-s14139" xml:space="preserve">Or ſi l’on conſidere une infinité de quarrés qui rem-<lb/>pliſſent le vuide de la voûte, tous ces quarrés auront leurs an-<lb/>gles dans les quarts de cercles F C, F A, F B, F D, & </s> <s xml:id="echoid-s14140" xml:space="preserve">leurs <lb/>côtés ſeront des lignes comme G H & </s> <s xml:id="echoid-s14141" xml:space="preserve">I K, tirées d’un quart <lb/>de cercle à l’autre parallélement aux côtés A D ou D B, & </s> <s xml:id="echoid-s14142" xml:space="preserve">la <lb/>moitié de toutes les diagonales, comme E A & </s> <s xml:id="echoid-s14143" xml:space="preserve">L M ſeront <lb/>les ordonnées d’un quart de cercle A F E. </s> <s xml:id="echoid-s14144" xml:space="preserve">Or comme la ligne <lb/>E F ou E A qui marque la hauteur de la voûte, exprime la <lb/>ſomme de tous ces quarrés, il s’enſuit que les ordonnées E A <lb/>& </s> <s xml:id="echoid-s14145" xml:space="preserve">L M ſervant de demi-diagonales à ces quarrés, l’on trou-<lb/>vera la valeur de tous ces quarrés, comme on trouve celles <lb/>des ordonnées d’un quart de cercle; </s> <s xml:id="echoid-s14146" xml:space="preserve">mais nous avons vu <lb/>(art. </s> <s xml:id="echoid-s14147" xml:space="preserve">821), que la valeur des quarrés des ordonnées d’un quart <lb/>de cercle ſe connoiſſoit en multipliant la plus grande ordon-<lb/>née E A par les deux tiers de la ligne E F: </s> <s xml:id="echoid-s14148" xml:space="preserve">il faudra donc, pour <lb/>trouver la ſolidité du corps A F B, multiplier le quarré A B, <lb/>qui lui ſert de baſe, par les deux tiers de la ligne E F, qui en <lb/>exprime la hauteur.</s> <s xml:id="echoid-s14149" xml:space="preserve"/> </p> <div xml:id="echoid-div1125" type="float" level="2" n="6"> <note position="right" xlink:label="note-0495-01" xlink:href="note-0495-01a" xml:space="preserve">Figure 260. <lb/>& 261.</note> </div> <p> <s xml:id="echoid-s14150" xml:space="preserve">838. </s> <s xml:id="echoid-s14151" xml:space="preserve">Si la voûte étoit ſur des pieds-droits, qui compoſaſſent <lb/> <anchor type="note" xlink:label="note-0495-02a" xlink:href="note-0495-02"/> enſemble un priſme, & </s> <s xml:id="echoid-s14152" xml:space="preserve">que ce priſme fût de ſix côtés, le <lb/>corps qui formeroit le vuide de la voûte auroit une figure <lb/>comme G H I K, formée auſſi par demi-cercles: </s> <s xml:id="echoid-s14153" xml:space="preserve">& </s> <s xml:id="echoid-s14154" xml:space="preserve">comme ce <lb/>corps ſeroit compoſé d’une quantité infinie de polygones ſem-<lb/>blables, de même que celui que nous venons de voir eſt com-<lb/>poſé de quarrés, ſi l’on conſidere le quart de cercle I K G, <lb/>l’on verra que toutes les ordonnées, comme O P & </s> <s xml:id="echoid-s14155" xml:space="preserve">Q R de ce <lb/>quart de cercle, ſervent de rayons aux polygones, dont le ſolide <lb/>eſt compoſé: </s> <s xml:id="echoid-s14156" xml:space="preserve">mais ces polygones étant tous ſemblables, & </s> <s xml:id="echoid-s14157" xml:space="preserve"><lb/>dans la raiſon des quarrés de leurs rayons (art. </s> <s xml:id="echoid-s14158" xml:space="preserve">492), l’on en <lb/>trouvera la valeur, comme on trouve celle des quarrés de leurs <lb/>rayons, c’eſt-à-dire, en multipliant la ſuperficie du plus grand <lb/>polygone par les deux tiers de la ligne qui en exprime la quan-<lb/>tité. </s> <s xml:id="echoid-s14159" xml:space="preserve">Ainſi pour trouver la valeur du ſolide G I H, il faut mul-<lb/>tiplier la baſe G H par les deux tiers de la perpendiculaire I K.</s> <s xml:id="echoid-s14160" xml:space="preserve"/> </p> <div xml:id="echoid-div1126" type="float" level="2" n="7"> <note position="right" xlink:label="note-0495-02" xlink:href="note-0495-02a" xml:space="preserve">Figure 261.</note> </div> <p> <s xml:id="echoid-s14161" xml:space="preserve">839. </s> <s xml:id="echoid-s14162" xml:space="preserve">Mais ſi au lieu de demi-cercles, c’étoit des demi-ellipſes<pb o="430" file="0496" n="510" rhead="NOUVEAU COURS"/> A B C & </s> <s xml:id="echoid-s14163" xml:space="preserve">D B E, qui partageaſſent la voûte, on trouveroit de <lb/>même la valeur du vuide, en multipliant la baſe A C par les <lb/>deux tiers de l’axe B F: </s> <s xml:id="echoid-s14164" xml:space="preserve">car ſi le plan A C eſt un quarré, tous <lb/>ceux qui compoſeront le ſolide ſeront auſſi des quarrés: </s> <s xml:id="echoid-s14165" xml:space="preserve">donc <lb/>les demi-diagonales ſeront les ordonnées K L & </s> <s xml:id="echoid-s14166" xml:space="preserve">M N du quart <lb/>d’ellipſe H G I ou F B C: </s> <s xml:id="echoid-s14167" xml:space="preserve">& </s> <s xml:id="echoid-s14168" xml:space="preserve">comme l’on trouve la valeur de <lb/>tous les quarrés des ordonnées d’un quart d’ellipſe, comme <lb/>on trouve celles des ordonnées d’un quart de cercle (art. </s> <s xml:id="echoid-s14169" xml:space="preserve">798), <lb/>c’eſt-à-dire en multipliant le quarré de la plus grande ordon-<lb/>née H I par les deux tiers de la ligne G H, il s’enſuit que l’on <lb/>trouvera toujours la ſolidité d’une voûte quelconque, ſoit que <lb/>ſes arrêtes ſe trouvent être des ellipſes, ſoit qu’elles ſoient <lb/>ſeulement des quart de cercles. </s> <s xml:id="echoid-s14170" xml:space="preserve">Cela vient de ce que l’on doit <lb/>toujours déterminer la ſolidité d’un corps, dont les élément <lb/>croiſſent dans la raiſon des quarrés des ordonnées d’une el-<lb/>lipſe ou d’un quart de cercle, en multipliant le plus grand élé-<lb/>ment qui ſert de baſe par les deux tiers de la hauteur, quelle <lb/>que ſoit d’ailleurs la figure du polygone qui ſert de baſe régu-<lb/>liere ou irréguliere.</s> <s xml:id="echoid-s14171" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14172" xml:space="preserve">840. </s> <s xml:id="echoid-s14173" xml:space="preserve">Il eſt encore une autre eſpece de voûte, que l’on nomme <lb/>voûte en bourlet, parce qu’en effet le vuide de cette voûte reſ-<lb/>ſemble aſſez à un bourlet; </s> <s xml:id="echoid-s14174" xml:space="preserve">& </s> <s xml:id="echoid-s14175" xml:space="preserve">pour en donner une idée, con-<lb/>ſidérez les figures 264 & </s> <s xml:id="echoid-s14176" xml:space="preserve">265, dont la premiere eſt le plan <lb/>d’une Tour, où l’on voit dans le milieu un pilier A B, ſur le-<lb/>quel repoſe une voûte, qui répond auſſi aux murs de la Tour; <lb/></s> <s xml:id="echoid-s14177" xml:space="preserve">de ſorte que de quelque ſens qu’on puiſſe prendre le profil de <lb/>cette Tour, il ſera toujours ſemblable à la figure 265. </s> <s xml:id="echoid-s14178" xml:space="preserve">Or <lb/>comme la voûte regne autour du pilier A B E, il faut pour la <lb/>toiſer, commencer par meſurer la maſſe H I C D, tant pleine <lb/>que vuide, qui eſt un cylindre qui a pour baſe un cercle, <lb/>dont C D eſt le diametre, & </s> <s xml:id="echoid-s14179" xml:space="preserve">H C la hauteur.</s> <s xml:id="echoid-s14180" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14181" xml:space="preserve">Préſentement pour trouver le vuide qu’il faut déduire de ce <lb/>cylindre, il faut chercher la ſuperficie du demi-cercle C M A, <lb/>& </s> <s xml:id="echoid-s14182" xml:space="preserve">la multiplier par la circonférence du cercle, qui ſera moyenne <lb/>arithmétique entre les circonférences de la Tour & </s> <s xml:id="echoid-s14183" xml:space="preserve">du pilier, <lb/>c’eſt-à-dire entre les circonférences qui auront pour rayons <lb/>A F & </s> <s xml:id="echoid-s14184" xml:space="preserve">F C; </s> <s xml:id="echoid-s14185" xml:space="preserve">& </s> <s xml:id="echoid-s14186" xml:space="preserve">retranchant ce produit-ci du précédent, on <lb/>aura la valeur de la voûte.</s> <s xml:id="echoid-s14187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14188" xml:space="preserve">Comme le bourlet eſt compoſé d’autant de demi-cercles <lb/>que l’eſpace qui eſt entre les deux circonférences C O D Q &</s> <s xml:id="echoid-s14189" xml:space="preserve"> <pb o="431" file="0497" n="511" rhead="DE MATHÉMATIQUE. Liv. XII."/> A N B P contient de lignes, comme A C & </s> <s xml:id="echoid-s14190" xml:space="preserve">N O, qui ſervent <lb/>de diametre aux demi-cercles, il s’enſuit que la ligne qui ex-<lb/>primera la ſomme de tous les élémens qui compoſent la cou-<lb/>ronne, c’eſt-à-dire la ſomme de toutes les lignes A C & </s> <s xml:id="echoid-s14191" xml:space="preserve">N O, <lb/>marquera auſſi la ſomme de tous les demi-cercles qui compo-<lb/>ſent le bourlet. </s> <s xml:id="echoid-s14192" xml:space="preserve">Or comme cette ligne n’eſt autre choſe qu’une <lb/>circonférence G H moyenne arithmétique entre les deux <lb/>C O D Q & </s> <s xml:id="echoid-s14193" xml:space="preserve">A N B P, qui renferment la couronne, il s’enſuit <lb/>qu’il faut multiplier le demi-cercle, qui auroit pour diametre <lb/>C A par la circonférence G H, pour avoir la valeur du bourlet.</s> <s xml:id="echoid-s14194" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14195" xml:space="preserve">A l’égard du revêtement de la Tour, l’on voit que pour en <lb/>trouver la ſolidité, il faut ôter de la valeur du cône tronqué, <lb/>dont R S T X ſeroit la coupe, le cylindre qui auroit pour dia-<lb/>metre du cercle de ſa baſe la ligne H I, & </s> <s xml:id="echoid-s14196" xml:space="preserve">pour hauteur la ligne <lb/>H Z, afin d’avoir la différence, qui ſera ce qu’on demande.</s> <s xml:id="echoid-s14197" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14198" xml:space="preserve">841. </s> <s xml:id="echoid-s14199" xml:space="preserve">On peut être ſouvent dans le cas de toiſer la ſuper-<lb/>ficie des voûtes dont nous venons d’examiner la ſolidité: </s> <s xml:id="echoid-s14200" xml:space="preserve">c’eſt <lb/>pourquoi il eſt à propos de ſçavoir la maniere dont il faudroit <lb/>s’y prendre ſi l’on avoit de pareilles ſurfaces courbes à me-<lb/>ſurer. </s> <s xml:id="echoid-s14201" xml:space="preserve">La méthode que je vais expliquer ici ne peut s’appli-<lb/>quer qu’aux voûtes telles que A B C, dont la baſe eſt un po-<lb/> <anchor type="note" xlink:label="note-0497-01a" xlink:href="note-0497-01"/> lygone régulier, & </s> <s xml:id="echoid-s14202" xml:space="preserve">dont la hauteur B F eſt égale au rayon G F, <lb/>mené du centre F du polygone régulier qui ſert de baſe, per-<lb/>pendiculairement au côté A E. </s> <s xml:id="echoid-s14203" xml:space="preserve">Si l’on pouvoit trouver le <lb/>moyen de toiſer par une méthode générale & </s> <s xml:id="echoid-s14204" xml:space="preserve">facile la ſurface <lb/>d’un ellipſoïde, la méthode que nous allons propoſer s’appli-<lb/>queroit avec la même facilité aux voûtes ſurbaiſſées & </s> <s xml:id="echoid-s14205" xml:space="preserve">ſur-<lb/>montées. </s> <s xml:id="echoid-s14206" xml:space="preserve">En général on dit qu’une voûte quelconque eſt en <lb/>plein cintre, lorſque la hauteur B F ou la perpendiculaire <lb/>abaiſſée du ſommet ſur le plan de la baſe eſt égale à la ligne <lb/>menée du centre F de la baſe où tombe la perpendiculaire B F, <lb/>au milieu de chaque côté du polygone régulier, comme eſt ici <lb/>la ligne F G. </s> <s xml:id="echoid-s14207" xml:space="preserve">Si cette ligne B F eſt plus grande ou plus petite <lb/>que G F, la voûte eſt appellée ſurmontée ou ſurbaiſſée. </s> <s xml:id="echoid-s14208" xml:space="preserve">Le principe <lb/>que nous allons expliquer a ceci d’avantageux, que quoiqu’on <lb/>ne puiſſe l’appliquer qu’aux voûtes en plein cintres, on trouve <lb/>encore par ſon moyen la ſurface d’une voûte fort commune, <lb/>à laquelle on a donné le nom de voûte d’arrête. </s> <s xml:id="echoid-s14209" xml:space="preserve">Lafigure 254, <lb/>planche 17, repréſente une voûte d’arrête. </s> <s xml:id="echoid-s14210" xml:space="preserve">Nous ferons voir <lb/>auſſi la maniere de toiſer la ſolidité de cette voûte, en ne fai-<lb/>ſant uſage que des principes précédens.</s> <s xml:id="echoid-s14211" xml:space="preserve"/> </p> <div xml:id="echoid-div1127" type="float" level="2" n="8"> <note position="right" xlink:label="note-0497-01" xlink:href="note-0497-01a" xml:space="preserve">Figure 262.</note> </div> <pb o="432" file="0498" n="512" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1129" type="section" level="1" n="835"> <head xml:id="echoid-head1011" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s14212" xml:space="preserve">842. </s> <s xml:id="echoid-s14213" xml:space="preserve">Suppoſant toujours la voûte en plein ceintre, en arc <lb/>de cloître, comme celle qui eſt repréſentée par la figure 262, <lb/>nous appellerons chaque portion de la ſurface courbe de la <lb/>voûte, telle que A B E, un pan de voûte: </s> <s xml:id="echoid-s14214" xml:space="preserve">ainſi dans la même <lb/>figure, la voûte propoſée eſt une voûte à quatre pans. </s> <s xml:id="echoid-s14215" xml:space="preserve">En gé-<lb/>néral, une voûte en arc de cloître & </s> <s xml:id="echoid-s14216" xml:space="preserve">en plein ceintre, aura <lb/>toujours autant de pans que le polygone régulier qui lui ſert <lb/>de baſe a de côtés.</s> <s xml:id="echoid-s14217" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1130" type="section" level="1" n="836"> <head xml:id="echoid-head1012" xml:space="preserve">PROPOSITION XIX. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s14218" xml:space="preserve">843. </s> <s xml:id="echoid-s14219" xml:space="preserve">La ſuperficie courbe A B E d’un pan de voûte quelconque <lb/> <anchor type="note" xlink:label="note-0498-01a" xlink:href="note-0498-01"/> eſt double du triangle qui lui ſert de baſe.</s> <s xml:id="echoid-s14220" xml:space="preserve"/> </p> <div xml:id="echoid-div1130" type="float" level="2" n="1"> <note position="left" xlink:label="note-0498-01" xlink:href="note-0498-01a" xml:space="preserve">Figure 262.</note> </div> <p> <s xml:id="echoid-s14221" xml:space="preserve">Soit repréſenté par 2a le côté du polygone régulier qui ſert <lb/>de baſe à notre voûte, & </s> <s xml:id="echoid-s14222" xml:space="preserve">par b la perpendiculaire G F abaiſſée <lb/>du centre F du polygone ſur ſon côté A E, laquelle (art. </s> <s xml:id="echoid-s14223" xml:space="preserve">841) <lb/>doit être égale à la hauteur B F de la voûte, puiſqu’on la ſup-<lb/>poſe en plein ceintre; </s> <s xml:id="echoid-s14224" xml:space="preserve">la ſurface du triangle A F E qui ſert de <lb/>baſe à la portion A B F E de la voûte ſera a b: </s> <s xml:id="echoid-s14225" xml:space="preserve">& </s> <s xml:id="echoid-s14226" xml:space="preserve">pour avoir le <lb/>ſolide de cette portion de voûte, il faudra, ſuivant l’art. </s> <s xml:id="echoid-s14227" xml:space="preserve">837, <lb/>multiplier le plus grand élément ou le triangle A F E par les <lb/>deux tiers de B F; </s> <s xml:id="echoid-s14228" xml:space="preserve">ce qui donnera pour la ſolidité du corps <lb/>A B F E {2ab<emph style="sub">2</emph>/3}.</s> <s xml:id="echoid-s14229" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14230" xml:space="preserve">Préſentement je fais attention que l’on pourroit conſidérer <lb/>la ſolidité de ce corps d’une autre maniere, en le concevant <lb/>comme étant compoſé d’une infinité de petits cônes, tels que <lb/>F G, F g, F h, qui ont tous leur ſommet au point F, & </s> <s xml:id="echoid-s14231" xml:space="preserve">dont <lb/>les baſes ſont répandues uniformément ſur la ſurface ou le pan <lb/>de voûte A B E. </s> <s xml:id="echoid-s14232" xml:space="preserve">Il eſt aiſé de voir que de tous ces cônes il n’y <lb/>a que ceux qui ſont diſpoſés ſur le quart de cercle qui puiſſent <lb/>être droits, & </s> <s xml:id="echoid-s14233" xml:space="preserve">que tous les autres ſont néceſſairement obliques <lb/>& </s> <s xml:id="echoid-s14234" xml:space="preserve">différemment inclinés, quoiqu’ils aient tous la même hau-<lb/>teur F G. </s> <s xml:id="echoid-s14235" xml:space="preserve">Ainſi pour avoir la ſolidité de la portion de voûte <lb/>A B F E conſidérée de cette maniere, il faudra multiplier la <lb/>ſomme des baſes de tous ces petits cônes, qui n’eſt autre choſe <lb/>que la ſurface du pan de voûte A B E, par le tiers du rayon F G:</s> <s xml:id="echoid-s14236" xml:space="preserve"> <pb o="433" file="0499" n="513" rhead="DE MATHÉMATIQUE. Liv. XII."/> donc en déſignant cette ſurface par S, on aura le ſolide du <lb/>corps A B F E = S x {a/3}. </s> <s xml:id="echoid-s14237" xml:space="preserve">D’ailleurs, nous venons de voir que <lb/>le même ſolide eſt exprimé par {2/3}a<emph style="sub">2</emph>b, en le conſidérant com-<lb/>poſé d’élémens triangulaires, tels que I L K qui croiſſent com-<lb/>me les quarrés des ordonnées L H au quart de cercle B H G: <lb/></s> <s xml:id="echoid-s14238" xml:space="preserve">on aura donc S x {a/3} = {2/3}a<emph style="sub">2</emph>b, & </s> <s xml:id="echoid-s14239" xml:space="preserve">en diviſant par {1/3}a, S = 2ab; </s> <s xml:id="echoid-s14240" xml:space="preserve"><lb/>d’où il ſuit évidemment que le pan de voûte A B E eſt double <lb/>du triangle correſpondant A F E qui lui ſert de baſe.</s> <s xml:id="echoid-s14241" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14242" xml:space="preserve">Nota. </s> <s xml:id="echoid-s14243" xml:space="preserve">Il faut remarquer que ſelon la figure où la baſe A DC E <lb/>eſt un quarré, la ſurface du triangle eſt aa, parce que la per-<lb/>pendiculaire F G ſe trouve, par la propriété du quarré, égale à <lb/>la moitié A G du côté A E. </s> <s xml:id="echoid-s14244" xml:space="preserve">Comme cela n’eſt qu’accidentel, <lb/>& </s> <s xml:id="echoid-s14245" xml:space="preserve">que notre démonſtration doit s’entendre d’un polygone quel-<lb/>conque, il étoit à propos de ne point ſuppoſer la perpendicu-<lb/>laire G F = A G, pour que la propoſition fût démontrée dans <lb/>toute ſa généralité.</s> <s xml:id="echoid-s14246" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1132" type="section" level="1" n="837"> <head xml:id="echoid-head1013" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s14247" xml:space="preserve">844. </s> <s xml:id="echoid-s14248" xml:space="preserve">Il ſuit delà que la ſurface d’une voûte en arc de cloître <lb/>en plein cintre eſt toujours double de la ſurface du polygone <lb/> <anchor type="note" xlink:label="note-0499-01a" xlink:href="note-0499-01"/> régulier qui lui ſert de baſe: </s> <s xml:id="echoid-s14249" xml:space="preserve">ainſi ſuppoſant que la ligne D K <lb/>perpendiculaire au côté G N de l’exagone, ſoit égale à la ligne <lb/>I K, menée du ſommet I de la voûte perpendiculairement à <lb/>la baſe, au centre K de cette même baſe, la ſurface de cette <lb/>voûte ſera double de celle de l’exagone M N G L O H qui lui <lb/>ſert de baſe, puiſque chaque pan N I M, N I G ſera double du <lb/>triangle correſpondant N K M, N K G.</s> <s xml:id="echoid-s14250" xml:space="preserve"/> </p> <div xml:id="echoid-div1132" type="float" level="2" n="1"> <note position="right" xlink:label="note-0499-01" xlink:href="note-0499-01a" xml:space="preserve">Figure 261.</note> </div> </div> <div xml:id="echoid-div1134" type="section" level="1" n="838"> <head xml:id="echoid-head1014" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s14251" xml:space="preserve">845. </s> <s xml:id="echoid-s14252" xml:space="preserve">Il ſuit de cette propoſition, que la ſurface d’une demi-<lb/>ſphere eſt double du cercle qui lui ſert de baſe; </s> <s xml:id="echoid-s14253" xml:space="preserve">enſorte que <lb/>la propoſition que nous avons démontré ſur la ſuperficie de <lb/>la ſphere devient un corollaire trés-ſimple de celle-ci; </s> <s xml:id="echoid-s14254" xml:space="preserve">car <lb/>puiſque notre démonſtration eſt applicable à tous ies poiy-<lb/>gones réguliers, elle eſt auſſi applicable au cercle. </s> <s xml:id="echoid-s14255" xml:space="preserve">En effet, <lb/>on peut concevoir la ſurface de la ſphere comme compoſée <lb/>d’une infinité de petits triangles curvilignes qui ont leur ſom-<lb/>met au pôle de cette demi-ſphere, & </s> <s xml:id="echoid-s14256" xml:space="preserve">qui vont ſe terminer à <lb/>la circonférence, leſquels ſont tous, par la propoſition pré- <pb o="434" file="0500" n="514" rhead="NOUVEAU COURS"/> ſente, doubles des petits triangles correſpondans dans le cer-<lb/>cle qui lui ſert de baſe.</s> <s xml:id="echoid-s14257" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1135" type="section" level="1" n="839"> <head xml:id="echoid-head1015" xml:space="preserve"><emph style="sc">Scholie</emph>.</head> <p> <s xml:id="echoid-s14258" xml:space="preserve">846. </s> <s xml:id="echoid-s14259" xml:space="preserve">On peut faire uſage de la propoſition précédente pour <lb/>trouver la ſuperficie des voûtes d’arrêtes, telle que celle qui <lb/>eſt repréſentée par la figure 254 (planche 17). </s> <s xml:id="echoid-s14260" xml:space="preserve">Mais avant que <lb/>de chercher la ſuperficie de ces ſortes de voûtes, il eſt à propos <lb/>de rechercher de quelle maniere elles peuvent être formées; <lb/></s> <s xml:id="echoid-s14261" xml:space="preserve">c’eſt ce que nous allons examiner dans les articles ſuivans, <lb/>après quoi il nous ſera facile de déterminer leur ſurface, & </s> <s xml:id="echoid-s14262" xml:space="preserve"><lb/>leur ſolidité par la même occaſion.</s> <s xml:id="echoid-s14263" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14264" xml:space="preserve">847. </s> <s xml:id="echoid-s14265" xml:space="preserve">A E D C F B eſt un demi-cylindre droit, dont la baſe <lb/> <anchor type="note" xlink:label="note-0500-01a" xlink:href="note-0500-01"/> eſt un paralléogramme rectangle A D C B. </s> <s xml:id="echoid-s14266" xml:space="preserve">Le côté A D eſt <lb/> <anchor type="note" xlink:label="note-0500-02a" xlink:href="note-0500-02"/> diviſé en deux également en K, & </s> <s xml:id="echoid-s14267" xml:space="preserve">de ce point on a tiré aux <lb/>angles B, C les lignes droites K B, K C. </s> <s xml:id="echoid-s14268" xml:space="preserve">Par ces lignes & </s> <s xml:id="echoid-s14269" xml:space="preserve">la <lb/>ligne E K perpendiculaire au plan de la baſe renſermée dans <lb/>le plan du demi-cercle A E D, il faut concevoir deux plans <lb/>coupans K E I B, K E H C qui ſeront néceſſairement perpendi-<lb/>culaires au plan de la baſe. </s> <s xml:id="echoid-s14270" xml:space="preserve">Il eſt viſible que ces plans retran-<lb/>chent du demi-cylindre ou berceau deux corps égaux A K E B, <lb/>D K E C qui ſont dans le cas de ceux que nous venons d’exa-<lb/>miner dans tout ce qui précéde, dont on pourra trouver la <lb/>ſolidité, en multipliant chaque triangle qui lui ſert de baſe <lb/>par les deux tiers du rayon A K, & </s> <s xml:id="echoid-s14271" xml:space="preserve">dont on aura la ſurface en <lb/>doublant les mêmes triangles égaux A K B, D K C. </s> <s xml:id="echoid-s14272" xml:space="preserve">La corps <lb/>E K B F C terminé en coin du côté de la ligne E K, eſt évi-<lb/>demment égal à ce qui reſte du cylindre, après en avoir ôté les <lb/>deux corps A K E B, E K D C: </s> <s xml:id="echoid-s14273" xml:space="preserve">donc puiſque l’on peut toiſer <lb/>ces deux corps, ainſi que le demi-cylindre, on aura auſſi la <lb/>ſolidité du corps E K B F C. </s> <s xml:id="echoid-s14274" xml:space="preserve">De même la ſurface courbe de <lb/>ce même corps eſt égale à celle du demi-cylindre, après en <lb/>avoir ôté celles des corps A K E B, D K E C: </s> <s xml:id="echoid-s14275" xml:space="preserve">donc puiſque la <lb/>ſuperficie courbe de ces deux corps peut être déterminée, on <lb/>peut auſſi trouver celle du corps E K B F C.</s> <s xml:id="echoid-s14276" xml:space="preserve"/> </p> <div xml:id="echoid-div1135" type="float" level="2" n="1"> <note position="left" xlink:label="note-0500-01" xlink:href="note-0500-01a" xml:space="preserve">Pl. XVII.</note> <note position="left" xlink:label="note-0500-02" xlink:href="note-0500-02a" xml:space="preserve">Figure 255.</note> </div> <p> <s xml:id="echoid-s14277" xml:space="preserve">848. </s> <s xml:id="echoid-s14278" xml:space="preserve">Cela poſé, une voûte d’arrête telle que celle qui eſt <lb/>repréſentée par la figure 254, n’eſt autre choſe que différens <lb/>corps R G E L D, R G F I E, tous égaux entr’eux, & </s> <s xml:id="echoid-s14279" xml:space="preserve">formés de <lb/>la même maniere que le corps E K B F C de la figure 255, <lb/>leſquels ſe touchent tous dans les ſurfaces planes qui forment <pb o="435" file="0501" n="515" rhead="DE MATHÉMATIQUE. Liv. XII."/> leurs côtés qui ſeront toujours des quart d’ellipſe, & </s> <s xml:id="echoid-s14280" xml:space="preserve">ſont <lb/>tous terminés à une même ligne perpendiculaire au plan de la <lb/>voûte. </s> <s xml:id="echoid-s14281" xml:space="preserve">Il eſt viſible que tous ces corps doivent être parfaite-<lb/>ment égaux, que leurs cercles F I E, E L D doivent auſſi être <lb/>égaux, ainſi que les triangles qui leur ſervent de baſe. </s> <s xml:id="echoid-s14282" xml:space="preserve">On voit <lb/>par-là que la ſuperficie & </s> <s xml:id="echoid-s14283" xml:space="preserve">la ſolidité ſe réduit à trouver la ſur-<lb/>face & </s> <s xml:id="echoid-s14284" xml:space="preserve">la ſolidité du corps E K B F C de la figure 255.</s> <s xml:id="echoid-s14285" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14286" xml:space="preserve">849. </s> <s xml:id="echoid-s14287" xml:space="preserve">Soit le rayon A K ou E K = a; </s> <s xml:id="echoid-s14288" xml:space="preserve">la ligne A B qui me-<lb/>ſure la longueur du cylindre ſoit égale à b: </s> <s xml:id="echoid-s14289" xml:space="preserve">pour trouver la <lb/>ſurface de ce corps, je chercherai d’abord celle du cylindre. <lb/></s> <s xml:id="echoid-s14290" xml:space="preserve">Je commence par déterminer la demi-circonférence B F C par <lb/>la proportion ſuivante, 7 : </s> <s xml:id="echoid-s14291" xml:space="preserve">22 :</s> <s xml:id="echoid-s14292" xml:space="preserve">: a : </s> <s xml:id="echoid-s14293" xml:space="preserve">{22/7}a; </s> <s xml:id="echoid-s14294" xml:space="preserve">puiſque le rapport du <lb/>rayon à la demi-circonférence eſt le même que celui du dia-<lb/>metre à la circonférence. </s> <s xml:id="echoid-s14295" xml:space="preserve">Multipliant cette demi-circonfé-<lb/>rence par b, j’aurai {22/7}ab pour la ſurface du demi-cylindre: </s> <s xml:id="echoid-s14296" xml:space="preserve"><lb/>ôtant de cette ſuperficie celles des corps A K E B, D K E C, <lb/>leſquelles ſont égales enſemble au rectangle A B, on aura pour <lb/>la ſuperficie du corps E K B F C, {22/7}ab - 2ab = {22/7}ab - {14/7}ab <lb/>= {18/7}ab; </s> <s xml:id="echoid-s14297" xml:space="preserve">d’où il ſuit que cette ſurface eſt égale à ab + {1/7}ab, <lb/>c’eſt-à-dire égale à la baſe, plus {1/7} de la même baſe A B C D : </s> <s xml:id="echoid-s14298" xml:space="preserve"><lb/>donc pour avoir la ſurface d’une voûte d’arrête en plein cintre, <lb/>comme celle de la figure 254, & </s> <s xml:id="echoid-s14299" xml:space="preserve">dont la baſe eſt un polygone <lb/>régulier, il faut à cette même baſe ajouter un ſeptieme.</s> <s xml:id="echoid-s14300" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14301" xml:space="preserve">850. </s> <s xml:id="echoid-s14302" xml:space="preserve">Pour la ſolidité du même corps, je cherche la ſurface <lb/>du demi-cercle B F C, en multipliant la demi-circonférence <lb/>{22/7} a par la moitié du rayon; </s> <s xml:id="echoid-s14303" xml:space="preserve">ce qui me donne {11/7}a<emph style="sub">2</emph>: </s> <s xml:id="echoid-s14304" xml:space="preserve">ſi je mul-<lb/>tiplie ce produit par b, j’aurai la ſolidité du demi-cylindre <lb/>qui ſera {11/7}a<emph style="sub">2</emph>b. </s> <s xml:id="echoid-s14305" xml:space="preserve">Préſentement je cherche la ſolidité des deux <lb/>corps égaux A K E B, E K D C, qui eſt {2/3}a<emph style="sub">2</emph>b: </s> <s xml:id="echoid-s14306" xml:space="preserve">donc la ſolidité <lb/>du corps E K B F C ſera {11/7}a<emph style="sub">2</emph>b - {2/3}a<emph style="sub">2</emph>b, ou en réduiſant au <lb/>même dénominateur √{33/21} - {14/21}\x{0020} x a<emph style="sub">2</emph>b = {19/21}a<emph style="sub">2</emph>b; </s> <s xml:id="echoid-s14307" xml:space="preserve">d’où il ſuit que <lb/>ce ſolide eſt au demi-cylindre A E D C F B :</s> <s xml:id="echoid-s14308" xml:space="preserve">: 19 : </s> <s xml:id="echoid-s14309" xml:space="preserve">33 : </s> <s xml:id="echoid-s14310" xml:space="preserve">donc <lb/>ce même corps ſera les {19/33} du même demi-cylindre. </s> <s xml:id="echoid-s14311" xml:space="preserve">Pour ap-<lb/>pliquer ce que nous venons de dire au toiſé du ſolide d’une <lb/>voûte d’arrête, dont la baſe eſt un polygone régulier, il fau-<lb/>dra chercher la ſolidité du demi-cylindre, qui auroit pour baſe <lb/>un rectangle formé ſur le côté E D du polygone, & </s> <s xml:id="echoid-s14312" xml:space="preserve">la perpen-<lb/>diculaire G S abaiſſée du centre du polygone ſur le côté D E, <lb/>& </s> <s xml:id="echoid-s14313" xml:space="preserve">enſuite prendre les {19/33} de ce ſolide autant de fois que le poly-<lb/>gone de la baſe aura de côtés.</s> <s xml:id="echoid-s14314" xml:space="preserve"/> </p> <pb o="436" file="0502" n="516" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s14315" xml:space="preserve">851. </s> <s xml:id="echoid-s14316" xml:space="preserve">Il faut bien remarquer que quoiqu’on ne puiſſe pas <lb/>trouver par notre méthode la ſuperficie d’une voûte d’arrête <lb/>ſurbaiſſée ou ſurmontée, cependant on détermineroit avec <lb/>la derniere facilité le ſolide de ces ſortes de voûtes dans ces <lb/>deux cas. </s> <s xml:id="echoid-s14317" xml:space="preserve">Je laiſſe aux Commençans le plaiſir d’en trouver eux-<lb/>mêmes la démonſtration.</s> <s xml:id="echoid-s14318" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14319" xml:space="preserve">Comme ces ſortes de voûtes ſont ordinairement remplies <lb/>de maçonnerie du côté des toits des Egliſes ou autres endroits <lb/>où elles ſe trouvent; </s> <s xml:id="echoid-s14320" xml:space="preserve">on toiſera la ſolidité du priſme droit qui <lb/>auroit même baſe & </s> <s xml:id="echoid-s14321" xml:space="preserve">même hauteur, & </s> <s xml:id="echoid-s14322" xml:space="preserve">du tout on déduira la <lb/>ſolidité des voûtes, ſelon la méthode que nous venons d’ex-<lb/>pliquer.</s> <s xml:id="echoid-s14323" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14324" xml:space="preserve">Il eſt aiſé de voir qu’il ne nous a pas été poſſible de parler <lb/>de la ſuperficie de ces ſortes de voûtes dans l’article de la <lb/>meſure des ſurfaces, parce que la connoiſſance de ces mêmes <lb/>ſurfaces ne peut ſe déduire que de la ſolidité de ces voûtes, <lb/>au moins dans la méthode que j’ai ſuivie ici.</s> <s xml:id="echoid-s14325" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1137" type="section" level="1" n="840"> <head xml:id="echoid-head1016" style="it" xml:space="preserve">Application de la Géométrie à la maniere de toiſer le revêtement <lb/>d’une Fortification.</head> <p> <s xml:id="echoid-s14326" xml:space="preserve">852. </s> <s xml:id="echoid-s14327" xml:space="preserve">Quand on trace une fortification, il y a une ligne qui <lb/>regne tout autour des ouvrages, que l’on nomme magiſtrale, <lb/>qui ſert à donner les longueurs que doivent avoir les parties <lb/>de la fortification; </s> <s xml:id="echoid-s14328" xml:space="preserve">& </s> <s xml:id="echoid-s14329" xml:space="preserve">cette ligne eſt celle qui eſt repréſentée <lb/>par le cordon du revêtement d’un ouvrage: </s> <s xml:id="echoid-s14330" xml:space="preserve">par exemple, ſi l’on <lb/>dit qu’une face de baſtion a 50 toiſes, cela doit s’entendre de-<lb/>puis une extrêmité du cordon de cette face juſqu’à l’autre; </s> <s xml:id="echoid-s14331" xml:space="preserve">ou, <lb/>ce qui eſt la même choſe, depuis une extrêmité juſqu’à l’autre <lb/>de l’entablement de la muraille de la face.</s> <s xml:id="echoid-s14332" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14333" xml:space="preserve">Préſentement pour meſurer le revêtement du baſtion re-<lb/>préſenté dans la figure 266, conſidérez-en le profil, dont les <lb/> <anchor type="note" xlink:label="note-0502-01a" xlink:href="note-0502-01"/> dimenſions ont été priſes ſelon le profil général de M. </s> <s xml:id="echoid-s14334" xml:space="preserve">de <lb/>Vauban, pour le revêtement ordinaire d’un rempart, qui <lb/>auroit 30 pieds, depuis la retraite A G des fondemens juſqu’à <lb/>la hauteur C H du cordon: </s> <s xml:id="echoid-s14335" xml:space="preserve">& </s> <s xml:id="echoid-s14336" xml:space="preserve">comme la partie D E F G n’a <lb/>point de talud, nous n’en parlerons point ici, parce qu’elle eſt <lb/>facile à meſurer; </s> <s xml:id="echoid-s14337" xml:space="preserve">nous conſidérerons ſeulement la muraille, <lb/>depuis la retraite juſqu’au cordon; </s> <s xml:id="echoid-s14338" xml:space="preserve">& </s> <s xml:id="echoid-s14339" xml:space="preserve">faiſant auſſi abſtraction <lb/>des contre-forts, il faut, à cauſe des pyramides tronquées qui <pb o="437" file="0503" n="517" rhead="DE MATHÉMATIQUE. Liv. XII."/> ſe rencontrent aux angles des points A & </s> <s xml:id="echoid-s14340" xml:space="preserve">D, abaiſſer les per-<lb/>pendiculaires A B & </s> <s xml:id="echoid-s14341" xml:space="preserve">D E, & </s> <s xml:id="echoid-s14342" xml:space="preserve">meſurer la ſuperficie du trapeze <lb/>A B C G du profil par la longueur A D de la face, priſe le long <lb/>des contre-forts, & </s> <s xml:id="echoid-s14343" xml:space="preserve">le produit ſera regardé comme le revête-<lb/>ment de la face: </s> <s xml:id="echoid-s14344" xml:space="preserve">venant enſuite dans l’angle flanquant I, l’on <lb/>tirera une perpendiculaire G H, de ſorte qu’elle correſponde <lb/>dans l’angle K du pied de la muraille; </s> <s xml:id="echoid-s14345" xml:space="preserve">& </s> <s xml:id="echoid-s14346" xml:space="preserve">ayant auſſi abaiſſé <lb/>la perpendiculaire C A, l’on multipliera le profil précédent <lb/>par la longueur H A ou G C du flanc, & </s> <s xml:id="echoid-s14347" xml:space="preserve">l’on fera de même <lb/>pour toiſer la courtine & </s> <s xml:id="echoid-s14348" xml:space="preserve">les autres parties où l’on aura re-<lb/>tranché les pyramides des angles.</s> <s xml:id="echoid-s14349" xml:space="preserve"/> </p> <div xml:id="echoid-div1137" type="float" level="2" n="1"> <note position="left" xlink:label="note-0502-01" xlink:href="note-0502-01a" xml:space="preserve">Figure 266. <lb/>& 267.</note> </div> <p> <s xml:id="echoid-s14350" xml:space="preserve">Pour connoître la valeur de ces pyramides tronquées, je <lb/> <anchor type="note" xlink:label="note-0503-01a" xlink:href="note-0503-01"/> conſidere que celle qui eſt à l’angle de l’épaule & </s> <s xml:id="echoid-s14351" xml:space="preserve">à l’angle ſail-<lb/>lant, reſſemblent aſſez à la figure 270. </s> <s xml:id="echoid-s14352" xml:space="preserve">Ainſi connoiſſant les <lb/>deux plans V T & </s> <s xml:id="echoid-s14353" xml:space="preserve">Q R, je meſure cette pyramide tronquée <lb/>comme à l’ordinaire, & </s> <s xml:id="echoid-s14354" xml:space="preserve">ſuppoſant qu’elle ſoit celle de l’angle <lb/>flanqué, je me garde bien de la prendre auſſi pour celle de <lb/>l’angle de l’épaule, parce qu’elles ſont différentes en ſolidité <lb/>c’eſt pourquoi je meſure cette derniere, comme je viens de <lb/>faire la précédente.</s> <s xml:id="echoid-s14355" xml:space="preserve"/> </p> <div xml:id="echoid-div1138" type="float" level="2" n="2"> <note position="right" xlink:label="note-0503-01" xlink:href="note-0503-01a" xml:space="preserve">Figure 220.</note> </div> <p> <s xml:id="echoid-s14356" xml:space="preserve">Quant à ce qui nous reſte à meſurer dans l’angle flanquant I, <lb/> <anchor type="note" xlink:label="note-0503-02a" xlink:href="note-0503-02"/> je conſidere la figure 269, comme étant cette partie-là déta-<lb/>chée, qui reſſembleroit à un priſme, ſi le vuide B C E H G <lb/>étoit rempli: </s> <s xml:id="echoid-s14357" xml:space="preserve">ſuppoſant donc qu’il le ſoit, je cherche la valeur <lb/>du priſme A F G, de laquelle je ſouſtrais celle de la pyramide <lb/>K M I, que je ſuppoſe être égale au vuide B E G, & </s> <s xml:id="echoid-s14358" xml:space="preserve">la diffé-<lb/>rence donne la partie que je cherche.</s> <s xml:id="echoid-s14359" xml:space="preserve"/> </p> <div xml:id="echoid-div1139" type="float" level="2" n="3"> <note position="right" xlink:label="note-0503-02" xlink:href="note-0503-02a" xml:space="preserve">Figure 268. <lb/>& 269.</note> </div> <p> <s xml:id="echoid-s14360" xml:space="preserve">853. </s> <s xml:id="echoid-s14361" xml:space="preserve">Ce ſeroit peu de choſe que de toiſer le revêtement <lb/> <anchor type="note" xlink:label="note-0503-03a" xlink:href="note-0503-03"/> d’une fortification, s’il étoit toujours compoſé de lignes droites, <lb/>comme dans cette figure; </s> <s xml:id="echoid-s14362" xml:space="preserve">mais il y a bien d’autres difficultés, <lb/>quand il faut toiſer le revêtement des parties des baſtions à <lb/>orillons, comme celle du baſtion repréſenté dans la figure 271. <lb/></s> <s xml:id="echoid-s14363" xml:space="preserve">Cependant comme les articles 854, 855 ont été rapportés ex-<lb/>près pour en faciliter l’intelligence, nous allons faire enſorte <lb/>d’en rendre les opérations aiſées.</s> <s xml:id="echoid-s14364" xml:space="preserve"/> </p> <div xml:id="echoid-div1140" type="float" level="2" n="4"> <note position="right" xlink:label="note-0503-03" xlink:href="note-0503-03a" xml:space="preserve">Figure 271.</note> </div> <p> <s xml:id="echoid-s14365" xml:space="preserve">La figure 275 repréſente le flanc d’un baſtion à orillon, <lb/> <anchor type="note" xlink:label="note-0503-04a" xlink:href="note-0503-04"/> dont la largeur A B marque l’épaiſſeur du revêtement au cor-<lb/> <anchor type="note" xlink:label="note-0503-05a" xlink:href="note-0503-05"/> don, qui eſt toujours de 5 pieds, & </s> <s xml:id="echoid-s14366" xml:space="preserve">la largeur B C marque le <lb/>talud I du revêtement, qui eſt ici de 6 pieds; </s> <s xml:id="echoid-s14367" xml:space="preserve">de ſorte que toute <lb/>la largeur A C marque l’épaiſſeur du revêtement ſur la retraite, <pb o="438" file="0504" n="518" rhead="NOUVEAU COURS"/> qui ſera de 11 pieds, & </s> <s xml:id="echoid-s14368" xml:space="preserve">la ligne F K I G D E la magiſtrale. <lb/></s> <s xml:id="echoid-s14369" xml:space="preserve">Pour ſçavoir comment il faut s’y prendre pour toiſer l’orillon <lb/>G S D, nous allons voir premiérement de quelle façon il a été <lb/>tracé, afin de connoître l’angle G H D, & </s> <s xml:id="echoid-s14370" xml:space="preserve">le rayon H D, <lb/>dont nous aurons beſoin.</s> <s xml:id="echoid-s14371" xml:space="preserve"/> </p> <div xml:id="echoid-div1141" type="float" level="2" n="5"> <note position="right" xlink:label="note-0503-04" xlink:href="note-0503-04a" xml:space="preserve">Pl. XX.</note> <note position="right" xlink:label="note-0503-05" xlink:href="note-0503-05a" xml:space="preserve">Figure 275.</note> </div> <p> <s xml:id="echoid-s14372" xml:space="preserve">L’on fçait que pour tracer l’orillon, ſelon la méthode de <lb/>M. </s> <s xml:id="echoid-s14373" xml:space="preserve">de Vauban, l’on diviſe le flanc F D en trois parties égales, <lb/>& </s> <s xml:id="echoid-s14374" xml:space="preserve">que la troiſieme partie G D devient la corde d’une portion <lb/>de cercle qui forme l’orillon, & </s> <s xml:id="echoid-s14375" xml:space="preserve">que pour décrire cette por-<lb/>tion de cercle, l’on éleve ſur le milieu de la partie G D une <lb/>perpendiculaire I H, & </s> <s xml:id="echoid-s14376" xml:space="preserve">une autre D H ſur l’extrêmité D E de <lb/>la face du baſtion, & </s> <s xml:id="echoid-s14377" xml:space="preserve">que ces deux perpendiculaires venant ſe <lb/>rencontrer au point H, donnent le centre de l’orillon, ou <lb/>autrement de l’arc G V D, dont le rayon eſt la perpendicu-<lb/>laire D H.</s> <s xml:id="echoid-s14378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14379" xml:space="preserve">Cela poſé, ſi avec les rayons H B, H G, H Q l’on décrit <lb/> <anchor type="note" xlink:label="note-0504-01a" xlink:href="note-0504-01"/> trois cercles, & </s> <s xml:id="echoid-s14380" xml:space="preserve">que l’on conſidere la figure 273, l’on verra <lb/>que ces trois cercles compoſent un cône tronqué, dans le mi-<lb/>lieu duquel eſt un cylindre, & </s> <s xml:id="echoid-s14381" xml:space="preserve">le plan B Y étant le profil de <lb/> <anchor type="note" xlink:label="note-0504-02a" xlink:href="note-0504-02"/> l’orillon, la ligne G Q dans l’une & </s> <s xml:id="echoid-s14382" xml:space="preserve">l’autre figure marquera le <lb/>talud du revêtement; </s> <s xml:id="echoid-s14383" xml:space="preserve">la ligne G B, ſon épaiſſeur à l’endroit <lb/>du cordon, & </s> <s xml:id="echoid-s14384" xml:space="preserve">la ligne H G, le demi-diametre de l’orillon, qui <lb/>eſt la même choſe que H D. </s> <s xml:id="echoid-s14385" xml:space="preserve">Or comme le revêtement de l’o-<lb/>rillon eſt un ſecteur de cône tronqué, après en avoir ôté le <lb/>cylindre, qui eſt dans le milieu, & </s> <s xml:id="echoid-s14386" xml:space="preserve">que la grandeur de ce ſec-<lb/>teur eſt déterminée par l’angle G H D, voici comment on <lb/>pourra connoître la valeur des lignes dont nous avons beſoin <lb/>pour meſurer ce ſecteur.</s> <s xml:id="echoid-s14387" xml:space="preserve"/> </p> <div xml:id="echoid-div1142" type="float" level="2" n="6"> <note position="left" xlink:label="note-0504-01" xlink:href="note-0504-01a" xml:space="preserve">Figure 273. <lb/>& 274.</note> <note position="left" xlink:label="note-0504-02" xlink:href="note-0504-02a" xml:space="preserve">On n’a re-<lb/>préſenté que <lb/>la moitié du <lb/>cônetronqué, <lb/>afin de ména-<lb/>ger l’eſpace de <lb/>la planche.</note> </div> <p> <s xml:id="echoid-s14388" xml:space="preserve">On a vu (art. </s> <s xml:id="echoid-s14389" xml:space="preserve">741) que l’angle de l’épaule F D E étoit de <lb/>117 degrés 39 minutes: </s> <s xml:id="echoid-s14390" xml:space="preserve">par conſéquent ſi l’on en ſouſtrait <lb/>l’angle droit H D B, il reſtera 27 degrés 39 minutes pour l’an-<lb/>gle I H D du triangle rectangle H L D. </s> <s xml:id="echoid-s14391" xml:space="preserve">Ainſi l’angle L D H <lb/>ſera de 62 degrés 21 minutes: </s> <s xml:id="echoid-s14392" xml:space="preserve">& </s> <s xml:id="echoid-s14393" xml:space="preserve">comme on a trouvé auſſi <lb/>(art. </s> <s xml:id="echoid-s14394" xml:space="preserve">541) que le flanc F D étoit de 27 toiſes 2 pieds, la ligne <lb/>L D en étant la ſixieme partie, ſera de 4 toiſes 3 pieds 4 pouces. <lb/></s> <s xml:id="echoid-s14395" xml:space="preserve">Or comme du triangle L H D l’on connoît les trois angles & </s> <s xml:id="echoid-s14396" xml:space="preserve"><lb/>le côte L D, il ſera facile de connoître le côté D H, que l’on <lb/>trouvera de 5 toiſes 9 pouces. </s> <s xml:id="echoid-s14397" xml:space="preserve">Cela étant, on connoîtra toutes <lb/>les lignes de la figure; </s> <s xml:id="echoid-s14398" xml:space="preserve">car le demi-diametre H G étant de <lb/>5 toiſes 9 pouces, & </s> <s xml:id="echoid-s14399" xml:space="preserve">la ligne G B de 5 pieds, le rayon H B du <pb o="439" file="0505" n="519" rhead="DE MATHÉMATIQUE. Liv. XII."/> cylindre ſera de 5 toiſes 1 pied 9 pouces, & </s> <s xml:id="echoid-s14400" xml:space="preserve">le talud G Q étant <lb/>de 6 pieds, le demi-diametre H Q de la baſe du cône tronqué <lb/>ſera de 6 toiſes 9 pouces, & </s> <s xml:id="echoid-s14401" xml:space="preserve">l’axe H Z exprimant la hauteur du <lb/>revêtement, ſera de 5 toiſes: </s> <s xml:id="echoid-s14402" xml:space="preserve">ainſi l’on connoît tout ce qu’il <lb/>faut pour meſurer le cône tronqué & </s> <s xml:id="echoid-s14403" xml:space="preserve">le cylindre qui eſt dans <lb/>le milieu.</s> <s xml:id="echoid-s14404" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14405" xml:space="preserve">Ayant donc meſuré le cône tronqué & </s> <s xml:id="echoid-s14406" xml:space="preserve">le cylindre, on re-<lb/>tranchera la valeur du cylindre de celle du cône tronqué, pour <lb/>avoir le fragment qui en fait la différence: </s> <s xml:id="echoid-s14407" xml:space="preserve">& </s> <s xml:id="echoid-s14408" xml:space="preserve">comme le re-<lb/>vêtement de l’orillon eſt un ſecteur de ce fragment, l’on en <lb/>cherchera la valeur, en fuivant ce qu’on a vu dans l’art. </s> <s xml:id="echoid-s14409" xml:space="preserve">820, <lb/>c’eſt-à-dire, que connoiſſant l’angle G H D, qui eſt de 124 <lb/>degrés 42 minutes, l’on dira: </s> <s xml:id="echoid-s14410" xml:space="preserve">Si 360 degrés m’ont donné tant <lb/>pour la valeur du cône tronqué, après en avoir ôté le cylindre, <lb/>que me donneront 124 degrés 42 minutes pour le ſecteur, ou <lb/>autrement pour la valeur du revêtement de l’orillon, qui ſe <lb/>trouvera, en faiſant le calcul des parties que l’on vient d’in-<lb/>diquer.</s> <s xml:id="echoid-s14411" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14412" xml:space="preserve">854. </s> <s xml:id="echoid-s14413" xml:space="preserve">Avant que de chercher à toiſer le flanc concave KI, il <lb/> <anchor type="note" xlink:label="note-0505-01a" xlink:href="note-0505-01"/> faut être prévenu que pour le tracer on a prolongé la ligne de <lb/>défenſe S F de la longueur F K de 5 toiſes pour faire la briſure, <lb/>& </s> <s xml:id="echoid-s14414" xml:space="preserve">que par l’angle flanqué S, & </s> <s xml:id="echoid-s14415" xml:space="preserve">le point G l’on a tiré la ligne <lb/>S I, pour avoir la partie G I auſſi de 5 toiſes; </s> <s xml:id="echoid-s14416" xml:space="preserve">& </s> <s xml:id="echoid-s14417" xml:space="preserve">enſuite on a <lb/>tiré la ligne K I, ſur laquelle on fait un triangle équilatéral <lb/>K P I, pour avoir le point P, qui a ſervi de centre pour décrire <lb/>avec le rayon P K l’arc K I, avec le rayon P N l’arc N O, & </s> <s xml:id="echoid-s14418" xml:space="preserve"><lb/>avec le rayon P L l’arc R M.</s> <s xml:id="echoid-s14419" xml:space="preserve"/> </p> <div xml:id="echoid-div1143" type="float" level="2" n="7"> <note position="right" xlink:label="note-0505-01" xlink:href="note-0505-01a" xml:space="preserve">Figure 272. <lb/>& 275.</note> </div> <p> <s xml:id="echoid-s14420" xml:space="preserve">Préſentement la premiere difficulté eſt d’avoir la valeur du <lb/>rayon P K, que l’on trouvera pourtant en conſidérant qu’on <lb/>connoît l’angle S F G de So degrés 47 minutes par l’art. </s> <s xml:id="echoid-s14421" xml:space="preserve">741 <lb/>qui nous a donné auſſi la ligne E F de 82 toiſes, à laquelle <lb/>ajoutant la ligne S E, c’eſt-à-dire la face du baſtion, qui eſt <lb/>de 50 toiſes, on aura toute la ligne S E F de 132 toiſes: </s> <s xml:id="echoid-s14422" xml:space="preserve">& </s> <s xml:id="echoid-s14423" xml:space="preserve"><lb/>comme la ligne F G eſt les deux tiers du flanc E D, que nous <lb/>avons trouvé de 27 toiſes 2 pieds, elle ſera donc de 18 toiſes <lb/>1 pied 4 pouces. </s> <s xml:id="echoid-s14424" xml:space="preserve">Or comme du triangle S F G on connoît les <lb/>côtés F S & </s> <s xml:id="echoid-s14425" xml:space="preserve">F G avec l’angle compris, on trouvera par leur <lb/>moyen que l’angle F S G eſt de 8 degrés, & </s> <s xml:id="echoid-s14426" xml:space="preserve">que le côté eſt de <lb/>126 toiſes 5 pieds; </s> <s xml:id="echoid-s14427" xml:space="preserve">& </s> <s xml:id="echoid-s14428" xml:space="preserve">ſi au côté S F on ajoute la ligne F K de <lb/>5 toiſes, & </s> <s xml:id="echoid-s14429" xml:space="preserve">au côté S G la ligne G I auſſi de 5 toiſes, l’on aura <lb/> <lb/> <lb/> <lb/> <lb/> <pb o="440" file="0506" n="520" rhead="NOUVEAU COURS"/> un autre triangle K S I, dont on connoîtra le côté S K de 137 <lb/>toiſes, le côté S I de 131 toiſes 5 pieds, & </s> <s xml:id="echoid-s14430" xml:space="preserve">l’angle K S I de <lb/>8 degrés, avec leſquels on trouvera la ligne K I de 18 toiſes <lb/>4 pieds quelque choſe; </s> <s xml:id="echoid-s14431" xml:space="preserve">& </s> <s xml:id="echoid-s14432" xml:space="preserve">comme cette ligne eſt égale au rayon <lb/>P K, il ſera donc auſſi de 18 toiſes 4 pieds.</s> <s xml:id="echoid-s14433" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14434" xml:space="preserve">Si l’on conſidere bien le revêtement du flanc concave K I, <lb/> <anchor type="note" xlink:label="note-0506-01a" xlink:href="note-0506-01"/> on verra qu’il n’eſt autre choſe qu’un ſecteur du cylindre, dans <lb/>le milieu duquel il y auroit un vuide en forme de cône tronqué, <lb/>comme dans l’art. </s> <s xml:id="echoid-s14435" xml:space="preserve">820; </s> <s xml:id="echoid-s14436" xml:space="preserve">& </s> <s xml:id="echoid-s14437" xml:space="preserve">pour le mieux comprendre, ima-<lb/>ginons que X V eſt la moitié d’un cylindre, dont le rayon P N <lb/>du cercle eſt le même que celui de l’arc N O du flanc, & </s> <s xml:id="echoid-s14438" xml:space="preserve">que <lb/>le rayon P K étant de 18 toiſes 4 pieds, ſi on y ajoute la ligne <lb/>K N, qui marque l’épaiſſeur de la muraille au cordon, & </s> <s xml:id="echoid-s14439" xml:space="preserve">qui <lb/>eſt par conſéquent de 5 pieds, on aura la ligne P N de 17 toiſes <lb/>3 pieds: </s> <s xml:id="echoid-s14440" xml:space="preserve">ſi donc de la ligne P K on retranche la ligne K L, <lb/>qui marque le talud de la muraille, qui eſt de 6 pieds, l’on aura <lb/>la ligne P L de 17 toiſes 4 pieds; </s> <s xml:id="echoid-s14441" xml:space="preserve">& </s> <s xml:id="echoid-s14442" xml:space="preserve">ſi la ligne N V eſt égale à <lb/>la hauteur du revêtement, c’eſt-à-dire de 5 toiſes, le trapeze <lb/>K L V N ſera le profil du revêtement: </s> <s xml:id="echoid-s14443" xml:space="preserve">ainſi comme l’on con-<lb/>noît le rayon P N du cylindre, le demi-diametre P K du plus <lb/>grand cercle du cône tronqué, & </s> <s xml:id="echoid-s14444" xml:space="preserve">le demi-diametre P L du <lb/>plus petit cercle du même cône, & </s> <s xml:id="echoid-s14445" xml:space="preserve">de plus l’axe P p de 5 toiſes; <lb/></s> <s xml:id="echoid-s14446" xml:space="preserve">on a tout ce qu’il faut pour meſurer la ſolidité du cylindre <lb/>X V & </s> <s xml:id="echoid-s14447" xml:space="preserve">celle du cône tronqué. </s> <s xml:id="echoid-s14448" xml:space="preserve">Ayant donc trouvé ces ſolidités, <lb/>on ſouſtraira celle du cône tronqué de celle du cylindre, pour <lb/>avoir la différence, qui étant une fois trouvée, l’on dira: </s> <s xml:id="echoid-s14449" xml:space="preserve">Si <lb/>360 m’ont donné tant pour la différence du cylindre au cône <lb/>tronqué, que me donneront 60 degrés, valeur de l’angle <lb/>N O P pour la ſolidité du ſecteur de la partie du cylindre, <lb/>après en avoir ôté le cône tronqué, & </s> <s xml:id="echoid-s14450" xml:space="preserve">ce qu’on trouvera ſera <lb/>au juſte la valeur du revêtement du flanc concave. </s> <s xml:id="echoid-s14451" xml:space="preserve">Quant à <lb/>la briſure F K, & </s> <s xml:id="echoid-s14452" xml:space="preserve">au revers G I de l’orillon, ce ſont des par-<lb/>ties trop aiſées à toiſer, pour avoir beſoin d’explication.</s> <s xml:id="echoid-s14453" xml:space="preserve"/> </p> <div xml:id="echoid-div1144" type="float" level="2" n="8"> <note position="left" xlink:label="note-0506-01" xlink:href="note-0506-01a" xml:space="preserve">Figure 272, <lb/>273 & 274.</note> </div> <p> <s xml:id="echoid-s14454" xml:space="preserve">855. </s> <s xml:id="echoid-s14455" xml:space="preserve">La maniere de toiſer l’arrondiſſement d’une contreſ-<lb/> <anchor type="note" xlink:label="note-0506-02a" xlink:href="note-0506-02"/> carpe, eſt encore une opération qui a auſſi ſes difficultés: </s> <s xml:id="echoid-s14456" xml:space="preserve">mais <lb/>comme cette partie eſt la même que celle du flanc concave, <lb/>ſi on a bien entrendu ce que j’ai dit ci-devant, je ne crois pas <lb/>qu’on ſe trouve embarraſſé. </s> <s xml:id="echoid-s14457" xml:space="preserve">Cependant comme je ne veux rien <lb/>laiſſer à deviner, conſidérez que pour toiſer la maçonnerie de <lb/>la contreſcarpe de la figure 278, on s’y prendra comme on a <pb o="441" file="0507" n="521" rhead="DE MATHÉMATIQUE. Liv. XII."/> fait pour le baſtion de la figure 266, c’eſt-à-dire que faiſant <lb/>abſtraction des contre-forts, on multipliera la ſuperficie de la <lb/>maçonnerie par la longueur de la contreſcarpe rectiligne, & </s> <s xml:id="echoid-s14458" xml:space="preserve"><lb/>qu’on meſurera auſſi les pyramides tronquées, qui ſe trouve-<lb/>ront dans les angles rentrans; </s> <s xml:id="echoid-s14459" xml:space="preserve">& </s> <s xml:id="echoid-s14460" xml:space="preserve">pour l’arrondiſſement, on s’y <lb/>prendra comme il ſuit.</s> <s xml:id="echoid-s14461" xml:space="preserve"/> </p> <div xml:id="echoid-div1145" type="float" level="2" n="9"> <note position="left" xlink:label="note-0506-02" xlink:href="note-0506-02a" xml:space="preserve">Figure 278.</note> </div> <p> <s xml:id="echoid-s14462" xml:space="preserve">856. </s> <s xml:id="echoid-s14463" xml:space="preserve">Suppoſant que l’arc A C B marque le pied de la mu-<lb/> <anchor type="note" xlink:label="note-0507-01a" xlink:href="note-0507-01"/> raille dans le foſſé, l’arc D F G le ſommet, & </s> <s xml:id="echoid-s14464" xml:space="preserve">l’arc H I K avec <lb/>le précédent l’épaiſſeur au ſommet, & </s> <s xml:id="echoid-s14465" xml:space="preserve">l’intervalle C F le talud, <lb/>on commencera par chercher la valeur de la corde A B, que <lb/>nous ſuppoſerons de 20 toiſes, & </s> <s xml:id="echoid-s14466" xml:space="preserve">celle de la fleche L C, qui <lb/>ſera, par exemple, de 4, afin de connoître le diametre de <lb/>l’arc A C B, qu’on trouvera, auſſi-bien que celui de tout autre <lb/>arc, en cherchant une troiſieme proportionnelle à la fleche L C, <lb/>& </s> <s xml:id="echoid-s14467" xml:space="preserve">à la moitié de la corde L A, c’eſt-à-dire à 4 & </s> <s xml:id="echoid-s14468" xml:space="preserve">à 10: </s> <s xml:id="echoid-s14469" xml:space="preserve">cette <lb/>troiſieme proportionnelle, qui eſt ici de 25 toiſes, ſera la va-<lb/>leur du diametre qu’on demande.</s> <s xml:id="echoid-s14470" xml:space="preserve"/> </p> <div xml:id="echoid-div1146" type="float" level="2" n="10"> <note position="right" xlink:label="note-0507-01" xlink:href="note-0507-01a" xml:space="preserve">Figure 278.</note> </div> <p> <s xml:id="echoid-s14471" xml:space="preserve">857. </s> <s xml:id="echoid-s14472" xml:space="preserve">La raiſon de ceci s’entendra aiſément, en conſidé-<lb/> <anchor type="note" xlink:label="note-0507-02a" xlink:href="note-0507-02"/> rant que l’arc A C B de la figure n 276 eſt le même que le précé-<lb/>dent; </s> <s xml:id="echoid-s14473" xml:space="preserve">& </s> <s xml:id="echoid-s14474" xml:space="preserve">on remarquera qu’ayant achevé le cercle, la demi-<lb/>corde L B eſt moyenne proportionnelle entre la fleche C L & </s> <s xml:id="echoid-s14475" xml:space="preserve"><lb/>la partie L M du diametre; </s> <s xml:id="echoid-s14476" xml:space="preserve">& </s> <s xml:id="echoid-s14477" xml:space="preserve">qu’ayant trouvé la ligne L M <lb/>troiſieme proportionnelle à C L & </s> <s xml:id="echoid-s14478" xml:space="preserve">L B, on n’a qu’à l’ajouter à <lb/>la fleche C L, pour avoir le diametre C M.</s> <s xml:id="echoid-s14479" xml:space="preserve"/> </p> <div xml:id="echoid-div1147" type="float" level="2" n="11"> <note position="right" xlink:label="note-0507-02" xlink:href="note-0507-02a" xml:space="preserve">Figure 276.</note> </div> <p> <s xml:id="echoid-s14480" xml:space="preserve">Comme nous avons beſoin de connoître auſſi la quantité de <lb/>degrés que contient l’arc A C B, ſi on tire les rayons N B & </s> <s xml:id="echoid-s14481" xml:space="preserve"><lb/>N A du centre, l’on aura le triangle A B N, dont on connoît <lb/>le côté A B de 20 toiſes, & </s> <s xml:id="echoid-s14482" xml:space="preserve">les côtés N B & </s> <s xml:id="echoid-s14483" xml:space="preserve">N A chacun de <lb/>14 toiſes 3 pieds: </s> <s xml:id="echoid-s14484" xml:space="preserve">il ſera donc facile de connoître l’angle <lb/>A N B, que l’on trouvera de 90 degrés 44 minutes.</s> <s xml:id="echoid-s14485" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14486" xml:space="preserve">Préſentement ſi l’on conſidere le profil de la contreſcarpe <lb/>dans la figure 281, on verra que reſſemblant à celui du flanc <lb/>concave, l’arrondiſſement du foſſé eſt un ſecteur de cylindre, <lb/>duquel on a ôté un cône tronqué, dont l’axe commun ſeroit <lb/>la ligne O P. </s> <s xml:id="echoid-s14487" xml:space="preserve">Or ſi la hauteur F R ou O P eſt de 18 pieds, <lb/>& </s> <s xml:id="echoid-s14488" xml:space="preserve">l’épaiſſeur F I de 3, le talud C R de 4, le rayon P C étant <lb/>de 14 toiſes 3 pieds, le rayon O F ſera de 15 toiſes 1 pied, & </s> <s xml:id="echoid-s14489" xml:space="preserve"><lb/>le rayon O I ſera de 15 toiſes 4 pieds: </s> <s xml:id="echoid-s14490" xml:space="preserve">& </s> <s xml:id="echoid-s14491" xml:space="preserve">comme on connoît <lb/>toutes les lignes du cylindre, qui auroient pour plan généra-<lb/>teur le rectangle P I, & </s> <s xml:id="echoid-s14492" xml:space="preserve">celles du cône tronqué, qui auroient <pb o="442" file="0508" n="522" rhead="NOUVEAU COURS"/> pour plan générateur le trapézoïde P O F D, ſi on cherche la <lb/>ſolidité de l’un & </s> <s xml:id="echoid-s14493" xml:space="preserve">de l’autre, & </s> <s xml:id="echoid-s14494" xml:space="preserve">qu’on ôte celle du cône tron-<lb/>qué de celle du cylindre, on aura la différence qui nous don-<lb/>nera la ſolidité que nous cherchons, en diſant: </s> <s xml:id="echoid-s14495" xml:space="preserve">Si 360 degrés <lb/>m’ont donné cette différence, que me donneront 90 degrés <lb/>44 minutes pour la valeur de l’arrondiſſement.</s> <s xml:id="echoid-s14496" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14497" xml:space="preserve">Je n’ai rien dit juſqu’ici ſur la maniere de toiſer les contre-<lb/>forts, parce qu’ils ne ſont autre choſe que des parallélepipedes, <lb/>dont la ſolidité ſe trouve en multipliant la baſe par la hauteur.</s> <s xml:id="echoid-s14498" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1149" type="section" level="1" n="841"> <head xml:id="echoid-head1017" xml:space="preserve">PROPOSITION XX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14499" xml:space="preserve">858. </s> <s xml:id="echoid-s14500" xml:space="preserve">Maniere de meſurer la ſolidité de l’onglet d’un batardeau. <lb/></s> </p> <p> <s xml:id="echoid-s14501" xml:space="preserve">Quand les foſſés d’une fortification ſont inondés, on y fait <lb/>ordinairement aux endroits les plus convenables des batardeaux <lb/>de maçonnerie, pour retenir les eaux ou pour les lâcher, ſelon <lb/>le beſoin qu’on en a. </s> <s xml:id="echoid-s14502" xml:space="preserve">Pour connoître ce batardeau, conſidérez <lb/>la figure 277, qui fait voir que cet ouvrage n’eſt autre choſe <lb/>qu’un corps de maçonnerie, dont le profil A B C D E marque <lb/>que le deſſus B C D eſt en dos d’âne pour l’écoulement des eaux <lb/>de pluie, & </s> <s xml:id="echoid-s14503" xml:space="preserve">pour empêcher qu’un homme ne puiſſe paſſer <lb/>deſſus: </s> <s xml:id="echoid-s14504" xml:space="preserve">cependant comme les ſoldats pourroient, en deſcen-<lb/>dant du rempart avec une corde, paſſer le foſſé en s’acheva-<lb/>lant ſur cette chappe, on fait, pour y mettre empêchement, <lb/>une tourelle dans le milieu, qui s’oppoſe abſolument au paſ-<lb/>ſage. </s> <s xml:id="echoid-s14505" xml:space="preserve">Pour toiſer ce batardeau, on commence par meſurer la <lb/>ſuperficie du profil A B C D E, qu’on multiplie par toute la <lb/>largeur du foſſé en cet endroit; </s> <s xml:id="echoid-s14506" xml:space="preserve">enſuite on cherche la ſolidité <lb/>du cylindre F I K G, auſſi-bien que celle de ſa couverture, qui <lb/>eſt quelquefois un cône I L K, ou une demi-ſphere. </s> <s xml:id="echoid-s14507" xml:space="preserve">Juſques-là <lb/>tout eſt facile; </s> <s xml:id="echoid-s14508" xml:space="preserve">mais ce qui embarraſſe preſque tous les Ingé-<lb/>nieurs, c’eſt de toiſer les deux fragmens, comme F H G, de <lb/>la tourelle, qui ſont à droite & </s> <s xml:id="echoid-s14509" xml:space="preserve">à gauche, comme on peut les <lb/>voir encore mieux en X & </s> <s xml:id="echoid-s14510" xml:space="preserve">Z de la figure 282, qui eſt un profil <lb/>de la tourelle & </s> <s xml:id="echoid-s14511" xml:space="preserve">du batardeau.</s> <s xml:id="echoid-s14512" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14513" xml:space="preserve">Ce problême me fut propoſé par pluſieurs Ingénieurs, qui <lb/> <anchor type="note" xlink:label="note-0508-02a" xlink:href="note-0508-02"/> déſiroient d’en avoir la ſolution. </s> <s xml:id="echoid-s14514" xml:space="preserve">Je la cherchai, & </s> <s xml:id="echoid-s14515" xml:space="preserve">la trouvai <lb/>de pluſieurs manieres; </s> <s xml:id="echoid-s14516" xml:space="preserve">je pris tant de plaiſir à y travailler, que <lb/>je cherchai même pluſieurs choſes fort curieuſes à ſon occa- <pb o="443" file="0509" n="523" rhead="DE MATHÉMATIQUE. Liv. XII."/> ſion; </s> <s xml:id="echoid-s14517" xml:space="preserve">entr’autres de ſçavoir quelle eſt la quadrature de la ſur-<lb/>face de l’onglet, c’eſt-à-dire trouver un rectangle égal à ſa <lb/>ſurface: </s> <s xml:id="echoid-s14518" xml:space="preserve">& </s> <s xml:id="echoid-s14519" xml:space="preserve">comme je crois qu’on ſera bien aiſe de ſçavoir ce <lb/>qu’on peut dire de plus intéreſſant là-deſſus, on n’a qu’à exa-<lb/>miner ce qui ſuit.</s> <s xml:id="echoid-s14520" xml:space="preserve"/> </p> <div xml:id="echoid-div1149" type="float" level="2" n="1"> <note position="left" xlink:label="note-0508-02" xlink:href="note-0508-02a" xml:space="preserve">Figure 282.</note> </div> <p> <s xml:id="echoid-s14521" xml:space="preserve">Comme l’axe du cylindre qui compoſe la tourelle répond <lb/> <anchor type="note" xlink:label="note-0509-01a" xlink:href="note-0509-01"/> ſur l’arrête de la cape du batardeau, cette arrête partage la <lb/>cape du cylindre en deux également; </s> <s xml:id="echoid-s14522" xml:space="preserve">de ſorte que chaque <lb/>demi-cercle devient une des faces N Q M de l’onglet. </s> <s xml:id="echoid-s14523" xml:space="preserve">Or ſi <lb/>l’on conſidere ce ſolide comme compoſé d’une quantité infinie <lb/>de triangles rectangles, tels que P O Q, qui ont tous pour <lb/>baſe les ordonnées Q O, R S, T V, des quarts de cercles O Q N <lb/>& </s> <s xml:id="echoid-s14524" xml:space="preserve">O P M, on verra que tous ces triangles étant ſemblables, <lb/>ils ſont dans la même raiſon que les quarrés de leurs baſes; </s> <s xml:id="echoid-s14525" xml:space="preserve">& </s> <s xml:id="echoid-s14526" xml:space="preserve"><lb/>ne prenant que les triangles qui compoſent la moitié Q N O P <lb/>de l’onglet, il s’enſuit qu’on en trouvera la valeur, comme <lb/>on trouve celle des quarrés de leurs baſes, ou autrement comme <lb/>on trouve celle des quarrés des ordonnées d’un quart de cer-<lb/>cle (art. </s> <s xml:id="echoid-s14527" xml:space="preserve">569); </s> <s xml:id="echoid-s14528" xml:space="preserve">mais nous ſçavons que pour trouver la valeur <lb/>de tous ces quarrés, il faut multiplier celui de la plus grande <lb/>ordonnée O Q par les deux tiers de la ligne O N, qui en ex-<lb/>prime la quantité: </s> <s xml:id="echoid-s14529" xml:space="preserve">ſil faudra donc, pour trouver la valeur de <lb/>tous les triangles, multiplier le plus grand triangle P O Q par <lb/>les deux tiers de la ligne O N: </s> <s xml:id="echoid-s14530" xml:space="preserve">mais comme ceci ne donne <lb/>que la moitié de la ſolidité de l’onglet, il faudra donc, pour <lb/>l’avoir toute entiere, multiplier le triangle P O Q par les deux <lb/>tiers du diametre M N.</s> <s xml:id="echoid-s14531" xml:space="preserve"/> </p> <div xml:id="echoid-div1150" type="float" level="2" n="2"> <note position="right" xlink:label="note-0509-01" xlink:href="note-0509-01a" xml:space="preserve">Figure 279.</note> </div> <p> <s xml:id="echoid-s14532" xml:space="preserve">Suppoſant que cet onglet-ci ſoit le même que celui qui eſt <lb/> <anchor type="note" xlink:label="note-0509-02a" xlink:href="note-0509-02"/> en X, le triangle O P Q ſera le même que A B C: </s> <s xml:id="echoid-s14533" xml:space="preserve">par conſé-<lb/>quent ſi la ligne C A eſt de 5 pieds, & </s> <s xml:id="echoid-s14534" xml:space="preserve">le diametre A D de 9, <lb/>la ligne B C ſera de 4 & </s> <s xml:id="echoid-s14535" xml:space="preserve">demi, & </s> <s xml:id="echoid-s14536" xml:space="preserve">la ſuperficie du triangle <lb/>A B C ſera de 11 pieds 3 pouces, qui étant multipliés par les <lb/>deux tiers du diametre A D, c’eſt-à-dire par 6, donnera 67 pieds <lb/>& </s> <s xml:id="echoid-s14537" xml:space="preserve">demi cubes pour la ſolidité de l’onglet X.</s> <s xml:id="echoid-s14538" xml:space="preserve"/> </p> <div xml:id="echoid-div1151" type="float" level="2" n="3"> <note position="right" xlink:label="note-0509-02" xlink:href="note-0509-02a" xml:space="preserve">Figure 279 <lb/>& 282.</note> </div> <p> <s xml:id="echoid-s14539" xml:space="preserve">Si on imagine l’onglet coupé par une quantité de plans, <lb/>qui paſſant par le centre B du demi-cercle, aillent tomber ſur <lb/>la circonférence A F D, c’eſt-à-dire perpendiculairement ſur <lb/>la ſurface de l’onglet, ces plans partageront l’onglet en une <lb/>infinité de petites pyramides, qui auront toutes pour hauteur <lb/>commune le rayon du demi-cercle, & </s> <s xml:id="echoid-s14540" xml:space="preserve">leurs baſes dans la ſur- <pb o="444" file="0510" n="524" rhead="NOUVEAU COURS"/> face de l’onglet. </s> <s xml:id="echoid-s14541" xml:space="preserve">Mais comme toutes ces pyramides, priſes <lb/>enſemble, ſont égales à une ſeule qui auroit pour baſe la <lb/>ſomme de toutes les baſes, c’eſt-à-dire la ſurface de l’onglet, <lb/>& </s> <s xml:id="echoid-s14542" xml:space="preserve">pour hauteur ſon rayon, il s’enſuit qu’on trouvera encore <lb/>la ſolidité de l’onglet, en multipliant ſa ſurface par le tiers du <lb/>rayon.</s> <s xml:id="echoid-s14543" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14544" xml:space="preserve">859. </s> <s xml:id="echoid-s14545" xml:space="preserve">Préſentement je dis que la ſurface de l’onglet X eſt <lb/>égale à un rectangle, qui auroit pour baſe le diametre B D ou <lb/>M N de l’onglet, & </s> <s xml:id="echoid-s14546" xml:space="preserve">pour hauteur, la hauteur même de l’on-<lb/>glet, c’eſt-à-dire la ligne B A.</s> <s xml:id="echoid-s14547" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14548" xml:space="preserve">Si l’on nomme la ligne B A, a; </s> <s xml:id="echoid-s14549" xml:space="preserve">le rayon C B ou C D, b; <lb/></s> <s xml:id="echoid-s14550" xml:space="preserve">le diametre B D ſera 2b. </s> <s xml:id="echoid-s14551" xml:space="preserve">Cela poſé, il faut faire voir que B D <lb/>x B A (2ba) eſt égal à la ſurface de l’onglet.</s> <s xml:id="echoid-s14552" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14553" xml:space="preserve">Conſidérez que la ſuperficie du triangle A B C eſt {ab/2}, & </s> <s xml:id="echoid-s14554" xml:space="preserve">que <lb/>ſi on multiplie cette quantité par les deux tiers du diametre <lb/>B D, c’eſt-à-dire par {4b/3}, l’on aura {4abb/6} pour la ſolidité de l’on-<lb/>glet: </s> <s xml:id="echoid-s14555" xml:space="preserve">mais comme ce produit peut être regardé comme le pro-<lb/>duit de la ſurface de l’onglet par le tiers du rayon, il s’enſuit <lb/>que diviſant {4abb/6} par {b/3}, le quotient ſera néceſſairement la ſur-<lb/>face de l’onglet. </s> <s xml:id="echoid-s14556" xml:space="preserve">Si l’on fait la diviſion, on trouvera que ce <lb/>quotient eſt 2ab = B D x B A; </s> <s xml:id="echoid-s14557" xml:space="preserve">ce qui fait voir que la ſurface <lb/>de l’onglet eſt égale au rectangle que nous avons dit.</s> <s xml:id="echoid-s14558" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14559" xml:space="preserve">Ceci rentre dans la propoſition que nous avons donnée ſur <lb/>la ſuperficie des voûtes en plein cintre, & </s> <s xml:id="echoid-s14560" xml:space="preserve">ſur leur ſolidité; <lb/></s> <s xml:id="echoid-s14561" xml:space="preserve">l’onglet que nous venons de meſurer pouvant être regardé <lb/>comme un double pan de voûte, dont chacun auroit la même <lb/>hauteur, & </s> <s xml:id="echoid-s14562" xml:space="preserve">pour baſe le triangle B F I.</s> <s xml:id="echoid-s14563" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1153" type="section" level="1" n="842"> <head xml:id="echoid-head1018" style="it" xml:space="preserve">Principe général pour meſurer les ſurfaces & les ſolides.</head> <p> <s xml:id="echoid-s14564" xml:space="preserve">860. </s> <s xml:id="echoid-s14565" xml:space="preserve">Rien ne fait mieux connoître la beauté de la Géomé-<lb/>trie, que la fécondité de ſes principes qui ſemblent, à l’envi, <lb/>ouvrir de nouveaux chemins pour parvenir à la même choſe; <lb/></s> <s xml:id="echoid-s14566" xml:space="preserve">témoin les belles découvertes qu’on a faites de notre tems, <lb/>parmi leſquelles en voici une qui eſt trop intéreſſante pour la <lb/>refuſer à ceux dont le principal objet, en étudiant la Géomé-<lb/>trie, eſt de ſçavoir meſurer les corps; </s> <s xml:id="echoid-s14567" xml:space="preserve">mais comme ſa con-<lb/>noiſſance dépend de certaines choſes dont nous n’avons point <pb o="445" file="0511" n="525" rhead="DE MATHÉMATIQUE. Liv. XII."/> parlé juſqu’ici, nous allons faire enſorte de ne rien laiſſer à <lb/>deviner.</s> <s xml:id="echoid-s14568" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1154" type="section" level="1" n="843"> <head xml:id="echoid-head1019" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s14569" xml:space="preserve">861. </s> <s xml:id="echoid-s14570" xml:space="preserve">L’on nomme centre de gravité d’une ligne droite, un <lb/>point par lequel cette ligne étant ſuſpendue, toutes ſes parties <lb/>ſont en équilibre: </s> <s xml:id="echoid-s14571" xml:space="preserve">car quoiqu’une ligne ſoit regardée comme <lb/>n’ayant aucune peſanteur, cela n’empêche pas que la diffé-<lb/>rence de ſes parties ne ſoit conſidérée comme un obſtacle à <lb/>l’équilibre. </s> <s xml:id="echoid-s14572" xml:space="preserve">Ainſi la ligne A B étant diviſée en deux également <lb/>au point C, ce point eſt pris pour celui d’équilibre, c’eſt-à-dire <lb/>pour l’endroit par lequel cette ligne étant ſuſpendue, les parties <lb/>égales C A & </s> <s xml:id="echoid-s14573" xml:space="preserve">C B ſeront en équilibre, parce que n’étant pas <lb/>plus longues l’une que l’autre, il n’y a point de raiſon pour que <lb/>l’extrêmité A ſoit plus ſollicitée à ſe mouvoir que l’extrêmité <lb/>D: </s> <s xml:id="echoid-s14574" xml:space="preserve">& </s> <s xml:id="echoid-s14575" xml:space="preserve">quand cela eſt ainſi à l’égard d’un plan, ce point eſt ap-<lb/>pellé le centre de gravité du plan: </s> <s xml:id="echoid-s14576" xml:space="preserve">car quoique le plan, auſſi-<lb/>bien que la ligne, ſoit conſidéré ſans peſanteur, cela n’em-<lb/>pêche pas qu’on ne regarde encore ſes parties comme pouvant <lb/>être un obſtacle à leur équilibre.</s> <s xml:id="echoid-s14577" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14578" xml:space="preserve">862. </s> <s xml:id="echoid-s14579" xml:space="preserve">Par exemple, ſi l’on a un rectangle A B, & </s> <s xml:id="echoid-s14580" xml:space="preserve">qu’on tire <lb/> <anchor type="note" xlink:label="note-0511-01a" xlink:href="note-0511-01"/> les diagonales A B & </s> <s xml:id="echoid-s14581" xml:space="preserve">C D, le point E où elles ſe coupent en <lb/> <anchor type="note" xlink:label="note-0511-02a" xlink:href="note-0511-02"/> ſera le centre de gravité, parce que ſi ce plan étoit poſé ſur un <lb/>pivot fort aigu qui répondît à l’endroit E, il n’y auroit point <lb/>de raiſon pour que le plan inclinât plus du côté D B que du <lb/>côté A C, ni du côté A D, plutôt que du côté C B.</s> <s xml:id="echoid-s14582" xml:space="preserve"/> </p> <div xml:id="echoid-div1154" type="float" level="2" n="1"> <note position="right" xlink:label="note-0511-01" xlink:href="note-0511-01a" xml:space="preserve">Pl. XXI.</note> <note position="right" xlink:label="note-0511-02" xlink:href="note-0511-02a" xml:space="preserve">Figure 283.</note> </div> <p> <s xml:id="echoid-s14583" xml:space="preserve">Comme les ſurfaces circulaires ſont formées par la circon-<lb/> <anchor type="note" xlink:label="note-0511-03a" xlink:href="note-0511-03"/> volution uniforme d’une ligne droite, & </s> <s xml:id="echoid-s14584" xml:space="preserve">que les ſolides cir-<lb/>culaires ſont formés par la circonvolution d’un plan, c’eſt la <lb/>valeur de ces ſurfaces & </s> <s xml:id="echoid-s14585" xml:space="preserve">de ces ſolides qu’on ſe propoſe de <lb/>trouver ici, moyennant la connoiſſance du centre de gravité <lb/>de la ligne génératrice, & </s> <s xml:id="echoid-s14586" xml:space="preserve">celui du plan générateur: </s> <s xml:id="echoid-s14587" xml:space="preserve">car ſi le <lb/>point C eſt le centre de gravité de la ligne A B, & </s> <s xml:id="echoid-s14588" xml:space="preserve">qu’on éleve <lb/>à cet endroit la perpendiculaire C D, nous ferons voir que <lb/>(la ligne A B ayant fait une circonvolution autour de la ligne <lb/>E F, qui ſera appellée axe, & </s> <s xml:id="echoid-s14589" xml:space="preserve">qui eſt auſſi perpendiculaire ſur <lb/>D C), la ſurface que décrira la ligne A B, ſera égale à un rec-<lb/>tangle, qui auroit pour baſe la ligne A B, & </s> <s xml:id="echoid-s14590" xml:space="preserve">pour hauteur <lb/>une ligne égale à la circonférence, qui auroit pour rayon la <lb/>ligne D C, qui exprime la diſtance du centre de gravité C à <pb o="446" file="0512" n="526" rhead="NOUVEAU COURS"/> l’axe; </s> <s xml:id="echoid-s14591" xml:space="preserve">& </s> <s xml:id="echoid-s14592" xml:space="preserve">que ſi du centre de gravité E l’on abaiſſe une perpen-<lb/>diculaire E F ſur le côté C B, & </s> <s xml:id="echoid-s14593" xml:space="preserve">qu’on faſſe faire une circon-<lb/>volution au rectangle A B ſur le côté C B (que nous nomme-<lb/>rons auſſi axe), le corps que décrira le plan, ſera égal à un pa-<lb/>rallélepipede qui auroit pour baſe ce plan même, & </s> <s xml:id="echoid-s14594" xml:space="preserve">pour <lb/>hauteur une ligne égale à la circonférence du cercle, qui au-<lb/>roit pour rayon la ligne E F; </s> <s xml:id="echoid-s14595" xml:space="preserve">ce que nous rendrons général <lb/>pour meſurer toutes les ſurfaces dont on pourra connoître les <lb/>centres de gravité de leurs lignes génératrices, & </s> <s xml:id="echoid-s14596" xml:space="preserve">pour me-<lb/>ſurer tous les ſolides dont on pourra connoître le centre de <lb/>gravité de leur plan générateur.</s> <s xml:id="echoid-s14597" xml:space="preserve"/> </p> <div xml:id="echoid-div1155" type="float" level="2" n="2"> <note position="right" xlink:label="note-0511-03" xlink:href="note-0511-03a" xml:space="preserve">Figure 285 <lb/>& 283.</note> </div> </div> <div xml:id="echoid-div1157" type="section" level="1" n="844"> <head xml:id="echoid-head1020" xml:space="preserve">PROPOSITION XXI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14598" xml:space="preserve">863. </s> <s xml:id="echoid-s14599" xml:space="preserve">Connoiſſant le centre de gravité d’une ligne droite A B, <lb/> <anchor type="note" xlink:label="note-0512-01a" xlink:href="note-0512-01"/> trouver la valeur de la ſurface qu’elle décrira, aprés avoir fait une <lb/>circonvolution autour de l’axe E F.</s> <s xml:id="echoid-s14600" xml:space="preserve"/> </p> <div xml:id="echoid-div1157" type="float" level="2" n="1"> <note position="left" xlink:label="note-0512-01" xlink:href="note-0512-01a" xml:space="preserve">Figure 285 <lb/>& 286.</note> </div> <p> <s xml:id="echoid-s14601" xml:space="preserve">Je dis qu’il faut multiplier la ligne A B par la circonférence <lb/>du cercle, qui auroit pour rayon D C, & </s> <s xml:id="echoid-s14602" xml:space="preserve">qu’on aura la ſurface <lb/>que l’on demande: </s> <s xml:id="echoid-s14603" xml:space="preserve">car comme cette ligne décrira un cylindre <lb/>G B, & </s> <s xml:id="echoid-s14604" xml:space="preserve">que pour trouver la ſurface de ce cylindre, il faut <lb/>multiplier le cercle du rayon F B de la baſe par la hauteur A B <lb/>du cylindre, il s’enſuit que la ligne D C étant égale à F B, <lb/>les circonférences de ces lignes ſeront auſſi égales, & </s> <s xml:id="echoid-s14605" xml:space="preserve">que par <lb/>conſéquent le produit de la ligne A B par la circonférence du <lb/>rayon D C, ſera égal à la ſurface qu’on demande.</s> <s xml:id="echoid-s14606" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14607" xml:space="preserve">864. </s> <s xml:id="echoid-s14608" xml:space="preserve">Mais ſi la ligne A B, au lieu d’être parallele à l’axe <lb/> <anchor type="note" xlink:label="note-0512-02a" xlink:href="note-0512-02"/> E F étoit oblique, comme eſt, par exemple, la ligne G H: <lb/></s> <s xml:id="echoid-s14609" xml:space="preserve">je dis qu’ayant fait une circonvolution à l’entour de l’axe E F, <lb/>la ſurface qu’elle décrira ſera encore égale au rectangle com-<lb/>pris ſous la même ligne G H, & </s> <s xml:id="echoid-s14610" xml:space="preserve">ſous la circonférence du cer-<lb/>cle, qui auroit pour rayon la ligne D C, tirée du centre de <lb/>gravité C perpendiculaire ſur l’axe E F.</s> <s xml:id="echoid-s14611" xml:space="preserve"/> </p> <div xml:id="echoid-div1158" type="float" level="2" n="2"> <note position="left" xlink:label="note-0512-02" xlink:href="note-0512-02a" xml:space="preserve">Figure 287 <lb/>& 258.</note> </div> <p> <s xml:id="echoid-s14612" xml:space="preserve">Comme cette ligne aura décrit la ſurface I H d’un cône <lb/>tronqué, & </s> <s xml:id="echoid-s14613" xml:space="preserve">que la ligne D C eſt moyenne arithmétique entre <lb/>E G & </s> <s xml:id="echoid-s14614" xml:space="preserve">F H, la circonférence qui auroit pour rayon D C <lb/>ſera moyenne arithmétique entre les circonférences des <lb/>rayons E G & </s> <s xml:id="echoid-s14615" xml:space="preserve">F H: </s> <s xml:id="echoid-s14616" xml:space="preserve">mais comme ces circonférences ſervent <lb/>de côtés paralleles au trapézoïde qui auroit pour hauteur la <pb o="447" file="0513" n="527" rhead="DE MATHÉMATIQUE. Liv. XII."/> ligne G H, & </s> <s xml:id="echoid-s14617" xml:space="preserve">que ce trapeze eſt égal à la ſurface du cône <lb/>tronqué, il s’enſuit que le rectangle compris ſous G H, & </s> <s xml:id="echoid-s14618" xml:space="preserve">la <lb/>circonférence du cercle, qui auroit pour rayon D C, eſt égal <lb/>à la ſurface décrite par la ligne G H.</s> <s xml:id="echoid-s14619" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14620" xml:space="preserve">865. </s> <s xml:id="echoid-s14621" xml:space="preserve">Enfin ſi la ligne génératrice venoit rencontrer, comme <lb/> <anchor type="note" xlink:label="note-0513-01a" xlink:href="note-0513-01"/> E K, l’axe E F, je dis encore que ſi elle fait une circonvolution <lb/>à l’entour de l’axe E F, la ſurface qu’elle décrira ſera égale au <lb/>rectangle compris ſous la même ligne E K, & </s> <s xml:id="echoid-s14622" xml:space="preserve">ſous la cir-<lb/>conférence du cercle qui auroit pour rayon D C.</s> <s xml:id="echoid-s14623" xml:space="preserve"/> </p> <div xml:id="echoid-div1159" type="float" level="2" n="3"> <note position="right" xlink:label="note-0513-01" xlink:href="note-0513-01a" xml:space="preserve">Figure 289 <lb/>& 290.</note> </div> <p> <s xml:id="echoid-s14624" xml:space="preserve">Si l’on fait attention que la ligne génératrice aura décrit la <lb/>ſurface du cône L E K, on verra que cette ſurface étant égale <lb/>au rectangle compris ſous le côté E K, & </s> <s xml:id="echoid-s14625" xml:space="preserve">ſous la moitié de <lb/>la circonférence du cercle L K (art. </s> <s xml:id="echoid-s14626" xml:space="preserve">547), la ligne D C étant <lb/>moitié du rayon F K, la circonférence dont elle ſera le rayon <lb/>ſera auſſi moitié de L K, & </s> <s xml:id="echoid-s14627" xml:space="preserve">que par conſéquent le rectangle <lb/>compris ſous la ligne génératrice E K, & </s> <s xml:id="echoid-s14628" xml:space="preserve">ſous la circonfé-<lb/>rence du cercle, qui auroit pour rayon D C, ſera égale à la <lb/>ſurface qu’elle aura décrite.</s> <s xml:id="echoid-s14629" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1161" type="section" level="1" n="845"> <head xml:id="echoid-head1021" xml:space="preserve">PROPOSITION XXII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14630" xml:space="preserve">866. </s> <s xml:id="echoid-s14631" xml:space="preserve">Si l’on a une demi-circonférence E B F, & </s> <s xml:id="echoid-s14632" xml:space="preserve">que le point <lb/> <anchor type="note" xlink:label="note-0513-02a" xlink:href="note-0513-02"/> C ſoit le centre de gravité, je dis que cette demi-circonférence ayant <lb/>fait une circonvolution ſur l’axe E F, la ſurface qu’elle décrira, <lb/>qui ſera celle d’une ſphere, ſera égale au rectangle compris ſous <lb/>une ligne égale à la demi-circonférence E B F, & </s> <s xml:id="echoid-s14633" xml:space="preserve">ſous celle qui <lb/>ſeroit égale à la circonférence dont la ligne C D ſeroit le rayon.</s> <s xml:id="echoid-s14634" xml:space="preserve"/> </p> <div xml:id="echoid-div1161" type="float" level="2" n="1"> <note position="right" xlink:label="note-0513-02" xlink:href="note-0513-02a" xml:space="preserve">Figure 294.</note> </div> <p> <s xml:id="echoid-s14635" xml:space="preserve">Comme il faut connoître le centre de gravité C par rap-<lb/>port aux autres parties de la figure, on ſçaura que la ligne <lb/>C D, qui en détermine la poſition par rapport au centre du <lb/>demi-cercle, doit être quatrieme proportionnelle à la demi-<lb/>circonférence E B F, au diametre E F, & </s> <s xml:id="echoid-s14636" xml:space="preserve">au demi-diametre <lb/>D F. </s> <s xml:id="echoid-s14637" xml:space="preserve">Ainſi ayant nommé la demi-circonférence a; </s> <s xml:id="echoid-s14638" xml:space="preserve">le dia-<lb/>metre E F, b; </s> <s xml:id="echoid-s14639" xml:space="preserve">le demi-diametre D F ſera {b/2}; </s> <s xml:id="echoid-s14640" xml:space="preserve">& </s> <s xml:id="echoid-s14641" xml:space="preserve">par conſé-<lb/>quent on aura a : </s> <s xml:id="echoid-s14642" xml:space="preserve">b :</s> <s xml:id="echoid-s14643" xml:space="preserve">: {b/2} : </s> <s xml:id="echoid-s14644" xml:space="preserve">{bb/2a}, qui fait voir que {bb/2a} eſt égal à la <lb/>ligne D C: </s> <s xml:id="echoid-s14645" xml:space="preserve">mais comme nous avons beſoin de la circonfé-<lb/>rence de la ligne D C, on la trouvera, en diſant: </s> <s xml:id="echoid-s14646" xml:space="preserve">Comme le <pb o="448" file="0514" n="528" rhead="NOUVEAU COURS"/> rayon D F ({b/2}) eſt à ſa circonférence (2a), ainſi le rayon D C <lb/>({bb/2a}) eſt à ſa circonférence: </s> <s xml:id="echoid-s14647" xml:space="preserve">c’eſt pourquoi multipliant le ſe-<lb/>cond terme par le troiſieme, & </s> <s xml:id="echoid-s14648" xml:space="preserve">diviſant le produit par le pre-<lb/>mier (art. </s> <s xml:id="echoid-s14649" xml:space="preserve">206), on trouvera le quatrieme, qui ſera 2b.</s> <s xml:id="echoid-s14650" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14651" xml:space="preserve">Comme 2b eſt la circonférence du rayon D C, ſi on la mul-<lb/>tiplie par la demi-circonférence E B F (a), l’on aura 2ab pour <lb/>la ſurface que la demi-circonférence aura décrite; </s> <s xml:id="echoid-s14652" xml:space="preserve">ce qui eſt <lb/>évident: </s> <s xml:id="echoid-s14653" xml:space="preserve">car comme cette ſurface eſt ici celle d’une ſphere, <lb/>& </s> <s xml:id="echoid-s14654" xml:space="preserve">que la ſurface d’une ſphere eſt égale au produit du diametre <lb/>du grand cercle par la circonférence du même cercle (art. </s> <s xml:id="echoid-s14655" xml:space="preserve">574), <lb/>toute la circonférence étant ici 2a, & </s> <s xml:id="echoid-s14656" xml:space="preserve">le diametre b, la ſurface <lb/>ſera toujours 2ab.</s> <s xml:id="echoid-s14657" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1163" type="section" level="1" n="846"> <head xml:id="echoid-head1022" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s14658" xml:space="preserve">Je viens d’en dire aſſez pour faire voir que dès qu’on aura le <lb/>centre de gravité d’une ligne droite ou courbe, on trouvera <lb/>toujours la ſurface dont elle aura été la génératrice, & </s> <s xml:id="echoid-s14659" xml:space="preserve">que <lb/>rien au monde ne ſeroit plus beau que ce principe, ſi on avoit <lb/>autant de facilité à trouver le centre de gravité de ces lignes, <lb/>qu’on en a à trouver la valeur des ſurfaces qu’elles décrivent. <lb/></s> <s xml:id="echoid-s14660" xml:space="preserve">Ainſi ayant ſatisfait à mon premier deſſein, je vais remplir le <lb/>ſecond, en montrant comment on peut auſſi, par les centres <lb/>de gravité des plans générateurs, trouver la ſolidité des corps <lb/>qu’ils auront décrits.</s> <s xml:id="echoid-s14661" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1164" type="section" level="1" n="847"> <head xml:id="echoid-head1023" xml:space="preserve">PROPOSITION XXIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14662" xml:space="preserve">867. </s> <s xml:id="echoid-s14663" xml:space="preserve">Si l’on a un rectangle A F, qui faſſe une circonvolution <lb/> <anchor type="note" xlink:label="note-0514-01a" xlink:href="note-0514-01"/> autour de l’axe E F, je dis que la ſolidité du corps qu’il décrira <lb/>ſera égale au produit du plan A F par la circonférence, qui auroit <lb/>pour rayon la ligne C D, tirée du centre de gravité C, perpendi-<lb/>culaire ſur l’axe E F.</s> <s xml:id="echoid-s14664" xml:space="preserve"/> </p> <div xml:id="echoid-div1164" type="float" level="2" n="1"> <note position="left" xlink:label="note-0514-01" xlink:href="note-0514-01a" xml:space="preserve">Figure 284.</note> </div> <p> <s xml:id="echoid-s14665" xml:space="preserve">Comme ce ſolide ſera un cylindre, nous ſuppoſerons que <lb/>c’eſt le cylindre A G: </s> <s xml:id="echoid-s14666" xml:space="preserve">ainſi nommant l’axe E F, a; </s> <s xml:id="echoid-s14667" xml:space="preserve">la ligne <lb/>A E, b; </s> <s xml:id="echoid-s14668" xml:space="preserve">la ligne C D ſera {b/2}, puiſqu’elle eſt la moitié de A E; <lb/></s> <s xml:id="echoid-s14669" xml:space="preserve">& </s> <s xml:id="echoid-s14670" xml:space="preserve">ſi l’on nomme la circonférence du rayon E A, c; </s> <s xml:id="echoid-s14671" xml:space="preserve">celle du <lb/>rayon C D ſera {c/2}.</s> <s xml:id="echoid-s14672" xml:space="preserve"/> </p> <pb file="0515" n="529"/> <pb file="0515a" n="530"/> <figure> <image file="0515a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0515a-01"/> </figure> <pb file="0516" n="531"/> <pb file="0517" n="532"/> <pb file="0517a" n="533"/> <figure> <image file="0517a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0517a-01"/> </figure> <pb file="0518" n="534"/> <pb file="0519" n="535"/> <pb file="0519a" n="536"/> <figure> <image file="0519a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0519a-01"/> </figure> <pb file="0520" n="537"/> <pb file="0521" n="538"/> <pb file="0521a" n="539"/> <figure> <image file="0521a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0521a-01"/> </figure> <pb file="0522" n="540"/> <pb file="0523" n="541"/> <pb file="0523a" n="542"/> <figure> <image file="0523a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0523a-01"/> </figure> <pb file="0524" n="543"/> <pb file="0525" n="544"/> <pb file="0525a" n="545"/> <figure> <image file="0525a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0525a-01"/> </figure> <pb file="0526" n="546"/> <pb o="449" file="0527" n="547" rhead="DE MATHÉMATIQUE. Liv. XII."/> <p> <s xml:id="echoid-s14673" xml:space="preserve">Cela poſé, A E x E F (ab) ſera la valeur du plan généra-<lb/>teur, qui étant multiplié par la circonférence du rayon C D <lb/>({c/2}), doit être {abc/2} pour la valeur du ſolide, formé par la cir-<lb/>convolution du plan A F; </s> <s xml:id="echoid-s14674" xml:space="preserve">ce qui eſt évident: </s> <s xml:id="echoid-s14675" xml:space="preserve">car comme ce <lb/>ſolide, ou autrement le cylindre A G, eſt égal au produit du <lb/>cercle de ſa baſe par l’axe E F (art. </s> <s xml:id="echoid-s14676" xml:space="preserve">812), on voit que la ſu-<lb/>perficie de ce cercle étant {bc/2}, ſi on la multiplie par l’axe E F, <lb/>on aura encore {abc/2}.</s> <s xml:id="echoid-s14677" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1166" type="section" level="1" n="848"> <head xml:id="echoid-head1024" xml:space="preserve">PROPOSITION XXIV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14678" xml:space="preserve">868. </s> <s xml:id="echoid-s14679" xml:space="preserve">Si l’on a un triangle iſoſcele E B F, dont le centre de <lb/> <anchor type="note" xlink:label="note-0527-01a" xlink:href="note-0527-01"/> gravité ſoit le point C, je dis que ſi ce triangle fait une circon-<lb/>volution autour de l’axe E F, le ſolide qu’il décrira ſera égal au <lb/>produit du plan générateur par la circonférence du cercle, qui au-<lb/>roit pour rayon la ligne C D, tirée du centre de gravité perpendi-<lb/>culaire ſur l’axe.</s> <s xml:id="echoid-s14680" xml:space="preserve"/> </p> <div xml:id="echoid-div1166" type="float" level="2" n="1"> <note position="right" xlink:label="note-0527-01" xlink:href="note-0527-01a" xml:space="preserve">Figure 291 <lb/>& 292.</note> </div> <p> <s xml:id="echoid-s14681" xml:space="preserve">Remarquez que le ſolide I K G H qu’aura décrit le triangle <lb/>E B F, eſt compoſé de deux cônes K G H & </s> <s xml:id="echoid-s14682" xml:space="preserve">K I H, & </s> <s xml:id="echoid-s14683" xml:space="preserve">qu’il <lb/>s’agit de faire voir que le produit du plan E B F, par la cir-<lb/>conférence du rayon C D, eſt égal à ces deux cônes: </s> <s xml:id="echoid-s14684" xml:space="preserve">mais pour <lb/>cela, il faut être prévenu que le centre de gravité du triangle <lb/>iſoſcele eſt un point tel que C, pris dans la perpendiculaire <lb/>B D à une diſtance C D de la baſe, qui eſt le tiers de la per-<lb/>pendiculaire. </s> <s xml:id="echoid-s14685" xml:space="preserve">Ainſi nommant la ligne E F, a; </s> <s xml:id="echoid-s14686" xml:space="preserve">la ligne B D, b; <lb/></s> <s xml:id="echoid-s14687" xml:space="preserve">& </s> <s xml:id="echoid-s14688" xml:space="preserve">c la circonférence dont elle ſeroit le rayon, C D étant le <lb/>tiers de B D, la circonférence dont elle ſeroit le rayon ſera{c/3}.</s> <s xml:id="echoid-s14689" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14690" xml:space="preserve">Cela poſé, le triangle E B F ſera {ab/2}, qui étant multiplié <lb/>par {c/3}, l’on aura {abc/6} pour la valeur du ſolide K G H I; </s> <s xml:id="echoid-s14691" xml:space="preserve">ce qui <lb/>eſt évident: </s> <s xml:id="echoid-s14692" xml:space="preserve">car ſi l’on cherche par la voie ordinaire la ſolidité <lb/>du cône K G H, dont le plan générateur eſt le triangle E B D, <lb/>la ligne B D étant le rayon du cercle de la baſe, ſa valeur ſera <lb/>{bc/2}, qui étant multipliée par le tiers de la ligne E D (art. </s> <s xml:id="echoid-s14693" xml:space="preserve">556), <pb o="450" file="0528" n="548" rhead="NOUVEAU COURS"/> ou par la ſixieme partie de E F ({a/6}), donnera {abc/12} pour la va-<lb/>leur du cône; </s> <s xml:id="echoid-s14694" xml:space="preserve">& </s> <s xml:id="echoid-s14695" xml:space="preserve">par conſéquent {2abc/12}, ou bien {abc/6} pour la va-<lb/>leur des deux cônes, c’eſt-à-dire du ſolide K G H I, qui ſe <lb/>trouve la même que la précédente.</s> <s xml:id="echoid-s14696" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14697" xml:space="preserve">869. </s> <s xml:id="echoid-s14698" xml:space="preserve">Mais ſi le triangle E B F faiſoit une circonvolution <lb/>autour de l’axe L M, il décrira un ſolide d’une autre figure, <lb/>dont le rapport avec le précédent ſera comme la ligne B C <lb/>eſt à la ligne C D: </s> <s xml:id="echoid-s14699" xml:space="preserve">car pour trouver la valeur de ce ſolide, il <lb/>faudra multiplier le plan E B F par la circonférence du cercle, <lb/>qui auroit pour rayon B C: </s> <s xml:id="echoid-s14700" xml:space="preserve">& </s> <s xml:id="echoid-s14701" xml:space="preserve">comme l’un & </s> <s xml:id="echoid-s14702" xml:space="preserve">l’autre ſolide <lb/>aura pour baſe le même plan E B F, ils ſeront dans la même <lb/>raiſon que leurs hauteurs, c’eſt-à-dire dans la raiſon des cir-<lb/>conférences des rayons B C & </s> <s xml:id="echoid-s14703" xml:space="preserve">C D, qui ſont dans la même <lb/>raiſon que ces rayons.</s> <s xml:id="echoid-s14704" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14705" xml:space="preserve">L’on peut remarquer encore qu’ayant un triangle rectangle <lb/>E B D, qui faſſe une circonvolution autour du côté E D, il <lb/>décrira un cône dont on trouvera la valeur, en multipliant <lb/>le triangle B E D par la circonférence du cercle, qui auroit <lb/>pour rayon la ligne C D égale au tiers de la baſe B D: </s> <s xml:id="echoid-s14706" xml:space="preserve">car mul-<lb/>tipliant B D (b) par la moitié de E D ({a/4}), l’on aura {ab/4} pour <lb/>la ſuperficie du triangle, qui étant multiplié par {c/d}, donnera {abc/12}.</s> <s xml:id="echoid-s14707" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14708" xml:space="preserve">Et ſi le triangle E B D faiſoit une circonvolution autour de <lb/> <anchor type="note" xlink:label="note-0528-01a" xlink:href="note-0528-01"/> l’axe H B, il décriroit l’entonnoir F G B D E, qui ſeroit dou-<lb/>ble du cône: </s> <s xml:id="echoid-s14709" xml:space="preserve">car comme le cône & </s> <s xml:id="echoid-s14710" xml:space="preserve">l’entonnoir ont le même <lb/>plan générateur, ils ſeront dans la raiſon des circonférences <lb/>décrites par le centre de gravité C: </s> <s xml:id="echoid-s14711" xml:space="preserve">& </s> <s xml:id="echoid-s14712" xml:space="preserve">comme le rayon B C <lb/>eſt double de C D, l’entonnoir ſera double du cône; </s> <s xml:id="echoid-s14713" xml:space="preserve">ce qui <lb/>fait voir qu’un cône eſt le tiers d’un cylindre de même baſe <lb/>& </s> <s xml:id="echoid-s14714" xml:space="preserve">de même hauteur.</s> <s xml:id="echoid-s14715" xml:space="preserve"/> </p> <div xml:id="echoid-div1167" type="float" level="2" n="2"> <note position="left" xlink:label="note-0528-01" xlink:href="note-0528-01a" xml:space="preserve">Figure 295.</note> </div> <p> <s xml:id="echoid-s14716" xml:space="preserve">870. </s> <s xml:id="echoid-s14717" xml:space="preserve">Enfin ſi l’on avoit un triangle B A D, dont le point C <lb/> <anchor type="note" xlink:label="note-0528-02a" xlink:href="note-0528-02"/> fût le centre de gravité du triangle double de celui-ci, que l’on <lb/>prolongeât la ligne A D indéfiniment juſqu’aux points E & </s> <s xml:id="echoid-s14718" xml:space="preserve">F, <lb/>& </s> <s xml:id="echoid-s14719" xml:space="preserve">que l’on fît faire une circonvolution au triangle B A D au-<lb/>tour de l’axe G F, le ſolide qu’il décriroit ſeroit égal au pro-<lb/>duit du plan B A D par la circonférence du cercle, qui auroit <lb/>pour rayon la ligne C F, qui eſt la diſtance du centre de gra- <pb o="451" file="0529" n="549" rhead="DE MATHÉMATIQUE. Liv. XII."/> vité C à l’axe F G; </s> <s xml:id="echoid-s14720" xml:space="preserve">& </s> <s xml:id="echoid-s14721" xml:space="preserve">ſi le triangle, au lieu de faire une cir-<lb/>convolution autour de l’axe G F, en faiſoit une autre autour <lb/>de l’axe H E, le ſolide qu’il décriroit ſeroit égal au produit du <lb/>plan A B D par la circonférence du cercle, qui auroit pour <lb/>rayon la ligne C E, tirée du centre de gravité à l’axe, & </s> <s xml:id="echoid-s14722" xml:space="preserve">ces <lb/>deux ſolides ſeroient dans la raiſon des rayons C F & </s> <s xml:id="echoid-s14723" xml:space="preserve">C E.</s> <s xml:id="echoid-s14724" xml:space="preserve"/> </p> <div xml:id="echoid-div1168" type="float" level="2" n="3"> <note position="left" xlink:label="note-0528-02" xlink:href="note-0528-02a" xml:space="preserve">Figure 293.</note> </div> <p> <s xml:id="echoid-s14725" xml:space="preserve">Je laiſſe au lecteur le plaiſir d’en chercher la démonſtra-<lb/>tion; </s> <s xml:id="echoid-s14726" xml:space="preserve">& </s> <s xml:id="echoid-s14727" xml:space="preserve">je me contenterai de dire ſeulement que le ſolide, <lb/>formé par la circonvolution du triangle A B D autour de l’axe <lb/>G F, eſt ſemblable à celui dont nous avons parlé dans l’arti-<lb/>cle 820, c’eſt-à-dire qu’il fait la différence d’un cylindre, <lb/>duquel on auroit ôté un cône tronqué; </s> <s xml:id="echoid-s14728" xml:space="preserve">& </s> <s xml:id="echoid-s14729" xml:space="preserve">que le ſolide, formé <lb/>par la circonvolution du triangle A B D autour de l’axe H E, <lb/>eſt auſſi ſemblable à celui de l’art. </s> <s xml:id="echoid-s14730" xml:space="preserve">819, c’eſt-à-dire qu’il fait la <lb/>différence d’un cône tronqué, duquel on auroit ôté un cylin-<lb/>dre: </s> <s xml:id="echoid-s14731" xml:space="preserve">& </s> <s xml:id="echoid-s14732" xml:space="preserve">comme la maniere de trouver la valeur de ces ſolides <lb/>de la façon que je viens de dire, eſt plus aiſée que celle des <lb/>articles 819, 820, l’on pourra s’en ſervir pour toiſer la ma-<lb/>çonnerie, compriſe par le talud de l’orillon, du flanc concave, <lb/>& </s> <s xml:id="echoid-s14733" xml:space="preserve">de l’arrondiſſement de la contreſcarpe.</s> <s xml:id="echoid-s14734" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1170" type="section" level="1" n="849"> <head xml:id="echoid-head1025" xml:space="preserve">PROPOSITION XXV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14735" xml:space="preserve">871. </s> <s xml:id="echoid-s14736" xml:space="preserve">Si on a un demi-cercle E B F, dont le centre de gravité <lb/> <anchor type="note" xlink:label="note-0529-01a" xlink:href="note-0529-01"/> ſoit le point I, & </s> <s xml:id="echoid-s14737" xml:space="preserve">que de ce point l’on abaiſſe la perpendiculaire <lb/>I D, je dis que le ſolide formé par la circonvolution du demi-<lb/>cercle E B F autour de l’axe E F, qui ſera une ſphere, ſera égal au <lb/>produit du plan E B F par la circonférence du cercle, qui auroit <lb/>pour rayon la ligne I D.</s> <s xml:id="echoid-s14738" xml:space="preserve"/> </p> <div xml:id="echoid-div1170" type="float" level="2" n="1"> <note position="right" xlink:label="note-0529-01" xlink:href="note-0529-01a" xml:space="preserve">Figure 294.</note> </div> <p> <s xml:id="echoid-s14739" xml:space="preserve">Il faut être prévenu que la ligne I D, qui marque la diſ-<lb/>tance du centre de gravité I au centre D du demi-cercle, <lb/>eſt une quatrieme proportionnelle à la moitié de la circonfé-<lb/>rence E B F au rayon D E, & </s> <s xml:id="echoid-s14740" xml:space="preserve">aux deux tiers du même rayon. <lb/></s> <s xml:id="echoid-s14741" xml:space="preserve">Ainſi nommant la demi-circonférence E B F, a; </s> <s xml:id="echoid-s14742" xml:space="preserve">le rayon <lb/>D E, b; </s> <s xml:id="echoid-s14743" xml:space="preserve">la moitié de la circonférence E B F ſera {a/2}; </s> <s xml:id="echoid-s14744" xml:space="preserve">& </s> <s xml:id="echoid-s14745" xml:space="preserve">les deux <lb/>tiers du rayon D E ſeront {2b/3}: </s> <s xml:id="echoid-s14746" xml:space="preserve">on trouvera la ligne D I, en di- <pb o="452" file="0530" n="550" rhead="NOUVEAU COURS"/> ſant: </s> <s xml:id="echoid-s14747" xml:space="preserve">Comme {a/2} eſt à b, ainſi {2b/3} eſt à D I, qui ſera {4bb/3a}: </s> <s xml:id="echoid-s14748" xml:space="preserve">& </s> <s xml:id="echoid-s14749" xml:space="preserve">com-<lb/>me nous avons beſoin de la circonférence du rayon D I, on <lb/>dira: </s> <s xml:id="echoid-s14750" xml:space="preserve">Si le rayon D E (b) donne 2a pour ſa circonférence, <lb/>que donnera le rayon D I ({4bb/3a}) pour ſa circonférence, qui <lb/>ſera {8abb/3a}, ou bien {8bb/3}? </s> <s xml:id="echoid-s14751" xml:space="preserve">Or ſi l’on multiplie cette circonférence <lb/>par la valeur du demi-cercle E B F ({ab/2}), l’on aura {8abb/6}, ou <lb/>bien {4abb/3} pour la valeur du ſolide; </s> <s xml:id="echoid-s14752" xml:space="preserve">ce qui eſt aiſé à prouver: <lb/></s> <s xml:id="echoid-s14753" xml:space="preserve">car comme une ſphere eſt égale au produit de quatre fois ſon <lb/>grand cercle par le tiers du rayon (art. </s> <s xml:id="echoid-s14754" xml:space="preserve">568 & </s> <s xml:id="echoid-s14755" xml:space="preserve">570), la ſu-<lb/>perficie du demi-cercle étant {ab/2}, celle de tout le cercle ſera ab, <lb/>qui étant multipliée par 4, donnera 4ab pour la valeur des <lb/>quatre cercles; </s> <s xml:id="echoid-s14756" xml:space="preserve">& </s> <s xml:id="echoid-s14757" xml:space="preserve">ſi l’on multiplie cette quantité par le tiers <lb/>du rayon, c’eſt-à-dire par {b/3}, l’on aura {4abb/3} pour la valeur de <lb/>la ſphere, qui eſt la même que celle que nous venons de <lb/>trouver.</s> <s xml:id="echoid-s14758" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14759" xml:space="preserve">Mais ſi le demi-cercle E B F faiſoit une circonvolution au-<lb/>tour de la tangente G A, parallele au diametre E F, il décri-<lb/>roit un ſolide, dont on trouvera la valeur, en multipliant le <lb/>demi-cercle par la circonférence, qui auroit pour rayon la <lb/>ligne I B, qui eſt la diſtance du centre de gravité I à l’axe <lb/>G A, & </s> <s xml:id="echoid-s14760" xml:space="preserve">ſi le demi-cercle fait encore une circonvolution au-<lb/>tour de l’axe A H perpendiculaire à E F, il décrira une eſpece <lb/>de bourlet, dont on trouvera la valeur, en multipliant le <lb/>demi-cercle par la circonférence du rayon I K, ou du rayon <lb/>D F, qui eſt la même choſe; </s> <s xml:id="echoid-s14761" xml:space="preserve">& </s> <s xml:id="echoid-s14762" xml:space="preserve">pour lors le ſolide décrit par <lb/>le demi-cercle autour de l’axe E F, ſera au ſolide décrit au-<lb/>tour de l’axe G A, comme le rayon I D eſt au rayon I B, & </s> <s xml:id="echoid-s14763" xml:space="preserve"><lb/>le ſolide formé par la circonvolution du demi-cercle autour <lb/>de l’axe E F, ſera à celui qui aura été formé par une circon-<lb/>volution du même demi-cercle autour de l’axe A H, comme <lb/>le rayon I D eſt au rayon I K ou D F.</s> <s xml:id="echoid-s14764" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1172" type="section" level="1" n="850"> <head xml:id="echoid-head1026" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s14765" xml:space="preserve">Je n’ai point donné la maniere de trouver les centres de <pb o="453" file="0531" n="551" rhead="DE MATHEMATIQUE. Liv. XII."/> gravité, parce que c’eût été m’écarter de mon ſujet, n’ayant <lb/>eu en vue que d’exercer l’eſprit des Commençans, & </s> <s xml:id="echoid-s14766" xml:space="preserve">leur <lb/>faire ſentir le prix de ce principe général, par le moyen duquel <lb/>on peut, indépendamment de ce que nous avons enſeigné <lb/>dans le huitieme Livre de la premiere Partie, réſoudre une <lb/>quantité de problêmes, dès qu’on a les centres de gravité des <lb/>figures génératrices, que l’on ne peut trouver d’une façon gé-<lb/>nérale, qu’avec le ſecours du calcul intégral: </s> <s xml:id="echoid-s14767" xml:space="preserve">cependant on <lb/>peut voir ce qu’en a dit M. </s> <s xml:id="echoid-s14768" xml:space="preserve">Ozanam dans ſon Traité de <lb/>Méchanique, où il trouve les centres de gravité de pluſieurs <lb/>figures par la Géométrie ordinaire.</s> <s xml:id="echoid-s14769" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1173" type="section" level="1" n="851"> <head xml:id="echoid-head1027" xml:space="preserve">Fin du douzieme Livre.</head> <figure> <image file="0531-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0531-01"/> </figure> <pb o="454" file="0532" n="552"/> <figure> <image file="0532-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0532-01"/> </figure> </div> <div xml:id="echoid-div1174" type="section" level="1" n="852"> <head xml:id="echoid-head1028" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head1029" xml:space="preserve">LIVRE TREIZIEME, <lb/>Où l’on applique la Géométrie à la diviſion des Champs, <lb/>& à l’uſage du Compas de proportion.</head> <head xml:id="echoid-head1030" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14770" xml:space="preserve">872. </s> <s xml:id="echoid-s14771" xml:space="preserve">DIviſer un triangle en autant de parties égales qu’on vou-<lb/> <anchor type="note" xlink:label="note-0532-01a" xlink:href="note-0532-01"/> dra, par des lignes tirées de l’angle oppoſé à la baſe.</s> <s xml:id="echoid-s14772" xml:space="preserve"/> </p> <div xml:id="echoid-div1174" type="float" level="2" n="1"> <note position="left" xlink:label="note-0532-01" xlink:href="note-0532-01a" xml:space="preserve">Figure 296.</note> </div> <p> <s xml:id="echoid-s14773" xml:space="preserve">Pour diviſer un triangle A B C en trois parties égales par <lb/>des lignes tirées de l’angle oppoſé à la baſe, il faut diviſer la <lb/>baſe A C en trois parties égales aux points D & </s> <s xml:id="echoid-s14774" xml:space="preserve">E, tirer les <lb/>lignes B D & </s> <s xml:id="echoid-s14775" xml:space="preserve">B E, & </s> <s xml:id="echoid-s14776" xml:space="preserve">le triangle ſera diviſé en trois triangles <lb/>égaux, puiſque ces triangles ont des baſes égales, & </s> <s xml:id="echoid-s14777" xml:space="preserve">qu’ils ont <lb/>la même hauteur.</s> <s xml:id="echoid-s14778" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1176" type="section" level="1" n="853"> <head xml:id="echoid-head1031" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14779" xml:space="preserve">873. </s> <s xml:id="echoid-s14780" xml:space="preserve">Diviſer un triangle en deux parties égales par une ligne <lb/> <anchor type="note" xlink:label="note-0532-02a" xlink:href="note-0532-02"/> tirée d’un point donné ſur un des côtés du triangle.</s> <s xml:id="echoid-s14781" xml:space="preserve"/> </p> <div xml:id="echoid-div1176" type="float" level="2" n="1"> <note position="left" xlink:label="note-0532-02" xlink:href="note-0532-02a" xml:space="preserve">Figure 297.</note> </div> <p> <s xml:id="echoid-s14782" xml:space="preserve">L’on demande qu’on diviſe le triangle A B C en deux par-<lb/>ties égales par une ligne tirée du point D, parce que l’on ſup-<lb/>poſe que ce triangle eſt un champ, ſur le bord duquel eſt un <pb o="455" file="0533" n="553" rhead="NOUVEAU COURS DE MATH. Liv. XIII."/> lieu avantageux au point D, qui doit être commun à chacun <lb/>de ceux qui auront part au champ.</s> <s xml:id="echoid-s14783" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14784" xml:space="preserve">Pour réſoudre ce problême, il faut diviſer la baſe A C en <lb/>deux parties égales au point E, & </s> <s xml:id="echoid-s14785" xml:space="preserve">tirer de ce point les lignes <lb/>E B & </s> <s xml:id="echoid-s14786" xml:space="preserve">E D; </s> <s xml:id="echoid-s14787" xml:space="preserve">puis du point B tirer la ligne B F parallele à D E; <lb/></s> <s xml:id="echoid-s14788" xml:space="preserve">enfin tirer la ligne E D, qui diviſera le triangle en deux par-<lb/>ties égales B D F A & </s> <s xml:id="echoid-s14789" xml:space="preserve">D F C.</s> <s xml:id="echoid-s14790" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14791" xml:space="preserve">Pour prouver cette opération, conſidérez que le triangle <lb/>A B E eſt la moitié de tout le triangle A B C; </s> <s xml:id="echoid-s14792" xml:space="preserve">& </s> <s xml:id="echoid-s14793" xml:space="preserve">qu’à cauſe <lb/>des paralleles B F & </s> <s xml:id="echoid-s14794" xml:space="preserve">D E, le triangle B F D eſt égal au trian-<lb/>gle B E F; </s> <s xml:id="echoid-s14795" xml:space="preserve">d’où il s’enſuit que le triangle O F E, que l’on a <lb/>retranché du triangle B E A, eſt égal au triangle O D B, que <lb/>l’on a retranché du triangle E B C: </s> <s xml:id="echoid-s14796" xml:space="preserve">ce qui fait voir que le tra-<lb/>peze B D F A eſt égal au triangle F D C.</s> <s xml:id="echoid-s14797" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1178" type="section" level="1" n="854"> <head xml:id="echoid-head1032" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14798" xml:space="preserve">874. </s> <s xml:id="echoid-s14799" xml:space="preserve">Diviſer un triangle en trois parties égales par des lignes <lb/> <anchor type="note" xlink:label="note-0533-01a" xlink:href="note-0533-01"/> tirées d’un point pris ſur un de ſes côtés.</s> <s xml:id="echoid-s14800" xml:space="preserve"/> </p> <div xml:id="echoid-div1178" type="float" level="2" n="1"> <note position="right" xlink:label="note-0533-01" xlink:href="note-0533-01a" xml:space="preserve">Figure 298.</note> </div> <p> <s xml:id="echoid-s14801" xml:space="preserve">Pour diviſer le triangle A B C en trois parties égales par des <lb/>lignes tirées du point D, il faut partager le côté A C en trois <lb/>parties égales aux points E & </s> <s xml:id="echoid-s14802" xml:space="preserve">F; </s> <s xml:id="echoid-s14803" xml:space="preserve">enſuite tirer la ligne D B, à <lb/>laquelle il faut mener des points E & </s> <s xml:id="echoid-s14804" xml:space="preserve">F les paralleles E H & </s> <s xml:id="echoid-s14805" xml:space="preserve"><lb/>F G: </s> <s xml:id="echoid-s14806" xml:space="preserve">& </s> <s xml:id="echoid-s14807" xml:space="preserve">ſi l’on tire du point D les lignes D G & </s> <s xml:id="echoid-s14808" xml:space="preserve">D H, on aura <lb/>le triangle diviſé en trois parties égales A H D, D H B G, <lb/>& </s> <s xml:id="echoid-s14809" xml:space="preserve">D G C.</s> <s xml:id="echoid-s14810" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14811" xml:space="preserve">Pour le prouver, il ne faut que tirer les lignes B E & </s> <s xml:id="echoid-s14812" xml:space="preserve">B F, <lb/>qui diviſeront le triangle en trois autres triangles égaux. </s> <s xml:id="echoid-s14813" xml:space="preserve">Or <lb/>comme le triangle A B E eſt égal au triangle A H D, à cauſe <lb/>des paralleles H E & </s> <s xml:id="echoid-s14814" xml:space="preserve">B D: </s> <s xml:id="echoid-s14815" xml:space="preserve">on verra par la même raiſon que le <lb/>triangle D G C eſt égal au triangle B F C, & </s> <s xml:id="echoid-s14816" xml:space="preserve">que par conſé-<lb/>quent ils ſont chacun le tiers de toute la figure.</s> <s xml:id="echoid-s14817" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1180" type="section" level="1" n="855"> <head xml:id="echoid-head1033" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14818" xml:space="preserve">875. </s> <s xml:id="echoid-s14819" xml:space="preserve">Diviſer un triangle en trois parties égales par des lignes <lb/>tirées dans les trois angles.</s> <s xml:id="echoid-s14820" xml:space="preserve"/> </p> <pb o="456" file="0534" n="554" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s14821" xml:space="preserve">On demande un point dans le triangle A B C, duquel ayant <lb/> <anchor type="note" xlink:label="note-0534-01a" xlink:href="note-0534-01"/> tiré des lignes dans les angles, elles diviſent le triangle en trois <lb/> <anchor type="note" xlink:label="note-0534-02a" xlink:href="note-0534-02"/> parties égales.</s> <s xml:id="echoid-s14822" xml:space="preserve"/> </p> <div xml:id="echoid-div1180" type="float" level="2" n="1"> <note position="left" xlink:label="note-0534-01" xlink:href="note-0534-01a" xml:space="preserve">Pl. XXII.</note> <note position="left" xlink:label="note-0534-02" xlink:href="note-0534-02a" xml:space="preserve">Figure 299.</note> </div> <p> <s xml:id="echoid-s14823" xml:space="preserve">Pour réſoudre le problême, il faut faire la ligne A F égale <lb/>au tiers de la baſe A C, du point F tirer la ligne F E parallele <lb/>au côté A B, & </s> <s xml:id="echoid-s14824" xml:space="preserve">diviſer la parallele F E en deux également au <lb/>point D, ce point ſera celui qu’on cherche: </s> <s xml:id="echoid-s14825" xml:space="preserve">car ayant tiré dans <lb/>les angles du triangle les lignes D B, D A & </s> <s xml:id="echoid-s14826" xml:space="preserve">D C, elles diviſe-<lb/>ront le triangle en trois parties égales.</s> <s xml:id="echoid-s14827" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14828" xml:space="preserve">Pour le prouver, je tire la ligne B F, qui me donne le <lb/>triangle B A F, qui eſt le tiers de toute la figure: </s> <s xml:id="echoid-s14829" xml:space="preserve">& </s> <s xml:id="echoid-s14830" xml:space="preserve">comme <lb/>ce triangle eſt égal au triangle A D B, à cauſe des paralleles, <lb/>il s’enſuit que ce dernier triangle eſt auſſi le tiers de la figure: <lb/></s> <s xml:id="echoid-s14831" xml:space="preserve">& </s> <s xml:id="echoid-s14832" xml:space="preserve">comme les triangles A D C & </s> <s xml:id="echoid-s14833" xml:space="preserve">B D C ſont égaux entr’eux, <lb/>comme il eſt facile de le voir, il s’enſuit que le problême eſt <lb/>réſolu.</s> <s xml:id="echoid-s14834" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1182" type="section" level="1" n="856"> <head xml:id="echoid-head1034" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14835" xml:space="preserve">876. </s> <s xml:id="echoid-s14836" xml:space="preserve">Diviſer un triangle en deux parties égales par des lignes <lb/> <anchor type="note" xlink:label="note-0534-03a" xlink:href="note-0534-03"/> tirées d’un point donné à volonté dans la ſuperficie du triangle.</s> <s xml:id="echoid-s14837" xml:space="preserve"/> </p> <div xml:id="echoid-div1182" type="float" level="2" n="1"> <note position="left" xlink:label="note-0534-03" xlink:href="note-0534-03a" xml:space="preserve">Figure 300.</note> </div> <p> <s xml:id="echoid-s14838" xml:space="preserve">Pour diviſer en deux également le triangle A B C par des <lb/>lignes tirées du point donné F, il faut diviſer la baſe A C en <lb/>deux également au point D, & </s> <s xml:id="echoid-s14839" xml:space="preserve">tirer la ligne D F, à laquelle il <lb/>faut mener une parallele B E; </s> <s xml:id="echoid-s14840" xml:space="preserve">après quoi l’on n’aura qu’à tirer <lb/>les lignes E F & </s> <s xml:id="echoid-s14841" xml:space="preserve">F B pour avoir la figure A B F E égale à la <lb/>figure B F E C.</s> <s xml:id="echoid-s14842" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14843" xml:space="preserve">Pour le prouver, tirez la ligne B D, & </s> <s xml:id="echoid-s14844" xml:space="preserve">conſidérez qu’à cauſe <lb/>des paralleles le triangle B F E eſt égal au triangle B D E, & </s> <s xml:id="echoid-s14845" xml:space="preserve"><lb/>que par conſéquent ce qu’on a retranché d’une part eſt égal à <lb/>ce que l’on a ajouté de l’autre dans les deux triangles A B D <lb/>& </s> <s xml:id="echoid-s14846" xml:space="preserve">D B C.</s> <s xml:id="echoid-s14847" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1184" type="section" level="1" n="857"> <head xml:id="echoid-head1035" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14848" xml:space="preserve">877. </s> <s xml:id="echoid-s14849" xml:space="preserve">Diviſer un triangle en deux parties égales par une ligne <lb/> <anchor type="note" xlink:label="note-0534-04a" xlink:href="note-0534-04"/> parallele à la baſe.</s> <s xml:id="echoid-s14850" xml:space="preserve"/> </p> <div xml:id="echoid-div1184" type="float" level="2" n="1"> <note position="left" xlink:label="note-0534-04" xlink:href="note-0534-04a" xml:space="preserve">Figure 301.</note> </div> <p> <s xml:id="echoid-s14851" xml:space="preserve">Pour diviſer le triangle A B C par une ligne D E parallele à <pb o="457" file="0535" n="555" rhead="DE MATHÉMATIQUE. Liv. XIII."/> la baſe, il faut partager en deux également l’un des autres <lb/>côtés, par exemple, le côté B C; </s> <s xml:id="echoid-s14852" xml:space="preserve">puis chercher une moyenne <lb/>proportionnelle entre tout le côté B C, & </s> <s xml:id="echoid-s14853" xml:space="preserve">ſa moitié B F: </s> <s xml:id="echoid-s14854" xml:space="preserve">& </s> <s xml:id="echoid-s14855" xml:space="preserve"><lb/>ſuppoſant que la ligne B E ſoit égale à la moyenne, que l’on <lb/>aura trouvée, on n’aura qu’à mener du point E la parallele <lb/>E D à la baſe A C, pour avoir réſolu le problême.</s> <s xml:id="echoid-s14856" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14857" xml:space="preserve">Pour le prouver, faites attention que les lignes B C, B E, <lb/>B F étant proportionnelles, il y aura même raiſon du quarré <lb/>fait ſur la ligne B C au quarré fait ſur la ligne B E, que de <lb/>la premiere ligne B C à la derniere B F (art. </s> <s xml:id="echoid-s14858" xml:space="preserve">497). </s> <s xml:id="echoid-s14859" xml:space="preserve">Or comme <lb/>les triangles ſont dans la même raiſon que les quarrés de leurs <lb/>côtés homologues, le triangle B A C ſera double du triangle <lb/>B D E, puiſque le quarré du côté B C eſt double du quarré <lb/>du côté B E, à cauſe que la ligne B C eſt double de la ligne B F.</s> <s xml:id="echoid-s14860" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14861" xml:space="preserve">Si l’on vouloit diviſer un triangle en trois parties égales par <lb/>des lignes tirées paralleles à la baſe, il faudroit chercher d’a-<lb/>bord une moyenne proportionnelle entre l’un des côtés du <lb/>triangle, & </s> <s xml:id="echoid-s14862" xml:space="preserve">les deux tiers du même côté; </s> <s xml:id="echoid-s14863" xml:space="preserve">& </s> <s xml:id="echoid-s14864" xml:space="preserve">ayant déterminé <lb/>la longueur de cette moyenne ſur le côté qu’on aura diviſé, <lb/>l’on tirera une parallele de l’extrêmité de cette ligne à la baſe: <lb/></s> <s xml:id="echoid-s14865" xml:space="preserve">on aura un triangle intérieur, qui ſera les deux tiers de celui <lb/>qu’on veut partager en trois: </s> <s xml:id="echoid-s14866" xml:space="preserve">& </s> <s xml:id="echoid-s14867" xml:space="preserve">ſi l’on diviſe le rectangle qui <lb/>contient les deux tiers du grand, en deux également, comme <lb/>on vient de le faire dans la propoſition précédente, tout le <lb/>triangle ſe trouvera diviſé en trois parties égales.</s> <s xml:id="echoid-s14868" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1186" type="section" level="1" n="858"> <head xml:id="echoid-head1036" xml:space="preserve">PROPOSITION VII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14869" xml:space="preserve">878. </s> <s xml:id="echoid-s14870" xml:space="preserve">Diviſer un Trapézoïde en deux parties égales par une <lb/> <anchor type="note" xlink:label="note-0535-01a" xlink:href="note-0535-01"/> ligne parallele à la baſe.</s> <s xml:id="echoid-s14871" xml:space="preserve"/> </p> <div xml:id="echoid-div1186" type="float" level="2" n="1"> <note position="right" xlink:label="note-0535-01" xlink:href="note-0535-01a" xml:space="preserve">Figure 302.</note> </div> <p> <s xml:id="echoid-s14872" xml:space="preserve">Pour diviſer le trapézoïde A B C D par une ligne parallele à <lb/>la baſe, il faut prolonger les deux côtés A B & </s> <s xml:id="echoid-s14873" xml:space="preserve">D C pour <lb/>qu’ils ſe rencontrent au point G, puis élever ſur l’extrêmité <lb/>G la perpendiculaire G H égale à la ligne G B; </s> <s xml:id="echoid-s14874" xml:space="preserve">tirer la ligne <lb/>H A, & </s> <s xml:id="echoid-s14875" xml:space="preserve">décrire ſur cette ligne un demi-cercle, dont il faudra <lb/>diviſer la circonférence en deux également au point I; </s> <s xml:id="echoid-s14876" xml:space="preserve">& </s> <s xml:id="echoid-s14877" xml:space="preserve"><lb/>ayant tiré la ligne I H, on fera G E égal à I H: </s> <s xml:id="echoid-s14878" xml:space="preserve">& </s> <s xml:id="echoid-s14879" xml:space="preserve">ſi par le <lb/>point E l’on mene la parallele E F à la baſe A D, je dis qu’elle <lb/>diviſera le trapézoïde en deux parties égales.</s> <s xml:id="echoid-s14880" xml:space="preserve"/> </p> <pb o="458" file="0536" n="556" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s14881" xml:space="preserve">Pour le prouver, je conſidere que la ligne H A eſt le côté <lb/>du quarré, qui vaut la ſomme des quarrés B G & </s> <s xml:id="echoid-s14882" xml:space="preserve">G A; </s> <s xml:id="echoid-s14883" xml:space="preserve">& </s> <s xml:id="echoid-s14884" xml:space="preserve">que <lb/>la ligne I H eſt le côté d’un quarré qui vaut la moitié du <lb/>quarré H A: </s> <s xml:id="echoid-s14885" xml:space="preserve">par conſéquent le quarré I H ou G E eſt moyenne <lb/>arithmétique entre les quarrés G A & </s> <s xml:id="echoid-s14886" xml:space="preserve">G B. </s> <s xml:id="echoid-s14887" xml:space="preserve">Et comme les <lb/>triangles ſemblables ſont dans la même raiſon que les quarrés <lb/>de leurs côtés homologues, il s’enſuit que les quarrés des côtés <lb/>G B, G E, G A étant en progreſſion arithmétique, les trian-<lb/>gles G B C, G E F, G A D ſont en proportion arithmétique, <lb/>par conſéquent ſe ſurpaſſent également; </s> <s xml:id="echoid-s14888" xml:space="preserve">& </s> <s xml:id="echoid-s14889" xml:space="preserve">comme les gran-<lb/>deurs dont ils ſont ſurpaſſés, ne ſont autre choſe que le tra-<lb/>pézoïde E C, & </s> <s xml:id="echoid-s14890" xml:space="preserve">A F, je conclus que ces trapézoïdes ſont <lb/>égaux, & </s> <s xml:id="echoid-s14891" xml:space="preserve">que par conſéquent le problême eſt réſolu.</s> <s xml:id="echoid-s14892" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1188" type="section" level="1" n="859"> <head xml:id="echoid-head1037" xml:space="preserve">PROPOSITION VIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14893" xml:space="preserve">879. </s> <s xml:id="echoid-s14894" xml:space="preserve">Diviſer un trapeze en deux également par une ligne pa-<lb/> <anchor type="note" xlink:label="note-0536-01a" xlink:href="note-0536-01"/> rallele à l’un de ſes côtés.</s> <s xml:id="echoid-s14895" xml:space="preserve"/> </p> <div xml:id="echoid-div1188" type="float" level="2" n="1"> <note position="left" xlink:label="note-0536-01" xlink:href="note-0536-01a" xml:space="preserve">Figure 303.</note> </div> <p> <s xml:id="echoid-s14896" xml:space="preserve">Pour diviſer le trapeze A B C D par une ligne parallele au <lb/>côté A B, il faut prolonger les côtés B C & </s> <s xml:id="echoid-s14897" xml:space="preserve">A D, tant qu’ils <lb/>ſe rencontrent au point G; </s> <s xml:id="echoid-s14898" xml:space="preserve">puis réduire le trapeze en triangle <lb/>pour avoir le point F: </s> <s xml:id="echoid-s14899" xml:space="preserve">après quoi on diviſera la baſe A F du <lb/>triangle A B F en deux également au point H; </s> <s xml:id="echoid-s14900" xml:space="preserve">on cherchera <lb/>une moyenne proportionnelle entre A G & </s> <s xml:id="echoid-s14901" xml:space="preserve">H G, qui ſera, <lb/>par exemple, I G; </s> <s xml:id="echoid-s14902" xml:space="preserve">& </s> <s xml:id="echoid-s14903" xml:space="preserve">ſi du point I l’on mene la ligne I K pa-<lb/>rallele à A B, elle diviſera le trapeze en deux parties égales <lb/>A B K I & </s> <s xml:id="echoid-s14904" xml:space="preserve">I K C D.</s> <s xml:id="echoid-s14905" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14906" xml:space="preserve">Pour le prouver, remarquez que les triangles A B G & </s> <s xml:id="echoid-s14907" xml:space="preserve"><lb/>I K G ſont ſemblables, & </s> <s xml:id="echoid-s14908" xml:space="preserve">qu’étant dans la même raiſon que <lb/>les quarrés de leurs côtés homologues, ils ſeront comme les <lb/>lignes A G & </s> <s xml:id="echoid-s14909" xml:space="preserve">H G (art. </s> <s xml:id="echoid-s14910" xml:space="preserve">497). </s> <s xml:id="echoid-s14911" xml:space="preserve">Or comme les triangles A B G <lb/>& </s> <s xml:id="echoid-s14912" xml:space="preserve">H B G ont la même hauteur, ils ſeront dans la même rai-<lb/>ſon que leurs baſes, & </s> <s xml:id="echoid-s14913" xml:space="preserve">auront par conſéquent même raiſon <lb/>que les lignes A G & </s> <s xml:id="echoid-s14914" xml:space="preserve">H G; </s> <s xml:id="echoid-s14915" xml:space="preserve">d’où il s’enſuit que le triangle I K G <lb/>eſt égal au triangle H B G. </s> <s xml:id="echoid-s14916" xml:space="preserve">Cela poſé, ſi l’on retranche de part <lb/>& </s> <s xml:id="echoid-s14917" xml:space="preserve">d’autre la figure H O K G qui eſt commune à ces deux trian-<lb/>gles, il reſtera le triangle O I H égal au triangle O B K: </s> <s xml:id="echoid-s14918" xml:space="preserve">mais <lb/>comme le triangle B A H eſt égal à la moitié du trapeze, il <lb/>s’enſuit que la figure A I K B eſt auſſi égale à la moitié du tra- <pb o="459" file="0537" n="557" rhead="DE MATHÉMATIQUE. Liv. XIII."/> peze, & </s> <s xml:id="echoid-s14919" xml:space="preserve">que par conſéquent la ligne I K le partage en deux <lb/>également.</s> <s xml:id="echoid-s14920" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1190" type="section" level="1" n="860"> <head xml:id="echoid-head1038" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14921" xml:space="preserve">880. </s> <s xml:id="echoid-s14922" xml:space="preserve">Diviſer un trapézoïde en trois parties égales.</s> <s xml:id="echoid-s14923" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figurè 304.</note> <p> <s xml:id="echoid-s14924" xml:space="preserve">Cette propoſition eſt peu conſidérable, mais elle eſt miſe <lb/>ici pour ſervir d’introduction aux ſuivantes. </s> <s xml:id="echoid-s14925" xml:space="preserve">Ainſi conſidérant <lb/>le trapézoïde A C, qu’on propoſe à diviſer en trois parties <lb/>égales, on verra qu’il ne faut que diviſer les côtés B C & </s> <s xml:id="echoid-s14926" xml:space="preserve">A D <lb/>en trois parties égales, & </s> <s xml:id="echoid-s14927" xml:space="preserve">tirer les lignes G E & </s> <s xml:id="echoid-s14928" xml:space="preserve">H F, qui <lb/>donneront les figures égales A G, E H, F C, puiſqu’elles ſont <lb/>compoſées chacune de deux triangles égaux.</s> <s xml:id="echoid-s14929" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1191" type="section" level="1" n="861"> <head xml:id="echoid-head1039" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14930" xml:space="preserve">881. </s> <s xml:id="echoid-s14931" xml:space="preserve">Diviſer un trapeze en deux parties égales.</s> <s xml:id="echoid-s14932" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Figure 305.</note> <p> <s xml:id="echoid-s14933" xml:space="preserve">Pour diviſer le trapeze A B C D en deux parties égales, il <lb/>faut du point B tirer la ligne B H parallele à A D, & </s> <s xml:id="echoid-s14934" xml:space="preserve">diviſer <lb/>les lignes B H & </s> <s xml:id="echoid-s14935" xml:space="preserve">A D en deux parties égales aux points G & </s> <s xml:id="echoid-s14936" xml:space="preserve">F; <lb/></s> <s xml:id="echoid-s14937" xml:space="preserve">enſuite tirer les lignes G C & </s> <s xml:id="echoid-s14938" xml:space="preserve">G F, qui donneront la figure <lb/>C B A F G égale à la figure C G F D, qui ſont chacune moitié <lb/>du trapeze: </s> <s xml:id="echoid-s14939" xml:space="preserve">car par l’opération le trapézoïde A G eſt égal au <lb/>trapezoïde G D, & </s> <s xml:id="echoid-s14940" xml:space="preserve">le triangle B C G eſt égal au triangle G C H.</s> <s xml:id="echoid-s14941" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14942" xml:space="preserve">Mais pour que les deux parties du trapeze fuſſent plus régu-<lb/>lieres, il ſeroit à propos que les lignes de diviſion C G & </s> <s xml:id="echoid-s14943" xml:space="preserve">G F <lb/>ne fiſſent qu’une ligne droite. </s> <s xml:id="echoid-s14944" xml:space="preserve">Or ſi l’on tire à la ligne F C <lb/>la parallele G E, on n’aura qu’à tirer de E en F pour avoir le <lb/>trapeze diviſé en deux parties égales par la ſeule ligne E F, <lb/>comme on le peut voir par les triangles F G C & </s> <s xml:id="echoid-s14945" xml:space="preserve">F E C, qui <lb/>ſont renfermés entre les mêmes paralleles.</s> <s xml:id="echoid-s14946" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1192" type="section" level="1" n="862"> <head xml:id="echoid-head1040" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14947" xml:space="preserve">882. </s> <s xml:id="echoid-s14948" xml:space="preserve">Diviſer un trapeze en deux parties égales par une ligne <lb/> <anchor type="note" xlink:label="note-0537-03a" xlink:href="note-0537-03"/> tirée d’un de ſes angles.</s> <s xml:id="echoid-s14949" xml:space="preserve"/> </p> <div xml:id="echoid-div1192" type="float" level="2" n="1"> <note position="right" xlink:label="note-0537-03" xlink:href="note-0537-03a" xml:space="preserve">Figure 306.</note> </div> <p> <s xml:id="echoid-s14950" xml:space="preserve">L’on demande qu’on diviſe le trapeze A B C D en deux par-<lb/>ties égales par une ligne tirée de l’angle B.</s> <s xml:id="echoid-s14951" xml:space="preserve"/> </p> <pb o="460" file="0538" n="558" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s14952" xml:space="preserve">Pour réſoudre ce problême, tirez les diagonales A C & </s> <s xml:id="echoid-s14953" xml:space="preserve"><lb/>B D, & </s> <s xml:id="echoid-s14954" xml:space="preserve">diviſez la premiere A C en deux parties égales au <lb/>point E, & </s> <s xml:id="echoid-s14955" xml:space="preserve">de ce point menez la ligne E F parallele à B D; <lb/></s> <s xml:id="echoid-s14956" xml:space="preserve">& </s> <s xml:id="echoid-s14957" xml:space="preserve">ſi vous tirez une ligne de l’angle B au point F, elle diviſera <lb/>le trapeze en deux parties égales.</s> <s xml:id="echoid-s14958" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14959" xml:space="preserve">Pour le démontrer, conſidérez qu’ayant tiré les lignes E B <lb/>& </s> <s xml:id="echoid-s14960" xml:space="preserve">E D, elles donnent les triangles A E D & </s> <s xml:id="echoid-s14961" xml:space="preserve">E C D égaux en-<lb/>tr’eux, auſſi-bien que les triangles A B E & </s> <s xml:id="echoid-s14962" xml:space="preserve">E B C. </s> <s xml:id="echoid-s14963" xml:space="preserve">Cela étant, <lb/>le trapeze ſe trouve diviſé en deux parties égales par les lignes <lb/>E B & </s> <s xml:id="echoid-s14964" xml:space="preserve">E D: </s> <s xml:id="echoid-s14965" xml:space="preserve">& </s> <s xml:id="echoid-s14966" xml:space="preserve">comme les triangles qui ſont renfermés entre <lb/>les mêmes paralleles nous donnent E B O égal à O F D, il <lb/>s’enſuit que la ſeule ligne B F diviſe le trapeze en deux égale-<lb/>ment.</s> <s xml:id="echoid-s14967" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1194" type="section" level="1" n="863"> <head xml:id="echoid-head1041" xml:space="preserve">PROPOSITION XII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14968" xml:space="preserve">883. </s> <s xml:id="echoid-s14969" xml:space="preserve">Diviſer un trapézoïde en deux parties égales par une <lb/> <anchor type="note" xlink:label="note-0538-01a" xlink:href="note-0538-01"/> ligne tirée d’un point pris ſur l’un de ſes côtés.</s> <s xml:id="echoid-s14970" xml:space="preserve"/> </p> <div xml:id="echoid-div1194" type="float" level="2" n="1"> <note position="left" xlink:label="note-0538-01" xlink:href="note-0538-01a" xml:space="preserve">Figure 307.</note> </div> <p> <s xml:id="echoid-s14971" xml:space="preserve">Pour diviſer en deux également le trapézoïde A B C D par <lb/>une ligne tirée du point H, il faut commencer par réduire le <lb/>trapézoïde en triangle, en tirant à la diagonale B D la parallele <lb/>C F, afin d’avoir le point F pour tirer la ligne F B, qui don-<lb/>nera le triangle A B F égal au trapézoïde. </s> <s xml:id="echoid-s14972" xml:space="preserve">Cela poſé, il faut <lb/>diviſer la baſe A F du triangle en deux également au point E, <lb/>& </s> <s xml:id="echoid-s14973" xml:space="preserve">tirer la ligne B E, pour avoir le triangle A B E, qui ſera la <lb/>moitié du trapézoïde. </s> <s xml:id="echoid-s14974" xml:space="preserve">Préſentement il faut tirer la ligne B H, <lb/>& </s> <s xml:id="echoid-s14975" xml:space="preserve">lui mener du point E la parallele E G; </s> <s xml:id="echoid-s14976" xml:space="preserve">& </s> <s xml:id="echoid-s14977" xml:space="preserve">ſi on tire la ligne <lb/>H G, elle diviſera le trapézoïde en deux également.</s> <s xml:id="echoid-s14978" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s14979" xml:space="preserve">Pour le démontrer, faites attention qu’à cauſe des paralleles, <lb/>les triangles O H E & </s> <s xml:id="echoid-s14980" xml:space="preserve">O B G ſont égaux, & </s> <s xml:id="echoid-s14981" xml:space="preserve">que par conſé-<lb/>quent la figure A B G H eſt égale à la moitié du trapézoïde, <lb/>puiſqu’elle eſt égale au triangle A B E.</s> <s xml:id="echoid-s14982" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1196" type="section" level="1" n="864"> <head xml:id="echoid-head1042" xml:space="preserve">PROPOSITION XIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s14983" xml:space="preserve">884. </s> <s xml:id="echoid-s14984" xml:space="preserve">Diviſerun pentagone en trois parties égales par des lignes <lb/> <anchor type="note" xlink:label="note-0538-02a" xlink:href="note-0538-02"/> tirées d’un de ſes angles.</s> <s xml:id="echoid-s14985" xml:space="preserve"/> </p> <div xml:id="echoid-div1196" type="float" level="2" n="1"> <note position="left" xlink:label="note-0538-02" xlink:href="note-0538-02a" xml:space="preserve">Figure 308.</note> </div> <p> <s xml:id="echoid-s14986" xml:space="preserve">Pour diviſer en trois parties égales le pentagone A B C D E <lb/>par les lignes tirées de l’angle C, il faut commencer par ré- <pb o="461" file="0539" n="559" rhead="DE MATHÉMATIQUE. Liv. XIII."/> duire le pentagone en triangle; </s> <s xml:id="echoid-s14987" xml:space="preserve">& </s> <s xml:id="echoid-s14988" xml:space="preserve">cela, en tirant aux lignes <lb/>C A & </s> <s xml:id="echoid-s14989" xml:space="preserve">C E les paralleles B F & </s> <s xml:id="echoid-s14990" xml:space="preserve">D G, & </s> <s xml:id="echoid-s14991" xml:space="preserve">en menant des lignes <lb/>du point C au point F, & </s> <s xml:id="echoid-s14992" xml:space="preserve">du même point C au point G, qui <lb/>donneront le triangle F C G égal au pentagone, comme on le <lb/>peut connoître facilement. </s> <s xml:id="echoid-s14993" xml:space="preserve">Après cela, ſi l’on diviſe la baſe <lb/>F G en trois parties égales aux points H & </s> <s xml:id="echoid-s14994" xml:space="preserve">I, on n’aura plus <lb/>qu’à tirer les lignes C H & </s> <s xml:id="echoid-s14995" xml:space="preserve">C I pour avoir le triangle H C I, <lb/>qui ſera le tiers du triangle F C G, par conſéquent du penta-<lb/>gone, & </s> <s xml:id="echoid-s14996" xml:space="preserve">il ſe trouvera que les parties H A B C & </s> <s xml:id="echoid-s14997" xml:space="preserve">I C D E ſe-<lb/>ront égales entr’elles, & </s> <s xml:id="echoid-s14998" xml:space="preserve">ſeront par conſéquent chacune le tiers <lb/>du pentagone.</s> <s xml:id="echoid-s14999" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1198" type="section" level="1" n="865"> <head xml:id="echoid-head1043" style="it" xml:space="preserve">Application de la Géométrie à l’uſage du Compas de proportion.</head> <p> <s xml:id="echoid-s15000" xml:space="preserve">De tous les inſtrumens de Mathématique, il n’y en a point <lb/>dont l’uſage ſoit ſi univerſel que celui qu’on nomme compas de <lb/>proportion; </s> <s xml:id="echoid-s15001" xml:space="preserve">car il facilite la pratique de toute la théorie de la <lb/>Géométrie: </s> <s xml:id="echoid-s15002" xml:space="preserve">par exemple, la ligne des parties égales ſert à di-<lb/>viſer une ligne, ſelon une raiſon donnée, & </s> <s xml:id="echoid-s15003" xml:space="preserve">à trouver des <lb/>troiſiemes & </s> <s xml:id="echoid-s15004" xml:space="preserve">quatriemes proportionnelles: </s> <s xml:id="echoid-s15005" xml:space="preserve">la ligne des cordes <lb/>tient lieu de rapporteur, puiſque par ſon moyen l’on peut <lb/>connoître la valeur des angles, & </s> <s xml:id="echoid-s15006" xml:space="preserve">en déterminer de quelque <lb/>quantité de degrés qu’on voudra: </s> <s xml:id="echoid-s15007" xml:space="preserve">la ligne des polygones ſert <lb/>à diviſer un cercle en une quantité de parties égales, pour y <lb/>inſcrire des polygones: </s> <s xml:id="echoid-s15008" xml:space="preserve">par le moyen de la ligne des plans, <lb/>l’on trouve les côtés des figures ſemblables qu’on veut aug-<lb/>menter ou diminuer ſelon les raiſons données: </s> <s xml:id="echoid-s15009" xml:space="preserve">enfin la ligne <lb/>des ſolides, qui peut paſſer pour la plus conſidérable du compas <lb/>de proportion, ſert à trouver deux moyennes proportionnelles <lb/>entre deux lignes données, à diminuer & </s> <s xml:id="echoid-s15010" xml:space="preserve">augmenter les ſolides <lb/>ſemblables, ſelon les raiſons que l’on voudra. </s> <s xml:id="echoid-s15011" xml:space="preserve">Ce ſont toutes <lb/>ces propriétés que nous allons enſeigner ici, en commençant <lb/>par les lignes de parties égales.</s> <s xml:id="echoid-s15012" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1199" type="section" level="1" n="866"> <head xml:id="echoid-head1044" xml:space="preserve">PROPOSITION XIV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15013" xml:space="preserve">885. </s> <s xml:id="echoid-s15014" xml:space="preserve">Diviſer une ligne droite en tant de parties égales qu’on <lb/> <anchor type="note" xlink:label="note-0539-01a" xlink:href="note-0539-01"/> voudra.</s> <s xml:id="echoid-s15015" xml:space="preserve"/> </p> <div xml:id="echoid-div1199" type="float" level="2" n="1"> <note position="right" xlink:label="note-0539-01" xlink:href="note-0539-01a" xml:space="preserve">Figure 309.</note> </div> <p> <s xml:id="echoid-s15016" xml:space="preserve">L’on trouvera marqué d’un côté ſur chaque jambe du compas <pb o="462" file="0540" n="560" rhead="NOUVEAU COURS"/> de proportion une ligne que l’on verra nommée parties égales, <lb/>parce qu’elles ſervent effectivement à diviſer les lignes droites <lb/>en parties égales: </s> <s xml:id="echoid-s15017" xml:space="preserve">& </s> <s xml:id="echoid-s15018" xml:space="preserve">pour faire voir comment on s’en ſert, <lb/>nous ſuppoſerons qu’on veut diviſer la ligne H I en neuf par-<lb/>ties, pour faire, par exemple, l’échelle d’un plan: </s> <s xml:id="echoid-s15019" xml:space="preserve">pour cela, <lb/>il faut avec le compas ordinaire, prendre la longueur de la <lb/>ligne H I, & </s> <s xml:id="echoid-s15020" xml:space="preserve">ouvrir le compas de proportion, de maniere que <lb/>les pointes du compas ordinaire puiſſent être poſées dans les <lb/>points de la ligne des parties égales, où l’on verra marqué 90, <lb/>qui ſera, par exemple, les points D & </s> <s xml:id="echoid-s15021" xml:space="preserve">E. </s> <s xml:id="echoid-s15022" xml:space="preserve">Préſentement laiſ-<lb/>ſant le compas de proportion ouvert, il faut, avec le compas <lb/>ordinaire, prendre l’intervalle des points où l’on verra le nom-<lb/>bre 10, qui ſera, par exemple, l’intervalle F G. </s> <s xml:id="echoid-s15023" xml:space="preserve">Or ſi vous <lb/>portez préſentement le compas ainſi ouvert ſur la ligne H I, <lb/>vous trouverez que ſon ouverture ſera la neuvieme partie de <lb/>cette même ligne,</s> </p> <p> <s xml:id="echoid-s15024" xml:space="preserve">Pour le démontrer, conſidérez que les triangles A F G & </s> <s xml:id="echoid-s15025" xml:space="preserve"><lb/>A D E ſont ſemblables, & </s> <s xml:id="echoid-s15026" xml:space="preserve">que par conſéquent il y aura même <lb/>raiſon de A F à A D, que de F G à D E. </s> <s xml:id="echoid-s15027" xml:space="preserve">Or comme A F eſt la <lb/>neuvieme partie de A D, F G ſera la neuvieme partie de D E.</s> <s xml:id="echoid-s15028" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1201" type="section" level="1" n="867"> <head xml:id="echoid-head1045" xml:space="preserve">PROPOSITION XV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15029" xml:space="preserve">886. </s> <s xml:id="echoid-s15030" xml:space="preserve">Trouver une troiſieme proportionnelle à deux lignes données.</s> <s xml:id="echoid-s15031" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 310.</note> <p> <s xml:id="echoid-s15032" xml:space="preserve">Pour trouver une troiſieme proportionnelle à deux lignes <lb/>données F & </s> <s xml:id="echoid-s15033" xml:space="preserve">G, il faut prendre la premiere F avec le compas <lb/>ordinaire, & </s> <s xml:id="echoid-s15034" xml:space="preserve">la porter ſur la ligne des parties égales, comme <lb/>ſi elle occupoit, par exemple, la diſtance depuis A juſqu’en D; <lb/></s> <s xml:id="echoid-s15035" xml:space="preserve">enſuite prendre la ſeconde G, & </s> <s xml:id="echoid-s15036" xml:space="preserve">la porter depuis A juſqu’en B. </s> <s xml:id="echoid-s15037" xml:space="preserve"><lb/>Il faut après cela ouvrir le compas de proportion d’une gran-<lb/>deur telle que la diſtance D E (des deux nombres égaux qui <lb/>correſpondent aux points D & </s> <s xml:id="echoid-s15038" xml:space="preserve">E) ſoit égale à la ligne G. </s> <s xml:id="echoid-s15039" xml:space="preserve"><lb/>Préſentement ſi l’on prend la diſtance B C, c’eſt-à-dire l’in-<lb/>tervalle du chiffre, qui eſt au point B à celui qui lui correſpond <lb/>au point C, l’on aura la troiſieme proportionnelle que l’on <lb/>cherche, qui ſera, par exemple, H.</s> <s xml:id="echoid-s15040" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15041" xml:space="preserve">Pour le prouver, conſidérez que les triangles A B C & </s> <s xml:id="echoid-s15042" xml:space="preserve">E A D <lb/>ſont ſemblables, & </s> <s xml:id="echoid-s15043" xml:space="preserve">que la ligne A B étant égale à la ligne D E, <lb/>l’on aura A D : </s> <s xml:id="echoid-s15044" xml:space="preserve">D E : </s> <s xml:id="echoid-s15045" xml:space="preserve">: </s> <s xml:id="echoid-s15046" xml:space="preserve">A B : </s> <s xml:id="echoid-s15047" xml:space="preserve">B C; </s> <s xml:id="echoid-s15048" xml:space="preserve">par conſéquent {.</s> <s xml:id="echoid-s15049" xml:space="preserve">./.</s> <s xml:id="echoid-s15050" xml:space="preserve">.} F. </s> <s xml:id="echoid-s15051" xml:space="preserve">G. </s> <s xml:id="echoid-s15052" xml:space="preserve">H.</s> <s xml:id="echoid-s15053" xml:space="preserve"/> </p> <pb o="463" file="0541" n="561" rhead="DE MATHÉMATIQUE. Liv. XIII."/> </div> <div xml:id="echoid-div1202" type="section" level="1" n="868"> <head xml:id="echoid-head1046" xml:space="preserve">PROPOSITION XVI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15054" xml:space="preserve">887. </s> <s xml:id="echoid-s15055" xml:space="preserve">Trouver une quatrieme proportionnelle àtrois lignes données. <lb/></s> </p> <p> <s xml:id="echoid-s15056" xml:space="preserve">Pour trouver une quatrieme proportionnelle aux trois li-<lb/>gnes données A, B, C, il faut prendre la ligne A, & </s> <s xml:id="echoid-s15057" xml:space="preserve">la porter <lb/>avec le compas ordinaire ſur la ligne des parties égales, enſorte <lb/>qu’elle occupe l’intervalle E F; </s> <s xml:id="echoid-s15058" xml:space="preserve">puis porter la ſeconde B de-<lb/>puis le point F juſqu’au point correſpondant G: </s> <s xml:id="echoid-s15059" xml:space="preserve">enfin il faut <lb/>prendre la troiſieme C, enſorte qu’elle occupe l’eſpace E H, <lb/>& </s> <s xml:id="echoid-s15060" xml:space="preserve">l’intervalle du point H à celui qui lui correſpond en I, ſera la <lb/>quatrieme proportionnelle, comme eſt, par exemple, la ligne D.</s> <s xml:id="echoid-s15061" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15062" xml:space="preserve">Pour le prouver, remarquez que les triangles E F G & </s> <s xml:id="echoid-s15063" xml:space="preserve">E H I <lb/>ſont ſemblables, & </s> <s xml:id="echoid-s15064" xml:space="preserve">par conſéquent l’on aura EF:</s> <s xml:id="echoid-s15065" xml:space="preserve">FG:</s> <s xml:id="echoid-s15066" xml:space="preserve">:EH:</s> <s xml:id="echoid-s15067" xml:space="preserve">HI, <lb/>ou bien A : </s> <s xml:id="echoid-s15068" xml:space="preserve">B :</s> <s xml:id="echoid-s15069" xml:space="preserve">: C : </s> <s xml:id="echoid-s15070" xml:space="preserve">D.</s> <s xml:id="echoid-s15071" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1203" type="section" level="1" n="869"> <head xml:id="echoid-head1047" xml:space="preserve"><emph style="sc">Usage de la ligne des</emph> <emph style="sc">Polygones</emph>.</head> <head xml:id="echoid-head1048" xml:space="preserve">PROPOSITION XVII. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s15072" xml:space="preserve">888. </s> <s xml:id="echoid-s15073" xml:space="preserve">Inſcrire un polygone dans un cercle. <lb/></s> <s xml:id="echoid-s15074" xml:space="preserve"> <anchor type="note" xlink:label="note-0541-02a" xlink:href="note-0541-02"/> & </s> <s xml:id="echoid-s15075" xml:space="preserve">313. ront à décrire l’octogone. <pb o="464" file="0542" n="562" rhead="NOUVEAU COURS"/> cercle, que vous voulez réduire en polygone.</s></p> </div> <div xml:id="echoid-div1204" type="section" level="1" n="870"> <head xml:id="echoid-head1049" xml:space="preserve">PROPOSITION XVIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15076" xml:space="preserve">889. </s> <s xml:id="echoid-s15077" xml:space="preserve">Décrire un polygone régulier ſur une ligne donnée. fois la ligne K L ſur la circonférence du cercle.</s></p> </div> <div xml:id="echoid-div1205" type="section" level="1" n="871"> <head xml:id="echoid-head1050" xml:space="preserve"><emph style="sc">Usage de la ligne des</emph> <emph style="sc">Cordes</emph>.</head> <head xml:id="echoid-head1051" xml:space="preserve">PROPOSITION XIX. <lb/><emph style="sc">Probleme</emph>.</head> <p><s xml:id="echoid-s15078" xml:space="preserve"><anchor type="note" xlink:label="note-0542-01a" xlink:href="note-0542-01"/> de degrés qu’on voudra.</s></p> <div xml:id="echoid-div1205" type="float" level="2" n="1"> <note position="left" xlink:label="note-0542-01" xlink:href="note-0542-01a" xml:space="preserve">Figure 312 <lb/>& 314.</note> </div> <p> <s xml:id="echoid-s15079" xml:space="preserve">Si l’on vouloit prendre ſur la circonférence du cercle H un <lb/>arc de 70 degrés, il faudra avec le compas ordinaire, porter <lb/>ſur la ligne des cordes aux endroits marqués 60 la grandeur ou <lb/>le rayon H I: </s> <s xml:id="echoid-s15080" xml:space="preserve">ainſi ſuppoſant que l’angle A B C eſt formé par les <lb/>lignes des cordes du compas de proportion, de maniere que <lb/>l’on ait ouvert la grandeur D E égale au rayon H I, l’on pren-<lb/>dra l’intervalle de F en G, que je ſuppoſe être de 70 en 70, <lb/>& </s> <s xml:id="echoid-s15081" xml:space="preserve">la ligne F G ſera la corde de 70 degrés, qu’on n’aura qu’à <lb/>porter ſur la circonférence du cercle, pour avoir l’arc M I qu’on <lb/>demande.</s> <s xml:id="echoid-s15082" xml:space="preserve"/> </p> <pb o="465" file="0543" n="563" rhead="DE MATHÉMATIQUE. Liv. XIII."/> </div> <div xml:id="echoid-div1207" type="section" level="1" n="872"> <head xml:id="echoid-head1052" xml:space="preserve">PROPOSITION XX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15083" xml:space="preserve">891. </s> <s xml:id="echoid-s15084" xml:space="preserve">Un angle étant donné ſur le papier, en trouver la va-<lb/>leur par le moyen de la ligne des cordes.</s> <s xml:id="echoid-s15085" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15086" xml:space="preserve">Pour connoître la valeur d’un angle A B C, il faut, du point <lb/>B, comme centre, décrire l’arc A C d’une ouverture de compas <lb/>indéterminée; </s> <s xml:id="echoid-s15087" xml:space="preserve">enſuite prendre le rayon B C, & </s> <s xml:id="echoid-s15088" xml:space="preserve">ouvrir le <lb/>compas de proportion, de maniere que l’intervalle de 60 en <lb/>60, marqué ſur la ligne des cordes, ſoit égal au rayon. </s> <s xml:id="echoid-s15089" xml:space="preserve">Pré-<lb/>ſentement ſi on prend avec le compas la corde A C, & </s> <s xml:id="echoid-s15090" xml:space="preserve">qu’on <lb/>la porte ſur la ligne des cordes, de façon qu’il convienne <lb/>dans deux points également éloignés du centre, les nombres <lb/>qui correſpondront à ces points, donneront la valeur de l’an-<lb/>gle: </s> <s xml:id="echoid-s15091" xml:space="preserve">ainſi ſuppoſant que ce ſoit de 50 en 50, l’on connoîtra <lb/>que l’angle A B C eſt de 50 degrés.</s> <s xml:id="echoid-s15092" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1208" type="section" level="1" n="873"> <head xml:id="echoid-head1053" xml:space="preserve">PROPOSITION XXI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15093" xml:space="preserve">892. </s> <s xml:id="echoid-s15094" xml:space="preserve">Connoiſſant la quantité de degrés d’un arc de cercle, trouver <lb/> <anchor type="note" xlink:label="note-0543-01a" xlink:href="note-0543-01"/> ſon rayon.</s> <s xml:id="echoid-s15095" xml:space="preserve"/> </p> <div xml:id="echoid-div1208" type="float" level="2" n="1"> <note position="right" xlink:label="note-0543-01" xlink:href="note-0543-01a" xml:space="preserve">Figure 314 <lb/>& 315.</note> </div> <p> <s xml:id="echoid-s15096" xml:space="preserve">Si l’on a un arc de cercle B A de 50 degrés, & </s> <s xml:id="echoid-s15097" xml:space="preserve">qu’on veuille <lb/>connoître le rayon du cercle de cet arc, il faudra prendre avec <lb/>le compas la corde B A, & </s> <s xml:id="echoid-s15098" xml:space="preserve">la porter ſur la ligne des cordes <lb/>pour ouvrir le compas de proportion de 50 en 50: </s> <s xml:id="echoid-s15099" xml:space="preserve">par exemple, <lb/>ſi les points D & </s> <s xml:id="echoid-s15100" xml:space="preserve">E correſpondent au nombre 50, il faut faire <lb/>l’intervalle D E égal à la corde B A; </s> <s xml:id="echoid-s15101" xml:space="preserve">& </s> <s xml:id="echoid-s15102" xml:space="preserve">ſi après cela l’on prend <lb/>l’intervalle F G de 60 en 60, elle ſera le rayon que l’on de-<lb/>mande, c’eſt-à-dire que la ligne F G ſera égale au demi-dia-<lb/>metre C B.</s> <s xml:id="echoid-s15103" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1210" type="section" level="1" n="874"> <head xml:id="echoid-head1054" xml:space="preserve">PROPOSITION XXII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15104" xml:space="preserve">893. </s> <s xml:id="echoid-s15105" xml:space="preserve">Ouvrir le compas de proportion de maniere que les lignes <lb/> <anchor type="note" xlink:label="note-0543-02a" xlink:href="note-0543-02"/> des cordes faſſent tel angle que l’on voudra, ſuppoſant que les <lb/>lignes A B & </s> <s xml:id="echoid-s15106" xml:space="preserve">C B ſoient celles des cordes; </s> <s xml:id="echoid-s15107" xml:space="preserve">on demande de faire <lb/>avec elles un angle de 70 degrés.</s> <s xml:id="echoid-s15108" xml:space="preserve"/> </p> <div xml:id="echoid-div1210" type="float" level="2" n="1"> <note position="right" xlink:label="note-0543-02" xlink:href="note-0543-02a" xml:space="preserve">Figure 314.</note> </div> <pb o="466" file="0544" n="564" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s15109" xml:space="preserve">Il faut prendre avec le compas ordinaire l’intervalle qu’il <lb/>y a du centre B au point F ou G, que je ſuppoſe être de 70 <lb/>degrés; </s> <s xml:id="echoid-s15110" xml:space="preserve">puis porter les pointes du compas ainſi ouvert dans les <lb/>points de 60 en 60: </s> <s xml:id="echoid-s15111" xml:space="preserve">par exemple, ſi les points D & </s> <s xml:id="echoid-s15112" xml:space="preserve">E ſont <lb/>ceux de 60 en 60, il faut faire la diſtance D E égale à l’inter-<lb/>valle B F, & </s> <s xml:id="echoid-s15113" xml:space="preserve">les lignes des cordes formeront l’angle A B C de <lb/>70 degrés.</s> <s xml:id="echoid-s15114" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1212" type="section" level="1" n="875"> <head xml:id="echoid-head1055" xml:space="preserve">PROPOSITION XXIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15115" xml:space="preserve">894. </s> <s xml:id="echoid-s15116" xml:space="preserve">Le compas de proportion étant ouvert d’une grandeur quel-<lb/> <anchor type="note" xlink:label="note-0544-01a" xlink:href="note-0544-01"/> conque, connoître la valeur de l’angle formé par les lignes des <lb/>cordes.</s> <s xml:id="echoid-s15117" xml:space="preserve"/> </p> <div xml:id="echoid-div1212" type="float" level="2" n="1"> <note position="left" xlink:label="note-0544-01" xlink:href="note-0544-01a" xml:space="preserve">Figure 314.</note> </div> <p> <s xml:id="echoid-s15118" xml:space="preserve">Si l’on veut ſçavoir la valeur de l’angle A B C, formé par <lb/>les lignes des cordes, l’on n’aura qu’à prendre avec le compas <lb/>ordinaire l’intervalle de 60 en 60, puis la porter ſur l’une des <lb/>cordes, en commençant du centre, l’on trouvera la quantité <lb/>de degrés que contient l’angle: </s> <s xml:id="echoid-s15119" xml:space="preserve">ainſi les points D & </s> <s xml:id="echoid-s15120" xml:space="preserve">E étant <lb/>ſuppoſés ceux de 60, l’on prendra la ligne D E pour la porter <lb/>ſur B F; </s> <s xml:id="echoid-s15121" xml:space="preserve">& </s> <s xml:id="echoid-s15122" xml:space="preserve">ſi l’on voit que le point F correſpond à un nom-<lb/>bre, par exemple, de 70, l’on verra par-là que l’angle A B C <lb/>eſt de 70 degrés.</s> <s xml:id="echoid-s15123" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1214" type="section" level="1" n="876"> <head xml:id="echoid-head1056" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s15124" xml:space="preserve">Comme l’on applique quelquefois des pinnules aux extrê-<lb/>mités des cordes du compas de proportion, pour prendre des <lb/>angles ſur le terrein, on peut en former de telle ouverture <lb/>que l’on voudra, puiſque par ces deux propoſitions l’on peut <lb/>faire un angle quelconque avec les lignes des cordes, & </s> <s xml:id="echoid-s15125" xml:space="preserve">qu’on <lb/>peut d’ailleurs connoître la valeur des angles qu’elles peuvent <lb/>former.</s> <s xml:id="echoid-s15126" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1215" type="section" level="1" n="877"> <head xml:id="echoid-head1057" xml:space="preserve"><emph style="sc">Usage de la ligne des</emph> <emph style="sc">Plans</emph>.</head> <head xml:id="echoid-head1058" xml:space="preserve">PROPOSITION XXIV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15127" xml:space="preserve">895. </s> <s xml:id="echoid-s15128" xml:space="preserve">Faire un quarré qui ſoit à un autre ſelon une raiſon donnée. <lb/></s> <s xml:id="echoid-s15129" xml:space="preserve"> <anchor type="note" xlink:label="note-0544-02a" xlink:href="note-0544-02"/> & 321. <pb o="467" file="0545" n="565" rhead="DE MATHÉMATIQUE. Liv. XIII."/> C D, que de 5 à 2.</s></p> </div> <div xml:id="echoid-div1216" type="section" level="1" n="878"> <head xml:id="echoid-head1059" xml:space="preserve">PROPOSITION XXV. <lb/><emph style="sc">Probleme</emph>.</head> <p><s xml:id="echoid-s15130" xml:space="preserve"><anchor type="note" xlink:label="note-0545-01a" xlink:href="note-0545-01"/> & 321. quarré C D.</s></p> </div> <div xml:id="echoid-div1217" type="section" level="1" n="879"> <head xml:id="echoid-head1060" xml:space="preserve">PROPOSITION XXVI. <lb/><emph style="sc">Probleme</emph>.</head> <p><s xml:id="echoid-s15131" xml:space="preserve"><anchor type="note" xlink:label="note-0545-02a" xlink:href="note-0545-02"/> des plans forment un angle droit.</s></p> <div xml:id="echoid-div1217" type="float" level="2" n="1"> <note position="right" xlink:label="note-0545-02" xlink:href="note-0545-02a" xml:space="preserve">Figure 317.</note> </div> <p> <s xml:id="echoid-s15132" xml:space="preserve">Pour faire un angle droit tel que B A C avec les deux lignes <lb/>des plans, il faut avec le compas ordinaire prendre l’intervalle <lb/>du centre à un nombre quelconque D, qui ſera, par exemple, <lb/>20, puis ouvrir le compas de proportion, de maniere que l’in-<lb/>tervalle des points (qui correſpondront à la moitié de ce nom-<lb/>bre) ſoit égal à la longueur A D: </s> <s xml:id="echoid-s15133" xml:space="preserve">ainſi prenant les nombres <lb/>10 & </s> <s xml:id="echoid-s15134" xml:space="preserve">10, qui ſeront moitié de 20, l’on n’aura qu’à faire l’in-<lb/>tervalle F G égal à la diſtance A D, & </s> <s xml:id="echoid-s15135" xml:space="preserve">les lignes des plans <lb/>A B & </s> <s xml:id="echoid-s15136" xml:space="preserve">A C formeront un angle droit.</s> <s xml:id="echoid-s15137" xml:space="preserve"/> </p> <pb o="468" file="0546" n="566" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1219" type="section" level="1" n="880"> <head xml:id="echoid-head1061" xml:space="preserve">PROPOSITION XXVII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15138" xml:space="preserve">898. </s> <s xml:id="echoid-s15139" xml:space="preserve">Faire un quarré égal à deux autres donnés.</s> <s xml:id="echoid-s15140" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 318 <lb/>& 321.</note> <p> <s xml:id="echoid-s15141" xml:space="preserve">Pour faire un quarré qui ſoit égal aux deux autres A B & </s> <s xml:id="echoid-s15142" xml:space="preserve"><lb/>C D, il faut ouvrir le compas de proportion, de maniere que <lb/>les lignes des plans forment un angle droit, comme eſt l’angle <lb/>E F G; </s> <s xml:id="echoid-s15143" xml:space="preserve">puis prendre ſur la ligne F E la longueur F I égale au <lb/>côté A B, & </s> <s xml:id="echoid-s15144" xml:space="preserve">bien retenir le nombre où l’extrêmité I viendra <lb/>aboutir: </s> <s xml:id="echoid-s15145" xml:space="preserve">enſuite il faut prendre de même la longueur F H <lb/>égale au côté C D de l’autre quarré & </s> <s xml:id="echoid-s15146" xml:space="preserve">la diſtance de H en I, <lb/>qui ſera, par exemple, celle de 18 en 5, ſera le côté du quarré <lb/>égal aux deux quarrés propoſés.</s> <s xml:id="echoid-s15147" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1220" type="section" level="1" n="881"> <head xml:id="echoid-head1062" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s15148" xml:space="preserve">Comme toutes les figures ſemblables ſont dans la même <lb/>raiſon que les quarrés de leurs côtés homologues, l’on pourra <lb/>faire les mêmes opérations pour les triangles, les polygones <lb/>& </s> <s xml:id="echoid-s15149" xml:space="preserve">les cercles que l’on a faits dans les propoſitions précédentes <lb/>pour les quarrés.</s> <s xml:id="echoid-s15150" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1221" type="section" level="1" n="882"> <head xml:id="echoid-head1063" xml:space="preserve"><emph style="sc">Usage de la ligne des</emph> <emph style="sc">Solides</emph>.</head> <head xml:id="echoid-head1064" xml:space="preserve">PROPOSITION XXVIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15151" xml:space="preserve">899. </s> <s xml:id="echoid-s15152" xml:space="preserve">Faire un cube qui ſoit à un autre ſelon une raiſon donnée.</s> <s xml:id="echoid-s15153" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Figure 319 <lb/>& 322.</note> <p> <s xml:id="echoid-s15154" xml:space="preserve">Si l’on veut avoir un cube qui ſoit au cube A B, comme <lb/>3 eſt à 7, il faut commencer par prendre avec le compas or-<lb/>dinaire le côté A B, & </s> <s xml:id="echoid-s15155" xml:space="preserve">le porter ſur la ligne des ſolides, de <lb/>maniere qu’il correſponde aux points 7 & </s> <s xml:id="echoid-s15156" xml:space="preserve">7: </s> <s xml:id="echoid-s15157" xml:space="preserve">ainſi ſuppoſant <lb/>que l’intervalle des points K & </s> <s xml:id="echoid-s15158" xml:space="preserve">L ſoit celui du nombre 7, l’on <lb/>n’aura plus qu’à prendre l’intervalle I H de 3 en 3 pour avoir le <lb/>côté du cube que l’on demande. </s> <s xml:id="echoid-s15159" xml:space="preserve">Ainſi faiſant C D égal à H I, <lb/>il y aura même raiſon du cube A B au cube C D, que de 7 à 3.</s> <s xml:id="echoid-s15160" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1222" type="section" level="1" n="883"> <head xml:id="echoid-head1065" xml:space="preserve">PROPOSITION XXIX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15161" xml:space="preserve">900. </s> <s xml:id="echoid-s15162" xml:space="preserve">Trouver le rapport qui eſt entre deux cubes.</s> <s xml:id="echoid-s15163" xml:space="preserve"/> </p> <pb o="469" file="0547" n="567" rhead="DE MATHÉMATIQUE. Liv. XIII."/> <p> <s xml:id="echoid-s15164" xml:space="preserve">Pour trouver le rapport qui eſt entre deux cubes quelconques <lb/> <anchor type="note" xlink:label="note-0547-01a" xlink:href="note-0547-01"/> C D & </s> <s xml:id="echoid-s15165" xml:space="preserve">A B, il faut prendre le côté C D du plus petit cube, <lb/>& </s> <s xml:id="echoid-s15166" xml:space="preserve">ouvrir le compas de proportion, enſorte que l’intervalle H I, <lb/>pris vers le centre, ſoit égal à ce côté. </s> <s xml:id="echoid-s15167" xml:space="preserve">Après cela, l’on pren-<lb/>dra le côté A B pour le porter en un endroit, comme K L, <lb/>dont l’intervalle lui ſoit égal, & </s> <s xml:id="echoid-s15168" xml:space="preserve">le rapport que l’on trouvera <lb/>entre les nombres qui ſeront marqués aux points I & </s> <s xml:id="echoid-s15169" xml:space="preserve">K, ſera <lb/>le même que celui du cube C D au cube A B.</s> <s xml:id="echoid-s15170" xml:space="preserve"/> </p> <div xml:id="echoid-div1222" type="float" level="2" n="1"> <note position="right" xlink:label="note-0547-01" xlink:href="note-0547-01a" xml:space="preserve">Figure 319 <lb/>& 322.</note> </div> </div> <div xml:id="echoid-div1224" type="section" level="1" n="884"> <head xml:id="echoid-head1066" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s15171" xml:space="preserve">Comme tous les ſolides ſemblables ſont dans la même raiſon <lb/>que les cubes de leurs côtés homologues, il s’enſuit que l’on <lb/>pourra faire à l’égard des cylindres, des cônes, des pyramides, <lb/>& </s> <s xml:id="echoid-s15172" xml:space="preserve">des ſpheres, les mêmes opérations que l’on vient de faire <lb/>pour les cubes, comme dans les propoſitions précédentes.</s> <s xml:id="echoid-s15173" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1225" type="section" level="1" n="885"> <head xml:id="echoid-head1067" xml:space="preserve"><emph style="sc">Application de la</emph> <emph style="sc">Geometrie a l’</emph><emph style="sc">Artillerie</emph>.</head> <head xml:id="echoid-head1068" xml:space="preserve">PROPOSITION XXX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15174" xml:space="preserve">901. </s> <s xml:id="echoid-s15175" xml:space="preserve">Faire l’analyſe de l’alliage du métail dont on fait les pieces <lb/>de canon.</s> <s xml:id="echoid-s15176" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15177" xml:space="preserve">Pour connoître l’utilité de ce problême, il faut être prévenu <lb/>que le métail dont on fait les pieces d’Artillerie de fonte, eſt <lb/>compoſé de roſette, que l’on appelle communément cuivre <lb/>rouge, & </s> <s xml:id="echoid-s15178" xml:space="preserve">d’étain fin d’Angleterre; </s> <s xml:id="echoid-s15179" xml:space="preserve">& </s> <s xml:id="echoid-s15180" xml:space="preserve">comme il doit y avoir <lb/>une proportion entre la roſette & </s> <s xml:id="echoid-s15181" xml:space="preserve">l’étain qui compoſent le <lb/>métail, les Fondeurs les plus expérimentés ſuivent celle de <lb/>100 à 12, c’eſt-à-dire que ſur 100 livres de roſette ils mettent <lb/>12 livres d’étain.</s> <s xml:id="echoid-s15182" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15183" xml:space="preserve">Or comme il arrive tous les jours que dans les Fonderies on <lb/>fond des pieces qui ſont hors d’état de ſervir pour en faire de <lb/>nouvelles, & </s> <s xml:id="echoid-s15184" xml:space="preserve">que les Fondeurs ſont embarraſſés pour ſçavoir <lb/>ſi le métail eſt conforme à l’alliage qu’ils ſuivent, pour qu’il <lb/>ne ſoit ni trop aigre ni trop doux; </s> <s xml:id="echoid-s15185" xml:space="preserve">voici comment on pourra <lb/>connoître au juſte la quantité de roſette & </s> <s xml:id="echoid-s15186" xml:space="preserve">d’étain qui compoſe <lb/>le métail des pieces.</s> <s xml:id="echoid-s15187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15188" xml:space="preserve">C’eſt une choſe démontrée par l’expérience, & </s> <s xml:id="echoid-s15189" xml:space="preserve">dont la rai-<lb/>ſon phyſique eſt facile à appercevoir, que les métaux perdent <pb o="470" file="0548" n="568" rhead="NOUVEAU COURS"/> de leur peſanteur lorſqu’ils ſont dans l’eau: </s> <s xml:id="echoid-s15190" xml:space="preserve">par exemple, ſi <lb/>l’on attache à une balance romaine un morceau de plomb pe-<lb/>ſant 48 livres, l’on verra que le corps étant mis dans l’eau, <lb/>de ſorte qu’il en ſoit environné de toutes parts, au lieu de peſer <lb/>48 livres, n’en peſera que 44, parce que le plomb perd dans <lb/>l’eau la douzieme partie de ſon poids, ainſi des autres métaux <lb/>qui perdent plus ou moins, ſelon qu’ils ſont plus ou moins pe-<lb/>ſans. </s> <s xml:id="echoid-s15191" xml:space="preserve">Mais comme nous avons beſoin de connoître ici ce que <lb/>perdent l’étain & </s> <s xml:id="echoid-s15192" xml:space="preserve">la roſette, l’on ſçaura que l’étain perd la <lb/>ſeptieme partie de ſon poids, & </s> <s xml:id="echoid-s15193" xml:space="preserve">que la roſette n’en perd que <lb/>la neuvieme partie.</s> <s xml:id="echoid-s15194" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15195" xml:space="preserve">Cela poſé, pour connoître la quantité de roſette & </s> <s xml:id="echoid-s15196" xml:space="preserve">d’étain <lb/>qui ſe trouve dans une piece de 24 livres de balle, qui peſe en-<lb/>viron 5200 livres, il faut avoir un morceau de la piece, qui <lb/>ſera, par exemple, un de ſes tronçons, & </s> <s xml:id="echoid-s15197" xml:space="preserve">le peſer bien exac-<lb/>tement; </s> <s xml:id="echoid-s15198" xml:space="preserve">& </s> <s xml:id="echoid-s15199" xml:space="preserve">ſuppoſant qu’il peſe 163 livres, on le peſera en-<lb/>ſuite dans l’eau, pour voir combien il perd de ſa peſanteur, <lb/>& </s> <s xml:id="echoid-s15200" xml:space="preserve">nous ſuppoſerons qu’il en perd 19 livres.</s> <s xml:id="echoid-s15201" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15202" xml:space="preserve">Préſentement il faut conſidérer le métail comme étant tout <lb/>de roſette, afin de voir, ſelon cette ſuppoſition, combien il <lb/>perd de ſa peſanteur, & </s> <s xml:id="echoid-s15203" xml:space="preserve">l’on trouvera qu’il perd {164/9}; </s> <s xml:id="echoid-s15204" xml:space="preserve">& </s> <s xml:id="echoid-s15205" xml:space="preserve">con-<lb/>ſidérant auſſi le métail comme étant tout étain, l’on cher-<lb/>chera combien il perd de ſa peſanteur, & </s> <s xml:id="echoid-s15206" xml:space="preserve">l’on trouvera qu’il <lb/>perd {163/7}: </s> <s xml:id="echoid-s15207" xml:space="preserve">ainſi ſi l’on nomme a la peſanteur du métail, b ſa <lb/>perte, c la perte du poids du métail, s’il étoit tout de roſette, <lb/>d la perte du même poids, s’il étoit tout étain, l’on aura <lb/>a = 163, b = 19, c = {163/9}, d = {163/7}; </s> <s xml:id="echoid-s15208" xml:space="preserve">& </s> <s xml:id="echoid-s15209" xml:space="preserve">nommant x la quan-<lb/>tité de roſette qui eſt dans le métail, & </s> <s xml:id="echoid-s15210" xml:space="preserve">y la quantité d’étain, <lb/>voici comment on trouvera la valeur de ces deux inconnues.</s> <s xml:id="echoid-s15211" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15212" xml:space="preserve">Il faut commencer par faire deux proportions, en diſant: <lb/></s> <s xml:id="echoid-s15213" xml:space="preserve">Comme a, poids du métail conſidéré comme roſette eſt à c, <lb/>perte de ce poids de roſette, ainſi x, qui eſt la quantité de <lb/>roſette inconnue, eſt à la perte du poids de la même roſette <lb/>inconnue; </s> <s xml:id="echoid-s15214" xml:space="preserve">ce qui donne a:</s> <s xml:id="echoid-s15215" xml:space="preserve">c:</s> <s xml:id="echoid-s15216" xml:space="preserve">:x:</s> <s xml:id="echoid-s15217" xml:space="preserve">{cx/a}; </s> <s xml:id="echoid-s15218" xml:space="preserve">& </s> <s xml:id="echoid-s15219" xml:space="preserve">faiſant la même choſe <lb/>pour l’étain, l’on dira: </s> <s xml:id="echoid-s15220" xml:space="preserve">Comme a, poids du métail conſidéré <lb/>comme étain eſt à d, perte de ce poids d’étain, ainſi y, va <lb/>leur de la quantité inconnue, eſt à la perte de cette quantité <lb/>d’étain, qui donnera encore cette proportion a:</s> <s xml:id="echoid-s15221" xml:space="preserve">d:</s> <s xml:id="echoid-s15222" xml:space="preserve">:y:</s> <s xml:id="echoid-s15223" xml:space="preserve">{dy/a}.</s> <s xml:id="echoid-s15224" xml:space="preserve"/> </p> <pb o="471" file="0549" n="569" rhead="DE MATHÉMATIQUE. Liv. XIII."/> <p> <s xml:id="echoid-s15225" xml:space="preserve">Mais comme l’on a trouvé {cx/a} pour la perte du poids de la <lb/>roſette qui eſt dans le métail, & </s> <s xml:id="echoid-s15226" xml:space="preserve">{dy/a} pour la perte du poids d’é-<lb/>tain, qui eſt auſſi dans le métail, & </s> <s xml:id="echoid-s15227" xml:space="preserve">que ces deux quantités <lb/>font enſemble la perte du poids du métail: </s> <s xml:id="echoid-s15228" xml:space="preserve">l’on aura donc <lb/>cette équation {cx/a} + {dy/a} = b; </s> <s xml:id="echoid-s15229" xml:space="preserve">& </s> <s xml:id="echoid-s15230" xml:space="preserve">comme x & </s> <s xml:id="echoid-s15231" xml:space="preserve">y repréſentent <lb/>la roſette & </s> <s xml:id="echoid-s15232" xml:space="preserve">l’étain qui compoſent le métail, l’on pourra en-<lb/>core former cette équation x + y = a; </s> <s xml:id="echoid-s15233" xml:space="preserve">& </s> <s xml:id="echoid-s15234" xml:space="preserve">dégageant une de <lb/>ces deux inconnues, qui ſera, par exemple x, l’on aura <lb/>x = a - y; </s> <s xml:id="echoid-s15235" xml:space="preserve">& </s> <s xml:id="echoid-s15236" xml:space="preserve">ſubſtituant la valeur de x dans l’équation <lb/>{cx/a} + {dy/a} = b, il viendra {ac - yc + dy/a} = b, ou bien c + {dy-yc/a} = b. <lb/></s> <s xml:id="echoid-s15237" xml:space="preserve">Or ſi l’on fait paſſer c du premier membre dans le ſecond, & </s> <s xml:id="echoid-s15238" xml:space="preserve"><lb/>que l’on multiplie les deux membres par a, il viendra dy - yc <lb/>= ab - ac, qui étant diviſé par d - c, donne y = {ab - ac/d - c}, où <lb/>y eſt égal à des quantités connues: </s> <s xml:id="echoid-s15239" xml:space="preserve">par conſéquent ſi l’on met <lb/>dans l’équation x = a - y la valeur de y, l’on aura x = a <lb/>- {ab + ac/d - e} = {ad + ab/d - c}, qui donne auſſi la valeur de x.</s> <s xml:id="echoid-s15240" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15241" xml:space="preserve">Or pour connoître y en nombres, je conſidere qu’il eſt égal <lb/>à ab - ac diviſé par d - c: </s> <s xml:id="echoid-s15242" xml:space="preserve">& </s> <s xml:id="echoid-s15243" xml:space="preserve">comme b - c eſt multiplié par <lb/>a, je ſouſtrais de 19 de b {163/9} valeur de c, & </s> <s xml:id="echoid-s15244" xml:space="preserve">le reſte eſt {8/9}, que <lb/>je multiplie par 163, qui eſt la valeur de a pour avoir {1304/9}, que <lb/>je diviſe par {363/7} - {163/9} valeur de d - c, qui eſt {416/63}; </s> <s xml:id="echoid-s15245" xml:space="preserve">la diviſion <lb/>étant faite, l’on trouvera 28 pour la valeur de y: </s> <s xml:id="echoid-s15246" xml:space="preserve">& </s> <s xml:id="echoid-s15247" xml:space="preserve">cher-<lb/>chant de même la valeur de x, l’on trouvera qu’elle eſt de <lb/>135; </s> <s xml:id="echoid-s15248" xml:space="preserve">ce qui fait voir qu’il y a 135 livres de roſette, & </s> <s xml:id="echoid-s15249" xml:space="preserve">28 livres <lb/>d’étain dans le morceau de métail.</s> <s xml:id="echoid-s15250" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15251" xml:space="preserve">Pour ſçavoir préſentement la quantité d’étain qu’il y a dans <lb/>la piece de canon, il faut dire: </s> <s xml:id="echoid-s15252" xml:space="preserve">Si dans 163 livres de métail il <lb/>y a 28 livres d’étain, combien y en aura-t-il dans 5200 livres, <lb/>poids de la piece? </s> <s xml:id="echoid-s15253" xml:space="preserve">l’on trouvera qu’il y en a environ 894 livres, <lb/>& </s> <s xml:id="echoid-s15254" xml:space="preserve">par conſéquent il y a 4306 livres de roſette.</s> <s xml:id="echoid-s15255" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15256" xml:space="preserve">Mais comme la raiſon de 4306 livres à 894 n’eſt pas égale à <lb/>celle de 100 à 12, parce que nous avons ſuppoſé qu’il y avoit <lb/>dans le métail beaucoup plus d’étain qu’il n’en falloit, il ſera <lb/>facile de ſçavoir combien il faut ajouter de roſette pour que <lb/>l’alliage ſoit bien fait, en diſant: </s> <s xml:id="echoid-s15257" xml:space="preserve">Si pour 12 livres d’étain il <lb/>faut 100 livres de roſette, combien en faudra-t-il pour 894 <pb o="472" file="0550" n="570" rhead="NOUVEAU COURS"/> livres. </s> <s xml:id="echoid-s15258" xml:space="preserve">On trouvera qu’il en faut 7450 livres; </s> <s xml:id="echoid-s15259" xml:space="preserve">& </s> <s xml:id="echoid-s15260" xml:space="preserve">comme il y <lb/>en a déja 4306 livres, il faudra en ajouter 3144 livres.</s> <s xml:id="echoid-s15261" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15262" xml:space="preserve">Si l’on a pluſieurs pieces à refondre en même-tems, l’on <lb/>cherchera par la regle précédente ce qui manque à chacune de <lb/>roſette ou d’étain, afin que l’alliage ſoit dans la raiſon de <lb/>100 à 12.</s> <s xml:id="echoid-s15263" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1226" type="section" level="1" n="886"> <head xml:id="echoid-head1069" xml:space="preserve">PROPOSITION XXXI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15264" xml:space="preserve">902. </s> <s xml:id="echoid-s15265" xml:space="preserve">Trouver le calibre des boulets & </s> <s xml:id="echoid-s15266" xml:space="preserve">des pieces de canon.</s> <s xml:id="echoid-s15267" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15268" xml:space="preserve">Pour trouver le calibre des boulets de telle peſanteur que <lb/>l’on voudra, il faut ſçavoir d’abord le diametre d’un boulet <lb/>de même métail d’un poids déterminé, comme, par exem-<lb/>ple, celui d’une livre de fer coulé, qui eſt d’un pouce 10 lignes <lb/>8 points, & </s> <s xml:id="echoid-s15269" xml:space="preserve">conſidérer le diametre comme étant diviſé en un <lb/>grand nombre de petites parties égales, comme en 500 (pour <lb/>que dans le calcul on puiſſe négliger les reſtes), enſuite cuber <lb/>la valeur du diametre en petites parties, pour avoir 125000000 <lb/>pour ſon cube, que nous regarderons ici comme le boulet <lb/>même, parce que les boulets étant des ſpheres, ils ſont dans <lb/>la même raiſon que les cubes de leurs diametres: </s> <s xml:id="echoid-s15270" xml:space="preserve">c’eſt pour-<lb/>quoi ſi l’on veut avoir le diametre d’un boulet de 24, l’on <lb/>n’aura qu’à multiplier le cube d’un boulet d’une livre, c’eſt-à-<lb/>dire 125000000 par 24 pour avoir 3000000000, qui ſera le <lb/>cube du diametre du boulet de 24, puiſqu’il eſt 24 fois plus grand <lb/>que l’autre. </s> <s xml:id="echoid-s15271" xml:space="preserve">Ainſi en extrayant la racine cube de 3000000000, <lb/>l’on aura 1442 petites parties, que l’on pourra changer en <lb/>pouces, lignes & </s> <s xml:id="echoid-s15272" xml:space="preserve">points, en diſant: </s> <s xml:id="echoid-s15273" xml:space="preserve">Si 500 petites parties don-<lb/>nent un pouce 10 lignes 8 points pour le diametre du boulet <lb/>d’une livre, combien donneront 1442 petites parties pour le <lb/>diametre du boulet de 24. </s> <s xml:id="echoid-s15274" xml:space="preserve">On trouvera, après la regle faite, <lb/>que le diametre eſt de 5 pouces 5 lignes, & </s> <s xml:id="echoid-s15275" xml:space="preserve">un peu plus de <lb/>4 points.</s> <s xml:id="echoid-s15276" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15277" xml:space="preserve">Si l’on veut avoir le diametre de tout autre boulet, par <lb/>exemple, celui de 16, l’on fera comme on a fait pour celui <lb/>de 24, avec cette différence, qu’au lieu de multiplier 125000000 <lb/>par 24, il faudra le multiplier par 16, afin d’avoir le cube du <lb/>diametre du boulet qu’on cherche: </s> <s xml:id="echoid-s15278" xml:space="preserve">& </s> <s xml:id="echoid-s15279" xml:space="preserve">l’on pourra ſur ce prin-<lb/>cipe calculer une table pour tous les autres boulets.</s> <s xml:id="echoid-s15280" xml:space="preserve"/> </p> <pb o="473" file="0551" n="571" rhead="DE MATHÉMATIQUE. Liv. XIII."/> <p> <s xml:id="echoid-s15281" xml:space="preserve">Mais comme l’on a beſoin de connoître particuliérement <lb/>les diametres des boulets pour faire les coquilles dans leſquelles <lb/>on coule le fer qui doit les former, & </s> <s xml:id="echoid-s15282" xml:space="preserve">que la plûpart pour-<lb/>roient ſe trouver embarraſſés, s’ils ne connoiſſoient pas le dia-<lb/>metre du boulet d’une livre, ou s’ils ſoupçonnoient qu’il ne fût <lb/>pas aſſez juſte pour ſervir de baſe à une regle générale, en ce <lb/>cas l’on pourra faire couler un boulet de tel diametre que l’on <lb/>voudra, comme de 3 pouces, ſans s’embarraſſer de ſa peſan-<lb/>teur qu’après qu’il ſera fondu, parce que pour lors on le peſera <lb/>bien exactement; </s> <s xml:id="echoid-s15283" xml:space="preserve">& </s> <s xml:id="echoid-s15284" xml:space="preserve">ſuppoſant qu’on a trouvé qu’il peſe 5 livres <lb/>& </s> <s xml:id="echoid-s15285" xml:space="preserve">demie, l’on réduira ſon diametre en petites parties pour le <lb/>cuber, & </s> <s xml:id="echoid-s15286" xml:space="preserve">enſuite l’on dira: </s> <s xml:id="echoid-s15287" xml:space="preserve">Si 5 livres & </s> <s xml:id="echoid-s15288" xml:space="preserve">demie donnent tant <lb/>de petites parties pour le cube du diametre de ſon boulet, com-<lb/>bien une livre donnera-t’elle de petites parties pour le cube de <lb/>ſon diametre: </s> <s xml:id="echoid-s15289" xml:space="preserve">& </s> <s xml:id="echoid-s15290" xml:space="preserve">lorſqu’on aura trouvé ce que l’on cherche, <lb/>on en extraira la racine cube, qui donnera en petites parties <lb/>la valeur du diametre du boulet d’une livre, qu’il ſera facile <lb/>de réduire en pouces, lignes, &</s> <s xml:id="echoid-s15291" xml:space="preserve">c. </s> <s xml:id="echoid-s15292" xml:space="preserve">ſçachant que le diametre du <lb/>premier boulet eſt de 3 pouces.</s> <s xml:id="echoid-s15293" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15294" xml:space="preserve">Pour trouver le diametre des pieces, l’on ſçaura qu’il ne <lb/>differe que de peu de choſe de celui de leurs boulets; </s> <s xml:id="echoid-s15295" xml:space="preserve">& </s> <s xml:id="echoid-s15296" xml:space="preserve">comme <lb/>cette différence, qui eſt ce qu’on appelle vent du boulet, n’eſt <lb/>pas la même pour toutes les pieces, il ſuffira de ſçavoir le dia-<lb/>metre de la piece d’une livre, pour trouver celui de tous les au-<lb/>tres: </s> <s xml:id="echoid-s15297" xml:space="preserve">& </s> <s xml:id="echoid-s15298" xml:space="preserve">comme le diametre eſt d’un pouce 11 lignes 6 points, <lb/>parce que le boulet de cette piece a environ une ligne de <lb/>vent, on ſuppoſera, comme on a fait pour ſon boulet, que <lb/>le diametre de la piece eſt diviſé en 500 parties; </s> <s xml:id="echoid-s15299" xml:space="preserve">& </s> <s xml:id="echoid-s15300" xml:space="preserve">voulant <lb/>trouver celui de la piece de 24, l’on cubera 500 pour multiplier <lb/>le produit par 24, dont on extraira la racine cube, qui eſt en-<lb/>core 1442, dont on pourra connoître la valeur en pouces, <lb/>lignes, &</s> <s xml:id="echoid-s15301" xml:space="preserve">c. </s> <s xml:id="echoid-s15302" xml:space="preserve">en diſant: </s> <s xml:id="echoid-s15303" xml:space="preserve">Si 500 donnent un pouce 11 lignes <lb/>6 points pour le diametre de ſa piece d’une livre, combien <lb/>donneront 1442 pour le diametre de la piece de 24: </s> <s xml:id="echoid-s15304" xml:space="preserve">on trou-<lb/>vera que ce diametre eſt de 5 pouces 7 lignes 9 points.</s> <s xml:id="echoid-s15305" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1227" type="section" level="1" n="887"> <head xml:id="echoid-head1070" xml:space="preserve">PROPOSITION XXXII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15306" xml:space="preserve">903. </s> <s xml:id="echoid-s15307" xml:space="preserve">Trouver le diametre des cylindres ſervant à meſurer la poudre.</s> <s xml:id="echoid-s15308" xml:space="preserve"/> </p> <pb o="474" file="0552" n="572" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s15309" xml:space="preserve">L’on ne ſe ſert preſque jamais de balances dans les magaſins <lb/>& </s> <s xml:id="echoid-s15310" xml:space="preserve">dans les Arcenaux pour meſurer la poudre que l’on diſtribue <lb/>aux troupes, ſoit pour des détachemens ou pour tout autre <lb/>ſujet, parce qu’il faudroit trop de tems pour en faire la diſtri-<lb/>bution: </s> <s xml:id="echoid-s15311" xml:space="preserve">on ſe ſert, au lieu de balances, de certaines meſures de <lb/>fer blanc ou de cuivre, de figure cylindrique, qui contiennent <lb/>plus ou moins de livres de poudre, ou de parties de livres. </s> <s xml:id="echoid-s15312" xml:space="preserve">Or <lb/>comme ſouvent l’on eſt obligé de faire faire de ces meſures, <lb/>& </s> <s xml:id="echoid-s15313" xml:space="preserve">qu’on ne peut, ſans le ſecours de la Géométrie, ſçavoir les <lb/>dimenſions qu’il faut leur donner pour contenir une quantité <lb/>de poudre quelconque, voici une regle générale qui pourra <lb/>ſervir pour trouver le diametre de toutes les meſures que l’on <lb/>voudra: </s> <s xml:id="echoid-s15314" xml:space="preserve">mais comme il faut que ces meſures ſoient ſembla-<lb/>bles pour que la regle puiſſe convenir à toutes également, nous <lb/>ſuppoſerons que ces meſures étant cylindriques, la hauteur du <lb/>cylindre eſt égale au diametre du cercle qui lui ſert de baſe.</s> <s xml:id="echoid-s15315" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15316" xml:space="preserve">Cela poſé, étant prévenu qu’une meſure cylindrique, dont <lb/>le diametre eſt de 3 pouces, contient 4 livres de poudre, l’on <lb/>trouvera le diametre d’une meſure pour autant de livres que <lb/>l’on voudra: </s> <s xml:id="echoid-s15317" xml:space="preserve">par exemple, pour 10 livres, en diſant: </s> <s xml:id="echoid-s15318" xml:space="preserve">Si 4 livres <lb/>de poudre donne 125 pouces pour le cube du diametre de ſa <lb/>meſure, combien donneront 10 livres de poudre? </s> <s xml:id="echoid-s15319" xml:space="preserve">l’on trou-<lb/>vera 312 pouces & </s> <s xml:id="echoid-s15320" xml:space="preserve">demi cubes, dont il faudra extraire la ra-<lb/>cine qui ſera de 6 pouces 8 lignes 9 points, qui eſt la grandeur <lb/>qu’il faut donner au diametre de la meſure de 10 livres, qui <lb/>doit avoir auſſi la même hauteur: </s> <s xml:id="echoid-s15321" xml:space="preserve">il en ſera de même pour <lb/>telle autre meſure que l’on voudra.</s> <s xml:id="echoid-s15322" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15323" xml:space="preserve">Mais ſi l’on ignore le diametre d’une meſure pour une <lb/>certaine quantité de poudre, & </s> <s xml:id="echoid-s15324" xml:space="preserve">ſi l’on n’a aucun terme de la <lb/>proportion connue, dans ce cas il faut faire faire une me-<lb/>ſure à laquelle on donnera le diametre que l’on voudra, & </s> <s xml:id="echoid-s15325" xml:space="preserve">on <lb/>la remplira de poudre, aſin de ſçavoir ce qu’elle contient; </s> <s xml:id="echoid-s15326" xml:space="preserve">& </s> <s xml:id="echoid-s15327" xml:space="preserve"><lb/>ſçachant ce qu’elle contient, & </s> <s xml:id="echoid-s15328" xml:space="preserve">la valeur du diametre, l’on ſe <lb/>ſervira de la regle précédente pour trouver le diametre de <lb/>toutes les autres meſures, faiſant attention que ces meſures <lb/>ne peuvent avoir lieu que pour la poudre dont les grains ſont <lb/>approchans de même groſſeur que ſont ceux de la poudre à <lb/>canon: </s> <s xml:id="echoid-s15329" xml:space="preserve">car ſi les grains étoient plus fins, les meſures contien-<lb/>droient moins de poudre en peſanteur.</s> <s xml:id="echoid-s15330" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15331" xml:space="preserve">L’on voit que cette regle eſt établie ſur ce que les cylindres <pb o="475" file="0553" n="573" rhead="DE MATHÉMATIQUE. Liv. XIII."/> ſemblables ſont dans la même raiſon que les cubes de leurs <lb/>diametres. </s> <s xml:id="echoid-s15332" xml:space="preserve">Or comme les meſures dont il s’agit ici ſont ſup-<lb/>poſées avoir une hauteur égale à leur diametre, elles ſeront <lb/>donc ſemblables, & </s> <s xml:id="echoid-s15333" xml:space="preserve">par conſéquent leurs ſolidités, qui ne ſont <lb/>autre choſe que la quantité de poudre qu’elles contiennent, <lb/>ſeront dans la raiſon des cubes des diametres.</s> <s xml:id="echoid-s15334" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15335" xml:space="preserve">Mais ſi l’on vouloit avoir des meſures, dont la hauteur fût <lb/>plus grande ou plus petite que le diametre de la baſe (que nous <lb/>nommerons meſure irréguliere), il faudroit chercher le dia-<lb/>metre de la meſure pour la quantité de poudre que l’on veut <lb/>que cette meſure contienne, comme ſi cette meſure devoit <lb/>être réguliere, c’eſt-à-dire que le diametre fût égal à la hau-<lb/>teur; </s> <s xml:id="echoid-s15336" xml:space="preserve">enſuite cuber le diametre, & </s> <s xml:id="echoid-s15337" xml:space="preserve">diviſer le produit par la <lb/>hauteur de la meſure irréguliere, & </s> <s xml:id="echoid-s15338" xml:space="preserve">le quotient ſera la va-<lb/>leur du quarré du diametre de cette meſure. </s> <s xml:id="echoid-s15339" xml:space="preserve">Après cela, ſi <lb/>l’on extrait la racine quarrée de cette quantité, l’on aura le <lb/>diametre du cercle qui doit ſervir de baſe à la meſure que l’on <lb/>cherche.</s> <s xml:id="echoid-s15340" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15341" xml:space="preserve">Comme les cercles ſont dans la raiſon des quarrés de leurs <lb/>diametres, l’on pourra prendre à la place des cercles les quarrés <lb/>de leurs diametres. </s> <s xml:id="echoid-s15342" xml:space="preserve">Or comme les cylindres ſont égaux, lorſ-<lb/>que leurs hauteurs & </s> <s xml:id="echoid-s15343" xml:space="preserve">leurs baſes, ou les quarrés des diametres <lb/>de leurs baſes ſont réciproques, nommant a le diametre de la <lb/>baſe du cylindre régulier, a ſera auſſi ſa hauteur; </s> <s xml:id="echoid-s15344" xml:space="preserve">& </s> <s xml:id="echoid-s15345" xml:space="preserve">nommant <lb/>b la hauteur du cylindre irrégulier, & </s> <s xml:id="echoid-s15346" xml:space="preserve">x le diametre de ſa baſe, <lb/>il faut, pour que le cylindre régulier ſoit égal à l’irrégulier, <lb/>que b : </s> <s xml:id="echoid-s15347" xml:space="preserve">a :</s> <s xml:id="echoid-s15348" xml:space="preserve">: aa : </s> <s xml:id="echoid-s15349" xml:space="preserve">xx, d’où l’on tire bxx = aaa, ou bien xx = {aaa/b}, <lb/>ou encore x = √{aaa/b}\x{0020} = a √{b/a}\x{0020}, qui fait voir la raiſon de la regle <lb/>précédente.</s> <s xml:id="echoid-s15350" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15351" xml:space="preserve">Ce que nous venons de dire à l’égard des meſures pour la <lb/>poudre, ſe peut appliquer à toutes autres meſures cylindriques <lb/>pour telles choſes que ce ſoit.</s> <s xml:id="echoid-s15352" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1228" type="section" level="1" n="888"> <head xml:id="echoid-head1071" xml:space="preserve">PROPOSITION XXXIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15353" xml:space="preserve">904. </s> <s xml:id="echoid-s15354" xml:space="preserve">Trouver quelle longueur doivent avoir les pieces de canon <lb/>par rapport à leurs calibres.</s> <s xml:id="echoid-s15355" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15356" xml:space="preserve">Les extrêmités dans leſquelles on eſt tombé pour régler la <pb o="476" file="0554" n="574" rhead="NOUVEAU COURS"/> longueur des pieces de canon, en faiſant celles de même cali-<lb/>bre, tantôt fort longues, tantôt fort courtes, m’ont fait penſer <lb/>qu’il devoit y avoir une longueur pour les pieces cylindriques <lb/>de chaque calibre, qui étoit telle, qu’avec la charge ordinaire <lb/>le boulet reçût la plus grande vîteſſe que l’impulſion de la pou-<lb/>dre eſt capable de lui donner; </s> <s xml:id="echoid-s15357" xml:space="preserve">& </s> <s xml:id="echoid-s15358" xml:space="preserve">ſi pour la connoître l’on eſt <lb/>obligé de conſidérer les effets de la poudre dans le canon, <lb/>voici, à mon avis, ce que l’on peut dire de plus plauſible ſur <lb/>ce ſujet.</s> <s xml:id="echoid-s15359" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15360" xml:space="preserve">Comme l’on ne peut douter que plus il y a de poudre en-<lb/> <anchor type="note" xlink:label="note-0554-01a" xlink:href="note-0554-01"/> flammée dans un canon, & </s> <s xml:id="echoid-s15361" xml:space="preserve">plus le boulet reçoit de mouve-<lb/> <anchor type="note" xlink:label="note-0554-02a" xlink:href="note-0554-02"/> ment, nous ſuppoſerons que l’on a mis pour la charge de la <lb/>piece D G la quantité de poudre D E. </s> <s xml:id="echoid-s15362" xml:space="preserve">Cela poſé, auſſi-tôt que <lb/>le feu de l’amorce ſe ſera introduit au point A de la lumiere, <lb/>les premiers grains de poudre enflammés raréfieront l’air qu’ils <lb/>contiennent, & </s> <s xml:id="echoid-s15363" xml:space="preserve">celui dont ils ſont environnés, & </s> <s xml:id="echoid-s15364" xml:space="preserve">écarte-<lb/>ront à la ronde tout ce qui leur fera obſtacle, & </s> <s xml:id="echoid-s15365" xml:space="preserve">ſucceſſivement <lb/>la poudre continuant à s’enflammer, elle occupera un bien plus <lb/>grand volume qu’auparavant; </s> <s xml:id="echoid-s15366" xml:space="preserve">& </s> <s xml:id="echoid-s15367" xml:space="preserve">agiſſant avec beaucoup de <lb/>violence à droite & </s> <s xml:id="echoid-s15368" xml:space="preserve">à gauche du point A, & </s> <s xml:id="echoid-s15369" xml:space="preserve">particuliérement <lb/>du côté où elle trouvera moins de réfiſtance, qui eſt celui du <lb/>boulet qu’elle chaſſera du côté de la bouche, avec une grande <lb/>quantité de poudre, qui n’aura pas encore eu le tems de s’en-<lb/>flammer, & </s> <s xml:id="echoid-s15370" xml:space="preserve">la vîteſſe du boulet augmentant dans la même <lb/>raiſon du volume de la poudre enflammée, il ſe trouvera dans <lb/>un inſtant chaſſé en G pour ſortir de la piece. </s> <s xml:id="echoid-s15371" xml:space="preserve">Or ſi dans le <lb/>tems que le boulet a parcouru l’eſpace E G, la poudre qui l’ac-<lb/>compagnoit n’a pu être enflammée entiérement, il en ſortira <lb/>une quantité F avec le boulet, qui s’écartera comme du petit <lb/>plomb, au lieu que ſi la piece avoit été plus longue que je ne <lb/>la ſuppoſe ici, le boulet ayant à parcourir un plus grand eſ-<lb/>pace, la poudre qui a été chaſſée avec lui auroit eu le tems <lb/>de s’enflammer, & </s> <s xml:id="echoid-s15372" xml:space="preserve">par conſéquent auroit été capable d’un plus <lb/>grand effort: </s> <s xml:id="echoid-s15373" xml:space="preserve">ainſi l’on peut conclure que la proportion qu’il <lb/>doit y avoir entre D E & </s> <s xml:id="echoid-s15374" xml:space="preserve">D G, c’eſt-à-dire entre la charge & </s> <s xml:id="echoid-s15375" xml:space="preserve"><lb/>la longueur de la piece, doit être telle que la poudre acheve <lb/>de s’enflammer entiérement à l’inſtant que le boulet ſort de la <lb/>piece; </s> <s xml:id="echoid-s15376" xml:space="preserve">d’où il ſuit qu’un canon qui eſt chargé plus qu’il ne <lb/>faut, ne chaſſe pas pour cela ſon boulet plus loin, & </s> <s xml:id="echoid-s15377" xml:space="preserve">même <lb/>au contraire, puiſque plus il y aura de parties entre la poudre <pb o="477" file="0555" n="575" rhead="DE MATHÉMATIQUE. Liv. XIII."/> agiſſante & </s> <s xml:id="echoid-s15378" xml:space="preserve">le boulet, moins il recevra de mouvement: </s> <s xml:id="echoid-s15379" xml:space="preserve">& </s> <s xml:id="echoid-s15380" xml:space="preserve"><lb/>cela eſt ſi vrai, que ſi au lieu d’un bouchon de fourrage ordi-<lb/>naire entre la poudre & </s> <s xml:id="echoid-s15381" xml:space="preserve">le boulet, l’on en mettoit cinq ou <lb/>ſix, l’on s’appercevroit viſiblement que la portée ne ſeroit pas <lb/>ſi longue que s’il n’y en avoit qu’un, comme j’en ai fait l’ex-<lb/>périence: </s> <s xml:id="echoid-s15382" xml:space="preserve">car le boulet ne recevant de mouvement que par <lb/>l’impulſion que la poudre a imprimée au premier bouchon, <lb/>celui-ci ne peut le communiquer aux autres, pour aller juſ-<lb/>qu’au boulet, ſans l’altérer; </s> <s xml:id="echoid-s15383" xml:space="preserve">ce qui fait qu’il s’en faut de beau-<lb/>coup que le boulet n’ait autant de vîteſſe que s’il avoit reçu <lb/>ſon impulſion immédiatement de la poudre même. </s> <s xml:id="echoid-s15384" xml:space="preserve">Ainſi le <lb/>trop de poudre fera le même effet que s’il y avoit trop de <lb/>bourre.</s> <s xml:id="echoid-s15385" xml:space="preserve"/> </p> <div xml:id="echoid-div1228" type="float" level="2" n="1"> <note position="left" xlink:label="note-0554-01" xlink:href="note-0554-01a" xml:space="preserve">Pl. XXIII.</note> <note position="left" xlink:label="note-0554-02" xlink:href="note-0554-02a" xml:space="preserve">Figure 323.</note> </div> <p> <s xml:id="echoid-s15386" xml:space="preserve">Mais ſi au lieu d’une piece trop courte nous en ſuppoſons <lb/>une trop longue, comme L O, il n’y a point de doute, quoi-<lb/>qu’elle ſoit de même calibre que la précédente, & </s> <s xml:id="echoid-s15387" xml:space="preserve">chargée <lb/>avec la même quantité de poudre, qu’elle ne porte pas ſi loin <lb/>que ſi elle étoit d’une juſte longueur: </s> <s xml:id="echoid-s15388" xml:space="preserve">car ſuppoſant que la <lb/>poudre L M faiſant ſon effet, ait pouſſé le boulet juſqu’au <lb/>point N, qui eſt l’endroit où elle auroit achevé de s’enflammer <lb/>entiérement, il eſt certain que ſi le boulet a encore à parcourir <lb/>l’eſpace N O, il ſortira avec moins de violence de l’endroit O, <lb/>que s’il étoit parti d’abord de l’endroit N: </s> <s xml:id="echoid-s15389" xml:space="preserve">car dans le tems que <lb/>le reſte de la poudre acheve de s’enflammer vers N, la flamme <lb/>de celle qui a commencé vers la culaſſe ſe dilate, & </s> <s xml:id="echoid-s15390" xml:space="preserve">l’air raréſié <lb/>s’amortiſſant de ce côté-là, il n’y a plus que celui qui eſt vers <lb/>N, qui fait impreſſion ſur le boulet; </s> <s xml:id="echoid-s15391" xml:space="preserve">de ſorte que ſi la piece <lb/>étoit aſſez longue pour que l’impulſion de la poudre fût entiére-<lb/>ment amortie à l’inſtant que le boulet eſt prêt à ſortir de la piece, <lb/>il pourroit arriver que l’air que le boulet auroit chaſſé avec beau-<lb/>coup de violence, cherchant à rentrer dans la piece, le repouſ-<lb/>ſeroit vers la culaſſe; </s> <s xml:id="echoid-s15392" xml:space="preserve">ce qui arriveroit ſans doute, ſi à l’inſtant <lb/>que le feu a pris à la poudre, l’on pouvoit boucher la lumiere <lb/>avec aſſez de promptitude, pour empêcher que l’air que le <lb/>boulet chaſſe ne ſoit remplacé par celui qui s’introduiroit <lb/>par-là.</s> <s xml:id="echoid-s15393" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15394" xml:space="preserve">Puiſque les pieces d’une trop grande longueur font moins <lb/>d’effet que les autres, il ne faut donc plus s’étonner ſi la cou-<lb/>levrine de Nancy (contre l’opinion commune) a moins de <lb/>portée que les pieces de même calibre, comme M. </s> <s xml:id="echoid-s15395" xml:space="preserve">Dumez <pb o="478" file="0556" n="576" rhead="NOUVEAU COURS"/> l’a obſervé dans les épreuves qu’il a faites à Dunkerque.</s> <s xml:id="echoid-s15396" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15397" xml:space="preserve">Ce raiſonnement fait voir que la charge doit dépendre de <lb/>la longueur de la piece, & </s> <s xml:id="echoid-s15398" xml:space="preserve">la longueur de la piece de la force <lb/>de la charge: </s> <s xml:id="echoid-s15399" xml:space="preserve">mais comme pour de groſſes charges il faudroit <lb/>de longues pieces, dont le ſervice & </s> <s xml:id="echoid-s15400" xml:space="preserve">le tranſport ſouffriroient <lb/>bien des difficultés, joint à la grande conſommation de pou-<lb/>dre que l’on ſeroit obligé de faire; </s> <s xml:id="echoid-s15401" xml:space="preserve">comme il ſemble que la <lb/>méthode de charger (comme on le pratique ordinairement) <lb/>les pieces à la moitié du poids du boulet eſt la meilleure, il faut, <lb/>en comptant là-deſſus, chercher quelle doit être la longueur <lb/>d’une piece par rapport à un calibre quelconque, parce qu’après <lb/>cela l’on peut établir des regles pour connoître la longueur de <lb/>tous les calibres imaginables. </s> <s xml:id="echoid-s15402" xml:space="preserve">Je crois que le plus ſûr moyen <lb/>pour parvenir à cette connoiſſance, eſt de faire un canon fort <lb/>long, dont le calibre ſeroit, par exemple, de 8 livres, & </s> <s xml:id="echoid-s15403" xml:space="preserve">le <lb/>charger à la moitié du poids de ſon boulet, puis le tirer de but <lb/>en blanc, pour voir ſa portée: </s> <s xml:id="echoid-s15404" xml:space="preserve">& </s> <s xml:id="echoid-s15405" xml:space="preserve">comme l’on ſuppoſe que la <lb/>piece eſt plus longue qu’elle ne doit être, on la ſciera pour la <lb/>diminuer d’un calibre, & </s> <s xml:id="echoid-s15406" xml:space="preserve">on tirera un autre coup pour voir <lb/>de combien elle aura porté plus loin que le premier; </s> <s xml:id="echoid-s15407" xml:space="preserve">& </s> <s xml:id="echoid-s15408" xml:space="preserve">conti-<lb/>nuant toujours à raccourcir la piece, en la diminuant de quel-<lb/>ques pouces, ſur la fin l’on arrivera à un point où la piece, <lb/>pour être un peu trop courte, portera moins loin qu’aupara-<lb/>vant; </s> <s xml:id="echoid-s15409" xml:space="preserve">& </s> <s xml:id="echoid-s15410" xml:space="preserve">conſidérant la longueur moyenne entre celle du der-<lb/>nier coup & </s> <s xml:id="echoid-s15411" xml:space="preserve">le pénultieme, l’on aura au juſte la longueur de <lb/>la piece par rapport à ſa charge, pour que la poudre ſoit ca-<lb/>pable du plus grand effet qu’il eſt poſſible avec la même quan-<lb/>tité de poudre.</s> <s xml:id="echoid-s15412" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15413" xml:space="preserve">Cependant comme ce que je propoſe ici pourroit peut-être <lb/>n’avoir pas ſes partiſans, quoique le ſujet ſoit aſſez de conſé-<lb/>quence pour prendre toutes ces meſures, voici encore ce que <lb/>l’on pourroit faire.</s> <s xml:id="echoid-s15414" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15415" xml:space="preserve">Comme l’expérience fait voir tous les jours que les petites <lb/>pieces portent plus loin à proportion que les groſſes, puiſque, <lb/>ſelon les épreuves qu’en a faites M. </s> <s xml:id="echoid-s15416" xml:space="preserve">Dumez, il a trouvé que <lb/>nos pieces de France chargées aux deux tiers de la peſanteur <lb/>du boulet, & </s> <s xml:id="echoid-s15417" xml:space="preserve">pointées à 45 degrés, portoient,</s> </p> <pb o="479" file="0557" n="577" rhead="DE MATHÉMATIQUE. Liv. XIII."/> <note position="right" xml:space="preserve"># { # la piece de 24 à # 2250 toiſes. <lb/># # de 16 à # 2020 <lb/>Premiérement, # # de 12 à # 1870 <lb/># # de 8 à # 1660 <lb/># # & la piece de 4 à # 1520; <lb/></note> <p> <s xml:id="echoid-s15418" xml:space="preserve">Ce qui me fait croire que la longueur des petites pieces eſt <lb/>mieux proportionnée par rapport à leurs calibres, que celle <lb/>des groſſes: </s> <s xml:id="echoid-s15419" xml:space="preserve">ainſi ſuppoſant qu’une piece de canon de 4, qui a <lb/>ordinairement 6 pieds de longueur dans l’ame, ſoit bien pro-<lb/>portionnée, voici comment on pourra trouver la longueur <lb/>des pieces de tel calibre que l’on voudra.</s> <s xml:id="echoid-s15420" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15421" xml:space="preserve">Conſidérant A C comme étant la longueur de l’ame d’une <lb/> <anchor type="note" xlink:label="note-0557-02a" xlink:href="note-0557-02"/> piece de 4; </s> <s xml:id="echoid-s15422" xml:space="preserve">A B l’eſpace qu’occupe la poudre dans le canon; </s> <s xml:id="echoid-s15423" xml:space="preserve">& </s> <s xml:id="echoid-s15424" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0557-03a" xlink:href="note-0557-03"/> H K la longueur de la piece de 24, que je cherche, & </s> <s xml:id="echoid-s15425" xml:space="preserve">H I l’eſ-<lb/>pace qu’occupe ſa charge; </s> <s xml:id="echoid-s15426" xml:space="preserve">je fais attention que la poudre agiſ-<lb/>ſant dans la piece de 4 & </s> <s xml:id="echoid-s15427" xml:space="preserve">dans la piece de 24, dans la raiſon <lb/>de la quantité qu’il s’en trouve dans l’une & </s> <s xml:id="echoid-s15428" xml:space="preserve">dans l’autre (en <lb/>faiſant abſtraction des forces unies), il faut, afin que le boulet <lb/>de l’une & </s> <s xml:id="echoid-s15429" xml:space="preserve">de l’autre piece parte dans le moment que la poudre <lb/>eſt entiérement allumée, qu’il y ait même raiſon du cylindre <lb/>A B au cylindre A C, que du cylindre H I au cylindre H K: <lb/></s> <s xml:id="echoid-s15430" xml:space="preserve">& </s> <s xml:id="echoid-s15431" xml:space="preserve">comme je puis prendre à la place des cylindres A B & </s> <s xml:id="echoid-s15432" xml:space="preserve">H I <lb/>la quantité de poudre qu’ils contiennent, & </s> <s xml:id="echoid-s15433" xml:space="preserve">à la place des cy-<lb/>lindres A C & </s> <s xml:id="echoid-s15434" xml:space="preserve">H K le cube de leurs axes, puiſqu’ils doivent <lb/>être ſemblables, l’on pourra (pour trouver la longueur H K) <lb/>dire: </s> <s xml:id="echoid-s15435" xml:space="preserve">Si deux livres de poudre, qui eſt la charge de la piece de <lb/>4, donne 216 pour le cube de ſon axe, combien donneront <lb/>12 livres de poudre, qui eſt la charge de la piece de 24, pour le <lb/>cube de l’axe de la même piece? </s> <s xml:id="echoid-s15436" xml:space="preserve">l’on trouvera 1296 pieds <lb/>cubes, dont la racine cube eſt 11 pieds, moins très-peu de <lb/>choſe: </s> <s xml:id="echoid-s15437" xml:space="preserve">ainſi l’on voit que l’ame de la piece de 24, pour être <lb/>proportionnée à ſa charge par rapport à celle de 4, doit avoir <lb/>11 pieds de longueur; </s> <s xml:id="echoid-s15438" xml:space="preserve">& </s> <s xml:id="echoid-s15439" xml:space="preserve">comme l’ame de ces mêmes pieces <lb/>n’a ordinairement qu’environ 9 pieds: </s> <s xml:id="echoid-s15440" xml:space="preserve">ſelon ce principe, elles <lb/>ſont trop courtes de 2 pieds.</s> <s xml:id="echoid-s15441" xml:space="preserve"/> </p> <div xml:id="echoid-div1229" type="float" level="2" n="2"> <note position="right" xlink:label="note-0557-02" xlink:href="note-0557-02a" xml:space="preserve">Pl. XXIII.</note> <note position="right" xlink:label="note-0557-03" xlink:href="note-0557-03a" xml:space="preserve">Figure 323.</note> </div> <p> <s xml:id="echoid-s15442" xml:space="preserve">L’on pourra trouver de même la longueur de toutes les au-<lb/>tres pieces, lorſqu’elles auront leurs chambres cylindriques: <lb/></s> <s xml:id="echoid-s15443" xml:space="preserve">car ſi elles étoient autrement, il faudroit prendre d’autres <lb/>meſures.</s> <s xml:id="echoid-s15444" xml:space="preserve"/> </p> <pb o="480" file="0558" n="578" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s15445" xml:space="preserve">Les pieces dont on ſe ſert ordinairement n’étant point d’une <lb/>longueur proportionnée à celle de la piece de 4, & </s> <s xml:id="echoid-s15446" xml:space="preserve">comme il <lb/>n’y a point d’apparence qu’on les fonde toutes exprès pour les <lb/>y faire convenir, il faut, puiſque la charge d’une piece dépend <lb/>de ſa longueur, comme la longueur dépend de la charge, faire <lb/>voir comment on peut trouver la charge de toutes les pieces, en <lb/>connoiſſant le calibre & </s> <s xml:id="echoid-s15447" xml:space="preserve">la longueur. </s> <s xml:id="echoid-s15448" xml:space="preserve">Comme les ames des <lb/>pieces qui ne ſont point ſemblables, ſont dans la raiſon com-<lb/>poſée des quarrés des diametres des pieces & </s> <s xml:id="echoid-s15449" xml:space="preserve">des axes des <lb/>mêmes pieces, ſi l’on multiplie le quarré du diametre de cha-<lb/>que piece par l’axe, l’on pourra trouver la charge qui convient <lb/>aux pieces, puiſque ces charges doivent être dans la raiſon des <lb/>produits des quarrés des diametres des pieces, par les axes des <lb/>mêmes pieces. </s> <s xml:id="echoid-s15450" xml:space="preserve">Ainſi voulant ſçavoir la charge d’une piece de <lb/>24 ordinaire, dont l’ame a 9 pieds de longueur; </s> <s xml:id="echoid-s15451" xml:space="preserve">j’ai recours <lb/>à la piece de 4, pour en prendre le diametre, qui eſt 3 pouces, <lb/>que je quarre pour en multiplier le quarré par la longueur de <lb/>l’axe, qui eſt 6 pieds, dont le produit eſt 54; </s> <s xml:id="echoid-s15452" xml:space="preserve">enſuite je quarre <lb/>le diametre de la piece de 24, qui donne 29 pouces 9 lignes <lb/>6 points, que je multiplie par l’axe, qui eſt 9, & </s> <s xml:id="echoid-s15453" xml:space="preserve">le produit <lb/>eſt 268. </s> <s xml:id="echoid-s15454" xml:space="preserve">Après cela, je fais une Regle de Trois, en diſant: <lb/></s> <s xml:id="echoid-s15455" xml:space="preserve">Si 54, produit du quarré du diametre de la piece de 4 par ſon <lb/>axe, donne deux livres pour ſa charge, combien donneront <lb/>268, produit du quarré du diametre de la piece de 24 par ſon <lb/>axe, pour la charge de la même piece? </s> <s xml:id="echoid-s15456" xml:space="preserve">l’on trouvera 10 livres <lb/>moins quelque petite choſe, qui fait voir que les pieces de 24, <lb/>dont l’ame à 9 pieds de longueur, doivent être chargées à <lb/>10 livres de poudre, quand la piece de 4 ſera chargée à la moitié <lb/>de ſon boulet.</s> <s xml:id="echoid-s15457" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15458" xml:space="preserve">De la même façon, ſi l’on veut ſçavoir quelle doit être la <lb/>charge de la coulevrine de Nancy, par rapport à la piece de <lb/>4, chargée à la moitié de ſon boulet, il faut être prévenu que <lb/>cette piece eſt de 18 livres de balle, que ſon diametre eſt de <lb/>5 pouces 1 ligne 6 points, & </s> <s xml:id="echoid-s15459" xml:space="preserve">que la longueur de ſon axe eſt <lb/>de 20 pieds: </s> <s xml:id="echoid-s15460" xml:space="preserve">ainſi faiſant la regle, on trouvera qu’elle doit <lb/>être chargée à 20 livres de poudre.</s> <s xml:id="echoid-s15461" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15462" xml:space="preserve">Mais comme ſon métail ne réſiſteroit peut-être pas à une <lb/>charge auſſi forte que celle-ci, il n’y a qu’à voir la longueur <lb/>qui lui convient pour la charge de la moitié de ſon boulet, <lb/>c’eſt-à-dire pour 9 livres de poudre, en diſant: </s> <s xml:id="echoid-s15463" xml:space="preserve">Si 2 livres de <pb o="481" file="0559" n="579" rhead="DE MATHÉMATIQUE. Liv. XIII."/> poudre, qui eſt la charge de la piece de 4, donnent 216 pour <lb/>le cube de ſon axe, que donneront 9 livres de poudre, qui eſt <lb/>la charge d’une piece de 18, pour le cube de ſon axe, que l’on <lb/>trouvera de 972, dont la racine cube eſt environ 9 pieds <lb/>11 pouces, qui eſt la longueur que devroit avoir l’ame de la <lb/>coulevrine, pour être bien proportionnée? </s> <s xml:id="echoid-s15464" xml:space="preserve">Ainſi l’on con-<lb/>noîtra que cette piece eſt environ de 10 pieds plus longue qu’elle <lb/>ne devroit être.</s> <s xml:id="echoid-s15465" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15466" xml:space="preserve">905. </s> <s xml:id="echoid-s15467" xml:space="preserve">Depuis 1723 que j’ai écrit ce diſcours, j’ai fait des <lb/>épreuves pour ſçavoir quelle étoit la charge des pieces de dif-<lb/>férens calibre en uſage en France pour chaſſer le boulet à la <lb/>plus grande diſtance, ou pour battre en breche avec le plus de <lb/>violence qu’il eſt poſſible, afin que, partant de ce point, on pût <lb/>la diminuer ſelon les occaſions, & </s> <s xml:id="echoid-s15468" xml:space="preserve">jamais l’augmenter. </s> <s xml:id="echoid-s15469" xml:space="preserve">J’ai <lb/>fait mes premieres épreuves à l’Ecole de la Fere, dans le mois <lb/>d’Octobre 1739, en préſence de Meſſieurs les Officiers d’Ar-<lb/>tillerie, en chargeant chaque piece de 8, de 12, de 16, & </s> <s xml:id="echoid-s15470" xml:space="preserve">de <lb/>24, avec des charges qui alloient en augmentant par gradation <lb/>d’une demi-livre de poudre, en commençant par une charge <lb/>égale à la huitieme partie de la peſanteur du boulet, & </s> <s xml:id="echoid-s15471" xml:space="preserve">finiſ-<lb/>ſoient par celle des deux tiers de la même peſanteur. </s> <s xml:id="echoid-s15472" xml:space="preserve">L’on <lb/>tiroit de ſuite quatre coups avec la même charge, dont on pre-<lb/>noit enſuite la portée moyenne. </s> <s xml:id="echoid-s15473" xml:space="preserve">J’entends que le premier coup <lb/>pour la piece de 16 a été chargée de deux livres de poudre, que <lb/>la ſeconde charge a été de deux livres & </s> <s xml:id="echoid-s15474" xml:space="preserve">demie, la troiſieme <lb/>de trois livres, la quatrieme de trois livres & </s> <s xml:id="echoid-s15475" xml:space="preserve">demie, ainſi de <lb/>ſuite juſqu’à dix livres & </s> <s xml:id="echoid-s15476" xml:space="preserve">demie, qui eſt à peu près les deux <lb/>tiers de 16, peſanteur du boulet. </s> <s xml:id="echoid-s15477" xml:space="preserve">On en a uſé de même pour <lb/>les pieces des autres calibres toutes pointées ſous l’angle de <lb/>4 degrés formé par la direction de l’ame avec l’horizon.</s> <s xml:id="echoid-s15478" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15479" xml:space="preserve">Ayant meſuré bien exactement toutes les portées de ces <lb/>pieces pour chaque charge différente, j’ai reconnu que celle <lb/>qui produiroit le plus grand effet, c’eſt-à-dire qui chaſſoit le <lb/>boulet à la plus grande diſtance, étoit à peu près égale au tiers <lb/>de la peſanteur du même boulet, & </s> <s xml:id="echoid-s15480" xml:space="preserve">que tout ce que l’on em-<lb/>ployoit de poudre au-delà étoit en pure perte, parce qu’elle <lb/>ne s’enflammoit qu’après que le boulet étoit ſorti de la piece; <lb/></s> <s xml:id="echoid-s15481" xml:space="preserve">il eſt vrai que plus l’on met de poudre dans un canon, plus la <lb/>détonnation eſt forte, ce qui arrive également quand l’on tire <lb/>ſans boulet: </s> <s xml:id="echoid-s15482" xml:space="preserve">par conſéquent ces expériences ont fait voir que <pb o="482" file="0560" n="580" rhead="NOUVEAU COURS"/> pour le plus grand effet il falloit charger la piece de 8 de trois <lb/>livres de poudre, celle de 12 de quatre, celle de 16 de cinq & </s> <s xml:id="echoid-s15483" xml:space="preserve"><lb/>demie, & </s> <s xml:id="echoid-s15484" xml:space="preserve">celle de 24 de huit à neuf livres.</s> <s xml:id="echoid-s15485" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15486" xml:space="preserve">Ces épreuves ayant été conteſtées avec beaucoup de cha-<lb/>leur de la part de ceux qui ne les avoient point vues, la Cour <lb/>ordonna qu’elles fuſſent répétées à Metz, en préſence de M. </s> <s xml:id="echoid-s15487" xml:space="preserve">le <lb/>Maréchal de Belle-Iſle, qui étoit chargé de la part du Roi de <lb/>veiller à leur exactitude, pour être en état d’en rendre compte <lb/>à Sa Majeſté: </s> <s xml:id="echoid-s15488" xml:space="preserve">elles eurent le même ſuccès qu’à la Fere, ayant <lb/>auſſi reconnu qu’il falloit environ le tiers de la peſanteur du <lb/>boulet pour la charge la plus forte; </s> <s xml:id="echoid-s15489" xml:space="preserve">mais on s’en eſt tenu à <lb/>neuf livres pour celle du plus grand effet des pieces de 24.</s> <s xml:id="echoid-s15490" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15491" xml:space="preserve">Dans le mois d’Août de la même année, l’on a encore ré-<lb/>pété ces épreuves à Strasbourg, mais avec des circonſtances <lb/>propres à les rendre plus exactes. </s> <s xml:id="echoid-s15492" xml:space="preserve">L’on s’eſt ſervi d’une piece <lb/>de 24 bien conditionnée, que l’on a pointée ſous l’angle de <lb/>45 degrés & </s> <s xml:id="echoid-s15493" xml:space="preserve">maintenue inébranlable, on ne s’eſt ſervi que <lb/>de boulets bien calibrés & </s> <s xml:id="echoid-s15494" xml:space="preserve">bien ébarbés. </s> <s xml:id="echoid-s15495" xml:space="preserve">L’on verra dans le <lb/>Traité du Jet des Bombes, que le canon tiré ſous l’angle de <lb/>45 degrés ſe trouve dirigé de la maniere la plus convenable <lb/>pour faire des épreuves deſtinées à juger de l’effet des différentes <lb/>charges, parce que les portées des boulets qui partent ſous <lb/>une direction au deſſus ou au deſſous de 45 degrés, ſont plus <lb/>courtes avec une même charge que ne ſont celles des boulets <lb/>qui ſuivent la direction de l’ame pointée ſous cet angle; </s> <s xml:id="echoid-s15496" xml:space="preserve">d’où il <lb/>ſuit que les plus grandes portées ne doivent être attribuées <lb/>qu’à la force de la poudre, & </s> <s xml:id="echoid-s15497" xml:space="preserve">non pas aux accidens qui ne <lb/>peuvent que lesraccourcir.</s> <s xml:id="echoid-s15498" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15499" xml:space="preserve">L’on a employé un nombre de charges en progreſſion arith-<lb/>métique, tirées de ſuite en augmentant d’une livre pour <lb/>chacune, en commençant par huit livres, & </s> <s xml:id="echoid-s15500" xml:space="preserve">finiſſant par <lb/>vingt-quatre. </s> <s xml:id="echoid-s15501" xml:space="preserve">L’on a reconnu que la charge de neuf livres de <lb/>poudre avoit chaſſé le boulet à 2500 toiſes, & </s> <s xml:id="echoid-s15502" xml:space="preserve">que toutes les <lb/>autres charges plus fortes, juſqu’à celle de vingt-quatre, n’a-<lb/>voit jamais chaſſé le boulet plus loin, au grand étonnement <lb/>de ceux qui en avoient douté. </s> <s xml:id="echoid-s15503" xml:space="preserve">Le lendemain de cette pre-<lb/>miere ſéance, l’on a répété les mêmes épreuves avec les mê-<lb/>mes charges; </s> <s xml:id="echoid-s15504" xml:space="preserve">mais au lieu de commencer par huit livres de <lb/>poudre, & </s> <s xml:id="echoid-s15505" xml:space="preserve">finir par vingt-quatre, l’on a tiré le premier coup <lb/>à vingt-quatre livres, & </s> <s xml:id="echoid-s15506" xml:space="preserve">le dernier à huit, en ſuivant la même <pb o="483" file="0561" n="581" rhead="DE MATHEMATIQUE. Liv. XIII."/> progreſſion des nombres naturels dans un ordre renverſé, & </s> <s xml:id="echoid-s15507" xml:space="preserve"><lb/>jamais les fortes charges ne l’ont emporté ſur celle de neuf <lb/>livres.</s> <s xml:id="echoid-s15508" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15509" xml:space="preserve">Comme je n’ai point eu de part à ces dernieres épreuves, <lb/>elles ne peuvent être ſuſpectées, ainſi elles conſtatent de la <lb/>maniere la plus évidente, que la plus forte charge du canon <lb/>doit être à peu près le tiers de la peſanteur du boulet.</s> <s xml:id="echoid-s15510" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15511" xml:space="preserve">L’on trouvera dans l’Hiſtoire de l’Académie Royale des <lb/>Sciences de l’année 1757, un Mémoire que j’y ai lu ſur la charge <lb/>du plus grand effet du canon, & </s> <s xml:id="echoid-s15512" xml:space="preserve">qui répand un plus grand <lb/>jour ſur cette matiere que je n’ai fait juſqu’ici: </s> <s xml:id="echoid-s15513" xml:space="preserve">on pourra y avoir <lb/>recours, ſi on le juge à propos.</s> <s xml:id="echoid-s15514" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15515" xml:space="preserve">906. </s> <s xml:id="echoid-s15516" xml:space="preserve">Il y a encore une difficulté touchant les armes à feu, <lb/>qui eſt de ſçavoir à quel endroit doit être poſée la lumiere, <lb/>pour que la poudre faſſe un plus grand effet, & </s> <s xml:id="echoid-s15517" xml:space="preserve">je ne crois pas <lb/>que l’on ſe ſoit déterminé là-deſſus: </s> <s xml:id="echoid-s15518" xml:space="preserve">les uns diſent qu’il faut <lb/>la placer dans le milieu de la longueur de la chambre, parce <lb/>que la poudre s’enflamme à la ronde, & </s> <s xml:id="echoid-s15519" xml:space="preserve">en bien plus grande <lb/>quantité: </s> <s xml:id="echoid-s15520" xml:space="preserve">les autres ſont d’une opinion contraire, & </s> <s xml:id="echoid-s15521" xml:space="preserve">veulent <lb/>qu’elle ſoit placée à l’extrêmité de la chambre contre la cu-<lb/>laſſe, diſant pour leur raiſon que la piece n’a pas tant de recul. <lb/></s> <s xml:id="echoid-s15522" xml:space="preserve">Ces deux raiſonnemens ſont également vrais; </s> <s xml:id="echoid-s15523" xml:space="preserve">cependant <lb/>comme les reſſorts de la poudre, auſſi-bien que tous les autres <lb/>reſſorts, n’agiſſent avec plus ou moins de violence, qu’autant <lb/>que les corps qui leur réſiſtent cedent plus ou moins vîte, il s’en-<lb/>ſuit que quand une arme à feu n’a preſque point de recul, <lb/>c’eſt une marque que la poudre a trouvé ſi peu de réſiſtance <lb/>pour chaſſer la balle, qu’elle n’a eu beſoin que de ſon pre-<lb/>mier effort, au lieu que ſi elle trouve beaucoup de réſiſtance <lb/>vers la culaſſe & </s> <s xml:id="echoid-s15524" xml:space="preserve">du côté de la balle, tous ſes efforts ſe déban-<lb/>deront en même tems, quoique le recul ſoit plus grand, la <lb/>balle ira bien plus loin que ſi le canon n’avoit point eu de <lb/>recul: </s> <s xml:id="echoid-s15525" xml:space="preserve">ainſi la lumiere étant placée dans le milieu de la cham-<lb/>bre, les reſſorts agiront en bien plus grande quantité dans le <lb/>même tems, que ſi elle étoit contre la culaſſe, où ces mêmes <lb/>reſſorts ne peuvent agir que ſucceſſivement, puiſque la poudre <lb/>s’enflamme ainſi; </s> <s xml:id="echoid-s15526" xml:space="preserve">& </s> <s xml:id="echoid-s15527" xml:space="preserve">ſi le boulet vient à partir dès que la poudre <lb/>commence à s’enflammer, il arrivera encore qu’une grande <lb/>partie ſera chaſſée hors de la piece ſans faire aucun effet: </s> <s xml:id="echoid-s15528" xml:space="preserve">ainſi <lb/>il me ſemble que la lumiere placée dans le milieu de la <pb o="484" file="0562" n="582" rhead="NOUVEAU COURS"/> chambre, convient beaucoup mieux que partout ailleurs: </s> <s xml:id="echoid-s15529" xml:space="preserve">car <lb/>comme le canon ne recule qu’avec peine, à cauſe de la peſan-<lb/>teur de la machine & </s> <s xml:id="echoid-s15530" xml:space="preserve">du frottement de l’affût contre la plate-<lb/>forme, il ſe fait une réaction d’une grande partie de poudre <lb/>qui agit contre la culaſſe, qui vient augmenter l’impulſion de <lb/>celle qui chaſſe le boulet.</s> <s xml:id="echoid-s15531" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15532" xml:space="preserve">Je crois qu’il ne ſera pas ici mal-à-propos de déſabuſer ceux qui <lb/>croient que le boulet, en ſortant de la piece, s’éleve au deſſus <lb/>de la même piece, & </s> <s xml:id="echoid-s15533" xml:space="preserve">qui penſent qu’après avoir décrit une <lb/>courbe, il reprend une direction horizontale, pour en décrire <lb/>après cela une autre: </s> <s xml:id="echoid-s15534" xml:space="preserve">la plûpart ſont ſi opiniâtres à ſoutenir <lb/>cette erreur, qu’on a beau leur dire que la peſanteur du boulet, <lb/>bien loin de permettre qu’il puiſſe s’élever au deſſus de l’axe de <lb/>la piece, l’emporte au deſſous, dès l’inſtant même qu’il ſort, <lb/>& </s> <s xml:id="echoid-s15535" xml:space="preserve">lui fait décrire une courbe, qui à la vérité eſt d’abord fort <lb/>approchante de la ligne droite, mais qui devient ſenſible à <lb/>meſure qu’il s’éloigne de la piece; </s> <s xml:id="echoid-s15536" xml:space="preserve">& </s> <s xml:id="echoid-s15537" xml:space="preserve">une preuve à laquelle ils <lb/>ont tous recours pour ſoutenir leur opinion, c’eſt, diſent-ils, que <lb/>quand on tire après une piece de gibier à la chaſſe, il faut tirer <lb/>un peu au deſſous de l’animal, pour gagner la diſtance dont <lb/>la balle s’eſt élevée au deſſus du canon: </s> <s xml:id="echoid-s15538" xml:space="preserve">mais comme cette <lb/>raiſon ne vaut abſolument rien, en voici l’unique cauſe.</s> <s xml:id="echoid-s15539" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15540" xml:space="preserve">Si l’on attache un canon de fuſil ſur une petite planche, & </s> <s xml:id="echoid-s15541" xml:space="preserve"><lb/>qu’aux deux côtés de cette planche on y mette deux tourillons, <lb/>enſorte que le canon ſoit en équilibre ſur ces tourillons, comme <lb/>le bras d’une balance, on verra que l’ayant chargé à balle, ſi <lb/>l’on tire au deſſus de l’horizon, la partie de la poudre qui agira <lb/>contre la culaſſe, & </s> <s xml:id="echoid-s15542" xml:space="preserve">qui cauſe ordinairement le recul, fera <lb/>baiſſer la culaſſe, & </s> <s xml:id="echoid-s15543" xml:space="preserve">par conſéquent lever le bout du canon: <lb/></s> <s xml:id="echoid-s15544" xml:space="preserve">& </s> <s xml:id="echoid-s15545" xml:space="preserve">comme cela ſe fera avant même que la balle ſoit ſortie du <lb/>canon, il arrivera qu’elle ira au deſſus de l’objet vers lequel <lb/>on avoit pointé, parce qu’en ſortant elle ira ſelon la direction <lb/>de l’ame, & </s> <s xml:id="echoid-s15546" xml:space="preserve">non pas ſelon celle du rayon viſuel, qui ne ſera <lb/>plus la même à cauſe du dérangement de la culaſſe. </s> <s xml:id="echoid-s15547" xml:space="preserve">Or ſi l’on <lb/>fait attention que le fuſil entre les mains du chaſſeur fait le <lb/>même effet que je viens de dire, l’on verra que quand on <lb/>veut pointer juſte, il faut pointer au deſſous de l’objet.</s> <s xml:id="echoid-s15548" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15549" xml:space="preserve">Cependant ce qui fait qu’il ſemble que le boulet à une cer-<lb/>taine diſtance s’éleve au deſſus de la piece, c’eſt que la ſurface <lb/>extérieure de la piece n’étant point parallele avec l’ame, le <pb o="485" file="0563" n="583" rhead="DE MATHÉMATIQUE. Liv. XIII."/> boulet emporté avec beaucoup de violence, approche fort <lb/>pendant un tems de la direction de l’ame: </s> <s xml:id="echoid-s15550" xml:space="preserve">& </s> <s xml:id="echoid-s15551" xml:space="preserve">comme cette <lb/>direction ſe coupe avec celle de la ſurface de la piece, de ces <lb/>deux lignes prolongées, celle de l’ame paſſe au deſſus de la <lb/>ſurface: </s> <s xml:id="echoid-s15552" xml:space="preserve">& </s> <s xml:id="echoid-s15553" xml:space="preserve">ſi le boulet ſuit encore à peu près la direction de <lb/>l’ame au-delà de la ſection des deux lignes, il arrive en effet <lb/>que le boulet eſt au deſſus de la ſurface de la piece, mais non <lb/>pas au deſſus de la direction de l’ame prolongée; </s> <s xml:id="echoid-s15554" xml:space="preserve">& </s> <s xml:id="echoid-s15555" xml:space="preserve">il y a <lb/>même apparence que des Fondeurs ont eu égard à l’obliquité <lb/>de la ſurface de la piece par rapport à l’ame, afin de rectifier <lb/>la ligne courbe pour tirer de but en blanc.</s> <s xml:id="echoid-s15556" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1231" type="section" level="1" n="889"> <head xml:id="echoid-head1072" xml:space="preserve">PROPOSITION XXXIV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s15557" xml:space="preserve">907. </s> <s xml:id="echoid-s15558" xml:space="preserve">Trouver la maniere de connoître le nombre de boulets <lb/>qui ſont en pile.</s> <s xml:id="echoid-s15559" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15560" xml:space="preserve">Les boulets de canon & </s> <s xml:id="echoid-s15561" xml:space="preserve">les bombes qui ſont dans les Arce-<lb/>naux, ſont ordinairement rangés en pile; </s> <s xml:id="echoid-s15562" xml:space="preserve">ces piles ſont de <lb/>trois ſortes: </s> <s xml:id="echoid-s15563" xml:space="preserve">il y en a qui ont pour baſe un quarré, que l’on <lb/>nomme piles quarrées, comme dans la figure 324, d’autres un <lb/> <anchor type="note" xlink:label="note-0563-01a" xlink:href="note-0563-01"/> triangle, que l’on nomme piles triangulaires, comme dans la <lb/>figure 325, & </s> <s xml:id="echoid-s15564" xml:space="preserve">d’autres un parallélogramme, comme dans la <lb/>figure 326, que l’on nomme piles oblongues. </s> <s xml:id="echoid-s15565" xml:space="preserve">Or comme la <lb/>maniere de compter ces boulets dépend d’un calcul qui eſt <lb/>différent, ſelon la figure de la pile, en voici la méthode.</s> <s xml:id="echoid-s15566" xml:space="preserve"/> </p> <div xml:id="echoid-div1231" type="float" level="2" n="1"> <note position="right" xlink:label="note-0563-01" xlink:href="note-0563-01a" xml:space="preserve">Figure 324, <lb/>325 & 327.</note> </div> <p> <s xml:id="echoid-s15567" xml:space="preserve">Avant toutes choſes, il faut conſidérer que les faces de la <lb/>pile quarrée & </s> <s xml:id="echoid-s15568" xml:space="preserve">de la pile triangulaire ſont toujours des trian-<lb/>gles, dont les trois côtés ſont égaux, & </s> <s xml:id="echoid-s15569" xml:space="preserve">que ces triangles étant <lb/>formés par des boulets, ils compoſent une progreſſion arith-<lb/>métique, qui commence par l’unité, c’eſt-à-dire par le boulet <lb/>qui eſt au ſommet de la pile, & </s> <s xml:id="echoid-s15570" xml:space="preserve">que le plus grand terme de <lb/>la progreſſion eſt la baſe du triangle. </s> <s xml:id="echoid-s15571" xml:space="preserve">Et comme nous ſerons <lb/>obligés de connoître la quantité de boulets contenue dans une <lb/>face, que nous nommerons dans la ſuite triangle arithmétique, <lb/>voici comment on les pourra compter d’une maniere fort <lb/>aiſée.</s> <s xml:id="echoid-s15572" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15573" xml:space="preserve">Pour ſçavoir combien il y a de boulets dans le triangle <lb/>A B C, il faut compter combien il s’en trouve dans le côté <lb/>A C, ajouter à ce nombre l’unité, enſuite multiplier cette <pb o="486" file="0564" n="584" rhead="NOUVEAU COURS"/> quantité par la moitié du côté A B ou A C, qui eſt la même <lb/>choſe, & </s> <s xml:id="echoid-s15574" xml:space="preserve">le produit donnera le nombre des boulets contenus <lb/>dans le triangle: </s> <s xml:id="echoid-s15575" xml:space="preserve">ainſi le côté A C étant de ſix boulets, ſi j’a-<lb/>joute à ce nombre l’unité pour avoir 7, & </s> <s xml:id="echoid-s15576" xml:space="preserve">que je les multiplie <lb/>par la moitié de A B ou de A C, qui eſt 3, le produit ſera 21, <lb/>qui eſt le nombre des boulets que l’on cherche. </s> <s xml:id="echoid-s15577" xml:space="preserve">Il en ſera de <lb/>même pour tous les autres triangles arithmétiques.</s> <s xml:id="echoid-s15578" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15579" xml:space="preserve">La raiſon de ceci eſt que dans une progreſſion arithmétique, <lb/>a. </s> <s xml:id="echoid-s15580" xml:space="preserve">a + e, a + 2e, a + 3e, a + 4e, a + 5e, dont les termes <lb/>ſe ſurpaſſent d’une quantité e, la ſomme des deux termes a + e <lb/>& </s> <s xml:id="echoid-s15581" xml:space="preserve">a + 4e également éloignés des extrêmes, eſt égale à la <lb/>ſomme des extrêmes a & </s> <s xml:id="echoid-s15582" xml:space="preserve">a + 5e, ou à celle des deux autres <lb/>termes quelconques auſſi également éloignés des extrêmes, <lb/>puiſque la ſomme des uns & </s> <s xml:id="echoid-s15583" xml:space="preserve">des autres donne 2a + 5e; </s> <s xml:id="echoid-s15584" xml:space="preserve">mais <lb/>il y a la moitié autant de fois 2a + 5e (qui eſt la ſomme des <lb/>extrêmes) qu’il y a de termes dans la progreſſion: </s> <s xml:id="echoid-s15585" xml:space="preserve">donc pour <lb/>avoir la valeur de tous les termes d’une progreſſion arithmé-<lb/>tique, qui commence par l’unité, ou par tout autre nombre, <lb/>il faut multiplier le premier & </s> <s xml:id="echoid-s15586" xml:space="preserve">le dernier terme par la moitié <lb/>du nombre qui exprime la quantité des termes: </s> <s xml:id="echoid-s15587" xml:space="preserve">c’eſt pourquoi <lb/>nous avons ajouté le premier terme A C avec le dernier B, <lb/>& </s> <s xml:id="echoid-s15588" xml:space="preserve">nous avons multiplié la ſomme par la moitié du côté A B, <lb/>c’eſt-à-dire par la moitié du nombre des termes de la pro-<lb/>greſſion pour avoir les boulets du triangle.</s> <s xml:id="echoid-s15589" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15590" xml:space="preserve">Prévenu de ceci, il faut encore conſidérer que ſi l’on a une <lb/>quantité de boulets qui forment par leurs arrangemens un <lb/>priſme triangulaire D E H G F, ſoutenu par un plan incliné <lb/> <anchor type="note" xlink:label="note-0564-01a" xlink:href="note-0564-01"/> IK, dont la baſe ſoit le triangle E G H, ce priſme étant <lb/>coupé par un plan E F, parallele à la baſe, ſe trouvera diviſé <lb/>en deux parties, dont l’une, comme D E F, ſera le tiers de <lb/>tout le priſme, & </s> <s xml:id="echoid-s15591" xml:space="preserve">l’autre, comme E F G H, en ſera les deux <lb/>tiers; </s> <s xml:id="echoid-s15592" xml:space="preserve">car la partie E D F eſt une pyramide triangulaire, qui a <lb/>pour baſe le triangle oppoſé à E G H, & </s> <s xml:id="echoid-s15593" xml:space="preserve">pour hauteur la hau-<lb/>teur D E du priſme: </s> <s xml:id="echoid-s15594" xml:space="preserve">par conſéquent la partie E F G H, qui <lb/>eſt auſſi une pyramide, qui a pour baſe un quarré, en ſera les <lb/>deux tiers. </s> <s xml:id="echoid-s15595" xml:space="preserve">Mais il faut remarquer que le plan E F partage un <lb/>triangle de boulet, tel que E F G, qui ſe rencontre dans la <lb/>coupe; </s> <s xml:id="echoid-s15596" xml:space="preserve">ce qui rendra les deux pyramides imparfaites, quand <lb/>on les conſidérera compoſées de boulets: </s> <s xml:id="echoid-s15597" xml:space="preserve">car comme le plan <lb/>E F paſſe par tiers de chaque boulet L, il faudra donner à la <pb o="487" file="0565" n="585" rhead="DE MATHEMATIQUE. Liv. XIII."/> pyramide triangulaire D E F les deux tiers de la quantité des <lb/>boulets du triangle arithmétique qui ſe rencontre dans la <lb/>coupe E F. </s> <s xml:id="echoid-s15598" xml:space="preserve">De même pour rendre réguliere la pyramide quarrée <lb/>E F G H, il faudra lui donner le tiers du même triangle arith-<lb/>métique. </s> <s xml:id="echoid-s15599" xml:space="preserve">Or ſi l’on ſuppoſe que l’on a détaché du priſme la <lb/>pyramide quarrée E F G H pour tenir lieu de la pyramide <lb/>A B C Q, & </s> <s xml:id="echoid-s15600" xml:space="preserve">que la pyramide triangulaire D E F qui reſte ſoit <lb/> <anchor type="note" xlink:label="note-0565-01a" xlink:href="note-0565-01"/> regardée comme la pyramide M N O P, on pourra donc dire <lb/>que la pyramide A B C Q eſt plus grande que les deux tiers du <lb/>priſme qui auroit pour baſe le triangle A B C, qui eſt la même <lb/>choſe que E G H, & </s> <s xml:id="echoid-s15601" xml:space="preserve">pour hauteur le côté A B, qui eſt la même <lb/>choſe que D E, du tiers du triangle A B C, qui eſt la même <lb/>que celui qui ſe trouve dans la coupe E F.</s> <s xml:id="echoid-s15602" xml:space="preserve"/> </p> <div xml:id="echoid-div1232" type="float" level="2" n="2"> <note position="left" xlink:label="note-0564-01" xlink:href="note-0564-01a" xml:space="preserve">Figure 327.</note> <note position="right" xlink:label="note-0565-01" xlink:href="note-0565-01a" xml:space="preserve">Figure 324 <lb/>& 325.</note> </div> <p> <s xml:id="echoid-s15603" xml:space="preserve">Enfin l’on pourroit dire auſſi que la pyramide M N O P ſera <lb/>plus grande que le tiers du priſme, qui auroit pour baſe le <lb/>triangle M N O, qui eſt le même que E G H, & </s> <s xml:id="echoid-s15604" xml:space="preserve">pour hau-<lb/>teur le côté M N, qui eſt le même que E D, des deux tiers du <lb/>triangle M N O, qui eſt le même que le triangle arithmétique <lb/>qui ſe rencontre dans la coupe E F.</s> <s xml:id="echoid-s15605" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15606" xml:space="preserve">D’où il s’enſuit, 1°. </s> <s xml:id="echoid-s15607" xml:space="preserve">que pour trouver la quantité de boulets <lb/>contenue dans une pile quarrée A B C Q, il faut d’abord cher-<lb/>cher le nombre de ceux qui ſont contenus dans le triangle <lb/>arithmétique A B C, & </s> <s xml:id="echoid-s15608" xml:space="preserve">le multiplier par les deux tiers du côté <lb/>A B ou A C, & </s> <s xml:id="echoid-s15609" xml:space="preserve">ajouter au produit le tiers du triangle A B C.</s> <s xml:id="echoid-s15610" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15611" xml:space="preserve">908. </s> <s xml:id="echoid-s15612" xml:space="preserve">Ainſi le côté A C étant de 6, je commence par trou-<lb/>ver le triangle A B C, en ajoutant l’unité au nombre 6 pour <lb/>avoir 7, que je multiplie par la moitié du côté A B, qui eſt 3, <lb/>& </s> <s xml:id="echoid-s15613" xml:space="preserve">le produit donne 21, que je multiplie par les deux tiers du <lb/>côté A B, qui eſt 4, pour avoir 84 au produit, auquel ajoutant <lb/>le tiers du triangle arithmétique A B C, qui eſt 7, il vient 91 <lb/>pour le nombre des boulets de la pile.</s> <s xml:id="echoid-s15614" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15615" xml:space="preserve">909. </s> <s xml:id="echoid-s15616" xml:space="preserve">L’on pourra donc dire auſſi que pour trouver le nom-<lb/>bre de boulets contenus dans la pile triangulaire M N O P, il <lb/>faut multiplier le triangle M N O par le tiers du côté M N, <lb/>& </s> <s xml:id="echoid-s15617" xml:space="preserve">ajouter au produit les deux tiers du nombre de boulets <lb/>contenus dans le triangle M N O: </s> <s xml:id="echoid-s15618" xml:space="preserve">ainſi le côté N O étant en-<lb/>core de 6, le triangle arithmétique ſera de 21, qui étant mul-<lb/>tiplié par le tiers du côté M N, qui eſt 2, l’on aura 42, aux-<lb/>quels ajoutant les deux tiers du triangle, qui eſt 14, l’on aura <lb/>56 pour le nombre de boulets contenus dans cette pile.</s> <s xml:id="echoid-s15619" xml:space="preserve"/> </p> <pb o="488" file="0566" n="586" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s15620" xml:space="preserve">A l’égard de la pile oblongue, il eſt fort facile d’en con-<lb/> <anchor type="note" xlink:label="note-0566-01a" xlink:href="note-0566-01"/> noître la quantité de boulets: </s> <s xml:id="echoid-s15621" xml:space="preserve">car comme elle eſt compoſée <lb/>d’un priſme triangulaire R S T V, & </s> <s xml:id="echoid-s15622" xml:space="preserve">d’une pyramide quarrée <lb/>V T X Y, l’on voit qu’il n’y a d’abord qu’à chercher la quantité <lb/>de boulets contenue dans une pyramide quarrée, qui auroit <lb/>pour côté X Y ou V X; </s> <s xml:id="echoid-s15623" xml:space="preserve">enſuite ajouter à la valeur de cette <lb/>pyramide celle du priſme R S T V, que l’on trouvera en mul-<lb/>tipliant le triangle X T V ou celui de la coupe T V, qui eſt la <lb/>même choſe, par la quantité de boulets R T qui ſe trouve au <lb/>ſommet de la pile moins une unité; </s> <s xml:id="echoid-s15624" xml:space="preserve">quand je dis moins une <lb/>unité, c’eſt qu’on doit faire attention que le premier boulet T, <lb/>avec le triangle arithmétique T V, qui lui correſpond, appar-<lb/>tient entiérement à la pyramide T V X Y, & </s> <s xml:id="echoid-s15625" xml:space="preserve">par conſéquent il <lb/>doit être ſupprimé de la quantité R T.</s> <s xml:id="echoid-s15626" xml:space="preserve"/> </p> <div xml:id="echoid-div1233" type="float" level="2" n="3"> <note position="left" xlink:label="note-0566-01" xlink:href="note-0566-01a" xml:space="preserve">Figure 326.</note> </div> <p> <s xml:id="echoid-s15627" xml:space="preserve">Ainſi ſuppoſant que le côté X Y ou T X ſoit de 9, j’ajoute <lb/>1 à 9 pour avoir 10, que je multiplie par la moitié de 9; </s> <s xml:id="echoid-s15628" xml:space="preserve">ou, <lb/>ce qui eſt la même choſe, 9 par la moitié de 10, qui eſt 5, le <lb/>produit ſera 45 pour la quantité de boulets du triangle X T Y, <lb/>que je multiplie par les deux tiers de 9, c’eſt-à-dire par 6, & </s> <s xml:id="echoid-s15629" xml:space="preserve"><lb/>il vient 270 pour le produit, auquel j’ajoute le tiers du triangle, <lb/>qui eſt 15, & </s> <s xml:id="echoid-s15630" xml:space="preserve">le tout fait 285 pour la pyramide. </s> <s xml:id="echoid-s15631" xml:space="preserve">Or ſuppoſant <lb/>auſſi que R T ſoit de 15 boulets, je multiplie 15 moins 1, qui <lb/>eſt 14, par le triangle arithmétique, qui eſt 45, & </s> <s xml:id="echoid-s15632" xml:space="preserve">il vient 630 <lb/>pour le nombre de boulets du priſme R S T V, qui étant ajouté <lb/>avec ceux de la pyramide, l’on trouvera 715 boulets dans la <lb/>pyramide oblongue.</s> <s xml:id="echoid-s15633" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15634" xml:space="preserve">910. </s> <s xml:id="echoid-s15635" xml:space="preserve">Comme il n’y a rien de plus commode pour l’imagi-<lb/>nation que les formules qui nous indiquent par leurs expreſſions <lb/>ce que nous avons à faire dans tous les cas imaginables, nous <lb/>allons donner une formule très-ſimple, par le moyen de la-<lb/>quelle on pourra trouver le nombre des boulets ou des bombes <lb/>rangés en piles, ſoit que ces piles ſoient diſpoſées en forme <lb/>priſmatique, comme dans la figure 326, ſoit qu’elles ſoient <lb/>en pyramide quarrée ou en pyramide triangulaire. </s> <s xml:id="echoid-s15636" xml:space="preserve">Notre for-<lb/>mule peut s’appliquer à tous ces cas: </s> <s xml:id="echoid-s15637" xml:space="preserve">car il eſt évident que <lb/>pour connoître le nombre de boulets compris dans la pile de la <lb/>figure 326, il faut, comme nous l’avons dit, décompoſer cette <lb/>pile en deux corps, dont l’un eſt le priſme triangulaire RQXYT, <lb/>lequel n’a aucune difficulté, & </s> <s xml:id="echoid-s15638" xml:space="preserve">dont l’autre eſt une pyramide <lb/>qui a même nombre de rangs que le priſme triangulaire, ou <pb o="489" file="0567" n="587" rhead="DE MATHÉMATIQUE. Liv. XIII."/> qui a autant de rangs qu’il y a de boulets dans le côté R Q du <lb/>triangle S R Q.</s> <s xml:id="echoid-s15639" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15640" xml:space="preserve">Il n’eſt pas moins viſible que cette pile eſt la ſomme des <lb/>quarrés d’autant de nombres depuis l’unité qu’il y a de boulets <lb/>dans le côté R Q: </s> <s xml:id="echoid-s15641" xml:space="preserve">ainſi ſi l’on a 9 boulets, la pyramide ſera <lb/>égale à la ſomme des quarrés des neuf premiers nombres, <lb/>1, 2, 3, 4, &</s> <s xml:id="echoid-s15642" xml:space="preserve">c. </s> <s xml:id="echoid-s15643" xml:space="preserve">Tout ſe réduira donc à trouver la ſomme des <lb/>quarrés de tant de nombres naturels que l’on voudra. </s> <s xml:id="echoid-s15644" xml:space="preserve">Sur quoi <lb/>je remarque que tous les quarrés des nombres naturels réſul-<lb/>tent de l’addition des termes de deux ſuites égales des nom-<lb/>bres triangulaires, diſpoſées de maniere que la premiere ait <lb/>un terme de plus que la ſeconde.</s> <s xml:id="echoid-s15645" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">1, # 3, # 6, # 10, # 15, # 21, # 28, # 36, # &c. <lb/># 1, # 3, # 6, # 10, # 15, # 21, # 28, # &c. <lb/>1, # 4, # 9, # 16, # 25, # 36, # 49, # 64, # &c. <lb/></note> <p> <s xml:id="echoid-s15646" xml:space="preserve">Par exemple, ſi l’on diſ-<lb/>poſe ces deux ſuites, com-<lb/>me on voit ici, & </s> <s xml:id="echoid-s15647" xml:space="preserve">que <lb/>l’on les ajoute terme par <lb/>terme, il eſt évident qu’il en réſultera la ſuite des quarrés des <lb/>nombres naturels que l’on voit au deſſous. </s> <s xml:id="echoid-s15648" xml:space="preserve">Ainſi tout ſe réduit à <lb/>trouver la ſomme des quarrés de tant de termes que l’on voudra <lb/>de la ſuite des nombres naturels: </s> <s xml:id="echoid-s15649" xml:space="preserve">car de cette maniere on pourra <lb/>trouver le nombre des boulets contenus dans une pile trian-<lb/>gulaire & </s> <s xml:id="echoid-s15650" xml:space="preserve">dans une pyramide quarrée quelconque. </s> <s xml:id="echoid-s15651" xml:space="preserve">La pyra-<lb/>mide triangulaire ſe trouvera, en ſommant autant de termes <lb/>qu’il y a de boulets dans le côté du triangle M N O, & </s> <s xml:id="echoid-s15652" xml:space="preserve">la py-<lb/>ramide quarrée ſe trouvera, en ſommant d’abord un nombre <lb/>de termes de la ſuite des nombres triangulaires égal au nom-<lb/>bre de boulets contenus dans le côté B C du triangle B C Q, <lb/> <anchor type="note" xlink:label="note-0567-02a" xlink:href="note-0567-02"/> & </s> <s xml:id="echoid-s15653" xml:space="preserve">en ſommant un nombre de termes de la même ſuite trian-<lb/>gulaire diminué de l’unité, la ſomme de ces deux premieres <lb/>ſera la ſomme des boulets de la pyramide quarrée. </s> <s xml:id="echoid-s15654" xml:space="preserve">Voici la <lb/>formule que j’ai trouvée: </s> <s xml:id="echoid-s15655" xml:space="preserve">Si m eſt égal au nombre de boulets <lb/>contenus dans le côté M O du triangle M N O, la ſomme des <lb/>boulets ſera {m<emph style="sub">3</emph> + 3m<emph style="sub">2</emph> + 2m/6}, par exemple, dans notre figure <lb/>m = 6: </s> <s xml:id="echoid-s15656" xml:space="preserve">donc on aura {216 + 108 + 12/6} = 56, c’eſt le nombre que <lb/>l’on a trouvé (art. </s> <s xml:id="echoid-s15657" xml:space="preserve">907). </s> <s xml:id="echoid-s15658" xml:space="preserve">Si la pyramide eſt une pyramide <lb/>quarrée, on pourra trouver le nombre des boulets par la même <lb/>formule. </s> <s xml:id="echoid-s15659" xml:space="preserve">Si m = 6, on aura pour la premiere ſomme 56, & </s> <s xml:id="echoid-s15660" xml:space="preserve"><lb/>pour la ſeconde, en faiſant m = 5, c’eſt-à dire en prenant la <pb o="490" file="0568" n="588" rhead="NOUVEAU COURS DE MATH. Liv. XIII."/> ſomme des mêmes nombres triangulaires, diminuée d’un <lb/>terme, on aura {125 + 75 + 10/6} = 35, dont la ſomme, avec 56, <lb/>fait 91, comme on l’a déja trouvé à l’art. </s> <s xml:id="echoid-s15661" xml:space="preserve">906. </s> <s xml:id="echoid-s15662" xml:space="preserve">J’ai trouvé cette <lb/>formule, en recherchant les propriétés des nombres triangu-<lb/>laires; </s> <s xml:id="echoid-s15663" xml:space="preserve">mais comme la théorie ſeroit peut-être un peu difficile <lb/>pour des Commençans, je me contente de donner la formule <lb/>qui eſt aſſez ſimple, pour qu’on puiſſe s’en reſſouvenir dans <lb/>tous les cas poſſibles. </s> <s xml:id="echoid-s15664" xml:space="preserve">Il faut bien remarquer que par cette for-<lb/>mule, on pourra ſommer autant de termes que l’on voudra de <lb/>la ſuite des quarrés des nombres naturels.</s> <s xml:id="echoid-s15665" xml:space="preserve"/> </p> <div xml:id="echoid-div1234" type="float" level="2" n="4"> <note position="right" xlink:label="note-0567-02" xlink:href="note-0567-02a" xml:space="preserve">Figure 324.</note> </div> <p> <s xml:id="echoid-s15666" xml:space="preserve">911. </s> <s xml:id="echoid-s15667" xml:space="preserve">Suivant ces principes, on peut aiſément déduire une <lb/>formule pour ſommer tant de nombres quarrés que l’on vou-<lb/>dra: </s> <s xml:id="echoid-s15668" xml:space="preserve">pour cela, il n’y a qu’à faire dans la formule m = m - 1, <lb/>& </s> <s xml:id="echoid-s15669" xml:space="preserve">ajouter ce qui en viendra à la même formule, la ſomme <lb/>ſera une formule propre à ſommer tant de nombres quarrés <lb/>que l’on voudra: </s> <s xml:id="echoid-s15670" xml:space="preserve">cette ſubſtitution donne <lb/>{m<emph style="sub">3</emph> - 3m<emph style="sub">2</emph> + 3m - 1 + 3m<emph style="sub">2</emph> - 6m + 3 + 2m - 2/6} = {m<emph style="sub">3</emph> - m/6}, qui étant <lb/>jointe avec {m<emph style="sub">3</emph> + 3m<emph style="sub">2</emph> + 2m/6}, donnera {2m<emph style="sub">3</emph> + 3m<emph style="sub">2</emph> + m/6} = {m<emph style="sub">3</emph>/3} + {1/2} m<emph style="sub">2</emph> <lb/>+ {1/6} m. </s> <s xml:id="echoid-s15671" xml:space="preserve">Il eſt à propos de ſe ſervir de cette formule pour trouver <lb/>les nombres des boulets rangés en pyramide quarrée, puiſ-<lb/>que l’on trouve la ſomme demandée par une ſeule opération, <lb/>au lieu que par l’autre formule il faut néceſſairement en faire <lb/>deux. </s> <s xml:id="echoid-s15672" xml:space="preserve">Par exemple, ſi le nombre des rangs de boulets eſt 6, <lb/>en faiſant m = 6 dans cette derniere formule, on aura {216/3} + 18 <lb/>+ 1 = 91, comme on l’avoit trouvé ci-devant. </s> <s xml:id="echoid-s15673" xml:space="preserve">Cette formule <lb/>pour ſommer les nombres quarrés eſt démontrée, en admettant <lb/>celle que nous avons donnée pour ſommer les nombres trian-<lb/>gulaires.</s> <s xml:id="echoid-s15674" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1236" type="section" level="1" n="890"> <head xml:id="echoid-head1073" style="it" xml:space="preserve">Fin du treizieme Livre.</head> <figure> <image file="0568-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0568-01"/> </figure> <pb o="491" file="0569" n="589"/> <figure> <image file="0569-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0569-01"/> </figure> </div> <div xml:id="echoid-div1237" type="section" level="1" n="891"> <head xml:id="echoid-head1074" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head1075" xml:space="preserve">LIVRE QUATORZIEME. <lb/>Du mouvement des Corps, & du jet des Bombes.</head> <p style="it"> <s xml:id="echoid-s15675" xml:space="preserve">LE principal objet que je me ſuis propoſé dans le Traité du <lb/>Mouvement que je donne ici, a été d’enſeigner l’art de jetter <lb/>les bombes. </s> <s xml:id="echoid-s15676" xml:space="preserve">Il eſt vrai que je ne commence pas d’abord par-là, parce <lb/>qu’il m’a paru qu’il étoit bon de donner une connoiſſance du choc <lb/>des corps, afin d’en tirer quelques principes qui nous ſerviront <lb/>beaucoup dans la méchanique. </s> <s xml:id="echoid-s15677" xml:space="preserve">Je pourrois dire la même choſe du <lb/>chapitre du mouvement, parce qu’il me donnera auſſi lieu dans la <lb/>méchanique d’expliquer pluſieurs choſes qui n’auroient pu être en-<lb/>tendues ſans une connoiſſance de la chûte des corps: </s> <s xml:id="echoid-s15678" xml:space="preserve">d’ailleurs il <lb/>eſt abſolument néceſſaire à ceux qui veulent s’attacher aux Ma-<lb/>thématiques & </s> <s xml:id="echoid-s15679" xml:space="preserve">à la Phyſique, pour expliquer quantité de choſes <lb/>curieuſes dans l’Artillerie, de ſcavoir les principales regles du <lb/>choc & </s> <s xml:id="echoid-s15680" xml:space="preserve">du mouvement des corps: </s> <s xml:id="echoid-s15681" xml:space="preserve">ainſi ce Traité contient trois <lb/>chapitres; </s> <s xml:id="echoid-s15682" xml:space="preserve">le premier traite du choc des corps, le ſecond des <lb/>regles du mouvement, & </s> <s xml:id="echoid-s15683" xml:space="preserve">le troiſieme de la théorie & </s> <s xml:id="echoid-s15684" xml:space="preserve">de la pratique <lb/>du jet des bombes.</s> <s xml:id="echoid-s15685" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s15686" xml:space="preserve">A l’égard du jet des bombes, je ne vois pas que les Bombar-<lb/>diers ſe ſoient mis beaucoup en peine de ſcavoir s’il y avoit des <lb/>regles certaines ſur ce ſujet, dans la penſée où ils ont toujours été <lb/>qu’il n’y avoit que la ſeule pratique qui puiſſe ſervir au Bombar-<lb/>dier, pour lui faire jetter des bombes avec ſuccès; </s> <s xml:id="echoid-s15687" xml:space="preserve">& </s> <s xml:id="echoid-s15688" xml:space="preserve">cela vient <lb/>ſans doute de ce que la plûpart n’ayant aucune connoiſſance des <pb o="492" file="0570" n="590" rhead="NOUVEAU COURS"/> Mathématiques ni de la Phyſique, ne peuvent point s’imaginer <lb/>qu’il eſt poſſible de donner des loix des effets de la poudre, au <lb/>caprice de laquelle ils attribuent les fautes qu’ils font. </s> <s xml:id="echoid-s15689" xml:space="preserve">J’avoue <lb/>qu’il y a tant de choſes qui concourent dans la charge d’un mortier <lb/>à déranger tout ce que les regles & </s> <s xml:id="echoid-s15690" xml:space="preserve">l’attention du Bombardier le <lb/>plus adroit ſont en état de faire, qu’il y auroit de la témérité à <lb/>croire qu’on peut jetter des bombes dans un endroit comme ſi on <lb/>les y portoit avec la main. </s> <s xml:id="echoid-s15691" xml:space="preserve">Mais ce qu’il y a de ſûr, c’eſt que <lb/>ſi un Bombardier avoit aſſez d’attention, en chargeant ſon mortier, <lb/>pour en examiner le défaut, & </s> <s xml:id="echoid-s15692" xml:space="preserve">pour faire enſorte de charger tou-<lb/>jours également, les regles ſeroient d’un uſage excellent, puiſ-<lb/>que l’on n’auroit pour chaſſer des bombes à une diſtance quel-<lb/>conque, qu’à en tirer une avec la charge que l’on aura jugé à pro-<lb/>pos, & </s> <s xml:id="echoid-s15693" xml:space="preserve">à un degré d’élevation à volonté, pour connoître l’éléva-<lb/>tion qu’il convient de donner au mortier, pour jetter les autres <lb/>bombes à la diſtance qu’on demande. </s> <s xml:id="echoid-s15694" xml:space="preserve">Mais ceux qui n’ont que la <lb/>pratique, ſoutiennent qu’il eſt impoſſible de pouvoir obſerver cette <lb/>préciſion dans la maniere de charger également: </s> <s xml:id="echoid-s15695" xml:space="preserve">car, diſent-ils, <lb/>l’inégalité des grains de poudre, ſoit dans leur groſſeur ou dans <lb/>les matieres qui la compoſent, fait que la même quantité pour <lb/>chaque charge produit des effets différens; </s> <s xml:id="echoid-s15696" xml:space="preserve">ce qui peut venir auſſi <lb/>de la part de la terre avec laquelle on remplit la chambre, qui <lb/>peut être plus ou moins refoulée une fois que l’autre: </s> <s xml:id="echoid-s15697" xml:space="preserve">d’ailleurs <lb/>les bombes qui ne ſont point toutes bien calibrées & </s> <s xml:id="echoid-s15698" xml:space="preserve">d’égale pe-<lb/>ſanteur, & </s> <s xml:id="echoid-s15699" xml:space="preserve">ſouvent mal coulées, la plate-forme qui ſe dérange preſ-<lb/>que à chaque coup que l’on tire, ſont autant de ſujets qui prouvent <lb/>que moralement il n’eſt pas poſſible de jamais tirer des bombes <lb/>comme il faut. </s> <s xml:id="echoid-s15700" xml:space="preserve">Mais quoiqu’on puiſſe remédier à tout ceci quand <lb/>on voudra y bien prendre garde, il n’y a point de doute qu’un <lb/>Bombardier expérimenté d’ailleurs dans ſon métier, & </s> <s xml:id="echoid-s15701" xml:space="preserve">qui ſcaura <lb/>l’art de jetter les bombes, ne ſoit plus ſûr de ſon fait que celui <lb/>qui n’a que la ſimple pratique: </s> <s xml:id="echoid-s15702" xml:space="preserve">car s’il s’apperçoit que ſon premier <lb/>& </s> <s xml:id="echoid-s15703" xml:space="preserve">ſon ſecond coup ne jettent point la bombe où il veut qu’elle <lb/>tombe, il pourra ſe corriger, au lieu que ce dernier tâtonnera en <lb/>augmentant ou diminuant la poudre ou les degrés pendant un tems <lb/>conſidérable; </s> <s xml:id="echoid-s15704" xml:space="preserve">& </s> <s xml:id="echoid-s15705" xml:space="preserve">quoiqu’on diſe que c’eſt le pur hazard qui gou-<lb/>verne l’action du mortier, l’expérience m’a fait voir que quand on <lb/>vouloit apporter tous ſes ſoins à charger également, & </s> <s xml:id="echoid-s15706" xml:space="preserve">à poſer l’affût <lb/>toujours dans le même endroit de la plate-forme, & </s> <s xml:id="echoid-s15707" xml:space="preserve">les tourrillons <lb/>dans la même ſituation ſur l’affût, il étoit très-poſſible de tirer <pb o="493" file="0571" n="591" rhead="DE MATHÉMATIQUE. Liv. XIV."/> quantité de bombes toujours à peu près dans le même endroit. </s> <s xml:id="echoid-s15708" xml:space="preserve">Qu’on <lb/>revienne donc de l’opinion où l’on eſt, que les regles pour jetter <lb/>les bombes ne peuvent être d’aucun ſecours, puiſque ſi l’on a ſoin de <lb/>charger bien également, & </s> <s xml:id="echoid-s15709" xml:space="preserve">que l’on ſe ſerve des bombes à peu près <lb/>de même poids, l’on n’aura plus lieu de douter de la certitude de <lb/>ces regles.</s> <s xml:id="echoid-s15710" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s15711" xml:space="preserve">Après cela on peut dire qu’il y a ſi peu de Bombardiers qui <lb/>ſe ſoient attachés à ſçavoir ces regles, & </s> <s xml:id="echoid-s15712" xml:space="preserve">encore moins à les prati-<lb/>quer, que certainement il y a plus de préjugé que de connoiſſance <lb/>dans leur fait; </s> <s xml:id="echoid-s15713" xml:space="preserve">& </s> <s xml:id="echoid-s15714" xml:space="preserve">quand ils pourroient s’en paſſer pour jetter des <lb/>bombes dans un endroit de niveau avec la batterie, après en avoir <lb/>tiré un grand nombre d’inutiles, comme cela arrive toujours, com-<lb/>ment s’y prendroient-ils pour en jetter dans quelque fortereſſe fort <lb/>élevée, comme ſur un rocher eſcarpé, au pied duquel ſeroit la bat-<lb/>terie, ou bien ſi la batterie étoit un lieu fort élevé, pour en jetter <lb/>dans un fond? </s> <s xml:id="echoid-s15715" xml:space="preserve">Il n’y a point de Bombardier, que je ſçache, à <lb/>qui l’expérience ait donné quelque pratique pour cela, d’autant <lb/>plus qu’ils ne regardent point ces deux cas comme problématiques. <lb/></s> <s xml:id="echoid-s15716" xml:space="preserve">Enfin il réſulte de tout ce qui vient d’être dit, que jamais on ne <lb/>parviendra à jetter des bombes à une diſtance donnée, que l’on ne <lb/>ſçache les regles qui ſont établies pour cela, & </s> <s xml:id="echoid-s15717" xml:space="preserve">qu’on n’ait aſſez <lb/>d’expérience pour prévoir tous les accidens auxquels le mortier & </s> <s xml:id="echoid-s15718" xml:space="preserve">la <lb/>bombe ſont ſujets.</s> <s xml:id="echoid-s15719" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1238" type="section" level="1" n="892"> <head xml:id="echoid-head1076" xml:space="preserve">CHAPITRE PREMIER. <lb/>Du Choc des Corps. <lb/><emph style="sc">Définitions</emph>. <lb/>I.</head> <p> <s xml:id="echoid-s15720" xml:space="preserve">912. </s> <s xml:id="echoid-s15721" xml:space="preserve">LE mouvement d’un corps eſt le tranſport de ce corps <lb/>d’un lieu dans un autre. </s> <s xml:id="echoid-s15722" xml:space="preserve">Le mouvement eſt réel, lorſque le <lb/>corps parcourt lui-même, en vertu d’une force qui lui a été ap-<lb/>pliquée, les parties de l’étendue compriſes entre les deux termes <lb/>du mouvement, qui ſont le point de départ & </s> <s xml:id="echoid-s15723" xml:space="preserve">le point d’ar-<lb/>rivée. </s> <s xml:id="echoid-s15724" xml:space="preserve">Tel eſt le mouvement d’une boule que l’on a jettée ſur <lb/>un plan horizontal. </s> <s xml:id="echoid-s15725" xml:space="preserve">Le mouvement eſt relatif ou reſpectif, lorſ-<lb/>que le corps paſſe d’un lieu en un autre par le moyen d’un <pb o="494" file="0572" n="592" rhead="NOUVEAU COURS"/> corps en mouvement, quoiqu’il ſoit lui-même en repos. </s> <s xml:id="echoid-s15726" xml:space="preserve">Tel <lb/>eſt le mouvement d’un homme dans un bateau. </s> <s xml:id="echoid-s15727" xml:space="preserve">Dans le mou-<lb/>vement d’un corps, il y a cinq choſes à conſidérer, le corps <lb/>mis en mouvement, la force motrice, l’eſpace parcouru, le <lb/>tems du mouvement, la direction de ce mouvement.</s> <s xml:id="echoid-s15728" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1239" type="section" level="1" n="893"> <head xml:id="echoid-head1077" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s15729" xml:space="preserve">913. </s> <s xml:id="echoid-s15730" xml:space="preserve">On appelle force motrice tout ce qui peut mouvoir un <lb/>corps. </s> <s xml:id="echoid-s15731" xml:space="preserve">Un corps en mouvement eſt lui-même une force mo-<lb/>trice: </s> <s xml:id="echoid-s15732" xml:space="preserve">car l’expérience nous apprend qu’il peut lui-même en <lb/>mettre un autre en mouvement. </s> <s xml:id="echoid-s15733" xml:space="preserve">Pour eſtimer une force mo-<lb/>trice, il faut connoître la maſſe du corps mis en mouvement, <lb/>l’eſpace que ce corps a parcouru, & </s> <s xml:id="echoid-s15734" xml:space="preserve">le tems pendant lequel <lb/>il a parcouru cet eſpace.</s> <s xml:id="echoid-s15735" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1240" type="section" level="1" n="894"> <head xml:id="echoid-head1078" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s15736" xml:space="preserve">914. </s> <s xml:id="echoid-s15737" xml:space="preserve">La vîteſſe d’un corps eſt le plus ou le moins de chemin <lb/>qu’il fait pendant un certain tems, lorſque quelque cauſe l’a <lb/>mis en mouvement; </s> <s xml:id="echoid-s15738" xml:space="preserve">d’autres ont défini la vîteſſe, le rapport <lb/>de l’eſpace au tems. </s> <s xml:id="echoid-s15739" xml:space="preserve">En effet, pour avoir une idée de la vîteſſe <lb/>d’un mobile, il ne ſuffit pas de connoître ſeulement l’eſpace <lb/>qu’il a parcouru, ou le tems qu’il a été en mouvement, mais il <lb/>faut connoître pendant quel tems il a parcouru un eſpace dé-<lb/>terminé. </s> <s xml:id="echoid-s15740" xml:space="preserve">Par exemple, on ne peut pas dire qu’un homme ait <lb/>fait une grande diligence, parce qu’il a parcouru dix lieues: <lb/></s> <s xml:id="echoid-s15741" xml:space="preserve">mais cette même vîteſſe eſt connue, lorſqu’on ſçait qu’il les a <lb/>faites pendant cinq heures.</s> <s xml:id="echoid-s15742" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1241" type="section" level="1" n="895"> <head xml:id="echoid-head1079" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s15743" xml:space="preserve">915. </s> <s xml:id="echoid-s15744" xml:space="preserve">La vîteſſe d’un corps eſt uniforme ou variable, elle ſe <lb/>nomme uniforme, lorſque dans des tems égaux elle fait par-<lb/>courir des eſpaces égaux, & </s> <s xml:id="echoid-s15745" xml:space="preserve">elle ſe nomme variable, lorſque <lb/>dans des tems égaux elle fait parcourir des eſpaces inégaux. <lb/></s> <s xml:id="echoid-s15746" xml:space="preserve">Les vîteſſes uniformes ou variables ſont entr’elles comme les <lb/>eſpaces qu’elles font parcourir en des tems égaux. </s> <s xml:id="echoid-s15747" xml:space="preserve">Si l’une dans <lb/>une minute fait parcourir dix toiſes, & </s> <s xml:id="echoid-s15748" xml:space="preserve">l’autre 20 dans le <lb/>même-tems, ces deux vîteſſes ſont entr’elles comme 10 & </s> <s xml:id="echoid-s15749" xml:space="preserve">20, <lb/>c’eſt-à-dire que la derniere eſt double de la ſeconde.</s> <s xml:id="echoid-s15750" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1242" type="section" level="1" n="896"> <head xml:id="echoid-head1080" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s15751" xml:space="preserve">916. </s> <s xml:id="echoid-s15752" xml:space="preserve">La direction d’un corps eſt la détermination de ſon <pb o="495" file="0573" n="593" rhead="DE MATHÉMATIQUE. Liv. XIV."/> mouvement, ſuivant une certaine ligne qu’il tend à parcourir <lb/>en vertu de la force qui lui a été communiquée, & </s> <s xml:id="echoid-s15753" xml:space="preserve">qu’il dé-<lb/>crit effectivement, ſi rien ne le détourne de cette ligne.</s> <s xml:id="echoid-s15754" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1243" type="section" level="1" n="897"> <head xml:id="echoid-head1081" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s15755" xml:space="preserve">917. </s> <s xml:id="echoid-s15756" xml:space="preserve">Comme il eſt évident qu’un corps ne peut aller par <lb/>deux chemins différens, lorſque pluſieurs forces concourent <lb/>par leurs actions réunies à le mettre en mouvement, le mou-<lb/>vement s’appelle mouvement compoſé, & </s> <s xml:id="echoid-s15757" xml:space="preserve">la direction que ſuit <lb/>le corps eſt appellée direction moyenne.</s> <s xml:id="echoid-s15758" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1244" type="section" level="1" n="898"> <head xml:id="echoid-head1082" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s15759" xml:space="preserve">918. </s> <s xml:id="echoid-s15760" xml:space="preserve">Les corps dont on conſidere le mouvement, ſont durs <lb/>ou fluides: </s> <s xml:id="echoid-s15761" xml:space="preserve">il y en a auſſi qui ont du reſſort, & </s> <s xml:id="echoid-s15762" xml:space="preserve">d’autres qui <lb/>n’en ont pas.</s> <s xml:id="echoid-s15763" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1245" type="section" level="1" n="899"> <head xml:id="echoid-head1083" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s15764" xml:space="preserve">919. </s> <s xml:id="echoid-s15765" xml:space="preserve">On appelle corps dur celui dont les parties ne ſe divi-<lb/>ſent pas aiſément, & </s> <s xml:id="echoid-s15766" xml:space="preserve">qui étant diviſées ne ſe réuniſſent point <lb/>facilement, comme une pierre.</s> <s xml:id="echoid-s15767" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1246" type="section" level="1" n="900"> <head xml:id="echoid-head1084" xml:space="preserve">IX.</head> <p> <s xml:id="echoid-s15768" xml:space="preserve">920. </s> <s xml:id="echoid-s15769" xml:space="preserve">On appelle corps fluide celui dont les parties ſe divi-<lb/>ſent aiſément, & </s> <s xml:id="echoid-s15770" xml:space="preserve">leſquelles étant diviſées ſe réuniſſent facile-<lb/>ment, comme l’eau.</s> <s xml:id="echoid-s15771" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1247" type="section" level="1" n="901"> <head xml:id="echoid-head1085" xml:space="preserve">X.</head> <p> <s xml:id="echoid-s15772" xml:space="preserve">921. </s> <s xml:id="echoid-s15773" xml:space="preserve">On appelle corps ſans reſſort celui qui à la rencontre <lb/>d’un autre, ne change point de figure, ou s’il en change, ne <lb/>ſe rétablit point dans ſa premiere figure.</s> <s xml:id="echoid-s15774" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1248" type="section" level="1" n="902"> <head xml:id="echoid-head1086" xml:space="preserve">XI.</head> <p> <s xml:id="echoid-s15775" xml:space="preserve">922. </s> <s xml:id="echoid-s15776" xml:space="preserve">On appelle corps à reſſort celui qui à la rencontre d’un <lb/>autre, change de figure dans le choc, & </s> <s xml:id="echoid-s15777" xml:space="preserve">enſuite ſe rétablit <lb/>comme auparavant.</s> <s xml:id="echoid-s15778" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15779" xml:space="preserve">Nota. </s> <s xml:id="echoid-s15780" xml:space="preserve">Nous n’examinerons dansce Traité que les corps durs <lb/>ſans reſſort; </s> <s xml:id="echoid-s15781" xml:space="preserve">à l’égard des autres, nous en parlerons aux en-<lb/>droits qu’il conviendra.</s> <s xml:id="echoid-s15782" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1249" type="section" level="1" n="903"> <head xml:id="echoid-head1087" xml:space="preserve"><emph style="sc">Demandes</emph>. <lb/>I.</head> <p> <s xml:id="echoid-s15783" xml:space="preserve">923. </s> <s xml:id="echoid-s15784" xml:space="preserve">L’on demande qu’il ſoit regardé comme inconteſtable <pb o="496" file="0574" n="594" rhead="NOUVEAU COURS"/> que lorſque deux corps ſe rencontrent dans des directions dia-<lb/>métralement oppoſées, ils ſe communiquent mutuellement <lb/>leur mouvement, & </s> <s xml:id="echoid-s15785" xml:space="preserve">qu’un corps perd autant de ſon mouve-<lb/>ment qu’il en communique à un autre.</s> <s xml:id="echoid-s15786" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1250" type="section" level="1" n="904"> <head xml:id="echoid-head1088" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s15787" xml:space="preserve">924. </s> <s xml:id="echoid-s15788" xml:space="preserve">Que lorſque deux corps ſans reſſort ſe rencontrent, <lb/>ils ne ſe repouſſent point l’un l’autre, & </s> <s xml:id="echoid-s15789" xml:space="preserve">que le plus fort em-<lb/>porte le plus foible dans ſa même détermination.</s> <s xml:id="echoid-s15790" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1251" type="section" level="1" n="905"> <head xml:id="echoid-head1089" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s15791" xml:space="preserve">925. </s> <s xml:id="echoid-s15792" xml:space="preserve">Il ſuit delà que lorſqu’un corps a plus de force qu’un <lb/>autre, il pouſſe devant lui celui qui eſt le plus foible, & </s> <s xml:id="echoid-s15793" xml:space="preserve">que <lb/>ces deux corps peuvent être regardés comme s’ils n’en fai-<lb/>ſoient plus qu’un, qui les vaut tous deux.</s> <s xml:id="echoid-s15794" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1252" type="section" level="1" n="906"> <head xml:id="echoid-head1090" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s15795" xml:space="preserve">926. </s> <s xml:id="echoid-s15796" xml:space="preserve">On ſuppoſe encore que les corps ſe meuvent dans un <lb/>milieu, qui ne réſiſte point à leurs mouvemens; </s> <s xml:id="echoid-s15797" xml:space="preserve">de ſorte que <lb/>ſi un corps parcourt 4 toiſes dans la premiere minute de ſon <lb/>mouvement, il continuera de parcourir 4 toiſes dans chaque <lb/>minute.</s> <s xml:id="echoid-s15798" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1253" type="section" level="1" n="907"> <head xml:id="echoid-head1091" xml:space="preserve"><emph style="sc">Axiome</emph>.</head> <p> <s xml:id="echoid-s15799" xml:space="preserve">927. </s> <s xml:id="echoid-s15800" xml:space="preserve">Les effets ſont proportionnels à leurs cauſes.</s> <s xml:id="echoid-s15801" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1254" type="section" level="1" n="908"> <head xml:id="echoid-head1092" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s15802" xml:space="preserve">928. </s> <s xml:id="echoid-s15803" xml:space="preserve">Il ſuit delà que ſi l’on a deux corps égaux A & </s> <s xml:id="echoid-s15804" xml:space="preserve">C, qui <lb/> <anchor type="note" xlink:label="note-0574-01a" xlink:href="note-0574-01"/> étant mis en mouvement, parcourent en même tems les eſ-<lb/> <anchor type="note" xlink:label="note-0574-02a" xlink:href="note-0574-02"/> paces A B & </s> <s xml:id="echoid-s15805" xml:space="preserve">C D, ces deux corps ont reçu des degrés de <lb/>vîteſſe, qui ſont dans la raiſon des mêmes eſpaces A B & </s> <s xml:id="echoid-s15806" xml:space="preserve">C D; <lb/></s> <s xml:id="echoid-s15807" xml:space="preserve">puiſque les degrés de viteſſe de ces corps peuvent être pris <lb/>pour les cauſes, & </s> <s xml:id="echoid-s15808" xml:space="preserve">les eſpaces parcourus pour les effets.</s> <s xml:id="echoid-s15809" xml:space="preserve"/> </p> <div xml:id="echoid-div1254" type="float" level="2" n="1"> <note position="left" xlink:label="note-0574-01" xlink:href="note-0574-01a" xml:space="preserve">Pl. XXIII.</note> <note position="left" xlink:label="note-0574-02" xlink:href="note-0574-02a" xml:space="preserve">Figure 329.</note> </div> </div> <div xml:id="echoid-div1256" type="section" level="1" n="909"> <head xml:id="echoid-head1093" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s15810" xml:space="preserve">Comme les corps que l’on fait rouler ſur un plan parcourent <lb/>des lignes droites (pourvu qu’une ſeule force les ait mis en <lb/>mouvement), nous prendrons dans la ſuite des lignes droites <lb/>pour exprimer non ſeulement le chemin que ces corps par-<lb/>courent, ou auront à parcourir, mais encore pour exprimer <pb o="497" file="0575" n="595" rhead="DE MATHÉMATIQUE. Liv. XIV."/> les degrés de force qu’on leur aura communiqué: </s> <s xml:id="echoid-s15811" xml:space="preserve">nous ſuppo-<lb/>ſerons auſſi que les corps dont nous parlerons ſeront de figure <lb/>ſphérique.</s> <s xml:id="echoid-s15812" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1257" type="section" level="1" n="910"> <head xml:id="echoid-head1094" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s15813" xml:space="preserve">929. </s> <s xml:id="echoid-s15814" xml:space="preserve">Si deux corps ſemblables de même matiere & </s> <s xml:id="echoid-s15815" xml:space="preserve">égaux ſont <lb/>mus avec des vîteſſes inégales, l’effort du corps qui aura le plus de <lb/>vîteſſe ſera plus grand ſur le corps qu’il rencontrera, que celui dont <lb/>la vîteſſe ſera plus petite.</s> <s xml:id="echoid-s15816" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1258" type="section" level="1" n="911"> <head xml:id="echoid-head1095" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s15817" xml:space="preserve">Si l’on ſuppoſe que de deux corps égaux l’un ait une vîteſſe <lb/>double de l’autre, je dis que ces deux corps venant à frapper <lb/>un autre corps, celui qui aura la vîteſſe double, le frappera <lb/>avec deux fois plus de force que l’autre: </s> <s xml:id="echoid-s15818" xml:space="preserve">car les effets étant <lb/>proportionnés à leurs cauſes (art. </s> <s xml:id="echoid-s15819" xml:space="preserve">927 & </s> <s xml:id="echoid-s15820" xml:space="preserve">928) ſi l’on prend les <lb/>vîteſſes pour les cauſes, & </s> <s xml:id="echoid-s15821" xml:space="preserve">les chocs pour les effets, le corps <lb/>qui aura deux fois plus de vîteſſe que l’autre, agira avec deux <lb/>fois plus de force contre celui qu’il rencontrera.</s> <s xml:id="echoid-s15822" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1259" type="section" level="1" n="912"> <head xml:id="echoid-head1096" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s15823" xml:space="preserve">930. </s> <s xml:id="echoid-s15824" xml:space="preserve">Si deux corps inégaux & </s> <s xml:id="echoid-s15825" xml:space="preserve">de même matiere ſont pouſſés <lb/>avec des vîteſſes égales, le plus grand corps fera plus d’impreſſion <lb/>ſur le corps qu’il rencontrera que le plus petit.</s> <s xml:id="echoid-s15826" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1260" type="section" level="1" n="913"> <head xml:id="echoid-head1097" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s15827" xml:space="preserve">Si l’on ſuppoſe deux corps, l’un de quatre livres, & </s> <s xml:id="echoid-s15828" xml:space="preserve">l’autre <lb/>de deux livres, il eſt conſtant que ſi ces deux corps ont des de-<lb/>grés de vîteſſe égaux, le plus grand aura deux fois plus de <lb/>force que le plus petit: </s> <s xml:id="echoid-s15829" xml:space="preserve">car ſi l’on ſuppoſe le corps de quatre <lb/>livres diviſé en deux également, l’on aura deux autres corps, <lb/>dont chacun ſera égal à celui de deux livres; </s> <s xml:id="echoid-s15830" xml:space="preserve">& </s> <s xml:id="echoid-s15831" xml:space="preserve">comme ils <lb/>auront la même vîteſſe que celui de deux livres, la force de <lb/>chacun en particulier ſera égale à celle du plus petit: </s> <s xml:id="echoid-s15832" xml:space="preserve">ainſi ces <lb/>deux corps n’en faiſant qu’un, la force du plus grand corps <lb/>ſera par conſéquent double de celle du plus petit.</s> <s xml:id="echoid-s15833" xml:space="preserve"/> </p> <pb o="498" file="0576" n="596" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1261" type="section" level="1" n="914"> <head xml:id="echoid-head1098" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s15834" xml:space="preserve">931. </s> <s xml:id="echoid-s15835" xml:space="preserve">Il ſuit des deux théorêmes précédens que la force d’un <lb/>corps, qu’on peut appeller auſſi quantité de mouvement de ce <lb/>corps, ne dépend pas ſeulement de ſa vîteſſe, mais encore de <lb/>ſa maſſe: </s> <s xml:id="echoid-s15836" xml:space="preserve">c’eſt pourquoi l’on connoîtra toujours la quantité <lb/>de mouvement de deux ou de pluſieurs corps, en multipliant <lb/>la maſſe de chacun par ſa vîteſſe. </s> <s xml:id="echoid-s15837" xml:space="preserve">Pour ſe convaincre de cette <lb/>vîteſſe, imaginons deux corps, dont l’un ait trois parties de <lb/>maſſe & </s> <s xml:id="echoid-s15838" xml:space="preserve">4 degrés de vîteſſe, & </s> <s xml:id="echoid-s15839" xml:space="preserve">l’autre cinq parties de maſſe <lb/>& </s> <s xml:id="echoid-s15840" xml:space="preserve">6 degrés de vîteſſe, & </s> <s xml:id="echoid-s15841" xml:space="preserve">nommons f la force qui eſt en état <lb/>de donner un degré de vîteſſe à un corps qui n’aura qu’une <lb/>partie de maſſe, puiſque les effets ſont proportionnés aux <lb/>cauſes, celle qui ſera en état de donner quatre degrés de vîteſſe <lb/>ſera 4f. </s> <s xml:id="echoid-s15842" xml:space="preserve">Si le corps devient trois fois plus grand, & </s> <s xml:id="echoid-s15843" xml:space="preserve">qu’il faille <lb/>lui donner encore 4 degrés de vîteſſe, il n’eſt pas moins évi-<lb/>dent que la force devient 3 x 4f ou 12f. </s> <s xml:id="echoid-s15844" xml:space="preserve">Par la même raiſon, <lb/>puiſque les degrés de vîteſſe ſont égaux, en appellant toujours <lb/>f celle qui peut donner un degré de vîteſſe à une partie du ſe-<lb/>cond corps, 6f ſera celle qui eſt capable de lui en donner <lb/>6 degrés, & </s> <s xml:id="echoid-s15845" xml:space="preserve">ſi le corps devient cinq fois plus gros, il faudra <lb/>une force cinq fois plus grande: </s> <s xml:id="echoid-s15846" xml:space="preserve">donc la force qui lui donne <lb/>cette même vîteſſe ſera 5 x 6f ou 30f: </s> <s xml:id="echoid-s15847" xml:space="preserve">donc les quantités de <lb/>mouvement de ces corps, ou les forces qui les ont miſes en <lb/>mouvement ſeront entr’elles comme 12f eſt à 30f, ou comme <lb/>12 à 30, c’eſt-à-dire comme les produits des maſſes par les <lb/>vîteſſes. </s> <s xml:id="echoid-s15848" xml:space="preserve">Ainſi ayant deux corps, que nous nommerons a & </s> <s xml:id="echoid-s15849" xml:space="preserve">b, <lb/>nommant c la vîteſſe du premier, & </s> <s xml:id="echoid-s15850" xml:space="preserve">d la vîteſſe du ſecond, <lb/>a c ſera la quantité de mouvement de l’un, & </s> <s xml:id="echoid-s15851" xml:space="preserve">b d la quantité <lb/>de mouvement de l’autre.</s> <s xml:id="echoid-s15852" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1262" type="section" level="1" n="915"> <head xml:id="echoid-head1099" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s15853" xml:space="preserve">932. </s> <s xml:id="echoid-s15854" xml:space="preserve">Il ſuit encore delà que connoiſſant la quantité de <lb/>mouvement d’un corps & </s> <s xml:id="echoid-s15855" xml:space="preserve">ſa maſſe, en diviſant la quantité <lb/>de mouvement par la maſſe, l’on aura au quotient la vîteſſe; <lb/></s> <s xml:id="echoid-s15856" xml:space="preserve">& </s> <s xml:id="echoid-s15857" xml:space="preserve">que diviſant de même la quantité de mouvement par la <lb/>vîteſſe, le quotient donnera la maſſe.</s> <s xml:id="echoid-s15858" xml:space="preserve"/> </p> <pb o="499" file="0577" n="597" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1263" type="section" level="1" n="916"> <head xml:id="echoid-head1100" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s15859" xml:space="preserve">933. </s> <s xml:id="echoid-s15860" xml:space="preserve">Si deux corps ont des maſſes & </s> <s xml:id="echoid-s15861" xml:space="preserve">des vîteſſes qui ſoient en <lb/>raiſon réciproque, ces deux corps auront une même quantité de <lb/>mouvement.</s> <s xml:id="echoid-s15862" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1264" type="section" level="1" n="917"> <head xml:id="echoid-head1101" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s15863" xml:space="preserve">Par ce qui précede, la force d’un corps ou ſa quantité de mou-<lb/>vement dépend de ces deux choſes, ſa maſſe & </s> <s xml:id="echoid-s15864" xml:space="preserve">ſa vîteſſe, <lb/>c’eſt-à-dire, eſt en raiſon compoſée de la maſſe & </s> <s xml:id="echoid-s15865" xml:space="preserve">de la vîteſſe, <lb/>ou comme le produit de ſa maſſe par ſa vîteſſe: </s> <s xml:id="echoid-s15866" xml:space="preserve">par hypotheſe, <lb/>la maſſe du premier eſt à celle du ſecond, comme la vîteſſe <lb/>du même ſecond eſt à celle du premier: </s> <s xml:id="echoid-s15867" xml:space="preserve">donc les quantités <lb/>de mouvemens ou les forces de ces deux corps ſont égales. <lb/></s> <s xml:id="echoid-s15868" xml:space="preserve">Ainſi nommant a la maſſe du premier, & </s> <s xml:id="echoid-s15869" xml:space="preserve">c ſa vîteſſe; </s> <s xml:id="echoid-s15870" xml:space="preserve">b la <lb/>maſſe du ſecond, & </s> <s xml:id="echoid-s15871" xml:space="preserve">d ſa vîteſſe, on aura a : </s> <s xml:id="echoid-s15872" xml:space="preserve">b :</s> <s xml:id="echoid-s15873" xml:space="preserve">: d : </s> <s xml:id="echoid-s15874" xml:space="preserve">c. </s> <s xml:id="echoid-s15875" xml:space="preserve">Donc <lb/>a c = b d. </s> <s xml:id="echoid-s15876" xml:space="preserve">C. </s> <s xml:id="echoid-s15877" xml:space="preserve">Q. </s> <s xml:id="echoid-s15878" xml:space="preserve">F. </s> <s xml:id="echoid-s15879" xml:space="preserve">D.</s> <s xml:id="echoid-s15880" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1265" type="section" level="1" n="918"> <head xml:id="echoid-head1102" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s15881" xml:space="preserve">934. </s> <s xml:id="echoid-s15882" xml:space="preserve">Il ſuit delà que ſi l’on a deux corps A & </s> <s xml:id="echoid-s15883" xml:space="preserve">B, dont les <lb/> <anchor type="note" xlink:label="note-0577-01a" xlink:href="note-0577-01"/> maſſes ſoient réciproques aux vîteſſes, ces deux corps venant <lb/>à ſe rencontrer ſuivant des directions diamétralement oppo-<lb/>ſées, ſe choqueront également, & </s> <s xml:id="echoid-s15884" xml:space="preserve">qu’ils demeureront tous <lb/>les deux en repos au moment qu’ils ſe ſeront choqués: </s> <s xml:id="echoid-s15885" xml:space="preserve">car <lb/>ſuppoſant que le corps A ſoit de 4 livres, & </s> <s xml:id="echoid-s15886" xml:space="preserve">ſa vîteſſe ſoit de <lb/>12 degrés, que le corps B ſoit de 6 livres, & </s> <s xml:id="echoid-s15887" xml:space="preserve">ſa vîteſſe de 8 de-<lb/>grés; </s> <s xml:id="echoid-s15888" xml:space="preserve">la maſſe du corps A, qui eſt 4, étant multipliée par ſa <lb/>vîteſſe, qui eſt 12, donnera 48 pour la quantité de mouve-<lb/>ment du corps A. </s> <s xml:id="echoid-s15889" xml:space="preserve">De même, ſi l’on multiplie la maſſe du <lb/>corps B, qui eſt 6, par ſa vîteſſe, qui eſt 8, ſa quantité de <lb/>mouvement ſera encore 48: </s> <s xml:id="echoid-s15890" xml:space="preserve">ils viendront donc ſe choquer <lb/>avec des forces égales & </s> <s xml:id="echoid-s15891" xml:space="preserve">diamétralement oppoſées; </s> <s xml:id="echoid-s15892" xml:space="preserve">le corps <lb/>A choquera donc autant le corps B, que le corps B choquera <lb/>le corps A: </s> <s xml:id="echoid-s15893" xml:space="preserve">ainſi ils demeureront en repos, puiſque l’un ne <lb/>fera pas plus d’effort que l’autre, & </s> <s xml:id="echoid-s15894" xml:space="preserve">qu’il n’y a pas de raiſon <lb/>pour que l’un l’emporte ſur l’autre.</s> <s xml:id="echoid-s15895" xml:space="preserve"/> </p> <div xml:id="echoid-div1265" type="float" level="2" n="1"> <note position="right" xlink:label="note-0577-01" xlink:href="note-0577-01a" xml:space="preserve">Figure 330.</note> </div> <p> <s xml:id="echoid-s15896" xml:space="preserve">Cette égalité entre deux forces ou quantités de mouvemens <lb/>qui agiſſent ſuivant des directions diamétralement oppoſées, ſe <lb/>nomme équilibre. </s> <s xml:id="echoid-s15897" xml:space="preserve">Ainſi pour qu’il y ait équilibre entre deux ou un <pb o="500" file="0578" n="598" rhead="NOUVEAU COURS"/> plus grand nombre de forces qui agiſſent ſuivant des directions <lb/>quelconques, il faut qu’on puiſſe les réduire à deux forces égales <lb/>& </s> <s xml:id="echoid-s15898" xml:space="preserve">directement oppoſées.</s> <s xml:id="echoid-s15899" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1267" type="section" level="1" n="919"> <head xml:id="echoid-head1103" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s15900" xml:space="preserve">935. </s> <s xml:id="echoid-s15901" xml:space="preserve">Il ſuit encore delà que ſi deux corps égaux avec des vîteſſes <lb/>égales, viennent à ſe rencontrer dans des lignes de direction <lb/>diamétralement oppoſées, ils ſeront en équilibre à l’inſtant <lb/>du choc, puiſqu’ils auront chacun une même quantité de mou-<lb/>vement.</s> <s xml:id="echoid-s15902" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1268" type="section" level="1" n="920"> <head xml:id="echoid-head1104" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s15903" xml:space="preserve">936. </s> <s xml:id="echoid-s15904" xml:space="preserve">Lorſque deux corps ſans reſſort ſe meuvent dans la même <lb/>détermination, & </s> <s xml:id="echoid-s15905" xml:space="preserve">vers un même côté, le corps qui a le plus de <lb/>vîteſſe ayant rencontré celui qui en a moins, & </s> <s xml:id="echoid-s15906" xml:space="preserve">ces deux corps <lb/>allant enſemble, ils auront une quantité de mouvement égale à la <lb/>ſomme de celles qu’ils avoient avant le choc.</s> <s xml:id="echoid-s15907" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1269" type="section" level="1" n="921"> <head xml:id="echoid-head1105" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s15908" xml:space="preserve">Si ces deux corps ſe meuvent d’un même côté, il n’y aura <lb/>rien d’oppoſé qui puiſſe détruire leur mouvement: </s> <s xml:id="echoid-s15909" xml:space="preserve">c’eſt pour-<lb/>quoi ils conſerveront après le choc la même quantité de mou-<lb/>vement qu’ils avoient avant le choc: </s> <s xml:id="echoid-s15910" xml:space="preserve">car ſi celui qui a le plus <lb/>de mouvement en communique à celui qui en a moins, cette <lb/>quantité de mouvement reſte dans ce dernier. </s> <s xml:id="echoid-s15911" xml:space="preserve">Or ces deux <lb/>corps étant conſidérés comme n’en faiſant qu’un ſeul (art. </s> <s xml:id="echoid-s15912" xml:space="preserve">925) <lb/>après le choc; </s> <s xml:id="echoid-s15913" xml:space="preserve">il s’enſuit que leur quantité de mouvement eſt <lb/>la ſomme de celles qu’ils avoient avant le choc.</s> <s xml:id="echoid-s15914" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1270" type="section" level="1" n="922"> <head xml:id="echoid-head1106" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s15915" xml:space="preserve">937. </s> <s xml:id="echoid-s15916" xml:space="preserve">Il ſuit delà que connoiſſant la quantité de mouvement <lb/>de deux corps, qui n’en font plus qu’un, après s’être rencon-<lb/>trés, l’on trouvera la vîteſſe en diviſant la quantité de mouve-<lb/>ment par la ſomme des maſſes; </s> <s xml:id="echoid-s15917" xml:space="preserve">& </s> <s xml:id="echoid-s15918" xml:space="preserve">que connoiſſant la vîteſſe, <lb/>l’on trouvera la ſomme des maſſes, en diviſant la quantité de <lb/>mouvement par la vîteſſe.</s> <s xml:id="echoid-s15919" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1271" type="section" level="1" n="923"> <head xml:id="echoid-head1107" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s15920" xml:space="preserve">938. </s> <s xml:id="echoid-s15921" xml:space="preserve">Par conſéquent ſi l’on a deux corps égaux mus ſur une <pb o="501" file="0579" n="599" rhead="DE MATHÉMATIQUE. Liv. XIV."/> même ligne de direction, & </s> <s xml:id="echoid-s15922" xml:space="preserve">que l’un ſoit en repos, & </s> <s xml:id="echoid-s15923" xml:space="preserve">l’autre <lb/>en mouvement, celui qui eſt en mouvement venant à ren-<lb/>contrer celui qui eſt en repos (ces deux corps n’en faiſant plus <lb/>qu’un), il lui comumniquera la moitié de la vîteſſe qu’il avoit <lb/>avant le choc; </s> <s xml:id="echoid-s15924" xml:space="preserve">puiſque pour avoir cette vîteſſe, il faut diviſer <lb/>la quantité de mouvement par une maſſe double: </s> <s xml:id="echoid-s15925" xml:space="preserve">enſin ſi le <lb/>corps mobile en rencontre un autre en repos, dont la maſſe <lb/>ſoit triple de la ſienne, ſa vîteſſe ne ſera plus que d’un quart, <lb/>ainſi des autres.</s> <s xml:id="echoid-s15926" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15927" xml:space="preserve">En général ſoit u la vîteſſe du premier corps, & </s> <s xml:id="echoid-s15928" xml:space="preserve">m ſa maſſe, <lb/>v la vîteſſe du ſecond corps, & </s> <s xml:id="echoid-s15929" xml:space="preserve">M ſa maſſe. </s> <s xml:id="echoid-s15930" xml:space="preserve">Soit V la vîteſſe <lb/>après le choc, on aura, ſuivant ce que nous venons de voir, <lb/>V = {m u + Mv/m + M}. </s> <s xml:id="echoid-s15931" xml:space="preserve">On pourra par cette formule déterminer la <lb/>vîteſſe V dans tous les cas poſſibles, quel que ſoit le rapport de <lb/>m à M, & </s> <s xml:id="echoid-s15932" xml:space="preserve">de u à v. </s> <s xml:id="echoid-s15933" xml:space="preserve">Suppoſons, par exemple, u = o, & </s> <s xml:id="echoid-s15934" xml:space="preserve">m = M, <lb/>on aura V = {Mv/2M} = {v/2}; </s> <s xml:id="echoid-s15935" xml:space="preserve">c’eſt ce que nous venons de voir.</s> <s xml:id="echoid-s15936" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1272" type="section" level="1" n="924"> <head xml:id="echoid-head1108" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s15937" xml:space="preserve">939. </s> <s xml:id="echoid-s15938" xml:space="preserve">Si deux corps ſe meuvent dans un ſens directement oppoſé <lb/>ſur une même direction, ces deux corps venant à ſe rencontrer, & </s> <s xml:id="echoid-s15939" xml:space="preserve"><lb/>n’en faiſant plus qu’un, la quantité de mouvement de ces corps <lb/>ſera la différence des quantités de mouvement que les deux corps <lb/>avoient avant le choc.</s> <s xml:id="echoid-s15940" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1273" type="section" level="1" n="925"> <head xml:id="echoid-head1109" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s15941" xml:space="preserve">Si ces deux corps ſe meuvent dans des déterminations di-<lb/>rectement oppoſées, ils tendront mutuellement à s’arrêter; </s> <s xml:id="echoid-s15942" xml:space="preserve">de <lb/>ſorte que s’ils avoient des forces égales, ils demeureroient en <lb/>repos après le choc; </s> <s xml:id="echoid-s15943" xml:space="preserve">ainſi le plus fort perd autant de ſa force <lb/>que le plus foible en a. </s> <s xml:id="echoid-s15944" xml:space="preserve">Il ne reſte donc pour mouvoir ces deux <lb/>corps après leur choc, que la différence de leurs forces, ou de <lb/>leur quantité de mouvement; </s> <s xml:id="echoid-s15945" xml:space="preserve">mais ces deux corps étant con-<lb/>ſidérés comme n’en faiſant plus qu’un, ſa quantité de mou-<lb/>vement ſera la différence de celles des deux corps avant le <lb/>choc.</s> <s xml:id="echoid-s15946" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1274" type="section" level="1" n="926"> <head xml:id="echoid-head1110" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s15947" xml:space="preserve">940. </s> <s xml:id="echoid-s15948" xml:space="preserve">Il ſuit delà que pour trouver la vîteſſe de ces corps <pb o="502" file="0580" n="600" rhead="NOUVEAU COURS"/> après leur choc, il faut diviſer la différence des quantités <lb/>de mouvement qu’ils avoient avant le choc, par la ſomme de <lb/>leurs maſſes, & </s> <s xml:id="echoid-s15949" xml:space="preserve">le quotient donnera cette vîteſſe, laquelle <lb/>ſera dans la détermination du corps qui avoit la plus grande <lb/>quantité de mouvement avant le choc: </s> <s xml:id="echoid-s15950" xml:space="preserve">donc la formule gé-<lb/>nérale pour déterminer la vîteſſe des corps après le choc, ſoit <lb/>dans une même direction ou dans des directions diamétrale-<lb/>ment oppoſées, ſera V = {mu ± Mv/m + M}.</s> <s xml:id="echoid-s15951" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1275" type="section" level="1" n="927"> <head xml:id="echoid-head1111" xml:space="preserve">CHAPITRE II. <lb/>Du mouvement des Corps jettés. <lb/><emph style="sc">Définitions</emph>. <lb/>I.</head> <p> <s xml:id="echoid-s15952" xml:space="preserve">941. </s> <s xml:id="echoid-s15953" xml:space="preserve">SI un corps ſe meut pendant un certain tems, lequel <lb/>tems ſoit diviſé en pluſieurs parties égales, nous appellerons <lb/>chacune de ces petites parties moment ou inſtant.</s> <s xml:id="echoid-s15954" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1276" type="section" level="1" n="928"> <head xml:id="echoid-head1112" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s15955" xml:space="preserve">942. </s> <s xml:id="echoid-s15956" xml:space="preserve">Si un corps reçoit dans chaque inſtant une augmenta-<lb/>tion égale de vîteſſe, cette vîteſſe ſera nommée accélerée; </s> <s xml:id="echoid-s15957" xml:space="preserve">& </s> <s xml:id="echoid-s15958" xml:space="preserve"><lb/>ſi au contraire un corps à chaque inſtant perd des degrés égaux <lb/>de vîteſſe, cette vîteſſe ſera nommée retardée. </s> <s xml:id="echoid-s15959" xml:space="preserve">La vîteſſe d’un <lb/>corps qui tombe eſt une vîteſſe accélerée, parce que la peſan-<lb/>teur agit à chaque inſtant ſur lui, & </s> <s xml:id="echoid-s15960" xml:space="preserve">lui communique des <lb/>degrés égaux de vîteſſe. </s> <s xml:id="echoid-s15961" xml:space="preserve">Par une raiſon contraire, la vîteſſe <lb/>d’un corps jetté de bas en haut eſt une vîteſſe retardée, puiſ-<lb/>que la peſanteur ôte à chaque inſtant des degrés égaux de <lb/>vîteſſe. </s> <s xml:id="echoid-s15962" xml:space="preserve">Si les degrés de vîteſſe reçus ou perdus à chaque inſ-<lb/>tant ne ſont pas égaux entr’eux, mais varient ſuivant des rap-<lb/>ports conſtans, ces vîteſſes ſont appellées variables accélerées <lb/>ou variables retardées.</s> <s xml:id="echoid-s15963" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1277" type="section" level="1" n="929"> <head xml:id="echoid-head1113" xml:space="preserve"><emph style="sc">Axiome</emph> I.</head> <p> <s xml:id="echoid-s15964" xml:space="preserve">943. </s> <s xml:id="echoid-s15965" xml:space="preserve">Un corps en mouvement ou en repos eſt toujours le <lb/>même corps; </s> <s xml:id="echoid-s15966" xml:space="preserve">il eſt encore le même quelle que ſoit la détermi-<lb/>nation de ſon mouvement & </s> <s xml:id="echoid-s15967" xml:space="preserve">ſa quantité.</s> <s xml:id="echoid-s15968" xml:space="preserve"/> </p> <pb o="503" file="0581" n="601" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1278" type="section" level="1" n="930"> <head xml:id="echoid-head1114" xml:space="preserve"><emph style="sc">Axiome</emph> II.</head> <p> <s xml:id="echoid-s15969" xml:space="preserve">944. </s> <s xml:id="echoid-s15970" xml:space="preserve">Le corps de lui-même ou de ſa nature eſt tout-à-fait <lb/>indifférent au mouvement ou au repos, & </s> <s xml:id="echoid-s15971" xml:space="preserve">par conſéquent ce <lb/>corps étant une fois mis en mouvement, il y reſtera toujours <lb/>juſqu’à ce que quelque cauſe le lui ait ôté; </s> <s xml:id="echoid-s15972" xml:space="preserve">& </s> <s xml:id="echoid-s15973" xml:space="preserve">réciproquement <lb/>un corps une fois en repos, ne ſe mettra jamais de lui-même <lb/>en mouvement.</s> <s xml:id="echoid-s15974" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1279" type="section" level="1" n="931"> <head xml:id="echoid-head1115" xml:space="preserve"><emph style="sc">Axiome</emph> III.</head> <p> <s xml:id="echoid-s15975" xml:space="preserve">945. </s> <s xml:id="echoid-s15976" xml:space="preserve">Le corps de ſoi ou de ſa nature eſt tout-à-fait indif-<lb/>férent à quelque détermination, ou à quelque vîteſſe que ce <lb/>puiſſe être, & </s> <s xml:id="echoid-s15977" xml:space="preserve">par conſéquent ce corps ne changera jamais de <lb/>lui-même ni la vîteſſe, ni la détermination qu’il a eu en der-<lb/>nier lieu.</s> <s xml:id="echoid-s15978" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s15979" xml:space="preserve">946. </s> <s xml:id="echoid-s15980" xml:space="preserve">Nous venons de voir qu’un corps ne peut être en mou-<lb/>vement ſans une cauſe qui l’y ait mis, & </s> <s xml:id="echoid-s15981" xml:space="preserve">que ſi rien ne s’oppoſe <lb/>à ſon mouvement, il y ſera éternellement. </s> <s xml:id="echoid-s15982" xml:space="preserve">Dans la même <lb/>ſuppoſition que rien ne s’oppoſe à ſon mouvement, ſi petite ou <lb/>ſi grande que ſoit la force motrice, il eſt évident que la durée <lb/>du mouvement ſeroit éternelle. </s> <s xml:id="echoid-s15983" xml:space="preserve">On pourroit donc en appa-<lb/>rence inférer delà que la plus petite force comme la plus grande <lb/>produiroit un effet infini en durée, & </s> <s xml:id="echoid-s15984" xml:space="preserve">croire que les forces <lb/>mêmes ſont infinies, ſuivant notre axiome, qui dit que les <lb/>effets ſont proportionnels aux cauſes. </s> <s xml:id="echoid-s15985" xml:space="preserve">Pour n’être pas ſéduits <lb/>par ce ſophiſme, il faut, 1°. </s> <s xml:id="echoid-s15986" xml:space="preserve">diſtinguer la durée du mouve-<lb/>ment du plus ou moins d’eſpace que la force motrice fait par-<lb/>courir au corps dans un tems fini. </s> <s xml:id="echoid-s15987" xml:space="preserve">2°. </s> <s xml:id="echoid-s15988" xml:space="preserve">Faire attention que, <lb/>dans l’hypotheſe, que rien ne s’oppoſe au mouvement du corps, <lb/>la durée infinie de ce mouvement ne vient pas directement de <lb/>la force motrice, mais bien de l’indifférence du même corps <lb/>au mouvement ou au repos; </s> <s xml:id="echoid-s15989" xml:space="preserve">d’où il ſuit évidemment que les <lb/>effets des cauſes ſeront toujours finis & </s> <s xml:id="echoid-s15990" xml:space="preserve">proportionnels à ces <lb/>cauſes, puiſque les effets ne ſeront que le plus ou le moins d’eſ-<lb/>pace parcouru dans un tems donné.</s> <s xml:id="echoid-s15991" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1280" type="section" level="1" n="932"> <head xml:id="echoid-head1116" xml:space="preserve"><emph style="sc">Demande</emph>.</head> <p> <s xml:id="echoid-s15992" xml:space="preserve">947. </s> <s xml:id="echoid-s15993" xml:space="preserve">L’on demande qu’il ſoit accordé que la peſanteur de <lb/>quelque cauſe qu’elle puiſſe provenir, preſſe toujours le corps <lb/>avec une même force pour le faire deſcendre.</s> <s xml:id="echoid-s15994" xml:space="preserve"/> </p> <pb o="504" file="0582" n="602" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1281" type="section" level="1" n="933"> <head xml:id="echoid-head1117" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s15995" xml:space="preserve">948. </s> <s xml:id="echoid-s15996" xml:space="preserve">Si rien ne s’oppoſoit au mouvement des corps jettés; <lb/></s> <s xml:id="echoid-s15997" xml:space="preserve">chacun de ces corps conſerveroit toujours avec une vîteſſe égale le <lb/>mouvement qu’il auroit réçu, & </s> <s xml:id="echoid-s15998" xml:space="preserve">ſuivant toujours une même ligne <lb/>droite.</s> <s xml:id="echoid-s15999" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1282" type="section" level="1" n="934"> <head xml:id="echoid-head1118" xml:space="preserve"><emph style="sc">DÉMONSTRATION</emph>.</head> <p> <s xml:id="echoid-s16000" xml:space="preserve">Comme un corps ne peut jamais de lui-même ſe mettre en <lb/>repos, ni changer ſa détermination ou la vîteſſe qu’il a reçue <lb/>(art. </s> <s xml:id="echoid-s16001" xml:space="preserve">944 & </s> <s xml:id="echoid-s16002" xml:space="preserve">945), il s’enſuit que, ſi rien ne s’oppoſoit à cette <lb/>vîteſſe, le corps conſerveroit perpétuellement ſon mouve-<lb/>ment, & </s> <s xml:id="echoid-s16003" xml:space="preserve">avec une vîteſſe toujours égale, & </s> <s xml:id="echoid-s16004" xml:space="preserve">ſuivroit toujours <lb/>une même ligne droite. </s> <s xml:id="echoid-s16005" xml:space="preserve">C. </s> <s xml:id="echoid-s16006" xml:space="preserve">Q. </s> <s xml:id="echoid-s16007" xml:space="preserve">F. </s> <s xml:id="echoid-s16008" xml:space="preserve">D.</s> <s xml:id="echoid-s16009" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1283" type="section" level="1" n="935"> <head xml:id="echoid-head1119" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16010" xml:space="preserve">949. </s> <s xml:id="echoid-s16011" xml:space="preserve">Donc le mouvement tel qu’il eſt de la part de la puiſ-<lb/>ſance qui meut, ſoit horizontalement, ſoit obliquement, ſoit <lb/>verticalement, ſeroit perpétuel & </s> <s xml:id="echoid-s16012" xml:space="preserve">égal, en allant toujours de <lb/>même côté, ſi l’air ne réſiſtoit pas au corps, & </s> <s xml:id="echoid-s16013" xml:space="preserve">ſi ſa peſanteur <lb/>ne le faiſoit pas toujours deſcendre en bas; </s> <s xml:id="echoid-s16014" xml:space="preserve">de ſorte que le <lb/>mouvement, préciſément comme il eſt de la part du mobile, <lb/>doit être conſidéré comme égal, perpétuel, & </s> <s xml:id="echoid-s16015" xml:space="preserve">toujours diviſé <lb/>vers le même côté où le corps eſt pouſſé.</s> <s xml:id="echoid-s16016" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1284" type="section" level="1" n="936"> <head xml:id="echoid-head1120" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16017" xml:space="preserve">950. </s> <s xml:id="echoid-s16018" xml:space="preserve">De même, ſi immédiatement après qu’un corps a <lb/>acquis une certaine vîteſſe en tombant, l’action de la pe-<lb/>ſanteur venoit à ceſſer tout-à-fait, & </s> <s xml:id="echoid-s16019" xml:space="preserve">que l’air ne réſiſtât <lb/>point, ce corps néanmoins continueroit de ſe mouvoir avec la <lb/>même vîteſſe qu’il auroit reçue en dernier lieu, conſervant <lb/>toujours également cette même vîteſſe, & </s> <s xml:id="echoid-s16020" xml:space="preserve">ſuivant toujours la <lb/>même ligne droite.</s> <s xml:id="echoid-s16021" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1285" type="section" level="1" n="937"> <head xml:id="echoid-head1121" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s16022" xml:space="preserve">951. </s> <s xml:id="echoid-s16023" xml:space="preserve">Donc puiſque l’action de la peſanteur ne nuit point à <lb/>la vîteſſe d’un corps qui tombe, ſi l’air, ni autre choſe ne s’y <lb/>oppoſoit, la vîteſſe que la peſanteur cauſeroit au corps dans <lb/>le premier inſtant, ſubſiſteroit dans le ſecond inſtant avec une <pb o="505" file="0583" n="603" rhead="DE MATHÉMATIQUE. Liv. XIV."/> pareille vîteſſe cauſée par la même peſanteur; </s> <s xml:id="echoid-s16024" xml:space="preserve">par la même <lb/>raiſon les vîteſſes des deux premiers inſtans ſubſiſteroient avec <lb/>celles du troiſieme inſtant; </s> <s xml:id="echoid-s16025" xml:space="preserve">& </s> <s xml:id="echoid-s16026" xml:space="preserve">ainſi les vîteſſes de tous ces pre-<lb/>miers inſtans ſubſiſteroient avec les vîteſſes que ce même corps <lb/>recevroit dans chacun des inſtans ſuivans, ou bien (ce qui eſt <lb/>la même choſe) lorſqu’un corps tombe, ce corps reçoit des <lb/>parties égales de vîteſſe dans des tems égaux, en ſuppoſant <lb/>que l’action de la peſanteur eſt uniforme, & </s> <s xml:id="echoid-s16027" xml:space="preserve">négligeant la <lb/>réſiſtance de l’air.</s> <s xml:id="echoid-s16028" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1286" type="section" level="1" n="938"> <head xml:id="echoid-head1122" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s16029" xml:space="preserve">952. </s> <s xml:id="echoid-s16030" xml:space="preserve">Un corps qui tombe reçoit des degrés égaux de vîteſſe <lb/>dans des tems égaux; </s> <s xml:id="echoid-s16031" xml:space="preserve">de ſorte que dans le ſecond inſtant il a une <lb/>vîteſſe double de celle qu’il avoit dans le premier inſtant de ſa chûte, <lb/>& </s> <s xml:id="echoid-s16032" xml:space="preserve">dans le troiſieme il en a une triple, & </s> <s xml:id="echoid-s16033" xml:space="preserve">ainſi des autres.</s> <s xml:id="echoid-s16034" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1287" type="section" level="1" n="939"> <head xml:id="echoid-head1123" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16035" xml:space="preserve">Puiſqu’un corps qui tombe eſt continuellement pouſſé en <lb/>bas, par l’action de ſa peſanteur, qui eſt toujours la même <lb/>(art. </s> <s xml:id="echoid-s16036" xml:space="preserve">951), il s’enſuit que la peſanteur doit donner à ce corps, <lb/>à chaque inſtant de ſa chûte des degrés égaux de vîteſſe: </s> <s xml:id="echoid-s16037" xml:space="preserve">donc <lb/>puiſque les degrés de vîteſſe que le corps a reçus en premier <lb/>lieu ſubſiſtent entiérement avec ceux qu’il auroit reçus en der-<lb/>nier lieu (art. </s> <s xml:id="echoid-s16038" xml:space="preserve">951), le corps en tombant ſe trouve avoir au-<lb/>tant de degrés de vîteſſe, cauſés par ſa peſanteur, qu’il s’eſt <lb/>écoulé de momens depuis le commencement de ſa chûte juſ-<lb/>qu’au moment que l’on compte: </s> <s xml:id="echoid-s16039" xml:space="preserve">donc ce corps aura à la fin <lb/>du ſecond inſtant une vîteſſe double de celle du premier, au <lb/>troiſieme inſtant une vîteſſe triple, &</s> <s xml:id="echoid-s16040" xml:space="preserve">c. </s> <s xml:id="echoid-s16041" xml:space="preserve">C. </s> <s xml:id="echoid-s16042" xml:space="preserve">Q. </s> <s xml:id="echoid-s16043" xml:space="preserve">F. </s> <s xml:id="echoid-s16044" xml:space="preserve">D.</s> <s xml:id="echoid-s16045" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1288" type="section" level="1" n="940"> <head xml:id="echoid-head1124" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s16046" xml:space="preserve">953. </s> <s xml:id="echoid-s16047" xml:space="preserve">Il ſuit delà que les degrés de vîteſſe qu’un corps a ac-<lb/>quis à la fin de chaque inſtant de chûte, ſont comme les tems <lb/>qui ſe ſont écoulés depuis le commencement de ſa chûte: </s> <s xml:id="echoid-s16048" xml:space="preserve">donc <lb/>puiſque les inſtans écoulés depuis le premier moment de la <lb/>chûte ſont en progreſſion arithmétique, les degrés de vîteſſe <lb/>acquis à la fin de ces tems ſont auſſi en progreſſion arithmé-<lb/>tique.</s> <s xml:id="echoid-s16049" xml:space="preserve"/> </p> <pb o="506" file="0584" n="604" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1289" type="section" level="1" n="941"> <head xml:id="echoid-head1125" xml:space="preserve"><emph style="sc">Demande</emph>.</head> <p> <s xml:id="echoid-s16050" xml:space="preserve">954. </s> <s xml:id="echoid-s16051" xml:space="preserve">On demande qu’il ſoit permis de repréſenter les tems <lb/>par des lignes; </s> <s xml:id="echoid-s16052" xml:space="preserve">ce qui ne doit faire aucune difficulté: </s> <s xml:id="echoid-s16053" xml:space="preserve">car ayant <lb/>repréſenté une minute par une ligne d’un pouce, je repréſen-<lb/>terai deux ou trois minutes par des lignes de deux ou de trois <lb/>pouces. </s> <s xml:id="echoid-s16054" xml:space="preserve">Par cette ſuppoſition, on ne prétend pas que les tems <lb/>& </s> <s xml:id="echoid-s16055" xml:space="preserve">les lignes ſoient des quantités de même nature, mais bien <lb/>que les dernieres ſont des expreſſions propres à repréſenter les <lb/>différens rapports des premiers.</s> <s xml:id="echoid-s16056" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1290" type="section" level="1" n="942"> <head xml:id="echoid-head1126" xml:space="preserve">PROPOSITION III. <lb/>T <emph style="sc">HÉOREME</emph>.</head> <p style="it"> <s xml:id="echoid-s16057" xml:space="preserve">955. </s> <s xml:id="echoid-s16058" xml:space="preserve">Si deux corps égaux A & </s> <s xml:id="echoid-s16059" xml:space="preserve">B ſont en mouvement pendant <lb/> <anchor type="note" xlink:label="note-0584-01a" xlink:href="note-0584-01"/> un même-tems, l’un d’une vîteſſe uniforme, l’autre d’un mouvement <lb/>uniformément accéléré, tel que le dernier degré de la vîteſſe acquiſe <lb/>ſoit égal à la vîteſſe conſtante du corps qui ſe meut uniformément, <lb/>l’eſpace parcouru par le premier ſera double de l’eſpace parcouru par <lb/>le ſecond.</s> <s xml:id="echoid-s16060" xml:space="preserve"/> </p> <div xml:id="echoid-div1290" type="float" level="2" n="1"> <note position="left" xlink:label="note-0584-01" xlink:href="note-0584-01a" xml:space="preserve">Figure 331.</note> </div> </div> <div xml:id="echoid-div1292" type="section" level="1" n="943"> <head xml:id="echoid-head1127" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16061" xml:space="preserve">Soit repréſenté le tems du mouvement par A C, & </s> <s xml:id="echoid-s16062" xml:space="preserve">ſuppo-<lb/>ſons-le partagé en un nombre infini d’inſtans égaux. </s> <s xml:id="echoid-s16063" xml:space="preserve">Si pen-<lb/>dant un de ces inſtans le corps qui ſe meut d’un mouvement <lb/>uniforme parcoure C D pendant le tems A C, il parcourra <lb/>autant de fois C D qu’il y a d’inſtans dans le tems du mouve-<lb/>ment, ou qu’il y a de points dans A C: </s> <s xml:id="echoid-s16064" xml:space="preserve">donc le rectangle <lb/>A C x C D, repréſentera l’eſpace parcouru pendant le tems <lb/>A C par le corps, dont le mouvement eſt uniforme. </s> <s xml:id="echoid-s16065" xml:space="preserve">Préſen-<lb/>tement voyons quel ſera l’eſpace que parcourra le mobile qui <lb/>ſe meut d’un mouvement uniformément accéléré, pendant le <lb/>même tems A C, en ſuppoſant que la vîteſſe qu’il a acquiſe à <lb/>la fin du dernier inſtant du tems A C eſt auſſi repréſentée par <lb/>C D. </s> <s xml:id="echoid-s16066" xml:space="preserve">Cela poſé, par le dernier corollaire, puiſque les vîteſſes <lb/>ſont comme les tems, à la fin du tems A B, c’eſt-à-dire dans <lb/>l’inſtant B, il parcourra une ligne B E qui ſera à C D, comme <lb/>A C : </s> <s xml:id="echoid-s16067" xml:space="preserve">A B : </s> <s xml:id="echoid-s16068" xml:space="preserve">donc la ſomme des eſpaces parcourus par le corps <lb/>qui eſt mu d’un mouvement uniformément accéléré, ſera re-<lb/>préſentée par la ſomme des élémens du triangle A C B : </s> <s xml:id="echoid-s16069" xml:space="preserve">donc <lb/>l’eſpace total parcouru pendant le tems A B n’eſt pas différent <pb o="507" file="0585" n="605" rhead="DE MATHÉMATIQUE. Liv. XIV."/> de cette ſomme, & </s> <s xml:id="echoid-s16070" xml:space="preserve">ſera repréſenté par A C x {C D/2} : </s> <s xml:id="echoid-s16071" xml:space="preserve">donc le pre-<lb/>mier mobile parcourt dans le même-tems un eſpace double du <lb/>ſecond. </s> <s xml:id="echoid-s16072" xml:space="preserve">C. </s> <s xml:id="echoid-s16073" xml:space="preserve">Q. </s> <s xml:id="echoid-s16074" xml:space="preserve">F. </s> <s xml:id="echoid-s16075" xml:space="preserve">D.</s> <s xml:id="echoid-s16076" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1293" type="section" level="1" n="944"> <head xml:id="echoid-head1128" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16077" xml:space="preserve">956. </s> <s xml:id="echoid-s16078" xml:space="preserve">Puiſque les deux corps ſont égaux, on peut n’en ſup-<lb/>poſer qu’un ſeul; </s> <s xml:id="echoid-s16079" xml:space="preserve">d’où il ſuit que ſi un même corps eſt mu <lb/>d’un mouvement uniforme pendant un certain tems, & </s> <s xml:id="echoid-s16080" xml:space="preserve">que <lb/>dans un tems égal il ait acquis d’un mouvement uniformé-<lb/>ment accéléré une vîteſſe égale à celle du mouvement uni-<lb/>forme, l’eſpace qu’il aura parcouru dans le premier cas ſera <lb/>double de celui qui a été parcouru dans le ſecond.</s> <s xml:id="echoid-s16081" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1294" type="section" level="1" n="945"> <head xml:id="echoid-head1129" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16082" xml:space="preserve">957. </s> <s xml:id="echoid-s16083" xml:space="preserve">Donc les eſpaces parcourus dans un mouvement uni-<lb/>formément accéléré ſont entr’eux comme les quarrés des tems, <lb/>à commencer de l’inſtant que le corps a été mis en mouve-<lb/>ment: </s> <s xml:id="echoid-s16084" xml:space="preserve">car il eſt évident que les triangles A B E, A C D qui <lb/>repréſentent les eſpaces parcourus pendant les tems A B, A C <lb/>étant ſemblables, ſont comme les quarrés des tems A B & </s> <s xml:id="echoid-s16085" xml:space="preserve">A C.</s> <s xml:id="echoid-s16086" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1295" type="section" level="1" n="946"> <head xml:id="echoid-head1130" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s16087" xml:space="preserve">958. </s> <s xml:id="echoid-s16088" xml:space="preserve">Puiſque les tems ſont comme les vîteſſes (art. </s> <s xml:id="echoid-s16089" xml:space="preserve">953), <lb/>& </s> <s xml:id="echoid-s16090" xml:space="preserve">que les eſpaces parcourus depuis le premier inſtant du mou-<lb/>vement ſont comme les quarrés des tems, ils ſeront auſſi en-<lb/>tr’eux comme les quarrés des vîteſſes acquiſes. </s> <s xml:id="echoid-s16091" xml:space="preserve">Ainſi nommant <lb/>L une longueur parcourue depuis le point du repos; </s> <s xml:id="echoid-s16092" xml:space="preserve">T, le tems <lb/>employé à la parcourir; </s> <s xml:id="echoid-s16093" xml:space="preserve">V, la vîteſſe acquiſe à la fin de ces <lb/>tems; </s> <s xml:id="echoid-s16094" xml:space="preserve">& </s> <s xml:id="echoid-s16095" xml:space="preserve">l, une autre longueur parcourue depuis le point de <lb/>repos; </s> <s xml:id="echoid-s16096" xml:space="preserve">t, le tems employé à la parcourir; </s> <s xml:id="echoid-s16097" xml:space="preserve">u, la vîteſſe acquiſe <lb/>à la fin de ce tems, l’on aura L : </s> <s xml:id="echoid-s16098" xml:space="preserve">l :</s> <s xml:id="echoid-s16099" xml:space="preserve">: T T : </s> <s xml:id="echoid-s16100" xml:space="preserve">tt, ou bien L : </s> <s xml:id="echoid-s16101" xml:space="preserve">l <lb/>:</s> <s xml:id="echoid-s16102" xml:space="preserve">: V V : </s> <s xml:id="echoid-s16103" xml:space="preserve">uu.</s> <s xml:id="echoid-s16104" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1296" type="section" level="1" n="947"> <head xml:id="echoid-head1131" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s16105" xml:space="preserve">959. </s> <s xml:id="echoid-s16106" xml:space="preserve">Puiſque l’on a L : </s> <s xml:id="echoid-s16107" xml:space="preserve">l :</s> <s xml:id="echoid-s16108" xml:space="preserve">: V V : </s> <s xml:id="echoid-s16109" xml:space="preserve">uu, ſi on extrait la racine <lb/>quarrée de chaque terme, on aura √L\x{0020} : </s> <s xml:id="echoid-s16110" xml:space="preserve">√l\x{0020} :</s> <s xml:id="echoid-s16111" xml:space="preserve">: V : </s> <s xml:id="echoid-s16112" xml:space="preserve">u; </s> <s xml:id="echoid-s16113" xml:space="preserve">ce qui <lb/>fait voir que dans le mouvement accéléré, on peut exprimer <lb/>les vîteſſes par les racines des longueurs parcourues depuis le <lb/>point de repos. </s> <s xml:id="echoid-s16114" xml:space="preserve">Il faut s’appliquer à comprendre ceci pour n’être <lb/>point arrêté dans la ſuite.</s> <s xml:id="echoid-s16115" xml:space="preserve"/> </p> <pb o="508" file="0586" n="606" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1297" type="section" level="1" n="948"> <head xml:id="echoid-head1132" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s16116" xml:space="preserve">960. </s> <s xml:id="echoid-s16117" xml:space="preserve">Comme dans la chûte des corps graves la peſanteur <lb/>agit à chaque inſtant pour les approcher du centre de la terre, <lb/>qu’elle leur communique à chaque inſtant des degrés égaux <lb/>de vîteſſe (au moins cette ſuppoſition ne peut cauſer aucune <lb/>erreur en les conſidérant à des diſtances peu conſidérables de <lb/>la ſurface de la terre, même de quelques lieues); </s> <s xml:id="echoid-s16118" xml:space="preserve">il s’enſuit <lb/>que les eſpaces parcourus par un corps qui tombe librement, <lb/>à compter du point de repos, ſont comme les quarrés des <lb/>inſtans écoulés depuis le repos.</s> <s xml:id="echoid-s16119" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1298" type="section" level="1" n="949"> <head xml:id="echoid-head1133" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s16120" xml:space="preserve">961. </s> <s xml:id="echoid-s16121" xml:space="preserve">Il ſuit delà que les eſpaces qu’un corps parcourt en tom-<lb/>bant pendant destems égaux, ſont entr’eux comme la ſuite des <lb/>nombres impairs 1, 3, 5, 7, 9, &</s> <s xml:id="echoid-s16122" xml:space="preserve">c. </s> <s xml:id="echoid-s16123" xml:space="preserve">Imaginons que dans le premier <lb/>inſtant de ſa chûte le corps ait parcouru une toiſe. </s> <s xml:id="echoid-s16124" xml:space="preserve">Comme <lb/>cette vîteſſe a été acquiſe par degrés, & </s> <s xml:id="echoid-s16125" xml:space="preserve">que d’ailleurs il la <lb/>conſerve dans tous les inſtans ſuivans; </s> <s xml:id="echoid-s16126" xml:space="preserve">dans le ſecond inſtant, <lb/>en vertu de ce premier degré de vîteſſe, le corps parcourra un <lb/>eſpace double, c’eſt-à-dire 2 toiſes (art. </s> <s xml:id="echoid-s16127" xml:space="preserve">955), mais la peſan-<lb/>teur a toujours agi de la même maniere; </s> <s xml:id="echoid-s16128" xml:space="preserve">donc elle aura fait <lb/>parcourir au corps une toiſe de plus dans ce ſecond inſtant: <lb/></s> <s xml:id="echoid-s16129" xml:space="preserve">il aura donc parcouru 3 toiſes. </s> <s xml:id="echoid-s16130" xml:space="preserve">De même avec les deux degrés <lb/>de vîteſſe qu’il poſſede, dans le troiſieme inſtant il parcourra <lb/>4 toiſes, & </s> <s xml:id="echoid-s16131" xml:space="preserve">en vertu du nouveau degré que la peſanteur lui <lb/>communique par ſon action, il parcourra encore une toiſe: </s> <s xml:id="echoid-s16132" xml:space="preserve"><lb/>donc dans le troiſieme inſtant il aura parcouru 5 toiſes, & </s> <s xml:id="echoid-s16133" xml:space="preserve">ainſi <lb/>des autres. </s> <s xml:id="echoid-s16134" xml:space="preserve">Donc les eſpaces qu’un corps qui tombe parcourt <lb/>pendant des tems égaux ſont comme les nombres 1, 3, 5, 7, 9, &</s> <s xml:id="echoid-s16135" xml:space="preserve">c; </s> <s xml:id="echoid-s16136" xml:space="preserve"><lb/>d’où il ſuit encore de nouveau que les eſpaces parcourus depuis <lb/>le premier inſtant de la chûte, ſont comme les quarrés des <lb/>inſtans qui ſe ſont écoulés; </s> <s xml:id="echoid-s16137" xml:space="preserve">puiſqu’en ajoutant continuelle-<lb/>ment les nombres impairs depuis l’unité, il en réſulte les nom-<lb/>bres quarrés: </s> <s xml:id="echoid-s16138" xml:space="preserve">car il eſt évident que 1 = 1, 4 = 1 + 3, <lb/>9 = 1 + 3 + 5, 16 = 1 + 3 + 5 + 7, &</s> <s xml:id="echoid-s16139" xml:space="preserve">c.</s> <s xml:id="echoid-s16140" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1299" type="section" level="1" n="950"> <head xml:id="echoid-head1134" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s16141" xml:space="preserve">962. </s> <s xml:id="echoid-s16142" xml:space="preserve">Il ſuit encore delà que ſi un corps, après avoir parcouru <lb/>un certain eſpace pendant un certain nombre d’inſtans, ve- <pb o="509" file="0587" n="607" rhead="DE MATHÉMATIQUE. Liv. XIV."/> noit à être abandonné tout d’un coup de la force de la peſan-<lb/>teur, il continueroit néan moins à ſe mouvoir avec une vîteſſe <lb/>uniforme égale à celle que la peſanteur lui a communiquée dans <lb/>le premier tems, & </s> <s xml:id="echoid-s16143" xml:space="preserve">par conſéquent pendant un tems égal à <lb/>celui de la deſcente, il parcourroit toujours un eſpace double <lb/>de celui qu’il a parcouru pendant tout le tems que la peſan-<lb/>teur a agi ſur lui.</s> <s xml:id="echoid-s16144" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1300" type="section" level="1" n="951"> <head xml:id="echoid-head1135" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s16145" xml:space="preserve">963. </s> <s xml:id="echoid-s16146" xml:space="preserve">Il ſuit encore delà que ſi l’on jette un corps de bas en <lb/>haut, ſuivant une direction perpendiculaire ou oblique à l’ho-<lb/>rizon, le corps ſera mu d’un mouvement uniformément re-<lb/>tardé: </s> <s xml:id="echoid-s16147" xml:space="preserve">car il eſt évident que dans cette ſuppoſition la peſan-<lb/>teur étant oppoſée en tout ou en partie au mouvement de pro-<lb/>jection de ce corps, doit lui ôter à chaque inſtant des degrés <lb/>égaux de vîteſſe, & </s> <s xml:id="echoid-s16148" xml:space="preserve">par conſéquent au bout d’un certain tems, <lb/>lorſque la peſanteur aura détruit toute la force que le mobile <lb/>avoit pour s’élever perpendiculairement, il commencera à <lb/>tomber, & </s> <s xml:id="echoid-s16149" xml:space="preserve">paſſera ſucceſſivement par tous les degrés poſſibles <lb/>d’accélération, juſqu’à ce qu’il ſoit arrivé à quelque corps qui <lb/>l’arrête entiérement.</s> <s xml:id="echoid-s16150" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1301" type="section" level="1" n="952"> <head xml:id="echoid-head1136" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <p> <s xml:id="echoid-s16151" xml:space="preserve">964. </s> <s xml:id="echoid-s16152" xml:space="preserve">Donc la vîteſſe qu’un corps a acquiſe en tombant <lb/>d’une certaine hauteur, eſt égale à celle qui auroit pu le faire <lb/>monter à cette hauteur; </s> <s xml:id="echoid-s16153" xml:space="preserve">ou, ce qui revient au même, ſi l’on <lb/>jette un corps de bas en haut avec une force égale à celle <lb/>qu’il a acquiſe en tombant d’une certaine hauteur, cette force <lb/>ſera capable de le faire remonter à la même hauteur; </s> <s xml:id="echoid-s16154" xml:space="preserve">d’où il <lb/>ſuit encore que les eſpaces parcourus par un corps pouſſé de <lb/>bas en haut, ſeront comme les nombres impairs, pris dans un <lb/>ordre renverſé, ſi les tems ſont égaux.</s> <s xml:id="echoid-s16155" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1302" type="section" level="1" n="953"> <head xml:id="echoid-head1137" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s16156" xml:space="preserve">965. </s> <s xml:id="echoid-s16157" xml:space="preserve">Donc ſi l’on modifie la force de la peſanteur d’une <lb/>maniere conſtante, les eſpaces que cette force modifiée fera <lb/>parcourir à un corps quelconque, ſeront toujours ſuivant les <lb/>loix générales de la peſanteur. </s> <s xml:id="echoid-s16158" xml:space="preserve">Par exemple, un corps qui <lb/>tombe le long d’un plan incliné à l’horizon, ne gliſſe ſur le <lb/>plan qu’en conſéquence des loix de la peſanteur qui l’oblige <pb o="510" file="0588" n="608" rhead="NOUVEAU COURS"/> toujours à deſcendre: </s> <s xml:id="echoid-s16159" xml:space="preserve">donc il doit parcourir des eſpaces qui <lb/>ſoient dans la raiſon des quarrés des tems, à compter depuis <lb/>le premier inſtant du mouvement. </s> <s xml:id="echoid-s16160" xml:space="preserve">Si dans l’expérience on ne <lb/>trouvoit pas cette loi avec toute la préciſion poſſible, il n’y a <lb/>que le frottement & </s> <s xml:id="echoid-s16161" xml:space="preserve">la réſiſtance du milieu dans lequel ſe fait <lb/>le mouvement qui pourroit en altérer la juſteſſe, ce qui ne <lb/>conclud rien contre les principes que nous venons d’établir, <lb/>puiſque nous n’avons pas eu égard à ces circonſtances.</s> <s xml:id="echoid-s16162" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1303" type="section" level="1" n="954"> <head xml:id="echoid-head1138" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s16163" xml:space="preserve">966. </s> <s xml:id="echoid-s16164" xml:space="preserve">Comme la théorie de la peſanteur a une application <lb/>directe & </s> <s xml:id="echoid-s16165" xml:space="preserve">immédiate à la projection des corps; </s> <s xml:id="echoid-s16166" xml:space="preserve">que l’on ne <lb/>peut entendre celle du jet des bombes ſans être convaincu des <lb/>vérités que nous venons d’établir, & </s> <s xml:id="echoid-s16167" xml:space="preserve">que d’ailleurs il y auroit <lb/>une infinité de peſanteurs poſſibles, capables de faire parcourir <lb/>aux corps ſoumis à leur action des eſpaces qui ſeroient entr’eux <lb/>comme les quarrés des tems depuis le premier inſtant du mou-<lb/>vement, ou comme les nombres impairs depuis l’unité, en ſup-<lb/>poſant les tems égaux, c’eſt à l’expérience à décider quelle eſt <lb/>la force de la peſanteur auprès de la ſurface de la terre: </s> <s xml:id="echoid-s16168" xml:space="preserve">car <lb/>dans la ſuppoſition même que cette force augmentât ou di-<lb/>minuât à raiſon de ſes différentes diſtances de la terre, ſui-<lb/>vant un rapport quelconque, les diſtances auxquelles on peut <lb/>jetter les corps, même les plus grandes, ne ſont pas aſſez con-<lb/>ſidérables pour que l’on puiſſe appercevoir des variations dans <lb/>l’action de la peſanteur.</s> <s xml:id="echoid-s16169" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16170" xml:space="preserve">967. </s> <s xml:id="echoid-s16171" xml:space="preserve">On a reconnu par l’expérience qu’un corps qui tombe <lb/>parcourt 15 pieds dans la premiere ſeconde de ſa chûte, qu’il <lb/>en parcourt 45 dans la ſeconde, 75 dans la troiſieme, & </s> <s xml:id="echoid-s16172" xml:space="preserve">ainſi <lb/>de ſuite: </s> <s xml:id="echoid-s16173" xml:space="preserve">donc la peſanteur eſt une force capable de faire par-<lb/>courir 30 pieds dans une ſeconde à tout corps qui auroit été <lb/>ſoumis à ſon action pendant le même-tems, puiſque les 15 <lb/>pieds n’ont été parcourus que d’un mouvement accéléré, & </s> <s xml:id="echoid-s16174" xml:space="preserve"><lb/>qu’il s’agit ici d’un mouvement uniforme. </s> <s xml:id="echoid-s16175" xml:space="preserve">De même ſi la pe-<lb/>ſanteur a agi pendant 3 ſecondes, elle fera parcourir au corps <lb/>270 pieds pendant le même-tems, & </s> <s xml:id="echoid-s16176" xml:space="preserve">par conſéquent 90 pieds <lb/>dans une ſeconde. </s> <s xml:id="echoid-s16177" xml:space="preserve">Or il eſt viſible que les vîteſſes 30 & </s> <s xml:id="echoid-s16178" xml:space="preserve">90 pieds <lb/>par ſeconde, ſont comme les tems pendant leſquels le mobile <lb/>eſt tombé.</s> <s xml:id="echoid-s16179" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16180" xml:space="preserve">968. </s> <s xml:id="echoid-s16181" xml:space="preserve">Tout nous prouve cette prodigieuſe augmentation de <pb o="511" file="0589" n="609" rhead="DE MATHÉMATIQUE. Liv. XIV."/> vîteſſe à raiſon des racines quarrées des hauteurs d’où les corps <lb/>tombent: </s> <s xml:id="echoid-s16182" xml:space="preserve">car ſi la vîteſſe d’un corps qui tombe de 3 pieds étoit <lb/>égale à celle d’un corps qui tombe de 60 pieds de haut, il n’y <lb/>auroit pas plus de danger à tomber d’un troiſieme étage qu’à <lb/>tomber de deux ou trois pieds de haut, puiſque l’on ne frappe <lb/>la terre qu’à raiſon de la vîteſſe avec laquelle on tombe. </s> <s xml:id="echoid-s16183" xml:space="preserve">De <lb/>même il n’y a perſonne qui ne ſçache qu’une pierre nous bleſſe <lb/>d’autant plus qu’elle tombe de plus haut.</s> <s xml:id="echoid-s16184" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1304" type="section" level="1" n="955"> <head xml:id="echoid-head1139" style="it" xml:space="preserve">Digreſſion ſur les variations de la peſanteur.</head> <p> <s xml:id="echoid-s16185" xml:space="preserve">969. </s> <s xml:id="echoid-s16186" xml:space="preserve">Nous avons déja inſinué que la peſanteur pouvoit <lb/>bien n’être pas une force conſtante & </s> <s xml:id="echoid-s16187" xml:space="preserve">égale, à toutes les diſ-<lb/>tances différentes de notre globe, quoiqu’on la regarde <lb/>comme telle dans les diſtances médiocres; </s> <s xml:id="echoid-s16188" xml:space="preserve">c’eſt ce qui arrive <lb/>en effet. </s> <s xml:id="echoid-s16189" xml:space="preserve">M. </s> <s xml:id="echoid-s16190" xml:space="preserve">Newton a démontré le premier que cette force <lb/>décroît en raiſon inverſe des quarrés des diſtances, enſorte <lb/>que la force qui, combinée avec une force de projection, <lb/>retient la lune dans ſon orbite, en l’écartant continuelle-<lb/>ment de la tangente qu’elle tend à décrire, n’eſt autre choſe <lb/>que la peſanteur diminuée en raiſon inverſe du quarré des <lb/>diſtances. </s> <s xml:id="echoid-s16191" xml:space="preserve">Il prouve qu’en vertu de la peſanteur, la lune <lb/>dans une minute s’écarte de 15 pieds de la tangente au point <lb/>de ſon orbite où elle étoit au commencement de cette mi-<lb/>nute: </s> <s xml:id="echoid-s16192" xml:space="preserve">donc pour comparer la force de la peſanteur près de <lb/>la lune à celle de la peſanteur près de la terre, il faut voir <lb/>ce qu’elle fera décrire dans une ſeconde, en ſuppoſant tou-<lb/>jours que les eſpaces parcourus ſont comme les quarrés des <lb/>tems, & </s> <s xml:id="echoid-s16193" xml:space="preserve">faiſant l’eſpace de 15 pieds égal à l’unité. </s> <s xml:id="echoid-s16194" xml:space="preserve">Une mi-<lb/>nute vaut 60 ſecondes: </s> <s xml:id="echoid-s16195" xml:space="preserve">donc les quarrés des tems ſeront 3600 <lb/>& </s> <s xml:id="echoid-s16196" xml:space="preserve">1; </s> <s xml:id="echoid-s16197" xml:space="preserve">faiſant cette proportion 3600 : </s> <s xml:id="echoid-s16198" xml:space="preserve">1 :</s> <s xml:id="echoid-s16199" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s16200" xml:space="preserve">{1/3600}, ce qua-<lb/>trieme terme ſera l’eſpace parcouru près de la lune dans une <lb/>ſeconde de tems: </s> <s xml:id="echoid-s16201" xml:space="preserve">donc les eſpaces que les corps parcourent <lb/>dans une ſeconde près de la lune, & </s> <s xml:id="echoid-s16202" xml:space="preserve">près de la terre, en vertu <lb/>de la peſanteur, ſont comme {1/3600} : </s> <s xml:id="echoid-s16203" xml:space="preserve">1. </s> <s xml:id="echoid-s16204" xml:space="preserve">Mais les diſtances des <lb/>corps qui ſont ſur notre globe & </s> <s xml:id="echoid-s16205" xml:space="preserve">de la lune au centre de la <lb/>terre ſont 1 & </s> <s xml:id="echoid-s16206" xml:space="preserve">60, parce que la diſtance moyenne de la lune <lb/>à la terre eſt de 60 rayons de la terre: </s> <s xml:id="echoid-s16207" xml:space="preserve">ces quarrés ſont 1 & </s> <s xml:id="echoid-s16208" xml:space="preserve"><lb/>3600, qui ſont préciſément dans la raiſon inverſe des forces <lb/>ou eſpaces parcourus {1/3600} & </s> <s xml:id="echoid-s16209" xml:space="preserve">1, puiſque {1/3600} : </s> <s xml:id="echoid-s16210" xml:space="preserve">1 :</s> <s xml:id="echoid-s16211" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s16212" xml:space="preserve">3600. <lb/></s> <s xml:id="echoid-s16213" xml:space="preserve">C. </s> <s xml:id="echoid-s16214" xml:space="preserve">Q. </s> <s xml:id="echoid-s16215" xml:space="preserve">F. </s> <s xml:id="echoid-s16216" xml:space="preserve">D.</s> </p> <pb o="512" file="0590" n="610" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s16217" xml:space="preserve">On a auſſi découvert que la peſanteur varie ſelon les <lb/>diſtances à l’équateur, enſorte qu’elle va en augmentant de <lb/>l’équateur vers les poles, & </s> <s xml:id="echoid-s16218" xml:space="preserve">réciproquement. </s> <s xml:id="echoid-s16219" xml:space="preserve">On s’eſt ap-<lb/>perçu de cette variation, en obſervant qu’un pendule qui bat <lb/>les ſecondes à Paris, c’eſt-à-dire qui fait ſoixante vibrations <lb/>dans une minute, en faiſoit un plus petit nombre vers l’é-<lb/>quateur; </s> <s xml:id="echoid-s16220" xml:space="preserve">d’où on a conclu avec certitude que la peſanteur <lb/>eſt moindre vers l’équateur que vers les poles, puiſque les <lb/>vibrations des pendules, qui ne ſont que des effets de la pe-<lb/>ſanteur, ſont plus lentes vers l’équateur que vers les poles. <lb/></s> <s xml:id="echoid-s16221" xml:space="preserve">Cette diminution de peſanteur eſt cauſée par le mouvement <lb/>de rotation de la terre autour de ſon axe, duquel il réſulte <lb/>une force centrifuge plus grande vers l’équateur que vers les <lb/>poles.</s> <s xml:id="echoid-s16222" xml:space="preserve">Tout ce que nous venons de voir ſur les variations de la pe-<lb/>ſanteur n’empêche pas qu’on ne la doive regarder comme une <lb/>force conſtante, puiſque ces variations ne peuvent être ſen-<lb/>ſibles dans les plus grandes diſtances auxquelles on puiſſe jetter <lb/>les corps. </s> <s xml:id="echoid-s16223" xml:space="preserve">Quoique ces vérités n’aient pas une application di-<lb/>recte au jet des bombes, qui doit faire le principal objet de <lb/>l’Ingénieur, je n’ai pas cru cependant devoir les ſupprimer, <lb/>parce qu’elles ſont trop belles pour qu’il ſoit permis à un <lb/>homme de ſcience de les ignorer, & </s> <s xml:id="echoid-s16224" xml:space="preserve">que de plus il eſt à pro-<lb/>pos que l’on ſçache quels ſont les changemens qui peuvent <lb/>altérer les loix que nous venons d’établir, & </s> <s xml:id="echoid-s16225" xml:space="preserve">quels ſont ceux <lb/>qui ne peuvent produire le même effet.</s> <s xml:id="echoid-s16226" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1305" type="section" level="1" n="956"> <head xml:id="echoid-head1140" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16227" xml:space="preserve">970. </s> <s xml:id="echoid-s16228" xml:space="preserve">Un corps eſt tombé perpendiculairement pendant quatre <lb/>ſecondes; </s> <s xml:id="echoid-s16229" xml:space="preserve">on demande l’eſpace que la peſanteur lui a fait parcourir.</s> <s xml:id="echoid-s16230" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1306" type="section" level="1" n="957"> <head xml:id="echoid-head1141" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s16231" xml:space="preserve">Soit appellé x cet eſpace inconnu; </s> <s xml:id="echoid-s16232" xml:space="preserve">puiſque les eſpaces par-<lb/>courus dans des tems différens, depuis le commencement du <lb/>mouvement, ſont comme les quarrés des tems (art. </s> <s xml:id="echoid-s16233" xml:space="preserve">958), & </s> <s xml:id="echoid-s16234" xml:space="preserve"><lb/>que d’ailleurs on ſçait par expérience qu’un corps parcourt <lb/>15 pieds dans la premiere ſeconde; </s> <s xml:id="echoid-s16235" xml:space="preserve">on aura cette proportion, <lb/>1 : </s> <s xml:id="echoid-s16236" xml:space="preserve">16 :</s> <s xml:id="echoid-s16237" xml:space="preserve">: 15 : </s> <s xml:id="echoid-s16238" xml:space="preserve">x = {15 x 16/1} = 240 pieds. </s> <s xml:id="echoid-s16239" xml:space="preserve">C. </s> <s xml:id="echoid-s16240" xml:space="preserve">Q. </s> <s xml:id="echoid-s16241" xml:space="preserve">F. </s> <s xml:id="echoid-s16242" xml:space="preserve">T. </s> <s xml:id="echoid-s16243" xml:space="preserve">& </s> <s xml:id="echoid-s16244" xml:space="preserve">D.</s> <s xml:id="echoid-s16245" xml:space="preserve"/> </p> <pb o="513" file="0591" n="611" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1307" type="section" level="1" n="958"> <head xml:id="echoid-head1142" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16246" xml:space="preserve">971. </s> <s xml:id="echoid-s16247" xml:space="preserve">Un corps a parcouru en tombant par la ſeule force de la <lb/>peſanteur un eſpace de 375 pieds, on demande le tems qu’il lui a <lb/>fallu pour les parcourir.</s> <s xml:id="echoid-s16248" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1308" type="section" level="1" n="959"> <head xml:id="echoid-head1143" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s16249" xml:space="preserve">Soit x le tems cherché; </s> <s xml:id="echoid-s16250" xml:space="preserve">puiſque les eſpaces parcourus ſont <lb/>comme les quarrés des tems (art. </s> <s xml:id="echoid-s16251" xml:space="preserve">957), on aura cette propor-<lb/>tion, 15 : </s> <s xml:id="echoid-s16252" xml:space="preserve">375 :</s> <s xml:id="echoid-s16253" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s16254" xml:space="preserve">xx: </s> <s xml:id="echoid-s16255" xml:space="preserve">donc xx ={375/15}=25, d’où l’on tire <lb/>x=5, c’eſt-à-dire que le corps a été 5 ſecondes de tems en <lb/>mouvement. </s> <s xml:id="echoid-s16256" xml:space="preserve">C. </s> <s xml:id="echoid-s16257" xml:space="preserve">Q. </s> <s xml:id="echoid-s16258" xml:space="preserve">F. </s> <s xml:id="echoid-s16259" xml:space="preserve">T. </s> <s xml:id="echoid-s16260" xml:space="preserve">& </s> <s xml:id="echoid-s16261" xml:space="preserve">D.</s> <s xml:id="echoid-s16262" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16263" xml:space="preserve">972. </s> <s xml:id="echoid-s16264" xml:space="preserve">Comme dans le jet des bombes le mobile ſe trouve <lb/>entre deux forces, l’une de projection & </s> <s xml:id="echoid-s16265" xml:space="preserve">ſimplement motrice, <lb/>c’eſt la force de la poudre, l’autre accélératrice ou retardatrice <lb/>conſtante; </s> <s xml:id="echoid-s16266" xml:space="preserve">c’eſt la force de la peſanteur; </s> <s xml:id="echoid-s16267" xml:space="preserve">ſuivant que le corps <lb/>deſcend ou monte, quelle que ſoit ſa direction, & </s> <s xml:id="echoid-s16268" xml:space="preserve">que d’ailleurs <lb/>abſtraction faite des réſiſtances de l’air à ces deux forces, on <lb/>ne peut trouver la courbe que le mobile décrit pendant ſon <lb/>mouvement, ſans ſuppoſer qu’il ſatisfait à chacune de ces deux <lb/>forces à la fois, comme s’il n’avoit été pouſſé que par l’une ou <lb/>l’autre ſéparément. </s> <s xml:id="echoid-s16269" xml:space="preserve">Nous allons démontrer cette propoſition, <lb/>que l’on appelle le mouvement compoſé.</s> <s xml:id="echoid-s16270" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1309" type="section" level="1" n="960"> <head xml:id="echoid-head1144" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s16271" xml:space="preserve">973. </s> <s xml:id="echoid-s16272" xml:space="preserve">On appelle ſimplement force motrice, toute force qui <lb/>n’eſt appliquée à un corps que pendant un ſeul inſtant. </s> <s xml:id="echoid-s16273" xml:space="preserve">Tout <lb/>corps dur en mouvement eſt une force motrice par rapport à <lb/>celui qu’il rencontre, car il ne lui eſt appliqué que pendant <lb/>l’inſtant du choc.</s> <s xml:id="echoid-s16274" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16275" xml:space="preserve">974. </s> <s xml:id="echoid-s16276" xml:space="preserve">Si deux ou pluſieurs forces motrices ſont appliquées <lb/>dans un même inſtant à un même corps pour le mouvoir, <lb/>chacune ſuivant ſa direction, on les appelle forces compoſantes. <lb/></s> <s xml:id="echoid-s16277" xml:space="preserve">La force qu’elles donnent eſt appellée réſultante.</s> <s xml:id="echoid-s16278" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1310" type="section" level="1" n="961"> <head xml:id="echoid-head1145" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s16279" xml:space="preserve">975. </s> <s xml:id="echoid-s16280" xml:space="preserve">Si un corps K eſt pouſſé à la fois par deux forces M, N <lb/> <anchor type="note" xlink:label="note-0591-01a" xlink:href="note-0591-01"/> ſimplement motrices, & </s> <s xml:id="echoid-s16281" xml:space="preserve">capables de lui faire parcourir dans le même <pb o="514" file="0592" n="612" rhead="NOUVEAU COURS"/> tems, ſuivant leurs directions, l’une A B, & </s> <s xml:id="echoid-s16282" xml:space="preserve">l’autre A C; </s> <s xml:id="echoid-s16283" xml:space="preserve">je dis, <lb/>1°. </s> <s xml:id="echoid-s16284" xml:space="preserve">que le corps par l’effort compoſé de ces deux puiſſances, dé-<lb/>crira d’un mouvement uniforme la diagonale A D du parallélo-<lb/>gramme formé ſur leurs directions; </s> <s xml:id="echoid-s16285" xml:space="preserve">2°. </s> <s xml:id="echoid-s16286" xml:space="preserve">qu’il parcourra cette même <lb/>diagonale pendant le même-tems qu’il auroit parcouru le côté A B <lb/>ou le côté A C, ſi l’une des deux forces ſeulement eût agi ſur lui.</s> <s xml:id="echoid-s16287" xml:space="preserve"/> </p> <div xml:id="echoid-div1310" type="float" level="2" n="1"> <note position="right" xlink:label="note-0591-01" xlink:href="note-0591-01a" xml:space="preserve">Figure 333.</note> </div> </div> <div xml:id="echoid-div1312" type="section" level="1" n="962"> <head xml:id="echoid-head1146" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16288" xml:space="preserve">Concevons deux regles infiniment minces M A B, N A C, <lb/>chacune égale en peſanteur au corps K, elles-mêmes en mou-<lb/>vement, & </s> <s xml:id="echoid-s16289" xml:space="preserve">dirigées, l’une ſuivant A B, & </s> <s xml:id="echoid-s16290" xml:space="preserve">l’autre ſuivant A C, <lb/>avec des vîteſſes qui leur font parcourir le double des lignes <lb/>A C & </s> <s xml:id="echoid-s16291" xml:space="preserve">A B, que les puiſſances M & </s> <s xml:id="echoid-s16292" xml:space="preserve">N font parcourir au corps <lb/>K. </s> <s xml:id="echoid-s16293" xml:space="preserve">Ces regles venant à choquer le corps K, que l’on ſuppoſe <lb/>en repos, lui communiqueront chacune la moitié de leurs <lb/>vîteſſes, ſuivant les loix des corps à reſſort, & </s> <s xml:id="echoid-s16294" xml:space="preserve">par conſéquent <lb/>elles font préciſément ſur ce corps un effet égal à celui qu’au-<lb/>roient fait les puiſſances M, N que nous avons ſuppoſées agir <lb/>ſur lui, & </s> <s xml:id="echoid-s16295" xml:space="preserve">elles ne ſont pas différentes de ces mêmes forces. <lb/></s> <s xml:id="echoid-s16296" xml:space="preserve">Cela poſé, le corps K doit décrire la diagonale A D dans le <lb/>même-tems qu’il auroit décrit la ligne A B ou la ligne A C, <lb/>s’il n’eût été pouſſé que par une ſeule puiſſance M ou N. </s> <s xml:id="echoid-s16297" xml:space="preserve">Pour <lb/>s’en convaincre, on remarquera que les regles ne choquent le <lb/>corps qu’autant qu’il eſt néceſſaire pour qu’il ne puiſſe s’op-<lb/>poſer au mouvement qui leur reſte, lequel eſt la moitié de <lb/>celui qu’ils avoient avant le choc. </s> <s xml:id="echoid-s16298" xml:space="preserve">Il faut de plus remarquer <lb/>que les regles ne faiſant que gliſſer l’une ſur l’autre, ne peu-<lb/>vent détruire mutuellement le mouvement qui leur reſte: </s> <s xml:id="echoid-s16299" xml:space="preserve"><lb/>donc elles ſont mues avec des vîteſſes qui leur font parcourir <lb/>les lignes A B, A C dans le tems que les puiſſances M, N au-<lb/>roient fait parcourir au corps K les mêmes lignes. </s> <s xml:id="echoid-s16300" xml:space="preserve">Enfin on <lb/>fera attention que dans ce même-tems leur interſection mu-<lb/>tuelle décrit la diagonale A D: </s> <s xml:id="echoid-s16301" xml:space="preserve">car il eſt évident que lorſque <lb/>la regle A B eſt venue en E F, la regle A C a parcouru une <lb/>partie proportionnelle de l’eſpace A B, & </s> <s xml:id="echoid-s16302" xml:space="preserve">ſe trouve par con-<lb/>ſéquent en G H; </s> <s xml:id="echoid-s16303" xml:space="preserve">d’où il ſuit évidemment que l’interſection I <lb/>eſt un point de la diagonale: </s> <s xml:id="echoid-s16304" xml:space="preserve">donc pour que le corps K ne <lb/>s’oppoſe point au mouvement de ces regles, il ſuffit qu’il ſoit <lb/>venu d’une égale vîteſſe de K en I, c’eſt-à-dire qu’il ait par-<lb/>couru K I dans le tems que les regles ont parcouru les eſpaces <pb o="515" file="0593" n="613" rhead="DE MATHÉMATIQUE. Liv. XIV."/> A E, A H: </s> <s xml:id="echoid-s16305" xml:space="preserve">donc il arrivera en D dans le tems que les regles <lb/>ſeront venues en B D & </s> <s xml:id="echoid-s16306" xml:space="preserve">en C D. </s> <s xml:id="echoid-s16307" xml:space="preserve">D’ailleurs il eſt viſible, <lb/>comme nous l’avons déja fait remarquer, que ces regles ſont <lb/>égales aux puiſſances M & </s> <s xml:id="echoid-s16308" xml:space="preserve">N, puiſqu’elles communiquent la <lb/>même vîteſſe au corps K, ſuivant les loix des corps durs: <lb/></s> <s xml:id="echoid-s16309" xml:space="preserve">donc le corps décrira la diagonale A D dans le même-tems <lb/>qu’il eût parcouru A B ou A C, s’il n’eût été pouſſé que par <lb/>l’une des forces M ou N. </s> <s xml:id="echoid-s16310" xml:space="preserve">C. </s> <s xml:id="echoid-s16311" xml:space="preserve">Q. </s> <s xml:id="echoid-s16312" xml:space="preserve">F. </s> <s xml:id="echoid-s16313" xml:space="preserve">D.</s> <s xml:id="echoid-s16314" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16315" xml:space="preserve">On peut encore concevoir le mouvement compoſé d’une <lb/>autre maniere. </s> <s xml:id="echoid-s16316" xml:space="preserve">Imaginons que le corps K eſt mu ſur une regle <lb/>A B, & </s> <s xml:id="echoid-s16317" xml:space="preserve">que dans le même-tems qu’il parcourt la longueur de <lb/>la regle, une force emporte cette regle le long de A C en lui <lb/>faiſant parcourir A C. </s> <s xml:id="echoid-s16318" xml:space="preserve">Il eſt évident que dans ce mouvement <lb/>le corps K décrit encore la diagonale A D, puiſque les eſ-<lb/>paces entiers A B, A C & </s> <s xml:id="echoid-s16319" xml:space="preserve">leurs parties proportionnelles ſont <lb/>décrits dans des tems égaux. </s> <s xml:id="echoid-s16320" xml:space="preserve">Donc, &</s> <s xml:id="echoid-s16321" xml:space="preserve">c.</s> <s xml:id="echoid-s16322" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16323" xml:space="preserve">On pourroit craindre dans cette derniere démonſtration, <lb/>que la ſuppoſition que nous avons faite que le corps eſt mu ſur <lb/>la ligne A B, & </s> <s xml:id="echoid-s16324" xml:space="preserve">que cette ligne eſt emportée ſur A C parallé-<lb/>lement à elle-même, ne changeât quelque choſe dans la force <lb/>N qui meut le corps de A en C. </s> <s xml:id="echoid-s16325" xml:space="preserve">Pour prévenir cette objec-<lb/>tion, nous remarquerons, avec M. </s> <s xml:id="echoid-s16326" xml:space="preserve">Varignon, que lorſque <lb/>deux corps ſont mus d’une égale vîteſſe, comme dans notre <lb/>hypotheſe, cette même vîteſſe les mettant dans l’impoſſibilité <lb/>de s’aider ou de ſe nuire réciproquement, la force qui meut <lb/>chacun ſéparément, eſt toujours la même; </s> <s xml:id="echoid-s16327" xml:space="preserve">d’où il ſuit que la <lb/>force qui fait parcourir A C au corps K eſt toujours la même, <lb/>ſoit qu’il ſoit emporté ſur la regle A B, ou que la regle ſoit <lb/>ſupprimée; </s> <s xml:id="echoid-s16328" xml:space="preserve">moyennant quoi on peut regarder cette derniere <lb/>démonſtration comme une des plus ſatisfaiſantes.</s> <s xml:id="echoid-s16329" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16330" xml:space="preserve">Au reſte comme cette propoſition ne ſe trouve ici que re-<lb/>lativement au jet des bombes, & </s> <s xml:id="echoid-s16331" xml:space="preserve">pour faire voir qu’un corps <lb/>qui eſt entre deux puiſſances, prend une direction par laquelle <lb/>il ſatisfait à l’impulſion de chacune des forces qui agiſſent ſur <lb/>lui, nous donnerons encore une démonſtration plus lumineuſe <lb/>& </s> <s xml:id="echoid-s16332" xml:space="preserve">plus convaincante de cette même propoſition dans le Traité <lb/>de Méchanique qui doit ſuivre. </s> <s xml:id="echoid-s16333" xml:space="preserve">Comme cette propoſition eſt <lb/>de la derniere importance dans tout ce qui a rapport à la com-<lb/>poſition & </s> <s xml:id="echoid-s16334" xml:space="preserve">la décompoſition des forces, on doit, autant qu’il <lb/>eſt poſſible, s’appliquer à bien éclaircir les principes.</s> <s xml:id="echoid-s16335" xml:space="preserve"/> </p> <pb o="516" file="0594" n="614" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1313" type="section" level="1" n="963"> <head xml:id="echoid-head1147" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16336" xml:space="preserve">976. </s> <s xml:id="echoid-s16337" xml:space="preserve">Donc la force réſultante, ſuivant la diagonale A D, <lb/>eſt à l’une des compoſantes M ou N, comme A D: </s> <s xml:id="echoid-s16338" xml:space="preserve">A B, ou <lb/>comme AD: </s> <s xml:id="echoid-s16339" xml:space="preserve">AC: </s> <s xml:id="echoid-s16340" xml:space="preserve">car les forces qui meuvent des corps égaux <lb/>ſont comme les eſpaces parcourus en même-tems.</s> <s xml:id="echoid-s16341" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1314" type="section" level="1" n="964"> <head xml:id="echoid-head1148" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16342" xml:space="preserve">977. </s> <s xml:id="echoid-s16343" xml:space="preserve">Donc le corps ſatisfait aux deux puiſſances en même <lb/>tems, comme s’il n’avoit été pouſſé que par l’une des deux: <lb/></s> <s xml:id="echoid-s16344" xml:space="preserve">car il eſt évident que lorſqu’il eſt au point D, il ſe trouve <lb/>éloigné de la ligne A B d’une quantité B D = A C, & </s> <s xml:id="echoid-s16345" xml:space="preserve">réci-<lb/>proquement il ſe trouve éloigné de la direction A C d’une <lb/>quantité D C = A B.</s> <s xml:id="echoid-s16346" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1315" type="section" level="1" n="965"> <head xml:id="echoid-head1149" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s16347" xml:space="preserve">978. </s> <s xml:id="echoid-s16348" xml:space="preserve">Donc la force que le corps a, ſuivant la diagonale, eſt <lb/>capable de faire équilibre avec les compoſantes, ſi elle eſt dirigée <lb/>dans un ſens contraire, c’eſt-à-dire que ſi l’on pouſſe un corps de <lb/>D vers A avec une vîteſſe capable de lui faire parcourir A D <lb/>dans un certain tems, ce corps arrêtera avec la force qu’il a, <lb/>dans cette hypotheſe, celle des puiſſances capables de lui faire <lb/>parcourir A B & </s> <s xml:id="echoid-s16349" xml:space="preserve">A C dans le même-tems, puiſque l’effort ré-<lb/>ſultant de ces deux puiſſances appliquées dans le même inſtant <lb/>à ce corps, ne peuvent lui faire parcourir que la diagonale <lb/>avec la même vîteſſe.</s> <s xml:id="echoid-s16350" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1316" type="section" level="1" n="966"> <head xml:id="echoid-head1150" xml:space="preserve"><emph style="sc">Corollaire</emph>. IV.</head> <p> <s xml:id="echoid-s16351" xml:space="preserve">979. </s> <s xml:id="echoid-s16352" xml:space="preserve">Donc ſi l’on a une force quelconque, on pourra tou-<lb/>jours la regarder comme la réſultante de deux autres forces, & </s> <s xml:id="echoid-s16353" xml:space="preserve"><lb/>ſuppoſer qu’elle eſt la diagonale du parallélogramme formé ſur <lb/>les directions de ces deux nouvelles forces; </s> <s xml:id="echoid-s16354" xml:space="preserve">d’où il ſuit encore <lb/>qu’il y a une infinité de forces dans leſquelles on peut décom-<lb/>poſer une force quelconque, puiſqu’une ligne peut être dia-<lb/>gonale d’une infinité de parallélogrammes différens. </s> <s xml:id="echoid-s16355" xml:space="preserve">L’état de <lb/>la queſtion ou les conditions du problême, déterminent ordi-<lb/>nairement quelles ſont les forces dans leſquelles on doit dé-<lb/>compoſer une force donnée. </s> <s xml:id="echoid-s16356" xml:space="preserve">On en verra des exemples dans <lb/>la méchanique.</s> <s xml:id="echoid-s16357" xml:space="preserve"/> </p> <pb o="517" file="0595" n="615" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1317" type="section" level="1" n="967"> <head xml:id="echoid-head1151" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s16358" xml:space="preserve">980. </s> <s xml:id="echoid-s16359" xml:space="preserve">Donc ſi un corps ſe trouve à la fois ſoumis à l’action de <lb/>deux forces accélératrices ou retardatrices conſtantes, il dé-<lb/>crira encore la diagonale du parallélogramme formé ſur les di-<lb/>rections de ces forces: </s> <s xml:id="echoid-s16360" xml:space="preserve">car ces forces ne ſont que des forces <lb/>motrices qui renouvellent leur action à chaque inſtant; </s> <s xml:id="echoid-s16361" xml:space="preserve">& </s> <s xml:id="echoid-s16362" xml:space="preserve"><lb/>comme les degrés d’augmentation ou de diminution ſont pro-<lb/>portionnels dans tous les inſtans du mouvement pour chaque <lb/>force, il s’enſuit que la ligne décrite par le mobile doit être <lb/>une ligne droite.</s> <s xml:id="echoid-s16363" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1318" type="section" level="1" n="968"> <head xml:id="echoid-head1152" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s16364" xml:space="preserve">981. </s> <s xml:id="echoid-s16365" xml:space="preserve">Si les deux forces ne gardent pas un certain rapport <lb/>conſtant pendant chaque inſtant du mouvement, la ligne dé-<lb/>crite par le mobile ne peut être qu’une ligne courbe; </s> <s xml:id="echoid-s16366" xml:space="preserve">cepen-<lb/>dant toujours telle qu’il ſatisfait durant chaque inſtant du <lb/>mouvement à chacune des deux forces à la fois, comme s’il <lb/>n’avoit été ſoumis qu’à l’une des deux.</s> <s xml:id="echoid-s16367" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1319" type="section" level="1" n="969"> <head xml:id="echoid-head1153" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s16368" xml:space="preserve">982. </s> <s xml:id="echoid-s16369" xml:space="preserve">Donc ſi l’une des forces eſt variable, & </s> <s xml:id="echoid-s16370" xml:space="preserve">l’autre conſ-<lb/>tante, la ligne décrite par le corps ſoumis à l’action de ces <lb/>deux forces ſera néceſſairement une ligne courbe; </s> <s xml:id="echoid-s16371" xml:space="preserve">d’où il ſuit <lb/>& </s> <s xml:id="echoid-s16372" xml:space="preserve">du corollaire précédent, que l’on peut ramener la théorie <lb/>des courbes à celles du mouvement; </s> <s xml:id="echoid-s16373" xml:space="preserve">& </s> <s xml:id="echoid-s16374" xml:space="preserve">réciproquement con-<lb/>noître quel rapport les forces motrices doivent avoir entr’elles <lb/>pour faire décrire à un corps une courbe déterminée.</s> <s xml:id="echoid-s16375" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16376" xml:space="preserve">Tout ce que nous venons de voir eſt ſuffiſant pour con-<lb/>noître la courbe décrite par un corps ſoumis à l’action d’une <lb/>force motrice, & </s> <s xml:id="echoid-s16377" xml:space="preserve">à celle de la peſanteur, abſtraction faite des <lb/>réſiſtances de l’air & </s> <s xml:id="echoid-s16378" xml:space="preserve">des différentes circonſtances qui peuvent <lb/>altérer la préciſion des regles que nous allons établir. </s> <s xml:id="echoid-s16379" xml:space="preserve">Il ſuffira <lb/>de dire que les expériences du vuide démontrent que les corps <lb/>tomberoient avec la même vîteſſe, quels que ſoient leurs maſſes <lb/>& </s> <s xml:id="echoid-s16380" xml:space="preserve">leurs volumes, ſi l’air ne réſiſtoit pas à leur mouvement. <lb/></s> <s xml:id="echoid-s16381" xml:space="preserve">Si l’on vouloit avoir égard à cette réſiſtance, il faut déter-<lb/>miner auparavant la réſiſtance des fluides aux corps en mou-<lb/>vement, à raiſon de leurs volumes, de leurs maſſes & </s> <s xml:id="echoid-s16382" xml:space="preserve">de <lb/>leurs vîteſſes. </s> <s xml:id="echoid-s16383" xml:space="preserve">Ainſi l’on voit que nous ne pouvons actuelle- <pb o="518" file="0596" n="616" rhead="NOUVEAU COURS"/> ment déterminer la courbe que les corps décrivent en montant <lb/>ou en deſcendant, ſuivant une ligne oblique à l’horizon, qu’en <lb/>faiſant abſtraction des réſiſtances de l’air, puiſque l’air eſt un <lb/>fluide, & </s> <s xml:id="echoid-s16384" xml:space="preserve">que nous n’avons pas encore donné la théorie de la <lb/>réſiſtance des fluides.</s> <s xml:id="echoid-s16385" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1320" type="section" level="1" n="970"> <head xml:id="echoid-head1154" xml:space="preserve">CHAPITRE III.</head> <head xml:id="echoid-head1155" style="it" xml:space="preserve">De la théorie & de la pratique du Jet des Bombes pour ſervir à <lb/>l’intelligence de la conſtruction & de l’uſage d’un inſtrument <lb/>univerſel pour le jet des bombes.</head> <p> <s xml:id="echoid-s16386" xml:space="preserve">983. </s> <s xml:id="echoid-s16387" xml:space="preserve"><emph style="sc">TOus</emph> ceux qui tirent des bombes ſçavent en général <lb/>que la bombe décrit une courbe, en allant du mortier au lieu <lb/>où elle tombe; </s> <s xml:id="echoid-s16388" xml:space="preserve">mais il faut encore ſçavoir quelle eſt cette <lb/>courbe, afin d’établir quelques regles qui ſervent de principes <lb/>dans la pratique, en conſéquence des propriétés de la courbe <lb/>décrite pendant le mouvement. </s> <s xml:id="echoid-s16389" xml:space="preserve">Nous allons démontrer que <lb/>la courbe décrite non ſeulement par une bombe, mais par <lb/>tout corps, quelle que ſoit ſa direction parallele ou oblique à <lb/>l’horizon, eſt toujours une parabole. </s> <s xml:id="echoid-s16390" xml:space="preserve">On ſuppoſe encore ici <lb/>comme dans ce qui précéde, que l’air ne fait aucune réſiſtance, <lb/>ſoit à la force d’impulſion, ſoit à celle de la peſanteur. </s> <s xml:id="echoid-s16391" xml:space="preserve">Si la di-<lb/>rection du projectile eſt verticale ou perpendiculaire à l’ho-<lb/>rizon, tout le monde ſçait que le corps doit décrire une ligne <lb/>droite, ainſi il n’eſt pas queſtion de cette direction dans le cas <lb/>préſent.</s> <s xml:id="echoid-s16392" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1321" type="section" level="1" n="971"> <head xml:id="echoid-head1156" xml:space="preserve"><emph style="sc">Demande</emph></head> <p> <s xml:id="echoid-s16393" xml:space="preserve">984. </s> <s xml:id="echoid-s16394" xml:space="preserve">On demande qu’on puiſſe regarder la force de la pou-<lb/>dre comme une force capable de faire parcourir au corps jetté <lb/>des eſpaces égaux. </s> <s xml:id="echoid-s16395" xml:space="preserve">Cette demande eſt une ſuite immédiate de <lb/>l’hypotheſe préſente qu’on n’a pas égard à la réſiſtance de l’air; <lb/></s> <s xml:id="echoid-s16396" xml:space="preserve">d’ailleurs la force de la poudre eſt une force ſimplement mo-<lb/>trice, qui n’agit ſur le corps que dans un certain tems, que <lb/>l’on peut regarder comme un inſtant par rapport à la durée du <lb/>mouvement.</s> <s xml:id="echoid-s16397" xml:space="preserve"/> </p> <pb o="519" file="0597" n="617" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1322" type="section" level="1" n="972"> <head xml:id="echoid-head1157" xml:space="preserve">PROPOSITION VII. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s16398" xml:space="preserve">985. </s> <s xml:id="echoid-s16399" xml:space="preserve">Si un corps A eſt pouſſé par une force motrice ſuivant une <lb/> <anchor type="note" xlink:label="note-0597-01a" xlink:href="note-0597-01"/> direction parallele ou oblique à l’horizon, je dis que par l’effort <lb/>compoſé du mouvement d’impulſion & </s> <s xml:id="echoid-s16400" xml:space="preserve">de la peſanteur, il décrira <lb/>une parabole.</s> <s xml:id="echoid-s16401" xml:space="preserve"/> </p> <div xml:id="echoid-div1322" type="float" level="2" n="1"> <note position="right" xlink:label="note-0597-01" xlink:href="note-0597-01a" xml:space="preserve">Figure 134 <lb/>& 135.</note> </div> </div> <div xml:id="echoid-div1324" type="section" level="1" n="973"> <head xml:id="echoid-head1158" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16402" xml:space="preserve">Quelle que ſoit la direction de la force motrice, le corps A <lb/>ſe trouvera entre deux forces, l’une conſtante, c’eſt la force <lb/>de la poudre, l’autre accélératrice conſtante, c’eſt celle de la <lb/>peſanteur: </s> <s xml:id="echoid-s16403" xml:space="preserve">donc (art. </s> <s xml:id="echoid-s16404" xml:space="preserve">975) il doit ſatisfaire dans le même <lb/>tems à chacune de ces deux forces, comme s’il n’avoit été <lb/>ſoumis qu’à l’une des deux. </s> <s xml:id="echoid-s16405" xml:space="preserve">En vertu de la force d’impulſion, <lb/>il parcourt, dans des tems égaux, des eſpaces égaux A E, E G, <lb/>G I, I B, & </s> <s xml:id="echoid-s16406" xml:space="preserve">en vertu de la peſanteur, il parcourt à la fin de <lb/>chacun de ces tems des eſpaces E F, G H, I K, B D, qui ſont <lb/>comme les quarrés des tems écoulés depuis le premier inſ-<lb/>tant du mouvement. </s> <s xml:id="echoid-s16407" xml:space="preserve">Cela poſé, puiſque les eſpaces A E, A G, <lb/>A I croiſſent en progreſſion arithmétique, & </s> <s xml:id="echoid-s16408" xml:space="preserve">que les tems <lb/>croiſſent dans la même proportion; </s> <s xml:id="echoid-s16409" xml:space="preserve">& </s> <s xml:id="echoid-s16410" xml:space="preserve">que d’ailleurs les eſ-<lb/>paces parcourus à la fin de chacun de ces tems, à compter du <lb/>premier inſtant, ſont comme les quarrés des tems; </s> <s xml:id="echoid-s16411" xml:space="preserve">ces mê-<lb/>mes eſpaces E F, G H, I K, B D ſeront auſſi entr’eux comme <lb/>les quarré des lignes A E, A G, A I, A B proportionnelles au <lb/>tems; </s> <s xml:id="echoid-s16412" xml:space="preserve">& </s> <s xml:id="echoid-s16413" xml:space="preserve">prenant au lieu des lignes A E, A G, A I, leurs paral-<lb/>leles L F, M H, N K, & </s> <s xml:id="echoid-s16414" xml:space="preserve">de même au lieu des lignes E F, G H, <lb/>I K, leurs paralleles A I, A M, A N, on aura, par ce qu’on vient <lb/>de voir L F<emph style="sub">2</emph>: </s> <s xml:id="echoid-s16415" xml:space="preserve">M H<emph style="sub">2</emph>: </s> <s xml:id="echoid-s16416" xml:space="preserve">N K<emph style="sub">2</emph>:</s> <s xml:id="echoid-s16417" xml:space="preserve">: A L: </s> <s xml:id="echoid-s16418" xml:space="preserve">A M: </s> <s xml:id="echoid-s16419" xml:space="preserve">N N; </s> <s xml:id="echoid-s16420" xml:space="preserve">d’où il ſuit <lb/>que la courbe A F D eſt une parabole, puiſque les quarrés des <lb/>ordonnées ſont entr’eux comme leurs abſciſſes. </s> <s xml:id="echoid-s16421" xml:space="preserve">C. </s> <s xml:id="echoid-s16422" xml:space="preserve">Q. </s> <s xml:id="echoid-s16423" xml:space="preserve">F. </s> <s xml:id="echoid-s16424" xml:space="preserve">D.</s> <s xml:id="echoid-s16425" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1325" type="section" level="1" n="974"> <head xml:id="echoid-head1159" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16426" xml:space="preserve">986. </s> <s xml:id="echoid-s16427" xml:space="preserve">Comme la peſanteur n’eſt pas un inſtant ſans agir ſur <lb/>le projectile, quelle que ſoit ſa direction, il eſt évident qu’elle <lb/>le détourne de cette ligne dès le premier inſtant du mouve-<lb/>ment: </s> <s xml:id="echoid-s16428" xml:space="preserve">donc la ligne A B qui exprime la direction de la force <lb/>motrice, eſt tangente à la parabole.</s> <s xml:id="echoid-s16429" xml:space="preserve"/> </p> <pb o="520" file="0598" n="618" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1326" type="section" level="1" n="975"> <head xml:id="echoid-head1160" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16430" xml:space="preserve">987. </s> <s xml:id="echoid-s16431" xml:space="preserve">Si la direction de la force motrice eſt parallele à l’ho-<lb/>rizon, la verticale A O menée par le point A ſera l’axe de la <lb/>parabole, & </s> <s xml:id="echoid-s16432" xml:space="preserve">le point A eſt le ſommet de la courbe. </s> <s xml:id="echoid-s16433" xml:space="preserve">Si la direc-<lb/>tion eſt oblique, la ligne A O menée par le même point A ſera <lb/> <anchor type="note" xlink:label="note-0598-01a" xlink:href="note-0598-01"/> un diametre. </s> <s xml:id="echoid-s16434" xml:space="preserve">Si le corps eſt pouſſé de A vers B, le point H déter-<lb/>miné par la verticale, menée par le milieu de G B, ſera le plus <lb/>haut où le corps puiſſe s’élever; </s> <s xml:id="echoid-s16435" xml:space="preserve">s’il eſt pouſſé de A vers Q, le <lb/>point A ſera le plus haut où il puiſſe ſe trouver dans le mouve-<lb/>ment.</s> <s xml:id="echoid-s16436" xml:space="preserve"/> </p> <div xml:id="echoid-div1326" type="float" level="2" n="1"> <note position="left" xlink:label="note-0598-01" xlink:href="note-0598-01a" xml:space="preserve">Figure 335.</note> </div> </div> <div xml:id="echoid-div1328" type="section" level="1" n="976"> <head xml:id="echoid-head1161" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s16437" xml:space="preserve">988. </s> <s xml:id="echoid-s16438" xml:space="preserve">Les paraboles décrite par un même mobile ont d’au-<lb/>tant plus d’étendue que la force motrice eſt plus grande ſous <lb/>la même inclinaiſon: </s> <s xml:id="echoid-s16439" xml:space="preserve">car l’étendue dépend de la force motrice <lb/>& </s> <s xml:id="echoid-s16440" xml:space="preserve">de l’inclinaiſon de la direction de cette même force à l’ho-<lb/>rizon.</s> <s xml:id="echoid-s16441" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1329" type="section" level="1" n="977"> <head xml:id="echoid-head1162" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s16442" xml:space="preserve">989. </s> <s xml:id="echoid-s16443" xml:space="preserve">La ligne A B, direction de la force motrice, eſt nom-<lb/>mée la ligne de projection; </s> <s xml:id="echoid-s16444" xml:space="preserve">la ligne B D élevée du point D de <lb/>l’horizon où le corps tombe perpendiculairement juſqu’à la <lb/>ligne de projection eſt nommée ligne de chûte. </s> <s xml:id="echoid-s16445" xml:space="preserve">La ligne A D <lb/>menée du point d’où le corps part juſqu’au point où il arrive <lb/>ſur l’horizon, eſt appellée ligne de but. </s> <s xml:id="echoid-s16446" xml:space="preserve">Si cette ligne eſt ho-<lb/>rizontale, comme dans la figure 335, on l’appelle amplitude <lb/>de la parabole; </s> <s xml:id="echoid-s16447" xml:space="preserve">cette ligne détermine l’étendue du jet, & </s> <s xml:id="echoid-s16448" xml:space="preserve">c’eſt <lb/>pour cela qu’on l’appelle amplitude.</s> <s xml:id="echoid-s16449" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1330" type="section" level="1" n="978"> <head xml:id="echoid-head1163" xml:space="preserve"><emph style="sc">Principe</emph> <emph style="sc">Fondamental</emph>.</head> <p> <s xml:id="echoid-s16450" xml:space="preserve">990. </s> <s xml:id="echoid-s16451" xml:space="preserve">Comme les étendues des paraboles décrites par un <lb/>même mobile dépendent de la force qui a mis le mobile en <lb/>mouvement; </s> <s xml:id="echoid-s16452" xml:space="preserve">pour ramener cette force à quelques meſures <lb/>fixes & </s> <s xml:id="echoid-s16453" xml:space="preserve">déterminées, les Géometres, après Galilée, ſont con-<lb/>venus d’eſtimer les forces par les hauteurs, dont il auroit fallu <lb/>que le même corps tombât pour acquérir la vîteſſe qu’on lui <lb/>ſuppoſe: </s> <s xml:id="echoid-s16454" xml:space="preserve">car comme un mobile en tombant acquiert à chaque <lb/>inſtant de nouveaux degrés de vîteſſe, il n’y a point de vîteſſe <lb/>ſi grande qu’on puiſſe imaginer, à laquelle le même mobile <lb/>ne puiſſe arriver, puiſque l’on peut ſuppoſer la hauteur dont <lb/>il eſt tombé auſſi grande que l’on voudra.</s> <s xml:id="echoid-s16455" xml:space="preserve"/> </p> <pb o="521" file="0599" n="619" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1331" type="section" level="1" n="979"> <head xml:id="echoid-head1164" xml:space="preserve">PROPOSITION VIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16456" xml:space="preserve">991. </s> <s xml:id="echoid-s16457" xml:space="preserve">Connoiſſant la ligne de projection A B, ſuppoſée parallele <lb/> <anchor type="note" xlink:label="note-0599-01a" xlink:href="note-0599-01"/> à l’horizon, & </s> <s xml:id="echoid-s16458" xml:space="preserve">la ligne de chûte B F de la parabole A E F décrite <lb/>par un mobile quelconque, on demande de quelle hauteur ce mobile <lb/>doit tomber pour avoir à la fin de ſa chûte une vîteſſe avec laquelle <lb/>il puiſſe parcourir la ligne A B d’un mouvement uniforme, dans le <lb/>même tems que la peſanteur lui fera parcourir B F ou A G.</s> <s xml:id="echoid-s16459" xml:space="preserve"/> </p> <div xml:id="echoid-div1331" type="float" level="2" n="1"> <note position="right" xlink:label="note-0599-01" xlink:href="note-0599-01a" xml:space="preserve">Figure 336.</note> </div> </div> <div xml:id="echoid-div1333" type="section" level="1" n="980"> <head xml:id="echoid-head1165" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s16460" xml:space="preserve">Soit x la hauteur d’où le corps doit tomber pour avoir la <lb/>vîteſſe demandée; </s> <s xml:id="echoid-s16461" xml:space="preserve">ſoit T le tems que le corps emploie à par-<lb/>courir B F en vertu de ſa peſanteur; </s> <s xml:id="echoid-s16462" xml:space="preserve">ſoit fait de plus B F = a, <lb/>& </s> <s xml:id="echoid-s16463" xml:space="preserve">A B = 2b. </s> <s xml:id="echoid-s16464" xml:space="preserve">La vîteſſe que la peſanteur a communiquée au <lb/>corps à la fin de ſa chûte par B F eſt telle qu’elle lui fait par-<lb/>courir 2a ou 2B F dans le tems T (art. </s> <s xml:id="echoid-s16465" xml:space="preserve">962): </s> <s xml:id="echoid-s16466" xml:space="preserve">la vîteſſe qui <lb/>doit être acquiſe par la hauteur inconnue x eſt telle qu’elle <lb/>fait parcourir au même corps l’eſpace 2b ou A B dans le même <lb/>tems T: </s> <s xml:id="echoid-s16467" xml:space="preserve">d’ailleurs (art. </s> <s xml:id="echoid-s16468" xml:space="preserve">959) les vîteſſes acquiſes par diffé-<lb/>rentes hauteurs ſont comme les racines quarrées de ces hau-<lb/>teurs, qui ſont a & </s> <s xml:id="echoid-s16469" xml:space="preserve">x: </s> <s xml:id="echoid-s16470" xml:space="preserve">on aura donc cette proportion, <lb/>√a\x{0020}: </s> <s xml:id="echoid-s16471" xml:space="preserve">√x\x{0020}:</s> <s xml:id="echoid-s16472" xml:space="preserve">: 2a: </s> <s xml:id="echoid-s16473" xml:space="preserve">2b:</s> <s xml:id="echoid-s16474" xml:space="preserve">: a: </s> <s xml:id="echoid-s16475" xml:space="preserve">b; </s> <s xml:id="echoid-s16476" xml:space="preserve">d’où l’on tire a√x\x{0020}=b√a\x{0020}: </s> <s xml:id="echoid-s16477" xml:space="preserve">éle-<lb/>vant tout au quarré, on aura a<emph style="sub">2</emph>x = b<emph style="sub">2</emph>a, d’où l’on déduit x <lb/>= {b<emph style="sub">2</emph>/a}: </s> <s xml:id="echoid-s16478" xml:space="preserve">donc on aura cette proportion, a: </s> <s xml:id="echoid-s16479" xml:space="preserve">b:</s> <s xml:id="echoid-s16480" xml:space="preserve">: b: </s> <s xml:id="echoid-s16481" xml:space="preserve">x. </s> <s xml:id="echoid-s16482" xml:space="preserve">Pour conſ-<lb/>truire cette valeur de x, du point G au point D milieu de la <lb/>ligne A B, on menera une ligne G D; </s> <s xml:id="echoid-s16483" xml:space="preserve">on élevera au point D, <lb/>la perpendiculaire C D à cette ligne, juſqu’à ce qu’elle ren-<lb/>contre la ligne A G, prolongée en C; </s> <s xml:id="echoid-s16484" xml:space="preserve">je dis que la ligne A C <lb/>eſt égale à x, c’eſt-à-dire que cette ligne eſt la hauteur dont le <lb/>corps doit tomber pour avoir la vîteſſe demandée: </s> <s xml:id="echoid-s16485" xml:space="preserve">car à cauſe <lb/>des triangles ſemblables G A D, D A C, on aura A G (b): <lb/></s> <s xml:id="echoid-s16486" xml:space="preserve">A D (b):</s> <s xml:id="echoid-s16487" xml:space="preserve">: A D (b): </s> <s xml:id="echoid-s16488" xml:space="preserve">A C ({bb/a}). </s> <s xml:id="echoid-s16489" xml:space="preserve">C. </s> <s xml:id="echoid-s16490" xml:space="preserve">Q. </s> <s xml:id="echoid-s16491" xml:space="preserve">F. </s> <s xml:id="echoid-s16492" xml:space="preserve">T. </s> <s xml:id="echoid-s16493" xml:space="preserve">& </s> <s xml:id="echoid-s16494" xml:space="preserve">D.</s> <s xml:id="echoid-s16495" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1334" type="section" level="1" n="981"> <head xml:id="echoid-head1166" xml:space="preserve">Suite du Problême précédent.</head> <p> <s xml:id="echoid-s16496" xml:space="preserve">992. </s> <s xml:id="echoid-s16497" xml:space="preserve">Si l’on veut ſçavoir de quelle hauteur le mobile doit <lb/>tomber pour acquérir une vîteſſe capable de lui faire parcourir <pb o="522" file="0600" n="620" rhead="NOUVEAU COURS"/> la ligne oblique G D dans un tems égal à la moitié de celui <lb/>qu’il auroit employé à tomber par A G: </s> <s xml:id="echoid-s16498" xml:space="preserve">on fera de même la <lb/>hauteur inconnue = y; </s> <s xml:id="echoid-s16499" xml:space="preserve">la ligne G D connue = d; </s> <s xml:id="echoid-s16500" xml:space="preserve">la hauteur <lb/>A G = a; </s> <s xml:id="echoid-s16501" xml:space="preserve">& </s> <s xml:id="echoid-s16502" xml:space="preserve">l’on dira: </s> <s xml:id="echoid-s16503" xml:space="preserve">La vîteſſe acquiſe par la hauteur A G <lb/>eſt à la vîteſſe acquiſe par la hauteur inconnue y, comme l’eſ-<lb/>pace A G qu’elle fait parcourir uniformément pendant la moitié <lb/>du tems de la chûte par A G, eſt à l’eſpace G D qui doit être <lb/>parcouru pendant le même-tems, ſelon l’hypotheſe: </s> <s xml:id="echoid-s16504" xml:space="preserve">& </s> <s xml:id="echoid-s16505" xml:space="preserve">comme <lb/>d’ailleurs les vîteſſes ſont comme les racines quarrées des hau-<lb/>teurs, on aura cette proportion, √a\x{0020}: </s> <s xml:id="echoid-s16506" xml:space="preserve">√y\x{0020}:</s> <s xml:id="echoid-s16507" xml:space="preserve">: a: </s> <s xml:id="echoid-s16508" xml:space="preserve">d: </s> <s xml:id="echoid-s16509" xml:space="preserve">donc en <lb/>élevant tout au quarré a: </s> <s xml:id="echoid-s16510" xml:space="preserve">y:</s> <s xml:id="echoid-s16511" xml:space="preserve">: a<emph style="sub">2</emph>: </s> <s xml:id="echoid-s16512" xml:space="preserve">dd: </s> <s xml:id="echoid-s16513" xml:space="preserve">donc y = {a<emph style="sub">2</emph>d<emph style="sub">2</emph>/aa} ou {dd/a}: <lb/></s> <s xml:id="echoid-s16514" xml:space="preserve">donc la ligne G C eſt la hauteur que l’on demande; </s> <s xml:id="echoid-s16515" xml:space="preserve">car à cauſe <lb/>des triangles rectangles ſemblables G A D, G D C, on a <lb/>A G (a): </s> <s xml:id="echoid-s16516" xml:space="preserve">G D (d):</s> <s xml:id="echoid-s16517" xml:space="preserve">: G D (d): </s> <s xml:id="echoid-s16518" xml:space="preserve">G C ({dd/a}). </s> <s xml:id="echoid-s16519" xml:space="preserve">C. </s> <s xml:id="echoid-s16520" xml:space="preserve">Q. </s> <s xml:id="echoid-s16521" xml:space="preserve">F. </s> <s xml:id="echoid-s16522" xml:space="preserve">T. </s> <s xml:id="echoid-s16523" xml:space="preserve">& </s> <s xml:id="echoid-s16524" xml:space="preserve">D.</s> <s xml:id="echoid-s16525" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1335" type="section" level="1" n="982"> <head xml:id="echoid-head1167" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s16526" xml:space="preserve">993. </s> <s xml:id="echoid-s16527" xml:space="preserve">Puiſque le mobile avec la vîteſſe acquiſe par la chûte <lb/> <anchor type="note" xlink:label="note-0600-01a" xlink:href="note-0600-01"/> C G parcourt G D dans la moitié du tems qu’il emploie à <lb/>parcourir A G en tombant; </s> <s xml:id="echoid-s16528" xml:space="preserve">pendant un tems quadruple il <lb/>parcourra une ligne quadruple G E: </s> <s xml:id="echoid-s16529" xml:space="preserve">donc dans le double du <lb/>tems de la chûte par A G il parcourra d’un mouvement uni-<lb/>forme la ligne G E quadruple de G D. </s> <s xml:id="echoid-s16530" xml:space="preserve">Mais dans le même <lb/>tems double de celui de la deſcente par A G, la peſanteur fera <lb/>parcourir un eſpace quadruple de A G, à compter depuis le <lb/>premier inſtant de la chûte; </s> <s xml:id="echoid-s16531" xml:space="preserve">d’où il ſuit que ſi un mobile eſt <lb/>pouſſé ſuivant une direction oblique G D avec la force acquiſe <lb/>par le diametre C G, il parcourra d’un mouvement uniforme <lb/>la ligne G E quadruple de G D dans le même tems que la pe-<lb/>ſanteur lui feroit parcourir d’un mouvement accéléré la ver-<lb/>ticale E F auſſi quadruple de G A, comme il eſt évident par ce <lb/>qu’on vient de dire, & </s> <s xml:id="echoid-s16532" xml:space="preserve">à cauſe des triangles ſemblables G A D, <lb/>G E N.</s> <s xml:id="echoid-s16533" xml:space="preserve"/> </p> <div xml:id="echoid-div1335" type="float" level="2" n="1"> <note position="left" xlink:label="note-0600-01" xlink:href="note-0600-01a" xml:space="preserve">Figure 337.</note> </div> </div> <div xml:id="echoid-div1337" type="section" level="1" n="983"> <head xml:id="echoid-head1168" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s16534" xml:space="preserve">994. </s> <s xml:id="echoid-s16535" xml:space="preserve">Toute ligne comme C G parcourue par un mobile <lb/>pour acquérir un degré de force capable de lui faire décrire <lb/>une parabole déterminée, eſt appellée la ligne de hauteur.</s> <s xml:id="echoid-s16536" xml:space="preserve"/> </p> <pb o="523" file="0601" n="621" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1338" type="section" level="1" n="984"> <head xml:id="echoid-head1169" xml:space="preserve">PROPOSITION IX. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s16537" xml:space="preserve">995. </s> <s xml:id="echoid-s16538" xml:space="preserve">Le parametre d’une parabole décrite par un mobile eſt <lb/> <anchor type="note" xlink:label="note-0601-01a" xlink:href="note-0601-01"/> quadruple de la ligne de hauteur.</s> <s xml:id="echoid-s16539" xml:space="preserve"/> </p> <div xml:id="echoid-div1338" type="float" level="2" n="1"> <note position="right" xlink:label="note-0601-01" xlink:href="note-0601-01a" xml:space="preserve">Figure 336.</note> </div> </div> <div xml:id="echoid-div1340" type="section" level="1" n="985"> <head xml:id="echoid-head1170" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16540" xml:space="preserve">Ce problême renferme deux cas; </s> <s xml:id="echoid-s16541" xml:space="preserve">car le corps eſt projetté <lb/>horizontalement comme dans la figure 336, ou ſuivant une <lb/>ligne oblique à l’horizon, comme dans la figure 337. </s> <s xml:id="echoid-s16542" xml:space="preserve">Nous <lb/>l’allons démontrer dans l’un & </s> <s xml:id="echoid-s16543" xml:space="preserve">l’autre cas.</s> <s xml:id="echoid-s16544" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16545" xml:space="preserve">1°. </s> <s xml:id="echoid-s16546" xml:space="preserve">Si le mobile eſt projetté horizontalement, ſuivant la <lb/>ligne A B, l’ordonnée G F eſt égale à la ligne A B, & </s> <s xml:id="echoid-s16547" xml:space="preserve">partant <lb/>égale à 2 A D. </s> <s xml:id="echoid-s16548" xml:space="preserve">Par la propriété de la parabole, le quarré de <lb/>G F eſt égal au produit de ſon abſciſſe A G par le parametre, <lb/>ainſi nous aurons G F<emph style="sub">2</emph> ou 4 A D<emph style="sub">2</emph>=A G x 4 A C: </s> <s xml:id="echoid-s16549" xml:space="preserve">mais à cauſe <lb/>des triangles ſemblables D A G, C A D, on a A G: </s> <s xml:id="echoid-s16550" xml:space="preserve">A D :</s> <s xml:id="echoid-s16551" xml:space="preserve">: A D: </s> <s xml:id="echoid-s16552" xml:space="preserve">A C; <lb/></s> <s xml:id="echoid-s16553" xml:space="preserve">donc A D<emph style="sub">2</emph> = A G x A C: </s> <s xml:id="echoid-s16554" xml:space="preserve">donc 4 A D<emph style="sub">2</emph>, ou G F<emph style="sub">2</emph> = A G x 4 A C: </s> <s xml:id="echoid-s16555" xml:space="preserve"><lb/>donc le quadruple de A C ou de la ligne de hauteur eſt égalau <lb/>parametre. </s> <s xml:id="echoid-s16556" xml:space="preserve">C. </s> <s xml:id="echoid-s16557" xml:space="preserve">Q. </s> <s xml:id="echoid-s16558" xml:space="preserve">F. </s> <s xml:id="echoid-s16559" xml:space="preserve">1°. </s> <s xml:id="echoid-s16560" xml:space="preserve">D.</s> <s xml:id="echoid-s16561" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16562" xml:space="preserve">2°. </s> <s xml:id="echoid-s16563" xml:space="preserve">Si la ligne de projection G E eſt oblique à l’horizon, <lb/> <anchor type="note" xlink:label="note-0601-02a" xlink:href="note-0601-02"/> on remarquera d’abord que la ligne de projection E G étant <lb/>tangente à la parabole décrite en G, la ligne H I parallele à <lb/>G B ſera ordonnée au diametre G I; </s> <s xml:id="echoid-s16564" xml:space="preserve">& </s> <s xml:id="echoid-s16565" xml:space="preserve">comme, par hypotheſe, <lb/>G B eſt double de G D; </s> <s xml:id="echoid-s16566" xml:space="preserve">I H = G B ſera auſſi double de G D. <lb/></s> <s xml:id="echoid-s16567" xml:space="preserve">Mais à cauſe des triangles ſemblables G A D, G D C, on <lb/>aura G A: </s> <s xml:id="echoid-s16568" xml:space="preserve">G D:</s> <s xml:id="echoid-s16569" xml:space="preserve">: G D: </s> <s xml:id="echoid-s16570" xml:space="preserve">G C: </s> <s xml:id="echoid-s16571" xml:space="preserve">donc G D<emph style="sub">2</emph> = G A x G C, & </s> <s xml:id="echoid-s16572" xml:space="preserve"><lb/>partant 4 G D<emph style="sub">2</emph> ou I H<emph style="sub">2</emph> = G A x 4 G C = G I x 4 G C, puiſque <lb/>G A = G I. </s> <s xml:id="echoid-s16573" xml:space="preserve">C. </s> <s xml:id="echoid-s16574" xml:space="preserve">Q. </s> <s xml:id="echoid-s16575" xml:space="preserve">F. </s> <s xml:id="echoid-s16576" xml:space="preserve">2°. </s> <s xml:id="echoid-s16577" xml:space="preserve">D.</s> <s xml:id="echoid-s16578" xml:space="preserve"/> </p> <div xml:id="echoid-div1340" type="float" level="2" n="1"> <note position="right" xlink:label="note-0601-02" xlink:href="note-0601-02a" xml:space="preserve">Figure 337.</note> </div> </div> <div xml:id="echoid-div1342" type="section" level="1" n="986"> <head xml:id="echoid-head1171" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16579" xml:space="preserve">996. </s> <s xml:id="echoid-s16580" xml:space="preserve">Il ſuit delà que ſi on éleve ſur la ligne de projection <lb/>G E une perpendiculaire E M, qui aille rencontrer la ligne de <lb/>hauteur G C prolongée en M, M G ſera le parametre du dia-<lb/>metre G K: </s> <s xml:id="echoid-s16581" xml:space="preserve">car les triangles G C D & </s> <s xml:id="echoid-s16582" xml:space="preserve">G M E étant ſembla-<lb/>bles, on aura G D: </s> <s xml:id="echoid-s16583" xml:space="preserve">G E:</s> <s xml:id="echoid-s16584" xml:space="preserve">: G C: </s> <s xml:id="echoid-s16585" xml:space="preserve">G M: </s> <s xml:id="echoid-s16586" xml:space="preserve">donc puiſque G E eſt <lb/>quadruple de G D, G M ſera auſſi quadruple de G C.</s> <s xml:id="echoid-s16587" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1343" type="section" level="1" n="987"> <head xml:id="echoid-head1172" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16588" xml:space="preserve">997. </s> <s xml:id="echoid-s16589" xml:space="preserve">Il ſuit encore delà que connoiſſant le parametre d’une <pb o="524" file="0602" n="622" rhead="NOUVEAU COURS"/> parabole décrite par un mobile, on connoîtra aiſément de <lb/>quelle hauteur le mobile doit tomber pour acquérir une force <lb/>capable de lui faire décrire la parabole à laquelle appartient <lb/>le parametre, puiſque cette hauteur ſera toujours le quart du <lb/>parametre.</s> <s xml:id="echoid-s16590" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1344" type="section" level="1" n="988"> <head xml:id="echoid-head1173" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s16591" xml:space="preserve">998. </s> <s xml:id="echoid-s16592" xml:space="preserve">Comme avec un même parametre on peut décrire une <lb/>infinité de paraboles différentes, lorſque l’angle des ordonnées <lb/>avec le diametre n’eſt pas déterminé; </s> <s xml:id="echoid-s16593" xml:space="preserve">il s’enſuit qu’avec une <lb/>même force un corps projetté peut décrire une infinité de pa-<lb/>raboles différentes: </s> <s xml:id="echoid-s16594" xml:space="preserve">ces courbes varieront ſuivant les diffé-<lb/>rens angles de la ligne de projection avec l’horizon.</s> <s xml:id="echoid-s16595" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1345" type="section" level="1" n="989"> <head xml:id="echoid-head1174" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s16596" xml:space="preserve">999. </s> <s xml:id="echoid-s16597" xml:space="preserve">Il ſuit encore delà que ces trois lignes, le parametre <lb/>M G, la ligne de projection G E, & </s> <s xml:id="echoid-s16598" xml:space="preserve">la ligne de chûte E F, ſont <lb/>en progreſſion géométrique: </s> <s xml:id="echoid-s16599" xml:space="preserve">car à cauſe des triangles ſem-<lb/>blables M G E, E G F, on aura M G: </s> <s xml:id="echoid-s16600" xml:space="preserve">G E:</s> <s xml:id="echoid-s16601" xml:space="preserve">: G E: </s> <s xml:id="echoid-s16602" xml:space="preserve">E F: </s> <s xml:id="echoid-s16603" xml:space="preserve">donc <lb/>deux lignes quelconques étant connues, on trouvera toujours <lb/>la troiſieme.</s> <s xml:id="echoid-s16604" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1346" type="section" level="1" n="990"> <head xml:id="echoid-head1175" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s16605" xml:space="preserve">1000. </s> <s xml:id="echoid-s16606" xml:space="preserve">Les lignes de chûte E F étant perpendiculaires à l’ho-<lb/>rizon, elles formeront, avec les lignes de projection G E, des <lb/>triangles rectangles G E F qui ſeront ſemblables aux triangles <lb/>G M E, leſquels auront tous pour hypotheſe le parametre <lb/>M G; </s> <s xml:id="echoid-s16607" xml:space="preserve">d’où il ſuit que toutes les lignes de projections poſſi-<lb/>bles pour une même force ſont renfermées dans un demi-<lb/>cercle G E M.</s> <s xml:id="echoid-s16608" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1347" type="section" level="1" n="991"> <head xml:id="echoid-head1176" xml:space="preserve"><emph style="sc">Observation</emph>.</head> <p> <s xml:id="echoid-s16609" xml:space="preserve">1001. </s> <s xml:id="echoid-s16610" xml:space="preserve">Il faut bien s’attacher à concevoir la raiſon pour la-<lb/>quelle on a ſuppoſé que la force de projection eſt telle qu’elle <lb/>fait parcourir au corps d’un mouvement uniforme la ligne <lb/>G D dans la moitié du tems que le corps employeroit à par-<lb/>courir A G. </s> <s xml:id="echoid-s16611" xml:space="preserve">Pour cela, on remarquera que dans le tems que <lb/>le corps parcourt G B, la peſanteur qui a toujours agi ſur lui <lb/>a fait parcourir l’eſpace B H = A G; </s> <s xml:id="echoid-s16612" xml:space="preserve">de même dans le tems <lb/>que la force de projection lui auroit fait parcourir B E = G B, <lb/>la peſanteur lui fera parcourir E F, quadruple de A G, & </s> <s xml:id="echoid-s16613" xml:space="preserve">par <lb/>conſéquent il ſe trouvera en F ſur l’horizontale G F.</s> <s xml:id="echoid-s16614" xml:space="preserve"/> </p> <pb o="525" file="0603" n="623" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1348" type="section" level="1" n="992"> <head xml:id="echoid-head1177" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s16615" xml:space="preserve">1002. </s> <s xml:id="echoid-s16616" xml:space="preserve">Il ſuit delà que dans un tems double de la deſcente <lb/>par A G, le corps projetté ſuivant la ligne G D avec la vîteſſe <lb/>acquiſe par G C décrira la parabole G H F, & </s> <s xml:id="echoid-s16617" xml:space="preserve">de plus que la <lb/>vîteſſe qu’il a lorſqu’il eſt en F eſt égale à celle qu’il auroit <lb/>acquiſe par A G: </s> <s xml:id="echoid-s16618" xml:space="preserve">car il eſt viſible que le ſommet H de la pa-<lb/>rabole décrite eſt au milieu de la ligne B L, double de A G.</s> <s xml:id="echoid-s16619" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1349" type="section" level="1" n="993"> <head xml:id="echoid-head1178" xml:space="preserve">PROPOSITION X. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16620" xml:space="preserve">1003. </s> <s xml:id="echoid-s16621" xml:space="preserve">Etant donnée la ligne de but G F, l’angle M G E for-<lb/> <anchor type="note" xlink:label="note-0603-01a" xlink:href="note-0603-01"/> mé par le parametre M G, & </s> <s xml:id="echoid-s16622" xml:space="preserve">la direction G E du mortier, & </s> <s xml:id="echoid-s16623" xml:space="preserve">l’an-<lb/>gle E G F formé par la direction du mortier & </s> <s xml:id="echoid-s16624" xml:space="preserve">la ligne de but G F, <lb/>trouver le parametre M G, la ligne de projection G E, & </s> <s xml:id="echoid-s16625" xml:space="preserve">la ligne <lb/>de chûte E F.</s> <s xml:id="echoid-s16626" xml:space="preserve"/> </p> <div xml:id="echoid-div1349" type="float" level="2" n="1"> <note position="right" xlink:label="note-0603-01" xlink:href="note-0603-01a" xml:space="preserve">Figure 339, <lb/>340 & 341.</note> </div> <p> <s xml:id="echoid-s16627" xml:space="preserve">Conſidérez que les lignes M G & </s> <s xml:id="echoid-s16628" xml:space="preserve">E F étant paralleles, les <lb/>angles alternes M G E & </s> <s xml:id="echoid-s16629" xml:space="preserve">G E F ſont égaux; </s> <s xml:id="echoid-s16630" xml:space="preserve">& </s> <s xml:id="echoid-s16631" xml:space="preserve">que connoiſ-<lb/>ſant l’un, on connoîtra l’autre: </s> <s xml:id="echoid-s16632" xml:space="preserve">& </s> <s xml:id="echoid-s16633" xml:space="preserve">qu’ainſi l’on connoît dans <lb/>le triangle G E F le côté G F avec les angles E G F & </s> <s xml:id="echoid-s16634" xml:space="preserve">G E F; <lb/></s> <s xml:id="echoid-s16635" xml:space="preserve">& </s> <s xml:id="echoid-s16636" xml:space="preserve">que par conſéquent on trouvera par la Trigonométrie la <lb/>ligne de projection G E, & </s> <s xml:id="echoid-s16637" xml:space="preserve">la ligne de chûte E F: </s> <s xml:id="echoid-s16638" xml:space="preserve">mais E F: </s> <s xml:id="echoid-s16639" xml:space="preserve"><lb/>E G:</s> <s xml:id="echoid-s16640" xml:space="preserve">: E G: </s> <s xml:id="echoid-s16641" xml:space="preserve">G M (art. </s> <s xml:id="echoid-s16642" xml:space="preserve">999). </s> <s xml:id="echoid-s16643" xml:space="preserve">Ainſi l’on voit que cherchant <lb/>une troiſieme proportionnelle à la ligne de chûte & </s> <s xml:id="echoid-s16644" xml:space="preserve">à la ligne <lb/>de projection, l’on aura auſſi le parametre.</s> <s xml:id="echoid-s16645" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1351" type="section" level="1" n="994"> <head xml:id="echoid-head1179" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s16646" xml:space="preserve">1004. </s> <s xml:id="echoid-s16647" xml:space="preserve">Il ſuit delà que ſi l’on jette une bombe avec un mor-<lb/>tier, ſelon telle inclinaiſon que l’on voudra, pour trouver le <lb/>parametre de toutes les paraboles décrites par le même mo-<lb/>bile toujours pouſſé avec la même force, il n’y a qu’à ob-<lb/>ſerver l’angle d’inclinaiſon du mortier, & </s> <s xml:id="echoid-s16648" xml:space="preserve">meſurer la diſtance <lb/>où la bombe ſera tombée, puiſque le reſte ſe trouve après aiſé-<lb/>ment.</s> <s xml:id="echoid-s16649" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16650" xml:space="preserve">Suppoſons, par exemple, que l’on ait meſuré l’angle E G F <lb/> <anchor type="note" xlink:label="note-0603-02a" xlink:href="note-0603-02"/> d’inclinaiſon du mortier avec la ligne de but G F, que nous <lb/>ſuppoſerons de 500 toiſes; </s> <s xml:id="echoid-s16651" xml:space="preserve">& </s> <s xml:id="echoid-s16652" xml:space="preserve">qu’on ait auſſi meſuré l’angle <lb/>M G E que fait la même ligne de direction avec la verticale <lb/>M G ou le parametre. </s> <s xml:id="echoid-s16653" xml:space="preserve">On connoîtra donc trois choſes dans <pb o="526" file="0604" n="624" rhead="NOUVEAU COURS"/> le triangle E G F, ſçavoir la ligne de but G F, l’angle E G F, <lb/>& </s> <s xml:id="echoid-s16654" xml:space="preserve">l’angle G E F égal à ſon alterne M G E: </s> <s xml:id="echoid-s16655" xml:space="preserve">donc on connoîtra <lb/>les lignes, E F qui eſt la ligne de chûte, E G qui eſt celle de <lb/>direction; </s> <s xml:id="echoid-s16656" xml:space="preserve">& </s> <s xml:id="echoid-s16657" xml:space="preserve">par conſéquent on trouvera le parametre par cette <lb/>proportion, E F: </s> <s xml:id="echoid-s16658" xml:space="preserve">E G:</s> <s xml:id="echoid-s16659" xml:space="preserve">: E G: </s> <s xml:id="echoid-s16660" xml:space="preserve">M G. </s> <s xml:id="echoid-s16661" xml:space="preserve">Ainſi l’on ſçaura de quelle <lb/>hauteur le corps a dû tomber pour acquérir une force égale à <lb/>celle que lui a communiqué la charge de poudre dont il eſt <lb/>queſtion, en prenant le quart du parametre (art. </s> <s xml:id="echoid-s16662" xml:space="preserve">995). </s> <s xml:id="echoid-s16663" xml:space="preserve">D’ail-<lb/>leurs avec le même parametre on peut décrire une infinité de <lb/>paraboles, ſelon l’angle des ordonnées au diametre: </s> <s xml:id="echoid-s16664" xml:space="preserve">donc <lb/>ces obſervations ſuffiſent pour déterminer le parametre.</s> <s xml:id="echoid-s16665" xml:space="preserve"/> </p> <div xml:id="echoid-div1351" type="float" level="2" n="1"> <note position="right" xlink:label="note-0603-02" xlink:href="note-0603-02a" xml:space="preserve">Figure 342.</note> </div> </div> <div xml:id="echoid-div1353" type="section" level="1" n="995"> <head xml:id="echoid-head1180" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s16666" xml:space="preserve">Nous allons donner des ſolutions géométriques & </s> <s xml:id="echoid-s16667" xml:space="preserve">analyti-<lb/>ques de pluſieurs problêmes qui ont rapport au jet des bombes, <lb/>pour nous préparer à faire les mêmes choſes dans la pratique <lb/>avec un inſtrument univerſel, dont la conſtruction & </s> <s xml:id="echoid-s16668" xml:space="preserve">l’uſage <lb/>dépendent de ce que l’on va voir: </s> <s xml:id="echoid-s16669" xml:space="preserve">ainſi il ne faut pas que ceux <lb/>qui étudieront ce Traité, s’inquiétent ſi on ne les conduit pas <lb/>d’abord à la pratique, puiſqu’ils trouveront dans la ſuite de <lb/>quoi ſe contenter.</s> <s xml:id="echoid-s16670" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1354" type="section" level="1" n="996"> <head xml:id="echoid-head1181" xml:space="preserve">PROPOSITION XI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16671" xml:space="preserve">1005. </s> <s xml:id="echoid-s16672" xml:space="preserve">Trouver quelle élevation il faut donner à un mortier pour <lb/> <anchor type="note" xlink:label="note-0604-01a" xlink:href="note-0604-01"/> jetter une bombe à tel endroit que l’on voudra, pourvu que cet en-<lb/>droit ſoit de niveau avec la batterie.</s> <s xml:id="echoid-s16673" xml:space="preserve"/> </p> <div xml:id="echoid-div1354" type="float" level="2" n="1"> <note position="left" xlink:label="note-0604-01" xlink:href="note-0604-01a" xml:space="preserve">Figure 342.</note> </div> <p> <s xml:id="echoid-s16674" xml:space="preserve">Le mortier étant ſuppoſé au point G, & </s> <s xml:id="echoid-s16675" xml:space="preserve">le point F étant <lb/>celui où l’on veut jetter la bombe, nous ſuppoſerons que la <lb/>ligne G M, élevée perpendiculairement ſur G F, eſt le parametre <lb/>de projection. </s> <s xml:id="echoid-s16676" xml:space="preserve">Cela poſé, on le diviſera en deux également au <lb/>point A; </s> <s xml:id="echoid-s16677" xml:space="preserve">& </s> <s xml:id="echoid-s16678" xml:space="preserve">de ce point comme centre, on décrira un demi-<lb/>cercle, & </s> <s xml:id="echoid-s16679" xml:space="preserve">ſur le point F de la ligne horizontale G H, on éle-<lb/>vera la perpendiculaire F E, qui coupera le demi-cercle au <lb/>point E. </s> <s xml:id="echoid-s16680" xml:space="preserve">Or ſi l’on tire du point G aux points E les lignes <lb/>G E, je dis que le mortier pointé, ſelon l’une ou l’autre de ces <lb/>directions, jettera la bombe au point F.</s> <s xml:id="echoid-s16681" xml:space="preserve"/> </p> <pb o="527" file="0605" n="625" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1356" type="section" level="1" n="997"> <head xml:id="echoid-head1182" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16682" xml:space="preserve">Nous avons fait voir (art. </s> <s xml:id="echoid-s16683" xml:space="preserve">997) que le parametre, la ligne <lb/>de projection, & </s> <s xml:id="echoid-s16684" xml:space="preserve">la ligne de chûte étoient trois proportion-<lb/>nelles: </s> <s xml:id="echoid-s16685" xml:space="preserve">ainſi pour prouver que la ligne G E eſt la ligne de pro-<lb/>jection, il n’y a qu’à prouver qu’elle eſt moyenne proportion-<lb/>nelle entre le parametre M G & </s> <s xml:id="echoid-s16686" xml:space="preserve">la ligne de chûte correſpon-<lb/>dante E F. </s> <s xml:id="echoid-s16687" xml:space="preserve">Or ſi l’on tire les lignes M E, l’on aura les trian-<lb/>gles ſemblables M G E & </s> <s xml:id="echoid-s16688" xml:space="preserve">G E F; </s> <s xml:id="echoid-s16689" xml:space="preserve">car ils ont chacun un angle <lb/>droit, & </s> <s xml:id="echoid-s16690" xml:space="preserve">les angles G M E & </s> <s xml:id="echoid-s16691" xml:space="preserve">E G F ont chacun pour meſure <lb/>la moitié de l’arc G I E: </s> <s xml:id="echoid-s16692" xml:space="preserve">par conſéquent l’on a M G: </s> <s xml:id="echoid-s16693" xml:space="preserve">G E:</s> <s xml:id="echoid-s16694" xml:space="preserve">: <lb/>G E : </s> <s xml:id="echoid-s16695" xml:space="preserve">E F.</s> <s xml:id="echoid-s16696" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16697" xml:space="preserve">Mais ſi la perpendiculaire élevée ſur le point F, au lieu de <lb/> <anchor type="note" xlink:label="note-0605-01a" xlink:href="note-0605-01"/> couper le cercle, ne faiſoit que le toucher en un ſeul point <lb/>E; </s> <s xml:id="echoid-s16698" xml:space="preserve">je dis que la ligne G E ſera encore l’inclinaiſon du mor-<lb/>tier; </s> <s xml:id="echoid-s16699" xml:space="preserve">puiſqu’à cauſe des triangles ſemblables M G E & </s> <s xml:id="echoid-s16700" xml:space="preserve">G E F, <lb/>l’on aura M G : </s> <s xml:id="echoid-s16701" xml:space="preserve">G E :</s> <s xml:id="echoid-s16702" xml:space="preserve">: G E : </s> <s xml:id="echoid-s16703" xml:space="preserve">E F.</s> <s xml:id="echoid-s16704" xml:space="preserve"/> </p> <div xml:id="echoid-div1356" type="float" level="2" n="1"> <note position="right" xlink:label="note-0605-01" xlink:href="note-0605-01a" xml:space="preserve">Figure 343.</note> </div> <p> <s xml:id="echoid-s16705" xml:space="preserve">Enſin ſi l’on ſuppoſe que le point donné ſoit l’endroit C, & </s> <s xml:id="echoid-s16706" xml:space="preserve"><lb/>que la perpendiculaire C D ne rencontre pas le cercle, je dis <lb/>que le problême eſt impoſſible; </s> <s xml:id="echoid-s16707" xml:space="preserve">puiſque G D, qui eſt ſuppoſé <lb/>la ligne de projection, ne peut pas être moyenne proportion-<lb/>nelle entre le parametre M G & </s> <s xml:id="echoid-s16708" xml:space="preserve">la ligne de chûte D C: </s> <s xml:id="echoid-s16709" xml:space="preserve">car <lb/>pour cela il faudroit qu’elle fût un côté commun aux deux <lb/>triangles ſemblables M G E & </s> <s xml:id="echoid-s16710" xml:space="preserve">G D C; </s> <s xml:id="echoid-s16711" xml:space="preserve">ce qui ne peut arriver, <lb/>tant que la pointe D ſera hors du cercle.</s> <s xml:id="echoid-s16712" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1358" type="section" level="1" n="998"> <head xml:id="echoid-head1183" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16713" xml:space="preserve">1006. </s> <s xml:id="echoid-s16714" xml:space="preserve">Il ſuit delà que lorſque la perpendiculaire E F coupe <lb/>le cercle, le problême a deux ſolutions, & </s> <s xml:id="echoid-s16715" xml:space="preserve">que par conſéquent <lb/>on peut jetter une bombe en un même endroit par deux che-<lb/>mins différens: </s> <s xml:id="echoid-s16716" xml:space="preserve">car les arcs M E & </s> <s xml:id="echoid-s16717" xml:space="preserve">G E étant égaux, lorſque <lb/>le mortier ſera pointé à un degré d’élevation par un angle au-<lb/>tant au deſſus qu’au deſſous du quart de cercle, la bombe <lb/>ira également loin: </s> <s xml:id="echoid-s16718" xml:space="preserve">mais comme les angles M G E n’ont pour <lb/>meſure que les moitiés des arcs M E, & </s> <s xml:id="echoid-s16719" xml:space="preserve">que c’eſt toujours avec <lb/>la verticale M G & </s> <s xml:id="echoid-s16720" xml:space="preserve">les lignes de projections G E, que l’on con-<lb/>ſidere l’élevation du mortier; </s> <s xml:id="echoid-s16721" xml:space="preserve">l’on voit que cet angle ſera tou-<lb/>jours plus petit qu’un droit, & </s> <s xml:id="echoid-s16722" xml:space="preserve">qu’on pourra pointer le mortier <lb/>également au deſſus ou au deſſous de 45 degrés pour chaſſer la <lb/>bombe en un même endroit.</s> <s xml:id="echoid-s16723" xml:space="preserve"/> </p> <pb o="528" file="0606" n="626" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1359" type="section" level="1" n="999"> <head xml:id="echoid-head1184" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16724" xml:space="preserve">1007. </s> <s xml:id="echoid-s16725" xml:space="preserve">Comme le problême eſt toujours poſſible, ſoit que <lb/>la ligne E F coupe ou touche le cercle, l’on voit que lorſqu’elle <lb/>touchera le cercle, la bombe ſera chaſſée le plus loin qu’il eſt <lb/>poſſible avec la même charge, puiſque la ligne de but G F <lb/>eſt la plus grande de toutes celles qui peuvent être renfermées <lb/>entre le parametre & </s> <s xml:id="echoid-s16726" xml:space="preserve">la ligne de chûte. </s> <s xml:id="echoid-s16727" xml:space="preserve">Or comme l’angle <lb/>M G E a pour meſure la moitié du demi-cercle M E, l’on peur <lb/>dire que de toutes les bombes qui ſeront tirées avec une même <lb/>charge, celle qui ira le plus loin, ſera celle qui aura été tirée <lb/>ſous un angle de 45 degrés.</s> <s xml:id="echoid-s16728" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1360" type="section" level="1" n="1000"> <head xml:id="echoid-head1185" xml:space="preserve">PROPOSITION XII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16729" xml:space="preserve">1008. </s> <s xml:id="echoid-s16730" xml:space="preserve">Trouver quelle élevation il faut donner à un mortier <lb/>pour chaſſer une bombe à une diſtance donnée, en ſuppoſant que la <lb/>batterie n’eſt pas de niveau avec l’endroit où l’on veut jetter la <lb/>bombe, c’eſt-à-dire en ſuppoſant que cet endroit eſt beaucoup plus <lb/>élevé ou plus bas que la batterie.</s> <s xml:id="echoid-s16731" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16732" xml:space="preserve">Le point G étant ſuppoſé l’endroit du mortier, & </s> <s xml:id="echoid-s16733" xml:space="preserve">le point <lb/>F celui où l’on veut jetter la bombe, lequel ſera plus élevé <lb/>que la batterie, comme dans la figure 344, ou plus bas que la <lb/>batterie, comme dans la figure 345, il faut ſur la ligne ho-<lb/>rizontale G H élever la perpendiculaire G M égale au para-<lb/>metre de la charge du mortier, parce que je ſuppoſe que l’on <lb/>a fait une épreuve pour trouver ce parametre, comme il a été <lb/>dit, art. </s> <s xml:id="echoid-s16734" xml:space="preserve">1001; </s> <s xml:id="echoid-s16735" xml:space="preserve">enſuite l’on élevera la perpendiculaire G A ſur <lb/>la ligne du plan G L, & </s> <s xml:id="echoid-s16736" xml:space="preserve">l’on fera l’angle A M G égal à l’angle <lb/>A G M; </s> <s xml:id="echoid-s16737" xml:space="preserve">& </s> <s xml:id="echoid-s16738" xml:space="preserve">du point A, comme centre, l’on décrira la portion <lb/>de cercle M E G: </s> <s xml:id="echoid-s16739" xml:space="preserve">du point donné F l’on menera la ligne <lb/>F E parallele au parametre M G; </s> <s xml:id="echoid-s16740" xml:space="preserve">& </s> <s xml:id="echoid-s16741" xml:space="preserve">cette ligne venant couper <lb/>le cercle aux points E, je dis que ſi l’on tire les lignes G E, <lb/>qu’elles détermineront l’élevation qu’il faut donner au mortier <lb/>pour jetter la bombe au point F dans l’un & </s> <s xml:id="echoid-s16742" xml:space="preserve">l’autre cas.</s> <s xml:id="echoid-s16743" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1361" type="section" level="1" n="1001"> <head xml:id="echoid-head1186" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16744" xml:space="preserve">M G étantle parametre, G E la ligne de projection, & </s> <s xml:id="echoid-s16745" xml:space="preserve">E F, <pb o="529" file="0607" n="627" rhead="DE MATHÉMATIQUE. Liv. XIV."/> la ligne de chûte, il faut prouver, comme on l’a fait ci-devant, <lb/>que M G : </s> <s xml:id="echoid-s16746" xml:space="preserve">G E :</s> <s xml:id="echoid-s16747" xml:space="preserve">: G E : </s> <s xml:id="echoid-s16748" xml:space="preserve">E F. </s> <s xml:id="echoid-s16749" xml:space="preserve">Pour cela, conſidérez que les <lb/>triangles M G E & </s> <s xml:id="echoid-s16750" xml:space="preserve">G E F ſont ſemblables: </s> <s xml:id="echoid-s16751" xml:space="preserve">car comme la ligne <lb/>G F eſt perpendiculaire au rayon A G, l’angle E G F ſera égal <lb/>à l’angle G M E, puiſqu’ils ont chacun pour meſure la moitié <lb/>de l’arc G I E: </s> <s xml:id="echoid-s16752" xml:space="preserve">d’ailleurs à cauſe des paralleles M G & </s> <s xml:id="echoid-s16753" xml:space="preserve">E F <lb/>les angles M G E & </s> <s xml:id="echoid-s16754" xml:space="preserve">G E F ſont égaux, étant alternes: </s> <s xml:id="echoid-s16755" xml:space="preserve">ainſi <lb/>l’on aura M G : </s> <s xml:id="echoid-s16756" xml:space="preserve">G E :</s> <s xml:id="echoid-s16757" xml:space="preserve">: G E : </s> <s xml:id="echoid-s16758" xml:space="preserve">E F, ce qui fait voir que l’angle <lb/>M G E eſt celui qu’il faut que le mortier faſſe avec la verticale <lb/>pour chaſſer la bombe au point F. </s> <s xml:id="echoid-s16759" xml:space="preserve">C. </s> <s xml:id="echoid-s16760" xml:space="preserve">Q. </s> <s xml:id="echoid-s16761" xml:space="preserve">F. </s> <s xml:id="echoid-s16762" xml:space="preserve">D.</s> <s xml:id="echoid-s16763" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16764" xml:space="preserve">Pour ne pas répéter les mêmes choſes, nous avons compris <lb/>les deux cas précédens dans une même démonſtration: </s> <s xml:id="echoid-s16765" xml:space="preserve">mais <lb/>il ſeroit bon que les Commençans répétaſſent deux fois la dé-<lb/>monſtration précédente, pour ne conſidérer qu’une des deux <lb/>figures 344 & </s> <s xml:id="echoid-s16766" xml:space="preserve">345 à la fois.</s> <s xml:id="echoid-s16767" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1362" type="section" level="1" n="1002"> <head xml:id="echoid-head1187" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s16768" xml:space="preserve">1009. </s> <s xml:id="echoid-s16769" xml:space="preserve">Il arrivera dans les deux cas du problême précédent <lb/>ce que nous avons dit (art. </s> <s xml:id="echoid-s16770" xml:space="preserve">1006) à l’occaſion des bombes jet-<lb/>tées à un endroit de niveau avec la batterie, qui eſt que ſi la <lb/>parallele E F touche le cercle, au lieu de le couper, la portée <lb/>de la bombe ſera la plus grande de toutes celles qu’on peut <lb/>jetter avec la même charge; </s> <s xml:id="echoid-s16771" xml:space="preserve">& </s> <s xml:id="echoid-s16772" xml:space="preserve">que ſi la parallele E F ne tou-<lb/>choit ni ne coupoit le cercle, que le problême ſeroit impoſ-<lb/>ſible; </s> <s xml:id="echoid-s16773" xml:space="preserve">ce qui a été ſuffiſamment expliqué ailleurs (art. </s> <s xml:id="echoid-s16774" xml:space="preserve">1005), <lb/>pour n’avoir pas beſoin d’en faire voir encore la raiſon.</s> <s xml:id="echoid-s16775" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1363" type="section" level="1" n="1003"> <head xml:id="echoid-head1188" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s16776" xml:space="preserve">1010. </s> <s xml:id="echoid-s16777" xml:space="preserve">Il eſt bon que l’on ſçache que dans la pratique ordi-<lb/>naire du jet des bombes, l’on pointe toujours le mortier ſous <lb/>l’angle qui donne la plus grande ligne de chûte E F, afin que la <lb/>bombe tombant de plus haut, acquiere par ſa peſanteur un <lb/>degré de force capable de produire plus de dommage ſur les <lb/>édifices où elle tombe; </s> <s xml:id="echoid-s16778" xml:space="preserve">mais quand on eſt près d’un ouvrage <lb/>de fortification que l’on veut labourer par les bombes, pour <lb/>le mettre plutôt en état de l’attaquer, l’on pointe le mortier <lb/>ſous l’angle de la petite ligne de chûte E F, afin que la bombe <lb/>paſſant par le chemin le plus court, ne donne pas le tems à <lb/>ceux qui ſont dans l’ouvrage de ſe garantir des éclats.</s> <s xml:id="echoid-s16779" xml:space="preserve"/> </p> <pb o="530" file="0608" n="628" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1364" type="section" level="1" n="1004"> <head xml:id="echoid-head1189" xml:space="preserve">PROPOSITION XIII. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16780" xml:space="preserve">1011. </s> <s xml:id="echoid-s16781" xml:space="preserve">La ligne de but G F, l’angle qu’elle fait avec la verti-<lb/> <anchor type="note" xlink:label="note-0608-01a" xlink:href="note-0608-01"/> cale G M, & </s> <s xml:id="echoid-s16782" xml:space="preserve">la charge du mortier etant donnee, trouver l’angle <lb/>d’élévation ſous lequel il faut pointer le mortier pour qu’elle tombe <lb/>au point F.</s> <s xml:id="echoid-s16783" xml:space="preserve"/> </p> <div xml:id="echoid-div1364" type="float" level="2" n="1"> <note position="left" xlink:label="note-0608-01" xlink:href="note-0608-01a" xml:space="preserve">Figure 345.</note> </div> </div> <div xml:id="echoid-div1366" type="section" level="1" n="1005"> <head xml:id="echoid-head1190" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s16784" xml:space="preserve">Soit nommée a la ligne de but G F; </s> <s xml:id="echoid-s16785" xml:space="preserve">comme la charge du <lb/>mortier eſt auſſi connue, & </s> <s xml:id="echoid-s16786" xml:space="preserve">que d’ailleurs on ſuppoſe que l’ex-<lb/>périence a déterminé la force qu’une telle charge peut donner <lb/>à la bombe, il s’enſuit qu’on connoît le parametre de la para-<lb/>bole qu’elle doit décrire; </s> <s xml:id="echoid-s16787" xml:space="preserve">puiſque ce même parametre eſt qua-<lb/>druple de la ligne de hauteur, dont le projectile à dû tomber <lb/>pour acquérir une force égale à celle qu’il reçoit de la poudre; <lb/></s> <s xml:id="echoid-s16788" xml:space="preserve">ſoit p ce parametre. </s> <s xml:id="echoid-s16789" xml:space="preserve">Comme l’on connoît l’angle M G F dela <lb/>verticale avec la ligne de but G F, on connoîtra auſſi l’angle <lb/>de cette derniere ligne avec l’horizontale: </s> <s xml:id="echoid-s16790" xml:space="preserve">donc au triangle <lb/>rectangle G H F on connoîtra le côté H F & </s> <s xml:id="echoid-s16791" xml:space="preserve">le côté G H. </s> <s xml:id="echoid-s16792" xml:space="preserve"><lb/>Nous nommerons H F, d; </s> <s xml:id="echoid-s16793" xml:space="preserve">& </s> <s xml:id="echoid-s16794" xml:space="preserve">partant G H ſera √aa-dd\x{0020}: </s> <s xml:id="echoid-s16795" xml:space="preserve"><lb/>enfin la ligne E H qui détermine l’angle d’inclinaiſon demandé <lb/>ſera nommée x. </s> <s xml:id="echoid-s16796" xml:space="preserve">Cela poſé, il eſt viſible, à cauſe du triangle rec-<lb/>tangle G H E, que la ligne de projection G E eſt √aa-dd+xx\x{0020}; </s> <s xml:id="echoid-s16797" xml:space="preserve"><lb/>d’ailleurs la ligne de chûte E F = d + x, & </s> <s xml:id="echoid-s16798" xml:space="preserve">comme ces deux <lb/>lignes ſont en progreſſion avec le parametre (art. </s> <s xml:id="echoid-s16799" xml:space="preserve">999), on <lb/>aura E F: </s> <s xml:id="echoid-s16800" xml:space="preserve">G E :</s> <s xml:id="echoid-s16801" xml:space="preserve">: G E : </s> <s xml:id="echoid-s16802" xml:space="preserve">G M, ou d + x: </s> <s xml:id="echoid-s16803" xml:space="preserve">√aa-dd+xx\x{0020} :</s> <s xml:id="echoid-s16804" xml:space="preserve">: <lb/>√aa - dd + xx\x{0020}: </s> <s xml:id="echoid-s16805" xml:space="preserve">p; </s> <s xml:id="echoid-s16806" xml:space="preserve">d’où l’on tire px+pd=aa-dd+xx. </s> <s xml:id="echoid-s16807" xml:space="preserve"><lb/>Or donnant cette équation, & </s> <s xml:id="echoid-s16808" xml:space="preserve">la réſolvant ſuivant les regles <lb/>ordinaires, on aura d’abord xx -px=dd+pd-aa; </s> <s xml:id="echoid-s16809" xml:space="preserve">& </s> <s xml:id="echoid-s16810" xml:space="preserve">en-<lb/>ſuite x={1/2} p ± √dd+pd+{1/4}pp-aa\x{0020}. </s> <s xml:id="echoid-s16811" xml:space="preserve">Nous allons faire <lb/>voir que ces deux valeurs de x, conſtruites ſuivant la formule <lb/>algébrique, ſont préciſément les mêmes que celles que nous <lb/>avons ci-devant déterminé géométriquement.</s> <s xml:id="echoid-s16812" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16813" xml:space="preserve">Dans les conſtructions précédentes, on a d’abord ſur la <lb/>ligne G F élevée une perpendiculaire indéfinie G A; </s> <s xml:id="echoid-s16814" xml:space="preserve">au point <lb/>B, milieu du parametre M G, on a élevé une autre perpen-<lb/>diculaire B A qui coupe la premiere en A. </s> <s xml:id="echoid-s16815" xml:space="preserve">Du point A comme <pb o="531" file="0609" n="629" rhead="DE MATHÉMATIQUE. Liv. XIV."/> centre avec le rayon A G, on a décrite une portion de cercle <lb/>qui a déterminé ſur la verticale F E les points E, E qui don-<lb/>nent deux inclinaiſons différentes pour jetter la bombe en F. <lb/></s> <s xml:id="echoid-s16816" xml:space="preserve">Ainſi il faut faire voir que des deux lignes E F, E F, la plus pe-<lb/>tite eſt {1/2}p - √dd+pd+{1/4}pp-aa\x{0020}, & </s> <s xml:id="echoid-s16817" xml:space="preserve">la plus grande <lb/>{1/2}p+√dd+pd+{1/4}pp-aa\x{0020}. </s> <s xml:id="echoid-s16818" xml:space="preserve">Par conſtruction les triangles <lb/>G H F, G B A ſont ſemblables, & </s> <s xml:id="echoid-s16819" xml:space="preserve">donnent GH : </s> <s xml:id="echoid-s16820" xml:space="preserve">HF :</s> <s xml:id="echoid-s16821" xml:space="preserve">: GB : </s> <s xml:id="echoid-s16822" xml:space="preserve">BA, <lb/>& </s> <s xml:id="echoid-s16823" xml:space="preserve">analytiquement √aa-dd\x{0020}: </s> <s xml:id="echoid-s16824" xml:space="preserve">d :</s> <s xml:id="echoid-s16825" xml:space="preserve">: {p/2} : </s> <s xml:id="echoid-s16826" xml:space="preserve">{pd/2√aa-dd\x{0020}}: </s> <s xml:id="echoid-s16827" xml:space="preserve">doncle rayon <lb/>A G ſera égal à √{p<emph style="sub">2</emph>d<emph style="sub">2</emph>/4a<emph style="sub">2</emph>-4dd}+{pp/4}\x{0020}. </s> <s xml:id="echoid-s16828" xml:space="preserve">Je fais enſuite attention que <lb/>pour avoir E H (x) il faut déterminer A O parallele à B G, & </s> <s xml:id="echoid-s16829" xml:space="preserve"><lb/>terminé en O à la ligne D E parallele à G H. </s> <s xml:id="echoid-s16830" xml:space="preserve">O E = D E-D O <lb/>=G H- G N ou G H - A B, & </s> <s xml:id="echoid-s16831" xml:space="preserve">analytiquement O E = <lb/>√aa-dd\x{0020}-{pd/2√aa-dd\x{0020}}; </s> <s xml:id="echoid-s16832" xml:space="preserve">& </s> <s xml:id="echoid-s16833" xml:space="preserve">à cauſe du triangle rectangle A O E, <lb/>A O = √A E<emph style="sub">2</emph> - O E<emph style="sub">2</emph>\x{0020}: </s> <s xml:id="echoid-s16834" xml:space="preserve">donc l’expreſſion algébrique de A O <lb/>ſera √{p<emph style="sub">2</emph>d<emph style="sub">2</emph>/4aa - 4dd} + {pp/4} - aa + dd - {p<emph style="sub">2</emph>d<emph style="sub">2</emph>/4aa-4dd} + pd\x{0020}; </s> <s xml:id="echoid-s16835" xml:space="preserve">ce qui ſe <lb/>réduit à √dd+pd+{pp/4}-aa\x{0020}: </s> <s xml:id="echoid-s16836" xml:space="preserve">donc E H (x), ou B G-B D <lb/>={1/2}p-√dd+pd+{1/4}pp-aa\x{0020}; </s> <s xml:id="echoid-s16837" xml:space="preserve">d’où il ſuit évidemment <lb/>que la conſtruction géométrique eſt parfaitement d’accord <lb/>avec l’analyſe, & </s> <s xml:id="echoid-s16838" xml:space="preserve">qu’elle nous donne les mêmes ſolutions. </s> <s xml:id="echoid-s16839" xml:space="preserve"><lb/>C. </s> <s xml:id="echoid-s16840" xml:space="preserve">Q. </s> <s xml:id="echoid-s16841" xml:space="preserve">F. </s> <s xml:id="echoid-s16842" xml:space="preserve">T. </s> <s xml:id="echoid-s16843" xml:space="preserve">& </s> <s xml:id="echoid-s16844" xml:space="preserve">D.</s> <s xml:id="echoid-s16845" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1367" type="section" level="1" n="1006"> <head xml:id="echoid-head1191" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s16846" xml:space="preserve">1012. </s> <s xml:id="echoid-s16847" xml:space="preserve">Il ſuit delà, comme nous l’avons déja remarqué, que <lb/>le problême aura toujours deux ſolutions, tant que le radical <lb/>√dd+pd+{1/4}pp-aa\x{0020} ſera quelque choſe. </s> <s xml:id="echoid-s16848" xml:space="preserve">2°. </s> <s xml:id="echoid-s16849" xml:space="preserve">Il eſt évident <lb/>que dans le cas où {1/4}pp+pd+dd=aa, le problême ne <lb/>peut avoir qu’une ſolution; </s> <s xml:id="echoid-s16850" xml:space="preserve">il n’eſt pas moins évident que la <lb/>ligne E F devenue pour lors F I touche le cercle au ſeul point I, <lb/>puiſque l’expreſſion dd+pd+{1/4}pp, eſt le quarré de {1/2}p+d, <lb/>qui eſt égale à F I, & </s> <s xml:id="echoid-s16851" xml:space="preserve">qu’il n’ya a que dans le cas où {1/2}p+d=a, <lb/>que cette ligne eſt tangente. </s> <s xml:id="echoid-s16852" xml:space="preserve">Enfin ſi {1/4}pp+pd+dd, eſt plus <lb/>grand que aa, le problême ſera impoſſible, & </s> <s xml:id="echoid-s16853" xml:space="preserve">l’on en con-<lb/>clueroit qu’il faut augmenter la charge du mortier.</s> <s xml:id="echoid-s16854" xml:space="preserve"/> </p> <pb o="532" file="0610" n="630" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1368" type="section" level="1" n="1007"> <head xml:id="echoid-head1192" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s16855" xml:space="preserve">1013. </s> <s xml:id="echoid-s16856" xml:space="preserve">Si la ligne de but G F au lieu d’être au deſſous de <lb/>l’horizon étoit au deſſus, la formule ſerviroit toujours à <lb/>faire connoître les angles d’inclinaiſons demandés; </s> <s xml:id="echoid-s16857" xml:space="preserve">il n’y <lb/>auroit qu’à faire F H = - d, & </s> <s xml:id="echoid-s16858" xml:space="preserve">la formule deviendroit <lb/>x = {1/2}p ± √{1/4}pp-pd+dd-aa\x{0020}, ſur laquelle on feroit les <lb/>mêmes raiſonnemens que ſur la premiere. </s> <s xml:id="echoid-s16859" xml:space="preserve">La conſtruction de <lb/>cette formule revient encore à celle de la figure 344.</s> <s xml:id="echoid-s16860" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1369" type="section" level="1" n="1008"> <head xml:id="echoid-head1193" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s16861" xml:space="preserve">1014. </s> <s xml:id="echoid-s16862" xml:space="preserve">Enfin ſi la ligne de but eſt horizontale, on tirera <lb/>encore de cette formule la conſtruction de la premiere figure, <lb/>en faiſant d=o, d’où l’on tire x={1/2}p± √{1/4}pp-aa\x{0020}. </s> <s xml:id="echoid-s16863" xml:space="preserve">Ainſi <lb/>la formule que nous venons de donner renferme tous les cas <lb/>poſſibles.</s> <s xml:id="echoid-s16864" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1370" type="section" level="1" n="1009"> <head xml:id="echoid-head1194" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s16865" xml:space="preserve">1015. </s> <s xml:id="echoid-s16866" xml:space="preserve">On remarquera encore que dans toutes les poſitions <lb/>poſſibles de la ligne de but avec la ligne horizontale, la plus <lb/>grande partie du jet ſera toujours celle qui eſt déterminée par <lb/> <anchor type="note" xlink:label="note-0610-01a" xlink:href="note-0610-01"/> {1/4}pp=aa, ou par {1/4}pp+dd=aa+pd, ou par aa={1/4}pp <lb/> <anchor type="note" xlink:label="note-0610-02a" xlink:href="note-0610-02"/> +pd+dd, parce que dans tous ces cas la ligne de chûte eſt <lb/> <anchor type="note" xlink:label="note-0610-03a" xlink:href="note-0610-03"/> la tangente de la portion de cercle G I M, & </s> <s xml:id="echoid-s16867" xml:space="preserve">que cette tan-<lb/>gente détermine la plus grande portée du jet.</s> <s xml:id="echoid-s16868" xml:space="preserve"/> </p> <div xml:id="echoid-div1370" type="float" level="2" n="1"> <note position="left" xlink:label="note-0610-01" xlink:href="note-0610-01a" xml:space="preserve">Figure 343.</note> <note position="left" xlink:label="note-0610-02" xlink:href="note-0610-02a" xml:space="preserve">Figure 344.</note> <note position="left" xlink:label="note-0610-03" xlink:href="note-0610-03a" xml:space="preserve">Figure 345.</note> </div> </div> <div xml:id="echoid-div1372" type="section" level="1" n="1010"> <head xml:id="echoid-head1195" xml:space="preserve">PROPOSITION XIV <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16869" xml:space="preserve">1016. </s> <s xml:id="echoid-s16870" xml:space="preserve">Conſtruction d’un inſtrument univerſel pour jetter les <lb/> <anchor type="note" xlink:label="note-0610-04a" xlink:href="note-0610-04"/> bombes ſur toutes ſortes de plans.</s> <s xml:id="echoid-s16871" xml:space="preserve"/> </p> <div xml:id="echoid-div1372" type="float" level="2" n="1"> <note position="left" xlink:label="note-0610-04" xlink:href="note-0610-04a" xml:space="preserve">Figure 346.</note> </div> <p> <s xml:id="echoid-s16872" xml:space="preserve">On fera un cercle de cuivre ou de quelqu’autre matiere ſo-<lb/>lide & </s> <s xml:id="echoid-s16873" xml:space="preserve">polie, & </s> <s xml:id="echoid-s16874" xml:space="preserve">on diviſera ſa circonférence en 360 parties <lb/>égales ou degrés: </s> <s xml:id="echoid-s16875" xml:space="preserve">on appliquera à un de ſes points G une <lb/>regle fixe G N, qui le touche au point G, & </s> <s xml:id="echoid-s16876" xml:space="preserve">qui ſoit égale à <lb/>ſon diametre G B. </s> <s xml:id="echoid-s16877" xml:space="preserve">On diviſera cette regle en un grand nombre <lb/>de parties égales, comme en 200 parties; </s> <s xml:id="echoid-s16878" xml:space="preserve">& </s> <s xml:id="echoid-s16879" xml:space="preserve">on y attachera <lb/>un filet avec un plomb D, enſorte néanmoins que le filet puiſſe <lb/>couler le long de la regle, en s’approchant ou s’éloignant du <lb/>point G. </s> <s xml:id="echoid-s16880" xml:space="preserve">On expliquera l’uſage de cet inſtrument dans les pro-<lb/>blêmes ſuivans.</s> <s xml:id="echoid-s16881" xml:space="preserve"/> </p> <pb o="533" file="0611" n="631" rhead="DE MATHÉMATIQUE. Liv. XIV."/> </div> <div xml:id="echoid-div1374" type="section" level="1" n="1011"> <head xml:id="echoid-head1196" style="it" xml:space="preserve">Uſage de l’inſtrument univerſel pour le jet des bombes.</head> <head xml:id="echoid-head1197" xml:space="preserve">PROPOSITION XV. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16882" xml:space="preserve">1017. </s> <s xml:id="echoid-s16883" xml:space="preserve">Trouver par le moyen de l’inſtrument univerſel, quelle <lb/> <anchor type="note" xlink:label="note-0611-01a" xlink:href="note-0611-01"/> hauteur il faut donner à un mortier pour jetter une bombe à une <lb/>diſtance donnée, ſuppoſant que le lieu où l’on veut la jetter ſoit de <lb/>niveau avec la batterie.</s> <s xml:id="echoid-s16884" xml:space="preserve"/> </p> <div xml:id="echoid-div1374" type="float" level="2" n="1"> <note position="right" xlink:label="note-0611-01" xlink:href="note-0611-01a" xml:space="preserve">Figure 339.</note> </div> <p> <s xml:id="echoid-s16885" xml:space="preserve">Pour réſoudre ce problême, il faut commencer par faire <lb/>une épreuve, en jettant une bombe avec la charge qu’on ſe <lb/>propoſe de tirer, quiſera, par exemple, de deux livres de pou-<lb/>dre; </s> <s xml:id="echoid-s16886" xml:space="preserve">& </s> <s xml:id="echoid-s16887" xml:space="preserve">ſuppoſant que la bombe a été jettée à 400 toiſes, ſous <lb/>un angle que l’on aura pris à volonté, qui ſera, ſi l’on veut <lb/>de 30 degrés, il faut chercher le parametre: </s> <s xml:id="echoid-s16888" xml:space="preserve">ainſi l’angle <lb/>MGE étant de 30 degrés, l’angle GEF ſera auſſi de 30 de-<lb/>grés, parce que la ligne de chûte E F eſt parallele au parametre <lb/>MG: </s> <s xml:id="echoid-s16889" xml:space="preserve">& </s> <s xml:id="echoid-s16890" xml:space="preserve">comme l’angle E G F eſt de 60 degrés, & </s> <s xml:id="echoid-s16891" xml:space="preserve">qu’on con-<lb/>noît la ligne F G de 400 toiſes, l’on trouvera, par la Trigo-<lb/>nométrie, que la ligne de chûte E F eſt de 693 toiſes, & </s> <s xml:id="echoid-s16892" xml:space="preserve">que la <lb/>ligne de projection G E eſt de 800 toiſes. </s> <s xml:id="echoid-s16893" xml:space="preserve">Or cherchant une <lb/>troiſieme proportionnelle à 693 & </s> <s xml:id="echoid-s16894" xml:space="preserve">à 800 toiſes, l’on trouvera <lb/>qu’elle eſt de 923 toiſes, qui eſt la valeur du parametre G M.</s> <s xml:id="echoid-s16895" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16896" xml:space="preserve">Cela poſé, ſi l’on veut ſçavoir à quels degrés d’élevation <lb/> <anchor type="note" xlink:label="note-0611-02a" xlink:href="note-0611-02"/> il faut pointer le mortier pour chaſſer une bombe à 250 toiſes <lb/>avec une charge de deux livres de poudre, il faut faire une <lb/>regle de Trois, en diſant: </s> <s xml:id="echoid-s16897" xml:space="preserve">Si 923 toiſes, valeur du parametre, <lb/>donnent 250 toiſes pour la diſtance donnée, combien donne-<lb/>ront 200, valeur du diametre de l’inſtrument, c’eſt-à-dire <lb/>valeur de la ligne NG, pour le nombre de ſes parties que je <lb/>cherche, qu’on trouvera de 54.</s> <s xml:id="echoid-s16898" xml:space="preserve"/> </p> <div xml:id="echoid-div1375" type="float" level="2" n="2"> <note position="right" xlink:label="note-0611-02" xlink:href="note-0611-02a" xml:space="preserve">Figure 346.</note> </div> <p> <s xml:id="echoid-s16899" xml:space="preserve">Préſentement il faut mettre la regle N G parfaitement de <lb/>niveau, & </s> <s xml:id="echoid-s16900" xml:space="preserve">faire gliſſer le filet K D juſqu’au nombre 54, & </s> <s xml:id="echoid-s16901" xml:space="preserve">le <lb/>filet venant à couper la circonférence du cercle de l’inſtru-<lb/>ment aux deux endroits C, marquera que le problême a deux <lb/>ſolutions, & </s> <s xml:id="echoid-s16902" xml:space="preserve">qu’il doit être pointé ſous un angle moitié du <lb/>nombre des degrés compris dans les arcs G C. </s> <s xml:id="echoid-s16903" xml:space="preserve">Or comme le <lb/>plus grand eſt de 148 degrés, & </s> <s xml:id="echoid-s16904" xml:space="preserve">que le plus petit eſt de 32 de-<lb/>grés, prenant leurs moitiés, qui ſont 74 & </s> <s xml:id="echoid-s16905" xml:space="preserve">16, le mottier <pb o="534" file="0612" n="632" rhead="NOUVEAU COURS"/> pointé à l’une ou l’autre de ces élevations, chaſſera la bombe <lb/>à la diſtance propoſée.</s> <s xml:id="echoid-s16906" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1377" type="section" level="1" n="1012"> <head xml:id="echoid-head1198" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16907" xml:space="preserve">Pour faciliter la démonſtration de la pratique précédente, <lb/>nous ſuppoſerons que la ligne G F eſt la diſtance donnée, c’eſt-<lb/>à-dire qu’elle vaut 250 toiſes, & </s> <s xml:id="echoid-s16908" xml:space="preserve">que la perpendiculaire G M <lb/>eſt le parametre que l’on a trouvé. </s> <s xml:id="echoid-s16909" xml:space="preserve">Or ſi l’on décrit un demi-<lb/> <anchor type="note" xlink:label="note-0612-01a" xlink:href="note-0612-01"/> cercle M E G, & </s> <s xml:id="echoid-s16910" xml:space="preserve">que l’on mene la ligne F E parallele à G M, <lb/>& </s> <s xml:id="echoid-s16911" xml:space="preserve">que l’on tire les lignes G E aux points où cette parallele <lb/>coupe le cercle, l’on aura les angles M G E de l’élévation du <lb/>mortier, pour jetter la bombe au point F, comme on l’a dé-<lb/>montré ci-devant (art. </s> <s xml:id="echoid-s16912" xml:space="preserve">1000). </s> <s xml:id="echoid-s16913" xml:space="preserve">Préſentement ſi l’on imagine <lb/>que la regle N G de l’inſtrument ſoit miſe d’alignement avec <lb/>la ligne de but G F, & </s> <s xml:id="echoid-s16914" xml:space="preserve">que les diametres M G & </s> <s xml:id="echoid-s16915" xml:space="preserve">G B ſoient <lb/>auſſi d’alignement, & </s> <s xml:id="echoid-s16916" xml:space="preserve">que le filet K D ſoit encore à l’endroit <lb/>où on l’a poſé, c’eſt-à-dire au point 54, l’on aura, ſelon la <lb/>pratique du problême, G M : </s> <s xml:id="echoid-s16917" xml:space="preserve">G F :</s> <s xml:id="echoid-s16918" xml:space="preserve">: G B : </s> <s xml:id="echoid-s16919" xml:space="preserve">G K, parce qu’on <lb/>peut prendre ici le diametre G B pour la longueur de la regle <lb/>G N, ces deux lignes étant égales. </s> <s xml:id="echoid-s16920" xml:space="preserve">Cela étant, à cauſe de la <lb/>proportion, la perpendiculaire K D coupera le demi-cercle <lb/>G C B, de la même façon que la perpendiculaire F E coupe le <lb/>demi-cercle M E G: </s> <s xml:id="echoid-s16921" xml:space="preserve">ainſi les lignes E G & </s> <s xml:id="echoid-s16922" xml:space="preserve">G C n’en faiſant <lb/>qu’une ſeule E C, comme les lignes M G & </s> <s xml:id="echoid-s16923" xml:space="preserve">G B, l’arc M E <lb/>ſera égal à l’arc C B ou G C, qui eſt la même choſe: </s> <s xml:id="echoid-s16924" xml:space="preserve">ainſi ces <lb/>arcs ſeront de 32 degrés; </s> <s xml:id="echoid-s16925" xml:space="preserve">& </s> <s xml:id="echoid-s16926" xml:space="preserve">comme l’angle M G E n’a pour <lb/>meſure que la moitié de l’arc M E, il ne vaudra que 16 degrés, <lb/>qui eſt l’élevation qu’il faudra donner au mortier, ſi l’on veut <lb/>pointer au deſſous de 45 degrés: </s> <s xml:id="echoid-s16927" xml:space="preserve">ainſi l’on voit que l’on trouve, <lb/>par le moyen de l’inſtrument, les mêmes choſes que l’on a <lb/>trouvées ci-devant (art. </s> <s xml:id="echoid-s16928" xml:space="preserve">1008) avecle demi-cercle M E G. <lb/></s> <s xml:id="echoid-s16929" xml:space="preserve">C. </s> <s xml:id="echoid-s16930" xml:space="preserve">Q. </s> <s xml:id="echoid-s16931" xml:space="preserve">F. </s> <s xml:id="echoid-s16932" xml:space="preserve">D.</s> <s xml:id="echoid-s16933" xml:space="preserve"/> </p> <div xml:id="echoid-div1377" type="float" level="2" n="1"> <note position="left" xlink:label="note-0612-01" xlink:href="note-0612-01a" xml:space="preserve">Figure 347.</note> </div> </div> <div xml:id="echoid-div1379" type="section" level="1" n="1013"> <head xml:id="echoid-head1199" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s16934" xml:space="preserve">1018. </s> <s xml:id="echoid-s16935" xml:space="preserve">Il ſuit delà que lorſque le filet K D, au lieu de cou-<lb/> <anchor type="note" xlink:label="note-0612-02a" xlink:href="note-0612-02"/> per le demi-cercle G C B, ne fait que le toucher en C, le <lb/>mortier pointé ſous la moitié de l’arc G C, qui eſt 45 degrés, <lb/>chaſſera la bombe le plus loin qu’il eſt poſſible avec la même <lb/>charge, puiſque pour lors la ligne E F touchera auſſi le demi-<lb/>cercle M E G; </s> <s xml:id="echoid-s16936" xml:space="preserve">enfin que ſi le filet K D ne touchoit ni ne cou- <pb o="535" file="0613" n="633" rhead="DE MATHÉMATIQUE. Liv. XIV."/> poit le cercle, le problême ſera impoſſible; </s> <s xml:id="echoid-s16937" xml:space="preserve">puiſque dans ce <lb/>cas la ligne E F ne peut pas toucher non plus, ni couper le <lb/>demi-cercle M E G.</s> <s xml:id="echoid-s16938" xml:space="preserve"/> </p> <div xml:id="echoid-div1379" type="float" level="2" n="1"> <note position="left" xlink:label="note-0612-02" xlink:href="note-0612-02a" xml:space="preserve">Figure 348.</note> </div> </div> <div xml:id="echoid-div1381" type="section" level="1" n="1014"> <head xml:id="echoid-head1200" xml:space="preserve">PROPOSITION XVI. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s16939" xml:space="preserve">1019. </s> <s xml:id="echoid-s16940" xml:space="preserve">Trouver quelle élevation il faut donner au mortier pour <lb/> <anchor type="note" xlink:label="note-0613-01a" xlink:href="note-0613-01"/> chaſſer une bombe à une diſtance donnée, ſuppoſant que l’endroit <lb/>où l’on veut jetter la bombe ſoit plus haut ou plus has que la bat-<lb/>terie, & </s> <s xml:id="echoid-s16941" xml:space="preserve">cela en ſe ſervant de l’inſtrument univerſel.</s> <s xml:id="echoid-s16942" xml:space="preserve"/> </p> <div xml:id="echoid-div1381" type="float" level="2" n="1"> <note position="right" xlink:label="note-0613-01" xlink:href="note-0613-01a" xml:space="preserve">Figure 349 <lb/>& 350.</note> </div> <p> <s xml:id="echoid-s16943" xml:space="preserve">Suppoſant que de l’endroit G, où ſeroit une batterie de <lb/>mortiers, on veuille jetter des bombes à l’endroit F beaucoup <lb/>plus élevé ou plus bas que le plan de la batterie, il faut com-<lb/>commencer par chercher, en ſe ſervant de la Trigonométrie, <lb/>la diſtance horizontale G H, qui eſt l’amplitude de la para-<lb/>bole; </s> <s xml:id="echoid-s16944" xml:space="preserve">& </s> <s xml:id="echoid-s16945" xml:space="preserve">connoiſſant le parametre de la charge dont on vou-<lb/>dra ſe ſervir, que je ſuppoſe être le même que celui du pro-<lb/>blême précédent, c’eſt-à-dire de 923 toiſes, la charge étant <lb/>encore de deux livres de poudre, l’on dira: </s> <s xml:id="echoid-s16946" xml:space="preserve">comme le para-<lb/>metre eſt à la diſtance G H, ainſi la longueur G N de la regle <lb/>diviſée en 200 parties eſt à la longueur G K, qui donnera un <lb/>nombre de ces parties. </s> <s xml:id="echoid-s16947" xml:space="preserve">Or ſuppoſant qu’on a trouvé 60 parties, <lb/>l’on fera gliſſer le filet K D ſur le nombre 60, où il faudra le <lb/>tenir fixe; </s> <s xml:id="echoid-s16948" xml:space="preserve">enſuite on appuyera le cercle de l’inſtrument ſur un <lb/>endroit où il puiſſe être ſtable; </s> <s xml:id="echoid-s16949" xml:space="preserve">& </s> <s xml:id="echoid-s16950" xml:space="preserve">l’ayant mis bien verticale-<lb/>ment, on viſera le long de la regle N G le lieu donné F, & </s> <s xml:id="echoid-s16951" xml:space="preserve"><lb/>le filet K D coupera le cercle aux points C, où il déterminera <lb/>les arcs C G: </s> <s xml:id="echoid-s16952" xml:space="preserve">& </s> <s xml:id="echoid-s16953" xml:space="preserve">ſi l’on prend la moitié du nombre des degrés <lb/>contenus dans l’un ou l’autre de ces arcs, l’on aura la valeur <lb/>de l’angle que doit faire le mortier avec la verticale pour jetter <lb/>la bombe au point F.</s> <s xml:id="echoid-s16954" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1383" type="section" level="1" n="1015"> <head xml:id="echoid-head1201" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16955" xml:space="preserve">Ayant élevé ſur la ligne horizontale G H la perpendicu-<lb/> <anchor type="note" xlink:label="note-0613-02a" xlink:href="note-0613-02"/> laire G M égale au parametre, & </s> <s xml:id="echoid-s16956" xml:space="preserve">ſur le plan G F la perpendi-<lb/>culaire G A, on fera l’angle A M G égal à l’angle A G M, & </s> <s xml:id="echoid-s16957" xml:space="preserve"><lb/>du point A on décrira une portion de cercle M E G, & </s> <s xml:id="echoid-s16958" xml:space="preserve">du <lb/>point F on menera la ligne F E parallele à G M, qui coupera <lb/>le cercle aux points E, auxquels menant les lignes G E, l’on <pb o="536" file="0614" n="634" rhead="NOUVEAU COURS"/> aura les directions G E qu’il faut donner au mortier pour jetter <lb/>une bombe à l’endroit F (art. </s> <s xml:id="echoid-s16959" xml:space="preserve">1008). </s> <s xml:id="echoid-s16960" xml:space="preserve">Or ſi on place l’inſtru-<lb/>ment de maniere que la regle N G ſoit d’alignement avec le <lb/>diametre G O, & </s> <s xml:id="echoid-s16961" xml:space="preserve">que le filet K D ſoit toujours à l’endroit où <lb/>on l’a poſé dans l’opération, l’on verra que le demi-cercle <lb/>G C B eſt coupé par la perpendiculaire K D de la même façon <lb/>que le demi-cercle O E G eſt coupé par la perpendiculaire E F; <lb/></s> <s xml:id="echoid-s16962" xml:space="preserve">ce qui ſe prouve aſſez de ſoi-même, ſans qu’il ſoit beſoin de <lb/>répéter ce qui a déja été dit ailleurs à ce ſujet.</s> <s xml:id="echoid-s16963" xml:space="preserve"/> </p> <div xml:id="echoid-div1383" type="float" level="2" n="1"> <note position="right" xlink:label="note-0613-02" xlink:href="note-0613-02a" xml:space="preserve">Figure 351 <lb/>& 352.</note> </div> </div> <div xml:id="echoid-div1385" type="section" level="1" n="1016"> <head xml:id="echoid-head1202" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s16964" xml:space="preserve">Comme l’on peut ſe ſervir de la Trigonométrie pour jetter <lb/>des bombes par une méthode toute différente de celle que nous <lb/>venons d’enſeigner, voici deux propoſitions dont on pourra <lb/>faire uſage dans les occaſions où l’on n’auroit pas d’inſtrumens <lb/>tel que celui dont nous venons de parler; </s> <s xml:id="echoid-s16965" xml:space="preserve">il eſt vrai que tout <lb/>ce que nous allons enſeigner ne peut avoir lieu que lorſque <lb/>l’objet où l’on veut jetter les bombes eſt de niveau avec la bat-<lb/>terie; </s> <s xml:id="echoid-s16966" xml:space="preserve">mais comme cela ſe rencontre preſque toujours, je ne <lb/>me ſuis pas ſoucié de donner une méthode pour en jetter dans <lb/>un lieu qui ſeroit plus bas ou plus haut que la batterie, parce <lb/>que les opérations m’ont paru trop longues par la Trigono-<lb/>métrie. </s> <s xml:id="echoid-s16967" xml:space="preserve">Il faut remarquer que nous allons ſuppoſer dans les <lb/>propoſitions ſuivantes, que le mortier fait ſon angle d’éléva-<lb/>tion avec la ligne horizontale, quoique dans la pratique l’on <lb/>pourra, ſi l’on veut, le former avec la verticale.</s> <s xml:id="echoid-s16968" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1386" type="section" level="1" n="1017"> <head xml:id="echoid-head1203" xml:space="preserve">PROPOSITION XVII. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s16969" xml:space="preserve">1020. </s> <s xml:id="echoid-s16970" xml:space="preserve">Si l’on tire deux bombes avec la même charge à diffé-<lb/>rentes élévations de mortier, je dis que la portée de la premiere <lb/>bombe ſera à celle de la ſeconde, comme le ſinus d’un angle double <lb/>de l’élévation du mortier pour la premiere bombe, eſt au ſinus de <lb/>l’angle double de l’élévation pour la ſeconde.</s> <s xml:id="echoid-s16971" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s16972" xml:space="preserve">Ayant élevé ſur l’extrêmité B de la ligne horizontale B P <lb/> <anchor type="note" xlink:label="note-0614-01a" xlink:href="note-0614-01"/> une perpendiculaire B N à volonté, on la diviſera en deux éga-<lb/>lement au point M, pour décrire le demi-cercle N G B; </s> <s xml:id="echoid-s16973" xml:space="preserve">en-<lb/>ſuite ayant tiré les lignes B G & </s> <s xml:id="echoid-s16974" xml:space="preserve">B K, pour marquer les deux <lb/>inclinaiſons différentes du mortier, on les prolongera de ma- <pb o="537" file="0615" n="635" rhead="DE MATHÉMATIQUE. Liv. XIV."/> niere que K A ſoit égal à K B, & </s> <s xml:id="echoid-s16975" xml:space="preserve">que G D ſoit égal à B G, & </s> <s xml:id="echoid-s16976" xml:space="preserve"><lb/>des extrêmités A & </s> <s xml:id="echoid-s16977" xml:space="preserve">D, l’on abaiſſera les perpendiculaires A C <lb/>& </s> <s xml:id="echoid-s16978" xml:space="preserve">D E ſur la ligne horizontale B P; </s> <s xml:id="echoid-s16979" xml:space="preserve">enſuite ſi par le point K <lb/>l’on mene la ligne I L parallele à B C, l’on aura I K égal à K L, <lb/>& </s> <s xml:id="echoid-s16980" xml:space="preserve">A L égal à L C, à cauſe des paralleles I B & </s> <s xml:id="echoid-s16981" xml:space="preserve">A C: </s> <s xml:id="echoid-s16982" xml:space="preserve">ainſi I K <lb/>ſera moitié de B C; </s> <s xml:id="echoid-s16983" xml:space="preserve">& </s> <s xml:id="echoid-s16984" xml:space="preserve">menant auſſi par le point G la ligne <lb/>F H parallele à B E, l’on aura encore F G égal à G H, & </s> <s xml:id="echoid-s16985" xml:space="preserve">par <lb/>conſéquent F G ſera la moitié de B E.</s> <s xml:id="echoid-s16986" xml:space="preserve"/> </p> <div xml:id="echoid-div1386" type="float" level="2" n="1"> <note position="left" xlink:label="note-0614-01" xlink:href="note-0614-01a" xml:space="preserve">Figure 353.</note> </div> </div> <div xml:id="echoid-div1388" type="section" level="1" n="1018"> <head xml:id="echoid-head1204" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s16987" xml:space="preserve">Conſidérez que l’angle D B E ayant pour meſure la moitié <lb/>de l’arc G O B, la ligne G F étant le ſinus de l’angle G M B, <lb/>elle ſera le ſinus d’un angle double de l’angle D B E, & </s> <s xml:id="echoid-s16988" xml:space="preserve">que de <lb/>même l’angle A B C ayant pour meſure la moitié de l’arc <lb/>K G B, la ligne K I étant le ſinus de cet arc, ou bien de ſon <lb/>complément, qui eſt la même choſe, elle ſera le ſinus d’un <lb/>angle double de l’angle A B C. </s> <s xml:id="echoid-s16989" xml:space="preserve">Or la ligne B C étant double de <lb/>I K, & </s> <s xml:id="echoid-s16990" xml:space="preserve">la ligne B E double de F G, l’on aura donc B C: </s> <s xml:id="echoid-s16991" xml:space="preserve">B E <lb/>:</s> <s xml:id="echoid-s16992" xml:space="preserve">: IK: </s> <s xml:id="echoid-s16993" xml:space="preserve">F G. </s> <s xml:id="echoid-s16994" xml:space="preserve">Mais ſi à la place des demi-amplitudes B C & </s> <s xml:id="echoid-s16995" xml:space="preserve">B E, <lb/>l’on prend les amplitudes entieres B Q & </s> <s xml:id="echoid-s16996" xml:space="preserve">B P, c’eſt-à-dire la <lb/>portée entiere de chaque bombe, l’on aura comme B Q, portée <lb/>de la premiere bombe, eſt à B P portée de la ſeconde, ainſi <lb/>I K, ſinus de l’angle double de l’élévation de la premiere, eſt <lb/>à F G, ſinus de l’angle double de l’élévation de la ſeconde. <lb/></s> <s xml:id="echoid-s16997" xml:space="preserve">C. </s> <s xml:id="echoid-s16998" xml:space="preserve">Q. </s> <s xml:id="echoid-s16999" xml:space="preserve">F. </s> <s xml:id="echoid-s17000" xml:space="preserve">D.</s> <s xml:id="echoid-s17001" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1389" type="section" level="1" n="1019"> <head xml:id="echoid-head1205" xml:space="preserve"><emph style="sc">Application</emph>.</head> <p> <s xml:id="echoid-s17002" xml:space="preserve">1021. </s> <s xml:id="echoid-s17003" xml:space="preserve">Pourtirer des bombes avec une même charge à quelle <lb/>diſtance l’on voudra, il faut commencer par faire une épreuve: <lb/></s> <s xml:id="echoid-s17004" xml:space="preserve">cette épreuve ſe fera, par exemple, en chargeant le mortier à <lb/>deux livres de poudre, & </s> <s xml:id="echoid-s17005" xml:space="preserve">en le pointant à 45 degrés, qui eſt <lb/>l’élévation où le mortier chaſſera le plus loin avec cette charge, <lb/>comme nous l’avons déja dit: </s> <s xml:id="echoid-s17006" xml:space="preserve">après avoir tiré la bombe, on <lb/>meſurera exactement la diſtance du mortier à l’endroit où elle <lb/>ſera tombée, que je ſuppoſe qu’on aura trouvée de 800 toiſes. </s> <s xml:id="echoid-s17007" xml:space="preserve"><lb/>Cela étant fait, ſi l’on veut ſçavoir quelle élévation il faut <lb/>donner à un mortier pour envoyer une bombe à 500 toiſes, <lb/>pour la trouver il faut faire une Regle de Trois, dont le pre-<lb/>mier terme ſoit 800 toiſes, qui eſt la diſtance connue, le ſe-<lb/>cond 500 toiſes, qui eſt la diſtance où l’on veut envoyer la <pb o="538" file="0616" n="636" rhead="NOUVEAU COURS"/> bombe, le troiſieme le ſinus d’un angle double de 45 degrés, <lb/>qui eſt 100000. </s> <s xml:id="echoid-s17008" xml:space="preserve">La regle étant faite, l’on trouvera 62500, <lb/>qui eſt le ſinus d’un angle double de celui que l’on cherche: <lb/></s> <s xml:id="echoid-s17009" xml:space="preserve">après l’avoir trouvé dans la Table, l’on verra qu’il correſpond <lb/>à 38 degrés 40 minutes, dont la moitié eſt 19 degrés 20 mi-<lb/>nutes, qui eſt la valeur de l’angle que doit faire le mortier <lb/>avec l’horizon pour jetter une bombe à 500 toiſes.</s> <s xml:id="echoid-s17010" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1390" type="section" level="1" n="1020"> <head xml:id="echoid-head1206" xml:space="preserve">PROPOSITION XVIII. <lb/><emph style="sc">Théoreme.</emph></head> <p style="it"> <s xml:id="echoid-s17011" xml:space="preserve">1022. </s> <s xml:id="echoid-s17012" xml:space="preserve">Si l’on tire deux bombes à différens degrés d’élévations <lb/>avec la même charge, il y aura même raiſon du ſinus de l’angle <lb/>double de la premiere élévation au ſinus du double de la ſeconde, <lb/>que de la portée de la premiere élévation à la portée de la ſeconde.</s> <s xml:id="echoid-s17013" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1391" type="section" level="1" n="1021"> <head xml:id="echoid-head1207" xml:space="preserve"><emph style="sc">DÉMONSTRATION.</emph></head> <p> <s xml:id="echoid-s17014" xml:space="preserve">L’angle A B C étant celui de la premiere élévation du mor-<lb/> <anchor type="note" xlink:label="note-0616-01a" xlink:href="note-0616-01"/> tier, & </s> <s xml:id="echoid-s17015" xml:space="preserve">l’angle D B E celui de la ſeconde, l’on aura encore <lb/>I K : </s> <s xml:id="echoid-s17016" xml:space="preserve">F G :</s> <s xml:id="echoid-s17017" xml:space="preserve">: B C : </s> <s xml:id="echoid-s17018" xml:space="preserve">B E, ou bien I K : </s> <s xml:id="echoid-s17019" xml:space="preserve">F G :</s> <s xml:id="echoid-s17020" xml:space="preserve">: B Q : </s> <s xml:id="echoid-s17021" xml:space="preserve">B P, qui fait <lb/>voir que I K, ſinus d’un angle double de l’angle A B C, eſt à <lb/>la ligne F G, ſinus d’un angle double de l’angle D B E, comme <lb/>la premiere portée B Q eſt à la ſeconde B P.</s> <s xml:id="echoid-s17022" xml:space="preserve"/> </p> <div xml:id="echoid-div1391" type="float" level="2" n="1"> <note position="left" xlink:label="note-0616-01" xlink:href="note-0616-01a" xml:space="preserve">Figure 353.</note> </div> </div> <div xml:id="echoid-div1393" type="section" level="1" n="1022"> <head xml:id="echoid-head1208" xml:space="preserve"><emph style="sc">Application.</emph></head> <p> <s xml:id="echoid-s17023" xml:space="preserve">1023. </s> <s xml:id="echoid-s17024" xml:space="preserve">On peut, par le moyen de cette propoſition, ſçavoir à <lb/>quelle diſtance du mortier une bombe ira tomber, ayant fait <lb/>une épreuve comme nous l’avons dit ci-devant.</s> <s xml:id="echoid-s17025" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17026" xml:space="preserve">Suppoſons donc qu’une bombe a été tirée par un angle de <lb/>40 degrés, & </s> <s xml:id="echoid-s17027" xml:space="preserve">qu’elle ait été chaſſée à 1000 toiſes avec une cer-<lb/>taine charge, on demande à quelle diſtance ira la bombe avec <lb/>la même charge, le mortier étant pointé à 25 degrés, il faut <lb/>faire une Regle de Trois, dont le premier terme ſoit le ſinus <lb/>d’un angle double de 40 degrés, c’eſt-à-dire le ſinus de 80 de-<lb/>grés, qui eſt 98480, & </s> <s xml:id="echoid-s17028" xml:space="preserve">le fecond le ſinus d’un angle double de ce-<lb/>lui qu’on veut donner au mortier; </s> <s xml:id="echoid-s17029" xml:space="preserve">comme cet angle a été pro-<lb/>poſé de 25 degrés, on prendra donc le ſinus de 50 degrés, qui <lb/>eſt 76604, & </s> <s xml:id="echoid-s17030" xml:space="preserve">le troiſieme terme la diſtance où la bombe a été <lb/>chaſſée à 40 degrés, que nous avons ſuppoſé de 1000 toiſes, <lb/>la regle étant faite, l’on trouvera pour quatrieme terme 777 <pb o="539" file="0617" n="637" rhead="DE MATHÉMATIQUE. Liv. XIV."/> <lb/>toiſes, qui eſt la diſtance du mortier à l’endroit où tombera <lb/>la bombe, ayant été tirée ſous un angle de 25 degrés.</s> <s xml:id="echoid-s17031" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1394" type="section" level="1" n="1023"> <head xml:id="echoid-head1209" xml:space="preserve">PROPOSITION XIX. <lb/><emph style="sc">Probleme.</emph></head> <p style="it"> <s xml:id="echoid-s17032" xml:space="preserve">1024. </s> <s xml:id="echoid-s17033" xml:space="preserve">Connoiſſant l’amplitude d’une parabole décrite par une <lb/> <anchor type="note" xlink:label="note-0617-01a" xlink:href="note-0617-01"/> bombe, ſcavoir quelle eſt la hauteur où la bombe s’eſt élevée au <lb/>deſſus de l’horizon.</s> <s xml:id="echoid-s17034" xml:space="preserve"/> </p> <div xml:id="echoid-div1394" type="float" level="2" n="1"> <note position="right" xlink:label="note-0617-01" xlink:href="note-0617-01a" xml:space="preserve">Figure 353.</note> </div> <p> <s xml:id="echoid-s17035" xml:space="preserve">Nous ſervant de la figure précédente, où l’on a ſuppoſé <lb/>que la ligne B A marquoit l’élévation du mortier, l’on peut <lb/>dire que cette ligne eſt la tangente de la parabole B L Q; </s> <s xml:id="echoid-s17036" xml:space="preserve">& </s> <s xml:id="echoid-s17037" xml:space="preserve"><lb/>qu’ainſi la ſoutagente A C ſera double de l’abſciſſe L C (art. </s> <s xml:id="echoid-s17038" xml:space="preserve">614), <lb/>qui eſt ici la hauteur où la bombe aura été élevée ſous l’angle <lb/>A B C. </s> <s xml:id="echoid-s17039" xml:space="preserve">Suppoſant cet angle de 70 degrés, l’amplitude B Q <lb/>de 300 toiſes, la demi-amplitude B C ſera de 150 toiſes: </s> <s xml:id="echoid-s17040" xml:space="preserve">ainſi <lb/>dans le triangle A B C l’on connoît l’angle A B C de 70 degrés, <lb/>le côté B C de 150 toiſes, & </s> <s xml:id="echoid-s17041" xml:space="preserve">l’angle droit B C A: </s> <s xml:id="echoid-s17042" xml:space="preserve">ainſi par le <lb/>calcul ordinaire de la Trigonometrie, l’on trouvera le côté <lb/>A C de 412 toiſes, dont la moitié, qui eſt 206 toiſes, ſera la <lb/>valeur de la ligne L C, c’eſt-à-dire la hauteur où la bombe ſe <lb/>ſera élevée.</s> <s xml:id="echoid-s17043" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1396" type="section" level="1" n="1024"> <head xml:id="echoid-head1210" xml:space="preserve">PROPOSITION XX. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s17044" xml:space="preserve">1025. </s> <s xml:id="echoid-s17045" xml:space="preserve">Connoiſſant la hauteur où une bombe s’eſt élevée, ſçavoir <lb/>la peſanteur ou le degré de mouvement qu’elle a acquis en tombant <lb/>par ſon mouvement accéléré.</s> <s xml:id="echoid-s17046" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17047" xml:space="preserve">Suppoſant qu’une bombe de 12 pouces ſoit tombée de 206 <lb/>toiſes de hauteur, ſa vîteſſe ſera exprimée par la racine quarrée <lb/>de ſa chûte (art. </s> <s xml:id="echoid-s17048" xml:space="preserve">959), c’eſt-à-dire par la racine quarrée de <lb/>206, qui eſt 14 {1/3}. </s> <s xml:id="echoid-s17049" xml:space="preserve">Cela poſé, l’on ſçait que la force ou la quan-<lb/>tité du mouvement d’un corps, eſt le produit de ſa maſſe par <lb/>ſa vîteſſe (art. </s> <s xml:id="echoid-s17050" xml:space="preserve">931). </s> <s xml:id="echoid-s17051" xml:space="preserve">Or comme les bombes de 12 pouces <lb/>peſent environ 140 livres, l’on peut regarder cette quantité <lb/>comme la valeur de la maſſe, qui étant multipliée par la <lb/>vîteſſe, qui eſt 14 {1/3}, donnera 2006 pour la quantité de mou-<lb/>vemens, ou la force de la bombe.</s> <s xml:id="echoid-s17052" xml:space="preserve"/> </p> <pb o="540" file="0618" n="638" rhead="NOUVEAU COURS DE MATH.LIV. XIV."/> </div> <div xml:id="echoid-div1397" type="section" level="1" n="1025"> <head xml:id="echoid-head1211" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s17053" xml:space="preserve">1026. </s> <s xml:id="echoid-s17054" xml:space="preserve">Par les deux problêmes précédens, l’on voit qu’il eſt <lb/>facile de ſçavoir la force des bombes qui ſont chaſſées ſous dif-<lb/>férens degrés d’élévations, puiſque connoiſſant leurs ampli-<lb/>tudes, on connoîtra les hauteurs où elles ſe ſont élevées, & </s> <s xml:id="echoid-s17055" xml:space="preserve"><lb/>par conſéquent leur vîteſſe, qu’il ne faudra que multiplier par <lb/>la peſanteur des bombes de mêmes ou de différens calibres, <lb/>pour avoir des produits, dont les rapports ſeront les mêmes <lb/>que ceux des forces que les bombes auront acquiſes en tom-<lb/>bant. </s> <s xml:id="echoid-s17056" xml:space="preserve">Ainſi l’on peut ſçavoir quel degré d’élévation il fau-<lb/>droit donner à un mortier de 8 pouces, pour que la bombe de <lb/>ſon calibre tombant ſur un éaifice, comme, par exemple, ſur <lb/>un magaſin à poudre, fît autant de dommage qu’une bombe <lb/>de 12 pouces, qui auroit été jettée ſous un angle de direction <lb/>moindre que celui de la bombe de 8 pouces, cette derniere <lb/>devant acquérir, par la hauteur de ſa chûte, ce qu’elle a de <lb/>moins en peſanteur que celle de 12 pouces. </s> <s xml:id="echoid-s17057" xml:space="preserve">Ceci eſt non ſeu-<lb/>lement curieux, mais peut encore avoir ſon utilité dans l’at-<lb/>taque des places.</s> <s xml:id="echoid-s17058" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1398" type="section" level="1" n="1026"> <head xml:id="echoid-head1212" style="it" xml:space="preserve">Fin du quatorzieme Livre.</head> <figure> <image file="0618-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0618-01"/> </figure> <pb o="541" file="0619" n="639"/> <figure> <image file="0619-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0619-01"/> </figure> </div> <div xml:id="echoid-div1399" type="section" level="1" n="1027"> <head xml:id="echoid-head1213" xml:space="preserve">NOUVEAU COURS DE MATHÉMATIQUE.</head> <head xml:id="echoid-head1214" xml:space="preserve">LIVRE QUINZIEME, <lb/>Qui traite de la Méchanique Statique.</head> <p style="it"> <s xml:id="echoid-s17059" xml:space="preserve">DE toutes les parties des Mathématiques néceſſaires à un Ingé-<lb/>nieur, après les élémens de Géométrie, il n’y en a pas de plus <lb/>importante que celle que nous allons traiter. </s> <s xml:id="echoid-s17060" xml:space="preserve">On l’appelle mécha-<lb/>nique ſtatique, parce que nous y conſidérons les machines en <lb/>repos, ou plutôt en équilibre avec les fardeaux ou les poids qu’on <lb/>veut enlever par leur moyen. </s> <s xml:id="echoid-s17061" xml:space="preserve">On peut aiſément ſe convaincre de <lb/>l’avantage qu’il y a de conſidérer ainſi les machines ſimples ou <lb/>compoſées: </s> <s xml:id="echoid-s17062" xml:space="preserve">car ſi l’on connoît la force qui eſt capable de faire équi-<lb/>libre avec les puiſſances qui leur ſont appliquées, on ſçait, dès-là <lb/>même, celles qui ſont capables de les ſurmonter, en cas qu’il faille <lb/>vaincre l’équilibre. </s> <s xml:id="echoid-s17063" xml:space="preserve">En un mot, on eſt en état d’apprécier les forces <lb/>par les réſiſtances qu’elles ont à vaincre; </s> <s xml:id="echoid-s17064" xml:space="preserve">de déterminer les ſitua-<lb/>tions & </s> <s xml:id="echoid-s17065" xml:space="preserve">les directions les plus avantageuſes, ſuivant leſquelles on <lb/>doit appliquer les forces motrices aux machines dont on fait uſage. <lb/></s> <s xml:id="echoid-s17066" xml:space="preserve">Tout nous invite à découvrir les principes des effets que nous <lb/>voyons exécuter tous les jours. </s> <s xml:id="echoid-s17067" xml:space="preserve">N’y eût-il que la curioſité, on ne <lb/>peut s’empêcher de voir avec étonnement un homme, dont la force <lb/>ordinaire eſt très-petite, faire équilibre à l’aide d’une ſimple ma-<lb/>chine, avec des fardeaux de pluſieurs milliers, & </s> <s xml:id="echoid-s17068" xml:space="preserve">ſouvent les mettre <lb/>en mouvement. </s> <s xml:id="echoid-s17069" xml:space="preserve">Bientôt, lorſqu’on a reconnu les vraies cauſes <lb/>d’effets auſſi ſurprenans, on devient, pour ainſi dire, maître des <pb o="542" file="0620" n="640" rhead="NOUVEAU COURS"/> mouvemens, à l’aide de la théorie établie ſur ces mêmes principes: <lb/></s> <s xml:id="echoid-s17070" xml:space="preserve">leur combinaiſon nous découvre une infinité d’avantages particu-<lb/>liers, applicables aux Arts & </s> <s xml:id="echoid-s17071" xml:space="preserve">aux différentes ſituations dans leſ-<lb/>quelles on peut ſe trouver. </s> <s xml:id="echoid-s17072" xml:space="preserve">Quoique le génie de la méchanique, <lb/>ainſi que les autres talens, ſoit un don particulier, qui ſemble d’a-<lb/>bord dépendre beaucoup plus d’une heureuſe diſpoſition des organes <lb/>qui nous rend inventifs, que des regles générales, il faut cependant <lb/>regarder comme une vérité inconteſtable, que toutes choſes égales <lb/>d’ailleurs, celui qui poſſede les principes du mouvement & </s> <s xml:id="echoid-s17073" xml:space="preserve">de la <lb/>ſtatique, eſt beaucoup plus propre que tout autre à l’exécution d’un <lb/>grand nombre de manœuvres qui paroîtroient quelquefois imprati-<lb/>cables: </s> <s xml:id="echoid-s17074" xml:space="preserve">il ſçaura combiner avec certitude, calculer les forces des <lb/>machines qu’on lui préſentera, & </s> <s xml:id="echoid-s17075" xml:space="preserve">s’épargnera mille tâtonnemens <lb/>inutiles, mais inévitables pour ceux qui ne ſont pas inſtruits <lb/>comme lui. </s> <s xml:id="echoid-s17076" xml:space="preserve">Il eſt bon de prévenir ici, & </s> <s xml:id="echoid-s17077" xml:space="preserve">de combattre deux erreurs <lb/>groſſieres, dans leſquelles tombent la plûpart de ceux qui s’appli-<lb/>quent à la méchanique ſans en connoître les loix. </s> <s xml:id="echoid-s17078" xml:space="preserve">Ayant obſervé <lb/>la prodigieuſe augmentation des forces dans certaines machines, <lb/>ils s’imaginent pouvoir les augmenter à leur gré, en multipliant <lb/>les leviers ou les roues. </s> <s xml:id="echoid-s17079" xml:space="preserve">Ce qui ſeroit vrai dans un état parfait & </s> <s xml:id="echoid-s17080" xml:space="preserve"><lb/>dans la métaphyſique de la méchanique, devient faux par l’aug-<lb/>mentation des frottemens qui ſont inévitables dans les machines, <lb/>telles que celles dont on eſt obligé de faire uſage. </s> <s xml:id="echoid-s17081" xml:space="preserve">Une erreur à peu <lb/>près ſemblable, & </s> <s xml:id="echoid-s17082" xml:space="preserve">qui a toujours ſa ſource dans l’ignorance, eſt <lb/>celle de certaines perſonnes qui ayant exécuté une machine en petit, <lb/>en concluent avec la derniere aſſurance qu’elle doit produire les <lb/>mêmes effets en grand. </s> <s xml:id="echoid-s17083" xml:space="preserve">Ils ne font pas attention que les corps <lb/>ſemblables croiſſant en peſanteur dans la raiſon des cubes des di-<lb/>menſions homologues, les frottemens croiſſent dans la même raiſon; </s> <s xml:id="echoid-s17084" xml:space="preserve"><lb/>ce qui eſt cauſe que dans certaines machines la force ſur laquelle ils <lb/>comptent pour produire l’effet qu’ils annoncent, eſt employée toute <lb/>entiere, & </s> <s xml:id="echoid-s17085" xml:space="preserve">ſouvent n’eſt pas encore capable de vaincre les frotte-<lb/>mens. </s> <s xml:id="echoid-s17086" xml:space="preserve">Il eſt bien vrai qu’une machine qui produit certain effet en <lb/>grand, en produira un proportionnel en petit; </s> <s xml:id="echoid-s17087" xml:space="preserve">mais le réciproque <lb/>n’eſt jamais vrai: </s> <s xml:id="echoid-s17088" xml:space="preserve">ainſi il faut toujours compter ſur une augmen-<lb/>tation conſidérable de forces dans les machines que l’on exécute. </s> <s xml:id="echoid-s17089" xml:space="preserve"><lb/>Les meilleures ſont celles où cette augmentation pardeſſus la pro-<lb/>portion du modele avec la machine en grand ſe trouve être la plus <lb/>petite, toutes choſes égales d’ailleurs. </s> <s xml:id="echoid-s17090" xml:space="preserve">Il y a encore un troiſieme <lb/>défaut dans ceux qui ignorent la ſtatique, & </s> <s xml:id="echoid-s17091" xml:space="preserve">qui cependant ont un <pb o="543" file="0621" n="641" rhead="DE MATHÉMATIQUE. Liv. XV."/> goût décidé pour cette partie. </s> <s xml:id="echoid-s17092" xml:space="preserve">Mais on commence à craindre le ridi-<lb/>cule, en cherchant le mouvement perpétuel, & </s> <s xml:id="echoid-s17093" xml:space="preserve">l’on ſe perſuade <lb/>aiſément, quand on n’eſt pas entêté de ſes idées, qu’il y a des recher-<lb/>ches plus dignes de notre attention, après les efforts inutiles de <lb/>ceux qui ont voulu trouver la ſolution de ce problême, qui eſt or-<lb/>dinairement l’écueil des mauvais Méchaniciens, comme la qua-<lb/>drature du cercle eſt celui des médiocres Géometres.</s> <s xml:id="echoid-s17094" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1400" type="section" level="1" n="1028"> <head xml:id="echoid-head1215" xml:space="preserve">CHAPITRE PREMIER, <lb/>Dans lequel on donne l’introduction à la Méchanique. <lb/><emph style="sc">Définitions</emph>.</head> <head xml:id="echoid-head1216" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s17095" xml:space="preserve">1027. </s> <s xml:id="echoid-s17096" xml:space="preserve">L A méchanique eſt une ſcience qui conſidere les rap-<lb/>ports qui ſe rencontrent entre les puiſſances ou les forces ap-<lb/>pliquées ſur les corps pour les mouvoir par le moyen des ma-<lb/>chines. </s> <s xml:id="echoid-s17097" xml:space="preserve">Ainſi la méchanique priſe en général, eſt la ſcience <lb/>des mouvemens, & </s> <s xml:id="echoid-s17098" xml:space="preserve">par conſéquent la comparaiſon des maſſes <lb/>des corps, celle de leur vîteſſe, fait néceſſairement partie de <lb/>la méchanique, enviſagée ſous ce point de vue.</s> <s xml:id="echoid-s17099" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1401" type="section" level="1" n="1029"> <head xml:id="echoid-head1217" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s17100" xml:space="preserve">1028. </s> <s xml:id="echoid-s17101" xml:space="preserve">Si l’on détermine les rapports qui doivent ſe trouver <lb/>entre un certain nombre de puiſſances, pour leurs forces abſo-<lb/>lues, & </s> <s xml:id="echoid-s17102" xml:space="preserve">leurs directions, afin de produire l’équilibre, la mé-<lb/>chanique en général devient une partie déterminée, & </s> <s xml:id="echoid-s17103" xml:space="preserve">ſe <lb/>nomme méchanique ſtatique: </s> <s xml:id="echoid-s17104" xml:space="preserve">ſon objet eſt de mettre les <lb/>forces en équilibre, ou ſi elles y ſont, de déterminer les raiſons <lb/>qui concourent à l’établir.</s> <s xml:id="echoid-s17105" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1402" type="section" level="1" n="1030"> <head xml:id="echoid-head1218" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s17106" xml:space="preserve">1029. </s> <s xml:id="echoid-s17107" xml:space="preserve">Nous avons déja dit que toute force mouvante ou <lb/>puiſſance, eſt ce qui peut mouvoir un corps, & </s> <s xml:id="echoid-s17108" xml:space="preserve">par conſéquent <lb/>les corps en mouvement ſont des forces motrices, puiſqu’il <lb/>eſt démontré par l’expérience qu’ils peuvent faire mouvoir les <lb/>autres. </s> <s xml:id="echoid-s17109" xml:space="preserve">Nous n’examinons pas ici ſi cette propriété eſt atta-<lb/>chée eſſentiellement aux corps en mouvement, ou ſi elle ne <lb/>dépend que de la volonté de Dieu qui a établi la communica-<lb/>tion du mouvement par le moyen du choc.</s> <s xml:id="echoid-s17110" xml:space="preserve"/> </p> <pb o="544" file="0622" n="642" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1403" type="section" level="1" n="1031"> <head xml:id="echoid-head1219" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s17111" xml:space="preserve">1030. </s> <s xml:id="echoid-s17112" xml:space="preserve">L’équilibre eſt l’état d’un corps en repos, tiré par plu-<lb/>ſieurs forces, qui tendent à le mouvoir. </s> <s xml:id="echoid-s17113" xml:space="preserve">Un corps ſuſpendu, <lb/>au moyen d’un cordon, eſt en équilibre, & </s> <s xml:id="echoid-s17114" xml:space="preserve">tire autant le cor-<lb/>don de haut en bas, qu’il eſt lui-même tiré de bas en haut par <lb/>ce même cordon. </s> <s xml:id="echoid-s17115" xml:space="preserve">Cette machine nous préſente la maniere <lb/>dont ſe fait l’équilibre, & </s> <s xml:id="echoid-s17116" xml:space="preserve">nous montre qu’en général il ne <lb/>peut y avoir d’équilibre qu’entre deux forces égales, & </s> <s xml:id="echoid-s17117" xml:space="preserve">directe-<lb/>ment oppoſées. </s> <s xml:id="echoid-s17118" xml:space="preserve">Si donc il y a plus de deux forces en équi-<lb/>libre appliquées à un même corps, ce que l’on a à faire eſt <lb/>de déterminer, par le moyen d’une force connue, comment <lb/>toutes les autres, dont les directions ſont données, ſe compo-<lb/>ſent en une ſeule égale & </s> <s xml:id="echoid-s17119" xml:space="preserve">directement oppoſée à la premiere, <lb/>afin de produire l’équilibre.</s> <s xml:id="echoid-s17120" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1404" type="section" level="1" n="1032"> <head xml:id="echoid-head1220" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s17121" xml:space="preserve">1031. </s> <s xml:id="echoid-s17122" xml:space="preserve">On appelle poids, l’effort qui ſollicite les corps à <lb/>deſcendre au centre de la terre. </s> <s xml:id="echoid-s17123" xml:space="preserve">Dans les corps de même ma-<lb/>tiere, les poids ſont proportionnels aux volumes, & </s> <s xml:id="echoid-s17124" xml:space="preserve">par con-<lb/>ſéquent ſe déterminent par les regles de la Géométrie. </s> <s xml:id="echoid-s17125" xml:space="preserve">Comme <lb/>les meſures doivent être homologues aux choſes dont elles <lb/>ſont la meſure, les poids naturellement doivent ſe meſurer <lb/>par des poids. </s> <s xml:id="echoid-s17126" xml:space="preserve">Celui auquel on rapporte les autres, eſt regardé <lb/>comme l’unité, quoiqu’il puiſſe contenir un nombre indéfini <lb/>de parties égales: </s> <s xml:id="echoid-s17127" xml:space="preserve">ainſi la livre, qui eſt la meſure ordinaire <lb/>des poids, eſt regardée comme l’unité, quoiqu’elle contienne <lb/>ſeize parties égales, qui ſervent à meſurer les corps d’un moin-<lb/>dre poids, & </s> <s xml:id="echoid-s17128" xml:space="preserve">ainſi des autres meſures plus petites.</s> <s xml:id="echoid-s17129" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17130" xml:space="preserve">1032. </s> <s xml:id="echoid-s17131" xml:space="preserve">Il y a deux manieres de repréſenter une force. </s> <s xml:id="echoid-s17132" xml:space="preserve">La <lb/>premiere & </s> <s xml:id="echoid-s17133" xml:space="preserve">la plus naturelle, eſt d’exprimer l’effort dont elle <lb/>eſt capable par les poids auxquels elle peut faire équilibre. <lb/></s> <s xml:id="echoid-s17134" xml:space="preserve">Ainſi une puiſſance capable de ſoutenir un poids de 20 livres <lb/>eſt une force de 20 livres. </s> <s xml:id="echoid-s17135" xml:space="preserve">Mais comme il s’agit moins des <lb/>forces abſolues que des rapports qu’elles ont entr’elles, les <lb/>Géometres ſont convenus de déſigner les forces par des lignes. </s> <s xml:id="echoid-s17136" xml:space="preserve"><lb/>Ainſi ayant repréſenté une force de 4 livres par une ligne d’une <lb/>certaine longueur, une force triple ou quadruple, c’eſt-à-dire <lb/>de 12 ou de 16 livres, ſera repréſentée par une ligne triple ou <lb/>quadruple de la premiere. </s> <s xml:id="echoid-s17137" xml:space="preserve">Dans la théorie du mouvement, <lb/>nous avons déterminé les forces par les eſpaces qu’elles font <pb o="545" file="0623" n="643" rhead="DE MATHÉMATIQUE. Liv. XV."/> parcourir à des corps égaux en tems égaux: </s> <s xml:id="echoid-s17138" xml:space="preserve">ainſi les lignes <lb/>qui repréſentoient les forces motrices, ſont les expreſſions na-<lb/>turelles de ces forces. </s> <s xml:id="echoid-s17139" xml:space="preserve">Ici ces lignes déſignent les rapports qui <lb/>ſe trouvent entre les poids des forces appliquées aux corps. <lb/></s> <s xml:id="echoid-s17140" xml:space="preserve">Au reſte de quelque maniere que l’on les conſidere, on verra <lb/>que cela revient toujours au même.</s> <s xml:id="echoid-s17141" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1405" type="section" level="1" n="1033"> <head xml:id="echoid-head1221" xml:space="preserve">VI.</head> <p> <s xml:id="echoid-s17142" xml:space="preserve">1033. </s> <s xml:id="echoid-s17143" xml:space="preserve">La ligne de direction d’une puiſſance eſt la ligne ſui-<lb/>vant laquelle elle tend à mouvoir le corps en cas qu’elle ſoit <lb/>ſeule, & </s> <s xml:id="echoid-s17144" xml:space="preserve">que le corps cede à ſon impreſſion. </s> <s xml:id="echoid-s17145" xml:space="preserve">Une même force <lb/>ne peut pas agir ſuivant pluſieurs directions à la fois: </s> <s xml:id="echoid-s17146" xml:space="preserve">ainſi <lb/>une force ſeule qui tire un corps ne peut le mouvoir que ſui-<lb/>vant une ligne droite: </s> <s xml:id="echoid-s17147" xml:space="preserve">il faut encore remarquer que l’on ne <lb/>doit meſurer l’effort d’une force appliquée à un corps que par <lb/>la réſiſtance qu’elle éprouve de la part de ce corps: </s> <s xml:id="echoid-s17148" xml:space="preserve">car ſi le <lb/>corps a une maſſe de dix livres, il ne détruit qu’une force de <lb/>dix livres dans la puiſſance qui agit ſur lui, quand même elle <lb/>ſeroit en état de vaincre une force de 100 livres; </s> <s xml:id="echoid-s17149" xml:space="preserve">c’eſt ce que <lb/>les Méchaniciens entendent par ce principe général: </s> <s xml:id="echoid-s17150" xml:space="preserve">l’action <lb/>eſt égale à la réaction.</s> <s xml:id="echoid-s17151" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1406" type="section" level="1" n="1034"> <head xml:id="echoid-head1222" xml:space="preserve">VII.</head> <p> <s xml:id="echoid-s17152" xml:space="preserve">1034. </s> <s xml:id="echoid-s17153" xml:space="preserve">On appelle machines tous les inſtrumens propres à <lb/>faire mouvoîr ou à arrêter le mouvement des corps; </s> <s xml:id="echoid-s17154" xml:space="preserve">il y en a <lb/>de ſimples & </s> <s xml:id="echoid-s17155" xml:space="preserve">de compoſées.</s> <s xml:id="echoid-s17156" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17157" xml:space="preserve">1035. </s> <s xml:id="echoid-s17158" xml:space="preserve">Les machines ſimples ſont au nombre de ſix, ſçavoir, <lb/>le levier, la roue dans ſon aiſſieu, la poulie, le plan incliné, le <lb/>coin, & </s> <s xml:id="echoid-s17159" xml:space="preserve">la vis.</s> <s xml:id="echoid-s17160" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17161" xml:space="preserve">1036. </s> <s xml:id="echoid-s17162" xml:space="preserve">Les machines compoſées ſont ſans nombre, & </s> <s xml:id="echoid-s17163" xml:space="preserve">dépen-<lb/>dent des différentes combinaiſons de celles-ci, priſes en tel <lb/>nombre qu’on le juge à propos.</s> <s xml:id="echoid-s17164" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1407" type="section" level="1" n="1035"> <head xml:id="echoid-head1223" xml:space="preserve">VIII.</head> <p> <s xml:id="echoid-s17165" xml:space="preserve">1037. </s> <s xml:id="echoid-s17166" xml:space="preserve">On appelle centre de gravité d’un corps un point par <lb/>lequel ce corps étant ſuſpendu, demeure en équilibre dans toutes <lb/>les ſituations imaginables. </s> <s xml:id="echoid-s17167" xml:space="preserve">Il ſuit delà que la puiſſance qui eſt <lb/>appliquée à ce point, arrête tous les efforts de la peſanteur des <lb/>parties qui compoſent ce corps: </s> <s xml:id="echoid-s17168" xml:space="preserve">donc on peut concevoir que <lb/>cette même peſanteur eſt réunie à ce point. </s> <s xml:id="echoid-s17169" xml:space="preserve">Nous verrons par <lb/>la ſuite la maniere de déterminer les centres de gravité des <pb o="546" file="0624" n="644" rhead="NOUVEAU COURS"/> principales figures & </s> <s xml:id="echoid-s17170" xml:space="preserve">des principaux corps qui peuvent être <lb/>mis en uſage. </s> <s xml:id="echoid-s17171" xml:space="preserve">Cette recherche ne peut être que trés-utile dans <lb/>la méchanique.</s> <s xml:id="echoid-s17172" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1408" type="section" level="1" n="1036"> <head xml:id="echoid-head1224" xml:space="preserve"><emph style="sc">Axiome</emph>.</head> <p> <s xml:id="echoid-s17173" xml:space="preserve">1038. </s> <s xml:id="echoid-s17174" xml:space="preserve">Le poids d’un corps agit avec la même force dans <lb/>tous les points de ſa direction. </s> <s xml:id="echoid-s17175" xml:space="preserve">Concevons un corps attaché à <lb/>une corde flexible, & </s> <s xml:id="echoid-s17176" xml:space="preserve">faiſons abſtraction du poids de cette <lb/>corde: </s> <s xml:id="echoid-s17177" xml:space="preserve">il eſt évident que ce corps tire toujours autant le point <lb/>auquel il eſt attaché par cette corde, quelle que ſoit la longueur <lb/>de la corde. </s> <s xml:id="echoid-s17178" xml:space="preserve">Cela ſuppoſe que la force qui pouſſe le corps au <lb/>centre de la terre eſt toujours la même. </s> <s xml:id="echoid-s17179" xml:space="preserve">Cette fauſſe ſuppoſition <lb/>ne peut être ſenſible dans les plus grandes diſtances, relative-<lb/>ment aux machines.</s> <s xml:id="echoid-s17180" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1409" type="section" level="1" n="1037"> <head xml:id="echoid-head1225" style="it" xml:space="preserve"><emph style="sc">Avertissement</emph>.</head> <p> <s xml:id="echoid-s17181" xml:space="preserve">Après avoir conſidéré dans le Traité du mouvement le pa-<lb/>rallélogramme des forces, c’eſt-à-dire la compoſition des forces, <lb/>pour déterminer la vîteſſe que les forces compoſantes procu-<lb/>rent au mobile, nous allons reprendre la même queſtion par <lb/>rapport à la méchanique ſtatique, c’eſt-à-dire conſidérer quelle <lb/>eſt la force capable de faire équilibre avec les forces com-<lb/>poſantes.</s> <s xml:id="echoid-s17182" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1410" type="section" level="1" n="1038"> <head xml:id="echoid-head1226" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s17183" xml:space="preserve">1039. </s> <s xml:id="echoid-s17184" xml:space="preserve">Si un corps K eſt pouſſé à la fois par deux puiſſances <lb/> <anchor type="note" xlink:label="note-0624-01a" xlink:href="note-0624-01"/> égales repréſentées par les côtés A B, A C d’un quarré A B D C, <lb/>& </s> <s xml:id="echoid-s17185" xml:space="preserve">dirigées ſuivant ces mêmes côtés, je dis qu’il décrira la diago-<lb/>nale A D du même quarré dans le tems qu’il eût décrit le côté A C, <lb/>s’il n’avoit été pouſſé que par une ſeule force.</s> <s xml:id="echoid-s17186" xml:space="preserve"/> </p> <div xml:id="echoid-div1410" type="float" level="2" n="1"> <note position="left" xlink:label="note-0624-01" xlink:href="note-0624-01a" xml:space="preserve">Figure 354.</note> </div> </div> <div xml:id="echoid-div1412" type="section" level="1" n="1039"> <head xml:id="echoid-head1227" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s17187" xml:space="preserve">Il eſt d’abord évident que le corps doit ſe mouvoir ſur la <lb/>diagonale A D : </s> <s xml:id="echoid-s17188" xml:space="preserve">car ne pouvant aller que par un ſeul chemin, <lb/>& </s> <s xml:id="echoid-s17189" xml:space="preserve">ſe trouvant entre deux forces motrices entiérement égales, <lb/>il n’y a pas de raiſon pour qu’il s’approche plutôt de l’une que <lb/>de l’autre; </s> <s xml:id="echoid-s17190" xml:space="preserve">ce qui arriveroit, s’il décrivoit toute autre ligne que <lb/>la diagonale. </s> <s xml:id="echoid-s17191" xml:space="preserve">Pour ſçavoir préſentement ſi la force réſultante <lb/>eſt repréſentée par la même diagonale, je nomme x cette <lb/>même réſultante, dont la longueur eſt inconnue, & </s> <s xml:id="echoid-s17192" xml:space="preserve">je fais <pb o="547" file="0625" n="645" rhead="DE MATHÉMATIQUE. Liv. XV."/> attention que ſi des deux forces égales A C, A B il en réſulte <lb/>une ſeule force x: </s> <s xml:id="echoid-s17193" xml:space="preserve">pareillement ces mêmes forces peuvent être <lb/>regardées comme les réſultantes, chacune de deux forces <lb/>égales, diſpoſées de la même maniere qu’elles le ſont elles-<lb/>mêmes par rapport à la réſultante x, & </s> <s xml:id="echoid-s17194" xml:space="preserve">proportionnelles à ces <lb/>mêmes forces: </s> <s xml:id="echoid-s17195" xml:space="preserve">mais ces forces font un angle de 45 degrés <lb/>avec la diagonale: </s> <s xml:id="echoid-s17196" xml:space="preserve">donc leurs compoſantes doivent être priſes, <lb/>deux ſur la ligne E A F perpendiculaire à la diagonale, & </s> <s xml:id="echoid-s17197" xml:space="preserve">deux <lb/>autres ſur la diagonale elle-même. </s> <s xml:id="echoid-s17198" xml:space="preserve">On aura donc cette pro-<lb/>portion, la réſultante x eſt à ſa compoſante A B, que je nom-<lb/>merai a, comme la même compoſante A B, priſe pour réſul-<lb/>tante des forces A E & </s> <s xml:id="echoid-s17199" xml:space="preserve">A G, eſt à ſa compoſante A E; </s> <s xml:id="echoid-s17200" xml:space="preserve">d’où <lb/>l’on tire cette analogie, x : </s> <s xml:id="echoid-s17201" xml:space="preserve">a :</s> <s xml:id="echoid-s17202" xml:space="preserve">: a : </s> <s xml:id="echoid-s17203" xml:space="preserve">{aa/x} = A E. </s> <s xml:id="echoid-s17204" xml:space="preserve">On démon-<lb/>trera de même que la force A F eſt auſſi égale à {aa/x}: </s> <s xml:id="echoid-s17205" xml:space="preserve">donc au <lb/>lieu des deux forces égales A B, A C, on aura quatre nou-<lb/>velles forces égales, dont deux A E, A F ſont directement op-<lb/>poſées, & </s> <s xml:id="echoid-s17206" xml:space="preserve">ſe détruiſent par conſéquent, & </s> <s xml:id="echoid-s17207" xml:space="preserve">deux autres ſont <lb/>toutes les deux dirigées ſuivant A D, & </s> <s xml:id="echoid-s17208" xml:space="preserve">ſont chacune repré-<lb/>ſentées par {aa/x}. </s> <s xml:id="echoid-s17209" xml:space="preserve">Mais ces deux forces tirant dans le même ſens, <lb/>ſont les ſeules qui forment la réſultante inconnue x, à laquelle <lb/>elles ſont égales. </s> <s xml:id="echoid-s17210" xml:space="preserve">On aura donc cette équation {2aa/x} = x, ou en <lb/>multipliant par x, 2aa = xx; </s> <s xml:id="echoid-s17211" xml:space="preserve">d’où il ſuit évidemment que la <lb/>réſultante eſt non ſeulement dirigée ſuivant la diagonale, <lb/>mais encore égale à cette même diagonale. </s> <s xml:id="echoid-s17212" xml:space="preserve">C. </s> <s xml:id="echoid-s17213" xml:space="preserve">Q. </s> <s xml:id="echoid-s17214" xml:space="preserve">F. </s> <s xml:id="echoid-s17215" xml:space="preserve">D.</s> <s xml:id="echoid-s17216" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1413" type="section" level="1" n="1040"> <head xml:id="echoid-head1228" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s17217" xml:space="preserve">1040. </s> <s xml:id="echoid-s17218" xml:space="preserve">Comme toute ligne peut être regardée comme la <lb/>diagonale d’un quarré, il s’enſuit qu’au lieu d’une ſeule force <lb/>on peut prendre deux forces compoſantes, repréſentées par <lb/>les côtés du quarré, dont l’expreſſion de la premiere eſt dia-<lb/>gonale: </s> <s xml:id="echoid-s17219" xml:space="preserve">car ces deux nouvelles forces ne produiront pas d’autre <lb/>effet ſur le mobile que celui qui réſultoit de la premiere.</s> <s xml:id="echoid-s17220" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1414" type="section" level="1" n="1041"> <head xml:id="echoid-head1229" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s17221" xml:space="preserve">1041. </s> <s xml:id="echoid-s17222" xml:space="preserve">Il ſuit delà que ſi un corps eſt tiré ou pouſſé à la fois <lb/>par deux forces motrices, repréſentées & </s> <s xml:id="echoid-s17223" xml:space="preserve">dirigées ſuivant les <lb/>côtés A B, A C d’un parallélogramme rectangle, il décrira, <lb/>par l’effort compoſé des deux puiſſances, la diagonale A D du <pb o="548" file="0626" n="646" rhead="NOUVEAU COURS"/> même parallélogramme, dans le tems qu’il eût décrit l’un ou <lb/>l’autre des côtés A B, A C, s’il n’cût été pouſſé que par une <lb/>ſeule force M ou N.</s> <s xml:id="echoid-s17224" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1415" type="section" level="1" n="1042"> <head xml:id="echoid-head1230" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s17225" xml:space="preserve">Par le corollaire précédent, on peut décompoſer les deux <lb/> <anchor type="note" xlink:label="note-0626-01a" xlink:href="note-0626-01"/> forces A C, A B, chacune en deux autres qui ſoient les côtés <lb/>du quarré, dont ces mêmes lignes ſont les diagonales: </s> <s xml:id="echoid-s17226" xml:space="preserve">de <lb/>plus, il eſt évident que la ligne A E qui diviſe l’angle droit <lb/>en deux angles égaux, doit réunir deux de ces quatre forces <lb/>dans leſquelles nous décompoſons les premieres compoſantes <lb/>A B & </s> <s xml:id="echoid-s17227" xml:space="preserve">A C; </s> <s xml:id="echoid-s17228" xml:space="preserve">mais il eſt aiſé de voir que le corps ne peut pas <lb/>ſuivre la ligne A E: </s> <s xml:id="echoid-s17229" xml:space="preserve">car pour cela il faudroit que les forces <lb/>A H, A I, directement oppoſées, fuſſent égales, ce qui eſt <lb/>impoſſible, puiſque les lignes ou les forces qu’elles repréſentent <lb/>ſont dans la raiſon des lignes ou forces A C, A B, qui ſont iné-<lb/>gales (par hypotheſe): </s> <s xml:id="echoid-s17230" xml:space="preserve">donc tandis que le corps ſera pouſſé <lb/>par la ſomme des forces A F, A G, dirigées ſur la même ligne, <lb/>il y aura encore une force repréſentée par A K, différence des <lb/>forces directement oppoſées A H, A I. </s> <s xml:id="echoid-s17231" xml:space="preserve">Pour déterminer toutes <lb/>ces forces, nous nommerons A B, a; </s> <s xml:id="echoid-s17232" xml:space="preserve">A C, b; </s> <s xml:id="echoid-s17233" xml:space="preserve">on aura A F ou <lb/>A H = √{1/2} b<emph style="sub">2</emph>\x{0020}, & </s> <s xml:id="echoid-s17234" xml:space="preserve">de même A I ou A G = √{1/2} aa\x{0020}: </s> <s xml:id="echoid-s17235" xml:space="preserve">donc A E <lb/>ou A F + A G = √{1/2} a<emph style="sub">2</emph>\x{0020} + √{1/2} b<emph style="sub">2</emph>\x{0020}; </s> <s xml:id="echoid-s17236" xml:space="preserve">& </s> <s xml:id="echoid-s17237" xml:space="preserve">A K ou A I - A H <lb/>= √{1/2} a<emph style="sub">2</emph>\x{0020} - √{1/2} b<emph style="sub">2</emph>\x{0020}. </s> <s xml:id="echoid-s17238" xml:space="preserve">Préſentement voyons ſi cette force A K, <lb/>appliquée perpendiculairement en E, eſt capable de ramener <lb/>le corps à l’extrêmité D de la diagonale A D; </s> <s xml:id="echoid-s17239" xml:space="preserve">ſi cela eſt, il <lb/>faut que l’angle A E D étant rectangle, on ait A D<emph style="sub">2</emph> = A K<emph style="sub">2</emph> <lb/>+ (A F + A G)<emph style="sub">2</emph>. </s> <s xml:id="echoid-s17240" xml:space="preserve">J’éleve donc D E ou A K au quarré, & </s> <s xml:id="echoid-s17241" xml:space="preserve">j’ai <lb/>{1/2} a<emph style="sub">2</emph> - 2 √{1/4} a<emph style="sub">2</emph> b<emph style="sub">2</emph>\x{0020} + {1/2} b<emph style="sub">2</emph> pour le quarré de √{1/2} a<emph style="sub">2</emph>\x{0020} - √{1/2} b<emph style="sub">2</emph>\x{0020}. <lb/></s> <s xml:id="echoid-s17242" xml:space="preserve">J’éleve de même A E ou A F + A G = √{1/2} a<emph style="sub">2</emph>\x{0020} + √{1/2} b<emph style="sub">2</emph>\x{0020} au <lb/>quarré, & </s> <s xml:id="echoid-s17243" xml:space="preserve">j’ai {1/2} a<emph style="sub">2</emph> + 2 √{1/4} a<emph style="sub">2</emph> b<emph style="sub">2</emph>\x{0020} + {1/2} b<emph style="sub">2</emph>: </s> <s xml:id="echoid-s17244" xml:space="preserve">ajoutant les deux quarrés <lb/>enſemble, la ſomme eſt a<emph style="sub">2</emph> + b<emph style="sub">2</emph>; </s> <s xml:id="echoid-s17245" xml:space="preserve">d’où il ſuit que pendant que <lb/>les efforts conjoints A F, A G font décrire au mobile la ligne <lb/>A E égale à leur ſomme, la force A K ou D E, ramene le corps <lb/>à l’extrêmité de la diagonale: </s> <s xml:id="echoid-s17246" xml:space="preserve">Donc dans ce cas des forces <lb/>inégales dirigées ſuivant les côtés du parallélogramme rectan-<lb/>gle & </s> <s xml:id="echoid-s17247" xml:space="preserve">repréſentées par ces côtés, le corps décrit encore la dia-<lb/>gonale. </s> <s xml:id="echoid-s17248" xml:space="preserve">C. </s> <s xml:id="echoid-s17249" xml:space="preserve">Q. </s> <s xml:id="echoid-s17250" xml:space="preserve">F. </s> <s xml:id="echoid-s17251" xml:space="preserve">D.</s> <s xml:id="echoid-s17252" xml:space="preserve"/> </p> <div xml:id="echoid-div1415" type="float" level="2" n="1"> <note position="left" xlink:label="note-0626-01" xlink:href="note-0626-01a" xml:space="preserve">Figure 355.</note> </div> <pb o="549" file="0627" n="647" rhead="DE MATHÉMATIQUE. Liv. XV."/> </div> <div xml:id="echoid-div1417" type="section" level="1" n="1043"> <head xml:id="echoid-head1231" xml:space="preserve"><emph style="sc">Observation</emph>.</head> <p> <s xml:id="echoid-s17253" xml:space="preserve">1042. </s> <s xml:id="echoid-s17254" xml:space="preserve">On pourroit craindre d’être tombé dans un paralo-<lb/>giſme, parce que nous démontrons que le corps, entre les <lb/>forces A E & </s> <s xml:id="echoid-s17255" xml:space="preserve">A K, qui ſont les côtés d’un parallélogramme <lb/>rectangle, & </s> <s xml:id="echoid-s17256" xml:space="preserve">qui repréſentent les forces qui agiſſent ſur lui, <lb/>décrit la diagonale A D du nouveau parallélogramme; </s> <s xml:id="echoid-s17257" xml:space="preserve">ce qui <lb/>ſemble être préciſément l’état de la queſtion. </s> <s xml:id="echoid-s17258" xml:space="preserve">Mais il eſt aiſé de <lb/>ſe convaincre que quelles que ſoient les forces dans leſquelles <lb/>on décompoſe les premieres A B, A C, la réſultante eſt néceſ-<lb/>ſairement égale à la diagonale: </s> <s xml:id="echoid-s17259" xml:space="preserve">c’eſt ce que nous allons faire <lb/>en peu de mots. </s> <s xml:id="echoid-s17260" xml:space="preserve">Soit x la réſultante, dirigée ſuivant A E, qui <lb/>fait un angle quelconque avec la ligne A C, & </s> <s xml:id="echoid-s17261" xml:space="preserve">ſoient faits <lb/>A B = a, A C = b; </s> <s xml:id="echoid-s17262" xml:space="preserve">puiſque les forces a & </s> <s xml:id="echoid-s17263" xml:space="preserve">b font parcourir x, <lb/>deux forces proportionnelles à a & </s> <s xml:id="echoid-s17264" xml:space="preserve">b feront parcourir A B, <lb/>pourvu qu’elles ſoient diſpoſées de la même maniere que les <lb/>lignes A B & </s> <s xml:id="echoid-s17265" xml:space="preserve">A C le ſont par rapport à A E; </s> <s xml:id="echoid-s17266" xml:space="preserve">ce qui arrivera ſi <lb/>l’on prend l’une A G ſur A E, & </s> <s xml:id="echoid-s17267" xml:space="preserve">l’autre ſur la ligne A I per-<lb/>pendiculaire à la diagonale: </s> <s xml:id="echoid-s17268" xml:space="preserve">car A B fait avec A E le même <lb/>angle que A G fait avec A B, & </s> <s xml:id="echoid-s17269" xml:space="preserve">A C fait avec A E le même <lb/>angle que A I fait avec A B. </s> <s xml:id="echoid-s17270" xml:space="preserve">Donc les forces dirigées, ſuivant <lb/>ces lignes, ſont diſpoſées à l’égard de A B, comme A B & </s> <s xml:id="echoid-s17271" xml:space="preserve"><lb/>A C le ſont à l’égard de A E: </s> <s xml:id="echoid-s17272" xml:space="preserve">de même puiſque les forces a & </s> <s xml:id="echoid-s17273" xml:space="preserve">b <lb/>font parcourir A E ou x, deux forces proportionnelles, & </s> <s xml:id="echoid-s17274" xml:space="preserve">diſ-<lb/>poſées de la même maniere à l’égard de A E, feront parcourir <lb/>A C; </s> <s xml:id="echoid-s17275" xml:space="preserve">ce qui arrivera, ſi l’on prend l’une ſur A E, & </s> <s xml:id="echoid-s17276" xml:space="preserve">l’autre <lb/>ſur A H, auſſi perpendiculaire à A E. </s> <s xml:id="echoid-s17277" xml:space="preserve">On aura donc ces quatre <lb/>proportions, A E : </s> <s xml:id="echoid-s17278" xml:space="preserve">A B :</s> <s xml:id="echoid-s17279" xml:space="preserve">: A B : </s> <s xml:id="echoid-s17280" xml:space="preserve">A G, ou x : </s> <s xml:id="echoid-s17281" xml:space="preserve">a :</s> <s xml:id="echoid-s17282" xml:space="preserve">: a : </s> <s xml:id="echoid-s17283" xml:space="preserve">{a a/x} = A G <lb/>A E : </s> <s xml:id="echoid-s17284" xml:space="preserve">A C :</s> <s xml:id="echoid-s17285" xml:space="preserve">: A B : </s> <s xml:id="echoid-s17286" xml:space="preserve">A I, ou x : </s> <s xml:id="echoid-s17287" xml:space="preserve">b :</s> <s xml:id="echoid-s17288" xml:space="preserve">: a : </s> <s xml:id="echoid-s17289" xml:space="preserve">{a b/x} = A I; </s> <s xml:id="echoid-s17290" xml:space="preserve">& </s> <s xml:id="echoid-s17291" xml:space="preserve">encore A E <lb/>: </s> <s xml:id="echoid-s17292" xml:space="preserve">A C :</s> <s xml:id="echoid-s17293" xml:space="preserve">: A C : </s> <s xml:id="echoid-s17294" xml:space="preserve">A F, ou x : </s> <s xml:id="echoid-s17295" xml:space="preserve">b :</s> <s xml:id="echoid-s17296" xml:space="preserve">: b : </s> <s xml:id="echoid-s17297" xml:space="preserve">{b b/x} = A F; </s> <s xml:id="echoid-s17298" xml:space="preserve">& </s> <s xml:id="echoid-s17299" xml:space="preserve">enſin A E : </s> <s xml:id="echoid-s17300" xml:space="preserve">A B <lb/>:</s> <s xml:id="echoid-s17301" xml:space="preserve">: A C : </s> <s xml:id="echoid-s17302" xml:space="preserve">A L, ou x : </s> <s xml:id="echoid-s17303" xml:space="preserve">a :</s> <s xml:id="echoid-s17304" xml:space="preserve">: b : </s> <s xml:id="echoid-s17305" xml:space="preserve">{a b/x} = A L : </s> <s xml:id="echoid-s17306" xml:space="preserve">donc au lieu des deux <lb/>forces A B & </s> <s xml:id="echoid-s17307" xml:space="preserve">A C nous en avons quatre, A F, A G, A L, A I, <lb/>dont les deux dernieres ſont égales, & </s> <s xml:id="echoid-s17308" xml:space="preserve">directement oppoſées, <lb/>puiſque nous avons trouvé pour A L & </s> <s xml:id="echoid-s17309" xml:space="preserve">pour A I {a b/x}, & </s> <s xml:id="echoid-s17310" xml:space="preserve">dontles <lb/>deux premieres ſont dirigées ſur la même ligne A E, & </s> <s xml:id="echoid-s17311" xml:space="preserve">par <lb/>conſéquent concourent ſeules à produire A E; </s> <s xml:id="echoid-s17312" xml:space="preserve">ce qui donne <pb o="550" file="0628" n="648" rhead="NOUVEAU COURS"/> {a a/x} + {b b/x} = x; </s> <s xml:id="echoid-s17313" xml:space="preserve">d’où l’on tire a a + b b = x x: </s> <s xml:id="echoid-s17314" xml:space="preserve">ce qui prouve <lb/>invinciblement que la réſultante des quatre nouvelles forces <lb/>ou des deux compoſantes eſt néceſſairement égale à la dia-<lb/>gonale. </s> <s xml:id="echoid-s17315" xml:space="preserve">Mais ces quatre forces dans leſquelles on a décom-<lb/>poſées les deux premieres A B & </s> <s xml:id="echoid-s17316" xml:space="preserve">A C, font préciſément le <lb/>même effet que les forces A H, A F, A I, A G, dans leſquelles <lb/>nous avions d’abord décompoſé les forces M & </s> <s xml:id="echoid-s17317" xml:space="preserve">N, en regar-<lb/>dant les lignes A B & </s> <s xml:id="echoid-s17318" xml:space="preserve">A C comme les diagonales des quarrés <lb/>G I, F H: </s> <s xml:id="echoid-s17319" xml:space="preserve">donc il eſt inconteſtablement démontré que la force <lb/>D E ou A K a dû ramener le corps K ſur la diagonale A D.</s> <s xml:id="echoid-s17320" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1418" type="section" level="1" n="1044"> <head xml:id="echoid-head1232" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s17321" xml:space="preserve">1043. </s> <s xml:id="echoid-s17322" xml:space="preserve">Donc ſi l’on a une force quelconque, on pourra, ſi <lb/>on le juge à propos, la décompoſer en deux autres forces per-<lb/>pendiculaires entr’elles, & </s> <s xml:id="echoid-s17323" xml:space="preserve">la regarder comme la réſultante ou <lb/>la diagonale d’un parallélogramme rectangle, dont les côtés <lb/>expriment les forces réſultantes qui l’ont produites; </s> <s xml:id="echoid-s17324" xml:space="preserve">ſeulement <lb/>il faut bien remarquer que comme une même ligne peut être <lb/>diagonale d’une inſinité de parallélogrammes rectangles diffé-<lb/>rens, il ne faut pas la décompoſer au hazard, mais examiner <lb/>la décompoſition la plus analogue à l’état de la queſtion. </s> <s xml:id="echoid-s17325" xml:space="preserve">On <lb/>en va voir un exemple dans le corollaire ſuivant.</s> <s xml:id="echoid-s17326" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1419" type="section" level="1" n="1045"> <head xml:id="echoid-head1233" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s17327" xml:space="preserve">1044. </s> <s xml:id="echoid-s17328" xml:space="preserve">Il ſuit encore delà que ſi un corps eſt pouſſé à la fois <lb/> <anchor type="note" xlink:label="note-0628-01a" xlink:href="note-0628-01"/> par deux forces M, N, repréſentées par les côtés A C, A B <lb/>d’un parallélogramme obliqu’angle ou obtuſangle, & </s> <s xml:id="echoid-s17329" xml:space="preserve">dirigées <lb/>ſuivant les mêmes côtés, il décrira encore la diagonale A D <lb/>dans le tems qu’il eût décrit l’une ou l’autre des lignes A B, <lb/>A C, en n’obéiſſant qu’à une ſeule force M ou N. </s> <s xml:id="echoid-s17330" xml:space="preserve">Pour s’en <lb/>convaincre, du point C ſur la diagonale A D, ſoit abaiſſée la <lb/>ligne C F, & </s> <s xml:id="echoid-s17331" xml:space="preserve">formé le parallélogramme A F C H; </s> <s xml:id="echoid-s17332" xml:space="preserve">pareille-<lb/>ment du point B ſoit abaiſſée la perpendiculaire B G à la dia-<lb/>gonale A D, & </s> <s xml:id="echoid-s17333" xml:space="preserve">ſoit achevé le parallélogramme rectangle <lb/>A I B G: </s> <s xml:id="echoid-s17334" xml:space="preserve">les lignes A H, A F, A I, A G feront le même effet <lb/>que les forces A C, A B (art. </s> <s xml:id="echoid-s17335" xml:space="preserve">1043). </s> <s xml:id="echoid-s17336" xml:space="preserve">De plus, les forces repré-<lb/>ſentées par A H, A I ſont évidemment égales, & </s> <s xml:id="echoid-s17337" xml:space="preserve">directement <lb/>oppoſées, puiſqu’elles meſurent les hauteurs des triangles <lb/>égaux A B D, A C D: </s> <s xml:id="echoid-s17338" xml:space="preserve">donc il ne reſte pour mouvoir le corps <pb o="551" file="0629" n="649" rhead="DE MATHÉMATIQUE. Liv. XV."/> que les forces conſpirantes A F, A G, dirigées ſuivant la dia-<lb/>gonale: </s> <s xml:id="echoid-s17339" xml:space="preserve">reſte donc à faire voir que leur ſomme eſt égal à la <lb/>diagonale; </s> <s xml:id="echoid-s17340" xml:space="preserve">ce qui eſt évident, à cauſe des triangles rectangles <lb/>égaux & </s> <s xml:id="echoid-s17341" xml:space="preserve">ſemblables B G D, C F A, qui donnent G D = A F: <lb/></s> <s xml:id="echoid-s17342" xml:space="preserve">donc encore dans ce cas le corps décrit la diagonale A D du <lb/>parallélogramme, formé ſur la direction des forces compo-<lb/>ſantes. </s> <s xml:id="echoid-s17343" xml:space="preserve">C. </s> <s xml:id="echoid-s17344" xml:space="preserve">Q. </s> <s xml:id="echoid-s17345" xml:space="preserve">F. </s> <s xml:id="echoid-s17346" xml:space="preserve">3°. </s> <s xml:id="echoid-s17347" xml:space="preserve">D.</s> <s xml:id="echoid-s17348" xml:space="preserve"/> </p> <div xml:id="echoid-div1419" type="float" level="2" n="1"> <note position="left" xlink:label="note-0628-01" xlink:href="note-0628-01a" xml:space="preserve">Figure 356.</note> </div> </div> <div xml:id="echoid-div1421" type="section" level="1" n="1046"> <head xml:id="echoid-head1234" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s17349" xml:space="preserve">1045. </s> <s xml:id="echoid-s17350" xml:space="preserve">Donc quel que ſoit l’angle de direction des forces <lb/>compoſantes, le corps pouſſé par deux forces décrira toujours <lb/>la diagonale du parallélogramme formé ſur ces directions. </s> <s xml:id="echoid-s17351" xml:space="preserve">Et <lb/>de plus, ſi l’on oppoſe au corps dans la direction de la diago-<lb/>nale une force repréſentée par cette même ligne, cette force <lb/>fera équilibre avec les deux autres; </s> <s xml:id="echoid-s17352" xml:space="preserve">puiſque de ces deux forces <lb/>il n’en réſulte qu’une force égale à celle de la diagonale, & </s> <s xml:id="echoid-s17353" xml:space="preserve">à <lb/>laquelle on ſuppoſe la nouvelle force directement oppoſée.</s> <s xml:id="echoid-s17354" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1422" type="section" level="1" n="1047"> <head xml:id="echoid-head1235" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s17355" xml:space="preserve">1046. </s> <s xml:id="echoid-s17356" xml:space="preserve">Donc plus l’angle de direction des puiſſances com-<lb/>poſantes ſera petit, plus auſſi la ligne ou force réſultante ſera <lb/>grande pour les mêmes forces compoſantes: </s> <s xml:id="echoid-s17357" xml:space="preserve">de maniere que <lb/>le corps ſe mouvra avec la ſomme des compoſantes dans le <lb/>cas où ces deux forces ſont dirigées ſur une même ligne; </s> <s xml:id="echoid-s17358" xml:space="preserve">& </s> <s xml:id="echoid-s17359" xml:space="preserve"><lb/>réciproquement plus cet angle ſera obtus, plus la force réſul-<lb/>tante ſera petite; </s> <s xml:id="echoid-s17360" xml:space="preserve">enſorte que dans le cas où cet angle devien-<lb/>droit égal à deux droits, les forces ſe détruiſent réciproque-<lb/>ment, & </s> <s xml:id="echoid-s17361" xml:space="preserve">le corps eſt emporté dans la direction de la plus forte <lb/>puiſſance, & </s> <s xml:id="echoid-s17362" xml:space="preserve">reſte en repos, ſi les forces compoſantes ſont <lb/>égales.</s> <s xml:id="echoid-s17363" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1423" type="section" level="1" n="1048"> <head xml:id="echoid-head1236" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s17364" xml:space="preserve">1047. </s> <s xml:id="echoid-s17365" xml:space="preserve">Il ſuit encore delà que les trois forces étant repré-<lb/>ſentées par les lignes A B, A C, A D, elles le ſont auſſi par <lb/>les lignes A B, A D, B D qui forment le triangle A B D: </s> <s xml:id="echoid-s17366" xml:space="preserve">donc <lb/>elles ſont entr’elles comme les ſinus des angles du triangle <lb/>A B D, puiſque dans tout triangle, les côtés ſont entr’eux <lb/>comme les côtés oppoſés à ces angles. </s> <s xml:id="echoid-s17367" xml:space="preserve">On aura donc B D: <lb/></s> <s xml:id="echoid-s17368" xml:space="preserve">A B : </s> <s xml:id="echoid-s17369" xml:space="preserve">A D :</s> <s xml:id="echoid-s17370" xml:space="preserve">: ſin. </s> <s xml:id="echoid-s17371" xml:space="preserve">B A D: </s> <s xml:id="echoid-s17372" xml:space="preserve">ſin. </s> <s xml:id="echoid-s17373" xml:space="preserve">A D B: </s> <s xml:id="echoid-s17374" xml:space="preserve">ſin. </s> <s xml:id="echoid-s17375" xml:space="preserve">A B D; </s> <s xml:id="echoid-s17376" xml:space="preserve">mais à cauſe <lb/>des paralleles C D, A B, l’angle C A D = l’angle A D B, l’angle <pb o="552" file="0630" n="650" rhead="NOUVEAU COURS"/> B A D = l’angle B A D, & </s> <s xml:id="echoid-s17377" xml:space="preserve">le ſinus de l’angle A B D eſt le <lb/>même que celui de l’angle A B C: </s> <s xml:id="echoid-s17378" xml:space="preserve">donc on aura cette propor-<lb/>tion, A B : </s> <s xml:id="echoid-s17379" xml:space="preserve">A C : </s> <s xml:id="echoid-s17380" xml:space="preserve">A D :</s> <s xml:id="echoid-s17381" xml:space="preserve">: ſin. </s> <s xml:id="echoid-s17382" xml:space="preserve">C A D: </s> <s xml:id="echoid-s17383" xml:space="preserve">ſin. </s> <s xml:id="echoid-s17384" xml:space="preserve">B A D: </s> <s xml:id="echoid-s17385" xml:space="preserve">ſin. </s> <s xml:id="echoid-s17386" xml:space="preserve">B A C; <lb/></s> <s xml:id="echoid-s17387" xml:space="preserve">d’où il ſuit que chaque puiſſance eſt repréſentée par le ſinus <lb/>de l’angle formé par les directions des deux puiſſances que l’on <lb/>ne compare pas.</s> <s xml:id="echoid-s17388" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1424" type="section" level="1" n="1049"> <head xml:id="echoid-head1237" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s17389" xml:space="preserve">1048. </s> <s xml:id="echoid-s17390" xml:space="preserve">Il ſuit encore delà que ſi l’on a trois forces repré-<lb/>ſentées par les lignes P, Q, R à mettre en équilibre, il n’y a <lb/>qu’à former un triangle A B D avec ces trois lignes ou leurs <lb/>égales, & </s> <s xml:id="echoid-s17391" xml:space="preserve">achevant enſuite le parallélogramme A B D C, les <lb/>lignes A B, A C, A D diſpoſées comme elles ſe trouveront par <lb/>la conſtruction du parallélogramme, détermineront les ſitua-<lb/>tions reſpectives des puiſſances données dans le cas de l’équi-<lb/>libre.</s> <s xml:id="echoid-s17392" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1425" type="section" level="1" n="1050"> <head xml:id="echoid-head1238" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <p> <s xml:id="echoid-s17393" xml:space="preserve">1049. </s> <s xml:id="echoid-s17394" xml:space="preserve">De plus, comme chaque côté A B, A D, B D du <lb/>triangle A B D peut être pris pour la diagonale du parallélo-<lb/>gramme à conſtruire, il s’enſuit que trois forces pourront re-<lb/>cevoir trois diſpoſitions différentes, & </s> <s xml:id="echoid-s17395" xml:space="preserve">toutes les trois propres <lb/>à produire l’équilibre.</s> <s xml:id="echoid-s17396" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1426" type="section" level="1" n="1051"> <head xml:id="echoid-head1239" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s17397" xml:space="preserve">1050. </s> <s xml:id="echoid-s17398" xml:space="preserve">Il ſuit encore delà que ſi l’on donne un nombre quel-<lb/>conque de forces déterminées de grandeur & </s> <s xml:id="echoid-s17399" xml:space="preserve">de poſition qui <lb/>tirent toutes dans un même plan, & </s> <s xml:id="echoid-s17400" xml:space="preserve">qui ſont appliquées au <lb/>même corps, on pourra toujours déterminer la réſultante de <lb/>toutes ces forces, ſoit pour ſa direction, ſoit pour ſa quantité <lb/>de force. </s> <s xml:id="echoid-s17401" xml:space="preserve">Pour cela on commencera par chercher la réſultante <lb/>de deux forces quelconques; </s> <s xml:id="echoid-s17402" xml:space="preserve">enſuite on cherchera la réſul-<lb/>tante de cette nouvelle force équivalente aux deux autres, & </s> <s xml:id="echoid-s17403" xml:space="preserve"><lb/>d’une troiſieme; </s> <s xml:id="echoid-s17404" xml:space="preserve">ce qui réduira trois forces à une ſeule: </s> <s xml:id="echoid-s17405" xml:space="preserve">on <lb/>continuera le même procédé juſqu’à ce que l’on n’ait plus qu’une <lb/>ſeule force, & </s> <s xml:id="echoid-s17406" xml:space="preserve">alors la réſultante derniere ſera celle qu’on <lb/>demande.</s> <s xml:id="echoid-s17407" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1427" type="section" level="1" n="1052"> <head xml:id="echoid-head1240" xml:space="preserve"><emph style="sc">Scholie</emph>.</head> <p> <s xml:id="echoid-s17408" xml:space="preserve">1051. </s> <s xml:id="echoid-s17409" xml:space="preserve">Il ſeroit aiſé de déduire encore un grand nombre de <lb/>corollaires de cette propoſition, & </s> <s xml:id="echoid-s17410" xml:space="preserve">l’on peut dire même que <pb o="553" file="0631" n="651" rhead="DE MATHÉMATIQUE. Liv. XV."/> toute la théorie de la méchanique ſtatique n’eſt qu’une ſuite <lb/>de conſéquences déduites de ce principe. </s> <s xml:id="echoid-s17411" xml:space="preserve">Il étoit donc de la <lb/>derniere importance de le démontrer dans toute la rigueur <lb/>poſſible: </s> <s xml:id="echoid-s17412" xml:space="preserve">peut-être la démonſtration que j’apporte paroîtra-<lb/>t’elle un peu longue; </s> <s xml:id="echoid-s17413" xml:space="preserve">mais on ſentira bientôt que cette lon-<lb/>gueur eſt pardonnable, ſi l’on veut approfondir les démonſ-<lb/>trations de pluſieurs Auteurs. </s> <s xml:id="echoid-s17414" xml:space="preserve">Il ne leur eſt pas bien difficile <lb/>de démontrer que le corps décrit la diagonale, lorſqu’ils ont <lb/>tellement combiné les forces motrices ou tractives, que le <lb/>corps eſt néceſſairement obligé de ſe mouvoir en diagonale. <lb/></s> <s xml:id="echoid-s17415" xml:space="preserve">Ce n’eſt pas là l’état de la queſtion. </s> <s xml:id="echoid-s17416" xml:space="preserve">Il faut, comme le dit <lb/>M. </s> <s xml:id="echoid-s17417" xml:space="preserve">d’Alembert, laiſſer le corps libre de choiſir telle direction <lb/>qu’il voudra, & </s> <s xml:id="echoid-s17418" xml:space="preserve">faire voir enſuite que cette direction doit ſe <lb/>trouver abſolument ſur la diagonale, & </s> <s xml:id="echoid-s17419" xml:space="preserve">que la force réſultante <lb/>doit être repréſentée par cette même diagonale; </s> <s xml:id="echoid-s17420" xml:space="preserve">c’eſt ce que <lb/>je crois avoir fait dans lesart. </s> <s xml:id="echoid-s17421" xml:space="preserve">1042 & </s> <s xml:id="echoid-s17422" xml:space="preserve">1043. </s> <s xml:id="echoid-s17423" xml:space="preserve">Dans ce dernier, la <lb/>direction eſt priſe au hazard, & </s> <s xml:id="echoid-s17424" xml:space="preserve">je démontre que la force réſul-<lb/>tante eſt exprimée par la diagonale, quelle que ſoit la direction de <lb/>cette réſultante; </s> <s xml:id="echoid-s17425" xml:space="preserve">d’où il ſuit que puiſque les quatre forces dont il <lb/>eſt queſtion dans cet article, produiſent une diagonale, les quatre <lb/>dont il étoit queſtion dans le précédent, & </s> <s xml:id="echoid-s17426" xml:space="preserve">qui ſont, ainſi que <lb/>les quatre premieres équivalentes aux deux forces motrices <lb/>M & </s> <s xml:id="echoid-s17427" xml:space="preserve">N, doivent auſſi produire une diagonale; </s> <s xml:id="echoid-s17428" xml:space="preserve">d’où il ſuit <lb/> <anchor type="note" xlink:label="note-0631-01a" xlink:href="note-0631-01"/> que le corps ne peut pas décrire la ligne AE; </s> <s xml:id="echoid-s17429" xml:space="preserve">ce qui fixe par <lb/>conſéquent la direction du corps ſur la diagonale. </s> <s xml:id="echoid-s17430" xml:space="preserve">J’ai auſſi <lb/>ſuppoſé deux forces ſimplement motrices: </s> <s xml:id="echoid-s17431" xml:space="preserve">car ſi la propoſition <lb/>eſt vraie dans ce cas, elle le ſera auſſi dans le cas des forces <lb/>tractives, parce que l’on peut regarder la force qui meut un <lb/>corps, après que la force motrice a agi ſur lui dans un inſtant, <lb/>comme une force tractive.</s> <s xml:id="echoid-s17432" xml:space="preserve"/> </p> <div xml:id="echoid-div1427" type="float" level="2" n="1"> <note position="right" xlink:label="note-0631-01" xlink:href="note-0631-01a" xml:space="preserve">Figure 355.</note> </div> </div> <div xml:id="echoid-div1429" type="section" level="1" n="1053"> <head xml:id="echoid-head1241" xml:space="preserve">CHAPITRE II,</head> <head xml:id="echoid-head1242" style="it" xml:space="preserve">Où l’on fait voir le rapport des puiſſances qui ſoutiennent des <lb/>poids avec des cordes.</head> <p> <s xml:id="echoid-s17433" xml:space="preserve">1052. </s> <s xml:id="echoid-s17434" xml:space="preserve"><emph style="sc">Comme</emph> nous avons conſidéré dans le Traité du Mou-<lb/>vement la théorie des corps qui ſe choquent ou qui ſe rencon-<lb/>trent, celle des corps jettés ſelon des directions perpendicu- <pb o="554" file="0632" n="652" rhead="NOUVEAU COURS"/> laires, obliques ou paralleles à l’horizon; </s> <s xml:id="echoid-s17435" xml:space="preserve">il ſemble que, pour <lb/>ſuivre un ordre dans la méchanique, dont l’objet eſt de con-<lb/>ſidérer en équilibre les corps qui tendent naturellement à ſe <lb/>mouvoir, il eſt néceſſaire d’expliquer, avant toutes choſes, <lb/>ce qui a le plus de rapport avec ce qui précéde immédiatement: <lb/></s> <s xml:id="echoid-s17436" xml:space="preserve">or ce ſera ſans doute la théorie des corps ſoutenus par des <lb/>puiſſances qui ſont en équilibre avec ces corps dans toutes les <lb/>ſituations qu’on peut leur donner; </s> <s xml:id="echoid-s17437" xml:space="preserve">& </s> <s xml:id="echoid-s17438" xml:space="preserve">c’eſt ce qu’on ſe propoſe <lb/>d’enſeigner dans ce ſecond chapitre, parce qu’après cela nous <lb/>ferons voir dans le troiſieme les poids qui tendent à rouler ſur <lb/>des plans inclinés, & </s> <s xml:id="echoid-s17439" xml:space="preserve">le rapport de leur peſanteur avec les <lb/>puiſſances qui les ſoutiennent en repos.</s> <s xml:id="echoid-s17440" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1430" type="section" level="1" n="1054"> <head xml:id="echoid-head1243" xml:space="preserve">PROPOSITION.</head> <head xml:id="echoid-head1244" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s17441" xml:space="preserve">1053. </s> <s xml:id="echoid-s17442" xml:space="preserve">Si les deux puiſſances P & </s> <s xml:id="echoid-s17443" xml:space="preserve">Q ſoutiennent un poids R <lb/> <anchor type="note" xlink:label="note-0632-01a" xlink:href="note-0632-01"/> tendant à ſuivre la direction BR, je dis que ces deux puiſſances <lb/> <anchor type="note" xlink:label="note-0632-02a" xlink:href="note-0632-02"/> ſeront en équilibre entr’elles, ſi elles ſont en raiſon réciproque des <lb/>perpendiculaires BC & </s> <s xml:id="echoid-s17444" xml:space="preserve">BG, tirées d’un des points B de la direc-<lb/>tion BR ſur les directions FP & </s> <s xml:id="echoid-s17445" xml:space="preserve">FQ, c’eſt-à-dire que P : </s> <s xml:id="echoid-s17446" xml:space="preserve">Q <lb/>:</s> <s xml:id="echoid-s17447" xml:space="preserve">: BG: </s> <s xml:id="echoid-s17448" xml:space="preserve">BC.</s> <s xml:id="echoid-s17449" xml:space="preserve"/> </p> <div xml:id="echoid-div1430" type="float" level="2" n="1"> <note position="left" xlink:label="note-0632-01" xlink:href="note-0632-01a" xml:space="preserve">Pl. XXVII.</note> <note position="left" xlink:label="note-0632-02" xlink:href="note-0632-02a" xml:space="preserve">Figure 360.</note> </div> </div> <div xml:id="echoid-div1432" type="section" level="1" n="1055"> <head xml:id="echoid-head1245" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s17450" xml:space="preserve">Pour que ces deux puiſſances faſſent équilibre entr’elles, il <lb/>faut qu’elles ſoient comme les côtés FE & </s> <s xml:id="echoid-s17451" xml:space="preserve">FD d’un parallé-<lb/>logramme, dont la diagonale BF exprimeroit la force ou la <lb/>peſanteur du poids R, parce que pour lors le poids R étant <lb/>pris pour la puiſſance réſiftante, il ſera en équilibre avec les <lb/>deux puiſſances agiſſantes, parce qu’il ſe trouvera de part & </s> <s xml:id="echoid-s17452" xml:space="preserve"><lb/>d’autre une égalité de force; </s> <s xml:id="echoid-s17453" xml:space="preserve">mais prenant BD à la place de <lb/>EF, nous aurons les côtés BD & </s> <s xml:id="echoid-s17454" xml:space="preserve">DF du triangle BDF, qui <lb/>feront dans la raiſon des puiſſances P & </s> <s xml:id="echoid-s17455" xml:space="preserve">Q; </s> <s xml:id="echoid-s17456" xml:space="preserve">& </s> <s xml:id="echoid-s17457" xml:space="preserve">comme les <lb/>côtés BD & </s> <s xml:id="echoid-s17458" xml:space="preserve">DF ſont auſſi dans la raiſon des ſinus de leurs <lb/>angles oppoſés, qui ne ſont autre choſe que les perpendicu-<lb/>laires BC & </s> <s xml:id="echoid-s17459" xml:space="preserve">BG, l’on aura donc P : </s> <s xml:id="echoid-s17460" xml:space="preserve">Q :</s> <s xml:id="echoid-s17461" xml:space="preserve">: BC : </s> <s xml:id="echoid-s17462" xml:space="preserve">BG. <lb/></s> <s xml:id="echoid-s17463" xml:space="preserve">C. </s> <s xml:id="echoid-s17464" xml:space="preserve">Q. </s> <s xml:id="echoid-s17465" xml:space="preserve">F. </s> <s xml:id="echoid-s17466" xml:space="preserve">D.</s> <s xml:id="echoid-s17467" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17468" xml:space="preserve">De même ſi d’un point D de la direction F Q l’on tire les <lb/> <anchor type="note" xlink:label="note-0632-03a" xlink:href="note-0632-03"/> perpendiculaires DG & </s> <s xml:id="echoid-s17469" xml:space="preserve">DC ſur les directions BR & </s> <s xml:id="echoid-s17470" xml:space="preserve">FP, l’on <lb/>aura le rapport de la puiſſance P au poids Q, étant en raiſon <lb/>réciproque des perpendiculaires DC & </s> <s xml:id="echoid-s17471" xml:space="preserve">DG: </s> <s xml:id="echoid-s17472" xml:space="preserve">car à cauſe que <pb o="555" file="0633" n="653" rhead="DE MATHÉMATIQUE. Liv. XV."/> ces perpendiculaires ſont les ſinus des angles oppoſés aux côtés <lb/>BF & </s> <s xml:id="echoid-s17473" xml:space="preserve">BD du triangle BDF, l’on aura BD : </s> <s xml:id="echoid-s17474" xml:space="preserve">BF :</s> <s xml:id="echoid-s17475" xml:space="preserve">: DG : </s> <s xml:id="echoid-s17476" xml:space="preserve">DC, <lb/>ou bien P : </s> <s xml:id="echoid-s17477" xml:space="preserve">R :</s> <s xml:id="echoid-s17478" xml:space="preserve">: DG : </s> <s xml:id="echoid-s17479" xml:space="preserve">DC.</s> <s xml:id="echoid-s17480" xml:space="preserve"/> </p> <div xml:id="echoid-div1432" type="float" level="2" n="1"> <note position="left" xlink:label="note-0632-03" xlink:href="note-0632-03a" xml:space="preserve">Figure 361.</note> </div> <p> <s xml:id="echoid-s17481" xml:space="preserve">Enfin ſi du point E, pris dans la direction de la puiſſance <lb/> <anchor type="note" xlink:label="note-0633-01a" xlink:href="note-0633-01"/> P, l’on abaiſſe les perpendiculaires EG & </s> <s xml:id="echoid-s17482" xml:space="preserve">EC ſur les direc-<lb/>tions des puiſſances R & </s> <s xml:id="echoid-s17483" xml:space="preserve">Q, l’on aura encore Q : </s> <s xml:id="echoid-s17484" xml:space="preserve">R :</s> <s xml:id="echoid-s17485" xml:space="preserve">: EG : </s> <s xml:id="echoid-s17486" xml:space="preserve">EC.</s> <s xml:id="echoid-s17487" xml:space="preserve"/> </p> <div xml:id="echoid-div1433" type="float" level="2" n="2"> <note position="right" xlink:label="note-0633-01" xlink:href="note-0633-01a" xml:space="preserve">Figure 362.</note> </div> </div> <div xml:id="echoid-div1435" type="section" level="1" n="1056"> <head xml:id="echoid-head1246" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s17488" xml:space="preserve">1054. </s> <s xml:id="echoid-s17489" xml:space="preserve">Il ſuit delà que ſi l’on ſuppoſe que le poids R diminue <lb/> <anchor type="note" xlink:label="note-0633-02a" xlink:href="note-0633-02"/> continuellement, les deux puiſſances P & </s> <s xml:id="echoid-s17490" xml:space="preserve">Q demeurant les <lb/>mêmes, la diagonale BF du parallélogramme ED, diminuera <lb/>à proportion du corps R. </s> <s xml:id="echoid-s17491" xml:space="preserve">Or comme les côtés FD & </s> <s xml:id="echoid-s17492" xml:space="preserve">FE de-<lb/>meureront les mêmes, l’angle EFD augmentera, parce que <lb/>les puiſſances P & </s> <s xml:id="echoid-s17493" xml:space="preserve">Q deſcendront, & </s> <s xml:id="echoid-s17494" xml:space="preserve">le poids R remontera: <lb/></s> <s xml:id="echoid-s17495" xml:space="preserve">mais tant que le poids R ſera d’une grandeur finie, la diago-<lb/>nale BF ſera toujours une ligne finie, & </s> <s xml:id="echoid-s17496" xml:space="preserve">pourra toujours for-<lb/>mer le parallélogramme ED, & </s> <s xml:id="echoid-s17497" xml:space="preserve">par conſéquent les directions <lb/>FP & </s> <s xml:id="echoid-s17498" xml:space="preserve">FQ formeront toujours un angle en F.</s> <s xml:id="echoid-s17499" xml:space="preserve"/> </p> <div xml:id="echoid-div1435" type="float" level="2" n="1"> <note position="right" xlink:label="note-0633-02" xlink:href="note-0633-02a" xml:space="preserve">Figure 363.</note> </div> </div> <div xml:id="echoid-div1437" type="section" level="1" n="1057"> <head xml:id="echoid-head1247" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s17500" xml:space="preserve">1055. </s> <s xml:id="echoid-s17501" xml:space="preserve">Il ſuit delà qu’une corde ne peut jamais être tendue <lb/>en ligne droite que par une puiſſance infinie: </s> <s xml:id="echoid-s17502" xml:space="preserve">car ſon poids, <lb/>quelque petit qu’on le ſuppoſe, ſera toujours d’une grandeur <lb/>finie, & </s> <s xml:id="echoid-s17503" xml:space="preserve">peut être regardé, étant réuni en un ſeul point, <lb/>comme le poids R attaché à quelqu’un des points F de la même <lb/>corde.</s> <s xml:id="echoid-s17504" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1438" type="section" level="1" n="1058"> <head xml:id="echoid-head1248" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s17505" xml:space="preserve">1056. </s> <s xml:id="echoid-s17506" xml:space="preserve">Si des points E & </s> <s xml:id="echoid-s17507" xml:space="preserve">D l’on abaiſſe les perpendiculaires <lb/> <anchor type="note" xlink:label="note-0633-03a" xlink:href="note-0633-03"/> EG & </s> <s xml:id="echoid-s17508" xml:space="preserve">DH ſur la direction BR, & </s> <s xml:id="echoid-s17509" xml:space="preserve">qu’on acheve les parallé-<lb/>logrammes rectangles GI & </s> <s xml:id="echoid-s17510" xml:space="preserve">HK, l’on aura les côtés EI & </s> <s xml:id="echoid-s17511" xml:space="preserve"><lb/>IE, qui repréſenteront deux forces égales à la force EF, & </s> <s xml:id="echoid-s17512" xml:space="preserve"><lb/>les deux côtés FK & </s> <s xml:id="echoid-s17513" xml:space="preserve">KD, qui exprimeront auſſi deux forces <lb/>égales à DF (art. </s> <s xml:id="echoid-s17514" xml:space="preserve">1045); </s> <s xml:id="echoid-s17515" xml:space="preserve">mais IF & </s> <s xml:id="echoid-s17516" xml:space="preserve">FK ſont deux forces <lb/>égales qui ne ſoutiennent aucune partie du poids R: </s> <s xml:id="echoid-s17517" xml:space="preserve">ainſi la <lb/>partie du poids que ſoutient la puiſſance Q, ſera exprimée <lb/>par DK, & </s> <s xml:id="echoid-s17518" xml:space="preserve">la partie du poids que ſoutient la puiſſance P, <lb/>ſera exprimée par EI. </s> <s xml:id="echoid-s17519" xml:space="preserve">Il s’enſuit donc que les parties du poids <lb/>R que ſoutiennent les puiſſances P & </s> <s xml:id="echoid-s17520" xml:space="preserve">Q, ſont l’une à l’autre, <lb/>comme EI eſt à DK, ou comme GF eſt à HF: </s> <s xml:id="echoid-s17521" xml:space="preserve">mais comme <pb o="556" file="0634" n="654" rhead="NOUVEAU COURS"/> BH eſt égal à GF, BF exprimera toute la peſanteur du poids: <lb/></s> <s xml:id="echoid-s17522" xml:space="preserve">ainſi l’on aura donc P : </s> <s xml:id="echoid-s17523" xml:space="preserve">R :</s> <s xml:id="echoid-s17524" xml:space="preserve">: EI, ou G F : </s> <s xml:id="echoid-s17525" xml:space="preserve">BF; </s> <s xml:id="echoid-s17526" xml:space="preserve">& </s> <s xml:id="echoid-s17527" xml:space="preserve">de l’autre <lb/>part Q : </s> <s xml:id="echoid-s17528" xml:space="preserve">R :</s> <s xml:id="echoid-s17529" xml:space="preserve">: D K ou HF : </s> <s xml:id="echoid-s17530" xml:space="preserve">BF.</s> <s xml:id="echoid-s17531" xml:space="preserve"/> </p> <div xml:id="echoid-div1438" type="float" level="2" n="1"> <note position="right" xlink:label="note-0633-03" xlink:href="note-0633-03a" xml:space="preserve">Figure 364.</note> </div> </div> <div xml:id="echoid-div1440" type="section" level="1" n="1059"> <head xml:id="echoid-head1249" xml:space="preserve"><emph style="sc">Corollaire</emph>. IV.</head> <p> <s xml:id="echoid-s17532" xml:space="preserve">1057. </s> <s xml:id="echoid-s17533" xml:space="preserve">Mais ſi la puiſſance Q étoit dans la ligne horizontale <lb/> <anchor type="note" xlink:label="note-0634-01a" xlink:href="note-0634-01"/> ED, & </s> <s xml:id="echoid-s17534" xml:space="preserve">que la puiſſance P fût au deſſus de l’horizontale, cette <lb/>puiſſance ſoutiendra elle ſeule tout le poids R: </s> <s xml:id="echoid-s17535" xml:space="preserve">car ayant achevé <lb/>le parallélogramme rectangle BE, la perpendiculaire HE ex-<lb/>primera la partie du poids R, que porte la puiſſance P; </s> <s xml:id="echoid-s17536" xml:space="preserve">mais <lb/>HE eſt égal à la diagonale BF, qui exprime toute la pe-<lb/>ſanteur du poids: </s> <s xml:id="echoid-s17537" xml:space="preserve">ainſi la puiſſance P ſoutiendra tout le <lb/>poids.</s> <s xml:id="echoid-s17538" xml:space="preserve"/> </p> <div xml:id="echoid-div1440" type="float" level="2" n="1"> <note position="left" xlink:label="note-0634-01" xlink:href="note-0634-01a" xml:space="preserve">Figure 365.</note> </div> </div> <div xml:id="echoid-div1442" type="section" level="1" n="1060"> <head xml:id="echoid-head1250" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s17539" xml:space="preserve">1058. </s> <s xml:id="echoid-s17540" xml:space="preserve">Mais ſi la puiſſance Q étoit au deſſous de l’horizon-<lb/> <anchor type="note" xlink:label="note-0634-02a" xlink:href="note-0634-02"/> tale HL, & </s> <s xml:id="echoid-s17541" xml:space="preserve">la puiſſance P au deſſus, il arrivera que la puiſ-<lb/>ſance P ſoutiendra non ſeulement tout le poids R, mais en-<lb/>core la partie du poids que ſoutiendroit la puiſſance Q, ſi <lb/>elle étoit autant au deſſus de l’horizontale HL, comme elle <lb/>ſe trouve ici au deſſous: </s> <s xml:id="echoid-s17542" xml:space="preserve">car ayant formé les parallélogrammes <lb/>rectangles IH & </s> <s xml:id="echoid-s17543" xml:space="preserve">GK, la ligne EH exprimera ce que porte la <lb/>puiſſance P, & </s> <s xml:id="echoid-s17544" xml:space="preserve">la ligne FK exprimera l’effort que fait la puiſ-<lb/>ſance Q. </s> <s xml:id="echoid-s17545" xml:space="preserve">Or comme FK eſt égal à IB, il s’enſuit que E H <lb/>ou IF eſt compoſé de B F & </s> <s xml:id="echoid-s17546" xml:space="preserve">de BI, c’eſt-à-dire de B F, qui <lb/>exprime la peſanteur du poids, & </s> <s xml:id="echoid-s17547" xml:space="preserve">de BI qui eſt la partie du <lb/>poids R que ſoutiendra la puiſſance Q, ſi elle étoit autant au <lb/>deſſus de l’horizontale H L qu’elle eſt au deſſous: </s> <s xml:id="echoid-s17548" xml:space="preserve">ce qui fait <lb/>voir que la puiſſance P ſoutient plus que la peſanteur du <lb/>poids R.</s> <s xml:id="echoid-s17549" xml:space="preserve"/> </p> <div xml:id="echoid-div1442" type="float" level="2" n="1"> <note position="left" xlink:label="note-0634-02" xlink:href="note-0634-02a" xml:space="preserve">Figure 366.</note> </div> </div> <div xml:id="echoid-div1444" type="section" level="1" n="1061"> <head xml:id="echoid-head1251" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s17550" xml:space="preserve">1059. </s> <s xml:id="echoid-s17551" xml:space="preserve">Enfin il ſuit delà que ſi l’on a un corps peſant HI, <lb/> <anchor type="note" xlink:label="note-0634-03a" xlink:href="note-0634-03"/> ſoutenu par deux puiſſances P & </s> <s xml:id="echoid-s17552" xml:space="preserve">Q, ces deux puiſſances ſeront <lb/>en équilibre, ſi elles ſont en raiſon réciproques des perpendi-<lb/>culaires F G & </s> <s xml:id="echoid-s17553" xml:space="preserve">F C, tirées d’un des points de la direction B F <lb/>ſur celles des puiſſances P & </s> <s xml:id="echoid-s17554" xml:space="preserve">Q: </s> <s xml:id="echoid-s17555" xml:space="preserve">car ſi l’on ſuppoſe que toute <lb/>la peſanteur du corps H I ſoit ramaſſée autour de ſon centre <lb/>de gravité F pour ſormer le poids R, il faudra, pour ſoutenir <lb/>ce poids, que P ſoit à Q, comme B E eſt à B D, ou comme <lb/>F D eſt à B D. </s> <s xml:id="echoid-s17556" xml:space="preserve">Or comme les ſinus des angles dans le triangle <pb o="557" file="0635" n="655" rhead="DE MATHÉMATIQUE. Liv. XV."/> FBD ſont dans la même raiſon que leurs côtés oppoſés, FG <lb/>étant le ſinus de l’angle FBG, & </s> <s xml:id="echoid-s17557" xml:space="preserve">FC le ſinus de l’angle BFD, <lb/>puiſqu’il eſt celui de ſon alterne CBF, l’on aura FD : </s> <s xml:id="echoid-s17558" xml:space="preserve">BD :</s> <s xml:id="echoid-s17559" xml:space="preserve">: <lb/>FG : </s> <s xml:id="echoid-s17560" xml:space="preserve">FC, ou bien BE : </s> <s xml:id="echoid-s17561" xml:space="preserve">BD :</s> <s xml:id="echoid-s17562" xml:space="preserve">: FG : </s> <s xml:id="echoid-s17563" xml:space="preserve">FC; </s> <s xml:id="echoid-s17564" xml:space="preserve">par conſéquent <lb/>P : </s> <s xml:id="echoid-s17565" xml:space="preserve">Q :</s> <s xml:id="echoid-s17566" xml:space="preserve">: FG : </s> <s xml:id="echoid-s17567" xml:space="preserve">FC.</s> <s xml:id="echoid-s17568" xml:space="preserve"/> </p> <div xml:id="echoid-div1444" type="float" level="2" n="1"> <note position="left" xlink:label="note-0634-03" xlink:href="note-0634-03a" xml:space="preserve">Figure 367.</note> </div> <p> <s xml:id="echoid-s17569" xml:space="preserve">Mais ſi le corps peſant HI étoit appuyé par une de ſes ex-<lb/>trêmités H, & </s> <s xml:id="echoid-s17570" xml:space="preserve">ſoutenu ſeulement à l’extrêmité I par la puiſ-<lb/>ſance Q, cette puiſſance Q ſera au poids R, comme BD eſt <lb/>à BF; </s> <s xml:id="echoid-s17571" xml:space="preserve">& </s> <s xml:id="echoid-s17572" xml:space="preserve">comme ces lignes ſont les côtés du triangle BFD, <lb/>elles ſeront dans la raiſon des ſinus des angles BFD & </s> <s xml:id="echoid-s17573" xml:space="preserve">BDF, <lb/>qui ſont les perpendiculaires EG & </s> <s xml:id="echoid-s17574" xml:space="preserve">EC; </s> <s xml:id="echoid-s17575" xml:space="preserve">ce qui fait voir que <lb/>la puiſſance Q eſt au poids R dans la raiſon réciproque des <lb/>perpendiculaires E C & </s> <s xml:id="echoid-s17576" xml:space="preserve">E G, tirées d’un des points E de la di-<lb/>rection de la puiſſance P ſur celles des puiſſances Q & </s> <s xml:id="echoid-s17577" xml:space="preserve">R.</s> <s xml:id="echoid-s17578" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1446" type="section" level="1" n="1062"> <head xml:id="echoid-head1252" xml:space="preserve">CHAPITRE III.</head> <head xml:id="echoid-head1253" style="it" xml:space="preserve">Du Plan incliné.</head> <head xml:id="echoid-head1254" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s17579" xml:space="preserve">1060. </s> <s xml:id="echoid-s17580" xml:space="preserve"><emph style="sc">On</emph> appelle plan incliné toute ſuperficie inclinée à <lb/>l’horizon, le long de laquelle on fait mouvoir un poids. </s> <s xml:id="echoid-s17581" xml:space="preserve">Ce <lb/>plan peut toujours être exprimé par l’hypoténuſe d’un triangle <lb/>rectangle.</s> <s xml:id="echoid-s17582" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1447" type="section" level="1" n="1063"> <head xml:id="echoid-head1255" xml:space="preserve">PROPOSITION.</head> <head xml:id="echoid-head1256" xml:space="preserve"><emph style="sc">Theoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s17583" xml:space="preserve">1061. </s> <s xml:id="echoid-s17584" xml:space="preserve">Si une puiſſance Q ſoutient un poids ſphérique P par une <lb/> <anchor type="note" xlink:label="note-0635-01a" xlink:href="note-0635-01"/> ligne de direction D E, parallele au plan incliné A B, je dis, <lb/> <anchor type="note" xlink:label="note-0635-02a" xlink:href="note-0635-02"/> 1°. </s> <s xml:id="echoid-s17585" xml:space="preserve">que la puiſſance ſera au poids, comme la hauteur du plan in-<lb/>cliné eſt à ſa longueur, c’eſt-à-dire que Q : </s> <s xml:id="echoid-s17586" xml:space="preserve">P :</s> <s xml:id="echoid-s17587" xml:space="preserve">: BC : </s> <s xml:id="echoid-s17588" xml:space="preserve">BA.</s> <s xml:id="echoid-s17589" xml:space="preserve"/> </p> <div xml:id="echoid-div1447" type="float" level="2" n="1"> <note position="right" xlink:label="note-0635-01" xlink:href="note-0635-01a" xml:space="preserve">Pl. <emph style="sc">XXVIII</emph>.</note> <note position="right" xlink:label="note-0635-02" xlink:href="note-0635-02a" xml:space="preserve">Figure 369.</note> </div> <p style="it"> <s xml:id="echoid-s17590" xml:space="preserve">2°. </s> <s xml:id="echoid-s17591" xml:space="preserve">Que ſi le poids eſt ſoutenu par une puiſſance Q, qui tire <lb/> <anchor type="note" xlink:label="note-0635-03a" xlink:href="note-0635-03"/> ſelon une direction DE, parallele à la baſe AC du plan, la puiſ-<lb/>ſance ſera au poids comme la hauteur du plan eſt à la longueur <lb/>de ſa baſe, c’eſt-à-dire que Q : </s> <s xml:id="echoid-s17592" xml:space="preserve">P :</s> <s xml:id="echoid-s17593" xml:space="preserve">: BC : </s> <s xml:id="echoid-s17594" xml:space="preserve">AC.</s> <s xml:id="echoid-s17595" xml:space="preserve"/> </p> <div xml:id="echoid-div1448" type="float" level="2" n="2"> <note position="right" xlink:label="note-0635-03" xlink:href="note-0635-03a" xml:space="preserve">Figure 370.</note> </div> </div> <div xml:id="echoid-div1450" type="section" level="1" n="1064"> <head xml:id="echoid-head1257" xml:space="preserve"><emph style="sc">Démonstration du premier cas</emph>.</head> <p> <s xml:id="echoid-s17596" xml:space="preserve">Si l’on tire la ligne DF perpendiculaire ſur le plan incliné <lb/> <anchor type="note" xlink:label="note-0635-04a" xlink:href="note-0635-04"/> AB, cette ligne ſera la direction de la puiſſance réſiſtante: </s> <s xml:id="echoid-s17597" xml:space="preserve">&</s> <s xml:id="echoid-s17598" xml:space="preserve"> <pb o="558" file="0636" n="656" rhead="NOUVEAU COURS"/> faiſant le parallélogramme I G, le côté D G exprimera une <lb/>des puiſſances agiſſantes, & </s> <s xml:id="echoid-s17599" xml:space="preserve">le côté D I l’autre puiſſance agiſ-<lb/>ſante, & </s> <s xml:id="echoid-s17600" xml:space="preserve">ces deux puiſſances agiſſantes enſemble ſeront en <lb/>équilibre avec la puiſſance réſiſtante D F; </s> <s xml:id="echoid-s17601" xml:space="preserve">mais ces deux puiſ-<lb/>ſances étant l’une à l’autre comme D G eſt à D I, ſeront <lb/>comme les côtés I F & </s> <s xml:id="echoid-s17602" xml:space="preserve">I D du triangle rectangle D I F; </s> <s xml:id="echoid-s17603" xml:space="preserve">& </s> <s xml:id="echoid-s17604" xml:space="preserve"><lb/>comme ce triangle eſt ſemblable au triangle A B C, l’on aura <lb/>I F, ou D G : </s> <s xml:id="echoid-s17605" xml:space="preserve">I D :</s> <s xml:id="echoid-s17606" xml:space="preserve">: B C : </s> <s xml:id="echoid-s17607" xml:space="preserve">B A, ou bien Q : </s> <s xml:id="echoid-s17608" xml:space="preserve">P :</s> <s xml:id="echoid-s17609" xml:space="preserve">: B C : </s> <s xml:id="echoid-s17610" xml:space="preserve">B A.</s> <s xml:id="echoid-s17611" xml:space="preserve"/> </p> <div xml:id="echoid-div1450" type="float" level="2" n="1"> <note position="right" xlink:label="note-0635-04" xlink:href="note-0635-04a" xml:space="preserve">Figure 369.</note> </div> </div> <div xml:id="echoid-div1452" type="section" level="1" n="1065"> <head xml:id="echoid-head1258" xml:space="preserve"><emph style="sc">Démonstration du second cas</emph>.</head> <p> <s xml:id="echoid-s17612" xml:space="preserve">1062. </s> <s xml:id="echoid-s17613" xml:space="preserve">Si la direction D E de la puiſſance Q eſt parallele à <lb/> <anchor type="note" xlink:label="note-0636-01a" xlink:href="note-0636-01"/> la baſe A C du plan incliné, il ſera facile de prouver que <lb/>Q : </s> <s xml:id="echoid-s17614" xml:space="preserve">P :</s> <s xml:id="echoid-s17615" xml:space="preserve">: B C : </s> <s xml:id="echoid-s17616" xml:space="preserve">C A: </s> <s xml:id="echoid-s17617" xml:space="preserve">car ſi la ligne D F eſt perpendiculaire ſur <lb/>A B, elle exprimera encore la puiſſance réſiſtante; </s> <s xml:id="echoid-s17618" xml:space="preserve">& </s> <s xml:id="echoid-s17619" xml:space="preserve">ſi l’on <lb/>fait le parallélogramme rectangle I G, l’on aura Q : </s> <s xml:id="echoid-s17620" xml:space="preserve">P :</s> <s xml:id="echoid-s17621" xml:space="preserve">: D G : </s> <s xml:id="echoid-s17622" xml:space="preserve">D I. <lb/></s> <s xml:id="echoid-s17623" xml:space="preserve">Or ſi à la place du D G on prend I F, l’on aura les côtés I F <lb/>& </s> <s xml:id="echoid-s17624" xml:space="preserve">I D du triangle rectangle D I F, qui ſeront comme Q eſt à <lb/>P : </s> <s xml:id="echoid-s17625" xml:space="preserve">& </s> <s xml:id="echoid-s17626" xml:space="preserve">comme ce triangle eſt ſemblable au triangle A C B, l’on <lb/>aura F I : </s> <s xml:id="echoid-s17627" xml:space="preserve">I D :</s> <s xml:id="echoid-s17628" xml:space="preserve">: B C : </s> <s xml:id="echoid-s17629" xml:space="preserve">C A, ou bien Q : </s> <s xml:id="echoid-s17630" xml:space="preserve">P :</s> <s xml:id="echoid-s17631" xml:space="preserve">: B C : </s> <s xml:id="echoid-s17632" xml:space="preserve">C A.</s> <s xml:id="echoid-s17633" xml:space="preserve"/> </p> <div xml:id="echoid-div1452" type="float" level="2" n="1"> <note position="left" xlink:label="note-0636-01" xlink:href="note-0636-01a" xml:space="preserve">Figure 370.</note> </div> <p> <s xml:id="echoid-s17634" xml:space="preserve">1063. </s> <s xml:id="echoid-s17635" xml:space="preserve">Mais ſi la ligne de direction D E de la puiſſance Q <lb/> <anchor type="note" xlink:label="note-0636-02a" xlink:href="note-0636-02"/> n’étoit point parallele au plan incliné A B, ni à ſa baſe A C, <lb/>& </s> <s xml:id="echoid-s17636" xml:space="preserve">que cependant la puiſſance & </s> <s xml:id="echoid-s17637" xml:space="preserve">le poids fuſſent en équilibre, <lb/>en ce cas la puiſſance ſera au poids dans la raiſon réciproque <lb/>des perpendiculaires F I & </s> <s xml:id="echoid-s17638" xml:space="preserve">F L : </s> <s xml:id="echoid-s17639" xml:space="preserve">car ayant fait le parallélo-<lb/>gramme K G, l’on aura toujours Q : </s> <s xml:id="echoid-s17640" xml:space="preserve">P :</s> <s xml:id="echoid-s17641" xml:space="preserve">: D G : </s> <s xml:id="echoid-s17642" xml:space="preserve">D K, ou G F; <lb/></s> <s xml:id="echoid-s17643" xml:space="preserve">mais les côtés D G & </s> <s xml:id="echoid-s17644" xml:space="preserve">G F du triangle G D F ſont comme les <lb/>ſinus de leurs angles oppoſés, qui ſont les perpendiculaires <lb/>E I & </s> <s xml:id="echoid-s17645" xml:space="preserve">F L : </s> <s xml:id="echoid-s17646" xml:space="preserve">ainſi l’on aura D G : </s> <s xml:id="echoid-s17647" xml:space="preserve">G F ou D K :</s> <s xml:id="echoid-s17648" xml:space="preserve">: F I : </s> <s xml:id="echoid-s17649" xml:space="preserve">F L, ou bien <lb/>Q : </s> <s xml:id="echoid-s17650" xml:space="preserve">P :</s> <s xml:id="echoid-s17651" xml:space="preserve">: F I : </s> <s xml:id="echoid-s17652" xml:space="preserve">F L. </s> <s xml:id="echoid-s17653" xml:space="preserve">L’on trouvera comme dans les propoſitions <lb/>précédentes le rapport de chacune des puiſſances agiſſantes <lb/>P & </s> <s xml:id="echoid-s17654" xml:space="preserve">Q à la réſiſtance R, qui eſt l’effort que le poids P fait <lb/>contre le plan A B.</s> <s xml:id="echoid-s17655" xml:space="preserve"/> </p> <div xml:id="echoid-div1453" type="float" level="2" n="2"> <note position="left" xlink:label="note-0636-02" xlink:href="note-0636-02a" xml:space="preserve">Figure 371.</note> </div> </div> <div xml:id="echoid-div1455" type="section" level="1" n="1066"> <head xml:id="echoid-head1259" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s17656" xml:space="preserve">1064. </s> <s xml:id="echoid-s17657" xml:space="preserve">Il ſuit delà que ſi deux corps P & </s> <s xml:id="echoid-s17658" xml:space="preserve">Q ſe ſoutiennent <lb/> <anchor type="note" xlink:label="note-0636-03a" xlink:href="note-0636-03"/> mutuellement ſur des plans diverſement inclinés par des lignes <lb/>R P & </s> <s xml:id="echoid-s17659" xml:space="preserve">R Q, paralles à ces plans, ils ſeront entr’eux comme <lb/>les longueurs des plans, c’eſt-à-dire que P : </s> <s xml:id="echoid-s17660" xml:space="preserve">Q :</s> <s xml:id="echoid-s17661" xml:space="preserve">: B A : </s> <s xml:id="echoid-s17662" xml:space="preserve">B C : </s> <s xml:id="echoid-s17663" xml:space="preserve">car <lb/>comme B D eſt la hauteur commune des deux plans, la puiſ-<lb/>ſance qui ſeroit en R ne fera pas plus d’effort pour ſoutenir <pb o="559" file="0637" n="657" rhead="DE MATHÉMATIQUE. Liv. XV."/> <lb/>le poids, que pour ſoutenir le poids Q, c’eſt-à-dire qu’elle <lb/>pourroit être la puiſſance commune: </s> <s xml:id="echoid-s17664" xml:space="preserve">ainſi comme le rapport <lb/>de la puiſſance R à la hauteur D B, eſt le même pour chaque <lb/>plan incliné, le rapport des plans & </s> <s xml:id="echoid-s17665" xml:space="preserve">des poids ſera auſſi le <lb/>même.</s> <s xml:id="echoid-s17666" xml:space="preserve"/> </p> <div xml:id="echoid-div1455" type="float" level="2" n="1"> <note position="left" xlink:label="note-0636-03" xlink:href="note-0636-03a" xml:space="preserve">Figure 371.</note> </div> </div> <div xml:id="echoid-div1457" type="section" level="1" n="1067"> <head xml:id="echoid-head1260" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s17667" xml:space="preserve">1065. </s> <s xml:id="echoid-s17668" xml:space="preserve">De même ſi deux poids P & </s> <s xml:id="echoid-s17669" xml:space="preserve">Q ſe ſoutiennent mu-<lb/> <anchor type="note" xlink:label="note-0637-01a" xlink:href="note-0637-01"/> tuellement ſur des plans diverſement inclinés par des lignes <lb/>de directions paralleles aux baſes, ces deux poids ſeront en-<lb/>tr’eux comme les longueurs des baſes, c’eſt-à-dire que P : </s> <s xml:id="echoid-s17670" xml:space="preserve">Q :</s> <s xml:id="echoid-s17671" xml:space="preserve">: <lb/>D A : </s> <s xml:id="echoid-s17672" xml:space="preserve">D C : </s> <s xml:id="echoid-s17673" xml:space="preserve">car comme B D eſt la hauteur commune des deux <lb/>plans, la puiſſance R pourra devenir commune pour les deux <lb/>poids. </s> <s xml:id="echoid-s17674" xml:space="preserve">Ainſi comme le rapport de la hauteur B D à la puiſſance <lb/>de part & </s> <s xml:id="echoid-s17675" xml:space="preserve">d’autre ſera le même, le rapport des poids & </s> <s xml:id="echoid-s17676" xml:space="preserve">des <lb/>baſes ſera auſſi le même.</s> <s xml:id="echoid-s17677" xml:space="preserve"/> </p> <div xml:id="echoid-div1457" type="float" level="2" n="1"> <note position="right" xlink:label="note-0637-01" xlink:href="note-0637-01a" xml:space="preserve">Figure 373.</note> </div> </div> <div xml:id="echoid-div1459" type="section" level="1" n="1068"> <head xml:id="echoid-head1261" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s17678" xml:space="preserve">1066. </s> <s xml:id="echoid-s17679" xml:space="preserve">Il ſuit encore delà que lorſqu’une puiſſance Q tire <lb/> <anchor type="note" xlink:label="note-0637-02a" xlink:href="note-0637-02"/> ou pouſſe un poids P par une ligne de direction parallele au <lb/>plan, la puiſſance eſt au poids comme le ſinus B C de l’angle <lb/>d’inclinaiſon B A C du plan eſt au ſinus total A B, & </s> <s xml:id="echoid-s17680" xml:space="preserve">que <lb/>par conſéquent la puiſſance eſt toujours moindre que le poids.</s> <s xml:id="echoid-s17681" xml:space="preserve"/> </p> <div xml:id="echoid-div1459" type="float" level="2" n="1"> <note position="right" xlink:label="note-0637-02" xlink:href="note-0637-02a" xml:space="preserve">Figure 369.</note> </div> </div> <div xml:id="echoid-div1461" type="section" level="1" n="1069"> <head xml:id="echoid-head1262" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s17682" xml:space="preserve">1067. </s> <s xml:id="echoid-s17683" xml:space="preserve">Enfin l’on peut dire encore que lorſqu’une puiſſance <lb/> <anchor type="note" xlink:label="note-0637-03a" xlink:href="note-0637-03"/> Q tire ou pouſſe un poids P par une ligne de direction parallele <lb/>à la baſe A C du plan incliné, la puiſſance eſt au poids, comme <lb/>le ſinus B C de l’angle d’inclinaiſon B A C eſt au ſinus A C de <lb/>ſon complément A B C; </s> <s xml:id="echoid-s17684" xml:space="preserve">ce qui fait voir que la puiſſance eſt <lb/>égale au poids, lorſque l’angle d’inclinaiſon eſt de 45 degrés, <lb/>& </s> <s xml:id="echoid-s17685" xml:space="preserve">qu’elle eſt plus grande que le poids, lorſque l’angle d’in-<lb/>clinaiſon eſt au deſſus de 45 degrés.</s> <s xml:id="echoid-s17686" xml:space="preserve"/> </p> <div xml:id="echoid-div1461" type="float" level="2" n="1"> <note position="right" xlink:label="note-0637-03" xlink:href="note-0637-03a" xml:space="preserve">Figure 370.</note> </div> <figure> <image file="0637-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0637-01"/> </figure> <pb o="560" file="0638" n="658" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1463" type="section" level="1" n="1070"> <head xml:id="echoid-head1263" xml:space="preserve">CHAPITRE IV. <lb/>Du Levier. <lb/><emph style="sc">Definitions</emph>.</head> <p> <s xml:id="echoid-s17687" xml:space="preserve">1068. </s> <s xml:id="echoid-s17688" xml:space="preserve">LEvier eſt une verge inflexible conſidérée ſans peſan-<lb/>teur, à trois points de laquelle il y a trois puiſſances appliquées, <lb/>deux deſquelles, qui ſont les agiſſantes, agiſſent d’un certain <lb/>ſens, & </s> <s xml:id="echoid-s17689" xml:space="preserve">ont leurs directions dans un même plan; </s> <s xml:id="echoid-s17690" xml:space="preserve">& </s> <s xml:id="echoid-s17691" xml:space="preserve">la troi-<lb/>ſieme, qui eſt la réſiſtante, agit d’un ſens directement oppoſé <lb/>aux deux autres, entre leſquelles elle eſt toujours.</s> <s xml:id="echoid-s17692" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1464" type="section" level="1" n="1071"> <head xml:id="echoid-head1264" xml:space="preserve">PROPOSITION. <lb/><emph style="sc">Théoreme</emph>.</head> <p> <s xml:id="echoid-s17693" xml:space="preserve">1069. </s> <s xml:id="echoid-s17694" xml:space="preserve">Deux puiſſances P & </s> <s xml:id="echoid-s17695" xml:space="preserve">Q que l’on compare, ſeront en <lb/>équilibre, ſi elles ſont en raiſon réciproque des perpendiculaires <lb/>D G & </s> <s xml:id="echoid-s17696" xml:space="preserve">D H, tirées du point d’appui D ſur les lignes de direc-<lb/>tions C A & </s> <s xml:id="echoid-s17697" xml:space="preserve">C B des puiſſances P & </s> <s xml:id="echoid-s17698" xml:space="preserve">Q : </s> <s xml:id="echoid-s17699" xml:space="preserve">ainſi il faut prouver que <lb/>P : </s> <s xml:id="echoid-s17700" xml:space="preserve">Q :</s> <s xml:id="echoid-s17701" xml:space="preserve">: D H : </s> <s xml:id="echoid-s17702" xml:space="preserve">D G.</s> <s xml:id="echoid-s17703" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1465" type="section" level="1" n="1072"> <head xml:id="echoid-head1265" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s17704" xml:space="preserve">Si du point D l’on tire les lignes D E, D F paralleles aux <lb/>lignes de directions C A, C B, l’on aura un parallélogramme <lb/>E F, dont la diagonale C D exprimera la force de la puiſſance <lb/>qui réſiſte aux deux puiſſances P & </s> <s xml:id="echoid-s17705" xml:space="preserve">Q; </s> <s xml:id="echoid-s17706" xml:space="preserve">le côté C E exprimera <lb/>la force de la puiſſance P, & </s> <s xml:id="echoid-s17707" xml:space="preserve">le côté C F celle de la puiſſance <lb/>Q : </s> <s xml:id="echoid-s17708" xml:space="preserve">ainſi l’on aura P : </s> <s xml:id="echoid-s17709" xml:space="preserve">Q :</s> <s xml:id="echoid-s17710" xml:space="preserve">: E C, ou D F : </s> <s xml:id="echoid-s17711" xml:space="preserve">F C; </s> <s xml:id="echoid-s17712" xml:space="preserve">mais dans le <lb/>triangle D C F, l’on ſçait que les ſinus des angles ſont dans la <lb/>même raiſon que leurs côtés oppoſés: </s> <s xml:id="echoid-s17713" xml:space="preserve">l’on aura donc le côté <lb/>D F eſt au côté C F, comme le ſinus de l’angle D C F eſt au <lb/>ſinus de l’angle C D F. </s> <s xml:id="echoid-s17714" xml:space="preserve">Or comme D H eſt le ſinus de l’angle <lb/>D C F, & </s> <s xml:id="echoid-s17715" xml:space="preserve">que D G eſt le ſinus de l’angle C D F, puiſqu’il eſt <lb/>celui de l’angle alterne E C D, ſi à la place de D F on prend <lb/>E C, l’on aura E C : </s> <s xml:id="echoid-s17716" xml:space="preserve">F C :</s> <s xml:id="echoid-s17717" xml:space="preserve">: H D : </s> <s xml:id="echoid-s17718" xml:space="preserve">D G, & </s> <s xml:id="echoid-s17719" xml:space="preserve">ſi au lieu de E C & </s> <s xml:id="echoid-s17720" xml:space="preserve"><lb/>F C l’on prend les puiſſances P & </s> <s xml:id="echoid-s17721" xml:space="preserve">Q, l’on aura encore <lb/>P : </s> <s xml:id="echoid-s17722" xml:space="preserve">Q :</s> <s xml:id="echoid-s17723" xml:space="preserve">: D H : </s> <s xml:id="echoid-s17724" xml:space="preserve">D G. </s> <s xml:id="echoid-s17725" xml:space="preserve">C. </s> <s xml:id="echoid-s17726" xml:space="preserve">Q. </s> <s xml:id="echoid-s17727" xml:space="preserve">F. </s> <s xml:id="echoid-s17728" xml:space="preserve">D.</s> <s xml:id="echoid-s17729" xml:space="preserve"/> </p> <pb file="0639" n="659"/> <pb file="0639a" n="660"/> <figure> <image file="0639a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0639a-01"/> </figure> <pb file="0640" n="661"/> <pb file="0641" n="662"/> <pb file="0641a" n="663"/> <figure> <image file="0641a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0641a-01"/> </figure> <pb file="0642" n="664"/> <pb file="0643" n="665"/> <pb file="0643a" n="666"/> <figure> <image file="0643a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0643a-01"/> </figure> <pb file="0644" n="667"/> <pb file="0645" n="668"/> <pb file="0645a" n="669"/> <figure> <image file="0645a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0645a-01"/> </figure> <pb file="0646" n="670"/> <pb file="0647" n="671"/> <pb file="0647a" n="672"/> <figure> <image file="0647a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0647a-01"/> </figure> <pb file="0648" n="673"/> <pb file="0649" n="674"/> <pb file="0649a" n="675"/> <figure> <image file="0649a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0649a-01"/> </figure> <pb file="0650" n="676"/> <pb o="561" file="0651" n="677" rhead="DE MATHÉMATIQUE. Liv. XV."/> </div> <div xml:id="echoid-div1466" type="section" level="1" n="1073"> <head xml:id="echoid-head1266" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s17730" xml:space="preserve">1070. </s> <s xml:id="echoid-s17731" xml:space="preserve">Il eſt clair que ſi le point C s’éloignoit de plus en plus <lb/>des trois points A, D, B, de ſorte que les directions A C, <lb/>D C, B C des trois puiſſances P, R, Q, devinſſent enfin pa-<lb/>ralleles, elles ſeront perpendiculaires ou obliques; </s> <s xml:id="echoid-s17732" xml:space="preserve">ſi elles ſont <lb/>obliques, l’on aura encore P : </s> <s xml:id="echoid-s17733" xml:space="preserve">Q :</s> <s xml:id="echoid-s17734" xml:space="preserve">: D H : </s> <s xml:id="echoid-s17735" xml:space="preserve">D G : </s> <s xml:id="echoid-s17736" xml:space="preserve">car les lignes <lb/> <anchor type="note" xlink:label="note-0651-01a" xlink:href="note-0651-01"/> D H & </s> <s xml:id="echoid-s17737" xml:space="preserve">D G ſont des perpendiculaires tirées ſur les lignes de <lb/>directions des puiſſances P & </s> <s xml:id="echoid-s17738" xml:space="preserve">Q; </s> <s xml:id="echoid-s17739" xml:space="preserve">de plus à cauſe des triangles <lb/>ſemblables D A G & </s> <s xml:id="echoid-s17740" xml:space="preserve">D B H, l’on pourra à la place des lignes <lb/>D H, D G, prendre les lignes D B & </s> <s xml:id="echoid-s17741" xml:space="preserve">D A, d’où l’on tire <lb/>P : </s> <s xml:id="echoid-s17742" xml:space="preserve">Q :</s> <s xml:id="echoid-s17743" xml:space="preserve">: D B : </s> <s xml:id="echoid-s17744" xml:space="preserve">D A; </s> <s xml:id="echoid-s17745" xml:space="preserve">c’eſt-à-dire que deux puiſſances appliquées <lb/>aux extrêmités des bras d’un levier, ſont en équilibre, lorſ-<lb/>qu’ayant leurs directions paralleles, elles ſont en raiſon réci-<lb/>proque des bras du levier, c’eſt-à-dire ſi P : </s> <s xml:id="echoid-s17746" xml:space="preserve">Q :</s> <s xml:id="echoid-s17747" xml:space="preserve">: D B : </s> <s xml:id="echoid-s17748" xml:space="preserve">D A.</s> <s xml:id="echoid-s17749" xml:space="preserve"/> </p> <div xml:id="echoid-div1466" type="float" level="2" n="1"> <note position="right" xlink:label="note-0651-01" xlink:href="note-0651-01a" xml:space="preserve">Figure 374 <lb/>& 375.</note> </div> </div> <div xml:id="echoid-div1468" type="section" level="1" n="1074"> <head xml:id="echoid-head1267" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s17750" xml:space="preserve">1071. </s> <s xml:id="echoid-s17751" xml:space="preserve">L’on peut remarquer ici en paſſant, que ſi deux puiſ-<lb/> <anchor type="note" xlink:label="note-0651-02a" xlink:href="note-0651-02"/> ſances portent un poids E, appliqué dans le milieu d’un levier, <lb/>elles ſeront également chargées; </s> <s xml:id="echoid-s17752" xml:space="preserve">car il y aura même raiſon de <lb/>P à Q, que de C B à C A : </s> <s xml:id="echoid-s17753" xml:space="preserve">mais comme C B eſt égal à C A, <lb/>la puiſſance P ſera égale à la puiſſance Q. </s> <s xml:id="echoid-s17754" xml:space="preserve">Et ſi au contraire <lb/>le poids F, eſt plus près de A que de B, comme le poids F, la <lb/>puiſſance P ſera plus chargée que la puiſſance Q, puiſque l’on <lb/>aura P : </s> <s xml:id="echoid-s17755" xml:space="preserve">Q :</s> <s xml:id="echoid-s17756" xml:space="preserve">: D B : </s> <s xml:id="echoid-s17757" xml:space="preserve">D A. </s> <s xml:id="echoid-s17758" xml:space="preserve">Ainſi d’autant le bras ſera plus grand <lb/>que le bras D A, d’autant la puiſſance P ſera plus chargée que <lb/>la puiſſance Q.</s> <s xml:id="echoid-s17759" xml:space="preserve"/> </p> <div xml:id="echoid-div1468" type="float" level="2" n="1"> <note position="right" xlink:label="note-0651-02" xlink:href="note-0651-02a" xml:space="preserve">Figure 377.</note> </div> </div> <div xml:id="echoid-div1470" type="section" level="1" n="1075"> <head xml:id="echoid-head1268" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s17760" xml:space="preserve">1072. </s> <s xml:id="echoid-s17761" xml:space="preserve">Mais ſi l’on a un levier A B, dont le point d’appui <lb/> <anchor type="note" xlink:label="note-0651-03a" xlink:href="note-0651-03"/> ſoit à une des extrêmités A, & </s> <s xml:id="echoid-s17762" xml:space="preserve">que de deux puiſſances appli-<lb/>quées aux points D & </s> <s xml:id="echoid-s17763" xml:space="preserve">B, l’une tire ſelon la direction D Q, <lb/>& </s> <s xml:id="echoid-s17764" xml:space="preserve">l’autre ſelon la direction B P en ſens contraires, ces deux <lb/>puiſſances ſeront encore en équilibre, ſi elles ſont en raiſon <lb/>réciproque des perpendiculaires A G & </s> <s xml:id="echoid-s17765" xml:space="preserve">A H, tirées du point <lb/>d’appui A ſur leurs lignes de directions: </s> <s xml:id="echoid-s17766" xml:space="preserve">car faiſant le parallé-<lb/>logramme E F, le côté C F exprimera la force de la puiſſance <lb/>P, & </s> <s xml:id="echoid-s17767" xml:space="preserve">la diagonale C D celle de la puiſſance Q, pour que ces <lb/>deux puiſſances ſoient en équilibre. </s> <s xml:id="echoid-s17768" xml:space="preserve">Et comme dans le triangle <lb/>C F D, les côtés C F & </s> <s xml:id="echoid-s17769" xml:space="preserve">C D ſont dans la raiſon des ſinus de <pb o="562" file="0652" n="678" rhead="NOUVEAU COURS"/> leurs angles oppoſés, l’on aura C F : </s> <s xml:id="echoid-s17770" xml:space="preserve">C D :</s> <s xml:id="echoid-s17771" xml:space="preserve">: A H : </s> <s xml:id="echoid-s17772" xml:space="preserve">A G, ou <lb/>bien P : </s> <s xml:id="echoid-s17773" xml:space="preserve">Q :</s> <s xml:id="echoid-s17774" xml:space="preserve">: A H : </s> <s xml:id="echoid-s17775" xml:space="preserve">A G.</s> <s xml:id="echoid-s17776" xml:space="preserve"/> </p> <div xml:id="echoid-div1470" type="float" level="2" n="1"> <note position="right" xlink:label="note-0651-03" xlink:href="note-0651-03a" xml:space="preserve">Figure 377.</note> </div> </div> <div xml:id="echoid-div1472" type="section" level="1" n="1076"> <head xml:id="echoid-head1269" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s17777" xml:space="preserve">1073. </s> <s xml:id="echoid-s17778" xml:space="preserve">L’on peut dire encore, comme dans le coroll. </s> <s xml:id="echoid-s17779" xml:space="preserve">I, que <lb/> <anchor type="note" xlink:label="note-0652-01a" xlink:href="note-0652-01"/> ſi le point C s’éloignoit de plus en plus à l’infini des points <lb/>D & </s> <s xml:id="echoid-s17780" xml:space="preserve">B, enſorte que les lignes de directions B P & </s> <s xml:id="echoid-s17781" xml:space="preserve">D Q de-<lb/>vinſſent paralleles & </s> <s xml:id="echoid-s17782" xml:space="preserve">perpendiculaires au levier A B, les puiſ-<lb/>ſances P & </s> <s xml:id="echoid-s17783" xml:space="preserve">Q demeureront toujours en équilibre: </s> <s xml:id="echoid-s17784" xml:space="preserve">car dans ce <lb/>cas la perpendiculaire A G deviendra égale à la longueur du <lb/>levier A B, & </s> <s xml:id="echoid-s17785" xml:space="preserve">la perpendiculaire A H égale au bras A D, & </s> <s xml:id="echoid-s17786" xml:space="preserve"><lb/>l’on aura encore P : </s> <s xml:id="echoid-s17787" xml:space="preserve">Q :</s> <s xml:id="echoid-s17788" xml:space="preserve">: A D : </s> <s xml:id="echoid-s17789" xml:space="preserve">A B.</s> <s xml:id="echoid-s17790" xml:space="preserve"/> </p> <div xml:id="echoid-div1472" type="float" level="2" n="1"> <note position="left" xlink:label="note-0652-01" xlink:href="note-0652-01a" xml:space="preserve">Figure 377 <lb/>& 378.</note> </div> </div> <div xml:id="echoid-div1474" type="section" level="1" n="1077"> <head xml:id="echoid-head1270" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s17791" xml:space="preserve">1074. </s> <s xml:id="echoid-s17792" xml:space="preserve">Par conſéquent ſi une puiſſance P ſoutient un poids <lb/> <anchor type="note" xlink:label="note-0652-02a" xlink:href="note-0652-02"/> Q à l’aide d’un levier A B, enſorte que le poids ſoit dans le <lb/>milieu D, le point d’appui à l’extrêmité A, & </s> <s xml:id="echoid-s17793" xml:space="preserve">la puiſſance à <lb/>l’extrêmité B, cette puiſſance ne ſoutiendra que la moitié du <lb/>poids Q; </s> <s xml:id="echoid-s17794" xml:space="preserve">car l’on aura P : </s> <s xml:id="echoid-s17795" xml:space="preserve">Q :</s> <s xml:id="echoid-s17796" xml:space="preserve">: A D : </s> <s xml:id="echoid-s17797" xml:space="preserve">A B : </s> <s xml:id="echoid-s17798" xml:space="preserve">ainſi A D étant la <lb/>moitié de A B, P ſera la moitié de Q.</s> <s xml:id="echoid-s17799" xml:space="preserve"/> </p> <div xml:id="echoid-div1474" type="float" level="2" n="1"> <note position="left" xlink:label="note-0652-02" xlink:href="note-0652-02a" xml:space="preserve">Figure 379.</note> </div> </div> <div xml:id="echoid-div1476" type="section" level="1" n="1078"> <head xml:id="echoid-head1271" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s17800" xml:space="preserve">1075. </s> <s xml:id="echoid-s17801" xml:space="preserve">Donc ſi le poids, au lieu d’être dans le milieu du le-<lb/>vier, étoit au point C plus près de A que de B, la puiſſance <lb/>ſera moins chargée qu’elle n’étoit auparavant: </s> <s xml:id="echoid-s17802" xml:space="preserve">car l’on aura <lb/>toujours P : </s> <s xml:id="echoid-s17803" xml:space="preserve">Q :</s> <s xml:id="echoid-s17804" xml:space="preserve">: A C : </s> <s xml:id="echoid-s17805" xml:space="preserve">A B. </s> <s xml:id="echoid-s17806" xml:space="preserve">Et comme A C eſt moindre que <lb/>C B, P ſera moindre que la moitié de Q.</s> <s xml:id="echoid-s17807" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1477" type="section" level="1" n="1079"> <head xml:id="echoid-head1272" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s17808" xml:space="preserve">1076. </s> <s xml:id="echoid-s17809" xml:space="preserve">Il ſuit delà que ſi la puiſſance étoit appliquée à un <lb/> <anchor type="note" xlink:label="note-0652-03a" xlink:href="note-0652-03"/> point quelconque D du levier A B, & </s> <s xml:id="echoid-s17810" xml:space="preserve">que le poids fût à l’ex-<lb/>trêmité B, la puiſſance & </s> <s xml:id="echoid-s17811" xml:space="preserve">le poids ſeront encore en équilibre, <lb/>s’il y a même raiſon de la puiſſance au poids, que du levier <lb/>A B au bras A D.</s> <s xml:id="echoid-s17812" xml:space="preserve"/> </p> <div xml:id="echoid-div1477" type="float" level="2" n="1"> <note position="left" xlink:label="note-0652-03" xlink:href="note-0652-03a" xml:space="preserve">Pl. XXIX. <lb/>Figure 380.</note> </div> </div> <div xml:id="echoid-div1479" type="section" level="1" n="1080"> <head xml:id="echoid-head1273" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s17813" xml:space="preserve">1077. </s> <s xml:id="echoid-s17814" xml:space="preserve">Si l’on a un levier A B, dont le point d’appui ſoit en <lb/> <anchor type="note" xlink:label="note-0652-04a" xlink:href="note-0652-04"/> E, deux poids P & </s> <s xml:id="echoid-s17815" xml:space="preserve">Q attachés aux extrêmités A & </s> <s xml:id="echoid-s17816" xml:space="preserve">B ſeront <lb/>en équilibre, s’ils ſont en raiſon réciproque des bras du levier, <lb/>c’eſt-à-dire ſi P : </s> <s xml:id="echoid-s17817" xml:space="preserve">Q :</s> <s xml:id="echoid-s17818" xml:space="preserve">: E B : </s> <s xml:id="echoid-s17819" xml:space="preserve">E A : </s> <s xml:id="echoid-s17820" xml:space="preserve">car nous avons démontré que <lb/>deux puiſſances dans cet état étoient en équilibre, ſi au lieu <pb o="563" file="0653" n="679" rhead="DE MATHÉMATIQUE. Liv. XV."/> des puiſſances l’on met des poids qui leur ſoient équivalens, <lb/>ils feront le même effet, & </s> <s xml:id="echoid-s17821" xml:space="preserve">ſeront par conſéquent en équi-<lb/>libre.</s> <s xml:id="echoid-s17822" xml:space="preserve"/> </p> <div xml:id="echoid-div1479" type="float" level="2" n="1"> <note position="left" xlink:label="note-0652-04" xlink:href="note-0652-04a" xml:space="preserve">Figure 381.</note> </div> </div> <div xml:id="echoid-div1481" type="section" level="1" n="1081"> <head xml:id="echoid-head1274" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s17823" xml:space="preserve">1078. </s> <s xml:id="echoid-s17824" xml:space="preserve">Il ſuit encore delà que ſi l’on a deux poids appliqués <lb/> <anchor type="note" xlink:label="note-0653-01a" xlink:href="note-0653-01"/> aux extrêmités d’un levier ou d’une balance, on pourra tou-<lb/>jours trouver le point d’appui, autour duquel les deux poids <lb/>ſeront en équilibre, en diſant: </s> <s xml:id="echoid-s17825" xml:space="preserve">Comme la ſomme de deux <lb/>poids P & </s> <s xml:id="echoid-s17826" xml:space="preserve">Q eſt à toute la longueur de la balance A B, ainſi <lb/>le poids P eſt à la longueur du bras B E, qui donnera le point <lb/>E pour le point d’appui.</s> <s xml:id="echoid-s17827" xml:space="preserve"/> </p> <div xml:id="echoid-div1481" type="float" level="2" n="1"> <note position="right" xlink:label="note-0653-01" xlink:href="note-0653-01a" xml:space="preserve">Figure 381.</note> </div> <p> <s xml:id="echoid-s17828" xml:space="preserve">Par la même raiſon connoiſſant les bras A E & </s> <s xml:id="echoid-s17829" xml:space="preserve">E B avec un <lb/>poids P, l’on trouvera toujours l’autre poids Q, en diſant: <lb/></s> <s xml:id="echoid-s17830" xml:space="preserve">comme le poids P eſt au bras E B, ainſi le bras A E eſt au <lb/>poids Q.</s> <s xml:id="echoid-s17831" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1483" type="section" level="1" n="1082"> <head xml:id="echoid-head1275" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <p> <s xml:id="echoid-s17832" xml:space="preserve">1079. </s> <s xml:id="echoid-s17833" xml:space="preserve">Il ſuit encore delà qu’ayant une verge A B d’une pe-<lb/> <anchor type="note" xlink:label="note-0653-02a" xlink:href="note-0653-02"/> ſanteur quelconque, on pourra trouver un point tel que F, <lb/>par lequel la verge étant ſuſpendue, elle ſoit en équilibre <lb/>avec le poids C : </s> <s xml:id="echoid-s17834" xml:space="preserve">car il n’y a qu’à diviſer la verge A B en deux <lb/>également au point D, & </s> <s xml:id="echoid-s17835" xml:space="preserve">ſuppoſer que ſa peſanteur eſt raſ-<lb/>ſemblée autour de ſon centre de gravité pour avoir le poids <lb/>E, enſuite chercher dans la verge A D, qui n’a plus de peſan-<lb/>teur, un point d’appui F, en diſant: </s> <s xml:id="echoid-s17836" xml:space="preserve">comme la ſomme des <lb/>deux poids C & </s> <s xml:id="echoid-s17837" xml:space="preserve">F eſt à la longueur A D, ainſi le poids E eſt au <lb/>au bras A F.</s> <s xml:id="echoid-s17838" xml:space="preserve"/> </p> <div xml:id="echoid-div1483" type="float" level="2" n="1"> <note position="right" xlink:label="note-0653-02" xlink:href="note-0653-02a" xml:space="preserve">Figure 382.</note> </div> </div> <div xml:id="echoid-div1485" type="section" level="1" n="1083"> <head xml:id="echoid-head1276" xml:space="preserve"><emph style="sc">Corollaire</emph> X.</head> <p> <s xml:id="echoid-s17839" xml:space="preserve">1080. </s> <s xml:id="echoid-s17840" xml:space="preserve">Enfin l’on peut dire qu’ayant deux poids C & </s> <s xml:id="echoid-s17841" xml:space="preserve">D ap-<lb/> <anchor type="note" xlink:label="note-0653-03a" xlink:href="note-0653-03"/> pliqués aux deux extrêmités d’une balance A B, à laquelle <lb/>on ſuppoſe une peſanteur, pour trouver un point d’appui, <lb/>autour duquel la peſanteur de la balance & </s> <s xml:id="echoid-s17842" xml:space="preserve">celle des poids <lb/>ſoient en équilibre, il faut d’abord chercher un point d’appui <lb/>tel que E, autour duquel les deux poids C & </s> <s xml:id="echoid-s17843" xml:space="preserve">D ſoient en équi-<lb/>libre, en faiſant abſtraction de la peſanteur de la balance; </s> <s xml:id="echoid-s17844" xml:space="preserve">en-<lb/>ſuite ſuppoſer que les poids C & </s> <s xml:id="echoid-s17845" xml:space="preserve">D ſont réunis dans le ſeul <lb/>poids G au centre de gravité E, & </s> <s xml:id="echoid-s17846" xml:space="preserve">que la peſanteur de la ba-<lb/>lance eſt auſſi réunie dans le poids F autour de ſon centre de <lb/>gravité H, & </s> <s xml:id="echoid-s17847" xml:space="preserve">regardant la longueur E H comme une balance <pb o="564" file="0654" n="680" rhead="NOUVEAU COURS"/> aux extrêmités de laquelle ſont les poids G & </s> <s xml:id="echoid-s17848" xml:space="preserve">F, on en cher-<lb/>chera le point d’appui, en diſant: </s> <s xml:id="echoid-s17849" xml:space="preserve">Comme la ſomme des deux <lb/>poids G & </s> <s xml:id="echoid-s17850" xml:space="preserve">F eſt à la longueur E H, ainſi le poids F eſt au bras <lb/>E I, qui donnera le point I, qui ſera celui autour duquel la <lb/>peſanteur de la balance & </s> <s xml:id="echoid-s17851" xml:space="preserve">celle des poids C & </s> <s xml:id="echoid-s17852" xml:space="preserve">D ſeront en <lb/>équilibre.</s> <s xml:id="echoid-s17853" xml:space="preserve"/> </p> <div xml:id="echoid-div1485" type="float" level="2" n="1"> <note position="right" xlink:label="note-0653-03" xlink:href="note-0653-03a" xml:space="preserve">Figure 383.</note> </div> </div> <div xml:id="echoid-div1487" type="section" level="1" n="1084"> <head xml:id="echoid-head1277" xml:space="preserve"><emph style="sc">Corollaire</emph> XI.</head> <p> <s xml:id="echoid-s17854" xml:space="preserve">1081. </s> <s xml:id="echoid-s17855" xml:space="preserve">Enfin ſi l’on a une verge ou balance A B d’une cer-<lb/> <anchor type="note" xlink:label="note-0654-01a" xlink:href="note-0654-01"/> taine peſanteur avec un poids I ſuſpendu à l’extrêmité A, & </s> <s xml:id="echoid-s17856" xml:space="preserve"><lb/>qu’on prenne le point C pour le point d’appui, & </s> <s xml:id="echoid-s17857" xml:space="preserve">que l’on <lb/>veuille trouver dans le bras C B un endroit où un poids tel <lb/>que H, aidé de la peſanteur de la balance, ſoit en équilibre <lb/>avec le poids I, il faut diviſer la balance A B en deux égale-<lb/>ment au point E, & </s> <s xml:id="echoid-s17858" xml:space="preserve">ſuppoſer que ſa peſanteur ſoit réunie <lb/>dans le point F; </s> <s xml:id="echoid-s17859" xml:space="preserve">enſuite chercher la partie du poids I, qui fera <lb/>équilibre avec le poids F, ou autrement avec la balance, en <lb/>diſant: </s> <s xml:id="echoid-s17860" xml:space="preserve">Comme le bras A C eſt au poids F, ainſi le bras C E <lb/>eſt à la partie du poids I qui doit faire l’équilibre, qui ſera, <lb/>par exemple, la partie K. </s> <s xml:id="echoid-s17861" xml:space="preserve">Préſentement pour trouver le point <lb/>G, où le poids H doit être ſuſpendu pour être en équilibre <lb/>avec ce qui reſte du poids I, qui eſt la partie L, il faut dire: <lb/></s> <s xml:id="echoid-s17862" xml:space="preserve">Comme le poids H eſt au bras A C, ainſi le poids L eſt au bras <lb/>C G, que l’on trouvera après avoir déterminé la peſanteur de la <lb/>balance A B, & </s> <s xml:id="echoid-s17863" xml:space="preserve">celles des poids I & </s> <s xml:id="echoid-s17864" xml:space="preserve">H.</s> <s xml:id="echoid-s17865" xml:space="preserve"/> </p> <div xml:id="echoid-div1487" type="float" level="2" n="1"> <note position="left" xlink:label="note-0654-01" xlink:href="note-0654-01a" xml:space="preserve">Figure 384.</note> </div> <p> <s xml:id="echoid-s17866" xml:space="preserve">L’on tire de ce corollaire le moyen de faire la balance <lb/>romaine, que l’on nomme auſſi peſon.</s> <s xml:id="echoid-s17867" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1489" type="section" level="1" n="1085"> <head xml:id="echoid-head1278" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s17868" xml:space="preserve">1082. </s> <s xml:id="echoid-s17869" xml:space="preserve">Il y a encore une autre maniere de démontrer l’é-<lb/> <anchor type="note" xlink:label="note-0654-02a" xlink:href="note-0654-02"/> quilibre dans les machines dont nous n’avons pas encore parlé, <lb/>mais qui s’entendra aiſément, ſi l’on ſe rappelle ce qui a été <lb/>enſeigné dans le Traité du Mouvement.</s> <s xml:id="echoid-s17870" xml:space="preserve"/> </p> <div xml:id="echoid-div1489" type="float" level="2" n="1"> <note position="left" xlink:label="note-0654-02" xlink:href="note-0654-02a" xml:space="preserve">Figure 385.</note> </div> <p> <s xml:id="echoid-s17871" xml:space="preserve">Par exemple, pour prouver que deux poids P & </s> <s xml:id="echoid-s17872" xml:space="preserve">Q attachés <lb/>aux extrêmités d’un levier A B, ſont en équilibre, s’ils ſont <lb/>en raiſon réciproque des bras E B & </s> <s xml:id="echoid-s17873" xml:space="preserve">E A, c’eſt-à-dire ſi P : </s> <s xml:id="echoid-s17874" xml:space="preserve">Q <lb/>:</s> <s xml:id="echoid-s17875" xml:space="preserve">: E B : </s> <s xml:id="echoid-s17876" xml:space="preserve">E A.</s> <s xml:id="echoid-s17877" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17878" xml:space="preserve">Conſidérez que le poids P ne peut ſe mouvoir qu’il ne faſſe <lb/>auſſi mouvoir le poids Q. </s> <s xml:id="echoid-s17879" xml:space="preserve">Or ſuppoſant que le poids P puiſſe <lb/>emporter le poids Q, dans le tems que le poids P décrira l’arc <pb o="565" file="0655" n="681" rhead="DE MATHÉMATIQUE. Liv. XV."/> A F, le poids Q décrira l’arc G B : </s> <s xml:id="echoid-s17880" xml:space="preserve">ainſi l’arc A F marquera la <lb/>vîteſſe du poids P, & </s> <s xml:id="echoid-s17881" xml:space="preserve">l’arc G B la vîteſſe du poids Q en tems <lb/>égaux. </s> <s xml:id="echoid-s17882" xml:space="preserve">Mais nous avons fait voir (art. </s> <s xml:id="echoid-s17883" xml:space="preserve">933) que deux corps <lb/>avoient une même quantité de force, lorſqu’ils avoient des <lb/>maſſes & </s> <s xml:id="echoid-s17884" xml:space="preserve">des vîteſſes réciproques: </s> <s xml:id="echoid-s17885" xml:space="preserve">ainſi ces deux poids auront <lb/>des forces égales, ſi P : </s> <s xml:id="echoid-s17886" xml:space="preserve">Q :</s> <s xml:id="echoid-s17887" xml:space="preserve">: G B : </s> <s xml:id="echoid-s17888" xml:space="preserve">A F. </s> <s xml:id="echoid-s17889" xml:space="preserve">Or, ſelon la ſup-<lb/>poſition, P : </s> <s xml:id="echoid-s17890" xml:space="preserve">Q :</s> <s xml:id="echoid-s17891" xml:space="preserve">: E B : </s> <s xml:id="echoid-s17892" xml:space="preserve">E A : </s> <s xml:id="echoid-s17893" xml:space="preserve">ainſi prenant E B & </s> <s xml:id="echoid-s17894" xml:space="preserve">E A à la <lb/>place de G B & </s> <s xml:id="echoid-s17895" xml:space="preserve">A F, qui ſont dans la même raiſon, l’on aura <lb/>P : </s> <s xml:id="echoid-s17896" xml:space="preserve">Q :</s> <s xml:id="echoid-s17897" xml:space="preserve">: E B : </s> <s xml:id="echoid-s17898" xml:space="preserve">E A : </s> <s xml:id="echoid-s17899" xml:space="preserve">par conſéquent ces deux poids ayant une <lb/>même force, lorſqu’ils ſont dans la raiſon réciproque des bras <lb/>du levier, demeureront en équilibre, puiſque l’un ne fera <lb/>pas plus d’effort pour ſe mouvoir que l’autre.</s> <s xml:id="echoid-s17900" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1491" type="section" level="1" n="1086"> <head xml:id="echoid-head1279" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s17901" xml:space="preserve">1083. </s> <s xml:id="echoid-s17902" xml:space="preserve">Il ſuit delà que ſi à la place du poids Q on ſuppoſe <lb/> <anchor type="note" xlink:label="note-0655-01a" xlink:href="note-0655-01"/> une puiſſance, cette puiſſance ſera encore en équilibre avec le <lb/>poids P, s’ils ſont en raiſon réciproque de leurs chemins ou <lb/>des vîteſſes, qu’ils ont en tems égaux, c’eſt-à-dire ſi la <lb/>puiſſance Q eſt au poids, comme le chemin ou la vîteſſe A F <lb/>du poids eſt au chemin ou à la vîteſſe G B de la puiſſance: </s> <s xml:id="echoid-s17903" xml:space="preserve">c’eſt <lb/>pourquoi lorſque l’on fera voir dans les machines que le che-<lb/>min de la puiſſance & </s> <s xml:id="echoid-s17904" xml:space="preserve">celui du poids ſont en raiſon réciproque <lb/>de la puiſſance & </s> <s xml:id="echoid-s17905" xml:space="preserve">du poids, on prouvera toujours que la puiſ-<lb/>ſance & </s> <s xml:id="echoid-s17906" xml:space="preserve">le poids ſont en équilibre.</s> <s xml:id="echoid-s17907" xml:space="preserve"/> </p> <div xml:id="echoid-div1491" type="float" level="2" n="1"> <note position="right" xlink:label="note-0655-01" xlink:href="note-0655-01a" xml:space="preserve">Figure 385.</note> </div> <p> <s xml:id="echoid-s17908" xml:space="preserve">Par exemple, pour prouver que ſi une puiſſance Q appli-<lb/> <anchor type="note" xlink:label="note-0655-02a" xlink:href="note-0655-02"/> quée à l’extrêmité d’un levier, ſoutient un poids P, que la puiſ-<lb/>ſance & </s> <s xml:id="echoid-s17909" xml:space="preserve">le poids ſeront en équilibre, ſi Q : </s> <s xml:id="echoid-s17910" xml:space="preserve">P :</s> <s xml:id="echoid-s17911" xml:space="preserve">: A F : </s> <s xml:id="echoid-s17912" xml:space="preserve">A B. <lb/></s> <s xml:id="echoid-s17913" xml:space="preserve">Imaginons que la puiſſance & </s> <s xml:id="echoid-s17914" xml:space="preserve">le poids ſe ſoient mus, enſorte <lb/>que le levier A B ait pris la ſituation A D, la vîteſſe de la puiſ-<lb/>ſance ſera l’arc D B, & </s> <s xml:id="echoid-s17915" xml:space="preserve">la vîteſſe du poids l’arc E F; </s> <s xml:id="echoid-s17916" xml:space="preserve">& </s> <s xml:id="echoid-s17917" xml:space="preserve">dans <lb/>l’état de l’équilibre, l’on aura Q : </s> <s xml:id="echoid-s17918" xml:space="preserve">P :</s> <s xml:id="echoid-s17919" xml:space="preserve">: E F : </s> <s xml:id="echoid-s17920" xml:space="preserve">D B, & </s> <s xml:id="echoid-s17921" xml:space="preserve">ſi à la place <lb/>des arcs l’on prend les rayons, l’on aura Q : </s> <s xml:id="echoid-s17922" xml:space="preserve">P :</s> <s xml:id="echoid-s17923" xml:space="preserve">: A F : </s> <s xml:id="echoid-s17924" xml:space="preserve">A B.</s> <s xml:id="echoid-s17925" xml:space="preserve"/> </p> <div xml:id="echoid-div1492" type="float" level="2" n="2"> <note position="right" xlink:label="note-0655-02" xlink:href="note-0655-02a" xml:space="preserve">Figure 386.</note> </div> </div> <div xml:id="echoid-div1494" type="section" level="1" n="1087"> <head xml:id="echoid-head1280" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s17926" xml:space="preserve">1084. </s> <s xml:id="echoid-s17927" xml:space="preserve">Comme nous n’avons point mis de différence entre <lb/>les leviers dont nous venons de faire mention, & </s> <s xml:id="echoid-s17928" xml:space="preserve">que cepen-<lb/>dant le point d’appui, ou la puiſſance réſiſtante change le <lb/>levier de nature, ſelon qu’il eſt placé différemment, nous <lb/>nommerons levier du premier genre celui qui a une puiſſance à <lb/>une extrêmité, un poids à l’autre, & </s> <s xml:id="echoid-s17929" xml:space="preserve">le point d’appui entre <pb o="566" file="0656" n="682" rhead="NOUVEAU COURS"/> les deux. </s> <s xml:id="echoid-s17930" xml:space="preserve">Nous nommerons levier du ſecond genre celui dont <lb/>le point d’appui eſt à une des extrêmités, une puiſſance à l’autre, <lb/>& </s> <s xml:id="echoid-s17931" xml:space="preserve">le poids entre les deux. </s> <s xml:id="echoid-s17932" xml:space="preserve">Enfin nous nommerons levier du <lb/>troiſieme genre celui dont le point d’appui eſt à une des extrê-<lb/>mités, le poids à l’autre, & </s> <s xml:id="echoid-s17933" xml:space="preserve">la puiſlance entre les deux.</s> <s xml:id="echoid-s17934" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17935" xml:space="preserve">Il y a encore une quatrieme ſorte de levier, qu’on appelle <lb/>levier recourbé. </s> <s xml:id="echoid-s17936" xml:space="preserve">Ce levier eſt nommé ainſi, parce qu’il fait un <lb/>angle au point d’appui; </s> <s xml:id="echoid-s17937" xml:space="preserve">ce qui lui a fait auſſi donner le nom <lb/>d’angulaire. </s> <s xml:id="echoid-s17938" xml:space="preserve">Ce levier ſe rapporte toujours au levier du pre-<lb/>mier genre, parce que la puiſſance eſt à une des extrêmités, <lb/>le poids à l’autre, & </s> <s xml:id="echoid-s17939" xml:space="preserve">le point d’appui entre deux.</s> <s xml:id="echoid-s17940" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1495" type="section" level="1" n="1088"> <head xml:id="echoid-head1281" xml:space="preserve">CHAPITRE V.</head> <head xml:id="echoid-head1282" style="it" xml:space="preserve">De la Roue dans ſon aiſſieu.</head> <head xml:id="echoid-head1283" xml:space="preserve"><emph style="sc">Definitions</emph>.</head> <p> <s xml:id="echoid-s17941" xml:space="preserve">1085. </s> <s xml:id="echoid-s17942" xml:space="preserve">LA roue dans ſon aiſſieu eſt une machine compoſée <lb/>d’une roue attachée par ſes rayons fixement à un cylindre, que <lb/>l’on nomme treuil, aux extrêmités duquel ſont des pivots de <lb/>fer, poſés ſur un affût, qui n’eſt autre autre choſe qu’un aſ-<lb/>ſemblage de pieces de bois, qui ſert à porter la roue & </s> <s xml:id="echoid-s17943" xml:space="preserve">ſon <lb/>aiſſieu.</s> <s xml:id="echoid-s17944" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17945" xml:space="preserve">La puiſſance s’applique ordinairement à la circonférence de <lb/>la roue, qu’elle fait tourner par le moyen des chevilles qui <lb/>ſont perpendiculaires à ſon plan, comme aux roues quiſervent <lb/>à tirer les pierres des carrieres: </s> <s xml:id="echoid-s17946" xml:space="preserve">pour le poids, il eſt toujours <lb/>attaché à une corde qui tourne autour du treuil.</s> <s xml:id="echoid-s17947" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1496" type="section" level="1" n="1089"> <head xml:id="echoid-head1284" xml:space="preserve">PROPOSITION.</head> <head xml:id="echoid-head1285" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s17948" xml:space="preserve">1086. </s> <s xml:id="echoid-s17949" xml:space="preserve">Si une puiſſance ſoutient un poids à l’aide d’une roue, <lb/>& </s> <s xml:id="echoid-s17950" xml:space="preserve">que cette puiſſance agiſſe par une ligne de direction tangente à <lb/>la roue, je dis que la puiſſance ſera au poids comme le rayon du <lb/>treuil eſt au rayon de la roue.</s> <s xml:id="echoid-s17951" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1497" type="section" level="1" n="1090"> <head xml:id="echoid-head1286" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s17952" xml:space="preserve">Pour prouver que ſi la puiſſance Q ſoutient le poids P en <pb o="567" file="0657" n="683" rhead="DE MATHÉMATIQUE. Liv. XV."/> équilibre, il y aura même raiſon de Q à P, que du rayon C B <lb/> <anchor type="note" xlink:label="note-0657-01a" xlink:href="note-0657-01"/> du treuil au rayon C A de la roue. </s> <s xml:id="echoid-s17953" xml:space="preserve">Remarquez que la ligne <lb/>droite A B peut être regardée comme un levier, dont le point <lb/>d’appui eſt au centre C du treuil, & </s> <s xml:id="echoid-s17954" xml:space="preserve">que la puiſſance Q étant <lb/>à une des extrêmités du levier, & </s> <s xml:id="echoid-s17955" xml:space="preserve">le poids à l’autre, l’on aura <lb/>dans l’état de l’équilibre Q : </s> <s xml:id="echoid-s17956" xml:space="preserve">P :</s> <s xml:id="echoid-s17957" xml:space="preserve">: C B : </s> <s xml:id="echoid-s17958" xml:space="preserve">C A.</s> <s xml:id="echoid-s17959" xml:space="preserve"/> </p> <div xml:id="echoid-div1497" type="float" level="2" n="1"> <note position="right" xlink:label="note-0657-01" xlink:href="note-0657-01a" xml:space="preserve">Figure 387.</note> </div> <p> <s xml:id="echoid-s17960" xml:space="preserve">Mais ſi la puiſſance, au lieu d’agir ſelon la direction A Q, <lb/>agiſſoit ſelon la direction D F, toujours tangente à la roue, <lb/>la puiſſance ſera encore au poids comme le rayon du treuil eſt <lb/>au rayon de la roue: </s> <s xml:id="echoid-s17961" xml:space="preserve">car l’angle D C B fait un levier recourbé, <lb/>dont les bras ſont les rayons C B & </s> <s xml:id="echoid-s17962" xml:space="preserve">C D. </s> <s xml:id="echoid-s17963" xml:space="preserve">Or ſi la puiſſance <lb/>agit par une ligne de direction D F perpendiculaire au bras <lb/>C D, elle fera le même effet à l’endroit D qu’à l’endroit A: <lb/></s> <s xml:id="echoid-s17964" xml:space="preserve">ainſi le levier recourbé tenant lieu du levier du premier genre <lb/>(art. </s> <s xml:id="echoid-s17965" xml:space="preserve">1084), l’on aura toujours Q : </s> <s xml:id="echoid-s17966" xml:space="preserve">P :</s> <s xml:id="echoid-s17967" xml:space="preserve">: C B : </s> <s xml:id="echoid-s17968" xml:space="preserve">C A, ou bien <lb/>Q : </s> <s xml:id="echoid-s17969" xml:space="preserve">P :</s> <s xml:id="echoid-s17970" xml:space="preserve">: C B : </s> <s xml:id="echoid-s17971" xml:space="preserve">C D. </s> <s xml:id="echoid-s17972" xml:space="preserve">C. </s> <s xml:id="echoid-s17973" xml:space="preserve">Q. </s> <s xml:id="echoid-s17974" xml:space="preserve">F. </s> <s xml:id="echoid-s17975" xml:space="preserve">D.</s> <s xml:id="echoid-s17976" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17977" xml:space="preserve">L’on peut encore démontrer ceci par le mouvement, en <lb/>conſidérant que lorſque la puiſſance a fait un tour de la roue, <lb/>le poids a fait un tour du treuil; </s> <s xml:id="echoid-s17978" xml:space="preserve">mais nous ſçavons que la <lb/>puiſſance & </s> <s xml:id="echoid-s17979" xml:space="preserve">le poids ſont en équilibre, lorſqu’ils ſont en raiſon <lb/>réciproque de leurs vîteſſes: </s> <s xml:id="echoid-s17980" xml:space="preserve">ainſi la circonférence de la roue <lb/>exprimant la vîteſſe de la puiſſance, & </s> <s xml:id="echoid-s17981" xml:space="preserve">la circonférence du <lb/>treuil celle du poids, la puiſſance ſera au poids comme la cir-<lb/>conférence du treuil eſt à la circonférence de la roue; </s> <s xml:id="echoid-s17982" xml:space="preserve">mais <lb/>prenant les rayons à la place des circonférences, puiſqu’ils <lb/>ſont en même raiſon, l’on aura la puiſſance eſt au poids <lb/>comme le rayon du treuil eſt au rayon de la roue.</s> <s xml:id="echoid-s17983" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1499" type="section" level="1" n="1091"> <head xml:id="echoid-head1287" xml:space="preserve">CHAPITRE VI.</head> <head xml:id="echoid-head1288" style="it" xml:space="preserve">De la Poulie.</head> <head xml:id="echoid-head1289" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s17984" xml:space="preserve">1087. </s> <s xml:id="echoid-s17985" xml:space="preserve">LA poulie eſt une roue de bois ou de métal, qui eſt atta-<lb/>chée à une écharpe ou chape de fer, qui embraſſe la poulie.</s> <s xml:id="echoid-s17986" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s17987" xml:space="preserve">Lorſque la poulie eſt attachée à l’endroit d’une machine <lb/>d’où elle ne bouge point, on la nomme poulie fixe; </s> <s xml:id="echoid-s17988" xml:space="preserve">& </s> <s xml:id="echoid-s17989" xml:space="preserve">lorſ-<lb/>qu’elle eſt attachée à un poids que l’on veut enlever, on la <lb/>nomme poulie mobile.</s> <s xml:id="echoid-s17990" xml:space="preserve"/> </p> <pb o="568" file="0658" n="684" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s17991" xml:space="preserve">Lorſque pluſieurs poulies ſont enfermées dans la même <lb/>chape, ſoit qu’elles ſoient poſées les unes au deſſus des autres, <lb/>ou les unes à côté des autres, on les nomme poulies mouflées, <lb/>leſquelles peuvent être toutes enſemble fixes ou mobiles.</s> <s xml:id="echoid-s17992" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1500" type="section" level="1" n="1092"> <head xml:id="echoid-head1290" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s17993" xml:space="preserve">1088. </s> <s xml:id="echoid-s17994" xml:space="preserve">Dans la théorie de la poulie, comme dans celle <lb/>de toutes les autres machines, l’on n’a point d’égard aux frot-<lb/>temens des cordages, ni à celui de la poulie ſur ſon aiſſieu: <lb/></s> <s xml:id="echoid-s17995" xml:space="preserve">cependant l’on peut dire que plus la poulie ſera grande & </s> <s xml:id="echoid-s17996" xml:space="preserve">l’axe <lb/>petit, & </s> <s xml:id="echoid-s17997" xml:space="preserve">moins il y aura de frottement.</s> <s xml:id="echoid-s17998" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1501" type="section" level="1" n="1093"> <head xml:id="echoid-head1291" xml:space="preserve">PROPOSITION.</head> <head xml:id="echoid-head1292" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s17999" xml:space="preserve">1089. </s> <s xml:id="echoid-s18000" xml:space="preserve">Si une puiſſance ſoutient un poids à l’aide d’une poulie, <lb/>dont la chape ſoit immobile, je dis, 1°. </s> <s xml:id="echoid-s18001" xml:space="preserve">que la puiſſance ſera égale <lb/>au poids. </s> <s xml:id="echoid-s18002" xml:space="preserve">2°. </s> <s xml:id="echoid-s18003" xml:space="preserve">Que ſi la chape eſt mobile, de ſorte que le poids qui <lb/>y ſeroit attaché, ſoit enlevé par la puiſſance, cette puiſſance ſera <lb/>la moitié du poids, lorſque la direction de la puiſſance & </s> <s xml:id="echoid-s18004" xml:space="preserve">celle du <lb/>poids ſeront paralleles.</s> <s xml:id="echoid-s18005" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1502" type="section" level="1" n="1094"> <head xml:id="echoid-head1293" xml:space="preserve"><emph style="sc">Démonstration du premier cas</emph>.</head> <p> <s xml:id="echoid-s18006" xml:space="preserve">Si l’on conſidere le diametre A B de la poulie, comme un <lb/> <anchor type="note" xlink:label="note-0658-01a" xlink:href="note-0658-01"/> levier du premier genre, puiſque le poids eſt à une extrêmité, <lb/>la puiſſance à l’autre, & </s> <s xml:id="echoid-s18007" xml:space="preserve">le point d’appui entre les deux, qui eſt <lb/>ici le point C. </s> <s xml:id="echoid-s18008" xml:space="preserve">Il faudra, pour que la puiſſance ſoit en équilibre <lb/>avec le poids, avoir cette proportion, Q : </s> <s xml:id="echoid-s18009" xml:space="preserve">P :</s> <s xml:id="echoid-s18010" xml:space="preserve">: C A : </s> <s xml:id="echoid-s18011" xml:space="preserve">C B. </s> <s xml:id="echoid-s18012" xml:space="preserve">Mais <lb/>comme l’on a C A égal à C B, puiſque ce ſont les rayons d’un <lb/>même cercle, l’on aura Q = P. </s> <s xml:id="echoid-s18013" xml:space="preserve">C. </s> <s xml:id="echoid-s18014" xml:space="preserve">Q. </s> <s xml:id="echoid-s18015" xml:space="preserve">F. </s> <s xml:id="echoid-s18016" xml:space="preserve">D.</s> <s xml:id="echoid-s18017" xml:space="preserve"/> </p> <div xml:id="echoid-div1502" type="float" level="2" n="1"> <note position="left" xlink:label="note-0658-01" xlink:href="note-0658-01a" xml:space="preserve">Figure 388.</note> </div> <p> <s xml:id="echoid-s18018" xml:space="preserve">Pour démontrer ceci par le mouvement, faites attention <lb/>que ſi la puiſſance Q tire de haut en bas, la corde B Q de la <lb/>longueur de deux pieds, cela ne ſe pourra faire ſans que le <lb/>poids P ne ſoit monté, d’autant que la puiſſance eſt deſcen-<lb/>due, c’eſt-à-dire de deux pieds; </s> <s xml:id="echoid-s18019" xml:space="preserve">mais dans l’état de l’équi-<lb/>libre, la puiſſance doit être au poids dans la raiſon réciproque <lb/>de la vîteſſe ou du chemin de la puiſſance & </s> <s xml:id="echoid-s18020" xml:space="preserve">du poids. </s> <s xml:id="echoid-s18021" xml:space="preserve">Et <lb/>comme la vîteſſe de l’une eſt égale à la vîteſſe de l’autre, la <lb/>force de l’une ſera égale à la force de l’autre.</s> <s xml:id="echoid-s18022" xml:space="preserve"/> </p> <pb o="569" file="0659" n="685" rhead="DE MATHÉMATIQUE. Liv. XV."/> </div> <div xml:id="echoid-div1504" type="section" level="1" n="1095"> <head xml:id="echoid-head1294" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s18023" xml:space="preserve">1090. </s> <s xml:id="echoid-s18024" xml:space="preserve">Il ſuit delà que les poulies fixes n’augmentent point <lb/>la force de la puiſſance, & </s> <s xml:id="echoid-s18025" xml:space="preserve">qu’elles ne ſervent qu’à changer les <lb/>directions, & </s> <s xml:id="echoid-s18026" xml:space="preserve">à diminuer le frottement, qui ſeroit très con-<lb/>ſidérable, ſi la corde ne tournoit pas avec la poulie, & </s> <s xml:id="echoid-s18027" xml:space="preserve">étoit <lb/>obligée de gliſſer ou de paſſer pardeſſus un cylindre immobile; <lb/></s> <s xml:id="echoid-s18028" xml:space="preserve">au lieu qu’il n’eſt preſque queſtion ici que du frottement qui <lb/>ſe fait de la poulie contre ſon aiſſieu, qui eſt bien plus petit <lb/>que celui que feroit la corde ſur le cylindre immobile, le <lb/>frottement de l’aiſſieu étant à celui du cylindre immobile, <lb/>comme le rayon de l’aiſſieu eſt à celui de la poulie; </s> <s xml:id="echoid-s18029" xml:space="preserve">ce qui <lb/>fait voir, comme nous l’avons déja dit, que plus la poulie eſt <lb/>grande, & </s> <s xml:id="echoid-s18030" xml:space="preserve">l’aiſſieu petit, moins il y aura de frottement.</s> <s xml:id="echoid-s18031" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1505" type="section" level="1" n="1096"> <head xml:id="echoid-head1295" xml:space="preserve"><emph style="sc">Démonstration du second cas</emph>.</head> <p> <s xml:id="echoid-s18032" xml:space="preserve">Si l’on ſuppoſe une poulie A B, au deſſous de laquelle paſſe <lb/> <anchor type="note" xlink:label="note-0659-01a" xlink:href="note-0659-01"/> une corde, dont l’un des bouts ſoit attaché à un endroit fixe <lb/>G, & </s> <s xml:id="echoid-s18033" xml:space="preserve">qu’à l’autre bout A E ſoit appliquée une puiſſance Q, <lb/>ou bien que l’autre bout de la corde paſſe au deſſus d’une <lb/>poulie D E, afin que la puiſſance étant en Q, & </s> <s xml:id="echoid-s18034" xml:space="preserve">tirant de <lb/>haut en bas, agiſſe plus commodément: </s> <s xml:id="echoid-s18035" xml:space="preserve">enfin que le poids P <lb/>ſoit attaché à l’écharpe C I, il faut prouver que la puiſſance ne <lb/>ſoutient que la moitié du poids.</s> <s xml:id="echoid-s18036" xml:space="preserve"/> </p> <div xml:id="echoid-div1505" type="float" level="2" n="1"> <note position="right" xlink:label="note-0659-01" xlink:href="note-0659-01a" xml:space="preserve">Figure 389.</note> </div> <p> <s xml:id="echoid-s18037" xml:space="preserve">Pour cela, faites attention que le diametre A B de la poulie <lb/>peut être regardé comme un levier du ſecond genre, dont le <lb/>point d’appui eſt à l’extrêmité B, la puiſſance à l’extrêmité A, <lb/>& </s> <s xml:id="echoid-s18038" xml:space="preserve">le poids dans le milieu. </s> <s xml:id="echoid-s18039" xml:space="preserve">Or ſi la puiſſance eſt en équilibre <lb/>avec le poids, l’on aura Q : </s> <s xml:id="echoid-s18040" xml:space="preserve">P :</s> <s xml:id="echoid-s18041" xml:space="preserve">: C B : </s> <s xml:id="echoid-s18042" xml:space="preserve">A B; </s> <s xml:id="echoid-s18043" xml:space="preserve">mais le rayon C B, <lb/>eſt la moitié du diametre A B: </s> <s xml:id="echoid-s18044" xml:space="preserve">donc la puiſſance Q ſera la <lb/>moitié du poids P.</s> <s xml:id="echoid-s18045" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18046" xml:space="preserve">Il faut remarquer que par ce qui a été démontré dans le <lb/>premier cas, la poulie D E ne fait autre choſe ici que faciliter <lb/>l’action de la puiſſance, puiſqu’elle n’aura pas plus de force <lb/>appliquée dans la partie E A de la corde, que dans la partie <lb/>D Q, comptant toujours pour rien le frottement dans la <lb/>poulie D E, comme dans la poulie A B.</s> <s xml:id="echoid-s18047" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18048" xml:space="preserve">On démontrera encore ceci par le mouvement, en conſi-<lb/>dérant que ſi la puiſſance a élevé le poids P de deux pieds, <lb/>chaque brin de corde G B & </s> <s xml:id="echoid-s18049" xml:space="preserve">E A ſera diminué de deux pieds:</s> <s xml:id="echoid-s18050" xml:space="preserve"> <pb o="570" file="0660" n="686" rhead="NOUVEAU COURS"/> ainſi la puiſſance Q ſera deſcendue de quatre pieds, ou pour <lb/>mieux dire, le brin D Q ſera augmenté de quatre pieds: </s> <s xml:id="echoid-s18051" xml:space="preserve">ainſi <lb/>le mouvement de la puiſſance ſera double de celui du poids; <lb/></s> <s xml:id="echoid-s18052" xml:space="preserve">par conſéquent le poids ſera double de la puiſſance, puiſque <lb/>dans l’état de l’équilibre, la puiſſance & </s> <s xml:id="echoid-s18053" xml:space="preserve">le poids ſont dans la <lb/>raiſon réciproque de leurs vîteſſes.</s> <s xml:id="echoid-s18054" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1507" type="section" level="1" n="1097"> <head xml:id="echoid-head1296" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s18055" xml:space="preserve">1091. </s> <s xml:id="echoid-s18056" xml:space="preserve">Il eſt à remarquer que ſi les brins A Q & </s> <s xml:id="echoid-s18057" xml:space="preserve">B G ne <lb/>ſont point paralleles, l’analogie précédente nc ſera plus la <lb/>même, c’eſt-à-dire que l’on n’aura pas Q : </s> <s xml:id="echoid-s18058" xml:space="preserve">P :</s> <s xml:id="echoid-s18059" xml:space="preserve">: B C : </s> <s xml:id="echoid-s18060" xml:space="preserve">A B; <lb/></s> <s xml:id="echoid-s18061" xml:space="preserve">mais que le rapport de la puiſſance au poids ſera dans la raiſon <lb/>réciproque des perpendiculaires tirées du point d’appui B ſur <lb/>les lignes de directions du poids & </s> <s xml:id="echoid-s18062" xml:space="preserve">de la puiſſance. </s> <s xml:id="echoid-s18063" xml:space="preserve">Or pre-<lb/>nant la ligne A H pour la direction de la puiſſance, & </s> <s xml:id="echoid-s18064" xml:space="preserve">la ligne <lb/>C I pour celle du poids, B C ſera une perpendiculaire tirée ſur <lb/>la direction C I du poids, & </s> <s xml:id="echoid-s18065" xml:space="preserve">B F ſera une perpendiculaire ſur <lb/>la direction A H de la puiſſance: </s> <s xml:id="echoid-s18066" xml:space="preserve">ainſi l’on aura Q : </s> <s xml:id="echoid-s18067" xml:space="preserve">P :</s> <s xml:id="echoid-s18068" xml:space="preserve">: B C : </s> <s xml:id="echoid-s18069" xml:space="preserve">B F. </s> <s xml:id="echoid-s18070" xml:space="preserve"><lb/>Ce qui eſt facile à entendre, ſi l’on a bien compris ce qui a <lb/>été enſeigné au ſujet du levier.</s> <s xml:id="echoid-s18071" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18072" xml:space="preserve">Mais comme plus la ligne B A eſt grande par rapport à la <lb/>ligne B C, plus la puiſſance eſt grande par rapport au poids <lb/>dans le levier du ſecond genre, il s’enſuit que la ligne B F <lb/>devenant plus petite que B A, lorſque les brins ne ſont pas <lb/>paralleles, la puiſſance n’a pas tant de force dans ce cas ci que <lb/>dans l’autre, & </s> <s xml:id="echoid-s18073" xml:space="preserve">par conſéquent il faut que les brins ſoient <lb/>paralleles, pour que la puiſſance agiſſe avec toute ſa force.</s> <s xml:id="echoid-s18074" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1508" type="section" level="1" n="1098"> <head xml:id="echoid-head1297" xml:space="preserve">CHAPITRE VII.</head> <head xml:id="echoid-head1298" style="it" xml:space="preserve">Du Coin.</head> <head xml:id="echoid-head1299" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s18075" xml:space="preserve">1092. </s> <s xml:id="echoid-s18076" xml:space="preserve">LE coin eſt une machine de fer ou de bois ſervant à <lb/>élever des corps à une petite hauteur, ou à fendre du bois, <lb/>qui eſt ſon principal uſage. </s> <s xml:id="echoid-s18077" xml:space="preserve">Sa figure eſt ordinairement iſoſ-<lb/>cele, quand il ſert à fendre du bois; </s> <s xml:id="echoid-s18078" xml:space="preserve">mais on ſuppoſe qu’elle <lb/>eſt rectangle, quand on s’en ſert pour élever un corps peſant.</s> <s xml:id="echoid-s18079" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18080" xml:space="preserve">On ſuppoſe en premier lieu que les faces A O & </s> <s xml:id="echoid-s18081" xml:space="preserve">B O du coin <pb o="571" file="0661" n="687" rhead="DE MATHÉMATIQUE. Liv. XV."/> ſont égales, & </s> <s xml:id="echoid-s18082" xml:space="preserve">que le bois eſt flexible; </s> <s xml:id="echoid-s18083" xml:space="preserve">de maniere qu’étant <lb/>commencé à fendre, & </s> <s xml:id="echoid-s18084" xml:space="preserve">le coin introduit par la force qui le <lb/>pouſſe dans la fente, les faces de la fente ſont pliées en ligne <lb/>courbe, & </s> <s xml:id="echoid-s18085" xml:space="preserve">que les faces du coin les pouſſent en deux points <lb/>I & </s> <s xml:id="echoid-s18086" xml:space="preserve">K, où il y a deux puiſſances égales, qui réſiſtent ſelon <lb/>des directions E C & </s> <s xml:id="echoid-s18087" xml:space="preserve">F C perpendiculaires aux faces du coin, <lb/>& </s> <s xml:id="echoid-s18088" xml:space="preserve">à celle des fentes qui repouſſent celles du coin, autant <lb/>qu’elles ſont pouſſées par le coin, parce que l’action eſt égale <lb/>à la réaction, en ſuppoſant que la tête du coin eſt frappée en <lb/>G par un maillet ou une force, dont la direction eſt perpen-<lb/>diculaire à A B, & </s> <s xml:id="echoid-s18089" xml:space="preserve">paſſe par l’angle A O B du coin qu’elle diviſe <lb/>en deux également, puiſque le coin eſt iſoſcele. </s> <s xml:id="echoid-s18090" xml:space="preserve">Or l’objet de <lb/>ceci eſt de prouver premiérement que dans l’inſtant de l’équi-<lb/>libre que le coin eſt enchâſſé, comme on vient de le dire, le <lb/>bois ne ſe fend point, mais il ſe ſeroit fendu, pour peu que la <lb/>force du coin eût été plus grande; </s> <s xml:id="echoid-s18091" xml:space="preserve">il faut prouver, dis-je, que <lb/>dans l’inſtant de l’équilibre les faces du coin pouſſant celles <lb/>des fentes en ſont également repouſſées; </s> <s xml:id="echoid-s18092" xml:space="preserve">ou, ce qui eſt la <lb/>même choſe, que les deux efforts qui ſe font en I & </s> <s xml:id="echoid-s18093" xml:space="preserve">en K <lb/>ſont égaux.</s> <s xml:id="echoid-s18094" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18095" xml:space="preserve">Pour cela ayant pris ſur G O, direction de la puiſſance R, <lb/>un point quelconque D, & </s> <s xml:id="echoid-s18096" xml:space="preserve">achevé le parallélogramme C E D F, <lb/>je dis qu’il a tous ſes côtés égaux: </s> <s xml:id="echoid-s18097" xml:space="preserve">car les triangles C I O, <lb/>C K O, rectangles en I & </s> <s xml:id="echoid-s18098" xml:space="preserve">en K, ſont égaux & </s> <s xml:id="echoid-s18099" xml:space="preserve">ſemblables, <lb/>puiſqueles angles C O I, C O K ſont égaux, & </s> <s xml:id="echoid-s18100" xml:space="preserve">par conſéquent <lb/>auſſi les angles O C I, O C K; </s> <s xml:id="echoid-s18101" xml:space="preserve">mais l’angle O C F eſt égal à <lb/>l’angle C D E, étant alternes: </s> <s xml:id="echoid-s18102" xml:space="preserve">donc l’angle O C I égalà O C K, <lb/>eſt égal à l’angle C D E, & </s> <s xml:id="echoid-s18103" xml:space="preserve">par conſéquent C E & </s> <s xml:id="echoid-s18104" xml:space="preserve">D E ſont <lb/>égales entr’elles, & </s> <s xml:id="echoid-s18105" xml:space="preserve">partant le parallélogramme E F a les quatre <lb/>côtés égaux; </s> <s xml:id="echoid-s18106" xml:space="preserve">mais dans l’état de l’équilibre, l’action du coin <lb/>ou la réſiſtance du bois en I, eſt à l’action du coin ou à la <lb/>réſiſtance du bois en K, comme C E, C F: </s> <s xml:id="echoid-s18107" xml:space="preserve">donc puiſque C E <lb/>& </s> <s xml:id="echoid-s18108" xml:space="preserve">C F ſont égaux, l’effort du coin en I eſt égal à l’effort du <lb/>coin en K: </s> <s xml:id="echoid-s18109" xml:space="preserve">nommant donc la force qui pouſſe le coin R, & </s> <s xml:id="echoid-s18110" xml:space="preserve"><lb/>l’effort du coin en I, P, l’effort en K ſera auſſi P.</s> <s xml:id="echoid-s18111" xml:space="preserve"/> </p> <pb o="572" file="0662" n="688" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1509" type="section" level="1" n="1099"> <head xml:id="echoid-head1300" xml:space="preserve">PROPOSITION.</head> <head xml:id="echoid-head1301" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s18112" xml:space="preserve">1093. </s> <s xml:id="echoid-s18113" xml:space="preserve">La force qui chaſſe le coin eſt à la réſiſtance du bois, <lb/> <anchor type="note" xlink:label="note-0662-01a" xlink:href="note-0662-01"/> comme la moitié de la tête du coin eſt à la longueur d’un de ſes <lb/>côtés: </s> <s xml:id="echoid-s18114" xml:space="preserve">ainſi il faut prouver, 1°. </s> <s xml:id="echoid-s18115" xml:space="preserve">que R : </s> <s xml:id="echoid-s18116" xml:space="preserve">2P :</s> <s xml:id="echoid-s18117" xml:space="preserve">: A G : </s> <s xml:id="echoid-s18118" xml:space="preserve">A O. </s> <s xml:id="echoid-s18119" xml:space="preserve">2°. </s> <s xml:id="echoid-s18120" xml:space="preserve">Que <lb/>ſi une puiſſance ſoutient un poids à l’aide d’un coin, la puiſſance <lb/>ſera au poids, comme la hauteur du coin eſt à ſa longueur.</s> <s xml:id="echoid-s18121" xml:space="preserve"/> </p> <div xml:id="echoid-div1509" type="float" level="2" n="1"> <note position="left" xlink:label="note-0662-01" xlink:href="note-0662-01a" xml:space="preserve">Figure 390.</note> </div> </div> <div xml:id="echoid-div1511" type="section" level="1" n="1100"> <head xml:id="echoid-head1302" xml:space="preserve"><emph style="sc">Démonstration du premier cas</emph>.</head> <p> <s xml:id="echoid-s18122" xml:space="preserve">Il eſt clair que les trois puiſſances P, P, R peuvent être re-<lb/>gardées comme agiſſantes contre le point C, où leurs direc-<lb/>tions concourent: </s> <s xml:id="echoid-s18123" xml:space="preserve">c’eſt pourquoi l’on a R: </s> <s xml:id="echoid-s18124" xml:space="preserve">P:</s> <s xml:id="echoid-s18125" xml:space="preserve">: C D: </s> <s xml:id="echoid-s18126" xml:space="preserve">C E <lb/>+ C F, ou C E + E D; </s> <s xml:id="echoid-s18127" xml:space="preserve">mais les triangles A B O, C D B <lb/>ſont ſemblables: </s> <s xml:id="echoid-s18128" xml:space="preserve">car les triangles A G O, C I O le ſont, ayant <lb/>chacun un angle droit aux points G & </s> <s xml:id="echoid-s18129" xml:space="preserve">I; </s> <s xml:id="echoid-s18130" xml:space="preserve">& </s> <s xml:id="echoid-s18131" xml:space="preserve">l’angle au point O <lb/>commun: </s> <s xml:id="echoid-s18132" xml:space="preserve">c’eſt pourquoi C D: </s> <s xml:id="echoid-s18133" xml:space="preserve">C E + D E, ou 2CE :</s> <s xml:id="echoid-s18134" xml:space="preserve">: AB : </s> <s xml:id="echoid-s18135" xml:space="preserve">AO <lb/>+ BO ou 2AO: </s> <s xml:id="echoid-s18136" xml:space="preserve">donc R : </s> <s xml:id="echoid-s18137" xml:space="preserve">2P :</s> <s xml:id="echoid-s18138" xml:space="preserve">: AB : </s> <s xml:id="echoid-s18139" xml:space="preserve">2AO, ou R : </s> <s xml:id="echoid-s18140" xml:space="preserve">2P :</s> <s xml:id="echoid-s18141" xml:space="preserve">: AG : </s> <s xml:id="echoid-s18142" xml:space="preserve">AO, <lb/>en diviſant par 2 les deux termes du deuxieme rapport. <lb/></s> <s xml:id="echoid-s18143" xml:space="preserve">C. </s> <s xml:id="echoid-s18144" xml:space="preserve">Q. </s> <s xml:id="echoid-s18145" xml:space="preserve">F. </s> <s xml:id="echoid-s18146" xml:space="preserve">D.</s> <s xml:id="echoid-s18147" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1512" type="section" level="1" n="1101"> <head xml:id="echoid-head1303" xml:space="preserve"><emph style="sc">Démonstration du second cas</emph>.</head> <p> <s xml:id="echoid-s18148" xml:space="preserve">Pour démontrer préſentement que ſi une puiſſance Q ſou-<lb/> <anchor type="note" xlink:label="note-0662-02a" xlink:href="note-0662-02"/> tient un poids à l’aide d’un coin A B C, la puiſſance eſt au <lb/> <anchor type="note" xlink:label="note-0662-03a" xlink:href="note-0662-03"/> poids, comme ſa hauteur B C eſt à ſa longueur C A, ſuppo-<lb/>ſons que le poids P ſoit retenu par une corde G D, attachée à <lb/>un point fixe D, & </s> <s xml:id="echoid-s18149" xml:space="preserve">qu’une puiſſance Q pouſſe le coin, enſorte <lb/>que de l’endroit où il étoit, il ſoit parvenu en F A; </s> <s xml:id="echoid-s18150" xml:space="preserve">pour lors <lb/>le poids P ſera monté au ſommet B du coin, ou au ſommet E, <lb/>qui eſt la même choſe: </s> <s xml:id="echoid-s18151" xml:space="preserve">alors le chemin de la puiſſance ſera ex-<lb/>primé par la ligne A C, & </s> <s xml:id="echoid-s18152" xml:space="preserve">le chemin du poids par la ligne <lb/>CB: </s> <s xml:id="echoid-s18153" xml:space="preserve">car la puiſſance a été de A en F; </s> <s xml:id="echoid-s18154" xml:space="preserve">ou, ce qui eſt la même <lb/>choſe, de C en A dans le même-tems que le poids eſt monté <lb/>de la hauteur B C ou E A; </s> <s xml:id="echoid-s18155" xml:space="preserve">mais dans l’état de l’équilibre, la <lb/>puiſſance & </s> <s xml:id="echoid-s18156" xml:space="preserve">le poids ſont dans la raiſon réciproque de leurs <lb/>vîteſſes: </s> <s xml:id="echoid-s18157" xml:space="preserve">donc l’on aura Q : </s> <s xml:id="echoid-s18158" xml:space="preserve">P :</s> <s xml:id="echoid-s18159" xml:space="preserve">: B C : </s> <s xml:id="echoid-s18160" xml:space="preserve">C A. </s> <s xml:id="echoid-s18161" xml:space="preserve">C. </s> <s xml:id="echoid-s18162" xml:space="preserve">Q. </s> <s xml:id="echoid-s18163" xml:space="preserve">F. </s> <s xml:id="echoid-s18164" xml:space="preserve">D.</s> <s xml:id="echoid-s18165" xml:space="preserve"/> </p> <div xml:id="echoid-div1512" type="float" level="2" n="1"> <note position="left" xlink:label="note-0662-02" xlink:href="note-0662-02a" xml:space="preserve">Pl. XXX.</note> <note position="left" xlink:label="note-0662-03" xlink:href="note-0662-03a" xml:space="preserve">Figure 391.</note> </div> </div> <div xml:id="echoid-div1514" type="section" level="1" n="1102"> <head xml:id="echoid-head1304" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s18166" xml:space="preserve">1094. </s> <s xml:id="echoid-s18167" xml:space="preserve">Il ſuit delà que plus la hauteur ou la tête du coin eſt <lb/>petite, plus la puiſſance a de force.</s> <s xml:id="echoid-s18168" xml:space="preserve"/> </p> <pb o="573" file="0663" n="689" rhead="DE MATHÉMATIQUE. Liv. XV."/> </div> <div xml:id="echoid-div1515" type="section" level="1" n="1103"> <head xml:id="echoid-head1305" xml:space="preserve">CHAPITRE VIII.</head> <head xml:id="echoid-head1306" style="it" xml:space="preserve">De la Vis.</head> <p> <s xml:id="echoid-s18169" xml:space="preserve">1095. </s> <s xml:id="echoid-s18170" xml:space="preserve">LA vis eſt de toutes les machines celle qui donne le <lb/>plus de force à la puiſſance pour élever ou pour preſſer un <lb/>corps, lorſque la puiſſance ſe ſert d’un levier pour la mettre <lb/>en mouvement; </s> <s xml:id="echoid-s18171" xml:space="preserve">& </s> <s xml:id="echoid-s18172" xml:space="preserve">quoique cette machine ſoit connue de tout <lb/>le monde, voici cependant de la façon qu’il faut la conce-<lb/>voir, afin de mieux entendre l’analogie que nous en ferons.</s> <s xml:id="echoid-s18173" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18174" xml:space="preserve">Ayant un cylindre A B C D, imaginons que ſa hauteur B D <lb/> <anchor type="note" xlink:label="note-0663-01a" xlink:href="note-0663-01"/> eſt diviſée en un nombre de parties égales, & </s> <s xml:id="echoid-s18175" xml:space="preserve">que par chaque <lb/>point de diviſion, comme F & </s> <s xml:id="echoid-s18176" xml:space="preserve">H, l’on a tiré des perpendicu-<lb/>laires F E & </s> <s xml:id="echoid-s18177" xml:space="preserve">H G à la ligne B D, & </s> <s xml:id="echoid-s18178" xml:space="preserve">que chaque perpendicu-<lb/>laire ſoit égale à la circonférence du cercle du cylindre, c’eſt-<lb/>à-dire qui auroit A B pour diametre. </s> <s xml:id="echoid-s18179" xml:space="preserve">Or ſi l’on tire des lignes <lb/>E B & </s> <s xml:id="echoid-s18180" xml:space="preserve">G F, l’on aura autant de triangles rectangles E B F & </s> <s xml:id="echoid-s18181" xml:space="preserve"><lb/>G F H, qu’il y a de parties égales dans la hauteur B D; </s> <s xml:id="echoid-s18182" xml:space="preserve">& </s> <s xml:id="echoid-s18183" xml:space="preserve">ſi <lb/>l’on roule tous ces triangles ſur le cylindre, le point E viendra <lb/>aboutir en F, & </s> <s xml:id="echoid-s18184" xml:space="preserve">le point G en H, & </s> <s xml:id="echoid-s18185" xml:space="preserve">toutes les hypoténuſes <lb/>E B & </s> <s xml:id="echoid-s18186" xml:space="preserve">G F ainſi roulés, formeront enſemble une ſpirale ſur <lb/>le cylindre, qui commencera en B, & </s> <s xml:id="echoid-s18187" xml:space="preserve">finira en D; </s> <s xml:id="echoid-s18188" xml:space="preserve">ou autre-<lb/>ment toutes ces hypoténuſes formeront les filets de la vis, & </s> <s xml:id="echoid-s18189" xml:space="preserve"><lb/>les hauteurs B F & </s> <s xml:id="echoid-s18190" xml:space="preserve">F H ſeront les intervalles de ces filets, que <lb/>l’on nomme pas de la vis: </s> <s xml:id="echoid-s18191" xml:space="preserve">ainſi l’on peut dire que la vis eſt <lb/>un cylindre enveloppé de triangles rectangles, dont les hypo-<lb/>ténuſes E B & </s> <s xml:id="echoid-s18192" xml:space="preserve">G F formeront les filets, les hauteurs B F & </s> <s xml:id="echoid-s18193" xml:space="preserve"><lb/>F H les pas de la vis, & </s> <s xml:id="echoid-s18194" xml:space="preserve">les baſes E F & </s> <s xml:id="echoid-s18195" xml:space="preserve">G H le contour du <lb/>cylindre.</s> <s xml:id="echoid-s18196" xml:space="preserve"/> </p> <div xml:id="echoid-div1515" type="float" level="2" n="1"> <note position="right" xlink:label="note-0663-01" xlink:href="note-0663-01a" xml:space="preserve">Figure 392.</note> </div> <p> <s xml:id="echoid-s18197" xml:space="preserve">L’écroue dans lequel entre la vis, eſt un autre cylindre <lb/>creux, dont le diametre eſt égal à celui de la vis, & </s> <s xml:id="echoid-s18198" xml:space="preserve">dont la <lb/>ſurface intérieure eſt compoſée de triangles rectangles égaux, <lb/>& </s> <s xml:id="echoid-s18199" xml:space="preserve">ſemblables à ceux qui ſont roulés ſur le cylindre pour for-<lb/>mer la vis: </s> <s xml:id="echoid-s18200" xml:space="preserve">c’eſt ainſi que les Géometres regardent la vis & </s> <s xml:id="echoid-s18201" xml:space="preserve"><lb/>ſon écroue.</s> <s xml:id="echoid-s18202" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18203" xml:space="preserve">Mais afin de tirer de la vis toute l’utilité qu’on en attend, <lb/>il faut entailler le cylindre entre les filets formés par les hy-<lb/>poténuſes des triangles rectangles d’une certaine profondeur, <pb o="574" file="0664" n="690" rhead="NOUVEAU COURS"/> & </s> <s xml:id="echoid-s18204" xml:space="preserve">diminuer le diametre de l’écroue d’une grandeur égale à la <lb/>profondeur des entrailles de la vis, & </s> <s xml:id="echoid-s18205" xml:space="preserve">faire les mêmes entailles <lb/>dans les creux de l’écroue, afin que la vis puiſſe entrer dedans, <lb/>& </s> <s xml:id="echoid-s18206" xml:space="preserve">y tourner librement: </s> <s xml:id="echoid-s18207" xml:space="preserve">ſi l’écroue eſt fixe en tournant la vis, <lb/>on la fait avancer, & </s> <s xml:id="echoid-s18208" xml:space="preserve">ſi c’eſt la vis qui eſt immobile, on fait <lb/>avancer l’écroue.</s> <s xml:id="echoid-s18209" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18210" xml:space="preserve">Il y a encore une autre ſorte de vis, que l’on nomme vis <lb/>ſans fin, qui n’entre point dans un écroue. </s> <s xml:id="echoid-s18211" xml:space="preserve">Elle eſt miſe en <lb/>mouvement par une manivelle, ou par une roue dentée, dont <lb/>les dents gliſſent le long des pas de la vis, comme on le verra <lb/>dans les machines compoſées.</s> <s xml:id="echoid-s18212" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1517" type="section" level="1" n="1104"> <head xml:id="echoid-head1307" xml:space="preserve">PROPOSITION.</head> <head xml:id="echoid-head1308" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s18213" xml:space="preserve">1096. </s> <s xml:id="echoid-s18214" xml:space="preserve">Si une puiſſance preſſe ou enleve un poids à l’aide d’une <lb/>vis, la puiſſance ſera au poids, comme la hauteur d’un des pas de <lb/>la vis eſt à la circonférence du cercle que décrira la puiſſance ap-<lb/>pliquée au levier, par le moyen duquel on meut la vis.</s> <s xml:id="echoid-s18215" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1518" type="section" level="1" n="1105"> <head xml:id="echoid-head1309" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s18216" xml:space="preserve">Si l’on ſuppoſe que l’écroue C D de la vis ſoit immobile ſur <lb/> <anchor type="note" xlink:label="note-0664-01a" xlink:href="note-0664-01"/> le plan G H, la vis E F étant miſe en mouvement, fera monter <lb/>le poids P qui eſt attaché à ſon extrêmité F, & </s> <s xml:id="echoid-s18217" xml:space="preserve">ſi la puiſſance <lb/>Q eſt appliquée à l’extrêmité B d’un levier A B, il faudra, <lb/>pour faire tourner la vis, qu’elle tourne elle-même. </s> <s xml:id="echoid-s18218" xml:space="preserve">Or dans <lb/>le tems qu’elle aura décrit une circonférence de cercle, dont <lb/>le rayon ſera A B, la vis aura auſſi fait un tour, & </s> <s xml:id="echoid-s18219" xml:space="preserve">ſera montée <lb/>de la hauteur d’un pas: </s> <s xml:id="echoid-s18220" xml:space="preserve">ainſi le chemin ou la vîteſſe de la puiſ-<lb/>ſance ſera exprimé par la circonférence I B, & </s> <s xml:id="echoid-s18221" xml:space="preserve">le chemin ou <lb/>la vîteſſe du poids par la hauteur d’un pas de la vis; </s> <s xml:id="echoid-s18222" xml:space="preserve">mais dans <lb/>l’état de l’équilibre, la puiſſance eſt au poids dans la raiſon <lb/>réciproque de la vîteſſe de l’une à celle de l’autre: </s> <s xml:id="echoid-s18223" xml:space="preserve">donc la <lb/>puiſſance Q eſt au poids P, comme la hauteur d’un pas de la <lb/>vis eſt à la circonférence décrite par la puiſſance Q. </s> <s xml:id="echoid-s18224" xml:space="preserve">C. </s> <s xml:id="echoid-s18225" xml:space="preserve">Q. </s> <s xml:id="echoid-s18226" xml:space="preserve">F. </s> <s xml:id="echoid-s18227" xml:space="preserve">D.</s> <s xml:id="echoid-s18228" xml:space="preserve"/> </p> <div xml:id="echoid-div1518" type="float" level="2" n="1"> <note position="left" xlink:label="note-0664-01" xlink:href="note-0664-01a" xml:space="preserve">Figure 392 <lb/>& 393.</note> </div> </div> <div xml:id="echoid-div1520" type="section" level="1" n="1106"> <head xml:id="echoid-head1310" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s18229" xml:space="preserve">1097. </s> <s xml:id="echoid-s18230" xml:space="preserve">Il ſuit delà que plus les pas de la vis ſeront ſerrés, & </s> <s xml:id="echoid-s18231" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s18232" xml:space="preserve">le levier long, plus la puiſſance aura de force. </s> <s xml:id="echoid-s18233" xml:space="preserve">Ainſi ſuppo-<lb/>ſant que les pas de la vis ne ſoient éloignés que de deux pouces, <pb o="575" file="0665" n="691" rhead="DE MATHÉMATIQUE. Liv. XV."/> & </s> <s xml:id="echoid-s18234" xml:space="preserve">que le levier ſoit de 6 pieds, ou autrement de 72 pouces, <lb/>la circonférence du cercle, dont il ſera le rayon, ſera de 452 <lb/>pouces: </s> <s xml:id="echoid-s18235" xml:space="preserve">ainſi la puiſſance ſera au poids, comme 2 eſt à 452, <lb/>ou bien comme 1 eſt à 226: </s> <s xml:id="echoid-s18236" xml:space="preserve">par conſéquent une puiſſance <lb/>d’une livre ſera en équilibre avec un poids de 226 livres.</s> <s xml:id="echoid-s18237" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18238" xml:space="preserve">Nous n’avons point eu d’égard ici au frottement, non plus <lb/>que dans les autres machines, quoiqu’il ſoit conſidérable.</s> <s xml:id="echoid-s18239" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1521" type="section" level="1" n="1107"> <head xml:id="echoid-head1311" xml:space="preserve">CHAPITRE IX.</head> <head xml:id="echoid-head1312" style="it" xml:space="preserve">Des Machines compoſées.</head> <p> <s xml:id="echoid-s18240" xml:space="preserve">1098. </s> <s xml:id="echoid-s18241" xml:space="preserve">NOUS avons déja dit que lorſque pluſieurs machines <lb/>ſimples de mêmes ou de différentes eſpeces, ſervent à faire <lb/>mouvoir un corps, la machine qui étoit compoſée de toutes <lb/>celles-là, ſe nommoit machine compoſée. </s> <s xml:id="echoid-s18242" xml:space="preserve">Or comme ces ſortes <lb/>de machines montrent parfaitement l’utilité que l’on tire des <lb/>méchaniques dans la pratique des Arts, nous allons faire voir <lb/>les propriétés de celles qui ſont le plus d’uſage.</s> <s xml:id="echoid-s18243" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18244" xml:space="preserve">1099. </s> <s xml:id="echoid-s18245" xml:space="preserve">Mais avant cela, il faut ſçavoir que l’effort d’un <lb/>homme qui agit en pouſſant ou tirant (comme font ceux qui <lb/>tournent au cabeſtan, & </s> <s xml:id="echoid-s18246" xml:space="preserve">qui tirent les charrettes), n’eſt<unsure/> que <lb/>d’environ 25 livres, & </s> <s xml:id="echoid-s18247" xml:space="preserve">que celle des chevaux qui agiſient de la <lb/>même maniere, n’eſt que de 175 livres, ou égale à celle de <lb/>ſept hommes, ce qu’on a connu par expérience.</s> <s xml:id="echoid-s18248" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18249" xml:space="preserve">1100. </s> <s xml:id="echoid-s18250" xml:space="preserve">Que l’effort d’un homme gui<unsure/> tire du haut en bas, <lb/>peut être d’environ 50 ou 60 liv@es, & </s> <s xml:id="echoid-s18251" xml:space="preserve">même davantage; </s> <s xml:id="echoid-s18252" xml:space="preserve">mais <lb/>il ne peut agir ſi long-tems. </s> <s xml:id="echoid-s18253" xml:space="preserve">il peut même être égal à ſon poids; <lb/></s> <s xml:id="echoid-s18254" xml:space="preserve">mais alors il ne pourroit agir.</s> <s xml:id="echoid-s18255" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18256" xml:space="preserve">1101. </s> <s xml:id="echoid-s18257" xml:space="preserve">Que l’effort d’un homme qui marche dans une roue <lb/>eſt égal à ſon poids.</s> <s xml:id="echoid-s18258" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18259" xml:space="preserve">1102. </s> <s xml:id="echoid-s18260" xml:space="preserve">Que dans la pratique il faut avoir égard aux frotte-<lb/>mens, qui ſont d’autant plus grands, que la machine eſt plus <lb/>compoſée; </s> <s xml:id="echoid-s18261" xml:space="preserve">aux groſſeurs des cordes qui alongent les rayons <lb/>des cylindres de leur demi-diametre; </s> <s xml:id="echoid-s18262" xml:space="preserve">à la groſſeur des cordes <lb/>qui augmentent auſſi le rayon du cylindre; </s> <s xml:id="echoid-s18263" xml:space="preserve">d<unsure/> la roideur des <lb/>mêmes cordes; </s> <s xml:id="echoid-s18264" xml:space="preserve">que ſi l’on fait faire pluſieurs tours à la corde, <lb/>le rayon du cylindre augmente à chaque tour du diametre de <lb/>la corde.</s> <s xml:id="echoid-s18265" xml:space="preserve"/> </p> <pb o="576" file="0666" n="692" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1522" type="section" level="1" n="1108"> <head xml:id="echoid-head1313" xml:space="preserve"><emph style="sc">Analogie des poulies mouflées</emph>.</head> <p style="it"> <s xml:id="echoid-s18266" xml:space="preserve">1103. </s> <s xml:id="echoid-s18267" xml:space="preserve">Si une puiſſance ſoutient un poids à l’aide de pluſieurs <lb/>poulies, je dis que la puiſſance eſt au poids, comme l’unité eſt au <lb/>double du nombre des poulies d’en bas, qui ſont toujours les poulies <lb/>mobiles.</s> <s xml:id="echoid-s18268" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1523" type="section" level="1" n="1109"> <head xml:id="echoid-head1314" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s18269" xml:space="preserve">Soit H G la moufle d’en haut, qui eſt celle qui doit être <lb/> <anchor type="note" xlink:label="note-0666-01a" xlink:href="note-0666-01"/> fixe, & </s> <s xml:id="echoid-s18270" xml:space="preserve">D K la moufle d’en bas, qui eſt celle qui doit hauſſer <lb/>& </s> <s xml:id="echoid-s18271" xml:space="preserve">enlever le poids, ſoit auſſi un des bouts de la corde atta-<lb/>ché à l’extrêmité G de la moufle d’en haut; </s> <s xml:id="echoid-s18272" xml:space="preserve">après avoir paſſé <lb/>au deſſus des poulies A, B, C, & </s> <s xml:id="echoid-s18273" xml:space="preserve">au deſſous des poulies D, E, F, <lb/>enſorte que ſon autre extrêmité ſoit le bout où eſt appliquée <lb/>la puiſſance. </s> <s xml:id="echoid-s18274" xml:space="preserve">Cela poſé, lorſque la puiſſance tire le bout de <lb/>la corde pour faire monter le poids, toutes les parties de la <lb/>corde tirent d’une égale force à la puiſſance Q; </s> <s xml:id="echoid-s18275" xml:space="preserve">c’eſt pourquoi <lb/>chacune des poulies d’en bas, D, E, F, porte une égale partie <lb/>du poids P, c’eſt-à-dire que chacune porte un tiers, parce qu’il <lb/>y a trois poulies. </s> <s xml:id="echoid-s18276" xml:space="preserve">Or ſi l’on conſidere que la poulie F eſt un <lb/>levier du ſecond genre, dont le point d’appui eſt en M, la <lb/>puiſſance en N, ou dans la direction N O ou R Q, qui eſt la <lb/>même choſe, & </s> <s xml:id="echoid-s18277" xml:space="preserve">le poids dans le milieu F, l’on aura que la <lb/>puiſſance eſt au poids comme M N eſt à M F, c’eſt-à-dire que <lb/>la puiſſance ſera la moitié du poids; </s> <s xml:id="echoid-s18278" xml:space="preserve">mais comme la poulie <lb/>ne ſoutient ici<unsure/> que le tiers du poids, la puiſſance n’en ſou-<lb/>tiendra que la ſixieme partie, puiſque P : </s> <s xml:id="echoid-s18279" xml:space="preserve">R :</s> <s xml:id="echoid-s18280" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s18281" xml:space="preserve">6, qui fait <lb/>voir que la raiſon de la puiſſance au poids, eſt comme l’unité <lb/>au double du nombre des poulies D, E, F.</s> <s xml:id="echoid-s18282" xml:space="preserve"/> </p> <div xml:id="echoid-div1523" type="float" level="2" n="1"> <note position="left" xlink:label="note-0666-01" xlink:href="note-0666-01a" xml:space="preserve">Figure 394.</note> </div> <p> <s xml:id="echoid-s18283" xml:space="preserve">1104. </s> <s xml:id="echoid-s18284" xml:space="preserve">Mais ſi l’on avoit une moufle E F immobile, dont les <lb/> <anchor type="note" xlink:label="note-0666-02a" xlink:href="note-0666-02"/> poulies A, B, C, D fuſſent miſes les unes à côté des autres, <lb/>& </s> <s xml:id="echoid-s18285" xml:space="preserve">une moufle mobile L M, dont les poulies G, H, I, K fuſſent <lb/>dans la même diſpoſition que celles d’en haut, & </s> <s xml:id="echoid-s18286" xml:space="preserve">qu’une corde <lb/>dont une des extrêmités ſeroit attachée en I, paſſât au deſſous <lb/>des poulies d’en bas, & </s> <s xml:id="echoid-s18287" xml:space="preserve">au deſſus des poulies d’en haut, tant <lb/>que l’autre bout étant parvenu à la derniere poulie A fût retenu <lb/>par une puiſſance Q, l’on verroit encore que cette puiſſance <lb/>eſt au poids, comme l’unité eſt au double du nombre des <lb/>poulies d’en bas: </s> <s xml:id="echoid-s18288" xml:space="preserve">ainſi comme il y a quatre poulies G, H, I, K, <lb/>l’on aura Q : </s> <s xml:id="echoid-s18289" xml:space="preserve">P :</s> <s xml:id="echoid-s18290" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s18291" xml:space="preserve">8.</s> <s xml:id="echoid-s18292" xml:space="preserve"/> </p> <div xml:id="echoid-div1524" type="float" level="2" n="2"> <note position="left" xlink:label="note-0666-02" xlink:href="note-0666-02a" xml:space="preserve">Figure 395.</note> </div> <pb o="577" file="0667" n="693" rhead="DE MATHÉMATIQUE. Liv. XV."/> </div> <div xml:id="echoid-div1526" type="section" level="1" n="1110"> <head xml:id="echoid-head1315" style="it" xml:space="preserve">Autre démonſtration par le mouvement.</head> <p> <s xml:id="echoid-s18293" xml:space="preserve">1105. </s> <s xml:id="echoid-s18294" xml:space="preserve">Pour prouver que Q : </s> <s xml:id="echoid-s18295" xml:space="preserve">P :</s> <s xml:id="echoid-s18296" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s18297" xml:space="preserve">6 dans la figure 394, ou <lb/> <anchor type="note" xlink:label="note-0667-01a" xlink:href="note-0667-01"/> que Q : </s> <s xml:id="echoid-s18298" xml:space="preserve">P :</s> <s xml:id="echoid-s18299" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s18300" xml:space="preserve">8 dans la figure 395, remarquez que pour que <lb/>le Poids P ſoit élevé par la puiſſance Q d’un pied, il faut que <lb/>chacune des cordes qui ſoutient le poids ſe raccourciſſe auſſi d’un <lb/>pied, & </s> <s xml:id="echoid-s18301" xml:space="preserve">qu’ainſi la puiſſance doit deſcendre d’autant de pieds <lb/>qu’il y a de brins de cordes qui ſe raccourciſſent: </s> <s xml:id="echoid-s18302" xml:space="preserve">mais il y a <lb/>deux fois autant de brins de corde qu’il y a de poulies mobiles; <lb/></s> <s xml:id="echoid-s18303" xml:space="preserve">ce qui fait voir que la vîteſſe du poids eſt à celle de la puiſſance, <lb/>comme l’unité eſt au double du nombre des poulies d’en bas, <lb/>& </s> <s xml:id="echoid-s18304" xml:space="preserve">par conſéquent la puiſſance & </s> <s xml:id="echoid-s18305" xml:space="preserve">le poids ſont en équilibre, <lb/>puiſqu’ils ſont en raiſon réciproque de leur vîteſſe.</s> <s xml:id="echoid-s18306" xml:space="preserve"/> </p> <div xml:id="echoid-div1526" type="float" level="2" n="1"> <note position="right" xlink:label="note-0667-01" xlink:href="note-0667-01a" xml:space="preserve">Figure 394.</note> </div> </div> <div xml:id="echoid-div1528" type="section" level="1" n="1111"> <head xml:id="echoid-head1316" style="it" xml:space="preserve">Application de l’effet des poulies aux manœuvres de l’Artillerie.</head> <p> <s xml:id="echoid-s18307" xml:space="preserve">1106. </s> <s xml:id="echoid-s18308" xml:space="preserve">De toutes les machines compoſées, il n’y en a pas <lb/> <anchor type="note" xlink:label="note-0667-02a" xlink:href="note-0667-02"/> qui ſoient plus en uſage pour les manœuvres de l’Artillerie, & </s> <s xml:id="echoid-s18309" xml:space="preserve"><lb/>pour celles qu’on pratique en général, pour élever facilement <lb/>des corps fort peſans, que la chevre. </s> <s xml:id="echoid-s18310" xml:space="preserve">Or pour faire voir ici l’effet <lb/>de la chevre A B C D, qui eſt équipée de deux poulies mouflées <lb/>immobiles E, F, & </s> <s xml:id="echoid-s18311" xml:space="preserve">de deux autres mobiles G, H, à la moufle <lb/>deſquelles eſt attachée une piece de canon peſant 4800 livres. <lb/></s> <s xml:id="echoid-s18312" xml:space="preserve">Conſidérez que ſi la puiſſance eſt appliquée à la corde E Q, <lb/>l’on aura Q : </s> <s xml:id="echoid-s18313" xml:space="preserve">P :</s> <s xml:id="echoid-s18314" xml:space="preserve">: 1 : </s> <s xml:id="echoid-s18315" xml:space="preserve">4; </s> <s xml:id="echoid-s18316" xml:space="preserve">ainſi la puiſſance ne ſoutiendra que la <lb/>quatrieme partie du poids, c’eſt-à-dire 1200 livres; </s> <s xml:id="echoid-s18317" xml:space="preserve">mais la <lb/>puiſſance, quand on ſe ſert d’une chevre, n’eſt jamais appli-<lb/>quée aux cordes, elle eſt toujours appliquée à un levier M O, <lb/>qui paſſe dans le treuil K L de la chevre. </s> <s xml:id="echoid-s18318" xml:space="preserve">Or ſi le treuil a un <lb/>pied de diametre, & </s> <s xml:id="echoid-s18319" xml:space="preserve">que le levier depuis l’axe du treuil juſqu à <lb/>l’endroit où eſt appliquée la puiſſance, ſoit de 5 pieds, ou au-<lb/>trement de 60 pouces, le rayon du treuil & </s> <s xml:id="echoid-s18320" xml:space="preserve">la longueur du <lb/>levier feront un levier du ſecond genre, dont le point d’appui <lb/>ſera au centre du treuil, la puiſſance à l’extrêmité O, & </s> <s xml:id="echoid-s18321" xml:space="preserve">le <lb/>poids à l’endroit I de la circonférence du treuil. </s> <s xml:id="echoid-s18322" xml:space="preserve">Si la puiſſance <lb/>ſoutient le poids en équilibre, il y aura même raiſon de cette <lb/>puiſſance au poids, que du rayon du treuil à la longueur du <lb/>levier, c’eſt-à-dire comme 6 pouces eſt à 60 pouces, ou bien <lb/>comme 1 eſt à 10; </s> <s xml:id="echoid-s18323" xml:space="preserve">mais à l’endroit I, le poids de 4800 eſt ré-<lb/>duit à 1200: </s> <s xml:id="echoid-s18324" xml:space="preserve">la puiſſance qui ſeroit appliquée au levier ne ſou- <pb o="578" file="0668" n="694" rhead="NOUVEAU COURS"/> tiendra donc que la dixieme partie de 1200 livres, qui eſt 120 <lb/>livres: </s> <s xml:id="echoid-s18325" xml:space="preserve">ainſi l’on voit qu’une puiſſance de 120 livres ſoutient, <lb/>par le moyen de la chevre, un poids de 4800 livres, & </s> <s xml:id="echoid-s18326" xml:space="preserve">qu’elle <lb/>en pourroit élever un beaucoup plus peſant avec une force <lb/>même moindre que celle qu’on lui a ſuppoſée ici, en augmen-<lb/>tant le nombre des poulies, & </s> <s xml:id="echoid-s18327" xml:space="preserve">la longueur du levier.</s> <s xml:id="echoid-s18328" xml:space="preserve"/> </p> <div xml:id="echoid-div1528" type="float" level="2" n="1"> <note position="right" xlink:label="note-0667-02" xlink:href="note-0667-02a" xml:space="preserve">Figure 396.</note> </div> </div> <div xml:id="echoid-div1530" type="section" level="1" n="1112"> <head xml:id="echoid-head1317" xml:space="preserve"><emph style="sc">Définitions</emph>.</head> <p> <s xml:id="echoid-s18329" xml:space="preserve">1107. </s> <s xml:id="echoid-s18330" xml:space="preserve">La machine ſimple à laquelle une puiſſance eſt immé-<lb/>diatement appliquée, & </s> <s xml:id="echoid-s18331" xml:space="preserve">qui donne le mouvement à toutes les <lb/>autres, eſt nommée la premiere; </s> <s xml:id="echoid-s18332" xml:space="preserve">celle ſur laquelle la premiere <lb/>agit, la ſeconde; </s> <s xml:id="echoid-s18333" xml:space="preserve">& </s> <s xml:id="echoid-s18334" xml:space="preserve">celle ſur laquelle la ſeconde agit, la troi-<lb/>ſieme, ainſi de ſuite.</s> <s xml:id="echoid-s18335" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1531" type="section" level="1" n="1113"> <head xml:id="echoid-head1318" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s18336" xml:space="preserve">1108. </s> <s xml:id="echoid-s18337" xml:space="preserve">Il ſuit delà que l’effet de la premiere machine eſt à la <lb/>cauſe qui fait agir la ſeconde, comme l’effet de la ſeconde eſt <lb/>à la cauſe qui fait agir la troiſieme, ainſi de ſuite juſqu’à la <lb/>derniere.</s> <s xml:id="echoid-s18338" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1532" type="section" level="1" n="1114"> <head xml:id="echoid-head1319" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s18339" xml:space="preserve">1109. </s> <s xml:id="echoid-s18340" xml:space="preserve">Il ſuit encore delà que dans les machines compoſées <lb/>le rapport de la puiſſance au poids eſt compoſé de l’effet de la <lb/>premiere machine à la cauſe qui fait agir la ſeconde, & </s> <s xml:id="echoid-s18341" xml:space="preserve">de <lb/>l’effet de la ſeconde à la cauſe qui fait agir la troiſieme, ainſi <lb/>de ſuite, juſqu’à la cauſe qui fait mouvoir le poids: </s> <s xml:id="echoid-s18342" xml:space="preserve">par exem-<lb/>ple, dans la chevre dont nous venons de parler, le rapport de <lb/>la puiſſance Q au poids P eſt compoſée de celui de 1 à 10, & </s> <s xml:id="echoid-s18343" xml:space="preserve"><lb/>de celui de 1 à 4: </s> <s xml:id="echoid-s18344" xml:space="preserve">ainſi multipliant les antécédens de ces rap-<lb/>ports les uns par les autres, & </s> <s xml:id="echoid-s18345" xml:space="preserve">les conſéquens auſſi les uns par <lb/>les autres, on aura {1/40} pour le rapport compoſé, qui eſt celui de <lb/>la puiſſance au poids, & </s> <s xml:id="echoid-s18346" xml:space="preserve">qui fait voir que la puiſſance eſt la <lb/>quarantieme partie du poids: </s> <s xml:id="echoid-s18347" xml:space="preserve">car {1/40} eſt la même choſe que <lb/>{120/4800}, qui eſt le rapport que nous avons trouvé.</s> <s xml:id="echoid-s18348" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1533" type="section" level="1" n="1115"> <head xml:id="echoid-head1320" style="it" xml:space="preserve"><emph style="sc">Des</emph> <emph style="sc">Roues dentées</emph>.</head> <head xml:id="echoid-head1321" xml:space="preserve"><emph style="sc">Definitions</emph>.</head> <p> <s xml:id="echoid-s18349" xml:space="preserve">1110. </s> <s xml:id="echoid-s18350" xml:space="preserve">Lorſqu’une machine eſt compoſée de pluſieurs roues, <lb/>il faut que toutes les roues ſoient dentées, excepté la premiere, <lb/>& </s> <s xml:id="echoid-s18351" xml:space="preserve">que toutes les lanternes ou pignons le ſoient auſſi, excepté <pb o="579" file="0669" n="695" rhead="DE MATHÉMATIQUE. Liv. XV."/> le dernier, qui doit être rond, afin que la corde qui enleve le <lb/>poids, s’entortille à l’entour; </s> <s xml:id="echoid-s18352" xml:space="preserve">il faut auſſi qu’il y ait à chaque <lb/>extrêmité des pivots des axes, pour pouvoir être ajuſtés dans <lb/>une eſpece d’affût, de maniere que la lanterne ou le pignon de <lb/>l’axe de la premiere roue engraine dans les dents de la ſeconde, <lb/>la lanterne ou le pignon de la deuxieme dans les dents de la troi-<lb/>ſieme, ainſi de ſuite juſqu’à la derniere. </s> <s xml:id="echoid-s18353" xml:space="preserve">Cette machine, ainſi <lb/>compoſée, eſt nommée machine des roues dentées, qui eſt pro-<lb/>pre pour élever de très-gros fardeaux, & </s> <s xml:id="echoid-s18354" xml:space="preserve">d’autant plus gros <lb/>& </s> <s xml:id="echoid-s18355" xml:space="preserve">plus peſans que les roues ſeroient en plus grand nombre.</s> <s xml:id="echoid-s18356" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1534" type="section" level="1" n="1116"> <head xml:id="echoid-head1322" xml:space="preserve"><emph style="sc">Analogie des</emph> <emph style="sc">Roues dentées</emph>.</head> <p style="it"> <s xml:id="echoid-s18357" xml:space="preserve">1111. </s> <s xml:id="echoid-s18358" xml:space="preserve">Ayant nommé f le rayon de la premiere roue, à la cir-<lb/> <anchor type="note" xlink:label="note-0669-01a" xlink:href="note-0669-01"/> conférence de laquelle eſt appliquée la puiſſance, a le rayon de ſon <lb/> <anchor type="note" xlink:label="note-0669-02a" xlink:href="note-0669-02"/> pignon, g le rayon de la ſeconde roue, b celui de ſon pignon, <lb/>h le rayon de la troiſieme roue, c celui de ſon pignon, k le <lb/>rayon de la quatrieme roue, d celui de ſon pignon, l le rayon <lb/>de la cinquieme roue, & </s> <s xml:id="echoid-s18359" xml:space="preserve">e celui de ſon pignon (qui n’eſt point <lb/>denté), il faut faire voir que le rapport de la puiſſance Q au poids <lb/>P, eſt comme le produit des rayons des aiſſieux au produit des <lb/>rayons des roues.</s> <s xml:id="echoid-s18360" xml:space="preserve"/> </p> <div xml:id="echoid-div1534" type="float" level="2" n="1"> <note position="right" xlink:label="note-0669-01" xlink:href="note-0669-01a" xml:space="preserve">Pl. XXXI.</note> <note position="right" xlink:label="note-0669-02" xlink:href="note-0669-02a" xml:space="preserve">Figure 398.</note> </div> <p> <s xml:id="echoid-s18361" xml:space="preserve">Si la premiere roue étoit ſeule, & </s> <s xml:id="echoid-s18362" xml:space="preserve">que la puiſſance enlevât <lb/>par ſon moyen le poids P, qui devroit pour cela être ſuſpendu <lb/>au pignon ou au treuil de cette roue, l’on auroit Q : </s> <s xml:id="echoid-s18363" xml:space="preserve">P :</s> <s xml:id="echoid-s18364" xml:space="preserve">: a : </s> <s xml:id="echoid-s18365" xml:space="preserve">f; <lb/></s> <s xml:id="echoid-s18366" xml:space="preserve">mais l’effet de la premiere roue, au lieu d’être employé à lever <lb/>un poids, eſt employé à faire tourner la ſeconde par le moyen <lb/>des dents de ſon pignon qui engraine dans les dents de la ſe-<lb/>conde roue; </s> <s xml:id="echoid-s18367" xml:space="preserve">d’où l’on voit que l’effet de la premiere roue eſt <lb/>la cauſe qui fait agir la ſeconde, parce que l’effet des dents <lb/>de ſon aiſſieu contre les dents de la ſeconde roue, eſt égal au <lb/>poids qu’elle pourroit enlever. </s> <s xml:id="echoid-s18368" xml:space="preserve">Il en eſt ainſi des autres. </s> <s xml:id="echoid-s18369" xml:space="preserve">Or ſi <lb/>l’on nomme l’effet de la premiere roue r, l’effet de la ſeconde <lb/>ſ, celui de la troiſieme t, & </s> <s xml:id="echoid-s18370" xml:space="preserve">celui de la quatrieme u, l’on aura <lb/>pour le premier rapport q : </s> <s xml:id="echoid-s18371" xml:space="preserve">r :</s> <s xml:id="echoid-s18372" xml:space="preserve">: a : </s> <s xml:id="echoid-s18373" xml:space="preserve">f, pour le ſecond r : </s> <s xml:id="echoid-s18374" xml:space="preserve">ſ :</s> <s xml:id="echoid-s18375" xml:space="preserve">: b : </s> <s xml:id="echoid-s18376" xml:space="preserve">g, <lb/>pour le troiſieme ſ : </s> <s xml:id="echoid-s18377" xml:space="preserve">t :</s> <s xml:id="echoid-s18378" xml:space="preserve">: c : </s> <s xml:id="echoid-s18379" xml:space="preserve">h, pour le quatrieme t : </s> <s xml:id="echoid-s18380" xml:space="preserve">u :</s> <s xml:id="echoid-s18381" xml:space="preserve">: d : </s> <s xml:id="echoid-s18382" xml:space="preserve">k, <lb/>enfin pour le cinquieme & </s> <s xml:id="echoid-s18383" xml:space="preserve">dernier rapport, u : </s> <s xml:id="echoid-s18384" xml:space="preserve">p :</s> <s xml:id="echoid-s18385" xml:space="preserve">: e : </s> <s xml:id="echoid-s18386" xml:space="preserve">l.</s> <s xml:id="echoid-s18387" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18388" xml:space="preserve">Préſentement ſi l’on multiplie ces cinq proportions terme <lb/>par terme, c’eſt-à-dire les antécédens par les antécédens, & </s> <s xml:id="echoid-s18389" xml:space="preserve"><lb/>les conſéquens par les conſéquens, l’on aura cette proportion, <pb o="580" file="0670" n="696" rhead="NOUVEAU COURS"/> q r ſ t u : </s> <s xml:id="echoid-s18390" xml:space="preserve">r ſ t u p :</s> <s xml:id="echoid-s18391" xml:space="preserve">: a b c d e : </s> <s xml:id="echoid-s18392" xml:space="preserve">f g h k l. </s> <s xml:id="echoid-s18393" xml:space="preserve">Et fi l’on diviſe <lb/> <anchor type="note" xlink:label="note-0670-01a" xlink:href="note-0670-01"/> les deux premiers termes par r ſ t u, l’on aura <lb/>Q : </s> <s xml:id="echoid-s18394" xml:space="preserve">P :</s> <s xml:id="echoid-s18395" xml:space="preserve">: a b c d e : </s> <s xml:id="echoid-s18396" xml:space="preserve">f g h k l; </s> <s xml:id="echoid-s18397" xml:space="preserve">d’où l’on tire cette ana-<lb/>logie pour toutes les machines compoſées des <lb/>roues dentées: </s> <s xml:id="echoid-s18398" xml:space="preserve">Si une puiſſance ſoutient un poids <lb/>à l’aide de pluſieurs roues, la puiſſance eſt au poids <lb/>comme le produit des rayons des pignons eſt au produit des rayons <lb/>des roues.</s> <s xml:id="echoid-s18399" xml:space="preserve"/> </p> <div xml:id="echoid-div1535" type="float" level="2" n="2"> <note position="right" xlink:label="note-0670-01" xlink:href="note-0670-01a" xml:space="preserve">q: r :: a: f <lb/>r: ſ :: b: g <lb/>ſ: t :: c: h <lb/>t: u :: d: k <lb/>u: p :: e: l <lb/></note> </div> </div> <div xml:id="echoid-div1537" type="section" level="1" n="1117"> <head xml:id="echoid-head1323" xml:space="preserve"><emph style="sc">Application</emph>.</head> <p> <s xml:id="echoid-s18400" xml:space="preserve">1112. </s> <s xml:id="echoid-s18401" xml:space="preserve">Pour faire voir la force immenſe qu’on peut donner <lb/>à une puiſſance, par le moyen des roues dentées, ſuppoſons <lb/>que la force de la puiſſance ſoit de 50 livres, & </s> <s xml:id="echoid-s18402" xml:space="preserve">que cette puiſ-<lb/>fance ſoit appliquée à la premiere roue d’une machine com-<lb/>poſée de cinq roues de chacune 12 pouces de rayon, parce <lb/>que nous les ſuppoſons égales, auſſi-bien que les pignons qui <lb/>ſeront, par exemple, d’un pouce de rayon. </s> <s xml:id="echoid-s18403" xml:space="preserve">Cela poſé, le rap-<lb/>port du rayon de chaque pignon au rayon de chaque roue, ſera <lb/>comme 1 eſt à 12: </s> <s xml:id="echoid-s18404" xml:space="preserve">ainſi le produit de tous les pignons ſera 1, <lb/>& </s> <s xml:id="echoid-s18405" xml:space="preserve">celui de tous les rayons des roues ſera 248832. </s> <s xml:id="echoid-s18406" xml:space="preserve">Or ſi l’on <lb/>veut ſçavoir quelle eſt la peſanteur du poids qu’une puiſſance <lb/>de 50 livres, que je ſuppoſe être la force d’un homme, pour-<lb/>roit enlever avec cette machine: </s> <s xml:id="echoid-s18407" xml:space="preserve">je conſidere que ſelon ce <lb/>qui vient d’être démontré, la puiſſance eſt au poids comme <lb/>le produit des rayons des pignons eſt au produit des rayons <lb/>des roues, & </s> <s xml:id="echoid-s18408" xml:space="preserve">que par conſéquent le produit des rayons des <lb/>pignons eſt au produit des rayons des roues, comme la puiſ-<lb/>ſance eſt au poids; </s> <s xml:id="echoid-s18409" xml:space="preserve">ainſi pour trouver le poids, je dis: </s> <s xml:id="echoid-s18410" xml:space="preserve">Si <lb/>1, produit des rayons des pignons, donne 248832 pour le pro-<lb/>duit des rayons des roues, que donnera la puiſſance de 50 livres <lb/>pour le poids qu’elle ſeroit capable d’enlever? </s> <s xml:id="echoid-s18411" xml:space="preserve">l’on trouvera <lb/>12441600, qui eſt le nombre de livres qu’un homme peut en-<lb/>lever avec une force moyenne, aidée d’une machine com-<lb/>poſée de cinq roues dentées.</s> <s xml:id="echoid-s18412" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1538" type="section" level="1" n="1118"> <head xml:id="echoid-head1324" style="it" xml:space="preserve"><emph style="sc">Du</emph> <emph style="sc">Cric</emph>.</head> <p> <s xml:id="echoid-s18413" xml:space="preserve">1113. </s> <s xml:id="echoid-s18414" xml:space="preserve">Le cric, dont l’uſage eſt ſi fréquent dans l’Artillerie, <lb/>fait encore voir combien les roues dentées augmentent la puiſ-<lb/>ſance, & </s> <s xml:id="echoid-s18415" xml:space="preserve">pour en calculer la force, conſidérez la figure 397 <lb/>qui repréſente à peu près les parties dont l’intérieur eſt com- <pb o="581" file="0671" n="697" rhead="DE MATHÉMATIQUE. Liv. XV."/> pofé, qui eſt mis en mouvement par la manivelle A B C, où <lb/> <anchor type="note" xlink:label="note-0671-01a" xlink:href="note-0671-01"/> eſt appliquée la puiſſance; </s> <s xml:id="echoid-s18416" xml:space="preserve">cette manivelle en tournant, fait <lb/> <anchor type="note" xlink:label="note-0671-02a" xlink:href="note-0671-02"/> tourner le petit pignon D, lequel étant engrainé dans la roue <lb/>E, la fait auſſi tourner. </s> <s xml:id="echoid-s18417" xml:space="preserve">Au centre de cette roue, eſt un autre <lb/>pignon F, qui fait monter le cric G H, pour enlever le fardeau. <lb/></s> <s xml:id="echoid-s18418" xml:space="preserve">Préſentement ſi l’on ſuppoſe que la manivelle A B (que nous <lb/>conſidérons ici comme le rayon d’une roue), ſoit de 15 pouces, <lb/>que le pignon D ait un pouce de rayon, la roue E, 12 pouces <lb/>auſſi de rayon, & </s> <s xml:id="echoid-s18419" xml:space="preserve">le pignon F deux, l’on connoîtra le rapport <lb/>de la puiſſance au poids qu’on peut enlever, en conſidérant le <lb/>rapport du produit des rayons des pignons au produit des rayons <lb/>des roues: </s> <s xml:id="echoid-s18420" xml:space="preserve">ainſi le produit des pignons ſera 2, & </s> <s xml:id="echoid-s18421" xml:space="preserve">le produit des <lb/>roues 180; </s> <s xml:id="echoid-s18422" xml:space="preserve">ce qui fait voir que la puiſſance ſera au poids, <lb/>comme 2 eſt à 180, ou bien comme l’unité eſt à 90. </s> <s xml:id="echoid-s18423" xml:space="preserve">Or ſi l’on <lb/>ſuppoſe que la puiſſance eſt 50, multipliant 50 par 90, l’on <lb/>aura 4500, qui eſt à peu près le poids qu’un homme peut en-<lb/>lever par le moyen d’un cric tel que celui que nous venons <lb/>d’expliquer : </s> <s xml:id="echoid-s18424" xml:space="preserve">& </s> <s xml:id="echoid-s18425" xml:space="preserve">ſi au lieu de deux roues il y en avoit davan-<lb/>tage, l’on voit qu’on peut avec le cric lever des fardeaux d’une <lb/>peſanteur immenſe.</s> <s xml:id="echoid-s18426" xml:space="preserve"/> </p> <div xml:id="echoid-div1538" type="float" level="2" n="1"> <note position="right" xlink:label="note-0671-01" xlink:href="note-0671-01a" xml:space="preserve">Pl. XXX.</note> <note position="right" xlink:label="note-0671-02" xlink:href="note-0671-02a" xml:space="preserve">Figure 397.</note> </div> </div> <div xml:id="echoid-div1540" type="section" level="1" n="1119"> <head xml:id="echoid-head1325" style="it" xml:space="preserve">De la Vis ſans fin, appliquée aux roues dentées.</head> <p> <s xml:id="echoid-s18427" xml:space="preserve">1114. </s> <s xml:id="echoid-s18428" xml:space="preserve">La vis ſans fin eſt encore une machine propre à aug-<lb/> <anchor type="note" xlink:label="note-0671-03a" xlink:href="note-0671-03"/> menter extrêmement la force de la puiſſance, ſurtout quand <lb/> <anchor type="note" xlink:label="note-0671-04a" xlink:href="note-0671-04"/> elle met en mouvement pluſieurs roues dentées. </s> <s xml:id="echoid-s18429" xml:space="preserve">Suppoſant <lb/>donc qu’on a une machine compoſée d’une vis ſans fin, & </s> <s xml:id="echoid-s18430" xml:space="preserve">de <lb/>trois roues, comme celle de la figure 399, pour ſçavoir le rap-<lb/>port de la puiſſance Q au poids P, je conſidere que la puiſſance <lb/>étant appliquée à une manivelle ou à un levier A B, fera tour-<lb/>ner la vis, qui mettra en mouvement la premiere roue, à <lb/>cauſe que les pas de la vis ſont engrainés avec les dents de la <lb/>premiere roue, dont les pignons qui s’engrainent avec les <lb/>dents de la ſeconde roue, la fera tourner auſſi, & </s> <s xml:id="echoid-s18431" xml:space="preserve">le pignon de <lb/>celle-ci la troiſieme roue, au pignon de laquelle eſt attaché le <lb/>poids.</s> <s xml:id="echoid-s18432" xml:space="preserve"/> </p> <div xml:id="echoid-div1540" type="float" level="2" n="1"> <note position="right" xlink:label="note-0671-03" xlink:href="note-0671-03a" xml:space="preserve">Pl. XXXI.</note> <note position="right" xlink:label="note-0671-04" xlink:href="note-0671-04a" xml:space="preserve">Figure 399.</note> </div> <p> <s xml:id="echoid-s18433" xml:space="preserve">Préſentement ſi l’on nomme n la circonférence du cercle, <lb/>qui auroit pour rayon le levier A C, a l’intervalle d’un pas de <lb/>la vis, f l’effet des filets contre les dents de la roue, g le <lb/>rayon de la premiere roue, b celui de ſon pignon, h le rayon <pb o="582" file="0672" n="698" rhead="NOUVEAU COURS"/> de la ſeconde roue, & </s> <s xml:id="echoid-s18434" xml:space="preserve">d le rayon de ſon pignon, k le rayon <lb/>de la troiſieme roue, & </s> <s xml:id="echoid-s18435" xml:space="preserve">c celui de ſon pignon, t l’effet de la <lb/>premiere roue, & </s> <s xml:id="echoid-s18436" xml:space="preserve">u l’effet de la ſeconde. </s> <s xml:id="echoid-s18437" xml:space="preserve">Voici comme il faut <lb/>raiſonner: </s> <s xml:id="echoid-s18438" xml:space="preserve">L’on ſçait que la puiſſance qui eſt appliquée au le-<lb/>vier d’une vis, eſt à l’effet de la vis, comme l’intervalle d’un <lb/>des pas de la vis eſt à la circonférence du cercle que décrit la <lb/>puiſſance, l’on aura donc cette proportion, q : </s> <s xml:id="echoid-s18439" xml:space="preserve">ſ :</s> <s xml:id="echoid-s18440" xml:space="preserve">: a : </s> <s xml:id="echoid-s18441" xml:space="preserve">n, & </s> <s xml:id="echoid-s18442" xml:space="preserve"><lb/> <anchor type="note" xlink:label="note-0672-01a" xlink:href="note-0672-01"/> l’effet de la premiere roue don-<lb/>nera encore ſ : </s> <s xml:id="echoid-s18443" xml:space="preserve">t :</s> <s xml:id="echoid-s18444" xml:space="preserve">: b : </s> <s xml:id="echoid-s18445" xml:space="preserve">g, l’effet <lb/>de la ſeconde t : </s> <s xml:id="echoid-s18446" xml:space="preserve">u :</s> <s xml:id="echoid-s18447" xml:space="preserve">: d : </s> <s xml:id="echoid-s18448" xml:space="preserve">h, & </s> <s xml:id="echoid-s18449" xml:space="preserve">celui <lb/>de la troiſieme, u : </s> <s xml:id="echoid-s18450" xml:space="preserve">p :</s> <s xml:id="echoid-s18451" xml:space="preserve">: c : </s> <s xml:id="echoid-s18452" xml:space="preserve">k. </s> <s xml:id="echoid-s18453" xml:space="preserve">Or <lb/>multipliant ces quatre propor-<lb/>tions, termes par termes, l’on <lb/>aura q ſ t u : </s> <s xml:id="echoid-s18454" xml:space="preserve">ſ t u p :</s> <s xml:id="echoid-s18455" xml:space="preserve">: a b c d : </s> <s xml:id="echoid-s18456" xml:space="preserve">h g n k, & </s> <s xml:id="echoid-s18457" xml:space="preserve">diviſant les deux premiers <lb/>termes par ſ t u, l’on aura Q : </s> <s xml:id="echoid-s18458" xml:space="preserve">P :</s> <s xml:id="echoid-s18459" xml:space="preserve">: a c d b : </s> <s xml:id="echoid-s18460" xml:space="preserve">h g n k; </s> <s xml:id="echoid-s18461" xml:space="preserve">d’où l’on tire <lb/>cette analogie : </s> <s xml:id="echoid-s18462" xml:space="preserve">Si une puiſſance enleve un poids à l’aide d’une <lb/>vis & </s> <s xml:id="echoid-s18463" xml:space="preserve">de pluſieurs roues dentées, la puiſſance ſera au poids comme <lb/>le produit de l’intervalle d’un des pas de la vis, par les rayons <lb/>des pignons des roues, eſt au produit de la circonférence qui décrit <lb/>la puiſſance par les rayons des roues.</s> <s xml:id="echoid-s18464" xml:space="preserve"/> </p> <div xml:id="echoid-div1541" type="float" level="2" n="2"> <note position="right" xlink:label="note-0672-01" xlink:href="note-0672-01a" xml:space="preserve">q : ſ :: a : n <lb/>ſ : t :: b : g <lb/>t : u :: d : h <lb/>u : p :: c : k <lb/>qſtu : ſtup :: acdb : hgnk <lb/></note> </div> </div> <div xml:id="echoid-div1543" type="section" level="1" n="1120"> <head xml:id="echoid-head1326" xml:space="preserve"><emph style="sc">Application</emph>.</head> <p> <s xml:id="echoid-s18465" xml:space="preserve">1115. </s> <s xml:id="echoid-s18466" xml:space="preserve">Pour ſçavoir quel eſt le poids qu’une puiſſance de <lb/>50 livres peut enlever par le moyen de la machine précédente, <lb/>nous ſuppoſerons que le rayon C A du cercle que décrit la puiſ-<lb/>ſance eſt de 10 {1/2} pouces; </s> <s xml:id="echoid-s18467" xml:space="preserve">par conſéquent la circonférence ſera <lb/>de 66 pouces: </s> <s xml:id="echoid-s18468" xml:space="preserve">de plus qu’un des pas de la vis eſt de 2 pouces, <lb/>que le rayon de la premiere roue eſt de 24 pouces, & </s> <s xml:id="echoid-s18469" xml:space="preserve">celui <lb/>de ſon pignon de 3, que le rayon de la ſeconde roue eſt de 20 <lb/>pouces, & </s> <s xml:id="echoid-s18470" xml:space="preserve">celui de ſon pignon de 2, enfin, que le rayon de <lb/>la troiſieme roue eſt de 18 pouces, & </s> <s xml:id="echoid-s18471" xml:space="preserve">celui de ſon pignon d’un <lb/>pouce & </s> <s xml:id="echoid-s18472" xml:space="preserve">demi. </s> <s xml:id="echoid-s18473" xml:space="preserve">Cela poſé, ſi l’on multiplie les rayons des pi-<lb/>gnons les uns par les autres, l’on aura 9 au produit, qui étant <lb/>multiplié par un des pas de la vis, qui eſt de 2 pouces, l’on aura <lb/>18 pour un des termes de la proportion; </s> <s xml:id="echoid-s18474" xml:space="preserve">& </s> <s xml:id="echoid-s18475" xml:space="preserve">multipliant auſſi <lb/>les rayons des roues les unes par les autres, & </s> <s xml:id="echoid-s18476" xml:space="preserve">enſuite le pro-<lb/>duit par la circonférence que décrira la puiſſance, l’on aura <lb/>570240 pour un autre terme de la proportion: </s> <s xml:id="echoid-s18477" xml:space="preserve">ainſi la puiſ-<lb/>ſance ſera au poids, comme 18 eſt à 570240, ou comme 1 <lb/>eſt à 31680. </s> <s xml:id="echoid-s18478" xml:space="preserve">L’on pourra donc dire comme 1 eſt à 31680, qui <pb o="583" file="0673" n="699" rhead="DE MATHÉMATIQUE. Liv. XV."/> eſt le rapport du produit des rayons des pignons par un pas de <lb/>la vis au produit des rayons des roues par la circonférence dé-<lb/>crite par la puiſſance : </s> <s xml:id="echoid-s18479" xml:space="preserve">ainſi 50, qui eſt la force de la puiſſance, <lb/>eſt au poids que cette puiſſance eſt capable d’enlever, l’on trou-<lb/>vera que ce poids eſt de 1584000 livres.</s> <s xml:id="echoid-s18480" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1544" type="section" level="1" n="1121"> <head xml:id="echoid-head1327" xml:space="preserve"><emph style="sc">Remarque.</emph></head> <p> <s xml:id="echoid-s18481" xml:space="preserve">Si un auſſi grand poids que celui que nous venons de trou-<lb/>ver peut être enlevé par la force moyenne d’un ſeul homme <lb/>avec une vis à trois roues ſeulement, ce n’eſt pas ſans raiſon <lb/>qu’ Archimede diſoit, pour faire voir juſqu’à quel point on pou-<lb/>voit augmenter la force de la puiſſance, que ſi on lui donnoit <lb/>un point fixe pour appuyer ſa machine, il ne ſeroit pas em-<lb/>barraſſé d’enlever toute la terre, malgré l’immenſité de ſon <lb/>poids. </s> <s xml:id="echoid-s18482" xml:space="preserve">Da punctum, & </s> <s xml:id="echoid-s18483" xml:space="preserve">movebo.</s> <s xml:id="echoid-s18484" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1545" type="section" level="1" n="1122"> <head xml:id="echoid-head1328" style="it" xml:space="preserve">Machine compoſée d’une roue, & d’un plan incliné.</head> <p> <s xml:id="echoid-s18485" xml:space="preserve">1116. </s> <s xml:id="echoid-s18486" xml:space="preserve">Ayant un plan incliné G H, dont la hauteur eſt G I, <lb/> <anchor type="note" xlink:label="note-0673-01a" xlink:href="note-0673-01"/> & </s> <s xml:id="echoid-s18487" xml:space="preserve">un poids P ſur ce plan, où il eſt retenu par une corde B P <lb/> <anchor type="note" xlink:label="note-0673-02a" xlink:href="note-0673-02"/> parallele à G H, dont un des bouts eſt attaché au treuil d’un <lb/>tourniquet, qui eſt mis en mouvement par une puiſſance Q, <lb/>appliquée à un des leviers A Q, A D ou A C, qui ſervent à <lb/>faire tourner le treuil pour attirer le poids P vers le ſommet G; <lb/></s> <s xml:id="echoid-s18488" xml:space="preserve">on demande quel eſt le rapport de la puiſſance au poids?</s> <s xml:id="echoid-s18489" xml:space="preserve"/> </p> <div xml:id="echoid-div1545" type="float" level="2" n="1"> <note position="right" xlink:label="note-0673-01" xlink:href="note-0673-01a" xml:space="preserve">Pl. XXXI.</note> <note position="right" xlink:label="note-0673-02" xlink:href="note-0673-02a" xml:space="preserve">Figure 401.</note> </div> <p> <s xml:id="echoid-s18490" xml:space="preserve">Ayant nommé G H, a; </s> <s xml:id="echoid-s18491" xml:space="preserve">G I, b; </s> <s xml:id="echoid-s18492" xml:space="preserve">le rayon du treuil, c; </s> <s xml:id="echoid-s18493" xml:space="preserve">& </s> <s xml:id="echoid-s18494" xml:space="preserve"><lb/>la longueur d’un des leviers A C, A Q ou A D, d; </s> <s xml:id="echoid-s18495" xml:space="preserve">& </s> <s xml:id="echoid-s18496" xml:space="preserve">l’effort <lb/>que fait la puiſſance qui ſeroit appliquée dans la direction P B <lb/>pour ſoutenir le poids P, ſ; </s> <s xml:id="echoid-s18497" xml:space="preserve">l’on aura par la propriété du plan <lb/>incliné, ſ: </s> <s xml:id="echoid-s18498" xml:space="preserve">p :</s> <s xml:id="echoid-s18499" xml:space="preserve">: b : </s> <s xml:id="echoid-s18500" xml:space="preserve">a, & </s> <s xml:id="echoid-s18501" xml:space="preserve">par la propriété de la roue, la puiſſance <lb/>Q ne ſoutenant que l’effort ſ de l’autre puiſſance q, l’on aura <lb/>Q : </s> <s xml:id="echoid-s18502" xml:space="preserve">ſ :</s> <s xml:id="echoid-s18503" xml:space="preserve">: c : </s> <s xml:id="echoid-s18504" xml:space="preserve">d. </s> <s xml:id="echoid-s18505" xml:space="preserve">Or multipliant les termes de ces deux propor-<lb/>tions, l’on aura Q x ſ : </s> <s xml:id="echoid-s18506" xml:space="preserve">pſ :</s> <s xml:id="echoid-s18507" xml:space="preserve">: bc : </s> <s xml:id="echoid-s18508" xml:space="preserve">ad, & </s> <s xml:id="echoid-s18509" xml:space="preserve">diviſant les deux pre-<lb/>miers termes de cette proportion par ſ, il viendra Q : </s> <s xml:id="echoid-s18510" xml:space="preserve">P :</s> <s xml:id="echoid-s18511" xml:space="preserve">: bc : </s> <s xml:id="echoid-s18512" xml:space="preserve">ad, <lb/>qui fait voir que la puiſſance eſt au poids, comme le produit <lb/>du rayon de l’aiſſieu par la hauteur du plan incliné eſt au pro-<lb/>duit du rayon de la roue ou de la longueur du levier par la lon-<lb/>gueur du plan incliné.</s> <s xml:id="echoid-s18513" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1547" type="section" level="1" n="1123"> <head xml:id="echoid-head1329" xml:space="preserve"><emph style="sc">Application</emph>.</head> <p> <s xml:id="echoid-s18514" xml:space="preserve">1117. </s> <s xml:id="echoid-s18515" xml:space="preserve">Il arrive fort ſouvent que pour tirer des corps peſans <pb o="584" file="0674" n="700" rhead="NOUVEAU COURS"/> d’une cave, comme ſont, par exemple, les muids de vin ou <lb/>d’eau-de-vie, l’on ſe ſert d’un tourniquet pour en faciliter le <lb/>tranſport: </s> <s xml:id="echoid-s18516" xml:space="preserve">ainſi ſi les marches de la cave ſont dans un même <lb/>plan, l’eſcalier pourroit être regardé comme un plan incliné. <lb/></s> <s xml:id="echoid-s18517" xml:space="preserve">Si donc la hauteur de ce plan incliné eſt à ſa longueur, comme <lb/>4 eſt à 6, & </s> <s xml:id="echoid-s18518" xml:space="preserve">qu’ayant un tourniquet à l’entrée de l’eſcalier, le <lb/>treuil ſoit, par exemple, de 6 pouces de rayon, & </s> <s xml:id="echoid-s18519" xml:space="preserve">le levier de <lb/>36 pouces de longueur, depuis le centre du treuil juſqu’à l’en-<lb/>droit où eſt appliquée la puiſſance; </s> <s xml:id="echoid-s18520" xml:space="preserve">& </s> <s xml:id="echoid-s18521" xml:space="preserve">qu’on veuille ſçavoir la <lb/>peſanteur du corps qu’une puiſſance de 50 livres peut ſoutenir <lb/>ou attirer à ſoi par le moyen du tourniquet, il faut commencer <lb/>par multiplier le rayon du treuil, qui eſt de 6 pouces, par la <lb/>hauteur du plan incliné, qui eſt de 4 pieds, ou qu’on peut <lb/>prendre pour telle, le produit ſera 24 pouces; </s> <s xml:id="echoid-s18522" xml:space="preserve">& </s> <s xml:id="echoid-s18523" xml:space="preserve">multipliant <lb/>la longueur du levier de 36 pouces par 6 pieds, le produit ſera <lb/>2592 : </s> <s xml:id="echoid-s18524" xml:space="preserve">ainſi la puiſſance ſera au poids qu’elle eſt capable de <lb/>ſoutenir, comme 24 eſt à 2592, ou comme 1 eſt à 108 : </s> <s xml:id="echoid-s18525" xml:space="preserve">ainſi <lb/>pour trouver le poids, il n’y a qu’à dire: </s> <s xml:id="echoid-s18526" xml:space="preserve">Si 1 donne 108, <lb/>combien donneront 50? </s> <s xml:id="echoid-s18527" xml:space="preserve">l’on trouvera 5400 livres pour le poids <lb/>que l’on cherche.</s> <s xml:id="echoid-s18528" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1548" type="section" level="1" n="1124"> <head xml:id="echoid-head1330" style="it" xml:space="preserve"><emph style="sc">De la</emph> <emph style="sc">Sonnette</emph>.</head> <p> <s xml:id="echoid-s18529" xml:space="preserve">1118. </s> <s xml:id="echoid-s18530" xml:space="preserve">Preſque toutes les machines compoſées augmentent <lb/> <anchor type="note" xlink:label="note-0674-01a" xlink:href="note-0674-01"/> la force de la puiſſance, excepté celle que l’on nomme com-<lb/>munément ſonnette, dont on ſe ſert pour enfoncer des pilots, <lb/>par le moyen d’un gros billot de bois, tel que A, que l’on <lb/>nomme mouton. </s> <s xml:id="echoid-s18531" xml:space="preserve">Ce mouton eſt attaché par deux mains de <lb/>fer ou crampons B, ſuſpendus à deux cordes qui paſſent ſur <lb/>des poulies G, & </s> <s xml:id="echoid-s18532" xml:space="preserve">à ces cordes ſont pluſieurs bouts O N, qui <lb/>ſont tirés tout à la fois par des hommes qui levent le mouton <lb/>vers G, & </s> <s xml:id="echoid-s18533" xml:space="preserve">le laiſſent tomber tout d’un coup ſur la tête du pilot <lb/>C F que l’on veut enfoncer, Mais comme il arrive qu’à meſure <lb/>que le pilot s’enfonce, le mouton tombe de plus haut, & </s> <s xml:id="echoid-s18534" xml:space="preserve">ac-<lb/>quiert par ſon accélération un plus grand degré de force; </s> <s xml:id="echoid-s18535" xml:space="preserve">voici <lb/>comme l’on pourra meſurer la force du mouton à chaque coup, <lb/>& </s> <s xml:id="echoid-s18536" xml:space="preserve">même ſçavoir combien il faudra de coups pour enfoncer un <lb/>pilot à refus de mouton.</s> <s xml:id="echoid-s18537" xml:space="preserve"/> </p> <div xml:id="echoid-div1548" type="float" level="2" n="1"> <note position="left" xlink:label="note-0674-01" xlink:href="note-0674-01a" xml:space="preserve">Figure 400.</note> </div> <p> <s xml:id="echoid-s18538" xml:space="preserve">Nous ſuppoſerons que le terrein dans lequel on veut enfon-<lb/>cer le pilot eſt homogene dans toutes ſes parties, & </s> <s xml:id="echoid-s18539" xml:space="preserve">qu’auſſitôt <lb/>que le bout de pilot eſt entré juſques un peu au deſſus de la <pb o="585" file="0675" n="701" rhead="DE MATHÉMATIQUE. Liv. XV."/> partie que l’on a taillée en pointe, le terrein dans lequel on <lb/>l’enfonce réſiſte toujours également, parce que l’on compte <lb/>pour rien le frottement de la terre qui entoure la ſurface du <lb/>pilot, qui ſe trouve de plus en plus couverte, à meſure que le <lb/>pilot enfonce.</s> <s xml:id="echoid-s18540" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18541" xml:space="preserve">Cela poſé, je ſuppoſe que le mouton A, après avoir été en-<lb/>levé juſqu’au plus haut de la ſonnette, ſe trouve éloigné de <lb/>3 pieds de la tête C du pilot, & </s> <s xml:id="echoid-s18542" xml:space="preserve">que l’ayant laiſſé tomber, le <lb/>pilot ſe ſoit enfoncé de 13 pouces, de ſorte que la tête ſera <lb/>deſcendue de C en D. </s> <s xml:id="echoid-s18543" xml:space="preserve">Or pour ſçavoir de combien le pilot <lb/>ſera enfoncé au ſecond coup, qui ſera plus fort que le pre-<lb/>mier, parce que le mouton, au lieu de tomber de H en C, tom-<lb/>bera de H en D; </s> <s xml:id="echoid-s18544" xml:space="preserve">je conſidere que la force ou la quantité de <lb/>mouvement d’un corps eſt le produit de ſa maſſe par ſa vîteſſe, <lb/>& </s> <s xml:id="echoid-s18545" xml:space="preserve">qu’ainſi la force du corps A, en tombant de H en C, ſera à <lb/>la force du même corps en tombant de H en D, comme le <lb/>produit de la peſanteur du corps A par la vîteſſe acquiſe de H <lb/>en C, eſt au produit de la peſanteur du même corps par la <lb/>vîteſſe acquiſe de H en D; </s> <s xml:id="echoid-s18546" xml:space="preserve">mais nous ſçavons que les vîteſſes <lb/>d’un corps qui tombe de différentes hauteurs, peuvent s’ex-<lb/>primer par les racines quarrées des eſpaces parcourus: </s> <s xml:id="echoid-s18547" xml:space="preserve">ainſi <lb/>nommant a la maſſe du corps A; </s> <s xml:id="echoid-s18548" xml:space="preserve">b l’eſpace parcouru H C; </s> <s xml:id="echoid-s18549" xml:space="preserve">& </s> <s xml:id="echoid-s18550" xml:space="preserve"><lb/>d l’eſpace parcouru H D, l’on aura √ b pour la vîteſſe acquiſe <lb/>de H en C, & </s> <s xml:id="echoid-s18551" xml:space="preserve">√ d pour la vîteſſe acquiſe de H en D: </s> <s xml:id="echoid-s18552" xml:space="preserve">ainfi la <lb/>force du corps A tombant en C & </s> <s xml:id="echoid-s18553" xml:space="preserve">en D, ſera comme a √ b eſt <lb/>à a √ d, ou bien comme √ b eſt à √ d. </s> <s xml:id="echoid-s18554" xml:space="preserve">Mais les effets étant <lb/>comme les cauſes, il s’enſuit que l’enfoncement du pilot au <lb/>premier coup ſera à l’enfoncement du pilot au ſecond coup, <lb/>comme la racine quarrée de l’eſpace parcouru par le mouton <lb/>au premier coup ſera à la racine quarrée de l’eſpace parcourn <lb/>au ſecond coup. </s> <s xml:id="echoid-s18555" xml:space="preserve">Or dans la ſuppoſition, l’eſpace parcouru dans <lb/>le premier coup eſt de 3 pieds, ou autrement de 36 pouces, <lb/>dont la racine ſera 6; </s> <s xml:id="echoid-s18556" xml:space="preserve">& </s> <s xml:id="echoid-s18557" xml:space="preserve">comme le pilot aura été enfoncé de <lb/>13 pouces, l’eſpace H D ſera de 49 pouces, dont la racine eſt <lb/>7. </s> <s xml:id="echoid-s18558" xml:space="preserve">Je dis donc, pour trouver l’enfoncement du pilot au ſecond <lb/>coup, ſi la vîteſſe 6 a donné 13 pour l’enfoncement du pilot <lb/>au premier coup, combien donnera la vîteſſe 7 pour l’enfon-<lb/>cement du pilot au ſecond coup? </s> <s xml:id="echoid-s18559" xml:space="preserve">l’on trouvera 15 & </s> <s xml:id="echoid-s18560" xml:space="preserve">{1/6}, qui <lb/>fait voir que le pilot ſera enfoncé au ſecond coup de 15 pouces <lb/>2 lignes, qui eſt la diſtance D E.</s> <s xml:id="echoid-s18561" xml:space="preserve"/> </p> <pb o="586" file="0676" n="702" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s18562" xml:space="preserve">Pour ſçavoir combien il ſera enfoncé au troiſieme coup, je <lb/> <anchor type="note" xlink:label="note-0676-01a" xlink:href="note-0676-01"/> conſidere que l’eſpace H E eſt de 64 & </s> <s xml:id="echoid-s18563" xml:space="preserve">{1/6}, dont la racine quarrée <lb/> <anchor type="note" xlink:label="note-0676-02a" xlink:href="note-0676-02"/> eſt 8, & </s> <s xml:id="echoid-s18564" xml:space="preserve">je dis encore: </s> <s xml:id="echoid-s18565" xml:space="preserve">Si la vîteſſe 6 donne 13 pour l’enfon-<lb/>cement du pilot au premier coup, combien donnera 8? </s> <s xml:id="echoid-s18566" xml:space="preserve">l’on <lb/>trouvera 17 pouces & </s> <s xml:id="echoid-s18567" xml:space="preserve">4 lignes, & </s> <s xml:id="echoid-s18568" xml:space="preserve">agiſſant toujours de même, <lb/>l’on trouvera que l’enfoncement du quatrieme coup ſera de <lb/>19 pouces 6 lignes, que celui du cinquieme ſera de 21 pouces <lb/>8 lignes, & </s> <s xml:id="echoid-s18569" xml:space="preserve">que celui du ſixieme ſera de 23 pouces 10 lignes: <lb/></s> <s xml:id="echoid-s18570" xml:space="preserve">ainſi l’on aura pour l’enfoncement du pilot à chaque coup les <lb/>ſix termes ſuivans, 13 pouces, 15 pouces, plus 2 lign. </s> <s xml:id="echoid-s18571" xml:space="preserve">17 + 4, <lb/>19 + 6, 21 + 8, 23 + 10, qui ſont tous en progreſſion arith-<lb/>métique, puiſqu’ils ſe ſurpaſſent de 2 pouces & </s> <s xml:id="echoid-s18572" xml:space="preserve">de 2 lignes; </s> <s xml:id="echoid-s18573" xml:space="preserve"><lb/>ils ſe ſurpaſſeroient même encore de quelques parties de point, <lb/>auxquelles je n’ai pas eu égard.</s> <s xml:id="echoid-s18574" xml:space="preserve"/> </p> <div xml:id="echoid-div1549" type="float" level="2" n="2"> <note position="left" xlink:label="note-0676-01" xlink:href="note-0676-01a" xml:space="preserve">Pl. XXXI.</note> <note position="left" xlink:label="note-0676-02" xlink:href="note-0676-02a" xml:space="preserve">Figure 400.</note> </div> <p> <s xml:id="echoid-s18575" xml:space="preserve">L’on ſera peut-être ſurpris de voir que les racines quarrées <lb/>des eſpaces parcourus par le mouton, ſont en progreſſion arith-<lb/>métique, de même que les quantités qui expriment l’enfonce-<lb/>ment du pilot à chaque coup; </s> <s xml:id="echoid-s18576" xml:space="preserve">mais cela ne peut arriver autre-<lb/>ment, comme on le va voir.</s> <s xml:id="echoid-s18577" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18578" xml:space="preserve">Si l’on a une progreſſion arithmétique {.</s> <s xml:id="echoid-s18579" xml:space="preserve">/.</s> <s xml:id="echoid-s18580" xml:space="preserve">} a. </s> <s xml:id="echoid-s18581" xml:space="preserve">b. </s> <s xml:id="echoid-s18582" xml:space="preserve">c. </s> <s xml:id="echoid-s18583" xml:space="preserve">d. </s> <s xml:id="echoid-s18584" xml:space="preserve">e. </s> <s xml:id="echoid-s18585" xml:space="preserve">f, dont <lb/>chaque terme marque le tems pendant lequel un corps tom-<lb/>bant de différentes hauteurs, a mis à parcourir différens eſ-<lb/>paces, & </s> <s xml:id="echoid-s18586" xml:space="preserve">que ces eſpaces ſoient, par exemple, g. </s> <s xml:id="echoid-s18587" xml:space="preserve">h. </s> <s xml:id="echoid-s18588" xml:space="preserve">i. </s> <s xml:id="echoid-s18589" xml:space="preserve">k. </s> <s xml:id="echoid-s18590" xml:space="preserve">l. </s> <s xml:id="echoid-s18591" xml:space="preserve">m, <lb/>ces eſpaces ſeront dans la raiſon des quarrés des tems, c’eſt-à-<lb/>dire comme aa, bb, cc, dd, ee, ff: </s> <s xml:id="echoid-s18592" xml:space="preserve">or ſi l’on extrait la ra-<lb/>cine quarrée de l’une & </s> <s xml:id="echoid-s18593" xml:space="preserve">l’autre de ces progreſſions, l’on aura <lb/>{.</s> <s xml:id="echoid-s18594" xml:space="preserve">/.</s> <s xml:id="echoid-s18595" xml:space="preserve">} a. </s> <s xml:id="echoid-s18596" xml:space="preserve">b. </s> <s xml:id="echoid-s18597" xml:space="preserve">c. </s> <s xml:id="echoid-s18598" xml:space="preserve">d. </s> <s xml:id="echoid-s18599" xml:space="preserve">e. </s> <s xml:id="echoid-s18600" xml:space="preserve">f pour les tems, & </s> <s xml:id="echoid-s18601" xml:space="preserve">√ g, √ h, √ i, √ k, √ l, √ m, <lb/>pour celles des eſpaces parcourus. </s> <s xml:id="echoid-s18602" xml:space="preserve">Or ſi les tems a, b, c, d, e, f <lb/>ſont en progreſſion arithmétique, les racines des eſpaces le <lb/>ſeront auſſi: </s> <s xml:id="echoid-s18603" xml:space="preserve">ainſi il n’eſt plus étonnant que ſi les tems que le <lb/>mouton met à tomber, ſont en progreſſion arithmétique, les <lb/>racines quarrées des eſpaces, qui ſont les vîteſſes acquiſes, le <lb/>ſoient auſſi: </s> <s xml:id="echoid-s18604" xml:space="preserve">mais les vîteſſes acquiſes pcuvent être regardées <lb/>comme les cauſes de l’enfoncement du pilot à chaque coup; <lb/></s> <s xml:id="echoid-s18605" xml:space="preserve">& </s> <s xml:id="echoid-s18606" xml:space="preserve">comme les effets ſont proportionnels à leurs cauſes, les <lb/>cauſes étant en proportion arithmétique, les effets le ſeront <lb/>auſſi; </s> <s xml:id="echoid-s18607" xml:space="preserve">ce qui fait que le pilot doit s’enfoncer plus au ſecond <lb/>coup qu’au premier, & </s> <s xml:id="echoid-s18608" xml:space="preserve">plus au troiſieme qu’au ſecond, dans <lb/>la raiſon d’une progreſſion arithmétique.</s> <s xml:id="echoid-s18609" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18610" xml:space="preserve">L’on peut tirer de ce qu’on vient de dire, la maniere de <pb o="587" file="0677" n="703" rhead="DE MATHÉMATIQUE. Liv. XV."/> connoître combien il faut donner de coups ſur un pilot pour <lb/>le faire entrer à refus de mouton: </s> <s xml:id="echoid-s18611" xml:space="preserve">car on n’a qu’à conſidérer <lb/>au premier coup de combien le pilot ſera enfoncé, & </s> <s xml:id="echoid-s18612" xml:space="preserve">regarder <lb/>cette quantité comme le premier terme d’une progreſſion arith-<lb/>métique. </s> <s xml:id="echoid-s18613" xml:space="preserve">Suppoſant donc que le mouton tombant de 3 pieds <lb/>de hauteur, le pilot ſe ſoit enfoncé de 12 pouces, & </s> <s xml:id="echoid-s18614" xml:space="preserve">ſuppo-<lb/>ſant auſſi qu’au ſecond coup le pilot ſe ſoit enfoncé de 14 <lb/>pouces, je regarde ce nombre comme le ſecond terme de la <lb/>progreſſion, & </s> <s xml:id="echoid-s18615" xml:space="preserve">comme la différence de ce terme-ci à l’autre <lb/>eſt 2, je vois que le troiſieme terme ſera 16, que le quatrieme <lb/>ſera 18, le cinquieme 20. </s> <s xml:id="echoid-s18616" xml:space="preserve">Or ſi j’ai un pilot, par exemple, de <lb/>12 pieds de longueur, cette longueur exprimera la valeur de <lb/>tous les termes de la progreſſion pris enſemble: </s> <s xml:id="echoid-s18617" xml:space="preserve">ainſi j’ajoute <lb/>les termes que je viens de trouver pour voir s’ils valent 144 <lb/>pouces; </s> <s xml:id="echoid-s18618" xml:space="preserve">& </s> <s xml:id="echoid-s18619" xml:space="preserve">comme il s’en faut beaucoup, je cherche encore <lb/>quelque terme, comme, par exemple 22, 24 & </s> <s xml:id="echoid-s18620" xml:space="preserve">26, qui font <lb/>avec les autres 152 pouces, qui ſurpaſſent la longueur du pilot <lb/>de 8 pouces; </s> <s xml:id="echoid-s18621" xml:space="preserve">& </s> <s xml:id="echoid-s18622" xml:space="preserve">comme ce ſont 8 termes qui m’ont donné <lb/>cette quantité, je vois qu’il faut 8 coups pour enfoncer le pilot <lb/>juſqu’au refus de mouton. </s> <s xml:id="echoid-s18623" xml:space="preserve">Au reſte l’on trouvera ce ſujet traité <lb/>encore plus exactement dans le premier volume de la ſeconde <lb/>Partie de l’Architecture Hydraulique, page 188.</s> <s xml:id="echoid-s18624" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1551" type="section" level="1" n="1125"> <head xml:id="echoid-head1331" style="it" xml:space="preserve">Application de la méchanique à la conſtruction des magaſins à <lb/>poudre.</head> <p> <s xml:id="echoid-s18625" xml:space="preserve">1119. </s> <s xml:id="echoid-s18626" xml:space="preserve">De tous les édifices militaires, il n’y en a point qui <lb/>ſoient d’une plus grande conſéquence que les magaſins à pou-<lb/>dre, & </s> <s xml:id="echoid-s18627" xml:space="preserve">qui demandent plus de précaution pour les bien conſ-<lb/>truire: </s> <s xml:id="echoid-s18628" xml:space="preserve">car comme on les fait toujours voûtés, il faut ſçavoir <lb/>quelles ſortes de voûtes conviennent le mieux, de la voûte <lb/>en plein ceintre, de celle qui eſt ſurbaiſſée, ou de celle qui eſt en <lb/>tiers point, pour être capable de réſiſter le plus à l’effort de la <lb/>bombe, quand elle tombe deſſus: </s> <s xml:id="echoid-s18629" xml:space="preserve">après cela, il faut ſçavoir <lb/>proportionner l’épaiſſeur des pieds droits, qui ſoutiennent les <lb/>voûtes au poids, à la pouſſée, & </s> <s xml:id="echoid-s18630" xml:space="preserve">à la grandeur des mêmes <lb/>voûtes.</s> <s xml:id="echoid-s18631" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18632" xml:space="preserve">L’opinion de la plûpart des Ingénieurs eſt partagée ſur la <lb/>maniere de voûter les magaſins à poudre; </s> <s xml:id="echoid-s18633" xml:space="preserve">les uns prétendent <lb/>que la voûte en plein ceintre eſt la meilleure de toutes, & </s> <s xml:id="echoid-s18634" xml:space="preserve">les <lb/>autres au contraire veulent que la voûte en tiers point ſoit <pb o="588" file="0678" n="704" rhead="NOUVEAU COURS"/> préférable à celle-ci. </s> <s xml:id="echoid-s18635" xml:space="preserve">Ce qu’il y a de certain, c’eſt que la voûte <lb/>en tiers point a moins de pouſſée que celle en plein ceintre, & </s> <s xml:id="echoid-s18636" xml:space="preserve"><lb/>celle en plein ceintre que celle qui eſt ſurbaiſſée; </s> <s xml:id="echoid-s18637" xml:space="preserve">ce que l’on <lb/>peut démontrer même géométriquement, & </s> <s xml:id="echoid-s18638" xml:space="preserve">ſans entrer dans <lb/>une grande théorie; </s> <s xml:id="echoid-s18639" xml:space="preserve">je vais faire voir comment la voûte en <lb/>plein ceintre a plus de pouſſée que celle en tiers point.</s> <s xml:id="echoid-s18640" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18641" xml:space="preserve">Conſidérez la figure 402, qui eſt le profil d’un magaſin à <lb/> <anchor type="note" xlink:label="note-0678-01a" xlink:href="note-0678-01"/> poudre, dont la voûte eſt en plein ceintre, & </s> <s xml:id="echoid-s18642" xml:space="preserve">la figure 403, <lb/>qui eſt un autre profil, dont la voûte eſt en tiers point: </s> <s xml:id="echoid-s18643" xml:space="preserve">dans <lb/>ces deux figures l’on a diviſé en deux également les arcs E D <lb/>& </s> <s xml:id="echoid-s18644" xml:space="preserve">L D par des lignes tirées de leurs centres. </s> <s xml:id="echoid-s18645" xml:space="preserve">Or ſi l’on con-<lb/>ſidere la partie ſupérieure B A G C de la voûte comme un coin <lb/>qui agit contre les pieds droits, & </s> <s xml:id="echoid-s18646" xml:space="preserve">contre les autres parties de <lb/>la voûte pour les écarter, l’on verra que plus l’angle A B C ſera <lb/>aigu, & </s> <s xml:id="echoid-s18647" xml:space="preserve">plus le coinaura de force par la loi des méchaniques, <lb/>ou bien ſi l’on regarde la ligne A B comme un plan incliné, <lb/>l’on verra encore que plus il ſera incliné, & </s> <s xml:id="echoid-s18648" xml:space="preserve">plus le corps G A B <lb/>qui tend à gliſſer deſſus aura de force pour deſcendre, puiſque <lb/>la peſanteur relative ſera moindre qu’elle ne le ſeroit, ſi le plan <lb/>incliné approchoit plus d’être horizontal. </s> <s xml:id="echoid-s18649" xml:space="preserve">Or dans la figure <lb/>403, ſi l’on regarde encore T Q R S comme un coin, l’on <lb/>verra que l’angle Q S R étant obtus, le coin fera moins d’ef-<lb/>fort pour écarter les parties R Z & </s> <s xml:id="echoid-s18650" xml:space="preserve">Q N, que dans la figure 402 <lb/>où l’angle du coin eſt droit; </s> <s xml:id="echoid-s18651" xml:space="preserve">& </s> <s xml:id="echoid-s18652" xml:space="preserve">ſi l’on conſidere de plus la ligne <lb/>Q P comme un plan incliné, l’on verra que l’étant beaucoup <lb/>moins que le plan A B, la partie T Q S n’aura pas tant de force <lb/>pour deſcendre que la partie G A B; </s> <s xml:id="echoid-s18653" xml:space="preserve">par conſéquent tous les <lb/>vouſſoirs qui compoſent la voûte en tiers point étant regardés <lb/>comme des coins, ou comme des corps qui tendent à gliſſer <lb/>ſucceſſivement ſur des plans inclinés, feront moins d’effort <lb/>que ceux de la voûte en plein ceintre; </s> <s xml:id="echoid-s18654" xml:space="preserve">d’où il s’enſuit que la <lb/>voûte en plein ceintre a plus de pouſſée que la voûte en tiers <lb/>point: </s> <s xml:id="echoid-s18655" xml:space="preserve">& </s> <s xml:id="echoid-s18656" xml:space="preserve">par un ſemblable raiſonnement, on fera voir que la <lb/>voûte ſurbaiſſée a plus de pouſſée que celle en plein ceintre.</s> <s xml:id="echoid-s18657" xml:space="preserve"/> </p> <div xml:id="echoid-div1551" type="float" level="2" n="1"> <note position="left" xlink:label="note-0678-01" xlink:href="note-0678-01a" xml:space="preserve">Figure 402 <lb/>& 403.</note> </div> <p> <s xml:id="echoid-s18658" xml:space="preserve">Un autre défaut de la voûte en plein ceintre, eſt qu’elle <lb/>oblige à faire le toît fort plat; </s> <s xml:id="echoid-s18659" xml:space="preserve">ce qui la rend moins capable <lb/>de réſiſter à la chûte des bombes, qui ne font point tant <lb/>d’effort quand le plan ſur lequel elles tombent eſt plus incliné, <lb/>parce qu’alors elles ne font que rouler ſans faire de dommage <lb/>conſidérable; </s> <s xml:id="echoid-s18660" xml:space="preserve">& </s> <s xml:id="echoid-s18661" xml:space="preserve">ſi l’on veut éviter ce défaut, au lieu de faire <pb o="589" file="0679" n="705" rhead="DE MATHÉMATIQUE. Liv. XV."/> le toît comme dans la figure 402, le faire comme dans la figure <lb/> <anchor type="note" xlink:label="note-0679-01a" xlink:href="note-0679-01"/> 404, c’eſt-à-dire plus roide, l’on eſt obligé de charger la voûte <lb/>à l’endroit de la clef, d’une maſſe de maçonnerie qui oblige <lb/>abſolument de faire les pieds droits plus épais: </s> <s xml:id="echoid-s18662" xml:space="preserve">d’ailleurs un <lb/>avantage de la voûte en tiers point, c’eſt que ſi l’on veut faire <lb/>un magaſin qui ne ſoit pas fort élevé, l’on peut commencer la <lb/>naiſſance de la voûte à 4 ou 5 pieds au deſſus du rez-de-chauſſée, <lb/>& </s> <s xml:id="echoid-s18663" xml:space="preserve">le magaſin eſt aſſez élevé, au lieu que le faiſant en plein <lb/>ceintre, il faut que les pieds droits aient au moins 8 ou 9 pieds <lb/>de hauteur; </s> <s xml:id="echoid-s18664" xml:space="preserve">ce qui oblige à les faire plus épais: </s> <s xml:id="echoid-s18665" xml:space="preserve">car il n’y a <lb/>point de doute qu’à meſure qu’on les fait plus élevés, il ne <lb/>faille leur donner plus d’épaiſſeur. </s> <s xml:id="echoid-s18666" xml:space="preserve">Enfin je pourrois rapporter <lb/>encore pluſieurs raiſons en faveur des voûtes en tiers point; <lb/></s> <s xml:id="echoid-s18667" xml:space="preserve">mais je crois que ce que j’en ai dit ſuffit pour faire voir com-<lb/>bien elles ſont à préférer à celles qui ſont en plein ceintre.</s> <s xml:id="echoid-s18668" xml:space="preserve"/> </p> <div xml:id="echoid-div1552" type="float" level="2" n="2"> <note position="right" xlink:label="note-0679-01" xlink:href="note-0679-01a" xml:space="preserve">Figure 402 <lb/>& 404.</note> </div> <p> <s xml:id="echoid-s18669" xml:space="preserve">Quoiqu’il ſoit preſque impoſſible de déterminer l’épaiſſeur <lb/>que doit avoir la voûte d’un magaſin à poudre pour être à l’é-<lb/>preuve de la bombe, puiſque les bombes ne ſont pas toutes d’é-<lb/>gale peſanteur, & </s> <s xml:id="echoid-s18670" xml:space="preserve">ſont ſujettes à tomber de différentes hau-<lb/>teurs, cela n’empêche point qu’on ne ſe ſoit déterminé à leur <lb/>donner 3 pieds d’épaiſſeur à l’endroit des reins, & </s> <s xml:id="echoid-s18671" xml:space="preserve">je crois que <lb/>cette épaiſſeur ſera ſuffiſante, quand le toit ne ſera point trop <lb/>plat.</s> <s xml:id="echoid-s18672" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18673" xml:space="preserve">Comme il m’a paru qu’il convenoit de donner une regle <lb/>pour déterminer l’angle que doit avoir le faîte du toit d’un <lb/>magaſin, afin qu’il ne ſoit ni trop obtus, ni trop aigu, voici <lb/>comme je m’y prends.</s> <s xml:id="echoid-s18674" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18675" xml:space="preserve">Suppoſant qu’on veuille faire un magaſin à poudre, dont la <lb/> <anchor type="note" xlink:label="note-0679-02a" xlink:href="note-0679-02"/> voûte ſoit en plein ceintre, je commence par déterminer la <lb/>largeur du magaſin, qui ſera, par exemple, la ligne A C, qui <lb/>doit ſervir de diametre au demi-cercle de la voûte; </s> <s xml:id="echoid-s18676" xml:space="preserve">enſuite j’é-<lb/>leve ſur le centre B la perpendiculaire B G, & </s> <s xml:id="echoid-s18677" xml:space="preserve">je diviſe en deux <lb/>également chaque quart de cercle A N & </s> <s xml:id="echoid-s18678" xml:space="preserve">N C par les lignes <lb/>B M & </s> <s xml:id="echoid-s18679" xml:space="preserve">B E; </s> <s xml:id="echoid-s18680" xml:space="preserve">je donne 3 pieds à chacune des lignes D E & </s> <s xml:id="echoid-s18681" xml:space="preserve"><lb/>L M, qui déterminent l’épaiſſeur des reins de la voûte, & </s> <s xml:id="echoid-s18682" xml:space="preserve">puis <lb/>du centre B je décris un demi-cercle à volonté, qui ſe trouve <lb/>diviſé en deux également par la perpendiculaire au point G, <lb/>& </s> <s xml:id="echoid-s18683" xml:space="preserve">dont le diametre eſt la ligne F I, je tire auſſi les cordes F G <lb/>& </s> <s xml:id="echoid-s18684" xml:space="preserve">G I, & </s> <s xml:id="echoid-s18685" xml:space="preserve">par les points E & </s> <s xml:id="echoid-s18686" xml:space="preserve">M je fais paſſer les paralleles O H <lb/>& </s> <s xml:id="echoid-s18687" xml:space="preserve">H K aux cordes qui ſont dans le demi-cercle, & </s> <s xml:id="echoid-s18688" xml:space="preserve">ces paral- <pb o="590" file="0680" n="706" rhead="NOUVEAU COURS"/> leles me donnent le toit O H K, qui forme un angle droit <lb/>en H, parce que l’angle H eſt égal à l’angle G: </s> <s xml:id="echoid-s18689" xml:space="preserve">ainſi ſans tâ-<lb/>tonner par cette méthode, il ſe trouvera toujours que l’angle <lb/>du faîte d’un magaſin à poudre ſera droit, & </s> <s xml:id="echoid-s18690" xml:space="preserve">cet angle me <lb/>paroît convenir mieux qu’un autre, parce qu’il tient un milieu <lb/>entre l’angle aigu & </s> <s xml:id="echoid-s18691" xml:space="preserve">l’angle obtus, qui conviennent moins <lb/>que celui-ci: </s> <s xml:id="echoid-s18692" xml:space="preserve">car l’angle obtus, comme je l’ai déja dit, rend <lb/>le toit trop plat, & </s> <s xml:id="echoid-s18693" xml:space="preserve">l’angle aigu charge trop la clef de la voûte <lb/>par le grand vuide qu’il laiſſe au deſſus de la clef, qu’on eſt <lb/>obligé de remplir de maçonnerie.</s> <s xml:id="echoid-s18694" xml:space="preserve"/> </p> <div xml:id="echoid-div1553" type="float" level="2" n="3"> <note position="right" xlink:label="note-0679-02" xlink:href="note-0679-02a" xml:space="preserve">Figure 404.</note> </div> <p> <s xml:id="echoid-s18695" xml:space="preserve">Pour tracer la voûte en tiers point, je ſuppoſe que les points <lb/> <anchor type="note" xlink:label="note-0680-01a" xlink:href="note-0680-01"/> V & </s> <s xml:id="echoid-s18696" xml:space="preserve">X marquent l’endroit où doit commencer la naiſſance <lb/>de la voûte, je tire une ligne de V en X, laquelle je diviſe en <lb/>quatre parties égales; </s> <s xml:id="echoid-s18697" xml:space="preserve">& </s> <s xml:id="echoid-s18698" xml:space="preserve">du point P comme centre, & </s> <s xml:id="echoid-s18699" xml:space="preserve">de l’in-<lb/>tervalle P V, je décris l’arc V Y, & </s> <s xml:id="echoid-s18700" xml:space="preserve">du point O & </s> <s xml:id="echoid-s18701" xml:space="preserve">de l’inter-<lb/>valle O X, je décris l’arc X Y, lequel forme avec le précédent <lb/>l’intradoſſe V Y X de la voûte: </s> <s xml:id="echoid-s18702" xml:space="preserve">après cela je diviſe chacun de <lb/>ces arcs en deux également, & </s> <s xml:id="echoid-s18703" xml:space="preserve">je tire les lignes O R & </s> <s xml:id="echoid-s18704" xml:space="preserve">P Q, <lb/>& </s> <s xml:id="echoid-s18705" xml:space="preserve">je donne à chacune des lignes A Q & </s> <s xml:id="echoid-s18706" xml:space="preserve">B R 3 pieds & </s> <s xml:id="echoid-s18707" xml:space="preserve">3 pouces, <lb/>& </s> <s xml:id="echoid-s18708" xml:space="preserve">puis je diviſe la perpendiculaire L Y en trois parties égales, <lb/>& </s> <s xml:id="echoid-s18709" xml:space="preserve">de l’extrêmité M de la premiere parties, je décris un demi-<lb/>cercle K T D, & </s> <s xml:id="echoid-s18710" xml:space="preserve">je tire, comme dans la figure précédente, <lb/>les cordes K N, N D, & </s> <s xml:id="echoid-s18711" xml:space="preserve">par les points Q & </s> <s xml:id="echoid-s18712" xml:space="preserve">R je fais paſſer <lb/>deux paralleles aux cordes qui forment le toît de la voûte, <lb/>dont l’angle du faîte eſt encore droit.</s> <s xml:id="echoid-s18713" xml:space="preserve"/> </p> <div xml:id="echoid-div1554" type="float" level="2" n="4"> <note position="left" xlink:label="note-0680-01" xlink:href="note-0680-01a" xml:space="preserve">Figure 403.</note> </div> <p> <s xml:id="echoid-s18714" xml:space="preserve">Si j’ai donné aux lignes A Q & </s> <s xml:id="echoid-s18715" xml:space="preserve">B R 3 pieds 3 pouces, c’eſt <lb/>parce qu’elles ſont au deſſous des reins de la voûte; </s> <s xml:id="echoid-s18716" xml:space="preserve">mais en <lb/>ſuivant ce qui vient d’être dit, l’épaiſſeur des reins de la voûte <lb/>ſe trouve dans leur plus foible avoir 3 pieds d’épaiſſeur: </s> <s xml:id="echoid-s18717" xml:space="preserve">vous <lb/>pouvez remarquer la différence de la maçonnerie qui ſe trouve <lb/>au deſſus de la clef de la voûte en tiers point, & </s> <s xml:id="echoid-s18718" xml:space="preserve">celle qui eſt <lb/>au deſſus de la voûte en plein ceintre, c’eſt-à-dire que l’une eſt <lb/>beaucoup moins chargée que l’autre; </s> <s xml:id="echoid-s18719" xml:space="preserve">car il n’y a que 6 pieds <lb/>de hauteur de maçonnerie au deſſus de la voûte en tiers point, <lb/>au lieu que dans celle en plein ceintre il y en a plus de 10: <lb/></s> <s xml:id="echoid-s18720" xml:space="preserve">c’eſt auſſi la raiſon pour laquelle les pieds droits de cette voûte <lb/>ſont bien moins épais que ceux de celles en plein ceintre, parce <lb/>que d’ailleurs ils ſont auſſi moins élevés.</s> <s xml:id="echoid-s18721" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18722" xml:space="preserve">Mais pour régler l’épaiſſeur des pieds droits, tant pour les <lb/>voûtes en tiers point, que pour les voûtes en plein ceintre, <pb o="591" file="0681" n="707" rhead="DE MATHÉMATIQUE. Liv. XV."/> j’ai jugé à propos de rapporter ici une Table que j’ai calculée, <lb/>pour proportionner préciſément l’épaiſſeur des pieds droits <lb/>des voûtes des magaſins à poudre par rapport à la largeur dans <lb/>œuvre qu’on peut leur donner, & </s> <s xml:id="echoid-s18723" xml:space="preserve">à l’élévation des mêmes <lb/>pieds droits, e’eſt-à-dire que j’ai cherché un juſte équilibre <lb/>entre leur réſiſtance & </s> <s xml:id="echoid-s18724" xml:space="preserve">l’effort des voûtes; </s> <s xml:id="echoid-s18725" xml:space="preserve">après quoi j’ai <lb/>augmenté la pouſſée d’un quart de ce qu’elle eſt effectivement <lb/>pour rendre les pieds droits capables de cette réſiſtance au <lb/>deſſus de l’équilibre: </s> <s xml:id="echoid-s18726" xml:space="preserve">j’ai fait abſtraction des contreforts que <lb/>l’on fait ordinairement pour ſoutenir les pieds droits, parce <lb/>qu’en quelque façon on pourroit s’en paſſer; </s> <s xml:id="echoid-s18727" xml:space="preserve">mais comme il <lb/>ſembleroit que ce ſeroit vouloir changer ce qui ſe pratique or-<lb/>dinairement, je laiſſe à la diſcrétion de ceux qui auront la <lb/>conduite de ces ſortes d’ouvrages, d’en faire autant qu’ils le <lb/>jugeront à propos, & </s> <s xml:id="echoid-s18728" xml:space="preserve">de leur donner les dimenſions qui leur <lb/>conviendront le mieux: </s> <s xml:id="echoid-s18729" xml:space="preserve">car quoiqu’il ſemble qu’après avoir <lb/>donné aux pieds droits des épaiſſeurs ſuffiſantes pour réſiſter à <lb/>la pouſſée des voûtes des magaſins, il ſoit inutile d’y ajouter <lb/>encore des contreforts, cela n’empêche pas qu’ils ne ſoient <lb/>très-bien placés, puiſqu’il convient même d’en faire aux murs <lb/>qui n’ont point de pouſſée.</s> <s xml:id="echoid-s18730" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18731" xml:space="preserve">Il me reſte à donner l’uſage de la Table ſuivante, que j’ai <lb/>calculée pour quatre ſortes de magaſins à poudre. </s> <s xml:id="echoid-s18732" xml:space="preserve">Dans la pre-<lb/>miere colonne l’on voit la largeur des magaſins, qui auroient <lb/>depuis 20 pieds juſqu’à 36 dans œuvre; </s> <s xml:id="echoid-s18733" xml:space="preserve">& </s> <s xml:id="echoid-s18734" xml:space="preserve">la colonne qui eſt à <lb/>côté, marque l’épaiſſeur qu’il faut donner aux pieds droits <lb/>des voûtes en plein ceintre de ces magaſins; </s> <s xml:id="echoid-s18735" xml:space="preserve">ſuppoſant d’ail-<lb/>leurs que tous les pieds droits de ces différens magaſins aient <lb/>toujours 9 pieds de hauteur depuis le rez-de-chauſſée juſqu’à la <lb/>naiſſance de la voûte. </s> <s xml:id="echoid-s18736" xml:space="preserve">Ainſi voulant ſçavoir quelle épaiſſeur il <lb/>faut donner au pied droit d’un magaſin, dont la largeur ſeroit <lb/>de 30 pieds, & </s> <s xml:id="echoid-s18737" xml:space="preserve">dont les pieds droits auroient 9 pieds de hau-<lb/>teur depuis la fondation juſqu’à la naiſſance de la voûte, je <lb/>cherche dans la premiere colonne le nombre 30, & </s> <s xml:id="echoid-s18738" xml:space="preserve">je vois qu’il <lb/>correſpond à 7 pieds 7 pouces, qui eſt l’épaiſſeur qu’il faudra <lb/>leur donner, pour que leur réſiſtance ſoit au deſſus de l’équi-<lb/>libre avec la pouſſée de la voûte d’un magaſin fait à l’épreuve <lb/>de la bombe.</s> <s xml:id="echoid-s18739" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18740" xml:space="preserve">La ſeconde Table fait voir l’épaiſſeur qu’il faut donner aux <lb/>pieds droits des voûtes des magaſins à poudre, qui ſeroient <pb o="592" file="0682" n="708" rhead="NOUVEAU COURS"/> faits en tiers point, en ſuppoſant que la naiſſance de la voûte <lb/>commence à 5 pieds au deſſus du rez-de-chauſſée, comme on <lb/>le voit marqué au ſecond profil; </s> <s xml:id="echoid-s18741" xml:space="preserve">& </s> <s xml:id="echoid-s18742" xml:space="preserve">cela pour toutes les lar-<lb/>geurs marquées dans la premiere colonne: </s> <s xml:id="echoid-s18743" xml:space="preserve">ainſi pour ſçavoir <lb/>l’épaiſſeur qu’il faut donner au pied droit d’une voûte en tiers <lb/>point d’un magaſin, dont la largeur dans œuvre ſeroit de 24 <lb/>pieds, & </s> <s xml:id="echoid-s18744" xml:space="preserve">dont les pieds droits en dedans ne ſont élevés que de <lb/>5 pieds au deſſus du rez-de-chauſſée, il faut chercher dans la <lb/>premiere colonne le nombre 24, & </s> <s xml:id="echoid-s18745" xml:space="preserve">l’on verra qu’il correſpond <lb/>à 5 pieds 10 pouces, qui eſt l’épaiſſeur que l’on cherche.</s> <s xml:id="echoid-s18746" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18747" xml:space="preserve">La troiſieme Table ſert pour régler l’épaiſſeur qu’il faut <lb/>donner aux pieds droits des magaſins, qui ont un étage ſou-<lb/>terrein, & </s> <s xml:id="echoid-s18748" xml:space="preserve">j’ai ſuppoſé, en la calculant, que la hauteur des <lb/>pieds droit ſeroit de 12 pieds depuis la retraite au deſſus de la <lb/>fondation juſqu’à la naiſſance de la voûte qui doit être en tiers <lb/>point.</s> <s xml:id="echoid-s18749" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18750" xml:space="preserve">Enfin la quatrieme Table a été calculée pour les pieds droits <lb/>des magaſins à poudre, qui auroient un étage pratiqué dans la <lb/>voûte au deſſus de celui du rez-de-chauſſée, & </s> <s xml:id="echoid-s18751" xml:space="preserve">la hauteur des <lb/>pieds droits a été ſuppoſée de 9 pieds pour tous les magaſins, <lb/>dont la largeur auroit depuis 20 juſqu’à 36 pieds dans œuvre, <lb/>& </s> <s xml:id="echoid-s18752" xml:space="preserve">dont les voûtes ſeroient en tiers point.</s> <s xml:id="echoid-s18753" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18754" xml:space="preserve">Le principe qui m’a ſervi à calculer cette Table, eſt une <lb/>ſuite d’un des plus beaux problêmes d’architecture, que peu <lb/>de perſonnes ſçavent, non pas même les plus fameux Archi-<lb/>tectes. </s> <s xml:id="echoid-s18755" xml:space="preserve">Ce problême eſt de ſçavoir donner au pied droit d’une <lb/>voûte une épaiſſeur qui met la pouſſée de la voûte en équilibre <lb/>avec la réſiſtance des pieds droits; </s> <s xml:id="echoid-s18756" xml:space="preserve">ou, ce qui a encore rapport <lb/>au même, ſçavoir quelle épaiſſeur il faut donner aux culées <lb/>des ponts, pour ſoutenir la pouſſée des arches. </s> <s xml:id="echoid-s18757" xml:space="preserve">Le P. </s> <s xml:id="echoid-s18758" xml:space="preserve">Derand <lb/>dans ſon Traité de la coupe des Pierres, M. </s> <s xml:id="echoid-s18759" xml:space="preserve">Blondel dans ſon <lb/>Cours d’Architecture, & </s> <s xml:id="echoid-s18760" xml:space="preserve">pluſieurs autres, ont prétendu donner <lb/>des regles là-deſſus; </s> <s xml:id="echoid-s18761" xml:space="preserve">mais leur principe eſt faux, en ce qu’ils <lb/>n’ont point d’égard à la hauteur des pieds droits, ni à l’épaiſ-<lb/>ſeur de la voûte. </s> <s xml:id="echoid-s18762" xml:space="preserve">M. </s> <s xml:id="echoid-s18763" xml:space="preserve">de la Hire en a donné une parfaite ſolu-<lb/>tion dans les Mémoires de l’Académie des Sciences de 1712. <lb/></s> <s xml:id="echoid-s18764" xml:space="preserve">J’aurois pu rapporter ſon Mémoire, & </s> <s xml:id="echoid-s18765" xml:space="preserve">en expliquer les endroits <lb/>qui m’ont paru obſcurs, mais je me ſuis contenté de conſtruire <lb/>la Table que je rapporte ici, & </s> <s xml:id="echoid-s18766" xml:space="preserve">que l’on trouvera expliquée à <lb/>fonds dans la Science des Ingénieurs.</s> <s xml:id="echoid-s18767" xml:space="preserve"/> </p> <pb o="593" file="0683" n="709" rhead="DE MATHÉMATIQUE. Liv. XV."/> </div> <div xml:id="echoid-div1556" type="section" level="1" n="1126"> <head xml:id="echoid-head1332" xml:space="preserve">TABLE</head> <head xml:id="echoid-head1333" style="it" xml:space="preserve">Pour régler l’épaiſſeur qu’il faut donner aux pieds droits des voûtes <lb/>des magaſins à poudre.</head> <note position="right" xml:space="preserve">Largeur \\ des Ma- \\ gaſins à \\ poudre. # # # Epaiſſeur des \\ pieds droits des \\ voûtes en plein \\ ceintre pour les \\ magaſins à un \\ étage. # # # Epaiſſeur des \\ pieds droits des \\ voûtes en tiers \\ point pour les \\ magaſins à un \\ étage. # # # Epaiſſeur des \\ pieds droits des \\ voûtes pour les \\ Magaſins qui \\ ont un étage \\ ſouterrein. # # # Epaiſſeur des pieds \\ droits pour les voû- \\ tes des magaſins \\ qui ont un étage au \\ deſſus du rez - de- \\ chauſſée. <lb/>pieds. # pie. # pou. # lig. # pie. # pou. # lig. # pie. # pou. # lig. # pieds. # pou. # lig. <lb/>20 # 5 # 10 # 0 # 5 # 2 # 0 # 7 # 0 # 0 # 5 # 5 # 6 <lb/>21 # 5 # 11 # 8 # 5 # 3 # 0 # 7 # 2 # 5 # 5 # 8 # 6 <lb/>22 # 6 # 2 # 2 # 5 # 5 # 6 # 7 # 4 # 10 # 5 # 10 # 6 <lb/>23 # 6 # 4 # 6 # 5 # 7 # 4 # 7 # 7 # 3 # 6 # 0 # 10 <lb/>24 # 6 # 6 # 0 # 5 # 10 # 0 # 7 # 9 # 9 # 6 # 2 # 6 <lb/>25 # 6 # 8 # 3 # 6 # 0 # 4 # 8 # 0 # 1 # 6 # 4 # 6 <lb/>26 # 6 # 10 # 0 # 6 # 2 # 0 # 8 # 2 # 6 # 6 # 5 # 11 <lb/>27 # 6 # 11 # 9 # 6 # 5 # 0 # 8 # 4 # 10 # 6 # 8 # 0 <lb/>28 # 7 # 2 # 6 # 6 # 8 # 0 # 8 # 7 # 3 # 6 # 10 # 3 <lb/>29 # 7 # 4 # 9 # 6 # 10 # 6 # 8 # 9 # 8 # 7 # 0 # 0 <lb/>30 # 7 # 7 # 0 # 7 # 1 # 0 # 9 # 0 # 1 # 7 # 2 # 9 <lb/>31 # 7 # 9 # 4 # 7 # 2 # 4 # 9 # 2 # 6 # 7 # 5 # 6 <lb/>32 # 7 # 11 # 10 # 7 # 4 # 9 # 9 # 5 # 11 # 7 # 8 # 0 <lb/>33 # 8 # 2 # 8 # 7 # 7 # 0 # 9 # 8 # 4 # 7 # 10 # 6 <lb/>34 # 8 # 3 # 11 # 7 # 9 # 4 # 9 # 10 # 9 # 8 # 2 # 0 <lb/>35 # 8 # 5 # 9 # 7 # 11 # 0 # 10 # 1 # 2 # 8 # 4 # 2 <lb/>36 # 8 # 8 # 0 # 8 # 0 # 0 # 10 # 3 # 7 # 8 # 6 # 0 <lb/></note> <p> <s xml:id="echoid-s18768" xml:space="preserve">Après avoir parlé des magaſins à poudre, je crois qu’on <lb/>verra avec plaiſir de quelle maniere ſe fait le choc des bombes <lb/>qui tombent ſur leurs voûtes, afin qu’on ſente la différence <lb/>qu’il y a de conſidérer les chofes comme elles nous paroiſſent, <lb/>ou telles qu’elles ſont en elles-mêmes, & </s> <s xml:id="echoid-s18769" xml:space="preserve">que les Mathémati-<lb/>ques donnent ſur ce ſujet des connoiſſances que la pratique <lb/>des plus habiles Bombardiers ne peut appercevoir.</s> <s xml:id="echoid-s18770" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1557" type="section" level="1" n="1127"> <head xml:id="echoid-head1334" style="it" xml:space="preserve">Application des principes de la méchanique au jet des bombes.</head> <p> <s xml:id="echoid-s18771" xml:space="preserve">1120. </s> <s xml:id="echoid-s18772" xml:space="preserve">Nous avons fait voir (art. </s> <s xml:id="echoid-s18773" xml:space="preserve">1118) que pour trouver la <lb/>force avec laquelle une bombe tomboit ſur un plan, il falloit <lb/>multiplier ſa peſanteur par la racine quarrée de la hauteur où <pb o="594" file="0684" n="710" rhead="NOUVEAU COURS"/> elle s’étoit élevée, & </s> <s xml:id="echoid-s18774" xml:space="preserve">nous avons agi comme ſi la bombe tom-<lb/>boit ſelon une direction perpendiculaire à l’horizon, & </s> <s xml:id="echoid-s18775" xml:space="preserve">comme <lb/>ſi le plan qu’elle choquoit étoit de niveau avec la batterie. <lb/></s> <s xml:id="echoid-s18776" xml:space="preserve">Mais comme les bombes ne tombent que rarement par des di-<lb/>rections perpendiculaires aux plans qu’elles rencontrent, & </s> <s xml:id="echoid-s18777" xml:space="preserve"><lb/>que le plus ſouvent elles tombent ſur des ſurfaces qui ſont plus <lb/>élevées que la batterie, le problême dont je viens de parler, <lb/>n’eſt pas abſolument juſte, parce qu’on y fait abſtraction des <lb/>deux circonſtances précédentes; </s> <s xml:id="echoid-s18778" xml:space="preserve">& </s> <s xml:id="echoid-s18779" xml:space="preserve">ſi on ne les a pas fait en-<lb/>trer, c’eſt qu’on n’étoit pas encore prévenu du principe de <lb/>méchanique expliqué ci-devant. </s> <s xml:id="echoid-s18780" xml:space="preserve">Mais comme il ne reſte plus <lb/>rien à deſirer à ce ſujet, voici comme il faut raiſonner.</s> <s xml:id="echoid-s18781" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18782" xml:space="preserve">Si la ligne A B marque l’élévation du mortier ſur le plan <lb/>horizontal A C, & </s> <s xml:id="echoid-s18783" xml:space="preserve">que la parabole A H D ait été décrite par <lb/>la bombe, la ligne A B qui va rencontrer l’axe prolongé de la <lb/>parabole, ſera la tangente de cette courbe menée du point A, <lb/> <anchor type="note" xlink:label="note-0684-01a" xlink:href="note-0684-01"/> & </s> <s xml:id="echoid-s18784" xml:space="preserve">la ligne B D ſera une autre tangente menée du point D: <lb/></s> <s xml:id="echoid-s18785" xml:space="preserve"> <anchor type="note" xlink:label="note-0684-02a" xlink:href="note-0684-02"/> mais quand un corps eſt jetté par une direction qui n’eſt pas <lb/>perpendiculaire à l’horizon, la direction ſelon laquelle ce corps <lb/>choque un plan, eſt marquée par la tangente menée par le <lb/>point de la parabole, où le corps rencontre le plan: </s> <s xml:id="echoid-s18786" xml:space="preserve">ainſi la <lb/>bombe qui aura décrit la parabole A H D, choquera le plan <lb/>A C, ſelon la direction B D; </s> <s xml:id="echoid-s18787" xml:space="preserve">mais comme cette ligne eſt <lb/>oblique au plan A C, ſi la force de la bombe eſt exprimée par <lb/>la ligne F D, elle ne choquera pas le plan avec toute la force <lb/>F D: </s> <s xml:id="echoid-s18788" xml:space="preserve">car ſi l’on abaiſſe F E perpendiculaire ſur A C, & </s> <s xml:id="echoid-s18789" xml:space="preserve">qu’on <lb/>faſſe le parallélogramme E G, la force F D ſera égale aux <lb/>forces F G & </s> <s xml:id="echoid-s18790" xml:space="preserve">F E (art. </s> <s xml:id="echoid-s18791" xml:space="preserve">1039) agiſſantes enſemble; </s> <s xml:id="echoid-s18792" xml:space="preserve">mais la <lb/>force F G parallele à l’horizon, n’agit point du tout ſur le plan <lb/>A C: </s> <s xml:id="echoid-s18793" xml:space="preserve">il n’y a donc que la force exprimée par F E, qui choque le <lb/>plan; </s> <s xml:id="echoid-s18794" xml:space="preserve">ce qui fait voir que le choc de la bombe, ſelon la direc-<lb/>tion B D, eſt au choc de la même bombe, ſelon la direction <lb/>perpendiculaire B I, comme F E eſt à F D, ou comme B I eſt <lb/>à B D, c’eſt-à-dire comme la ſoutangente eſt à la tangente, <lb/>ou bien comme la tangente de l’angle de l’élévation du mor-<lb/>tier eſt à la ſécante du même angle, ou encore comme le ſinus <lb/>de l’angle de l’élévation eſt au ſinus total: </s> <s xml:id="echoid-s18795" xml:space="preserve">ainſi ſuppoſant que <lb/>l’angle B A I ſoit de 50 degrés, l’on peut dire que le choc de la <lb/>bombe tombant, ſelon la direction perpendiculaire B I, eſt au <lb/>choc par la direction B D, comme 100000 eſt à 76604.</s> <s xml:id="echoid-s18796" xml:space="preserve"/> </p> <div xml:id="echoid-div1557" type="float" level="2" n="1"> <note position="left" xlink:label="note-0684-01" xlink:href="note-0684-01a" xml:space="preserve">Pl. XXXII.</note> <note position="left" xlink:label="note-0684-02" xlink:href="note-0684-02a" xml:space="preserve">Figure 404.</note> </div> <pb o="595" file="0685" n="711" rhead="DE MATHÉMATIQUE. Liv. XV."/> <p> <s xml:id="echoid-s18797" xml:space="preserve">A ne conſidérer que le choc des bombes qui tombent ſur un <lb/>plan horizontal, il ſemble que ce que l’on vient de dire ne <lb/>ſoit pas d’une grande utilité, parce que les bombes que l’on <lb/>jette dans les ouvrages, ſoit de la part des Aſſiégés ou des <lb/>Aſſiégeans, font toujours beaucoup plus d’effet par leurs <lb/>éclats, quand elles crevent, que par le poids de leur chûte; <lb/></s> <s xml:id="echoid-s18798" xml:space="preserve">& </s> <s xml:id="echoid-s18799" xml:space="preserve">ſi le poids avoit lieu dans ce cas-ci, ce ne ſeroit qu’à l’occa-<lb/>ſion des ſouterreins que l’on pratique dans les Places ſous les <lb/>remparts pour les différens uſages auxquels ils ſont propres: </s> <s xml:id="echoid-s18800" xml:space="preserve"><lb/>mais comme le choc d’une bombe mérite plus d’attention, <lb/>lorſqu’elle tombe ſur un édifice que les Aſſiégeans ont intérêt <lb/>de ruiner, comme un magaſin à poudre, dont il s’agit de percer <lb/>la voûte, qui eſt un plan incliné à l’horizon, c’eſt particulié-<lb/>rement la chûte des bombes dans ce cas-ci qu’il nous faut exa-<lb/>miner.</s> <s xml:id="echoid-s18801" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18802" xml:space="preserve">Si l’on a un mortier au point A pour jetter une bombe ſur <lb/> <anchor type="note" xlink:label="note-0685-01a" xlink:href="note-0685-01"/> le plan incliné K L, & </s> <s xml:id="echoid-s18803" xml:space="preserve">qu’on veuille ſçavoir quel eſt le choc <lb/>de la bombe, qui après avoir décrit la parabole A H D, vien-<lb/>droit tomber à un point D du plan incliné, je conſidere que <lb/>la bombe frappant le point D, agit ſelon ſa direction B D, <lb/>qui eſt une tangente menée par le point D de la parabole. </s> <s xml:id="echoid-s18804" xml:space="preserve">Or <lb/>ſi l’on prend la ligne F D pour exprimer la force de la bombe, <lb/>lorſqu’elle eſt prête à tomber ſur le plan incliné, cette force <lb/>étant oblique au plan, n’exprimera pas la force avec laquelle <lb/>la bombe choquera ce plan, mais ſeulement la force de la <lb/>bombe en elle-même: </s> <s xml:id="echoid-s18805" xml:space="preserve">& </s> <s xml:id="echoid-s18806" xml:space="preserve">ſi du point F l’on mene la ligne F E <lb/>perpendiculaire ſur K L, elle exprimera la force avec laquelle <lb/>la bombe choquera le plan incliné: </s> <s xml:id="echoid-s18807" xml:space="preserve">car faiſant le parallélo-<lb/>gramme G E, l’on aura les côtés F E & </s> <s xml:id="echoid-s18808" xml:space="preserve">F G, qui exprime-<lb/>ront deux forces, leſquelles agiſſant enſemble, ſeront égales <lb/>à la ſeule F D; </s> <s xml:id="echoid-s18809" xml:space="preserve">mais la force F G étant parallele au plan K L, <lb/>n’agit point du tout ſur ce plan. </s> <s xml:id="echoid-s18810" xml:space="preserve">Il n’y a donc que la ligne F E <lb/>qui exprime le choc de la bombe: </s> <s xml:id="echoid-s18811" xml:space="preserve">ainſi l’on peut dire que le <lb/>choc d’une bombe qui tombe obliquement ſur un plan incliné, <lb/>eſt au choc de la direction perpendiculaire, comme F E eſt à <lb/>F D, ou comme le ſinus de l’angle F D E eſt au ſinus total, <lb/>étant tombée de la même hauteur.</s> <s xml:id="echoid-s18812" xml:space="preserve"/> </p> <div xml:id="echoid-div1558" type="float" level="2" n="2"> <note position="right" xlink:label="note-0685-01" xlink:href="note-0685-01a" xml:space="preserve">Figure 405.</note> </div> <p> <s xml:id="echoid-s18813" xml:space="preserve">Si l’on vouloit ſçavoir quel eſt ce rapport, il faudroit cher-<lb/>cher l’angle F D E, que l’on trouvera en connoiſſant la valeur <lb/>de l’angle K D C, formé par l’horizon & </s> <s xml:id="echoid-s18814" xml:space="preserve">le plan incliné, de <pb o="596" file="0686" n="712" rhead="NOUVEAU COURS"/> plus l’angle d’inclinaiſon B A D du mortier, qui eſt égal à <lb/>B D A: </s> <s xml:id="echoid-s18815" xml:space="preserve">ainſi ſuppoſant l’angle B D A de 50 degrés, & </s> <s xml:id="echoid-s18816" xml:space="preserve">l’angle <lb/>K C D de 70; </s> <s xml:id="echoid-s18817" xml:space="preserve">ſi on les ajoute enſemble, l’on aura 120 degrés, <lb/>qui étant ſouſtraits de deux droits, la différence ſera 60 de-<lb/>grés pour la valeur de l’angle F D E, dont le ſinus eſt 86602, <lb/>par conſéquent le rapport du choc de la bombe, ſelon la di-<lb/>rection perpendiculaire, eſt à celle, ſelon la direction oblique <lb/>F D, comme 100000 eſt à 86602.</s> <s xml:id="echoid-s18818" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18819" xml:space="preserve">Tout le monde croit (& </s> <s xml:id="echoid-s18820" xml:space="preserve">l’on a raiſon dans un ſens) que <lb/>plus les bombes tombent de haut, & </s> <s xml:id="echoid-s18821" xml:space="preserve">plus le choc ſur le plan <lb/>qu’elles rencontrent eſt violent. </s> <s xml:id="echoid-s18822" xml:space="preserve">Cependant ceci n’eſt vrai <lb/>que quand le plan que la bombe rencontre eſt de niveau avec <lb/>la batterie, parce que tombant de fort haut, elle décrit ſur la <lb/>fin une ligne courbe, qui approche fort de la verticale; </s> <s xml:id="echoid-s18823" xml:space="preserve">mais <lb/>quand le plan eſt incliné à l’horizon, la chûte par la verticale <lb/>même eſt celle qui choque le plan incliné avec moins de vio-<lb/>lence que par toutes les autres directions poſſibles, qui ſeroient <lb/>entre l’horizontale & </s> <s xml:id="echoid-s18824" xml:space="preserve">la verticale, ſi les bombes tombent d’une <lb/>hauteur égale; </s> <s xml:id="echoid-s18825" xml:space="preserve">& </s> <s xml:id="echoid-s18826" xml:space="preserve">ce n’eſt que quand la tangente menée au <lb/>point de la parabole qui rencontre le plan incliné, eſt perpen-<lb/>diculaire à ce plan même, que la bombe choque avec toute <lb/>ſa force abſolue. </s> <s xml:id="echoid-s18827" xml:space="preserve">Or pour faire enſorte qu’une bombe tombe <lb/>ſur un plan incliné par une direction perpendiculaire, il faut <lb/>connoître l’angle d’inclinaiſon que forme le plan avec l’ho-<lb/>rizon, & </s> <s xml:id="echoid-s18828" xml:space="preserve">pointer le mortier ſous un angle qui ſoit égal au <lb/>complément de celui du plan incliné.</s> <s xml:id="echoid-s18829" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18830" xml:space="preserve">Par exemple, ſi ſur le plan incliné K L, on éleve la perpen-<lb/> <anchor type="note" xlink:label="note-0686-01a" xlink:href="note-0686-01"/> diculaire B D au point D, qui aille rencontrer la perpendicu-<lb/>laire B E, élevée dans le milieu de l’amplitude A D de la pa-<lb/>rabole, & </s> <s xml:id="echoid-s18831" xml:space="preserve">qu’on tire la ligne A B, l’angle B A D ſera celui <lb/>qu’il faut donner au mortier pour chaſſer la bombe au point <lb/>D; </s> <s xml:id="echoid-s18832" xml:space="preserve">mais cette angle eſt égal à l’angle B D E, lequel eſt com-<lb/>plément de l’angle K D C, puiſque B D K eſt droit: </s> <s xml:id="echoid-s18833" xml:space="preserve">donc <lb/>l’angle B A E, complément de l’angle d’inclinaiſon, eſt celui <lb/>qu’il faut donner au mortier, pour que la bombe choque le <lb/>plan incliné par une direction perpendiculaire au même plan.</s> <s xml:id="echoid-s18834" xml:space="preserve"/> </p> <div xml:id="echoid-div1559" type="float" level="2" n="3"> <note position="left" xlink:label="note-0686-01" xlink:href="note-0686-01a" xml:space="preserve">Figure 406.</note> </div> <p> <s xml:id="echoid-s18835" xml:space="preserve">Par cette théorie l’on pourroit déterminer quelle eſt la <lb/>charge, ou ſi l’on veut, quels ſont les degrés de force que doit <lb/>avoir un mortier, & </s> <s xml:id="echoid-s18836" xml:space="preserve">l’angle qu’il lui faut donner pour chaſſer <lb/>une bombe ſur un plan incliné, enſorte que la bombe choque <pb o="597" file="0687" n="713" rhead="DE MATHÉMATIQUE. Liv. XV."/> ce plan avec toute la force qu’il eſt poſſible; </s> <s xml:id="echoid-s18837" xml:space="preserve">démontrer même <lb/>que lorſque les racines quarrées des différentes hauteurs d’où <lb/>une bombe tombera ſur un plan incliné, ſeront réciproque-<lb/>ment proportionnelles aux ſinus des angles d’incidence formés <lb/>par les différentes directions des bombes, le choc ſera tou-<lb/>jours égal, & </s> <s xml:id="echoid-s18838" xml:space="preserve">une quantité d’autres choſes, qui à la vérité <lb/>ſont plus propres à exercer l’eſprit, qu’à être miſes en pratique. <lb/></s> <s xml:id="echoid-s18839" xml:space="preserve">C’eſt pourquoi je ne parlerai plus que de deux cas qui me reſtent <lb/>à expliquer; </s> <s xml:id="echoid-s18840" xml:space="preserve">ſçavoir quel eſt le choc des bombes qui ſeroient <lb/>tirées d’un lieu plus bas ou plus élevé que le plan incliné qu’elle <lb/>doit rencontrer: </s> <s xml:id="echoid-s18841" xml:space="preserve">& </s> <s xml:id="echoid-s18842" xml:space="preserve">comme ſçachant un de ces cas, il eſt aiſé <lb/>de concevoir l’autre, voici celui qui regarde le plan incliné <lb/>plus élevé que la batterie.</s> <s xml:id="echoid-s18843" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18844" xml:space="preserve">Si par les regles du jet des bombes l’on a trouvé l’angle B A I <lb/>pour donner au mortier une élévation convenable, afin de <lb/>jetter une bombe au point D d’un plan incliné K L, plus <lb/>élevé que l’horizon A P, l’on connoîtra l’amplitude A P de la <lb/>parabole A H P, & </s> <s xml:id="echoid-s18845" xml:space="preserve">par conſéquent ſon axe H I; </s> <s xml:id="echoid-s18846" xml:space="preserve">& </s> <s xml:id="echoid-s18847" xml:space="preserve">avant cela <lb/>on aura dû ſçavoir l’élévation D Q du point D ſur l’horizon <lb/>A P: </s> <s xml:id="echoid-s18848" xml:space="preserve">mais ſi la bombe au lieu de tomber en P, tombe en D, <lb/>menant D O parallele à P A, la vîteſſe de la bombe ſera ex-<lb/>primée par la racine quarrée de H N. </s> <s xml:id="echoid-s18849" xml:space="preserve">Or ſi l’on prend la ligne <lb/>F D pour exprimer cette force, & </s> <s xml:id="echoid-s18850" xml:space="preserve">que l’on tire la ligne F E <lb/>perpendiculaire au plan K L, le choc de la bombe au point D <lb/>ſera exprimé par la ligne F E, & </s> <s xml:id="echoid-s18851" xml:space="preserve">non pas par la ligne F D, <lb/>comme on vient de le voir. </s> <s xml:id="echoid-s18852" xml:space="preserve">Or le rapport du choc perpendi-<lb/>culaire au choc oblique étant comme F D eſt à F E, ou comme <lb/>le ſinus total eſt au ſinus de l’angle F D E, ſi l’on veut avoir <lb/>ce ſinus pour connoître en nombre le rapport de la ligne F D <lb/>à la ligne F E, il faut chercher la valeur de l’angle M O N, <lb/>formé par l’ordonnée O N & </s> <s xml:id="echoid-s18853" xml:space="preserve">la tangente O M, qui eſt l’angle <lb/>qu’il auroit fallu donner au mortier, ſi la bombe avoit été tirée <lb/>de l’endroit O, de niveau avec le point D. </s> <s xml:id="echoid-s18854" xml:space="preserve">Pour le trouver, <lb/>conſidérez que l’on connoît l’abſciſſe H N, qui eſt la diffé-<lb/>rence de H I à H D, & </s> <s xml:id="echoid-s18855" xml:space="preserve">que par conſéquent on connoîtra auſſi <lb/>la ſoutangente M N, qui eſt un des côtés du triangle rectangle <lb/>M N O; </s> <s xml:id="echoid-s18856" xml:space="preserve">& </s> <s xml:id="echoid-s18857" xml:space="preserve">comme pour trouver l’angle que nous cherchons, <lb/>il nous faut encore le côté O N, pour le trouver, l’on dira: <lb/></s> <s xml:id="echoid-s18858" xml:space="preserve">Comme l’abſciſſe H I eſt à l’abſciſſe H N, ainſi le quárré de <lb/>l’ordonnée A I eſt au quarré de l’ordonnée O N, que l’on trou- <pb o="598" file="0688" n="714" rhead="NOUVEAU COURS"/> vera par la regle de proportion, dont extrayant la racine <lb/>quarrée, l’on aura le côté O N, qui donnera avec le côté M N <lb/>l’angle M O N ou M D N ſon égal; </s> <s xml:id="echoid-s18859" xml:space="preserve">& </s> <s xml:id="echoid-s18860" xml:space="preserve">ſi l’on ajoute à cet angle <lb/>la valeur de l’angle E D C, formé par le plan incliné & </s> <s xml:id="echoid-s18861" xml:space="preserve">l’ho-<lb/>rizon, & </s> <s xml:id="echoid-s18862" xml:space="preserve">que l’on ôte la ſomme de ces deux angles de la va-<lb/>leur de deux droits, l’on aura pour la différence l’angle F D E, <lb/>dont le ſinus ſervira à déterminer le choc de la bombe au point <lb/>D, par rapport au ſinus total qui exprime la force abſolue.</s> <s xml:id="echoid-s18863" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18864" xml:space="preserve">L’on peut auſſi tirer de tout ceci des regles pour déterminer <lb/> <anchor type="note" xlink:label="note-0688-01a" xlink:href="note-0688-01"/> la force d’un boulet de canon, qui choqueroit une ſurface, tiré <lb/>des batteries différemment éloignées de cette ſurface: </s> <s xml:id="echoid-s18865" xml:space="preserve">par <lb/>exemple, ſi l’on a une ſurface verticale A B, & </s> <s xml:id="echoid-s18866" xml:space="preserve">que du point <lb/>C l’on tire un boulet, enſorte que l’ame de la piece ſoit pointée <lb/>ſelon la direction C D perpendiculaire à cette ſurface, le boulet, <lb/>au lieu de frapper au point D, frappera au point G, plus bas <lb/>que le point D, parce que ſa peſanteur lui fera décrire la pa-<lb/>rabole C P G, & </s> <s xml:id="echoid-s18867" xml:space="preserve">le choc du boulet ſe fera ſelon la direction <lb/>de la ligne I G tangente à la parabole au point G: </s> <s xml:id="echoid-s18868" xml:space="preserve">ainſi ce ſera <lb/>la ligne I K perpendiculaire à la ſurface qui exprimera le choc <lb/>du boulet, & </s> <s xml:id="echoid-s18869" xml:space="preserve">non pas la ligne I G, diagonale du parallélo-<lb/>gramme K L. </s> <s xml:id="echoid-s18870" xml:space="preserve">Or ſi le même boulet, au lieu d’être chaſſé du <lb/>point C, eſt chaſſé du point E, avec la même force, la diſtance <lb/>E F étant plus grande que C A, choquera la ſurface au point H <lb/>avec moins de force qu’il ne la choque au point G; </s> <s xml:id="echoid-s18871" xml:space="preserve">ce n’eſt <lb/>pas que cette plus grande diſtance lui ait rien fait perdre de <lb/>ſon degré de mouvement (ſi l’on compte pour rien la réſiſ-<lb/>tance de l’air); </s> <s xml:id="echoid-s18872" xml:space="preserve">mais c’eſt que la parabole E q H étant plus <lb/>grande que C P G, le point H où le boulet aura choqué la ſur-<lb/>face, ſera bien plus éloigné de F que le point G ne l’eſt de <lb/>D: </s> <s xml:id="echoid-s18873" xml:space="preserve">par conſéquent la tangente M H, que l’on menera à la pa-<lb/>rabole par le point H, ſera plus incliné à la ſurface A B, que <lb/>la tangente I G ne l’eſt à la même ſurface. </s> <s xml:id="echoid-s18874" xml:space="preserve">Or faiſant M H <lb/>égal à I G, ſi l’on mene la ligne M N perpendiculaire à la ſur-<lb/>face A B, elle ſera dans la même raiſon avec la perpendicu-<lb/>laire I K, comme le choc du boulet tiré de l’endroit E ſera à <lb/>celui du boulet tiré de l’endroit C, ou bien comme le ſinus <lb/>de l’angle M H N ſera au ſinus de l’angle I G K; </s> <s xml:id="echoid-s18875" xml:space="preserve">d’où il s’en-<lb/>ſuit que quand on bat avec le canon une ſurface de fort loin, <lb/>ce n’eſt pas que le boulet ait rien perdu de ſa force, qui fait <lb/>qu’il ne choque pas la ſurface avec autant de violence, que s’il <pb o="599" file="0689" n="715" rhead="DE MATHÉMATIQUE. Liv. XV."/> avoit été tiré de plus près, comme bien des gens le croient; <lb/></s> <s xml:id="echoid-s18876" xml:space="preserve">mais au contraire c’eſt que ne frappant la ſurface que par une <lb/>direction fort oblique, il n’agit pas avec autant d’effort, que <lb/>s’il la frappoit par une autre direction qui approchât plus d’être <lb/>perpendiculaire: </s> <s xml:id="echoid-s18877" xml:space="preserve">car ſi un boulet en ſortant de la piece ne <lb/>rencontroit pas des corps à qui il communique du mouve-<lb/>ment qu’il a reçu de l’impulſion de la poudre, que l’air ne <lb/>lui fît aucun empêchement, & </s> <s xml:id="echoid-s18878" xml:space="preserve">que la peſanteur du boulet ne <lb/>le fît pas tendre vers le centre de la terre, en un mot, qu’il <lb/>pût toujours aller en ligne droite, ſa force ſeroit toujours la <lb/>même, à quelque diſtance qu’il fût porté, puiſqu’il conſerve-<lb/>roit toujours le mouvement qu’il a reçu, s’il n’en perdoit à <lb/>meſure qu’il en communique aux corps qu’il rencontre, n’y <lb/>ayant point de raiſon que cela puiſſe être autrement.</s> <s xml:id="echoid-s18879" xml:space="preserve"/> </p> <div xml:id="echoid-div1560" type="float" level="2" n="4"> <note position="left" xlink:label="note-0688-01" xlink:href="note-0688-01a" xml:space="preserve">Figure 408 <lb/>& 409.</note> </div> <figure> <image file="0689-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0689-01"/> <caption xml:id="echoid-caption1" style="it" xml:space="preserve">Fin du Quinzieme Livre.</caption> </figure> <pb o="600" file="0690" n="716"/> <figure> <image file="0690-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0690-01"/> </figure> </div> <div xml:id="echoid-div1562" type="section" level="1" n="1128"> <head xml:id="echoid-head1335" xml:space="preserve">NOUVEAU COURS <lb/>DE <lb/>MATHÉMATIQUE.</head> <head xml:id="echoid-head1336" xml:space="preserve">LIVRE SEIZIEME.</head> <head xml:id="echoid-head1337" xml:space="preserve">De l’Hydroſtatique.</head> <p style="it"> <s xml:id="echoid-s18880" xml:space="preserve"><emph style="sc">NOus</emph> allons traiter dans le Livre ſuivant des propriétés des <lb/>fluides conſidérés par rapport à l’équilibre; </s> <s xml:id="echoid-s18881" xml:space="preserve">& </s> <s xml:id="echoid-s18882" xml:space="preserve">c’eſt ce que l’on en-<lb/>tend par le mot Hydroſtatique. </s> <s xml:id="echoid-s18883" xml:space="preserve">Cette partie, comme l’on voit, eſt <lb/>une ſuite de la méchanique, & </s> <s xml:id="echoid-s18884" xml:space="preserve">peut être regardée comme la plus <lb/>importante. </s> <s xml:id="echoid-s18885" xml:space="preserve">Il ſeroit à ſouhaiter qu’on pût établir une théorie auſſi <lb/>ſimple ſur les fluides que ſur les corps ſolides que nous avons con-<lb/>ſidérés dans le Livre précédent. </s> <s xml:id="echoid-s18886" xml:space="preserve">Mais on voit bientôt qu’il n’eſt <lb/>pas également facile de traiter cette partie comme la précédente, <lb/>quoique ce ſoit la même peſanteur qui agiſſe ſur les corps & </s> <s xml:id="echoid-s18887" xml:space="preserve">les <lb/>fluides pour les faire deſcendre au centre de la terre. </s> <s xml:id="echoid-s18888" xml:space="preserve">Il n’y a, pour <lb/>ainſi dire, que ce phénomene qui ſoit commun aux uns & </s> <s xml:id="echoid-s18889" xml:space="preserve">aux au-<lb/>tres, auquel on peut joindre celui de la force d’inertie, qui eſt tou-<lb/>jours proportionnelle aux maſſes. </s> <s xml:id="echoid-s18890" xml:space="preserve">Il ſemble que plus les parties ſu-<lb/>jettes aux mathématiques deviennent intéreſſantes, plus elles de-<lb/>viennent obſcures & </s> <s xml:id="echoid-s18891" xml:space="preserve">complïquées. </s> <s xml:id="echoid-s18892" xml:space="preserve">Dans la ſtatique, la ſeule pe-<lb/>ſanteur des corps reconnue comme une force conſtante, quelle que <lb/>ſoit la cauſe dont elle provient, a ſuffi pour démontrer géométri-<lb/>quement les propriétés des machines, & </s> <s xml:id="echoid-s18893" xml:space="preserve">le rapport néceſſaire entre <lb/>tant de forces qu’on voudra, dont les directions étoient déterminées, <lb/>abſtraction faite des frottemens. </s> <s xml:id="echoid-s18894" xml:space="preserve">La même peſanteur nous a pareil- <pb o="601" file="0691" n="717" rhead="DE MATHÉMATIQUE. Liv. XVI."/> lement conduit à la découverte des courbes que les projectiles décri-<lb/>vent, quelle que ſoit l’intenſité de cette force. </s> <s xml:id="echoid-s18895" xml:space="preserve">Une ſeule expérience a <lb/>fixé parmi toutes les courbes poſſibles celle qu’ils décrivent réelle-<lb/>ment en conſéquence de l’action de la peſanteur auprès de notre <lb/>globe. </s> <s xml:id="echoid-s18896" xml:space="preserve">Il n’en eſt pas de même dans l’hydroſtatique: </s> <s xml:id="echoid-s18897" xml:space="preserve">la peſanteur <lb/>ſeule ne peut nous faire connoître tout ce qui a rapport au choc des <lb/>fluides à la maniere dont ils produiſent l’équilibre entr’eux ou avec <lb/>les corps ſolides. </s> <s xml:id="echoid-s18898" xml:space="preserve">Une théorie complette des fluides demanderoit que <lb/>l’on connût au moins quel eſt le principe général de la fluidité, <lb/>& </s> <s xml:id="echoid-s18899" xml:space="preserve">enſuite l’eſſence des parties de chacun en particulier: </s> <s xml:id="echoid-s18900" xml:space="preserve">mais la <lb/>nature ne nous laiſſe pas approcher de ſi près de ſes ſecrets. </s> <s xml:id="echoid-s18901" xml:space="preserve">La pro-<lb/>priété commune à tous les fluides de ſe mettre de niveau, & </s> <s xml:id="echoid-s18902" xml:space="preserve">de preſſer <lb/>également en tous ſens, eſt certainement une ſuite du principe gé-<lb/>néral de la fluidité; </s> <s xml:id="echoid-s18903" xml:space="preserve">auſſi doit-elle être regardée comme le premier <lb/>principe de l’hydroſtatique; </s> <s xml:id="echoid-s18904" xml:space="preserve">& </s> <s xml:id="echoid-s18905" xml:space="preserve">c’eſt delà qu’il faut partir pour dé-<lb/>couvrir, par la voie la plus naturelle, les autres propriétés des <lb/>fluides, relativement à l’équilibre: </s> <s xml:id="echoid-s18906" xml:space="preserve">car il eſt aiſé de voir que cette <lb/>même propriété ne donne rien à connoître ſur la figure des parties <lb/>élémentaires de chacun en particulier. </s> <s xml:id="echoid-s18907" xml:space="preserve">Lorſqu’on aura déduit de <lb/>cette propriété tout ce que l’analyſe peut nous fournir de conſé-<lb/>quences, il faut avoir recours aux expériences ſur chaque fluide <lb/>pour connoître ce qui les différencie les uns des autres, & </s> <s xml:id="echoid-s18908" xml:space="preserve">leurs pe-<lb/>ſanteurs ſpécifiques. </s> <s xml:id="echoid-s18909" xml:space="preserve">Il y a en général trois parties que nous de-<lb/>vons conſidérer dans l’hydroſtatique, 1°. </s> <s xml:id="echoid-s18910" xml:space="preserve">l’équilibre d’une liqueur <lb/>homogene, enſuite celui des fluides étérogenes, & </s> <s xml:id="echoid-s18911" xml:space="preserve">enfin celui des <lb/>ſolides avec les mêmes fluides. </s> <s xml:id="echoid-s18912" xml:space="preserve">Il ſeroit inutile d’entrer ici dans le <lb/>détail des avantages qu’on peut retirer de cette théorie. </s> <s xml:id="echoid-s18913" xml:space="preserve">On ſera <lb/>plus en état de les reconnoître lorſqu’on aura étudié ce Traité. </s> <s xml:id="echoid-s18914" xml:space="preserve">Il <lb/>n’eſt pas moins eſſentiel d’examiner la percuſſion des fluides, les <lb/>loix de leurs chocs contre les ſolides, à proportion des vîteſſes & </s> <s xml:id="echoid-s18915" xml:space="preserve"><lb/>des maſſes, ou denſités. </s> <s xml:id="echoid-s18916" xml:space="preserve">Quoique l’on ait fait uſage des fluides <lb/>pour ſe procurer une infinité de commodités & </s> <s xml:id="echoid-s18917" xml:space="preserve">d’avantages, ſans <lb/>connoître à fonds tout ce que l’on a découvert depuis environ un <lb/>ſiecle & </s> <s xml:id="echoid-s18918" xml:space="preserve">demi, il ne s’enſuit pas qu’il ſoit inutile de multi-<lb/>plier continuellement ſes recherches ſur cette partie. </s> <s xml:id="echoid-s18919" xml:space="preserve">Plus on aura <lb/>d’expériences bien analyſées, plus on aura des vues intéreſſantes <lb/>pour les Arts & </s> <s xml:id="echoid-s18920" xml:space="preserve">le bien public qui leur eſt attaché, plus on ſera à <lb/>portée de connoître les défauts des machines qui ont été exécutées <lb/>en ce genre, & </s> <s xml:id="echoid-s18921" xml:space="preserve">d’y remédier. </s> <s xml:id="echoid-s18922" xml:space="preserve">Comme nous ne donnons ici que les <lb/>élémens de l’hydroſtatique & </s> <s xml:id="echoid-s18923" xml:space="preserve">de l’hydraulique, on pourra recourir <pb o="602" file="0692" n="718" rhead="NOUVEAU COURS"/> à ce que nous avons donné ſur cette matiere dans le premier volume <lb/>de l’Architecture Hydraulique; </s> <s xml:id="echoid-s18924" xml:space="preserve">& </s> <s xml:id="echoid-s18925" xml:space="preserve">ceux qui auront les élémens ſuf-<lb/>fiſans pour pouſſer plus loin leurs recherches, ne pourront mieux <lb/>faire que d’étudier l’Hydraudinamique de MM. </s> <s xml:id="echoid-s18926" xml:space="preserve">Bernouilli & </s> <s xml:id="echoid-s18927" xml:space="preserve"><lb/>d’Alembert.</s> <s xml:id="echoid-s18928" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1563" type="section" level="1" n="1129"> <head xml:id="echoid-head1338" xml:space="preserve">CHAPITRE PREMIER.</head> <head xml:id="echoid-head1339" style="it" xml:space="preserve">De l’équilibre & du mouvement des liqueurs.</head> <head xml:id="echoid-head1340" xml:space="preserve"><emph style="sc">Définition</emph> I.</head> <p> <s xml:id="echoid-s18929" xml:space="preserve">1121. </s> <s xml:id="echoid-s18930" xml:space="preserve">ON appelle fluides les corps dont les parties ſe divi-<lb/>ſent & </s> <s xml:id="echoid-s18931" xml:space="preserve">cedent au moindre effort, & </s> <s xml:id="echoid-s18932" xml:space="preserve">ſe réuniſſent enſuite avec <lb/>la même facilité. </s> <s xml:id="echoid-s18933" xml:space="preserve">Par exemple, l’air, la flamme, l’eau, le mer-<lb/>cure, & </s> <s xml:id="echoid-s18934" xml:space="preserve">les autres liqueurs ſont des fluides.</s> <s xml:id="echoid-s18935" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1564" type="section" level="1" n="1130"> <head xml:id="echoid-head1341" xml:space="preserve"><emph style="sc">Remarque</emph> I.</head> <p> <s xml:id="echoid-s18936" xml:space="preserve">1122. </s> <s xml:id="echoid-s18937" xml:space="preserve">Il faut bien remarquer que tout liquide eſt fluide, <lb/>mais le réciproque n’eſt pas vrai. </s> <s xml:id="echoid-s18938" xml:space="preserve">Pour en ſentir la différence, <lb/>il faut ſçavoir que l’on appelle liquide tout fluide dont la ſur-<lb/>face ſe met de niveau dans le vaſe qui le contient. </s> <s xml:id="echoid-s18939" xml:space="preserve">Or il eſt <lb/>viſible que cette propriété ne convient pas à la flamme. </s> <s xml:id="echoid-s18940" xml:space="preserve">On <lb/>entend par une ſurface de niveau celle dont tous les points <lb/>ſont à égale diſtance du centre de la terre. </s> <s xml:id="echoid-s18941" xml:space="preserve">Il faut encore bien <lb/>remarquer que parmi les fluides il y en a qui ont du reſſort, <lb/>& </s> <s xml:id="echoid-s18942" xml:space="preserve">d’autres qui n’en ont pas. </s> <s xml:id="echoid-s18943" xml:space="preserve">Par exemple, on ſçait que l’air <lb/>ſe dilate & </s> <s xml:id="echoid-s18944" xml:space="preserve">ſe comprime, enſorte que la compreſſion eſt plus <lb/>plus grande à proportion des poids, au lieu que juſqu’ici on <lb/>n’a pu parvenir à réduire une certaine quantité d’eau à un <lb/>moindre volume; </s> <s xml:id="echoid-s18945" xml:space="preserve">ce qui ſeroit pourtant poſſible ſi l’eau avoit <lb/>une force de reſſort.</s> <s xml:id="echoid-s18946" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1565" type="section" level="1" n="1131"> <head xml:id="echoid-head1342" xml:space="preserve"><emph style="sc">Remarque</emph> II.</head> <p> <s xml:id="echoid-s18947" xml:space="preserve">1123. </s> <s xml:id="echoid-s18948" xml:space="preserve">La plûpart des Auteurs qui ont écrit ſur la nature <lb/>des fluides font conſiſter l’eſſence de la fluidité dans un mou-<lb/>vement continuel & </s> <s xml:id="echoid-s18949" xml:space="preserve">réciproque de toutes les parties du fluide <lb/>dans toutes les directions poſſibles. </s> <s xml:id="echoid-s18950" xml:space="preserve">Ce mouvement leur paroît <lb/>néceſſaire pour expliquer la diſſolution de certains corps plon-<lb/>gés dans un fluide, dont toutes les parties ſont enſuite em-<lb/>prégnées du corps qui a été mis en diſſolution. </s> <s xml:id="echoid-s18951" xml:space="preserve">Je ne fçais pas <lb/>ſi ce mouvement continuel ne peut pas être regardé comme <pb o="603" file="0693" n="719" rhead="DE MATHÉMATIQUE. Liv. XVI."/> une ſuppoſition de convenance, même en admettant la ma-<lb/>tiere ſubtile de M. </s> <s xml:id="echoid-s18952" xml:space="preserve">Deſcartes, qui les traverſe continuellement: <lb/></s> <s xml:id="echoid-s18953" xml:space="preserve">car il faudroit, ce me ſemble, avoir démontré que cette ma-<lb/>tiere ſubtile ne peut ainſi paſſer par les milieux des fluides ſans <lb/>les mettre en mouvement; </s> <s xml:id="echoid-s18954" xml:space="preserve">ce qui eſt préciſément l’état de la <lb/>queſtion. </s> <s xml:id="echoid-s18955" xml:space="preserve">D’ailleurs il me paroît que pour expliquer d’une <lb/>maniere auſſi ſatisfaiſante les mêmes effets, on n’a beſoin que <lb/>d’une tendance au mouvement, commune à toutes les parties <lb/>ſoumiſes au poids de l’atmoſphere. </s> <s xml:id="echoid-s18956" xml:space="preserve">Quant aux différentes diſ-<lb/>ſolutions, ne peuvent-elles pas s’expliquer auſſi par la différence <lb/>des parties de chaque fluide en particulier? </s> <s xml:id="echoid-s18957" xml:space="preserve">Au reſte ce même <lb/>ſyſtême, qui n’eſt pas nouveau, pourroit nous mener à des <lb/>diſcuſſions trop longues & </s> <s xml:id="echoid-s18958" xml:space="preserve">étrangeres à notre objet. </s> <s xml:id="echoid-s18959" xml:space="preserve">Il nous <lb/>ſuffit d’avoir expliqué ici ce que nous entendons par fluide, <lb/>ſans vouloir définir la nature de toutes les parties de chaque <lb/>fluide en particulier, ce qui a plus de rapport à la chymie qu’à <lb/>l’hydraulique ou l’hydroſtatique.</s> <s xml:id="echoid-s18960" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1566" type="section" level="1" n="1132"> <head xml:id="echoid-head1343" xml:space="preserve"><emph style="sc">Définition</emph> II.</head> <p> <s xml:id="echoid-s18961" xml:space="preserve">1124. </s> <s xml:id="echoid-s18962" xml:space="preserve">On appelle peſanteur ſpécifique de deux ou de pluſieurs <lb/>fluides ou corps en général, le poids de chacun de ces corps <lb/>meſurés ſous un même volume. </s> <s xml:id="echoid-s18963" xml:space="preserve">Ainſi ſi un corps peſe 3 livres <lb/>le pouce cube, & </s> <s xml:id="echoid-s18964" xml:space="preserve">un autre deux livres le pouce cube, les pe-<lb/>ſanteurs ſpécifiques de ces corps ſont comme 3 à 2.</s> <s xml:id="echoid-s18965" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18966" xml:space="preserve">Quand les volumes ſont inégaux, il faut pour connoître <lb/>les peſanteurs ſpécifiques les réduire à un même volume: </s> <s xml:id="echoid-s18967" xml:space="preserve">Par <lb/>exemple, ſi un corps peſe 12 livres ſous un volume de 3 pouces <lb/>cubes, & </s> <s xml:id="echoid-s18968" xml:space="preserve">un autre 16 livres ſous un volume de deux pouces; <lb/></s> <s xml:id="echoid-s18969" xml:space="preserve">pour avoir les peſanteurs ſpécifiques de ces mêmes corps, il <lb/>faudra chercher le poids d’un pouce cube de chacun; </s> <s xml:id="echoid-s18970" xml:space="preserve">le pre-<lb/>mier donnera 4 livres le pouce cube, & </s> <s xml:id="echoid-s18971" xml:space="preserve">le ſecond 8: </s> <s xml:id="echoid-s18972" xml:space="preserve">mais ces <lb/>nombres ſont ceux qui viennent en diviſant les poids par les <lb/>volumes. </s> <s xml:id="echoid-s18973" xml:space="preserve">On peut donc dire en général que les peſanteurs ſpé-<lb/>cifiques de pluſieurs corps ſont comme les poids diviſés par les <lb/>volumes, ou en raiſon compoſée de la raiſon directe des poids <lb/>& </s> <s xml:id="echoid-s18974" xml:space="preserve">de la raiſon inverſe des volumes; </s> <s xml:id="echoid-s18975" xml:space="preserve">ce qu’il eſt fort aiſé de <lb/>reconnoître: </s> <s xml:id="echoid-s18976" xml:space="preserve">car il eſt évident que plus un corps aura de poids <lb/>ſous un même volume, plus ſa peſanteur ſera grande, & </s> <s xml:id="echoid-s18977" xml:space="preserve">plus <lb/>il aura de volume pour le même poids, moins il aura de pe-<lb/>ſanteur ſpécifique.</s> <s xml:id="echoid-s18978" xml:space="preserve"/> </p> <pb o="604" file="0694" n="720" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1567" type="section" level="1" n="1133"> <head xml:id="echoid-head1344" xml:space="preserve">III.</head> <p> <s xml:id="echoid-s18979" xml:space="preserve">1125. </s> <s xml:id="echoid-s18980" xml:space="preserve">Le plus ou le moins de poids ſous un même volume <lb/>s’appelle denſité: </s> <s xml:id="echoid-s18981" xml:space="preserve">ainſi l’on peut dire en général que les denſités <lb/>ſont comme les peſanteurs ſpécifiques. </s> <s xml:id="echoid-s18982" xml:space="preserve">Pour épargner de longs <lb/>raiſonnemens ſur les rapports des denſités des corps ou fluides, <lb/>nous ferons le poids du premier corps P, ſon volume V & </s> <s xml:id="echoid-s18983" xml:space="preserve">ſa <lb/>denſité D; </s> <s xml:id="echoid-s18984" xml:space="preserve">pareillement nous ferons p le poids du ſecond corps; <lb/></s> <s xml:id="echoid-s18985" xml:space="preserve">v ſon volume, & </s> <s xml:id="echoid-s18986" xml:space="preserve">d ſa denſité: </s> <s xml:id="echoid-s18987" xml:space="preserve">on aura D : </s> <s xml:id="echoid-s18988" xml:space="preserve">d : </s> <s xml:id="echoid-s18989" xml:space="preserve">: </s> <s xml:id="echoid-s18990" xml:space="preserve">{P/V} : </s> <s xml:id="echoid-s18991" xml:space="preserve">{p/v} : </s> <s xml:id="echoid-s18992" xml:space="preserve">donc <lb/>D : </s> <s xml:id="echoid-s18993" xml:space="preserve">d : </s> <s xml:id="echoid-s18994" xml:space="preserve">: </s> <s xml:id="echoid-s18995" xml:space="preserve">Pv : </s> <s xml:id="echoid-s18996" xml:space="preserve">pV; </s> <s xml:id="echoid-s18997" xml:space="preserve">d’où l’on tire D p V = dPv: </s> <s xml:id="echoid-s18998" xml:space="preserve">donc ſi l’on <lb/>ſuppoſe que les denſités ſoient égales entr’elles, on aura pV=Pv. </s> <s xml:id="echoid-s18999" xml:space="preserve"><lb/>Donc p : </s> <s xml:id="echoid-s19000" xml:space="preserve">P : </s> <s xml:id="echoid-s19001" xml:space="preserve">: </s> <s xml:id="echoid-s19002" xml:space="preserve">v : </s> <s xml:id="echoid-s19003" xml:space="preserve">V, c’eſt-à-dire que les poids ſont proportionnels <lb/>aux volumes.</s> <s xml:id="echoid-s19004" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19005" xml:space="preserve">1126. </s> <s xml:id="echoid-s19006" xml:space="preserve">Si l’on ſuppoſe les poids égaux; </s> <s xml:id="echoid-s19007" xml:space="preserve">ou, ce qui eſt la même <lb/>choſe, ſi les maſſes ſont égales, on aura D V = dv : </s> <s xml:id="echoid-s19008" xml:space="preserve">donc <lb/>D : </s> <s xml:id="echoid-s19009" xml:space="preserve">d : </s> <s xml:id="echoid-s19010" xml:space="preserve">: </s> <s xml:id="echoid-s19011" xml:space="preserve">v : </s> <s xml:id="echoid-s19012" xml:space="preserve">V, c’eſt-à-dire que les denſités ſont dans la raiſon <lb/>inverſe des volumes, ou réciproquement les volumes dans la raiſon <lb/>inverſe des denſités. </s> <s xml:id="echoid-s19013" xml:space="preserve">On déduit encore de la formule DpV=dPv; <lb/></s> <s xml:id="echoid-s19014" xml:space="preserve">V : </s> <s xml:id="echoid-s19015" xml:space="preserve">v : </s> <s xml:id="echoid-s19016" xml:space="preserve">: </s> <s xml:id="echoid-s19017" xml:space="preserve">Pd : </s> <s xml:id="echoid-s19018" xml:space="preserve">p D, c’eſt-à-dire que les volumes de deux corps ſont <lb/>dans la raiſon compoſée de la directe des poids & </s> <s xml:id="echoid-s19019" xml:space="preserve">de l’inverſe des <lb/>denſités; </s> <s xml:id="echoid-s19020" xml:space="preserve">ce qui eſt bien évident, puiſque plus les poids ſeront <lb/>grands, plus il faudra de volume; </s> <s xml:id="echoid-s19021" xml:space="preserve">& </s> <s xml:id="echoid-s19022" xml:space="preserve">que plus les denſités ſe-<lb/>ront grandes, moins le volume ſera conſidérable.</s> <s xml:id="echoid-s19023" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19024" xml:space="preserve">1127. </s> <s xml:id="echoid-s19025" xml:space="preserve">On peut auſſi conclure de la même formule que <lb/>p : </s> <s xml:id="echoid-s19026" xml:space="preserve">P : </s> <s xml:id="echoid-s19027" xml:space="preserve">: </s> <s xml:id="echoid-s19028" xml:space="preserve">dv : </s> <s xml:id="echoid-s19029" xml:space="preserve">DV, c’eſt-à-dire que les poids ſont en raiſon com-<lb/>poſée des directes des denſités & </s> <s xml:id="echoid-s19030" xml:space="preserve">des volumes; </s> <s xml:id="echoid-s19031" xml:space="preserve">ce qui eſt encore <lb/>bien évident, puiſque les poids croiſſent à proportion des vo-<lb/>lumes & </s> <s xml:id="echoid-s19032" xml:space="preserve">de la maſſe compriſe ſous chaque volume. </s> <s xml:id="echoid-s19033" xml:space="preserve">On dé-<lb/>duiroit encore un grand nombre de proportions de cette éga-<lb/>lité; </s> <s xml:id="echoid-s19034" xml:space="preserve">mais il ſuffit de la connoître pour y avoir recours au <lb/>beſoin.</s> <s xml:id="echoid-s19035" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1568" type="section" level="1" n="1134"> <head xml:id="echoid-head1345" xml:space="preserve">IV.</head> <p> <s xml:id="echoid-s19036" xml:space="preserve">1128. </s> <s xml:id="echoid-s19037" xml:space="preserve">Les fluides peuvent être élaſtiques ou non élaſtiques. <lb/></s> <s xml:id="echoid-s19038" xml:space="preserve">Un fluide eſt élaſtique, lorſqu’on peut réduire la même maſſe <lb/>à un moindre volume par la compreſſion, & </s> <s xml:id="echoid-s19039" xml:space="preserve">que le corps rem-<lb/>plit toujours le même volume, après que la compreſſion a <lb/>ceſſée. </s> <s xml:id="echoid-s19040" xml:space="preserve">De tous les fluides, nous ne connoiſſons que l’air qui <lb/>ait cette propriété, au moins n’eſt-elle pas ſenſible dans les <lb/>autres.</s> <s xml:id="echoid-s19041" xml:space="preserve"/> </p> <pb o="605" file="0695" n="721" rhead="DE MATHEMATIQUE. Liv. XVI."/> </div> <div xml:id="echoid-div1569" type="section" level="1" n="1135"> <head xml:id="echoid-head1346" xml:space="preserve">V.</head> <p> <s xml:id="echoid-s19042" xml:space="preserve">1129. </s> <s xml:id="echoid-s19043" xml:space="preserve">On dit que la ſurface d’un fluide eſt de niveau, lorſ-<lb/>que tous les points de cette ſurface ſont à égale diſtance du <lb/>centre de la terre.</s> <s xml:id="echoid-s19044" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1570" type="section" level="1" n="1136"> <head xml:id="echoid-head1347" xml:space="preserve">PROPOSITION I.</head> <head xml:id="echoid-head1348" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19045" xml:space="preserve">1130. </s> <s xml:id="echoid-s19046" xml:space="preserve">Si on verſe une liqueur dans un vaſe, ſa ſurface ſera de <lb/>niveau, & </s> <s xml:id="echoid-s19047" xml:space="preserve">toutes ſes parties en équilibre.</s> <s xml:id="echoid-s19048" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1571" type="section" level="1" n="1137"> <head xml:id="echoid-head1349" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19049" xml:space="preserve">Si quelque partie du fluide étoit plus élevée que les autres, <lb/>comme d’ailleurs il n’y a rien qui l’empêche de gliſſer ſur les <lb/>autres, elle cédera néceſſairement à l’effort de ſa peſanteur <lb/>qui la ſollicite à deſcendre vers le centre de la terre; </s> <s xml:id="echoid-s19050" xml:space="preserve">d’où il <lb/>ſuit évidemment que la ſurface du fluide ſera de niveau, parce <lb/>que l’on feroit le même raiſonnement pour toutes les parties <lb/>de la ſurface du même fluide. </s> <s xml:id="echoid-s19051" xml:space="preserve">Donc 1°. </s> <s xml:id="echoid-s19052" xml:space="preserve">&</s> <s xml:id="echoid-s19053" xml:space="preserve">c. </s> <s xml:id="echoid-s19054" xml:space="preserve">2°. </s> <s xml:id="echoid-s19055" xml:space="preserve">Je dis que <lb/>toutes les parties ſont en équilibre. </s> <s xml:id="echoid-s19056" xml:space="preserve">Pour cela, concevons le <lb/>fluide partagé en une infinité de tranches verticales d’un même <lb/>diametre, & </s> <s xml:id="echoid-s19057" xml:space="preserve">faiſons attention que toutes ces colonnes ſe con-<lb/>trebalancent mutuellement, puiſque chacune doit ſoutenir le <lb/>poids de tout le fluide environnant: </s> <s xml:id="echoid-s19058" xml:space="preserve">car ſi l’on ſuppoſe que <lb/>l’une de ces colonnes fût plus foible que l’effort des autres qui <lb/>l’environnent, le poids de ces mêmes colonnes l’obligeroit <lb/>de s’élever pour céder à leur impreſſion, juſqu’à ce que toutes <lb/>les autres fuſſent réunies; </s> <s xml:id="echoid-s19059" xml:space="preserve">mais n’étant plus ſoutenue par ces <lb/>mêmes colonnes, elle ſe diſtribueroit uniformément ſur toute <lb/>la ſurface, en ajoutant des poids égaux à chaque colonne en <lb/>particulier, & </s> <s xml:id="echoid-s19060" xml:space="preserve">il y auroit alors équilibre; </s> <s xml:id="echoid-s19061" xml:space="preserve">mais comme il y a <lb/>toujours même maſſe de fluide, & </s> <s xml:id="echoid-s19062" xml:space="preserve">que d’ailleurs le vaſe n’a <lb/>pas changé de forme; </s> <s xml:id="echoid-s19063" xml:space="preserve">il s’enſuit que cette colonne eſt rem-<lb/>placée par une autre qui lui eſt parfaitement égale, & </s> <s xml:id="echoid-s19064" xml:space="preserve">qui fait <lb/>équilibre avec les autres: </s> <s xml:id="echoid-s19065" xml:space="preserve">donc elle-même étoit auſſi en équi-<lb/>libre avec les colonnes environnantes. </s> <s xml:id="echoid-s19066" xml:space="preserve">Et comme on démon-<lb/>trera la même choſe de toutes les colonnes collatérales, il <lb/>s’enſuit que toutes les parties ſont en équilibre.</s> <s xml:id="echoid-s19067" xml:space="preserve"/> </p> <pb o="606" file="0696" n="722" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1572" type="section" level="1" n="1138"> <head xml:id="echoid-head1350" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19068" xml:space="preserve">1131. </s> <s xml:id="echoid-s19069" xml:space="preserve">Il ſuit delà que quelle que ſoit la figure du vaſe qui <lb/> <anchor type="note" xlink:label="note-0696-01a" xlink:href="note-0696-01"/> contient une liqueur, ſa ſurface ſera toujours de niveau, & </s> <s xml:id="echoid-s19070" xml:space="preserve"><lb/>toutes ſes parties en équilibre. </s> <s xml:id="echoid-s19071" xml:space="preserve">De plus, comme l’effort de <lb/>toutes les colonnes verticales eſt égal, il s’enſuit que la preſſion <lb/>de toutes ces colonnes ſur le fond du vaſe eſt égale au pro-<lb/>duit de la même baſe, par la hauteur de la plus grande colonne <lb/>verticale. </s> <s xml:id="echoid-s19072" xml:space="preserve">Pour s’en convaincre, imaginons un vaſe compoſé <lb/>de deux cylindres A B C D, E F G H unis enſemble, & </s> <s xml:id="echoid-s19073" xml:space="preserve">que <lb/>l’on a rempli d’eau juſqu’à la hauteur G H; </s> <s xml:id="echoid-s19074" xml:space="preserve">il eſt évident que <lb/>toutes les colonnes, comme L M qui répondent aux côtés A E, <lb/>F D, ſont dans un effort continuel contre les mêmes côtés <lb/>pour s’élever juſqu’au niveau G H de la liqueur: </s> <s xml:id="echoid-s19075" xml:space="preserve">car la colonne <lb/>I K étant plus grande que L M, fait effort contre cette liqueur <lb/>qui cherche à s’échapper par le côté F D; </s> <s xml:id="echoid-s19076" xml:space="preserve">& </s> <s xml:id="echoid-s19077" xml:space="preserve">cet effort eſt égal <lb/>à celui que feroit la colonne I N ſur la baſe du cylindre <lb/>E G H F, s’il étoit ſéparé de l’autre A B C D.</s> <s xml:id="echoid-s19078" xml:space="preserve"/> </p> <div xml:id="echoid-div1572" type="float" level="2" n="1"> <note position="left" xlink:label="note-0696-01" xlink:href="note-0696-01a" xml:space="preserve">Figure 411.</note> </div> </div> <div xml:id="echoid-div1574" type="section" level="1" n="1139"> <head xml:id="echoid-head1351" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19079" xml:space="preserve">1132. </s> <s xml:id="echoid-s19080" xml:space="preserve">De même ſi l’on a un vaſe de figure conique, & </s> <s xml:id="echoid-s19081" xml:space="preserve">dont <lb/> <anchor type="note" xlink:label="note-0696-02a" xlink:href="note-0696-02"/> les parois ſoient inclinés à l’horizon, comme les lignes B E, <lb/>C F, & </s> <s xml:id="echoid-s19082" xml:space="preserve">qu’on rempliſſe ce vaſe de liqueur, la preſſion du fluide <lb/>ſur la baſe E F ſera égale à celle du poids d’un fluide de même <lb/>peſanteur ſpécifique qui auroit même baſe, & </s> <s xml:id="echoid-s19083" xml:space="preserve">dont le volume <lb/>ſeroit égal au ſolide fait ſur cette baſe, & </s> <s xml:id="echoid-s19084" xml:space="preserve">la hauteur E Q: <lb/></s> <s xml:id="echoid-s19085" xml:space="preserve">car dans le vaſe E B A D C F, il y a autant de colonnes qu’il y <lb/>a de points dans la baſe; </s> <s xml:id="echoid-s19086" xml:space="preserve">de plus, chaque colonne preſſe cette <lb/>baſe avec une force égale à celle de la colonne G H: </s> <s xml:id="echoid-s19087" xml:space="preserve">donc la <lb/>ſomme des preſſions ſur la baſe eſt égale au produit de la même <lb/>baſe par la hauteur G H.</s> <s xml:id="echoid-s19088" xml:space="preserve"/> </p> <div xml:id="echoid-div1574" type="float" level="2" n="1"> <note position="left" xlink:label="note-0696-02" xlink:href="note-0696-02a" xml:space="preserve">Figure 412.</note> </div> <p> <s xml:id="echoid-s19089" xml:space="preserve">1133. </s> <s xml:id="echoid-s19090" xml:space="preserve">L’expérience a fait voir auſſi que telle direction qu’on <lb/>puiſſe donner à l’eau que l’on fait ſortir d’un vaſe par des trous <lb/>pratiqués ſur ſes côtés, la force eſt toujours la même pour des <lb/>trous horizontaux & </s> <s xml:id="echoid-s19091" xml:space="preserve">verticaux, pourvu que la hauteur du ni-<lb/>veau de l’eau au deſſus de ces trous ſoit égale.</s> <s xml:id="echoid-s19092" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1576" type="section" level="1" n="1140"> <head xml:id="echoid-head1352" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19093" xml:space="preserve">1134. </s> <s xml:id="echoid-s19094" xml:space="preserve">Il ſuit encore delà que l’on peut multiplier conſidé-<lb/> <anchor type="note" xlink:label="note-0696-03a" xlink:href="note-0696-03"/> rablement les forces par le moyen des fluides. </s> <s xml:id="echoid-s19095" xml:space="preserve">Suppoſons, par <pb o="607" file="0697" n="723" rhead="DE MATHÉMATIQUE. Liv. XVI."/> exemple, que par le moyen d’un tuyau I N, & </s> <s xml:id="echoid-s19096" xml:space="preserve">d’un piſton <lb/>placé en I, on preſſe la ſurface de l’eau avec une force de 10 <lb/>livres: </s> <s xml:id="echoid-s19097" xml:space="preserve">je dis que cette preſſion pourra faire équilibre avec un <lb/>poids de 100 livres, placé ſur un trou R, dont le diametre ſe-<lb/>roit dix fois plus grand que le trou N: </s> <s xml:id="echoid-s19098" xml:space="preserve">car puiſque ce trou eſt <lb/>dix fois plus grand, il y a dix fois plus de filets d’eau qui font <lb/>effort contre le poids: </s> <s xml:id="echoid-s19099" xml:space="preserve">de plus, chacun de ces filets étant égal <lb/>au nombre de filets qui ſont en N à une force de dix livres; <lb/></s> <s xml:id="echoid-s19100" xml:space="preserve">donc tous les filets enſemble font équilibre avec 100 livres.</s> <s xml:id="echoid-s19101" xml:space="preserve"/> </p> <div xml:id="echoid-div1576" type="float" level="2" n="1"> <note position="left" xlink:label="note-0696-03" xlink:href="note-0696-03a" xml:space="preserve">Figure 411.</note> </div> </div> <div xml:id="echoid-div1578" type="section" level="1" n="1141"> <head xml:id="echoid-head1353" xml:space="preserve"><emph style="sc">Corollaire</emph>. IV.</head> <p> <s xml:id="echoid-s19102" xml:space="preserve">1135. </s> <s xml:id="echoid-s19103" xml:space="preserve">Si la ſurface A D du vaſe cylindrique eſt cent fois <lb/>plus grande que l’ouverture du trou que je ſuppoſe en N, & </s> <s xml:id="echoid-s19104" xml:space="preserve"><lb/>qui eſt preſſé par un poids de dix livres, la ſurface de l’eau fera <lb/>un effort de mille livres pour écarter cette ſurface des parois <lb/>A B C D. </s> <s xml:id="echoid-s19105" xml:space="preserve">C’eſt par cette propriété, commune à tous les fluides, <lb/>que l’on rendra raiſon de certains effets qui paroiſſent très-ſur-<lb/>prenans, & </s> <s xml:id="echoid-s19106" xml:space="preserve">qui pourroient en impoſer à tout autre qu’à des <lb/>perſonnes inſtruites de ce que nous venons de voir. </s> <s xml:id="echoid-s19107" xml:space="preserve">Le ſouffle <lb/>d’un enfant ſuffit pour enlever des poids conſidérables, par <lb/>le moyen d’une ou de pluſieurs veſſies ſur leſquelles ces poids <lb/>ſont placés, & </s> <s xml:id="echoid-s19108" xml:space="preserve">dans leſquelles on introduit l’air par le moyen <lb/>d’un petit chalumeau. </s> <s xml:id="echoid-s19109" xml:space="preserve">Plus le diametre eſt petit, plus il a de <lb/>facilité à enlever les poids. </s> <s xml:id="echoid-s19110" xml:space="preserve">Tout ceci peut encore ſe démon-<lb/>trer par le principe des vîteſſes.</s> <s xml:id="echoid-s19111" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1579" type="section" level="1" n="1142"> <head xml:id="echoid-head1354" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s19112" xml:space="preserve">1136. </s> <s xml:id="echoid-s19113" xml:space="preserve">Tout ce que nous venons de voir eſt de la derniere <lb/>importance dans l’hydroſtatique: </s> <s xml:id="echoid-s19114" xml:space="preserve">auſſi eſt-il de la plus grande <lb/>conſéquence de bien ſaiſir le vrai de cette même propoſition, <lb/>que l’on exprime ordinairement ainſi: </s> <s xml:id="echoid-s19115" xml:space="preserve">Les preſſions des fluides <lb/>ſur les baſes des vaiſſeaux qui les contiennent ſont en raiſon des <lb/>baſes multipliées par les hauteurs. </s> <s xml:id="echoid-s19116" xml:space="preserve">On pourroit objecter à cela, <lb/>qu’il s’enſuivroit delà que ſi l’on a un vaſe conique, & </s> <s xml:id="echoid-s19117" xml:space="preserve">un vaſe <lb/>cylindrique de même baſe & </s> <s xml:id="echoid-s19118" xml:space="preserve">de même hauteur que le premier, <lb/>l’un & </s> <s xml:id="echoid-s19119" xml:space="preserve">l’autre remplis de la même liqueur, le poids de l’un doit <lb/>être égal au poids de l’autre, puiſqu’il ſemble que la preſſion <lb/>occaſionne le poids. </s> <s xml:id="echoid-s19120" xml:space="preserve">Mais on va voir que quoique les preſſions <lb/>contre les baſes ſoient égales, il ne s’enſuit pas que les poids <lb/>abſolus doivent changer. </s> <s xml:id="echoid-s19121" xml:space="preserve">Pour s’en convaincre, il n’y a qu’à <pb o="608" file="0698" n="724" rhead="NOUVEAU COURS"/> faire attention que quoique dans le tambour d’une montre la <lb/>force du reſſort qui bande la chaîne ſoit très-conſidérable, on <lb/>ne ſent pourtant rien de cet effort, qui eſt détruit par la réſiſ-<lb/>tance de la chaîne. </s> <s xml:id="echoid-s19122" xml:space="preserve">Il en eſt de même de chaque filet, quoiqu’il <lb/>faſſe un effort conſidérable contre la baſe inférieure du vaſe: <lb/></s> <s xml:id="echoid-s19123" xml:space="preserve">comme cet effort eſt détruit par la réſiſtance des parois ſupé-<lb/>rieurs, on ne doit porter que le poids de la ſomme des filets, <lb/>c’eſt-à-dire le poids du volume de fluide contenu dans le vaſe. </s> <s xml:id="echoid-s19124" xml:space="preserve"><lb/>Auſſi ſi l’on détruit cette réſiſtance réciproque des parois du <lb/>vaſe, en pratiquant un fond mobile, alors l’expérience eſt <lb/>d’accord avec la théorie, & </s> <s xml:id="echoid-s19125" xml:space="preserve">nous fait voir qu’il faut une force <lb/>égale à celle d’un ſolide qui auroit une baſe égale à celle du <lb/>vaſe, & </s> <s xml:id="echoid-s19126" xml:space="preserve">une hauteur égale à celle de la plus haute colonne. </s> <s xml:id="echoid-s19127" xml:space="preserve"><lb/>Voyez le premier volume de notre Architecture Hydraulique, <lb/>art. </s> <s xml:id="echoid-s19128" xml:space="preserve">352, page 141.</s> <s xml:id="echoid-s19129" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1580" type="section" level="1" n="1143"> <head xml:id="echoid-head1355" xml:space="preserve">PROPOSITION II.</head> <head xml:id="echoid-head1356" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19130" xml:space="preserve">1137. </s> <s xml:id="echoid-s19131" xml:space="preserve">Si l’on verſe une liqueur, par exemple, de l’eau dans un <lb/>tuyau recourbé ou ſiphon, je dis que la ſurface de cette liqueur ſe <lb/>mettra de niveau dans les deux branches du ſiphon.</s> <s xml:id="echoid-s19132" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1581" type="section" level="1" n="1144"> <head xml:id="echoid-head1357" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19133" xml:space="preserve">1°. </s> <s xml:id="echoid-s19134" xml:space="preserve">Si les deux branches du ſiphon ſont d’égale groſſeur, il <lb/> <anchor type="note" xlink:label="note-0698-01a" xlink:href="note-0698-01"/> eſt aiſé de prouver que la ſurface de la liqueur dans chaque <lb/>tuyau ſe trouvera renfermée dans une ligne droite horizontale <lb/>A B; </s> <s xml:id="echoid-s19135" xml:space="preserve">puiſque les colonnes de la liqueur contenues dans chaque <lb/>tuyau, ſe trouveront dans le même cas que ſi elles étoient <lb/>compriſes dans un vaſe, c’eſt-à-dire de ſe contre-balancer éga-<lb/>lement, ſans faire plus d’effort l’une que l’autre pour baiſſer ou <lb/>hauſſer: </s> <s xml:id="echoid-s19136" xml:space="preserve">car les côtés L M & </s> <s xml:id="echoid-s19137" xml:space="preserve">N O du tuyau font le même effet <lb/>pour contenir la liqueur, que le feroient les colonnes L M P Q <lb/>& </s> <s xml:id="echoid-s19138" xml:space="preserve">R N Q O, ſi les deux colonnes L H & </s> <s xml:id="echoid-s19139" xml:space="preserve">N K étoient, auſſi-<lb/>bien que les précédentes, renfermées dans un ſeul vaſe A H B K; <lb/></s> <s xml:id="echoid-s19140" xml:space="preserve">mais ſelon cette ſuppoſition, les colonnes L H & </s> <s xml:id="echoid-s19141" xml:space="preserve">N K ſeroient <lb/>en équilibre (art. </s> <s xml:id="echoid-s19142" xml:space="preserve">1130), & </s> <s xml:id="echoid-s19143" xml:space="preserve">auroient leur ſurface de niveau: </s> <s xml:id="echoid-s19144" xml:space="preserve"><lb/>par conſéquent ſi l’on ſupprime toutes les colonnes d’eau qui <lb/>ſeroient entre ces deux-ci, & </s> <s xml:id="echoid-s19145" xml:space="preserve">qu’à la place l’on ſubſtitue les <lb/>côtés L M & </s> <s xml:id="echoid-s19146" xml:space="preserve">N O du ſiphon, l’eau reſtera de niveau dans les <lb/>deux tuyaux. </s> <s xml:id="echoid-s19147" xml:space="preserve">C. </s> <s xml:id="echoid-s19148" xml:space="preserve">Q. </s> <s xml:id="echoid-s19149" xml:space="preserve">F. </s> <s xml:id="echoid-s19150" xml:space="preserve">D.</s> <s xml:id="echoid-s19151" xml:space="preserve"/> </p> <div xml:id="echoid-div1581" type="float" level="2" n="1"> <note position="left" xlink:label="note-0698-01" xlink:href="note-0698-01a" xml:space="preserve">Figure 413.</note> </div> <pb o="609" file="0699" n="725" rhead="DE MATHÉMATIQUE. Liv. XVI."/> </div> <div xml:id="echoid-div1583" type="section" level="1" n="1145"> <head xml:id="echoid-head1358" xml:space="preserve"><emph style="sc">Autre demonstration</emph>.</head> <p> <s xml:id="echoid-s19152" xml:space="preserve">Pour démontrer ceci par les vîteſſes, ſuppoſons que la ſur-<lb/>face A L ſoit deſcendue de A en C, par exemple, de 4 pouces: <lb/></s> <s xml:id="echoid-s19153" xml:space="preserve">cela étant, la ſurface N B ſera montée de N en E auſſi de <lb/>4 pouces, puiſque les deux tuyaux ſont d’égale groſſeur: </s> <s xml:id="echoid-s19154" xml:space="preserve">ainſi <lb/>la quantité de mouvement du fluide dans le premier tuyau eſt <lb/>égale à la quantité du mouvement du fluide dans le ſecond <lb/>tuyau: </s> <s xml:id="echoid-s19155" xml:space="preserve">par conſéquent ils ſont en équilibre, & </s> <s xml:id="echoid-s19156" xml:space="preserve">leurs ſurfaces <lb/>ſont de niveau. </s> <s xml:id="echoid-s19157" xml:space="preserve">C. </s> <s xml:id="echoid-s19158" xml:space="preserve">Q. </s> <s xml:id="echoid-s19159" xml:space="preserve">F. </s> <s xml:id="echoid-s19160" xml:space="preserve">D.</s> <s xml:id="echoid-s19161" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1584" type="section" level="1" n="1146"> <head xml:id="echoid-head1359" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19162" xml:space="preserve">1138. </s> <s xml:id="echoid-s19163" xml:space="preserve">Il ſuit delà que ſi l’on a un ſiphon, dont la groſſeur <lb/> <anchor type="note" xlink:label="note-0699-01a" xlink:href="note-0699-01"/> des branches ſoit inégale, la liqueur qui ſera verſée dans le <lb/>ſiphon ſe mettra encore de niveau dans les deux branches: </s> <s xml:id="echoid-s19164" xml:space="preserve">car <lb/>ſi, par exemple, la branche I K eſt trois fois plus groſſe que la <lb/>branche G H, il y aura trois fois plus de liqueur dans la groſſe <lb/>branche que dans la petite. </s> <s xml:id="echoid-s19165" xml:space="preserve">Or ſi l’on imagine que l’eau de cette <lb/>branche ſoit partagée en trois colonnes égales, il y en aura <lb/>une, comme, par exemple, O L P M, qui ſera en équilibre <lb/>avec celle du petit tuyau, puiſqu’on ſuppoſe qu’elles ont des <lb/>baſes égales. </s> <s xml:id="echoid-s19166" xml:space="preserve">Or étant en équilibre, leurs ſurfaces ſeront de <lb/>niveau; </s> <s xml:id="echoid-s19167" xml:space="preserve">mais la colonne O L P M eſt en équilibre avec la co-<lb/>lonne N L M F ou N F B K, & </s> <s xml:id="echoid-s19168" xml:space="preserve">par conſéquent de niveau en-<lb/>tr’elles: </s> <s xml:id="echoid-s19169" xml:space="preserve">elles ſeront donc auſſi de niveau avec la colonne du <lb/>petit tuyau.</s> <s xml:id="echoid-s19170" xml:space="preserve"/> </p> <div xml:id="echoid-div1584" type="float" level="2" n="1"> <note position="right" xlink:label="note-0699-01" xlink:href="note-0699-01a" xml:space="preserve">Figure 414.</note> </div> <p> <s xml:id="echoid-s19171" xml:space="preserve">Pour prouver ceci par les vîteſſes, conſidérez que ſi la ſur-<lb/>face de l’eau du petit tuyau eſt deſcendue de A en C de 3 pouces, <lb/>par exemple, elle ſera montée de B en E d’un pouce dans le <lb/>grand tuyau, puiſque la baſe du grand tuyau eſt triple de celle <lb/>du petit: </s> <s xml:id="echoid-s19172" xml:space="preserve">ainſi les vîteſſes ſeront réciproques à leurs maſſes, <lb/>& </s> <s xml:id="echoid-s19173" xml:space="preserve">par conſéquent l’eau ſera en équilibre de part & </s> <s xml:id="echoid-s19174" xml:space="preserve">d’autre, & </s> <s xml:id="echoid-s19175" xml:space="preserve"><lb/>les ſurfaces de niveau.</s> <s xml:id="echoid-s19176" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1586" type="section" level="1" n="1147"> <head xml:id="echoid-head1360" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19177" xml:space="preserve">1139. </s> <s xml:id="echoid-s19178" xml:space="preserve">Mais ſi le tuyau avoit une branche perpendiculaire à <lb/>l’horizon, & </s> <s xml:id="echoid-s19179" xml:space="preserve">l’autre inclinée comme dans le ſiphon A B C, la <lb/> <anchor type="note" xlink:label="note-0699-02a" xlink:href="note-0699-02"/> liqueur que l’on verſera dans l’un des tuyaux, ſe mettra en-<lb/>core de niveau dans l’autre: </s> <s xml:id="echoid-s19180" xml:space="preserve">car ſi les deux branches de ce <lb/>ſiphon ſont d’égale groſſeur, & </s> <s xml:id="echoid-s19181" xml:space="preserve">que la ligne E G paſſe par la <pb o="610" file="0700" n="726" rhead="NOUVEAU COURS"/> ſurface de la liqueur dans chaque tuyau, l’eau de la branche <lb/>perpendiculaire ſera à celle de la branche oblique, comme E B <lb/>eſt à B G; </s> <s xml:id="echoid-s19182" xml:space="preserve">mais l’eau de la branche inclinée n’agit pas ſur la <lb/>baſe B avec toute ſa peſanteur abſolue; </s> <s xml:id="echoid-s19183" xml:space="preserve">& </s> <s xml:id="echoid-s19184" xml:space="preserve">conſidérant que <lb/>cette liqueur eſt appuyée ſur un plan incliné, l’on pourra dire <lb/>que la peſanteur relative de la liqueur eſt à ſa peſanteur ab-<lb/>ſolue, comme la hauteur G D du plan incliné eſt à ſa lon-<lb/>gueur G B; </s> <s xml:id="echoid-s19185" xml:space="preserve">& </s> <s xml:id="echoid-s19186" xml:space="preserve">comme nous avons vu que les liqueurs de chaque <lb/>tuyau étoient comme E B eſt à B G, il s’enſuit que les hauteurs <lb/>E B & </s> <s xml:id="echoid-s19187" xml:space="preserve">G D étant égales, l’eau du ſiphon eſt en équilibre, & </s> <s xml:id="echoid-s19188" xml:space="preserve"><lb/>que par conſéquent elle eſt de niveau; </s> <s xml:id="echoid-s19189" xml:space="preserve">ce que l’on démontrera <lb/>encore, quand même les branches du ſiphon ſeroient d’inégale <lb/>groſſeur.</s> <s xml:id="echoid-s19190" xml:space="preserve"/> </p> <div xml:id="echoid-div1586" type="float" level="2" n="1"> <note position="right" xlink:label="note-0699-02" xlink:href="note-0699-02a" xml:space="preserve">Figure 415.</note> </div> </div> <div xml:id="echoid-div1588" type="section" level="1" n="1148"> <head xml:id="echoid-head1361" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19191" xml:space="preserve">1140. </s> <s xml:id="echoid-s19192" xml:space="preserve">Il ſuit encore delà que l’eau qui eſt dans le canal <lb/> <anchor type="note" xlink:label="note-0700-01a" xlink:href="note-0700-01"/> H S T P fait autant d’effort contre les côtés du même canal <lb/>pour s’échapper, que l’eau de chaque tuyau en fait ſur la baſe <lb/>T V, qui ſeroit celle du cylindre, parce que l’eau des petites <lb/>colonnes Q T R P tend à ſe mettre de niveau avec la ſurface <lb/>de la liqueur de chaque branche; </s> <s xml:id="echoid-s19193" xml:space="preserve">auſſi l’expérience montre-<lb/>t’elle que ſi l’on fait un petit trou vertical au canal d’un ſiphon, <lb/>elle monte preſqu’à la hauteur de l’eau des branches.</s> <s xml:id="echoid-s19194" xml:space="preserve"/> </p> <div xml:id="echoid-div1588" type="float" level="2" n="1"> <note position="left" xlink:label="note-0700-01" xlink:href="note-0700-01a" xml:space="preserve">Figure 414.</note> </div> </div> <div xml:id="echoid-div1590" type="section" level="1" n="1149"> <head xml:id="echoid-head1362" xml:space="preserve">PROPOSITION III.</head> <head xml:id="echoid-head1363" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19195" xml:space="preserve">1141. </s> <s xml:id="echoid-s19196" xml:space="preserve">Si l’on met dans les deux branches d’un ſiphon des li-<lb/>queurs de différentes peſanteurs, je dis que les hauteurs de ces li-<lb/>queurs dans les tuyaux, ſeront entr’elles dans la raiſon réciproque <lb/>de leur peſanteur ſpécifique.</s> <s xml:id="echoid-s19197" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1591" type="section" level="1" n="1150"> <head xml:id="echoid-head1364" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19198" xml:space="preserve">Si l’on verſe du mercure dans le ſiphon A B C H, il ſe mettra <lb/> <anchor type="note" xlink:label="note-0700-02a" xlink:href="note-0700-02"/> de niveau dans les deux branches, comme toutes les autres li-<lb/>queurs. </s> <s xml:id="echoid-s19199" xml:space="preserve">Or ſi l’on ſuppoſe que la ligne horizontale D E marque <lb/>le niveau du mercure, & </s> <s xml:id="echoid-s19200" xml:space="preserve">qu’enſuite l’on verſe de l’eau dans la <lb/>branche A B juſqu’à la hauteur G, il eſt évident que le mer-<lb/>cure de cette branche ceſſera d’être de niveau avec celui de <lb/>l’autre branche, auſſi-tôt qu’on y aura verſé de l’eau, & </s> <s xml:id="echoid-s19201" xml:space="preserve">que <lb/>s’il eſt deſcendu de D en I de 2 pouces dans la premiere bran- <pb o="611" file="0701" n="727" rhead="DE MATHÉMATIQUE. Liv. XVI."/> che, il ſera monté de E en F auſſi de 2 pouces dans la ſeconde. <lb/></s> <s xml:id="echoid-s19202" xml:space="preserve">Préſentement ſi l’on tire la ligne horizontale I L, l’on voit <lb/>évidemment que le mercure I B de la premiere branche eſt en <lb/>équilibre avec le mercure L C de la ſeconde. </s> <s xml:id="echoid-s19203" xml:space="preserve">Or ſi l’eau ſe <lb/>maintient en repos à la hauteur G, & </s> <s xml:id="echoid-s19204" xml:space="preserve">le mercure à la hauteur <lb/>F, il s’enſuit que l’eau G I eſt en équilibre avec le mercure F L, <lb/>ſi les branches du ſiphon ſont d’égale groſſeur, & </s> <s xml:id="echoid-s19205" xml:space="preserve">que d’au-<lb/>tant la colonne G I eſt plus haute que F L, d’autant la peſan-<lb/>teur ſpécifique du mercure eſt plus grande que celle de l’eau, <lb/>& </s> <s xml:id="echoid-s19206" xml:space="preserve">que par conſéquent la peſanteur ſpécifique de ces deux li-<lb/>queurs eſt en raiſon réciproque de leurs hauteurs.</s> <s xml:id="echoid-s19207" xml:space="preserve"/> </p> <div xml:id="echoid-div1591" type="float" level="2" n="1"> <note position="left" xlink:label="note-0700-02" xlink:href="note-0700-02a" xml:space="preserve">Figure 416.</note> </div> </div> <div xml:id="echoid-div1593" type="section" level="1" n="1151"> <head xml:id="echoid-head1365" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s19208" xml:space="preserve">1142. </s> <s xml:id="echoid-s19209" xml:space="preserve">Il ſuit delà que ſi une des branches A B du ſiphon étoit <lb/> <anchor type="note" xlink:label="note-0701-01a" xlink:href="note-0701-01"/> plus groſſe que l’autre D C, le mercure qui ſeroit dans la groſſe <lb/>branche, ſera encore en équilibre avec l’eau de la petite, ſi <lb/>après avoir tiré l’horizontale F G, la hauteur E F du mercure <lb/>eſt à la hauteur H K de l’eau dans la raiſon réciproque de la <lb/>peſanteur ſpécifique de ces deux liqueurs: </s> <s xml:id="echoid-s19210" xml:space="preserve">car ſi l’on imagine <lb/>une colonne L F de mercure, dont la baſe ſoit égale à celle du <lb/>tuyau D C, cette colonne ſera en équilibre avec la colonne <lb/>d’eau H K. </s> <s xml:id="echoid-s19211" xml:space="preserve">Or ſi le tuyau A B eſt cinq fois plus gros que D C, <lb/>la quantité de mercure E I contiendra cinq colonnes, comme <lb/>L F, qui ſeront toutes en équilibre entr’elles, auſſi-bien qu’avec <lb/>la colonne H K: </s> <s xml:id="echoid-s19212" xml:space="preserve">ainſi il en ſera de la propoſition précédente <lb/>pour l’équilibre des liqueurs différentes dans des tuyaux d’iné-<lb/>gale groſſeur, la même choſe que dans l’article 1137, ſoit que <lb/>la liqueur la plus peſante ſe trouve dans le gros tuyau, ou dans <lb/>le petit.</s> <s xml:id="echoid-s19213" xml:space="preserve"/> </p> <div xml:id="echoid-div1593" type="float" level="2" n="1"> <note position="right" xlink:label="note-0701-01" xlink:href="note-0701-01a" xml:space="preserve">Figure 417.</note> </div> </div> <div xml:id="echoid-div1595" type="section" level="1" n="1152"> <head xml:id="echoid-head1366" xml:space="preserve">PROPOSITION IV.</head> <head xml:id="echoid-head1367" xml:space="preserve"><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19214" xml:space="preserve">1143. </s> <s xml:id="echoid-s19215" xml:space="preserve">1°. </s> <s xml:id="echoid-s19216" xml:space="preserve">Si un corps dur eſt mis dans un fluide de même pe-<lb/> <anchor type="note" xlink:label="note-0701-02a" xlink:href="note-0701-02"/> ſanteur ſpécifique, il y demeurera entiérement plongé, à quelque <lb/>hauteur qu’il ſe trouve.</s> <s xml:id="echoid-s19217" xml:space="preserve"/> </p> <div xml:id="echoid-div1595" type="float" level="2" n="1"> <note position="right" xlink:label="note-0701-02" xlink:href="note-0701-02a" xml:space="preserve">Figure 418</note> </div> <p style="it"> <s xml:id="echoid-s19218" xml:space="preserve">2°. </s> <s xml:id="echoid-s19219" xml:space="preserve">S’il eſt d’une peſanteur ſpécifique plus grande que celle du <lb/>fluide, il ira au fond du vaiſſeau.</s> <s xml:id="echoid-s19220" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s19221" xml:space="preserve">3°. </s> <s xml:id="echoid-s19222" xml:space="preserve">S’il eſt d’une peſanteur ſpécifique moindre que celle du fluide, <lb/>il n’y aura qu’une partie du corps qui s’enfoncera, & </s> <s xml:id="echoid-s19223" xml:space="preserve">l’autre partie <lb/>reſtera au deſſus de la ſurface du fluide.</s> <s xml:id="echoid-s19224" xml:space="preserve"/> </p> <pb o="612" file="0702" n="728" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1597" type="section" level="1" n="1153"> <head xml:id="echoid-head1368" xml:space="preserve"><emph style="sc">Démonstration du premier cas</emph>.</head> <p> <s xml:id="echoid-s19225" xml:space="preserve">Si l’on a un vaſe A B C D, rempli de telle liqueur que l’on <lb/>voudra, par exemple, de l’eau, & </s> <s xml:id="echoid-s19226" xml:space="preserve">qu’on y plonge un corps E, <lb/>dont la peſanteur ſoit égale à celle du volume d’eau, dont il <lb/>occupe la place, il eſt conſtant que ce corps demeurera en <lb/>équilibre, c’eſt-à-dire en repos, ſans monter ni deſcendre, <lb/>quelque ſituation qu’on lui donne: </s> <s xml:id="echoid-s19227" xml:space="preserve">car il a autant de force que <lb/>le volume d’eau qui ſeroit à ſa place, pour tendre au centre de <lb/>la terre: </s> <s xml:id="echoid-s19228" xml:space="preserve">mais les parties de l’eau ſont en équilibre avec toutes <lb/>celles de la même eau qui les environne: </s> <s xml:id="echoid-s19229" xml:space="preserve">ainſi le corps E te-<lb/>nant lieu d’une certaine quantité d’eau, dont il occupe la place, <lb/>ſera en équilibre avec toute celle du vaiſſeau, & </s> <s xml:id="echoid-s19230" xml:space="preserve">demeu-<lb/>rera entiérement plongé & </s> <s xml:id="echoid-s19231" xml:space="preserve">en repos, à quelque hauteur qu’on <lb/>le mette. </s> <s xml:id="echoid-s19232" xml:space="preserve">C. </s> <s xml:id="echoid-s19233" xml:space="preserve">Q. </s> <s xml:id="echoid-s19234" xml:space="preserve">F. </s> <s xml:id="echoid-s19235" xml:space="preserve">D.</s> <s xml:id="echoid-s19236" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1598" type="section" level="1" n="1154"> <head xml:id="echoid-head1369" xml:space="preserve"><emph style="sc">Démonstration du second cas</emph>.</head> <p> <s xml:id="echoid-s19237" xml:space="preserve">Si le corps F plongé dans le même vaſe, eſt plus peſant que <lb/>le volume d’eau, dont il occupe la place, il eſt aiſé de conce-<lb/>voir qu’il deſcendra au fond de l’eau: </s> <s xml:id="echoid-s19238" xml:space="preserve">car il tendra avec plus <lb/>de force au centre de la terre qu’un pareil volume d’eau: </s> <s xml:id="echoid-s19239" xml:space="preserve">ainſi <lb/>il ne ſera plus en équilibre avec les autres parties de l’eau dont <lb/>il eſt environné, & </s> <s xml:id="echoid-s19240" xml:space="preserve">ira par conſéquent au fond du vaiſſeau. <lb/></s> <s xml:id="echoid-s19241" xml:space="preserve">C. </s> <s xml:id="echoid-s19242" xml:space="preserve">Q. </s> <s xml:id="echoid-s19243" xml:space="preserve">F. </s> <s xml:id="echoid-s19244" xml:space="preserve">D.</s> <s xml:id="echoid-s19245" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1599" type="section" level="1" n="1155"> <head xml:id="echoid-head1370" xml:space="preserve"><emph style="sc">Démonstration du troisieme cas</emph>.</head> <p> <s xml:id="echoid-s19246" xml:space="preserve">Si le corps G eſt plus léger qu’un pareil volume d’eau, l’on <lb/>voit évidemment qu’il doit arriver tout le contraire du cas pré-<lb/>cédent, c’eſt-à-dire qu’au lieu d’aller au fond de l’eau, il doit <lb/>nager ſur la ſurface, & </s> <s xml:id="echoid-s19247" xml:space="preserve">ne s’enfoncer qu’en partie dedans, qui <lb/>ſera, par exemple, la partie I K M N qui occupe un volume <lb/>d’eau égal en peſanteur à tout le corps G: </s> <s xml:id="echoid-s19248" xml:space="preserve">car ſi, par exemple, <lb/>ce corps ne peſe que la moitié d’un pareil volume d’eau, la <lb/>partie enfoncée ſera la moitié du corps, & </s> <s xml:id="echoid-s19249" xml:space="preserve">l’eau que cette <lb/>moitié occupe étant d’une égale peſanteur que tout le corps, <lb/>ils tendront également au centre de la terre, & </s> <s xml:id="echoid-s19250" xml:space="preserve">ſeront par <lb/>conſéquent en équilibre, quoique le corps ne ſoit pas entié-<lb/>rement plongé dans l’eau. </s> <s xml:id="echoid-s19251" xml:space="preserve">C. </s> <s xml:id="echoid-s19252" xml:space="preserve">Q. </s> <s xml:id="echoid-s19253" xml:space="preserve">F. </s> <s xml:id="echoid-s19254" xml:space="preserve">D.</s> <s xml:id="echoid-s19255" xml:space="preserve"/> </p> <pb o="613" file="0703" n="729" rhead="DE MATHÉMATIQUE. Liv. XVI."/> </div> <div xml:id="echoid-div1600" type="section" level="1" n="1156"> <head xml:id="echoid-head1371" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19256" xml:space="preserve">1144. </s> <s xml:id="echoid-s19257" xml:space="preserve">Il ſuit du premier cas, que ſi une puiſſance Q vou-<lb/>loit ſortir de l’eau un poids E attaché à une corde, ſi le poids <lb/>eſt égal à la peſanteur ſpécifique de l’eau, la puiſſance ne s’ap-<lb/>percevra de la peſanteur du poids, que lorſqu’il commen-<lb/>cera à ſortir de l’eau, puiſque tant qu’il ſera plongé dedans, <lb/>elle n’en ſoutiendra aucune partie; </s> <s xml:id="echoid-s19258" xml:space="preserve">& </s> <s xml:id="echoid-s19259" xml:space="preserve">c’eſt la raiſon qui fait <lb/>que lorſque l’on tire de l’eau d’un puits, la puiſſance ne fait <lb/>preſque point d’effort pour ſoutenir le vaiſſeau plein d’eau, <lb/>tant qu’il eſt plongé dedans, parce qu’elle ne ſoutient aucune <lb/>partie de l’eau qui eſt dans le vaiſſeau, & </s> <s xml:id="echoid-s19260" xml:space="preserve">que le vaiſſeau lui-<lb/>même, quand il eſt de bois, eſt à peu près égal à la peſanteur <lb/>ſpécifique de l’eau, au lieu qu’étant entiérement dehors, l’ef-<lb/>fort de la puiſſance devient égal au poids de l’eau & </s> <s xml:id="echoid-s19261" xml:space="preserve">de celui <lb/>du vaiſſeau.</s> <s xml:id="echoid-s19262" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1601" type="section" level="1" n="1157"> <head xml:id="echoid-head1372" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19263" xml:space="preserve">1145. </s> <s xml:id="echoid-s19264" xml:space="preserve">Il ſuit du ſecond cas, que ſi une puiſſance Q ſoutient <lb/>un corps O plongé dans l’eau, & </s> <s xml:id="echoid-s19265" xml:space="preserve">que la peſanteur ſpécifique <lb/>du corps ſoit plus grande que celle de l’eau, cette puiſſance ne <lb/>ſoutiendra qu’une partie de la peſanteur du corps, qui ſera la <lb/>différence de ſa peſanteur ſpécifique à celle du volume d’eau <lb/>dont il occupe la place; </s> <s xml:id="echoid-s19266" xml:space="preserve">parce que ce corps peſe moins dans <lb/>l’eau que dans l’air, du poids d’un pareil volume d’eau: </s> <s xml:id="echoid-s19267" xml:space="preserve">ainſi <lb/>l’on peut dire en général que les corps plus peſans que l’eau <lb/>perdent de leur peſanteur, lorſqu’ils ſont plongés dedans; </s> <s xml:id="echoid-s19268" xml:space="preserve">& </s> <s xml:id="echoid-s19269" xml:space="preserve"><lb/>cela dans la raiſon de la gravité ſpécifique du corps à celle de <lb/>l’eau, qui eſt un principe dont nous avons déja parlé dans <lb/>l’art. </s> <s xml:id="echoid-s19270" xml:space="preserve">901.</s> <s xml:id="echoid-s19271" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1602" type="section" level="1" n="1158"> <head xml:id="echoid-head1373" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19272" xml:space="preserve">1146. </s> <s xml:id="echoid-s19273" xml:space="preserve">Il ſuit du troiſieme cas, que quand un corps eſt plus <lb/>léger qu’un pareil volume d’eau, la peſanteur ſpécifique de <lb/>l’eau eſt à celle du corps, comme le volume de tout le corps <lb/>eſt à ſa partie enfoncée: </s> <s xml:id="echoid-s19274" xml:space="preserve">ainſi ſuppoſant que le corps G ſoit un <lb/>cube ou un parallelépipede, la peſanteur ſpécifique de l’eau <lb/>ſera à celle de ce corps, comme H K eſt à I K.</s> <s xml:id="echoid-s19275" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1603" type="section" level="1" n="1159"> <head xml:id="echoid-head1374" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s19276" xml:space="preserve">1147. </s> <s xml:id="echoid-s19277" xml:space="preserve">Il ſuit auſſi qu’un corps s’enfonce différemment dans <pb o="614" file="0704" n="730" rhead="NOUVEAU COURS"/> les liqueurs dont les peſanteurs ſpécifiques ſont différentes, <lb/>étant certain qu’il s’enfoncera davantage dans une liqueur <lb/>d’une certaine peſanteur ſpécifique, que dans une autre qui <lb/>ſeroit plus peſante: </s> <s xml:id="echoid-s19278" xml:space="preserve">par exemple, l’on voit qu’un vaiſſeau <lb/>chargé s’enſonce plus dans une riviere que dans la mer, parce <lb/>que l’eau des rivieres eſt moins peſante que celle de la mer: <lb/></s> <s xml:id="echoid-s19279" xml:space="preserve">ainſi il ne faut pas s’étonner s’il eſt arrivé quelquefois qu’un <lb/>vaiſſeau, après avoir cinglé heureuſement en pleine mer, s’eſt <lb/>perdu & </s> <s xml:id="echoid-s19280" xml:space="preserve">a coulé à fond en arrivant à l’embouchure de quelque <lb/>riviere d’eau douce.</s> <s xml:id="echoid-s19281" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1604" type="section" level="1" n="1160"> <head xml:id="echoid-head1375" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s19282" xml:space="preserve">1148. </s> <s xml:id="echoid-s19283" xml:space="preserve">L’on peut encore remarquer que quoique les métaux <lb/>ſoient plus peſans que l’eau, cela n’empêche pas qu’ils ne puiſ-<lb/>ſent nager ſur l’eau: </s> <s xml:id="echoid-s19284" xml:space="preserve">car s’ils compoſent des corps creux, dont <lb/>la peſanteur ſpécifique ſoit moindre que celle du volume d’eau <lb/>dont ils occupent la place, ils ſurnageront ſans couler à fond.</s> <s xml:id="echoid-s19285" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1605" type="section" level="1" n="1161"> <head xml:id="echoid-head1376" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s19286" xml:space="preserve">1149. </s> <s xml:id="echoid-s19287" xml:space="preserve">Nous avons déja dit dans l’art. </s> <s xml:id="echoid-s19288" xml:space="preserve">901, que les métaux <lb/>perdoient de leur peſanteur, lorſqu’ils étoient plongés dans <lb/>l’eau: </s> <s xml:id="echoid-s19289" xml:space="preserve">& </s> <s xml:id="echoid-s19290" xml:space="preserve">comme c’eſt ici l’endroit d’en faire voir la raiſon, <lb/>l’on remarquera qu’il n’y en a pas d’autre que celle qui fait <lb/>qu’un corps étant plongé dans l’eau, eſt plus léger qu’il n’étoit <lb/>dans l’air de toute la peſanteur ſpécifique de l’eau dont il oc-<lb/>cupe la place. </s> <s xml:id="echoid-s19291" xml:space="preserve">Ainſi l’on pourra toujours trouver la raiſon de <lb/>la peſanteur ſpécifique d’un métal avec celle de l’eau, ou de <lb/>toute autre liqueur, en peſant dans l’air avec des juſtes balances <lb/>une piece de métal; </s> <s xml:id="echoid-s19292" xml:space="preserve">enſuite on l’attachera à l’un des bras ou <lb/>baſſins de la balance avec un fil de ſoie, pour voir après que <lb/>le métal ſera plongé dans l’eau, combien il peſera de moins; <lb/></s> <s xml:id="echoid-s19293" xml:space="preserve">& </s> <s xml:id="echoid-s19294" xml:space="preserve">la différence ſera celle de la peſanteur ſpécifique de ce métal <lb/>à celle de l’eau.</s> <s xml:id="echoid-s19295" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19296" xml:space="preserve">C’eſt en ſuivant ce que l’on vient de dire, qu’on a trouvé <lb/>que l’or perd dans l’eau environ la dix-neuvieme partie de ſon <lb/>poids, le mercure la quinzieme, le plomb la douzieme, l’ar-<lb/>gent la dixieme, le cuivre la neuvieme, le fer la huitieme, & </s> <s xml:id="echoid-s19297" xml:space="preserve"><lb/>l’étain la ſeptieme.</s> <s xml:id="echoid-s19298" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19299" xml:space="preserve">En ſuivant le même principe, on peut ſçavoir auſſi le rap-<lb/>port des peſanteurs ſpécifiques des liqueurs entr’elles, & </s> <s xml:id="echoid-s19300" xml:space="preserve">des <pb o="615" file="0705" n="731" rhead="DE MATHÉMATIQUE. Liv. XVI."/> métaux entr’eux, & </s> <s xml:id="echoid-s19301" xml:space="preserve">par conſéquent des liqueurs avec les mé-<lb/>taux, par exemple, le rapport du poids d’un pouce cube d’or <lb/>avec celui d’un pouce cube de mercure; </s> <s xml:id="echoid-s19302" xml:space="preserve">& </s> <s xml:id="echoid-s19303" xml:space="preserve">c’eſt ainſi que l’on <lb/>a trouvé la peſanteur d’un pouce cube des métaux & </s> <s xml:id="echoid-s19304" xml:space="preserve">des li-<lb/>queurs contenus dans la Table ſuivante.</s> <s xml:id="echoid-s19305" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1606" type="section" level="1" n="1162"> <head xml:id="echoid-head1377" style="it" xml:space="preserve">Poids d’un pouce cube.</head> <note position="right" xml:space="preserve">Matieres. # on. # gros. # gr. # Matieres. # on. # gros. # gr. <lb/>Or. # 12 # 2 # 17 # Marbre blanc. # 1 # 6 # 0 <lb/>Mercure. # 8 # 6 # 8 # Pierre de taille. # 1 # 2 # 24 <lb/>Plomb. # 7 # 3 # 30 # Eau de Seine. # 0 # 5 # 12 <lb/># # # # Vin. # 0 # 5 # 5 <lb/></note> <note position="right" xml:space="preserve">Matieres. # on. # gros. # gr. # Matieres. # on. # gros. # gr. <lb/>Argent. # 6 # 5 # 26 # Cire. # 0 # 4 # 65 <lb/>Cuivre. # 5 # 6 # 36 # Huile. # 0 # 4 # 43 <lb/>Fer. # 5 # 1 # 27 # Chêne ſec. # 0 # 4 # 22 <lb/>Etain. # 4 # 6 # 14 # Noyer. # 0 # 3 # 6 <lb/></note> <p> <s xml:id="echoid-s19306" xml:space="preserve">L’on peut encore par ce principe meſurer la ſolidité d’un <lb/>corps irrégulier: </s> <s xml:id="echoid-s19307" xml:space="preserve">car ſi ce corps peſe 90 livres dans l’air, & </s> <s xml:id="echoid-s19308" xml:space="preserve">que <lb/>dans l’eau il n’en peſe que 80, c’eſt une marque que le volume <lb/>d’eau, dont il occupe la place, peſe 10 livres: </s> <s xml:id="echoid-s19309" xml:space="preserve">ainſi il ne s’agit <lb/>que de ſçavoir combien 10 livres d’eau valent de pouces cubes; <lb/></s> <s xml:id="echoid-s19310" xml:space="preserve">ce que l’on trouvera, en diſant: </s> <s xml:id="echoid-s19311" xml:space="preserve">Si 70 livres valent un pied <lb/>cube d’eau, ou 1728 pouces, combien vaudront 10 livres? </s> <s xml:id="echoid-s19312" xml:space="preserve"><lb/>l’on trouvera 246 pouces & </s> <s xml:id="echoid-s19313" xml:space="preserve">{6/7} pour la ſolidité du corps.</s> <s xml:id="echoid-s19314" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1607" type="section" level="1" n="1163"> <head xml:id="echoid-head1378" style="it" xml:space="preserve">Application des principes précédens à la navigation.</head> <p> <s xml:id="echoid-s19315" xml:space="preserve">1150. </s> <s xml:id="echoid-s19316" xml:space="preserve">Quand on fait des tranſports de munitions de guerre <lb/>par des bateaux, comme cela arrive ſouvent, lorſqu’on a la <lb/>commodité des rivieres ou des canaux, & </s> <s xml:id="echoid-s19317" xml:space="preserve">que ces munitions <lb/>peuvent être accompagnées de gros fardeaux: </s> <s xml:id="echoid-s19318" xml:space="preserve">par exemple, <lb/>comme du canon, des affûts, en un mot tout ce qui compoſe <lb/>un équipage d’Artillerie, & </s> <s xml:id="echoid-s19319" xml:space="preserve">qu’un Officier qui a un peu de dé-<lb/>tail, n’ignore pas le poids des munitions dont il eſt chargé, <lb/>il faut faire voir ici comme il pourra eſtimer la charge que les <lb/>bateaux peuvent porter, afin de ſçavoir combien il lui en fau-<lb/>dra, ſi l’on n’avoit égard qu’aux poids des munitions, ſans <lb/>s’embarraſſer du volume.</s> <s xml:id="echoid-s19320" xml:space="preserve"/> </p> <pb o="616" file="0706" n="732" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s19321" xml:space="preserve">Comme le pied cube d’eau douce peſe environ 70 livres, <lb/>& </s> <s xml:id="echoid-s19322" xml:space="preserve">qu’un pied cube de bois de chêne ne peſe qu’environ 58, <lb/>l’on voit qu’un bateau pourroit être rempli d’eau, ſans pour <lb/>cela couler à fond, parce que l’eau qui ſeroit dedans eſt en <lb/>équilibre avec celle du dehors, & </s> <s xml:id="echoid-s19323" xml:space="preserve">que la peſanteur ſpécifique <lb/>du bois qui compoſe le bateau, eſt plus petite que celle de <lb/>l’eau. </s> <s xml:id="echoid-s19324" xml:space="preserve">L’on peut donc mettre dans le bateau un poids équi-<lb/>valent à celui de l’eau qu’il peut contenir. </s> <s xml:id="echoid-s19325" xml:space="preserve">Or ſi l’on meſure la <lb/>capacité du bateau, & </s> <s xml:id="echoid-s19326" xml:space="preserve">qu’on la trouve, par e@emple, de <lb/>4000 pieds cubes, ce bateau pourra porter 4000 fois 70 livres, <lb/>parce que nous avons dit qu’un pied cube d’eau peſoit 70 livres: <lb/></s> <s xml:id="echoid-s19327" xml:space="preserve">ainſi le bateau portera 280000 livres; </s> <s xml:id="echoid-s19328" xml:space="preserve">mais comme l’uſage <lb/>ſur les ports de mer eſt d’eſtimer la charge des vaiſſeaux par <lb/>tonneaux, & </s> <s xml:id="echoid-s19329" xml:space="preserve">la charge des bateaux ſur les rivieres par quin-<lb/>taux, l’on ſçaura que le tonneau eſt un poids de 2000 livres, <lb/>& </s> <s xml:id="echoid-s19330" xml:space="preserve">que le quintal eſt un poids de 100 livres: </s> <s xml:id="echoid-s19331" xml:space="preserve">ainſi quand l’on <lb/>dit en terme de Marine, qu’un vaiſſeau porte 100 tonneaux, <lb/>ou eſt de 100 tonneaux, cela veut dire qu’il peut porter 200000 <lb/>livres, ou 2000 quintaux.</s> <s xml:id="echoid-s19332" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19333" xml:space="preserve">Nous avons déja dit que l’eau de la mer étoit plus peſante <lb/>que celle des rivieres; </s> <s xml:id="echoid-s19334" xml:space="preserve">& </s> <s xml:id="echoid-s19335" xml:space="preserve">comme on pourroit avoir beſoin de <lb/>connoître ſon poids, l’on ſçaura que le pied cube peſe 73 livres, <lb/>qui eſt 3 livres de plus par pied cube que l’eau douce.</s> <s xml:id="echoid-s19336" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19337" xml:space="preserve">Nous allons encore faire voir dans la propoſition ſuivante <lb/>un principe de l’équilibre des liqueurs, qui eſt plus curieux <lb/>qu’utile dans la pratique: </s> <s xml:id="echoid-s19338" xml:space="preserve">c’eſt pourquoi je n’en ai pas parlé <lb/>plutôt; </s> <s xml:id="echoid-s19339" xml:space="preserve">mais comme il ne conviendroit pas de le paſſer ſous <lb/>ſilence, voici de quoi il eſt queſtion.</s> <s xml:id="echoid-s19340" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1608" type="section" level="1" n="1164"> <head xml:id="echoid-head1379" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19341" xml:space="preserve">1151. </s> <s xml:id="echoid-s19342" xml:space="preserve">Si l’on a un vaſe plus gros par un bout que par l’autre <lb/>rempli d’une liqueur quelconque; </s> <s xml:id="echoid-s19343" xml:space="preserve">cette liqueur aura autant de force <lb/>pour ſortir par une ouverture égale à ſa baſe, que ſi cette ouverture <lb/>étoit égale à celle d’en haut.</s> <s xml:id="echoid-s19344" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1609" type="section" level="1" n="1165"> <head xml:id="echoid-head1380" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19345" xml:space="preserve">Si l’on a un vaſe comme dans la figure 411, plus large par <lb/> <anchor type="note" xlink:label="note-0706-01a" xlink:href="note-0706-01"/> la baſe B C que par le haut G H, il eſt aiſé de concevoir que <pb o="617" file="0707" n="733" rhead="DE MATHÉMATIQUE. Liv. XVI."/> l’eau qui peſe ſur la baſe B C fait autant d’effort, que ſi elle <lb/>étoit chargée de toute l’eau du volume B O P C: </s> <s xml:id="echoid-s19346" xml:space="preserve">car nous <lb/>avons fait voir que toutes les colonnes d’eau, comme L M <lb/>(art. </s> <s xml:id="echoid-s19347" xml:space="preserve">1137), tendoient à monter à la hauteur G H ou O P, qui <lb/>eſt la même choſ<unsure/> & </s> <s xml:id="echoid-s19348" xml:space="preserve">que l’effort qu’elle faiſoit étoit exprimé <lb/>par le poids de la petite colonne I N : </s> <s xml:id="echoid-s19349" xml:space="preserve">mais l’effort exprimé <lb/>par I N, ſe fait également à l’endroit M de la baſe qu’à l’en-<lb/>droit L, à cauſe de la mobilité reſpective des parties qui com-<lb/>poſent les colonnes d’eau; </s> <s xml:id="echoid-s19350" xml:space="preserve">& </s> <s xml:id="echoid-s19351" xml:space="preserve">toutes les colonnes, comme <lb/>L M, indépendamment de l’effort exprimé par I N, font en-<lb/>core effort de tout le poids de leur hauteur L M : </s> <s xml:id="echoid-s19352" xml:space="preserve">d’où il ſuit <lb/>que la colonne L M peſe autant ſur la baſe que la colonne I K, <lb/>& </s> <s xml:id="echoid-s19353" xml:space="preserve">que par conſéquent la baſe eſt autant preſſée par l’eau, qui <lb/>eſt dans le vaſe, que ſi elle étoit chargée de tout le volume <lb/>B O P C. </s> <s xml:id="echoid-s19354" xml:space="preserve">C. </s> <s xml:id="echoid-s19355" xml:space="preserve">Q. </s> <s xml:id="echoid-s19356" xml:space="preserve">F. </s> <s xml:id="echoid-s19357" xml:space="preserve">D.</s> <s xml:id="echoid-s19358" xml:space="preserve"/> </p> <div xml:id="echoid-div1609" type="float" level="2" n="1"> <note position="left" xlink:label="note-0706-01" xlink:href="note-0706-01a" xml:space="preserve">Figure 411.</note> </div> <p> <s xml:id="echoid-s19359" xml:space="preserve">Si le vaſe a ſes côtés inclinés, comme dans la figure <lb/> <anchor type="note" xlink:label="note-0707-01a" xlink:href="note-0707-01"/> 412, l’on démontrera de même que l’eau fait autant d’effort <lb/>ſur la baſe E F, que ſi elle étoit chargée de toute celle qui ſe-<lb/>roit contenue dans le volume cylindrique E Q R F, qui a pour <lb/>hauteur celle de l’eau du vaſe.</s> <s xml:id="echoid-s19360" xml:space="preserve"/> </p> <div xml:id="echoid-div1610" type="float" level="2" n="2"> <note position="right" xlink:label="note-0707-01" xlink:href="note-0707-01a" xml:space="preserve">Figure 412.</note> </div> <p> <s xml:id="echoid-s19361" xml:space="preserve">L’expérience prouve ceci encore mieux que tout le raiſon-<lb/>nement que l’on peut faire: </s> <s xml:id="echoid-s19362" xml:space="preserve">car ſi l’on a un vaſe plus large par <lb/>en bas que par en haut, & </s> <s xml:id="echoid-s19363" xml:space="preserve">que le fond ſoit fermé par un <lb/>piſton qui ait la liberté de ſe mouvoir, ſans cependant que <lb/>l’eau puiſſe ſe répandre; </s> <s xml:id="echoid-s19364" xml:space="preserve">l’on voit, dis-je, que la puiſſance qui <lb/>ſoutient ce piſton, a beſoin d’une force égale au poids de l’eau <lb/>qui ſeroit contenue dans ce vaſe, s’il étoit auſſi large par en <lb/>haut que par en bas, à cauſe de l’effort que les petites colonnes <lb/>d’eau font pour ſe mettre au niveau des plus grandes: </s> <s xml:id="echoid-s19365" xml:space="preserve">mais <lb/>quand l’eau vient à être gelée, & </s> <s xml:id="echoid-s19366" xml:space="preserve">que ces parties ne ſont plus <lb/>en mouvement, elles ne font plus d’effort contre les côtés <lb/>du vaſe, & </s> <s xml:id="echoid-s19367" xml:space="preserve">la puiſſance n’a plus beſoin d’une ſi grande force, <lb/>parce que pour lors elle ne ſoutient plus que la peſanteur abſo-<lb/>lue de l’eau gelée.</s> <s xml:id="echoid-s19368" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19369" xml:space="preserve">1152. </s> <s xml:id="echoid-s19370" xml:space="preserve">Mais ſi le vaiſſeau étoit plus large par en haut que par <lb/> <anchor type="note" xlink:label="note-0707-02a" xlink:href="note-0707-02"/> en bas, comme eſt le vaſe A B C D, ſi on le remplit de liqueur, <lb/>elle ne fera pas plus d’effort contre la baſe B D, que ſi la lar-<lb/>geur d’en haut étoit égale à celle d’en bas: </s> <s xml:id="echoid-s19371" xml:space="preserve">car ſi l’on imagine <lb/>le cylindre d’eau B D E F, il ſera aiſé de juger que comme l’eau <lb/>peſe perpendiculairement, il n’y a que celle qui eſt contenue <pb o="618" file="0708" n="734" rhead="NOUVEAU COURS"/> dans le cylindre qui fait effort contre la baſe B D, parce que <lb/>celle qui eſt contenue autour du cylindre, ne peſe pas ſur la <lb/>baſe, mais ſeulement ſur les côtés inclinés du vaſe.</s> <s xml:id="echoid-s19372" xml:space="preserve"/> </p> <div xml:id="echoid-div1611" type="float" level="2" n="3"> <note position="right" xlink:label="note-0707-02" xlink:href="note-0707-02a" xml:space="preserve">Figure 419.</note> </div> </div> <div xml:id="echoid-div1613" type="section" level="1" n="1166"> <head xml:id="echoid-head1381" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s19373" xml:space="preserve">1153. </s> <s xml:id="echoid-s19374" xml:space="preserve">Il ſuit de cette propoſition, que quelque forme que <lb/>puiſſent avoir pluſieurs vaiſſeaux perpendiculaires à l’horizon, <lb/>& </s> <s xml:id="echoid-s19375" xml:space="preserve">d’égales hauteurs, ſi ces vaiſſeaux ont des baſes égales, & </s> <s xml:id="echoid-s19376" xml:space="preserve"><lb/>qu’ils ſoient remplis d’eau, les baſes ſeront également chargées,</s> </p> </div> <div xml:id="echoid-div1614" type="section" level="1" n="1167"> <head xml:id="echoid-head1382" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s19377" xml:space="preserve">1154. </s> <s xml:id="echoid-s19378" xml:space="preserve">L’effort des liqueurs ſe meſure à la livre comme celui <lb/> <anchor type="note" xlink:label="note-0708-01a" xlink:href="note-0708-01"/> des poids dans la méchanique; </s> <s xml:id="echoid-s19379" xml:space="preserve">& </s> <s xml:id="echoid-s19380" xml:space="preserve">comme on peut ſçavoir la <lb/>peſanteur d’un pied cube de toutes ſortes de liqueurs, particu-<lb/>liérement de celui de l’eau, qui peſe 70 livres, l’on trouvera <lb/>toujours l’effort de l’eau ſur le fond d’un vaſe, en multipliant <lb/>la capacité du fond par la hauteur perpendiculaire de l’eau du <lb/>vaſe: </s> <s xml:id="echoid-s19381" xml:space="preserve">ainſi ayant un vaſe A B C perpendiculaire à l’horizon, <lb/>& </s> <s xml:id="echoid-s19382" xml:space="preserve">rempli d’eau juſqu’à l’ouverture A, voulant ſçavoir l’effort <lb/>que fait l’eau ſur la baſe B C, nous ſuppoſerons que cette baſe <lb/>vaut 4 pieds quarrés, & </s> <s xml:id="echoid-s19383" xml:space="preserve">que la hauteur perpendiculaire A D eſt <lb/>de 40 pieds: </s> <s xml:id="echoid-s19384" xml:space="preserve">ainſi multipliant 40 par 4, l’on aura 160 pieds <lb/>cubes, qui étant multipliés par 70 livres, qui eſt la peſanteur <lb/>d’un pied cube d’eau, il viendra 11200 livres, qui eſt l’effort <lb/>que l’eau du vaſe A B C fait ſur la baſe B C; </s> <s xml:id="echoid-s19385" xml:space="preserve">& </s> <s xml:id="echoid-s19386" xml:space="preserve">ce qu’il y a de <lb/>ſurprenant, c’eſt que ſi tout le vaſe ne contenoit qu’un pied <lb/>cube d’eau, qui eſt équivalent au poids de 70 livres, il faudroit <lb/>que la puiſſance Q qui voudroit ſoutenir le fond C D (ſuppo-<lb/>ſant qu’il fût détaché du reſte), eût une force de 11200 liv. <lb/></s> <s xml:id="echoid-s19387" xml:space="preserve">pour être en équilibre avec l’effort de l’eau ſur la baſe B C.</s> <s xml:id="echoid-s19388" xml:space="preserve"/> </p> <div xml:id="echoid-div1614" type="float" level="2" n="1"> <note position="left" xlink:label="note-0708-01" xlink:href="note-0708-01a" xml:space="preserve">Figure 420.</note> </div> <figure> <image file="0708-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0708-01"/> </figure> <pb o="619" file="0709" n="735" rhead="DE MATHÉMATIQUE. Liv. XVI."/> </div> <div xml:id="echoid-div1616" type="section" level="1" n="1168"> <head xml:id="echoid-head1383" xml:space="preserve">CHAPITRE II.</head> <head xml:id="echoid-head1384" style="it" xml:space="preserve">De la vîteſſe des fluides qui ſortent par des ouvertures faites aux <lb/>vaſes qui les contiennent.</head> <head xml:id="echoid-head1385" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19389" xml:space="preserve">1155. </s> <s xml:id="echoid-s19390" xml:space="preserve">Si l’on a un tuyau A B C D perpendiculaire à l’horizon, <lb/> <anchor type="note" xlink:label="note-0709-01a" xlink:href="note-0709-01"/> & </s> <s xml:id="echoid-s19391" xml:space="preserve">rempli d’une liqueur quelconque, la vîteſſe de cette liqueur par <lb/>l’ouverture C D de la baſe ſera exprimée par la racine quarrée <lb/>de la hauteur.</s> <s xml:id="echoid-s19392" xml:space="preserve"/> </p> <div xml:id="echoid-div1616" type="float" level="2" n="1"> <note position="right" xlink:label="note-0709-01" xlink:href="note-0709-01a" xml:space="preserve">Figure 421.</note> </div> </div> <div xml:id="echoid-div1618" type="section" level="1" n="1169"> <head xml:id="echoid-head1386" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19393" xml:space="preserve">Si l’on ſuppoſe d’abord que l’ouverture de la baſe eſt égale <lb/>à la même baſe du cylindre, il eſt viſible que rien ne s’oppo-<lb/>ſant au paſſage du fluide renfermé dans le vaſe, toutes les <lb/>parties de la tranche inférieure C D doivent partir avec la <lb/>même vîteſſe; </s> <s xml:id="echoid-s19394" xml:space="preserve">toute la difficulté conſiſte à ſçavoir quelle doit <lb/>être la vîteſſe de cette tranche au premier inſtant du mouve-<lb/>ment. </s> <s xml:id="echoid-s19395" xml:space="preserve">Je dis que cette vîteſſe eſt égale à celle qu’auroit acquiſe <lb/>la premiere tranche ſupérieure A B en tombant de la hauteur <lb/>B D. </s> <s xml:id="echoid-s19396" xml:space="preserve">Pour cela, faites attention que la vîteſſe d’un corps qui <lb/>tombe augmente à chaque inſtant dans le rapport des momens <lb/>qui ſe ſont écoulés, & </s> <s xml:id="echoid-s19397" xml:space="preserve">par conſéquent la force de ce corps, <lb/>que l’on peut toujours exprimer par des poids, augmente dans <lb/>le même rapport. </s> <s xml:id="echoid-s19398" xml:space="preserve">Cela poſé, ſi nous imaginons que le tems <lb/>eſt repréſenté par la hauteur A C, il y aura autant de tranches <lb/>égales à la premiere qui preſſeront la derniere, qu’il y a d’inſ-<lb/>tans pour la chûte de la premiere tranche A B E G: </s> <s xml:id="echoid-s19399" xml:space="preserve">donc cette <lb/>derniere tranche reçoit du côté du poids de la colonne qui la <lb/>preſſe une force égale à celle qu’elle auroit acquiſe en tombant <lb/>de B en D: </s> <s xml:id="echoid-s19400" xml:space="preserve">d’ailleurs cette force ſeroit exprimée par la racine <lb/>quarrée de la hauteur. </s> <s xml:id="echoid-s19401" xml:space="preserve">Donc, &</s> <s xml:id="echoid-s19402" xml:space="preserve">c. </s> <s xml:id="echoid-s19403" xml:space="preserve">C. </s> <s xml:id="echoid-s19404" xml:space="preserve">Q. </s> <s xml:id="echoid-s19405" xml:space="preserve">F. </s> <s xml:id="echoid-s19406" xml:space="preserve">D.</s> <s xml:id="echoid-s19407" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1619" type="section" level="1" n="1170"> <head xml:id="echoid-head1387" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19408" xml:space="preserve">1156. </s> <s xml:id="echoid-s19409" xml:space="preserve">2°. </s> <s xml:id="echoid-s19410" xml:space="preserve">Si le trou D fait à la baſe du vaſe qui renferme la li-<lb/> <anchor type="note" xlink:label="note-0709-02a" xlink:href="note-0709-02"/> queur, n’eſt pas égal à la même baſe, je dis que la vîteſſe, au ſortir <pb o="620" file="0710" n="736" rhead="NOUVEAU COURS"/> de cette ouverture, ſera encore exprimée par la racine quarrée de la <lb/>hauteur. </s> <s xml:id="echoid-s19411" xml:space="preserve">On ſuppoſe que le vaſe eſt toujours entretenu à la même <lb/>hauteur.</s> <s xml:id="echoid-s19412" xml:space="preserve"/> </p> <div xml:id="echoid-div1619" type="float" level="2" n="1"> <note position="right" xlink:label="note-0709-02" xlink:href="note-0709-02a" xml:space="preserve">Figure 422 <lb/>& 423.</note> </div> </div> <div xml:id="echoid-div1621" type="section" level="1" n="1171"> <head xml:id="echoid-head1388" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s19413" xml:space="preserve">Je conſidere d’abord que les quantités de fluides écoulées <lb/>ſont dans la raiſon des vîteſſes pourune même ouverture, étant <lb/>évident qu’une vîteſſe double donnera une maſſe double, & </s> <s xml:id="echoid-s19414" xml:space="preserve"><lb/>ainſi de ſuite. </s> <s xml:id="echoid-s19415" xml:space="preserve">Cela poſé, concevons deux vaſes A B C & </s> <s xml:id="echoid-s19416" xml:space="preserve">E F G <lb/>percés chacun à leur baſe d’une même ouverture, & </s> <s xml:id="echoid-s19417" xml:space="preserve">dont les <lb/>hauteurs ſoient différentes. </s> <s xml:id="echoid-s19418" xml:space="preserve">Il eſt clair que les vîteſſes ſeront <lb/>différentes, quel que ſoit leur rapport, & </s> <s xml:id="echoid-s19419" xml:space="preserve">par conſéquent les <lb/>maſſes ou quantités de fluides le ſeront auſſi dans le même <lb/>rapport. </s> <s xml:id="echoid-s19420" xml:space="preserve">Soit V la vîteſſe du premier vaſe, & </s> <s xml:id="echoid-s19421" xml:space="preserve">u celle du ſe-<lb/>cond; </s> <s xml:id="echoid-s19422" xml:space="preserve">M la maſſe de fluide écoulé dans un certain tems, & </s> <s xml:id="echoid-s19423" xml:space="preserve"><lb/>m celle du fluide écoulé par le ſecond vaſe dans le même tems; <lb/></s> <s xml:id="echoid-s19424" xml:space="preserve">on aura M: </s> <s xml:id="echoid-s19425" xml:space="preserve">m :</s> <s xml:id="echoid-s19426" xml:space="preserve">: V : </s> <s xml:id="echoid-s19427" xml:space="preserve">u; </s> <s xml:id="echoid-s19428" xml:space="preserve">donc m={uM/V}. </s> <s xml:id="echoid-s19429" xml:space="preserve">Soit F le poids de la <lb/>colonne A D, & </s> <s xml:id="echoid-s19430" xml:space="preserve">f celui de la colonne E H. </s> <s xml:id="echoid-s19431" xml:space="preserve">Ces poids ou co-<lb/>lonnes ſeront dans la raiſon des hauteurs, puiſqu’elles ont des <lb/>baſes égales, & </s> <s xml:id="echoid-s19432" xml:space="preserve">que le fluide eſt le même pour chaque vaſe: </s> <s xml:id="echoid-s19433" xml:space="preserve"><lb/>on aura donc F : </s> <s xml:id="echoid-s19434" xml:space="preserve">f :</s> <s xml:id="echoid-s19435" xml:space="preserve">: A D : </s> <s xml:id="echoid-s19436" xml:space="preserve">E G. </s> <s xml:id="echoid-s19437" xml:space="preserve">Deplus, les forces étant comme <lb/>les quantités de mouvement qu’elles produiſent, c’eſt-à-dire <lb/>comme les produits des maſſes ou quantités écoulées par les <lb/>vîteſſes, on aura encore F : </s> <s xml:id="echoid-s19438" xml:space="preserve">f :</s> <s xml:id="echoid-s19439" xml:space="preserve">: MV : </s> <s xml:id="echoid-s19440" xml:space="preserve">{Mu<emph style="sub">2</emph>/V}: </s> <s xml:id="echoid-s19441" xml:space="preserve">donc F: </s> <s xml:id="echoid-s19442" xml:space="preserve">f :</s> <s xml:id="echoid-s19443" xml:space="preserve">: M V<emph style="sub">2</emph> <lb/>: </s> <s xml:id="echoid-s19444" xml:space="preserve">Mu<emph style="sub">2</emph> :</s> <s xml:id="echoid-s19445" xml:space="preserve">: V<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19446" xml:space="preserve">u<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19447" xml:space="preserve">donc A D : </s> <s xml:id="echoid-s19448" xml:space="preserve">E G :</s> <s xml:id="echoid-s19449" xml:space="preserve">: V<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19450" xml:space="preserve">u<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19451" xml:space="preserve">doncenfin V: </s> <s xml:id="echoid-s19452" xml:space="preserve">u:</s> <s xml:id="echoid-s19453" xml:space="preserve">: <lb/>√AD\x{0020} : </s> <s xml:id="echoid-s19454" xml:space="preserve">√EG\x{0020}. </s> <s xml:id="echoid-s19455" xml:space="preserve">C. </s> <s xml:id="echoid-s19456" xml:space="preserve">Q. </s> <s xml:id="echoid-s19457" xml:space="preserve">F. </s> <s xml:id="echoid-s19458" xml:space="preserve">D.</s> <s xml:id="echoid-s19459" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1622" type="section" level="1" n="1172"> <head xml:id="echoid-head1389" xml:space="preserve"><emph style="sc">Remarque</emph>.</head> <p> <s xml:id="echoid-s19460" xml:space="preserve">1157. </s> <s xml:id="echoid-s19461" xml:space="preserve">On voit par-là que le principe que nous avons établi <lb/>précédemment devient général, c’eſt-à-dire que les vîteſſes <lb/>ſeront toujours exprimées par les racines quarrées des hau-<lb/>teurs, quelle que ſoit l’ouverture, égale ou plus petite que la baſe <lb/>du vaſe qui renferme le fluide; </s> <s xml:id="echoid-s19462" xml:space="preserve">& </s> <s xml:id="echoid-s19463" xml:space="preserve">quelle que ſoit d’ailleurs la <lb/>figure du vaſe droit ou oblique, pourvu qu’il ſoit entretenu <lb/>plein à la même hauteur. </s> <s xml:id="echoid-s19464" xml:space="preserve">C’eſt en vain qu’on a tenté d’expli-<lb/>quer ce principe par l’accélération des vîteſſes, cauſée par la <lb/>peſanteur. </s> <s xml:id="echoid-s19465" xml:space="preserve">La premiere tranche arrivée en bas du vaſe ne peut <lb/>pas avoir acquis de vîteſſe plus grande que celle de la derniere, <lb/>puiſqu’elle ne peut paſſer qu’aprés elle, & </s> <s xml:id="echoid-s19466" xml:space="preserve">ainſi de toutes les <pb o="621" file="0711" n="737" rhead="DE MATHÉMATIQUE. Liv. XVI."/> autres ſucceſſivement. </s> <s xml:id="echoid-s19467" xml:space="preserve">Il faut avoir recours à d’autres dé-<lb/>monſtrations, tirées de la maniere dont les fluides agiſſent <lb/>ſur leurs parties. </s> <s xml:id="echoid-s19468" xml:space="preserve">On eſt redevable à M. </s> <s xml:id="echoid-s19469" xml:space="preserve">Varignon de la dé-<lb/>monſtration complette que nous venons d’apporter. </s> <s xml:id="echoid-s19470" xml:space="preserve">Ce prin-<lb/>cipe pouvoit être regardé avant lui comme douteux, puiſque <lb/>l’on ne l’avoit point démontré par une raiſon convenable à la <lb/>nature des fluides, & </s> <s xml:id="echoid-s19471" xml:space="preserve">qu’au contraire on avoit eu recours à <lb/>des cauſes qui ne peuvent avoir lieu.</s> <s xml:id="echoid-s19472" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19473" xml:space="preserve">1158. </s> <s xml:id="echoid-s19474" xml:space="preserve">2°. </s> <s xml:id="echoid-s19475" xml:space="preserve">Dans le cas où l’ouverture eſt égale au diametre <lb/> <anchor type="note" xlink:label="note-0711-01a" xlink:href="note-0711-01"/> de la baſe, quelques Auteurs prouvent que la vîteſſe de l’eau, <lb/>au ſortir de cette baſe, doit être égale à la racine quarrée de <lb/>la hauteur, en conſidérant le fluide qui tombe tout entier dans <lb/>le même tems comme un morceau de glace. </s> <s xml:id="echoid-s19476" xml:space="preserve">Je vois bien que <lb/>dans cette hypotheſe, lorſque la tranche A B ſera venue en <lb/>C D, elle aura acquiſe une vîteſſe exprimée par la racine de <lb/>cette hauteur: </s> <s xml:id="echoid-s19477" xml:space="preserve">mais je ne vois nullement que la derniere tran-<lb/>che C D, au premier inſtant de la chûte, ait la même vîteſſe; <lb/></s> <s xml:id="echoid-s19478" xml:space="preserve">ce qui eſt pourtant l’état de la queſtion. </s> <s xml:id="echoid-s19479" xml:space="preserve">Ainſi cette preuve ne <lb/>peut être admiſe, d’autant plus qu’il n’y a aucune comparaiſon <lb/>à faire entre un corps cylindrique de glace & </s> <s xml:id="echoid-s19480" xml:space="preserve">une colonne de <lb/>fluide de même baſe & </s> <s xml:id="echoid-s19481" xml:space="preserve">de même hauteur. </s> <s xml:id="echoid-s19482" xml:space="preserve">La raiſon en eſt, <lb/>que dans ce cylindre de glace, la tranche C D étant attachée <lb/>fortement avec toute la maſſe, ne peut reſſentir l’impreſſion <lb/>des parties ſupérieures; </s> <s xml:id="echoid-s19483" xml:space="preserve">au lieu que cette impreſſion nulle dans <lb/>un ſolide, a néceſſairement lieu dans un fluide.</s> <s xml:id="echoid-s19484" xml:space="preserve"/> </p> <div xml:id="echoid-div1622" type="float" level="2" n="1"> <note position="right" xlink:label="note-0711-01" xlink:href="note-0711-01a" xml:space="preserve">Figure 421.</note> </div> </div> <div xml:id="echoid-div1624" type="section" level="1" n="1173"> <head xml:id="echoid-head1390" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19485" xml:space="preserve">1159. </s> <s xml:id="echoid-s19486" xml:space="preserve">Il ſuit delà que la vîteſſe d’un fluide, à la ſortie du <lb/> <anchor type="note" xlink:label="note-0711-02a" xlink:href="note-0711-02"/> vaſe qui le contient, eſt égale à celle qu’un corps auroit ac-<lb/>quis en tombant d’une hauteur égale à celle de la ſurface de <lb/>l’eau au deſſus du fond du vaſe: </s> <s xml:id="echoid-s19487" xml:space="preserve">car cette vîteſſe eſt auſſi ex-<lb/>primée par la racine quarrée de la hauteur.</s> <s xml:id="echoid-s19488" xml:space="preserve"/> </p> <div xml:id="echoid-div1624" type="float" level="2" n="1"> <note position="right" xlink:label="note-0711-02" xlink:href="note-0711-02a" xml:space="preserve">Figure 421.</note> </div> </div> <div xml:id="echoid-div1626" type="section" level="1" n="1174"> <head xml:id="echoid-head1391" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19489" xml:space="preserve">1160. </s> <s xml:id="echoid-s19490" xml:space="preserve">Nous venons de voir que ſi un fluide s’échappe par <lb/>une ouverture égale à celle de ſa baſe, la vîteſſe qu’il a eſt <lb/>égale à celle d’un corps qui ſeroit tombé librement de la hau-<lb/>teur de cette colonne de fluide. </s> <s xml:id="echoid-s19491" xml:space="preserve">De plus avec cette vîteſſe, le <lb/>corps dans la moitié du tems de la chûte par B D parcourt le <lb/>même eſpace B D: </s> <s xml:id="echoid-s19492" xml:space="preserve">donc avec la vîteſſe que lefluide a au ſortir <pb o="622" file="0712" n="738" rhead="NOUVEAU COURS"/> du vaſe, il lui faudra, pour vuider le vaſe entiérement, un <lb/>tems égal à la moitié du tems qu’un corps grave employeroit à <lb/>parcourir la même hauteur B D.</s> <s xml:id="echoid-s19493" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1627" type="section" level="1" n="1175"> <head xml:id="echoid-head1392" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19494" xml:space="preserve">1161. </s> <s xml:id="echoid-s19495" xml:space="preserve">Comme la vîteſſe eſt la même, lorſque le trou eſt <lb/> <anchor type="note" xlink:label="note-0712-01a" xlink:href="note-0712-01"/> plus petit que la baſe, il s’enſuit que dans la moitié du tems <lb/>qu’un corps mettroit à parcourir A C, il paſſera une quantité <lb/>d’eau égale à la colonne A D: </s> <s xml:id="echoid-s19496" xml:space="preserve">par conſéquent dans le tems <lb/>de la chûte, par A D, il ſortira une colonne double de la <lb/>même colonne A D, pourvu que le vaiſſeau ſoit toujours en-<lb/>tretenu plein à la même hauteur, pour conſerver l’égalité de <lb/>vîteſſe. </s> <s xml:id="echoid-s19497" xml:space="preserve">On peut donc dire en général, que la dépenſe d’un <lb/>tuyau ou réſervoir, pendant le tems qu’il faudroit à un corps pour <lb/>tomber de la hauteur du niveau de l’eau au deſſus du fond, eſt égale <lb/>à une colonne qui auroit pour baſe l’orifice, & </s> <s xml:id="echoid-s19498" xml:space="preserve">pour hauteur une <lb/>ligne égale à celle que le corps parcourroit uniformément pendant <lb/>le même tems avec la vîteſſe acquiſe, c’eſt-à-dire une colonne double.</s> <s xml:id="echoid-s19499" xml:space="preserve"/> </p> <div xml:id="echoid-div1627" type="float" level="2" n="1"> <note position="left" xlink:label="note-0712-01" xlink:href="note-0712-01a" xml:space="preserve">Figure 422.</note> </div> </div> <div xml:id="echoid-div1629" type="section" level="1" n="1176"> <head xml:id="echoid-head1393" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s19500" xml:space="preserve">1162. </s> <s xml:id="echoid-s19501" xml:space="preserve">Il ſuit encore delà que l’on peut aiſément connoître <lb/>la dépenſe d’un tuyau dans un certain tems, ſi l’on connoît <lb/>le diametre de l’ouverture, & </s> <s xml:id="echoid-s19502" xml:space="preserve">la hauteur de l’eau au deſſus du <lb/>fond, que nous ſuppoſons toujours la même. </s> <s xml:id="echoid-s19503" xml:space="preserve">Pour cela, il n’y <lb/>aura qu’à chercher le tems de la chûte d’un corps par la hau-<lb/>teur de l’eau au deſſus de la baſe, enſuite chercher combien <lb/>de fois ce tems eſt contenu dans le propoſé, & </s> <s xml:id="echoid-s19504" xml:space="preserve">multiplier <lb/>après par le quotient une colonne double de celle qui auroit <lb/>pour baſe l’orifice, & </s> <s xml:id="echoid-s19505" xml:space="preserve">pour hauteur celle de l’eau au deſſus de <lb/>l’orifice. </s> <s xml:id="echoid-s19506" xml:space="preserve">Ce procédé ſuit évidemment du corollaire précédent: <lb/></s> <s xml:id="echoid-s19507" xml:space="preserve">car puiſque dans le tems de la chûte, par la hauteur de l’eau, <lb/>il s’écoule une colonne double de la même hauteur, ſi le tems <lb/>donné eſt décuple du tems de la chûte par cette hauteur, il <lb/>s’écoulera une colonne dix fois double, ou vingt fois plus <lb/>grande que la propoſée, pourvu, comme on le ſuppoſe, que la <lb/>hauteur ſoit toujours la même.</s> <s xml:id="echoid-s19508" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1630" type="section" level="1" n="1177"> <head xml:id="echoid-head1394" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s19509" xml:space="preserve">1163. </s> <s xml:id="echoid-s19510" xml:space="preserve">Si l’on a des vaſes qui aient des hauteurs inégales, & </s> <s xml:id="echoid-s19511" xml:space="preserve"><lb/>des orifices auſſi différens, mais ſemblables, comme des cercles <pb o="623" file="0713" n="739" rhead="DE MATHÉMATIQUE. Liv. XVI."/> ou des quarrés, les quantités d’eau écoulées ou les dépenſes <lb/>ſeront dans la raiſon compoſée des racines quarrées des hau-<lb/>teurs, & </s> <s xml:id="echoid-s19512" xml:space="preserve">des quarrés des diametres, pourvu que ces mêmes <lb/>vaſes ſoient toujours entretenus à la même hauteur d’eau: <lb/></s> <s xml:id="echoid-s19513" xml:space="preserve">donc ſi l’on appelle D & </s> <s xml:id="echoid-s19514" xml:space="preserve">d les dépenſes, H & </s> <s xml:id="echoid-s19515" xml:space="preserve">h les hauteurs, <lb/>L & </s> <s xml:id="echoid-s19516" xml:space="preserve">l les largeurs ou diametres des orifices, on aura D: </s> <s xml:id="echoid-s19517" xml:space="preserve">d:</s> <s xml:id="echoid-s19518" xml:space="preserve">: <lb/>√ H x L<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19519" xml:space="preserve">√h x l<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19520" xml:space="preserve">donc D x √h x l<emph style="sub">2</emph> = d √ H x L<emph style="sub">2</emph>. </s> <s xml:id="echoid-s19521" xml:space="preserve">Si les <lb/>dépenſes ſont égales, on aura √ h x l<emph style="sub">2</emph> = √ H x L<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19522" xml:space="preserve">donc <lb/>√ H : </s> <s xml:id="echoid-s19523" xml:space="preserve">√h:</s> <s xml:id="echoid-s19524" xml:space="preserve">: l<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19525" xml:space="preserve">L<emph style="sub">2</emph>, c’eſt-à-dire que les orifices ſont dans la <lb/>raiſon réciproque des racines quarrées des hauteurs, ou des <lb/>vîteſſes qui ſont exprimées par ces racines. </s> <s xml:id="echoid-s19526" xml:space="preserve">Si au lieu des dé-<lb/>penſes on met les produits des baſes par les vîteſſes qui leur <lb/>ſont égaux, on aura V x L<emph style="sub">2</emph> x √h x l<emph style="sub">2</emph> = u x l<emph style="sub">2</emph> x √ H x L<emph style="sub">2</emph>, <lb/>ou V x √h = u√ H, d’où l’on tire comme ci-devant, V : </s> <s xml:id="echoid-s19527" xml:space="preserve">u <lb/>:</s> <s xml:id="echoid-s19528" xml:space="preserve">: √ H : </s> <s xml:id="echoid-s19529" xml:space="preserve">√h.</s> <s xml:id="echoid-s19530" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1631" type="section" level="1" n="1178"> <head xml:id="echoid-head1395" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s19531" xml:space="preserve">1164. </s> <s xml:id="echoid-s19532" xml:space="preserve">Comme l’eau contenue dans un vaſe fait un effort <lb/> <anchor type="note" xlink:label="note-0713-01a" xlink:href="note-0713-01"/> égal de tous côtés pour s’échapper, il ſuit encore delà que ſi <lb/>l’on a un vaſe, comme A D, rempli d’eau, toujours entre-<lb/>tenue à la même hauteur, & </s> <s xml:id="echoid-s19533" xml:space="preserve">qu’on pratique deux ouvertures <lb/>en B & </s> <s xml:id="echoid-s19534" xml:space="preserve">C, les vîteſſes de l’eau, à la ſortie, ſeront comme les <lb/>racines quarrées des hauteurs A B & </s> <s xml:id="echoid-s19535" xml:space="preserve">A C, ſoit que l’eau, à la <lb/>ſortie des ouvertures, ſoit pouſſée ſuivant une direction ver-<lb/>ticale, horizontale, ou inclinée à l’horizon. </s> <s xml:id="echoid-s19536" xml:space="preserve">Il eſt cependant <lb/>à remarquer que cela ne ſe trouve pas exactement vrai, c’eſt-<lb/>à-dire que les vîteſſes de l’eau, ſuivant des directions incli-<lb/>nées, ne ſont pas ſi grandes en ſortant, que ſelon des direc-<lb/>tions horizontales ou des directions verticales, lorſqu’elle <lb/>coule de haut en bas. </s> <s xml:id="echoid-s19537" xml:space="preserve">Cette différence vient de ce que les <lb/>parties de l’eau ne s’échappent pas ſi aiſément, ſuivant des <lb/>directions obliques, que ſuivant des directions horizontales, <lb/>ni ſi facilement ſelon des directions horizontales, que ſuivant <lb/>des directions verticales.</s> <s xml:id="echoid-s19538" xml:space="preserve"/> </p> <div xml:id="echoid-div1631" type="float" level="2" n="1"> <note position="right" xlink:label="note-0713-01" xlink:href="note-0713-01a" xml:space="preserve">Figure 424.</note> </div> </div> <div xml:id="echoid-div1633" type="section" level="1" n="1179"> <head xml:id="echoid-head1396" xml:space="preserve"><emph style="sc">Corollaire</emph> VII.</head> <p> <s xml:id="echoid-s19539" xml:space="preserve">1165. </s> <s xml:id="echoid-s19540" xml:space="preserve">Il ſuit encore delà que ſi l’eau ſort ſuivant une di-<lb/> <anchor type="note" xlink:label="note-0713-02a" xlink:href="note-0713-02"/> rection horizontale, le jet décrira une parabole, dont le ſom-<lb/>met ſera en B: </s> <s xml:id="echoid-s19541" xml:space="preserve">car nous avons démontré dans le Traité du <lb/>Mouvement, que ſi l’on a un demi-cercle A F C, dont le dia- <pb o="624" file="0714" n="740" rhead="NOUVEAU COURS"/> metre A C ſoit vertical, & </s> <s xml:id="echoid-s19542" xml:space="preserve">qu’on pouſſe un corps quelconque <lb/>ſuivant une direction B D avec une force exprimée par la ra-<lb/>cine de A B, qui eſt celle qu’il auroit acquiſe en tombant de A <lb/>en B, ce corps décrira une parabole B G E, dont l’amplitude <lb/>C E ſeroit double de la perpendiculaire B F: </s> <s xml:id="echoid-s19543" xml:space="preserve">donc ſi l’on con-<lb/>ſidere les parties de l’eau comme une infinité de petits corps <lb/>pouſſés ſuivant la direction B D avec une force exprimée par <lb/>la racine quarrée de A B, on verra qu’ils décrivent pareille-<lb/>ment la parabole B G E.</s> <s xml:id="echoid-s19544" xml:space="preserve"/> </p> <div xml:id="echoid-div1633" type="float" level="2" n="1"> <note position="right" xlink:label="note-0713-02" xlink:href="note-0713-02a" xml:space="preserve">Figure 425.</note> </div> <p> <s xml:id="echoid-s19545" xml:space="preserve">De même ſi l’eau ſort ſuivant une direction C G avec <lb/> <anchor type="note" xlink:label="note-0714-01a" xlink:href="note-0714-01"/> une vîteſſe exprimée par la racine quarrée de la hauteur A C, <lb/>que je ſuppoſe être celle du niveau de l’eau au deſſus de la baſe, <lb/>le jet décrira la parabole C E F, dont le ſommet ſera le point <lb/>E, puiſque nous avons fait voir que tout corps pouſſé ſuivant <lb/>une direction C G oblique à l’horizon, avec une force expri-<lb/>mée par √AC\x{0020}, qui eſt la force de l’eau à ſa ſortie, doit décrire. <lb/></s> <s xml:id="echoid-s19546" xml:space="preserve">une parabole.</s> <s xml:id="echoid-s19547" xml:space="preserve"/> </p> <div xml:id="echoid-div1634" type="float" level="2" n="2"> <note position="left" xlink:label="note-0714-01" xlink:href="note-0714-01a" xml:space="preserve">Figure 426.</note> </div> </div> <div xml:id="echoid-div1636" type="section" level="1" n="1180"> <head xml:id="echoid-head1397" xml:space="preserve"><emph style="sc">Corollaire</emph> VIII.</head> <p> <s xml:id="echoid-s19548" xml:space="preserve">1166. </s> <s xml:id="echoid-s19549" xml:space="preserve">Il ſuit encore delà que ſi l’on a un réſervoir A B C D, <lb/> <anchor type="note" xlink:label="note-0714-02a" xlink:href="note-0714-02"/> au bas duquel il y ait une ouverture D, & </s> <s xml:id="echoid-s19550" xml:space="preserve">un tuyau recourbé <lb/>à cette ouverture de D vers E, l’eau montera dans ce tuyau <lb/>D E avec la vîteſſe acquiſe juſqu’à la hauteur dont elle eſt <lb/>deſcendue: </s> <s xml:id="echoid-s19551" xml:space="preserve">car nous avons vu que ſi un corps eſt pouſſé avec <lb/>la force qu’il a acquiſe en tombant d’une certaine hauteur, il <lb/>doit remonter à la même hauteur. </s> <s xml:id="echoid-s19552" xml:space="preserve">Ce principe eſt d’un grand <lb/>uſage dans la conduite des eaux, & </s> <s xml:id="echoid-s19553" xml:space="preserve">dans les différentes diſtri-<lb/>butions. </s> <s xml:id="echoid-s19554" xml:space="preserve">Lorſqu’on veut ſçavoir ſi l’on peut mener de l’eau <lb/>d’un endroit à un autre, il faut d’abord s’aſſurer ſi celui où <lb/>ſe trouve la ſource eſt plus élevé que l’endroit où l’on veut la <lb/>conduire, ce que l’on reconnoîtra par un nivellement exact. <lb/></s> <s xml:id="echoid-s19555" xml:space="preserve">Si cette ſource eſt tant ſoit peu plus élevée que le lieu auquel <lb/>on veut conduire de l’eau, alors par le moyen des canaux <lb/>pratiqués entre les deux endroits, on peut ſe la procurer. </s> <s xml:id="echoid-s19556" xml:space="preserve">Sur <lb/>quoi il eſt à remarquer que lorſqu’il faut que l’eau monte <lb/>pour arriver au lieu de ſa deſtination, après avoir deſcendu, <lb/>comme cela peut arriver par l’inégalité du terrein qui ſe trouve <lb/>entre deux, il faut que la ſource ſoit de quelque choſe plus <lb/>élevée que le lieu où on conduit ſes eaux, ſans quoi l’on s’ex-<lb/>poſeroit à une dépenſe inutile, parce que pluſieurs cauſes con- <pb o="625" file="0715" n="741" rhead="DE MATHÉMATIQUE. Liv. XVI."/> courent à altérer la vîteſſe de l’eau dans le tuyau, & </s> <s xml:id="echoid-s19557" xml:space="preserve">par con-<lb/>ſéquent diminuent la force qu’elle a pour monter.</s> <s xml:id="echoid-s19558" xml:space="preserve"/> </p> <div xml:id="echoid-div1636" type="float" level="2" n="1"> <note position="left" xlink:label="note-0714-02" xlink:href="note-0714-02a" xml:space="preserve">Figure 427.</note> </div> </div> <div xml:id="echoid-div1638" type="section" level="1" n="1181"> <head xml:id="echoid-head1398" xml:space="preserve"><emph style="sc">Corollaire</emph> IX.</head> <p> <s xml:id="echoid-s19559" xml:space="preserve">1167. </s> <s xml:id="echoid-s19560" xml:space="preserve">C’eſt auſſi à peu près la même raiſon qui fait que <lb/> <anchor type="note" xlink:label="note-0715-01a" xlink:href="note-0715-01"/> dans un jet d’eau l’eau ne monte pas tout-à-fait à la même <lb/>hauteur de celle du réſervoir qui fournit le même jet. </s> <s xml:id="echoid-s19561" xml:space="preserve">L’air ré-<lb/>ſiſtant aux parties de l’eau à meſure qu’elles ſortent de l’aju-<lb/>tage, qui eſt en C, diminue leur vîteſſe, & </s> <s xml:id="echoid-s19562" xml:space="preserve">les empêche de <lb/>s’élever juſqu’à la ſurface du niveau de l’eau du réſervoir. <lb/></s> <s xml:id="echoid-s19563" xml:space="preserve">M. </s> <s xml:id="echoid-s19564" xml:space="preserve">Mariotte dans ſon Traité du Mouvement des Eaux a fait <lb/>pluſieurs obſervations pour ſçavoir ſuivant quel rapport dimi-<lb/>nuent les hauteurs auxquelles s’élevent différens jets qui ont <lb/>mêmes ajutages, & </s> <s xml:id="echoid-s19565" xml:space="preserve">des réſervoirs inégaux. </s> <s xml:id="echoid-s19566" xml:space="preserve">Il a trouvé que <lb/>cette diminution ſuivoit le rapport des racines quarrées des <lb/>hauteurs; </s> <s xml:id="echoid-s19567" xml:space="preserve">d’où l’on voit que ſi l’on ſçait la hauteur à laquelle <lb/>un jet d’eau s’éleve, & </s> <s xml:id="echoid-s19568" xml:space="preserve">de plus la hauteur du réſervoir qui la <lb/>fournit, on pourroit connoître, par une ſeule proportion, la <lb/>hauteur à laquelle un jet d’eau d’un réſervoir donné de hau-<lb/>teur, peut s’élever par un ajutage de même diametre. </s> <s xml:id="echoid-s19569" xml:space="preserve">De <lb/>plus, les dépenſes étant toujours à proportion des vîteſſes, il <lb/>s’enſuit que ſi l’on connoît la dépenſe d’un réſervoir d’une <lb/>hauteur donnée par un ajutage donné, on connoîtra auſſi la <lb/>dépenſe d’un autre réſervoir de hauteur auſſi donnée par telle <lb/>ouverture que ce ſoit auſſi donnée.</s> <s xml:id="echoid-s19570" xml:space="preserve"/> </p> <div xml:id="echoid-div1638" type="float" level="2" n="1"> <note position="right" xlink:label="note-0715-01" xlink:href="note-0715-01a" xml:space="preserve">Figure 428.</note> </div> <p> <s xml:id="echoid-s19571" xml:space="preserve">M. </s> <s xml:id="echoid-s19572" xml:space="preserve">Mariotte a trouvé qu’ayant un réſervoir toujours rempli <lb/>d’eau, & </s> <s xml:id="echoid-s19573" xml:space="preserve">dont la hauteur A B étoit de 13 pieds, & </s> <s xml:id="echoid-s19574" xml:space="preserve">le diametre <lb/>de l’ajutage de 3 lignes, il ſort pendant une minute, par le <lb/>même ajutage, 14 pintes, meſure de paris; </s> <s xml:id="echoid-s19575" xml:space="preserve">la pinte peſant <lb/>deux livres: </s> <s xml:id="echoid-s19576" xml:space="preserve">ainſi il n’en faut pas davantage pour réſoudre le <lb/>problême ſuivant.</s> <s xml:id="echoid-s19577" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1640" type="section" level="1" n="1182"> <head xml:id="echoid-head1399" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s19578" xml:space="preserve">1168. </s> <s xml:id="echoid-s19579" xml:space="preserve">Trouver la dépenſe d’un jet d’eau pendant une minute <lb/>par un ajutage de 4 lignes de diametre, l’eau du réſervoir étant de <lb/>40 pieds de hauteur.</s> <s xml:id="echoid-s19580" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1641" type="section" level="1" n="1183"> <head xml:id="echoid-head1400" xml:space="preserve"><emph style="sc">Solution</emph>.</head> <p> <s xml:id="echoid-s19581" xml:space="preserve">Nous ſçavons que lorſque les ajutages ſont égaux, la dé- <pb o="626" file="0716" n="742" rhead="NOUVEAU COURS"/> penſe des eaux eſt dans la raiſon des racines quarrées des dif-<lb/>férentes hauteurs de l’eau; </s> <s xml:id="echoid-s19582" xml:space="preserve">& </s> <s xml:id="echoid-s19583" xml:space="preserve">que quand les ajutages ſont dif-<lb/>férens, les dépenſes ſont dans la raiſon compoſée des racines <lb/>quarrées des hauteurs, & </s> <s xml:id="echoid-s19584" xml:space="preserve">des quarrés des diametres des aju-<lb/>tages: </s> <s xml:id="echoid-s19585" xml:space="preserve">ainſi en faiſant uſage de l’expérience de M. </s> <s xml:id="echoid-s19586" xml:space="preserve">Mariotte, <lb/>nous dirons: </s> <s xml:id="echoid-s19587" xml:space="preserve">ſi le produit du quarré de 3 lignes, qui eſt 9, par la <lb/>racine de 13, donne 14 pintes pour la dépenſe de l’eau pen-<lb/>dant une minute, combien donnera le produit du quarré du <lb/>diametre 4 de l’ajutage, qui eſt 16, par la racine quarrée de <lb/>40, pour la dépenſe que l’on demande pendant le même tems? <lb/></s> <s xml:id="echoid-s19588" xml:space="preserve">Le quatrieme terme de cette Regle de Trois fera trouver le <lb/>nombre de pintes que l’on cherche, C. </s> <s xml:id="echoid-s19589" xml:space="preserve">Q. </s> <s xml:id="echoid-s19590" xml:space="preserve">F. </s> <s xml:id="echoid-s19591" xml:space="preserve">D.</s> <s xml:id="echoid-s19592" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1642" type="section" level="1" n="1184"> <head xml:id="echoid-head1401" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s19593" xml:space="preserve">1169. </s> <s xml:id="echoid-s19594" xml:space="preserve">Si les tems n’étoient pas égaux, on pourroit toujours <lb/>trouver par une ſeule Regle de Trois la dépenſe pendant un <lb/>tems donné; </s> <s xml:id="echoid-s19595" xml:space="preserve">car les dépenſes ſont toujours dans la raiſon <lb/>compoſée des racines quarrées deshauteurs, des quarrés des dia-<lb/>metres, & </s> <s xml:id="echoid-s19596" xml:space="preserve">de la raiſon ſimple des tems: </s> <s xml:id="echoid-s19597" xml:space="preserve">enſorte que ſi l’on a <lb/>un réſervoir, dont la hauteur ſoit H, la dépenſe D par un <lb/>ajutage; </s> <s xml:id="echoid-s19598" xml:space="preserve">dont le diametre ſoit F, pendant un tems T & </s> <s xml:id="echoid-s19599" xml:space="preserve">un <lb/>autre réſervoir, dont la hauteur ſoit h, la dépenſe d par un <lb/>ajutage; </s> <s xml:id="echoid-s19600" xml:space="preserve">dont le diametre ſoit f, pendant un tems t on aura <lb/>cette proportion, D : </s> <s xml:id="echoid-s19601" xml:space="preserve">d :</s> <s xml:id="echoid-s19602" xml:space="preserve">: F F T √ H : </s> <s xml:id="echoid-s19603" xml:space="preserve">f f t √ h; </s> <s xml:id="echoid-s19604" xml:space="preserve">d’où l’on tire <lb/>D f f t √ h = d F F T √ H; </s> <s xml:id="echoid-s19605" xml:space="preserve">& </s> <s xml:id="echoid-s19606" xml:space="preserve">l’on peut faire uſage de cette for-<lb/>mule pour déterminer tous les cas qui ont rapport aux diffé-<lb/>rentes queſtions que l’on peut propoſer ſur les dépenſes des ré-<lb/>ſervoirs, ſelon les différentes combinaiſons des tems, des hau-<lb/>teurs, & </s> <s xml:id="echoid-s19607" xml:space="preserve">des diametres.</s> <s xml:id="echoid-s19608" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1643" type="section" level="1" n="1185"> <head xml:id="echoid-head1402" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19609" xml:space="preserve">1170. </s> <s xml:id="echoid-s19610" xml:space="preserve">Si un vaſe cylindrique plein d’eau ſe déſemplit par une <lb/> <anchor type="note" xlink:label="note-0716-01a" xlink:href="note-0716-01"/> ouverture D, beaucoup plus petite que le fond de la baſe, les <lb/> <anchor type="note" xlink:label="note-0716-02a" xlink:href="note-0716-02"/> quantités d’eau qui s’écouleront dans des tems égaux ſeront comme <lb/>les nombres impairs pris dans un ordre renverſé, c’eſt-à-dire comme <lb/>la ſuite des nombres 11, 9, 7, 5, &</s> <s xml:id="echoid-s19611" xml:space="preserve">c.</s> <s xml:id="echoid-s19612" xml:space="preserve"/> </p> <div xml:id="echoid-div1643" type="float" level="2" n="1"> <note position="left" xlink:label="note-0716-01" xlink:href="note-0716-01a" xml:space="preserve">Pl. XXXIII.</note> <note position="left" xlink:label="note-0716-02" xlink:href="note-0716-02a" xml:space="preserve">Figure 422.</note> </div> </div> <div xml:id="echoid-div1645" type="section" level="1" n="1186"> <head xml:id="echoid-head1403" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19613" xml:space="preserve">Concevons le vaſe coupé par des plans paralleles, dont les <pb o="627" file="0717" n="743" rhead="DE MATHÉMATIQUE. Liv. XVI."/> hauteurs C L, C M, C F, O N ſoient comme les quarrés 1, 4, <lb/>9, 16, &</s> <s xml:id="echoid-s19614" xml:space="preserve">c. </s> <s xml:id="echoid-s19615" xml:space="preserve">des nombres naturels 1, 2, 3, 4, &</s> <s xml:id="echoid-s19616" xml:space="preserve">c. </s> <s xml:id="echoid-s19617" xml:space="preserve">Quand l’eau <lb/>commencera à couler, ſa vîteſſe étant exprimée par la racine <lb/>quarrée de la hauteur, ſera comme 4; </s> <s xml:id="echoid-s19618" xml:space="preserve">& </s> <s xml:id="echoid-s19619" xml:space="preserve">quand le niveau de <lb/>l’eau ſera deſcendu en E F, la vîteſſe deviendra comme racine <lb/>de 9, qui eſt 3. </s> <s xml:id="echoid-s19620" xml:space="preserve">Pareillement lorſque le niveau ſera en M m, <lb/>ſa vîteſſe ſera comme 2, & </s> <s xml:id="echoid-s19621" xml:space="preserve">enfin lorſqu’elle ſera en L l, la <lb/>vîteſſe ſera exprimée par 1. </s> <s xml:id="echoid-s19622" xml:space="preserve">Préſentement faiſons attention <lb/>que les cylindres, dont les hauteurs ſont C L, C M, C F, C N, <lb/>ayant des baſes égales, ſont entr’eux comme les mêmes hau-<lb/>teurs, c’eſt-à-dire comme 1, 4, 9, 16. </s> <s xml:id="echoid-s19623" xml:space="preserve">Et ſi l’on ſuppoſe pour <lb/>un inſtant que la vîteſſe de N juſqu’en F a été uniforme, que <lb/>celle de F juſqu’en M l’a été auſſi, les quantités N F, F M, <lb/>M L, L C, écoulées pendant des tems égaux, leſquelles ne <lb/>ſont que les différences des cylindres 7, 5, 3, 1, ſont préciſé-<lb/>ment dans la raiſon inverſe des nombres impairs 1, 3, 5, 7. <lb/></s> <s xml:id="echoid-s19624" xml:space="preserve">Préſentement ſi l’on fait attention que quoique la vîteſſe de <lb/>N en F ait diminué continuellement, cependant on peut <lb/>trouver une vîteſſe moyenne, qui regardée & </s> <s xml:id="echoid-s19625" xml:space="preserve">ſuppoſée conſ-<lb/>tante, ait donné la même dépenſe, & </s> <s xml:id="echoid-s19626" xml:space="preserve">ainſi des autres; </s> <s xml:id="echoid-s19627" xml:space="preserve">il s’en-<lb/>ſuit néceſſairement que les quantités d’eaux écoulées pendant <lb/>des tems égaux, ſont comme les nombres 7, 5, 3, 1. </s> <s xml:id="echoid-s19628" xml:space="preserve">C. </s> <s xml:id="echoid-s19629" xml:space="preserve">Q. </s> <s xml:id="echoid-s19630" xml:space="preserve">F. </s> <s xml:id="echoid-s19631" xml:space="preserve">D.</s> <s xml:id="echoid-s19632" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1646" type="section" level="1" n="1187"> <head xml:id="echoid-head1404" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19633" xml:space="preserve">1171. </s> <s xml:id="echoid-s19634" xml:space="preserve">Il eſt aiſé de voir que dans ce cas le diametre de l’ou-<lb/>verture doit être beaucoup plus petit que celui de la baſe: </s> <s xml:id="echoid-s19635" xml:space="preserve">car <lb/>alors l’eau tomberoit comme une ſeule maſſe, de maniere que <lb/>les parties inférieures n’auroient pas plus de vîteſſe que les ſu-<lb/>périeures. </s> <s xml:id="echoid-s19636" xml:space="preserve">C’eſt ce que l’on peut remarquer aiſément par un <lb/>grand entonnoir qui ſe forme tout d’un coup à la ſurface de <lb/>l’eau, & </s> <s xml:id="echoid-s19637" xml:space="preserve">qui prouve inconteſtablement que l’eau du milieu <lb/>ſort avec une plus grande vîteſſe dans ce cas que dans les autres.</s> <s xml:id="echoid-s19638" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1647" type="section" level="1" n="1188"> <head xml:id="echoid-head1405" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19639" xml:space="preserve">1172. </s> <s xml:id="echoid-s19640" xml:space="preserve">Il ſuit encore delà que l’on peut connoître les quan-<lb/>tités d’eau écoulées pendant un certain tems donné, ſi l’on con-<lb/>noît le tems total qu’un vaſe a employé à ſe vuider. </s> <s xml:id="echoid-s19641" xml:space="preserve">Suppo-<lb/>ſons, par exemple (fig. </s> <s xml:id="echoid-s19642" xml:space="preserve">421) que le vaſe ait été ſix heures detems <lb/>à ſe déſemplir par une ouverture beaucoup plus petite que la <lb/>baſe. </s> <s xml:id="echoid-s19643" xml:space="preserve">Je conçois le vaſe coupé par 36 tranches égales entr’elles.</s> <s xml:id="echoid-s19644" xml:space="preserve"> <pb o="628" file="0718" n="744" rhead="NOUVEAU COURS"/> Cela fait, je diviſe encore le nombre 36 en ſix autres parties <lb/>inégales entr’elles, dont la premiere contienne 11 de ces parties <lb/>égales, la ſeconde 9, la troiſieme 7, & </s> <s xml:id="echoid-s19645" xml:space="preserve">ainſi de ſuite. </s> <s xml:id="echoid-s19646" xml:space="preserve">De cette<unsure/> <lb/>maniere on verra que dans la premiere heure de l’écoulement <lb/>il eſt ſorti du vaſe un cylindre égal à 11 parties égales, c’eſt-<lb/>à-dire les {11/36} de l’eau contenu dans le même vaſe: </s> <s xml:id="echoid-s19647" xml:space="preserve">à la deuxieme <lb/>heure il en ſera ſorti {9/36} ou {1/4}, & </s> <s xml:id="echoid-s19648" xml:space="preserve">ainſi des autres; </s> <s xml:id="echoid-s19649" xml:space="preserve">ce qui eſt bien <lb/>évident, puiſque la ſomme des nombres 11, 9, 7, 5, 3, 2, 1 <lb/>fait préciſément 36, & </s> <s xml:id="echoid-s19650" xml:space="preserve">que par le théorême dont il s’agit, les <lb/>quantités écoulées dans des tems égaux ſuivent le rapport des <lb/>mêmes nombres.</s> <s xml:id="echoid-s19651" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1648" type="section" level="1" n="1189"> <head xml:id="echoid-head1406" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19652" xml:space="preserve">1173. </s> <s xml:id="echoid-s19653" xml:space="preserve">Il ſuit delà que puiſque la vîteſſe de l’eau ſortant d’un <lb/>vaſe qui ſe vuide eſt continuellement retardée, ſi le vaſe a voit tou-<lb/>jours été entretenu à la même hauteur d’eau, comme la vîteſſe <lb/>auroit été toujours uniforme, auſſi, ſuivant la loi de Galilée, <lb/>la quantité d’eau écoulée uniformément pendant le tems que <lb/>le vaſe ſe déſemplit, ſera double de l’eau qui étoit dans le vaſe.</s> <s xml:id="echoid-s19654" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1649" type="section" level="1" n="1190"> <head xml:id="echoid-head1407" xml:space="preserve"><emph style="sc">Corollaire</emph>. IV.</head> <p> <s xml:id="echoid-s19655" xml:space="preserve">1174. </s> <s xml:id="echoid-s19656" xml:space="preserve">Il ſuit encore delà que ſi des vaſes qui ſe déſempliſ-<lb/>ſent ont des hauteurs & </s> <s xml:id="echoid-s19657" xml:space="preserve">des ouvertures égales, avec des baſes <lb/>inégales, les tems qu’ils mettront à ſe vuider entiérement <lb/>ſeront dans la raiſon des baſes: </s> <s xml:id="echoid-s19658" xml:space="preserve">car les tems que ces vaſes em-<lb/>ploient à ſe vuider ſont égaux au tems qu’il faudroit pour <lb/>qu’il s’écoulât, par un mouvement uniforme, une quantité <lb/>double de l’eau qui eſt dans chaque vaſe, en les ſuppoſant en-<lb/>tretenus toujours à la même hauteur (art. </s> <s xml:id="echoid-s19659" xml:space="preserve">1173). </s> <s xml:id="echoid-s19660" xml:space="preserve">Et dans ce <lb/>dernier cas, les tems des écoulemens ſont proportionnels aux <lb/>baſes: </s> <s xml:id="echoid-s19661" xml:space="preserve">donc auſſi lorſque les vaſes ſe déſempliſſent totalement, <lb/>les tems doivent ſuivre la même raiſon.</s> <s xml:id="echoid-s19662" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1650" type="section" level="1" n="1191"> <head xml:id="echoid-head1408" xml:space="preserve"><emph style="sc">Corollaire</emph> V.</head> <p> <s xml:id="echoid-s19663" xml:space="preserve">1175. </s> <s xml:id="echoid-s19664" xml:space="preserve">Si les vaſes ont toujours même hauteur, & </s> <s xml:id="echoid-s19665" xml:space="preserve">des baſes <lb/>avec des ouvertures inégales, il eſt évident que les tems qu’ils <lb/>mettront à ſe vuider ſeront dans la raiſon compoſée de la di-<lb/>recte des baſes, & </s> <s xml:id="echoid-s19666" xml:space="preserve">de l’inverſe des ouvertures ou des quarrés des <lb/>diametres de ces ouvertures, ſi elles ſont des figures ſemblables.</s> <s xml:id="echoid-s19667" xml:space="preserve"/> </p> <pb o="629" file="0719" n="745" rhead="DE MATHÉMATIQUE. Liv. XVI."/> </div> <div xml:id="echoid-div1651" type="section" level="1" n="1192"> <head xml:id="echoid-head1409" xml:space="preserve"><emph style="sc">Corollaire</emph> VI.</head> <p> <s xml:id="echoid-s19668" xml:space="preserve">1176. </s> <s xml:id="echoid-s19669" xml:space="preserve">Il n’eſt pas moins évident que les tems ſeront en-<lb/>core dans la raiſon compoſée de la directe des racines quarrées <lb/>des hauteurs, ſi ces hauteurs ſont inégales, de la directe des <lb/>baſes & </s> <s xml:id="echoid-s19670" xml:space="preserve">de l’inverſe des quarrés des diametres des ouvertures; <lb/></s> <s xml:id="echoid-s19671" xml:space="preserve">enſorte que ſi l’on appelle H la hauteur de l’eau dans un vaſe, <lb/>B la baſe du même vaſe, D le diametre de l’ouverture, & </s> <s xml:id="echoid-s19672" xml:space="preserve">T le <lb/>tems qu’il met à ſe vuider, pareillement h la hauteur de l’eau <lb/>dans un autre vaſe, d le diametre de l’ouverture, b ſa baſe, <lb/>& </s> <s xml:id="echoid-s19673" xml:space="preserve">t le tems qu’il emploie à ſe vuider, on aura T : </s> <s xml:id="echoid-s19674" xml:space="preserve">t :</s> <s xml:id="echoid-s19675" xml:space="preserve">: {B√ H/D D} : </s> <s xml:id="echoid-s19676" xml:space="preserve"><lb/>{b√b/d d}, ou T : </s> <s xml:id="echoid-s19677" xml:space="preserve">t :</s> <s xml:id="echoid-s19678" xml:space="preserve">: B d d √ H : </s> <s xml:id="echoid-s19679" xml:space="preserve">b D D √ h; </s> <s xml:id="echoid-s19680" xml:space="preserve">d’où l’on tire T b D D √ h <lb/>= t B d d √ H; </s> <s xml:id="echoid-s19681" xml:space="preserve">& </s> <s xml:id="echoid-s19682" xml:space="preserve">l’on ſe ſerviroit de cette formule comme <lb/>des précédentes.</s> <s xml:id="echoid-s19683" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1652" type="section" level="1" n="1193"> <head xml:id="echoid-head1410" xml:space="preserve">CHAPITRE III.</head> <head xml:id="echoid-head1411" style="it" xml:space="preserve">Du cours des rivieres, & du choc des fluides en mouvement contre <lb/>les ſurfaces des corps qu’elles rencontrent.</head> <head xml:id="echoid-head1412" xml:space="preserve"><emph style="sc">Definitions</emph>.</head> <head xml:id="echoid-head1413" xml:space="preserve">I.</head> <p> <s xml:id="echoid-s19684" xml:space="preserve">1177. </s> <s xml:id="echoid-s19685" xml:space="preserve">LE lit d’un fleuve ou d’une riviere eſt le canal dans <lb/>lequel il coule.</s> <s xml:id="echoid-s19686" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1653" type="section" level="1" n="1194"> <head xml:id="echoid-head1414" xml:space="preserve">II.</head> <p> <s xml:id="echoid-s19687" xml:space="preserve">1178. </s> <s xml:id="echoid-s19688" xml:space="preserve">Si l’on conçoit un plan vertical qui coupe cette riviere <lb/>dans toute ſon étendue en largeur, & </s> <s xml:id="echoid-s19689" xml:space="preserve">perpendiculairement à <lb/>ſon cours, la figure qui en réſulte eſt appellée profil ou ſection <lb/>du fleuve. </s> <s xml:id="echoid-s19690" xml:space="preserve">Comme la ligne du terrein qui termine cette figure <lb/>eſt aſſez irréguliere, on la réduit en rectangle pour avoir une <lb/>meſure plus aiſée à déterminer.</s> <s xml:id="echoid-s19691" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1654" type="section" level="1" n="1195"> <head xml:id="echoid-head1415" xml:space="preserve">PROPOSITION I. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19692" xml:space="preserve">1179. </s> <s xml:id="echoid-s19693" xml:space="preserve">Toute riviere ou fleuve qui n’eſt point arrêté dans ſon <lb/>mouvement eſt mu d’une vîteſſe accélérée.</s> <s xml:id="echoid-s19694" xml:space="preserve"/> </p> <pb o="630" file="0720" n="746" rhead="NOUVEAU COURS"/> </div> <div xml:id="echoid-div1655" type="section" level="1" n="1196"> <head xml:id="echoid-head1416" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19695" xml:space="preserve">Ou bien le fond du lit de la riviere eſt horizontal, ou bien <lb/> <anchor type="note" xlink:label="note-0720-01a" xlink:href="note-0720-01"/> il eſt incliné à l’horizon. </s> <s xml:id="echoid-s19696" xml:space="preserve">Dans le premier cas, on concevra <lb/>d’abord le lit du fleuve, dont la hauteur eſt B C, repréſenté <lb/>par la ligne C D, & </s> <s xml:id="echoid-s19697" xml:space="preserve">le fleuve diviſé en une infinité de tran-<lb/>ches paralleles. </s> <s xml:id="echoid-s19698" xml:space="preserve">Il eſt viſible que chacune de ces tranches coule <lb/>avec une vîteſſe égale à celle qu’elle auroit acquiſe en tom-<lb/>bant de la hauteur correſpondante; </s> <s xml:id="echoid-s19699" xml:space="preserve">car chaque tranche étant <lb/>preſſée par le poids des tranches ſupérieures, ſe trouve dans <lb/>le cas de l’art. </s> <s xml:id="echoid-s19700" xml:space="preserve">1155. </s> <s xml:id="echoid-s19701" xml:space="preserve">De plus, comme elle eſt toujours ſou-<lb/>miſe à l’action des tranches ſupérieures, il s’enſuit qu’elle ac-<lb/>quiert de nouveaux degrés de vîteſſe: </s> <s xml:id="echoid-s19702" xml:space="preserve">donc elle eſt mue d’un <lb/>mouvement accéléré. </s> <s xml:id="echoid-s19703" xml:space="preserve">Dans le ſecond cas, c’eſt-à-dire lorſque <lb/>le lit eſt incliné à l’horizon indépendamment de cette premiere <lb/>accélération, cauſée par la preſſion de chaque tranche ſur celle <lb/>qui eſt au deſſous, & </s> <s xml:id="echoid-s19704" xml:space="preserve">modifiée par l’inclinaiſon du lit de la <lb/>riviere; </s> <s xml:id="echoid-s19705" xml:space="preserve">toute la maſſe tombant ſur un plan incliné, acquiert <lb/>à chaque inſtant de nouveaux degrés de vîteſſe, comme les <lb/>corps qui tombent le long des plans inclinés C. </s> <s xml:id="echoid-s19706" xml:space="preserve">Q. </s> <s xml:id="echoid-s19707" xml:space="preserve">F. </s> <s xml:id="echoid-s19708" xml:space="preserve">D.</s> <s xml:id="echoid-s19709" xml:space="preserve"/> </p> <div xml:id="echoid-div1655" type="float" level="2" n="1"> <note position="left" xlink:label="note-0720-01" xlink:href="note-0720-01a" xml:space="preserve">Figure 429.</note> </div> </div> <div xml:id="echoid-div1657" type="section" level="1" n="1197"> <head xml:id="echoid-head1417" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19710" xml:space="preserve">1180. </s> <s xml:id="echoid-s19711" xml:space="preserve">Il ſuit delà que quelle que ſoit la poſition du lit d’un <lb/>fleuve, la vîteſſe ſera d’autant plus grande que le fleuve ſera <lb/>plus éloigné de ſa ſource; </s> <s xml:id="echoid-s19712" xml:space="preserve">parce que dans le cas d’un lit ho-<lb/>rizontal, chaque tranche aura agi d’autant plus qu’il y a plus <lb/>de diſtances entre le point où l’on examine la vîteſſe du fleuve, <lb/>& </s> <s xml:id="echoid-s19713" xml:space="preserve">la ſource du même fleuve; </s> <s xml:id="echoid-s19714" xml:space="preserve">& </s> <s xml:id="echoid-s19715" xml:space="preserve">dans le cas d’un lit incliné <lb/>à l’horizon, la hauteur de la ſource au deſſus du même point <lb/>ſera d’autant plus grande.</s> <s xml:id="echoid-s19716" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1658" type="section" level="1" n="1198"> <head xml:id="echoid-head1418" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19717" xml:space="preserve">1181. </s> <s xml:id="echoid-s19718" xml:space="preserve">Il ſuit delà que les vîteſſes de deux rivieres différentes <lb/>à leurs embouchures, en ſuppoſant leur pente égale, ſont d’au-<lb/>tant plus grandes que ces mêmes embouchures ſont plus éloi-<lb/>gnées de leurs ſources: </s> <s xml:id="echoid-s19719" xml:space="preserve">en général les vîteſſes des fleuves dé-<lb/>pendent de la pente de leur lit, de la hauteur de leurs eaux, <lb/>& </s> <s xml:id="echoid-s19720" xml:space="preserve">de la diſtance de ces mêmes eaux à la ſource.</s> <s xml:id="echoid-s19721" xml:space="preserve"/> </p> <pb o="631" file="0721" n="747" rhead="DE MATHEMATIQUE. Liv. XVI."/> </div> <div xml:id="echoid-div1659" type="section" level="1" n="1199"> <head xml:id="echoid-head1419" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19722" xml:space="preserve">1182. </s> <s xml:id="echoid-s19723" xml:space="preserve">Il ſuit encore delà que les vîteſſes des différentes <lb/>tranches ſont d’autant plus grandes qu’elles ſont plus proches <lb/>du fond. </s> <s xml:id="echoid-s19724" xml:space="preserve">Si cette vérité ne ſe trouve pas entiérement confir-<lb/>mée par l’expérience, cela vient de ce que le fond des rivieres <lb/>eſt toujours rempli de corps inégaux, dont le frottement, <lb/>avec les dernieres couches, ralentit néceſſairement le mou-<lb/>vement de ces mêmes couches. </s> <s xml:id="echoid-s19725" xml:space="preserve">De plus, il eſt viſible que les <lb/>vîteſſes de chaque tranche étant exprimées par les racines <lb/>quarrées des hauteurs, ces vîteſſes peuvent être repréſentées <lb/>par les ordonnées d’une parabole A M O P, puiſque l’on a L M; <lb/></s> <s xml:id="echoid-s19726" xml:space="preserve"> <anchor type="note" xlink:label="note-0721-01a" xlink:href="note-0721-01"/> N O : </s> <s xml:id="echoid-s19727" xml:space="preserve">D P :</s> <s xml:id="echoid-s19728" xml:space="preserve">: √A L\x{0020}; </s> <s xml:id="echoid-s19729" xml:space="preserve">√A N\x{0020}: </s> <s xml:id="echoid-s19730" xml:space="preserve">√A D\x{0020}.</s> <s xml:id="echoid-s19731" xml:space="preserve"/> </p> <div xml:id="echoid-div1659" type="float" level="2" n="1"> <note position="right" xlink:label="note-0721-01" xlink:href="note-0721-01a" xml:space="preserve">Figure 429.</note> </div> </div> <div xml:id="echoid-div1661" type="section" level="1" n="1200"> <head xml:id="echoid-head1420" xml:space="preserve"><emph style="sc">Définition</emph>.</head> <p> <s xml:id="echoid-s19732" xml:space="preserve">1183. </s> <s xml:id="echoid-s19733" xml:space="preserve">Si l’on conçoit une vîteſſe uniforme qui ſoit telle <lb/>qu’il s’écoule pendant le même tems la même quantité d’eau <lb/>que celle qui s’écoule par la ſomme des vîteſſes inégales: </s> <s xml:id="echoid-s19734" xml:space="preserve">cette <lb/>vîteſſe eſt appellée vîteſſe moyenne.</s> <s xml:id="echoid-s19735" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1662" type="section" level="1" n="1201"> <head xml:id="echoid-head1421" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19736" xml:space="preserve">1184. </s> <s xml:id="echoid-s19737" xml:space="preserve">Il ſuit delà que la vîteſſe moyenne eſt les deux tiers <lb/> <anchor type="note" xlink:label="note-0721-02a" xlink:href="note-0721-02"/> de la vîteſſe de la derniere tranche, dans le cas où la ſection <lb/>du fleuve eſt un parallélogramme: </s> <s xml:id="echoid-s19738" xml:space="preserve">car il eſt évident que les <lb/>quantités d’eau qui s’écoulent par chaque tranche ou élément <lb/>de la ſection, ſont proportionnelles à la largeur de ces élémens <lb/>& </s> <s xml:id="echoid-s19739" xml:space="preserve">aux vîteſſes: </s> <s xml:id="echoid-s19740" xml:space="preserve">mais dans l’hypotheſe préſente, toutes les lar-<lb/>geurs ſont égales, dont les quantités d’eau qui s’écoulent par <lb/>chaque tranche, ſuivent le rapport des vîteſſes, c’eſt-à-dire <lb/>qu’elles vont en diminuant comme les ordonnées d’une para-<lb/>bole qui auroit pour hauteur A D: </s> <s xml:id="echoid-s19741" xml:space="preserve">donc ſi D P exprime la <lb/>vîteſſe de la derniere tranche, la quantité d’eau écoulée par la <lb/>ſurface du parallélogramme ſera les deux tiers de celle qui ſe <lb/>ſeroit écoulée, ſi toutes les vîteſſes étoient égales: </s> <s xml:id="echoid-s19742" xml:space="preserve">donc pour <lb/>avoir la vîteſſe moyenne, il n’y a qu’à prendre les deux tiers <lb/>de la derniere vîteſſe D P: </s> <s xml:id="echoid-s19743" xml:space="preserve">car en multipliant la hauteur A D, <lb/>par cette vîteſſe on aura la même quantité d’eau écoulée.</s> <s xml:id="echoid-s19744" xml:space="preserve"/> </p> <div xml:id="echoid-div1662" type="float" level="2" n="1"> <note position="right" xlink:label="note-0721-02" xlink:href="note-0721-02a" xml:space="preserve">Figure 429.</note> </div> </div> <div xml:id="echoid-div1664" type="section" level="1" n="1202"> <head xml:id="echoid-head1422" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19745" xml:space="preserve">1185. </s> <s xml:id="echoid-s19746" xml:space="preserve">Il ſuit encore delà que la vîteſſe moyenne varie ſelon <pb o="632" file="0722" n="748" rhead="NOUVEAU COURS"/> les différentes figures de la ſection de la riviere; </s> <s xml:id="echoid-s19747" xml:space="preserve">& </s> <s xml:id="echoid-s19748" xml:space="preserve">la regle <lb/>générale pour la trouver eſt de diviſer la quantité d’eau écou-<lb/>lée par la hauteur: </s> <s xml:id="echoid-s19749" xml:space="preserve">cette opération eſt la plus aiſée. </s> <s xml:id="echoid-s19750" xml:space="preserve">Celle qui <lb/>demande plus d’adreſſe eſt de trouver la quantité d’eau écoulée <lb/>pendant un certain tems, en faiſant uſage de ce principe, que <lb/>les quantités d’eau qui s’écoulent ſont en raiſon compoſée de <lb/>la directe des racines quarrées des hauteurs, & </s> <s xml:id="echoid-s19751" xml:space="preserve">de la directe <lb/>des élémens de la ſection. </s> <s xml:id="echoid-s19752" xml:space="preserve">Ceux qui auront connoiſſance du <lb/>calcul différentiel, pourront voir dans l’Architecture Hydrau-<lb/>lique différentes ſolutions de ce problême, & </s> <s xml:id="echoid-s19753" xml:space="preserve">pourront trou-<lb/>ver les vîteſſes moyennes correſpondantes par le moyen du <lb/>principe que j’expoſe ici.</s> <s xml:id="echoid-s19754" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1665" type="section" level="1" n="1203"> <head xml:id="echoid-head1423" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19755" xml:space="preserve">1186. </s> <s xml:id="echoid-s19756" xml:space="preserve">Il ſuit delà que la vîteſſe moyenne répond aux {4/9} de la hau-<lb/>teur A D: </s> <s xml:id="echoid-s19757" xml:space="preserve">car en ſuppoſant que N O ſoit cette vîteſſe, on aura <lb/>N O = {2/3} D P: </s> <s xml:id="echoid-s19758" xml:space="preserve">donc D P<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19759" xml:space="preserve">{4/9} D P<emph style="sub">2</emph>:</s> <s xml:id="echoid-s19760" xml:space="preserve">: A D : </s> <s xml:id="echoid-s19761" xml:space="preserve">{4/9} {A D x D P<emph style="sub">2</emph>/D P<emph style="sub">2</emph>} = {4/9} A D: <lb/></s> <s xml:id="echoid-s19762" xml:space="preserve">donc ſi l’on connoît la hauteur A D, & </s> <s xml:id="echoid-s19763" xml:space="preserve">la largeur de la ſec-<lb/>tion, que nous ſuppoſons parallélogrammique, avec la quan-<lb/>tité d’eau écoulée dans un certain tems, on connoîtra la vîteſſe <lb/>de la derniere tranche comme il ſuit. </s> <s xml:id="echoid-s19764" xml:space="preserve">Soit q la quantité d’eau <lb/>écoulée par cette ſection dans une minute; </s> <s xml:id="echoid-s19765" xml:space="preserve">a, la hauteur A D; </s> <s xml:id="echoid-s19766" xml:space="preserve"><lb/>on aura {2/3} √ a pour la vîteſſe moyenne (art. </s> <s xml:id="echoid-s19767" xml:space="preserve">1184): </s> <s xml:id="echoid-s19768" xml:space="preserve">donc la <lb/>vîteſſe de la derniere tranche eſt connue; </s> <s xml:id="echoid-s19769" xml:space="preserve">puiſque celle-ci en <lb/>eſt les deux tiers: </s> <s xml:id="echoid-s19770" xml:space="preserve">on fera donc {2/3} : </s> <s xml:id="echoid-s19771" xml:space="preserve">1 :</s> <s xml:id="echoid-s19772" xml:space="preserve">: {q/a}:</s> <s xml:id="echoid-s19773" xml:space="preserve">{3/2} {q/a}, c’eſt-à-dire que <lb/>l’on connoîtra la vîteſſe de la derniere tranche, en diviſant <lb/>le triple de la quantité d’eau écoulée par le double de la hau-<lb/>teur.</s> <s xml:id="echoid-s19774" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1666" type="section" level="1" n="1204"> <head xml:id="echoid-head1424" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s19775" xml:space="preserve">1187. </s> <s xml:id="echoid-s19776" xml:space="preserve">Il ſuit delà que ſi l’on connoît la vîteſſe de la derniere <lb/>tranche & </s> <s xml:id="echoid-s19777" xml:space="preserve">la vîteſſe moyenne avec la quantité d’eau qui s’eſt <lb/>écoulée, on connoîtra auſſi la hauteur de la ſection, & </s> <s xml:id="echoid-s19778" xml:space="preserve">partant <lb/>dans ce cas, comme dans le précédent, on déterminera faci-<lb/>lement le parametre de la parabole.</s> <s xml:id="echoid-s19779" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s19780" xml:space="preserve">Du choc des fluides contre les ſolides en repos ou en mouvement.</s> <s xml:id="echoid-s19781" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s19782" xml:space="preserve">1188. </s> <s xml:id="echoid-s19783" xml:space="preserve">Dans le choc des fluides, comme dans celui des ſo- <pb o="633" file="0723" n="749" rhead="DE MATHÉMATIQUE. Liv. XVI."/> lides, pour en eſtimer la force, il faut avoir égard à la denſité <lb/>& </s> <s xml:id="echoid-s19784" xml:space="preserve">à la vîteſſe du fluide dont elle dépend: </s> <s xml:id="echoid-s19785" xml:space="preserve">mais comme les fluides <lb/>agiſſent tout autrement que les ſolides, auſſi les loix de leur <lb/>choc ne ſont pas les mêmes; </s> <s xml:id="echoid-s19786" xml:space="preserve">la principale différence conſiſte <lb/>en ce que lorſqu’un corps ſolide vient en choquer un autre, il n’y <lb/>a que la ſurface antérieure de ce ſolide qui frappe le premier, <lb/>au lieu que dans les fluides toutes les lames élémentaires vien-<lb/>nent frapper chacune avec la même vîteſſe.</s> <s xml:id="echoid-s19787" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1667" type="section" level="1" n="1205"> <head xml:id="echoid-head1425" xml:space="preserve">PROPOSITION II. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19788" xml:space="preserve">1189. </s> <s xml:id="echoid-s19789" xml:space="preserve">Si un fluide choque avec différentes vîteſſes des ſurfaces <lb/>égales, expoſées perpendiculairement à ſon courant, les forces du <lb/>choc ſeront comme les quarrés des vîteſſes.</s> <s xml:id="echoid-s19790" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1668" type="section" level="1" n="1206"> <head xml:id="echoid-head1426" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19791" xml:space="preserve">Puiſque les ſurfaces ſont égales, & </s> <s xml:id="echoid-s19792" xml:space="preserve">chacune perpendiculaire <lb/>au courant ou à la direction du fluide, le nombre des filets <lb/>qui agiſſent contre elles eſt le même: </s> <s xml:id="echoid-s19793" xml:space="preserve">il eſt donc évident que <lb/>le choc du courant contre ces ſurfaces ſeroit égal, ſi les vîteſſes <lb/>étoient égales; </s> <s xml:id="echoid-s19794" xml:space="preserve">& </s> <s xml:id="echoid-s19795" xml:space="preserve">la différence ne peut venir que de l’inéga-<lb/>lité des vîteſſes. </s> <s xml:id="echoid-s19796" xml:space="preserve">Il faut donc faire voir que le rapport de ces <lb/>forces eſt celui des quarrés des vîteſſes. </s> <s xml:id="echoid-s19797" xml:space="preserve">Pour cela, ſuppoſons <lb/>que la premiere vîteſſe ſoit 1, & </s> <s xml:id="echoid-s19798" xml:space="preserve">la ſeconde 3: </s> <s xml:id="echoid-s19799" xml:space="preserve">donc dans le <lb/>même tems le plan oppoſé à la plus grande vîteſſe eſt frappé <lb/>trois fois davantage, puiſque la maſſe d’eau eſt trois fois plus <lb/>grande; </s> <s xml:id="echoid-s19800" xml:space="preserve">& </s> <s xml:id="echoid-s19801" xml:space="preserve">de plus, comme chaque partie de cette maſſe d’eau <lb/>égale à celle qui a un degré de vîteſſe, a (par hypotheſe) une <lb/>vîteſſe triple, la ſurface qui lui eſt oppoſée recevra donc trois <lb/>fois plus de mouvement de chacune de ces trois parties: </s> <s xml:id="echoid-s19802" xml:space="preserve">donc <lb/>la quantité de mouvement reçue, où la force du choc ſera ex-<lb/>primée par 9, quarré de 3; </s> <s xml:id="echoid-s19803" xml:space="preserve">pendant que le choc, contre la pre-<lb/>miere ſurface, ne ſera exprimé que par 1, quarré de la pre-<lb/>miere vîteſſe: </s> <s xml:id="echoid-s19804" xml:space="preserve">donc les forces du choc d’un fluide de même <lb/>denſité ſont comme les quarrés des vîteſſes, contre des ſur-<lb/>faces égales, & </s> <s xml:id="echoid-s19805" xml:space="preserve">expoſées perpendiculairement à ſon courant. <lb/></s> <s xml:id="echoid-s19806" xml:space="preserve">C. </s> <s xml:id="echoid-s19807" xml:space="preserve">Q. </s> <s xml:id="echoid-s19808" xml:space="preserve">F. </s> <s xml:id="echoid-s19809" xml:space="preserve">D.</s> <s xml:id="echoid-s19810" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1669" type="section" level="1" n="1207"> <head xml:id="echoid-head1427" xml:space="preserve"><emph style="sc">Scholie</emph>.</head> <p> <s xml:id="echoid-s19811" xml:space="preserve">1190. </s> <s xml:id="echoid-s19812" xml:space="preserve">Il faut bien remarquer que l’on ſuppoſe ici que toutes <pb o="634" file="0724" n="750" rhead="NOUVEAU COURS"/> les tranches horizontales du fluide ont une même vîteſſe: </s> <s xml:id="echoid-s19813" xml:space="preserve">ſi <lb/>cela n’étoit pas, il faudroit connoître le rapport ſuivant lequel <lb/>elles augmentent ou diminuent, pour déterminer les vîteſſes <lb/>moyennes; </s> <s xml:id="echoid-s19814" xml:space="preserve">& </s> <s xml:id="echoid-s19815" xml:space="preserve">les forces du courant contre ces ſurfaces ſeroient <lb/>entr’elles comme les quarrés de ces vîteſſes moyennes.</s> <s xml:id="echoid-s19816" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1670" type="section" level="1" n="1208"> <head xml:id="echoid-head1428" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19817" xml:space="preserve">1191. </s> <s xml:id="echoid-s19818" xml:space="preserve">Si les vîteſſes des fluides étant inégales, on ſuppoſe <lb/>de plus que les fluides ont des denſités inégales, & </s> <s xml:id="echoid-s19819" xml:space="preserve">qu’ils cho-<lb/>quent perpendiculairement des ſurfaces inégales; </s> <s xml:id="echoid-s19820" xml:space="preserve">alors il eſt <lb/>évident que les forces du choc contre ces ſurfaces eſt en raiſon <lb/>compoſée des ſurfaces choquées, des denſités des fluides & </s> <s xml:id="echoid-s19821" xml:space="preserve">des <lb/>quarrés des vîteſſes: </s> <s xml:id="echoid-s19822" xml:space="preserve">donc ſi l’on appelle F la force du premier <lb/>fluide, V ſa vîteſſe, D ſa denſité, & </s> <s xml:id="echoid-s19823" xml:space="preserve">S la ſurface qu’il rencon-<lb/>tre; </s> <s xml:id="echoid-s19824" xml:space="preserve">& </s> <s xml:id="echoid-s19825" xml:space="preserve">pareillement f la force d’un ſecond fluide, dont la den-<lb/>ſité eſt d, la vîteſſe v, & </s> <s xml:id="echoid-s19826" xml:space="preserve">qu’il rencontre une ſurface s; </s> <s xml:id="echoid-s19827" xml:space="preserve">on aura <lb/>cette analogie, F : </s> <s xml:id="echoid-s19828" xml:space="preserve">f :</s> <s xml:id="echoid-s19829" xml:space="preserve">: S V<emph style="sub">2</emph> D : </s> <s xml:id="echoid-s19830" xml:space="preserve">s u<emph style="sub">2</emph> d, d’où l’on tire cette éga-<lb/>lité, F s u<emph style="sub">2</emph>d = fS V<emph style="sub">2</emph> D qui pourra ſervir à déterminer les <lb/>différens rapports des forces du choc, ſelon les rapports des <lb/>denſités des vîteſſes & </s> <s xml:id="echoid-s19831" xml:space="preserve">des ſurfaces: </s> <s xml:id="echoid-s19832" xml:space="preserve">toujours dans le cas où <lb/>les tranches horizontales du fluide ont la même vîteſſe, comme <lb/>nous le ſuppoſons ici: </s> <s xml:id="echoid-s19833" xml:space="preserve">on avertira lorſque nous ferons d’autres <lb/>ſuppoſitions.</s> <s xml:id="echoid-s19834" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1671" type="section" level="1" n="1209"> <head xml:id="echoid-head1429" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19835" xml:space="preserve">1192. </s> <s xml:id="echoid-s19836" xml:space="preserve">Si les ſurfaces expoſées aux différens fluides ont des <lb/>vîteſſes particulieres, il eſt évident que dans le cas où ces vîteſſes <lb/>ſeroient vers le même point, elles doivent être moindres que <lb/>celles des fluides, & </s> <s xml:id="echoid-s19837" xml:space="preserve">alors les forces des fluides contre ces ſur-<lb/>faces ſeront en raiſon compoſée de la denſité de ces mêmes <lb/>fluides, de celle des ſurfaces & </s> <s xml:id="echoid-s19838" xml:space="preserve">des quarrés des différences des <lb/>vîteſſes de chaque fluide à la ſurface choquée. </s> <s xml:id="echoid-s19839" xml:space="preserve">Si les ſurfaces <lb/>choquées ont des vîteſſes particulieres, & </s> <s xml:id="echoid-s19840" xml:space="preserve">directement oppo-<lb/>ſées à celle des fluides; </s> <s xml:id="echoid-s19841" xml:space="preserve">les forces du choc contre ces ſurfaces <lb/>ſeront dans la raiſon compoſée des ſurfaces choquées, des <lb/>quarrés des ſommes des vîteſſes du fluide, & </s> <s xml:id="echoid-s19842" xml:space="preserve">de la ſurface <lb/>contre laquelle il choque, & </s> <s xml:id="echoid-s19843" xml:space="preserve">des denſités de ces mêmes <lb/>fluides.</s> <s xml:id="echoid-s19844" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1672" type="section" level="1" n="1210"> <head xml:id="echoid-head1430" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19845" xml:space="preserve">1193. </s> <s xml:id="echoid-s19846" xml:space="preserve">Si le fluide eſt ſuppoſé en repos, & </s> <s xml:id="echoid-s19847" xml:space="preserve">que la ſurface <pb o="635" file="0725" n="751" rhead="DE MATHÉMATIQUE. Liv. XVI."/> meuve dans ce même fluide avec une certaine vîteſſe, les ré-<lb/>ſiſtances qu’elle éprouvera ſeront comme les quarrés des vîteſſes: <lb/></s> <s xml:id="echoid-s19848" xml:space="preserve">car il eſt évident que c’eſt préciſément la même choſe de ſup-<lb/>poſer le fluide en repos, & </s> <s xml:id="echoid-s19849" xml:space="preserve">la ſurface en mouvement, ou la ſur-<lb/>face en repos, choquée par un fluide qui auroit la même <lb/>vîteſſe.</s> <s xml:id="echoid-s19850" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1673" type="section" level="1" n="1211"> <head xml:id="echoid-head1431" xml:space="preserve">PROPOSITION III. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19851" xml:space="preserve">1194. </s> <s xml:id="echoid-s19852" xml:space="preserve">Si deux ſurfaces égales ſont expoſées à un courant, dont <lb/>toutes les tranches horizontales ſont ſuppoſées avoir la même vîteſſe, <lb/>l’une perpendiculairement, & </s> <s xml:id="echoid-s19853" xml:space="preserve">l’autre obliquement au même fluide, <lb/>les chocs du fluide contre ces ſurfaces ſeront comme le quarré du <lb/>ſinus total au quarré du ſinus de l’angle d’inclinaiſon.</s> <s xml:id="echoid-s19854" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1674" type="section" level="1" n="1212"> <head xml:id="echoid-head1432" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19855" xml:space="preserve">Soit une ſurface T V, expoſée au courant, repréſenté dans <lb/> <anchor type="note" xlink:label="note-0725-01a" xlink:href="note-0725-01"/> cette figure, & </s> <s xml:id="echoid-s19856" xml:space="preserve">perpendiculaire à ce même courant; </s> <s xml:id="echoid-s19857" xml:space="preserve">& </s> <s xml:id="echoid-s19858" xml:space="preserve">ſoit <lb/>une autre ſurface T M inclinée à la direction du fluide, & </s> <s xml:id="echoid-s19859" xml:space="preserve">que <lb/>l’on ſuppoſe égale à la précédente; </s> <s xml:id="echoid-s19860" xml:space="preserve">ayant décrit l’arc M V, <lb/>& </s> <s xml:id="echoid-s19861" xml:space="preserve">abaiſſé la perpendiculaire M Q ſur T V, il eſt viſible que <lb/>T Q ſera le ſinus de l’angle d’inclinaiſon T M V: </s> <s xml:id="echoid-s19862" xml:space="preserve">il faut donc <lb/>démontrer que le choc du fluide contre T V eſt au choc du <lb/>fluide contre T M, comme le quarré T V<emph style="sub">2</emph> du ſinus total eſt <lb/>au quarré T Q<emph style="sub">2</emph> du ſinus de l’angle d’inclinaiſon.</s> <s xml:id="echoid-s19863" xml:space="preserve"/> </p> <div xml:id="echoid-div1674" type="float" level="2" n="1"> <note position="right" xlink:label="note-0725-01" xlink:href="note-0725-01a" xml:space="preserve">Figure 430.</note> </div> <p> <s xml:id="echoid-s19864" xml:space="preserve">On peut concevoir le fluide compoſé d’une infinité de lames <lb/>horizontales qui choquent toutes avec la même force. </s> <s xml:id="echoid-s19865" xml:space="preserve">Cela <lb/>poſé, il eſt évident que la force du choc dépend de la maniere <lb/>dont chacune agit directement, ou obliquement, & </s> <s xml:id="echoid-s19866" xml:space="preserve">du nom-<lb/>bre de ces tranches; </s> <s xml:id="echoid-s19867" xml:space="preserve">il n’eſt pas moins viſible que le nombre <lb/>des tranches qui choquent la ſurface T V eſt au nombre des <lb/>tranches qui choquent la ſurface T M, comme T V eſt à T Q. <lb/></s> <s xml:id="echoid-s19868" xml:space="preserve">Mais les tranches qui frappent la ſurface T M ne la choquent <lb/>pas directement, puiſque cette ſurface eſt oblique au courant: </s> <s xml:id="echoid-s19869" xml:space="preserve"><lb/>ainſi la force du choc contre cette ſurface doit encore dimi-<lb/>nuer dans la raiſon du ſinus total au ſinus de l’angle d’incli-<lb/>naiſon: </s> <s xml:id="echoid-s19870" xml:space="preserve">car ſi l’on ſuppoſe que la force abſolue d’une lame ſoit <lb/>repréſentée par P F, égale au ſinus total, cette force doit <lb/>ſe décompoſer en deux autres, l’une P H parallele au plan <lb/>T M, & </s> <s xml:id="echoid-s19871" xml:space="preserve">l’autre perpendiculaire F H: </s> <s xml:id="echoid-s19872" xml:space="preserve">or il eſt évident que <pb o="636" file="0726" n="752" rhead="NOUVEAU COURS"/> P F : </s> <s xml:id="echoid-s19873" xml:space="preserve">F H :</s> <s xml:id="echoid-s19874" xml:space="preserve">: T M : </s> <s xml:id="echoid-s19875" xml:space="preserve">T Q :</s> <s xml:id="echoid-s19876" xml:space="preserve">: T V : </s> <s xml:id="echoid-s19877" xml:space="preserve">T Q, à cauſe des triangles <lb/>ſemblables P H F, T M Q: </s> <s xml:id="echoid-s19878" xml:space="preserve">donc la force du choc contre T V <lb/>eſt à celle contre T M, comme T V<emph style="sub">2</emph>: </s> <s xml:id="echoid-s19879" xml:space="preserve">T Q, c’eſt-à-dire <lb/>comme le quarré du ſinus total eſt au ſinus de l’angle d’incli-<lb/>naiſon. </s> <s xml:id="echoid-s19880" xml:space="preserve">C.</s> <s xml:id="echoid-s19881" xml:space="preserve">Q.</s> <s xml:id="echoid-s19882" xml:space="preserve">F.</s> <s xml:id="echoid-s19883" xml:space="preserve">D.</s> <s xml:id="echoid-s19884" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1676" type="section" level="1" n="1213"> <head xml:id="echoid-head1433" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s19885" xml:space="preserve">1195. </s> <s xml:id="echoid-s19886" xml:space="preserve">Si le nombre des filets étoit égal de part & </s> <s xml:id="echoid-s19887" xml:space="preserve">d’autre, <lb/>ce qui arriveroit ſi la ſurface T M étoit prolongée en N juſ-<lb/>qu’à la ligne horizontale N V, alors l’inégalité du choc ne <lb/>vient que de l’obliquité du fluide, & </s> <s xml:id="echoid-s19888" xml:space="preserve">par conſéquent le choc <lb/>contre T V eſt au choc contre T N, comme le ſinus total au <lb/>ſinus de l’angle d’inclinaiſon, car la vîteſſe eſt la même, & </s> <s xml:id="echoid-s19889" xml:space="preserve"><lb/>comme la hauteur eſt auſſi la même, il y a même nombre de <lb/>tranches qui choquent les ſurfaces T V, T N, que l’on ſuppoſe <lb/>d’ailleurs avoir une largeur égale.</s> <s xml:id="echoid-s19890" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1677" type="section" level="1" n="1214"> <head xml:id="echoid-head1434" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s19891" xml:space="preserve">1196. </s> <s xml:id="echoid-s19892" xml:space="preserve">Il ſuit encore delà que les chocs de deux fluides dif-<lb/>férens en denſité contre des ſurfaces inégales, & </s> <s xml:id="echoid-s19893" xml:space="preserve">inégalemenc <lb/>inclinées, ſont dans la raiſon compoſée des quarrés des ſinus <lb/>des angles d’inclinaiſons, des denſités, & </s> <s xml:id="echoid-s19894" xml:space="preserve">des ſurfaces expo-<lb/>ſées à ces différens fluides, & </s> <s xml:id="echoid-s19895" xml:space="preserve">des quarrés des vîteſſes: </s> <s xml:id="echoid-s19896" xml:space="preserve">car les <lb/>ſinus de chacun des angles d’inclinaiſon meſurent le nom-<lb/>bre de lames horizontales qui choquent les ſurfaces propo-<lb/>ſées; </s> <s xml:id="echoid-s19897" xml:space="preserve">ils meſurent auſſi l’intenſité du choc, ſelon le plus ou <lb/>le moins d’inclinaiſon de ces ſurfaces: </s> <s xml:id="echoid-s19898" xml:space="preserve">donc les chocs ſonc <lb/>1°. </s> <s xml:id="echoid-s19899" xml:space="preserve">comme les quarrés des ſinus des angles d’inclinaiſon. <lb/></s> <s xml:id="echoid-s19900" xml:space="preserve">2°. </s> <s xml:id="echoid-s19901" xml:space="preserve">Il eſt évident que plus elles ſeront grandes, plus il y aura <lb/>de tranches qui les choqueront. </s> <s xml:id="echoid-s19902" xml:space="preserve">3°. </s> <s xml:id="echoid-s19903" xml:space="preserve">Il eſt encore viſible qu’à <lb/>vîteſſe égale plus les fluides ſeront denſes, plus le choc ſera <lb/>grand, à cauſe de la maſſe, toujours proportionnelle aux den-<lb/>ſités; </s> <s xml:id="echoid-s19904" xml:space="preserve">4°. </s> <s xml:id="echoid-s19905" xml:space="preserve">les chocs ſeront comme les quarrés des vîteſſes; </s> <s xml:id="echoid-s19906" xml:space="preserve">car <lb/>nous avons démontré (art. </s> <s xml:id="echoid-s19907" xml:space="preserve">1188) que les chocs ſuivent ce <lb/>rapport pour les fluides. </s> <s xml:id="echoid-s19908" xml:space="preserve">Donc ſi l’on appelle D la denſité d’un <lb/>fluide, V la vîteſſe comme à toutes les tranches (hyp.)</s> <s xml:id="echoid-s19909" xml:space="preserve">, S le <lb/>ſinus de l’angle d’inclinaiſon du plan, dont la ſurface eſt re-<lb/>préſentée par E, & </s> <s xml:id="echoid-s19910" xml:space="preserve">F la force du fluide contre cette ſurface; </s> <s xml:id="echoid-s19911" xml:space="preserve"><lb/>pareillement ſi l’on nomme d la denſité d’un autre fluide, <lb/>dont la vîteſſe eſt v, & </s> <s xml:id="echoid-s19912" xml:space="preserve">qui choque un plan, dont le ſinus <pb o="637" file="0727" n="753" rhead="DE MATHÉMATIQUE. Liv. XVI."/> d’inclinaiſon eſt s, & </s> <s xml:id="echoid-s19913" xml:space="preserve">dont la ſurface eſt e; </s> <s xml:id="echoid-s19914" xml:space="preserve">que l’on appellef, <lb/>du choc réſultant contre cette ſurface, on aura F : </s> <s xml:id="echoid-s19915" xml:space="preserve">f :</s> <s xml:id="echoid-s19916" xml:space="preserve">: D V<emph style="sub">2</emph>S<emph style="sub">2</emph> E <lb/>: </s> <s xml:id="echoid-s19917" xml:space="preserve">d v<emph style="sub">2</emph>s<emph style="sub">2</emph>e; </s> <s xml:id="echoid-s19918" xml:space="preserve">d’où l’on tire F d v<emph style="sub">2</emph>s<emph style="sub">2</emph>e = fD V<emph style="sub">2</emph>S<emph style="sub">2</emph> E. </s> <s xml:id="echoid-s19919" xml:space="preserve">On pourra dé-<lb/>duire de cette propoſition, & </s> <s xml:id="echoid-s19920" xml:space="preserve">de la formule qui a été conſ-<lb/>truite ſur ce que l’on vient de démontrer tout ce dont on <lb/>pourra avoir beſoin dans les différentes circonſtances qui peu-<lb/>vent avoir lieu dans le choc des fluides contre des ſurfaces en <lb/>repos. </s> <s xml:id="echoid-s19921" xml:space="preserve">On pourroit même l’appliquer au choc des fluides contre <lb/>des ſurfaces en mouvement, & </s> <s xml:id="echoid-s19922" xml:space="preserve">expoſées obliquement au cou-<lb/>rant, en prenant pour les vîteſſes V & </s> <s xml:id="echoid-s19923" xml:space="preserve">v la différence ou la <lb/>ſomme des vîteſſes du plan & </s> <s xml:id="echoid-s19924" xml:space="preserve">du fluide, ſelon que ces vîteſſes <lb/>ont des directions dans le même ſens, ou dans des ſens op-<lb/>poſés.</s> <s xml:id="echoid-s19925" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1678" type="section" level="1" n="1215"> <head xml:id="echoid-head1435" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s19926" xml:space="preserve">1197. </s> <s xml:id="echoid-s19927" xml:space="preserve">Pour faire voir quelques applications de cette formule, <lb/>nous ſuppoſerons que les vîteſſes ſont proportionnelles aux <lb/>denſités & </s> <s xml:id="echoid-s19928" xml:space="preserve">aux ſurfaces qu’elles rencontrent, c’eſt-à-dire que <lb/>V : </s> <s xml:id="echoid-s19929" xml:space="preserve">v :</s> <s xml:id="echoid-s19930" xml:space="preserve">: D : </s> <s xml:id="echoid-s19931" xml:space="preserve">d, & </s> <s xml:id="echoid-s19932" xml:space="preserve">que V : </s> <s xml:id="echoid-s19933" xml:space="preserve">v :</s> <s xml:id="echoid-s19934" xml:space="preserve">: E : </s> <s xml:id="echoid-s19935" xml:space="preserve">e : </s> <s xml:id="echoid-s19936" xml:space="preserve">donc en multipliant par <lb/>ordre, on aura V<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19937" xml:space="preserve">v<emph style="sub">2</emph> :</s> <s xml:id="echoid-s19938" xml:space="preserve">: D E : </s> <s xml:id="echoid-s19939" xml:space="preserve">de : </s> <s xml:id="echoid-s19940" xml:space="preserve">donc en ſubſtituant ces <lb/>rapports dans la proportion F : </s> <s xml:id="echoid-s19941" xml:space="preserve">f :</s> <s xml:id="echoid-s19942" xml:space="preserve">: D V<emph style="sub">2</emph>S<emph style="sub">2</emph>E : </s> <s xml:id="echoid-s19943" xml:space="preserve">dv<emph style="sub">2</emph>s<emph style="sub">2</emph>e, on aura <lb/>celle-ci, F : </s> <s xml:id="echoid-s19944" xml:space="preserve">f :</s> <s xml:id="echoid-s19945" xml:space="preserve">: D<emph style="sub">2</emph>E<emph style="sub">2</emph>S<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19946" xml:space="preserve">d<emph style="sub">2</emph>e<emph style="sub">2</emph>s<emph style="sub">2</emph>, ou F : </s> <s xml:id="echoid-s19947" xml:space="preserve">f :</s> <s xml:id="echoid-s19948" xml:space="preserve">: V<emph style="sub">4</emph>S<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19949" xml:space="preserve">v<emph style="sub">4</emph>s<emph style="sub">2</emph>, c’eſt-<lb/>à-dire que les forces ſont comme les produits des quarrés des <lb/>denſités, des ſurfaces, & </s> <s xml:id="echoid-s19950" xml:space="preserve">des ſinus des angles d’inclinaiſon, ou <lb/>dans la raiſon compoſée des quatriemes puiſſances des vîteſſes <lb/>& </s> <s xml:id="echoid-s19951" xml:space="preserve">des quarrés des ſinus des angles d’inclinaiſon.</s> <s xml:id="echoid-s19952" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1679" type="section" level="1" n="1216"> <head xml:id="echoid-head1436" xml:space="preserve"><emph style="sc">Corollaire</emph> IV.</head> <p> <s xml:id="echoid-s19953" xml:space="preserve">1198. </s> <s xml:id="echoid-s19954" xml:space="preserve">Si les vîteſſes ſont réciproques aux racines quarrées <lb/>des eſpaces, & </s> <s xml:id="echoid-s19955" xml:space="preserve">les denſités réciproques aux quarrés des ſinus <lb/>des angles d’inclinaiſon, les forces du choc ſeront égales: <lb/></s> <s xml:id="echoid-s19956" xml:space="preserve">car puiſque V : </s> <s xml:id="echoid-s19957" xml:space="preserve">v :</s> <s xml:id="echoid-s19958" xml:space="preserve">: √E\x{0020} : </s> <s xml:id="echoid-s19959" xml:space="preserve">√E\x{0020}, on a V<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19960" xml:space="preserve">v<emph style="sub">2</emph> :</s> <s xml:id="echoid-s19961" xml:space="preserve">: e : </s> <s xml:id="echoid-s19962" xml:space="preserve">E : </s> <s xml:id="echoid-s19963" xml:space="preserve">donc <lb/>V<emph style="sub">2</emph>E = v<emph style="sub">2</emph>e: </s> <s xml:id="echoid-s19964" xml:space="preserve">donc F : </s> <s xml:id="echoid-s19965" xml:space="preserve">f :</s> <s xml:id="echoid-s19966" xml:space="preserve">: D S<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19967" xml:space="preserve">ds<emph style="sub">2</emph>; </s> <s xml:id="echoid-s19968" xml:space="preserve">mais par hypotheſe <lb/>D : </s> <s xml:id="echoid-s19969" xml:space="preserve">d :</s> <s xml:id="echoid-s19970" xml:space="preserve">: s<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19971" xml:space="preserve">S<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19972" xml:space="preserve">donc D S<emph style="sub">2</emph> = ds<emph style="sub">2</emph> : </s> <s xml:id="echoid-s19973" xml:space="preserve">donc F = f, & </s> <s xml:id="echoid-s19974" xml:space="preserve">ainſi des <lb/>autres cas qu’il ſeroit inutile de détailler ici davantage. </s> <s xml:id="echoid-s19975" xml:space="preserve">C’eſtaux <lb/>Commençans à s’exercer à trouver eux-mêmes des ſuppoſitions, <lb/>pour connoître par la formule ce qui doit arriver dans ces <lb/>circonſtances; </s> <s xml:id="echoid-s19976" xml:space="preserve">mais ce qu’ils doivent étudier avec le plus de <lb/>ſoin, ce ſont les raiſons métaphyſiques des réſultats qu’ils ti-<lb/>reront de la formule, ſans quoi ils les auront auſſitôt oubliés <pb o="638" file="0728" n="754" rhead="NOUVEAU COURS"/> que découverts, & </s> <s xml:id="echoid-s19977" xml:space="preserve">contracteroient la pernicieuſe habitude de <lb/>ne raiſonner que par formules, lorſqu’ils ſont en état de le <lb/>faire par le jugement.</s> <s xml:id="echoid-s19978" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1680" type="section" level="1" n="1217"> <head xml:id="echoid-head1437" xml:space="preserve"><emph style="sc">Scholie</emph>.</head> <p> <s xml:id="echoid-s19979" xml:space="preserve">1199. </s> <s xml:id="echoid-s19980" xml:space="preserve">Dans ce qui précede, nous avons ſuppoſé que toutes les <lb/>tranches du fluide qui choque une ſurface perpendiculaire ou <lb/>oblique à ſon courant, étoient toutes mues avec une égale <lb/>vîteſſe; </s> <s xml:id="echoid-s19981" xml:space="preserve">mais comme il y a un grand nombre de cas où les <lb/>vîteſſes des tranches ne ſont pas égales, & </s> <s xml:id="echoid-s19982" xml:space="preserve">ſuivent différens <lb/>rapports, nous allons examiner dans la propoſition ſuivante <lb/>quelles doivent être les forces du choc, lorſque les vîteſſes de <lb/>chaque tranche ſont comme les racines quarrées des hauteurs, <lb/>comme cela arrive dans les rivieres & </s> <s xml:id="echoid-s19983" xml:space="preserve">autres courans qui ont <lb/>une certaine profondeur.</s> <s xml:id="echoid-s19984" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1681" type="section" level="1" n="1218"> <head xml:id="echoid-head1438" xml:space="preserve">PROPOSITION IV. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s19985" xml:space="preserve">1200. </s> <s xml:id="echoid-s19986" xml:space="preserve">Si deux ſurfaces égales ſont expoſées au courant d’un <lb/> <anchor type="note" xlink:label="note-0728-01a" xlink:href="note-0728-01"/> fluide, dont toutes les tranches ont différentes vîteſſes qui ſuivent <lb/>la progreſſion des racines des hauteurs, & </s> <s xml:id="echoid-s19987" xml:space="preserve">que l’une de ces ſurfaces <lb/>ſoit expoſée perpendiculairement, & </s> <s xml:id="echoid-s19988" xml:space="preserve">l’autre obliquement au même <lb/>fluide, le choc contre la premiere eſt au choc ſur la ſeconde ſur-<lb/>face, comme le cube du ſinus total eſt au cube de celui de l’angle <lb/>d’inclinaiſon.</s> <s xml:id="echoid-s19989" xml:space="preserve"/> </p> <div xml:id="echoid-div1681" type="float" level="2" n="1"> <note position="left" xlink:label="note-0728-01" xlink:href="note-0728-01a" xml:space="preserve">Figure 432.</note> </div> </div> <div xml:id="echoid-div1683" type="section" level="1" n="1219"> <head xml:id="echoid-head1439" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s19990" xml:space="preserve">Suppoſons que les lignes égales A B, A F repréſentent le <lb/>profil de chacune de ces ſurfaces, l’une A B perpendiculaire à <lb/>la direction du fluide, & </s> <s xml:id="echoid-s19991" xml:space="preserve">l’autre A F oblique au même fluide; <lb/></s> <s xml:id="echoid-s19992" xml:space="preserve">A B ſera le ſinus total, & </s> <s xml:id="echoid-s19993" xml:space="preserve">A G le ſinus de l’angle d’inclinaiſon: </s> <s xml:id="echoid-s19994" xml:space="preserve"><lb/>de plus, comme on ſuppoſe que les vîteſſes croiſſent comme <lb/>les racines quarrées des hauteurs, il eſt évident que la plus <lb/>grande vîteſſe des tranches qui répondent au plan oblique A F <lb/>ſera exprimée par √A G\x{0020}, & </s> <s xml:id="echoid-s19995" xml:space="preserve">la plus grande vîteſſe qui réponde <lb/>au plan perpendiculaire A B ſera exprimée par √A B\x{0020}. </s> <s xml:id="echoid-s19996" xml:space="preserve">On ſçait <lb/>par ce qui précéde, que le choc de ces différentes tranches <lb/>contre les ſurfaces qu’elles rencontrent perpendiculairement, <lb/>eſt comme le produit de ces ſurfaces par les quarrés des vîteſſes <pb o="639" file="0729" n="755" rhead="DE MATHÉMATIQUE. Liv. XVI."/> moyennes, leſquelles ſont comme les racines quarrées des hau-<lb/>teurs correſpondantes A B, A G, dont elles ſont les {2/3} (art, 1184). <lb/></s> <s xml:id="echoid-s19997" xml:space="preserve">Ainſi en appellant F le choc du fluide contre A B, f celui du <lb/>même fluide contre A G, on aura F : </s> <s xml:id="echoid-s19998" xml:space="preserve">f :</s> <s xml:id="echoid-s19999" xml:space="preserve">: A B x A B : </s> <s xml:id="echoid-s20000" xml:space="preserve">A G x A G : </s> <s xml:id="echoid-s20001" xml:space="preserve"><lb/>car les ſurfaces ayant une même largeur, ſont comme les hau-<lb/>teurs A B & </s> <s xml:id="echoid-s20002" xml:space="preserve">A G, & </s> <s xml:id="echoid-s20003" xml:space="preserve">de plus les quarrés des vîteſſes moyennes <lb/>correſpondantes ſont comme A B & </s> <s xml:id="echoid-s20004" xml:space="preserve">A G, puiſque A B eſt le <lb/>quarré de √A B\x{0020}, & </s> <s xml:id="echoid-s20005" xml:space="preserve">A G celui de √A G\x{0020}.</s> <s xml:id="echoid-s20006" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20007" xml:space="preserve">Préſentement pour avoir le choc des tranches meſurées par <lb/>A G contre la ſurface oblique A F, il faut faire attention que <lb/>le choc direct eſt au choc oblique, comme le ſinus total eſt au <lb/>ſinus de l’angle d’inclinaiſon, ou comme A B eſt à A G : </s> <s xml:id="echoid-s20008" xml:space="preserve">donc <lb/>en appellant ſ la force du choc oblique, on aura f : </s> <s xml:id="echoid-s20009" xml:space="preserve">ſ :</s> <s xml:id="echoid-s20010" xml:space="preserve">: A B : </s> <s xml:id="echoid-s20011" xml:space="preserve">A G; <lb/></s> <s xml:id="echoid-s20012" xml:space="preserve">mais nous avions ci-devant F: </s> <s xml:id="echoid-s20013" xml:space="preserve">f :</s> <s xml:id="echoid-s20014" xml:space="preserve">: A B<emph style="sub">2</emph> : </s> <s xml:id="echoid-s20015" xml:space="preserve">A G<emph style="sub">2</emph>: </s> <s xml:id="echoid-s20016" xml:space="preserve">donc en mul-<lb/>tipliant par ordre, & </s> <s xml:id="echoid-s20017" xml:space="preserve">diviſant les deux premiers termes par f, <lb/>il viendra F : </s> <s xml:id="echoid-s20018" xml:space="preserve">ſ :</s> <s xml:id="echoid-s20019" xml:space="preserve">: A B<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20020" xml:space="preserve">A G<emph style="sub">3</emph>; </s> <s xml:id="echoid-s20021" xml:space="preserve">d’où il ſuit évidemment que <lb/>dans cette hypotheſe, les forces du courant contre les ſurfaces <lb/>égales A B & </s> <s xml:id="echoid-s20022" xml:space="preserve">A F ſont comme les cubes du ſinus total & </s> <s xml:id="echoid-s20023" xml:space="preserve">celui <lb/>de l’angle d’inclinaiſon. </s> <s xml:id="echoid-s20024" xml:space="preserve">C.</s> <s xml:id="echoid-s20025" xml:space="preserve">Q.</s> <s xml:id="echoid-s20026" xml:space="preserve">F.</s> <s xml:id="echoid-s20027" xml:space="preserve">D.</s> <s xml:id="echoid-s20028" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1684" type="section" level="1" n="1220"> <head xml:id="echoid-head1440" xml:space="preserve"><emph style="sc">Corollaire</emph> I.</head> <p> <s xml:id="echoid-s20029" xml:space="preserve">1201. </s> <s xml:id="echoid-s20030" xml:space="preserve">Si l’on a une autre inclinaiſon pour le même plan, <lb/>comme A K, on aura encore F : </s> <s xml:id="echoid-s20031" xml:space="preserve">φ :</s> <s xml:id="echoid-s20032" xml:space="preserve">: A B : </s> <s xml:id="echoid-s20033" xml:space="preserve">A L<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20034" xml:space="preserve">donc les forces <lb/>contre la même ſurface différemment inclinée dans un fluide <lb/>homogene, ſont comme les cubes des ſinus des angles d’in-<lb/>clinaiſon: </s> <s xml:id="echoid-s20035" xml:space="preserve">car il eſt évident que puiſque l’on a F : </s> <s xml:id="echoid-s20036" xml:space="preserve">ſ :</s> <s xml:id="echoid-s20037" xml:space="preserve">: A B<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20038" xml:space="preserve">A G<emph style="sub">3</emph>, <lb/>& </s> <s xml:id="echoid-s20039" xml:space="preserve">que les antécédens de ces deux proportions ſont les mêmes, <lb/>les conſéquens doivent auſſi former une proportion: </s> <s xml:id="echoid-s20040" xml:space="preserve">donc <lb/>ſ: </s> <s xml:id="echoid-s20041" xml:space="preserve">φ :</s> <s xml:id="echoid-s20042" xml:space="preserve">: A G<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20043" xml:space="preserve">A L<emph style="sub">3</emph>.</s> <s xml:id="echoid-s20044" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1685" type="section" level="1" n="1221"> <head xml:id="echoid-head1441" xml:space="preserve"><emph style="sc">Corollaire</emph> II.</head> <p> <s xml:id="echoid-s20045" xml:space="preserve">1202. </s> <s xml:id="echoid-s20046" xml:space="preserve">Si les ſurfaces ſont inégales & </s> <s xml:id="echoid-s20047" xml:space="preserve">différemment inclinées <lb/>dans un fluide de même denſité, les forces du fluide contre ces <lb/>mêmes ſurfaces, ſont comme les produits de ces ſurfaces par <lb/>les produits des cubes des ſinus des angles d’inclinaiſon, par <lb/>les quarrés des vîteſſes moyennes correſpondantes.</s> <s xml:id="echoid-s20048" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20049" xml:space="preserve">Pour démontrer ce corollaire, ſoient repréſentées les ſur-<lb/> <anchor type="note" xlink:label="note-0729-01a" xlink:href="note-0729-01"/> faces inégales par les lignes A F, af, & </s> <s xml:id="echoid-s20050" xml:space="preserve">ſoient priſes les lignes <lb/>A B = A F, & </s> <s xml:id="echoid-s20051" xml:space="preserve">ab = af, chacune perpendiculaire au courant. <lb/></s> <s xml:id="echoid-s20052" xml:space="preserve">Soit F la force qui agit perpendiculairement contre la ſurface <pb o="640" file="0730" n="756" rhead="NOUVEAU COURS"/> A B, V la vîteſſe moyenne, & </s> <s xml:id="echoid-s20053" xml:space="preserve">aa la ſurface repréſentée par <lb/>A B ou A F; </s> <s xml:id="echoid-s20054" xml:space="preserve">ſoit pareillement f la force du courant contre la <lb/>ſurface ab; </s> <s xml:id="echoid-s20055" xml:space="preserve">v la vîteſſe moyenne correſpondante, & </s> <s xml:id="echoid-s20056" xml:space="preserve">b b la ſur-<lb/>face repréſentée par ab ou af ſon égale. </s> <s xml:id="echoid-s20057" xml:space="preserve">Soit de plus R le ſinus <lb/>total, S le ſinus d’inclinaiſon du plan A F, & </s> <s xml:id="echoid-s20058" xml:space="preserve">s le ſinus d’in-<lb/>clinaiſon du plan af; </s> <s xml:id="echoid-s20059" xml:space="preserve">on aura par la préſente propoſition <lb/>R<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20060" xml:space="preserve">S<emph style="sub">3</emph> :</s> <s xml:id="echoid-s20061" xml:space="preserve">: F ou (aa x V<emph style="sub">2</emph>) : </s> <s xml:id="echoid-s20062" xml:space="preserve">{aa x V<emph style="sub">2</emph> x S<emph style="sub">3</emph>/R<emph style="sub">3</emph>}, & </s> <s xml:id="echoid-s20063" xml:space="preserve">ce quatrieme terme eſt <lb/>la force qui agit contre la ſurface oblique A F; </s> <s xml:id="echoid-s20064" xml:space="preserve">puiſque la force <lb/>qui agit contre A B eſt préſentée par le produit de la ſurface par <lb/>le quarré de la vîteſſe. </s> <s xml:id="echoid-s20065" xml:space="preserve">De même on a R<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20066" xml:space="preserve">s<emph style="sub">3</emph> :</s> <s xml:id="echoid-s20067" xml:space="preserve">: f ou (b b x vv) : <lb/></s> <s xml:id="echoid-s20068" xml:space="preserve">{bb x vv x s<emph style="sub">3</emph>/R<emph style="sub">3</emph>}. </s> <s xml:id="echoid-s20069" xml:space="preserve">Les forces qui agiſſent contre les ſurfaces inégales <lb/>& </s> <s xml:id="echoid-s20070" xml:space="preserve">différemment inclinées, ſeront donc comme les deux der-<lb/>niers termes de ces deux proportions qui les expriment: </s> <s xml:id="echoid-s20071" xml:space="preserve">donc <lb/>ſi l’on appelle ces forces f & </s> <s xml:id="echoid-s20072" xml:space="preserve">φ, on aura f : </s> <s xml:id="echoid-s20073" xml:space="preserve">φ :</s> <s xml:id="echoid-s20074" xml:space="preserve">: {aa x V<emph style="sub">2</emph> x S<emph style="sub">3</emph>/R<emph style="sub">3</emph>}: </s> <s xml:id="echoid-s20075" xml:space="preserve"><lb/>{bb x vv x s<emph style="sub">3</emph>/R<emph style="sub">3</emph>} :</s> <s xml:id="echoid-s20076" xml:space="preserve">: aa x V<emph style="sub">2</emph> x S<emph style="sub">3</emph> : </s> <s xml:id="echoid-s20077" xml:space="preserve">bb x vv x s<emph style="sub">3</emph>. </s> <s xml:id="echoid-s20078" xml:space="preserve">C.</s> <s xml:id="echoid-s20079" xml:space="preserve">Q.</s> <s xml:id="echoid-s20080" xml:space="preserve">F.</s> <s xml:id="echoid-s20081" xml:space="preserve">D.</s> <s xml:id="echoid-s20082" xml:space="preserve"/> </p> <div xml:id="echoid-div1685" type="float" level="2" n="1"> <note position="right" xlink:label="note-0729-01" xlink:href="note-0729-01a" xml:space="preserve">Figure 432 <lb/>& 433.</note> </div> </div> <div xml:id="echoid-div1687" type="section" level="1" n="1222"> <head xml:id="echoid-head1442" xml:space="preserve"><emph style="sc">Corollaire</emph> III.</head> <p> <s xml:id="echoid-s20083" xml:space="preserve">1203. </s> <s xml:id="echoid-s20084" xml:space="preserve">Si les denſités D & </s> <s xml:id="echoid-s20085" xml:space="preserve">d ſont inégales, il faudroit en-<lb/>core multiplier les deux derniers termes des proportions pré-<lb/>cédentes par les mêmes denſités, pour avoir le rapport des <lb/>forces qu’ils expriment. </s> <s xml:id="echoid-s20086" xml:space="preserve">On pourroit de cette proportion dé-<lb/>duire une formule générale, pour déterminer tous les cas qui <lb/>ont rapport aux différentes ſuppoſitions que l’on peut faire <lb/>dans l’hypotheſe préſente; </s> <s xml:id="echoid-s20087" xml:space="preserve">mais il ſeroit inutile d’entrer dans <lb/>le détail de tous ces cas particuliers, que l’on ne doit recher-<lb/>cher que lorſque l’on en a beſoin.</s> <s xml:id="echoid-s20088" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1688" type="section" level="1" n="1223"> <head xml:id="echoid-head1443" xml:space="preserve"><emph style="sc">Remarque</emph> I.</head> <p> <s xml:id="echoid-s20089" xml:space="preserve">1204. </s> <s xml:id="echoid-s20090" xml:space="preserve">Il faut bien remarquer que dans cette propoſition & </s> <s xml:id="echoid-s20091" xml:space="preserve"><lb/>tous ſes corollaires, on a ſuppoſé que les ſurfaces, par rapport <lb/>auxquelles on eſtimoit le choc des fluides où elles étoient plon-<lb/>gées, répondent toutes à la même tranche ſupérieure, que l’on <lb/>ſuppoſe être la premiere du fluide, ſans quoi le théorême ne <lb/>ſeroit pas vrai, & </s> <s xml:id="echoid-s20092" xml:space="preserve">alors on parviendroit aiſément à fixer le <lb/>choc, en déterminant la vîteſſe moyenne comme nous l’avons <lb/>fait (art. </s> <s xml:id="echoid-s20093" xml:space="preserve">1184). </s> <s xml:id="echoid-s20094" xml:space="preserve">On remarquera encore que l’on pourroit trou-<lb/>ver le choc des fluides de même ou de différentes denſités, <pb o="641" file="0731" n="757" rhead="DE MATHÉMATIQUE. Liv. XVI."/> contre des ſurfaces différemment inclinées, en ſuppoſant, par <lb/>exemple, que les vîteſſes de ces tranches croiſſent comme les <lb/>hauteurs. </s> <s xml:id="echoid-s20095" xml:space="preserve">Je me ſuis arrêté à la premiere hypotheſe, parce que <lb/>c’eſt celle qui a lieu dans la nature des fluides.</s> <s xml:id="echoid-s20096" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1689" type="section" level="1" n="1224"> <head xml:id="echoid-head1444" xml:space="preserve"><emph style="sc">Remarque</emph> II.</head> <p> <s xml:id="echoid-s20097" xml:space="preserve">1205. </s> <s xml:id="echoid-s20098" xml:space="preserve">M. </s> <s xml:id="echoid-s20099" xml:space="preserve">Mariotte ayant fait pluſieurs expériences pour me-<lb/>ſurer le choc de l’eau, a trouvé que l’eau ayant un pied de <lb/>vîteſſe par ſeconde, fait un effort d’une livre & </s> <s xml:id="echoid-s20100" xml:space="preserve">demie contre <lb/>une ſurface d’un pied quarré. </s> <s xml:id="echoid-s20101" xml:space="preserve">Or pour ſe ſervir de cette expé-<lb/>rience à l’égard du choc que l’eau fait contre une ſurface, il <lb/>faut avoir une pendule ou une montre qui marque les minutes <lb/>bien exactement; </s> <s xml:id="echoid-s20102" xml:space="preserve">enſuite attacher au bout d’un fil de ſoie un <lb/>corps fort leger, comme, par exemple, un morceau de liege, <lb/>qu’il faudra faire ſurnager dans le milieu du courant de l’eau, <lb/>marquer un piquet à l’endroit où le corps aura commencé à <lb/>ſuivre le courant, & </s> <s xml:id="echoid-s20103" xml:space="preserve">faire enſorte d’accompagner ce corps le <lb/>long du bord de l’eau; </s> <s xml:id="echoid-s20104" xml:space="preserve">& </s> <s xml:id="echoid-s20105" xml:space="preserve">quand on aura parcouru une longueur <lb/>raiſonnable, on prendra garde combien il ſe ſera écoulé de <lb/>minutes depuis le moment qu’on ſera parti juſqu’à l’endroit <lb/>où l’on aura ceſſé d’accompagner ce corps; </s> <s xml:id="echoid-s20106" xml:space="preserve">& </s> <s xml:id="echoid-s20107" xml:space="preserve">ſuppoſant qu’on <lb/>ait mis 3 minutes, on meſurera bien exactement le chemin <lb/>qu’à fait le corps pendant ce tems, que je ſuppoſe être, par <lb/>exemple, de 120 toiſes. </s> <s xml:id="echoid-s20108" xml:space="preserve">Orpour ſçavoir le chemin que le corps <lb/>a parcouru pendant une ſeconde, je multiplie 60 par 3, pour <lb/>avoir 180 ſecondes (parce qu’une minute vaut 60 ſecondes), <lb/>& </s> <s xml:id="echoid-s20109" xml:space="preserve">voulant connoître la vîteſſe de l’eau pendant une ſeconde, <lb/>je réduis les toiſes en pieds pour avoir 720 pieds, que je diviſe <lb/>enſuite par 180 ſecondes, qui donnent 4 au quotient: </s> <s xml:id="echoid-s20110" xml:space="preserve">ainſi la <lb/>vîteſſe de l’eau pendant une ſeconde ſera de 4 pieds.</s> <s xml:id="echoid-s20111" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1690" type="section" level="1" n="1225"> <head xml:id="echoid-head1445" xml:space="preserve">PROPOSITION V. <lb/><emph style="sc">Probleme</emph>.</head> <p style="it"> <s xml:id="echoid-s20112" xml:space="preserve">1206. </s> <s xml:id="echoid-s20113" xml:space="preserve">Connoiſſant la vîteſſe de l’eau, trouver le choc de cette <lb/>eau contre une ſurface donnée.</s> <s xml:id="echoid-s20114" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20115" xml:space="preserve">Nous ſervant de l’expérience de M. </s> <s xml:id="echoid-s20116" xml:space="preserve">Mariotte, rapportée dans <lb/>la remarque précédente, on demande quel eſt le choc de l’eau <lb/>contre une ſurface de 20 pieds quarrés, en ſuppoſant que cette <pb o="642" file="0732" n="758" rhead="NOUVEAU COURS"/> eau a 4 pieds de vîteſſe par ſeconde. </s> <s xml:id="echoid-s20117" xml:space="preserve">Pour cela, il faut ſe rap-<lb/>peller que les chocs de l’eau avec des vîteſſes différentes contre <lb/>des ſurfaces inégales & </s> <s xml:id="echoid-s20118" xml:space="preserve">perpendiculaires au courant, ſont <lb/>comme les produits des quarrés des vîteſſes par les ſurfaces <lb/>oppoſées. </s> <s xml:id="echoid-s20119" xml:space="preserve">L’on pourra donc dire: </s> <s xml:id="echoid-s20120" xml:space="preserve">Si le quarré d’une ſeconde, <lb/>qui eſt 1, multiplié par une ſurface d’un pied, qui eſt encore 1, <lb/>donne une livre & </s> <s xml:id="echoid-s20121" xml:space="preserve">demie pour l’effort de l’eau contre la ſur-<lb/>face d’un pied quarré, que donnera le produit du quarré de la <lb/>vîteſſe de 4, qui eſt 16, par la ſurface de 20 pieds quarrés, qui <lb/>eſt 320 pour le choc de l’eau contre la ſurface de 20 pieds? </s> <s xml:id="echoid-s20122" xml:space="preserve">l’on <lb/>trouvera 480: </s> <s xml:id="echoid-s20123" xml:space="preserve">ce qui fait voir que la ſurface doit faire un effort <lb/>de 480 livres, pour être en équilibre avec le choc de l’eau.</s> <s xml:id="echoid-s20124" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1691" type="section" level="1" n="1226"> <head xml:id="echoid-head1446" xml:space="preserve"><emph style="sc">Application</emph>.</head> <p> <s xml:id="echoid-s20125" xml:space="preserve">1207. </s> <s xml:id="echoid-s20126" xml:space="preserve">Si l’on vouloit trouver l’effort de l’eau contre les aubes <lb/>d’un moulin, expoſées perpendiculairement à ſon courant, il <lb/>faut connoître d’abord la vîteſſe de l’eau, & </s> <s xml:id="echoid-s20127" xml:space="preserve">la grandeur des <lb/>aubes: </s> <s xml:id="echoid-s20128" xml:space="preserve">ainſi ſuppoſant que la vîteſſe de l’eau ſoit de 5 pieds <lb/>par ſeconde, & </s> <s xml:id="echoid-s20129" xml:space="preserve">les aubes de 6 pieds quarrés, l’on dira: </s> <s xml:id="echoid-s20130" xml:space="preserve">Si le <lb/>produit du quarré de la vîteſſe d’un pied par un pied quarré, <lb/>fait un effort d’une livre & </s> <s xml:id="echoid-s20131" xml:space="preserve">demie en une ſeconde, que fera le <lb/>produit du quarré de la vîteſſe de 5 pieds par la ſurface de 6 <lb/>pieds? </s> <s xml:id="echoid-s20132" xml:space="preserve">l’on trouvera pour l’effort que l’on cherche 225 livres.</s> <s xml:id="echoid-s20133" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1692" type="section" level="1" n="1227"> <head xml:id="echoid-head1447" xml:space="preserve">PROPOSITION VI. <lb/><emph style="sc">Théoreme</emph>.</head> <p style="it"> <s xml:id="echoid-s20134" xml:space="preserve">1208. </s> <s xml:id="echoid-s20135" xml:space="preserve">Si l’on a un vaiſſeau rempli d’eau, qui ſoit toujours en-<lb/> <anchor type="note" xlink:label="note-0732-01a" xlink:href="note-0732-01"/> tretenu à la même hauteur, je dis que les chocs de l’eau, à la ſortie <lb/>de deux ajutages égaux, ſeront dans la raiſon des hauteurs de l’eau <lb/>au deſſus du centre des deux ajutages.</s> <s xml:id="echoid-s20136" xml:space="preserve"/> </p> <div xml:id="echoid-div1692" type="float" level="2" n="1"> <note position="left" xlink:label="note-0732-01" xlink:href="note-0732-01a" xml:space="preserve">Figure 434.</note> </div> </div> <div xml:id="echoid-div1694" type="section" level="1" n="1228"> <head xml:id="echoid-head1448" xml:space="preserve"><emph style="sc">Démonstration</emph>.</head> <p> <s xml:id="echoid-s20137" xml:space="preserve">Si le vaiſſeau A B C D eſt rempli d’eau, & </s> <s xml:id="echoid-s20138" xml:space="preserve">qu’elle ſorte par <lb/>les deux ajutages E & </s> <s xml:id="echoid-s20139" xml:space="preserve">F, les vîteſſes de l’eau ſeront comme <lb/>√B E\x{0020} eſt à √B F\x{0020}; </s> <s xml:id="echoid-s20140" xml:space="preserve">& </s> <s xml:id="echoid-s20141" xml:space="preserve">ſi les ajutages ſont égaux, les quantités <lb/>d’eau qui ſortiront dans le même tems, ſeront encore comme <lb/>√B E\x{0020} eſt à √B F\x{0020}: </s> <s xml:id="echoid-s20142" xml:space="preserve">mais ces quantités d’eau peuvent être re-<lb/>gardées comme les maſſes, & </s> <s xml:id="echoid-s20143" xml:space="preserve">les racines de B E & </s> <s xml:id="echoid-s20144" xml:space="preserve">B F comme <lb/>leurs vîteſſes: </s> <s xml:id="echoid-s20145" xml:space="preserve">par conſéquent le choc dont l’eau ſera capable, <pb o="643" file="0733" n="759" rhead="DE MATHÉMATIQUE. Liv. XVI."/> à la ſortie des deux ajutages, ſera égal au produit de √BE\x{0020} <lb/>x √BE\x{0020} & </s> <s xml:id="echoid-s20146" xml:space="preserve">à √BF\x{0020} x √BF\x{0020}, c’eſt-à-dire comme le quarré des <lb/>racines des hauteurs de l’eau au deſſus du centre des ajutages; <lb/></s> <s xml:id="echoid-s20147" xml:space="preserve">mais ces deux produits ne ſont autre choſe que B E & </s> <s xml:id="echoid-s20148" xml:space="preserve">B F: </s> <s xml:id="echoid-s20149" xml:space="preserve"><lb/>par conſéquent les chocs de l’eau, à la ſortie des ajutages égaux, <lb/>ſont comme les hauteurs de l’eau au deſſus du centre des aju-<lb/>tages.</s> <s xml:id="echoid-s20150" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1695" type="section" level="1" n="1229"> <head xml:id="echoid-head1449" xml:space="preserve"><emph style="sc">Corollaire</emph>.</head> <p> <s xml:id="echoid-s20151" xml:space="preserve">1209. </s> <s xml:id="echoid-s20152" xml:space="preserve">Il ſuit delà que ſi les ajutages ſont de différentes gran-<lb/>deurs, les chocs de l’eau à leurs ſorties, ſeront comme les pro-<lb/>duits des quarrés des diametres des ajutages par la hauteur de <lb/>l’eau qui répond à leur centre, s’ils ſont circulaires; </s> <s xml:id="echoid-s20153" xml:space="preserve">mais s’ils <lb/>ſont de toute autre figure, il faudra multiplier leur capacité <lb/>par la hauteur de l’eau qui répond au centre.</s> <s xml:id="echoid-s20154" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1696" type="section" level="1" n="1230"> <head xml:id="echoid-head1450" xml:space="preserve">DISCOURS</head> <p style="it"> <s xml:id="echoid-s20155" xml:space="preserve"><emph style="sc">Sur la nature et les proprietés de l’</emph><emph style="sc">Air</emph>, <lb/>pour ſervir d’introduction à la Phyſique, ſervant auſſi à rendre <lb/>raiſon de l’effet des machines hydrauliques.</s> <s xml:id="echoid-s20156" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20157" xml:space="preserve">Quoique les Anciens nous aient laiſſé beaucoup de belles <lb/>connoiſſances, il ſemble qu’on pourroit leur reprocher de n’a-<lb/>voir point aſſez étudié la nature, ſurtout quand on fait réflexion <lb/>aux idées fauſſes qu’ils avoient de l’air: </s> <s xml:id="echoid-s20158" xml:space="preserve">ce n’eſt pourtant pas <lb/>manque qu’ils n’aient eu aſſez de tems pour en découvrir les <lb/>propriétés; </s> <s xml:id="echoid-s20159" xml:space="preserve">mais apparemment qu’il en étoit de ceci comme <lb/>d’une infinité d’autres choſes qui étoient reſervées aux décou-<lb/>vertes de notre tems: </s> <s xml:id="echoid-s20160" xml:space="preserve">& </s> <s xml:id="echoid-s20161" xml:space="preserve">pour ne parler que de l’air, nous al-<lb/>lons faire voir qu’il a de la peſanteur, qu’il a du reſſort, & </s> <s xml:id="echoid-s20162" xml:space="preserve"><lb/>qu’il eſt capable d’être condenſé & </s> <s xml:id="echoid-s20163" xml:space="preserve">dilaté.</s> <s xml:id="echoid-s20164" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20165" xml:space="preserve">Avant M. </s> <s xml:id="echoid-s20166" xml:space="preserve">Deſcartes & </s> <s xml:id="echoid-s20167" xml:space="preserve">M. </s> <s xml:id="echoid-s20168" xml:space="preserve">Paſcal, ſi l’on demandoit aux <lb/>Philoſophes pourquoi, en tirant le piſton d’une ſeringue ou <lb/>d’une pompe, l’eau monte & </s> <s xml:id="echoid-s20169" xml:space="preserve">ſuit comme ſi elle adhéroit; <lb/></s> <s xml:id="echoid-s20170" xml:space="preserve">pourquoi quand on remplit d’eau un ſiphon, & </s> <s xml:id="echoid-s20171" xml:space="preserve">qu’on met <lb/>chaque jambe dans un vaiſſeau plein d’eau, ſi un des vaiſſeaux <lb/>eſt un peu plus élevé que l’autre, l’eau monte par le ſiphon, <lb/>ſort du vaiſſeau qui eſt le plus élevé, pour deſcendre dans celui <pb o="644" file="0734" n="760" rhead="NOUVEAU COURS"/> qui eſt un peu plus bas, tant que toute l’eau de celui d’en <lb/>haut ſoit entrée dans celui d’en bas; </s> <s xml:id="echoid-s20172" xml:space="preserve">ils répondoient que la <lb/>nature avoit de l’horreur pour le vuide, ou bien que la nature <lb/>abhorroit le vuide, comme ſi elle étoit capable de paſſion, <lb/>pour avoir de l’horreur pour quelque choſe: </s> <s xml:id="echoid-s20173" xml:space="preserve">car à leur ſens ils <lb/>parloient comme ſi la nature faiſoit de grands efforts pour éviter <lb/>le vuide, quoiqu’on voie parfaitement qu’elle ne fait aucune <lb/>choſe pour l’éviter, ni pour le rechercher, & </s> <s xml:id="echoid-s20174" xml:space="preserve">que le vuide ou <lb/>le plein lui ſont fort indifférens.</s> <s xml:id="echoid-s20175" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20176" xml:space="preserve">Il eſt bien vrai que l’eau monte dans une pompe, quand il <lb/>n’y a point de jour par où l’air puiſſe entrer, & </s> <s xml:id="echoid-s20177" xml:space="preserve">qu’ainſi il y <lb/>auroit du vuide, ſi l’eau ne ſuivoit pas le piſton, & </s> <s xml:id="echoid-s20178" xml:space="preserve">même qu’elle <lb/>n’y monte pas, quand il y a des fentes par où l’air peut entrer <lb/>pour la remplir. </s> <s xml:id="echoid-s20179" xml:space="preserve">De même ſi l’on fait une petite ouverture au <lb/>haut d’un ſiphon, par où l’air puiſſe s’introduire, l’eau de cha-<lb/>que branche tombe dans ſon vaiſſeau, & </s> <s xml:id="echoid-s20180" xml:space="preserve">le tout demeure en <lb/>repos: </s> <s xml:id="echoid-s20181" xml:space="preserve">d’où l’on a conclu que la nature avoit de l’horreur pour <lb/>le vuide, puiſqu’auſſitôt qu’il n’y avoit point d’air dans un <lb/>tuyau, l’eau montoit d’elle-même, & </s> <s xml:id="echoid-s20182" xml:space="preserve">que l’air ſurvenant, l’eau <lb/>ſe remettroit dans ſon premier état; </s> <s xml:id="echoid-s20183" xml:space="preserve">ce qui a fait croire qu’elle <lb/>n’y montoit que pour empêcher le vuide.</s> <s xml:id="echoid-s20184" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20185" xml:space="preserve">Mais ſi l’on fait voir que ces effets (de même que pluſieurs <lb/>autres que nous expliquerons dans la ſuite) ne ſont cauſés <lb/>que par la peſanteur de l’air, on n’aura plus lieu de douter <lb/>que la nature n’a point d’horreur pour le vuide, qu’elle ſuit les <lb/>loix de la méchanique, auſſi-bien par rapport à l’air, que par <lb/>rapport aux liqueurs de différentes peſanteurs, & </s> <s xml:id="echoid-s20186" xml:space="preserve">que ce qu’on <lb/>peut dire de l’air n’eſt qu’une ſuite des principes que l’on a dé-<lb/>montrés dans le Traité précédent.</s> <s xml:id="echoid-s20187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20188" xml:space="preserve">Pour être convaincu de la peſanteur de l’air par une expé-<lb/>rience dont il eſt aiſé de ſe convaincre, prenez un tuyau de <lb/>verre de 20 ou 24 pouces, bien bouché par une de ſes extrê-<lb/>mités, après qu’on l’aura rempli de mercure; </s> <s xml:id="echoid-s20189" xml:space="preserve">bouchez enſuite <lb/>le bout qui eſt ouvert avec le doigt, & </s> <s xml:id="echoid-s20190" xml:space="preserve">ſoutenez le tuyau per-<lb/>pendiculairement, enſorte que le bout ouvert ſoit en bas: </s> <s xml:id="echoid-s20191" xml:space="preserve">ſi <lb/>vous plongez dans un vaſe où il y aura du mercure le bout <lb/>que vous aurez bouché avec le doigt, & </s> <s xml:id="echoid-s20192" xml:space="preserve">qu’après cela vous <lb/>laiſſiez la liberté au mercure de deſcendre, vous verrez que <lb/>bien loin qu’il retombe dans le vaſe pour ſe mêler avec l’autre, <lb/>il demeurera ſuſpendu de lui-même. </s> <s xml:id="echoid-s20193" xml:space="preserve">La raiſon de cet effet vient <pb o="645" file="0735" n="761" rhead="DE MATHÉMATIQUE. Liv. XVI."/> de la peſanteur de l’air, qui preſſe le mercure qui eſt dans le <lb/>vaſe, & </s> <s xml:id="echoid-s20194" xml:space="preserve">qui ne preſſe pas celui qui eſt dans le tuyau, qui eſt <lb/>moins peſant qu’une colonne d’air qui aura la même baſe: </s> <s xml:id="echoid-s20195" xml:space="preserve">ainſi <lb/>c’eſt le poids de l’air qui force le mercure de reſter dans le <lb/>tuyau; </s> <s xml:id="echoid-s20196" xml:space="preserve">& </s> <s xml:id="echoid-s20197" xml:space="preserve">pour en être plus certain, il n’y a qu’à ouvrir le <lb/>bout d’en haut qu’on a bouché, & </s> <s xml:id="echoid-s20198" xml:space="preserve">auſſitôt vous verrez le mer-<lb/>cure deſcendre, & </s> <s xml:id="echoid-s20199" xml:space="preserve">ſe mêler avec celui qui eſt dans le vaſe.</s> <s xml:id="echoid-s20200" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20201" xml:space="preserve">Si l’on prend encore un tuyau de 20 ou de 24 pouces, rempli <lb/>de mercure, bouché par une de ſes extrêmités, & </s> <s xml:id="echoid-s20202" xml:space="preserve">que l’autre <lb/>extrêmité ſoit recourbée, vous verrez que le mercure, quoi-<lb/>que le tuyau ne ſoit pas plongé dans un vaſe, ſe maintiendra <lb/>ſuſpendu ſans ſortir par le bout recourbé, à cauſe que le poids <lb/>de l’air qui peſe ſur le mercure du bout recourbé, eſt plus pe-<lb/>ſant que le mercure qui eſt dans le tuyau.</s> <s xml:id="echoid-s20203" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20204" xml:space="preserve">Si au lieu d’un tuyau de 20 ou 24 pouces l’on ſe ſert d’un <lb/>qui ait 25 ou 26 pieds, & </s> <s xml:id="echoid-s20205" xml:space="preserve">qu’au lieu de le remplir de mercure, <lb/>on le rempliſſe d’eau, l’on verra que l’eau demeurera ſuſpen-<lb/>due comme le mercure, quoique le tuyau ſoit plus grand: <lb/></s> <s xml:id="echoid-s20206" xml:space="preserve">car comme l’eau eſt beaucoup plus legere que le mercure, on <lb/>en mettra une bien plus grande hauteur dans un tuyau que de <lb/>mercure: </s> <s xml:id="echoid-s20207" xml:space="preserve">car nous ſçavons que les hauteurs de différentes li-<lb/>queurs ſont comme les poids des mêmes liqueurs.</s> <s xml:id="echoid-s20208" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20209" xml:space="preserve">Cependant quoique la peſanteur de l’air ſoutienne ſuſpendus <lb/>le mercure & </s> <s xml:id="echoid-s20210" xml:space="preserve">l’eau dans des tuyaux de la grandeur que nous <lb/>venons de dire, il ne faut pas croire que ſi l’on rempliſſoit <lb/>d’eau un tuyau qui auroit beaucoup plus de 25 ou 26 pieds, <lb/>comme, par exemple, de 40 pieds, que l’eau y demeurera <lb/>toute ſuſpendue: </s> <s xml:id="echoid-s20211" xml:space="preserve">car l’air ne peut pas ſoutenir un plus grand <lb/>poids que le ſien; </s> <s xml:id="echoid-s20212" xml:space="preserve">& </s> <s xml:id="echoid-s20213" xml:space="preserve">c’eſt par le moyen des tuyaux remplis de <lb/>mercure ou d’eau que l’on meſure la peſanteur de l’air, comme <lb/>on le va voir.</s> <s xml:id="echoid-s20214" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20215" xml:space="preserve">Si l’on a un tuyau de verre de 40 pouces, que l’on rempliſſe <lb/>de mercure, enſorte qu’il y ait toujours une de ſes extrêmités <lb/>bouchée, & </s> <s xml:id="echoid-s20216" xml:space="preserve">que l’autre bout auquel on aura mis le doigt, ſoit <lb/>plongé dans un vaſe où il y ait du mercure, ou que ce bout <lb/>ſoit ſeulement recourbé, & </s> <s xml:id="echoid-s20217" xml:space="preserve">qu’on le ſoutienne perpendiculai-<lb/>rement dans l’air ou dans le mercure, car cela ne fait rien; <lb/></s> <s xml:id="echoid-s20218" xml:space="preserve">l’on verra qu’auſſitôt qu’on aura ôté le doigt qu’on avoit ap-<lb/>pliqué ſur le bout ouvert, le mercure baiſſera tant qu’il ſera <lb/>parvenu à la hauteur de 28 pouces, qui eſt la hauteur où une <pb o="646" file="0736" n="762" rhead="NOUVEAU COURS"/> colonne de mercure eſt en équilibre avec la colonne d’air qui <lb/>lui répond.</s> <s xml:id="echoid-s20219" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20220" xml:space="preserve">Si l’on prend un tuyau de 40 pieds, conditionné comme <lb/>ceux dont nous avons parlé, l’on verra que l’ayant rempli <lb/>d’eau, elle deſcendra tant qu’elle ſoit à la hauteur de 31 pieds, <lb/>parce qu’une pareille colonne d’eau eſt en équilibre avec celle <lb/>de l’air qui lui répond, ou bien avec une colonne de vif-argent <lb/>de 28 pouces: </s> <s xml:id="echoid-s20221" xml:space="preserve">mais comme nous ſçavons qu’un pied cube d’eau <lb/>peſe 72 livres, ſi l’on multiplie 31 par 72, l’on aura 2232, qui <lb/>eſt la quantité de livres que peſe une colonne d’air, qui auroit <lb/>un pied quarré de baſe, & </s> <s xml:id="echoid-s20222" xml:space="preserve">pour hauteur celle de l’atmoſphere<anchor type="note" xlink:href="" symbol="*"/>.</s> <s xml:id="echoid-s20223" xml:space="preserve"/> </p> <note symbol="*" position="left" xml:space="preserve">L’on nom-<lb/>me atmoſ-<lb/>phere l’éten-<lb/>due de l’air <lb/>quieſt renfer-<lb/>mé dans le <lb/>tourbillon de <lb/>la terre.</note> <p> <s xml:id="echoid-s20224" xml:space="preserve">Cette épreuve eſt encore confirmée par les pompes aſpi-<lb/>rantes & </s> <s xml:id="echoid-s20225" xml:space="preserve">les ſeringues: </s> <s xml:id="echoid-s20226" xml:space="preserve">car auſſitôt qu’on tire le piſton d’une <lb/>pompe, l’eau ſuit le piſton; </s> <s xml:id="echoid-s20227" xml:space="preserve">& </s> <s xml:id="echoid-s20228" xml:space="preserve">ſi l’on continue à lever le piſton, <lb/>l’eau ſuivra toujours, mais non pas à la hauteur que l’on vou-<lb/>dra, puiſqu’elle ne paſſe pas 31 pieds: </s> <s xml:id="echoid-s20229" xml:space="preserve">car auſſitôt qu’on veut <lb/>la tirer plus haut, le piſton ne tire plus l’eau, & </s> <s xml:id="echoid-s20230" xml:space="preserve">elle demeure <lb/>immobile & </s> <s xml:id="echoid-s20231" xml:space="preserve">ſuſpendue à cette hauteur, où elle ſe trouve en <lb/>équilibre avec le poids de l’air qui peſe au dehors du tuyau <lb/>ſur l’eau qui l’environne. </s> <s xml:id="echoid-s20232" xml:space="preserve">L’on peut remarquer ici, pour dé-<lb/>ſabuſer ceux qui croient que l’eau monte dans les pompes, <lb/>parce que la nature a de l’horreur pour le vuide, que quand on <lb/>a hauſſé le piſton au-delà de 31 pieds, l’eau demeure à cette <lb/>hauteur, & </s> <s xml:id="echoid-s20233" xml:space="preserve">il ſe trouve un intervalle entre l’eau & </s> <s xml:id="echoid-s20234" xml:space="preserve">le piſton, <lb/>où il n’y a point, ou très-peu d’air que l’eau ne peut remplir, <lb/>ne pouvant être pouſſée plus haut par l’air extérieur. </s> <s xml:id="echoid-s20235" xml:space="preserve">Si nos <lb/>Philoſophes avoient pris garde à cela, ils auroient ſans doute <lb/>été fort étonnés de voir que la nature ceſſe d’avoir de l’horreur <lb/>pour le vuide au-delà de 31 pieds de hauteur, & </s> <s xml:id="echoid-s20236" xml:space="preserve">ils auroient <lb/>pu l’accuſer d’avoir du caprice, puiſqu’à une certaine hauteur <lb/>elle ne peut ſupporter le vuide, & </s> <s xml:id="echoid-s20237" xml:space="preserve">qu’après cela le vuide lui <lb/>devient indifférent.</s> <s xml:id="echoid-s20238" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20239" xml:space="preserve">Si l’on ſe ſert d’une ſeringue longue de 3 pieds ou de 3 pieds <lb/>& </s> <s xml:id="echoid-s20240" xml:space="preserve">demi, l’on verra encore que mettant le bout du tuyau, qui <lb/>eſt ouvert dans un vaſe de vif-argent, qu’en tirant le piſton, <lb/>le vif-argent montera à la hauteur de 28 pouces, & </s> <s xml:id="echoid-s20241" xml:space="preserve">qu’inutile-<lb/>ment on levera le piſton pour faire monter le vif-argent plus <lb/>haut, qu’il demeurera toujours à la hauteur qui le met en équi-<lb/>libre avec le poids de l’air: </s> <s xml:id="echoid-s20242" xml:space="preserve">ainſi l’eau, le vif-argent & </s> <s xml:id="echoid-s20243" xml:space="preserve">l’air <lb/>demeurent en équilibre, quand les hauteurs ſont entr’elles <pb o="647" file="0737" n="763" rhead="DE MATHÉMATIQUE. Liv. XVI."/> comme leurs poids; </s> <s xml:id="echoid-s20244" xml:space="preserve">& </s> <s xml:id="echoid-s20245" xml:space="preserve">cela de quelque groſſeur que ſoient les <lb/>tuyaux, parce que les liqueurs ne peſent pas ſelon la grandeur <lb/>de leurs baſes, mais ſelon leurs hauteurs.</s> <s xml:id="echoid-s20246" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20247" xml:space="preserve">Pour expliquer comme la peſanteur de l’air fait monter l’eau <lb/>dans les ſiphons, nous ſuppoſerons un ſiphon dont une des jam-<lb/>bes ſoit environ haute d’un pied, & </s> <s xml:id="echoid-s20248" xml:space="preserve">l’autre d’un pied un pouce. <lb/></s> <s xml:id="echoid-s20249" xml:space="preserve">Si on le remplit d’eau, & </s> <s xml:id="echoid-s20250" xml:space="preserve">qu’on bouche bien les deux ouver-<lb/>tures, pour qu’elle ne puiſſe pas ſortir, & </s> <s xml:id="echoid-s20251" xml:space="preserve">qu’après cela l’on ait <lb/>deux vaiſſeaux, dont l’un ſoit un peu plus élevé que l’autre, <lb/>& </s> <s xml:id="echoid-s20252" xml:space="preserve">que le plus élevé ſoit rempli d’eau, mettant la plus courte <lb/>jambe du ſiphon dans le vaiſſeau plus élevé, & </s> <s xml:id="echoid-s20253" xml:space="preserve">la plus longue <lb/>dans celui qui eſt un peu plus bas, la courte jambe trempant <lb/>dans l’eau, auſſitôt qu’on aura débouché les ouvertures, l’eau <lb/>qui eſt dedans, au lieu de deſcendre, cherchera à monter: </s> <s xml:id="echoid-s20254" xml:space="preserve">car <lb/>l’eau qui eſt dans les deux vaiſſeaux étant preſſée par l’air, & </s> <s xml:id="echoid-s20255" xml:space="preserve"><lb/>non pas celle qui eſt dans le ſiphon, la forcera d’y entrer pour <lb/>monter bien plus haut, s’il ſe pouvoit, puiſqu’elle ne montera <lb/>que d’un pied, au lieu que le poids de l’air eſt capable de la <lb/>faire monter de 31 pieds.</s> <s xml:id="echoid-s20256" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20257" xml:space="preserve">D’où il arrive que l’eau de chaque jambe étant pouſſée au <lb/>haut du ſiphon, elle ſe combat à cet endroit; </s> <s xml:id="echoid-s20258" xml:space="preserve">de ſorte qu’il <lb/>faut que celle qui a le plus de force l’emporte ſur celle qui en a <lb/>moins: </s> <s xml:id="echoid-s20259" xml:space="preserve">mais comme l’air a plus de hauteur d’un pouce ſur le <lb/>vaiſſeau plus bas que ſur le vaiſſeau plus élevé, il pouſſe en <lb/>haut l’eau de la longue jambe plus fortement que celle qui eſt <lb/>dans l’autre; </s> <s xml:id="echoid-s20260" xml:space="preserve">d’où il ſemble d’abord que l’eau doit être pouſſée <lb/>de la plus longue jambe dans la plus courte; </s> <s xml:id="echoid-s20261" xml:space="preserve">mais le poids de <lb/>l’eau de chaque jambe, quoiqu’il réſiſte à l’air, ne réſiſte pas <lb/>également: </s> <s xml:id="echoid-s20262" xml:space="preserve">car comme l’eau de la longue jambe a plus de hau-<lb/>teur d’un pouce que celle de la petite, elle réſiſte plus forte-<lb/>ment de la force que lui donne la hauteur d’un pouce d’eau. <lb/></s> <s xml:id="echoid-s20263" xml:space="preserve">Or elle n’eſt pouſſée en haut plus que celle de l’autre jambe, <lb/>que par la hauteur d’un pouce d’air; </s> <s xml:id="echoid-s20264" xml:space="preserve">mai le pouce d’eau qui <lb/>eſt dans la plus longue jambe, a plus de force pour deſcendre <lb/>que le pouce d’air n’en a pour le faire monter, puiſqu’un pouce <lb/>d’eau eſt plus peſant qu’un pouce d’air: </s> <s xml:id="echoid-s20265" xml:space="preserve">ainſi l’eau de la plus <lb/>courte jambe eſt pouſſée en haut avec plus de force que celle <lb/>de la plus grande; </s> <s xml:id="echoid-s20266" xml:space="preserve">ce qui fait qu’elle monte pour paſſer dans <lb/>l’autre vaiſſeau, & </s> <s xml:id="echoid-s20267" xml:space="preserve">continuera à monter tant qu’il y aura de <lb/>l’eau dans le vaiſſeau qui lui répond.</s> <s xml:id="echoid-s20268" xml:space="preserve"/> </p> <pb o="648" file="0738" n="764" rhead="NOUVEAU COURS"/> <p> <s xml:id="echoid-s20269" xml:space="preserve">C’eſt ainſi que toute l’eau du vaiſſeau le plus élevé, montera <lb/>& </s> <s xml:id="echoid-s20270" xml:space="preserve">ſe rendra dans le plus bas, tant que la branche du ſiphon, <lb/>qui y trempe, ſera au deſſous d’une hauteur de 31 pieds: </s> <s xml:id="echoid-s20271" xml:space="preserve">car <lb/>comme nous l’avons dit, le poids de l’air peut bien hauſſer & </s> <s xml:id="echoid-s20272" xml:space="preserve"><lb/>tenir ſuſpendue l’eau à cette hauteur; </s> <s xml:id="echoid-s20273" xml:space="preserve">mais dès que la branche <lb/>qui trempe dans le vaiſſeau élevé excédera cette hauteur, il <lb/>arrivera que le ſiphon ne fera plus ſon effet, j’entends que l’eau <lb/>du vaiſſeau élevé ne montera plus en haut du ſiphon pour ſe <lb/>rendre dans l’autre, parce que le poids de l’air ne peut pas l’é-<lb/>lever au-delà de 31 pieds; </s> <s xml:id="echoid-s20274" xml:space="preserve">de ſorte que l’eau ſe diviſera au haut <lb/>du ſiphon, & </s> <s xml:id="echoid-s20275" xml:space="preserve">tombera de chaque jambe dans ſon vaiſſeau, <lb/>juſqu’à ce qu’elle ſoit reſtée à la hauteur de 31 pieds au deſſus <lb/>de chaque vaiſſeau, où elle demeurera en repos ſuſpendue à <lb/>cette hauteur par le poids de l’air qui la contre-peſe.</s> <s xml:id="echoid-s20276" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20277" xml:space="preserve">Il arrive pluſieurs autres choſes dans la nature, que les An-<lb/>ciens ont toujours attribuées à l’horreur du vuide, mais qui <lb/>n’ont cependant d’autre cauſe que la peſanteur de l’air: </s> <s xml:id="echoid-s20278" xml:space="preserve">par <lb/>exemple, ſi deux corps fort polis ſont appliqués l’un contre <lb/>l’autre, l’on trouve une extrême réſiſtance à les ſéparer, & </s> <s xml:id="echoid-s20279" xml:space="preserve"><lb/>cette réſiſtance même eſt ſi grande, que l’on a cru qu’il n’y <lb/>avoit point de force humaine qui puiſſe les déſunir. </s> <s xml:id="echoid-s20280" xml:space="preserve">Cepen-<lb/>dant ſi l’on fait attention que n’y ayant point d’air entre ces <lb/>deux corps, ſi l’on tient celui d’en haut avec la main, il doit <lb/>arriver que celui d’en bas demeurera ſuſpendu, puiſqu’il eſt <lb/>preſſé par tout le poids de l’air qui le touche par deſſous, & </s> <s xml:id="echoid-s20281" xml:space="preserve"><lb/>qui fait qu’on ne peut les ſéparer qu’on n’emploie une force <lb/>plus grande que celle du poids de l’air; </s> <s xml:id="echoid-s20282" xml:space="preserve">tellement que ſi ces <lb/>deux corps ſont, par exemple, chacun d’un pied cube, & </s> <s xml:id="echoid-s20283" xml:space="preserve">qu’ils <lb/>en aient la figure, ils ſeront preſſés l’un contre l’autre par une <lb/>force de 2232 livres, qui eſt le poids d’une colonne d’air, qui <lb/>auroit un pied quarré de baſe: </s> <s xml:id="echoid-s20284" xml:space="preserve">ainſi pour vaincre la force de <lb/>l’air, afin de ſéparer ces deux corps, il faut employer une <lb/>force plus grande que celle de 2232 livres, & </s> <s xml:id="echoid-s20285" xml:space="preserve">pour lors ces <lb/>deux corps ſe déſuniront ſans aucune difficulté, puiſqu’il im-<lb/>porte fort peu à la nature qu’ils ſoient ſéparés ou non.</s> <s xml:id="echoid-s20286" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20287" xml:space="preserve">L’expérience nous fait voir encore qu’un ſoufflet, dont toutes <lb/>les ouvertures ſont bien bouchées, eſt très-difficile à ouvrir, <lb/>trouvant de la réſiſtance, comme ſi les aîles étoient collées: </s> <s xml:id="echoid-s20288" xml:space="preserve">ſi <lb/>on demande la cauſe de cet effet, on n’en trouvera pas d’autre <lb/>que celle de la peſanteur de l’air; </s> <s xml:id="echoid-s20289" xml:space="preserve">car comme il preſſe les aîles <pb o="649" file="0739" n="765" rhead="DE MATHÉMATIQUE. Liv. XVI."/> du ſoufflet, ſans pouvoir s’introduire dedans, l’on ne peut <lb/>lever une des aîles ſans lever auſſi toute la maſſe de l’air qui <lb/>eſt au deſſus, qui réſiſtera d’autant plus, que les aîles du ſoufflet <lb/>auront de capacité, tellement que ſi elles avoient un pied & </s> <s xml:id="echoid-s20290" xml:space="preserve"><lb/>demi de ſuperficie, il faudroit une force plus grande que celle <lb/>de 3348 livres, qui eſt égale au poids de l’air qui répond à un <lb/>plan d’un pied & </s> <s xml:id="echoid-s20291" xml:space="preserve">demi de ſuperficie; </s> <s xml:id="echoid-s20292" xml:space="preserve">mais dès que l’on fait <lb/>une ouverture au ſoufflet, l’air qui entre dedans fait équilibre <lb/>avec celui de dehors; </s> <s xml:id="echoid-s20293" xml:space="preserve">& </s> <s xml:id="echoid-s20294" xml:space="preserve">l’on ne trouve plus de difficulté à <lb/>l’ouvrir.</s> <s xml:id="echoid-s20295" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20296" xml:space="preserve">De même ſi l’on demande pourquoi en mettant la bouche <lb/>ſur l’eau, elle monte lorſque l’on aſpire, comme cela arrive auſſi <lb/>avec un chalumeau de paille, il n’y a qu’à conſidérer que l’eau <lb/>eſt preſſée de toutes parts par le poids de l’air, excepté à l’en-<lb/>droit de la bouche, où le chalumeau eſt appliqué, parce qu’en <lb/>aſpirant il arrive que les muſcles de la reſpiration élevant la poi-<lb/>trine, font la capacité du dedans plus grande; </s> <s xml:id="echoid-s20297" xml:space="preserve">ce qui donne <lb/>à l’air du dedans plus de place à remplir qu’il n’avoit aupara-<lb/>vant, & </s> <s xml:id="echoid-s20298" xml:space="preserve">lui donne moins de force pour empêcher l’eau d’en-<lb/>trer dans la bouche, que l’air du dehors n’en a pour l’y faire <lb/>monter: </s> <s xml:id="echoid-s20299" xml:space="preserve">ce qui devient le même cas que celui qui fait que <lb/>l’eau monte dans les pompes & </s> <s xml:id="echoid-s20300" xml:space="preserve">dans les ſeringues.</s> <s xml:id="echoid-s20301" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20302" xml:space="preserve">Comme la peſanteur de l’air n’eſt pas toujours la même, & </s> <s xml:id="echoid-s20303" xml:space="preserve"><lb/>qu’elle varie ſelon qu’il eſt plus ou moins chargé de vapeurs, <lb/>ſes effets varient auſſi continuellement dans un même lieu; </s> <s xml:id="echoid-s20304" xml:space="preserve">& </s> <s xml:id="echoid-s20305" xml:space="preserve"><lb/>c’eſt ce qu’on remarque par le barometre, où le mercure s’é-<lb/>leve quelquefois qu deſſus de 28 pouces, & </s> <s xml:id="echoid-s20306" xml:space="preserve">quelquefois deſ-<lb/>cend & </s> <s xml:id="echoid-s20307" xml:space="preserve">ſe met au deſſous; </s> <s xml:id="echoid-s20308" xml:space="preserve">quelque tems après il remonte, & </s> <s xml:id="echoid-s20309" xml:space="preserve"><lb/>toujours dans une viciſſitude continuelle qui ſuit celle de l’air. <lb/></s> <s xml:id="echoid-s20310" xml:space="preserve">La même choſe arrive par conſéquent dans les pompes où l’eau <lb/>monte quelquefois dans un tems à 31 pieds & </s> <s xml:id="echoid-s20311" xml:space="preserve">demi, puis elle <lb/>revient à 31 pieds, puis elle baiſſe, & </s> <s xml:id="echoid-s20312" xml:space="preserve">n’eſt plus qu’à la hau-<lb/>teur de 30 pieds & </s> <s xml:id="echoid-s20313" xml:space="preserve">quelques pouces, étant aſſujetties, comme <lb/>le barometre, aux différentes peſanteurs de l’air.</s> <s xml:id="echoid-s20314" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20315" xml:space="preserve">Comme l’air ſur les montagnes fort élevées, ne peſe pas <lb/>tant que ſur le bord de la mer, que nous prendrons pour le <lb/>lieu le plus bas de la terre, l’expérience fait voir que les pom-<lb/>pes qui ſont ſur les lieux fort élevés ne font pas monter l’eau <lb/>ſi haut; </s> <s xml:id="echoid-s20316" xml:space="preserve">l’on a même remarqué que ſur une montagne élevée <lb/>de 600 toiſes, l’eau, au lieu de monter à 31 pieds, comme nous <pb o="650" file="0740" n="766" rhead="NOUVEAU COURS"/> l’avons dit, ne montoit qu’à 26 pieds quelques pouces: </s> <s xml:id="echoid-s20317" xml:space="preserve">le même <lb/>changement arrive dans les lieux qui ſont fort bas, où l’eau <lb/>monte quelquefois juſqu’à 32 ou 33 pieds; </s> <s xml:id="echoid-s20318" xml:space="preserve">mais ces change-<lb/>mens s’obſervent bien mieux avec le barometre, qui peut ſer-<lb/>vir non ſeulement à connoître la peſanteur de l’air dans les <lb/>lieux différemment élevés, mais encore à meſurer la hauteur <lb/>des montagnes, & </s> <s xml:id="echoid-s20319" xml:space="preserve">même celle de l’atmoſphere.</s> <s xml:id="echoid-s20320" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20321" xml:space="preserve">Car ſi on eſt au pied d’une montagne, & </s> <s xml:id="echoid-s20322" xml:space="preserve">que le mercure à <lb/>cet endroit ſoit élevé de 28 pouces, l’on verra qu’à meſure que <lb/>l’on montera pour en gagner le ſommet, le mercure au lieu de <lb/>reſter à la hauteur de 28 pouces, baiſſera, parce qu’étant ſou-<lb/>tenu par une moindre colonne d’air, il faut néceſſairement <lb/>qu’il baiſſe pour ſe mettre en équilibre avec cette colonne: <lb/></s> <s xml:id="echoid-s20323" xml:space="preserve">ainſi il demeure ſuſpendu à une hauteur d’autant moindre, <lb/>qu’on le porte à un lieu plus élevé de ſorte que s’il étoit poſſi-<lb/>ble d’aller juſqu’qu haut de l’atmoſphere pour en ſortir entiére-<lb/>ment dehors, le vif-argent tomberoit, ſans qu’il en reſtât au-<lb/>cune partie, puiſqu’il n’y auroit plus aucun air pour le contre-<lb/>peſer.</s> <s xml:id="echoid-s20324" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20325" xml:space="preserve">L’on a fait pluſieurs belles expériences ſur la peſanteur de <lb/>l’air. </s> <s xml:id="echoid-s20326" xml:space="preserve">La premiere a été faite ſur une des plus hautes montagnes <lb/>d’Auvergne, proche Clermont, que l’on nomme la montagne <lb/>du Puy de Dome, & </s> <s xml:id="echoid-s20327" xml:space="preserve">a fait voir qu’ayant un tuyau plein de <lb/>mercure, bouché par un bout, & </s> <s xml:id="echoid-s20328" xml:space="preserve">recourbé par l’autre, le mer-<lb/>cure étant à la hauteur de 26 pouces 5 lignes au pied de la mon-<lb/>tagne, que partant delà pour aller au ſommet, à 10 toiſes le <lb/>mercure étoit deſcendu d’une ligne, qu’à 20 toiſes il étoit deſ-<lb/>cendu de 2 lignes, qu’à 100 toiſes il étoit deſcendu de 9 lignes, <lb/>& </s> <s xml:id="echoid-s20329" xml:space="preserve">qu’étant monté de 500 toiſes, il étoit deſcendu de 3 pouces <lb/>10 lignes; </s> <s xml:id="echoid-s20330" xml:space="preserve">& </s> <s xml:id="echoid-s20331" xml:space="preserve">l’on a trouvé qu’en deſcendant, pour venir au <lb/>pied de la montagne, à chaque endroit où le mercure étoit <lb/>deſcendu, il eſt remonté à la même hauteur, & </s> <s xml:id="echoid-s20332" xml:space="preserve">s’eſt retrouvé <lb/>à 26 pouces 5 lignes, au pied de la montagne, à l’endroit d’où <lb/>l’on étoit parti. </s> <s xml:id="echoid-s20333" xml:space="preserve">Il ne faut pas être ſurpris ſi, après avoir dit ail <lb/>leurs que la hauteur du mercure étoit ordinairement de 28 <lb/>pouces pour être en équilibre avec l’air, on ne la trouve <lb/>que de 26 pouces 5 lignes au plus bas lieu de la montagne du <lb/>Puy de Dome, c’eſt que cet endroit-là eſt apparemment plus <lb/>élevé que le bord de la mer, où effectivement le mercure eſt à <lb/>la hauteur de 28 pouces: </s> <s xml:id="echoid-s20334" xml:space="preserve">mais quand le barometre ſe trouve <pb o="651" file="0741" n="767" rhead="DE MATHÉMATIQUE. Liv. XVI."/> <lb/>dans un lieu plus élevé que le bord de la mer, le mercure eſt <lb/>toujours au deſſous de 28 pouces, ſelon que la colonne d’air <lb/>qui y répond, eſt moindre que ſur le bord de la mer.</s> <s xml:id="echoid-s20335" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20336" xml:space="preserve">Ceux qui ne raiſonnent pas ont de la peine à s’imaginer <lb/>que l’air ait de la peſanteur, parce qu’ils n’en ſentent pas le <lb/>poids; </s> <s xml:id="echoid-s20337" xml:space="preserve">mais ſi on leur fait remarquer qu’un animal qui eſt dans <lb/>l’eau a la liberté de ſe mouvoir ſans ſentir le poids de l’eau, à <lb/>cauſe qu’il en eſt preſſé également de toutes parts, ils ne s’é-<lb/>tonneront plus ſi on ne s’apperçoit pas du poids de l’air qui <lb/>nous preſſe auſſi également de toutes parts, & </s> <s xml:id="echoid-s20338" xml:space="preserve">qui eſt en équi-<lb/>libre avec celui que nous avons dans les poulmons & </s> <s xml:id="echoid-s20339" xml:space="preserve">dans le <lb/>ſang, & </s> <s xml:id="echoid-s20340" xml:space="preserve">avec celui qui eſt généralement répandu par tout le <lb/>corps.</s> <s xml:id="echoid-s20341" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20342" xml:space="preserve">Si l’on a cru ſi long tems que l’air étoit léger, c’eſt parce <lb/>que les anciens Auteurs l’ont dit, & </s> <s xml:id="echoid-s20343" xml:space="preserve">que ceux qui font profeſ-<lb/>ſion de les croire, les ſuivoient aveuglément, aux dépens même <lb/>de la vérité & </s> <s xml:id="echoid-s20344" xml:space="preserve">de la raiſon: </s> <s xml:id="echoid-s20345" xml:space="preserve">l’on a même été ſi éloigné de <lb/>penſer que la peſanteur de l’air fût la cauſe de l’élévation de <lb/>l’eau dans les pompes, qu’on a cru qu’il ſuffiſoit de tirer l’air <lb/>avec un piſton pour faire monter l’eau auſſi haut que l’on vou-<lb/>droit, & </s> <s xml:id="echoid-s20346" xml:space="preserve">qu’on pouvoit faire paſſer l’eau d’une riviere par deſ-<lb/>ſus une montagne pour la faire rendre dans le vallon oppoſé, <lb/>pourvu qu’il ſoit un peu plus bas que la riviere, par le moyen <lb/>d’un ſiphon placé ſur la montagne, dont l’une des jambes ré-<lb/>pondroit dans la riviere, puiſque pour cela il ne faudroit que <lb/>pomper l’air du ſiphon, & </s> <s xml:id="echoid-s20347" xml:space="preserve">il n’y a pas plus de 100 ans que l’on <lb/>étoit dans cette erreur.</s> <s xml:id="echoid-s20348" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20349" xml:space="preserve">L’air a encore la propriété de pouvoir être extrêmement <lb/>condenſé & </s> <s xml:id="echoid-s20350" xml:space="preserve">dilaté, & </s> <s xml:id="echoid-s20351" xml:space="preserve">de conſerver toujours une vertu de reſ-<lb/>ſort, par laquelle il fait effort pour repouſſer les corps qui le <lb/>preſſent, juſqu’à ce qu’il ait repris ſon exiſtence naturelle. <lb/></s> <s xml:id="echoid-s20352" xml:space="preserve">L’air ſe dilate auſſi très-facilement par la chaleur, & </s> <s xml:id="echoid-s20353" xml:space="preserve">ſe con-<lb/>denſe par le froid, comme on le remarque dans le thermo-<lb/>metre, où l’on voit que l’air qui eſt dans l’eſprit de vin fait <lb/>monter cette liqueur à vue d’œil dans le tuyau, quand on l’ap-<lb/>proche du feu, ou quand le ſoleil donne deſſus; </s> <s xml:id="echoid-s20354" xml:space="preserve">& </s> <s xml:id="echoid-s20355" xml:space="preserve">au contraire <lb/>on s’apperçoit qu’elle baiſſe beaucoup, quand il fait ſort froid, <lb/>ou quand on met le tuyau dans l’eau froide.</s> <s xml:id="echoid-s20356" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20357" xml:space="preserve">L’air qui eſt proche de la ſurface de la terre, eſt fort con-<lb/>denſé, parce qu’il n’a pas ſon étendue naturelle: </s> <s xml:id="echoid-s20358" xml:space="preserve">car puiſque <pb o="652" file="0742" n="768" rhead="NOUVEAU COURS"/> <lb/>celui qui eſt au deſſus eſt peſant, & </s> <s xml:id="echoid-s20359" xml:space="preserve">qu’il a une vertu de reſſort, <lb/>celui que nous reſpirons étant chargé du poids de tout l’atmoſ-<lb/>phere, eſt plus condenſé que celui qui eſt tout au haut: </s> <s xml:id="echoid-s20360" xml:space="preserve">par <lb/>conſéquent celui qui eſt entre ces deux extrêmités, doit être <lb/>moins condenſé que celui qui touche la terre, & </s> <s xml:id="echoid-s20361" xml:space="preserve">moins di-<lb/>laté que celui qui eſt au haut de l’atmoſphere. </s> <s xml:id="echoid-s20362" xml:space="preserve">Mais pour avoir <lb/>une idée claire de ceci, ſuppoſons un grand amas de laine <lb/>cardée de la hauteur de 80 ou 100 toiſes; </s> <s xml:id="echoid-s20363" xml:space="preserve">il eſt conſtant que la <lb/>laine qui eſt en bas étant chargée de toute la peſanteur de celle <lb/>qu’elle porte, ne ſera pas ſi étendue que celle qui eſt tout au <lb/>haut, & </s> <s xml:id="echoid-s20364" xml:space="preserve">celle qui eſt dans le milieu ne ſera pas ſi comprimée <lb/>que celle qui eſt au deſſous, ni ſi étendue que celle qui eft au <lb/>deſſus. </s> <s xml:id="echoid-s20365" xml:space="preserve">Or ſi l’on prend une poignée de la laine qui eſt en <lb/>bas, & </s> <s xml:id="echoid-s20366" xml:space="preserve">qu’on la porte au deſſus, en la tenant toujours preſ-<lb/>fée de la même façon qu’elle l’étoit dans l’endroit d’où on l’a <lb/>tirée, elle s’élargira d’elle-même, & </s> <s xml:id="echoid-s20367" xml:space="preserve">prendra la même éten-<lb/>due que celle qui eſt tout en haut; </s> <s xml:id="echoid-s20368" xml:space="preserve">& </s> <s xml:id="echoid-s20369" xml:space="preserve">au contraire ſi on prend <lb/>dans la main de celle qui eſt en haut, en lui laiſſant ſon éten-<lb/>due naturelle, ſans la preſſer aucunement, l’on verra que la <lb/>mettant ſous celle qui eſt en bas, elle ſe comprimera de la <lb/>même façon que celle qui eſt en bas. </s> <s xml:id="echoid-s20370" xml:space="preserve">L’on peut dire la même <lb/>choſe de l’air: </s> <s xml:id="echoid-s20371" xml:space="preserve">car ſi l’on prend une veſſie bien ſéche, ſoufflée <lb/>à la moitié de la groſſeur qu’elle devroit avoir, ſi on l’avoit <lb/>bien remplie d’air, ſi après l’avoir bien fermée, on la porte <lb/>au haut d’une montagne fort élevée, l’on verra qu’à meſure <lb/>que l’on montera, la veſſie deviendra plus enflée qu’elle n’é-<lb/>toit auparavant, & </s> <s xml:id="echoid-s20372" xml:space="preserve">lorſqu’on ſera parvenu au ſommet, on la <lb/>verra ronde & </s> <s xml:id="echoid-s20373" xml:space="preserve">toute auſſi enflée qu’elle eût été au pied de la <lb/>montagne, ſi on l’avoit ſoufflée autant qu’on fait ordinaire-<lb/>ment pour la rendre ſphérique. </s> <s xml:id="echoid-s20374" xml:space="preserve">Cependant il eſt à remarquer <lb/>que l’air qui eſt dans la veſſie eſt toujours le même qu’il étoit <lb/>au pied de la montagne, n’étant point augmenté ni diminué; <lb/></s> <s xml:id="echoid-s20375" xml:space="preserve">tout le changement qui lui eſt arrivé, c’eſt de s’être dilaté <lb/>conſidérablement, c’eſt-à-dire qu’il occupe un bien plus grand <lb/>eſpace qu’auparavant; </s> <s xml:id="echoid-s20376" xml:space="preserve">& </s> <s xml:id="echoid-s20377" xml:space="preserve">il eſt à préſumer que ſi on avoit porté <lb/>cette veſſie au haut d’une montagne beaucoup plus élevée que <lb/>celle que je ſuppoſe ici, l’air ſe ſeroit dilaté juſqu’au point de <lb/>crever la veſſie par la force de ſon reſſort. </s> <s xml:id="echoid-s20378" xml:space="preserve">La raiſon de cette <lb/>dilatation vient ſans doute de ce que l’air qu’on a mis dans la <lb/>veſſie au pied de la montagne, étant preſſé par le poids de l’air <pb o="653" file="0743" n="769" rhead="DE MATHÉMATIQUE. Liv. XVI."/> <lb/>extérieur, celui de dedans n’a pas plus de liberté de prendre <lb/>ſon étendue naturelle que celui de dehors, puiſqu’ils ſont éga-<lb/>lement chargés du poids de l’atmoſphere; </s> <s xml:id="echoid-s20379" xml:space="preserve">mais quand la veſſie <lb/>ſe trouve au haut de la montagne, l’air qui eſt à cette hauteur <lb/>n’étant point ſi chargé que celui d’en bas, ne preſſe pas tant <lb/>les corps qu’il environne; </s> <s xml:id="echoid-s20380" xml:space="preserve">ce qui fait que celui qui eſt dans la <lb/>veſſie ne trouvant pas une ſi grande réſiſtance pour s’étendre <lb/>qu’auparavant, ſe dilate & </s> <s xml:id="echoid-s20381" xml:space="preserve">occupe un bien plus grand eſpace <lb/>que celui où il étoit renfermé dans le lieu d’où on l’a ſorti.</s> <s xml:id="echoid-s20382" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20383" xml:space="preserve">Il arrive tout le contraire, ſi on remplit, autant qu’il eſt <lb/>poſſible, une veſſie au ſommet d’une haute montagne: </s> <s xml:id="echoid-s20384" xml:space="preserve">car ſi <lb/>l’on deſcend pour venir dans un lieu beaucoup plus bas, l’on <lb/>voit que la veſſie de bien tendue qu’elle étoit auparavant, de-<lb/>vient flaſque & </s> <s xml:id="echoid-s20385" xml:space="preserve">molle à meſure que l’on deſcend, tant qu’il <lb/>ne paroît preſque pas qu’elle ait été enflée; </s> <s xml:id="echoid-s20386" xml:space="preserve">ce qui ne peut man-<lb/>quer d’arriver par les raiſons que nous venons de dire: </s> <s xml:id="echoid-s20387" xml:space="preserve">car <lb/>l’air qui eſt dans la veſſie ſe trouvant comprimé de tous côtés <lb/>par celui qui l’environne, qui eſt beaucoup plus peſant que <lb/>ſur la montagne, il eſt forcé de ſe ramaſſer, c’eſt-à-dire de <lb/>ſe condenſer pour occuper un plus petit eſpace que celui qu’il <lb/>tenoit dans l’endroit d’où on l’a tiré.</s> <s xml:id="echoid-s20388" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20389" xml:space="preserve">C’eſt ſans doute à la dilatation & </s> <s xml:id="echoid-s20390" xml:space="preserve">à la condenſation que l’air <lb/>prend, quand il eſt porté dans un lieu plus élevé ou plus bas <lb/>que celui d’où il eſt ſorti, qu’on doit attribuer l’incommo-<lb/>dité que reſſentent ceux que le beſoin conduit ſur des hautes <lb/>montagnes: </s> <s xml:id="echoid-s20391" xml:space="preserve">car comme ils ont dans les poulmons & </s> <s xml:id="echoid-s20392" xml:space="preserve">dans le <lb/>ſang un air plus condenſé que celui de l’endroit où ils ſe trou-<lb/>vent, les chairs n’étant plus preſſées ſi fortement par l’air que <lb/>de coutume, laiſſent à celui qui eſt dans le corps la liberté de <lb/>ſe dilater; </s> <s xml:id="echoid-s20393" xml:space="preserve">ce qui ne peut ſe faire ſans déranger le tempéra-<lb/>ment de ceux à qui cela arrive. </s> <s xml:id="echoid-s20394" xml:space="preserve">L’on pourra expliquer par un <lb/>raiſonnement tout contraire à celui-ci la peine que reſſentent <lb/>ceux qui d’un lieu haut viennent habiter un lieu bas.</s> <s xml:id="echoid-s20395" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20396" xml:space="preserve">La raréfraction de l’air eſt très-conſidérable par les conſé-<lb/>quences que l’on a tirées de pluſieurs expériences; </s> <s xml:id="echoid-s20397" xml:space="preserve">& </s> <s xml:id="echoid-s20398" xml:space="preserve">M. <lb/></s> <s xml:id="echoid-s20399" xml:space="preserve">Mariotte, qui en a fait plus que perſonne, fait voir qu’un cer-<lb/>tain volume d’air, que nous reſpirons, peut ſe raréfier de 4000 <lb/>fois pour être dans ſon étendue naturelle, c’eſt-à-dire que s’il <lb/>étoit poſſible de porter un pied cube d’air de deſſus la ſurface <lb/>de la terre au haut de l’atmoſphere, il occuperoit un eſpace <pb o="654" file="0744" n="770" rhead="NOUVEAU COURS"/> <lb/>de 4000 pieds cubes, & </s> <s xml:id="echoid-s20400" xml:space="preserve">peut-être même d’une bien plus grande <lb/>étendue. </s> <s xml:id="echoid-s20401" xml:space="preserve">Si cette eſtimation approche de la vérité, il en ſera <lb/>la même choſe de la raréfraction de l’air naturel, c’eſt-à-dire <lb/>de l’air qui eſt au haut de l’atmoſphere, ſur la ſurface de la terre, <lb/>que lorſqu’il ſera comprimé par l’air du dehors; </s> <s xml:id="echoid-s20402" xml:space="preserve">il occupera un <lb/>volume quatre mille fois plus petit, pour devenir ſemblable à <lb/>celui que nous reſpirons: </s> <s xml:id="echoid-s20403" xml:space="preserve">mais comme l’expérience fait voir <lb/>que celui-ci peut être extrêmement condenſé, celui du haut <lb/>de l’atmoſphere qui ſe ſeroit condenſé de quatre mille fois, <lb/>pour devenir pareil au nôtre, peut donc l’être bien davantage <lb/>de quatre mille fois, pour devenir auſſi ſerré que le nôtre peut <lb/>être réduit.</s> <s xml:id="echoid-s20404" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20405" xml:space="preserve">Nous avons fait voir que quand on portoit un barometre du <lb/>pied d’une montagne au ſommet, à meſure que l’on mon-<lb/>toit, le mercure baiſſoit pour ſe mettre en équilibre avec la <lb/>colonne d’air, qui devient d’autant moindre, que la montagne <lb/>eſt plus élevée; </s> <s xml:id="echoid-s20406" xml:space="preserve">& </s> <s xml:id="echoid-s20407" xml:space="preserve">en parlant de l’expérience qui a été faite <lb/>ſur le Puy de Dome, nous avons dit qu’étant monté de 10 <lb/>toiſes, le mercure étoit deſcendu d’une ligne; </s> <s xml:id="echoid-s20408" xml:space="preserve">qu’étant monté <lb/>de 20 toiſes, il étoit deſcendu de 2 lignes; </s> <s xml:id="echoid-s20409" xml:space="preserve">qu’étant monté de <lb/>100 toiſes, il étoit deſcendu de 9 lignes; </s> <s xml:id="echoid-s20410" xml:space="preserve">enfin qu’étant monté <lb/>de 500 toiſes, il étoit deſcendu de 8 pouces 10 lignes, ou au-<lb/>trement de 46 lignes, où l’on peut remarquer que la diminu-<lb/>tion du mercure n’eſt pas dans la raiſon des différentes hau-<lb/>teurs où le barometre a été porté ſur la montagne: </s> <s xml:id="echoid-s20411" xml:space="preserve">car pour <lb/>que cela fût ainſi, il faudroit qu’à 100 toiſes le mercure fût <lb/>deſcendu de 10 lignes, & </s> <s xml:id="echoid-s20412" xml:space="preserve">qu’à 500 toiſes il fût deſcendu de <lb/>50 lignes: </s> <s xml:id="echoid-s20413" xml:space="preserve">pour lors l’on auroit deux progreſſions arithméti-<lb/>ques, l’une pour le barometre, & </s> <s xml:id="echoid-s20414" xml:space="preserve">l’autre pour les différentes <lb/>hauteurs ſur leſquelles il ſeroit porté; </s> <s xml:id="echoid-s20415" xml:space="preserve">les termes de la pre-<lb/>miere progreſſion ſe ſurpaſſeroient d’une unité, & </s> <s xml:id="echoid-s20416" xml:space="preserve">les termes <lb/>de la ſeconde ſe ſurpaſſeroient de 10 toiſes; </s> <s xml:id="echoid-s20417" xml:space="preserve">ce qui ſeroit fort <lb/>commode pour meſurer la hauteur des montagnes & </s> <s xml:id="echoid-s20418" xml:space="preserve">celle de <lb/>l’atmoſphere, puiſque le mercure deſcendant d’une ligne de <lb/>10 toiſes en 10 toiſes, l’on n’auroit qu’à obſerver de combien <lb/>de lignes il ſeroit deſcendu en allant du pied de la montagne <lb/>au ſommet; </s> <s xml:id="echoid-s20419" xml:space="preserve">enſuite multiplier cette quantité de lignes par <lb/>10 toiſes, & </s> <s xml:id="echoid-s20420" xml:space="preserve">le produit donneroit la hauteur de la montagne <lb/>au deſſus du vallon qui ſeroit au pied: </s> <s xml:id="echoid-s20421" xml:space="preserve">de même pour ſçavoir <lb/>la hauteur de l’atmoſphere, il n’y auroit qu’à multiplier 356 <pb o="655" file="0745" n="771" rhead="DE MATHÉMATIQUE. Liv. XVI."/> <lb/>lignes, qui eſt la hauteur du mercure ſur le bord de la mer, <lb/>par 10 toiſes, l’on auroit 3360 toiſes pour la hauteur de l’at-<lb/>moſphere: </s> <s xml:id="echoid-s20422" xml:space="preserve">mais comme la peſanteur de l’air ne ſuit point une <lb/>ſemblable progreſſion, & </s> <s xml:id="echoid-s20423" xml:space="preserve">qu’elle en ſuit une autre toute diffé-<lb/>rente, voici ce que MM. </s> <s xml:id="echoid-s20424" xml:space="preserve">Caſſini & </s> <s xml:id="echoid-s20425" xml:space="preserve">Maraldi ont fait pour la <lb/>trouver, que j’ai tiré des Mémoires de l’Académie Royale des <lb/>Sciences de l’année 1703.</s> <s xml:id="echoid-s20426" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20427" xml:space="preserve">Ils prirent d’abord géométriquement la hauteur des monta-<lb/>gnes qui ſe trouverent ſur le chemin de la Méridienne; </s> <s xml:id="echoid-s20428" xml:space="preserve">& </s> <s xml:id="echoid-s20429" xml:space="preserve"><lb/>quand ils purent ſe tranſporter juſqu’au haut, ils obſerverent <lb/>quelle étoit la deſcente du barometre. </s> <s xml:id="echoid-s20430" xml:space="preserve">Ils avoient fait le même <lb/>jour, lorſqu’il avoit été poſſible, une obſervation du baro-<lb/>metre ſur le bord de la mer, ou dans un lieu dont ils con-<lb/>noiſſoient l’élévation ſur le niveau de la mer, où en tout cas <lb/>ils ne pouvoient manquer de trouver à leur retour des obſer-<lb/>vations perpétuelles du barometre qu’on fait à l’Obſervatoire, <lb/>que l’on ſçait être plus haut que la mer de 46 toiſes.</s> <s xml:id="echoid-s20431" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20432" xml:space="preserve">Par les comparaiſons des différentes hauteurs des monta-<lb/>gnes, avec les différentes deſcentes du mercure ſur ces mon-<lb/>tagnes, ces Meſſieurs jugerent que la progreſſion, ſuivant la-<lb/>quelle les colonnes d’air qui répondoient à une ligne de mer-<lb/>cure, qui vont en augmentant des hauteurs, quand on deſ-<lb/>cend de la montagne, pouvoient être telles que la premiere <lb/>colonne ayant 61 pieds, la ſeconde en eût 62, la troiſieme <lb/>63, & </s> <s xml:id="echoid-s20433" xml:space="preserve">ainſi toujours de ſuite, du moins juſqu’à la hauteur <lb/>d’une demi-lieue; </s> <s xml:id="echoid-s20434" xml:space="preserve">car ils n’avoient pas obſervé ſur des monta-<lb/>gnes plus élevées.</s> <s xml:id="echoid-s20435" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20436" xml:space="preserve">En obſervant cette progreſſion, ils retrouverent toujours, <lb/>à quelques toiſes près, par la deſcente du mercure ſur une <lb/>montagne, la même hauteur de cette montagne qu’ils avoient <lb/>eue immédiatement après l’opération géométrique.</s> <s xml:id="echoid-s20437" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20438" xml:space="preserve">On peut donc, en admettant cette progreſſion, meſurer par <lb/>un barometre, qu’on portera ſur une montagne, combien <lb/>elle ſera élevée ſur le niveau de la mer, pourvu qu’on puiſſe <lb/>ſçavoir à quelle hauteur étoit à peu près en même tems le ba-<lb/>rometre ſur le bord de la mer, ou dans un lieu dont l’éléva-<lb/>tion au deſſus de la mer ſoit connu; </s> <s xml:id="echoid-s20439" xml:space="preserve">& </s> <s xml:id="echoid-s20440" xml:space="preserve">cette méthode réuſſira <lb/>le plus ſouvent, quand même la montagne ſeroit fort éloi-<lb/>gnée de la mer; </s> <s xml:id="echoid-s20441" xml:space="preserve">que ſi cette progreſſion régnoit dans tout l’at-<lb/>moſphere, il ſeroit bien facile d’en trouver la hauteur: </s> <s xml:id="echoid-s20442" xml:space="preserve">car les <pb o="656" file="0746" n="772" rhead="NOUVEAU COURS"/> 28 pouces de mercure étant la même choſe que 336 lignes, on <lb/>auroit une progreſſion arithmétique de 336 termes, dont la <lb/>différence ſeroit l’unité, & </s> <s xml:id="echoid-s20443" xml:space="preserve">le premier terme de 61: </s> <s xml:id="echoid-s20444" xml:space="preserve">mais <lb/>comme l’on n’eſt pas ſûr que la peſanteur de l’air ſuive une ſem-<lb/>blable progreſſion, le principe paroît trop incertain pour qu’on <lb/>puiſſe en rien conclure pour la hauteur de l’atmoſphere, qui <lb/>ne ſe trouveroit que de ſix lieues & </s> <s xml:id="echoid-s20445" xml:space="preserve">demie, ſelon cette pro-<lb/>greſſion, au lieu que M. </s> <s xml:id="echoid-s20446" xml:space="preserve">Mariotte a fait voir par une nou-<lb/>velle maniere de calculer la hauteur de l’atmoſphere, qu’elle <lb/>avoit environ 25 lieues, qui eſt la hauteur que tous les Phy-<lb/>ſiciens lui donnent préſentement: </s> <s xml:id="echoid-s20447" xml:space="preserve">mais la progreſſion précé-<lb/>dente peut être fort utile pour meſurer la hauteur d’une mon-<lb/>tagne qui ne paſſe point 1200 toiſes.</s> <s xml:id="echoid-s20448" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1697" type="section" level="1" n="1231"> <head xml:id="echoid-head1451" style="it" xml:space="preserve">Fin du Cours de Mathématique.</head> <pb file="0747" n="773"/> <pb file="0747a" n="774"/> <figure> <image file="0747a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0747a-01"/> </figure> <pb file="0748" n="775"/> <pb file="0749" n="776"/> <pb file="0749a" n="777"/> <figure> <image file="0749a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0749a-01"/> </figure> <pb file="0750" n="778"/> <pb file="0751" n="779"/> <pb file="0751a" n="780"/> <figure> <image file="0751a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0751a-01"/> </figure> <pb file="0752" n="781"/> <pb file="0753" n="782"/> <pb file="0753a" n="783"/> <figure> <image file="0753a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0753a-01"/> </figure> <pb file="0754" n="784"/> <pb file="0755" n="785"/> <pb file="0755a" n="786"/> <figure> <image file="0755a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0755a-01"/> </figure> <pb file="0756" n="787"/> <pb file="0757" n="788"/> <pb file="0757a" n="789"/> <figure> <image file="0757a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0757a-01"/> </figure> <pb file="0758" n="790"/> <pb file="0759" n="791"/> <pb file="0759a" n="792"/> <figure> <image file="0759a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0759a-01"/> </figure> <pb file="0760" n="793"/> <pb file="0761" n="794"/> <pb file="0761a" n="795"/> <figure> <image file="0761a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0761a-01"/> </figure> <pb file="0762" n="796"/> <pb file="0763" n="797"/> </div> <div xml:id="echoid-div1698" type="section" level="1" n="1232"> <head xml:id="echoid-head1452" style="it" xml:space="preserve">EXTR AIT des Regiſtres de l’Académie Royale des Sciences, <lb/>du 27 Janvier 1725.</head> <p> <s xml:id="echoid-s20449" xml:space="preserve"><emph style="sc">Le</emph>s Révérends Peres Sébaſtien & </s> <s xml:id="echoid-s20450" xml:space="preserve">Reneau, & </s> <s xml:id="echoid-s20451" xml:space="preserve">Meſſieurs Saurin, de Mairan <lb/>& </s> <s xml:id="echoid-s20452" xml:space="preserve">Chevalier qui avoient été nommés pour examiner un Ouvrage, préſenté <lb/>par M. </s> <s xml:id="echoid-s20453" xml:space="preserve">Belidor, Profeſſeur Royal de Mathématiques aux Ecoles d’Artillerie <lb/>de la Fere, & </s> <s xml:id="echoid-s20454" xml:space="preserve">intitulé Nouveau Cours de Mathématiques à l’uſage de l’ Ar-<lb/>tillerie & </s> <s xml:id="echoid-s20455" xml:space="preserve">du Génie, en ayant fait leur rapport, la Compagnie a jugé que <lb/>puiſque l’Auteur avoit recueilli avec choix & </s> <s xml:id="echoid-s20456" xml:space="preserve">avec ordre des diverſes par-<lb/>ties des Mathématiques les principales connoiſſances qui pouvoient ap-<lb/>partenir au Génie & </s> <s xml:id="echoid-s20457" xml:space="preserve">au ſervice de l’Artillerie, qu’il avoit rendu toutes ſes <lb/>démonſtrations plus nettes & </s> <s xml:id="echoid-s20458" xml:space="preserve">plus courtes, en y employant l’Algebre, dont <lb/>il donne les premiers élémens, & </s> <s xml:id="echoid-s20459" xml:space="preserve">qu’il faiſoit voir l’uſage des connoiſſances <lb/>qu’il donnoit, en les appliquant à des exemples conſidérables, tirés du Génie <lb/>même & </s> <s xml:id="echoid-s20460" xml:space="preserve">de l’Artillerie, il avoit bien rempli les vues qu’il s’étoit propoſées, <lb/>& </s> <s xml:id="echoid-s20461" xml:space="preserve">qu’on ne pouvoit trop louer ſon zele pour le progrés de l’Ecole à laquelle <lb/>il a voué ſes ſoins & </s> <s xml:id="echoid-s20462" xml:space="preserve">ſes travaux: </s> <s xml:id="echoid-s20463" xml:space="preserve">en foi de quoi j’ai ſigné le préſent Cer-<lb/>tificat. </s> <s xml:id="echoid-s20464" xml:space="preserve">A Paris ce 29 Janvier 1725.</s> <s xml:id="echoid-s20465" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20466" xml:space="preserve">FONTENELLE, Secr. </s> <s xml:id="echoid-s20467" xml:space="preserve">Pr. </s> <s xml:id="echoid-s20468" xml:space="preserve">de l’Ac. </s> <s xml:id="echoid-s20469" xml:space="preserve">Royale des Sciences.</s> <s xml:id="echoid-s20470" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1699" type="section" level="1" n="1233"> <head xml:id="echoid-head1453" style="it" xml:space="preserve">APPROBATION DU CENSEUR ROYAL.</head> <p> <s xml:id="echoid-s20471" xml:space="preserve">J’<emph style="sc">AI</emph> lu, par ordre de Monſeigneur le Chancelier, la nouvelle édition <lb/>du Cours de Mathématique de M. </s> <s xml:id="echoid-s20472" xml:space="preserve"><emph style="sc">Belidor</emph>. </s> <s xml:id="echoid-s20473" xml:space="preserve">Cet Ouvrage a été, dès le <lb/>commencement, bien reçu du Public; </s> <s xml:id="echoid-s20474" xml:space="preserve">il a été enſeigné avec ſuccès dans <lb/>les Ecoles d’Artillerie. </s> <s xml:id="echoid-s20475" xml:space="preserve">Les nouvelles augmentations dont on l’a enri-<lb/>chi, rendent cette édition très-complette, & </s> <s xml:id="echoid-s20476" xml:space="preserve">beaucoup ſupérieure aux <lb/>anciennes. </s> <s xml:id="echoid-s20477" xml:space="preserve">Fait à Paris, ce 5 Juin 1757.</s> <s xml:id="echoid-s20478" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20479" xml:space="preserve">MONTCARVILLE, Lecteur & </s> <s xml:id="echoid-s20480" xml:space="preserve">Profeſſeur Royal.</s> <s xml:id="echoid-s20481" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div1700" type="section" level="1" n="1234"> <head xml:id="echoid-head1454" style="it" xml:space="preserve">PRIVILEGE DU ROI.</head> <p> <s xml:id="echoid-s20482" xml:space="preserve"><emph style="sc">Louis</emph>, par la grace de Dieu, Roi de France & </s> <s xml:id="echoid-s20483" xml:space="preserve">de Navarre: </s> <s xml:id="echoid-s20484" xml:space="preserve">à nos <lb/>amés & </s> <s xml:id="echoid-s20485" xml:space="preserve">féaux Conſeillers, les Gens tenant nos Cours de Parlement, Maîtres <lb/>des Requêtes ordinaires de notre Hôtel, Grand Conſeil, Prevôt de Paris; <lb/></s> <s xml:id="echoid-s20486" xml:space="preserve">Baillifs, Sénéchaux, leurs Lieutenans Civils, & </s> <s xml:id="echoid-s20487" xml:space="preserve">autres nos Juſti-<lb/>ciers qu’il appartiendra: </s> <s xml:id="echoid-s20488" xml:space="preserve"><emph style="sc">Salut</emph>. </s> <s xml:id="echoid-s20489" xml:space="preserve">Notre amé <emph style="sc">Charles</emph>-<emph style="sc">Antoine</emph> <lb/><emph style="sc">Jombert</emph>, Notre Libraire à Paris, Nous a fait expoſer qu’il déſireroit <lb/>faire imprimer & </s> <s xml:id="echoid-s20490" xml:space="preserve">réimprimer les <emph style="sc">Œuvres de</emph> M. </s> <s xml:id="echoid-s20491" xml:space="preserve"><emph style="sc">Belidor</emph>; </s> <s xml:id="echoid-s20492" xml:space="preserve"><lb/>ſçavoir, le Cours de Mathématique; </s> <s xml:id="echoid-s20493" xml:space="preserve">la Science des Ingénieurs; </s> <s xml:id="echoid-s20494" xml:space="preserve">le Bombar-<lb/>dier François, & </s> <s xml:id="echoid-s20495" xml:space="preserve">l’Architecture Hydraulique, s’il Nous plaiſoit lui ac-<lb/>corder nos Lettres de privilege pour ce néceſſaires. </s> <s xml:id="echoid-s20496" xml:space="preserve">A <emph style="sc">C E S C A U S E S</emph>, <lb/>voulant favorablement traiter l’Expoſant, nous lui avons permis & </s> <s xml:id="echoid-s20497" xml:space="preserve"><lb/>permettons par ces Préſentes, de faire imprimer & </s> <s xml:id="echoid-s20498" xml:space="preserve">réimprimer leſdits <lb/>Ouvrages, autant de fois que bon lui ſemblera, & </s> <s xml:id="echoid-s20499" xml:space="preserve">de les vendre, faire <pb file="0764" n="798"/> vendre & </s> <s xml:id="echoid-s20500" xml:space="preserve">débiter par tout notre Royaume, pendant le tems de dix an-<lb/>nées conſécutives, à compter du jour de la date des Préſentes. </s> <s xml:id="echoid-s20501" xml:space="preserve">Faiſons <lb/>défenſes à tous Imprimeurs, Libraires, & </s> <s xml:id="echoid-s20502" xml:space="preserve">autres perſonnes, de quelque <lb/>qualité & </s> <s xml:id="echoid-s20503" xml:space="preserve">condition qu’elles ſoient, d’en introduire d’impreſſion étran-<lb/>gere dans aucun lieu de notre obéiſſance: </s> <s xml:id="echoid-s20504" xml:space="preserve">comme auſſi d’imprimer ou faire <lb/>imprimer, vendre, faire vendre, débiter ni contrefaire leſdits Ouvrages, <lb/>ni d’en faire aucun extrait, ſous quelque prétexte que ce ſoit d’augment-<lb/>tation, correction, changemens ou autres, ſans la permiſſion expreſſe & </s> <s xml:id="echoid-s20505" xml:space="preserve"><lb/>par écrit dudit Expoſant, ou de ceux qui auront droit de lui, à peine de <lb/>confiſcation des exemplaires contrefaits, de trois mille livres d’amende <lb/>contre chacun des contrevenans, dont un tiers à Nous, un tiers à l’Hôtel-<lb/>Dieu de Paris, & </s> <s xml:id="echoid-s20506" xml:space="preserve">l’autre tiers audit Expoſant, ou à celui qui aura droit <lb/>de lui, & </s> <s xml:id="echoid-s20507" xml:space="preserve">de tous dépens, dommages & </s> <s xml:id="echoid-s20508" xml:space="preserve">intérêts: </s> <s xml:id="echoid-s20509" xml:space="preserve">à la charge que ces Pré-<lb/>ſentes ſeront enrégiſtrées tout au long ſur le Regiſtre de la Communauté <lb/>des Imprimeurs & </s> <s xml:id="echoid-s20510" xml:space="preserve">Libraires de Paris, dans trois mois de la date d’i-<lb/>celles; </s> <s xml:id="echoid-s20511" xml:space="preserve">que l’impreſſion & </s> <s xml:id="echoid-s20512" xml:space="preserve">réimpreſſion deſdits Ouvrages ſera faite dans <lb/>notre Royaume, & </s> <s xml:id="echoid-s20513" xml:space="preserve">non ailleurs, en bon papier & </s> <s xml:id="echoid-s20514" xml:space="preserve">beaux caracteres, con-<lb/>formément à la feuille imprimée, attachée pour modele ſous le contre-<lb/>ſcel des Préſentes: </s> <s xml:id="echoid-s20515" xml:space="preserve">que l’impétrant ſe conformera en tout aux Réglemens de <lb/>la Librairie, & </s> <s xml:id="echoid-s20516" xml:space="preserve">notamment à celui du 10 Avril 1725; </s> <s xml:id="echoid-s20517" xml:space="preserve">& </s> <s xml:id="echoid-s20518" xml:space="preserve">qu’avant de <lb/>l’expoſer en vente, les manuſcrits & </s> <s xml:id="echoid-s20519" xml:space="preserve">imprimés qui auront ſervi de copie <lb/>à l’impreſſion & </s> <s xml:id="echoid-s20520" xml:space="preserve">réimpreſſion deſdits Ouvrages, ſeront remis dans le même <lb/>état où l’approbation y aura été donnée, ès mains de notre très - cher & </s> <s xml:id="echoid-s20521" xml:space="preserve"><lb/>féal Chevalier, Chancelier de France, le Sieur <emph style="sc">DE</emph> <emph style="sc">Lamoignon</emph>, & </s> <s xml:id="echoid-s20522" xml:space="preserve">qu’il en <lb/>ſera enſuite remis deux exemplaires de chacun dans notre Bibliotheque <lb/>publique, un dans celle de notre Château du Louvre, & </s> <s xml:id="echoid-s20523" xml:space="preserve">un dans celle de <lb/>notre très - cher & </s> <s xml:id="echoid-s20524" xml:space="preserve">féal Chevalier Chancelier de France le Sieur <emph style="sc">DE</emph> <emph style="sc">La</emph>-<lb/><emph style="sc">MOIGNON</emph>, & </s> <s xml:id="echoid-s20525" xml:space="preserve">un dans celle de notre très-cher & </s> <s xml:id="echoid-s20526" xml:space="preserve">féal Chevalier Garde des <lb/>Sceaux de France le Sieur de <emph style="sc">Machault</emph>, Commandeur de nos Ordres; <lb/></s> <s xml:id="echoid-s20527" xml:space="preserve">le tout à peine de nullité des Préſentes: </s> <s xml:id="echoid-s20528" xml:space="preserve">du contenu deſquelles vous man-<lb/>dons & </s> <s xml:id="echoid-s20529" xml:space="preserve">enjoignons de faire jouir ledit Expoſant, ou ſes ayans cauſe, plei-<lb/>nement & </s> <s xml:id="echoid-s20530" xml:space="preserve">paiſiblement, ſans ſouffrir qu’il leur ſoit fait aucun trouble <lb/>ou empêchement. </s> <s xml:id="echoid-s20531" xml:space="preserve">Voulons que la copie des Préſentes, qui ſera imprimée <lb/>tout au long au commencement ou à la fin deſdits Ouvrages, ſoit tenue <lb/>pour duement ſignifiée, & </s> <s xml:id="echoid-s20532" xml:space="preserve">qu’aux copies collationnées par l’un de nos <lb/>amés & </s> <s xml:id="echoid-s20533" xml:space="preserve">féaux Conſeillers Secretaires, foi ſoit ajoutée comme à l’Original. </s> <s xml:id="echoid-s20534" xml:space="preserve"><lb/>Commandons au premier notre Huiſſier ou Sergent ſur ce requis, de faire <lb/>pour l’exécution d’icelles tous actes requis & </s> <s xml:id="echoid-s20535" xml:space="preserve">néceſſaires, ſans demander <lb/>autre permiſſion, & </s> <s xml:id="echoid-s20536" xml:space="preserve">nonobſtant clameur de haro, Charte Normande, & </s> <s xml:id="echoid-s20537" xml:space="preserve"><lb/>Lettres à ce contraires; </s> <s xml:id="echoid-s20538" xml:space="preserve">car tel eſt notre plaiſir. </s> <s xml:id="echoid-s20539" xml:space="preserve"><emph style="sc">Donné</emph> à Verſailles le <lb/>vingt-unieme jour du mois d’Août, l’an de grace mil ſept cent cinquante-<lb/>deux, & </s> <s xml:id="echoid-s20540" xml:space="preserve">de notre regne le trente-ſeptieme. </s> <s xml:id="echoid-s20541" xml:space="preserve">Par le Roi en ſon Conſeil.</s> <s xml:id="echoid-s20542" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20543" xml:space="preserve">SAINSON.</s> <s xml:id="echoid-s20544" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s20545" xml:space="preserve">Regiſtrè ſur le Regiſtre XIII. </s> <s xml:id="echoid-s20546" xml:space="preserve">de la Chambre Royale des Libraires & </s> <s xml:id="echoid-s20547" xml:space="preserve"><lb/>Imprimeurs de Paris, N°. </s> <s xml:id="echoid-s20548" xml:space="preserve">19, fol. </s> <s xml:id="echoid-s20549" xml:space="preserve">12, conformément aux anciens Régle-<lb/>mens, confirmés par celui du yingt-huit Féyrier 1723. </s> <s xml:id="echoid-s20550" xml:space="preserve">A Paris le 29 Août <lb/>mil ſept cent cinquante-deux.</s> <s xml:id="echoid-s20551" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20552" xml:space="preserve">HERISANT, Adjoint.</s> <s xml:id="echoid-s20553" xml:space="preserve"/> </p> <pb file="0765" n="799"/> <pb file="0766" n="800"/> <pb file="0767" n="801"/> <pb file="0768" n="802"/> <pb file="0769" n="803"/> <pb file="0770" n="804"/> </div></text> </echo>