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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:1YZKBTHR.xml</dcterms:identifier> <dcterms:creator identifier="GND:117647993">Casati, Paolo</dcterms:creator> <dcterms:title xml:lang="it">Fabrica, et uso del compasso di proportione, dove insegna à gli artefici il modo di fare in esso le necessarie divisioni, e con varij problemi ...</dcterms:title> <dcterms:date xsi:type="dcterms:W3CDTF">1685</dcterms:date> <dcterms:language xsi:type="dcterms:ISO639-3">ita</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> <log>changed "potrà ſimilmentə diuidere" to "potrà ſimilmente diuidere" (approx. line 328)</log> </metadata> <text xml:lang="it" type="free"> <div xml:id="echoid-div1" type="section" level="1" n="1"><pb file="0001" n="1"/> <pb file="0002" n="2"/> <handwritten/> <pb file="0003" n="3"/> </div> <div xml:id="echoid-div2" type="section" level="1" n="2"> <head xml:id="echoid-head1" xml:space="preserve">FABRICA ET VSO <lb/>Del Compaſſo di Proportione, <lb/>Doue inſegna à gli ARTEFICI il modo di fare in eſſo <lb/>le neceſſarie diuiſioni, <lb/>E con varij Problemi vſuali moſtra l’vtilità <lb/>di queſto Stromento, <lb/>PAOLO CASATI <lb/>DELLA COMPAGNIA DI GIESV', <lb/>Dando le ragioni, & apportando le dimoſtrationi di tutte le <lb/>operationi nella Fabrica, e nell Vſo.</head> <head xml:id="echoid-head2" xml:space="preserve">OPERA VTILE</head> <p> <s xml:id="echoid-s1" xml:space="preserve">Non ſolo à Geometri, Agrimenſori, Architetti ciuih<unsure/>, e militari, Pittori, Scoltori, <lb/>& </s> <s xml:id="echoid-s2" xml:space="preserve">à tutti quelli, che vſano del Diſſegno, mà anche à Bombardieri, <lb/>Sergenti di Battaglia, Mercanti, & </s> <s xml:id="echoid-s3" xml:space="preserve">altri, per molte operationi <lb/>Aritmetiche, fatte con grandiſſima facilità,<unsure/> <lb/>Accreſciuta notabilmente in queſta ſeconda Editione dal medeſimo Autore.</s> <s xml:id="echoid-s4" xml:space="preserve"><unsure/></s> </p> <figure> <image file="0003-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0003-01"/> </figure> </div> <div xml:id="echoid-div3" type="section" level="1" n="3"> <head xml:id="echoid-head3" xml:space="preserve">IN BOLOGNA, Per Gioſeffo Longhi 1685. Conlic.<unsure/> de’Superiori.</head> <pb file="0004" n="4"/> <handwritten/> <pb file="0005" n="5"/> </div> <div xml:id="echoid-div4" type="section" level="1" n="4"> <head xml:id="echoid-head4" xml:space="preserve">Franciſcus Bellhomus Societatis Ieſu in Pro-<lb/>uincia Veneta Præpoſitus Prouincialis.</head> <p style="it"> <s xml:id="echoid-s5" xml:space="preserve">OPuſculum, cui titulus eſt, Fabrica, & </s> <s xml:id="echoid-s6" xml:space="preserve">Vſo <lb/>del Compaſſo di Proportione, &</s> <s xml:id="echoid-s7" xml:space="preserve">c. </s> <s xml:id="echoid-s8" xml:space="preserve">à P. <lb/></s> <s xml:id="echoid-s9" xml:space="preserve">Paulo Caſato Societat<unsure/>is noſtræ compoſitum, tres viri <lb/>graues, ac docti eiuſdem noſtræ Societatis perlegerunt, <lb/>& </s> <s xml:id="echoid-s10" xml:space="preserve">in lucem edi poſſe iudicarunt. </s> <s xml:id="echoid-s11" xml:space="preserve">Quare facultate mi-<lb/>hiconceſſa ab Adm. </s> <s xml:id="echoid-s12" xml:space="preserve">Reuer. </s> <s xml:id="echoid-s13" xml:space="preserve">P. </s> <s xml:id="echoid-s14" xml:space="preserve">Ioanne Paulo Oliva Vi-<lb/>cario Generali poteſtatem facio, vt imprimatur, ſi <lb/>alijs, ad quos ſpectat, it a viſumfuerit. </s> <s xml:id="echoid-s15" xml:space="preserve">Bononiæ die <lb/>26. </s> <s xml:id="echoid-s16" xml:space="preserve">Octobris 1662.</s> <s xml:id="echoid-s17" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s18" xml:space="preserve">Franciſcus Bellhomus.</s> <s xml:id="echoid-s19" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s20" xml:space="preserve">Locus + Sigilli.</s> <s xml:id="echoid-s21" xml:space="preserve"/> </p> <pb file="0006" n="6"/> <p> <s xml:id="echoid-s22" xml:space="preserve">V.</s> <s xml:id="echoid-s23" xml:space="preserve">D. </s> <s xml:id="echoid-s24" xml:space="preserve">Fulgentius Orighetus Rector Pœniten-<lb/>tiariæ, pro Illuſtriſsimo, & </s> <s xml:id="echoid-s25" xml:space="preserve">Reuerendiſsimo <lb/>D. </s> <s xml:id="echoid-s26" xml:space="preserve">Ioſepho Muſotto Vicario Capitulari.</s> <s xml:id="echoid-s27" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div5" type="section" level="1" n="5"> <head xml:id="echoid-head5" style="it" xml:space="preserve">Reimprimatur.</head> <p> <s xml:id="echoid-s28" xml:space="preserve">Fr. </s> <s xml:id="echoid-s29" xml:space="preserve">Vincentius Vbaldinus Vicarius Generalis <lb/>S. </s> <s xml:id="echoid-s30" xml:space="preserve">Officij Bonon. </s> <s xml:id="echoid-s31" xml:space="preserve">Ordinis Prædicat.</s> <s xml:id="echoid-s32" xml:space="preserve"/> </p> <pb file="0007" n="7"/> </div> <div xml:id="echoid-div6" type="section" level="1" n="6"> <head xml:id="echoid-head6" xml:space="preserve"><emph style="bf">TAVOLA</emph></head> <head xml:id="echoid-head7" xml:space="preserve">De’ Capi contenuti in queſto Trattato.</head> <note style="it" position="right" xml:space="preserve"> <lb/>CApo 1. Checoſa ſia il Compaſſo di Proportione, & in che ſia fondato. # Pag. 4. <lb/>Capo 2. Come ſi diuida il Compaſſo di Proportione per le ſemplici longhezze di <lb/># linee rette, & vſo di queſia linea Aritmetica. # 7 <lb/>Queſt. 1. Come ſi troui la partè determinat a in numeri d’vna linea data. # 10 <lb/>Queſt. 2. Come ad vna linea data ſi troui vna maggiore nella proportione determinata <lb/># in numeri. # 17 <lb/>Queſt. 3. Come ſi troui vna Quarta Proportionale, e ſi continui vna proportione. # 19 <lb/>Queſt 4. Come lo Stromento ſerua di ſcala vniuerſale per qualſiu@glia diſſegno. # 21 <lb/>Queſt. 5. Date du@ linee trouare la loro proportione<unsure/> in numeri. # 24 <lb/>Queſt. 6. Dati gli Aſsi d’vn’ Ellipſi, deſcriuere la ſuæ<unsure/> circonferenza. # 27 <lb/>Queſt. 7. Come potiamo ſeruirci dello Stromento di Proportione, in vece delle Tauole <lb/># Trigonometriche, per la ſolutione di molti Triangoli. # 29 <lb/>Queſt. 8. Come ſerua per la Proſpettiua lo Stromento. # 31 <lb/>Queſt. 9. Come potiamo valerci dello Stromento per pratticaï in Numeri la regola <lb/># del Trè, ò Aurea, che vogliamo dire. # 34 <lb/>Queſt. 10. Come d’vna linea data ſi poſſano prendere particelle piccioliſſime, quante <lb/># l<unsure/>e ne vorranno. # 51 <lb/>Capo 3. Come s’habbia à diuider’il Compaſſo di Proportione per le Superficie piane, & <lb/># vſo di queſta linea Geometrica. # 54 <lb/>Queſt. 1. Data vna ſigura regolare, come ſi poſſa deſcriuerne vn’ altra della ſteſſa ſpe-<lb/># cie nella proportione, che ſi deſidera. # 67 <lb/>Queſt. 2. Data vnafigura irregolare, come ſi poſſa deſcriuerne vna ſimile nella brama-<lb/># ta proportione. # 74 <lb/>Queſt. 3. Data vna linea in vn piano, come s’habbia à trouare la grandezza della linea, <lb/># che le corriſponde in vn’altro piano ſimile nella data proportione. # 77 <lb/>Queſt. 4. Data due figure piane ſimili trouar la loro proportione. # 82 <lb/>Queſt. 5. Data due, ò piu figure piane ſimili, trouarne vna ſimile vguale à tutte quel-<lb/># le inſieme. # 85 <lb/>Queſt. 6. Data due figure piane ſimili, e diſuguali, trouar’vna figura ſimile vguale alla <lb/># loro differenza. # 86 <lb/>Queſt. 7. Data due linee, come poſſa trouarſi la terza proportionale. # 87 <lb/>Queſt. 8. Come ſi troui vna media proportionale tra due linee date, e ſi faccia vn Qua-<lb/># drato vguale ad vna figura rettilinea. # 89 <lb/>Queſt. 9. Deſcriuere con facilità vna Parabola. # 90 <lb/>Queſt. 10 Data vna Parabola in vn Cono dato, trouar vn Quadrato à lei vguale. # 91 <lb/>Queſt. 11. Data due linee vguali, che ſi tagliano per mezzo obliquamente, deſcriuere in- <pb file="0008" n="8"/> # torno ad eſſe vn’ Ellipſi. # 92 <lb/>Queſt. 12. Data vna portione di Ouato trouar il reſtante delſuo diametro. # 94 <lb/>Queſt. 13. Dalli due diametri d’vn’ Ellipſi trouar l’area. # 96 <lb/>Queſt. 14. Dato vn numero, trouare la ſua radice quadratæ<unsure/>. # 97 <lb/>Capo 4. Come s’habbia à diuidere lo Stromento per i Corpi ſolidi; & vſo di queſta li-<lb/># nea Cubica. # 105 <lb/>Queſt. 1. Tradue linee date, come ſi trouino due medie continuamente proportionali: <lb/># ouero tra due numeri dati. # 113 <lb/>Queſt. 2. Come ſi poſſa ad vna linea data applicar’ vn ſolido rettangolo vguale ad vn <lb/># Cubo dato. # 116 <lb/>Queſt. 3. Dato vn ſolido, come s’habbia à trouarne vn’ altro ſimile nella data propor-<lb/># tione. # 118 <lb/>Queſt. 4. Dati due Corpi ſimili, come ſi conoſca la loro proportione. # 125 <lb/>Queſt. 5. Come ſi poſſafar’vn Cono vguale ad vn Cilindro dato, e che habbiano li dia-<lb/># metri delle baſi, e gl’Aſſi proportionali. # 128 <lb/>Queſt. 6. Come ſi troui vna Sfera vguale ad vn Cilindro dato. # 130 <lb/>Queſt. 7. Data vna Parabola, trouare la proportione di due ſegmenti terminati ad vn <lb/># medeſimo punto. # 132 <lb/>Queſt. 8. Data vna Parabola terminata, tagliata da vna linea parallela, trouar la pro-<lb/>portione delle parti nelle qualli è diuiſa. # 133 <lb/>Queſt. 9. Come d’vn numero dato ſi troui la Radice Cubica. # 134 <lb/>Capo 5. Come s’habbia à notare nello Stromento la Proportione de’Metalli; & vſo di <lb/># queſta linea Metallica. # 145 <lb/>Queſt. 1. Come ſi poſſa cauare la proportione delle grauità ſpecifiche di due, ò più <lb/># corpi. # 151 <lb/>Queſt. 2. Dato vn corpo, la cui grandezza, e grauità ſiano note, come ſi poſſa trouarne <lb/># vn’altro d’altra materia, che in grauità habbia la proportione data. # 154 <lb/>Queſt. 3. Come ſi poſſa trouare la grandezza di qualſiuoglia peſo, conoſcendone vn’ al-<lb/># tro d’altra materia. # 159 <lb/>Capo 6. In qual maniera s’habbiano à notare nello Stromento li Gradi del Circolo: & <lb/># vſo di tal linea. # 160 <lb/>Queſt. 1. Come ſi poſſa deſcriuer’vn’ angolo di quantità determinata. # 165 <lb/>Queſt. 2. Come ſi conoſca lagrandezza, e quantitd d’vn’ angolo dato. # 168 <lb/>Queſt. 3. Come con lo Stromento ſi poſſa pratticare tuttala Trigonometria ſenza Ta-<lb/># uole. # 171 <lb/>Queſt. 4. Trouar in numeri la proportione di due rette con l’aiuto delle Tauole de’Se-<lb/># ni. # 175 <lb/>Queſt. 5. Trouar in piccoli numeri i ſeni de’ gradi del quadrante. # 177 <lb/>Queſt. 6. Data vna linea corda d’vn’arco di determinata quantità, come ſi troui il ſuo <lb/># circolo. # 179 <lb/>Queſt. 7. Come ſi poſſa prendere qualſiuoglia parte determinata del circolo, c deſcriue-<lb/># re qualſiuoglia figura regolare. # 181 <pb file="0009" n="9"/> Queſt. 8. Dato il diametro d’vna sfera, come ſi trouila ſuperficie sferica, e la ſolidità <lb/># di qualſiuoglia ſegmento di detta sfera, conoſciuto nella quantità de’gradi d’vn circolo <lb/># maſsimo perpendicolare al piano della baſe di detto ſegmento. # 183 <lb/>Queſt. 9. Data in gradila circonferenza d’vn ſegmento di circolo, come ſi troui l’area <lb/># di detto ſegmento. # 189 <lb/>Capo 7. Come nello Stromento s’habbiano à ſegnare i lati delle figure regolari; vſo di <lb/># queſta linea de’Poligoni. # 191 <lb/>Queſt. 1. Come data vna linea ſi poſſa farne vna figura Regolare, qual più piace, ò de-<lb/># ſcriuere l’angolo d’vna figura Regolare, di quelle, che ſon ſegnate nello Stromento. # 196 <lb/>Queſt. 2. Data vna figura regolare, come ſe le poſſa circoſcriuere, ò inſcriuer’ vn cir-<lb/># colo. # 198 <lb/>Queſt. 3. Dato vn’arco, come ſi poſſa facilmente trouare in eſſo la quantità d’vn grado, <lb/># & altre parti del circolo non ſegnate nella linea de’ poligoni. # 199 <lb/>Queſt. 4. Come ſi conoſcala proportione de’lati delli poligoni deſoritti nello ſteſſo circo-<lb/># lo; e poi anche la proportione delli ſteſſi poligoni. # 203 <lb/>Queſt. 5. Dato vn Poligono regolare, trouarne vn’altro à lui vguale. # 206 <lb/>Capo 8. In qual maniera s’habbia à ſegnare nello Stromento la linea d’vguaglianza tra <lb/># piani regolari diſſomigliante, & vſo di queſta linea trasformatoria. # 207 <lb/>Queſt. 1. Data vna figura regolare, trasformatoria in vn’altra vguale di più, ò meno <lb/># lati. # 211 <lb/>Queſt. 2. Data vna figura regolare trouarne vn’altra regolare diuerſa, à cui habbia la <lb/># data Proportione. # 212 <lb/>Queſt. 3. Date due figure regolari diuerſe, conoſcere, che proportione habbiano trà di <lb/># loro. # 213 <lb/>Queſt. 4. Data l’area d’vn poligono regolare, trouaril ſuo lato. # 214 <lb/>Queſt. 5. Dati due poligoni regolari diuerſi vguali, trouare la proportione de’ circoli, <lb/># ne’quali eſſi ſi deſcriuono. # 215 <lb/>Queſt. 6. Data vna figuraregolare far’vn circolo a lei vguale, e dato vn circolo far vn <lb/># quadrato vguale. # 215 <lb/>Queſt. 7. Date due figure regolari diſſimili, e diſuguali, farne vna vguale à tutte due, e <lb/># diſſomigliante. # 216 <lb/>Queſt. 8. Dati due poligoni regolari diſſimili, e diſuguali, trouar’vn’altra figura diſſimi-<lb/># le, che ſia vguale alla loro differenza. # 217 <lb/>Capo 9. In qual maniera habbia à ſegnarſi la linea de’corpi regolari, & vſo di queſta <lb/># linea. # 218 <lb/>Queſt. 1. Conoſciuto il diametro d’vna sfera, come ſi poſſa formar’vn cubo, ò altro ſoli-<lb/># do regolare, che capiſca in eſſa. # 223 <lb/>Queſt. 2. Data vna piramide trouar la sſera, che contenga vn’ altra piramide in data <lb/># proportione. # 223 <lb/>Queſt. 3. Dato il diametro della sfera trouar la proportione de’corpi regolari inſcrit-<lb/># ti. # 224 <lb/>Queſt. 4. Data vna sfera trouar i lati de’corpi ordinati circoſcritti. # 227 <pb file="0010" n="10"/> Queſt. 5. Come dato vn corpo regolare ſi trasformi in vn’altro, chegli ſia vguale. # 228 <lb/>Capo 10. Come ſi poſſa diuidere vna linea, che ſerua per quadrare tutti i Segmenti del <lb/># Circolo, efigure inſcritte: & vſo di queſt a linea Quadratrice. # 231 <lb/>Queſt. 1. Se due Circoli diſuguali ſi tagliano, come ſi troui la quantità dell’area, in cui <lb/># communicano, e la lunula che reſta. # 236 <lb/>Queſt. 2. Dato vn trapezio in vn Circolo, eſegmento di circolo, trouare la ſua quanti-<lb/># tà. # 239 <lb/>Queſt. 3. Dato vn ſegmento di circolo, ò troppo grande, ò troppo piccolo, come ſi deb-<lb/># ba operare per trouar la linea, che dia il quadrato vguale al ſegmento. # 240 <lb/>Queſt. 4. Data vna portione di Circolo trouare la ſua grandezza in miſura determina-<lb/># ta. # 242 <lb/>Queſt. 5 Dato vn Segmento di Circolo, trouare la proportione, che il Segmento hà ad <lb/># vn dato Triangolo, che in eſſo capiſce. # 244 <lb/>Capo Vltimo. Come ſi poſſano con gran facilità fabricare molti Compaſſi di proportione <lb/># altrigrandi, altri piccoli. # 246 <lb/>Conchiuſione. # 248 <lb/></note> <figure> <image file="0010-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0010-01"/> </figure> <pb o="1" file="0011" n="11"/> <figure> <image file="0011-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0011-01"/> </figure> </div> <div xml:id="echoid-div7" type="section" level="1" n="7"> <head xml:id="echoid-head8" xml:space="preserve">DELLA FABRICA, <lb/>ETVSO</head> <head xml:id="echoid-head9" xml:space="preserve">Del Compaſſo di Proportione.</head> <p> <s xml:id="echoid-s33" xml:space="preserve">IO non pretendo di ſcriuere coſa nuoua, mà <lb/>impiegarmi in materia vtile. </s> <s xml:id="echoid-s34" xml:space="preserve">Ciò che del-<lb/>l’Organo ſi dice eſſer’vn Compendio de gli <lb/>Stromenti Muſicali à cagione deìla molti-<lb/>plicità, e varia combinatione de’regiſtri, <lb/>che contiene, parmi poſſa vgualmente dirſi <lb/>del Compaſſo di Proportione, cioè, che ſia vn Compendio <lb/>di molti ſtromenti Geometrici inuentati per la facilità di <lb/>molte operationi, poiche contiene varietà di linee diuerſa-<lb/>mente diuiſe, e ſeruendo variamente conforme alla diuerſa <lb/>apertura di detto Compaſſo, comprende vna grand’vniuer-<lb/>ſalità d’operationi. </s> <s xml:id="echoid-s35" xml:space="preserve">Mà alcuni ſi trouano prouiſti di ſimile <lb/>Stromento fabricato con grand’accuratezza, e politezza in <lb/>Francia, ò in Fiandra, à quali però non ſerue più che vna bel-<lb/>la pittura nella lor galeria, il cui vſo finiſce, con eſſer’attenta-<lb/>mente rimirata: </s> <s xml:id="echoid-s36" xml:space="preserve">eſſendoche ne conoſcono le linee, che vi <lb/>ſono notate, ſe non forſi quanto dalle parole aggiunte à ciaſ- <pb o="2" file="0012" n="12"/> cuna linea intendono<unsure/> qualche coſa, ne ſanno ſeruirſi del detto <lb/>Stromento. </s> <s xml:id="echoid-s37" xml:space="preserve">Altri poi ſono, che veramente ſariano capaci <lb/>di ſeruirſene con loro grand’vtilità, e piacere; </s> <s xml:id="echoid-s38" xml:space="preserve">mà la difficoltà <lb/>di far venire da paeſi ſtranierilo Stromento, e l’ignoranza <lb/>de’noſtri Artefici Italiani, quali (per alrro capaci di farlo <lb/>molto eſſattamente) non ſanno fabricarlo, è cagione, che <lb/>manchino di tal commodità. </s> <s xml:id="echoid-s39" xml:space="preserve">Quindi è, che à gl’vni, & </s> <s xml:id="echoid-s40" xml:space="preserve">à <lb/>gl’altri deſiderando di far coſa vtile, acciò e chi l hà ſappia <lb/>ſeruirſene, e chi ne manca poſſa facilmente prouederſene, mi <lb/>ſon riſoluto in primo luogo di moſtrar' il modo, con cui hab-<lb/>biano à diuiderſi le linee, che in queſto Stromento s’hanno à <lb/>deſcriuere; </s> <s xml:id="echoid-s41" xml:space="preserve">le quali diuiſioni, ò ſi potranno ſare da gli ſteſſi <lb/>Artefici, ò chi non ſi fidaſſe della lor diligenza, potrà farle <lb/>egli ſteſſo, doppo che dall’Art fice fatto ſarà tutto il mate-<lb/>riale dello Stromento; </s> <s xml:id="echoid-s42" xml:space="preserve">nel che non ſitroua tale difficoltà, che <lb/>non poſſa con poco trauaglio trouarſi Artefice, che lo faccia. <lb/></s> <s xml:id="echoid-s43" xml:space="preserve">Dipoi alla deſcrittione di ciaſcuna linea ſoggiungo in alcune <lb/>queſtioni l’vſo dello Stromento con tal linea. </s> <s xml:id="echoid-s44" xml:space="preserve">Dalle quali <lb/>queſtioni ciaſcuno colſno ingegno potra trouarne dell’altre, <lb/>& </s> <s xml:id="echoid-s45" xml:space="preserve">ampliare l’vſo dello Stromento; </s> <s xml:id="echoid-s46" xml:space="preserve">poiche io pretendo di <lb/>ſcriuere breuemente inſieme, e moſtrare la ſtrada à quei, che <lb/>non la ſanno.</s> <s xml:id="echoid-s47" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s48" xml:space="preserve">Da ciò ſi vede per qual cagione io habbia ſcritto in forma <lb/>ſemplice, & </s> <s xml:id="echoid-s49" xml:space="preserve">in lingua Italiana: </s> <s xml:id="echoid-s50" xml:space="preserve">eſſendo che così era conue-<lb/>niente di fare à chi voleua eſſer’inteſo dalli noſtri Artefici Ita-<lb/>liani: </s> <s xml:id="echoid-s51" xml:space="preserve">Oltre che eſſendo molti, iquali non hanno l’vſo della <lb/>lingua Latina così famigliare, e pure affettionandoſi alle co-<lb/>ſe Mattematiche, ſpenderiano vtilmente molto tempo, che <lb/>loro sfugge otioſamente, hò deſiderato di far loro in ciò coſa <lb/>grata, mentre non ſono ritirati dalla lettione di queſta Ope-<lb/>retta dalla qualità dell’Idioma.</s> <s xml:id="echoid-s52" xml:space="preserve"/> </p> <pb o="3" file="0013" n="13"/> <p> <s xml:id="echoid-s53" xml:space="preserve">E ſe ad alcuno pareſſe ſuperflua queſta mia fatica;</s> <s xml:id="echoid-s54" xml:space="preserve">eſſendo <lb/>che di queſto Stromento è ſtato ſcritto da altri; </s> <s xml:id="echoid-s55" xml:space="preserve">ſappia, che <lb/>tal’obiettione à me ancora è venuta in mente prima di met-<lb/>tetmi à ſcriuere queſti fogli; </s> <s xml:id="echoid-s56" xml:space="preserve">e quello che più mi ritraeua, era <lb/>il dubbio probabiliſsimo d’incontrar mi à dire molte coſe der-<lb/>te da altri, e ſoggiacer’alla riprenſione d’hauer copiato. </s> <s xml:id="echoid-s57" xml:space="preserve">Mà <lb/>finalmente mi ſon laſciato vincere dal deſiderio non di mia <lb/>lode, mà dell’altrui vtilità; </s> <s xml:id="echoid-s58" xml:space="preserve">tenendo per certo, che sì come <lb/>non oſtante ſia ſtato ſcritto da altri di queſta Mareria, ad ogni <lb/>modo io non hò hauuto forruna di vedere mai alcun’Autore, <lb/>fuorche il Galilei, di cui nel 1642. </s> <s xml:id="echoid-s59" xml:space="preserve">ventidue anni prima di <lb/>ſcriuere queſt’Operetta, nella Libreria noſtra del Collegio <lb/>Romano mi capitò vn picciolo libretto di queſta Materia, da <lb/>me allhora poco inteſo; </s> <s xml:id="echoid-s60" xml:space="preserve">così à molti altri poteua accadere ſi-<lb/>mile diſgratia, che non capitaſſe loro alle mani alcuno di que’ <lb/>buoni Autori; </s> <s xml:id="echoid-s61" xml:space="preserve">e perciò capitando loro queſta mia Operetta, <lb/>ne potranno trarre qualche vtilità. </s> <s xml:id="echoid-s62" xml:space="preserve">Oltre che vediamo da <lb/>tanti Huomini ſaggi eſſerſi ſpiegati gli medeſimi ſei primi li-<lb/>bri d’Euclide, e pur niuno ſi ſtima inutile, portandoſi con ciò <lb/>qualche maggior facilità a’principianti: </s> <s xml:id="echoid-s63" xml:space="preserve">e così per la ſteſſa <lb/>cagione hò creduto non eſſer queſta mia fatica ſuper flua, <lb/>mentre non ſcriuo per Mattematici prouetti, ma per <lb/>principianti, e poco eſperti nelle coſe della Geo-<lb/>metria. </s> <s xml:id="echoid-s64" xml:space="preserve">E per queſto per lo più cito le <lb/>propoſitioni d’Euclide, con le quali <lb/>ſi dimoſtrano le coſe, <lb/>che vado dicendo. <lb/></s> <s xml:id="echoid-s65" xml:space="preserve">§ § § § <lb/>§ §</s> </p> <pb o="4" file="0014" n="14"/> </div> <div xml:id="echoid-div8" type="section" level="1" n="8"> <head xml:id="echoid-head10" style="it" xml:space="preserve">CAPO PRIMO.</head> <head xml:id="echoid-head11" style="it" xml:space="preserve">Che coſa ſia il Compaſſo di Proportione, & in che ſia <lb/>fondato.</head> <p> <s xml:id="echoid-s66" xml:space="preserve">IL<unsure/> Compaſſo di Proportione non è altro, che vno Stro-<lb/>mento compoſto di due regole piane, e diritte di ma-<lb/>teria ſolida (ò ſia legno, ò ottone, ò argento) nell’vna delle <lb/>due eſtremità vnite inſieme in modo, che ſi poſſino allargar, <lb/>e ſtringere sì, che riſtrette ſi combacino, & </s> <s xml:id="echoid-s67" xml:space="preserve">allargate ſi ſten-<lb/>dano à formar vna ſola regola diritta. </s> <s xml:id="echoid-s68" xml:space="preserve">Che ſe bene non è <lb/>aſſoluta mente neceſſario, che poſſano tanto allargarſi, ò ſtrin-<lb/>gerſi, ad ogni modo così riuſcirà più vtile lo Stromento.</s> <s xml:id="echoid-s69" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s70" xml:space="preserve">Si chiama Compaſſo, perche il ſuo vſo è con allargarlo, ò <lb/>ſtringerlo à ſomiglianza del Compaſſo, con cui ſi deſcriuo-<lb/>no i circoli maggiori, ò minori. </s> <s xml:id="echoid-s71" xml:space="preserve">Si dice poi di Proportione, <lb/>perche ſerue à trouar linee nella proportione, che ſi deſidera.</s> <s xml:id="echoid-s72" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s73" xml:space="preserve">Dal centro dunque, circa di cui ſi muouono le due regole <lb/>(il quale conuien che ſia accuratiſſimamente ſegnato nella <lb/>ſuperfieie dello Stromento, e ſi troua nell’interſettione delli <lb/>lati interiori delle due regole, prolongati con linee occulte, <lb/>e ſottiliſſime, baſtando poi ſegnare viſibilmente ſolamente <lb/>il punto, che corriſponde al centro) ſi tira ſopra ciaſcheduna <lb/>regola vna linea retta, e queſta ſi diuide con la deſiderata pro-<lb/>portione; </s> <s xml:id="echoid-s74" xml:space="preserve">auuertendo, che l’vna, el’altra linea ſia vguale, e <lb/>ſimilmente diuiſa. </s> <s xml:id="echoid-s75" xml:space="preserve">E ciò fatto, s’hà lo Stromento, di cui hab-<lb/>biam biſogno per poter diuidere ſimilmente qualunque altra <lb/>linea, che non ſia maggiore della diſtanza, che è trà li due <lb/>eſtremi punti delle linee deſcritte sù le regole, quando ſtanno <lb/>diſteſe, e fanno vna regola ſola.</s> <s xml:id="echoid-s76" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s77" xml:space="preserve">Siano dunque le due regole AB, AC, congionte nel pun- <pb o="5" file="0015" n="15" rhead="Fondamento."/> to A, circa di cui, come intorno à centro, ſi poſſano girare; <lb/></s> <s xml:id="echoid-s78" xml:space="preserve">e ſul piano della regola AB tiriſi dal centro A, vna linea ret-<lb/>ta AE, e ſimilmente ſul piano dell’altra regola ſitiri dall’iſteſ-<lb/>ſo centro la retta vgnale<unsure/> all’AE. </s> <s xml:id="echoid-s79" xml:space="preserve">Se queſte due linee AE, AL <lb/>ſaranno ſimilmente diuiſe, qualunque linea, che non ſia mag-<lb/>giore della diſtanzatra E, L, quando ſono le due regole di-<lb/>ſteſe in vna ſola, ſi potrà ſimilmente diuidere. </s> <s xml:id="echoid-s80" xml:space="preserve">Come ſe per <lb/>eſſempio AE, & </s> <s xml:id="echoid-s81" xml:space="preserve">AL ſono ſimilmente diuiſe in H, & </s> <s xml:id="echoid-s82" xml:space="preserve">I, ſia vna <lb/>linea, che ſia la diſtanza EL; </s> <s xml:id="echoid-s83" xml:space="preserve">ſe ſi pigliarà la diſtanza HI, e <lb/>ſi traſportarà nella linea data, queſta ſarà diuiſa nella ſteſſa <lb/>proportione, che è diuiſa la linea AE in H. </s> <s xml:id="echoid-s84" xml:space="preserve">E perche le due <lb/>regole congiunte in A ſi puonno allargar, e ſtringere, ſi vede, <lb/>che tutte le linee, le quali poſſono capire trà la minima, e la <lb/>maſſima diſtanza di E, & </s> <s xml:id="echoid-s85" xml:space="preserve">L, tutte ſi poſſono diuidere nella <lb/>ſteſſa proportione di AE diuiſa in H. </s> <s xml:id="echoid-s86" xml:space="preserve">Dal che ſi raccoglie, <lb/>che quanto più lunghe ſaranno le regole AB, AC, anche <lb/>maggiore ſarà l’vſo loro per la diuiſione di linee molto <lb/>maggiori.</s> <s xml:id="echoid-s87" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s88" xml:space="preserve">Auuertaſi però, che, ſe bene ſin’hora non s’è parlato che <lb/>di diuiſioue di linea retta, non è, che à queſt’vſo ſolamente ſi <lb/>riſtringa il Compaſſo di Proportione, di cui parliamo; </s> <s xml:id="echoid-s89" xml:space="preserve">mà ciò <lb/>s’è detto per più facile intelligenza de gl’ineſperti: </s> <s xml:id="echoid-s90" xml:space="preserve">poiche <lb/>più à baſſo ſi ſpiegaranno gl’vſi molto maggiori, che per vna <lb/>ſemplice diuiſione. </s> <s xml:id="echoid-s91" xml:space="preserve">Quindi è, che per eſſer più obuio, e com-<lb/>mune l’vſo di queſto Stromento per le diuiſioni, è anche chia-<lb/>mat@ da molti Stromento delle Parti; </s> <s xml:id="echoid-s92" xml:space="preserve">ſe ben’<unsure/>il vocabolo di <lb/>Compaſſo, ò Siromento di Proportione pare più proprio, perche <lb/>comptende più vniuerſalmente il fine, à cuiſerue.</s> <s xml:id="echoid-s93" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s94" xml:space="preserve">Hor’acciò s’intenda fondamentalmente l’vſo di queſto <lb/>Stromento, e veggaſi, come quellc due diſtanze EL, & </s> <s xml:id="echoid-s95" xml:space="preserve">HI <pb o="6" file="0016" n="16" rhead="CAPO I."/> hanno trà di ſe la proportione di AE, & </s> <s xml:id="echoid-s96" xml:space="preserve">AH, ſia nella ſecon-<lb/>da figura il triangolo Iſoſcele AEL, e prendaſi AH vguale <lb/>alla AI, e tiriſi la linea HI. </s> <s xml:id="echoid-s97" xml:space="preserve">E' manifeſto, che <lb/> <anchor type="figure" xlink:label="fig-0016-01a" xlink:href="fig-0016-01"/> li due triangoli AEL, AHI ſono ſimili; </s> <s xml:id="echoid-s98" xml:space="preserve">perche <lb/>gl’angoli HI, ſon vguali trà di ſe (per la 5. </s> <s xml:id="echoid-s99" xml:space="preserve">del <lb/>1.) </s> <s xml:id="echoid-s100" xml:space="preserve">e ciaſcuno è la metà<unsure/> del complemento dell’ <lb/>angolo A, à due angoli retti (per la 32. </s> <s xml:id="echoid-s101" xml:space="preserve">del 1) <lb/>e per la ſteſſa ragione anche ciaſcuno de gli an-<lb/>goli E, & </s> <s xml:id="echoid-s102" xml:space="preserve">L è la metà dello ſteſſo complemen-<unsure/> <lb/>to. </s> <s xml:id="echoid-s103" xml:space="preserve">Dunque l’angolo I è vguale all’ angolo L, <lb/>e l’angolo H vguale all angolo E: </s> <s xml:id="echoid-s104" xml:space="preserve">dunque li due triangoli <lb/>A H I, AEL ſono equiangoli; </s> <s xml:id="echoid-s105" xml:space="preserve">dunque (per la 4 del 6.) </s> <s xml:id="echoid-s106" xml:space="preserve">ſono <lb/>ilati proportionali circa gl’angoli vguali; </s> <s xml:id="echoid-s107" xml:space="preserve">dunque come AE <lb/>ad EL, così AH à HI, e permutando come AE ad AH, così <lb/>EL à HI. </s> <s xml:id="echoid-s108" xml:space="preserve">Se dunque HI ſi trasferirà ſopra la EL, e ſia EK ſa-<lb/>rà la EL diuiſa in K proportionalmente alla diuiſione di <lb/>AE in H.</s> <s xml:id="echoid-s109" xml:space="preserve"/> </p> <div xml:id="echoid-div8" type="float" level="2" n="1"> <figure xlink:label="fig-0016-01" xlink:href="fig-0016-01a"> <image file="0016-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0016-01"/> </figure> </div> <p> <s xml:id="echoid-s110" xml:space="preserve">E queſta è la dimoſtrazione generale, qualunque ſia la pro-<lb/>portione, in cuiſia diuiſa la linea retta tirata ſul piano delle <lb/>regole dello Stromento. </s> <s xml:id="echoid-s111" xml:space="preserve">E perche varie aſſai puonno eſſere le <lb/>proportioni, nelle quali ſi può diuidere vna linea, così ſopra <lb/>la ſteſſa faccia della regola dello Stromento ſi tirano diuerſe <lb/>linee variamente diuiſe, acciò le ſteſſe due regole vengano à <lb/>ſeruirci per tanti Stromenti, quante linee ſono tirate in vna <lb/>delle ſudette regole. </s> <s xml:id="echoid-s112" xml:space="preserve">Sì che tutto l’ artificio di queſto Stro-<lb/>mento conſiſte in mettere ſopra le ſue regole quelle propor-<lb/>tioni, con cui ſi può deſiderare d’hauer altre linee in propor-<lb/>tioni ſimili; </s> <s xml:id="echoid-s113" xml:space="preserve">ancorche quelle linee non foſſero commenſura-<lb/>bili alle linee deſcritte nello Stromento.</s> <s xml:id="echoid-s114" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s115" xml:space="preserve">Da quel che s’è detto è manifeſto, che li due ttiangoli AEL, <pb file="0017" n="17"/> <pb file="0017a" n="18" rhead="Capo Secondo."/> <anchor type="figure" xlink:label="fig-0017a-01a" xlink:href="fig-0017a-01"/> <pb file="0018" n="19"/> <pb o="7" file="0019" n="20" rhead="Fondamento."/> AHI, deuono eſſere nell’ iſteſſo piano; </s> <s xml:id="echoid-s116" xml:space="preserve">onde ſe la linea AE <lb/>foſſe ſopra vna ſupei ficie incuruata, non procederebbe la di-<lb/>moſtrazione: </s> <s xml:id="echoid-s117" xml:space="preserve">Perciò ſi vede, quanto ſia neceſſario, che le re-<lb/>gole ſiano così ben’aggiuſtate e ſode, che ne in ſe ſteſſe facil-<lb/>mente s’incuruino, & </s> <s xml:id="echoid-s118" xml:space="preserve">anche allargate ſi conſeruino nell’iſteſſo <lb/>piano Deuono poi eſſere ciaſcuna tanto larghe, che vi poſ-<lb/>ſa capire tutta la moltitudine delle linee, che vi ſi vorranno <lb/>tirare, ſenza confuſione, & </s> <s xml:id="echoid-s119" xml:space="preserve">in modo, che li numeri notati alli <lb/>punti delle diuiſioni ſi poſſano commodamẽte oſſeruare ſen-<lb/>za pericolo d’errore, con prender’il numero corriſpondente <lb/>ad vn punto per vn’altro.</s> <s xml:id="echoid-s120" xml:space="preserve"/> </p> <div xml:id="echoid-div9" type="float" level="2" n="2"> <figure xlink:label="fig-0017a-01" xlink:href="fig-0017a-01a"> <image file="0017a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0017a-01"/> </figure> </div> <p> <s xml:id="echoid-s121" xml:space="preserve">Auuertaſi eſſer neceſſario nell’o perationi prendere col Cõ-<lb/>paſſo accuratamente la lunghezza delle linee, e perciò con-<lb/>uiene, che le ſue punte ſiano ben’ acute: </s> <s xml:id="echoid-s122" xml:space="preserve">e ſe tali non foſſero, <lb/>ſi potranno alle gambe del Compaſſo con ſottili cordicelle <lb/>da liuto legare ſtrettamente due aghi da cucire, le cui punte <lb/>ſono ſottiliſſime, & </s> <s xml:id="echoid-s123" xml:space="preserve">acute, quanto baſta ad ogni più accurata <lb/>operatione.</s> <s xml:id="echoid-s124" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div11" type="section" level="1" n="9"> <head xml:id="echoid-head12" xml:space="preserve">CAPO SECONDO.</head> <head xml:id="echoid-head13" style="it" xml:space="preserve">Come ſi diuida il Compaſſo di Proportione per le ſemplici lunghezze <lb/>di linee Rette, & vſo di queſta linea Aritmetica.</head> <p> <s xml:id="echoid-s125" xml:space="preserve">IL primo, e più facile vſo di queſto Stromento è in ordine <lb/>alle ſemplici lung hezze di linee Rette perciò da queſte ſi <lb/>comincia. </s> <s xml:id="echoid-s126" xml:space="preserve">Si tir<unsure/>ano dunque dal centro A due linee rette AE, <lb/>AL, e queſte ſi diuidono nelle più minute parti vguali, che ſi <lb/>può, ſalua la diſtintione neceſſaria, per non confonderſi nel <lb/>numerarle, & </s> <s xml:id="echoid-s127" xml:space="preserve">hauuto riſguardo alla lunghezza delle regole. <lb/></s> <s xml:id="echoid-s128" xml:space="preserve">E quì fà di<unsure/> meſtieri apportarui tutta la diligenza, per poter <pb o="8" file="0020" n="21" rhead="CAPO II."/> dipoi ſeruirſene con ſieurezza. </s> <s xml:id="echoid-s129" xml:space="preserve">Communemente ſi diuide in <lb/>cento parti, sì perche queſta è diuiſione ſofficiente, sì perche <lb/>dentro queſto numero ſi trouano quelle proportioni, che <lb/>communemente ſono vſuali, potendoſi maſſime tutte ridur-<lb/>re à ragione di centeſime, perle operationi Mecaniche, alle <lb/>quali ſeruono gli Stromenti. </s> <s xml:id="echoid-s130" xml:space="preserve">Mà ſelo Stromento foſſe aſſai <lb/>lungo, ſi potrà diuidere in 150. </s> <s xml:id="echoid-s131" xml:space="preserve">ouero in 200. </s> <s xml:id="echoid-s132" xml:space="preserve">particelle. <lb/></s> <s xml:id="echoid-s133" xml:space="preserve">E perche queſta linea è talmente diuiſa, che le diſtanze dal <lb/>centro A vanno ſempre creſcendo con vgual differenza, co-<lb/>me le progreſſioni Aritmetiche hanno vguali gl’incrementi, <lb/>ò decrementi de’ſuoi termini, perciò queſta linea diuiſa in <lb/>particelle vguali, con ragione ſi può chiamare linea Arit-<lb/>metica.</s> <s xml:id="echoid-s134" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s135" xml:space="preserve">Diuidaſi dunque la linea AE (ele diuiſioni fatte in queſta <lb/>ſi traſportino nella A L) con vn ben’ acuto, e ſodo compaſ-<lb/>ſo in due parti vguali; </s> <s xml:id="echoid-s136" xml:space="preserve">e ciaſcuna ſarà di 50. </s> <s xml:id="echoid-s137" xml:space="preserve">particelle cente-<lb/>ſime, onde al punto della diuiſione ſi noti il numero 50. </s> <s xml:id="echoid-s138" xml:space="preserve">Di-<lb/>poitutta la linea AE ſi diuida in cinque parti vguali, e ciaſcu-<lb/>na ſarà di 20. </s> <s xml:id="echoid-s139" xml:space="preserve">particelle: </s> <s xml:id="echoid-s140" xml:space="preserve">onde doueranno ſegnarſi con li nu-<lb/>meri 20. </s> <s xml:id="echoid-s141" xml:space="preserve">40. </s> <s xml:id="echoid-s142" xml:space="preserve">60. </s> <s xml:id="echoid-s143" xml:space="preserve">80. </s> <s xml:id="echoid-s144" xml:space="preserve">Cosìhauutaſi la diſtanza trà 40. </s> <s xml:id="echoid-s145" xml:space="preserve">e 50. <lb/></s> <s xml:id="echoid-s146" xml:space="preserve">shà la decima parte ditutta la linea AE, e con queſta comin-<lb/>ciando da A ſi ſegnano di dieci in dieci: </s> <s xml:id="echoid-s147" xml:space="preserve">con che anche ſi pro. </s> <s xml:id="echoid-s148" xml:space="preserve"><lb/>ua, ſe le prime diu ſioni furono accuratamente fatte. </s> <s xml:id="echoid-s149" xml:space="preserve">Simil-<lb/>mente ſe vna di queſte decime ſi diuide per metà (ouero ſe ne <lb/>piglino trè decime, e ſi diuidano per metà) s’<unsure/>hauranno le di-<lb/>uiſioni di cinque in cinque, e la linea AE ſarà diuiſa in 20. </s> <s xml:id="echoid-s150" xml:space="preserve"><lb/>parti vguali. </s> <s xml:id="echoid-s151" xml:space="preserve">E sì come le decime furono notate col numero, <lb/>& </s> <s xml:id="echoid-s152" xml:space="preserve">vna lineetta traſuerſale, così la metà delle decine ſi nota <lb/>con vna ſola lineetta più piccola, acciò ſubito ſi poſſa cono-<lb/>ſcere, e numerare le particelle, le altre poi ſi ſegnano con <pb o="9" file="0021" n="22" rhead="Linea Aritmetica."/> ſoli punti. </s> <s xml:id="echoid-s153" xml:space="preserve">Finalmente ciaſcuna di queſte parti venteſime ſi <lb/>diuide in cinque particelle vguali, e ſarà tutta la linea AE di-<lb/>uiſa in cento particelle vguali.</s> <s xml:id="echoid-s154" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s155" xml:space="preserve">E perche forſi il diuider’ vna di quelle parti venteſime <lb/>in cinque particelle vguali riuſcirebbe aſſai difficile, pigliſi da <lb/>A ſin a 30. </s> <s xml:id="echoid-s156" xml:space="preserve">e ſia la linea RS diuiſa in ſei di quelle parti vente-<lb/>ſime. </s> <s xml:id="echoid-s157" xml:space="preserve">Tutta la RS <lb/> <anchor type="figure" xlink:label="fig-0021-01a" xlink:href="fig-0021-01"/> ſi diuida in cinque <lb/>parti vguali, il che <lb/>ſi farà applicando <lb/>la RS all’interuallo 100. </s> <s xml:id="echoid-s158" xml:space="preserve">100. </s> <s xml:id="echoid-s159" xml:space="preserve">come più à baſſo ſi dirà, e <unsure/>’l’in-<lb/>teruallo 20. </s> <s xml:id="echoid-s160" xml:space="preserve">20. </s> <s xml:id="echoid-s161" xml:space="preserve">s’applichi alla linea RS in a, b, c, d; </s> <s xml:id="echoid-s162" xml:space="preserve">poiche <lb/>la diſtanza tra il numero 5. </s> <s xml:id="echoid-s163" xml:space="preserve">& </s> <s xml:id="echoid-s164" xml:space="preserve">il punto a, ſarà appunto la <lb/>quinta parte di tutta quella venteſima della linea AE: </s> <s xml:id="echoid-s165" xml:space="preserve">Il che <lb/>è manifeſto, perche RS è particelle 30; </s> <s xml:id="echoid-s166" xml:space="preserve">R a, che è quinto <lb/>di RS, è particelle 6; </s> <s xml:id="echoid-s167" xml:space="preserve">dunque la diſtanza di 5, & </s> <s xml:id="echoid-s168" xml:space="preserve">a, è la <lb/>trenteſima di tutta la RS, e così la centeſima di AE.</s> <s xml:id="echoid-s169" xml:space="preserve"/> </p> <div xml:id="echoid-div11" type="float" level="2" n="1"> <figure xlink:label="fig-0021-01" xlink:href="fig-0021-01a"> <image file="0021-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0021-01"/> </figure> </div> <p> <s xml:id="echoid-s170" xml:space="preserve">Ora per prouare ſe ſia giuſta la diuiſione, ſi prenda R a, e <lb/>ſe replicata cade nel 60. </s> <s xml:id="echoid-s171" xml:space="preserve">ella è giuſta, e ſegnarà tutti li punti <lb/>numerati dal 6. </s> <s xml:id="echoid-s172" xml:space="preserve">Così preſa 5 b ſi replichi, e ſe è giuſta, comin-<lb/>ciando da A centro, caderà nel 70. </s> <s xml:id="echoid-s173" xml:space="preserve">& </s> <s xml:id="echoid-s174" xml:space="preserve">in tutti li numeri molti-<lb/>plici di 7. </s> <s xml:id="echoid-s175" xml:space="preserve">Così 10 c, darà 8, & </s> <s xml:id="echoid-s176" xml:space="preserve">i ſuoi moltiplici, cadendo pre-<lb/>ciſamente in 80: </s> <s xml:id="echoid-s177" xml:space="preserve">e così anche 15 d, darà 9. </s> <s xml:id="echoid-s178" xml:space="preserve">& </s> <s xml:id="echoid-s179" xml:space="preserve">i ſuoi moltipli-<lb/>ci, cadendo nel 90. </s> <s xml:id="echoid-s180" xml:space="preserve">Et in queſta maniera traportando li ſu-<lb/>detti interualli non ſolo dalli punti delle decime, mà anche <lb/>dalle loro metà, come da 5. </s> <s xml:id="echoid-s181" xml:space="preserve">15. </s> <s xml:id="echoid-s182" xml:space="preserve">25. </s> <s xml:id="echoid-s183" xml:space="preserve">&</s> <s xml:id="echoid-s184" xml:space="preserve">c. </s> <s xml:id="echoid-s185" xml:space="preserve">ſi verranno à ſegnar <lb/>tutti i punti della linea AE con molta aggiuſtatezza, ò ſe furo-<lb/>no già ſegnati, ſi conoſcerà la buona diuiſione.</s> <s xml:id="echoid-s186" xml:space="preserve"/> </p> <pb o="10" file="0022" n="23" rhead="CAPO II."/> </div> <div xml:id="echoid-div13" type="section" level="1" n="10"> <head xml:id="echoid-head14" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head15" xml:space="preserve">Come ſi troua la parte determinata in numeri <lb/>d’ vna linea data.</head> <p> <s xml:id="echoid-s187" xml:space="preserve">SIa data la linea MN longhezza della Cortina in vn diſ-<lb/>ſegno di qualche Fortezza, e volendoſi prendeie la dif-<lb/>feſa dal quinto della Cortina, ſi cerchi la ſua <lb/> <anchor type="figure" xlink:label="fig-0022-01a" xlink:href="fig-0022-01"/> quinta parte. </s> <s xml:id="echoid-s188" xml:space="preserve">Allarghiſi lo Stromento in mo-<lb/>do, che la diſtanza 100. </s> <s xml:id="echoid-s189" xml:space="preserve">100. </s> <s xml:id="echoid-s190" xml:space="preserve">ſia la MN: </s> <s xml:id="echoid-s191" xml:space="preserve">poi <lb/>eſſendo 20. </s> <s xml:id="echoid-s192" xml:space="preserve">la quinta parte di 100. </s> <s xml:id="echoid-s193" xml:space="preserve">ſi pigli la <lb/>diſtanza 20. </s> <s xml:id="echoid-s194" xml:space="preserve">20, ritenendo la ſteſſa apertura <lb/>dello ſtromento, e queſta ſara la MO quinta <lb/>parte cercata di MN. </s> <s xml:id="echoid-s195" xml:space="preserve">Mà ſe la linea foſſe tale, <lb/>che la parte cercata foſſe molto piccola, ſi <lb/>prenda l’interuallo del reſto: </s> <s xml:id="echoid-s196" xml:space="preserve">come nella figu-<lb/>ra antecedente; </s> <s xml:id="echoid-s197" xml:space="preserve">ſe della linea RS ſi deſidera <lb/>la parte trenteſima, s’applichi RS all’ interual-<lb/>lo 30. </s> <s xml:id="echoid-s198" xml:space="preserve">30. </s> <s xml:id="echoid-s199" xml:space="preserve">& </s> <s xml:id="echoid-s200" xml:space="preserve">à quell’a pertura ſi prenda l’inter-<lb/>uallo 29. </s> <s xml:id="echoid-s201" xml:space="preserve">29. </s> <s xml:id="echoid-s202" xml:space="preserve">& </s> <s xml:id="echoid-s203" xml:space="preserve">il Compaſſo tagliando 29 par-<lb/>ti della linea RS, laſcierà vna trenteſima. </s> <s xml:id="echoid-s204" xml:space="preserve">Preſo <lb/>dipoi l’interuallo 28 28. </s> <s xml:id="echoid-s205" xml:space="preserve">e queſto applicato al-<lb/>la linea RS, laſcierà due trenteſime, e così di <lb/>mano in mano. </s> <s xml:id="echoid-s206" xml:space="preserve">Se bene fatta la prima ope-<lb/>ratione, ſe l’interuallo Si è di parti 29, vgua-<lb/>le à queſto ſia R e, ſimilmente di parti 29: </s> <s xml:id="echoid-s207" xml:space="preserve">la <lb/>diſtanza i e è di particelle 28: </s> <s xml:id="echoid-s208" xml:space="preserve">queſta dunque <lb/>applicata da S, darà S u parti 28: </s> <s xml:id="echoid-s209" xml:space="preserve">così u e ſarà <lb/>parti27.</s> <s xml:id="echoid-s210" xml:space="preserve"><unsure/> e perciò queſta applicata da S, darà S @ <lb/>di parti 27; </s> <s xml:id="echoid-s211" xml:space="preserve">e cosi dell’altre.</s> <s xml:id="echoid-s212" xml:space="preserve"/> </p> <div xml:id="echoid-div13" type="float" level="2" n="1"> <figure xlink:label="fig-0022-01" xlink:href="fig-0022-01a"> <image file="0022-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0022-01"/> </figure> </div> <pb o="11" file="0023" n="24" rhead="Linea Aritmetica."/> <p> <s xml:id="echoid-s213" xml:space="preserve">Che ſe ſi cercaſſe tal parte, la quale non foſſe preciſamente <lb/>nel numero 100; </s> <s xml:id="echoid-s214" xml:space="preserve">pigliſi vn’altro numero, che habbia tal par-<lb/>e, e ſopra di quello ſi ponga la longhezza MN, e poi il nu-<lb/>mero, che ſarà la parte cercata del numero preſo, darà la lon-<lb/>ghezza cercata. </s> <s xml:id="echoid-s215" xml:space="preserve">Per cagion d’eſſempio ſi deſideri della data <lb/>linea MN vna parte, che ſia quattro vndecime. </s> <s xml:id="echoid-s216" xml:space="preserve">Non ſi po<unsure/>-<lb/>tendo il 100 diuidere giuſtamente per 11, prendo vn nume. <lb/></s> <s xml:id="echoid-s217" xml:space="preserve">10<unsure/> qualſiuoglia, che ſia numerato dall’ 11; </s> <s xml:id="echoid-s218" xml:space="preserve">eſia 88. </s> <s xml:id="echoid-s219" xml:space="preserve">Apro lo <lb/>Stromento in modo, che MN ſia la diſtanza di 88; </s> <s xml:id="echoid-s220" xml:space="preserve">e perche <lb/>l’vndecima parte di 88 è 8, queſto replico quattro volte, e <lb/>32 ſono quattro vndecime: </s> <s xml:id="echoid-s221" xml:space="preserve">piglio dunque la diſtanza 32. </s> <s xml:id="echoid-s222" xml:space="preserve"><lb/>32, & </s> <s xml:id="echoid-s223" xml:space="preserve">è MR quattro vndecime di MN. </s> <s xml:id="echoid-s224" xml:space="preserve">Vn’ altra maniera <lb/>di trouar vna parte aſſai piccola, vedrai nel capo 7, queſtio-<lb/>ne 3. </s> <s xml:id="echoid-s225" xml:space="preserve">nel fine.</s> <s xml:id="echoid-s226" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s227" xml:space="preserve">Di quì ſi vede, che data vna linea maggiore, ſe ne può tro-<lb/>uar vna minore in qualſiuoglia proportione di quelle, che <lb/>con numeri ſi ponno eſprimere, pigliando dentro à 100 dne <lb/>numeri nella data proportione; </s> <s xml:id="echoid-s228" xml:space="preserve">& </s> <s xml:id="echoid-s229" xml:space="preserve">applicata la linea data al <lb/>maggiore di queſti due numeri, il minor numero darà la line@ <lb/>minore cercata. </s> <s xml:id="echoid-s230" xml:space="preserve">E ſe per auuentura li due numeri eſprimen-<lb/>ti la proportione foſſero tali, che eccedeſlero il 100, ſi ridu-<lb/>cano à centeſime; </s> <s xml:id="echoid-s231" xml:space="preserve">che per l’operatione Mecanica vi ſarà po-<lb/>chiſſimo sbaglio. </s> <s xml:id="echoid-s232" xml:space="preserve">Il che ſi fà (per ricordarlo alli meno prat-<lb/>tici) moltiplicando per 100 il Conſeguente della Proportio-<lb/>ne, & </s> <s xml:id="echoid-s233" xml:space="preserve">diuidendo il prodotto per l’Antecedente; </s> <s xml:id="echoid-s234" xml:space="preserve">e s’haurà la <lb/>proportione eſpreſia con due noui termini, il maggior de’ <lb/>quali ſarà il 100. </s> <s xml:id="echoid-s235" xml:space="preserve">& </s> <s xml:id="echoid-s236" xml:space="preserve">il minore, che ſi cerca, ſarà il Quotiente, <lb/>che riſulta da cotal diuiſione. </s> <s xml:id="echoid-s237" xml:space="preserve">Sia per cagion d’ eſſempio la <lb/>medeſima linea MN, e ſe ne cerchi vna minore, ò parte di <lb/>MN in tal proportione, che ſiano come 3, 22 {8/50}, che è quan- <pb o="12" file="0024" n="25" rhead="CAPO II."/> to dire come 150 à 108. </s> <s xml:id="echoid-s238" xml:space="preserve">Moltiplico 108 per 100, & </s> <s xml:id="echoid-s239" xml:space="preserve">è <lb/>10800, queſto diuido per 150, ene viene 72. </s> <s xml:id="echoid-s240" xml:space="preserve">Applico dun-<lb/>que la linea data al 100. </s> <s xml:id="echoid-s241" xml:space="preserve">100, e la diſtanza 72. </s> <s xml:id="echoid-s242" xml:space="preserve">72, mi dà <lb/>MX, che è quello, che ſi cercaua. </s> <s xml:id="echoid-s243" xml:space="preserve">In queſto eſſempio però, <lb/>perche 150, e 108 ſono a mbidue pari, baſta diuidere ciaſcu-<lb/>no per metà, e ne’ numeri 75, e 54 s’eſprime la ſteſſa propor-<lb/>tione; </s> <s xml:id="echoid-s244" xml:space="preserve">onde applicando MN à 75. </s> <s xml:id="echoid-s245" xml:space="preserve">75. </s> <s xml:id="echoid-s246" xml:space="preserve">la diſtanza 54. </s> <s xml:id="echoid-s247" xml:space="preserve">54 da-<lb/>rà l’iſteſſa MX.</s> <s xml:id="echoid-s248" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s249" xml:space="preserve">Mà ſe la linea data foſſ@ così lunga, che ò non haueſſimo <lb/>Compaſſo così grande, che baſtaſſe à prenderla tutta, per ap-<lb/>plicarla al noſtro stromento, ò lo Stromento foſſe così picco-<lb/>lo, che allargato non poteſſe capire tutta la linea data; </s> <s xml:id="echoid-s250" xml:space="preserve">Al-<lb/>lhora vna cotal linea ſi diuida per mezo, e ſe ancora riuſciſſe <lb/>troppo lunga, la metà ſi diuida di nuouo per mezo, e s’haurà <lb/>la quarta parte, e queſta quarta parte s’applichi allo Stro-<lb/>mento, come s ella foſſe la linea propoſta, e ſi cerchi la parte <lb/>determinata come ſopra; </s> <s xml:id="echoid-s251" xml:space="preserve">e poi queſta replicata tante volte, in <lb/>quante parti è ſtata diuiſa la linea data, ſarà la parte, che ſi de-<lb/>ſidera: </s> <s xml:id="echoid-s252" xml:space="preserve">onde ſe ſolo ſi diuiſe in due queſta parte trouata, ſi rad-<lb/>doppia, e ſe quella fù diuiſa in quattro, queſta ſi replica quat-<lb/>tro volte, perche le parti con i moltiplici han la ſteſſa pro-<lb/>portione (per la 15. </s> <s xml:id="echoid-s253" xml:space="preserve">del 5.) </s> <s xml:id="echoid-s254" xml:space="preserve">Così figurandoci vna linea lunga <lb/>300 determinate particelle, ſi prende la ſua quarta parte, che <lb/>ſia 75. </s> <s xml:id="echoid-s255" xml:space="preserve">e s’applichi allo Stromento 75. </s> <s xml:id="echoid-s256" xml:space="preserve">75, e ſe ſi vogliono <lb/>due terzi di tutta la data linea (che ſono 200) ſi prendano li <lb/>due terzi di 75, che ſono 50. </s> <s xml:id="echoid-s257" xml:space="preserve">e perche la linea tutta fù diuiſa <lb/>in quattro, ſi replichi queſta linea trouata tra 50. </s> <s xml:id="echoid-s258" xml:space="preserve">50 quat-<lb/>tro volte, e ſaranno appunto li due terzi della linea data, <lb/>cioè 200; </s> <s xml:id="echoid-s259" xml:space="preserve">poiche come 50 à 75, così 200 à 300.</s> <s xml:id="echoid-s260" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s261" xml:space="preserve">Che ſe dalla linea data ſi doueſſe cauar vna parte denomi- <pb o="13" file="0025" n="26" rhead="Linea Aritmeticæ."/> nata da vn numero Primo maggiore del 100, che è il ma ſſi-<lb/>mo della linea dello Stromento, tiriſi vn’ altra linea arbitra-<lb/>ria, che faccia angolo con la linea data; </s> <s xml:id="echoid-s262" xml:space="preserve">& </s> <s xml:id="echoid-s263" xml:space="preserve">in quella prendaſi <lb/>ſeparatamente l’ecceſſo ſopra il 100, e poi il 100, con hauer <lb/>data allo Stromento quell’apertura, che più piacerà. </s> <s xml:id="echoid-s264" xml:space="preserve">Dipoi <lb/>congionti gli eſtremi con vna linea, ſi tiri à queſta dall’ eſtre-<lb/>mo della prima diuiſione vna parallela; </s> <s xml:id="echoid-s265" xml:space="preserve">& </s> <s xml:id="echoid-s266" xml:space="preserve">ſi hauerà l’intento. <lb/></s> <s xml:id="echoid-s267" xml:space="preserve">Sia data la li-<lb/> <anchor type="figure" xlink:label="fig-0025-01a" xlink:href="fig-0025-01"/> nea BC della <lb/>quale diuiſa in <lb/>parti 111, ſi <lb/>vogliano 11 <lb/>parti. </s> <s xml:id="echoid-s268" xml:space="preserve">Tiriſi ad <lb/>arbitrio la li-<lb/>nea CA, & </s> <s xml:id="echoid-s269" xml:space="preserve">a-<lb/>perto arbitra-<lb/>riamente lo Stromento, prendaſi l’interuallo 11. </s> <s xml:id="echoid-s270" xml:space="preserve">11, e ſia <lb/>CE: </s> <s xml:id="echoid-s271" xml:space="preserve">@ndi la diſtanza 100. </s> <s xml:id="echoid-s272" xml:space="preserve">100, e ſia EA. </s> <s xml:id="echoid-s273" xml:space="preserve">Dunque CA è di <lb/>parti 111. </s> <s xml:id="echoid-s274" xml:space="preserve">Congiongaſi AB, & </s> <s xml:id="echoid-s275" xml:space="preserve">à queſta linea ſi tiri parallela <lb/>la EF; </s> <s xml:id="echoid-s276" xml:space="preserve">e così delle 111 parti ditutta la BC, ne ſaranno 11 la <lb/>parte CF: </s> <s xml:id="echoid-s277" xml:space="preserve">poiche come CE à CA, così CF a CB. </s> <s xml:id="echoid-s278" xml:space="preserve">L’iſteſſo <lb/>s’intenda, ſe l’ecceſſo ſopra 100 non doueſſe eſſere la parte <lb/>cercata; </s> <s xml:id="echoid-s279" xml:space="preserve">mà per eſempio ſi voleſſero 58 delle 111. </s> <s xml:id="echoid-s280" xml:space="preserve">Fatta <lb/>CA di 111, prendaſi in eſſa CH 58 parti come ſopra, e tirata <lb/>la parallela HI, ſi hauerà l’intento, cioè IC 58.</s> <s xml:id="echoid-s281" xml:space="preserve"/> </p> <div xml:id="echoid-div14" type="float" level="2" n="2"> <figure xlink:label="fig-0025-01" xlink:href="fig-0025-01a"> <image file="0025-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0025-01"/> </figure> </div> <p> <s xml:id="echoid-s282" xml:space="preserve">Mà forſi per gli Artefici, che per lo più cercano vna parte <lb/>aliquota, ò più parti aliquote non maggiori delle decime, <lb/>tornarà commodo vn’altra ſorte di linea Aritmetica, in cui <lb/>ſiano notate le parti aliquote ſin alle decime; </s> <s xml:id="echoid-s283" xml:space="preserve">come ſe ſi pren-<lb/>da la ST, & </s> <s xml:id="echoid-s284" xml:space="preserve">in eſſa ſi noti la ſua metà, il terzo, il quarto, e <pb o="14" file="0026" n="27" rhead="CAPO II."/> così di mano in mano ſin alla decima; </s> <s xml:id="echoid-s285" xml:space="preserve">e per maggior com-<lb/>modità dell operare, parimente, ſi notino le frattioni non <lb/>equiualenti ad vn’altra parte aliquota, ò ad v@’ altra frattio-<lb/>ne; </s> <s xml:id="echoid-s286" xml:space="preserve">e queſte frattioni ſi notino al ſuo punto con due numeri, <lb/>cioè col ſuo Numeratore, e ſuo De-<lb/>nominatore: </s> <s xml:id="echoid-s287" xml:space="preserve">Così ſi deue notare; </s> <s xml:id="echoid-s288" xml:space="preserve">mà <lb/> <anchor type="figure" xlink:label="fig-0026-01a" xlink:href="fig-0026-01"/> non {6/10}, che à quella ſono vguali; </s> <s xml:id="echoid-s289" xml:space="preserve">mà <lb/>non {4/6}, ò,, e così de gli altri. </s> <s xml:id="echoid-s290" xml:space="preserve">Solo <lb/>deue auuertirſi di mettere li numeri <lb/>con tal diſtintione, che non generino <lb/>confuſione, onde vno ſi prenda per <lb/>vn’altro. </s> <s xml:id="echoid-s291" xml:space="preserve">Nella ſteſſa maniera ſia diui-<lb/>ſa, e notata la SV totalmente vguale <lb/>alla ST. </s> <s xml:id="echoid-s292" xml:space="preserve">Non conſegliarei però di <lb/>mettere queſta linea (la quale però <lb/>chiamaſi Diuiſoria) ſopra dello Stro-<lb/>mento, in cui deuono metterſi le al-<lb/>tre linee, delle quali ſi dirà più auan-<lb/>ti; </s> <s xml:id="echoid-s293" xml:space="preserve">à fine che li numeri di queſta line@ <lb/>non ſi confondano con quelli d’altre <lb/>linee vicine; </s> <s xml:id="echoid-s294" xml:space="preserve">Mà ſarei di parere, che <lb/>ſi metteſſe queſta in vno Stromento <lb/>particolare, maſſime, che gli Artefici <lb/>più ordinarij non hanno biſogno di <lb/>quell’altre linee, e di queſta puonno <lb/>grandemente giouarſi.</s> <s xml:id="echoid-s295" xml:space="preserve"/> </p> <div xml:id="echoid-div15" type="float" level="2" n="3"> <figure xlink:label="fig-0026-01" xlink:href="fig-0026-01a"> <image file="0026-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0026-01"/> </figure> </div> <p> <s xml:id="echoid-s296" xml:space="preserve">L’vſo di queſta linea è manifeſto; <lb/></s> <s xml:id="echoid-s297" xml:space="preserve">perche poſta la linea da diuiderſi, ò di <lb/>cui ſi voglia vna parte determinata, <lb/>nell’ eſtremità alli punti 1. </s> <s xml:id="echoid-s298" xml:space="preserve">1, l’inter- <pb o="15" file="0027" n="28" rhead="Linea Aritmetica."/> uallo corriſpondente alla parte cercata ſubito la darà. </s> <s xml:id="echoid-s299" xml:space="preserve">Che ſe <lb/>la linea data ſoſse troppo lunga, ſi tagli per mezo, ò in quat-<lb/>tro parti, e conla meta, ò il quarto applicato alli punti 1. </s> <s xml:id="echoid-s300" xml:space="preserve">1. <lb/></s> <s xml:id="echoid-s301" xml:space="preserve">ſi operi come ſopra; </s> <s xml:id="echoid-s302" xml:space="preserve">poiche la parte trouata dourà raddop-<lb/>piarſi, ò quadruplicarſi per hauere la parte da principio cer-<lb/>cata. </s> <s xml:id="echoid-s303" xml:space="preserve">Così potrebbono i Legnaiuoli in vn gran Compaſſo <lb/>dilegno, computando le ſue punte nella lunghezza, deſcri-<lb/>uere le ſudette parti; </s> <s xml:id="echoid-s304" xml:space="preserve">perchecon detto Compaſſo preſa la <lb/>lunghezza della linea da diuiderſi, ſubito gl’interualli notati <lb/>sù le gambe del Compaſſo lot darebbono la parte cercata.</s> <s xml:id="echoid-s305" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s306" xml:space="preserve">Potrà anche queſta linea Diuiſoria ſeruire à Moltiplicar, e <lb/>Diuidere qualſiuoglia numero, il cui Moltiplicatore, ò Diui-<lb/>ſore ſia vn numero in eſſa notato. </s> <s xml:id="echoid-s307" xml:space="preserve">L’operatione è fondata ſo-<lb/>pra la verità nota à gli Aritmetici, che nella moltiplicatione <lb/>l’Vnità al Moltiplicatore hà la ſteſſa proportione, che il Mol-<lb/>tiplicato al Prodotto, e nella Diuiſione l’iſteſſa proportione <lb/>ha il Diuiſore all’Vnità, che ha<unsure/> il Diuiſo al Quotiente; </s> <s xml:id="echoid-s308" xml:space="preserve">eſſendo <lb/>manifeſto, che tante volte l’vnità è contenuta dal Moltiplica. <lb/></s> <s xml:id="echoid-s309" xml:space="preserve">tore, ò dal Diuiſore, quante volte il Moltiplicato è contenu. </s> <s xml:id="echoid-s310" xml:space="preserve"><lb/>to dal Prodotto, ò il Quotiente dal Diuiſo. </s> <s xml:id="echoid-s311" xml:space="preserve">Or habbiaſi vna <lb/>Scala di parti minutiſſime, la quale à molti vſi può ſeruire, & </s> <s xml:id="echoid-s312" xml:space="preserve"><lb/>in eſſa ſi prenda con vn Compaſſo vn numero di particelle <lb/>corriſpondente al numero dato da moltiplicarſi: </s> <s xml:id="echoid-s313" xml:space="preserve">ſe il Molti-<lb/>plicatore è numero intiero, quella grandezza di linea preſa <lb/>col Compaſſo, ſi applichi all’ interuallo della parte aliquor<unsure/>a <lb/>denominata da tal numero; </s> <s xml:id="echoid-s314" xml:space="preserve">come ſe foſſe 7, ſi applichi alli <lb/>Punti 7. </s> <s xml:id="echoid-s315" xml:space="preserve">7. </s> <s xml:id="echoid-s316" xml:space="preserve">Dipoi prenda@i nell’eſtremità l’interuallo 1. </s> <s xml:id="echoid-s317" xml:space="preserve">1, & </s> <s xml:id="echoid-s318" xml:space="preserve"><lb/>applicato alla Scala ſodetta, ſi trouarà nel numero delle par-<lb/>ticelle eſpreſſo il numero Prodotto, eſſendo che il primo in-<lb/>terual@o al ſecondo, per la coſtruttione, è come {1/7} ad 1, cioè, <pb o="16" file="0028" n="29" rhead="CAPO II."/> come 1 à 7: </s> <s xml:id="echoid-s319" xml:space="preserve">dunque le particelle applicatelal primo interual-<lb/>lo ſono come 1 à 7 in riguardo delle particelle trouate col ſe-<lb/>condo interuallo, cioè il Moltiplicato al Prodotto. </s> <s xml:id="echoid-s320" xml:space="preserve">Così do-<lb/>uendoſi moltiplicar 14 per 7; </s> <s xml:id="echoid-s321" xml:space="preserve">piglio nella Scala 14 particel-<lb/>le, & </s> <s xml:id="echoid-s322" xml:space="preserve">allargo lo Stromento tanto, che le poſſi applicare al 7. <lb/></s> <s xml:id="echoid-s323" xml:space="preserve">7; </s> <s xml:id="echoid-s324" xml:space="preserve">quindi prendo l’interuallo 1. </s> <s xml:id="echoid-s325" xml:space="preserve">1, & </s> <s xml:id="echoid-s326" xml:space="preserve">applicatolo alla Scala <lb/>trouo parti 08; </s> <s xml:id="echoid-s327" xml:space="preserve">e tanto ſi fà moltiplicando 14. </s> <s xml:id="echoid-s328" xml:space="preserve">pe 7.</s> <s xml:id="echoid-s329" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s330" xml:space="preserve">Mà ſe il Moltiplicatore foſſe vnode’<unsure/> otti notat sù lo Stro-<lb/>me ito, deue operarſi differentemente; </s> <s xml:id="echoid-s331" xml:space="preserve">cioè il numero Molti.</s> <s xml:id="echoid-s332" xml:space="preserve"><unsure/> <lb/>plicando ſi applica alli punti 1. </s> <s xml:id="echoid-s333" xml:space="preserve">1; </s> <s xml:id="echoid-s334" xml:space="preserve">e l’interuallo del rotto da-<lb/>to darà il Prodotto. </s> <s xml:id="echoid-s335" xml:space="preserve">Così volendo moltiplicar l’iſteſſo 14 per <lb/>{6/7}, applico il numero dato all’interuallo eſtremo 1. </s> <s xml:id="echoid-s336" xml:space="preserve">1; </s> <s xml:id="echoid-s337" xml:space="preserve">e l’inter-<lb/>uallo {6/7}. </s> <s xml:id="echoid-s338" xml:space="preserve">{6/7} darà nella ſcala 12, che è il numero Prodotto, eſ-<lb/>ſendo come l’V nità à {6/7}, così 14 à 12.</s> <s xml:id="echoid-s339" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s340" xml:space="preserve">Similmente nella Diuiſione prendo nella Scala il numero <lb/>dato da diuiderſi, & </s> <s xml:id="echoid-s341" xml:space="preserve">allargo lo Stromento sì, che capiſca trà <lb/>l’eſtremità 1. </s> <s xml:id="echoid-s342" xml:space="preserve">1; </s> <s xml:id="echoid-s343" xml:space="preserve">dipoi all’interuallo corriſpondente al nume-<lb/>ro intiero del Diuiſore trouo la linea, che sù la Scala dà il <lb/>Quotienre. </s> <s xml:id="echoid-s344" xml:space="preserve">Habbiaſi à diuidere 176 per 8: </s> <s xml:id="echoid-s345" xml:space="preserve">Nella ſcala pren-<lb/>do 176, e l’applico allo Stromento in 1. </s> <s xml:id="echoid-s346" xml:space="preserve">1: </s> <s xml:id="echoid-s347" xml:space="preserve">all’interuallo 8. <lb/></s> <s xml:id="echoid-s348" xml:space="preserve">8; </s> <s xml:id="echoid-s349" xml:space="preserve">trouo tal linea, che sù la Scala mi dà 22: </s> <s xml:id="echoid-s350" xml:space="preserve">poiche come 1 <lb/>ad {1/8}, cioè come il Diuiſore 8 à 1, cosi il Diuiſo 176 à 22 <lb/>Quotiente.</s> <s xml:id="echoid-s351" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s352" xml:space="preserve">Mà ſe il Diuiſore foſſe vn Rotto delli notati, à quell’inter-<lb/>uallo douria applicarſi il numero Diuiſo, perche l’interuallo <lb/>1. </s> <s xml:id="echoid-s353" xml:space="preserve">1 daria il Quotiente cercato, à cui il diuiſo hauerebbe la <lb/>ſteſſa proportione, che hà il Diuiſore all’ Vnità. </s> <s xml:id="echoid-s354" xml:space="preserve">Habbiaſi à <lb/>diuidere 176 per {2/3}: </s> <s xml:id="echoid-s355" xml:space="preserve">preſa dalla Scala la lunghezza di parti <lb/>176, l’applico alli punti {2/3}. </s> <s xml:id="echoid-s356" xml:space="preserve"><emph style="sub">3</emph><unsure/>: </s> <s xml:id="echoid-s357" xml:space="preserve">dipoi l’interuallo 1. </s> <s xml:id="echoid-s358" xml:space="preserve">1, tra-<lb/>portato sù la Scala darà il Quotiente 264: </s> <s xml:id="echoid-s359" xml:space="preserve">poiche veramen- <pb o="17" file="0029" n="30" rhead="Linea Aritmetica."/> teil Rotto {2/3} ſi contiene 264 volte nel numero 176, e co-<lb/>me il Diuiſore {2/3} all’ vnità, così il Diuiſo 176, al Quotien-<lb/>te 264.</s> <s xml:id="echoid-s360" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div17" type="section" level="1" n="11"> <head xml:id="echoid-head16" xml:space="preserve">QVESTIONE SECONDA. <lb/>Come ad una linea data ſi troua una maggiore nella proportione <lb/>determinata in numeri.</head> <p> <s xml:id="echoid-s361" xml:space="preserve">LI due numeri, co’quali s’eſprime la proportione deter-<lb/>minata ſe ſoſſero aſſai piccioli, ſi moltiplichino per <lb/>qualſiuoglia numero tale, che il prodotto dalla moltiplica-<lb/>tione per il maggiore non ecceda 100. </s> <s xml:id="echoid-s362" xml:space="preserve">Poi ſi piglino queſti <lb/>due prodotti come Antecedente, e Conſeguente della Pro-<lb/>portione, e la linea data s’applichi nello Stromento al nume-<unsure/> <lb/>ro minore, poicheil numero maggiore darà la lunghezza del-<lb/>la linea cercata. </s> <s xml:id="echoid-s363" xml:space="preserve">Sia la figura prima della queſtione prece-<lb/>dente, data la linea H, la quale debba ad vn’altra linea hauer <lb/>la proportione di 3 à 7. </s> <s xml:id="echoid-s364" xml:space="preserve">Moltiplico così il 3 come il 7 per 10, <lb/>e ſono 30, e 70. </s> <s xml:id="echoid-s365" xml:space="preserve">Allargo lo Stromento, & </s> <s xml:id="echoid-s366" xml:space="preserve">applico la linea H <lb/>alla diſtanza 30, 30; </s> <s xml:id="echoid-s367" xml:space="preserve">e poi ritenendo lo Stromento così allar-<lb/>gato, prendo la diſtanza 70. </s> <s xml:id="echoid-s368" xml:space="preserve">70, e ſarà la linea MN cercata. <lb/></s> <s xml:id="echoid-s369" xml:space="preserve">In queſta maniera ſe foſſe data in diſſegno vna fronte huma-<lb/>na, quanto è dal mezo doue finiſcono le ſopraciglia ſin alla <lb/>radice de’capegli, ſi trouerà la lunghezza della faccia, piglian-<lb/>do vna linea trè volte maggiore: </s> <s xml:id="echoid-s370" xml:space="preserve">E perche la faccia è la de-<lb/>cima parte, come ſcriue Vitruuio lib. </s> <s xml:id="echoid-s371" xml:space="preserve">3. </s> <s xml:id="echoid-s372" xml:space="preserve">cap, 1. </s> <s xml:id="echoid-s373" xml:space="preserve">ò come altri <lb/>vogliono, la nona parte di tutta la giuſta ſtatura humana, data <lb/>la fronte ſi pigli vna lina, che ſia 30, ouero 27 volte maggio. </s> <s xml:id="echoid-s374" xml:space="preserve"><lb/>re, e ſi haurà l’altezza del corpo proportionato.</s> <s xml:id="echoid-s375" xml:space="preserve"/> </p> <pb o="18" file="0030" n="31" rhead="CAPO II."/> <p> <s xml:id="echoid-s376" xml:space="preserve">Che ſe la linea dataf ſſe così grande, che non capiſſe com-<lb/>modamente nell’apertura dello Stromento, operiſi come s’è <lb/>detto nel fine della queſtione precedente; </s> <s xml:id="echoid-s377" xml:space="preserve">cioè pigliſi vna ſua <lb/>pa@@e aliquor<unsure/>a, econ eſſa s’operi al modo detto; </s> <s xml:id="echoid-s378" xml:space="preserve">poiche que-<lb/>ſta linea trouata, e replicata tante volte, in quante parti la li-<lb/>nea data fù diuiſa, ſarà appunto la linea cercata.</s> <s xml:id="echoid-s379" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s380" xml:space="preserve">Se finalmente la proportione foſle determinata in numeri <lb/>ambidue maggiori di 100. </s> <s xml:id="echoid-s381" xml:space="preserve">riducaſi à denominatione di cen-<lb/>teſime, facendo come il Conſeguente maggiore all’ Antece-<lb/>dente, minore nella Proportione data, così 100 ad vn’ altro <lb/>numero, e con queſti due vltimi s’operi, applicando la linea <lb/>data al numero minore trouato, e la diſtanza 100. </s> <s xml:id="echoid-s382" xml:space="preserve">100, darà <lb/>la linea cercata. </s> <s xml:id="echoid-s383" xml:space="preserve">Mà ſe de’ numeri eſprimenti la proportio-<lb/>ne, ſol’il maggiore eccedeſſe 100, baſterà, applicata la linea <lb/>data al numero minore, pigliare per la linea cercata prima la <lb/>diſtanza 100. </s> <s xml:id="echoid-s384" xml:space="preserve">100, poi la diſtanza del reſto del numero, e di <lb/>queſte due diſtanze farne vna ſola linea.</s> <s xml:id="echoid-s385" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s386" xml:space="preserve">Così per eſſempio habbiamo dato il Semidiametro d’vn <lb/>cerchio, e vogliamo vna linea retta proſſimamente vguale al-<lb/>la Semicirconferenza. </s> <s xml:id="echoid-s387" xml:space="preserve">Sappiamo per la Dottrina d’Archi-<lb/>mede, che la Circonferenza al Diametro (l’iſteſſo è delle loro <lb/>metà) è minore, che la tripla è dieci ſettanteſime, mà maggio-<lb/>re, che la t@ipla è dieci ſettantuneſime. </s> <s xml:id="echoid-s388" xml:space="preserve">Sì che la prima pro-<lb/>portione di 7 à 22, la ſeconda di 71 à 223. </s> <s xml:id="echoid-s389" xml:space="preserve">Sia dunque il ſe-<lb/>midiametro dato la linea B, la quale applicata al 7. </s> <s xml:id="echoid-s390" xml:space="preserve">7, ouero <lb/>14. </s> <s xml:id="echoid-s391" xml:space="preserve">14, darà nelli 22. </s> <s xml:id="echoid-s392" xml:space="preserve">22, ouero 44. </s> <s xml:id="echoid-s393" xml:space="preserve">44, la linea C vn poco <lb/>maggiore della vera Semicirconferenza. </s> <s xml:id="echoid-s394" xml:space="preserve">Per hauer poil’al-<lb/>tra proportione applichiſi la linea B alli 71. </s> <s xml:id="echoid-s395" xml:space="preserve">71, e poi per li <lb/>223, pigliſi due volte 100. </s> <s xml:id="echoid-s396" xml:space="preserve">100, e poi 23. </s> <s xml:id="echoid-s397" xml:space="preserve">23. </s> <s xml:id="echoid-s398" xml:space="preserve">e ſarà vna li-<lb/>nea di 223 particelle, delle quali B ne hà 71, così poco dif- <pb o="19" file="0031" n="32" rhead="Linea Aritmetica."/> ferente dalla linea C, che riuſcirà inſenſibile la <lb/> <anchor type="figure" xlink:label="fig-0031-01a" xlink:href="fig-0031-01"/> differenza. </s> <s xml:id="echoid-s399" xml:space="preserve">Mà ſe la linea B foſſe ſtata mol-<lb/>to maggiore, allhora ſaria riuſcita queſta ſe-<lb/>conda linea minore di C, con differenza tale, <lb/>che per hauer la Semicirconferenza proſſima <lb/>alla vera, ſi douria à queſta minore di C ag-<lb/>giungerela metà della accennata differenza.</s> <s xml:id="echoid-s400" xml:space="preserve"/> </p> <div xml:id="echoid-div17" type="float" level="2" n="1"> <figure xlink:label="fig-0031-01" xlink:href="fig-0031-01a"> <image file="0031-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0031-01"/> </figure> </div> </div> <div xml:id="echoid-div19" type="section" level="1" n="12"> <head xml:id="echoid-head17" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head18" xml:space="preserve">Come ſi troui vna Quarta Proportionale, <lb/>e ſi continui vna Proportione.</head> <p> <s xml:id="echoid-s401" xml:space="preserve">QVando ſon date trè linee, & </s> <s xml:id="echoid-s402" xml:space="preserve">alla Terza <lb/>ſi cerca vna Quarta, che ſia nella pro-<lb/>portione della Prima alla Seconda, <lb/>ſenza che ſia eſpreſſa in numeri la proportio-<lb/>ne, ſi traſporta la Prima dal centro dello Stro-<lb/>mento A ſopral’vno, e l’altro lato; </s> <s xml:id="echoid-s403" xml:space="preserve">eſe non <lb/>cade preciſamente ſopra alcuno de’ <unsure/>punti ſe-<lb/>gnati, baſta leggiermente con la punta del Compaſſ<unsure/>o tagliar <lb/>à trauerſo la linea tra l’vn punto, el’altro, tanto che ſi poſſa <lb/>riconoſcere. </s> <s xml:id="echoid-s404" xml:space="preserve">Poi s’allarghi lo Stromento tanto, che trà li <lb/>due punti già ſegnati con la punta del Compaſſo capiſca la fe-<lb/>conda delle linee date. </s> <s xml:id="echoid-s405" xml:space="preserve">Finalmente la Terza ſi traſporti ſi-<lb/>milmente dal centro A ſopra l’vno, el’altro lato, e ſi ſegni il <lb/>ſuo termine; </s> <s xml:id="echoid-s406" xml:space="preserve">poiche la diſtanza trà queſti due punti vltima-<lb/>mente ſegnati è la Quarta Proportionale, che ſi cerca.</s> <s xml:id="echoid-s407" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s408" xml:space="preserve">Siano date trè linee 1. </s> <s xml:id="echoid-s409" xml:space="preserve">2 3. </s> <s xml:id="echoid-s410" xml:space="preserve">e ſi cerchi la Quarta nella pro-<lb/>portione della prima alla Seconda. </s> <s xml:id="echoid-s411" xml:space="preserve">Traſporto la Prima ſo- <pb o="20" file="0032" n="33" rhead="CAPO II."/> <anchor type="figure" xlink:label="fig-0032-01a" xlink:href="fig-0032-01"/> pra l’vno, e l’altro lato dello Stromento dal centro A, eſe-<lb/>gno le linee laterali nelli punti R, S: </s> <s xml:id="echoid-s412" xml:space="preserve">Dipoi lo Stromento <lb/>tanto s’allarga, chela Seconda capiſca nella diſtanza RS. </s> <s xml:id="echoid-s413" xml:space="preserve">Il<unsure/> <lb/>che fattto applico la Terza sù l’vno, e l’altro lato, e ſegnati li <lb/>punti T, V, piendo la diſtanza T, V, & </s> <s xml:id="echoid-s414" xml:space="preserve">è la Quarta propor-<lb/>tionale cercata. </s> <s xml:id="echoid-s415" xml:space="preserve">La dimoſtrazione è manifeſta dalla ſecon-<lb/>da figura.</s> <s xml:id="echoid-s416" xml:space="preserve"/> </p> <div xml:id="echoid-div19" type="float" level="2" n="1"> <figure xlink:label="fig-0032-01" xlink:href="fig-0032-01a"> <image file="0032-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0032-01"/> </figure> </div> <pb o="21" file="0033" n="34" rhead="Linea Aritmetica."/> <p> <s xml:id="echoid-s417" xml:space="preserve">Di quì appariſce, come date due linee ſi poffa trouar la.</s> <s xml:id="echoid-s418" xml:space="preserve"><unsure/> <lb/>Terza in Proportione continua, e così di mano in mano: </s> <s xml:id="echoid-s419" xml:space="preserve">eſ-<lb/>ſendo che di trè continuamente proportionali, la Seconda hà <lb/>ragione di Conſeguente, e d Antecedente; </s> <s xml:id="echoid-s420" xml:space="preserve">e perciò la diſtan-<lb/>za l<unsure/>i traſporta dal centro A dello Stromento ſopra de’lati, co-<lb/>me s’ella foſſe vna Terza per trouar la Quarta. </s> <s xml:id="echoid-s421" xml:space="preserve">Così ſia data <lb/>la linea AB diuiſa in D, e ſi debba tagliar in proportione con-<lb/>tinua, come AB ad AD, così AD ad vn’altra. </s> <s xml:id="echoid-s422" xml:space="preserve">Piglio sù lo <lb/>Stromento AB, AC vguali alla data AB, l’allargo tanto che <lb/>capiſca la Seconda trà BC. </s> <s xml:id="echoid-s423" xml:space="preserve">Poi traſporto la diſtanza BC <lb/>in AD, AE, ela diſtanza DE è la Terza proportionale; </s> <s xml:id="echoid-s424" xml:space="preserve">qua-<lb/>le traſportata in AF, AG dà la diftanza FG Quarta propor-<lb/>tionale: </s> <s xml:id="echoid-s425" xml:space="preserve">Così FG trasferita in AH, AI dà la Quinta HI; </s> <s xml:id="echoid-s426" xml:space="preserve">& </s> <s xml:id="echoid-s427" xml:space="preserve">HI <lb/>applicata in AK, AL dà la Seſta KL e così di mano in mano. <lb/></s> <s xml:id="echoid-s428" xml:space="preserve">Onde trasferite le diuiſioni F, H K, O, sù la linea data AB, <lb/>queſta ſarà diuiſa, come ſi cercaua, e come AB ad AD, così <lb/>AD ad AE, cosi AE ad AH, così AH ad AK, & </s> <s xml:id="echoid-s429" xml:space="preserve">AK ad AO.</s> <s xml:id="echoid-s430" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s431" xml:space="preserve">La ragione di ciò è chiara, per quello, che s’è moſtrato <lb/>nelcap I. </s> <s xml:id="echoid-s432" xml:space="preserve">eſſendo come AB à BC (intendanſi tirate le line <lb/>BC, DE, &</s> <s xml:id="echoid-s433" xml:space="preserve">c.) </s> <s xml:id="echoid-s434" xml:space="preserve">così AD, cioè BC à DE cioè AF; </s> <s xml:id="echoid-s435" xml:space="preserve">dunque <lb/>AB, AD, AF ſono continuamente proportionali.</s> <s xml:id="echoid-s436" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div21" type="section" level="1" n="13"> <head xml:id="echoid-head19" xml:space="preserve">QVESTIONE QVARTA.</head> <head xml:id="echoid-head20" xml:space="preserve">Come lo Stromento ſerua di Scala vniuerſale per qualſiuoglia <lb/>diſſegno<unsure/>.</head> <p> <s xml:id="echoid-s437" xml:space="preserve">SI trouano alle volte diſſegni già fatti, ne v’è aggiunta la <lb/>Scala per poter ridurre tutt<unsure/>e le linee ad vna mif<unsure/>ura Ho-<lb/>mogenea: </s> <s xml:id="echoid-s438" xml:space="preserve">altre volte s’hà à far qualche diſſegno, & </s> <s xml:id="echoid-s439" xml:space="preserve">il douer <pb o="22" file="0034" n="35" rhead="CAPO II."/> à ciaſcuno far la ſua Scala particolare, è fatica aſſai noioſa; <lb/></s> <s xml:id="echoid-s440" xml:space="preserve">perciò lo Stromento di Proportione ſeruirà diScala vniuer-<lb/>fale, ò ſiano fatti li diſſegni, ò da farſi.</s> <s xml:id="echoid-s441" xml:space="preserve"/> </p> <figure> <image file="0034-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0034-01"/> </figure> <p> <s xml:id="echoid-s442" xml:space="preserve">Primieramente, ſia data la Campagna diſſegnata ne’ſuoi <lb/>termini A B C D E F, di cui ſi deſidera ſapere la grandezza. <lb/></s> <s xml:id="echoid-s443" xml:space="preserve">Se vno de’lati è conoſciuto in miſura, s’applichi quella linea <lb/>al numero corriſpondente nello Stromento: </s> <s xml:id="echoid-s444" xml:space="preserve">Come ſe il lato <lb/>AF ſi ſapeſſe eſſere paſſi 79. </s> <s xml:id="echoid-s445" xml:space="preserve">la lunghezza AF s’applichià 79. </s> <s xml:id="echoid-s446" xml:space="preserve"><lb/>79, el’altre linee tutte applicate allo Stromento, ritenuta la <pb o="23" file="0035" n="36" rhead="Linea Aritmetica."/> primiera appertura moſtreranno di quanti paſſi ſiano; </s> <s xml:id="echoid-s447" xml:space="preserve">& </s> <s xml:id="echoid-s448" xml:space="preserve">ope. <lb/></s> <s xml:id="echoid-s449" xml:space="preserve">rando conforme alli precetti della Geodeſia, ſi verrà à troua-<lb/>re la grandezza di tutta la Campagna. </s> <s xml:id="echoid-s450" xml:space="preserve">Et acciò chinon è <lb/>prattico, poſſa quì apprendere la forma, piacemi di moſtra-<lb/>re, come ſi tirino le linee per cauarne poi la grandezza dell’ <lb/>area.</s> <s xml:id="echoid-s451" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s452" xml:space="preserve">Dal punto A alla linea A B tiriſi la perpendicolare AG: <lb/></s> <s xml:id="echoid-s453" xml:space="preserve">poſcia dall angolo più baſſo E ſi tira la EH perpendicolare al-<lb/>la AG; </s> <s xml:id="echoid-s454" xml:space="preserve">che perciò EH vien ad eſſer parallela alla AB (per la <lb/>28. </s> <s xml:id="echoid-s455" xml:space="preserve">del primo) è doppo queſto dall angolo p ù interno, che <lb/>quì è B ſi tira la linea BI parallela alla AH: </s> <s xml:id="echoid-s456" xml:space="preserve">onde ſi hà il paral-<lb/>lelogrammo A I.</s> <s xml:id="echoid-s457" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s458" xml:space="preserve">Doppo queſto dall’angolo D ſi tirino due linee DK, DL <lb/>perpendicolarialle linee BI, & </s> <s xml:id="echoid-s459" xml:space="preserve">EI, ſopra le qualicadono; </s> <s xml:id="echoid-s460" xml:space="preserve">e ſi <lb/>hà il piccolo Rettangolo KL. </s> <s xml:id="echoid-s461" xml:space="preserve">E perche reſta il Trapezio BK <lb/>DC, tiriſi la linea DB, che lo diuide in due Triangoli. </s> <s xml:id="echoid-s462" xml:space="preserve">Si che <lb/>dall’area cauati li parallelogrammi, reſtano li Triangoli: </s> <s xml:id="echoid-s463" xml:space="preserve">Ne’ <lb/>quali ſe non v’è angolo Retto, tiriſi da vn’angolo al lato op. <lb/></s> <s xml:id="echoid-s464" xml:space="preserve">poſto vna perpendicolare. </s> <s xml:id="echoid-s465" xml:space="preserve">Così li Triangoli BKD, DLE, <lb/>EHG per eſſer rettangoli, non han biſogno d’altra perpen-<lb/>dicolare, come ne’Triangoli, AGF, BCD, fà di meſtieri ti-<lb/>rare le perpendicolati GN, DM.</s> <s xml:id="echoid-s466" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s467" xml:space="preserve">Ora ſe vno de’lati è conoſciuto, come AF paſſi 79 aperto <lb/>lo Stromento in modo, che trà 79, e 79 capiſca la linea AF, <lb/>ritengaſi la ſteſſa apertura, & </s> <s xml:id="echoid-s468" xml:space="preserve">applicando ciaſcuna linea ſi tro-<lb/>uerà la ſua grandezza. </s> <s xml:id="echoid-s469" xml:space="preserve">Ma per non prenderſi fatica ſouer-<lb/>chia, baſta nelli parallelogrammi prendere la miſura de’due <lb/>lati, che fanno l’angolo Retto; </s> <s xml:id="echoid-s470" xml:space="preserve">e queſti moltiplicati inſieme <lb/>danno l’area de’ſudetti parall logrammi. </s> <s xml:id="echoid-s471" xml:space="preserve">Nelli Triangoli <lb/>poi fi piglia la miſura della perpendicolare, e della baſe, ſopra <lb/> <pb o="24" file="0036" n="37" rhead="CAPO II."/> di cui ella cade; </s> <s xml:id="echoid-s472" xml:space="preserve">e moltiplicata la Perpendicolare per la metà <lb/>della baſe, ſi hà l’area del triangolo (per la 41. </s> <s xml:id="echoid-s473" xml:space="preserve">del 1<unsure/>.) </s> <s xml:id="echoid-s474" xml:space="preserve">E ridot-<lb/>te in vna ſomma tutte queſte aree, danno tutta l’area della <lb/>Campagna diſſegnata.</s> <s xml:id="echoid-s475" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s476" xml:space="preserve">Quindi ſi caua, che ſe il dato diſſegno foſſe Topografia di <lb/>paeſe non tanto grande, che ſenſibilmente s’allontanaſſe dal-<lb/>l’eſſer piano, con ogni facilità ſi potrà conoſcere la diſtanza <lb/>d’vn luogo dall’altro, purche vna qualche diſtanza ſia nota, <lb/>ſeruendo queſta per dar vna deter minata apertura allo Stro-<lb/>mento: </s> <s xml:id="echoid-s477" xml:space="preserve">come facilmente ſi raccoglie da ciò, che s’è detto <lb/>ſin’hora.</s> <s xml:id="echoid-s478" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s479" xml:space="preserve">Mà per traportar vn diſegno di grande in piccolo, ò di <lb/>piccolo in grande, non è di meſtieri dir altra coſa più parti-<lb/>colare, poiche ciò è manifeſto da ciò che ſi è detto nella <lb/>queſtione antecedente, non eſſendo queſto altra coſa, che <lb/>trouare la Quarta proportionale.</s> <s xml:id="echoid-s480" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div22" type="section" level="1" n="14"> <head xml:id="echoid-head21" xml:space="preserve">QVESTIONE QVINTA.</head> <head xml:id="echoid-head22" xml:space="preserve">Date due linee trouare la loro proportione in numeri.</head> <p> <s xml:id="echoid-s481" xml:space="preserve">E’Vero, che non tutte le linee ſono trà di loro commen-<lb/>ſurabili, ne hanno la proportione, che ſi poſſa eſpri-<lb/>mere con numeri, come è manifeſto dalla Geometria, e dal <lb/>libro Decimo d’Euclide; </s> <s xml:id="echoid-s482" xml:space="preserve">ad ogni modo per le operationi Me-<lb/>caniche, alle volte ci baſta ſapere, quali ſiano que’ numeri, <lb/>che più da vicino eſprimono la proportione, ò almeno li ter-<lb/>mini (per dir così) eſtrinſeci della proportione, cioè quelli, <lb/>che ſono immediatamente maggiori, & </s> <s xml:id="echoid-s483" xml:space="preserve">immediatamente mi-<lb/>nori del douere; </s> <s xml:id="echoid-s484" xml:space="preserve">tra’ quali prendendoſi il mezo Aritmetico <pb o="25" file="0037" n="38" rhead="Linea Aritmetica."/> ſi hà quel che ſi cerca, per quanto ſi può hauere Fiſica-<lb/>mente.</s> <s xml:id="echoid-s485" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s486" xml:space="preserve">Ora per operare più ſpeditamente in que-<lb/> <anchor type="figure" xlink:label="fig-0037-01a" xlink:href="fig-0037-01"/> ſta occaſione, ſarà bene hauer due Compaſ-<lb/>ſi, co’quali ſi prenda iſquiſitamente la lun-<lb/>ghezza (ò ſe foſſero troppo lunghe, la metà, <lb/>ò altra parte aliquota) di ciaſcuna delle date <lb/>linee, acciò variandoſi l’apertura dello Stro-<lb/>mento, ſi ritenga ſempre nelli due Compaſſi <lb/>aperti la ſteſſa lunghezza delle linee date <lb/>da poterſi applicar allo Stromento.</s> <s xml:id="echoid-s487" xml:space="preserve"/> </p> <div xml:id="echoid-div22" type="float" level="2" n="1"> <figure xlink:label="fig-0037-01" xlink:href="fig-0037-01a"> <image file="0037-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0037-01"/> </figure> </div> <p> <s xml:id="echoid-s488" xml:space="preserve">Siano dunque date le due linee C, B, la cui <lb/>proportione in numeri ſi cerca. </s> <s xml:id="echoid-s489" xml:space="preserve">Prendaſi <lb/>con vn Compaſſo accuratamente la lunghez-<lb/>za di C, e con l’altro Compaſſo quella di B, <lb/>dipoi s’applichi la lunghezza di C al 100, <lb/>100, e con la lunghezza di B ſi vegga ſopra <lb/>qual numero dello Stromento aperto ella ca-<lb/>da, e ſia per cagion d’eſſempio ſu’l 32, 32; </s> <s xml:id="echoid-s490" xml:space="preserve">e <lb/>diremo, che C à B hà la proportione di 100 à <lb/>32. </s> <s xml:id="echoid-s491" xml:space="preserve">Mà ſe la lunghezza di B foſſe minore del-<lb/>la diſtanza 32, 32, e maggiore della diſtanza <lb/>31, 31, diremo, che la proportione di 100 à 31 è maggior <lb/>della vera, e quella di 100 à 32 è minor della vera: </s> <s xml:id="echoid-s492" xml:space="preserve">onde eſ-<lb/>ſendo la differenza d’vna ſola centeſima parte di C, baſterà <lb/>per l’ordinario prendere la B per 31 {1/2}.</s> <s xml:id="echoid-s493" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s494" xml:space="preserve">Auanti però che ſi venga à queſto di prendere litermini <lb/>eſtrinſeci della proportione, cioè il maggior, & </s> <s xml:id="echoid-s495" xml:space="preserve">il minore, <lb/>conuien tentare in altri numeri, maſſime di quelli, che ſi chia-<lb/>mano Primi, cioè che non hanno altro numero, che li miſuri, <pb o="26" file="0038" n="39" rhead="CAPO II."/> & </s> <s xml:id="echoid-s496" xml:space="preserve">applicata ad eſſi la lunghezza di C, vedere ſe la lunghezza <lb/>di B ſi poſſa applicare preciſamente ad alcun numero dello <lb/>Stromento, ò al contrario applicata la B ad alcun numero <lb/>Primo, vedere ſe la C ſi poſſa applicare à qualche numero pre-<lb/>ciſamente nello Stromento. </s> <s xml:id="echoid-s497" xml:space="preserve">Quando dunque ſi troua inutile <lb/>ogni pruoua per hauer il numero preciſamente, allhora con-<lb/>uien oprare come di ſopra, prendendo il maggior, & </s> <s xml:id="echoid-s498" xml:space="preserve">il mino-<lb/>re. </s> <s xml:id="echoid-s499" xml:space="preserve">Et in tal caſo è meglio applicar la C al maſſimo numero <lb/>dello Stromento, cioè al 100, più toſto, che ad altro nume-<lb/>ro più piccolo, perche eſſendo la differenza de’due termini <lb/>trouati d’vna ſola centeſima, ſempre più s’accoſterà al vero, <lb/>che ſe ſi veniſſe ad adoprar vna differenza denominata da vn <lb/>numero minore di 100, eſſendo à tutti manifeſto, che è mi-<lb/>nor vna centeſima parte, che vna nouanteſima ſettima del <lb/>tutto.</s> <s xml:id="echoid-s500" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s501" xml:space="preserve">Mà per operar ancora più preciſamente in caſi ſimili, doue <lb/>non ſi poſſano hauere li numeri preciſi, meglio ſarà trouare la <lb/>differenza d’vna parte centeſima della linea minore B, perche <lb/>queſta è minor differenza, che vna centeſima della maggio-<lb/>re C, perche le parti hanno la proportione de’ Moltiplici, e<unsure/> <lb/>de gl’Intieri (per la 15, del 5.) </s> <s xml:id="echoid-s502" xml:space="preserve">e così c’accoſtaremo più al <lb/>vero. </s> <s xml:id="echoid-s503" xml:space="preserve">Tale dunque ſarà l’aperatione. </s> <s xml:id="echoid-s504" xml:space="preserve">La linea minore B, <lb/>s’applichi nello Stromento al 100. </s> <s xml:id="echoid-s505" xml:space="preserve">Poì la fteſſa B ſi caui dalla <lb/>maggiore C, quante volte ſi può, e ſiano per eſſempio trè <lb/>volte; </s> <s xml:id="echoid-s506" xml:space="preserve">ſi che reſta vna parte della C, minore della data B; </s> <s xml:id="echoid-s507" xml:space="preserve">e ſia <lb/>queſto reſtante IO. </s> <s xml:id="echoid-s508" xml:space="preserve">Onde di quali parti 100 è B, di tali 300 <lb/>è CI. </s> <s xml:id="echoid-s509" xml:space="preserve">Preſa dunque col Compaſſo la IO, & </s> <s xml:id="echoid-s510" xml:space="preserve">applicata allo <lb/>Stromento, trouo che è maggiore, che la diſtanza 14, 14 è <lb/>minore che trà 15. </s> <s xml:id="echoid-s511" xml:space="preserve">15. </s> <s xml:id="echoid-s512" xml:space="preserve">Sì che dico, che B à C, hà la propor-<lb/>tione maggiore di 100 à 315, e minore di 100 à 314; </s> <s xml:id="echoid-s513" xml:space="preserve">poiche <pb o="27" file="0039" n="40" rhead="Linea Aritmetica."/> la linea C è minore di 315, e maggiore di 314. </s> <s xml:id="echoid-s514" xml:space="preserve">E per il contra-<lb/>rio C à B hà la proportione minore di 315 à 100, e maggiore <lb/>di 314 à 100, come è manifeſto dalla 26. </s> <s xml:id="echoid-s515" xml:space="preserve">de l<unsure/> 5.</s> <s xml:id="echoid-s516" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s517" xml:space="preserve">Ora ſe f<unsure/>i farà come 315 à 100, così 100 à 31 {235/315}; </s> <s xml:id="echoid-s518" xml:space="preserve">c come <lb/>314 à 100, così 100 à 31 {266/314}; </s> <s xml:id="echoid-s519" xml:space="preserve">ſi vede chiaramente, che hab-<lb/>biamo li due Conſeguenti maggior, e minore della propor-<lb/>tione in termini più vicini trà di ſe, che non erano prima 31, e <lb/>32, mettendo la linea maggiore C per 100: </s> <s xml:id="echoid-s520" xml:space="preserve">poiche ridotte <lb/>le due frattioni allo ſteſſo Denominatore 98910, il Numera-<lb/>tore della prima ſarà 73790, quello della ſeconda 83790. <lb/></s> <s xml:id="echoid-s521" xml:space="preserve">Eridotti tutti gl’Intieri alla denominatione commune troua-<lb/>ta, ſarà la linea C 9891000, e la linea B ſarà maggiore di <lb/>3140000, e minore di 3150000; </s> <s xml:id="echoid-s522" xml:space="preserve">onde la differenza è di <lb/>10000 particelle di tutta la C; </s> <s xml:id="echoid-s523" xml:space="preserve">la qual differenza è minore, <lb/>che la centeſima parte della ſteſſa C; </s> <s xml:id="echoid-s524" xml:space="preserve">poiche queſta centeſi-<lb/>ma è delle particelle di C 98910.</s> <s xml:id="echoid-s525" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div24" type="section" level="1" n="15"> <head xml:id="echoid-head23" xml:space="preserve">QVESTIONE SESTA.</head> <head xml:id="echoid-head24" xml:space="preserve">Dati gli Aſsi d’ vn’ Ellipſi, deſcriuere la ſua <lb/>circonferenza.</head> <p> <s xml:id="echoid-s526" xml:space="preserve">SIa data la linea AB Aſſe maggiore, ela linea C Aſſe mi-<lb/>nore d’ vn’Ellipſi, e ſi voglia deſcriuere l’Ouato, di <lb/>cui ſono Aſſi. </s> <s xml:id="echoid-s527" xml:space="preserve">Primieramente per la Queſt. </s> <s xml:id="echoid-s528" xml:space="preserve">5. </s> <s xml:id="echoid-s529" xml:space="preserve">antecedente <lb/>ſi troui in numeri la loro proportione, e ſia per eſempio come <lb/>di 5 à 3. </s> <s xml:id="echoid-s530" xml:space="preserve">Dipoi circa AB come diametro ſi deſcriua vn circo-<lb/>lo: </s> <s xml:id="echoid-s531" xml:space="preserve">e dal punto eſtremo A ſi prendano di quà, e di là archi <lb/>vguali ad arbitrio AS, AR; </s> <s xml:id="echoid-s532" xml:space="preserve">AD, AF; </s> <s xml:id="echoid-s533" xml:space="preserve">AH, AI &</s> <s xml:id="echoid-s534" xml:space="preserve">c. </s> <s xml:id="echoid-s535" xml:space="preserve">e con <lb/>linee rette congionti li punti vgualmente diſtanti dall’eſtre- <pb o="28" file="0040" n="41" rhead="CAPO II."/> mità A, taglieranno il diametro AB ad angoli retti in O, M, <lb/>P &</s> <s xml:id="echoid-s536" xml:space="preserve">c. </s> <s xml:id="echoid-s537" xml:space="preserve">E così le linee per-<lb/>pendicolari alla AB ſa-<lb/> <anchor type="figure" xlink:label="fig-0040-01a" xlink:href="fig-0040-01"/> ranno parallele trà di lo-<lb/>ro, & </s> <s xml:id="echoid-s538" xml:space="preserve">ordinatamente <lb/>applicate così al diame-<lb/>tro del circolo, come all’ <lb/>Aſſe maggiore dell’ El-<lb/>lipſi.</s> <s xml:id="echoid-s539" xml:space="preserve"/> </p> <div xml:id="echoid-div24" type="float" level="2" n="1"> <figure xlink:label="fig-0040-01" xlink:href="fig-0040-01a"> <image file="0040-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0040-01"/> </figure> </div> <p> <s xml:id="echoid-s540" xml:space="preserve">Mettanſi dunque ciaſ-<lb/>cuna delle applicate nel <lb/>circolo ad vn numero <lb/>della linea Aritmetica, <lb/>che habbia vn’altro nu-<lb/>mero, à cui ella ſia come <lb/>5 à 3, come ſaria 50, 50; <lb/></s> <s xml:id="echoid-s541" xml:space="preserve">e 30, 30: </s> <s xml:id="echoid-s542" xml:space="preserve">perche il ſecon-<lb/>do interuallo 30, 30, darà l’Applicata dell’Ellipſi: </s> <s xml:id="echoid-s543" xml:space="preserve">Così OR <lb/>ad OV; </s> <s xml:id="echoid-s544" xml:space="preserve">MF ad MN; </s> <s xml:id="echoid-s545" xml:space="preserve">PI à PQ, e così ſuſſeguentemente, <lb/>ſaranno come 5 à 3, e pigliaraſſi ad OV vguale OG, & </s> <s xml:id="echoid-s546" xml:space="preserve">à <lb/>MN vguale MK &</s> <s xml:id="echoid-s547" xml:space="preserve">c. </s> <s xml:id="echoid-s548" xml:space="preserve">perche la linea tirata per li punti <lb/>Q, N, V, A, G, K, &</s> <s xml:id="echoid-s549" xml:space="preserve">c. </s> <s xml:id="echoid-s550" xml:space="preserve">ſarà Elliptica.</s> <s xml:id="echoid-s551" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s552" xml:space="preserve">Ciò ſi demoſtra, perche nell’ Ellipſi i Quadrati delle Ap-<lb/>plicate hanno la proportione delli rettangoli fatti dalli ſeg-<lb/>menti del diametro, à cui ſono Applicate: </s> <s xml:id="echoid-s553" xml:space="preserve">e nel circolo i <lb/>Quadrati delle perpendicolari OR, MF ſono vguali alli ret-<lb/>tangoli AOB, AMB fatti dalli ſteſſi ſegmenti: </s> <s xml:id="echoid-s554" xml:space="preserve">dunque co-<lb/>me il Quadrato di OV al Quadrato di MN, così il Quadra-<lb/>to di OR al Quadrato di MF. </s> <s xml:id="echoid-s555" xml:space="preserve">Dunque per la 22. </s> <s xml:id="echoid-s556" xml:space="preserve">del 6. </s> <s xml:id="echoid-s557" xml:space="preserve">co-<lb/>me OV ad MN, così OR ad MF, e permutando come OV <pb o="29" file="0041" n="42" rhead="Linea Aritmetica."/> ad OR, così MN ad MF; </s> <s xml:id="echoid-s558" xml:space="preserve">e perche OV ad OR per la co-<lb/>ſtruttione ſono come l’Aſſe maggiore AB all’Aſſe minore C, <lb/>cioè come le loro metà EX ad EL; </s> <s xml:id="echoid-s559" xml:space="preserve">dunque il Rettangolo <lb/>AEB al Rettangolo AOB è come il Quadrato della metà <lb/>dell’Aſſe minore al Quadrato dell’Applicata OV.</s> <s xml:id="echoid-s560" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div26" type="section" level="1" n="16"> <head xml:id="echoid-head25" style="it" xml:space="preserve">QVESTIONE SETTIMA.</head> <head xml:id="echoid-head26" style="it" xml:space="preserve">Come potiamo ſeruirci dello Stromento di Proportione, in vece <lb/>delle Tauole Trigonometriche, per la ſolutione <lb/>di molti Triangoli.</head> <p> <s xml:id="echoid-s561" xml:space="preserve">SE bene ciò appariſce aſſai chiaramente da ciò, che s’è <lb/>detto nella queſtione 4.</s> <s xml:id="echoid-s562" xml:space="preserve">ad ogni modo per maggior ſpie-<lb/> <anchor type="figure" xlink:label="fig-0041-01a" xlink:href="fig-0041-01"/> gatione è bene accennarlo quì <lb/>più particolarmente. </s> <s xml:id="echoid-s563" xml:space="preserve">Sia per <lb/>cagione d’eſſempio vna Torre, <lb/>la cui altezza, e diſtanza da noi, <lb/>deſideriamo di conoſcere. </s> <s xml:id="echoid-s564" xml:space="preserve">Pren-<lb/>daſi vn piano di qualunque ſor-<lb/>te, come ſaria vna tauola, MHC, <lb/>e ſi ponga in ſito verticale con la <lb/>Torre, di mode, che la linea ret-<lb/>ta del ſuo lato MH ſia parallela <lb/>all’Orizonte: </s> <s xml:id="echoid-s565" xml:space="preserve">poi collocato l’oc-<lb/>chio nel punto M, e riguardando la cima della Torre, ſia il <lb/>raggio viſuale la linea MB, la quale ſi ſegni. </s> <s xml:id="echoid-s566" xml:space="preserve">Fatto queſto, ſi <lb/>ritiri l’oſſeruatore più indietro, in modo però, che nella ſteſ-<lb/>ſa dirittura ſiano la Torre, & </s> <s xml:id="echoid-s567" xml:space="preserve">i luoghi delle due oſſeruationi: <lb/></s> <s xml:id="echoid-s568" xml:space="preserve">& </s> <s xml:id="echoid-s569" xml:space="preserve">in queſto ſecondo luogo di nuouo collocata la tauoletta <pb o="30" file="0042" n="43" rhead="CAPO II."/> MHC come prima, ſi noti il raggio viſuale MC, il quale ne-<lb/>ceſſariamente cade di ſotto di BM, douendo l’iſteſſa Torre in <lb/>ſito più lontano apparire ſotto angolo minore; </s> <s xml:id="echoid-s570" xml:space="preserve">e così CMH <lb/>deue eſſere minore di BMH: </s> <s xml:id="echoid-s571" xml:space="preserve">e ſe tutto ciò ſarà fatto accura-<lb/>tamente, habbiamo tutto ciò, che ci fà di meſtieri al noſtro <lb/>intento.</s> <s xml:id="echoid-s572" xml:space="preserve"/> </p> <div xml:id="echoid-div26" type="float" level="2" n="1"> <figure xlink:label="fig-0041-01" xlink:href="fig-0041-01a"> <image file="0041-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0041-01"/> </figure> </div> <p> <s xml:id="echoid-s573" xml:space="preserve">Tiriſi dunque in vn piano à parte la linea IN indefinita, e <lb/>dal puuto I ſi tiri vn’altra linea parimenti indefinita, mà che <lb/>faccia in I l’angolo vguale all’angolo CMH, che è il minore <lb/>delli due oſſeruati. </s> <s xml:id="echoid-s574" xml:space="preserve">Dipoi nella IN pigliſi il punto O arbi-<lb/>trariamente, e ſi faccia in O vn’altr’angolo vguale all’ angolo <lb/>BMH, che è il maggiore delli due oſſeruati. </s> <s xml:id="echoid-s575" xml:space="preserve">Et in tal manie-<lb/>ra IO rappreſenta la diſtanza delli due luoghi dell’ oſſerua-<lb/>tione; </s> <s xml:id="echoid-s576" xml:space="preserve">ele due linee OA, IA, che s’incontrano in A, rappre-<lb/>ſentano li due raggiviſuali, che ſi terminano nella cima della <lb/>Torre. </s> <s xml:id="echoid-s577" xml:space="preserve">E che s’incontrino in A, è manifeſto, perche li due <lb/>angoli AOI, AON ſon vguali à due retti (per la 13. </s> <s xml:id="echoid-s578" xml:space="preserve">del <lb/>lib. </s> <s xml:id="echoid-s579" xml:space="preserve">1.) </s> <s xml:id="echoid-s580" xml:space="preserve">l’angolo AIO è minore dell’angolo AON, per la con. <lb/></s> <s xml:id="echoid-s581" xml:space="preserve">ſtruttione, dunque li due AIO, AOI ſon minori di due retti; </s> <s xml:id="echoid-s582" xml:space="preserve"><lb/>dunque quelle due linee ſon conuergenti, e da quella parte <lb/>s’incontrano; </s> <s xml:id="echoid-s583" xml:space="preserve">e ciò ſi fà in A. </s> <s xml:id="echoid-s584" xml:space="preserve">Se dunque dal punto A, ſopra <lb/>la linea IN parallela all’Orizonte, ſi tirarà la perpendicola-<lb/>re AN, queſta ſarà l’altezza della Torre ſopra l’altezza dell’ <lb/>occhio dell’oſſeruatore, la quale ponendoſi IS, ò la ſua vgua-<lb/>le OR, ſarà tutta l’altezza della Tore AL, e la ſua diſtanza <lb/>ſarà ON, cioè RL.</s> <s xml:id="echoid-s585" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s586" xml:space="preserve">Ora portando ſopra dello Stromento la linea IO come <lb/>100, trouo per la queſtione precedente, che AN è 374, & </s> <s xml:id="echoid-s587" xml:space="preserve"><lb/>ON 328. </s> <s xml:id="echoid-s588" xml:space="preserve">Sì che eſſendo nota la diſtanza de’ due luoghi dell’ <lb/>oſſeruationi per cagion d’eſſempio di paſſi 18, trouo, che ſe <pb o="31" file="0043" n="44" rhead="Linea Aritmetica."/> IO 100 è paſſi 18, AN 374 è paſſi 67 {1<unsure/>/3} proſſimamente, & </s> <s xml:id="echoid-s589" xml:space="preserve"><lb/>ON 328 è paſſi 59. </s> <s xml:id="echoid-s590" xml:space="preserve">Se dunque all’altezza AN paſſi 67 {1/5} s’ag-<lb/>gionga l’altezza dell’ occhio ſopra il piano del piede della <lb/>Torre, per eſſempio di piedi Romani 6, ſarà tutta l’altezza <lb/>cercata AL di piedi 342 {2/3}, ela diſtanza cercata ON, ouero <lb/>RL di piedi 295.</s> <s xml:id="echoid-s591" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s592" xml:space="preserve">Di quì è manifeſto, che dato qualunque triangolo, ſi può <lb/>trouare la proportione de’ſuoi lati; </s> <s xml:id="echoid-s593" xml:space="preserve">e ſe vno di queſti è cono-<lb/>ſciuto in miſura determinata, ſi verrà anche in cognitione del-<lb/>la quantità de gl’altri due lati nella ſteſſa miſura.</s> <s xml:id="echoid-s594" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div28" type="section" level="1" n="17"> <head xml:id="echoid-head27" style="it" xml:space="preserve">QVESTIONE OTTAVA.</head> <head xml:id="echoid-head28" style="it" xml:space="preserve">Come ſerua per la Proſpettiua lo Stromento.</head> <p> <s xml:id="echoid-s595" xml:space="preserve">SIa l’occhio O, il punto della viſta C, in diſtanza di piedi <lb/>10 {1/2}; </s> <s xml:id="echoid-s596" xml:space="preserve">l’altezza dell’occhio OB piedi 6; </s> <s xml:id="echoid-s597" xml:space="preserve">à cuiè vguale <lb/> <anchor type="figure" xlink:label="fig-0043-01a" xlink:href="fig-0043-01"/> DC. </s> <s xml:id="echoid-s598" xml:space="preserve">AB è l’Orizonte. </s> <s xml:id="echoid-s599" xml:space="preserve">Non eſſendoui ſpatio nel Piano da-<lb/>te per tutte le diſtanze, così potraſſi operare con la ſola linea <lb/>DC, col Compaſſo di Proportione.</s> <s xml:id="echoid-s600" xml:space="preserve"/> </p> <div xml:id="echoid-div28" type="float" level="2" n="1"> <figure xlink:label="fig-0043-01" xlink:href="fig-0043-01a"> <image file="0043-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0043-01"/> </figure> </div> <pb o="32" file="0044" n="45" rhead="CAPO II."/> </div> <div xml:id="echoid-div30" type="section" level="1" n="18"> <head xml:id="echoid-head29" style="it" xml:space="preserve">Primo, Data la diſtanza dell’ oggetto, trouare in qual parallela <lb/>all’ Orizon@ale caſchi.</head> <p> <s xml:id="echoid-s601" xml:space="preserve">Prendaſi DC, e ſi metta ſul Compaſſo di Proportione al <lb/>numero corriſpondente alla diſtanza dell’ oggetto dall’ oc-<lb/>chio; </s> <s xml:id="echoid-s602" xml:space="preserve">e poi al numero corriſpondente alla diſtanza dell’oc-<lb/>chio dal Quadro, ſi trouera quanto ſotto al punto della viſta <lb/>C ſi debba tirare la cercata parallela. </s> <s xml:id="echoid-s603" xml:space="preserve">Sia la diſtanza dell’og-<lb/>getto BA piedi 28 {1/2}, & </s> <s xml:id="echoid-s604" xml:space="preserve">OC piedi 10 {1/2}. </s> <s xml:id="echoid-s605" xml:space="preserve">Metto la DC all’in-<lb/>teruallo 57, 57: </s> <s xml:id="echoid-s606" xml:space="preserve">e preſo l’interuallo 21, 21. </s> <s xml:id="echoid-s607" xml:space="preserve">mi viene CE, <lb/>per cu ſi tirarà la parallela EF. </s> <s xml:id="echoid-s608" xml:space="preserve">La ragione per la ſomiglian-<lb/>za de’ triangoli ADE, OCE è manifeſta, perche come AD <lb/>à OC, così DE à EC, e componendo come AD + OC(cioè <lb/>AB) à OC, così DC à CE.</s> <s xml:id="echoid-s609" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div31" type="section" level="1" n="19"> <head xml:id="echoid-head30" style="it" xml:space="preserve">Secondo, Data la lon@ananza dell’ oggetto dal piano Verticale, <lb/>in cui è l’Aſſe Viſuale, trouare il ſuo luogo nella <lb/>data diſtanza.</head> <p> <s xml:id="echoid-s610" xml:space="preserve">Prendaſi la CE, e ſi metta al numero dell’altezza dell’oc-<lb/>chio ſopra l’Orizonte; </s> <s xml:id="echoid-s611" xml:space="preserve">& </s> <s xml:id="echoid-s612" xml:space="preserve">al numero della diſtanza dell’ogget-<lb/>to dal mezzo, ſi hauerà l’interuallo douuto nella parallela tro-<lb/>uata. </s> <s xml:id="echoid-s613" xml:space="preserve">Sia dunque data la diſtanza di piedi 5.</s> <s xml:id="echoid-s614" xml:space="preserve">3′, come ſaria <lb/>DG. </s> <s xml:id="echoid-s615" xml:space="preserve">Perche CD è 6 piedi, intendaſi 60′. </s> <s xml:id="echoid-s616" xml:space="preserve">Dunque CE po-<lb/>ſta al 60. </s> <s xml:id="echoid-s617" xml:space="preserve">60, l’interuallo 53.</s> <s xml:id="echoid-s618" xml:space="preserve">53 darà EI. </s> <s xml:id="echoid-s619" xml:space="preserve">(ſe CE è troppo <lb/>piccola, prendaſi il triplo, e poi della linea trouata ſi prenda <lb/>la terza parte, e ſarà la EI). </s> <s xml:id="echoid-s620" xml:space="preserve">La ragione è, perche come <lb/>CD à DG, così CE à El.</s> <s xml:id="echoid-s621" xml:space="preserve"/> </p> <pb o="33" file="0045" n="46" rhead="Linea Aritmetica."/> </div> <div xml:id="echoid-div32" type="section" level="1" n="20"> <head xml:id="echoid-head31" style="it" xml:space="preserve">Terzo, Dato il luogo nel piano della Perſpettiua, data la diſtanza <lb/>dell’ occbio dal quadro, e data l’altezza perpendicolare <lb/>del corpo, trouar il punto doue ſi terminarà.</head> <p> <s xml:id="echoid-s622" xml:space="preserve">Sia il punto I il luogo nel piano della Perſpettiua: </s> <s xml:id="echoid-s623" xml:space="preserve">l’altez-<lb/> <anchor type="figure" xlink:label="fig-0045-01a" xlink:href="fig-0045-01"/> za data ſia di <lb/>piedi 15 {3/8}, cioè <lb/>BS; </s> <s xml:id="echoid-s624" xml:space="preserve">la diſtanza <lb/>dell’ occhio <lb/>OC piedi 10 {1/2}. <lb/></s> <s xml:id="echoid-s625" xml:space="preserve">Faciaſi come <lb/>CO ad SB, così <lb/>CH, cioè EI <lb/>data, ad I T. </s> <s xml:id="echoid-s626" xml:space="preserve"><lb/>Ora CO ad IB <lb/>è come 21 à <lb/>30 {3/4}; </s> <s xml:id="echoid-s627" xml:space="preserve">meſla <lb/>dunque la EI <lb/>all’ interuallo <lb/>21. </s> <s xml:id="echoid-s628" xml:space="preserve">21, l’inter-<lb/>uallo 30 {3/4}. </s> <s xml:id="echoid-s629" xml:space="preserve">30 {3/4} <lb/>darà la IT cercata.</s> <s xml:id="echoid-s630" xml:space="preserve"/> </p> <div xml:id="echoid-div32" type="float" level="2" n="1"> <figure xlink:label="fig-0045-01" xlink:href="fig-0045-01a"> <image file="0045-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0045-01"/> </figure> </div> <p> <s xml:id="echoid-s631" xml:space="preserve">Di qua ſi vede quanto facile ſarà trouare le conuerſe di <lb/>queſte trè propoſitioni. </s> <s xml:id="echoid-s632" xml:space="preserve">Primo, ſe ſi farà come CE à CD, <lb/>così OC à BA, s’haurà la diſtanza dell’oggetto. </s> <s xml:id="echoid-s633" xml:space="preserve">Secondo, <lb/>ſe come CE à EI, così CD à DS, s’haurà la diſtanza dall’aſse <lb/>viſuale. </s> <s xml:id="echoid-s634" xml:space="preserve">Terzo, ſe come EI à IT, così CO à BS, s’haurà di <lb/>quanta altezza perpendicolare ſia l’oggetto viſto in IT.</s> <s xml:id="echoid-s635" xml:space="preserve"/> </p> <pb o="34" file="0046" n="47" rhead="CAPO II."/> </div> <div xml:id="echoid-div34" type="section" level="1" n="21"> <head xml:id="echoid-head32" xml:space="preserve">QVESTIONE NONA.</head> <head xml:id="echoid-head33" style="it" xml:space="preserve">Come potiamo valerci dello Stromento per pratticar in Numeri <lb/>la Regola del Trè, ò Aurea, che vogliamo dire.</head> <p> <s xml:id="echoid-s636" xml:space="preserve">QVeſta prattica veramente non può riuſcire tanto preci-<lb/>ſa per ragione de’ Rotti, mà per gl’Intieri appariſce <lb/>faciliſſima, e preſta. </s> <s xml:id="echoid-s637" xml:space="preserve">Si pigli dal centro A dello <lb/>Stromento con vn Compaſſo la diſtanza ſin al punto corri-<lb/>ſpondente al ſecondo numero delli trè dati (ò per parlare più <lb/>vniuerſalmente, corriſpondente al numero, che è il Conſe-<lb/>guentetrà li dati) & </s> <s xml:id="echoid-s638" xml:space="preserve">à queſta diſtanza s’allarghilo Stromen-<lb/>to, applicandola al punto corriſpondente al numero, che è <lb/>Primo Antecedente della Proportione: </s> <s xml:id="echoid-s639" xml:space="preserve">perche all’incontro <lb/>del punto, che corriſponde al Terzo numero, ò al Secondo <lb/>Antecedente, ſi prenderà la diſtanza nello Stromento; </s> <s xml:id="echoid-s640" xml:space="preserve">e <lb/>queſta applieata dal Centro A ſopra la linea dello Stromento <lb/>moſtrerà il Quarto numero cercato.</s> <s xml:id="echoid-s641" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s642" xml:space="preserve">Sia per cagion d’eſſempio, ch’io habbia comprato 54 <lb/>braccia di panno per 36 zecchini; </s> <s xml:id="echoid-s643" xml:space="preserve">& </s> <s xml:id="echoid-s644" xml:space="preserve">vn’amico ne vorrebbe <lb/>hauere 2 I braccia; </s> <s xml:id="echoid-s645" xml:space="preserve">Quanto hà egli à pagare per ſua parte? <lb/></s> <s xml:id="echoid-s646" xml:space="preserve">Piglio col Compaſſo nello Stromento dal centro ſin al pun-<lb/>to 36; </s> <s xml:id="echoid-s647" xml:space="preserve">queſta diſtanza applico al 54. </s> <s xml:id="echoid-s648" xml:space="preserve">54. </s> <s xml:id="echoid-s649" xml:space="preserve">E ritenendo queſta <lb/>apertura piglio la diſtanza 21.</s> <s xml:id="echoid-s650" xml:space="preserve">21. </s> <s xml:id="echoid-s651" xml:space="preserve">Queſta traporto dal cen-<lb/>tro dello Strumento sù la linea, e vedendo che cade ſul pun-<lb/>to 14, dico al mio amico, toccali per ſua parte à pagare 14 <lb/>zecchini.</s> <s xml:id="echoid-s652" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s653" xml:space="preserve">La dimoſtratione di ciò è manifeſta, perche ſe di quali par-<lb/>ti 54 è AE, ditali 36 s’è preſa EL, dell’iſteſſa miſura hauen- <pb o="35" file="0047" n="48" rhead="Prattica in numeri della Regola del Trè."/> done AH 21, ſeguirà che HI applicata dal punto A alla li-<lb/>nea AE caderà in vn punto, che moſtrarà di <lb/> <anchor type="figure" xlink:label="fig-0047-01a" xlink:href="fig-0047-01"/> quante parti ella ſia in miſura homogenea al <lb/>termine ſuo corriſpondente, e caderà nel <lb/>punto 14.</s> <s xml:id="echoid-s654" xml:space="preserve"/> </p> <div xml:id="echoid-div34" type="float" level="2" n="1"> <figure xlink:label="fig-0047-01" xlink:href="fig-0047-01a"> <image file="0047-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0047-01"/> </figure> </div> <p> <s xml:id="echoid-s655" xml:space="preserve">E perche i’eſſempio poſto è della regola, <lb/>diretta, mettiamone vn’altro dell’@uerſa. </s> <s xml:id="echoid-s656" xml:space="preserve">Hò <lb/>vna laſtra d’argento lunga piedi 2 {1/2}, elarga <lb/>oncie 7: </s> <s xml:id="echoid-s657" xml:space="preserve">Vorei che l’orefice ne faceſſe vna <lb/>della ſteſſa groſſezza, mà larga oncie 10; </s> <s xml:id="echoid-s658" xml:space="preserve">Quanto dourà eſſer <lb/>longa? </s> <s xml:id="echoid-s659" xml:space="preserve">Quì è certo, che il Primo Antecedente deue eſſere <lb/>queſto numero, che è poſto nelterzo luogo, cioè il 10; </s> <s xml:id="echoid-s660" xml:space="preserve">ela <lb/>proportione ordinata ſarà come 10 à 7, così 30 (poiche <lb/>piedi 2 {1/2} ſono oncie 30) ad vn’altro. </s> <s xml:id="echoid-s661" xml:space="preserve">Preſa dal centro la di-<lb/>ſtanza ſin al punto 7 la colloco trà 10.</s> <s xml:id="echoid-s662" xml:space="preserve">10, e ritenuta la ſteſſa <lb/>apertura dello Stromento, prendo la diſtanzatrà 30. </s> <s xml:id="echoid-s663" xml:space="preserve">30; </s> <s xml:id="echoid-s664" xml:space="preserve">e <lb/>queſta diſtanza applicata alla linea dal centro, trouo, che ca-<lb/>de nel punto 21; </s> <s xml:id="echoid-s665" xml:space="preserve">e così dico, che la lunghezza cercata dourà <lb/>eſſere di oncie 21. </s> <s xml:id="echoid-s666" xml:space="preserve">Così d’vno ſquadrone diſoldati, che hà <lb/>60 di fronte, e 25 di fianco, volendo metterne 40 di fianco, <lb/>ſi cerca, quanti ſariano di fronte: </s> <s xml:id="echoid-s667" xml:space="preserve">la proportione ordinata <lb/>ſarà come 40 à 25, così 60, ad vn’ altro, & </s> <s xml:id="echoid-s668" xml:space="preserve">operando, come <lb/>s’è detto, ſi trouarà venire 37 difronte: </s> <s xml:id="echoid-s669" xml:space="preserve">vero è che ne auan-<lb/>zeranno 20: </s> <s xml:id="echoid-s670" xml:space="preserve">e perciò ſitrouerà che la punta del Compaſſo <lb/>caderà tra’l 37, e 38.</s> <s xml:id="echoid-s671" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s672" xml:space="preserve">Potrebbe occorrere, che li numeri foſſer ò troppo grandi, <lb/>ò troppo piccioli, ſi che ò non ſi trouaſſero per la ſua gran-<lb/>dezza nella linea ſegnata dello Stromento, che ſol arriua al <lb/>Ioo, ò non ſi poteſſero commodamente applicar all’apef<unsure/>-<lb/>tura dello Stromento per la ſua picciolezza. </s> <s xml:id="echoid-s673" xml:space="preserve">Se foſſero trop- <pb o="36" file="0048" n="49" rhead="CAPO II."/> pograndi, conuien diuiderli, e prenderne vna parte aliquo-<lb/>ta; </s> <s xml:id="echoid-s674" xml:space="preserve">ſe foſſero troppo piccioli, conuien pigliare li loro multi-<lb/>plici. </s> <s xml:id="echoid-s675" xml:space="preserve">E perche queſto può occorrere in più modi, per di-<lb/>ſtintione più chiara, ſarà bene parlar di ciaſcuno partico-<lb/>larmente.</s> <s xml:id="echoid-s676" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s677" xml:space="preserve">Primo delli trè numeri dati ſe ſolo il Secondo Antecedente <lb/>della Proportione è maggiore di 100, ſi prenda la ſua metà, ò <lb/>il terzo, e poi il numero trouato ſi raddoppij, ò ſi triplichi, e <lb/>s’haurà il quarto numero cercato. </s> <s xml:id="echoid-s678" xml:space="preserve">Per eſſempo, 24 perſone <lb/>in vn tal tempo conſumano 30 ſacchi di farina: </s> <s xml:id="echoid-s679" xml:space="preserve">in tempo <lb/>vguale 120 perſone quanta ne conſumeranno? </s> <s xml:id="echoid-s680" xml:space="preserve">La diſtanza <lb/>del centro ſin à 30, applicaſi trà 24. </s> <s xml:id="echoid-s681" xml:space="preserve">24; </s> <s xml:id="echoid-s682" xml:space="preserve">e perche 120 non <lb/>fi troua nella linea, prendo la ſua metà 60, ela diſtanza 60, <lb/>60, applicata alla linea, trouo eſſer 75; </s> <s xml:id="echoid-s683" xml:space="preserve">dunque queſta rad-<lb/>doppiata, dico richiederſi 150 ſacchi di farina per 120 <lb/>perſone.</s> <s xml:id="echoid-s684" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s685" xml:space="preserve">Secondo, ſe ſolo il Primo Antecedente, ò ſolo il Primo <lb/>Conſeguente, ò ambidue, ò l’vn, e l’altro Antecedente ſono <lb/>maggiori di 100; </s> <s xml:id="echoid-s686" xml:space="preserve">l’vno, el’Altro Antecedente, ò li primi An-<lb/>r<unsure/>ecedente, e Conſeguente, ſimilmente ſi diuidano, e con quel-<lb/>le parti s’operi, come quelle foſſero li termini dati. </s> <s xml:id="echoid-s687" xml:space="preserve">In vn <lb/>capitale di ſcudi 2000 s’è fatta perdita di ſcudi 1120; </s> <s xml:id="echoid-s688" xml:space="preserve">io che <lb/>cihaueuo per mia parte 75 ſcudi, quanto vengo à perdere? <lb/></s> <s xml:id="echoid-s689" xml:space="preserve">Perche li due priminumeri ſon troppo grandi, leuo à ciaſcuno <lb/>vn zero, e reſtano le loro decime parti 200, e 112: </s> <s xml:id="echoid-s690" xml:space="preserve">e perche <lb/>queſtiancora ſon troppo grandi, li diuido per metà, e ſono le <lb/>lor venteſime parti 100, e 56. </s> <s xml:id="echoid-s691" xml:space="preserve">Prendo dunque dal centro al <lb/>punto 56, e l’applico tra 100.</s> <s xml:id="echoid-s692" xml:space="preserve">100: </s> <s xml:id="echoid-s693" xml:space="preserve">poi trà 75.</s> <s xml:id="echoid-s694" xml:space="preserve">75 prendo la <lb/>diſtanza, & </s> <s xml:id="echoid-s695" xml:space="preserve">applicata alla linea dello Stromento, trouo ch’el-<lb/>la è 42; </s> <s xml:id="echoid-s696" xml:space="preserve">e perciò dico eſſer la perdita, che mi tocca di 42 ſcudi.</s> <s xml:id="echoid-s697" xml:space="preserve"/> </p> <pb o="37" file="0049" n="50" rhead="Prattica in numeri della Regola del Trè."/> <p> <s xml:id="echoid-s698" xml:space="preserve">Terzo, ſe tutti trè li numeri dati ſono maggiori di 100, <lb/>conuien diuiderli tuttitrè: </s> <s xml:id="echoid-s699" xml:space="preserve">E ciò ſi può far ò diuidendoli ſimil, <lb/>mente, come ſe 200 dà 150, che darà 160? </s> <s xml:id="echoid-s700" xml:space="preserve">perche, tutti di-<lb/>uiſi per metà, dico, ſe 100 dà 75, che darà 80? </s> <s xml:id="echoid-s701" xml:space="preserve">& </s> <s xml:id="echoid-s702" xml:space="preserve">applicati li <lb/>75 tra 100. </s> <s xml:id="echoid-s703" xml:space="preserve">100, la diſtanza 80. </s> <s xml:id="echoid-s704" xml:space="preserve">80, mi darà 60, e queſto <lb/>raddoppiato fà 120, che è quello che ſi cerca: </s> <s xml:id="echoid-s705" xml:space="preserve">Ouero ſi pon-<lb/>no diuidere ſimilmente ſolamente due, cioè ò li due Antece-<lb/>denti, ò il Primo Antecedente col ſuo Conſeguente, e di <lb/>quell’altro numero che reſta, prenderne quella parte che più <lb/>piacerà; </s> <s xml:id="echoid-s706" xml:space="preserve">poiche quello, che ſi trouarà, ſarà parte ſimile del <lb/>Quarto, che ſi cerca. </s> <s xml:id="echoid-s707" xml:space="preserve">Così ſtando nello ſteſſo eſſempio, ſe <lb/>200 dà 150, che darà 160? </s> <s xml:id="echoid-s708" xml:space="preserve">Piglio la metà del primo, e del <lb/>ſecondo 100 è 75, e del terzo 160 piglio la quarta parte 40, <lb/>& </s> <s xml:id="echoid-s709" xml:space="preserve">opro come prima, pigliando vltimamente la diſtanza trà <lb/>40, 40, e mi viene 30, il quale quadruplicato mi dà 120: <lb/></s> <s xml:id="echoid-s710" xml:space="preserve">ouero delli due Antecedenti propoſti 200, e 160. </s> <s xml:id="echoid-s711" xml:space="preserve">piglio la <lb/>metà 100, e 80, e del primo conſeguente 150 piglio la terza <lb/>parte 50, & </s> <s xml:id="echoid-s712" xml:space="preserve">oprando, come s’è più volte detto, trouo 40, il <lb/>qual’è la terza parte del numero cercato, cioè di 120.</s> <s xml:id="echoid-s713" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s714" xml:space="preserve">La ragione di queſto modo d’operare ſtà fondato nella 15, <lb/>& </s> <s xml:id="echoid-s715" xml:space="preserve">11 del lib. </s> <s xml:id="echoid-s716" xml:space="preserve">5. </s> <s xml:id="echoid-s717" xml:space="preserve">d’Euclide, cioè, che le parti hanno le propor-<lb/>tioni de’ ſuoi intieri, ele proportioni ſimili ad vna ſteſſa pro-<lb/>portione ſono ſimili trà di loro. </s> <s xml:id="echoid-s718" xml:space="preserve">E perciò ſe ſia come A al B, <lb/>così C al D, eſſendo {1/2} A al {1/2} B, come A al B, anche ſarà co-<lb/>me {1/2} A al {1/2} B, così C al D, eſſendo come C al D, così {1/3} C al {1/3} <lb/>D ſarà per conſeguenza, come {1/2} A al {1/2} B, così {1/3} C al {1/3} D. <lb/></s> <s xml:id="echoid-s719" xml:space="preserve">E perche ſe come A al B, così C al D, vale anche permutan-<lb/>do, come A al C, così B al D, ne ſeguirà con l’iſteſſo diſcorſo, <lb/>che come {1/2} A al {1/3} C, così {1/2} B al {1/3} D. </s> <s xml:id="echoid-s720" xml:space="preserve">Et in tal modo è manife-<lb/>ſta la ragione delle ſopraccennate operationi. </s> <s xml:id="echoid-s721" xml:space="preserve">E quello, che <pb o="38" file="0050" n="51" rhead="CAPO II."/> quì s’è detto de gl’Intieri riſpetto alle loro parti, così vale la <lb/>forma di diſcorrere delle parti, riſpetto de gl’Intieri, fatta ſo-<lb/>lo la conuerſione de’ter mini, per ciò che appreſſo ſi dirà de <lb/>gl’Intieri riſpetto de’ ſuoi moltiplici. </s> <s xml:id="echoid-s722" xml:space="preserve">Il che hò voluto così <lb/>breuemente accennare, per non replicar con tedio più volte <lb/>lo ſteſſo.</s> <s xml:id="echoid-s723" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s724" xml:space="preserve">Quarto, ſe ſolo il ſecondo Antecedente ſarà troppo picco-<lb/>lo, baſterà raddoppiarlo, ò triplicarlo, e ſeruirſi di queſto, co-<lb/>me ſe foſſe il vero Antecedente, perche del numero, che ſi tro-<lb/>uerà, dourà pigliarſi la metà, ò il terzo, per hauer il numero, <lb/>che ſi cerca. </s> <s xml:id="echoid-s725" xml:space="preserve">Per eſſempio. </s> <s xml:id="echoid-s726" xml:space="preserve">V na fontana, che getta l’acqua <lb/>fempre vniformemente, hà riempito vn vaſo capace di 54 <lb/>botti d’acqua in 23.</s> <s xml:id="echoid-s727" xml:space="preserve">ore, quant’ore ci vogliono per empir vno <lb/>capace di ſol 7 botti? </s> <s xml:id="echoid-s728" xml:space="preserve">Piglio dal centro ſin al punto 23. </s> <s xml:id="echoid-s729" xml:space="preserve">e <lb/>queſta diſtanza applico all interuallo 54. </s> <s xml:id="echoid-s730" xml:space="preserve">54. </s> <s xml:id="echoid-s731" xml:space="preserve">Dipoi perche <lb/>7.</s> <s xml:id="echoid-s732" xml:space="preserve">7. </s> <s xml:id="echoid-s733" xml:space="preserve">è troppo vicino, piglio la diſtanza 14. </s> <s xml:id="echoid-s734" xml:space="preserve">14. </s> <s xml:id="echoid-s735" xml:space="preserve">e queſta ap-<lb/>plicata dal centro cade ſul punto 6; </s> <s xml:id="echoid-s736" xml:space="preserve">onde perche il 7 ſi rad-<lb/>doppiò, prendo la metà di 6, e dico; </s> <s xml:id="echoid-s737" xml:space="preserve">che in 3 ore s’em pirà il <lb/>vaſo capace di ſol 7 botti. </s> <s xml:id="echoid-s738" xml:space="preserve">E’vero, che ciè qualche differen-<lb/>za, e non ſono preciſamente 3 ore, mà ſolo 2 {53/54}, il che nell’ <lb/>operatione, c’habbiamo per la mano, non è da conſiderarſi.</s> <s xml:id="echoid-s739" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s740" xml:space="preserve">Quinto, mà ſe ſolo il Primo Antecedente, ò ſolo il Primo <lb/>Conſeguente, ò ambidue, ò l’vn, el’altro Antecedente foſſero <lb/>troppo piccioli, tutti due gl’Antecedenti, ò li Primi Antece-<lb/>dente, e<unsure/> Conſeguente, ſimilmente ſi moltiplichino, raddop-<lb/>pino, ò triplichino, e s’opri, come ſe queſti foſſero li numeri <lb/>dati, perche ne verrà il numero cercato. </s> <s xml:id="echoid-s741" xml:space="preserve">Così s’io dico 7 mi <lb/>dà 10, che midarà 3? </s> <s xml:id="echoid-s742" xml:space="preserve">raddoppio il 7, & </s> <s xml:id="echoid-s743" xml:space="preserve">il 3, come troppo <lb/>piccioli, & </s> <s xml:id="echoid-s744" xml:space="preserve">opro, come ſe cercaſſi, 14 midà 10, che mi darà <lb/>6? </s> <s xml:id="echoid-s745" xml:space="preserve">e trouo, ch’è vn poco più di 4.</s> <s xml:id="echoid-s746" xml:space="preserve"/> </p> <pb o="39" file="0051" n="52" rhead="Prattica in numeri della Rego’a de’Trè."/> <p> <s xml:id="echoid-s747" xml:space="preserve">Seſto, ſe tutti trè li numeri dati ſono troppo piccioli, òtut. <lb/></s> <s xml:id="echoid-s748" xml:space="preserve">ti ſi moltiplichino vgualmente, & </s> <s xml:id="echoid-s749" xml:space="preserve">il numero, che ſitrouerà <lb/>dourà diuiderſi per il moltiplicatorepreſo, come ſe tutti ſi <lb/>raddoppiarono, ſi deue prendere la metà del trouato, per ha-<lb/>uer quello, che ſi cercaua, come è maniſeſto. </s> <s xml:id="echoid-s750" xml:space="preserve">Ouero due, <lb/>cioè ò li due Antecedenti, ò li due Primi termini ſi ponno <lb/>moltiplicare ſimilmente, e l’altro numero moltiplicar altri-<lb/>menti, perche quel che ſi trouerà, ſi dourà diuidere per il nu-<lb/>mero, che moltiplicò queſt’ vltimo. </s> <s xml:id="echoid-s751" xml:space="preserve">Per eſſempio: </s> <s xml:id="echoid-s752" xml:space="preserve">d’vn <lb/>drappo alto cinque quarte il Sarto me ne fece prendere brac-<lb/>cia 7 {1/2}, ora per far vna ſimil veſte d’vn drappo alto ſol 3 <lb/>quarte, quante braccia hò à comprarne? </s> <s xml:id="echoid-s753" xml:space="preserve">E’certo, che quì è <lb/>la proportione euerſa, cioè che le altezze, e le lunghezze ſo-<lb/>no reciprocamente proportionali, e come la ſeconda altezza <lb/>alla prima aitezza, così la prima lunghezza alla ſeconda lun-<lb/>ghezza, che ſi cerca: </s> <s xml:id="echoid-s754" xml:space="preserve">Sidice dunque, come 3 al 5, così 7 {1/2} ad <lb/>vn altro: </s> <s xml:id="echoid-s755" xml:space="preserve">quadruplico il 3, & </s> <s xml:id="echoid-s756" xml:space="preserve">il 7 {1/2}, eſono 12, e 30; </s> <s xml:id="echoid-s757" xml:space="preserve">duplico <lb/>il 5, & </s> <s xml:id="echoid-s758" xml:space="preserve">è 10. </s> <s xml:id="echoid-s759" xml:space="preserve">Oprodunque con queſtitrè numeri 12, 10, 30; </s> <s xml:id="echoid-s760" xml:space="preserve"><lb/>e preſadal centro la diſtanza ſin al punto 10, l’applico al 12. </s> <s xml:id="echoid-s761" xml:space="preserve"><lb/>12; </s> <s xml:id="echoid-s762" xml:space="preserve">e preſo l’interuallo 30. </s> <s xml:id="echoid-s763" xml:space="preserve">30, trouo eſſere 25. </s> <s xml:id="echoid-s764" xml:space="preserve">Ora perche <lb/>il 5 ſolo ſi duplicò, piglio la metà di 25, edico, che del ſecon-<lb/>do drappo me ne fan di meſtieri braccia 12 {1/2}. </s> <s xml:id="echoid-s765" xml:space="preserve">Equeſto ſteſ-<lb/>ſo haurei trouato, ſe haueſſi duplicato r<unsure/>utti trè li numeri; </s> <s xml:id="echoid-s766" xml:space="preserve">per-<lb/>che come 6 al 10, così 7 {1/2} al 12 {1/2}.</s> <s xml:id="echoid-s767" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s768" xml:space="preserve">Mà perche ſpeſſo occorre, che l’interuallo, che ſi troua, <lb/>non cade preciſamente ſul punto ſegnato da qualche numero <lb/>intiero, ſi potrà trouare la frattione, & </s> <s xml:id="echoid-s769" xml:space="preserve">auuicinarſi più al ve-<lb/>ro in queſto modo. </s> <s xml:id="echoid-s770" xml:space="preserve">Si prenda dal centro dello ſtro mento con <lb/>vn’altro Compaſſo la diſtanza ſin’al punto proſſimamente <lb/>maggiore, & </s> <s xml:id="echoid-s771" xml:space="preserve">il numero dital punto ſi moltiplichi, quanto ſi <pb o="40" file="0052" n="53" rhead="C A P O II."/> può, purch<unsure/>e non paſſi il 100, & </s> <s xml:id="echoid-s772" xml:space="preserve">allargato lo Stromento, à <lb/>queſto numero moltiplice s’applichi la lunghezza preſa con <lb/>queſto ſecondo Compaſſo; </s> <s xml:id="echoid-s773" xml:space="preserve">e poi ſi vegga in qual’ interuallo <lb/>capiſca la longhezza trouata col primo Compaſſo; </s> <s xml:id="echoid-s774" xml:space="preserve">perche la <lb/>frattione aderente all’intiero già conoſciuto, haurà per De-<lb/>nominatore il numero, che fù il moltiplicatore, e quanti pun-<lb/>ti ſi trouano mancare per giungcr à quella diſtanza maggio-<lb/>re, tanta deue eſſere la differenza tra’l<unsure/> Numeratore, & </s> <s xml:id="echoid-s775" xml:space="preserve">il De-<lb/>nominatore della frattione. </s> <s xml:id="echoid-s776" xml:space="preserve">Sia per eſſempio nell’o peratione <lb/>trouata vna tal lunghezza, che applicata dal centro cada tra <lb/>li punti 19, e 20; </s> <s xml:id="echoid-s777" xml:space="preserve">onde s’arguiſce, che il numero cercato è 19 <lb/>con vnafrattione. </s> <s xml:id="echoid-s778" xml:space="preserve">Ora con vn ſecondo Compaſſo preſala <lb/>diſtanza dal centro ſin’à 20, ſe applico queſta al 40. </s> <s xml:id="echoid-s779" xml:space="preserve">40, che <lb/>è duplo di 20, non mi può dare ſe non {1/2}, ſe al 60. </s> <s xml:id="echoid-s780" xml:space="preserve">60, che è <lb/>triplo, poſſo trouar li Terzi, ſe al 80. </s> <s xml:id="echoid-s781" xml:space="preserve">80, che è quadruplo, <lb/>trouerò li Quarti, e finalmente ſe al 100. </s> <s xml:id="echoid-s782" xml:space="preserve">100, che è quintu-<lb/>plo, trouerò li Quinti. </s> <s xml:id="echoid-s783" xml:space="preserve">Sia dunque applicata alli 100. </s> <s xml:id="echoid-s784" xml:space="preserve">100: <lb/></s> <s xml:id="echoid-s785" xml:space="preserve">e poi col primo Compaſſo, che daua quella miſura minore di <lb/>20, e maggiore di 19, veggoin qualinteruallo ſi poſſa appll-<lb/>care, etrouo che al 97. </s> <s xml:id="echoid-s786" xml:space="preserve">97, onde mancando 3 al 100 dico, <lb/>chela frattione aderente al 19 è {2/5}; </s> <s xml:id="echoid-s787" xml:space="preserve">ſe ſi foſſe applicata al99, <lb/>ſaria ſtato il numero cercato 19 {4/5}.</s> <s xml:id="echoid-s788" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s789" xml:space="preserve">La ragione di queſta operatione è, perche quelle 20 par-<lb/>ticelle applicate al 100. </s> <s xml:id="echoid-s790" xml:space="preserve">100, vengono come ad eſſere diuiſe <lb/>in 100 parti, cioè ciaſcuna ne’ſuoi quinti; </s> <s xml:id="echoid-s791" xml:space="preserve">ora ſe di quali 100 <lb/>parti ſono le 20, ditali 97 ſono quell’altre, è manifeſto; </s> <s xml:id="echoid-s792" xml:space="preserve">che à <lb/>queſte mancano {3/5} per arriuar à 20, ecosì ſono 19 {2/5}. </s> <s xml:id="echoid-s793" xml:space="preserve">Mà ſe <lb/>la diſtanza prima trouata foſſe ſtata maggiore di 24, e dal <lb/>centro ſin à 25 ſi foſſe applicata al 100. </s> <s xml:id="echoid-s794" xml:space="preserve">100, la frattione ſa-<lb/>ria di Quarti, e cadendo la diſtanza trouata ſul 97. </s> <s xml:id="echoid-s795" xml:space="preserve">97, ſaria <pb o="41" file="0053" n="54" rhead="Prattica in numeri della Regola delTrè."/> il numero cercato 24 {1/4}, poiche mancano {3/4}, per eſſere {100/4}, <lb/>cioè 25.</s> <s xml:id="echoid-s796" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s797" xml:space="preserve">Forſi riuſcirà ad alcuno più facile queſt’altro modo. </s> <s xml:id="echoid-s798" xml:space="preserve">Quan-<lb/>dola miſura trouata, e dalcentro applicata ſula linea dello <lb/>Stromento non cade in vn punto intiero, pigliſi con vn’altro <lb/>Compaſſo la miſura ſin al punto proſſimamente minore: </s> <s xml:id="echoid-s799" xml:space="preserve">& </s> <s xml:id="echoid-s800" xml:space="preserve">il <lb/>numero di tal punto moltiplicato, sì che non arriui à 100, s’a-<lb/>pra lo Stromento, & </s> <s xml:id="echoid-s801" xml:space="preserve">al punto, che cortiſponde al numero <lb/>moltiplicato, s’applichi la lunghezza preſa col ſecondo Com-<lb/>paſſo; </s> <s xml:id="echoid-s802" xml:space="preserve">poi applicatala miſura, che dà il primo Compaſſo, il <lb/>numero de’ punti, ehe eccedono quel moltiplicato, ſarà il <lb/>Numeratore della frattione, il cui Denominatore è quel che <lb/>fù il Moltiplicatore. </s> <s xml:id="echoid-s803" xml:space="preserve">Sia la miſura trouata maggiore di 17: <lb/></s> <s xml:id="echoid-s804" xml:space="preserve">Prendo con vn’altro Compaſſo dal centro ſinal punto 17; </s> <s xml:id="echoid-s805" xml:space="preserve">e <lb/>queſta diſtanza applico al numero 68. </s> <s xml:id="echoid-s806" xml:space="preserve">68, quadruplo del 17: </s> <s xml:id="echoid-s807" xml:space="preserve"><lb/>e perciò la frattione haurà il 4 per Denominatore: </s> <s xml:id="echoid-s808" xml:space="preserve">applicata <lb/>poi quella miſura trouata maggiore di 17, trouo che capiſce <lb/>al 71. </s> <s xml:id="echoid-s809" xml:space="preserve">71: </s> <s xml:id="echoid-s810" xml:space="preserve">e perciò dico, che eſſendo l’ecceſſo di 3 punti, la <lb/>frattione ſarà {3/4}, ecosì il numero, che ſi cercaua è 17 {3/4}.</s> <s xml:id="echoid-s811" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s812" xml:space="preserve">Laragione di queſto modo d’operare è, perche in quell’-<lb/>applicatione al numero quadruplo vengono le 17 vnità ad <lb/>eſſer diuiſe in tutti i ſuoi Quarti, che ſono 68; </s> <s xml:id="echoid-s813" xml:space="preserve">dunque ſe la <lb/>miſur a trouata hà di tali Quarti 71, ſarà il ſuo numero 17 {3/4}.</s> <s xml:id="echoid-s814" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s815" xml:space="preserve">Auuertaſi quì, che può occorrere, che la miſura tolta col <lb/>primo Compaſſo non poſſa applicarſi preciſamente a due <lb/>punti ſimili, come 71, e 71; </s> <s xml:id="echoid-s816" xml:space="preserve">ma ſolo a 71, e 72; </s> <s xml:id="echoid-s817" xml:space="preserve">& </s> <s xml:id="echoid-s818" xml:space="preserve">in tal caſo <lb/>èſegno, che è più di trè quarti: </s> <s xml:id="echoid-s819" xml:space="preserve">e ſe cade così preciſamente <lb/>ſu due punti 71, e 72, ſi può prendere per vna metà; </s> <s xml:id="echoid-s820" xml:space="preserve">ſe ca-<lb/>deſſe ſul 71, & </s> <s xml:id="echoid-s821" xml:space="preserve">alla metà del 72, ſi potria prendere per vn <lb/>Quarto. </s> <s xml:id="echoid-s822" xml:space="preserve">Ora mettiamo, che cada ſu li 71. </s> <s xml:id="echoid-s823" xml:space="preserve">72; </s> <s xml:id="echoid-s824" xml:space="preserve">e così oltre <pb o="42" file="0054" n="55" rhead="C A P O II."/> li {3/4}, v’èla metà d’vn Quarto, che è {1/8}, che aggiunto alli {3/4} ſo-<lb/>no in tutto {7/8}. </s> <s xml:id="echoid-s825" xml:space="preserve">Sefoſſe caduto alla metà del 72. </s> <s xml:id="echoid-s826" xml:space="preserve">era vn Quarto <lb/>d’vn Quarto, cioè {1/16}, ecosì tutta la frattionc {13/16}.</s> <s xml:id="echoid-s827" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s828" xml:space="preserve">E per non laſciare di ſpiegare anche meglio l’vſo di queſto <lb/>Stromento, per trouare con più preciſione le frattioni aggiun-<lb/>te agl’intieri, ſenza obligarcia prendere li numeri moltiplici, <lb/>maſſime, che bene ſpeſſo appena ſi ponno raddoppiare, ò tri-<lb/>plicare; </s> <s xml:id="echoid-s829" xml:space="preserve">perciò aggiungerò anche queſto modo d’operare. <lb/></s> <s xml:id="echoid-s830" xml:space="preserve">Preſo dunque, come ſi diſſe, con vn ſecondo Compaſſo dal <lb/>centro ſin al numero proſſimamente minore, s’apra lo Stro-<lb/>mento, e queſta diſtanza s’applichi a quell’interuallo, che più <lb/>piace, in maniera però, che poila diſtãza, che dà l’altro Com-<lb/>paſſo poſſa capire almeno tra 100. </s> <s xml:id="echoid-s831" xml:space="preserve">100; </s> <s xml:id="echoid-s832" xml:space="preserve">& </s> <s xml:id="echoid-s833" xml:space="preserve">il numero dital <lb/>interuallo ſarà il Denominatore della frattione. </s> <s xml:id="echoid-s834" xml:space="preserve">Di poi rite-<lb/>nuta l’a pertura medeſima dello Stromento, ſi vegga in qual <lb/>interuallo capiſca la prima miſura. </s> <s xml:id="echoid-s835" xml:space="preserve">Il numero de’ punti, che <lb/>queſto ſecondo interuallo è diſtante dal primo già coſtituito, <lb/>ſi moltiplichi per l’Intiero nu mero, che ſi preſe proſſimamen-<lb/>ce<unsure/> minore; </s> <s xml:id="echoid-s836" xml:space="preserve">e ciò per la molti plicatione ſi produce, ſarà il Nu-<lb/>meratore della frattione.</s> <s xml:id="echoid-s837" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s838" xml:space="preserve">Sia la miſura trouata maggiore di 6, ma minore di 7. </s> <s xml:id="echoid-s839" xml:space="preserve">Pren-<lb/>do dal centro ſin al 6, e queſta diſtanza applico ad arbitrio ad <lb/>wn numero, per eſſempio al 50. </s> <s xml:id="echoid-s840" xml:space="preserve">50:</s> <s xml:id="echoid-s841" xml:space="preserve">e perciò le parti della frat-<lb/>tione ſaranno cinquanteſime. </s> <s xml:id="echoid-s842" xml:space="preserve">Quindi applicata la miſura <lb/>trouata, veggo che cade ſul 53, 53. </s> <s xml:id="echoid-s843" xml:space="preserve">Dunque preſo l’ecceſſo <lb/>3, lo moltiplico per il numero intiero 6, e ſi fà 18, per nu-<lb/>meratoredella frattione; </s> <s xml:id="echoid-s844" xml:space="preserve">e perciò dico, che la miſura trouata <lb/>dà il nu mero cercato 6 {18/50}.</s> <s xml:id="echoid-s845" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s846" xml:space="preserve">La dimoſtratione di queſta operatione ſi vede dalla figura <lb/>preſente doue BC è parallela alla DE, e prendendoſi BF <pb o="43" file="0055" n="56" rhead="Fratticain numeri della Regola delTrè."/> <anchor type="figure" xlink:label="fig-0055-01a" xlink:href="fig-0055-01"/> vguale alla DE, e congiun-<lb/>gendoſi li punti E, F con <lb/>vna linea retta EF, viene <lb/>ad eſſer EF parallela alla <lb/>BD per la 33. </s> <s xml:id="echoid-s847" xml:space="preserve">del libro 1. <lb/></s> <s xml:id="echoid-s848" xml:space="preserve">Dunque per la 2. </s> <s xml:id="echoid-s849" xml:space="preserve">del lib. </s> <s xml:id="echoid-s850" xml:space="preserve">6. </s> <s xml:id="echoid-s851" xml:space="preserve">come AE ad EC, così BF à FC: </s> <s xml:id="echoid-s852" xml:space="preserve"><lb/>dunque il rettangolo fatto dalle due EC, BF, cioè DE, appli-<lb/>cato alla prima AE darà la FC: </s> <s xml:id="echoid-s853" xml:space="preserve">come appariſce dalla 16. </s> <s xml:id="echoid-s854" xml:space="preserve">del <lb/>lib. </s> <s xml:id="echoid-s855" xml:space="preserve">6. </s> <s xml:id="echoid-s856" xml:space="preserve">Se dunque DE è il numero 6. </s> <s xml:id="echoid-s857" xml:space="preserve">collocato ſu lo Stromen-<lb/>to nelli punti 50. </s> <s xml:id="echoid-s858" xml:space="preserve">50, cioè in AD, AE, ela miſura trouata BC <lb/>s’addatta alli punti B, & </s> <s xml:id="echoid-s859" xml:space="preserve">C 53. </s> <s xml:id="echoid-s860" xml:space="preserve">53, ſarà come AE 50, ad EC <lb/>3, così Bf, cioè DE 6 alla FC; </s> <s xml:id="echoid-s861" xml:space="preserve">e perciò EC 3 moltiplicando <lb/>DE 6 ſà 18 da diuiderſi per AE 50; </s> <s xml:id="echoid-s862" xml:space="preserve">onde il Quotiente {18/56} è la <lb/>FC da aggiungerſi alla BF, cioè alla De 6; </s> <s xml:id="echoid-s863" xml:space="preserve">ecosì tutta la BC <lb/>è 6 {18/50} numero cercato.</s> <s xml:id="echoid-s864" xml:space="preserve"/> </p> <div xml:id="echoid-div35" type="float" level="2" n="2"> <figure xlink:label="fig-0055-01" xlink:href="fig-0055-01a"> <image file="0055-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0055-01"/> </figure> </div> <p> <s xml:id="echoid-s865" xml:space="preserve">Di quì ſi vede, che ſe le due miſure preſe co’due Compaſſi, <lb/>come s’è detto, cadeſſero in tal apertura dello Stromento, che <lb/>non foſſero diſtanti, che vn punto ſolo, il Numeratore della <lb/>frattione ſarà il numero intiero preſo. </s> <s xml:id="echoid-s866" xml:space="preserve">Come per eſſempio, <lb/>ſe il numero è 27, & </s> <s xml:id="echoid-s867" xml:space="preserve">è applicato all’interuallo 43. </s> <s xml:id="echoid-s868" xml:space="preserve">43, e l’altra <lb/>miſura cade ſul 44. </s> <s xml:id="echoid-s869" xml:space="preserve">44, diremo, che il numero cercato è <lb/>27 {27/44}. </s> <s xml:id="echoid-s870" xml:space="preserve">Laragione è, perche l’vnità moltiplicando il 27 non <lb/>lo muta.</s> <s xml:id="echoid-s871" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s872" xml:space="preserve">Finalmente s’auuerta in queſto modo, che ſe la diſtanza <lb/>EC foſſe di molti punti, & </s> <s xml:id="echoid-s873" xml:space="preserve">il numero DE foſſe così grande, <lb/>che riuſciſſe difficile moltiplicarlo per EC così alla mente, ſi <lb/>dourà applicare la DE più vicina al centro A, che così la BC <lb/>riuſcirà più vicina alla DE, & </s> <s xml:id="echoid-s874" xml:space="preserve">EC ſarà numero minore.</s> <s xml:id="echoid-s875" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s876" xml:space="preserve">In vn’altra maniera potiamo ſeruirci di queſto stromento <lb/>per trouar il quarto numero proportionale ſenza applicar i <pb o="44" file="0056" n="57" rhead="C A P O II."/> numeri al lato dello Stromento, ma a gl’interualli: </s> <s xml:id="echoid-s877" xml:space="preserve">e poten-<lb/>doci ogni punto ſeruir per due, anche ſenza compaſſo molto <lb/>grande faremo ciò che deſideriamo. </s> <s xml:id="echoid-s878" xml:space="preserve">Per eſſempio 168 mi <lb/>dà 72, che coſa mi darà 63? </s> <s xml:id="echoid-s879" xml:space="preserve">Diuido li 168, & </s> <s xml:id="echoid-s880" xml:space="preserve">li 72 per me-<lb/>tà, e ſono 84, e 36. </s> <s xml:id="echoid-s881" xml:space="preserve">A qualunque apertura dello Stromento <lb/>prendo l’interuallo 84. </s> <s xml:id="echoid-s882" xml:space="preserve">84, con vn compaſſo, e col ſecondo <lb/>compaſſo alla ſteſſa apertura dello Stromento prendo 36, 36. <lb/></s> <s xml:id="echoid-s883" xml:space="preserve">Ritengo li Compaſſi così, & </s> <s xml:id="echoid-s884" xml:space="preserve">applico il primo compaſſo al <lb/>terzo numero dato, cioè à 63. </s> <s xml:id="echoid-s885" xml:space="preserve">63. </s> <s xml:id="echoid-s886" xml:space="preserve">allargando lo Stromento, <lb/>& </s> <s xml:id="echoid-s887" xml:space="preserve">a queſta apertura applicando il ſecondo compaſſo, trouo <lb/>che cade nell’interuallo 27. </s> <s xml:id="echoid-s888" xml:space="preserve">27. </s> <s xml:id="echoid-s889" xml:space="preserve">onde conchiudo, che il quar-<lb/>to numero cercato è 27. </s> <s xml:id="echoid-s890" xml:space="preserve">Queſta prattica è manifeſta per la <lb/>coſtruttione dello Stromento; </s> <s xml:id="echoid-s891" xml:space="preserve">perche di quali parti 84 era <lb/>la prima linea compreſa dal primo compaſlo, di tali 36 era <lb/>la ſeconda: </s> <s xml:id="echoid-s892" xml:space="preserve">ora preſa la prima di 63, la ſeconda viene ad eſ-<lb/>ſere di 27.</s> <s xml:id="echoid-s893" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s894" xml:space="preserve">Queſto modo d’operare moſtra vna grandiſſima facilità per <lb/>ſciogliere le queſtioni appartenenti al moltiplico de’capitali, <lb/>quando corrono intereſſi ſopra intereſſi, cioè che il frutto di <lb/>ciaſcun anno a capo d’anno s’accreſce al capitale: </s> <s xml:id="echoid-s895" xml:space="preserve">il che ſi fà, <lb/>eſſendo noto, quanto per cento ſia il frutto, perche ſe il 100 <lb/>guadagna nel primo anno per eſſempio 4. </s> <s xml:id="echoid-s896" xml:space="preserve">ſarà il capitale del <lb/>ſecondo anno 104; </s> <s xml:id="echoid-s897" xml:space="preserve">e così biſogna dire, ſe 100 a capo del pri-<lb/>mo anno dà 104, che coſa darà 104 a capo del ſecondo anno? <lb/></s> <s xml:id="echoid-s898" xml:space="preserve">e ſi troua, che dà 108 {16/100}. </s> <s xml:id="echoid-s899" xml:space="preserve">E poi ſeguitando all’ iſteſſo modo <lb/>a replicare la regola del Trè, ſe 100 dà 104, che coſa darà <lb/>108 {16/100} a capo del terzo anno? </s> <s xml:id="echoid-s900" xml:space="preserve">tante volte ſi replicherà, quan-<lb/>ti ſon gl’anni, che ſi laſcia il denaro a moltiplico. </s> <s xml:id="echoid-s901" xml:space="preserve">Il che, co-<lb/>me ſi vede, porta tempo, e fatica nel calcolo. </s> <s xml:id="echoid-s902" xml:space="preserve">Ma ſe le linee <lb/>Aritmetiche dello Stromento ſono accuratamente diuiſe, <pb o="45" file="0057" n="58" rhead="Prattica in numeri della Regola del Trè."/> queſta operatione ſi farà con pochiſſimo trauaglio.</s> <s xml:id="echoid-s903" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s904" xml:space="preserve">Sapendoſi quanto per cento ſi guadagna, prendaſi la metà <lb/>del 100, che è 50, ela metà del frutto annuo: </s> <s xml:id="echoid-s905" xml:space="preserve">& </s> <s xml:id="echoid-s906" xml:space="preserve">aperto lo <lb/>Stromento ad arbitrio, prendaſi l’interuallo 50. </s> <s xml:id="echoid-s907" xml:space="preserve">50, ma con-<lb/>ſeruiſi il compaſlo così aperto, come ſi preſe queſta prima <lb/>miſura, ouero ſi tiri vna linea vguale à tal’apertura, per hauer-<lb/>ne memoria, ouero ſi prenda queſta prima lunghezza vguale <lb/>ad vn numero determinato di punti preſi ſul lato dello Stro-<lb/>mento; </s> <s xml:id="echoid-s908" xml:space="preserve">e poi con vn’altro Compaſſo (ſe per altro in vno de’ <lb/>modi detti non ſi conſeruaſſe memoria della prima larghez-<lb/>za) eſſendo ancora lo Stromento allargato come prima, ſi <lb/>prenda l’interuallo corriſpondente alla metà del capitale, e <lb/>del frutto; </s> <s xml:id="echoid-s909" xml:space="preserve">e così ſe il frutto è 4 per 100, prendaſi 52. </s> <s xml:id="echoid-s910" xml:space="preserve">52, ſe <lb/>foſſe 6 per 100, prendaſi 53. </s> <s xml:id="echoid-s911" xml:space="preserve">53; </s> <s xml:id="echoid-s912" xml:space="preserve">e così de gl’altri. </s> <s xml:id="echoid-s913" xml:space="preserve">Queſta <lb/>larghezza vltima di Compaſſo per il ſecondo anno, di nuouo <lb/>s’applichi al 50. </s> <s xml:id="echoid-s914" xml:space="preserve">50, allargando lo Stromento, e di nuouo ſi <lb/>prenda il 52. </s> <s xml:id="echoid-s915" xml:space="preserve">52, ſe fù alli 4, ouero il 53. </s> <s xml:id="echoid-s916" xml:space="preserve">53, ſe fù alli 6 per <lb/>100. </s> <s xml:id="echoid-s917" xml:space="preserve">Di nuouo queſt’vltima lunghezza per ilterzo anno s’ap-<lb/>plichi al 50. </s> <s xml:id="echoid-s918" xml:space="preserve">50, con allargare lo Stromento, & </s> <s xml:id="echoid-s919" xml:space="preserve">al 52. </s> <s xml:id="echoid-s920" xml:space="preserve">52 s’ha-<lb/>urà la lunghezza conueniente al terzo anno; </s> <s xml:id="echoid-s921" xml:space="preserve">e così tante vol-<lb/>te, quanti ſon gl’anni, che ſi laſcia a moltiplico. </s> <s xml:id="echoid-s922" xml:space="preserve">Finalmente <lb/>ſi paragoni la prima larghezza, che fù preſa da principio con <lb/>queſt’vltima trouata; </s> <s xml:id="echoid-s923" xml:space="preserve">ela proportione di quella prima a <lb/>queſt’vltima è la proportione del capitale meſſo da principio <lb/>allo ſteſſo accreſciuto d’anno in anno, con i frutti, che diuen-<lb/>tarono capitale. </s> <s xml:id="echoid-s924" xml:space="preserve">Così ſe furono alli 4 per 100, troueremo che <lb/>li 100 in capo a dieci anni diuentano 148 {1/4} quaſi, cioè vn <lb/>poco più d’vn quinto: </s> <s xml:id="echoid-s925" xml:space="preserve">Onde dico, ſe in dieci anni 100 mi <lb/>danno 148 {1/4}, nello ſteſſo tempo vn capitale di dieci mila <lb/>ſcudi diuerrà 148 25.</s> <s xml:id="echoid-s926" xml:space="preserve"/> </p> <pb o="46" file="0058" n="59" rhead="C A P O II."/> <p> <s xml:id="echoid-s927" xml:space="preserve">In altra maniera ſi può operare ritenendo ſempre la me-<lb/>deſima apertura dello Stromento, ma prendendo nel ſuo lato <lb/>inumeri. </s> <s xml:id="echoid-s928" xml:space="preserve">Per eſſempio ſia al 4 per 100: </s> <s xml:id="echoid-s929" xml:space="preserve">prendaſi dal centro <lb/>A ſin al 52 la diſtanza, e queſta ſi metta tra 50, 50, e queſta <lb/>è l’apertura dello Stromento ſenza mutarla. </s> <s xml:id="echoid-s930" xml:space="preserve">Ora prendaſi <lb/>la metà del numero del capitale, e ſe è troppo grande, pren-<lb/>daſi vna parte aliquota di eſſo; </s> <s xml:id="echoid-s931" xml:space="preserve">come ſe foſſe il capitale 300 <lb/>Scudi, la ſua metà è 150, prendaſi 75, che è la 4. </s> <s xml:id="echoid-s932" xml:space="preserve">parte. </s> <s xml:id="echoid-s933" xml:space="preserve">E col <lb/>compaſſo preſo l’interuallo 75. </s> <s xml:id="echoid-s934" xml:space="preserve">75, mettaſi vna punta nel cen-<lb/>tro, e ſu li lati dello Stromento leggiermente ſi ſegni con l’al-<lb/>tra punta; </s> <s xml:id="echoid-s935" xml:space="preserve">prendaſi queſto interuallo tra li ſegni fatti, e di <lb/>nuouo dal centro ſi traporti, e ſegniſu li lati; </s> <s xml:id="echoid-s936" xml:space="preserve">e ciò tante volte <lb/>ſi replichi, quanti ſono gli anni: </s> <s xml:id="echoid-s937" xml:space="preserve">così ſe foſſero cinque anni, <lb/>ſi prendano cinque volte gl’interualli, e l’vltimo, cioè il quin-<lb/>to interuallo traportato dal centro ſul lato dello Stromento, <lb/>darà il numero cercato; </s> <s xml:id="echoid-s938" xml:space="preserve">e caderà proſſimamente al punto 91. <lb/></s> <s xml:id="echoid-s939" xml:space="preserve">Si che 75 ſcudi a capo di cinque anni danno 91 ſcudi proſſi-<lb/>mamente; </s> <s xml:id="echoid-s940" xml:space="preserve">e perche 75 è la quarta parte di 300, diremo che <lb/>300 ſcudi a capo di cinque anni ſaranno proſſimamente ſcu-<lb/>di 364. </s> <s xml:id="echoid-s941" xml:space="preserve">Di queſto modo d’operare la ragione è manifeſta, <lb/>perche ritenuta ſempre l’apertura medeſima dello Stromento <lb/>tutti i lati a gl’interualli ſono come 50 à 52, cioè 100 a 104; </s> <s xml:id="echoid-s942" xml:space="preserve"><lb/>e perche gl’interualli ſucceſſiuamente ſi traportano ſu li lati, <lb/>perciò ſempre ſi cõtinua la proportione iſteſſa di 100 a 104.</s> <s xml:id="echoid-s943" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s944" xml:space="preserve">Che ſe haueſſi curioſità di prouarlo col calcolo, ſe non <lb/>prenderai di volta in volta le frattioni proſſime alla vera <lb/>ora maggiori, ora minori, ma tutta la frattione intiera <lb/>(la quale è nel ſecondo anno di centeſime, nel terzo di dieci-<lb/>milleſime, e così ogn’anno aggiungendo due zeri al denomi-<lb/>natore) trouerai nel decimo anno vna frattione, che haurà <pb o="47" file="0059" n="60" rhead="Prattica in numeri della Regola del Trè."/> per denominatore l’vnità con diciotto zeri, & </s> <s xml:id="echoid-s945" xml:space="preserve">il numeratore <lb/>tale, che è proſſimo ad vn quarto d’vnità. </s> <s xml:id="echoid-s946" xml:space="preserve">E ſe cercaſſi<unsure/> per <lb/>vent’anni, l’vltimo denominatore ſaria di 38 zeri, ſempre due <lb/>meno del doppio del numero de gl’anni, eſſendo che per il <lb/>primo anno non ſi fà la diuiſione per 100, e per gli altri anni <lb/>ſi aggiongono ſempre due zeri al denominatore. </s> <s xml:id="echoid-s947" xml:space="preserve">In ſomma <lb/>(perche queſte coſe ſi ſcriuono per li meno eſperti) baſterà <lb/>per il fecondo anno moltiplicar il capitale col frutto in ſe ſteſ-<lb/>ſo, e per l’iſteſſo capitale col frutto, cioè per 104, ouero 105, <lb/>ò altro, moltiplicar di mano in mano i prodotti; </s> <s xml:id="echoid-s948" xml:space="preserve">e poi veden-<lb/>do quante volte hai fatto tal moltiplicatione, taglia dal nu-<lb/>mero vltimamente prodotto due volte altre tante figure; </s> <s xml:id="echoid-s949" xml:space="preserve">co-<lb/>me ſe hai fatto la moltiplicatione cinque volte, taglia alla de-<lb/>ſtra dieci figure, e queſte ſono il numeratore della frattione <lb/>aderente al numero d’intieri ſignificato dall’altre figure re-<lb/>ſtanti; </s> <s xml:id="echoid-s950" xml:space="preserve">e queſto ſaria il moltiplico del capitale fatto in 6 anni. <lb/></s> <s xml:id="echoid-s951" xml:space="preserve">Onde ſi vede eſſer quaſi vna progreſſione Geometrica, la cui <lb/>Radice è il capitale col frutto, cioè 104, &</s> <s xml:id="echoid-s952" xml:space="preserve">c. </s> <s xml:id="echoid-s953" xml:space="preserve">principiante <lb/>dall’vnità. </s> <s xml:id="echoid-s954" xml:space="preserve">E perciò in tal caſo conuiene trouar quella Pote-<lb/>ſtà, ò quel Grado della Progreſſione, il cui Eſponente è il nu-<lb/>mero de gl’anni (nel che ſe bene viſono alcuni compendij, v’è <lb/>però di molta fatica,) e trouato tal Grado della detta progreſ-<lb/>ſione, tagliarne, come s’è detto, le figure alla deſtra due meno <lb/>del doppio del numero di tal Grado, perche realmente il pri-<lb/>mo termine della progreſſione non è l’vnità, ma il 100. </s> <s xml:id="echoid-s955" xml:space="preserve">Il che <lb/>ſia detto per moſtrare di quanto compendio ſia l’vſo di que-<lb/>ſto Stromento, con cui preſtiſſimo ſi fà coſa per altro operoſa.</s> <s xml:id="echoid-s956" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s957" xml:space="preserve">Quindi volendo ſi ſa pere in quanto tempo raddoppiaraſſi <lb/>il Capitale, ſi piglia vna linea, & </s> <s xml:id="echoid-s958" xml:space="preserve">all’interuallo 50. </s> <s xml:id="echoid-s959" xml:space="preserve">50, ſia appli-<lb/>cata tal linea, dipoinel modo detto, conſiderato il frutto an- <pb o="48" file="0060" n="61" rhead="C A P O II."/> nuo, tante volte ſi replica l’operatione, ſin che ſi venga ad ha-<lb/>uer allargato il compaſſo, in modo che comprenda il doppio <lb/>della linea data da principio: </s> <s xml:id="echoid-s960" xml:space="preserve">e con quante operationi verrai <lb/>ad hauere tal linea doppia della data, tanti anni ſi ricercano <lb/>per raddoppiar il capitale.</s> <s xml:id="echoid-s961" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s962" xml:space="preserve">Dalle coſe dette ſi raccoglie anche il modo per tramutar <lb/>tra di ſe le ſpecie delle monete, eſſendo conoſciuto il lor valo-<lb/>re, riducendolo prima alla medeſima ſemplice denominatio-<lb/>ne; </s> <s xml:id="echoid-s963" xml:space="preserve">come ſe il valore d’vna ſpecie di moneta foſſe compoſto <lb/>di lire, e ſoldi, ſi riduce il valor d’ambidue in ſoldi, e così dell’ <lb/>altre denominationi di valore, e quando fatta queſta riduttio-<lb/>ne riuſciſſero i numeritroppo grandi, baſterà prendere, di <lb/>ambidue li numeri eſprimenti il valore, vna medeſima parte <lb/>aliquota. </s> <s xml:id="echoid-s964" xml:space="preserve">Per eſſem pio s’hanno a ridurre Ongari in Doppie; <lb/></s> <s xml:id="echoid-s965" xml:space="preserve">eſſendo il valor dell’Ongaro 17 giulij, quello della Doppia <lb/>30 giulij, è manifeſto, che 30 Ongari ſono 17 Doppie, per-<lb/>che l’iſteſſo numero ſi produce prendendoſi trenta volte il <lb/>17, e prendendoſi diciſette volte il 30. </s> <s xml:id="echoid-s966" xml:space="preserve">Dunque il numero de <lb/>gl’Ongari al numero delle Doppie ſarà reciprocamente co-<lb/>me il valor della Doppia al valore dell’Ongaro. </s> <s xml:id="echoid-s967" xml:space="preserve">Perciò aper-<lb/>to ad arbitrio lo Stromento, prendo con vn compaſſo l’inter-<lb/>uallo 30. </s> <s xml:id="echoid-s968" xml:space="preserve">30, e con vn’altro compaſſo l’interuallo 17. </s> <s xml:id="echoid-s969" xml:space="preserve">17. </s> <s xml:id="echoid-s970" xml:space="preserve"><lb/>Poſcia per ridurre vn numero d’Ongari in Doppie, applico <lb/>il primo compaſſo all’interuallo corriſpondente al numero <lb/>dato de gl’Ongari, & </s> <s xml:id="echoid-s971" xml:space="preserve">il ſecondo compaſſo con la ſua apertu-<lb/>ra caderà nel numero competente delle Doppie, ò ſe ſi foſſe <lb/>preſa vna parte aliquota del numero de gl’Ongari, s’haurà ſi-<lb/>mile parte del numero delle Doppie. </s> <s xml:id="echoid-s972" xml:space="preserve">Così ſe foſſero dati <lb/>180 Ongari, prendo la metà, che è 90, & </s> <s xml:id="echoid-s973" xml:space="preserve">appli<unsure/>co l’apertura <lb/>del primo compaſſo all’interuallo 90. </s> <s xml:id="echoid-s974" xml:space="preserve">90; </s> <s xml:id="echoid-s975" xml:space="preserve">& </s> <s xml:id="echoid-s976" xml:space="preserve">il ſecondo com- <pb o="49" file="0061" n="62" rhead="Pratticain numeri della Regola del Trè."/> paſſoapplicato, caderà al 51. </s> <s xml:id="echoid-s977" xml:space="preserve">51. </s> <s xml:id="echoid-s978" xml:space="preserve">Dunque conchiudo, che <lb/>90 Ongariſono Doppie 51, e perciò 180 Ongari ſono Dop-<lb/>pie 102. </s> <s xml:id="echoid-s979" xml:space="preserve">Per il contrario ſe voleſſi cambiar Doppie in Ongari, <lb/>al numero delle Doppie applico il ſecondo compaſſo, con <lb/>cui ſi preſe il valore delli Ongari; </s> <s xml:id="echoid-s980" xml:space="preserve">e l’altro compaſſo darà il <lb/>numero de gl’Ongari: </s> <s xml:id="echoid-s981" xml:space="preserve">Siano date Doppie 204, perche il nu-<lb/>mero è troppo grande, piglio la ſeſta parte, che è 34, & </s> <s xml:id="echoid-s982" xml:space="preserve">ap-<lb/>plico il ſecondo compaſſo con la ſua apertura all’interuallo <lb/>34. </s> <s xml:id="echoid-s983" xml:space="preserve">34, e poi l’altro compaſſo cadendo nell’interuallo 60. </s> <s xml:id="echoid-s984" xml:space="preserve">60, <lb/>moſtra, che ſi come il 34 era la ſeſta parte del numero delle <lb/>Doppie, così il 60 è il ſeſto delnumero de gl’Ongari, onde <lb/>Doppie 204 ſi cambiano in Ongari 360.</s> <s xml:id="echoid-s985" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s986" xml:space="preserve">Che ſe il valore è compoſto di diuerſe ſpecie, come in Vene-<lb/>tia lo Scudo è lire 9 ſoldi 6, & </s> <s xml:id="echoid-s987" xml:space="preserve">il Zecchino nuouo lire 17, con-<lb/>uien riſoluer tutto in ſoldi, ſi che lo Scudo è ſoldi 186, & </s> <s xml:id="echoid-s988" xml:space="preserve">il <lb/>Zecchino ſoldi 340, e perciò 340 Scudi ſono Zecchini 186, <lb/>e nella ſteſſa proportione ſono le parti aliquote ſimili. </s> <s xml:id="echoid-s989" xml:space="preserve">Onde <lb/>perche il 340, & </s> <s xml:id="echoid-s990" xml:space="preserve">il 186 ſon troppo grandi, ſi prende la lor <lb/>quarta parte 85, e 46 {1/2}, come ſe queſto foſſe il valore (pi-<lb/>gliandoſi adeſſo non più il valor in ſoldi, mà in groſſetti, eſſen-<lb/>done 85 groſſetti in vn Zecchino, e 46 {1/2} in vno Scudo) e ſi <lb/>opera come di ſopra.</s> <s xml:id="echoid-s991" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s992" xml:space="preserve">Auuertaſi in queſte operationi eſſere molto meglio, e più <lb/>ſicuro, quando quella prima apertura dello Stromento arbi-<lb/>traria ſi piglia aſſai grande, perche poi nelle ſeguenti opera-<lb/>tioni rieſce maggior diſtintione, ſenza pericolo di prender <lb/>vn’ intiero di più. </s> <s xml:id="echoid-s993" xml:space="preserve">Vero è che queſta operatione, come mec-<lb/>canica, non darà la preciſione della frattione aderente a gl’in-<lb/>tieri, mà queſta poi ſi troua, eſſendo aſſai hauer ſubito notitia <lb/>de gl’intieri con qualche facilità. </s> <s xml:id="echoid-s994" xml:space="preserve">Come nel propoſto eſſem- <pb o="50" file="0062" n="63" rhead="CAPO II."/> pio ſi vuol ſapere quanti Zecchini ci vogliono per far la ſom-<lb/>ma di cento ſcudi. </s> <s xml:id="echoid-s995" xml:space="preserve">Preſi gl’interualli 85, e 46 {1/2}, applico il <lb/>maggiore all’interuallo 100.</s> <s xml:id="echoid-s996" xml:space="preserve">100, che è il numero dato de gli <lb/>ſcudi, & </s> <s xml:id="echoid-s997" xml:space="preserve">il minore veggo eſſer più di 54, e meno di 55, onde <lb/>dico li 100 Scudi cambiarſi con Zecchini 54, & </s> <s xml:id="echoid-s998" xml:space="preserve">alcune lire <lb/>di più: </s> <s xml:id="echoid-s999" xml:space="preserve">E queſte ſi trouano paragonato inſieme il valore di <lb/>100 Scudi, e di 54 Zecchini, poiche la loro differenza è quel-<lb/>lo, che deue aggiungerſi alli 54 Zecchini trouati.</s> <s xml:id="echoid-s1000" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1001" xml:space="preserve">E queſto che s’è detto della traſmutatione delle monete <lb/>tra diloro, ſi deue intendere di tutte l’altre miſure, ò ſiano <lb/>dell’iſteſſo paeſe con diuerſe denominationi, o ſiano di paeſi <lb/>diuerſi con l’iſteſſa denominatione sì, ma con grandezze di. <lb/></s> <s xml:id="echoid-s1002" xml:space="preserve">uerſe; </s> <s xml:id="echoid-s1003" xml:space="preserve">perche hauutaſi la loro proportione, ſi tramutano con <lb/>proportione reciproca. </s> <s xml:id="echoid-s1004" xml:space="preserve">Così perche lo ſtadio Romano è <lb/>paſſi 125, & </s> <s xml:id="echoid-s1005" xml:space="preserve">il miglio paſſi 1000, mille ſtadij Romani ſono <lb/>125 miglia Romane: </s> <s xml:id="echoid-s1006" xml:space="preserve">e perche lo ſtadio Greco era di piedi <lb/>antichi Romani 600, elo ſtadio Aleſſandrino di piedi 720, è <lb/>manifeſto, che 600 ſtadij Aleſſandrini erano 720 ſtadij Gre-<lb/>ci: </s> <s xml:id="echoid-s1007" xml:space="preserve">Onde ſi vede correr quì la ſteſſa operatione, che s’è detta <lb/>per la traſmutatione delle monete.</s> <s xml:id="echoid-s1008" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1009" xml:space="preserve">Ma forſi troppo lungamente ci ſiamo fermati in moſtrare <lb/>queſto vſo dello Stromento di Proportione nella Regola del <lb/>Trè, per deſiderio d’eſſer meglio inteſi dalli principianti: </s> <s xml:id="echoid-s1010" xml:space="preserve">i <lb/>quali dalle coſe quì dette, potranno raccogliere ciò che deb-<lb/>ba farſi in caſi ſimili.</s> <s xml:id="echoid-s1011" xml:space="preserve"/> </p> <pb o="51" file="0063" n="64" rhead="Trouar Particelle piccioliſsime d’vna linea."/> </div> <div xml:id="echoid-div37" type="section" level="1" n="22"> <head xml:id="echoid-head34" style="it" xml:space="preserve">QVESTIONE DECIMA. <lb/>Come d’vna linea data ſi poſſano prendere particelle piccioliſsime <lb/>quante ſe ne voranno.</head> <p> <s xml:id="echoid-s1012" xml:space="preserve">QVeſta queſtione in ſoſtanza non è diſferente da quello, <lb/>che s’è detto nella prima, e ſeconda queſtione di que-<lb/>ſto capo ſecondo, ad ogni modo per facilità mag-<lb/>giore di chi non foſſe così prattico, ò non haueſſe così ben <lb/>compreſo, ciò che iui s’è detto, ſi conſidera quì la prattica <lb/>di trouare vna linea, che contenga vn determinato numero <lb/>di minute particelle d’vna linea data.</s> <s xml:id="echoid-s1013" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1014" xml:space="preserve">E quì conuien oſſeruare, che ſe bene la linea dello Stro-<lb/>mento non è attualmente diuiſa, che in 100 parti vguali, ad <lb/>ogni modo eſsẽdo all’occhio aſſai manifeſta la metà di ciaſcu-<lb/>na diqueſte centeſime, vien ad eſſere virtualmente ſegnata in <lb/>200 parti. </s> <s xml:id="echoid-s1015" xml:space="preserve">Quindi è, che ſe d’vna linea applicata all’interuallo <lb/>100. </s> <s xml:id="echoid-s1016" xml:space="preserve">100. </s> <s xml:id="echoid-s1017" xml:space="preserve">voleſſi hauere {157/200}, baſta ch’io cerchi l’interuallo <lb/>78 {1/2}. </s> <s xml:id="echoid-s1018" xml:space="preserve">78 {1/2}, perche ciaſcuna parte delle ſegnate nello Stro-<lb/>mento vale per due. </s> <s xml:id="echoid-s1019" xml:space="preserve">Così d’vna linea data ſe bramo hauere <lb/>{141/153} diuiſo per metà li 153, viene 76 {1/2}, & </s> <s xml:id="echoid-s1020" xml:space="preserve">a queſto interuallo <lb/>76 {1/2}. </s> <s xml:id="echoid-s1021" xml:space="preserve">76 {1/2} applicata la linea data, l’interuallo del numero, <lb/>che è la metà del 141, cioè 70 {1/2}. </s> <s xml:id="echoid-s1022" xml:space="preserve">70 {1/2}, mi darà la parte <lb/>che ſarà {141/153} della linea data.</s> <s xml:id="echoid-s1023" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1024" xml:space="preserve">Mà ſe voleſſi, che tali particelle non foſſero leuate, ma <lb/>aggiunte ad vna linea vguale, ò moltiplice alla data; </s> <s xml:id="echoid-s1025" xml:space="preserve">ſe bene <lb/>baſterebbe tirar vna linea indefinita, e da quella leuar vna <lb/>parte vguale, ò moltiplice alla data linea, & </s> <s xml:id="echoid-s1026" xml:space="preserve">a queſta parte <lb/>leuata aggiungere le ſudette particelle; </s> <s xml:id="echoid-s1027" xml:space="preserve">ad ogni modo alle <pb o="52" file="0064" n="65" rhead="CAPO II."/> volte per ragione, ò della picciolezza della <lb/> <anchor type="figure" xlink:label="fig-0064-01a" xlink:href="fig-0064-01"/> <anchor type="figure" xlink:label="fig-0064-02a" xlink:href="fig-0064-02"/> linea, ò del poco numero di dette particelle, <lb/>riuſcirebbe incommodo il prenderle ſepara-<lb/>tamente: </s> <s xml:id="echoid-s1028" xml:space="preserve">Perciò in tal occaſione applicata la <lb/>linea data al numero, che è la metà del deno-<lb/>minatore delle particelle, ſi intenderanno <lb/>gl’intieri vguali alla data linea riſoluti in ſimili <lb/>particelle, & </s> <s xml:id="echoid-s1029" xml:space="preserve">alla lor ſomma aggiunto il nu-<lb/>mero delle particelle: </s> <s xml:id="echoid-s1030" xml:space="preserve">ò più toſto intendaſi <lb/>vna ſola parte vguale alla linea data riſoluta <lb/>in tali particelle, con l’aggiunta del loro nu-<lb/>mero; </s> <s xml:id="echoid-s1031" xml:space="preserve">e la metà di tal ſomma darà il punto <lb/>nello Stromento, doue ſi trouerà la linea, che <lb/>ſi cerca.</s> <s xml:id="echoid-s1032" xml:space="preserve"/> </p> <div xml:id="echoid-div37" type="float" level="2" n="1"> <figure xlink:label="fig-0064-01" xlink:href="fig-0064-01a"> <image file="0064-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0064-01"/> </figure> <figure xlink:label="fig-0064-02" xlink:href="fig-0064-02a"> <image file="0064-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0064-02"/> </figure> </div> <p> <s xml:id="echoid-s1033" xml:space="preserve">Per eſſempio è data la linea H, e ne vor-<lb/>rei vna, che della detta linea foſſe 1 {71/100}. <lb/></s> <s xml:id="echoid-s1034" xml:space="preserve">Perche 100 è il denominatore delle particel-<lb/>le, applico la linea H all’interuallo 50. </s> <s xml:id="echoid-s1035" xml:space="preserve">50. </s> <s xml:id="echoid-s1036" xml:space="preserve"><lb/>Dipoi intendo quell’ altra linea nella parte <lb/>vguale alla H diuiſa in 100 particelle; </s> <s xml:id="echoid-s1037" xml:space="preserve">e perciò <lb/>tutta ſara {171/100} della H. </s> <s xml:id="echoid-s1038" xml:space="preserve">Dunque la metà di 171, <lb/>cioè l’interuallo 85 {1/2}. </s> <s xml:id="echoid-s1039" xml:space="preserve">85 {1/2}, mi darà nell’inde-<lb/>finita MN la parte MX, che ſarà 1 {71/100} della li-<lb/>nea H. </s> <s xml:id="echoid-s1040" xml:space="preserve">Che ſe haueſſi voluto vna linea, che <lb/>di detta linea H foſſe 4 {71/100}; </s> <s xml:id="echoid-s1041" xml:space="preserve">haurei in vna linea preſo trè vol-<lb/>te la lunghezza della H, & </s> <s xml:id="echoid-s1042" xml:space="preserve">a queſte haurei aggiunta queſta <lb/>trouata MX; </s> <s xml:id="echoid-s1043" xml:space="preserve">etutta la linea compoſta ſaria ſtata quella, che <lb/>ſi cercaua.</s> <s xml:id="echoid-s1044" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1045" xml:space="preserve">E queſto che s’è detto delle parti centeſime, s’intende, <lb/>quando la linea data non è così grande, che ſe ne poſſa pren- <pb o="53" file="0065" n="66" rhead="Trouar particelle piccioliſsime d’vna linea."/> der ò il quinto, ò il decimo, ò altra tal parte da poterſi com-<lb/>modamente applicar allo Stromento. </s> <s xml:id="echoid-s1046" xml:space="preserve">Poiche ſe la data linea <lb/>foſſe così grande, che ſe ne poteſſe prendere la quinta parte, <lb/>& </s> <s xml:id="echoid-s1047" xml:space="preserve">applicarla all’interuallo 100.</s> <s xml:id="echoid-s1048" xml:space="preserve">100, ſi potriano hauere le <lb/>milleſime, prendendo quel numero di milleſime, che auanza, <lb/>cauatine tutti li quinti del mille, cioè tutti li 200, & </s> <s xml:id="echoid-s1049" xml:space="preserve">applican-<lb/>do la metà del reſto all’interuallo, che gli corriſponde. </s> <s xml:id="echoid-s1050" xml:space="preserve">Come <lb/>ſe ſi voleſſero {792/1000} della linea; </s> <s xml:id="echoid-s1051" xml:space="preserve">queſta diuiſa in cinque parti, & </s> <s xml:id="echoid-s1052" xml:space="preserve"><lb/>applicato vn quinto d’eſſa all’interuallo 100.</s> <s xml:id="echoid-s1053" xml:space="preserve">100, cauo dal <lb/>792 trè volte il 200, e perciò prendo vna linea, che ſia trè <lb/>quinti della data, e queſta ſarà {600/1000}: </s> <s xml:id="echoid-s1054" xml:space="preserve">il reſto 192 applico all’ <lb/>interuallo della ſua metà, cioè a 96. </s> <s xml:id="echoid-s1055" xml:space="preserve">96, & </s> <s xml:id="echoid-s1056" xml:space="preserve">aggiunta alli detti <lb/>trè quinti la longhezza trouata in queſto interuallo, tutta ſarà <lb/>{792/1000} della data linea. </s> <s xml:id="echoid-s1057" xml:space="preserve">E queſta aggiunta al doppio della li-<lb/>nea data, farà vna lunghezza, che ſarà alla data come 2 {792/1000}. <lb/></s> <s xml:id="echoid-s1058" xml:space="preserve">E così dell’altre.</s> <s xml:id="echoid-s1059" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1060" xml:space="preserve">Nella ſteſſa maniera ſe la linea data foſſe così lunga, che la <lb/>ſua decima parte poteſſe commodamente applicarſi all’iter-<lb/>uallo 50. </s> <s xml:id="echoid-s1061" xml:space="preserve">50, commodiſſimamente ſi trouerà vn’altra linea in <lb/>proportione ſuperpartiente di milleſime; </s> <s xml:id="echoid-s1062" xml:space="preserve">perche eſſendo vna <lb/>decima della linea applicata al 50.</s> <s xml:id="echoid-s1063" xml:space="preserve">50, s’intende detta Deci-<lb/>ma diuiſa in 100; </s> <s xml:id="echoid-s1064" xml:space="preserve">e così tutta la linea in 1000. </s> <s xml:id="echoid-s1065" xml:space="preserve">Onde ogni me-<lb/>tà de’puntiſegnati nello Stromento, valendo vna centeſima <lb/>della Decima, vien ad eſſer {1/1000} della linea intiera. </s> <s xml:id="echoid-s1066" xml:space="preserve">Quindi ſe <lb/>della linea data, la cui Decima s’è applicata all’interuallo 50. <lb/></s> <s xml:id="echoid-s1067" xml:space="preserve">50, vorrò vn’altra linea, che ſia 1 {96/1000}, prendo il numeratore, <lb/>come ſe foſſe 196, e la ſua metâ 98 applico all’interuallo 98. </s> <s xml:id="echoid-s1068" xml:space="preserve"><lb/>98, e queſta lunghezza aggiungo à noue decime di tutta la <lb/>linea, poiche ne preſi vna da principio. </s> <s xml:id="echoid-s1069" xml:space="preserve">E generalmente in <lb/>queſto metodo d’operare, tutto il numero ſi butti in milleſi- <pb o="54" file="0066" n="67" rhead="CAPO II."/> me, e poi delle centenara, che ſono in tal numero, ſi prendo-<lb/>no tante decime della data linea, ma vnadi meno, e col reſto <lb/>s’operi come s’è detto. </s> <s xml:id="echoid-s1070" xml:space="preserve">Così ſi voglia vna linea, che ſia della <lb/>data 3 {240/1000}; </s> <s xml:id="echoid-s1071" xml:space="preserve">tutto è 3240 milleſime: </s> <s xml:id="echoid-s1072" xml:space="preserve">delle 32 centenara ne pi-<lb/>glio 31, ecosì replico la data linea trè volte, e v’aggiungo vna <lb/>decima: </s> <s xml:id="echoid-s1073" xml:space="preserve">del reſto 140 opro come s’è detto, & </s> <s xml:id="echoid-s1074" xml:space="preserve">aggiungo a <lb/>queſta linea di 31 decime della data l’interuallo 70. </s> <s xml:id="echoid-s1075" xml:space="preserve">70, che è <lb/>la metà di 140: </s> <s xml:id="echoid-s1076" xml:space="preserve">& </s> <s xml:id="echoid-s1077" xml:space="preserve">in tal modo ſarà la linea 3 {240/1000} della data.</s> <s xml:id="echoid-s1078" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div39" type="section" level="1" n="23"> <head xml:id="echoid-head35" xml:space="preserve">CAPO TERZO.</head> <head xml:id="echoid-head36" style="it" xml:space="preserve">Come s’habbia a diuider il Compaſſo di Proportione per le <lb/>Superficie Piane, & vſo di queſta linea Geometrica.</head> <p> <s xml:id="echoid-s1079" xml:space="preserve">POiche queſte coſe non ſi ſcriuono per huomini dotti, <lb/>conuien ricordar à quelli, che ſono men’eſperti, che fi-<lb/>gure ſimili ſon quelle, che tra di loro hanno gl’angoli vguali <lb/>(a benche gl’angoli di ciaſcuna ſiano tra di ſe diſuguali) & </s> <s xml:id="echoid-s1080" xml:space="preserve">i <lb/>lati, che fanno gl’angoli in vna, ſono proportionali alli lati, <lb/>che fanno gl’angoli vguali nell’altra figura; </s> <s xml:id="echoid-s1081" xml:space="preserve">come le definiſce <lb/>Euclide nel principio del libro 6, & </s> <s xml:id="echoid-s1082" xml:space="preserve">ilati, che nell’vna, e l’altra <lb/>figura ſi corriſpondono, ſi chiamano Lati Homologi. </s> <s xml:id="echoid-s1083" xml:space="preserve">In oltre <lb/>(come ſi dimoſtra nella 19. </s> <s xml:id="echoid-s1084" xml:space="preserve">e 20. </s> <s xml:id="echoid-s1085" xml:space="preserve">del lib. </s> <s xml:id="echoid-s1086" xml:space="preserve">6.) </s> <s xml:id="echoid-s1087" xml:space="preserve">così li triangoli, <lb/>come l’altre figure poligone ſimili, hanno trà di loro la pro-<lb/>portione duplicata, della proportione, che ſi troua trà li lati <lb/>Homologi; </s> <s xml:id="echoid-s1088" xml:space="preserve">cioè continuando la proportione de’ſudetti lati, <lb/>come il primo termine al terzo, così le figure trà di loro. </s> <s xml:id="echoid-s1089" xml:space="preserve">On-<lb/>de ſe per cagion d’eſſempio vn lato è la metà dell’altro, con-<lb/>uien continuare la proportione di 1 a 2, con vn terzo termi-<lb/>ne, eſarà 4; </s> <s xml:id="echoid-s1090" xml:space="preserve">e così la proportione di quelle due ſuperficie <pb file="0067" n="68"/> <pb file="0067a" n="69" rhead="Capo Terzo"/> <anchor type="figure" xlink:label="fig-0067a-01a" xlink:href="fig-0067a-01"/> <pb file="0068" n="70"/> <pb o="55" file="0069" n="71" rhead="Linea Geometrica."/> piane ſimili è come 1 a 4 Così ſe li lati foſſero come 2 a 3, queſta proportione ſi continua in tre termini, cioè 4, 6, 9, ele ſuperficie ſono trà di loro come 4 a 9: </s> <s xml:id="echoid-s1091" xml:space="preserve">e così di tutte l’altre.</s> <s xml:id="echoid-s1092" xml:space="preserve"/> </p> <div xml:id="echoid-div39" type="float" level="2" n="1"> <figure xlink:label="fig-0067a-01" xlink:href="fig-0067a-01a"> <image file="0067a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0067a-01"/> </figure> </div> <p> <s xml:id="echoid-s1093" xml:space="preserve">Ora ſicome nelli numeri, quando ſon trè minimi numeri <lb/>continuamente proportionali, li due eſtremi ſono numeri <lb/>quadrati, per il primo corollario della prop. </s> <s xml:id="echoid-s1094" xml:space="preserve">2. </s> <s xml:id="echoid-s1095" xml:space="preserve">del lib. </s> <s xml:id="echoid-s1096" xml:space="preserve">8. </s> <s xml:id="echoid-s1097" xml:space="preserve">e li <lb/>numeri piani ſimili hanno la proportione duplicata della pro-<lb/>portione de’lati Homologi, per la 18. </s> <s xml:id="echoid-s1098" xml:space="preserve">del lib. </s> <s xml:id="echoid-s1099" xml:space="preserve">8. </s> <s xml:id="echoid-s1100" xml:space="preserve">onde ne ſie-<lb/>gue, che li numeri piani ſimili hanno trà diloro la proportio-<lb/>ne de’Numeri Quadrati de’lati Homologi; </s> <s xml:id="echoid-s1101" xml:space="preserve">Così parimenti le <lb/>ſuperficie piane ſimili, hauendo la proportione duplicata de’ <lb/>lati Homologi, la qual proportione iſteſſa ſi troua trà li qua-<lb/>drati de’ſudetti lati Homologi, ſi dicono hauere trà di loro la <lb/>proportione delli quadrati de’lati homologi; </s> <s xml:id="echoid-s1102" xml:space="preserve">Eſe ben ſi potria <lb/>dire, che dette ſuperficie ſimili hanno la proportione de’trian-<lb/>goli ſimili, e ſimilmente poſti ſopra li detti lati Homologi; </s> <s xml:id="echoid-s1103" xml:space="preserve">ad <lb/>ogni modo per eſſer grande la varietà de’triangoli ſimili, che <lb/>ſopra detti lati ſi ponno intendere, perciò ſi dice più toſto, che <lb/>hanno la proportione de’quadrati di detti lati, poiche per la <lb/>vguaglianza de gl’angoli, e de’lati, che è nel quadrato, dato <lb/>vn lato, e conoſciuto tutto il quadrato.</s> <s xml:id="echoid-s1104" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1105" xml:space="preserve">Quindi è, che per conoſcere qual proportione habbiano <lb/>due figure ſimili, baſta conoſcere qual proportione habbiano <lb/>li quadrati de’loro lati Homolgi. </s> <s xml:id="echoid-s1106" xml:space="preserve">E per il contrario conoſciu-<lb/>ta la proportione de’quadrati, ſi manifeſtarà quella de’lati, la <lb/>qual è ſubduplicata di quella de’quadrati. </s> <s xml:id="echoid-s1107" xml:space="preserve">Onde ſe ſaranno <lb/>date due linee, e ſi deſiderino due quadrati nella proportio-<lb/>ne di dette due linee; </s> <s xml:id="echoid-s1108" xml:space="preserve">conuien trouar trà quelle vna media <lb/>proportionale, & </s> <s xml:id="echoid-s1109" xml:space="preserve">i quadrati della prima, e della ſeconda han-<lb/>no la proportione della prima alla terza: </s> <s xml:id="echoid-s1110" xml:space="preserve">e ciò che de’quadra- <pb o="56" file="0070" n="72" rhead="CAPO II."/> ti ſi dice, s’intenda anche delle figure ſimili, e ſimilmente po-<lb/>ſte ſopra la prima, e ſeconda linea delle trè continuamente <lb/>proportionali. </s> <s xml:id="echoid-s1111" xml:space="preserve">Perciò volendo ſopra vnalinea retta ſegnar <lb/>ilati di figure ſimili, le quali habbiano vna determinata pro-<lb/>portione, baſterà che ſopra detta linea ſi ſegnino i lati de’ <lb/>quadrati nella ſteſla proportione. </s> <s xml:id="echoid-s1112" xml:space="preserve">E queſti ſono facili a tro-<lb/>uarſi per la 47. </s> <s xml:id="echoid-s1113" xml:space="preserve">del Lib. </s> <s xml:id="echoid-s1114" xml:space="preserve">1.</s> <s xml:id="echoid-s1115" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1116" xml:space="preserve">Per venir dunque all’atto di ſegnar, e diuidere lo Stro-<lb/>mento per ſeruircene nelle ſuperficie piane, ſitiri dal centro <lb/>A, vna linea retta AZ; </s> <s xml:id="echoid-s1117" xml:space="preserve">& </s> <s xml:id="echoid-s1118" xml:space="preserve">vn’altra vguale A S: </s> <s xml:id="echoid-s1119" xml:space="preserve">le quali nonè <lb/>neceſſario ſegnare ſin ad A, ma baſterà, che comincino à ve-<lb/>derſi in F, e G; </s> <s xml:id="echoid-s1120" xml:space="preserve">in maniera tale però, che la diſtanza A F ſia <lb/>capace di 15 diuiſioni, caſo ch’ella foſſe {1/2} di tutta la AZ; </s> <s xml:id="echoid-s1121" xml:space="preserve">di <lb/>che ſi vedrà la ragione poco appreſſo.</s> <s xml:id="echoid-s1122" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1123" xml:space="preserve">Di poi la diſtanza A F dal punto F ſi vada replicando nel-<lb/>la linea A Z, in maniera, ch’ella venga diuiſa in parti vguali; <lb/></s> <s xml:id="echoid-s1124" xml:space="preserve">che quì non ponno commodamente eſſere più di 8. </s> <s xml:id="echoid-s1125" xml:space="preserve">Mà per <lb/>far più diuiſioni conuerrebbe, che lo Stromento foſſe più lun-<lb/>go. </s> <s xml:id="echoid-s1126" xml:space="preserve">Eciò che ſi dice della linea A Z, ſi faccia anche nella A S, <lb/>ſenza che habbiamo più di meſtieri diricordarlo. </s> <s xml:id="echoid-s1127" xml:space="preserve">Alli punti <lb/>notati ſi ſcriuano li numeri quadrati, intendendoſi nel punto <lb/>F 1, e cosìne gl’altri, 4, 9, 16, 25, 36, 49, 64, i quali ſono li <lb/>numeri quadrati di 2, 3, 4, 5, 6, 7, 8, conforme, che A 4 è <lb/>dupla di A F, & </s> <s xml:id="echoid-s1128" xml:space="preserve">A 9 è tripla della ſteſſa A F, e così dell’altre. </s> <s xml:id="echoid-s1129" xml:space="preserve"><lb/>E più volontieri da me ſi notano le diuiſioni di tal linea con <lb/>li ſo pradetti numeri quadrati, acciò quelli ſteſſi manifeſti-<lb/>no l’vſo dital Linea eſlere per le figure piane. </s> <s xml:id="echoid-s1130" xml:space="preserve">La ragione <lb/>poi di notare tali numeri è, perche eſſendo A 4 doppia di <lb/>A F, il quadrato di A 4 è quadruplo del quadrato di AF: </s> <s xml:id="echoid-s1131" xml:space="preserve">e <lb/>perche A 9 è tripla di AF, ilſuo quadrato è noncuplo, ecosì <lb/>de gl’altri.</s> <s xml:id="echoid-s1132" xml:space="preserve"/> </p> <pb o="57" file="0071" n="73" rhead="Linea Geometrica."/> <p> <s xml:id="echoid-s1133" xml:space="preserve">Volendoſi dunque notare ſu la linea AZ ilati de’quadrati, <lb/>che vanno creſcendo ſecondo l’ordine naturale de’numeri, ſi <lb/>vede che eſſendo dall’vnità al 4 la diſferenza 3, e dal 4 al 9 <lb/>la differenza 5, dal 9 al 16 la differenza 7, e così di mano in <lb/>mano aggiungendo li numeri diſpari, neceſſariamente ne ſie <lb/>gue, che delle ſette parti della linea F 64 la prima ſi diuide in <lb/>trè, la ſeconda in cinque, la terza in ſette, la quarta in noue, <lb/>la quintain vndici, la ſeſta in tredeci, e la ſettima in quindeci. <lb/></s> <s xml:id="echoid-s1134" xml:space="preserve">Perciò ſi diſſe, che la diſtanza AG, ò AF, che ſi piglia per il <lb/>lato del primo Quadrato, douea eſſer tanto lunga, che foſſe <lb/>capace di 15 diuiſioni. </s> <s xml:id="echoid-s1135" xml:space="preserve">Onde appariſce, che volendoſi pro-<lb/>ſeguire oltre 64, conuerrebbe che lo Stromento foſſe aſſai <lb/>più lungo, acciò la AF ſi pigliaſſe così grande, che vi ſi poteſ-<lb/>ſero commodamente notare tutte le diuiſioni neceſſarie per <lb/>l’vltima parte, le quali, come s’è accennato, vanno ſempre <lb/>creſcendo di moltitudine, conforme creſcono li numeri diſpa-<lb/>ri. </s> <s xml:id="echoid-s1136" xml:space="preserve">Quindi è, che riuſcendo queſte diuiſioni tra di loro diſu-<lb/>guali, & </s> <s xml:id="echoid-s1137" xml:space="preserve">in maniera, che la diſtanza dal centro A à ciaſcun-<lb/>punto non hà la proportione del numero, che gli corriſpon-<lb/>de, cioè A 1 ad A 2, nonè come à 2, anzi più toſto A 2 è tra <lb/>A 1, & </s> <s xml:id="echoid-s1138" xml:space="preserve">il ſuo duplo Media Proportionale di medietà Geo-<lb/>metrica; </s> <s xml:id="echoid-s1139" xml:space="preserve">perciò queſta linea in tal modo diuiſa può, e ſuole <lb/>da molti chiamarſi linea Geometrica, à differenza della pri-<lb/>ma, che habbiamo chiamato Aritmetica nel Capo prece-<lb/>dente.</s> <s xml:id="echoid-s1140" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1141" xml:space="preserve">Mà per fare nella linea AZ le diuiſioni per notar’i lati de’ <lb/>Quadrati moltiplici del Quadrato di AF, ſecondo l’ordine <lb/>naturale de’ numeri, è neceſſario ſopra vn piano (e ſarà otti-<lb/>ma vna laſtra dirame ben pulita, poiche in eſſa appariſcono <lb/>facilmenteli ſottiliſſimi ſegni, che ſi faranno colla punta del <pb o="58" file="0072" n="74" rhead="CAPO III."/> Compaſſo) tirar vna linea vguale alla AZ dello Stromento, & </s> <s xml:id="echoid-s1142" xml:space="preserve"><lb/>in eſſa prender AC vguale alla AF, dello Stromento, e queſta <lb/>replicarla in 4, 9, 16, &</s> <s xml:id="echoid-s1143" xml:space="preserve">c. </s> <s xml:id="echoid-s1144" xml:space="preserve">E per hauer poile altre diuiſioni, <lb/>dal punto A ſi tiri la perpendicolare AB vguale alla AC: </s> <s xml:id="echoid-s1145" xml:space="preserve">ma <lb/>auuertaſi di metter ogni diligenza per farla giuſtiſſimamen-<lb/>te perpendicolare, e preciſamente vguale alla AC; </s> <s xml:id="echoid-s1146" xml:space="preserve">perche <lb/>in vna di queſte due coſe, che ſi manchi, ridonda poinelle <lb/>diuiſioni non picciola imperfettione. </s> <s xml:id="echoid-s1147" xml:space="preserve">Perciò ſarà bene fare <lb/>la ſudetta perpendicolare più lunga del biſogno, acciò ſi poſ-<lb/>ſano far le pruoue più accertate, ſe l’angolo A ſia retto: </s> <s xml:id="echoid-s1148" xml:space="preserve">e tro-<lb/>uatoſiretto, allhora ſe ne taglia la AB vguale alla AC. </s> <s xml:id="echoid-s1149" xml:space="preserve">E ciò <lb/>fatto, tutto è preparato per le diuiſioni deſiderate.</s> <s xml:id="echoid-s1150" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1151" xml:space="preserve">Prendaſi dunque la diſtanza BC, e ſi traporti in AD, e ſarà <lb/>A D il lato del Quadrato duplo del Quadrato di AC; </s> <s xml:id="echoid-s1152" xml:space="preserve">come <lb/>appariſce dalla 47.</s> <s xml:id="echoid-s1153" xml:space="preserve">dellib. </s> <s xml:id="echoid-s1154" xml:space="preserve">1. </s> <s xml:id="echoid-s1155" xml:space="preserve">eſſendo vgualitra diſeilati AB, <lb/>A C. </s> <s xml:id="echoid-s1156" xml:space="preserve">Quindi preſa la diſtanza BD ſi traſporti in AE, e queſto <lb/>ſarà illato del quadrato tripolo del quadrato di AC; </s> <s xml:id="echoid-s1157" xml:space="preserve">perche <lb/>il quadrato di BD, cioè di AE è vguale alli quadrati di DA, <lb/>& </s> <s xml:id="echoid-s1158" xml:space="preserve">AB, cioè à trè quadrati di AB, cioè di AC. </s> <s xml:id="echoid-s1159" xml:space="preserve">E così ſuſſe-<lb/>guentemente pigliando la diſtanza B 4, e traſportandola dal <lb/>punto A, s’haurà il lato del quadrato quintuplo, & </s> <s xml:id="echoid-s1160" xml:space="preserve">in tal ma-<lb/>niera ſi procederà in ciaſcun punto, pigliando la diſtanza da <lb/>quello al punto B, e traportandola sù la linea, che ſi diuide.</s> <s xml:id="echoid-s1161" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1162" xml:space="preserve">E per non far molta fatica poco vtilmente, facendo diui-<lb/>ſioni non tanto aggiuſtate, ſi potranno di tanto in tanto nel <lb/>progreſſo far alcune proue per vedere, ſe le diuiſioni ſon fate <lb/>giuſtamente. </s> <s xml:id="echoid-s1163" xml:space="preserve">Ora perche A 4 è il doppio di AC, cioè AB, <lb/>preſoſi da principio, ne ſe ne può fiſicamente dubitare, pren-<lb/>deremo la diſtanza A 4, e poſto vn piede del compaſſo in B, <lb/>vedremo ſe l’altro piede cade giuſtamente in E, e ſarà ſegno, <pb o="59" file="0073" n="75" rhead="Linea Geometrica"/> che AE è preſa giuſtamente per il lato del triplo Quadrato. <lb/></s> <s xml:id="echoid-s1164" xml:space="preserve">E perche AE fù fatta vguale alla BD, ſarà anche ſegno, che <lb/>A D fù preſa con preciſione. </s> <s xml:id="echoid-s1165" xml:space="preserve">Mà per eſſaminar anche di van-<lb/>taggio ſe AD ſia giuſta, ella ſi replichi in H, ſi che AH ſia-<lb/>doppia di AD: </s> <s xml:id="echoid-s1166" xml:space="preserve">dunque il quadrato di AH è quadruplo del <lb/>quadrato di AD; </s> <s xml:id="echoid-s1167" xml:space="preserve">e perche il quadrato di AD ſi ſuppone du-<lb/>plo del quadrato di AC, ne ſeguirà, che il quadrato di AH ſia <lb/>ottuplo di quello di AC. </s> <s xml:id="echoid-s1168" xml:space="preserve">Dunque in H cade la diuiſione 8. </s> <s xml:id="echoid-s1169" xml:space="preserve"><lb/>Ora prendendoſi la diſtanza A 9, ſi traporti dal punto B in H, <lb/>poiche eſſendo BH lato del quadrato noncuplo, ſarà manife-<lb/>ſto, che AH è lato dell’ ottuplo, e per conſeguenza AD del <lb/>duplo, come ſi cercaua d’eſſaminare. </s> <s xml:id="echoid-s1170" xml:space="preserve">Che ſe in queſte proue <lb/>non ſi trouaſſero corriſponderſi li punti così preciſamente, di <lb/>nuouo s’eſſamini la rettitudine dell’angolo A, e l’vguaglianza <lb/>di AB con AC, & </s> <s xml:id="echoid-s1171" xml:space="preserve">emendate queſte ſi proceda auanti.</s> <s xml:id="echoid-s1172" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1173" xml:space="preserve">Trouati giuſti queſti punti eſſaminati, con eſſi ſe ne po-<lb/>tranno eſſaminare de gl’altri, ò anche da principio notare <lb/>con ſicurezza; </s> <s xml:id="echoid-s1174" xml:space="preserve">perche ſe AD replicata in H cade nel 8, repli-<lb/>cata di nuouo darà il lato del qnadrato noncuplo di AD, cioè <lb/>18, e di nuouo replicata darà il lato del ſedecuplo, cioè 32, e <lb/>preſa la quinta volta caderà nel termine del lato del Quadra-<lb/>to, che contiene 25 volte il Quadrato di AD, cioè 50 volte il <lb/>primo Quadrato di AC Così parimenti AE, che è 3 dupli-<lb/>cata darà 12, triplicata darà 27, quadruplicata 48. </s> <s xml:id="echoid-s1175" xml:space="preserve">Così A <lb/>5 duplicata darà 20, e triplicata 45. </s> <s xml:id="echoid-s1176" xml:space="preserve">A 6 duplicata darà 24, <lb/>e triplicata 54. </s> <s xml:id="echoid-s1177" xml:space="preserve">A 7 duplicata darà 28, e triplicata darà 63. <lb/></s> <s xml:id="echoid-s1178" xml:space="preserve">A 10 duplicata darà 40. </s> <s xml:id="echoid-s1179" xml:space="preserve">A 11 duplicata darà 44, e così del-<lb/>l’altre ſin’à A 15, che duplicate darà 60.</s> <s xml:id="echoid-s1180" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1181" xml:space="preserve">Per eſſaminare poi gl’altri punti, ſi prenda da vno di queſti <lb/>già certi, e determ<unsure/>inati la diſtanza ſin’à B, e s’applichi in A, <pb o="60" file="0074" n="76" rhead="CAPO III."/> e caderà nel punto proſſimamente maggiore; </s> <s xml:id="echoid-s1182" xml:space="preserve">di nuouo ſi <lb/>prenda dall’iſteſſo punto ſin’ad A, e s’applichi in B, e caderà <lb/>nel punto proſſimamente minore, ſe da principio s’oprò giu-<lb/>ſtamente. </s> <s xml:id="echoid-s1183" xml:space="preserve">Come per eſſempio, habbiamo certo il punto di <lb/>16, prendo la diſtanza B 16, e dourà darmi A 17; </s> <s xml:id="echoid-s1184" xml:space="preserve">e così A 16 <lb/>dourà dare B 15: </s> <s xml:id="echoid-s1185" xml:space="preserve">il che ſe ſarà, moſtrerà, che quando ſi preſe <lb/>B 14 per notare A 15, s’era oprato bene. </s> <s xml:id="echoid-s1186" xml:space="preserve">E così de gl’ altri.</s> <s xml:id="echoid-s1187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1188" xml:space="preserve">Vn’altra maniera aſſai facile per trouare ilati de’quadrati <lb/>ſi hà colbene-<lb/> <anchor type="figure" xlink:label="fig-0074-01a" xlink:href="fig-0074-01"/> ficio d’vn ſe-<lb/>micircolo de-<lb/>ſcritto ſopta <lb/>la lunghezza, <lb/>di cui deu’ eſ-<lb/>ſere la linea <lb/>Geometrica; <lb/></s> <s xml:id="echoid-s1189" xml:space="preserve">e ſia il ſemi-<lb/>circolo ſopra <lb/>la linea AZ.</s> <s xml:id="echoid-s1190" xml:space="preserve"/> </p> <div xml:id="echoid-div40" type="float" level="2" n="2"> <figure xlink:label="fig-0074-01" xlink:href="fig-0074-01a"> <image file="0074-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0074-01"/> </figure> </div> <p> <s xml:id="echoid-s1191" xml:space="preserve">Prendaſi il lato del primo quadrato in vna commoda di-<lb/>ſtanza dal centro dello ſtromento; </s> <s xml:id="echoid-s1192" xml:space="preserve">e ſia AF, la quale ſia ap-<lb/>plicata al ſemicircolo dall’eſtre mità del diametro A, e dal <lb/>punto F ſi tiri la perpendicolare FG, che prolongata in D ta-<lb/>gliarà il lato del rettangolo AC. </s> <s xml:id="echoid-s1193" xml:space="preserve">Ora la diſtanza AG ſi re-<lb/>plichi in H, I, K, &</s> <s xml:id="echoid-s1194" xml:space="preserve">c. </s> <s xml:id="echoid-s1195" xml:space="preserve">quante volte ci può capire; </s> <s xml:id="echoid-s1196" xml:space="preserve">e ſimilmen-<lb/>re la BD ſi replichi in E, O, &</s> <s xml:id="echoid-s1197" xml:space="preserve">c. </s> <s xml:id="echoid-s1198" xml:space="preserve">le quali ſono vguali alle prime. <lb/></s> <s xml:id="echoid-s1199" xml:space="preserve">Tirate dunque le linee EH, OI, &</s> <s xml:id="echoid-s1200" xml:space="preserve">c. </s> <s xml:id="echoid-s1201" xml:space="preserve">ſaranno tutte parallele <lb/>alla DG, e perciò perpendicolari al diametro AZ, eſega-<lb/>ran<unsure/>no la circonferenza in S, T, V, X, Y. </s> <s xml:id="echoid-s1202" xml:space="preserve">Dico che A Sè il lato <lb/>del quadrato duplo di AF, & </s> <s xml:id="echoid-s1203" xml:space="preserve">AT è lato dd triplo, e così di <pb o="61" file="0075" n="77" rhead="Linea Geometrica"/> mano in mano. </s> <s xml:id="echoid-s1204" xml:space="preserve">Onde ſe queſte linee AS, AT, &</s> <s xml:id="echoid-s1205" xml:space="preserve">c. </s> <s xml:id="echoid-s1206" xml:space="preserve">ſi trapor-<lb/>taranno ſu la linea Geometrica da diuiderſi, ſarà fatta la giu-<lb/>ſta diuiſione.</s> <s xml:id="echoid-s1207" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1208" xml:space="preserve">E che queſtiſian’i lati che ſi cercano, è manifeſto dall’ 8. <lb/></s> <s xml:id="echoid-s1209" xml:space="preserve">del 6. </s> <s xml:id="echoid-s1210" xml:space="preserve">perche AF è media proportionale trà AZ, & </s> <s xml:id="echoid-s1211" xml:space="preserve">AG, on-<lb/>de per la 17 del 6 il quadrato di AF è vguale al rettangolo di <lb/>A Z in AG. </s> <s xml:id="echoid-s1212" xml:space="preserve">Similmente per la ſteſſa ragione il quadrato di <lb/>A S è vguale al rettangolo di AZ in AH: </s> <s xml:id="echoid-s1213" xml:space="preserve">dunque li quadrati <lb/>di AF, & </s> <s xml:id="echoid-s1214" xml:space="preserve">AS, ſono come i rettangoli di AZ in AG, & </s> <s xml:id="echoid-s1215" xml:space="preserve">AZ in <lb/>A H. </s> <s xml:id="echoid-s1216" xml:space="preserve">Mà perche queſti rettangoli hanno la ſteſſa altezza <lb/>A Z, ſono per la prima del 6. </s> <s xml:id="echoid-s1217" xml:space="preserve">come le baſi AG, & </s> <s xml:id="echoid-s1218" xml:space="preserve">AH, e di <lb/>queſte la ſeconda è dupla della prima;</s> <s xml:id="echoid-s1219" xml:space="preserve">dunque anche il rettan-<lb/>golo di AZ, & </s> <s xml:id="echoid-s1220" xml:space="preserve">AH, cioè il quadrato di AS è doppio del ret-<lb/>tangolo di AZ in AG, cioè del quadrato di AF.</s> <s xml:id="echoid-s1221" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1222" xml:space="preserve">Così dimoſtraraſſi il rettangolo di AZ in AI, cioè il qua-<lb/>drato di A T, eſler triplo del rettangolo di A C in A G, cioè <lb/>del quadrato di A F, eſſendo che A I è tripla di A G. </s> <s xml:id="echoid-s1223" xml:space="preserve">E così <lb/>dituttigli altri. </s> <s xml:id="echoid-s1224" xml:space="preserve">Auuertaſi però, che per hauer il ſemicirco-<lb/>lo preparato conforme all’intento, baſterà ſegnare nella cir-<lb/>conferenza i punti doue ſi taglia dalla regola applicata alli <lb/>punti oppoſti del rettangolo A C, ſenzatirare le linee paral-<lb/>lele, ne meno le linee ſuttendenti gli archi; </s> <s xml:id="echoid-s1225" xml:space="preserve">perche baſtarà <lb/>prendere con il compaſſo le diſtanze A F, A S, A T, &</s> <s xml:id="echoid-s1226" xml:space="preserve">c. </s> <s xml:id="echoid-s1227" xml:space="preserve">etra-<lb/>portarle sù lo ſtromento.</s> <s xml:id="echoid-s1228" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1229" xml:space="preserve">Fatte sù la laſtra dirame queſte diuiſioni (le quali fatte vna <lb/>volta per vno ſtromento, ſeruiranno all’Artefice per molti <lb/>altri ſenza nuoua fatica) altro non reſta, che con diligenza-<lb/>traportarle sù la linea A Z dello ſtromento e nello ſteſſo tem-<lb/>po, che vna diuiſione ſi ſegna nell’A Z, ſi deue ſegnare nell’ <lb/>A S, acciò ciaſcuna ſia vgualmente preſa dal centro A. </s> <s xml:id="echoid-s1230" xml:space="preserve">E nel <pb o="62" file="0076" n="78" rhead="CAPO III."/> traportarle ſtimo ſarà più facile, eſicuro prender ſempre nel-<lb/>la linea la diſtanza di ciaſcun punto dall’A: </s> <s xml:id="echoid-s1231" xml:space="preserve">ſe forſi nel pro-<lb/>greſſo, quando conuien’allargar’ aſſai il compaſſo, non ſi giu-<lb/>dicaſſe di prendere le diſtanze da traportarſi da vn qualch’ al-<lb/>tro punto più vicino; </s> <s xml:id="echoid-s1232" xml:space="preserve">nel chel iſperienza inſegnerà a ciaſcu-<lb/>no ciò, che gli tornerà più a conto per la facilità d’operare, e <lb/>per la ſicurezza della preciſione, & </s> <s xml:id="echoid-s1233" xml:space="preserve">aggiuſtatezza neceſſaria <lb/>al fine preteſo. </s> <s xml:id="echoid-s1234" xml:space="preserve">Mà ſe tirate sù lo ſtromento le linee AZ, & </s> <s xml:id="echoid-s1235" xml:space="preserve"><lb/>AS, ti fidaſſi d’allargar lo ſtromento in modo, che foſſi ſicuro, <lb/>che le dette due linee faceſſero vn’angolo retto (il che cono-<lb/>ſcereſti con l’applicatione d’vna ſquadra giuſtiſſima, ouero <lb/>fatto vn quadrato d’vna linea vguale ad A F, allargaſſi lo ſtro-<lb/>mento in modo, che il diametro di detto quadrato foſſe l’in-<lb/>teruallo FG) in tal caſo, ſenza traportar le diuiſioni fatte pri-<lb/>ma in vna laſtra, ſi potriano far’ immediatamente nello ſteſ-<lb/>ſo ſtromento ritenuto in quella apertura, poiche è lo ſteſſo, <lb/>che ſe foſſe vna laſtra.</s> <s xml:id="echoid-s1236" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1237" xml:space="preserve">Se ben’il modo ſin’ora preſcritto per ſegnar’i lati de’qua-<lb/>drati è ſicuriſſimo, e Geometrico, e perciò il più preciſo; <lb/></s> <s xml:id="echoid-s1238" xml:space="preserve">nientedimeno ò gl’Arteficinon vorranno prenderſi tanta bri-<lb/>ga, la quale forſi ſtimeranno maggiore di quello, che real-<lb/>mente è, ò alcuno temerà, che quello traportare li punti del-<lb/>la laſtra sù lo ſtromento poſſa portar qualche variatione, ò <lb/>anche ſi vorrà con altro modo di operare prouare, quanto <lb/>preciſamente ſiano notati li punti in queſta linea quadratica, <lb/>ò Geometrica, che chiamar la vogliamo. </s> <s xml:id="echoid-s1239" xml:space="preserve">Perciò ecco vn’al-<lb/>tra forma mecanica, in cui ci ſeruirà la linea Aritmetica del <lb/>Capo precedente.</s> <s xml:id="echoid-s1240" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1241" xml:space="preserve">Queſto conſiſte in eſtrarre la Radice quadrata di ciaſcun <lb/>numero dall’ 1 ſin’al 64, come ſe foſſe quadrato: </s> <s xml:id="echoid-s1242" xml:space="preserve">e ſe ben’è <pb o="63" file="0077" n="79" rhead="Linea Geometrica"/> certo, che non eſſendo tutti quadrati, non hanno preciſamen-<lb/>te la Radice, ad ogni modo ſi può auuicinar’aſſai alla vera Ra-<lb/>dice, con inueſtigare in parti milleſime la frattione, che s’ag-<lb/>giunge al numero intiero. </s> <s xml:id="echoid-s1243" xml:space="preserve">Il che ſi fà con aggiunger’al nume-<lb/>ro, la cui radice quadrata ſi cerca, ſei zeri, poiche così verrà <lb/>vna radice di quattro figure, el’vltime trè ſaranno milleſime: <lb/></s> <s xml:id="echoid-s1244" xml:space="preserve">così per hauere la radice di 3, cauo la radice quadrata dal <lb/>3000000, e venendo 1732, dico la radice del 3 eſſer 1 {732/1000}. </s> <s xml:id="echoid-s1245" xml:space="preserve"><lb/>E così de gl’altri numeri, come nella tauoletta quì aggiunta <lb/>ſi può vedere; </s> <s xml:id="echoid-s1246" xml:space="preserve">in cui dirim petto à ciaſcun numero ſtà la ſua <lb/>radice, le cui trè vltime figure ſono milleſime parti dell’vnità. </s> <s xml:id="echoid-s1247" xml:space="preserve"><lb/>Mà perche nè meno ſi vien preciſamente nel numero delle <lb/>milleſime, perciò quando viſi dourebbe aggiunger qualche <lb/>coſa, s’è poſto il ſegno †; </s> <s xml:id="echoid-s1248" xml:space="preserve">come quando l’vltima figura è vn <lb/>poco troppo grande, e ſi douria leuar qualche coſa, s’è po-<lb/>ſto il ſegno-: </s> <s xml:id="echoid-s1249" xml:space="preserve">Tutta però la differenza dell’ aggiunger, ò <lb/>leuare non arriua ad vna milleſima; </s> <s xml:id="echoid-s1250" xml:space="preserve">onde ſi vede, che nell’o-<lb/>peratione ordinaria di ſtromento non molto grande non può <lb/>eſſer la differenza d’vna punta di compaſſo; </s> <s xml:id="echoid-s1251" xml:space="preserve">e perciò ſi può <lb/>adoperare francamente tutto il numero notato.</s> <s xml:id="echoid-s1252" xml:space="preserve"/> </p> <pb o="64" file="0078" n="80" rhead="CATO III."/> <note position="right" xml:space="preserve"> <lb/>######## Tauola de’ numeri con le ſue Radici Quadrate eſpreſſe \\ in particelle Milleſime dell’ Vnità. <lb/>Quad. # Radici # Quad. # Radici # Quad. # Radici # Quad. # Radici <lb/>1 # 1000 # 17 # 4123† # 33 # 5744† # 49 # 7000 <lb/>2 # 1415- # 18 # 4242† # 34 # 5830† # 50 # 7071† <lb/>3 # 1732† # 19 # 4359- # 35 # 5916† # 51 # 7142-<lb/>4 # 2000 # 20 # 4472† # 36 # 6000 # 52 # 7212-<lb/>5 # 2236† # 21 # 4582† # 37 # 6082† # 53 # 7280† <lb/>6 # 2450- # 22 # 4690† # 38 # 6164† # 54 # 7348† <lb/>7 # 2646- # 23 # 4796- # 39 # 6245- # 55 # 7416† <lb/>8 # 2828† # 24 # 4898† # 40 # 6324† # 56 # 7484-<lb/>9 # 3000 # 25 # 5000 # 41 # 6404- # 57 # 7550-<lb/>10 # 3162† # 26 # 5099† # 42 # 6480† # 58 # 7616-<lb/>11 # 3316† # 27 # 5169† # 43 # 6558- # 59 # 7682-<lb/>12 # 3465- # 28 # 5292- # 44 # 6633† # 60 # 7746-<lb/>13 # 3606- # 29 # 5386- # 45 # 6708† # 61 # 7810† <lb/>14 # 3742- # 30 # 5478- # 46 # 6782† # 62 # 7874† <lb/>15 # 3872† # 31 # 5568- # 47 # 6856- # 63 # 7937† <lb/>16 # 4000 # 32 # 5656† # 48 # 6928† # 64 # 8000 <lb/></note> <p> <s xml:id="echoid-s1253" xml:space="preserve">E per ſodisfar’al dubbio, che alcuno potria hauere, per <lb/>qual cagione potendoſi tutte le Radici notare vn poco mag-<lb/>giori, ò tutte vn poco minori, altre ſi ſiano notate maggiori <lb/>del douere col ſegno--, altre minori col ſegno †; </s> <s xml:id="echoid-s1254" xml:space="preserve">dico eſſerſi <lb/>ciò fatto, perche la radice vera è più vicina al numero ſegna-<lb/>to, che à quello, che foſſe minore, ò maggiore per vna mil-<lb/>leſima: </s> <s xml:id="echoid-s1255" xml:space="preserve">e pois’è hauuto riſguardo di far sì, che con queſta al-<lb/>ternatione ora di più, ora di meno ſi venga a conſeruare <lb/>quanto ſi può la giuſta miſura, la quale, aggiunte inſieme <lb/>quelle piccole, & </s> <s xml:id="echoid-s1256" xml:space="preserve">inſenſibili differenze, nel progreſſo verreb-<lb/>be ad alterarſi notabilmente.</s> <s xml:id="echoid-s1257" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1258" xml:space="preserve">Che ſe la lunghezza del lato del primo quadrato non foſſe <lb/>tale, che occorreſſe eſſer ſollecito delle parti milleſime, ba-<lb/>ſterà prenderele centeſime, laſciando l’vltima figura della <pb o="65" file="0079" n="81" rhead="Linea Geometrica"/> tauoletta, maſſime ſe haueſſe aggiunto il ſegno--, e foſſe <lb/>minore di 5: </s> <s xml:id="echoid-s1259" xml:space="preserve">e ſe queſt’vltima figura foſſe maggiore del 5, & </s> <s xml:id="echoid-s1260" xml:space="preserve"><lb/>haueſſe aggiunto il ſegno †, potrà accreſcerſi la penultima fi-<lb/>gura d’vn’ vnità. </s> <s xml:id="echoid-s1261" xml:space="preserve">Come per eſſempio, la radice di 2 è 1.</s> <s xml:id="echoid-s1262" xml:space="preserve">415, <lb/>baſterà prendere 141, cioè applicata AF all’interuallo 50. </s> <s xml:id="echoid-s1263" xml:space="preserve">50 <lb/>(come s’è detto nel Cap.</s> <s xml:id="echoid-s1264" xml:space="preserve">2. </s> <s xml:id="echoid-s1265" xml:space="preserve">Queſt. </s> <s xml:id="echoid-s1266" xml:space="preserve">9.) </s> <s xml:id="echoid-s1267" xml:space="preserve">pigliare l’interuallo del-<lb/>la metà di detto numero, cioè 70 {1/2}. </s> <s xml:id="echoid-s1268" xml:space="preserve">70 {1/2}, e queſta ſarà la <lb/>lunghezza di A 2, lato del quadrato duplo. </s> <s xml:id="echoid-s1269" xml:space="preserve">Per il contrario <lb/>la radice di 8 è 2828 †, perche l’vltima figura è 8 †, accre-<lb/>ſco la figura penultima 2 d’vn’vnità, onde ſia la radice in een-<lb/>teſime 283; </s> <s xml:id="echoid-s1270" xml:space="preserve">e così conſiderata queſta, come ſe foſſe 183, <lb/>prendo l’interuallo della metà 91 {1/2}. </s> <s xml:id="echoid-s1271" xml:space="preserve">91 {1/2}, e dal punto F tra-<lb/>portandolo, ſarà tutta la A 8 radice del quadrato ottu plo: </s> <s xml:id="echoid-s1272" xml:space="preserve">e <lb/>così de gi’altri. </s> <s xml:id="echoid-s1273" xml:space="preserve">Quando poi l’vltima figura foſſe maggiore <lb/>del 5, & </s> <s xml:id="echoid-s1274" xml:space="preserve">haueſſe il ſegno--, ouero minore del 5 col ſegno †, <lb/>ſi può ſicura mente prendere, come ſe non foſſe, ſenza peri-<lb/>colo disbaglio notabile, maſſime quando nella radice ante-<lb/>cedente ſi foſſe aggiunta l’vnità alla penultima figura nel mo-<lb/>do detto.</s> <s xml:id="echoid-s1275" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1276" xml:space="preserve">Mà ſe voleſſi ampliar l’vſo di queſta linea Geometrica à <lb/>numeri moltiplici delli numeri in eſſa ſegnati, cioè alli dop-<lb/>pij, triplici &</s> <s xml:id="echoid-s1277" xml:space="preserve">c. </s> <s xml:id="echoid-s1278" xml:space="preserve">baſterà nella AF, & </s> <s xml:id="echoid-s1279" xml:space="preserve">AG laſciate occulte, ſe-<lb/>gnare il lato de’ quadrati ſubmultiplici del quadrato di AF; <lb/></s> <s xml:id="echoid-s1280" xml:space="preserve">perche con vn compaſſo prendi la lunghezza AF, e quefta <lb/>applica all’interuallo 2. </s> <s xml:id="echoid-s1281" xml:space="preserve">2. </s> <s xml:id="echoid-s1282" xml:space="preserve">Dipoi ritenuta quella ſteſſa aper-<lb/>tura dello ſtromento, prendi l’interuallo FG, e queſto trapor-<lb/>tato dal punto A nelle linee AF, AG, ſegnerà il punto del la-<lb/>to del quadrato, che è la metà del quadrato di AF. </s> <s xml:id="echoid-s1283" xml:space="preserve">Nell’i-<lb/>ſteſſo modo la lunghezza AF applica all’ interuallo 33, e <lb/>l’interuallo FG darà la quãtità da ſegnarſi nelle line AF, AG, <pb o="66" file="0080" n="82" rhead="CAPO III."/> e ſarà il lato del quadrato, che è la terza parte del quadrato di <lb/>AF. </s> <s xml:id="echoid-s1284" xml:space="preserve">E così procedendo in altri numeri, ſe vorrai la quarta, <lb/>ò quinta, ò ſeſta parte del quadrato di AF. </s> <s xml:id="echoid-s1285" xml:space="preserve">Quindi è che cer-<lb/>cando illato d’vn quadrato, che ſia al quadrato dato di AF, <lb/>come 1 12 à 1, ſarà l’iſteſſo, che trouare quello, che ſia come <lb/>56 à {1/2} del quadrato AF; </s> <s xml:id="echoid-s1286" xml:space="preserve">ouero volendo vn quadrato, che ſia <lb/>come 147 à 1, ſarà l’iſteſſo, come ſe voleſſi quello, che è co-<lb/>me 49 à {1/3} del quadrato di AF. </s> <s xml:id="echoid-s1287" xml:space="preserve">Nel che ſarà vn gran compen-<lb/>dio nell’operare. </s> <s xml:id="echoid-s1288" xml:space="preserve">Noi però di fatto non habbiamo ſegnato <lb/>queſti punti delle parti del quadrato di AF, per sfuggire la <lb/>confuſione del Lettore, acciò nella figura vedendo li molti-<lb/>plici, eli ſubmoltiplici di AF, non prendeſſe gl’ vni in vece <lb/>de gl’altri.</s> <s xml:id="echoid-s1289" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1290" xml:space="preserve">E per non replicar più volte l’iſteſſo con tedio di chi legge, <lb/>auuerti, che queſto ſteſſo, che s’è detto del ſegnare le parti <lb/>del quadrato in queſta linea Geometrica, ſi potrà far’anche <lb/>nella linea cubica, di cui ſi parlerà nel Capo ſeguente, ado <lb/>prando l’iſteſſo modo per ſegnare nelle AH, AI i lati de’cubi <lb/>ſubmoltiplici. </s> <s xml:id="echoid-s1291" xml:space="preserve">Onde propoſta vna proportione moltiplice, <lb/>il cui termine maggiore ſupera il maſſimo ſegnato nello ſtro-<lb/>mento, diuidital numero per vno delli denominatori delle <lb/>parti notate, & </s> <s xml:id="echoid-s1292" xml:space="preserve">il quotiente darà l’intiero, che hà alla detta <lb/>parte l’iſteſſa proportione; </s> <s xml:id="echoid-s1293" xml:space="preserve">come appariſce eſſere 147 à 1, <lb/>come 49 à {1/2}.</s> <s xml:id="echoid-s1294" xml:space="preserve"/> </p> <pb o="67" file="0081" n="83" rhead="Linea Geometrica"/> </div> <div xml:id="echoid-div42" type="section" level="1" n="24"> <head xml:id="echoid-head37" style="it" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head38" style="it" xml:space="preserve">Data vna figura regolare, come ſi poſſa deſcriuerne vn’ altra <lb/>della ſteſſa ſpecie nella proportione, che ſi deſidera.</head> <p> <s xml:id="echoid-s1295" xml:space="preserve">FIgura Regolare ſi chiama quella, che hà ne’ſuoi termini, <lb/>da’ quali è compreſa, tutte le le parti vniformi; </s> <s xml:id="echoid-s1296" xml:space="preserve">perciò <lb/>quelle, che hanno molti lati, & </s> <s xml:id="echoid-s1297" xml:space="preserve">angoli, ſaranno Regolari, ſe <lb/>ſaranno Equilatere, & </s> <s xml:id="echoid-s1298" xml:space="preserve">Equiangole; </s> <s xml:id="echoid-s1299" xml:space="preserve">& </s> <s xml:id="echoid-s1300" xml:space="preserve">il Circolo ſe bene non <lb/>hà, propriamente parlando, nè lati, nè angoli, è però figura <lb/>regolare, perche le parti della circonferenza, che lo termina, <lb/>ſono vniformemente diſpoſte: </s> <s xml:id="echoid-s1301" xml:space="preserve">il che non ſi può dire dell’El-<lb/>lipſi, della Parabola, nè dell’Hiperbola, perche con tutto che <lb/>i termini di tali figure ſiano regolati da certe, e deter minate <lb/>conditioni, non ſono però in ogni ſua parte vniformi. </s> <s xml:id="echoid-s1302" xml:space="preserve">Quin-<lb/>diè, che delle Fortezze alcune ſi chiamano Regolari, perche <lb/>la figura, che ſi fortifica è Regolare, cioè Equilatera, & </s> <s xml:id="echoid-s1303" xml:space="preserve">Equi-<lb/>angola. </s> <s xml:id="echoid-s1304" xml:space="preserve">E ſe bene è manifeſto, che non tutte le linee della <lb/>fortificatione ſono trà loro vguali, eſſendo certo, che la faccia <lb/>del Baloardo, la ſpalla, ò fianco, ela cortina, ſono trà di loro <lb/>diſuguali: </s> <s xml:id="echoid-s1305" xml:space="preserve">ad ogni modo, perche tutte le cortine trà di loro, <lb/>tutte le ſpalle de’Baloardi trà di loro, e tutte le faccie trà di <lb/>loro ſono vguali, anche per queſto capo ſi puonno chiamar <lb/>Regolari, à diſferenza dell’Irregolari, doue le cortine ſono <lb/>trà di loro diſuguali, ele parti d’vn Baloardo non ſon’vguali <lb/>alle lor’homogenee d’vn’altro Baloardo. </s> <s xml:id="echoid-s1306" xml:space="preserve">Noi però quì par-<lb/>lando di figure Regolari, prendiamo quelle, che aſſoluta-<lb/>mente parlando ſon’Equilatere, & </s> <s xml:id="echoid-s1307" xml:space="preserve">Equiangole, conſiderãdo-<lb/>le aſſolutamente in ſe ſteſſe, e non come ordinate nel circolo.</s> <s xml:id="echoid-s1308" xml:space="preserve"/> </p> <pb o="68" file="0082" n="84" rhead="CAPO III."/> <p> <s xml:id="echoid-s1309" xml:space="preserve">Sia primiera-<lb/> <anchor type="figure" xlink:label="fig-0082-01a" xlink:href="fig-0082-01"/> mente data in nu-<lb/>meri la proportio-<lb/>ne, che deuono <lb/>hauere le due figu. <lb/></s> <s xml:id="echoid-s1310" xml:space="preserve">re regolari ſimili; </s> <s xml:id="echoid-s1311" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s1312" xml:space="preserve">applicato il lato <lb/>della figura data <lb/>al numero delle <lb/>linee Geometri-<lb/>che AZ, AS, l’in-<lb/>teruallo, che ſarà <lb/>al numero, che <lb/>corriſponde alla, <lb/>figura cercata, darà il lato, che ſi <lb/> <anchor type="figure" xlink:label="fig-0082-02a" xlink:href="fig-0082-02"/> deſidera. </s> <s xml:id="echoid-s1313" xml:space="preserve">Per cagione d’eſſempio, <lb/>ſia data la linea R lato dello ſpa-<lb/>tio, in cui ſtà ordinata vna Batta-<lb/>glia quadra di terreno, e voglia-<lb/>mo vn’altr’ area pur quadra, che <lb/>ſia il doppio, e quattro quinti della prima: </s> <s xml:id="echoid-s1314" xml:space="preserve">sì che la propor-<lb/>tione della prima alla ſeconda è di 5 à 14. </s> <s xml:id="echoid-s1315" xml:space="preserve">Applico dunque la <lb/>linea R all’interuallo 5. </s> <s xml:id="echoid-s1316" xml:space="preserve">5, e poil’interuallo 14. </s> <s xml:id="echoid-s1317" xml:space="preserve">14 mi darà <lb/>la linea Slato del quadrato, che ſi cerca.</s> <s xml:id="echoid-s1318" xml:space="preserve"/> </p> <div xml:id="echoid-div42" type="float" level="2" n="1"> <figure xlink:label="fig-0082-01" xlink:href="fig-0082-01a"> <image file="0082-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0082-01"/> </figure> <figure xlink:label="fig-0082-02" xlink:href="fig-0082-02a"> <image file="0082-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0082-02"/> </figure> </div> <p> <s xml:id="echoid-s1319" xml:space="preserve">La dimoſtratione di ciò non è punto diffe-<lb/> <anchor type="figure" xlink:label="fig-0082-03a" xlink:href="fig-0082-03"/> rente da quella, che s’apportò per fonda-<lb/>mento nel Capo 1. </s> <s xml:id="echoid-s1320" xml:space="preserve">Sia AH vguale all’A 5. <lb/></s> <s xml:id="echoid-s1321" xml:space="preserve">& </s> <s xml:id="echoid-s1322" xml:space="preserve">AE vguale all’A 14: </s> <s xml:id="echoid-s1323" xml:space="preserve">HI ſia la linea R, & </s> <s xml:id="echoid-s1324" xml:space="preserve"><lb/>EL la linea S. </s> <s xml:id="echoid-s1325" xml:space="preserve">Ora perchecome AH ad AE, <lb/>così HIad EL, come già ſi dimoſtrò, ſarà an- <pb o="69" file="0083" n="85" rhead="Linea Geometrica."/> che come il quadrato d’AH al quadrato d’AE, così il quadra-<lb/>to di HI, cioè di R, al quadrato d’EL, cioè di S, per la 22 del <lb/>lib. </s> <s xml:id="echoid-s1326" xml:space="preserve">6: </s> <s xml:id="echoid-s1327" xml:space="preserve">li due primi quadrati ſono come 5 à 14, per la conſtrut-<lb/>tione dello ſtromento; </s> <s xml:id="echoid-s1328" xml:space="preserve">dunque anche li quadrati di R, & </s> <s xml:id="echoid-s1329" xml:space="preserve">S <lb/>hanno la ſteſſa proportione.</s> <s xml:id="echoid-s1330" xml:space="preserve"/> </p> <div xml:id="echoid-div43" type="float" level="2" n="2"> <figure xlink:label="fig-0082-03" xlink:href="fig-0082-03a"> <image file="0082-03" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0082-03"/> </figure> </div> <p> <s xml:id="echoid-s1331" xml:space="preserve">Dalla ſteſſa propoſitione 22 dellib. </s> <s xml:id="echoid-s1332" xml:space="preserve">6 ſi dimoſtra, che qual <lb/>ſi voglia altra ſpecie di figure ſimili, e ſimilmente poſte ſopra <lb/>le due ſeconde linee R, & </s> <s xml:id="echoid-s1333" xml:space="preserve">S, ſiano di quanti lati, & </s> <s xml:id="echoid-s1334" xml:space="preserve">angoli eſ-<lb/>ſere ſi vogliano, hanno trà di loro la proportione de’quadra-<lb/>ti delle due prime linee ſegnate sù lo ſtromento: </s> <s xml:id="echoid-s1335" xml:space="preserve">E così ſe la <lb/>linea Sfoſſe data lato d’vn pentagono regolare da fortificarſi, <lb/>e voleſſimo metter’in diſſegno vn’altro pentagono minore <lb/>nella proportione di 14 à 10, applicata la linea S alli punti <lb/>14. </s> <s xml:id="echoid-s1336" xml:space="preserve">14, prendaſi la diſtanza 10. </s> <s xml:id="echoid-s1337" xml:space="preserve">10, e ſarà la linea T lato del <lb/>pentagono regolare, à cui mancano due ſettimi del maggiore <lb/>pentagono.</s> <s xml:id="echoid-s1338" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1339" xml:space="preserve">E perche ſpeſſo occorre, che douendoſi vn diſſegno tra-<lb/>portare digrande in piccolo ſecondo vna data proportione, <lb/>& </s> <s xml:id="echoid-s1340" xml:space="preserve">il lato dato è così grande, che non capiſce nello ſtromen-<lb/>to; </s> <s xml:id="echoid-s1341" xml:space="preserve">prendaſivna parte aliquota di detto lato, e con eſſa s’ope-<lb/>ri, come ſe foſſe il lato ſteſſo, perche ſi trouerà la parte ali-<lb/>quota ſimile del lato cercato; </s> <s xml:id="echoid-s1342" xml:space="preserve">come ſe la ſo pradetta linea S <lb/>foſſe la ſeſta parte dellato del pentagono maggiore, la linea T <lb/>trouata ſarà la ſeſta del minore. </s> <s xml:id="echoid-s1343" xml:space="preserve">Perche come S à T, così il <lb/>ſeſtuplo di S al ſeſtuplo di T, dunque per la 22 del 6, come il <lb/>pentagono di Sal pentagono di T, cioè come 14 à 10, così il <lb/>pentagono del ſeſtuplo di S, al pentagono del ſeſtuplo di T.</s> <s xml:id="echoid-s1344" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1345" xml:space="preserve">Per il contrario volendoſi ttaſportar’vn diſſegno d’vna fi-<lb/>gura regolare di piccolo in grande, può eſſer’il lato dato tale, <lb/>che non capiſca nell’interuallo del minore de’due numeri <pb o="70" file="0084" n="86" rhead="CAPO III."/> eſprimenti la proportione; </s> <s xml:id="echoid-s1346" xml:space="preserve">& </s> <s xml:id="echoid-s1347" xml:space="preserve">in tal caſo ſi trouino altri due <lb/>termini maggiori nella ſteſſa proportione: </s> <s xml:id="echoid-s1348" xml:space="preserve">Come pereſſem-<lb/>pio, ſi debba trouar’il lato d’vn poligono maggiore del poli-<lb/>gono dato nella proportione di 3 à 2. </s> <s xml:id="echoid-s1349" xml:space="preserve">Perche il lato S dato <lb/>non capiſce nell’interuallo 2. </s> <s xml:id="echoid-s1350" xml:space="preserve">2, in vece delli due numeri 2, e <lb/>3, prendo 14, e 21 nella ſteſſa proportione; </s> <s xml:id="echoid-s1351" xml:space="preserve">& </s> <s xml:id="echoid-s1352" xml:space="preserve">applicato il la-<lb/>to S al punto 14. </s> <s xml:id="echoid-s1353" xml:space="preserve">14, la diſtanza 21. </s> <s xml:id="echoid-s1354" xml:space="preserve">21, cioè la linea V ſarà <lb/>illato cercato del poligono ſeſquialtero del dato.</s> <s xml:id="echoid-s1355" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1356" xml:space="preserve">Ciò che de’poligoni regolari ſi dice, dee intenderſi anche <lb/>de’circoli, i quali per la 2 del lib. </s> <s xml:id="echoid-s1357" xml:space="preserve">12 ſono nella proportione <lb/>de’quadrati de’ſuoi diametri, e perche li quadrati de’ diametri <lb/>ſono quadrupli de’quadrati de’ſemidiametri, ſaranno anchei <lb/>circoli nella proportione de’quadrati delli ſemidiametri. </s> <s xml:id="echoid-s1358" xml:space="preserve">Sì <lb/>che volendo due circoli in vna determinata proportione, ba-<lb/>ſterà trouar’i lati de’quadrati nella ſteſſa proportione, e quel-<lb/>le linee ſaranno li ſemidiametri de’circoli nella bramata pro-<lb/>portione. </s> <s xml:id="echoid-s1359" xml:space="preserve">Sia data la forma per improntar’vna moneta d’ar-<lb/>gento; </s> <s xml:id="echoid-s1360" xml:space="preserve">e ſe ne vuol far vn’altra per improntar vna moneta, che <lb/>nella ſteſſa groſſezza ſia il doppio della prima. </s> <s xml:id="echoid-s1361" xml:space="preserve">Sia la linea R <lb/>il ſemidiametro della moneta ABC; </s> <s xml:id="echoid-s1362" xml:space="preserve">applico R al punto 5. </s> <s xml:id="echoid-s1363" xml:space="preserve">5, <lb/>e preſo l’interuallo 10. </s> <s xml:id="echoid-s1364" xml:space="preserve">10, trouo T ſenndiametro della mo-<lb/>neta DEF, che ſarà doppia della prima: </s> <s xml:id="echoid-s1365" xml:space="preserve">perche eſſendo am-<lb/>bidue della ſteſſa groſſezza, come ſi ſuppone, hanno la pro-<lb/>portione delle lor baſi circolari, per la 11 del lib. </s> <s xml:id="echoid-s1366" xml:space="preserve">12, e queſte <lb/>hanno la proportione de’quadrati delli loro ſemidiametri, co-<lb/>me s’è detto; </s> <s xml:id="echoid-s1367" xml:space="preserve">e tali quadrati ſono come 10 à 5, cioè vnodop-<lb/>pio dell’altro.</s> <s xml:id="echoid-s1368" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1369" xml:space="preserve">Di quì vedendoſi, che cauato il circolo minore del mag-<lb/>giore, reſta il cingolo, ò annello DEFABC vguale al circolo <lb/>minore ABC, perche egliè la metà del maggiore, ſi raccoglie <pb o="71" file="0085" n="87" rhead="Linea Geometrica"/> il modo di trouar’vna portione annulare, che habbia la bra-<lb/>mata proportione ad vn circolo dato, ò ad vn’altra portione <lb/>annulare. </s> <s xml:id="echoid-s1370" xml:space="preserve">Primieramente dal circolo ABC ſi voglia cauar’ <lb/>vna portione, che ſia {2/5} dello ſteſſo circolo. </s> <s xml:id="echoid-s1371" xml:space="preserve">Veggo, che ba-<lb/>ſta trouar’il ſemidiametro d’vn circolo, che ſia al dato circolo, <lb/>come 3 à 5, & </s> <s xml:id="echoid-s1372" xml:space="preserve">applicato il ſemidiametro dato al 5. </s> <s xml:id="echoid-s1373" xml:space="preserve">5, l’inter-<lb/>uallo 3. </s> <s xml:id="echoid-s1374" xml:space="preserve">3 midà il ſemidiametro del circolo HIK, che deſcrit-<lb/>to dallo ſteſſo centro laſcia il cingolo ABC, KHI, che è {2/5} del <lb/>dato circolo ABC.</s> <s xml:id="echoid-s1375" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1376" xml:space="preserve">Secondo. </s> <s xml:id="echoid-s1377" xml:space="preserve">E’dato il circolo HIK, e voglio trouar’vna por-<lb/>tione annulare, che lo contenga vna volta, e due terzi, cioè, <lb/>che ſia come 5 à 3, mà che le circonferenze, che la terminano <lb/>ſiano ambidue maggiori di quella del circolo dato. </s> <s xml:id="echoid-s1378" xml:space="preserve">Applico <lb/>il ſemidiametro dato al punto 3. </s> <s xml:id="echoid-s1379" xml:space="preserve">3. </s> <s xml:id="echoid-s1380" xml:space="preserve">E poi à mio piacere <lb/>prendo vn’interuallo di qualche punto maggiore, come ſaria <lb/>10. </s> <s xml:id="echoid-s1381" xml:space="preserve">10, econ queſto dallo ſteſſo centro deſcriuo la circonfe-<lb/>renza DEF. </s> <s xml:id="echoid-s1382" xml:space="preserve">Quindi ſe voglio l’altra circonferenza ancor <lb/>maggiore, perche il cingolo deue eſſere come 5 à 3, prendo <lb/>l’interuallo di cinque punti più diſtanti dal 10. </s> <s xml:id="echoid-s1383" xml:space="preserve">10, cioè 15. <lb/></s> <s xml:id="echoid-s1384" xml:space="preserve">15, edeſcritta la circonferenza LMN ſarà il cingolo LMNF-<lb/>DE al circolo HIK, come 5 à 3: </s> <s xml:id="echoid-s1385" xml:space="preserve">poiche il circolo LMN al cir-<lb/>colo HIK è come 15 à 3: </s> <s xml:id="echoid-s1386" xml:space="preserve">& </s> <s xml:id="echoid-s1387" xml:space="preserve">al circolo DEF, come 15 à 10, <lb/>dunque leuato DEF dal circolo LMN, quel che rimane è al <lb/>dato circolo HIK, come 5 à 3. </s> <s xml:id="echoid-s1388" xml:space="preserve">Mà ſe voglio, che la circonfe-<lb/>renza maggiore ſia DEF, prendo l’interuallo di cinque punti <lb/>minori del 10, & </s> <s xml:id="echoid-s1389" xml:space="preserve">è 5. </s> <s xml:id="echoid-s1390" xml:space="preserve">5; </s> <s xml:id="echoid-s1391" xml:space="preserve">onde la circonferenza ABC termi-<lb/>narà il cingolo DEFABC, che ſarà al dato circolo, come 5 à <lb/>3, come è manifeſto per lo ſteſſo diſcorſo.</s> <s xml:id="echoid-s1392" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1393" xml:space="preserve">Ora dal ſopradetto raccogliendoſi, come li due cingoli <lb/>AHBICK, & </s> <s xml:id="echoid-s1394" xml:space="preserve">LDMENF ſono come 2 à 5, è chiaro il modo <pb o="72" file="0086" n="88" rhead="CAPO III."/> di far due cingoli nella data proportione; </s> <s xml:id="echoid-s1395" xml:space="preserve">come ciaſcuno <lb/>ſenz’altro nuouo diſcorſo può per ſe ſteſſo raccoglier da quel <lb/>che ſin’ora s’è detto.</s> <s xml:id="echoid-s1396" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1397" xml:space="preserve">Nella ſteſſa maniera volendoſi vn circolo vguale à tutta la <lb/>ſuperficie sferica d’vn globo dato, poiche ſi sà da Archime-<lb/>de lib. </s> <s xml:id="echoid-s1398" xml:space="preserve">de Sph. </s> <s xml:id="echoid-s1399" xml:space="preserve">& </s> <s xml:id="echoid-s1400" xml:space="preserve">Cylind. </s> <s xml:id="echoid-s1401" xml:space="preserve">prop. </s> <s xml:id="echoid-s1402" xml:space="preserve">30. </s> <s xml:id="echoid-s1403" xml:space="preserve">che queſta è quadrupla <lb/>del circolo maſſimo di detta sfera, prendaſi il diametro del <lb/>globo dato, e pongaſi nella linea Geometrica all’interuallo <lb/>d’vn numero, di cui vi ſia il quadruplo come al 6.</s> <s xml:id="echoid-s1404" xml:space="preserve">6, e prendaſi <lb/>l’interuallo 24.</s> <s xml:id="echoid-s1405" xml:space="preserve">24, che darà il diametro del circolo vguale <lb/>alla ſuperficie sferica del globo. </s> <s xml:id="echoid-s1406" xml:space="preserve">ll che ſi può fare col ſolo rad-<lb/>doppiare il diametro del globo. </s> <s xml:id="echoid-s1407" xml:space="preserve">Quindi hauendoſi vn globo <lb/>piccolo, nella cui ſuperficie foſſero deſcritte le ſtelle, eſe ne <lb/>voleſſe far vn’altro, la cui ſuperficie foſſe ſette volte maggio-<lb/>re, acciò più diſtintamente compariſſero le ſtelle; </s> <s xml:id="echoid-s1408" xml:space="preserve">primiera-<lb/>mente trouiſi il diametro del circolo vguale alla data ſuperfi-<lb/>cie sferica, come ſi è detto; </s> <s xml:id="echoid-s1409" xml:space="preserve">dipoi queſto diametro trouato <lb/>ſi metta all’interuallo d’vn numero, a cui ſia nella linea Geo-<lb/>metrica notato vn’altro ſettuplo, come ſe ſi prendeſſe 4. </s> <s xml:id="echoid-s1410" xml:space="preserve">4, e <lb/>poi 28.</s> <s xml:id="echoid-s1411" xml:space="preserve">28, e queſto ſecondo interuallo darà il diametro d’vn <lb/>circolo vguale ad vna ſu perficie sferica ſettupla della ſuperfi-<lb/>cie data Perciò diuiſo tal diametro trouato in due parti <lb/>vguali, la ſua metà ſarà il diametro del globo di tal ſuper-<lb/>ficie.</s> <s xml:id="echoid-s1412" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1413" xml:space="preserve">Mà ſe la proportione, in cui ſi deuono formare li due po-<lb/>ligoni ſimili regolari foſſe eſpreſſa non in numeri, ma con li-<lb/>nee; </s> <s xml:id="echoid-s1414" xml:space="preserve">conuerrà trà le due linee eſprimenti la proportione tro-<lb/>uare vna Media proportionale, per la 13 del lib. </s> <s xml:id="echoid-s1415" xml:space="preserve">6, e ſegnate <lb/>ſottilmente le prime due delle trè continue proportionali sù <lb/>le linee Geometriche AZ, AS, (caſo che non cadeſlero in al- <pb o="73" file="0087" n="89" rhead="Linea Geometrica"/> cuno de’punti in eſſe notati) s’applichi il lato del dato poligo-<lb/>no all’interuallo, che gli corriſponde, maggiore, ò minore <lb/>che ſia, el’altro interuallo darà il lato cercato dell’altro poli-<lb/>gono. </s> <s xml:id="echoid-s1416" xml:space="preserve">Sia eſ-<lb/> <anchor type="figure" xlink:label="fig-0087-01a" xlink:href="fig-0087-01"/> preſſa la pro-<lb/>portione con <lb/>le due linee <lb/>AB, BC, queſte <lb/>ſi vniſcano in <lb/>vna, e tuta la <lb/>AC diuiſa per <lb/>metà in D, all’ <lb/>interuallo DA <lb/>ſi deſeriua il ſemicircolo AEC: </s> <s xml:id="echoid-s1417" xml:space="preserve">e dal punto B alzata la per-<lb/>pendicolare BE, ſarà la Media proportionale trale due date. <lb/></s> <s xml:id="echoid-s1418" xml:space="preserve">Dunque sù le linee Geometriche dello ſtromento AZ, AS, <lb/>cominciando dal centro A, ſi ſegnino ſottilmente colla punta <lb/>del Compaſſo le linee BE, & </s> <s xml:id="echoid-s1419" xml:space="preserve">AB: </s> <s xml:id="echoid-s1420" xml:space="preserve">e ſe illato dato deue eſſer <lb/>minore di quello, che ſi cerca, queſto s’applichi nello ſtro-<lb/>mento all’interuallo, doue furono ſegnati li ter mini della BE, <lb/>perche li ter mini della maggiore AB ſegnati nello ſtromen-<lb/>to, daranno l’interuallo per il lato maggiore. </s> <s xml:id="echoid-s1421" xml:space="preserve">La ragione di <lb/>queſta operatione è, perche come le linee ſegnate ne’lati, così <lb/>ſono gl’interualli de’loro eſtremi, come più volte s’è detto; </s> <s xml:id="echoid-s1422" xml:space="preserve"><lb/>dunque come i quadrati delle ſudette linee, così li quadrati de <lb/>gl’interualli, per la 22 dellib.</s> <s xml:id="echoid-s1423" xml:space="preserve">6. </s> <s xml:id="echoid-s1424" xml:space="preserve">Mà il quadrato di AB al qua-<lb/>drato di BE è come la linea AB alla BC, per la 20 dellib.</s> <s xml:id="echoid-s1425" xml:space="preserve">6; </s> <s xml:id="echoid-s1426" xml:space="preserve"><lb/>dunque anche i quadrati de gl’interualli, cioè li poligoni ſimi-<lb/>li, ſono come AB à BC; </s> <s xml:id="echoid-s1427" xml:space="preserve">come ſi cercaua.</s> <s xml:id="echoid-s1428" xml:space="preserve"/> </p> <div xml:id="echoid-div44" type="float" level="2" n="3"> <figure xlink:label="fig-0087-01" xlink:href="fig-0087-01a"> <image file="0087-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0087-01"/> </figure> </div> <p> <s xml:id="echoid-s1429" xml:space="preserve">Quì però deue auuertirſi, che queſta operatione non è alli.</s> <s xml:id="echoid-s1430" xml:space="preserve"> <pb o="74" file="0088" n="90" rhead="CAPO III."/> gata à queſta linea AZ diuiſa per le ſuperficie, mã trouata la <lb/>Media proportionale ſi può pratticare anche cõ la linea ſem-<lb/>plicemente diuiſa in parti vguali come nel Capo 2. </s> <s xml:id="echoid-s1431" xml:space="preserve">Dal che <lb/>ſi caua, che con quella ſola linea diuiſa vgualmente ſi puonno <lb/>far le operationi de’piani, ſe la proportione de’numeri s’eſpri-<lb/>me in linee nella ſteſſa proportione rationale, come s’è inſe-<lb/>gnato nella Queſt. </s> <s xml:id="echoid-s1432" xml:space="preserve">1. </s> <s xml:id="echoid-s1433" xml:space="preserve">e 2. </s> <s xml:id="echoid-s1434" xml:space="preserve">del Capo 2. </s> <s xml:id="echoid-s1435" xml:space="preserve">e poi tra queſte ſi pren-<lb/>da vna Media proportionale: </s> <s xml:id="echoid-s1436" xml:space="preserve">poiche traportate la prima, e <lb/>la ſeconda di queſte tre proportionali ſul lato dello ſtromen-<lb/>to, gl’interualli daranno ciò, che ſi cerca; </s> <s xml:id="echoid-s1437" xml:space="preserve">come dal già detto <lb/>è manifeſto. </s> <s xml:id="echoid-s1438" xml:space="preserve">Mà per leuar la briga di trouare la Medi<unsure/>a pro-<lb/>portionale, ſi fà queſt’altra diuiſione della linea AZ per i lati <lb/>de’quadrati commenſurabili.</s> <s xml:id="echoid-s1439" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1440" xml:space="preserve">Che ſe la proportione foſſe eſpreſſa con due figure rettili-<lb/>nee diſſimili, & </s> <s xml:id="echoid-s1441" xml:space="preserve">irregolari; </s> <s xml:id="echoid-s1442" xml:space="preserve">queſte, per la 14 dellib. </s> <s xml:id="echoid-s1443" xml:space="preserve">2, ſi ridu-<lb/>cano à quadrati; </s> <s xml:id="echoid-s1444" xml:space="preserve">e poi, come il lato d’vn quadrato al lato dell’-<lb/>altro quadrato, così ſi faccia il lato del poligono regolare da-<lb/>to, al lato cercato del poligono ſimile, che ſi deſidera.</s> <s xml:id="echoid-s1445" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div46" type="section" level="1" n="25"> <head xml:id="echoid-head39" style="it" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head40" style="it" xml:space="preserve">Data vna figur a irregolare, come ſi poſſa deſcriuere vna ſimile <lb/>nella bramata proport<unsure/>ione.</head> <p> <s xml:id="echoid-s1446" xml:space="preserve">DVe maniere ſi puonno tenere per venir all’ eſſecutione <lb/>di queſto problema. </s> <s xml:id="echoid-s1447" xml:space="preserve">La prima è, pigliando i lati del-<lb/>la figura data, etraportando ciaſcuno sù lo ſtromento al nu-<lb/>mero corriſpondente all’antecedente della data proportione, <lb/>e<unsure/>pigliando poi, per illato, che ſi cerca, l’interuallo, che dà il <lb/>numero, con cuis’eſprime il conſega<unsure/>ente di detta proportio.</s> <s xml:id="echoid-s1448" xml:space="preserve"> <pb o="75" file="0089" n="91" rhead="Linea Geometrica"/> ne; </s> <s xml:id="echoid-s1449" xml:space="preserve">auuertendo di far l’angolo ſul fine d’vna linea trouata <lb/>vguale all’angolo, che nell’iſteſſa poſitura gli corriſponde nel-<lb/>la figura data. </s> <s xml:id="echoid-s1450" xml:space="preserve">Sia vn Baloardo ABCDEF, e ſe ne voglia far’ <lb/>vn ſimile, ma ſia vn quarto più di capacità, & </s> <s xml:id="echoid-s1451" xml:space="preserve">ampiezza. <lb/></s> <s xml:id="echoid-s1452" xml:space="preserve">Dunque il Dato al Cercato, deue eſſere, come 4à 5. </s> <s xml:id="echoid-s1453" xml:space="preserve">ouero <lb/>come 16 à 20, come più tornerà <lb/> <anchor type="figure" xlink:label="fig-0089-01a" xlink:href="fig-0089-01"/> commodo eſprimere la propor-<lb/>tione con numeri maggiori, ò <lb/>minori.</s> <s xml:id="echoid-s1454" xml:space="preserve"/> </p> <div xml:id="echoid-div46" type="float" level="2" n="1"> <figure xlink:label="fig-0089-01" xlink:href="fig-0089-01a"> <image file="0089-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0089-01"/> </figure> </div> <p> <s xml:id="echoid-s1455" xml:space="preserve">Per tanto tirate le due linee <lb/>RF, FS, che facciano l’angolo <lb/>RFSvguale al’angolo AFE, per <lb/>la 23 del lib. </s> <s xml:id="echoid-s1456" xml:space="preserve">1, ſi prenda la mez-<lb/>za gola FA, es’applichi all’inter-<lb/> <anchor type="figure" xlink:label="fig-0089-02a" xlink:href="fig-0089-02"/> uallo 16. </s> <s xml:id="echoid-s1457" xml:space="preserve">16, poiche l’intetuallo <lb/>20. </s> <s xml:id="echoid-s1458" xml:space="preserve">20 darà FL, eperciò anche la <lb/>ſua vguale FM mezze gole del <lb/>Baloardo maggiore che s’hà à de-<lb/>ſcriuere. </s> <s xml:id="echoid-s1459" xml:space="preserve">Ciò fatto, dalli punti L, <lb/>& </s> <s xml:id="echoid-s1460" xml:space="preserve">M s’alzino due linee indefinite, <lb/>che facciano l’angolo FLI vguale <lb/>all’angolo FAB, el’angolo FMK vguale all’angolo FED; </s> <s xml:id="echoid-s1461" xml:space="preserve">& </s> <s xml:id="echoid-s1462" xml:space="preserve"><lb/>applicato il fianco AB all’interuallo 16.</s> <s xml:id="echoid-s1463" xml:space="preserve">16, ſi trouarà l’inter-<lb/>uallo 20.</s> <s xml:id="echoid-s1464" xml:space="preserve">20, che ſarà LI, & </s> <s xml:id="echoid-s1465" xml:space="preserve">il ſuo vguale MK fianchi del Ba-<lb/>loardo maggiore. </s> <s xml:id="echoid-s1466" xml:space="preserve">Quindi ſi faccia l’angolo I vguale all’an-<lb/>golo B, el’angolo K vguale all’angolo D, e le due linee IH, KH <lb/>s’incontreranno nel punto H; </s> <s xml:id="echoid-s1467" xml:space="preserve">e ſarà ſegno, che ſi ſia ben’opra-<lb/>to, ſe applicando BC all’interuallo 16.</s> <s xml:id="echoid-s1468" xml:space="preserve">16, l’interuallo 20.</s> <s xml:id="echoid-s1469" xml:space="preserve">20 <lb/>darà preciſamente IH.</s> <s xml:id="echoid-s1470" xml:space="preserve"/> </p> <div xml:id="echoid-div47" type="float" level="2" n="2"> <figure xlink:label="fig-0089-02" xlink:href="fig-0089-02a"> <image file="0089-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0089-02"/> </figure> </div> <p> <s xml:id="echoid-s1471" xml:space="preserve">E' dunque il Baloardo LIHKMF in proportione ſeſqui- <pb o="76" file="0090" n="92" rhead="CAPO III."/> quarta al Baloardo dato: </s> <s xml:id="echoid-s1472" xml:space="preserve">poiche, per la 20 del lib. </s> <s xml:id="echoid-s1473" xml:space="preserve">6. </s> <s xml:id="echoid-s1474" xml:space="preserve">più vol-<lb/>te mentouata, ſono nella duplicata proportione de’lati homo-<lb/>logi, cioè come i quadrati di detti lati: </s> <s xml:id="echoid-s1475" xml:space="preserve">ora perche il quadra-<lb/>to di AF, al quadrato di LF è come 16 à 20, cioè come 4 à 5, <lb/>anche il Baloardo dato al Baloardo fatto è come 4 à 5.</s> <s xml:id="echoid-s1476" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1477" xml:space="preserve">La ſeconda maniera è, con prender vn’angolo della figura, <lb/>e da quello tirar linee rette à tutti gl’angoli, che eſcano fuori <lb/>della figura data: </s> <s xml:id="echoid-s1478" xml:space="preserve">poiche trouata vna ſola linea sù lo ſtro-<lb/>mento, con ſolo tir ar linee parallele alli lati della data figura, <lb/>ſarà fatto ciò, che ſi cerca. </s> <s xml:id="echoid-s1479" xml:space="preserve">Sia dato lo ſteſſo Baloardo ABC-<lb/>DEF, e ſen’habbia à fare, come di ſopra, vno ſeſquiquarto. <lb/></s> <s xml:id="echoid-s1480" xml:space="preserve">Prendo il punto F, e tiro la Capitale FC, prolongandola an-<lb/>che fuori; </s> <s xml:id="echoid-s1481" xml:space="preserve">ſimilmente prolongo FB, FD, FA, FE. </s> <s xml:id="echoid-s1482" xml:space="preserve">Doppo di <lb/>che applico la Capitale FC all’interuallo 16.</s> <s xml:id="echoid-s1483" xml:space="preserve">16, e l’inter-<lb/>uallo 20.</s> <s xml:id="echoid-s1484" xml:space="preserve">20 mi dà FH Capitale del maggior Baloardo. </s> <s xml:id="echoid-s1485" xml:space="preserve">Ora <lb/>dal punto H tiro due parallele alle due faccie CB, CD, che <lb/>rincontrando le prolongate FB, FD in I, & </s> <s xml:id="echoid-s1486" xml:space="preserve">K, fanno le faccie <lb/>del nuouo Baloardo HI, HK, e ſimilmente dalli punti I, & </s> <s xml:id="echoid-s1487" xml:space="preserve">K <lb/>tirandoſi le IL, KM parallele alle BA, DE, s’hauranno li fian-<lb/>chi del Baloardo maggiore, e determinaranno le ſue mezze <lb/>gole LF, & </s> <s xml:id="echoid-s1488" xml:space="preserve">MF. </s> <s xml:id="echoid-s1489" xml:space="preserve">La dimoſtratione è la ſteſſa, che di ſopra, <lb/>per la 20 del lib. </s> <s xml:id="echoid-s1490" xml:space="preserve">6, eſſendo manifeſto per il paralleliſmo del-<lb/>le linee, che cosìl’vno, come l’altro Baloardo ſono riſoluti in <lb/>triangoli ſimili.</s> <s xml:id="echoid-s1491" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1492" xml:space="preserve">Fattoſi il diſſegno à queſto modo del maggiore intorno al <lb/>minore (l’iſteſſa for ma d’operare ſi tiene, quando data vna <lb/>figura maggiore, ſe ne voglia far vna minore) non è difficile il <lb/>traportarlo ſeparatamente, ò col Compaſſo di tre punte, ſo-<lb/>Prapplicandole alli punti FLI, & </s> <s xml:id="echoid-s1493" xml:space="preserve">alla linea FR applicando le <lb/>punte, che danno la diſtanza FL, poiche l’altra punta mo- <pb o="77" file="0091" n="93" rhead="Linea Geometrica"/> ſtra il punto I, per tirar la linea LI, e così di mano in mano: <lb/></s> <s xml:id="echoid-s1494" xml:space="preserve">Ouero col Compaſſo ordinario di due punte, col beneficio <lb/>de gl’archi, che ſi tagliano, cioè nella FR pigliaſi la FL, poi <lb/>all’interuallo LI ſi deſcriue vn’arco occulto, & </s> <s xml:id="echoid-s1495" xml:space="preserve">all’interuallo <lb/>FI ſe ne deſcriue vn’altro pur occulto, che tagliando il primo <lb/>in I, dà il punto per tirar la LI. </s> <s xml:id="echoid-s1496" xml:space="preserve">Similmente à gl’interualli IH, <lb/>& </s> <s xml:id="echoid-s1497" xml:space="preserve">FH altri due archi daranno nella lor’ interſettione il punto <lb/>H;</s> <s xml:id="echoid-s1498" xml:space="preserve">e nella ſteſſa maniera ſi trouerà il punto K, & </s> <s xml:id="echoid-s1499" xml:space="preserve">il punto M: </s> <s xml:id="echoid-s1500" xml:space="preserve"><lb/>e congiunti tali punticon linee, ſarà traportato il diſegno fat-<lb/>to intorno alla figura minore data.</s> <s xml:id="echoid-s1501" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div49" type="section" level="1" n="26"> <head xml:id="echoid-head41" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head42" style="it" xml:space="preserve">Data vna linea in vn piano, come s’habbia à trouarela grandezza <lb/>dellalinea, che le corriſponde in un’ altro piano ſimile <lb/>nella data proportione.</head> <p> <s xml:id="echoid-s1502" xml:space="preserve">OCcorre alcune volte, che eſſendo data vna ſuperficie <lb/>piana, in cui ſono deſcritte varie linee, ſenza prenderſi <lb/>la briga di deſcriuere tutta l’altra ſuperficie ſimile maggior, ò <lb/>minore nella data proportione, vorriamo ſapere, quanta <lb/>douria eſſere la grandezza d’vna linea, che in quella ſuperficie <lb/>da farſi corriſpondeſſe ad vna tal linea, che habbiamo nella <lb/>ſuperficie data. </s> <s xml:id="echoid-s1503" xml:space="preserve">L’operatione è facile, poiche baſterà nello <lb/>ſtromento prendere nella linea A Z li due numeri eſprimenti <lb/>la data proportione de’piani, & </s> <s xml:id="echoid-s1504" xml:space="preserve">applicata la data linea all’ in-<lb/>teruallo del numero congruente, l’interuallo dell’ altro nume-<lb/>ro darà la linea cercata.</s> <s xml:id="echoid-s1505" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1506" xml:space="preserve">Sia per cagion d’eſlempio dato in piccolo il diſſegno d’vn’ <lb/>Orologio à Sole, eſi voglia ſapere, quanto maggiore dourà <pb o="78" file="0092" n="94" rhead="CAPO III."/> eſſere lo ſtile d’vn’Orologio totalmente ſimile in vn’altro pia-<lb/>no dato maggiore. </s> <s xml:id="echoid-s1507" xml:space="preserve">Se non sò quanto maggiore, ſia queſto <lb/>ſecondo piano. </s> <s xml:id="echoid-s1508" xml:space="preserve">Prendo la lunghezza, ò la larghezza del da-<lb/>to Orologio, & </s> <s xml:id="echoid-s1509" xml:space="preserve">applicatala alla lunghezza, ò larghezza del <lb/>piano, in cui s’hà à deſcriuereil nuouo Orologio, veggo, che <lb/>proportione habbiano le lunghezze tra <lb/> <anchor type="figure" xlink:label="fig-0092-01a" xlink:href="fig-0092-01"/> loro, ò le larghezze tra loro (poiche è <lb/>tutto il medeſimo) e preſi li quadrati de’ <lb/>numeri eſprimenti la proportione di <lb/>dette lunghezze, ò larghezze, queſti <lb/>daranno la proportione de’ piani. </s> <s xml:id="echoid-s1510" xml:space="preserve">Così <lb/>ſe la lunghezza del diſſegno ſi contiene <lb/>ſei volte nella lunghezza del piano, le <lb/>ſuperficie de gl’Orologi ſaranno come <lb/>1 à 36. </s> <s xml:id="echoid-s1511" xml:space="preserve">Dunque prendo la lunghezza <lb/>dello ſtile A B nel diſſegno, e nello ſtromento l’applico all’ in-<lb/>teruallo 1. </s> <s xml:id="echoid-s1512" xml:space="preserve">1; </s> <s xml:id="echoid-s1513" xml:space="preserve">poiche l’interuallo 36. </s> <s xml:id="echoid-s1514" xml:space="preserve">36 mi darà CD lun-<lb/>ghezza dello ſtile per l’Orologio da deſcriuerſi nel piano, che <lb/>è 36 volte maggiore.</s> <s xml:id="echoid-s1515" xml:space="preserve"/> </p> <div xml:id="echoid-div49" type="float" level="2" n="1"> <figure xlink:label="fig-0092-01" xlink:href="fig-0092-01a"> <image file="0092-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0092-01"/> </figure> </div> <p> <s xml:id="echoid-s1516" xml:space="preserve">Egli è vero, che conoſciuta la proportione de’ lati delle ſu-<lb/>perficie, il trouar poi queſte linee ſi può fare per quello, che <lb/>s’è detto nel primo Capo, con la linea dello ſtromento diuiſa <lb/>in parti vguali per le linee ſemplici, poiche tali linee hanno <lb/>tra di loro la proportione de’lati delle figure ſimili; </s> <s xml:id="echoid-s1517" xml:space="preserve">Mà ſe ſia <lb/>data la proportione ſolamente de’ piani, e non quella de’lati, <lb/>conuien’ operare con queſta linea AZ dello ſtromento nel <lb/>modo detto: </s> <s xml:id="echoid-s1518" xml:space="preserve">e così ſe la proportione de’piani foſſe data, co-<lb/>me 1 à 24, la lunghezza dello ſtile douria eſſere CE, prenden-<lb/>doſi l’interuallo 24.</s> <s xml:id="echoid-s1519" xml:space="preserve">24.</s> <s xml:id="echoid-s1520" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1521" xml:space="preserve">La dimoſtratione di ciò, che s’è operato è, perche la pro- <pb o="79" file="0093" n="95" rhead="Linea Geometrica."/> portione, che vna linea hà ad vn’altra linea dello ſteſſo piano, <lb/>è l’iſteſſa con la proportione, che nell’altro piano ſimile han-<lb/>no le due linee homologe, e permutando &</s> <s xml:id="echoid-s1522" xml:space="preserve">c. </s> <s xml:id="echoid-s1523" xml:space="preserve">Dunque data <lb/>la proportione de’ piani ſimili, le linee homologe de’ detti <lb/>piani ſono tali, che li loro quadrati ſono nella proportione <lb/>de’piani dati. </s> <s xml:id="echoid-s1524" xml:space="preserve">Dunque pigliandoſi nello ſtromento tali due <lb/>linee, che li loro quadrati hanno la proportione de’ piani da-<lb/>ti, quella è la grandezza cercata della linea homologa alla li-<lb/>nea data.</s> <s xml:id="echoid-s1525" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1526" xml:space="preserve">Mà ſe occoreſſe, che la linea data foſſe così grande, che <lb/>nello ſtromento non capiſſe all’interuallo del numero, che le <lb/>corm<unsure/>ſponde ne’ termini della proportione data, prendaſi vn a <lb/>parte aliquota di detta linea, poiche l’interuallo dell’altro nu-<lb/>mero della proportione darà vna ſimile parte aliquota della <lb/>linea, che ſi cerca: </s> <s xml:id="echoid-s1527" xml:space="preserve">perche eſſendo le parti nella proportione <lb/>de’ſuoi intieri, per la 15 del lib. </s> <s xml:id="echoid-s1528" xml:space="preserve">5, anche i quadrati delle parti <lb/>hanno la proportione de’ quadrati de’ ſuoi intieri, per la 23 <lb/>dellib. </s> <s xml:id="echoid-s1529" xml:space="preserve">6. </s> <s xml:id="echoid-s1530" xml:space="preserve">Come ſe la proportione de’ piani doueſſe eſſere, <lb/>come 4 à 63, e la linea nel piano dato foſſe lunga vn palmo, <lb/>queſta non capirebbe nell’interuallo 4.</s> <s xml:id="echoid-s1531" xml:space="preserve">4; </s> <s xml:id="echoid-s1532" xml:space="preserve">prendaſi dunque tal <lb/>parte, che commodamente vi capiſca, e ſia la quinta parte; <lb/></s> <s xml:id="echoid-s1533" xml:space="preserve">queſta s’applichi all’interuallo 4 4, el’interuallo 63. </s> <s xml:id="echoid-s1534" xml:space="preserve">63 darà <lb/>la quinta parte della linea, che ſi cerca.</s> <s xml:id="echoid-s1535" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1536" xml:space="preserve">Che ſe alcuno de’termini della proportione foſſe eſpreſſo <lb/>con vn numero maggiore di quelli, che ſono notati nella li-<lb/>nea AZ, veggaſi s’egli ſi può diuidere per qualche numero <lb/>quadrato, e ſeruaſi del quotiente, per pigliar nello ſtromen-<lb/>to l’interuallo, che à tal numero corriſponde; </s> <s xml:id="echoid-s1537" xml:space="preserve">e poiqueſto in-<lb/>teruallo ſi replichi tante volte, quante vnità ſono nella radice <lb/>di quel numero quadrato, che ſeruì per diuiſore; </s> <s xml:id="echoid-s1538" xml:space="preserve">che così s’ha- <pb o="80" file="0094" n="96" rhead="CAPO III."/> urà tutta la linea cercata. </s> <s xml:id="echoid-s1539" xml:space="preserve">Per eſſempio, ſia dato il ſemidia-<lb/>metro d’vn circolo, e ſi deſideri il ſemidia metro d’vn’altro cir-<lb/>colo, che riſpetto al primo ſia come 2 {22/25} à 1. </s> <s xml:id="echoid-s1540" xml:space="preserve">la proportione <lb/>dunque è come 72 à 25. </s> <s xml:id="echoid-s1541" xml:space="preserve">Applico alli punti 25 25 il dato ſe-<lb/>midiametro; </s> <s xml:id="echoid-s1542" xml:space="preserve">e perche nella linea AZ dello ſtromento non <lb/>v’è il num. </s> <s xml:id="echoid-s1543" xml:space="preserve">72, diuido queſto per vn numero quadrato, come <lb/>per 9, la cui radice è 3: </s> <s xml:id="echoid-s1544" xml:space="preserve">evenendo il quotiente 8, prendo l’in-<lb/>teruallo 8.</s> <s xml:id="echoid-s1545" xml:space="preserve">8: </s> <s xml:id="echoid-s1546" xml:space="preserve">e perche 3 è radice del 9 diuiſore, triplico la li-<lb/>nea trouata all’ interuallo 8. </s> <s xml:id="echoid-s1547" xml:space="preserve">8, e cosìhò il ſemidiametro cer-<lb/>cato d’vn circolo, che ſarà al dato circolo, come 72 à 25. </s> <s xml:id="echoid-s1548" xml:space="preserve">La <lb/>ragione è, perche l’interuallo 8. </s> <s xml:id="echoid-s1549" xml:space="preserve">8 dà il raggio d’vn circolo, <lb/>che è al dato, come 8 à 25. </s> <s xml:id="echoid-s1550" xml:space="preserve">Mà il raggio triplo di quello, è <lb/>raggio d’vn circolo non cuplo; </s> <s xml:id="echoid-s1551" xml:space="preserve">dunque d’vn circolo, che è <lb/>come 72.</s> <s xml:id="echoid-s1552" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1553" xml:space="preserve">Similmente ſe ambidue li numeri foſſero troppo grandi, ne <lb/>ſi poteſſero diuidere per lo ſteſſo numero quadrato, baſterà <lb/>diuidere ciaſcuno per quello, che ſi può, edella linea data <lb/>prendere la parte, che dimoſtra la radice quadrata del Diui-<lb/>ſore del numero, che le corriſponde. </s> <s xml:id="echoid-s1554" xml:space="preserve">Per eſſempio nella fig. <lb/></s> <s xml:id="echoid-s1555" xml:space="preserve">15 la linea CD è in vna figura piana, e ſi cerca la grandezza <lb/>di quella, che le corriſponde in vn’altra figura piana, cheſia <lb/>alla data figura, come 99 à 80. </s> <s xml:id="echoid-s1556" xml:space="preserve">Diuido 80 per il quadrato di <lb/>2, che è 4, & </s> <s xml:id="echoid-s1557" xml:space="preserve">il quotiente è 20: </s> <s xml:id="echoid-s1558" xml:space="preserve">perciò diuiſa la CD per me-<lb/>tà (poiche 2 è la radice del Diuiſore) queſta metà applico <lb/>all’interuallo 20. </s> <s xml:id="echoid-s1559" xml:space="preserve">20. </s> <s xml:id="echoid-s1560" xml:space="preserve">Poi diuiſo il 99 per 9, il quotiente 11 <lb/>mi moſtra, che debbo prendere l’interuallo 11. </s> <s xml:id="echoid-s1561" xml:space="preserve">11, e perche <lb/>la radice del diuiſore è 3, triplico queſt’ interuallo, e ſarà ciò <lb/>che ſi cercaua. </s> <s xml:id="echoid-s1562" xml:space="preserve">La ragione è, perche l’interuallo 20. </s> <s xml:id="echoid-s1563" xml:space="preserve">20 è <lb/>l’interuallo 11. </s> <s xml:id="echoid-s1564" xml:space="preserve">11, dannoi lati de’quadrati, che ſonocome <lb/>20 à 11. </s> <s xml:id="echoid-s1565" xml:space="preserve">Dunque il primo lato duplicato è lato d’vn qua- <pb o="81" file="0095" n="97" rhead="Linea Geometrica."/> drato, che è quadruplo di 20, cioè come 80, & </s> <s xml:id="echoid-s1566" xml:space="preserve">il ſecondo la-<lb/>to triplicato è lato d’vn quadrato noncuplo di 11, cioè co-<lb/>me 99.</s> <s xml:id="echoid-s1567" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1568" xml:space="preserve">Se poi li due numeri eſprimenti la proportione del piano <lb/>ſono tali, che niuno d’eſſi ſi poſſa diuidere per alcuno de’nu-<lb/>meri quadrati, ſi riducano ad altri numeri, che proſſimamente <lb/>eſprimano la data proportione, ſe bene non tanto preciſa-<lb/>mente; </s> <s xml:id="echoid-s1569" xml:space="preserve">quando l’operatione Mecanica non richiede tanta ac-<lb/>curatezza. </s> <s xml:id="echoid-s1570" xml:space="preserve">Il che ſi fà prendendo ò il maſſimo numero, ò vno <lb/>de’maggiori di quelli, che ſono notati nello ſtromento, e que. <lb/></s> <s xml:id="echoid-s1571" xml:space="preserve">ſto moltiplicato per il minore delli due della proportione, il <lb/>prodotto diuiſo per l’altro numero, chereſta, cioè per il ter-<lb/>mine maggiore della proportione, il quotiente darà l’altro <lb/>numero, che ſarà il termine minore, con cui ſi eſprime la pro-<lb/>portione ridotta à queſta nuoua denominatione. </s> <s xml:id="echoid-s1572" xml:space="preserve">Per eſſem-<lb/>pio debbano eſſer due piani, che habbiano la proportione di <lb/>223 à 71: </s> <s xml:id="echoid-s1573" xml:space="preserve">prendo per nuouo termine maggiore 62, che mol-<lb/>tiplicato per il minore 71, produce 4402, il quale diuiſo per <lb/>il maggiore 223, dà per nuouo termine 19 {165/223}, che è quaſi <lb/>19 {3/4}: </s> <s xml:id="echoid-s1574" xml:space="preserve">onde prendendo l’interuallo vn poco minore di 20.</s> <s xml:id="echoid-s1575" xml:space="preserve">20, <lb/>s’haurà quanto baſta per operare fiſicamente. </s> <s xml:id="echoid-s1576" xml:space="preserve">Che ſe vi foſſe <lb/>di meſtieri di maggior preciſione, conuerrebbe in tal caſo <lb/>operare conforme alle regole della Geometria, trouando la <lb/>media proportionale tra due linee, che haueſſero la propor-<lb/>tione data de’piani, e quella media ſaria la lunghezza cercata <lb/>della linea.</s> <s xml:id="echoid-s1577" xml:space="preserve"/> </p> <pb o="82" file="0096" n="98" rhead="CAPO III."/> </div> <div xml:id="echoid-div51" type="section" level="1" n="27"> <head xml:id="echoid-head43" xml:space="preserve">QVESTIONE QVARTA.</head> <head xml:id="echoid-head44" style="it" xml:space="preserve">Date due figure piane ſimili trouar laloro proportione.</head> <p> <s xml:id="echoid-s1578" xml:space="preserve">NOn ſi vuol negare, che vi ſiano delle ſigure ſimili, la cui <lb/>proportione non ſi può eſprimere con numeri, come <lb/>quelle, che ſono incommenſurabili, & </s> <s xml:id="echoid-s1579" xml:space="preserve">hanno i lati homologi <lb/>incommenſurabili di lunghezza, e di potenza, come ſi parla <lb/>nellib. </s> <s xml:id="echoid-s1580" xml:space="preserve">10 d’Euclide. </s> <s xml:id="echoid-s1581" xml:space="preserve">Adogni modo, per la prattica, à cui ſer-<lb/>ue queſto ſtromento, baſterà trouare appreſſo di poco, qual <lb/>ſia la loro proportione. </s> <s xml:id="echoid-s1582" xml:space="preserve">E per far ciò, con due diſtinti com-<lb/>paſſi ſi prenda la lunghezza de’lati homologi delle figure, <lb/>cioè di quelli, che ſono frapoſti <lb/> <anchor type="figure" xlink:label="fig-0096-01a" xlink:href="fig-0096-01"/> fra gl’angoli ſimili, e poſta la li-<lb/>nea minore ad vn’interuallo, che <lb/>ſi ſtimerà più à propoſito, con-<lb/>forme à ciò che la prattica inſe-<lb/>gnarà, veggaſi sù qual’ interuallo <lb/>capiſca l’altra linea maggiore; </s> <s xml:id="echoid-s1583" xml:space="preserve">& </s> <s xml:id="echoid-s1584" xml:space="preserve"><lb/>inumeri, ne’quali caderà queſta <lb/>applicatione, eſprimeranno la <lb/>proportione. </s> <s xml:id="echoid-s1585" xml:space="preserve">Come per @ſſem-<lb/>pio, ſono dati li due Baloardi ſi-<lb/>mili, e ſi deſidera ſapere, che pro-<lb/>portione habbiano; </s> <s xml:id="echoid-s1586" xml:space="preserve">prendo con <lb/>due compaſſi la lunghezza delle <lb/>faccie CD, & </s> <s xml:id="echoid-s1587" xml:space="preserve">HK; </s> <s xml:id="echoid-s1588" xml:space="preserve">& </s> <s xml:id="echoid-s1589" xml:space="preserve">applicata <lb/>CD all’interuallo 24. </s> <s xml:id="echoid-s1590" xml:space="preserve">24, trouo, <lb/>che HK cade nell’interuallo 30. </s> <s xml:id="echoid-s1591" xml:space="preserve">30, onde cauo, che le lor’aree <lb/>ſono come 24 à 30, cioè come 4 à 5.</s> <s xml:id="echoid-s1592" xml:space="preserve"/> </p> <div xml:id="echoid-div51" type="float" level="2" n="1"> <figure xlink:label="fig-0096-01" xlink:href="fig-0096-01a"> <image file="0096-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0096-01"/> </figure> </div> <pb o="83" file="0097" n="99" rhead="Linea Geometrica."/> <p> <s xml:id="echoid-s1593" xml:space="preserve">Equì è da auuertire eſſer meglio applicare la linea minore <lb/>à tal’a pertura dello ſtromento, che la maggiore venga à ca-<lb/>dere verſo li numeri maggiori, perche eſſendo li punti delle <lb/>diuiſioni verſo il fine dello ſtromento tra diloro poco diſtanti, <lb/>ſi vien’anche à trouare più preciſamente l’interuallo capace <lb/>della maggiore, paſſandoſi dall’vn punto all’ altro con poca<unsure/> <lb/>differenza, doue che nelle parti dello ſtromento più vicine al <lb/>centro non è così <lb/> <anchor type="figure" xlink:label="fig-0097-01a" xlink:href="fig-0097-01"/> facile, che ſi affron-<lb/>ti preciſamente in <lb/>tal’apertura, che li <lb/>due Compaſſi ſi <lb/>poſſano giuſtamẽ. <lb/></s> <s xml:id="echoid-s1594" xml:space="preserve">te applicare a’pun-<lb/>ti, che ſi cercano. </s> <s xml:id="echoid-s1595" xml:space="preserve"><lb/>Così ſia il circolo <lb/>HIK la larghezza <lb/>d’vn cannello di <lb/>bronzo, per cui <lb/>vno riceue l’acqua <lb/>dal bottino d’vna <lb/>fontana; </s> <s xml:id="echoid-s1596" xml:space="preserve">& </s> <s xml:id="echoid-s1597" xml:space="preserve">il circo-<lb/>lo DEF ſia la lar-<lb/>ghezza d’vn’altro cannello, per <lb/>cui l’acqua della ſteſſa fontana ſi <lb/>deriua ad vn’altro: </s> <s xml:id="echoid-s1598" xml:space="preserve">ſi cerca la pro-<lb/>portionc dell’acqua, che ciaſcuno <lb/>riceue, quanto è per queſto capo. </s> <s xml:id="echoid-s1599" xml:space="preserve"><lb/>Prendo il ſemidiametro, ò il diametro del primo, e l’applico <lb/>all’interuallo 15. </s> <s xml:id="echoid-s1600" xml:space="preserve">15; </s> <s xml:id="echoid-s1601" xml:space="preserve">dipoi veggo doue cada il ſemidiametro, <pb o="84" file="0098" n="100" rhead="CAPO III."/> ò dia metro dell’altro, e trouo, che cade nel 50; </s> <s xml:id="echoid-s1602" xml:space="preserve">dunquc argo-<lb/>mento, che l’acqua ſi diuide trà queſti due nella proportione <lb/>di 15 à 50, cioè di 3 à 10.</s> <s xml:id="echoid-s1603" xml:space="preserve"/> </p> <div xml:id="echoid-div52" type="float" level="2" n="2"> <figure xlink:label="fig-0097-01" xlink:href="fig-0097-01a"> <image file="0097-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0097-01"/> </figure> </div> <p> <s xml:id="echoid-s1604" xml:space="preserve">Che ſe le linee date foſſero troppo lunghe, già dalle coſe <lb/>dette di ſopra ſi caua, in qual maniera poſſiamo ſeruirci delle <lb/>lor parti aliquote. </s> <s xml:id="echoid-s1605" xml:space="preserve">Se ſi piglia d’amendue la ſteſſa parte ali-<lb/>quota, come la metà, ò il terzo di ciaſcuna, li numeri in cui <lb/>cadono, eſprimono la proportione, perche la ſteſſa propor-<lb/>tione è de’quadrati de gl’intieri, e de’quadrati delle parti ſi-<lb/>mili. </s> <s xml:id="echoid-s1606" xml:space="preserve">Se vna linea è ſtata applicata intiera, e dell’altra s’è ap-<lb/>plicata vna parte, il numero in cui cade, ſi moltiplichi per il <lb/>quadrato del denominatore della parte; </s> <s xml:id="echoid-s1607" xml:space="preserve">come ſe la linea mi-<lb/>nore ſi foſſe applicata al 27. </s> <s xml:id="echoid-s1608" xml:space="preserve">27, e della maggiore preſa la <lb/>metà, cadeſſe nel 18. </s> <s xml:id="echoid-s1609" xml:space="preserve">18, perche il 2 è denominatore della <lb/>parte, cioè della metà, piglio il ſuo quadrato 4, e moltiplica-<lb/>to per eſſo il 18, trouo, che viene 72; </s> <s xml:id="echoid-s1610" xml:space="preserve">onde dico, che li piani <lb/>ſono come 27 à 72, cioè come 3 à 8. </s> <s xml:id="echoid-s1611" xml:space="preserve">Se in vece della metà <lb/>haueſſe preſo il terzo, e foſſe caduto nell’ interuallo 8. </s> <s xml:id="echoid-s1612" xml:space="preserve">8, per-<lb/>che 9 è quadrato del 3 denominatore della parte preſa, mol-<lb/>tiplicato 8 per 9, all’iſteſſo modo ſi ſaria trouato 72. </s> <s xml:id="echoid-s1613" xml:space="preserve">Se fi-<lb/>nalmente d’vna linea ſi foſſe preſa la metà, dell’altra il quin-<lb/>to, il num. </s> <s xml:id="echoid-s1614" xml:space="preserve">della prima ſi molti plicarebbe per 4, e quello del-<lb/>la ſeconda per 25, che ſonoi quadrati de’denominatori delle <lb/>parti preſe, & </s> <s xml:id="echoid-s1615" xml:space="preserve">i prodotti eſprimerebbono la proportione <lb/>cercata de’ piani ſimili.</s> <s xml:id="echoid-s1616" xml:space="preserve"/> </p> <pb o="85" file="0099" n="101" rhead="Linea Geometrica"/> </div> <div xml:id="echoid-div54" type="section" level="1" n="28"> <head xml:id="echoid-head45" xml:space="preserve">QVESTIONE QVINTA.</head> <head xml:id="echoid-head46" xml:space="preserve">Date due, ò piu figure piane ſimili, trouarne vna ſimile vguale <lb/>à tutte quelle inſieme.</head> <p> <s xml:id="echoid-s1617" xml:space="preserve">OCcorre alle volte hauer’alcune figure la ſomma delle <lb/>quali ſi vuol’hauere in vna ſola figura ſimile à quelle: <lb/></s> <s xml:id="echoid-s1618" xml:space="preserve">e ſe bene ciò ſi può pratticare, mediante la 47 del lib. </s> <s xml:id="echoid-s1619" xml:space="preserve">1, come <lb/>appariſce da ciò, che s’è detto nella deſcrittione di queſte <lb/>linee Geometriche; </s> <s xml:id="echoid-s1620" xml:space="preserve">ad ogni modo ſenz’altro trauaglio facil-<lb/>mente ſi troua il lato della figura, che ſi cerca mediante que-<lb/>ſto ſtromento. </s> <s xml:id="echoid-s1621" xml:space="preserve">Siano dati due, ò più pentagoni, per farne <lb/>vno ſimile vguale à tutti inſieme. </s> <s xml:id="echoid-s1622" xml:space="preserve">Prendo con tanti compaſ-<lb/>ſi, quante ſono le figure date, li lati di dette figure, e confor-<lb/>me alla Queſtione precedente trouo la proportione di dette <lb/>figure tra di loro: </s> <s xml:id="echoid-s1623" xml:space="preserve">e conſiderati i numeri eſprimenti la pro-<lb/>portione, li riduco in vna ſomma, & </s> <s xml:id="echoid-s1624" xml:space="preserve">il numero, che ne riſulta <lb/>è quello, à cui nelle linee Geometriche ſi deue prender l’in-<lb/>teruallo, per hauer’il lato del pentagono, che ſi cerca. </s> <s xml:id="echoid-s1625" xml:space="preserve">Così ſe <lb/>ſi è trouato, che la proportione delli dati due pentagoni è co-<lb/>me 7 à 10. </s> <s xml:id="echoid-s1626" xml:space="preserve">il pentagono vguale à tutti due ſarà come 17; </s> <s xml:id="echoid-s1627" xml:space="preserve">on-<lb/>de ritenuta quella ſteſſa apertura dello ſtromento, prendo <lb/>l’interuallo 17. </s> <s xml:id="echoid-s1628" xml:space="preserve">17, e queſto è illato del pentagono vguale al-<lb/>li due pentagoni dati.</s> <s xml:id="echoid-s1629" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1630" xml:space="preserve">Mà ſe eſſendo più di due le figure date, ò non haueſſi tanti <lb/>compaſſi, quante ſon quelle, ouero nella ſteſſa apertura di <lb/>ſtromento non ſi trouaſſe, che cadeſſero giuſtamente sù li <lb/>punti, ſi faccia così: </s> <s xml:id="echoid-s1631" xml:space="preserve">ſe ne prendano due di quelli, che caden-<lb/>do sù li punti moſtrano la proportione, e ſe ne troui vno <pb o="86" file="0100" n="102" rhead="CAPO III."/> vguale à quelli, come ſopra, & </s> <s xml:id="echoid-s1632" xml:space="preserve">è ſtato all’interuallo 17. </s> <s xml:id="echoid-s1633" xml:space="preserve">17. <lb/></s> <s xml:id="echoid-s1634" xml:space="preserve">Ritengo con vn compaſſo queſto interuallo, e con vn’altro <lb/>compaſſo prendo il lato del terzo pentagono dato, & </s> <s xml:id="echoid-s1635" xml:space="preserve">appli-<lb/>cando queſti due compaſſi alle linee Geometriche con altra <lb/>apertura di ſtromento, trouo la proportione loro, e cadano <lb/>per eſſem pio sù li punti 12. </s> <s xml:id="echoid-s1636" xml:space="preserve">12, e 13. </s> <s xml:id="echoid-s1637" xml:space="preserve">13: </s> <s xml:id="echoid-s1638" xml:space="preserve">dunque il pentago-<lb/>no vguale à queſti due ſarà come 25, & </s> <s xml:id="echoid-s1639" xml:space="preserve">all’interuallo 25. </s> <s xml:id="echoid-s1640" xml:space="preserve">25, <lb/>haurò il lato conueniente al pentagono vguale alli tre penta-<lb/>goni dati.</s> <s xml:id="echoid-s1641" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div55" type="section" level="1" n="29"> <head xml:id="echoid-head47" xml:space="preserve">QVESTIONE SESTA.</head> <head xml:id="echoid-head48" xml:space="preserve">Date due figure piane ſimili, e diſuguali, trouar’vna figura ſimile <lb/>vguale alla lor differenza.</head> <p> <s xml:id="echoid-s1642" xml:space="preserve">QVeſta operatione ſeguita per il conuerſo della prece-<lb/>dente, perche ſe vniti i numeri eſprimenti la propor-<lb/>tione ſi troua la ſomma, ſottratto il minore dal <lb/>maggiore ſi hà il reſiduo. </s> <s xml:id="echoid-s1643" xml:space="preserve">Dati dunque due Baloardi ſimili <lb/>nella figura della queſtione 4, ſe ne voglia far’vno vguale alla <lb/>lor differenza; </s> <s xml:id="echoid-s1644" xml:space="preserve">prendo in eſſi due lati homologi, per eſſem-<lb/>pio le mezze gole FE, FM, & </s> <s xml:id="echoid-s1645" xml:space="preserve">applicatele allo ſtromento nel-<lb/>le linee Geometriche, trouo, che cadono ne’ punti 16, e 20; <lb/></s> <s xml:id="echoid-s1646" xml:space="preserve">onde la proportione de’piani è nota; </s> <s xml:id="echoid-s1647" xml:space="preserve">ſottrago il 16 dal 20, <lb/>& </s> <s xml:id="echoid-s1648" xml:space="preserve">il reſiduo 4 mi moſtra, che all’interuallo 4. </s> <s xml:id="echoid-s1649" xml:space="preserve">4, haurò la mez-<lb/>za gola del Baloardo ſimile vguale alla loro differenza.</s> <s xml:id="echoid-s1650" xml:space="preserve"/> </p> <pb o="87" file="0101" n="103" rhead="Linea Geometrica"/> </div> <div xml:id="echoid-div56" type="section" level="1" n="30"> <head xml:id="echoid-head49" style="it" xml:space="preserve">QVESTIONE SETTIMA.</head> <head xml:id="echoid-head50" style="it" xml:space="preserve">Date due linee, come poſſa trouarſi la terza proportionale.</head> <p> <s xml:id="echoid-s1651" xml:space="preserve">SI piglino le lunghezze delle due linee date con due di-<lb/>ſtinti compaſſi, es’appplichino allo ſtromento nel mo-<lb/>do detto alla queſtione precedente: </s> <s xml:id="echoid-s1652" xml:space="preserve">e ſi oſſerui ſopra quali <lb/>numeri cadano. </s> <s xml:id="echoid-s1653" xml:space="preserve">Dipoi la lunghezza della prima s’applichi <lb/>nella linea Aritmetica, di cui ſi parlò nel Capo 2, al numero, <lb/>che le corriſponde; </s> <s xml:id="echoid-s1654" xml:space="preserve">perche l’interuallo, che nella ſteſla linea <lb/>Aritmetica darà l’altro numero corriſpondente nella linea <lb/>Geometrica, ſarà la terza proportionale, che ſi cerca.</s> <s xml:id="echoid-s1655" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1656" xml:space="preserve">Siano date due <lb/> <anchor type="figure" xlink:label="fig-0101-01a" xlink:href="fig-0101-01"/> linee T, V, alle <lb/>quali conuenga <lb/>trouare la terza <lb/>proportionale: <lb/></s> <s xml:id="echoid-s1657" xml:space="preserve">le applico nella <lb/>linea Geometri-<lb/>ca AZ, AS, etro-<lb/>uo, che T cade <lb/>nell’ interuallo <lb/>17. </s> <s xml:id="echoid-s1658" xml:space="preserve">17, & </s> <s xml:id="echoid-s1659" xml:space="preserve">V ca-<lb/>de nell’interuallo 33. </s> <s xml:id="echoid-s1660" xml:space="preserve">33. </s> <s xml:id="echoid-s1661" xml:space="preserve">Perciò nella linea Aritmetica A E, <lb/>AL della figura 1 applico la linea data T all’interuallo 17. </s> <s xml:id="echoid-s1662" xml:space="preserve">17, <lb/>el’interuallo 33. </s> <s xml:id="echoid-s1663" xml:space="preserve">33, nella ſteſſa linea darà la terza propor-<lb/>tionale X. </s> <s xml:id="echoid-s1664" xml:space="preserve">La dimoſtratione è manifeſta, perche di tre con-<lb/>tinue proportionali la proportione della prima alla terza è <lb/>duplicata della proportione della prima alla ſeconda, cioè <pb o="88" file="0102" n="104" rhead="CAPO III."/> come il quadrato della prima al quadrato della ſeconda, così <lb/>la prima alla terza. </s> <s xml:id="echoid-s1665" xml:space="preserve">Or eſſendo il quadrato di T al quadrato <lb/>di V, come 17 à 33, come moſtrò la linea Geometrica, & </s> <s xml:id="echoid-s1666" xml:space="preserve">eſ-<lb/>ſendo la T alla X, come 17 à 33, come s’è fatto con la linea <lb/>Aritmetica; </s> <s xml:id="echoid-s1667" xml:space="preserve">ne ſeguita, che la T alla X hà la proportione del <lb/>quadrato di D al quadrato di V, e perciò continua la propor-<lb/>tione della linea T alla linea V.</s> <s xml:id="echoid-s1668" xml:space="preserve"/> </p> <div xml:id="echoid-div56" type="float" level="2" n="1"> <figure xlink:label="fig-0101-01" xlink:href="fig-0101-01a"> <image file="0101-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0101-01"/> </figure> </div> <p> <s xml:id="echoid-s1669" xml:space="preserve">Quindi ſe ſarà dato il quadrato HO ſopra la linea HI, che <lb/>rappreſenta vn campo di terra; </s> <s xml:id="echoid-s1670" xml:space="preserve">e ſarà data la linea KL fianco <lb/>d’vn’ altro pezzo diterra, che debba eſſer’ vguale al detto <lb/>quadrato HO, ſi vede eſſer neceſſario trouar’vna Terza pro-<lb/>portionale, à fine, che ſi faccia il rettangolo vguale al qua-<lb/>drato, per la 17 del lib. </s> <s xml:id="echoid-s1671" xml:space="preserve">6. </s> <s xml:id="echoid-s1672" xml:space="preserve">Applico dunque le due linee HI, <lb/>KL alla linea Geometrica, e vego, che cadono ne gl’interual-<lb/>li quella 14. </s> <s xml:id="echoid-s1673" xml:space="preserve">14, queſta 49. </s> <s xml:id="echoid-s1674" xml:space="preserve">49. </s> <s xml:id="echoid-s1675" xml:space="preserve">Perciò nella linea Aritmeti-<lb/>ca applico la linea KL all’interuallo 49. </s> <s xml:id="echoid-s1676" xml:space="preserve">49, el’interuallo 14. <lb/></s> <s xml:id="echoid-s1677" xml:space="preserve">14 nella ſteſſa linea Aritmetica midà la KM, onde il rettan-<lb/>golo ML è vguale al quadrato HO.</s> <s xml:id="echoid-s1678" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1679" xml:space="preserve">Della ſteſſa maniera dato vn ſegmento di circolo, ſi troua-<lb/>rà il diametro di eſſo circolo: </s> <s xml:id="echoid-s1680" xml:space="preserve">poiche diuiſa la corda per mez-<lb/>zo, e tirata à perpendicolo vna linea indefinita, ſi ponga in <lb/>primo luogo l’altezza del ſegmento, nel ſecondo la metà del-<lb/>la corda, e trouiſi la terza proportionale: </s> <s xml:id="echoid-s1681" xml:space="preserve">e queſta aggionta <lb/>all’altezza del ſegmento, darà il diametro del circolo, come <lb/>appariſce dalla 13 del lib. </s> <s xml:id="echoid-s1682" xml:space="preserve">6.</s> <s xml:id="echoid-s1683" xml:space="preserve"/> </p> <pb o="89" file="0103" n="105" rhead="Linea Geometrica."/> </div> <div xml:id="echoid-div58" type="section" level="1" n="31"> <head xml:id="echoid-head51" style="it" xml:space="preserve">QVESTIONE OTTAVA.</head> <head xml:id="echoid-head52" style="it" xml:space="preserve">Come ſi troui vna media proportionale tra due linee date, <lb/>e ſi faccia vn Quadrato vguale ad vna figura <lb/>rettilinea.</head> <p> <s xml:id="echoid-s1684" xml:space="preserve">SE la proportione delle linee date è conoſciuta in nume. <lb/></s> <s xml:id="echoid-s1685" xml:space="preserve">ri, ſi applichi nella linea Geometrica vna delle date li-<lb/>nee all’interuallo d’vno de’numeri, ch’eſprimono la propor-<lb/>tione delle due linee eſtreme, poiche l’interuallo corriſpon-<lb/>dente all’altro di detti numeri darà la lunghezza della media <lb/>proportionale. </s> <s xml:id="echoid-s1686" xml:space="preserve">Mà ſe non ſi sà, che proportione habbiano <lb/>tra di loro le due linee eſtreme date, queſta ſi troui sù la linea<unsure/> <lb/>Aritmetica nel modo inſegnato alla Queſtione 5. </s> <s xml:id="echoid-s1687" xml:space="preserve">del Cap. </s> <s xml:id="echoid-s1688" xml:space="preserve">2, <lb/>e poi s’opri, come s’è detto.</s> <s xml:id="echoid-s1689" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1690" xml:space="preserve">Sia dato vn triangolo KSL nella fig. </s> <s xml:id="echoid-s1691" xml:space="preserve">della queſt. </s> <s xml:id="echoid-s1692" xml:space="preserve">antece-<lb/>dente, e ſi voglia vn quadrato, che gli ſia vguale. </s> <s xml:id="echoid-s1693" xml:space="preserve">Per quel-<lb/>lo, che ſi caua dalla 41. </s> <s xml:id="echoid-s1694" xml:space="preserve">del lib. </s> <s xml:id="echoid-s1695" xml:space="preserve">1, il ſudetto triangolo è vguale <lb/>al parallelogrammo rettangolo, che habbia la ſteſſa baſe, e <lb/>la metà dell’ altezza perpendicolare, ò la ſteſſa altezza è la <lb/>metà della baſe. </s> <s xml:id="echoid-s1696" xml:space="preserve">Dunque ſe ſi trouerà vna media proportio-<lb/>nale tra la baſe, e la metà dell’ altezza perpendicolare del <lb/>triangolo, queſta ſarà il lato del quadrato vguale al triango-<lb/>lo dato KSL, eſſendo che per la 17 del 6, il quadrato di quel-<lb/>la è vguale alrettangolo ſotto le due eſtreme. </s> <s xml:id="echoid-s1697" xml:space="preserve">Diuido dun-<lb/>que per metà l’altezza SL in R, e nella linea Aritmetica ap-<lb/>plicate KL, & </s> <s xml:id="echoid-s1698" xml:space="preserve">LR, trouo, che la prima è 49, la ſeconda 14: <lb/></s> <s xml:id="echoid-s1699" xml:space="preserve">perciò nella linea Geometrica applico KL all’ interuallo 49. </s> <s xml:id="echoid-s1700" xml:space="preserve"><lb/>49, e nella ſteſſa preſo l’<unsure/>interuallo 14. </s> <s xml:id="echoid-s1701" xml:space="preserve">14, dà la linea HI me- <pb o="90" file="0104" n="106" rhead="CAPO III."/> dia proportionale cercata, il cui quadrato HO è vguale al da-<lb/>to triangolo KSL. </s> <s xml:id="echoid-s1702" xml:space="preserve">E che HI ſia la media proportionale cer-<lb/>cata è manifeſto, perche per la coſtruttione dello ſtromento <lb/>il quadrato di KL al quadrato di HIè come 49 à 14, cioè co-<lb/>me la linea KL ad LR: </s> <s xml:id="echoid-s1703" xml:space="preserve">dunque eſſendo la proportione di KL <lb/>ad LR duplicata della proportione di KL ad HI, ſaranno <lb/>continuamente proportionali KL, HI, LR. </s> <s xml:id="echoid-s1704" xml:space="preserve">Che ſe la figura <lb/>ſia di molti lati, ſi riſolua in triangoli, & </s> <s xml:id="echoid-s1705" xml:space="preserve">in ciaſcheduno ſi tiri <lb/>la perpendicolare, etrouiſi il quadrato di ciaſcun triangolo, <lb/>e poi per la queſt. </s> <s xml:id="echoid-s1706" xml:space="preserve">5. </s> <s xml:id="echoid-s1707" xml:space="preserve">ſi troui il quadrato vguale à tutti queſti <lb/>quadrati.</s> <s xml:id="echoid-s1708" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div59" type="section" level="1" n="32"> <head xml:id="echoid-head53" style="it" xml:space="preserve">QVESTIONE NONA.</head> <head xml:id="echoid-head54" style="it" xml:space="preserve">Deſcriuere con facilità vna Parabola.</head> <p> <s xml:id="echoid-s1709" xml:space="preserve">EDimoſtrato, che nella Parabola li quadrati delle linee <lb/>Applicate al diametro ſono in tal proportione, quale <lb/>hanno le Saette (che ſono la parte del diametro intercetta <lb/>tra’l punto dell’ Applicatione, & </s> <s xml:id="echoid-s1710" xml:space="preserve">il Vertice della Parabola) <lb/>eſſendoche ciaſcun Quadrato delle Applicate è vguale al ret-<lb/>tangolo fatto dalla Saetta, e dal lato Retto; </s> <s xml:id="echoid-s1711" xml:space="preserve">e perciò hauen-<lb/>do tutti i rettangoli l’altezza medeſima, che è il lato Retto, <lb/>hanno la proportione delle baſi, cioè delle Saette.</s> <s xml:id="echoid-s1712" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1713" xml:space="preserve">Preſo dunque il Diametro della Parabola ſi diuida in quan-<lb/>te ſi vogliano parti vguali cominciando dal Vertice, e per i <lb/>punti delle diuiſioni ſi tirino linee parallele tra di loro, ò ſiano <lb/>perpendicolari al diametro, ò oblique, come più piacerà. <lb/></s> <s xml:id="echoid-s1714" xml:space="preserve">Dipoi prendaſi il lato Retto, ſe è dato, e tra eſſo, e la prima <lb/>Saetta, trouiſi vna Media proportionale, per la queſt. </s> <s xml:id="echoid-s1715" xml:space="preserve">8, e que- <pb o="91" file="0105" n="107" rhead="Linea Geometrica"/> ſta ſarà la grandezza della prima Applicata. </s> <s xml:id="echoid-s1716" xml:space="preserve">Ciò fatto, pon-<lb/>gaſi queſta prima Applicata tra li punti 1. </s> <s xml:id="echoid-s1717" xml:space="preserve">1. </s> <s xml:id="echoid-s1718" xml:space="preserve">della linea Geo. <lb/></s> <s xml:id="echoid-s1719" xml:space="preserve">metrica; </s> <s xml:id="echoid-s1720" xml:space="preserve">e poſ<unsure/>cia preſa la diſtanza 2. </s> <s xml:id="echoid-s1721" xml:space="preserve">2. </s> <s xml:id="echoid-s1722" xml:space="preserve">ſi ponga nella ſecon-<lb/>da parallela, e ſarà la ſeconda Applicata; </s> <s xml:id="echoid-s1723" xml:space="preserve">nella terza paral-<lb/>lela ſi metta la diſtanza 3. </s> <s xml:id="echoid-s1724" xml:space="preserve">3. </s> <s xml:id="echoid-s1725" xml:space="preserve">e ſarà la terza Applicata, e così <lb/>di mano in mano. </s> <s xml:id="echoid-s1726" xml:space="preserve">Finalmente la linea, che paſſarà per que-<lb/>ſti punti eſtremi delle Applicate, ſarà Parabolica.</s> <s xml:id="echoid-s1727" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1728" xml:space="preserve">Che ſe il lato Retto non è dato, prendaſi la prima Appli-<lb/>cata grande ad arbitrio, e ſi operi, come ſi è detto; </s> <s xml:id="echoid-s1729" xml:space="preserve">e ad vna <lb/>delle Saette, & </s> <s xml:id="echoid-s1730" xml:space="preserve">alla ſua Applicata trouandoſi per la queſt. </s> <s xml:id="echoid-s1731" xml:space="preserve">7. <lb/></s> <s xml:id="echoid-s1732" xml:space="preserve">la Terza Proportionale ſarà illato Retto di tal Parabola.</s> <s xml:id="echoid-s1733" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div60" type="section" level="1" n="33"> <head xml:id="echoid-head55" style="it" xml:space="preserve">QVESTIONE DECIMA.</head> <head xml:id="echoid-head56" style="it" xml:space="preserve">Data vna Parabola in vn Cono dato, trouar vn Quadrato <lb/>à lei vguale.</head> <p> <s xml:id="echoid-s1734" xml:space="preserve">SIa dato il Cono ABC, e dal punto D ſia fatta la Settione, <lb/>che genera la Parabola FDG. </s> <s xml:id="echoid-s1735" xml:space="preserve">Or eſſendo DE paralle-<lb/>la ad AB, come CA à CB, così <lb/> <anchor type="figure" xlink:label="fig-0105-01a" xlink:href="fig-0105-01"/> CD à CE, la quale perciò, per <lb/>la queſt. </s> <s xml:id="echoid-s1736" xml:space="preserve">3. </s> <s xml:id="echoid-s1737" xml:space="preserve">del capo 2, ſarà no-<lb/>ta. </s> <s xml:id="echoid-s1738" xml:space="preserve">E perche CB è diametro <lb/>del circolo BFCG, tagliata ad <lb/>angoli retti dalla ſettione FG, <lb/>perciò tra CE, & </s> <s xml:id="echoid-s1739" xml:space="preserve">EB ſi troui <lb/>la Media Proportionale, e ſarà <lb/>EG, conforme alla 13. </s> <s xml:id="echoid-s1740" xml:space="preserve">del 6. </s> <s xml:id="echoid-s1741" xml:space="preserve">Ora il Maſſimo Triangolo <lb/>della Parabola ha per baſe FG, e per altezza ED Aſſe della <lb/>Parabola, e perciò è vguale al rettangolo fatto da ED, EG.</s> <s xml:id="echoid-s1742" xml:space="preserve"> <pb o="92" file="0106" n="108" rhead="CAPO III."/> Dunque tra ED, EG ſi troui vna Media proportionale, e ſia <lb/>per cagione d’eſempio la linea H; </s> <s xml:id="echoid-s1743" xml:space="preserve">& </s> <s xml:id="echoid-s1744" xml:space="preserve">il quadrato di queſta ſa-<lb/>râ vguale al Triangolo maſſimo della Parabola FDG. </s> <s xml:id="echoid-s1745" xml:space="preserve">Final-<lb/>mente, perche dalle coſe dimoſtrate da Archimede la Para-<lb/>bola al ſuo maſſimo Triangolo è come 4 à 3, quella linea vl-<lb/>timamente trouata Hpongaſi nella linea Geometrica all’in-<lb/>teruallo 3. </s> <s xml:id="echoid-s1746" xml:space="preserve">3, e poi ſi prenda l’interuallo 4. </s> <s xml:id="echoid-s1747" xml:space="preserve">4: </s> <s xml:id="echoid-s1748" xml:space="preserve">che queſto darà <lb/>vna linea il cui quadrato è vguale alla Parabola data, eſſendo <lb/>anch’egli ſeſquiterzo del maſſimo Triangolo medeſimo.</s> <s xml:id="echoid-s1749" xml:space="preserve"/> </p> <div xml:id="echoid-div60" type="float" level="2" n="1"> <figure xlink:label="fig-0105-01" xlink:href="fig-0105-01a"> <image file="0105-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0105-01"/> </figure> </div> </div> <div xml:id="echoid-div62" type="section" level="1" n="34"> <head xml:id="echoid-head57" style="it" xml:space="preserve">QVESTIONE VNDECIMA.</head> <head xml:id="echoid-head58" style="it" xml:space="preserve">Date due linee vguali, che ſitagliano per mezzo obliquamēnte, <lb/>deſcriuere intorno ad eſſe vn’ Ellipſi.</head> <p> <s xml:id="echoid-s1750" xml:space="preserve">SIano le due linee AB, CD, che ſi tagliano per mezzo ob-<lb/>liquamente in E; </s> <s xml:id="echoid-s1751" xml:space="preserve">& </s> <s xml:id="echoid-s1752" xml:space="preserve">intorno ad eſſe habbiaſi à deſcriuer <lb/>vn’ Ellipſi, di cui elle ſono i diametri <lb/> <anchor type="figure" xlink:label="fig-0106-01a" xlink:href="fig-0106-01"/> coniugati vguali. </s> <s xml:id="echoid-s1753" xml:space="preserve">Prima ſi trouino gli <lb/>Aſſi: </s> <s xml:id="echoid-s1754" xml:space="preserve">il che breuemente ſi fà tirando le <lb/>linee AC, AD; </s> <s xml:id="echoid-s1755" xml:space="preserve">e queſte diuiſe vgual-<lb/>mente in F, e G, dal centro E ſi tirino <lb/>le linee EH, EI indefinite: </s> <s xml:id="echoid-s1756" xml:space="preserve">Queſte ſi di-<lb/>moſtra, che ſonogli Aſſi, perche eſſen-<lb/>do li punti D, A, C, eſtremità delli dia-<lb/>metri vguali dati nella circonferenza <lb/>dell’Ellipſi, così la linea AD, come la <lb/>AC ſono Applicate, quella al diame-<lb/>tro EI, e queſta al diametro EH. </s> <s xml:id="echoid-s1757" xml:space="preserve">Ora <lb/>perche AE è vguale ad EC, per l’hipo- <pb o="93" file="0107" n="109" rhead="Linea Geometrica"/> teſi, & </s> <s xml:id="echoid-s1758" xml:space="preserve">AF vguale à FC per la coſtruttione, e FE è commu-<lb/>ne, ſono li Triangoli AFE, CFE vguali, egli angoli poſti à <lb/>F ſono vguali, e perciò retti: </s> <s xml:id="echoid-s1759" xml:space="preserve">dunque il diametro EH è Aſſe. <lb/></s> <s xml:id="echoid-s1760" xml:space="preserve">Similmente ſi dimoſtra gli angolià Geſſer retti, cper conſe-<lb/>guenzail diametro EI eſſer Aſſe.</s> <s xml:id="echoid-s1761" xml:space="preserve"/> </p> <div xml:id="echoid-div62" type="float" level="2" n="1"> <figure xlink:label="fig-0106-01" xlink:href="fig-0106-01a"> <image file="0106-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0106-01"/> </figure> </div> <p> <s xml:id="echoid-s1762" xml:space="preserve">Per trouar il termine de gli Aſſi, dal punto A ſi tiri vna pa-<lb/>rallela all’altro diametro DC, la quale è Tangente dell’Ellip-<lb/>ſi, e taglia gli Aſſi in H, & </s> <s xml:id="echoid-s1763" xml:space="preserve">I. </s> <s xml:id="echoid-s1764" xml:space="preserve">Trouiſi dunque tra EF, & </s> <s xml:id="echoid-s1765" xml:space="preserve">EH, <lb/>la media Proportionale EL, per la queſt. </s> <s xml:id="echoid-s1766" xml:space="preserve">8, e queſto è il termi-<lb/>ne dell’ Aſſe maggiore: </s> <s xml:id="echoid-s1767" xml:space="preserve">e ſimilmente tra EG, & </s> <s xml:id="echoid-s1768" xml:space="preserve">EI trouiſi la <lb/>Media proportionale EK, & </s> <s xml:id="echoid-s1769" xml:space="preserve">è K termine dell’ Aſſe minore. <lb/></s> <s xml:id="echoid-s1770" xml:space="preserve">Tirata per tanto la KL è Applicata al diametro AB.</s> <s xml:id="echoid-s1771" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1772" xml:space="preserve">Ciò fatto, nel Diametro AB prendanſi quelli punti che ſi <lb/>vogliono M, P, & </s> <s xml:id="echoid-s1773" xml:space="preserve">altri, e ſi tirino linee parallele all’Applica-<lb/>ta KL, ò pure al diametro DC, che tutto torna allo ſteſſo. <lb/></s> <s xml:id="echoid-s1774" xml:space="preserve">E per hauere la quantità di queſte, ſi prenda, per la queſt. </s> <s xml:id="echoid-s1775" xml:space="preserve">8, <lb/>la media proportionale tra li due ſegmenti del diametro: </s> <s xml:id="echoid-s1776" xml:space="preserve">così <lb/>tra AM, MB ſia MN; </s> <s xml:id="echoid-s1777" xml:space="preserve">e tra AP, PB ſia PR, e così dell’altre: </s> <s xml:id="echoid-s1778" xml:space="preserve"><lb/>perche li punti N, R, &</s> <s xml:id="echoid-s1779" xml:space="preserve">c. </s> <s xml:id="echoid-s1780" xml:space="preserve">ſono anch’eſſi nella circonferenza <lb/>ſteſſa con gli altri. </s> <s xml:id="echoid-s1781" xml:space="preserve">Il che ſi dimoſtra, perche nell’ Ellipſi i <lb/>Quadrati delle Applicate ſono nella proportione delli Ret-<lb/>tangoli fattidalli ſegmenti del diametro, à cuiſono Applica-<lb/>te. </s> <s xml:id="echoid-s1782" xml:space="preserve">Onde come il rettangolo AOB al rettangolo AMB, così <lb/>il Quadrato OL al Quadrato MN: </s> <s xml:id="echoid-s1783" xml:space="preserve">e così in realtà ſono, eſ-<lb/>ſendoſi poſte OL, MN medie Proportionali.</s> <s xml:id="echoid-s1784" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1785" xml:space="preserve">E che li Quadrati delle Applicate all’vno de’Diametri con-<lb/>iugati vguali, ſiano vguali alli Rettangoli fatti dalli ſegmenti, <lb/>è manifeſto; </s> <s xml:id="echoid-s1786" xml:space="preserve">perche come il rettangolo AEB al Quadrato EC, <lb/>così il rettangolo AOB al Quadrato OL: </s> <s xml:id="echoid-s1787" xml:space="preserve">Mà ilrettangolo <lb/>AEB è vguale al Quadrato EC (eſſendo vguali le trè linee <pb o="94" file="0108" n="110" rhead="CAPO III."/> EA, EB, EC, per l’hipoteſi) dunque anche il rettangolo AOB <lb/>è vguale al Quadrato OL, & </s> <s xml:id="echoid-s1788" xml:space="preserve">AMB al Quadrato MN.</s> <s xml:id="echoid-s1789" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1790" xml:space="preserve">Auuertaſi dalli meno prattici, che tal modo di deſcriuere <lb/>l’Ellipſi con le Medie proportionali al modo ſodetto, conuie-<lb/>ne ſolo alli diametri coniugati vguali.</s> <s xml:id="echoid-s1791" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1792" xml:space="preserve">Nella maniera che ſi è deſcrita vna quarta parte dell’El-<lb/>lipſi, ſi fà il quadrante oppoſto; </s> <s xml:id="echoid-s1793" xml:space="preserve">e l’iſteſlo artificio ſi vſa con <lb/>gli altri quadranti; </s> <s xml:id="echoid-s1794" xml:space="preserve">il che non hò fatto in queſto eſempio per <lb/>isfuggire la confuſione delle linee. </s> <s xml:id="echoid-s1795" xml:space="preserve">Che poi HS, & </s> <s xml:id="echoid-s1796" xml:space="preserve">IZ ſiano <lb/>gli Aſſi, che ad angoli retti ſi tagliano in E, è maniſeſto; </s> <s xml:id="echoid-s1797" xml:space="preserve">per-<lb/>che da E vſcendo trè linee EA, EC, ED vguali, quello è cen-<lb/>tro del circolo, che paſſa per li punti eſtremi, onde CAD è an-<lb/>golo retto, eſſendo nel ſemicircolo; </s> <s xml:id="echoid-s1798" xml:space="preserve">e perciò AC, & </s> <s xml:id="echoid-s1799" xml:space="preserve">IE ſono <lb/>parallele, e l’angolo IEF è vguale all’angolo AFE retto, poi-<lb/>che tutti due inſieme ſi vguagliano à due retti.</s> <s xml:id="echoid-s1800" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div64" type="section" level="1" n="35"> <head xml:id="echoid-head59" style="it" xml:space="preserve">QVESTIONE DVODECIMA.</head> <head xml:id="echoid-head60" style="it" xml:space="preserve">Data vna portione di Ouato trouar il reſtante del ſuo <lb/>diametro.</head> <p> <s xml:id="echoid-s1801" xml:space="preserve">SIa data la portione Elliptica BAC, <lb/> <anchor type="figure" xlink:label="fig-0108-01a" xlink:href="fig-0108-01"/> in cui ſia tirata la retta BC, e diuiſa <lb/>per mezzo in D; </s> <s xml:id="echoid-s1802" xml:space="preserve">à queſta tiriſi parallela <lb/>vn’altra linea EF ſimilmente diuiſa in G. <lb/></s> <s xml:id="echoid-s1803" xml:space="preserve">Quindi per D, e G tirata la retta DA ſa-<lb/>rà parte del Diametro, di cui ſi cerca il re-<lb/>ſiduo DH. </s> <s xml:id="echoid-s1804" xml:space="preserve">Prendanſi le Applicate DC, <lb/>e FG, e la proportione de’ loro Quadrati <lb/>ſi troui nella linea Geometrica: </s> <s xml:id="echoid-s1805" xml:space="preserve">Dipoinella linea Aritmeti- <pb o="95" file="0109" n="111" rhead="Linea Geometrica"/> ca ſi troui la proportione delle linee GA, DA.</s> <s xml:id="echoid-s1806" xml:space="preserve"/> </p> <div xml:id="echoid-div64" type="float" level="2" n="1"> <figure xlink:label="fig-0108-01" xlink:href="fig-0108-01a"> <image file="0108-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0108-01"/> </figure> </div> <p> <s xml:id="echoid-s1807" xml:space="preserve">Ora, perche come il Quadrato di GF al Quadrato di DC, <lb/>così è il rettangolo AGH al rettangolo ADH; </s> <s xml:id="echoid-s1808" xml:space="preserve">perciò à fine di <lb/>trouare la DH, queſta ſi metta I℞ al modo gli Algebriſti. <lb/></s> <s xml:id="echoid-s1809" xml:space="preserve">Eſa<unsure/>ppongaſi, che GA ſia 3, e DA ſia 5: </s> <s xml:id="echoid-s1810" xml:space="preserve">dunque GD è 2: </s> <s xml:id="echoid-s1811" xml:space="preserve">e <lb/>così GH è 2 + I℞. </s> <s xml:id="echoid-s1812" xml:space="preserve">Dunque il rettangolo AGH è 6+3℞, & </s> <s xml:id="echoid-s1813" xml:space="preserve">il <lb/>rettangolo ADH è 5℞. </s> <s xml:id="echoid-s1814" xml:space="preserve">Quindiè, che trouatoſi il Quadrato <lb/>di GF eſſere 17, & </s> <s xml:id="echoid-s1815" xml:space="preserve">il Quadrato di DC 25 (per cagion d’eſſem-<lb/>pio) ſarà come 17à 25, così 6 + 3℞, à 5℞: </s> <s xml:id="echoid-s1816" xml:space="preserve">e per la 16 del 6, <lb/>ò 19 del 7. </s> <s xml:id="echoid-s1817" xml:space="preserve">ſaranno 85 ℞ vguali à 150 † 75℞, e leuate da ambe <lb/>le parti 75℞, reſtano 10℞ vguali à 150; </s> <s xml:id="echoid-s1818" xml:space="preserve">diuiſo 150 per 10, il <lb/>Quotiente 15 dà la quantità di vna Radice, cioè DH, che è <lb/>15 parti di quelle, che in DA ſono 5; </s> <s xml:id="echoid-s1819" xml:space="preserve">e tutto il diametro AH <lb/>è di parti 20.</s> <s xml:id="echoid-s1820" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1821" xml:space="preserve">Quindi per vedere ſe il diametro AH ſia Aſſe dell’Ellipſi, <lb/>oſſeruiſi, ſel’angolo CDA ſia retto, ò nò: </s> <s xml:id="echoid-s1822" xml:space="preserve">il che facilmente ſi <lb/>farà mettendo nella linea Geometrica la DC all’interuallo <lb/>25.</s> <s xml:id="echoid-s1823" xml:space="preserve">25, come ſi trouò; </s> <s xml:id="echoid-s1824" xml:space="preserve">e vedendo doue capiſca la DA, aggion-<lb/>ganſi queſti due Quadrati. </s> <s xml:id="echoid-s1825" xml:space="preserve">Dipoi tirata la retta AC anch’ella <lb/>applicata alla linea Geometrica, ritenuta la ſteſſa apertura <lb/>dello ſtromento, moſtrarà il ſuo Quadrato: </s> <s xml:id="echoid-s1826" xml:space="preserve">il quale ſe ſarà <lb/>vguale alla ſomma di que’due Quadrati, l’angolo CDA è ret-<lb/>to, per la 48 del 1: </s> <s xml:id="echoid-s1827" xml:space="preserve">ſe è maggiore, l’angolo è ottuſo per la <lb/>12 del 2, e ſe è minore l’angolo è acuto per la 13 del 2. </s> <s xml:id="echoid-s1828" xml:space="preserve">Se <lb/>dunque non è angolo retto, quel diametro non è Aſſe.</s> <s xml:id="echoid-s1829" xml:space="preserve"/> </p> <pb o="96" file="0110" n="112" rhead="CAPO III."/> </div> <div xml:id="echoid-div66" type="section" level="1" n="36"> <head xml:id="echoid-head61" style="it" xml:space="preserve">QVESTIONE DECIMATERZA.</head> <head xml:id="echoid-head62" style="it" xml:space="preserve">Dalli due diametri d’vn Ellipſi trouar l’area.</head> <p> <s xml:id="echoid-s1830" xml:space="preserve">PRimieramente ſi faccia come 14 à 11, così il Quadrato <lb/>del diametro maggiore ad vn’altro, e ſarà l’area del <lb/>circolo di detto diametro, per la 2. </s> <s xml:id="echoid-s1831" xml:space="preserve">di Archimede lib. </s> <s xml:id="echoid-s1832" xml:space="preserve">de di-<lb/>menſ. </s> <s xml:id="echoid-s1833" xml:space="preserve">circuli. </s> <s xml:id="echoid-s1834" xml:space="preserve">Dipoi per le coſe dimoſtrate dall’ iſteſſo Archi-<lb/>mede lib. </s> <s xml:id="echoid-s1835" xml:space="preserve">de Conoid. </s> <s xml:id="echoid-s1836" xml:space="preserve">& </s> <s xml:id="echoid-s1837" xml:space="preserve">Sphæroid. </s> <s xml:id="echoid-s1838" xml:space="preserve">prop 5. </s> <s xml:id="echoid-s1839" xml:space="preserve">Facciaſi come il <lb/>diametro maggiore al minore, così il Quadrato del diame-<lb/>tro maggiore ad vn’altro, e ſarà l’area dell’Ellipſi.</s> <s xml:id="echoid-s1840" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1841" xml:space="preserve">Perciò nelle linee Geometriche pongaſi la linea data, che <lb/>è maggior diametro dell’Ellipſi, all’interuallo 14. </s> <s xml:id="echoid-s1842" xml:space="preserve">14, e di poi <lb/>prendaſi l’interuallo 11. </s> <s xml:id="echoid-s1843" xml:space="preserve">11, e ſarà lato d’vn Quadrato vguale <lb/>al circolo di detto diametro.</s> <s xml:id="echoid-s1844" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1845" xml:space="preserve">Dipoi habbiaſi in numeri la proportione delli due Dia-<lb/>metri dati, e ſia per cagion d’eſſempio 15 a 13: </s> <s xml:id="echoid-s1846" xml:space="preserve">Dunque <lb/>quell’interuallo trouato tra 11. </s> <s xml:id="echoid-s1847" xml:space="preserve">11, ſi ponga tra 15. </s> <s xml:id="echoid-s1848" xml:space="preserve">15, poi-<lb/>che l’interuallo 13. </s> <s xml:id="echoid-s1849" xml:space="preserve">13, darà illato del Quadrato, che è vguale <lb/>all’area dell Ellipſi cercata.</s> <s xml:id="echoid-s1850" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1851" xml:space="preserve">Finalmente queſt’vltimo lato trouato ſi paragoni col dia-<lb/>metro maggiore dato, e sì come è noto il Quadrato di eſſo <lb/>diametro maggiore, cosìſarà noto il Quadraro del lato vlti-<lb/>mamente trouato, e per conſeguenza ſarà nota l’area dell’ <lb/>Ellipſi.</s> <s xml:id="echoid-s1852" xml:space="preserve"/> </p> <pb o="97" file="0111" n="113" rhead="Linea Geometrica"/> </div> <div xml:id="echoid-div67" type="section" level="1" n="37"> <head xml:id="echoid-head63" xml:space="preserve">QVESTIONE DECIMAQVARTA.</head> <head xml:id="echoid-head64" xml:space="preserve">Dato vn numero, trouare la ſuaradice quadrata.</head> <p> <s xml:id="echoid-s1853" xml:space="preserve">E’Vero, che non tutti li numeri ſono quadrati, e perciò <lb/>non hanno la radice preciſa, ad ogni modo, per le ope. <lb/></s> <s xml:id="echoid-s1854" xml:space="preserve">rationi Fiſiche, ci baſta la radice più vicina ne’numeri intieri, <lb/>e nel formare ſquadroni quadri di gente, non occorre ſaper <lb/>li rotti. </s> <s xml:id="echoid-s1855" xml:space="preserve">Mà perche tutti li numeri diſotto del 100. </s> <s xml:id="echoid-s1856" xml:space="preserve">ſono di <lb/>due ſole figure, perciò nello ſtromento non ſi trouerà imme-<lb/>diatamente, che la radice di numeri non maggiori di quattro <lb/>figure, perche vn numero ditre, ò quattro figure hà la radice <lb/>di due figure, mà ſe il numero habbia cinque, ò ſei figure, la <lb/>radice è di tre figure, come è manifeſto, & </s> <s xml:id="echoid-s1857" xml:space="preserve">allhora ſi richiede <lb/>qualch’altro artificio da ſpiegarſi. </s> <s xml:id="echoid-s1858" xml:space="preserve">Ora ſe è nota la proportio-<lb/>ne di due quadrati, la ſubduplicata è la proportione delle loro <lb/>radici, e così di quali parti è vna, ditali ſarà anche l’altra. </s> <s xml:id="echoid-s1859" xml:space="preserve">Per-<lb/>ciò dato vn numero, ſappiamo, che proportione habbia ad <lb/>vn’altro numero, preſi tutti due come quadrati nella linea <lb/>Geometrica. </s> <s xml:id="echoid-s1860" xml:space="preserve">E ſe ſarà nota la radice d’vno nella linea Arit-<lb/>metica, ſi manifeſterà anche l’altra radice in particelle ſimili. </s> <s xml:id="echoid-s1861" xml:space="preserve"><lb/>Quindi è, che dato vn numero d’alcune figure, ne piglio <lb/>vn’altro ad arbitrio, mà preciſamente quadrato, il quale ò <lb/>tutto intiero, ò gettati via li zeri, ſia tra li numeri ſegnati nella <lb/>linea Geometrica. </s> <s xml:id="echoid-s1862" xml:space="preserve">Et il numero dato ò tutto intiero, ò getta-<lb/>te via tante figure, quanti zeri ſi leuarono dal quadrato pre-<lb/>ciſo, lo prendo al ſuo interuallo nella linea Geometrica, allar-<lb/>gato lo ſtromento ad arbitrio: </s> <s xml:id="echoid-s1863" xml:space="preserve">e poi con vn’altro Compaſſo <lb/>prendo l’interuallo del numero preciſamente quadrato nel <pb o="98" file="0112" n="114" rhead="CAPO III."/> modo detto, tolto ad arbitrio. </s> <s xml:id="echoid-s1864" xml:space="preserve">Poſcia nella linea Aritmetica <lb/>applico queſto ſecondo interuallo al numero, che è radice co-<lb/>noſciuta del quadrato precilo, el’altro interuallo darà nella <lb/>linea Aritmetica la radice cercata.</s> <s xml:id="echoid-s1865" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1866" xml:space="preserve">Sia dato il numero di Soldati 5400, di cui deſidero la radice <lb/>quadrata per ſapere, quanti debbano eſſer per fronte, volen-<lb/>do far ſquadrone quadro di gente; </s> <s xml:id="echoid-s1867" xml:space="preserve">leuo li due zeri, & </s> <s xml:id="echoid-s1868" xml:space="preserve">aperto <lb/>lo ſtromento ad arbitrio, prendo nella linea Geometrica l’in-<lb/>teruallo 54. </s> <s xml:id="echoid-s1869" xml:space="preserve">54. </s> <s xml:id="echoid-s1870" xml:space="preserve">Eritenuta quell’apertura di ſtromento, pi-<lb/>glio nella ſteſſa linea l’interuallo d’vn numero preciſamente <lb/>quadrato, come 4.</s> <s xml:id="echoid-s1871" xml:space="preserve">9. </s> <s xml:id="echoid-s1872" xml:space="preserve">16, ò altro tale. </s> <s xml:id="echoid-s1873" xml:space="preserve">Sia preſo per eſſempio <lb/>l’interuallo 9. </s> <s xml:id="echoid-s1874" xml:space="preserve">9, la cui radice è nota eſſere 3. </s> <s xml:id="echoid-s1875" xml:space="preserve">Ora perche ſi <lb/>gettaron via due zeri dal numero dato 5400, s’intendono le-<lb/>uati due zeri anche dal 900; </s> <s xml:id="echoid-s1876" xml:space="preserve">ſono dunque li due quadrati ap-<lb/>plicati nella proportione di 900 à 5400; </s> <s xml:id="echoid-s1877" xml:space="preserve">e così la radice del <lb/>primo è 3 con vn zero, cioè 30. </s> <s xml:id="echoid-s1878" xml:space="preserve">l’interuallo dunque 9. </s> <s xml:id="echoid-s1879" xml:space="preserve">9 del-<lb/>la linea Geometrica applicato nella linea Aritmetica al 30. <lb/></s> <s xml:id="echoid-s1880" xml:space="preserve">30, l’apertura dell’altro Compaſſo, che daua 54. </s> <s xml:id="echoid-s1881" xml:space="preserve">54 nella li-<lb/>nea Geometrica, caderà nella linea Aritmetica all’interuallo <lb/>73. </s> <s xml:id="echoid-s1882" xml:space="preserve">73, e così dico la radice del numero 5400 eſſere 73, e <lb/>perciò eſſere 73 file di Soldati, ciaſcuna delle quali ne hà 73 <lb/>difronte.</s> <s xml:id="echoid-s1883" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1884" xml:space="preserve">L’iſteſſo ſarebbe, ſe in vece di prendere 9. </s> <s xml:id="echoid-s1885" xml:space="preserve">9 ſi foſſe preſo <lb/>25. </s> <s xml:id="echoid-s1886" xml:space="preserve">25, poiche quell’interuallo 25. </s> <s xml:id="echoid-s1887" xml:space="preserve">25 della linea Geometri-<lb/>ca applicato nella linea Aritmetica al 50. </s> <s xml:id="echoid-s1888" xml:space="preserve">50, ſimilmente <lb/>hauria dato l’intiero 73 per radice del 5400. </s> <s xml:id="echoid-s1889" xml:space="preserve">Mà perche <lb/>quell’interuallo è vn poco maggiore del 73. </s> <s xml:id="echoid-s1890" xml:space="preserve">73, è ſegno, che <lb/>al numero 73 và aggiunta vna frattione.</s> <s xml:id="echoid-s1891" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1892" xml:space="preserve">Mà ſe il numero dato foſſe ſtato 5486, ſaria ſtato bene in <lb/>vece di 54 prendere 55, poiche quel numero più s’accoſta <pb o="99" file="0113" n="115" rhead="Linea Geometricâ"/> al 5500, & </s> <s xml:id="echoid-s1893" xml:space="preserve">allhora la radice, che viene 74 è proſſima alla <lb/>vera: </s> <s xml:id="echoid-s1894" xml:space="preserve">il che deue farſi, quando ſi tagliano due figure, che paſ-<lb/>ſano la metà di 100, poiche in vece del numero intiero s’ope-<lb/>ra col ſubcentuplo.</s> <s xml:id="echoid-s1895" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1896" xml:space="preserve">Che ſeil numero<unsure/>, di cui ſi cerca la radice, foſſe piccolo in <lb/>modo, che nello ſtromento non ſi poteſſe facilmente prender <lb/>nella linea Aritmetica l’interuallo proprio, ſi prenda il decu-<lb/>plo, e ſi trouerà in decime la frattione attaccata all’intiero. <lb/></s> <s xml:id="echoid-s1897" xml:space="preserve">Come per eſſempio, cerco la radice di 18 piedi, che ſono l’a-<lb/>rea d’vn piano da ridurſi in quadro: </s> <s xml:id="echoid-s1898" xml:space="preserve">prendo nella linea Geo-<lb/>metrica l’interuallo 18. </s> <s xml:id="echoid-s1899" xml:space="preserve">18, e poi nella ſteſſa prendo l’inter-<lb/>uallo d’vn numero quadrato, per eſſem pio 49. </s> <s xml:id="echoid-s1900" xml:space="preserve">49, la cui ra-<lb/>dice è 7: </s> <s xml:id="echoid-s1901" xml:space="preserve">mà perche rieſce ò ſcommodo, ò impoſſibile met-<lb/>tere quell’interuallo nella linea Aritmetica al 7. </s> <s xml:id="echoid-s1902" xml:space="preserve">7, lo metto al <lb/>70. </s> <s xml:id="echoid-s1903" xml:space="preserve">70, e trouando, che il primo interuallo preſo cade quaſi <lb/>al 42 {1/2}. </s> <s xml:id="echoid-s1904" xml:space="preserve">42 {1/2}, poiche li 70 non erano ſe non 7, così li 40 non <lb/>ſono ſe non 4, & </s> <s xml:id="echoid-s1905" xml:space="preserve">il reſto dà li decimi d’vn’intero, perciò dico, <lb/>che la radice di piedi 18 è piedi 4 {1/4} quaſi, ma certo è più di <lb/>4 {1/5}, perche cade in vn’interuallo maggiore di 42. </s> <s xml:id="echoid-s1906" xml:space="preserve">42, cioè <lb/>maggiore di 4 {2/10}.</s> <s xml:id="echoid-s1907" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1908" xml:space="preserve">Occorrendo poi, che il numero foſſe ditre ſole figure, ò <lb/>anche di due, ma maggiore del maſſimo quadrato notato <lb/>nella linea Geometrica, prendaſi vna parte aliquota di eſſo <lb/>tale, che ſia minore del numero 64 maſſimo delli notati nel-<lb/>la linea: </s> <s xml:id="echoid-s1909" xml:space="preserve">e queſto interuallo s’applichiad vn’altro numero in <lb/>tal linea, il qual’habbi vn’altro così moltiplice, come tutto il <lb/>numero è moltiplice di quella parte preſa; </s> <s xml:id="echoid-s1910" xml:space="preserve">e queſto vltimo in-<lb/>teruallo del moltiplice ſarà l’interuallo, che nella linea Arit-<lb/>metica moſtrerà, quanti intieri, e quante decime habbia la <lb/>radice. </s> <s xml:id="echoid-s1911" xml:space="preserve">Per eſſempio, cerco la radice di 96: </s> <s xml:id="echoid-s1912" xml:space="preserve">perche è troppo <pb o="100" file="0114" n="116" rhead="CAPO III."/> grandeil numero, piglio la metà 48, e prendo nella linea Geo-<lb/>metrica l’interuallo 48. </s> <s xml:id="echoid-s1913" xml:space="preserve">48, e con vn’altro Compaſſo l’inter-<lb/>uallo per eſſempio 4. </s> <s xml:id="echoid-s1914" xml:space="preserve">4, la cui radice è 2, ma per commodità <lb/>nella linea Aritmetica s’applicherà all’interuallo 20. </s> <s xml:id="echoid-s1915" xml:space="preserve">20, onde <lb/>poi s’hauranno li decimi dell’vnità: </s> <s xml:id="echoid-s1916" xml:space="preserve">ſe ſi applicaſſe alla linea <lb/>Arit metica, l’interuallo preſo 48. </s> <s xml:id="echoid-s1917" xml:space="preserve">48 non hauriamo ſe non la <lb/>radice della metà del quadrato, & </s> <s xml:id="echoid-s1918" xml:space="preserve">eſſa caderebbe all’interual-<lb/>lo 69. </s> <s xml:id="echoid-s1919" xml:space="preserve">69, cioè la radice ſaria 6 {9/10}, onde per hauer la radice <lb/>del doppio quadrato, cioè di 96, conuerrebbe raddoppiare <lb/>la radice trouata, e tra 69 decime, e 138 decime trouare il <lb/>medio proportionale 9 {7/10}. </s> <s xml:id="echoid-s1920" xml:space="preserve">Mà per trouare ciò ſenza fatica di <lb/>calcolo in trouar queſto medio proportionale, prendo quell’-<lb/>apertura di compaſſo, che pigliaua l’interuallo 48. </s> <s xml:id="echoid-s1921" xml:space="preserve">48, e l’ap-<lb/>plico nella linea Geometrica all’interuallo 10. </s> <s xml:id="echoid-s1922" xml:space="preserve">10, e poi (per-<lb/>che 48 è la metà di 96) prendo l’interuallo del doppio di 10, <lb/>cioè 20. </s> <s xml:id="echoid-s1923" xml:space="preserve">20, e queſto applico alla linea Aritmetica, in cuil’a-<lb/>pertura dell’altro Compaſſo è applicata al 20. </s> <s xml:id="echoid-s1924" xml:space="preserve">20, e trouo, <lb/>che queſt’vltimo interuallo cade nel 97. </s> <s xml:id="echoid-s1925" xml:space="preserve">97, e quaſi nel 98. <lb/></s> <s xml:id="echoid-s1926" xml:space="preserve">98, onde conchiudo, chela radice del numero 96 è 9 {7/10}, e <lb/>quaſi 9 {8/10}.</s> <s xml:id="echoid-s1927" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1928" xml:space="preserve">E perche operando in tal maniera occorrerà, che l’interual-<lb/>lo vltimo da applicarſi alla linea Aritmetica ſarà tale, che non <lb/>capirà nell’interuallo dell’a pertura dello ſtromento, perciò ti-<lb/>riſi vna linea lunga quanto porta queſt’interuallo preſo nella <lb/>linea Geometrica: </s> <s xml:id="echoid-s1929" xml:space="preserve">e poi preſo nell’ Aritmetiche l’interuallo <lb/>100. </s> <s xml:id="echoid-s1930" xml:space="preserve">100, ſi leui dalla linea tirata; </s> <s xml:id="echoid-s1931" xml:space="preserve">il reſto della linea s’appli-<lb/>chi all’interuallo dell’ Aritmetiche, e s’haurà il numero da <lb/>aggiungerſi al 100: </s> <s xml:id="echoid-s1932" xml:space="preserve">tutte le decine ſaranno vnità, il reſto da-<lb/>rà i decimi dell’vnità. </s> <s xml:id="echoid-s1933" xml:space="preserve">Per eſſempio cerco la radice di 156: <lb/></s> <s xml:id="echoid-s1934" xml:space="preserve">perche è troppo grande, piglio la terza parte, che è 52, e nel- <pb o="101" file="0115" n="117" rhead="Linea Geometrica."/> le linee Geometriche prendo l’interuallo 52. </s> <s xml:id="echoid-s1935" xml:space="preserve">52, e con quell’ <lb/>apertura prendo l’interuallo d’vn numero quadrato, per eſ-<lb/>ſempio 4, la cui radice è 2, e queſto interuallo s’applicherà <lb/>nell’Aritmetiche al 20. </s> <s xml:id="echoid-s1936" xml:space="preserve">20. </s> <s xml:id="echoid-s1937" xml:space="preserve">Dipoi quell’apertura di compaſ-<lb/>ſo, che daua l’interuallo 52. </s> <s xml:id="echoid-s1938" xml:space="preserve">52, allargato lo ſtromento, la <lb/>metto nelle ſteſſe linee Geometriche ad vn numero, che hab-<lb/>bia il triplo, per eſſempio al 15. </s> <s xml:id="echoid-s1939" xml:space="preserve">15, e poi prendo il triplo, cioè <lb/>45. </s> <s xml:id="echoid-s1940" xml:space="preserve">45. </s> <s xml:id="echoid-s1941" xml:space="preserve">E queſto è l’interuallo, che darà la radice di 156. <lb/></s> <s xml:id="echoid-s1942" xml:space="preserve">Mâ<unsure/> perche applicato il ſecondo Compaſſo nelle linee Arit-<lb/>metiche, come ſi diſſe, al 20. </s> <s xml:id="echoid-s1943" xml:space="preserve">20, queſt’ altro interuallo non <lb/>ci capiſce; </s> <s xml:id="echoid-s1944" xml:space="preserve">perciò alla miſura di queſto interuallo tiro vna <lb/>linea, e preſo il maſſimo interuallo delle linee Aritmetiche <lb/>100, 100, lo taglio dalla linea deſcritta, e quel che auanza <lb/>della linea, l’applico allo ſtromento, e vedo, che cade all’in-<lb/>teruallo 24. </s> <s xml:id="echoid-s1945" xml:space="preserve">24: </s> <s xml:id="echoid-s1946" xml:space="preserve">onde conchiudo eſſere 124 decime, cioè <lb/>12 {4/10} la proſſima radice di 156.</s> <s xml:id="echoid-s1947" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1948" xml:space="preserve">Di quì ſi caua il modo ditrouar la radice quadrata anche <lb/>de’ numeri maggiori di quattro figure, perche ſe ſarà il num. <lb/></s> <s xml:id="echoid-s1949" xml:space="preserve">18412, dicui ſi cerchila radice quadrata, getto via le due <lb/>vltime figure 12, e del reſto 184 prendo la quarta parte, che <lb/>è 46, e nelle linee Geometriche prendo la diſtanza 46. </s> <s xml:id="echoid-s1950" xml:space="preserve">46, e <lb/>con vn’altro Compaſſo l’interuallo di qualche numero qua-<lb/>drato, per eſſempio 9. </s> <s xml:id="echoid-s1951" xml:space="preserve">9; </s> <s xml:id="echoid-s1952" xml:space="preserve">e così, come quello 46 è di centina-<lb/>ra, così anche queſto 9, onde ſono due quadrati 900, e 4600; </s> <s xml:id="echoid-s1953" xml:space="preserve"><lb/>e queſto è la quarta parte del numero propoſto, dunque ap-<lb/>plicando queſto interuallo ad vn numero, di cui ſi troui il <lb/>quadruplo, per eſſempio al 15. </s> <s xml:id="echoid-s1954" xml:space="preserve">15, l’interuallo 60. </s> <s xml:id="echoid-s1955" xml:space="preserve">60, ſarà <lb/>la radice del quadrato 18400. </s> <s xml:id="echoid-s1956" xml:space="preserve">Dunque applicato quell’in-<lb/>teruallo 9.</s> <s xml:id="echoid-s1957" xml:space="preserve">9, preſo da principio col ſecondo Compaſſo, alla <lb/>linea Aritmetica al punto 30. </s> <s xml:id="echoid-s1958" xml:space="preserve">30, l’altro Compaſſo con l’a- <pb o="102" file="0116" n="118" rhead="CAPO III."/> pertura dell’vltimo interuallo preſo darà nelle ſteſſe linee <lb/>Aritmetiche vn’interuallo maggiore dell’interuallo 100.</s> <s xml:id="echoid-s1959" xml:space="preserve">100. <lb/></s> <s xml:id="echoid-s1960" xml:space="preserve">Perciò da vna linea vguale à queſt’interuallo cauo l’interuallo <lb/>100.</s> <s xml:id="echoid-s1961" xml:space="preserve">100, & </s> <s xml:id="echoid-s1962" xml:space="preserve">applicato il reſto di detta linea, trouo, che <lb/>cade all’interuallo 35. </s> <s xml:id="echoid-s1963" xml:space="preserve">35, & </s> <s xml:id="echoid-s1964" xml:space="preserve">vn poco più; </s> <s xml:id="echoid-s1965" xml:space="preserve">onde conchiudo, <lb/>che la radice del numero propoſto 18412 è 135, e qualche <lb/>coſa di vantaggio.</s> <s xml:id="echoid-s1966" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1967" xml:space="preserve">Due coſe quì ſono da auuertire: </s> <s xml:id="echoid-s1968" xml:space="preserve">la prima è, che li 100 pun-<lb/>ti della linea Aritmetica potendoſi prendere per 200, ſi può <lb/>rendere più breue l’operatione, poiche applicandoſi all’inter-<lb/>uallo 15. </s> <s xml:id="echoid-s1969" xml:space="preserve">15, come ſe foſſe 30. </s> <s xml:id="echoid-s1970" xml:space="preserve">30, verrà l’altro interuallo alli <lb/>punti 67 {1/2}. </s> <s xml:id="echoid-s1971" xml:space="preserve">67 {1/2}, in circa, onde immediatamente ſi caua eſ-<lb/>ſer la radice 135 in circa, come prima. </s> <s xml:id="echoid-s1972" xml:space="preserve">La ſeconda è, che ſe <lb/>da principio ſi darà alle linee Geometriche l’apertura, pren-<lb/>dendo prima nella linea Aritmetica ſopra illato la lunghezza <lb/>corriſpondente al numero, che è radice del quadrato preciſo, <lb/>come di 30 punti, ò di 15, che s’intendano valer 30, e queſti <lb/>s’applichino al 9. </s> <s xml:id="echoid-s1973" xml:space="preserve">9, e poi preſo l’interuallo corriſpondente <lb/>del numero dato, queſto poi applicato allato dello ſtromen-<lb/>to sù la linea Aritmetica, ſi potranno hauer le frattioni ade-<lb/>renti nel modo, che s’è detto nel Capo 2. </s> <s xml:id="echoid-s1974" xml:space="preserve">queſt. </s> <s xml:id="echoid-s1975" xml:space="preserve">7. </s> <s xml:id="echoid-s1976" xml:space="preserve">verſo il <lb/>fine.</s> <s xml:id="echoid-s1977" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1978" xml:space="preserve">Seil numero dato foſſe così grande, che lidue numeri mol-<lb/>tiplicati inſieme, che lo producono, foſſero ambidue mag-<lb/>giori di quelli, cheſon notati nelle linee, ſe ne prendano tre, <lb/>che ſiano minori, e lo miſurino, moltiplicati tra di loro. </s> <s xml:id="echoid-s1979" xml:space="preserve">Per <lb/>eſſempio ſia il numero dato 604812, leuate le due vltime fi-<lb/>gure, reſta 6048, il quale ſi produce dal 72 per 84, niuno de’ <lb/>quali ſi troua notato nelle linee Geometriche. </s> <s xml:id="echoid-s1980" xml:space="preserve">Perciò pren-<lb/>do tre numeri, che inſieme moltiplicatilo producono, e ſono <pb o="103" file="0117" n="119" rhead="Linea Geometrica"/> 56. </s> <s xml:id="echoid-s1981" xml:space="preserve">9. </s> <s xml:id="echoid-s1982" xml:space="preserve">12. </s> <s xml:id="echoid-s1983" xml:space="preserve">Ecosì preſo l’interuallo 56. </s> <s xml:id="echoid-s1984" xml:space="preserve">56, deuo trouar’il la-<lb/>to del quadrato noncuplo, e perciò l’applico al 4. </s> <s xml:id="echoid-s1985" xml:space="preserve">4, il cui <lb/>noncuplo è 36, el’interuallo 36. </s> <s xml:id="echoid-s1986" xml:space="preserve">36 ſarà il lato del quadrato <lb/>noncuplo del primo. </s> <s xml:id="echoid-s1987" xml:space="preserve">E perche à queſto ſi deue trouar’il duo-<lb/>decuplo, applico queſto ſecondo interuallo al 5. </s> <s xml:id="echoid-s1988" xml:space="preserve">5, e piglio il <lb/>duodecuplo, che ſarà all’interuallo 60. </s> <s xml:id="echoid-s1989" xml:space="preserve">60, e con queſto ope-<lb/>rando nelle linee Aritmetiche, come s’è detto, trouo la ra-<lb/>dice quadrata del numero dato 604812 eſſere 777, e quaſi <lb/>778, poiche nella linea deſcritta ſi può leuare ſette volte <lb/>l’interuallo 100. </s> <s xml:id="echoid-s1990" xml:space="preserve">100, & </s> <s xml:id="echoid-s1991" xml:space="preserve">il reſtante è quaſi 78.</s> <s xml:id="echoid-s1992" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1993" xml:space="preserve">Mà cercando la Radice Quadrata d’vn Rotto, prendi nel-<lb/>le linee Geometriche li due interualli corriſpondenti al Nu-<lb/>meratore, & </s> <s xml:id="echoid-s1994" xml:space="preserve">al Denominatore: </s> <s xml:id="echoid-s1995" xml:space="preserve">dipoi traportali nelle linee <lb/>Aritmetiche, aprendo lo ſtromento in modo, che capiſca, <lb/>l’interuallo del numero, che vuoi ritenere; </s> <s xml:id="echoid-s1996" xml:space="preserve">poiche l’altro in-<lb/>teruallo nelle ſteſſe linee darà il numero cercato.</s> <s xml:id="echoid-s1997" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1998" xml:space="preserve">Sia il Rotto {4/9}, di cui ſi cerca la Radice Quadrata: </s> <s xml:id="echoid-s1999" xml:space="preserve">prendo <lb/>nelle linee Geometriche 4.</s> <s xml:id="echoid-s2000" xml:space="preserve">4, con vn Compaſſo, e con vn’al-<lb/>tro 9. </s> <s xml:id="echoid-s2001" xml:space="preserve">9. </s> <s xml:id="echoid-s2002" xml:space="preserve">Dipoi volendo ritener il Numeratore 4; </s> <s xml:id="echoid-s2003" xml:space="preserve">apro lo <lb/>ſtromento in modo, che l’interuallo del primo Compaſſo ſi <lb/>addatti alli punti 4.</s> <s xml:id="echoid-s2004" xml:space="preserve">4, nelle linee Aritmetiche; </s> <s xml:id="echoid-s2005" xml:space="preserve">poiche l’altro <lb/>Compaſſo ſi addattarà alli punti 6. </s> <s xml:id="echoid-s2006" xml:space="preserve">6: </s> <s xml:id="echoid-s2007" xml:space="preserve">onde dirò che la radi-<lb/>ce cercata è {4/6}, cioè {2/3}. </s> <s xml:id="echoid-s2008" xml:space="preserve">Ouero addattando il ſecondo Com-<lb/>paſſo, che corriſponde al Denominatore, alli punti 9. </s> <s xml:id="echoid-s2009" xml:space="preserve">9, tro-<lb/>uo che l’altro corriſponde alli 6. </s> <s xml:id="echoid-s2010" xml:space="preserve">6: </s> <s xml:id="echoid-s2011" xml:space="preserve">onde dirò, che la Radice <lb/>cercata è {6/9}. </s> <s xml:id="echoid-s2012" xml:space="preserve">E perche il 4, & </s> <s xml:id="echoid-s2013" xml:space="preserve">il 9 ſono interualli troppo pic-<lb/>coli, in lor vece ſi prendano li moltiplici, cioè 40, e 90, ò <lb/>qualſiuoglia altro. </s> <s xml:id="echoid-s2014" xml:space="preserve">II che molto più ſerue, quando il Rotto <lb/>dato non hà la Radice preciſa, poiche ſi trouarebbe la Radi-<lb/>ce più vicina alla vera. </s> <s xml:id="echoid-s2015" xml:space="preserve">Così cercando la Radice di {4/10} ſi tro- <pb o="104" file="0118" n="120" rhead="C A P O III."/> uarebbe ben ſi eſſer di denominatione maggiore di {4/6}, mà ſi <lb/>ſappia appreſſo di poco quanto maggiore; </s> <s xml:id="echoid-s2016" xml:space="preserve">mà applicandoſi <lb/>li Compaſſi al decuplo, ſi trouarà eſſer di denominatione <lb/>maggiore di {40/63}. </s> <s xml:id="echoid-s2017" xml:space="preserve">Quindi eſſendo il denominatore troppo pic-<lb/>colo, la frattione con lo ſteſſo Numeratore è maggiore del <lb/>douere.</s> <s xml:id="echoid-s2018" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2019" xml:space="preserve">Queſto modo dioperare è fondato nella regola per troua-<lb/>re tal Radice Aritmeticamente, la quale ſi approſſimi alla <lb/>vera; </s> <s xml:id="echoid-s2020" xml:space="preserve">cioè ſi moltiplica il Numeratore per il Denominatore: <lb/></s> <s xml:id="echoid-s2021" xml:space="preserve">del prodotto ſi caua la Radice Quadrata proſſima; </s> <s xml:id="echoid-s2022" xml:space="preserve">e queſta ſi <lb/>mette per Denominatore al Numeratore dato, ouero per <lb/>Numeratore al dato Denominatore. </s> <s xml:id="echoid-s2023" xml:space="preserve">Così per {4/10} ſi caua la <lb/>Radicc di 40 fatto dal 4 in 10, & </s> <s xml:id="echoid-s2024" xml:space="preserve">è 6 {4/13}: </s> <s xml:id="echoid-s2025" xml:space="preserve">onde la Radice proſ-<lb/>ſimamente è {52/82}, ouero {82/130}; </s> <s xml:id="echoid-s2026" xml:space="preserve">la prima è maggiore del douere, <lb/>eſſendo che quadrandoſi vien vna frattione maggiore di {4/10}; </s> <s xml:id="echoid-s2027" xml:space="preserve">la <lb/>ſeconda è minore del douere, perche quadrandoſi dà vna <lb/>frattione minore di {4/10}.</s> <s xml:id="echoid-s2028" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2029" xml:space="preserve">E’la ragione di queſto prendere la Media Proportionale <lb/>tra il Numeratore, & </s> <s xml:id="echoid-s2030" xml:space="preserve">il Denominatore dati, cauaſi dalla na-<lb/>tura delli Quadrati, cheſono nella duplicata proportione de’ <lb/>ſuoi lati. </s> <s xml:id="echoid-s2031" xml:space="preserve">Perciò volendoſi la Radice Quadrata d’vn Rotto, <lb/>ſi cerca vna frattione, il cui Numeratore ſia al Denominatore <lb/>nella proportione ſubduplicata del Numeratore al Denomi-<lb/>natore della frattione data. </s> <s xml:id="echoid-s2032" xml:space="preserve">E così ritenuto il primo Nume-<lb/>ratore, queſta Media Proportionale è il Denominatore; </s> <s xml:id="echoid-s2033" xml:space="preserve">e ſe <lb/>queſta ſi mette per Numeratore, reſta il primo Denomina-<lb/>tore.</s> <s xml:id="echoid-s2034" xml:space="preserve"/> </p> <pb o="105" file="0119" n="121" rhead="Linea Cubica"/> </div> <div xml:id="echoid-div68" type="section" level="1" n="38"> <head xml:id="echoid-head65" xml:space="preserve">CAPO QVARTO.</head> <head xml:id="echoid-head66" xml:space="preserve">Come s’habbia à diuidere lo Stromento per i corpi ſolidi: <lb/>& uſo di queſta linea Cubica.</head> <p> <s xml:id="echoid-s2035" xml:space="preserve">SI come le ſuperficie ſono terminate da linee, dalle quali <lb/>riceuono la denominatione, così li corpi ſolidi ſono ter-<lb/>minati da ſuperficie, e da queſte, ò per la qualità loro, ò per <lb/>la moltitudine vien denominata la figura ſolida; </s> <s xml:id="echoid-s2036" xml:space="preserve">perchc s’ella <lb/>è vna ſuperficie ſola in tutti i ſuoi punti vgualmente diſtante <lb/>dal centro, che s’intende nel mezzo della ſolidità del corpo, <lb/>ſarà quel corpo vna sfera; </s> <s xml:id="echoid-s2037" xml:space="preserve">ma ſe non hà queſta vgual diſtanza <lb/>dal centro, ſarà ben sì sferoidale la figura, ma non sfera; </s> <s xml:id="echoid-s2038" xml:space="preserve">tale <lb/>è la ſuperficie d’vn vouo, & </s> <s xml:id="echoid-s2039" xml:space="preserve">altre tali ò Elliptiche, ò Pſeudoel-<lb/>liptiche; </s> <s xml:id="echoid-s2040" xml:space="preserve">ma ſe ſono più ſuperficie terminanti il corpo di di-<lb/>uerſo genere, cioè altre ſuperficie piane, altre curue, & </s> <s xml:id="echoid-s2041" xml:space="preserve">incli-<lb/>nate à far’vn’angolo ſolido, dalla qualità delle ſuperficie ſi <lb/>denominarà il corpo, ò Cono, ò Cilindro, ò con altro nome <lb/>compoſto; </s> <s xml:id="echoid-s2042" xml:space="preserve">come li Conoidi Parabolici, ò Hiperbolici, &</s> <s xml:id="echoid-s2043" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s2044" xml:space="preserve">Que’ſolidi però, che più communemente ſi conſiderano, ſono <lb/>quelli, che hanno molte faccie, e ſon terminati da ſuperficie <lb/>piane; </s> <s xml:id="echoid-s2045" xml:space="preserve">e conforme al numero, e qualità di tali ſuperficie ſono <lb/>chiamati tali corpi, come ciaſcuno sà, e può facilmente vede-<lb/>re nelle definitioni del lib. </s> <s xml:id="echoid-s2046" xml:space="preserve">11. </s> <s xml:id="echoid-s2047" xml:space="preserve">d’Euclide.</s> <s xml:id="echoid-s2048" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2049" xml:space="preserve">Ora nella guiſa, che quelle ſuperficie ſi dicono ſimili, le <lb/>quali hanno vgual numero di linee, che le terminano, e tra <lb/>loro proportionali: </s> <s xml:id="echoid-s2050" xml:space="preserve">Così le figure ſolide ſimili (che tanto è, <lb/>quanto dire corpi ſimili) s’intendono eſſer quelle, che ſono <lb/>terminate da vgual numero di ſuperficie ſimili. </s> <s xml:id="echoid-s2051" xml:space="preserve">Onde ſe le <pb o="106" file="0120" n="122" rhead="C A P O IV."/> ſuperficie d’vn corpo ſaranno non ſolamente vguali di nu-<lb/>mero, ma anche di grandezza alle ſuperficie d’vn’altro cor-<lb/>po, tali due corpiſaranno vguali, e ſimili; </s> <s xml:id="echoid-s2052" xml:space="preserve">ma ſe le ſuperficie <lb/>vguali di numero, e diſuguali di grandezza ſono ſimili, li cor-<lb/>pi ſono ben sì ſimili, ma non vguali. </s> <s xml:id="echoid-s2053" xml:space="preserve">Di queſta maniera vn <lb/>cubo è ſimile all’altro cubo, perche così l’vno, come l’altro <lb/>hanno ſei faccie piane, e ciaſcheduna è quadrata; </s> <s xml:id="echoid-s2054" xml:space="preserve">e poiche <lb/>tutti li quadrati ſon ſimili, perciò anche li cubi ſono ſimili: </s> <s xml:id="echoid-s2055" xml:space="preserve">ma <lb/>ſe vn quadrato d’vno ſarà maggiore d’vn quadrato dell’altro, <lb/>ſaranno i cubi diſuguali. </s> <s xml:id="echoid-s2056" xml:space="preserve">Paragonando poi due Parallele pi-<lb/>pedi (chi non è così prattico de’vocaboli, s’imagini vna tra-<lb/>ue, vna tauola, ò coſa tale ben ſquadrata) hanno ben sì cia-<lb/>ſcuno ſei piani quadrilateri, de’quali li due oppoſti ſono pa-<lb/>ralleli, ma a fine che ſiano ſimili li Parallelepipedi, conuiene <lb/>che detti piani d’vno ſiano ſimili alli piani dell’altro. </s> <s xml:id="echoid-s2057" xml:space="preserve">Mà par-<lb/>lando de’Coni, e de’Cilindri, ſe bene potria dirſi eſſer tra loro <lb/>ſimili quelli, che hanno le baſi, e le ſuperficie Coniche, ò Ci-<lb/>lindriche ſimili; </s> <s xml:id="echoid-s2058" xml:space="preserve">ad ogni modo per eſſer più immediatamente <lb/>nota la lunghezza della lor baſe, e la lor’altezza perpendi-<lb/>colare, ò per parlar più generalmente, il lor’Aſſe, quelli ſono <lb/>Coni, ò Cilindri ſimili, che hanno gli aſſi, & </s> <s xml:id="echoid-s2059" xml:space="preserve">i diametri delle <lb/>baſi proportionali; </s> <s xml:id="echoid-s2060" xml:space="preserve">il che però ſi deue intendere con la mede-<lb/>ſima inclinatione dell’aſſe alla baſe, come è manifeſto, per-<lb/>che ſe vn’aſſe cadeſſe perpendicolare alla baſe, e l’altro aſſe <lb/>foſſe obliquo, con tutto, che dettiaſſi haueſſero nella lunghez-<lb/>za loro la proportione delli diametri delle baſi, non per tan-<lb/>to ſariano ſimilii Coni, ò Cilindri.</s> <s xml:id="echoid-s2061" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2062" xml:space="preserve">Permeſſe queſte coſe, per più chiara intelligenza, auuerto, <lb/>che nelle cofe ſeguenti prenderò il nome di Lati Homologi nel <lb/>ſenſo medeſimo, che s’è detto nel Capo precedente; </s> <s xml:id="echoid-s2063" xml:space="preserve">e per <pb o="107" file="0121" n="123" rhead="Linea Cubica"/> nome di Piani Homologi intenderò que’ piani, che ne’ due <lb/>corpi ſimili ſono ſimilmente poſti in ordine à gl’altri piani <lb/>delle figure, che terminano.</s> <s xml:id="echoid-s2064" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2065" xml:space="preserve">Eſſendo dunque l’vſo di queſto ſtromento di Proportione <lb/>in ordine alle figure ſimili, per poter’ in eſſo deſcriuere due li-<lb/>nee talmente diuiſe, che poſſano ſeruir’ al fine preteſo in or-<lb/>dine a’corpi ſolidi, conuien ſupporre ciò che nel lib. </s> <s xml:id="echoid-s2066" xml:space="preserve">11, e 12 <lb/>d’Euclide s’inſegna, cioè, che li ſolidi ſimili ſono nella tripli-<lb/>cata proportione de’lati homologi, come le sfere ſono nella <lb/>triplicata proportione de’ſuoi diametri. </s> <s xml:id="echoid-s2067" xml:space="preserve">Il che è quanto dire, <lb/>che dati due lati homologi di due corpi ſimili, ò due diametri <lb/>di due sfere, ſe ſi continuarà la proportione ſin’al quarto ter-<lb/>mine; </s> <s xml:id="echoid-s2068" xml:space="preserve">qual proportione hà il primo al quarto termine, tale è <lb/>d’vn ſolido all’altro, ò d’vna sfera all’<unsure/>altra. </s> <s xml:id="echoid-s2069" xml:space="preserve">Sì che date quat-<lb/>tro linee continuamente proportionali, come la prima alla <lb/>quarta, così il ſolido sù la prima al ſolido ſimile sù la ſeconda.</s> <s xml:id="echoid-s2070" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2071" xml:space="preserve">Quindiè, che data in linee la proportione, che debbano <lb/>hauere due ſolidi, conuiene tra quelle trouare due medie con-<lb/>tinuamente proportionali, per potere sù la prima, e sù la ſe-<lb/>conda fare li ſolidi ſimili, come auuertiti furono da Platone <lb/>quei di Delo, quando cercauano di raddoppiare l’ altare <lb/>d’Apolline (il qual’era ſtimato vno de’ ſette miracoli, per eſ-<lb/>ſer fatto tutto di ſole corna deſtre, ſenza eſſer’ incollate, ne le-<lb/>gate inſieme, come riferiſce Plutarco nel fine del libro De ſo-<lb/>lertia animalium) conforme all’Oracolo hauuto, & </s> <s xml:id="echoid-s2072" xml:space="preserve">eſſi in ve-<lb/>ce di raddoppiarlo, ne haueano fatto vno quattro volte mag-<lb/>giore del douere, come dice lo ſteſſo Plutarco nel libro de <lb/>Genio Socratis; </s> <s xml:id="echoid-s2073" xml:space="preserve">Et è aſſai noto appreſſo molti Scittori eſſere <lb/>queſta la famoſa duplicatione del Cubo, cioè l’inuentione di <lb/>due medie proportionali tra due eſtreme, l’vna delle quali ſia <lb/>doppia dell’altra.</s> <s xml:id="echoid-s2074" xml:space="preserve"/> </p> <pb o="108" file="0122" n="124" rhead="CAPO IV."/> <p> <s xml:id="echoid-s2075" xml:space="preserve">Varij ſono ſtati li tentatiui, evarie ſono le forme per tro-<lb/>uare mecanicamente queſte due medie proportionali; </s> <s xml:id="echoid-s2076" xml:space="preserve">e chi <lb/>vuole può vedere nell’ Annotationi di Guglielmo Filandro <lb/>ſopra il libro 9. </s> <s xml:id="echoid-s2077" xml:space="preserve">di Vitruuio cap. </s> <s xml:id="echoid-s2078" xml:space="preserve">3. </s> <s xml:id="echoid-s2079" xml:space="preserve">qual foſſe il Meſolabio <lb/>d’Eratoſtene; </s> <s xml:id="echoid-s2080" xml:space="preserve">nel Villalpando tom. </s> <s xml:id="echoid-s2081" xml:space="preserve">1. </s> <s xml:id="echoid-s2082" xml:space="preserve">part. </s> <s xml:id="echoid-s2083" xml:space="preserve">2. </s> <s xml:id="echoid-s2084" xml:space="preserve">lib. </s> <s xml:id="echoid-s2085" xml:space="preserve">1. </s> <s xml:id="echoid-s2086" xml:space="preserve">cap. </s> <s xml:id="echoid-s2087" xml:space="preserve">3. <lb/></s> <s xml:id="echoid-s2088" xml:space="preserve">prop. </s> <s xml:id="echoid-s2089" xml:space="preserve">12. </s> <s xml:id="echoid-s2090" xml:space="preserve">E nella Geometria di Renato di Chartes ſul prin-<lb/>ci pio del lib. </s> <s xml:id="echoid-s2091" xml:space="preserve">3. </s> <s xml:id="echoid-s2092" xml:space="preserve">trouerà, come perl’inuentione delle medie <lb/>proportionali, egli ſi ſerua d’vno Stromento da lui propoſto <lb/>nel principio del lib. </s> <s xml:id="echoid-s2093" xml:space="preserve">2. </s> <s xml:id="echoid-s2094" xml:space="preserve">Ma quanto appartiene al noſtro fine <lb/>preſente, meglio ſarà ſeruirci d’vna tauola di numeri, co’qua-<lb/>li ſi notaranno tanto preciſamente, quanto baſta, per l’ope-<lb/>rationi mecaniche, li punti richieſti in ordine alli ſolidi.</s> <s xml:id="echoid-s2095" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2096" xml:space="preserve">E perche tra li ſolidi il più conoſciuto, e facile ad hauerſi la <lb/>ſua miſura è il cubo, come quello, che hà le tre dimenſioni <lb/>di tal maniera vguali, che data la lunghezza d’vna ſua linea, e <lb/>queſta moltiplicata in ſe ſteſſa, ſe ſi moltiplica di nuouo il pro-<lb/>dotto per la medeſima, ſi fà nota la ſua ſolidità; </s> <s xml:id="echoid-s2097" xml:space="preserve">e date quat-<lb/>tro linee continuamente proportionali, come il cubo della <lb/>prima al cubo della ſeconda, così qual ſi voglia ſolido sù la <lb/>prima ad vn’altro ſolido ſimile sù la ſeconda, eſſendo che tan-<lb/>to i cubi, quanto quegl’ altri ſolidi ſono nella proportione <lb/>della linea prima alla quarta: </s> <s xml:id="echoid-s2098" xml:space="preserve">Perciò ſegnandoſi nello ſtro-<lb/>mento di Proportione i lati de’ cubi, che vanno creſcendo ſe-<lb/>condo la ſerie naturale de’numeri, ſi vengono ad hauere pari-<lb/>menti ſegnati i lati homologi di qualunque ſolidi ſimili. </s> <s xml:id="echoid-s2099" xml:space="preserve">Quin-<lb/>di è, che tal linea ſi chiama più toſto col nome ſpecifico di <lb/>Cubica, che col generico di Stereometrica; </s> <s xml:id="echoid-s2100" xml:space="preserve">sì perche tutti li <lb/>cubi ſono ſimili, sì anche perche riducendo le proportioni <lb/>a’numeri, ſi trouano le medie proportionali coll’eſtrattione <lb/>della radice cubica.</s> <s xml:id="echoid-s2101" xml:space="preserve"/> </p> <pb o="109" file="0123" n="125" rhead="Linea Cubica"/> <p> <s xml:id="echoid-s2102" xml:space="preserve">Sì che per formare la ſottoſcitta tauoletta, in cui ſi notano <lb/>le proportioni, che hà la radice di ciaſcun cubo alla radice del <lb/>primo cubo, conuiene tra li due numeri eſprimenti la propor-<lb/>tione de’ cubi trouare il primo de’ due medij proportionali; <lb/></s> <s xml:id="echoid-s2103" xml:space="preserve">perche queſto ſarà la radice del cubo, che hà al cubo del pri-<lb/>mo numero la proportione, che hà il quarto numero al pri-<lb/>mo, com’è manifeſto da quello, che delle linee s’è detto. </s> <s xml:id="echoid-s2104" xml:space="preserve"><lb/>E perche la maggior parte de’numeri non hà la radice cubica <lb/>preciſa, & </s> <s xml:id="echoid-s2105" xml:space="preserve">aggionger’à gl’intieri frattioni di diuerſe deno-<lb/>minationi, ſaria coſa, che nella prattica porterebbe molto di-<lb/>ſturbo, quindiè, che riuſcirà commodiſſimo intendere l’vni-<lb/>tà diuiſa in mille particelle, perche così tutte le frattioni ag-<lb/>giunte à gl’intieri ſaranno di milleſime;</s> <s xml:id="echoid-s2106" xml:space="preserve">e nel numero, che ver-<lb/>rà per radice, le tre vltime figure ſaranno numeratore delle <lb/>parti milleſime aggiunte à gl’ intieri ſignificati dal reſto delle <lb/>figure antecedenti nel modo detto nel Capo precedente, do-<lb/>ue ſi parlò delle radici de’ quadrati.</s> <s xml:id="echoid-s2107" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2108" xml:space="preserve">Sia dunque nella fig. </s> <s xml:id="echoid-s2109" xml:space="preserve">dello Stromento tirata dal centro del-<lb/>lo ſtromento la linea AL, ela AM, nella quale ſi prendano <lb/>AH, & </s> <s xml:id="echoid-s2110" xml:space="preserve">AI vguali, e perciò non è neceſſario, che queſte parti <lb/>AH, AI ſiano viſibili; </s> <s xml:id="echoid-s2111" xml:space="preserve">e s’intenda AH eſſer’ il lato del primo <lb/>cubo; </s> <s xml:id="echoid-s2112" xml:space="preserve">queſta ſi replichi quante volte ſi può, nelli numeri 8, e <lb/>27, in maniera, che A 8 è doppia, & </s> <s xml:id="echoid-s2113" xml:space="preserve">A 27 è tripla della lun-<lb/>ghezza AH. </s> <s xml:id="echoid-s2114" xml:space="preserve">E per queſto s’è notato nel ſecondo punto 8, e <lb/>nelterzo 27, per denotare, che il cubo di A 8 contiene otto <lb/>volte, & </s> <s xml:id="echoid-s2115" xml:space="preserve">il cubo di A 27 contiene ventiſette volte il cubo di <lb/>AH. </s> <s xml:id="echoid-s2116" xml:space="preserve">E ſe la linea AL foſſe più lunga, che ſi poteſſe vn’altra <lb/>volta replicare, nel quarto punto ſi notarebbe 64, percheil <lb/>cubo della linea quadrupla di AH, contiene 64 cubi di AH. <lb/></s> <s xml:id="echoid-s2117" xml:space="preserve">Ma perche ſi vede che tra 8, e 27, è molto più tra 27, e 64 <pb o="110" file="0124" n="126" rhead="CAPO IV."/> cadono molti numeri, onde dette parti deuon’ eſſer capaci di <lb/>molte diuiſioni, perciò s’è preſo da principio la linea AH vn <lb/>poco grandicella; </s> <s xml:id="echoid-s2118" xml:space="preserve">altrimenti non riuſcirebbe commoda la <lb/>diuiſione. </s> <s xml:id="echoid-s2119" xml:space="preserve">E queſta è la cagione, che non capirà ſe non circa <lb/>50 diuiſioni tutta la AL: </s> <s xml:id="echoid-s2120" xml:space="preserve">la quale in vno ſtromento più gran-<lb/>de, in cui poſſa prenderſi aſſai più lunga la AH, riuſcirà anche <lb/>capace di più numero di lati cubici.</s> <s xml:id="echoid-s2121" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2122" xml:space="preserve">Mà per ſegnare li lati de gl’altri cubi, e vedere, come ſi ſia <lb/>fatta la ſeguente tauoletta delle radici, conuien trouare tra <lb/>l’vnità, & </s> <s xml:id="echoid-s2123" xml:space="preserve">il numero di ciaſcun cubo il primo delli due medij <lb/>continuamente proportionali; </s> <s xml:id="echoid-s2124" xml:space="preserve">il che ſi fà moltiplicando il <lb/>quadrato del primo nel quarto numero; </s> <s xml:id="echoid-s2125" xml:space="preserve">e la radice cubica <lb/>del prodotto è il ſecondo numero, che ſi cerca. </s> <s xml:id="echoid-s2126" xml:space="preserve">Il fondamen-<lb/>to di ciò fare è, perche dati quattro termini continuamente <lb/>proportionali A, B, C, D, il piano fatto dalli due eſtremi A <lb/>in D, è eguale al piano fatto dalli due medij Bin C, per la <lb/>16 del 6, e 19 del 7. </s> <s xml:id="echoid-s2127" xml:space="preserve">Dunque li ſolidi fattì<unsure/> dalli due piani <lb/>detti, e dal primo termine, ſono vguali, e così il quadrato <lb/>del primo nel quarto A quadrato in D, e vguale al ſolido fatto <lb/>dallitre primi A in B in C. </s> <s xml:id="echoid-s2128" xml:space="preserve">E perche A, B, C, ſono continua-<lb/>mente proportionali, il piano fatto da gl’eſtremi, A in C, è <lb/>vguale al quadrato del medio, B quadrato per la 17 del 6, e <lb/>20 del 7, li ſolidi fatti da queſti due piani, e dal ſecondo ter-<lb/>mine B ſono vguali, e così A in B in C, cioè, come ſopra s’è <lb/>dimoſtrato, A quadrato in D, è vguale al cubo di B ſecondo <lb/>termine delli quattro. </s> <s xml:id="echoid-s2129" xml:space="preserve">Dunque eſſendo noti li due eſtremi, <lb/>moltiplicato il quadrato del primo nell’ altro eſtremo, il lato <lb/>cubico del prodotto è il ſecondo termine delli quattro con-<lb/>tinuamenre proportionali. </s> <s xml:id="echoid-s2130" xml:space="preserve">Nella ſteſſa maniera ſi dimoſtra, <lb/>che moltiplicato il quadrato del quarto termine nel primo, la <pb o="111" file="0125" n="127" rhead="Linea Cubica."/> radice cubica del prodotto è il terzo termine delli quattro.</s> <s xml:id="echoid-s2131" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2132" xml:space="preserve">Di quì ſi vede, che ſe il primo termine AH ſia 1000, & </s> <s xml:id="echoid-s2133" xml:space="preserve">il <lb/>ſuo doppio 2000, il quadrato del primo 1000000 moltipli-<lb/>cato per 2000, darà il ſolido 2000000000, la cui radice cu-<lb/>bica 1259 è il ſecondo termine delli quattro, & </s> <s xml:id="echoid-s2134" xml:space="preserve">è radice del <lb/>cubo doppio del cubo di AH. </s> <s xml:id="echoid-s2135" xml:space="preserve">Elo ſteſſo s’intende diqualſi-<lb/>uoglia altro numero: </s> <s xml:id="echoid-s2136" xml:space="preserve">onde baſterà à ciaſcun numeroal 3, al <lb/>4, al 9, &</s> <s xml:id="echoid-s2137" xml:space="preserve">c. </s> <s xml:id="echoid-s2138" xml:space="preserve">aggiunger noue zeri, perche così la radice cubica <lb/>ſarà di quattro figure, la prima delle quali moſtra, quante <lb/>volte ſi debba prender la linea AH, e le tre vltime figure mo-<lb/>ſtreranno, quante milleſime della ſteſſa AH ſi debbano di più <lb/>aggiungere. </s> <s xml:id="echoid-s2139" xml:space="preserve">Che ſe ſi foſſero per AH preſe ſolo le centeſi-<lb/>me, con aggiunger’ ad eſſa due zeri, allhora à gl’altri numeri <lb/>doueua aggiungerſi ſolamente ſei zeri, e la radice di tre ſigu-<lb/>re hauria con le due vltime moſtrato il numero delle cente-<lb/>ſime. </s> <s xml:id="echoid-s2140" xml:space="preserve">Ma perche volendo ſeruirci ſolo delle centeſime ſi <lb/>opera con più preciſione, conoſciuto il numero delle mil-<lb/>leſime, perciò nell’anneſſa tauolletta ſi ſon poſte le milleſi-<lb/>me, ſegnando le radici ſin’al cubo, che è cinquanta volte <lb/>maggiore del cubo di AH.</s> <s xml:id="echoid-s2141" xml:space="preserve"/> </p> <pb o="112" file="0126" n="128" rhead="CAPO IV."/> <note position="right" xml:space="preserve"> <lb/>######## Tauola de’numeri con le ſue Radici Cubicbe eſpreſſe \\ in particelle Milleſime dell’ Vnità. <lb/>Cubi # Radici # Cubi # Radici # Cubi # Radici # Cubi # Radici <lb/>1 # 1000 # 16 # 2520 - # 31 # 3142 - # 46 # 3583† <lb/>2 # 1259† # 17 # 2572 - # 32 # 3175 - # 47 # 3609 -<lb/>3 # 1442† # 18 # 2620† # 33 # 3208† # 48 # 3634† <lb/>4 # 11<unsure/>87† # 19 # 2664 - # 34 # 3240 - # 49 # 3660 -<lb/>5 # 1710 - # 20 # 2715 - # 35 # 3271† # 50 # 3684† <lb/>6 # 1817† # 21 # 2759 - # 3<unsure/>6 # 3301† <lb/>7 # 1913 # 22 # 2702† # 37 # 3332† <lb/>8 # 2000 # 23 # 2844 - # 38 # 3362 -<lb/>9 # 2080† # 24 # 2885 - # 39 # 3391† <lb/>10 # 2154† # 25 # 2924† # 40 # 3420 -<lb/>11 # 2224 - # 26 # 2962† # 41 # 3448† <lb/>12 # 2290 # 27 # 3000 # 42 # 3476† <lb/>13 # 2352 - # 28 # 3037 - # 43 # 3504 -<lb/>14 # 2410† # 29 # 3072† # 44 # 3530† <lb/>15 # 2466† # 30 # 3108 - # 45 # 3557 -<lb/></note> <p> <s xml:id="echoid-s2142" xml:space="preserve">Il modo di ſeruirſi di queſta Tauola per portare sùle linee <lb/>AL, AM le diuiſioni, eſſendo lo ſteſſo con quello, che s’è det-<lb/>to di ſopra nelle Radici de’Quadrati, non hà biſogno di più <lb/>lunga eſpoſitione. </s> <s xml:id="echoid-s2143" xml:space="preserve">E finita la diuiſione di tutta la linea, ſi po-<lb/>tranno notare tutte le decine, e con vna lineeta ſegnare la <lb/>metà delle decine, acciò con maggior facilità ſi poſſano pren-<lb/>deri punti corriſpondenti à que’ numeri che più piaceranno.</s> <s xml:id="echoid-s2144" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2145" xml:space="preserve">In queſta linea Cubica non potiamo hauere nel diuiderla <lb/>que’vantaggi compendioſi, che s’hebbero nella linea Geo-<lb/>metrica, raddo ppiando, ò triplicando i lati ſegnati; </s> <s xml:id="echoid-s2146" xml:space="preserve">perche il <lb/>lato doppio dà il cubo ottuplo, e così A 2 raddoppiata cade <lb/>nel punto 16, A 3 duplicata nel punto 24, A 4 nel punto 32, <lb/>A 5 nel 40, A 6 nel 48; </s> <s xml:id="echoid-s2147" xml:space="preserve">& </s> <s xml:id="echoid-s2148" xml:space="preserve">oltre di queſte niun’ altra ſi può <lb/>raddoppiare; </s> <s xml:id="echoid-s2149" xml:space="preserve">onde queſti ſoli punti ſi puonno eſſaminare.</s> <s xml:id="echoid-s2150" xml:space="preserve"/> </p> <pb o="113" file="0127" n="129" rhead="Linea Cubica"/> <p> <s xml:id="echoid-s2151" xml:space="preserve">Segnati di queſta maniera nelli lati dello Stromento i lati <lb/>de’eubi, che vanno creſcendo conforme alla ſerie naturale <lb/>de’numeri, è manifeſto per la dimoſtratione fondamentale <lb/>portata nel capo 1, che anche gl’interualli dello Stromento <lb/>allargato danno i lati de’Cubi, che ſono nella ſteſſa proportio-<lb/>ne indicata dalli numeri notati nello Stromento: </s> <s xml:id="echoid-s2152" xml:space="preserve">poiche eſ-<lb/>ſendo quattro linee proportionali (cioè li due lati nello Stro-<lb/>mento, e li due interualli loro corriſpondenti) i ſolidi ſimili <lb/>ſopra di eſſe ſono proportionali per la 37. </s> <s xml:id="echoid-s2153" xml:space="preserve">del lib. </s> <s xml:id="echoid-s2154" xml:space="preserve">11.</s> <s xml:id="echoid-s2155" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div69" type="section" level="1" n="39"> <head xml:id="echoid-head67" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head68" xml:space="preserve">Tra due linee date, come ſi trouino due medie continuamente <lb/>Proportionali: ouero t<unsure/>ra due numeri dati.</head> <p> <s xml:id="echoid-s2156" xml:space="preserve">SE la proportione delle due linee date non è conoſciuta in <lb/>numeri, ſi cerchi per la queſt. </s> <s xml:id="echoid-s2157" xml:space="preserve">5. </s> <s xml:id="echoid-s2158" xml:space="preserve">del capo 2, la quale tro-<lb/>uata, s’applichi nella linea cubica dello Stromento la prima <lb/>delle date linee all interuallo del numero, che le corriſponde, <lb/>perche l’interuallo dell’altro numero nella ſteſſa linea cubica, <lb/>darà la ſeconda delle quattro proportionali. </s> <s xml:id="echoid-s2159" xml:space="preserve">Di poi l’ altra <lb/>delle due date linee, allargando, ò ſtringendo lo Stromento, <lb/>s’applichi all’interuallo del numero, chele corriſponde, per-<lb/>che l’interuallo del numero corriſpondente all’ altra, darà la <lb/>terza delle Quattro Proportionali.</s> <s xml:id="echoid-s2160" xml:space="preserve"/> </p> <figure> <image file="0127-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0127-01"/> </figure> <p> <s xml:id="echoid-s2161" xml:space="preserve">Siano date due linee R, S, le quali <lb/>ſi troua, che hanno la proportione di <lb/>29 à 42; </s> <s xml:id="echoid-s2162" xml:space="preserve">applico la linea R all’inter-<lb/>uallo 29, 29 della linea cubica dello <lb/>Stromento, e ritenuta la ſteſſa aper- <pb o="114" file="0128" n="130" rhead="CAPO IV."/> tura, prendo l’interuallo 42. </s> <s xml:id="echoid-s2163" xml:space="preserve">42, e mi dà la linea A prima del-<lb/>le due medie. </s> <s xml:id="echoid-s2164" xml:space="preserve">Di poi applico la linea S all’interuallo 42, 42 <lb/>della linea cubica, e l’interuallo 29. </s> <s xml:id="echoid-s2165" xml:space="preserve">29, mi dàla linea B ſecon-<lb/>da delle due medie. </s> <s xml:id="echoid-s2166" xml:space="preserve">Onde le quattro R, A, B, S, ſono contin-<lb/>uamete Proportionali: </s> <s xml:id="echoid-s2167" xml:space="preserve">il che così ſi dimoſtra. </s> <s xml:id="echoid-s2168" xml:space="preserve">Il cubo di R <lb/>al cubo di A è come 29 à 42, per la coſtruttione dello ſtro-<lb/>mento, e per la propoitione, che gl’interualli preſi hanno <lb/>conilati dello ſtromento; </s> <s xml:id="echoid-s2169" xml:space="preserve">dunque la linea R alla linea A hà <lb/>la proportione ſubtriplicata di 29 à 42, cioè della linea R alla <lb/>linea S: </s> <s xml:id="echoid-s2170" xml:space="preserve">dunquetra R, & </s> <s xml:id="echoid-s2171" xml:space="preserve">S poſte due medie in continuata <lb/>proportione la linea A è la ſeconda proportionale. </s> <s xml:id="echoid-s2172" xml:space="preserve">Simil-<lb/>mente il cubo di S al cubo di B è nella proportione di 42 à <lb/>29, per la coſtruttione dello Stromento, & </s> <s xml:id="echoid-s2173" xml:space="preserve">applicatione fat-<lb/>ta: </s> <s xml:id="echoid-s2174" xml:space="preserve">dunque la linea S alla linea B, hà la proportione ſubtripli-<lb/>cata di 42 à 29, e per conuerſione B à S, hà la ſubtriplicata <lb/>di 29 à 42, cioè di R à S: </s> <s xml:id="echoid-s2175" xml:space="preserve">Eſſendo dunque la proportione di <lb/>R ad A, e quella di B ad S, ſubtriplicate della proportione di <lb/>R ad S, reſta che anche quella di A à B, ſia ſubtriplicata della <lb/>ſteſſa; </s> <s xml:id="echoid-s2176" xml:space="preserve">e perciò come R ad A, così A à B, così B à S.</s> <s xml:id="echoid-s2177" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2178" xml:space="preserve">L’iſteſſo ſi farà dati due numeri, tra’quali ſi voleſſero due <lb/>medij proportionali; </s> <s xml:id="echoid-s2179" xml:space="preserve">come per eſſempio tra 8, e 27. </s> <s xml:id="echoid-s2180" xml:space="preserve">A qual-<lb/>ſiuoglia apertura dello Stromento nella linea cubica, prendo <lb/>con due Compaſſi gl’interualli 8, 8, e 27, 27. </s> <s xml:id="echoid-s2181" xml:space="preserve">Dipoi trapor-<lb/>tando il primo interuallo ſu la linea Aritmetica all’interuallo <lb/>S, 8, applico l’altro Compaſſo, e veggo che cade nell’ inter-<lb/>uallo 12, 12; </s> <s xml:id="echoid-s2182" xml:space="preserve">onde dico, che il num. </s> <s xml:id="echoid-s2183" xml:space="preserve">12 è il ſecondo propor-<lb/>tionale. </s> <s xml:id="echoid-s2184" xml:space="preserve">Quindi ritenendo l’interuallo preſo con queſto ſe-<lb/>condo Compaſſo, l’applico nella ſteſſa linea Aritmetica al <lb/>punto 27, 27, ſtringendo lo Stromento, come fà di biſogno, <lb/>e conſiderando che l’interuallo preſo col primo Compaſſo, <pb o="115" file="0129" n="131" rhead="Linea Cubica."/> cade nel punto 18, 18, dico che il terzo proportionale è 18; <lb/></s> <s xml:id="echoid-s2185" xml:space="preserve">onde ſono continuatamente Proportionali 8. </s> <s xml:id="echoid-s2186" xml:space="preserve">12. </s> <s xml:id="echoid-s2187" xml:space="preserve">18. </s> <s xml:id="echoid-s2188" xml:space="preserve">27. </s> <s xml:id="echoid-s2189" xml:space="preserve">e tra <lb/>li due eſtremi propoſti, ſi ſono trouati due medij propor-<lb/>tionali.</s> <s xml:id="echoid-s2190" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2191" xml:space="preserve">E quì s’auuerta ciò che in altre occaſioni s’è detto, che ſe <lb/>non foſſe commodo applicare alla linea Aritmetica il Com-<lb/>paſſo con la ſua apertura preſa nella linea cubica, quella ſteſ-<lb/>ſa apertura s’applichi ad alcun numero moltiplice, ò ſubmol-<lb/>tiplice, poiche l’altro Compaſſo darà vn numero ſimilmente <lb/>moltiplice, ò ſubmoltiplice del numero, che ſi cerca. </s> <s xml:id="echoid-s2192" xml:space="preserve">Cosìſe <lb/>l’interuallo primo non ſi può applicare all’interuallo della li-<lb/>nea Aritmetica 8. </s> <s xml:id="echoid-s2193" xml:space="preserve">8, s’applichi al numero triplo 24. </s> <s xml:id="echoid-s2194" xml:space="preserve">24, per-<lb/>che così il ſecondo interuallo caderà nel 36. </s> <s xml:id="echoid-s2195" xml:space="preserve">36 triplo del 12, <lb/>che ſi cerca: </s> <s xml:id="echoid-s2196" xml:space="preserve">e ſe il ſecondo interuallo s’applicherà al numero <lb/>duplo 54. </s> <s xml:id="echoid-s2197" xml:space="preserve">54, il primo interuallo caderà nel 36. </s> <s xml:id="echoid-s2198" xml:space="preserve">36 duplo del <lb/>18, che ſi cerca.</s> <s xml:id="echoid-s2199" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2200" xml:space="preserve">Quando però li due numeri dati non ſono ſimili ſolidi, non <lb/>ſi troueranno li due medij proportionali preciſi, ma vi ſaran-<lb/>no aggiunte frattioni, che ſolo s’auuicineranno al vero ſenza <lb/>dar preciſione, come ſi può raccogliere dalla 19, e 21 del lib. <lb/></s> <s xml:id="echoid-s2201" xml:space="preserve">8, e per trouar tali frattioni, potremo valerci dell’ artificio <lb/>moſtrato nel Capo 2 alla Queſt. </s> <s xml:id="echoid-s2202" xml:space="preserve">7, quando le linee, ò apertu-<lb/>re del Compaſſo, che per lo ſteſſo ſi prendono, non cadono <lb/>preciſamente ne’ punti dello ſtromento.</s> <s xml:id="echoid-s2203" xml:space="preserve"/> </p> <pb o="116" file="0130" n="132" rhead="CAPO IV."/> </div> <div xml:id="echoid-div70" type="section" level="1" n="40"> <head xml:id="echoid-head69" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head70" xml:space="preserve">Come ſi poſſa ad vna linea data applicar’ vn ſolido rettangolo <lb/>vguale ad vn Cubo dato.</head> <p> <s xml:id="echoid-s2204" xml:space="preserve">HAuendo il corpo tre dimenſioni in Lunghezza, Lar-<lb/>ghezza, e Groſſezza, che altri chiamano Altezza, ò <lb/>Profondità, ſi dice, che vn ſolido ſia applicato ad vna linea <lb/>data, quando ſi ſuppone, che detta linea ſia vna delle ſue tre <lb/>dimenſioni, e ſi determina, quali, e quanto grandi ſiano l’al-<lb/>tre due dimenſioni dello ſteſſo corpo. </s> <s xml:id="echoid-s2205" xml:space="preserve">E per maggior facilità <lb/>di queſto eſſempio, maſſime che è conforme all’vſo più com-<lb/>mune, ſuppongo eſſer’ il ſolido, che deue applicarſi alla data <lb/> <anchor type="figure" xlink:label="fig-0130-01a" xlink:href="fig-0130-01"/> linea, rettangolo; <lb/></s> <s xml:id="echoid-s2206" xml:space="preserve">poiche poi ſopra la <lb/>ſteſſa baſe qualſiuo. </s> <s xml:id="echoid-s2207" xml:space="preserve"><lb/>glia parallelepipe-<lb/>do, che habbia la <lb/>ſteſſa altezza per-<lb/>pendicolare, gli ſa-<lb/>rà vguale, per la 30 <lb/>del lib. </s> <s xml:id="echoid-s2208" xml:space="preserve">11, e per <lb/>conſeguenza ſarà <lb/>vguale al dato cu-<lb/>bo.</s> <s xml:id="echoid-s2209" xml:space="preserve"/> </p> <div xml:id="echoid-div70" type="float" level="2" n="1"> <figure xlink:label="fig-0130-01" xlink:href="fig-0130-01a"> <image file="0130-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0130-01"/> </figure> </div> <p> <s xml:id="echoid-s2210" xml:space="preserve">Sia dunque dato <lb/>il cubo V T il cui <lb/>lato V S, e ſia datà<unsure/> <lb/>lalinea CD, la quale debba eſſere vna delle dimenſioni del ſo- <pb o="117" file="0131" n="133" rhead="Linea Cubica"/> lido rettangolo vguale al cubo dato. </s> <s xml:id="echoid-s2211" xml:space="preserve">In due maniere ciò ſi <lb/>può fare. </s> <s xml:id="echoid-s2212" xml:space="preserve">Primieramente con trouare alle linee CD, VS vna <lb/>terza proportionale E, perche il ſolido fatto da queſte tre, <lb/>cioè il ſolido C I H è vguale al dato cubo fatto dalla media <lb/>V S, per la 36 del lib. </s> <s xml:id="echoid-s2213" xml:space="preserve">11. </s> <s xml:id="echoid-s2214" xml:space="preserve">Secondariamente con trouare la <lb/>quarta proportionale, mettendo CD la prima, & </s> <s xml:id="echoid-s2215" xml:space="preserve">VS la ſe-<lb/>conda; </s> <s xml:id="echoid-s2216" xml:space="preserve">poiche il quadrato della prima con la quarta fanno vn <lb/>ſolido vguale al cubo della ſeconda. </s> <s xml:id="echoid-s2217" xml:space="preserve">Dunque con due Com-<lb/>paſſi prendendo le linee CD, & </s> <s xml:id="echoid-s2218" xml:space="preserve">VS, vedo nella linea cubica, <lb/>ſopra quali interualli cadano, etrouando, che cade la CD <lb/>nell’interuallo 29. </s> <s xml:id="echoid-s2219" xml:space="preserve">29, e la V S nell’interuallo 4. </s> <s xml:id="echoid-s2220" xml:space="preserve">4, applico la <lb/>CD nella linea Aritmetica al punto doppio del 29, cioè al <lb/>58. </s> <s xml:id="echoid-s2221" xml:space="preserve">58, & </s> <s xml:id="echoid-s2222" xml:space="preserve">all’interuallo 8. </s> <s xml:id="echoid-s2223" xml:space="preserve">8 doppio del 4 trouo la quarta <lb/>proportionale F. </s> <s xml:id="echoid-s2224" xml:space="preserve">Dunque della CD fatto il quadrato CM, <lb/>preſa DL vguale alla F quarta proportionale, ſarà il ſolido <lb/>CML vguale al cubo dato.</s> <s xml:id="echoid-s2225" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2226" xml:space="preserve">Così ſe foſſe dato vn pezzo di marmo ben ſquadrato, che <lb/>foſſe per ogni verſo ſette palmi, e da vn’altro gran pezzo di <lb/>marmo, che per vn verſo è 10 palmi, per l’altro 11, e per il <lb/>terzo 4 palmi, ſi doueſſe cauar’ vn pezzo vguale al primo, <lb/>ma quadro in vna delle faccie; </s> <s xml:id="echoid-s2227" xml:space="preserve">facilmente ſi cauerà in numeri, <lb/>quanta debba eſſer la groſſezza. </s> <s xml:id="echoid-s2228" xml:space="preserve">Primieramente ſi pigli il cu-<lb/>bo di 7, & </s> <s xml:id="echoid-s2229" xml:space="preserve">è il pezzo cubico dato 343 palmi ſolidi. </s> <s xml:id="echoid-s2230" xml:space="preserve">Dipoi il <lb/>pezzorozzo non può ſquadrarſi, che con hauer 10 palmi in <lb/>quadro, e così il quadrato di 10 è 100; </s> <s xml:id="echoid-s2231" xml:space="preserve">per il quale diuiden-<lb/>do il cubo 343, viene per la groſſezza cercata palmi 3 {43/100}. </s> <s xml:id="echoid-s2232" xml:space="preserve">Mà <lb/>ſe non ſapeſſi alcun numero, che miſuraſſe i lati de’ ſudetti <lb/>pezzi di marmo, prendo con vn Compaſſo tal parte aliquota <lb/>dellato del cubo, che poſſa commodamente capire ne gl’in-<lb/>terualli dello Stromento: </s> <s xml:id="echoid-s2233" xml:space="preserve">e ſimile parte aliquota prendo nel <pb o="118" file="0132" n="134" rhead="CAPO IV."/> lato mezzano dell’altro pezzo di marmo, per eſſempio la de-<lb/>cima parte. </s> <s xml:id="echoid-s2234" xml:space="preserve">Et applicando queſte due miſure à gl’interualli <lb/>della linea cubica, oſſeruo in quali numeri cadano; </s> <s xml:id="echoid-s2235" xml:space="preserve">perche la <lb/>proportione, che hauranno queſti due numeri, tale dourà ha-<lb/>uer’il lato mezzano oſſeruato alla linea della groſſezza, che <lb/>ſi cerca. </s> <s xml:id="echoid-s2236" xml:space="preserve">Laragione di queſta operatione è, perche eſſendo <lb/>le miſure preſe con i Compaſſi ciaſcuna la decima parte del <lb/>lato, il cubo di tal parte è vna milleſima di tutto il cubo di <lb/>quei lati intieri: </s> <s xml:id="echoid-s2237" xml:space="preserve">dunque li cubi delle parti hanno la propor-<lb/>tione de’cubi intieri. </s> <s xml:id="echoid-s2238" xml:space="preserve">Dunque per l’applicatione fatta allo <lb/>Stromento trouandoſi in numerila proportione de’ cubi, due <lb/>linee, che ſiano nella ſteſſa proportione di queſti numeri ſo-<lb/>no due eſtreme di quattro continuatamente proportionali: <lb/></s> <s xml:id="echoid-s2239" xml:space="preserve">Dunque anche le decuple di queſte ſono ſimilmente eſtreme <lb/>di quattro proportionali, delle quali la prima è il lato, di cui <lb/>ſi deue far’ il quadrato, la ſeconda è il lato del cubo dato, ela <lb/>quarta ſarà queſta trouata, la quale col quadrato della prima <lb/>farà vn ſolido vguale al cubo della ſeconda.</s> <s xml:id="echoid-s2240" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div72" type="section" level="1" n="41"> <head xml:id="echoid-head71" style="it" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head72" style="it" xml:space="preserve">Dato vn ſolido, come s’habbia à trouare vn’ altro ſimile <lb/>nella data proportione.</head> <p> <s xml:id="echoid-s2241" xml:space="preserve">POſſono li ſolidi eſſere Regolari, ò Irregolari; </s> <s xml:id="echoid-s2242" xml:space="preserve">Regolari, <lb/>quando tutte le linee, & </s> <s xml:id="echoid-s2243" xml:space="preserve">i piani del corpo ſono vguali <lb/>tra diloro; </s> <s xml:id="echoid-s2244" xml:space="preserve">Irregolari, quando non v’è queſta vguaglianza. <lb/></s> <s xml:id="echoid-s2245" xml:space="preserve">Nell’operatione v’è queſta ſola differenza, che ne’ Regolari <lb/>trouata vna linea, che habbia la douuta proportione con il la-<lb/>to del ſolido ſimile, non s’hà à cercar’ altra linea; </s> <s xml:id="echoid-s2246" xml:space="preserve">mà ne gl’Ir- <pb o="119" file="0133" n="135" rhead="Linea Cubica"/> regolari conuien far queſta operatione circa tutte le linee, <lb/>che concorrono alla coſtitutione dell’ angolo ſolido. </s> <s xml:id="echoid-s2247" xml:space="preserve">Ne lle <lb/>sfere baſta trouar’ il diametro, ma per li Coni, e Cilindri ſi. <lb/></s> <s xml:id="echoid-s2248" xml:space="preserve">mili conuien trouare il diametro della baſe, e l’aſſe.</s> <s xml:id="echoid-s2249" xml:space="preserve"/> </p> <figure> <image file="0133-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0133-01"/> </figure> <p> <s xml:id="echoid-s2250" xml:space="preserve">Se dunque il cor-<lb/>po dato è cubo, ò <lb/>altro de’ corpi Re-<lb/>golari, veggaſi con <lb/>quali numeri ſi e-<lb/>ſprima la propor-<lb/>tione data, & </s> <s xml:id="echoid-s2251" xml:space="preserve">il la-<lb/>to del corpo dato <lb/>ſi applichi nella li-<lb/>nea cubica all’ in-<lb/>teruallo del nume-<lb/>ro, che gli corri-<lb/>ſponde, e l’ inter-<lb/>uallo dell’ altro nu-<lb/>mero darà il lato, <lb/>che ſicerca. </s> <s xml:id="echoid-s2252" xml:space="preserve">Così ſe al cubo VST ſi debba farne vno, che <lb/>ſia {7/8} di quello, applico il lato V S all’interuallo 8. </s> <s xml:id="echoid-s2253" xml:space="preserve">8, e l’inter-<lb/>uallo 7. </s> <s xml:id="echoid-s2254" xml:space="preserve">7, mi darà il lato del cubo cercato. </s> <s xml:id="echoid-s2255" xml:space="preserve">Mà ſe foſſe dato <lb/>DAH ſolido di lati diſuguali, e conueniſſe farne vn ſimile, che <lb/>foſſe parimenti {7/8}, applico D I all’interuallo 8. </s> <s xml:id="echoid-s2256" xml:space="preserve">8, e l’ inter-<lb/>uallo 7. </s> <s xml:id="echoid-s2257" xml:space="preserve">7 dà il lato homologo RB. </s> <s xml:id="echoid-s2258" xml:space="preserve">Dipoi all’iſteſſo interual-<lb/>lo 8. </s> <s xml:id="echoid-s2259" xml:space="preserve">8 applico I A, e la diſtanza 7. </s> <s xml:id="echoid-s2260" xml:space="preserve">7 dà il lato homologo BK, <lb/>che col primo trouato faccia l’angolo R BK vguale all’angolo <lb/>D I A. </s> <s xml:id="echoid-s2261" xml:space="preserve">Finalmente allo ſteſſo interuallo 8. </s> <s xml:id="echoid-s2262" xml:space="preserve">8 applico IH, e la <lb/>diſtanza 7. </s> <s xml:id="echoid-s2263" xml:space="preserve">7 dà il terzo lato homologo B O, il quale con il <lb/>ſecondo trouato faccia l’angolo KBO vguale all’ angolo AIH:</s> <s xml:id="echoid-s2264" xml:space="preserve"> <pb o="120" file="0134" n="136" rhead="CAPO IV."/> e compiti tutti li parallelogrammi, ſarà fatto il corpo RKO <lb/>ſimile al dato DAH; </s> <s xml:id="echoid-s2265" xml:space="preserve">e che è à quello, come 7 à 8. </s> <s xml:id="echoid-s2266" xml:space="preserve">Che ſia <lb/>ſimile è chiaro, per l’vguaglianza de gl’angoli, circai quali <lb/>ſono i lati homologi, ciaſcuno preſo nello Stromento à gl’i-<lb/>ſteſſi interualli, e perciò nella medefima proportione; </s> <s xml:id="echoid-s2267" xml:space="preserve">onde <lb/>li piani RK, DA; </s> <s xml:id="echoid-s2268" xml:space="preserve">e li piani KO, AH, e RO, DH ſono ſimili. <lb/></s> <s xml:id="echoid-s2269" xml:space="preserve">E perche, per la 33 dellib. </s> <s xml:id="echoid-s2270" xml:space="preserve">11, li ſolidi ſimili ſono nella pro-<lb/>portione triplicata de’lati homologi, cioè nella proportione <lb/>de’cubi di detti lati homologi, eſſendo tali cubi, come 7 à 8, <lb/>per la coſtruttione dello Stromento, anche li ſolidi ſimili <lb/>RKO, DAH ſono come 7 à 8.</s> <s xml:id="echoid-s2271" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2272" xml:space="preserve">L’iſteſſo modo ſi dourà tenere ne’ Coni, e Cilindri ſimili, <lb/>ſeruendoſi de gl’interualli delli ſteſſi numeri peri diametri <lb/>delle baſi, e per gl’aſſi.</s> <s xml:id="echoid-s2273" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2274" xml:space="preserve">Così li Pittori, per eſprimere vn corpo, che ſia più picco-<lb/>lo di vn’ altro ſimile in data proportione, ſi ſeruiranno di que-<lb/>ſta linea cubica; </s> <s xml:id="echoid-s2275" xml:space="preserve">altrimenti ſe per far’vn dito la metà più pic-<lb/>colo, lo faceſſero la metà più corto, ſaria rappreſentato vn <lb/>dito otto volte minore: </s> <s xml:id="echoid-s2276" xml:space="preserve">perciò applicato il dito maggiore <lb/>all’interuallo 2. </s> <s xml:id="echoid-s2277" xml:space="preserve">2 di queſta linea cubica, l’interuallo 1. </s> <s xml:id="echoid-s2278" xml:space="preserve">1 darà <lb/>la lunghezza deſiderata; </s> <s xml:id="echoid-s2279" xml:space="preserve">e così dell’altre parti. </s> <s xml:id="echoid-s2280" xml:space="preserve">Quindi è, che <lb/>deuono auuertire li Pittori altra coſa eſſere far’vn Quadro la <lb/>metà più piccolo, altra coſa far le figure in eſſo la metà più <lb/>piccole: </s> <s xml:id="echoid-s2281" xml:space="preserve">perche l’impicciolire il Quadro è impicciolir’ vna <lb/>ſuperficie, doue che l’impicciolire le figure, è far corpi mi-<lb/>nori: </s> <s xml:id="echoid-s2282" xml:space="preserve">in quello ſerue la linea Geometrica, & </s> <s xml:id="echoid-s2283" xml:space="preserve">in queſto la <lb/>Cubica.</s> <s xml:id="echoid-s2284" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2285" xml:space="preserve">Così parimenti ſeruirà queſta linea Cubica alli Scultori, <lb/>& </s> <s xml:id="echoid-s2286" xml:space="preserve">alli Fonditori nel far le for me per Campane, Artiglierie, ò <lb/>coſe ſomiglianti, ſe voleſſero far’vna Statua, ò altra figura ſi- <pb o="121" file="0135" n="137" rhead="Linea Cubica."/> mile ad vna data. </s> <s xml:id="echoid-s2287" xml:space="preserve">Poiche ciaſcheduna parte applicata all’in-<lb/>teruallo conueniente, s’haurà la miſura corriſpondente nella <lb/>figura ſimile.</s> <s xml:id="echoid-s2288" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2289" xml:space="preserve">Mà commodiſſi mo riuſcirà queſto noſtro Compaſſo di <lb/>Proportione alli Bombardieri, per notar li diametri delle <lb/>palle, e dalla grandczza della bocca dell’ Artiglieria raccoglier <lb/>la loro portata, e formarne li ſuoi Calibri, ò Colibri, come <lb/>altri li chiamano; </s> <s xml:id="echoid-s2290" xml:space="preserve">e con ragione da molti ſi deplora l’ignoran-<lb/>za di molti di queſta profeſſione, che hanno Calibri ſpropo-<lb/>ſitatiſſimi; </s> <s xml:id="echoid-s2291" xml:space="preserve">mà con queſta linea Cubica fatta nel Compaſſo di <lb/>Proportione con qualche accuratezza, e diligenza, potrà cia-<lb/>ſcuno eſſaminare nel ſuo Calibre, ſe ſiano ben notati li diame-<lb/>tri; </s> <s xml:id="echoid-s2292" xml:space="preserve">e con ſomma facilità, e preſtezza potrà notare li diametri <lb/>delle palle di ferro, di piombo, di pietra à ragion di libre ò <lb/>communi di 12 oncie, ò, come in moltiluoghi s’ vſa, di 16. <lb/></s> <s xml:id="echoid-s2293" xml:space="preserve">oncie.</s> <s xml:id="echoid-s2294" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2295" xml:space="preserve">Habbiaſi noto il diametro d’vna palla, il cui peſo ſi sà, per <lb/>cagion d’eſſempio, di libre 7, queſto diametro ſi noti sù la <lb/>Regola, ò Calibre, e nella linea Cubica s’applichi all’inter-<lb/>uallo 7. </s> <s xml:id="echoid-s2296" xml:space="preserve">7; </s> <s xml:id="echoid-s2297" xml:space="preserve">perche ritenuta quell’apertura dello Stromento, <lb/>prendendo tutti gl’interuali da 1 ſin’ à 50, e traportandoli sù <lb/>la Regola, s’hauranno li diametri delle palle ſin’ à 50 libre di <lb/>peſo, della ſteſſa materia, di cui era quella, il cui diametro era <lb/>noto. </s> <s xml:id="echoid-s2298" xml:space="preserve">E queſto, che s’è fatto con vna palla di ferro, ſaputaſi <lb/>la proportione, che hà la pietra col ferro, ſi potrà fare con le <lb/>palle di pietra: </s> <s xml:id="echoid-s2299" xml:space="preserve">onde ſe la pietra, conſorme all’ opinione de’ <lb/>Bombardieri, è la terza parte del peſo del ferro in parità di <lb/>mole, conuerrà pigliar’vna linea, che ſia diametro d’vna sfe-<lb/>ra, la qual ſia tre volte tanto, quanto la palla di ferro nota di <lb/>libre 7, e ſarà il diametro della palla di pietra di libre 7, & </s> <s xml:id="echoid-s2300" xml:space="preserve">ap- <pb o="122" file="0136" n="138" rhead="CAPO IV."/> Plicato all’interuallo 7. </s> <s xml:id="echoid-s2301" xml:space="preserve">7, nella linea Cubica, all’iſteſſo modo <lb/>s’hauranno li diametri delle palle di pietra. </s> <s xml:id="echoid-s2302" xml:space="preserve">Ne differente <lb/>ſarà la forma per le palle di piombo, perche ſupponendoſi il <lb/>peſo del piombo ſeſquialtero à quello del ferro, ſi prenderà <lb/>il diametro della palla di piombo, di peſo vguale con quella <lb/>di ferro, che ſia diametro d’vna sfera, la qual fia {2/3} della pal. <lb/></s> <s xml:id="echoid-s2303" xml:space="preserve">la di ferro. </s> <s xml:id="echoid-s2304" xml:space="preserve">E finalmente per notare le palle à ragion d’oncie <lb/>16 per libra, auuerti che 4 libre da oncie 12 fanno 3 libre da <lb/>oncie 16 l’vna: </s> <s xml:id="echoid-s2305" xml:space="preserve">perciò prendi il diametro trouato di libre 4 <lb/>piccole, e notatolo ſopra vn lato della Regola, ò Calibre ſia il <lb/>diametro di libre 3 groſſe, e queſto diametro applicato nello <lb/>Stromento all’interuallo 3. </s> <s xml:id="echoid-s2306" xml:space="preserve">3, s’hauranno da gl’altri interualli <lb/>tutti li diametri delle palle à ragion di peſo d’oncie 16 per li-<lb/>bra. </s> <s xml:id="echoid-s2307" xml:space="preserve">Dal che ciaſcun vede, che queſti diametri ſon tali, che <lb/>ciaſcuno aggiunge vn terzo di peſo alle palle, che hanno la, <lb/>ſteſſa denominatione nella ſerie de’diametrià ragione d’oncie <lb/>12 per libra. </s> <s xml:id="echoid-s2308" xml:space="preserve">E così il diametro di 45 libre groſſe è il diame-<lb/>tro di libre 60 piccole, perche come 16 à 12, così 60 à 45.</s> <s xml:id="echoid-s2309" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2310" xml:space="preserve">E così ſi faccia rifleſſione, quanto più giuſti ſaranno com-<lb/>munemente li diametri delle palle notate, e preſe dal Com-<lb/>paſſo di Proportione ſegnato nella linea Cubica, come hab-<lb/>biamo detto in queſto Capo, che con la forma preſcritta da <lb/>Luigi Colliado nella ſua Prattica Manuale di Artiglieria trat-<lb/>tato 4 cap. </s> <s xml:id="echoid-s2311" xml:space="preserve">32, doue ciaſcuno potrà eſſaminare, quanto s’al-<lb/>lontani dalla preciſione. </s> <s xml:id="echoid-s2312" xml:space="preserve">E ſia per eſſempio ciò ch’ egli dice <lb/>per hauer’il diametro d’vna palla di due libre; </s> <s xml:id="echoid-s2313" xml:space="preserve">prendaſi, dice <lb/>egli, il diametro d’vna palla d’vna libra, e diuiſo in quattro <lb/>parti, vna ſe ne aggiunga, sì che il diametro di vna libra è co-<lb/>me 4, e quello di due è come 5; </s> <s xml:id="echoid-s2314" xml:space="preserve">li cubiſono 64, e 125, e pure <lb/>queſto, per eſſer doppio, douria eſſere 128, onde manca <pb o="123" file="0137" n="139" rhead="Linea Cubica"/> dalla preciſione {3/64}. </s> <s xml:id="echoid-s2315" xml:space="preserve">Mà nel noſtro Stromento il diametro di <lb/>vna palla d’vna libra è 1000, quello di due è 1259, il cubo di <lb/>queſto è 1995616979, il quale douria eſſere 2000000000, <lb/>e perciò manco della preciſione {4383021/1000000000}, doue che li {3/64} ridotti <lb/>alla ſſeſta denominatione, ſono {46875000/1000000000}, che è vna differenza <lb/>dieci volte maggiore di quella, che viene dal modo da noi <lb/>tenuto. </s> <s xml:id="echoid-s2316" xml:space="preserve">Cosìper il diametro della palla di 3 lib. </s> <s xml:id="echoid-s2317" xml:space="preserve">diuide in ſet-<lb/>te parti quello di due, & </s> <s xml:id="echoid-s2318" xml:space="preserve">vna di queſte aggiunge, onde il dia-<lb/>metro di due al diametro di tre libre è come 7 à 8; </s> <s xml:id="echoid-s2319" xml:space="preserve">il diametro <lb/>di due era {5/4} del primo diametro, dunque il diametro di tre <lb/>libre è {10/7} del primo diametro, com’è manifeſto, ſe le due pro-<lb/>portioni 4 à 5, e 7 à 8 ſi continuano in tre termine 28. </s> <s xml:id="echoid-s2320" xml:space="preserve">35. </s> <s xml:id="echoid-s2321" xml:space="preserve">40. <lb/></s> <s xml:id="echoid-s2322" xml:space="preserve">Dunque il diametro d’vna lib. </s> <s xml:id="echoid-s2323" xml:space="preserve">al diametro di tre libre è come <lb/>7 à 10: </s> <s xml:id="echoid-s2324" xml:space="preserve">il cubo di quello è 343, il cubo di queſto è 1000, e <lb/>pur’il triplo del primo è 1029; </s> <s xml:id="echoid-s2325" xml:space="preserve">sì che è minor del douere di <lb/>{29/343}, le qualiridotte ſono {84548104/1000000000}. </s> <s xml:id="echoid-s2326" xml:space="preserve">Mà nel noſtro Stromen-<lb/>to il diametro della palla di tre libre è 1442, il cui cubo <lb/>2998442888 mãca dal triplo cubo del primo 3000000000 <lb/>ſolamente di {1557112/1000000000}. </s> <s xml:id="echoid-s2327" xml:space="preserve">Dal che manifeſtamente appariſce, <lb/>quanto più accuratamente con qneſta maniera poſſano farſi <lb/>Calibri giuſtiſſimi, e con facilità grandiſſima, & </s> <s xml:id="echoid-s2328" xml:space="preserve">eſſaminare <lb/>igià fatti.</s> <s xml:id="echoid-s2329" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2330" xml:space="preserve">Mà ſe il Bombardiere haurà ſeco queſto Stromento di <lb/>Proportione, haurà ſeco vn Calibre vniuerſale per tutti i <lb/>Paeſi, ſecondo la diuerſità de’ peſi; </s> <s xml:id="echoid-s2331" xml:space="preserve">poiche conoſciuto il dia-<lb/>metro d’vna palla di determinato peſo di quel paeſe, ritenuta <lb/>quell’apertura dello Stromento, à cui tal diametro è applica-<lb/>to al numero corriſpondente alle libre del peſo, ſubito ſi co-<lb/>noſcerà il diametro di qual ſi voglia altra palla di tal materia <lb/>di qual ſi voglia peſo.</s> <s xml:id="echoid-s2332" xml:space="preserve"/> </p> <pb o="124" file="0138" n="140" rhead="CAPO IV."/> <p> <s xml:id="echoid-s2333" xml:space="preserve">Quindi volendo diametri di palle minori d’ vna libra, <lb/>metta il diametro d’vna libra al numero 12. </s> <s xml:id="echoid-s2334" xml:space="preserve">12, e potrà ha-<lb/>uer il diametro d’vna, due, e più oncie, & </s> <s xml:id="echoid-s2335" xml:space="preserve">anche minori dell’ <lb/>oncia, ſe trouato il diametro d’vn’ oncia ſi applichi ad vn nu. <lb/></s> <s xml:id="echoid-s2336" xml:space="preserve">mero capace della diuiſione cercata; </s> <s xml:id="echoid-s2337" xml:space="preserve">così mettendoſi al 50. </s> <s xml:id="echoid-s2338" xml:space="preserve"><lb/>50, ſi potrà hauer il diametro d’vna palla, che ſia {1/50} d’oncia.</s> <s xml:id="echoid-s2339" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2340" xml:space="preserve">Che ſe per auuentura la proportione, che deuono hauer’i <lb/>ſolidi ſimili foſſe eſpreſſa in numero maggiore del 50, che ſi <lb/>troua nella linea Cubica dello Stromento, come ſe la propor-<lb/>tione foſſe di 40 à 72, ſi riduca à minor termini, come di 10 <lb/>à 18, ouero di 5 à 9, e con queſti numeri ſi operi, comeſe in <lb/>eſſi foſſe data la proportione, poiche in realtà è la ſteſſa pro-<lb/>portione diuerſamente eſpreſſa. </s> <s xml:id="echoid-s2341" xml:space="preserve">Mà ſe li numeri della Pro-<lb/>portione non haueſſero alcuna commune miſura, come 49 <lb/>à 60, s’applichi il lato del ſolido dato all’interuallo 49. </s> <s xml:id="echoid-s2342" xml:space="preserve">49; <lb/></s> <s xml:id="echoid-s2343" xml:space="preserve">dipoi ritenuta quell’ apertura dello Stromento, diuiſo il 60 <lb/>per alcun numero, che lo miſuri, ſia per cagion d’eſſempio, il <lb/>12, che lo miſura per 5, prendo l’interuallo 12. </s> <s xml:id="echoid-s2344" xml:space="preserve">12, e conſer-<lb/>uo queſta lunghezza, la quale applico all’interuallo di qual-<lb/>che numero, che habbia tra’numeri della linea vn numero <lb/>quintuplo à cagione, che il 12 miſuraua per 5 il 60; </s> <s xml:id="echoid-s2345" xml:space="preserve">e per eſ-<lb/>ſempio l’applico al 7. </s> <s xml:id="echoid-s2346" xml:space="preserve">7; </s> <s xml:id="echoid-s2347" xml:space="preserve">Quindi al quintuplo di 7, cioè all’in-<lb/>teruallo 35. </s> <s xml:id="echoid-s2348" xml:space="preserve">35 haurò il lato del ſolido, che ſarà come 60 in, <lb/>riguardo del dato, che è 49. </s> <s xml:id="echoid-s2349" xml:space="preserve">E che ciò ſia, è chiaro dall’ope-<lb/>ratione, perche nella prima operatione ſi trouò il lato d’vn, <lb/>ſolido, che al 49 era come 12; </s> <s xml:id="echoid-s2350" xml:space="preserve">nella ſeconda operatione s’è <lb/>trouato il lato d’vn ſolido quintuplo di quello, e perciò pren-<lb/>dendoſi cinque volte il 12, vien’ad eſſere 60. </s> <s xml:id="echoid-s2351" xml:space="preserve">Così per hauer’ <lb/>il lato del ſolido, che ſia come 51 ad vn’ altro il cuilato s’ad-<lb/>datta all’interuallo 28. </s> <s xml:id="echoid-s2352" xml:space="preserve">28, prendo l’interuallo 3. </s> <s xml:id="echoid-s2353" xml:space="preserve">3: </s> <s xml:id="echoid-s2354" xml:space="preserve">queſto <pb o="125" file="0139" n="141" rhead="Linea Cubica"/> applico, aprendo lo Stromento, al punto 2. </s> <s xml:id="echoid-s2355" xml:space="preserve">2; </s> <s xml:id="echoid-s2356" xml:space="preserve">& </s> <s xml:id="echoid-s2357" xml:space="preserve">al 34. </s> <s xml:id="echoid-s2358" xml:space="preserve">34 <lb/>trouo la grandezza del lato di 51: </s> <s xml:id="echoid-s2359" xml:space="preserve">perche 34 contiene il 2 <lb/>dieciſette volte; </s> <s xml:id="echoid-s2360" xml:space="preserve">all’interuallo 2. </s> <s xml:id="echoid-s2361" xml:space="preserve">2 fù applicato il lato del ſoli-<lb/>do 3; </s> <s xml:id="echoid-s2362" xml:space="preserve">dunque il 3 preſo 17 volte dà 51. </s> <s xml:id="echoid-s2363" xml:space="preserve">Di quì appariſce, che <lb/>ſe il numero maggiore ſi miſura dall’ 8, preſo l’altro numero, <lb/>che lo miſura, e raddoppiato l’interuallo, ſarà il lato cercato; <lb/></s> <s xml:id="echoid-s2364" xml:space="preserve">Come ſe ſi voleſſe il lato di 96, il quale ſi miſura dal 12 per 8; </s> <s xml:id="echoid-s2365" xml:space="preserve"><lb/>preſo l’interuallo 12. </s> <s xml:id="echoid-s2366" xml:space="preserve">12, eraddoppiato, darà ciò, che ſi cer-<lb/>ca, perche illato doppio dà il cubo ottuplo, e così il 12 ottu-<lb/>plicato è 96.</s> <s xml:id="echoid-s2367" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2368" xml:space="preserve">Mà quando occorreſſe, che il numero maggiore di 50 foſ-<lb/>ſe numero primo, non miſurato da altro numero, che dall’<unsure/> <lb/>vnità, e per conſeguenza diſpari, come ſe foſſe 83, ſi potrà <lb/>ſenza pericolo di errore ſenſibile prendere la metà del nu-<lb/>mero all’interuallo 41 {1/2}. </s> <s xml:id="echoid-s2369" xml:space="preserve">41 {1/2}, e poi applicata queſta diſtan-<lb/>za al punto 25. </s> <s xml:id="echoid-s2370" xml:space="preserve">25, l’interuallo 50. </s> <s xml:id="echoid-s2371" xml:space="preserve">50 darà il lato cercato di <lb/>83: </s> <s xml:id="echoid-s2372" xml:space="preserve">perche ſe bene quel lato, che dà il 41’ preſo à occhio, <lb/>non è così preciſo, è però tanto poca la differenza, che per <lb/>l’operatione ſiſica non porta errore notabile.</s> <s xml:id="echoid-s2373" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div73" type="section" level="1" n="42"> <head xml:id="echoid-head73" style="it" xml:space="preserve">QVESTIONE QVARTA. <lb/>Dati due corpi ſimili, come ſi coneſca la loro proportione.</head> <p> <s xml:id="echoid-s2374" xml:space="preserve">COn due Compaſſi ſi prendano i due lati homologi, & </s> <s xml:id="echoid-s2375" xml:space="preserve"><lb/>applicati nella linea Cubica à gl’ interualli, ne’quali <lb/>caderanno con preciſione la maggiore che ſi potrà, i numeri, <lb/>che cortiſpondono eſprimeranno la pro portione. </s> <s xml:id="echoid-s2376" xml:space="preserve">E ſe i lati <lb/>de’ corpi dati foſſero troppo grandi per applicargli allo ſtro-<lb/>mento, ſi opericon vnalor parte aliquota ſimile, perche il ſo- <pb o="126" file="0140" n="142" rhead="CAPO IV."/> lido ſimile ſopra la parte del lato d’vno, hà al ſolido ſimile ſo-<lb/>pra parte ſimile dellato dell’altro la proportione, che hanno <lb/>tra di loro gl’intieri ſolidi ſimili ſopra i lati intieri.</s> <s xml:id="echoid-s2377" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2378" xml:space="preserve">Prendiamo l’eſſempio dalli Bombardieri, i quali danno il <lb/>vento alle palle dell’ artiglieria, cioè prendono le palle vn, <lb/>poco minori di quello, che richiede la bocca del pezzo, à fine <lb/>che mancando per auuentura, come ſpeſſo accade, la douuta <lb/>rotondità alla palla, non reſti impedita dal poterſi ſpinger à <lb/>baſſo, quanto conuiene, ò nello ſparare non incontraſſe con, <lb/>qualche piccola prominenza à ſerrar così giuſto, che perico-<lb/>laſſe il pezzo. </s> <s xml:id="echoid-s2379" xml:space="preserve">Due ſono le prattiche, che adoprano. </s> <s xml:id="echoid-s2380" xml:space="preserve">Pri-<lb/>mieramente prendono il diametro della bocca del pezzo, e <lb/>diuiſolo in 21 parti, ne danno 20 per il diametro della palla. <lb/></s> <s xml:id="echoid-s2381" xml:space="preserve">Ora per ſapere, che proportione habbia la palla, che real-<lb/>mente s’ adopra, à quella, che giuſtamente porta il pezzo, <lb/>s’ella foſſe iſquiſitamente polita, e liſcia; </s> <s xml:id="echoid-s2382" xml:space="preserve">prendaſi il diametro <lb/>dell’ anima del pezzo, e nella linea cubica dello ſtromento <lb/>s’applichi all’ interuallo di quel numero, che è il peſo della <lb/>palla, che lo denomina, e ſia vn cannone da 40, onde dourà <lb/>applicarſi all’interuallo 40. </s> <s xml:id="echoid-s2383" xml:space="preserve">40; </s> <s xml:id="echoid-s2384" xml:space="preserve">e poi ſi vegga à che interual-<lb/>lo ſi poſſa applicare il diametro della palla, ch’è {20/21} del diame-<lb/>tro del pezzo, e ſi trouerà, che cade tra li numeri 34, e 35, <lb/>onde ſi raccoglie, che tal palla non arriua à 35 libre di peſo, <lb/>mà è circa 34 {1/2}. </s> <s xml:id="echoid-s2385" xml:space="preserve">E cio ſi conferma, ſe delli due diametri 21, <lb/>e 20 ſi prendano i cubi 9261, & </s> <s xml:id="echoid-s2386" xml:space="preserve">8000: </s> <s xml:id="echoid-s2387" xml:space="preserve">& </s> <s xml:id="echoid-s2388" xml:space="preserve">eſſendo il primo <lb/>libre 40, ſi faccia come 9261 à 8000, così libre 40 à libre <lb/>34 {5/9}, & </s> <s xml:id="echoid-s2389" xml:space="preserve">in queſta maniera, ſe la portata del pezzo foſſe di <lb/>libre 50, dato il vento alla palla, con leuare al ſuo diametro <lb/>{1/31}, ſaria la palla ſolo di libre 43 {1/5} poco meno.</s> <s xml:id="echoid-s2390" xml:space="preserve"/> </p> <pb o="127" file="0141" n="143" rhead="Linea Cubica"/> <p> <s xml:id="echoid-s2391" xml:space="preserve">La ſeconda maniera è tale; </s> <s xml:id="echoid-s2392" xml:space="preserve">il cir-<lb/> <anchor type="figure" xlink:label="fig-0141-01a" xlink:href="fig-0141-01"/> colo CDAB ſia la bocca del pezzo, <lb/>e dal punto A s’applichi il ſemidia-<lb/>metro in AB, & </s> <s xml:id="echoid-s2393" xml:space="preserve">AD: </s> <s xml:id="echoid-s2394" xml:space="preserve">e preſo l’inter-<lb/>uallo DB, dal punto A ſi tagli il dia-<lb/>metro AC nel punto E; </s> <s xml:id="echoid-s2395" xml:space="preserve">& </s> <s xml:id="echoid-s2396" xml:space="preserve">del reſtan-<lb/>te EC ſi laſci vn terzo IC; </s> <s xml:id="echoid-s2397" xml:space="preserve">& </s> <s xml:id="echoid-s2398" xml:space="preserve">IA ſarà <lb/>il diametro della palla, à cui s’è dato <lb/>il vento. </s> <s xml:id="echoid-s2399" xml:space="preserve">Per ſaper dunque quanto meno peſi della giuſta <lb/>portata del pezzo, s’applichi nella linea cubica il diametro <lb/>AC al numero del peſo, che denomina il pezzo, per eſſempio <lb/>da 40, all’interuallo 40. </s> <s xml:id="echoid-s2400" xml:space="preserve">40; </s> <s xml:id="echoid-s2401" xml:space="preserve">e poi il numero dell’interuallo, in <lb/>cui cade il diametro AI manifeſtarà il peſo vero della palla <lb/>35. </s> <s xml:id="echoid-s2402" xml:space="preserve">Equeſto ſi confermarà, ſe preſo il diametro AC, come <lb/>200, trouerò tanto nella linea Aritmetica dello ſtromento, <lb/>quanto nelle Tauole Trigonometriche, che BD corda digr. <lb/></s> <s xml:id="echoid-s2403" xml:space="preserve">120, cioè AE è 173, e per conſeguenza EC 27, la cui terza <lb/>parte 9 è CI; </s> <s xml:id="echoid-s2404" xml:space="preserve">e perciò IE 18 aggiunta alla EA 173 da tutto il <lb/>diametro della palla AI 191, & </s> <s xml:id="echoid-s2405" xml:space="preserve">AC è 200; </s> <s xml:id="echoid-s2406" xml:space="preserve">i quali numeri <lb/>nella tauoletta poſta in queſto Capo ſono radici delli cubi 7, <lb/>& </s> <s xml:id="echoid-s2407" xml:space="preserve">8: </s> <s xml:id="echoid-s2408" xml:space="preserve">e così ſe 8 dà libre 40, 7 ne darà 35. </s> <s xml:id="echoid-s2409" xml:space="preserve">Come pure con <lb/>queſto metodo, ſe l’anima del pezzo foſſe capace di palla di <lb/>libre 50, datogli il vento, ſi trouerà, che ſarà ſolo di libre <lb/>43 {3/4}.</s> <s xml:id="echoid-s2410" xml:space="preserve"/> </p> <div xml:id="echoid-div73" type="float" level="2" n="1"> <figure xlink:label="fig-0141-01" xlink:href="fig-0141-01a"> <image file="0141-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0141-01"/> </figure> </div> <p> <s xml:id="echoid-s2411" xml:space="preserve">Dalle coſe dette ſi caua, come ſi poſſa <lb/> <anchor type="figure" xlink:label="fig-0141-02a" xlink:href="fig-0141-02"/> anche venir’in cognitione della ſolidità de’ <lb/>corpi vuoti, quando la vacuità di dentro è <lb/>capace d’vn corpo ſolido ſimile à quello <lb/>di tutto il vaſo ſe foſſe pieno. </s> <s xml:id="echoid-s2412" xml:space="preserve">Come nella <lb/>figura 20, ſe ſia dato il vaſo BEV, la cui vacuità ſi riem pireb- <pb o="128" file="0142" n="144" rhead="CAPO IV."/> be con vn corpo ſimile, e ſia la ſua bocca OI, in maniera che, <lb/>come DE ad EV, così OS ad SI, e come ED à DB, così SO ad <lb/>OT profondità della capacità del vaſo. </s> <s xml:id="echoid-s2413" xml:space="preserve">Applico il lato DE <lb/>all’interuallo 18. </s> <s xml:id="echoid-s2414" xml:space="preserve">18, e preſo col Compaſſo il lato OS, trouo, <lb/>che cade nell’interuallo 9.</s> <s xml:id="echoid-s2415" xml:space="preserve">9, onde argomento, che la ſolidità <lb/>del vaſo è tanta, quanta è la capacità ſua.</s> <s xml:id="echoid-s2416" xml:space="preserve"/> </p> <div xml:id="echoid-div74" type="float" level="2" n="2"> <figure xlink:label="fig-0141-02" xlink:href="fig-0141-02a"> <image file="0141-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0141-02"/> </figure> </div> </div> <div xml:id="echoid-div76" type="section" level="1" n="43"> <head xml:id="echoid-head74" xml:space="preserve">QV ESTIONE QVINTA.</head> <head xml:id="echoid-head75" xml:space="preserve">Come ſi poſſa far’vn Cono vguale ad vn Cilindro dato, e che <lb/>habbiano li diametri delle baſi, e gl’ Aſsi proportionali.</head> <p> <s xml:id="echoid-s2417" xml:space="preserve">OGni Cono paragonato con vn Cilindro, che habbia la <lb/>baſe, e l’aſſe, vguale alla baſe, & </s> <s xml:id="echoid-s2418" xml:space="preserve">all’ aſſe del Cono, è <lb/>la terza parte del Cilindro, per la 10 del lib. </s> <s xml:id="echoid-s2419" xml:space="preserve">12, e perciò da-<lb/>to il Cilindro, baſterà trouar’il diametro della baſe, e l’aſſe <lb/>d’vn ſimile Cilindro, che foſſe tre volte maggiore, perche il <lb/>Cono, che haurà queſto diametro della baſe, e queſto aſſe, <lb/>eſſendo la terza parte di queſto Cilindro triplo del primo, <lb/>ſarà vguale al primo Cilindro. </s> <s xml:id="echoid-s2420" xml:space="preserve">Ora perche li Cilindri ſimili <lb/>ſono nella triplicata proportione <lb/> <anchor type="figure" xlink:label="fig-0142-01a" xlink:href="fig-0142-01"/> delli diametri delle baſi, per la 12 <lb/>del lib. </s> <s xml:id="echoid-s2421" xml:space="preserve">12, cioè come i cubi di detti <lb/>diametri; </s> <s xml:id="echoid-s2422" xml:space="preserve">perciò applicato il dia-<lb/>metro del Cilindro dato AB à qual <lb/>ſi voglia numero della linea cubica, <lb/>come per eſſempio all’interuallo 6. <lb/></s> <s xml:id="echoid-s2423" xml:space="preserve">6@@ rendaſi il numero triplo (poiche <lb/>il Cilindro da farſi deue eſſer triplo) <lb/>e l’interuallo 18. </s> <s xml:id="echoid-s2424" xml:space="preserve">18, darà la linea <pb o="129" file="0143" n="145" rhead="Linea Cubica"/> <anchor type="figure" xlink:label="fig-0143-01a" xlink:href="fig-0143-01"/> EF diametro della baſe il cui centro è G. <lb/></s> <s xml:id="echoid-s2425" xml:space="preserve">Dipoi all’iſteſſo interuallo 6. </s> <s xml:id="echoid-s2426" xml:space="preserve">6, applica-<lb/>to l’aſſe CD del Cilindro dato, l’inter-<lb/>uallo 18. </s> <s xml:id="echoid-s2427" xml:space="preserve">18, darà l’aſſe GH; </s> <s xml:id="echoid-s2428" xml:space="preserve">e perciò il <lb/>Cilindro EIF è ſimile al Cilindro ADB, <lb/>eſſendo come AB ad EF diametri, co-<lb/>sì CD à GH aſſi; </s> <s xml:id="echoid-s2429" xml:space="preserve">& </s> <s xml:id="echoid-s2430" xml:space="preserve">eſſendo il cubo di EF triplo del cubo di <lb/>AB, per la conſtruttione dello ſtromento, anche il Cilindro <lb/>EIF è triplo del Cilindro dato ADB: </s> <s xml:id="echoid-s2431" xml:space="preserve">Dunque eſſendo il Ci-<lb/>lindro EIF triplo an che del Cono EHF ſopra la ſteſſa baſe <lb/>GEF, con la ſteſſa altezza GH ſarà il Cono EHF vguale al Ci-<lb/>lindro dato ADB, & </s> <s xml:id="echoid-s2432" xml:space="preserve">hauranno li diametri delle baſi, e gl’aſſi <lb/>proportionali, come s’era propoſto.</s> <s xml:id="echoid-s2433" xml:space="preserve"/> </p> <div xml:id="echoid-div76" type="float" level="2" n="1"> <figure xlink:label="fig-0142-01" xlink:href="fig-0142-01a"> <image file="0142-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0142-01"/> </figure> <figure xlink:label="fig-0143-01" xlink:href="fig-0143-01a"> <image file="0143-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0143-01"/> </figure> </div> <pb o="130" file="0144" n="146" rhead="CAPO IV."/> </div> <div xml:id="echoid-div78" type="section" level="1" n="44"> <head xml:id="echoid-head76" xml:space="preserve">QVESTIONE SESTA.</head> <head xml:id="echoid-head77" xml:space="preserve">Come ſi troui vna Sfera vguale ad vn Cilindro dato.</head> <p> <s xml:id="echoid-s2434" xml:space="preserve">SE foſſe data vna gran Colonna, e ſi voleſſe ſapere, quan-<lb/>to, ò quale douria eſſer’ il diametro d’vna sfera vguale <lb/>alla colonna (la quale ſuppongo eſſer’ vn cilindro retto, cioè, <lb/>che l’aſſe cade perpendicolare nella baſe; </s> <s xml:id="echoid-s2435" xml:space="preserve">ſe nò, facilmente ſi <lb/>ridurrà ad vn cilindro retto, che habbia l’iſteſſa baſe, e l’iſteſ-<lb/>ſa altezza perpendicolare, che ſia aſſe, come ſi raccoglie dal <lb/>Corollario della 11 del lib. </s> <s xml:id="echoid-s2436" xml:space="preserve">12) prendaſi il diametro della ba-<lb/>ſe, e l’altezza di tal cilindro; </s> <s xml:id="echoid-s2437" xml:space="preserve">ſi troui la lor proportione in <lb/>numeri, per la queſt. </s> <s xml:id="echoid-s2438" xml:space="preserve">5. </s> <s xml:id="echoid-s2439" xml:space="preserve">del cap. </s> <s xml:id="echoid-s2440" xml:space="preserve">2. </s> <s xml:id="echoid-s2441" xml:space="preserve">e nella linea cubica dello <lb/>ſtromento applicato il diametro all’ interuallo del numero, <lb/>che gli corriſponde, ſi prenda l’interuallo, che dà l’altro nu-<lb/>mero corriſpondente all’aſſe. </s> <s xml:id="echoid-s2442" xml:space="preserve">Queſta diſtanza trouata s’ap-<lb/>plichi nello ſtromento all’ interuallo 2. </s> <s xml:id="echoid-s2443" xml:space="preserve">2, poiche l’interuallo <lb/>3. </s> <s xml:id="echoid-s2444" xml:space="preserve">3 darà il diametro cercato della sfera vguale al cilindro. <lb/></s> <s xml:id="echoid-s2445" xml:space="preserve">E ſe gl’interualli 2. </s> <s xml:id="echoid-s2446" xml:space="preserve">2, e 3. </s> <s xml:id="echoid-s2447" xml:space="preserve">3 foſſero troppo piccolli, ſi pren-<lb/>dano li loro equemoltiplici in qualunque proportione. </s> <s xml:id="echoid-s2448" xml:space="preserve">Sia <lb/>nell’iſteſſa fig. </s> <s xml:id="echoid-s2449" xml:space="preserve">21 dato il cilindro EIF, à cui ſi voglia far’vna <lb/>sfera vguale; </s> <s xml:id="echoid-s2450" xml:space="preserve">ſi troua, che il diametro della baſe EF all’ aſſe <lb/>GH è come 91 à 200, cioè come 5 à 11, nella linea cubica <lb/>applico EF all’interuallo 5. </s> <s xml:id="echoid-s2451" xml:space="preserve">5, e l’interuallo 11. </s> <s xml:id="echoid-s2452" xml:space="preserve">11 mi dà la <lb/>linea R. </s> <s xml:id="echoid-s2453" xml:space="preserve">Applico la linea R all’interuallo 2. </s> <s xml:id="echoid-s2454" xml:space="preserve">2, e l’interuallo <lb/>3. </s> <s xml:id="echoid-s2455" xml:space="preserve">3 mi dà la linea S diametro della sfera MN vguale al dato <lb/>cilindro EIF.</s> <s xml:id="echoid-s2456" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2457" xml:space="preserve">Per dimoſtrare, che ciò ſia, prendaſi la linea R diametro, <lb/>& </s> <s xml:id="echoid-s2458" xml:space="preserve">aſſe del cilindro quadroto KPXL, & </s> <s xml:id="echoid-s2459" xml:space="preserve">in queſto cilindro s’in- <pb o="131" file="0145" n="147" rhead="Linea Cubica"/> tenda la sfera, il cui centro Q, e così il diametro della baſe <lb/>del cilindro KL, come l’altezza KP ſia vguale al diametro <lb/>della sfera. </s> <s xml:id="echoid-s2460" xml:space="preserve">Ora perche li cubi di EF, e di R ſono come 5, e <lb/>11, per la co ſtruttione dello ſtromento, la proportione di 5 <lb/>à 11, cioè di EF à GH, è triplicata della proportione de’lati, <lb/>cioè di EF à R; </s> <s xml:id="echoid-s2461" xml:space="preserve">dunque R è la ſeconda di quattro continuata-<lb/>mente proportionali, delle qualli EF è la prima, e GH la quar-<lb/>ta; </s> <s xml:id="echoid-s2462" xml:space="preserve">e ſia V la terza. </s> <s xml:id="echoid-s2463" xml:space="preserve">Dunque perche le baſi de’ cilindri EIF, <lb/>KPL ſono nella proportione duplicata de’ diameri EF, KL, <lb/>cioè R, le baſi di detti cilindri ſono come EF prima alla V <lb/>terza. </s> <s xml:id="echoid-s2464" xml:space="preserve">Mà come EF à V, così R à GH; </s> <s xml:id="echoid-s2465" xml:space="preserve">dunque come la ba-<lb/>ſe, il cui diametro EF, alla baſe, il cui diametro KL, così l’al-<lb/>tezza PK per la coſtruttione vguale alla linea R, all’altezza <lb/>GH. </s> <s xml:id="echoid-s2466" xml:space="preserve">Dunque, per la 15 del lib. </s> <s xml:id="echoid-s2467" xml:space="preserve">12, reciprocandoſi le baſi, e <lb/>l’altezze, i due cilindri EIF, KPL ſono vguali. </s> <s xml:id="echoid-s2468" xml:space="preserve">Dunque la <lb/>sfera QZOY, il cui diametro è la linea R vguale all’altezza <lb/>del cilindro, & </s> <s xml:id="echoid-s2469" xml:space="preserve">il cui circolo maſſime è vgualle alla baſe di det <lb/>to cilindro, è ſubſeſquialtera al cilindro, cioè come 2 à 3, per <lb/>il Manifeſto 9 del lib. </s> <s xml:id="echoid-s2470" xml:space="preserve">1. </s> <s xml:id="echoid-s2471" xml:space="preserve">de Sphæra; </s> <s xml:id="echoid-s2472" xml:space="preserve">& </s> <s xml:id="echoid-s2473" xml:space="preserve">Cylindro d’Archime-<lb/>de. </s> <s xml:id="echoid-s2474" xml:space="preserve">Dunque eſſendoſi preſa la linea R lato del cubo 2, e la <lb/>linea S lato del cubo 3, la sfera MN, il cui diametro è la li-<lb/>nea S è ſeſquialtera della sfera QZOY, il cui diametro è la li-<lb/>nea R. </s> <s xml:id="echoid-s2475" xml:space="preserve">Dunque così la sfera MN, come il cilindro KPL eſ-<lb/>ſendo ſeſquialteri della ſteſſa sfera Q Z O Y, ſono vguali; <lb/></s> <s xml:id="echoid-s2476" xml:space="preserve">dunque anche la sſera MN è vguale al dato cilindro EIF.</s> <s xml:id="echoid-s2477" xml:space="preserve"/> </p> <pb o="132" file="0146" n="148" rhead="CAPO IV."/> </div> <div xml:id="echoid-div79" type="section" level="1" n="45"> <head xml:id="echoid-head78" xml:space="preserve">QVESTIONE SETTIMA.</head> <head xml:id="echoid-head79" xml:space="preserve">Data vna Parabola, trouare la proportione di due ſegmenti <lb/>terminati ad vn medeſimo punto.</head> <p> <s xml:id="echoid-s2478" xml:space="preserve">SIa data la Parabola ABC, & </s> <s xml:id="echoid-s2479" xml:space="preserve">in eſſa due ſegmenti AFB, <lb/>e BC terminati nello ſteſl<unsure/>o punto B. </s> <s xml:id="echoid-s2480" xml:space="preserve">Si cerca la propor-<lb/>tione di queſti due ſegmenti. </s> <s xml:id="echoid-s2481" xml:space="preserve">Tiriſi <lb/> <anchor type="figure" xlink:label="fig-0146-01a" xlink:href="fig-0146-01"/> il Diametro BD: </s> <s xml:id="echoid-s2482" xml:space="preserve">il che ſi farà, ſe con-<lb/>gionte le eſtremità de’ ſegmenti con <lb/>la retta AC, à queſta dal punto B ſi <lb/>tirarà parallela la BG; </s> <s xml:id="echoid-s2483" xml:space="preserve">e così I’vna <lb/>come I’altra parallela diuiſe per <lb/>mezzo in H & </s> <s xml:id="echoid-s2484" xml:space="preserve">I; </s> <s xml:id="echoid-s2485" xml:space="preserve">la retta HI prodot-<lb/>ta ſin in F ſarà il diametro, à cui ſono Applicate HE, IB. <lb/></s> <s xml:id="echoid-s2486" xml:space="preserve">Dunque ſia BD parallela alla FH, e ſarà diametro, eſſendo <lb/>che nella Parabola tuttii diametri ſon paralleli all’ Aſſe. </s> <s xml:id="echoid-s2487" xml:space="preserve">Sì <lb/>che il diametro BD taglia la AC in E.</s> <s xml:id="echoid-s2488" xml:space="preserve"/> </p> <div xml:id="echoid-div79" type="float" level="2" n="1"> <figure xlink:label="fig-0146-01" xlink:href="fig-0146-01a"> <image file="0146-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0146-01"/> </figure> </div> <p> <s xml:id="echoid-s2489" xml:space="preserve">Ora perche li ſegmenti AFB, e BC hanno tra di loro la tri-<lb/>plicata proportione della linea AE all’EC, come dimoſtra <lb/>Gregorio di S. </s> <s xml:id="echoid-s2490" xml:space="preserve">Vicenzo lib. </s> <s xml:id="echoid-s2491" xml:space="preserve">5. </s> <s xml:id="echoid-s2492" xml:space="preserve">Quadr. </s> <s xml:id="echoid-s2493" xml:space="preserve">circ. </s> <s xml:id="echoid-s2494" xml:space="preserve">prop. </s> <s xml:id="echoid-s2495" xml:space="preserve">260; </s> <s xml:id="echoid-s2496" xml:space="preserve">mettaſi <lb/>la linea AE in qualſiuoglia interuallo della linea Cubica; </s> <s xml:id="echoid-s2497" xml:space="preserve">e <lb/>quell’interuallo, doue capirà la linea EC col numero oppo-<lb/>ſto dimoſtrarà la proportione delli due ſegmenti: </s> <s xml:id="echoid-s2498" xml:space="preserve">poiche eſ-<lb/>ſendo triplicata della proportione di AE ad EC, ſarà la me-<lb/>deſima delli Cubi di dette linee AE, EC.</s> <s xml:id="echoid-s2499" xml:space="preserve"/> </p> <pb o="133" file="0147" n="149" rhead="Linea Cubica"/> </div> <div xml:id="echoid-div81" type="section" level="1" n="46"> <head xml:id="echoid-head80" xml:space="preserve">QVESTIONE OTTAVA.</head> <head xml:id="echoid-head81" style="it" xml:space="preserve">Data vna Parabola terminata, tagliata da vna linea parallela, <lb/>trouar la proportione delle parti, nelle qualli è diuiſa.</head> <p> <s xml:id="echoid-s2500" xml:space="preserve">SIa data la Parabola DBE terminata dalla linea DE; </s> <s xml:id="echoid-s2501" xml:space="preserve">& </s> <s xml:id="echoid-s2502" xml:space="preserve"><lb/>à queſta ſia parallela la linea AC. </s> <s xml:id="echoid-s2503" xml:space="preserve">Si cerca la propor-<lb/>tione del ſegmento ABC al re-<lb/> <anchor type="figure" xlink:label="fig-0147-01a" xlink:href="fig-0147-01"/> ſtante DACE. </s> <s xml:id="echoid-s2504" xml:space="preserve">Diuiſe le due pa-<lb/>rallele in mezzo in F, c G, ſia ti-<lb/>rata la BG diametro della Para-<lb/>bola. </s> <s xml:id="echoid-s2505" xml:space="preserve">Ora perche le line BF, BG <lb/>ſono nella duplicata proportione <lb/>di AF à DG (eſſendo tra di loro <lb/>come ſi quadrati delle ordinatamente Applicate, alli quali <lb/>ſon vguali i Rettãgoli da eſſe ſaette & </s> <s xml:id="echoid-s2506" xml:space="preserve">illato Retto) cioè di tut-<lb/>te le intiere AC, DE; </s> <s xml:id="echoid-s2507" xml:space="preserve">la proportione del Triangoſo ABC, <lb/>al Triangolo DBE è compoſta della proportione delle baſi <lb/>AC, DE, e dell’ altezze BF, e BG, cioè è triplicara di AC <lb/>à DE.</s> <s xml:id="echoid-s2508" xml:space="preserve"/> </p> <div xml:id="echoid-div81" type="float" level="2" n="1"> <figure xlink:label="fig-0147-01" xlink:href="fig-0147-01a"> <image file="0147-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0147-01"/> </figure> </div> <p> <s xml:id="echoid-s2509" xml:space="preserve">Mà perche la Parabola ABC alla Parabola DBE è nella <lb/>proportione del ſuo Triangolo maſſimo ABC aſ Triangolo <lb/>maſſimo DBE; </s> <s xml:id="echoid-s2510" xml:space="preserve">dunque la Parabola ABC alla Parabola DBE <lb/>è nella triplicata proportione della linea AC alla linea DE. <lb/></s> <s xml:id="echoid-s2511" xml:space="preserve">Mettaſi dunque nella linea Cubica dello ſtromento à qualſi-<lb/>uoglia interuallo ſa linea DE, etrouiſi doue capiſca l’inter-<lb/>uallo AC, che ſarà manifeſta la proportione delle due Para-<lb/>bole: </s> <s xml:id="echoid-s2512" xml:space="preserve">e preſa la differenza trà di loro, ſarà manifeſta la pro-<lb/>portione delſegmento ABC al reſtante DACE.</s> <s xml:id="echoid-s2513" xml:space="preserve"/> </p> <pb o="134" file="0148" n="150" rhead="CAPO IV."/> </div> <div xml:id="echoid-div83" type="section" level="1" n="47"> <head xml:id="echoid-head82" xml:space="preserve">QVESTIONE NONA.</head> <head xml:id="echoid-head83" xml:space="preserve">Come d’vn numero dato ſi troui la Radice Cubica.</head> <p> <s xml:id="echoid-s2514" xml:space="preserve">APerto lo Stromento; </s> <s xml:id="echoid-s2515" xml:space="preserve">gl’interualli de’numeri nelle linee <lb/>cubiche danno i lati de’cubi, i qualli hanno tra diloro <lb/>la proportione eſpreſſa dalli numeri adiacenti. </s> <s xml:id="echoid-s2516" xml:space="preserve">Dunque ſe <lb/>detti lati s’applicheranno ad interualli delle linee Aritmeti-<lb/>che, ſi conoſcerà la proportione di detti lati; </s> <s xml:id="echoid-s2517" xml:space="preserve">la qual’è la ſub-<lb/>triplicata della proportione de’Cubi. </s> <s xml:id="echoid-s2518" xml:space="preserve">Dunque conoſciuta la <lb/>proportione di due cubi, & </s> <s xml:id="echoid-s2519" xml:space="preserve">il lato d’vno di eſſi, ſi conoſcerà <lb/>anche l’altro. </s> <s xml:id="echoid-s2520" xml:space="preserve">Quindi è, che applicato vn cubo ad vn nu-<lb/>mero delle linee cubiche, e preſo il lato d’vn’altro cubo co-<lb/>noſciuto nella ſua radice, & </s> <s xml:id="echoid-s2521" xml:space="preserve">applicata queſta all’interuallo <lb/>corriſpondente nelle linee Aritmetiche, l’altro lato del cubo <lb/>dato ſi conoſcerà, eſſendo applicato all’interuallo proportio-<lb/>nato delle linee ſteſſe Aritmetiche. </s> <s xml:id="echoid-s2522" xml:space="preserve">Perciò dato vn numero <lb/>preſo come cubo; </s> <s xml:id="echoid-s2523" xml:space="preserve">& </s> <s xml:id="echoid-s2524" xml:space="preserve">applicato alle linee cubiche (nel modo <lb/>proportionatamente, che ſi diſſe dell’ eſtrattione della radice <lb/>quadrata con le linee Geometriche) quelche reſta tagliate <lb/>via le tre vltime figure, e preſo l’interuallo d’vno de’numeri <lb/>cubi ſegnati nelle linee, cioè 8, ouero 27, radice de’ quali ſo-<lb/>no 2, e 3, e queſto poinelle ſinee Aritmetiche applicato al 20. <lb/></s> <s xml:id="echoid-s2525" xml:space="preserve">20, ouero al 30. </s> <s xml:id="echoid-s2526" xml:space="preserve">30, l’altro interuallo applicato alla ſteſſa li-<lb/>nea, darà la radice cubica cercata. </s> <s xml:id="echoid-s2527" xml:space="preserve">E la ragione, perche ſi<unsure/> <lb/>buttino via le tre vltime figure, è perche li cubi di 20, e di <lb/>30 ſono 8000, e 27000, e così gettate via le tre vltime figure, <lb/>reſta la proportione de’cubi eſpreſſa in numeri minori, che <lb/>ſono ſegnati nelle linee dello Stromento: </s> <s xml:id="echoid-s2528" xml:space="preserve">& </s> <s xml:id="echoid-s2529" xml:space="preserve">applicati poi <pb o="135" file="0149" n="151" rhead="Linea Cubica."/> gl’interualli alli 20, ouero 30, & </s> <s xml:id="echoid-s2530" xml:space="preserve">à numeri corriſpondenti, <lb/>vengono le radici cercate.</s> <s xml:id="echoid-s2531" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2532" xml:space="preserve">Cerchiſi la radice cubica del numero 14119; </s> <s xml:id="echoid-s2533" xml:space="preserve">gettate via <lb/>le tre ſigure 119, il reſto 14 applico all’interuallo 14. </s> <s xml:id="echoid-s2534" xml:space="preserve">14 del-<lb/>le linee cubiche: </s> <s xml:id="echoid-s2535" xml:space="preserve">poicon vn’altro Compaſſo prendo l’inter-<lb/>uallo 8. </s> <s xml:id="echoid-s2536" xml:space="preserve">8 nella ſteſſa apertura dello Stromento. </s> <s xml:id="echoid-s2537" xml:space="preserve">Poinelle li-<lb/>nee Aritmetiche applico queſto ſecondo interuallo preſo alli <lb/>punti 20<unsure/>. </s> <s xml:id="echoid-s2538" xml:space="preserve">20, che è la radice di 8000, e vedendo, che il primo <lb/>interuallo preſo applicato à queſte ſteſſe linee Aritmetiche <lb/>cade al 24. </s> <s xml:id="echoid-s2539" xml:space="preserve">24, e vn poco più; </s> <s xml:id="echoid-s2540" xml:space="preserve">dico, che la radice cubica del <lb/>dato numero 14119 è 24 con vna frattione aderente. </s> <s xml:id="echoid-s2541" xml:space="preserve">Che <lb/>ſe le tre vltime figure tagliate paſſano li 500, ſi può accreſcer <lb/>d’vn’vnità il numero, che reſta, poiche più s’accoſta al mille. <lb/></s> <s xml:id="echoid-s2542" xml:space="preserve">Così cercandoſi la radice di 19864, ſi può in vece del 19 <lb/>prendere il 20, & </s> <s xml:id="echoid-s2543" xml:space="preserve">operando come prima, ſi troua eſſer la ſua <lb/>radice 27, e poco più.</s> <s xml:id="echoid-s2544" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2545" xml:space="preserve">Mà ſe il numero reſtante foſſe maggiore del maſſimo no-<lb/>tato nelle linee cubiche, prendaſi vna parte aliquota tale, che <lb/>nelle linee cubiche ſiano due numeri così moltiplici l’vno <lb/>dell’altro, come il tutto è moltiplice della detta parte aliquo-<lb/>ta: </s> <s xml:id="echoid-s2546" xml:space="preserve">come ſe ſi prende la ſeſta parte, viſia vn numero ſeſtuplo <lb/>d’vn’altro. </s> <s xml:id="echoid-s2547" xml:space="preserve">Et in tali occaſioni è bene nel principio prendere <lb/>piccola apertura dello Stromento, per poter poi applicar <lb/>quell’interuallo preſo à numeri minori, come moſtrerà l’iſpe-<lb/>rienza. </s> <s xml:id="echoid-s2548" xml:space="preserve">Cerchiſi la radice cubica di 336212: </s> <s xml:id="echoid-s2549" xml:space="preserve">tagliate le tre <lb/>vltime figure, reſta 336, il qual’è troppo grande; </s> <s xml:id="echoid-s2550" xml:space="preserve">piglio <lb/>dunque la ſettima patte di 336, cioè 48, & </s> <s xml:id="echoid-s2551" xml:space="preserve">aperto lo Stro-<lb/>mento, prendo nelle linee cubiche l’interuallo 48. </s> <s xml:id="echoid-s2552" xml:space="preserve">48, e con <lb/>vn’altro Compaſſo l’interuallo 8, 8. </s> <s xml:id="echoid-s2553" xml:space="preserve">Mà perche il lato preſo <lb/>di 48 è ſolo il lato d’vn cubo ſubſettuplo del cubo dato, per- <pb o="136" file="0150" n="152" rhead="CATO IV."/> ciò cerco nella linea cubica due numeri, vno de’ qualiſia ſet-<lb/>tuplo dell’altro, eſono 5, e 35, perciò quell’ interuallo preſo <lb/>48. </s> <s xml:id="echoid-s2554" xml:space="preserve">48, allargando lo Stromento, lo metto alli punti 5. </s> <s xml:id="echoid-s2555" xml:space="preserve">5, & </s> <s xml:id="echoid-s2556" xml:space="preserve"><lb/>allhora prendo l’interuallo 35. </s> <s xml:id="echoid-s2557" xml:space="preserve">35, che è quello, che ſi cerca-<lb/>ua. </s> <s xml:id="echoid-s2558" xml:space="preserve">Quindi l’interuallo, che fù preſo tra 8. </s> <s xml:id="echoid-s2559" xml:space="preserve">8, applico nelle <lb/>linee Aritmetiche al 20. </s> <s xml:id="echoid-s2560" xml:space="preserve">20; </s> <s xml:id="echoid-s2561" xml:space="preserve">& </s> <s xml:id="echoid-s2562" xml:space="preserve">in quell’apertura di Stromen. <lb/></s> <s xml:id="echoid-s2563" xml:space="preserve">to trouando, che l’vltimo interuallo s’applica nelle dette linee <lb/>Aritmetiche alli punti 69. </s> <s xml:id="echoid-s2564" xml:space="preserve">69, & </s> <s xml:id="echoid-s2565" xml:space="preserve">vn poco più, dico, che la ra-<lb/>dice del numero 336212 è 69 con vna frattione.</s> <s xml:id="echoid-s2566" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2567" xml:space="preserve">Quando poi l’interuallo vltimo riuſciſſe così grande, che <lb/>foſſe maggiore dell’interuallo 100. </s> <s xml:id="echoid-s2568" xml:space="preserve">100 della linea Aritmeti-<lb/>ca, ſi deſcriue vna linea vguale à tal’ interuallo delle linee Cu-<lb/>biche vltimamente trouato, e cauatone la diſtanza 100. </s> <s xml:id="echoid-s2569" xml:space="preserve">100 <lb/>delle Aritmetiche, s’applica il reſto della linea, e ſi vede quan-<lb/>to di più vada aggiunto al 100. </s> <s xml:id="echoid-s2570" xml:space="preserve">Cerchiſi la radice cubica di <lb/>1840325, gettate le tre vltime figure, diuido il reſto 1840 <lb/>in quaranta parti, e trouo, che la ſua quaranteſima patte è <lb/>46. </s> <s xml:id="echoid-s2571" xml:space="preserve">Apro mediocremente lo Stromento, e prendo col primo <lb/>Compaſſo l’interuallo 46. </s> <s xml:id="echoid-s2572" xml:space="preserve">46, e col ſecondo Compaſſo l’in-<lb/>teruallo 8, 8. </s> <s xml:id="echoid-s2573" xml:space="preserve">Dipoi, perche il cubo 46. </s> <s xml:id="echoid-s2574" xml:space="preserve">46 và moltiplicato <lb/>40 volte, applico quell’interuallo preſo col primo Compaſ-<lb/>ſo all’interuallo 1. </s> <s xml:id="echoid-s2575" xml:space="preserve">1, e poi prendo l’interuallo 40, 40. </s> <s xml:id="echoid-s2576" xml:space="preserve">Et <lb/>operando poi, con hauer’ applicato l’interualo preſo col ſe-<lb/>condo Compaſſo alli punti 20. </s> <s xml:id="echoid-s2577" xml:space="preserve">20 delle linee Aritmetiche, <lb/>trouo, che eccede l’altro Compaſſo la maſſima diſtanza <lb/>100. </s> <s xml:id="echoid-s2578" xml:space="preserve">100: </s> <s xml:id="echoid-s2579" xml:space="preserve">perciò<unsure/> da vna linea deſcritta vguale all’vltimo in-<lb/>teruallo preſo col Compaſſo alli punti 40, 40 delle cubiche, <lb/>cauo l’interuallo 100. </s> <s xml:id="echoid-s2580" xml:space="preserve">100 dell’Aritmetiche, & </s> <s xml:id="echoid-s2581" xml:space="preserve">applico à <lb/>quello il reſto della linea deſcritta, e cadendo alli punti 22, <lb/>dico, che la radice cubica del numero dato 1840325, è 122 <lb/>con qualche frattione.</s> <s xml:id="echoid-s2582" xml:space="preserve"/> </p> <pb o="137" file="0151" n="153" rhead="Linea Cubica."/> <p> <s xml:id="echoid-s2583" xml:space="preserve">Quì pure nelnumero così grande, che due numeri, i quali <lb/>moltiplicati inſieme lo producono, ſono maggiori delli nota-<lb/>ti nella linea cubica dello ſtromento, ſe ne piglino 3, ò anche <lb/>quattro, dalla moltiplicatione de’quali vien prodotto il nu-<lb/>mero, chereſta, leuate le tre vltime figure, nel modo detto, <lb/>quando ſi parlò dell’eſtrattione della radice quadrata. </s> <s xml:id="echoid-s2584" xml:space="preserve">Così <lb/>cercando la radice cubica di 3600000, leuate le tre vltime <lb/>figure, reſta 3600, che ſi fà dal 60 per 60: </s> <s xml:id="echoid-s2585" xml:space="preserve">poſſo dunque <lb/>prendere tre numeri 15. </s> <s xml:id="echoid-s2586" xml:space="preserve">15. </s> <s xml:id="echoid-s2587" xml:space="preserve">16, e preſo l’interuallo 15. </s> <s xml:id="echoid-s2588" xml:space="preserve">15, <lb/>prender poiillato del cubo quindecuplo di queſto, applican-<lb/>do quell’interuallo al 3. </s> <s xml:id="echoid-s2589" xml:space="preserve">3, epoi prendendo i’interuallo 45. <lb/></s> <s xml:id="echoid-s2590" xml:space="preserve">45, & </s> <s xml:id="echoid-s2591" xml:space="preserve">hauuto queſto, s’hà à prender’il lato del cubo ſedecu-<lb/>plo, il che ſi farà applicando queſto ſecondo interuallo tro-<lb/>uato al 3. </s> <s xml:id="echoid-s2592" xml:space="preserve">3, e poi prendendo l’interuallo 48. </s> <s xml:id="echoid-s2593" xml:space="preserve">48, & </s> <s xml:id="echoid-s2594" xml:space="preserve">operan-<lb/>do con queſto nel modo detto, nelle linee Aritmetiche ſi tro-<lb/>ua, che la radice cubica di 3600000, ſarà 153 in circa.</s> <s xml:id="echoid-s2595" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2596" xml:space="preserve">Finalmente per i piccoli numeri s’opera ſenza tagliarne <lb/>alcuna figura; </s> <s xml:id="echoid-s2597" xml:space="preserve">e s’hanno l’intieri con le decime. </s> <s xml:id="echoid-s2598" xml:space="preserve">Cerco la ra-<lb/>dice del numero 47; </s> <s xml:id="echoid-s2599" xml:space="preserve">prendo l’interuallo 47. </s> <s xml:id="echoid-s2600" xml:space="preserve">47, & </s> <s xml:id="echoid-s2601" xml:space="preserve">anche 8. </s> <s xml:id="echoid-s2602" xml:space="preserve">8, <lb/>queſto ſecondo nelle linee Aritmetiche applico al 20. </s> <s xml:id="echoid-s2603" xml:space="preserve">20, e <lb/>l’altro cade nel 36. </s> <s xml:id="echoid-s2604" xml:space="preserve">36, poco più: </s> <s xml:id="echoid-s2605" xml:space="preserve">onde dico, che la radice cu-<lb/>bica di 47 è 3 {6/10}, poco più: </s> <s xml:id="echoid-s2606" xml:space="preserve">perche per radice di 8 douea, <lb/>prenderſi 2, e non 20; </s> <s xml:id="echoid-s2607" xml:space="preserve">dunque hauutiſi i decimi del cubo <lb/>preciſo, vengono li decimi del cubo dato non così preciſo. <lb/></s> <s xml:id="echoid-s2608" xml:space="preserve">Cerco la radice di 180, prendo il quinto 36, e l’interuallo 36. </s> <s xml:id="echoid-s2609" xml:space="preserve"><lb/>36 applico ad vn’altro numero, dicui ſia il quintuplo nelle <lb/>linee cubiche, per eſſempio al 5. </s> <s xml:id="echoid-s2610" xml:space="preserve">5, e poi prendo l’interuallo <lb/>quintuplo 25. </s> <s xml:id="echoid-s2611" xml:space="preserve">25. </s> <s xml:id="echoid-s2612" xml:space="preserve">Poi applicato l’interuallo 8. </s> <s xml:id="echoid-s2613" xml:space="preserve">8, preſo da <lb/>principio al 20. </s> <s xml:id="echoid-s2614" xml:space="preserve">20, delle linee Aritmetiche, trouo, che l vlti-<lb/>mo interuallo cade nelle linee Aritmetiche al 56. </s> <s xml:id="echoid-s2615" xml:space="preserve">56, e qua- <pb o="138" file="0152" n="154" rhead="C A P O IV."/> ſi 57. </s> <s xml:id="echoid-s2616" xml:space="preserve">57. </s> <s xml:id="echoid-s2617" xml:space="preserve">onde conchiudo, che la radice cubica di 180 è <lb/>5 {6/10} in circa.</s> <s xml:id="echoid-s2618" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2619" xml:space="preserve">Che ſe il numero dato non foſſe intiero, ma vn rotto, di cui <lb/>ſi cercaſſe la radice cubica; </s> <s xml:id="echoid-s2620" xml:space="preserve">ſarà ſacile il trouarla; </s> <s xml:id="echoid-s2621" xml:space="preserve">cioè nelle <lb/>linee cubiche applicando all’interuallo corriſpondente al nu-<lb/>mero, che ſi vuol ritenere (ò ſia il Numeratore, ò pure il De-<lb/>nominatore) iſ compaſſo con quell’apertura, che ſi vuole; </s> <s xml:id="echoid-s2622" xml:space="preserve">e <lb/>di poicon altro compaſſo prendendo l’interuallo riſpondente <lb/>all’altro numero della frattione data; </s> <s xml:id="echoid-s2623" xml:space="preserve">poiche nelle linee <lb/>Aritmetiche applicato il primo compaſſo al numero, che ſi <lb/>vuol ritenere della data frattione, ouero ad vn ſuo moltiplice, <lb/>(il che ſarà meglio, per hauer la radice più vicina alla preciſio-<lb/>ne) l’ltro compaſſo moſtrarà il numero cercato. </s> <s xml:id="echoid-s2624" xml:space="preserve">Sia per ca-<lb/>gione d’eſempio dato il roto {4/7}, di cui ſi vuole la radice cubi-<lb/>ca: </s> <s xml:id="echoid-s2625" xml:space="preserve">prendo nelle cubiche l’interuallo 4. </s> <s xml:id="echoid-s2626" xml:space="preserve">4. </s> <s xml:id="echoid-s2627" xml:space="preserve">(poiche voglio ri-<lb/>tener il Numeratore) e con altro compaſſo l’interuallo 7. </s> <s xml:id="echoid-s2628" xml:space="preserve">7. <lb/></s> <s xml:id="echoid-s2629" xml:space="preserve">Quindi applico il primo compaſſo nelle linee Aritmetiche al <lb/>decuplo di 4, cioè al 40, & </s> <s xml:id="echoid-s2630" xml:space="preserve">il ſecondo compaſſo caderà all’in-<lb/>teruallo 48. </s> <s xml:id="echoid-s2631" xml:space="preserve">48, poco più: </s> <s xml:id="echoid-s2632" xml:space="preserve">onde la radice ſarà proſſimamen-<lb/>te {40/48}, cioè proſſimamente {5/6}, il cui cubo {125/216} è poco maggiore <lb/>del cubo dato {4/7}. </s> <s xml:id="echoid-s2633" xml:space="preserve">Che ſe nelle linee cubiche prendo col primo <lb/>compaſſo l’interuallo 7. </s> <s xml:id="echoid-s2634" xml:space="preserve">7, e col ſecondo 4. </s> <s xml:id="echoid-s2635" xml:space="preserve">4, nelle Aritmeti-<lb/>che applico il primo compaſſo al 70. </s> <s xml:id="echoid-s2636" xml:space="preserve">70, & </s> <s xml:id="echoid-s2637" xml:space="preserve">il ſecondo cade <lb/>all’interuallo 58.</s> <s xml:id="echoid-s2638" xml:space="preserve">58. </s> <s xml:id="echoid-s2639" xml:space="preserve">onde la radice è proſſimamente {58/70}, cioè <lb/>{29/35}; </s> <s xml:id="echoid-s2640" xml:space="preserve">il cui cubo {24389/42875} è poco minore del cubo dato {4/7}. </s> <s xml:id="echoid-s2641" xml:space="preserve">La ragio-<lb/>ne di queſto modo di operare è manifeſta, perche cercando ſi <lb/>laradice cubica ad vn numero rotto, ſi cerca vna frattione, il <lb/>cui Numeratore al ſuo Denominatore habbia la propottio-<lb/>ne ſubtriplicata del Numeratore al Denominatore della data <lb/>frattione. </s> <s xml:id="echoid-s2642" xml:space="preserve">Ora per la conſtruttione dello ſtromento ſi hanno <pb o="139" file="0153" n="155" rhead="Linea Cubica"/> i lati de’cubi, che ſono nella ſubrriplicata proportione de gli <lb/>ſteſſi cubi; </s> <s xml:id="echoid-s2643" xml:space="preserve">dunque prendendo come cubi il Numeratore, & </s> <s xml:id="echoid-s2644" xml:space="preserve"><lb/>il Deno minatore, gl’interualli, che alli loro numeri corriſpon-<lb/>dono, ſono nella ſubtriplicata proportione; </s> <s xml:id="echoid-s2645" xml:space="preserve">e perciò eſami-<lb/>nata la loro quantità nelle linee Aritmetiche, ſi hauranno due <lb/>numeri nelſa ſubtriplicata proportione, come ſi cerca. </s> <s xml:id="echoid-s2646" xml:space="preserve">Per-<lb/>ciò à fine di cauare la ſudetta radice Cubica ſenza lo ſtromen-<lb/>to, baſtarà moltiplicar il quadrato del Numeratore 4, cioè <lb/>16, per il Denominatore 7, e dal prodotto cauata la radice <lb/>cubica ſarà la prima delle due medie proportionali tra 4, e 7, <lb/>e perciò Denominatore ſotto il Numeratore 4. </s> <s xml:id="echoid-s2647" xml:space="preserve">Ouero il <lb/>quadrato del Denominatore 7, cioè 49, ſi moltiplicarà per il <lb/>Numeratore 4, e dal prodotto la radice cubica ſarà la ſecon-<lb/>da delle due medie tra 4, e 7, e perciò Numeratore, a cui per <lb/>Denominatore ſi dà il 7.</s> <s xml:id="echoid-s2648" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2649" xml:space="preserve">In queſto luogo, come per aggiunta, mi perſuado non ſia <lb/>per eſſer diſcaro al mio Lettore, ſe proporrò vna maniera, <lb/>aſſai facile per trouar la radice cubica de’ numeri, almeno <lb/>molto vicina alla preciſione, della quale non ſi curano più <lb/>che tanto quelli, che cercano tali compendij, diſſi vicina alla <lb/>preciſione, non perche non ſi poſſa hauere la radice preciſa, <lb/>quando ella c’è, ma perche in alcuni numeri grandi, come ap-<lb/>preſſo ſi vedrà, non ſempre s’affronterà.</s> <s xml:id="echoid-s2650" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2651" xml:space="preserve">Per li numeri, che non ſiano maggiori di ſei figure, e per-<lb/>ciò la radice non è che di due figure, ſeruirà con ogni preciſio-<lb/>ne la ſeguente tauoletta, in cui nel capo di ciaſcun’ordine, <lb/>dou’è C 2. </s> <s xml:id="echoid-s2652" xml:space="preserve">C 3. </s> <s xml:id="echoid-s2653" xml:space="preserve">&</s> <s xml:id="echoid-s2654" xml:space="preserve">c. </s> <s xml:id="echoid-s2655" xml:space="preserve">ſi moſtra che, quando la prima nota del-<lb/>la radice è 2, ouero 3, ò qualunque altro numero, tutto quel-<lb/>lo, che ſi dourà cauare, è vno de’numeri poſti in quell’ordine <lb/>venendo à baſſo; </s> <s xml:id="echoid-s2656" xml:space="preserve">e nella prima colonna, doue ſon poſte le 9 <pb o="140" file="0154" n="156" rhead="C A P O IV."/> radici, corriſponde aſ<unsure/> numero la figura, che ſi deue aggiun-<lb/>ger’alla radice trouata da principio.</s> <s xml:id="echoid-s2657" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"> <lb/>R # C. # C. 1 # C. 2 # C. 3 # C. 4 # C 5 # C. 6 # C.7 # C.8 # C.9 <lb/>1 # 1 # 331 # 1261 # 2791 # 4921 # 7651 # 10981 # 14911 # 19441 # 24571 <lb/>2 # 8 # 728 # 2648 # 5768 # 10088 # 15608 # 22328 # 30348 # 39368 # 49688 <lb/>3 # 27 # 1197 # 4167 # 8937 # 15507 # 23877 # 34047 # 46017 # 59787 # 75357 <lb/>4 # 64 # 1744 # 5824 # 12304 # 21184 # 32464 # 46144 # 52224 # 80704 # 101584 <lb/>5 # 125 # 2375 # 7625 # 15875 # 27125 # 41375 # 58625 # 78875 # 102125 # 128375 <lb/>6 # 216 # 3096 # 9576 # 19656 # 33336 # 50616 # 71469 # 95976 # 124056 # 155736 <lb/>7 # 343 # 3913 # 11683 # 23653 # 39823 # 60193 # 84763 # 113523 # 146503 # 183673 <lb/>8 # 512 # 4832 # 13953 # 27872 # 46592 # 70112 # 98432 # 131552 # 169472 # 212192 <lb/>9 # 729 # 5859 # 16389 # 32319 # 53649 # 80379 # 112509 # 150039 # 192969 # 241299 <lb/></note> <p> <s xml:id="echoid-s2658" xml:space="preserve">Sia dato il numero 438976, da cui de-<lb/> <anchor type="note" xlink:label="note-0154-02a" xlink:href="note-0154-02"/> ueſi eſtrarre la radice cubica. </s> <s xml:id="echoid-s2659" xml:space="preserve">Noto li <lb/>punti ſotto il 6, el’8 al modo conſueto: </s> <s xml:id="echoid-s2660" xml:space="preserve">e <lb/>nel ſecondo ordine, che è de’cubi, trouo, <lb/>che il cubo proſſimamente minore di <lb/>438 è 343 cubo di 7; </s> <s xml:id="echoid-s2661" xml:space="preserve">dunque noto 7 per <lb/>radice, e leuo 343 dal 438, e reſta 95. </s> <s xml:id="echoid-s2662" xml:space="preserve">A queſte figure 95, <lb/>che ſon reſtate, aggiungo l’altre tre figure del numero dato, <lb/>& </s> <s xml:id="echoid-s2663" xml:space="preserve">è 95976.</s> <s xml:id="echoid-s2664" xml:space="preserve"/> </p> <div xml:id="echoid-div83" type="float" level="2" n="1"> <note position="right" xlink:label="note-0154-02" xlink:href="note-0154-02a" xml:space="preserve"> <lb/>438976 # 76 <lb/>343 <lb/>95976 <lb/>95976 <lb/>0 <lb/></note> </div> <p> <s xml:id="echoid-s2665" xml:space="preserve">Ora perche la radice trouata da principio è 7, cerco nell’ <lb/>ordine C. </s> <s xml:id="echoid-s2666" xml:space="preserve">7, venendo à baſſo vn numero vguale, ò proſſima-<lb/>mente minore del 95976, e lo trouo preciſamente à dirittu-<lb/>ra della radice 6 nella prima colonna: </s> <s xml:id="echoid-s2667" xml:space="preserve">perciò aggiungo il 6 <lb/>alla radice 7, e fatta l’eſtrattione, nulla rimane; </s> <s xml:id="echoid-s2668" xml:space="preserve">onde conchiu-<lb/>do, che il num. </s> <s xml:id="echoid-s2669" xml:space="preserve">dato 438976 è preciſamente cubo, e la ſua ra-<lb/>diee è 76.</s> <s xml:id="echoid-s2670" xml:space="preserve"/> </p> <pb o="141" file="0155" n="157" rhead="Linea Cubica"/> <p> <s xml:id="echoid-s2671" xml:space="preserve">Nell’iſteſſa maniera dato 749812, leuo dal 749 il cubo di <lb/>9, che è 729, e rimane 20. </s> <s xml:id="echoid-s2672" xml:space="preserve">Il numero, <lb/>che reſta è 20812. </s> <s xml:id="echoid-s2673" xml:space="preserve">Ora perche la radi-<lb/> <anchor type="note" xlink:label="note-0155-01a" xlink:href="note-0155-01"/> ce è 9, cerco nella colonna C. </s> <s xml:id="echoid-s2674" xml:space="preserve">9 vn nu-<lb/>mero proſſimamente minore, e niuno <lb/>ve n’è; </s> <s xml:id="echoid-s2675" xml:space="preserve">onde aggiungo il o alla radice, <lb/>che ſarà 90, e reſta per numeratore della frattione adiacente <lb/>il numero 20812; </s> <s xml:id="echoid-s2676" xml:space="preserve">e per denominatore al modo ſolito ſarà <lb/>il triplo della radice trouata, cioè 270, moſtipſicato per la-<lb/>ſteſſa radice, & </s> <s xml:id="echoid-s2677" xml:space="preserve">il prodotto 24300 ſarà il denominatore, oue-<lb/>ro moltiplicato per la radice accreſciuta dell’vnità, cioè per <lb/>91, & </s> <s xml:id="echoid-s2678" xml:space="preserve">il prodotto 24570 ſarà il denominatore, a cui per lo <lb/>più torna bene aggiungere l’vnità, onde ſia 24571, quello <lb/>dà la frattione maggiore, e queſto minore del douere.</s> <s xml:id="echoid-s2679" xml:space="preserve"/> </p> <div xml:id="echoid-div84" type="float" level="2" n="2"> <note position="right" xlink:label="note-0155-01" xlink:href="note-0155-01a" xml:space="preserve"> <lb/>749812 # # 20812 <lb/># 90 <lb/>729 # # 24570 <lb/>20812 <lb/></note> </div> <p> <s xml:id="echoid-s2680" xml:space="preserve">Mà ſe il numero dato foſſe 57649, le-<lb/>uo dal 57 il cubo di 3, che è 27, e reſta <lb/> <anchor type="note" xlink:label="note-0155-02a" xlink:href="note-0155-02"/> 30; </s> <s xml:id="echoid-s2681" xml:space="preserve">sì che il numero rimanente per la ſe-<lb/>conda operatione è 30649. </s> <s xml:id="echoid-s2682" xml:space="preserve">Cerco dun-<lb/>que neila colonna C. </s> <s xml:id="echoid-s2683" xml:space="preserve">3 vn numero proſ-<lb/>ſimamente minore di queſtc, che è rima-<lb/>ſto, e trouo 27872, quale cauo dal <lb/>30649, ereſta 2777. </s> <s xml:id="echoid-s2684" xml:space="preserve">E perche all’ in-<lb/>contro del ſudetto numero 27872 ſi troua la radice 8, ag-<lb/>giungo queſta al 3, & </s> <s xml:id="echoid-s2685" xml:space="preserve">è la radice del numero dato 38 con vna <lb/>frattione, il cui numeratore è quel 2777, che reſtò, & </s> <s xml:id="echoid-s2686" xml:space="preserve">il de-<lb/>nominatore è il triplo della radice 38 moltiplato per 39, per <lb/>hauer ſa frattione minore, ouero il triplo quadrato della ra-<lb/>dice 38, per hauer la frattione maggiore.</s> <s xml:id="echoid-s2687" xml:space="preserve"/> </p> <div xml:id="echoid-div85" type="float" level="2" n="3"> <note position="right" xlink:label="note-0155-02" xlink:href="note-0155-02a" xml:space="preserve"> <lb/>57649 # 38 <lb/>27 <lb/>30649 <lb/>27872 <lb/>2777 <lb/></note> </div> <p> <s xml:id="echoid-s2688" xml:space="preserve">La ragione di queſto modo d’operare è, perche i numeri <lb/>di ciaſcuna area della tauoletta ſono quelli, che ſi fanno dal <pb o="142" file="0156" n="158" rhead="CAPO IV."/> triplo quadrato del numero poſto in cima (preſo però come <lb/>numero decadico, cioè non 2, ma 20, e così de gl’altri) molti-<lb/>plicato nel numero laterale corriſpondente della radice, e di <lb/>più dal quadrato della radice poſ<unsure/>ta nella prima colonna nel <lb/>triplo del primo numero della radice preſo pure come deca-<lb/>dico, e di più dal cubo della detta ſeconda figura della radice. <lb/></s> <s xml:id="echoid-s2689" xml:space="preserve">Per eſſempio, ſotto il C. </s> <s xml:id="echoid-s2690" xml:space="preserve">3. </s> <s xml:id="echoid-s2691" xml:space="preserve">ſi troua corriſpondente alla radi-<lb/>ce laterale 3 il numero 8937. </s> <s xml:id="echoid-s2692" xml:space="preserve">Queſto ſi fà dal quadrato di 3 <lb/>(cioè dello 30 poſto in cima) preſo tre volte, & </s> <s xml:id="echoid-s2693" xml:space="preserve">è 2700, mol-<lb/>tiplicato per ſa ſeconda radice laterale 3, onde è 8100. </s> <s xml:id="echoid-s2694" xml:space="preserve">Di <lb/>più il triplo della prima radice, che era 3 (cioè 30) è 90, e <lb/>queſto ſi moltiplica per il quadrato della ſeconda radice 3, <lb/>cioè per 9, eſi fà 810. </s> <s xml:id="echoid-s2695" xml:space="preserve">Finaſ<unsure/>mente prendo il cubo della ſe-<lb/>conda figura della radice 3, cioè 27, & </s> <s xml:id="echoid-s2696" xml:space="preserve">aggiunti inſie me que-<lb/>ſti tre numeri ſolidi 8100, 810, 27, ſi fà la ſomma 8937: </s> <s xml:id="echoid-s2697" xml:space="preserve"><lb/>E queſto numero ſi dourâ ſempre cauare nella ſeconda ope-<lb/>ratione, quando la prima figura della radice ſarà 3, e la ſecon-<lb/>da ſarà parimenti 3. </s> <s xml:id="echoid-s2698" xml:space="preserve">L’iſteſſo s’intenda fatto in tutti gl’altri <lb/>numeri areali di queſta tauol<unsure/>etta. </s> <s xml:id="echoid-s2699" xml:space="preserve">Onde fatta la fatica vna <lb/>volta in far la tauoletta, rieſce poi facile l’operatione nel mo-<lb/>do detto.</s> <s xml:id="echoid-s2700" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2701" xml:space="preserve">Che ſe il numero dato ſarà maggiore di ſei figure, ſi diuida <lb/>per vn numero cubo, di cui ſia conoſciuta la radice, e del quo-<lb/>tiente rimaſto minore di ſette figure ſi caui nel modo predet-<lb/>to la radice; </s> <s xml:id="echoid-s2702" xml:space="preserve">poiche ſe queſta radice trouata ſi moltiplicarà <lb/>per la radice nota del cubo, che fù diuiſore, ſi produrrà la ra-<lb/>dice cercata del numero dato. </s> <s xml:id="echoid-s2703" xml:space="preserve">La ragione di ciè è manifeſta, <lb/>perche come l’vnità al diuiſore, così il<unsure/> quotiente al numero <lb/>diuiſo; </s> <s xml:id="echoid-s2704" xml:space="preserve">dunque eſſendo l’iſteſſa la lor proportione ſubtriplica-<lb/>ta, è ancho come la radice cubica dell’vnità alla radice cubi- <pb o="143" file="0157" n="159" rhead="Linea Cubica"/> ca del diuiſore, così la radice cubica del quotiente alla radice <lb/>cubica del numero diuiſo; </s> <s xml:id="echoid-s2705" xml:space="preserve">queſta dunque ſi fà con la molti-<lb/>plicatione delle radici cubiche del quotiente, che è trouata, e <lb/>del diuiſore, che ſi ſuppone nota. </s> <s xml:id="echoid-s2706" xml:space="preserve">Sia dato il numero 32001-<lb/>3504000, di cui ſicerca la radice cubica. </s> <s xml:id="echoid-s2707" xml:space="preserve">Mi è noto, come <lb/>ſuppongo, che 438976 è numero cubo, la cui radice è 76. <lb/></s> <s xml:id="echoid-s2708" xml:space="preserve">Prendo quel numero per diuiſore del numero dato, e mi vien <lb/>per quotiente 729000; </s> <s xml:id="echoid-s2709" xml:space="preserve">di queſto cerco la radice cubica nel <lb/>modo ſopradetto, etrouata eſſer 90, moltiplico 90 per 76 <lb/>radice del diuiſore, eſi produce 6840 radice cercata del nu-<lb/>mero dato. </s> <s xml:id="echoid-s2710" xml:space="preserve">Così ſia dato 128024064: </s> <s xml:id="echoid-s2711" xml:space="preserve">queſto diuido per <lb/>343 cubo del 7: </s> <s xml:id="echoid-s2712" xml:space="preserve">del quotiente 373248 trouo la radice eſſere <lb/>72; </s> <s xml:id="echoid-s2713" xml:space="preserve">e queſta moltiplicata per 7 radice del diuiſore, produce <lb/>504 radice cercata del numero dato.</s> <s xml:id="echoid-s2714" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2715" xml:space="preserve">Ma ſe vn numero ſarà così grande, che non ti ſia noto vn <lb/>cubo, che diuidendoſo laſci per quotiente meno di 7 figure, <lb/>diuidilo per qucl cubo, che ti è noto: </s> <s xml:id="echoid-s2716" xml:space="preserve">& </s> <s xml:id="echoid-s2717" xml:space="preserve">il quotiente troppo <lb/>grande diuidi ſimilmente per vn cubo noto, ſin che habbi vn <lb/>quotiente piccolo à tuo modo, dal quale poſſi cauar la radi-<lb/>ce: </s> <s xml:id="echoid-s2718" xml:space="preserve">dipoi queſta radice moltiplicata ſucceſſiuamente con le <lb/>radici de’cubi preſi per diuiſori, darà finaſmente la radice <lb/>cercata.</s> <s xml:id="echoid-s2719" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2720" xml:space="preserve">Di quì hai vn modo aſſai facile per cauare l<unsure/>a radice cubica <lb/>anche ſenza queſta tauoletta, ſe ſolamente ſaprai i primi no-<lb/>ue cubi, diuidendo per eſſi il tuo numero, ſin che reſti vn quo-<lb/>tiente minore di 4 figure, di cui ti ſarà nota la radice; </s> <s xml:id="echoid-s2721" xml:space="preserve">e queſta <lb/>poi moltiplica per tutte le radici de’cubi diuiſori. </s> <s xml:id="echoid-s2722" xml:space="preserve">Sia dato <lb/>lo ſteſſo numero poco prima poſto 128024064: </s> <s xml:id="echoid-s2723" xml:space="preserve">lo diuido <lb/>per 729 cubo del 9, & </s> <s xml:id="echoid-s2724" xml:space="preserve">il quotiente 175616 diuido di nuouo <lb/>per 343 cubo del 7, e viene il quotiente 512, la cui radice è <pb o="144" file="0158" n="160" rhead="CAPO IV."/> preciſamente 8. </s> <s xml:id="echoid-s2725" xml:space="preserve">Dunque moltiplicate infieme queſte tre ra-<lb/>dici 9, 7, 8, ſi produce dell’8 in 9 il 72, e queſto per il 7 dà 504 <lb/>radice del detto numero.</s> <s xml:id="echoid-s2726" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2727" xml:space="preserve">Dal che potrai anche inferire la facilità del ſeruirſi delli cu-<lb/>bi di 10, 100, 1000, &</s> <s xml:id="echoid-s2728" xml:space="preserve">c. </s> <s xml:id="echoid-s2729" xml:space="preserve">tagliando dal dato numero alla de-<lb/>ſtra tanti numeri ternarij di figure, che non reſtino più di tre <lb/>figure, delle quali prendi il cubo maggiore con la ſua radice, <lb/>e quel che auanza del numero reſtato aggiungi alle figure ta-<lb/>gliate, e ſerue per numeratore della frattione, il cui denomi-<lb/>natore ſarà il triplo quadrato della radice trouata, aggiunti <lb/>tanti zeri, quante figure tagſiaſti fuora: </s> <s xml:id="echoid-s2730" xml:space="preserve">Dipoi queſta radi-<lb/>ce trouata moltiplica per il 10, ouero 100, &</s> <s xml:id="echoid-s2731" xml:space="preserve">c. </s> <s xml:id="echoid-s2732" xml:space="preserve">conforme <lb/>tagliaſti fuora 3, ò 6, ò 9 figure, e ſi produrrà la radice cer-<lb/>cata; </s> <s xml:id="echoid-s2733" xml:space="preserve">è ben vero, che ſarà vn poco maggiore del douere, co-<lb/>me per il contrario, ſe haueſſi accreſciuto d’vn’vnità quel tri-<lb/>plo quadrato della radice, verrebbe vn poco minore del do-<lb/>uere. </s> <s xml:id="echoid-s2734" xml:space="preserve">Così ſia dato l’iſteſſo 128024064: </s> <s xml:id="echoid-s2735" xml:space="preserve">taglio ſei figure, <lb/>che è come diuiderlo per 1000000, cubo del 100, reſta <lb/>128 {024064/1000000}, da cui cauato 125 cubo di 5, reſta 3 con la frat-<lb/>tione: </s> <s xml:id="echoid-s2736" xml:space="preserve">Dunque, poiche 75 è il triplo quadrato di 5, la radi-<lb/>ce ſarà 53 {024064/1000000/75}, cioè 5 {3024064/75000000}, queſta radice moltiplicata <lb/>per 100 radice del cubo diuiſore, produce 504, con l’aggiun-<lb/>ta d’vna frattione, la quale fà il numeratore troppo grande, <lb/>che ſe in vece del 75 haueſſi preſo 76, ſaria venuto meno di <lb/>504, onde ſi caua douerſi prendere 504.</s> <s xml:id="echoid-s2737" xml:space="preserve"/> </p> <pb o="145" file="0159" n="161" rhead="Linea Metallica"/> </div> <div xml:id="echoid-div87" type="section" level="1" n="48"> <head xml:id="echoid-head84" style="it" xml:space="preserve">CAPO V.</head> <head xml:id="echoid-head85" style="it" xml:space="preserve">Come s’habbia à notare nello Stromento la Proportione de’Metalli; <lb/>& vſo di queſta linea Metallica.</head> <p> <s xml:id="echoid-s2738" xml:space="preserve">HAbbiamo ſin’ora nelle linee ſegnate sù lo Stromento, <lb/>riſguardato preci ſamente le grandezze, ò ſiano lun-<lb/>ghezza, ò aree, ò corpi, ſenza tener conto della materia; <lb/></s> <s xml:id="echoid-s2739" xml:space="preserve">Ora per cagion d’eſſempio, onde altri potrà à ſuo talento de-<lb/>ſcriuerne altre, conſideriamo le grandezze in materie deter-<lb/>minate in quanto ſi poſſono paragonar’inſieme, e ſiano li me-<lb/>talli, aggiungendoui la Cal<unsure/>amita, il Marmo, e la Pietra, per <lb/>hauer dieci materie da paragonar’inſieme. </s> <s xml:id="echoid-s2740" xml:space="preserve">In due maniere <lb/>ſi può inſtituire queſta comparatione, cioè nella grauità, eſ-<lb/>ſendo vguale la lor mole; </s> <s xml:id="echoid-s2741" xml:space="preserve">ouero nella mole, eſſendo vguale <lb/>illor peſo. </s> <s xml:id="echoid-s2742" xml:space="preserve">Mà perche hauere nello Stromento vna linea di-<lb/>uiſa nella proportione della grauità, è coſa, che non hà mol-<lb/>ta difficoltà, poiche è vna diuiſione di linea ſemplice, e tutte <lb/>le ſue operationi non ſolo ſi puonno facilmente fare con la li-<lb/>nea Aritmetica, hauuto riſguardo all<unsure/>a Tauoletta, che quì ſi <lb/>porrà, nella cui ſeconda colonna s’eſprimono le proportioni <lb/>delle grauità; </s> <s xml:id="echoid-s2743" xml:space="preserve">ma anche ſenza la Tauoletta ſi potranno caua-<lb/>re dallo Stromento nel modo, che quì à baſſo nella Queſt. </s> <s xml:id="echoid-s2744" xml:space="preserve">1. </s> <s xml:id="echoid-s2745" xml:space="preserve"><lb/>ſi dirà; </s> <s xml:id="echoid-s2746" xml:space="preserve">perciò è meglio hauer le proportioni de’lati cubici, <lb/>ouero delli diametri delle sfere, ch’eſſendo di diuerſa mate-<lb/>ria, ſono però di vgual<unsure/> peſo; </s> <s xml:id="echoid-s2747" xml:space="preserve">e queſto hauendo qualche diffi-<lb/>coltà, conuerrà quì ſpiegare, acciò ſi vegga il modo, che ſi de-<lb/>ue tenere; </s> <s xml:id="echoid-s2748" xml:space="preserve">poiche li meno prattici vi ci potriano prendere <lb/>non piccolo sbaglio.</s> <s xml:id="echoid-s2749" xml:space="preserve"/> </p> <pb o="146" file="0160" n="162" rhead="CAPO V."/> <p> <s xml:id="echoid-s2750" xml:space="preserve">Suppongo noto dalla Statica, che la ſpecie della grauità de’ <lb/>corpi paragonati inſieme ſi conoſce dal peſo di ciaſcuno nell’ <lb/>iſteſſo mezzo, in cui grauitano, eſſendo di mole vguali: </s> <s xml:id="echoid-s2751" xml:space="preserve">così <lb/>perche vna palla di ferro peſata nell’aria ſi troua eſſere libre <lb/>21, doue che vna di pietra della ſteſſa grandezza peſata pure <lb/>nell’aria, non è che libre 7, perciò diceſi, che il ferro è tre vol-<lb/>te più peſante della pietra. </s> <s xml:id="echoid-s2752" xml:space="preserve">In oltre ſuppongo ciò, che nella <lb/>Statica ſi dimoſtra, che le grauità ſpecifiche de’ corpi, e le lo-<lb/>ro moli ſono reciprocamente proportionali, cioè, come la <lb/>grauità ſpecifica del primo, alla grauità ſpecifica del ſecondo, <lb/>quando le mol<unsure/>i ſono vguali, così quando le grauità aſſolute <lb/>ſon’vguali, la mole del ſecondo alla mole del primo. </s> <s xml:id="echoid-s2753" xml:space="preserve">E per <lb/>ſtare nell’eſſempio propoſto del ferro, e della pietra, il ferro <lb/>è in ſpecie tre volte più peſante della pietra; </s> <s xml:id="echoid-s2754" xml:space="preserve">dunque quando <lb/>faranno due maſle, vna di ferro, e l’altra di pietra vguali di <lb/>peſo, la maſſa di pietra ſarà reciprocamente tre volte mag-<lb/>giore di quella di ferro. </s> <s xml:id="echoid-s2755" xml:space="preserve">Così perche in mole vguale il peſo <lb/>dell’oro è come 100, & </s> <s xml:id="echoid-s2756" xml:space="preserve">il peſo del rame è come 47 {1/3}, così in <lb/>peſo vguale la mole del rame ſarà come 100, ela mole dell’ <lb/>oro ſarà come 47 {1/3}; </s> <s xml:id="echoid-s2757" xml:space="preserve">ecosì di tutte l’altre grauità.</s> <s xml:id="echoid-s2758" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2759" xml:space="preserve">Quindiè, che conoſciuta la proportione, che hanno le gra-<lb/>uità ſpecifiche de’corpi propoſti, ſi verrà a trouar la propor-<lb/>tione della loro ſolidità, quando ſi ſuppongano di peſi vgua-<lb/>li, ſe ſi riuoltarà la proportione delle grauità in modo, che <lb/>quello, ch’era conſeguente nelle grauità, diuenga anteceden-<lb/>te della proportione nelle ſolidità. </s> <s xml:id="echoid-s2760" xml:space="preserve">Onde eſſendo li dieci cor-<lb/>pi propoſti nella grauità tali, che l’oro è il più peſante, e la <lb/>pietrail più leggiero, per il contrario, ſe ſi faranno dieci palle <lb/>di peſo vguale, quella di pietra è la più grande, e quella d’oro <lb/>la più pic cola.</s> <s xml:id="echoid-s2761" xml:space="preserve"/> </p> <pb o="147" file="0161" n="163" rhead="Linea Metallica"/> <p> <s xml:id="echoid-s2762" xml:space="preserve">E prima di paſſar’ auanti, mi conuien quì auuiſare, che ſi <lb/>troua appreſſo gl’ Autori qualche diuerſitànel determinare le <lb/>proportioni delle grauità ſpecifiche; </s> <s xml:id="echoid-s2763" xml:space="preserve">e ciò è potuto accadere <lb/>ſenza alcun errore, ò imperfettione nelle lor’ iſperienze, per-<lb/>che il ferro, ò l’argento, ò l’oro di tutte le miniere non è per-<lb/>fettamente ſimile, ne tuttii marmi ſono giuſtamente peſanti <lb/>à vn modo, e da queſta diuerſità de’ corpi oſſeruati hà potu-<lb/>to naſcere la diuerſità delle proportioni, che ſi ſono deter-<lb/>minate: </s> <s xml:id="echoid-s2764" xml:space="preserve">anzi deue auuertirſi, che ſi troua diuerſità di peſo nel <lb/>metallo coniato, e nel metallo fuſo, perche nel fonderlo non <lb/>ſi condenſa tanto, quanto nel<unsure/> batterlo per coniarlo, e così <lb/>nella ſteſſa mole ſi può trouare diuerſità di peſo tra argento, <lb/>& </s> <s xml:id="echoid-s2765" xml:space="preserve">argento tolto dalla ſteſſa miniera. </s> <s xml:id="echoid-s2766" xml:space="preserve">Mà purche ſi prenda la <lb/>proportione trouata da alcun’eſſatto, e diligente oſſeruatore, <lb/>tanto baſta; </s> <s xml:id="echoid-s2767" xml:space="preserve">perche nell’operatione fiſica, à cui ſerue que-<lb/>ſto Stromento di Proportione, di cui trattiamo, non può riu-<lb/>ſcir’errore notabile. </s> <s xml:id="echoid-s2768" xml:space="preserve">A me è piacciuta la proportione ap-<lb/>portata dal Merſennio ne’ſuoi Hidraulici, come quella, che <lb/>mettendo la grauità dell’oro, come 100, e paragonando con <lb/>eſſa l’altre grauità, moſtra alla prima aſſai intelligibiln ente <lb/>la loro proportione.</s> <s xml:id="echoid-s2769" xml:space="preserve"/> </p> <pb o="148" file="0162" n="164" rhead="CAPOV."/> <note position="right" xml:space="preserve"> <lb/>#### Tauola delle granità ſpecifiche d’ alcuni corpi, della ſolidità \\ delle sfere vgualmente peſanti, e loro diametri \\ in particelle milleſime. <lb/>Corpi # Grauità \\ ſpecifiche # Solidità \\ delle sfere, \\ ò de’cubi # Proportioni \\ de’ diametri, \\ ò lati cub. <lb/>Pietra # 14 # 100 # 4. 641 † <lb/>Marmo # 21 # 66 {2/3} # 4. 055 --<lb/>Calamita # 26 # 53 {11/13} # 3. 776 † <lb/>Stagno # 38 {1/4} # 36 {23/38} # 3. 320 † <lb/>Ferro # 42 # 33 {1/3} # 3. 218 † <lb/>Rame # 47 {1/3} # 29 {27/47} # 3. 094 --<lb/>Argento # 54 {1/2} # 25 {37/54} # 2. 950 † <lb/>Piombo # 60 {1/2} # 23 {17/321} # 2. 850 --<lb/>Argento viuo # 71 {1/2} # 19 {41/71} # 2. 695 † <lb/>Oro # 100 # 14 # 2. 410 † <lb/></note> <p> <s xml:id="echoid-s2770" xml:space="preserve">Or’ecco in qual maniera s’è fatta queſta Tauoletta, in cui <lb/>nella prima colonna ſono poſti i corpi per ordine, come van-<lb/>no creſcendo di grauità, e calando di mole; </s> <s xml:id="echoid-s2771" xml:space="preserve">nella ſeconda <lb/>ſono le grauità ſpeeifiche, cioè i peſi di detti corpi, quando <lb/>ſono di mole vguali; </s> <s xml:id="echoid-s2772" xml:space="preserve">nella terza la ſolidità delle sfere fatte di <lb/>ciaſcun corpo, sì che però ſiano di peſo vguali: </s> <s xml:id="echoid-s2773" xml:space="preserve">e quel che <lb/>delle sfere ſi dice, s’intende de’ cubi, e di qualſiuoglia altro <lb/>corpo ſimile, poiche tutti ſono nella triplicata proportione <pb o="149" file="0163" n="165" rhead="Linea Metallica"/> de’ lati homologi, come le sfere ſono nella triplicata propor-<lb/>portione de’diametri: </s> <s xml:id="echoid-s2774" xml:space="preserve">nella quarta poi ſono le proportioni <lb/>de’ diametri del@e sfere, ò lati de’ cubi: </s> <s xml:id="echoid-s2775" xml:space="preserve">Ecco, dico, in qual <lb/>maniera s’è fatta queſta Tauoletta. </s> <s xml:id="echoid-s2776" xml:space="preserve">Perche la grauità della <lb/>pietra è 14, e l’altra eſtrema dell’oro è 100, la mole della pie-<lb/>tra ſi pone 100, e quella dell’oro 14. </s> <s xml:id="echoid-s2777" xml:space="preserve">Dipoi paragonando la <lb/>pietra cel marmo, quella è in grauità 14, e queſto 21; </s> <s xml:id="echoid-s2778" xml:space="preserve">dunque <lb/>quella in mole è 21, e queſto 14, ma s’è poſta la mole della <lb/>pietra 100, dunque dico, ſe 21 dà 14, 100 danno 66 {2/3}, e <lb/>queſta ſarà la mole del marmo. </s> <s xml:id="echoid-s2779" xml:space="preserve">Nell’iſteſſa maniera s’ande-<lb/>rà paragonando la grauità della pietra con la grauità de gl’ <lb/>altri, e ſi farà reciprocamente tale la mole della pietra alla <lb/>mole di detti corpi. </s> <s xml:id="echoid-s2780" xml:space="preserve">E queſto compendioſamente ſi fà pi-<lb/>gliando il numero 1400, e diuidendolo per ciaſcun numero <lb/>delle grauità, cioè per 26 grauità della calamita, & </s> <s xml:id="echoid-s2781" xml:space="preserve">il quo-<lb/>tiente 53 {11/@@} è la mole della calamita; </s> <s xml:id="echoid-s2782" xml:space="preserve">per 38 {1/4} grauità dello <lb/>ſtagno, & </s> <s xml:id="echoid-s2783" xml:space="preserve">il quotiente 36 {23/38} è la mole dello ſtagno, ecosì de <lb/>gl’ altri.</s> <s xml:id="echoid-s2784" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2785" xml:space="preserve">E perche nello Stromento conuien notare la proportione <lb/>ſubtriplicata delle sfere, ò de’ cubi, perciò da ciaſcun nume-<lb/>ro delle ſolidità ſi caua la radice cubica, aggiungendo à cia-<lb/>ſcun numero noue zeri, à fine d’hauer la radice in parti mille-<lb/>ſime: </s> <s xml:id="echoid-s2786" xml:space="preserve">nel che s’è operato nella ſteſſa maniera, che nel Capo <lb/>4. </s> <s xml:id="echoid-s2787" xml:space="preserve">onde circa il modo di ſeruirci de’ numeri della quarta co-<lb/>lonna per notar le diuiſioni dello Stromento, non occorre re-<lb/>plicar ciò, che già di ſopra s’è detto.</s> <s xml:id="echoid-s2788" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2789" xml:space="preserve">Per venir dunque all’eſſecutione dal centro dello Stro-<lb/>mento, tiro le due @inee AP vguali; </s> <s xml:id="echoid-s2790" xml:space="preserve">e pongo, che A P ſia dia-<lb/>metro d’vna palla di pietra, il quale conforme alla Tauoletta <lb/>è 464 centeſime: </s> <s xml:id="echoid-s2791" xml:space="preserve">onde ſi può intendere tutta la linea diuiſa <pb o="150" file="0164" n="166" rhead="CAPO V."/> in 116 parti, ciaſcuna delle quali ſia {4/100}. </s> <s xml:id="echoid-s2792" xml:space="preserve">Quindi è, che pren-<lb/>dendo la metà della linea AP, ſarà di queſte parti 58; </s> <s xml:id="echoid-s2793" xml:space="preserve">e perciò <lb/>nella linea Aritmetica dello Stromento applico la metà di <lb/>AP all’interuallo 58. </s> <s xml:id="echoid-s2794" xml:space="preserve">58; </s> <s xml:id="echoid-s2795" xml:space="preserve">& </s> <s xml:id="echoid-s2796" xml:space="preserve">hò lo Stromento aperto per po-<lb/>ter ſegnare occultamente nella linea AP gl’intieri, che ſono 4. <lb/></s> <s xml:id="echoid-s2797" xml:space="preserve">Eſſendo dunque ciaſcuna di quelle 116 parti di {4/100}, vn’intiero <lb/>ne contiene 25: </s> <s xml:id="echoid-s2798" xml:space="preserve">onde prendendo l’interuallo 25.</s> <s xml:id="echoid-s2799" xml:space="preserve">25, dal pun-<lb/>to A, @o ſegno occulta mente nella linea AP, replicandolo ſo-<lb/>lo tre volte ne’punti a, b, c: </s> <s xml:id="echoid-s2800" xml:space="preserve">perche tanto baſta per il reſto <lb/>dell’operatione. </s> <s xml:id="echoid-s2801" xml:space="preserve">Sì che vna di queſte parti vltimamente <lb/>trouate è 100 di quelle particelle, delle quali tutta la AP è <lb/>464.</s> <s xml:id="echoid-s2802" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2803" xml:space="preserve">Dunque per hauer le parti centeſime in ordine à ſegnar <lb/>nella linea AP gl’altri diametri, la grandezza d’vna di queſte <lb/>parti vltimamente trouate per vn’intiero, applico nella ſteſſa <lb/>linea Aritmetica all’interual@o 50. </s> <s xml:id="echoid-s2804" xml:space="preserve">50; </s> <s xml:id="echoid-s2805" xml:space="preserve">eritenuto lo Stromen-<lb/>to nel@a ſteſſa apertura paſſo all’inueſtigatione de gl’ altri dia-<lb/>metri nel modo che nella Queſt. </s> <s xml:id="echoid-s2806" xml:space="preserve">10. </s> <s xml:id="echoid-s2807" xml:space="preserve">del Cap. </s> <s xml:id="echoid-s2808" xml:space="preserve">2. </s> <s xml:id="echoid-s2809" xml:space="preserve">ſi diſſe. </s> <s xml:id="echoid-s2810" xml:space="preserve">Così <lb/>perche il diametro della sfera di marmo è 405, prendo 105, <lb/>& </s> <s xml:id="echoid-s2811" xml:space="preserve">all’interuallo della metà cioè al 52 {1/2}. </s> <s xml:id="echoid-s2812" xml:space="preserve">52 {1/4} hò la parte da <lb/>aggiunger alli tre intieri, cioe dal punto c ſin’all’M; </s> <s xml:id="echoid-s2813" xml:space="preserve">e così di <lb/>quali parti AP è 464, di tali eſſendone Ac 300, e c M 105, <lb/>tutta @a AM è 405 diametro d’vna sfera di marmo di peſo <lb/>vguale alla sfera di pietra. </s> <s xml:id="echoid-s2814" xml:space="preserve">Così per la calamita alli due in-<lb/>tieri A b aggiungo l’interuallo della metà di 178, cioè di 89. <lb/></s> <s xml:id="echoid-s2815" xml:space="preserve">89, & </s> <s xml:id="echoid-s2816" xml:space="preserve">è b C; </s> <s xml:id="echoid-s2817" xml:space="preserve">onde AC è il diametro per la calamita: </s> <s xml:id="echoid-s2818" xml:space="preserve">E così <lb/>de gl’altri. </s> <s xml:id="echoid-s2819" xml:space="preserve">Similmente per l’argento, il cui diametro è 295, <lb/>prendo alla metà di 295 l’interuallo 97 {1/2}. </s> <s xml:id="echoid-s2820" xml:space="preserve">97 {1/2}, e l’aggiungo <lb/>ad vn intiero, cioè dal pnntoa, onde AA è il diametro di vna <lb/>sfera d’argento. </s> <s xml:id="echoid-s2821" xml:space="preserve">E nella iſteſſa maniera s’anderanno aggiun- <pb o="151" file="0165" n="167" rhead="Linea Metallica"/> gendo ne gl’altri ad vn intiero gl’interualli proportionati; </s> <s xml:id="echoid-s2822" xml:space="preserve">il <lb/>che già tante volte s’è detto, che non occorre replicarlo.</s> <s xml:id="echoid-s2823" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2824" xml:space="preserve">Quì auuerto che nello Stromento ſi ſon poſte le lettere ini-<lb/>tiatiue de’nomi ltaliani, e per l’argento viuo, già che hà otte-<lb/>nuto da’Chimici il nome di Mercurio fattogli già commune, <lb/>s’è poſta la lettera M, la qual’eſſendo la più vicina alla lettera <lb/>O, e ſapendoſi, che doppo l’oro l’argento viuo è il più peſan-<lb/>te, ogn’vno facilmente intende eſſere la M per l’argento vi-<lb/>uo. </s> <s xml:id="echoid-s2825" xml:space="preserve">Sarà poi lecito à qualſiuoglia Artefice porre quelle let-<lb/>tere, che più gli piacerà, purche ſiano tali, che ſi poſſa facil-<lb/>mente conoſcere qual nome dimoſtrino.</s> <s xml:id="echoid-s2826" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div88" type="section" level="1" n="49"> <head xml:id="echoid-head86" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head87" style="it" xml:space="preserve">Come ſi poſſa cauare la proportione delle grauità ſpecifiche <lb/>di due, ò più corpi.</head> <p> <s xml:id="echoid-s2827" xml:space="preserve">GI<unsure/>à s’è detto, che le grauità ſpecifiche ſono reciproca-<lb/>mente, come le moli, e grandezze delli peſi aſſoluta-<lb/>mente vguali; </s> <s xml:id="echoid-s2828" xml:space="preserve">onde è manifeſto, che hauendoſi nello Stro-<lb/>mento la proportione ſubtriplicata delle moli, queſta pro-<lb/>portione triplicata darà la proportione delle moli, e rouer-<lb/>ſciata ſarà proportione delle grauità ſpecifiche. </s> <s xml:id="echoid-s2829" xml:space="preserve">Si può dun-<lb/>que in due maniere operare. </s> <s xml:id="echoid-s2830" xml:space="preserve">Primieramente, allargando lo <lb/>Stromento, quanto piace, e prendendo con due Compaſſi <lb/>gl’interualli de’due corpi, la cui proportione delle grauit à ſpe-<lb/>ciſiche ſi cerca: </s> <s xml:id="echoid-s2831" xml:space="preserve">dipoi con la linea Aritmetica per la Queſt. </s> <s xml:id="echoid-s2832" xml:space="preserve">5. <lb/></s> <s xml:id="echoid-s2833" xml:space="preserve">del Cap. </s> <s xml:id="echoid-s2834" xml:space="preserve">2 ſi vegga, che proportione in numeri habbiano <lb/>quelli due interualli preſi: </s> <s xml:id="echoid-s2835" xml:space="preserve">li numeri ſi cubichino, e ſarà nota <lb/>la proportione cercata, ſe ſi riuolterà. </s> <s xml:id="echoid-s2836" xml:space="preserve">Per eſſempio voglio <pb o="152" file="0166" n="168" rhead="CAPO V."/> paragonar l’oro con la pietra, prendo gl’interualli dell’vno, <lb/>e dell’altra, e con la linea Aritmetica trouo alla pietra corri-<lb/>ſponder 100, & </s> <s xml:id="echoid-s2837" xml:space="preserve">all’oro 51, & </s> <s xml:id="echoid-s2838" xml:space="preserve">vn poco più, quaſi 52: </s> <s xml:id="echoid-s2839" xml:space="preserve">piglio <lb/>il cubo di 100, che è 1000000, & </s> <s xml:id="echoid-s2840" xml:space="preserve">il cubo di 51, che è 132651 <lb/>e dico, che l’oro alla pietra in mole vguale, è di peſo, come <lb/>1000000, à 132651 in circa, cioè come 100 à 13 {2651/10000}. </s> <s xml:id="echoid-s2841" xml:space="preserve">Mà <lb/>preſo il cubo di 52, che è 140608 trouo, che è come 100 à <lb/>14 {608/10000}, onde, poiche il 52 è ſtato preſo troppo grande, @e <lb/>grauità ſpecifice ſono come 100, e 14.</s> <s xml:id="echoid-s2842" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2843" xml:space="preserve">Secondariamente ſi può fare con più facilità, quando nello <lb/>stromento vi ſia la linea cubica; </s> <s xml:id="echoid-s2844" xml:space="preserve">poiche il primo modo pro-<lb/>poſto è buono, quando nello Stromento eſſendoui la ſinea <lb/>metallica non v’è la cubica. </s> <s xml:id="echoid-s2845" xml:space="preserve">Prendanſi come prima gl’inter-<lb/>ualli della linea metallica, e ſi vegga nella linea cubica, à quali <lb/>interualli s’addattino, & </s> <s xml:id="echoid-s2846" xml:space="preserve">i numeri della linea cubica moſtre-<lb/>ranno i termini della Proportione reciproca, poiche mo-<lb/>ſtrano la proportione delle grandezze. </s> <s xml:id="echoid-s2847" xml:space="preserve">Così l’interuallo FF <lb/>nella linea metallica corriſpondente al ferro portato sù la li-<lb/>nea cubica all’interuallo 13. </s> <s xml:id="echoid-s2848" xml:space="preserve">13, l’interuallo CC corriſpon-<lb/>dente alla calamita, cadendo nella linea cubica all’interuallo <lb/>21. </s> <s xml:id="echoid-s2849" xml:space="preserve">21, dimoſtra, che la mole della calamita alla mole del fer-<lb/>ro è come 21 à 13, e perciò reciprocamente la grauità del <lb/>ferro alla grauità della calamita è come 21 à 13.</s> <s xml:id="echoid-s2850" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2851" xml:space="preserve">La dimoſtratione è chiara: </s> <s xml:id="echoid-s2852" xml:space="preserve">perche gl’interualli CC, & </s> <s xml:id="echoid-s2853" xml:space="preserve">FF <lb/>ſono nella proportione di AC ad AF, per quello che s’è det-<lb/>to nel Capo 1; </s> <s xml:id="echoid-s2854" xml:space="preserve">dunque eſſendo queſte, per la conſtruttione <lb/>dello Stromento nella proportione ſubtriplicata delle gran-<lb/>dezze, anche gl’interualli CC, FF ſono nella ſteſſa propor-<lb/>tione ſubtriplicata; </s> <s xml:id="echoid-s2855" xml:space="preserve">dunque queſte portate come interualli <lb/>della linea cubica, ſono nella ſteſſa proportione, in cui ſono <pb o="153" file="0167" n="169" rhead="Linea Metallica"/> ilati cubici ſegnati nella ſteſſa linea cubica: </s> <s xml:id="echoid-s2856" xml:space="preserve">dunque i ſolidi de <lb/>gl’interualli CC, FF ſono nella proportione de’cubi de’ lati <lb/>cubici corriſpondenti; </s> <s xml:id="echoid-s2857" xml:space="preserve">e così i numeri eſprimenti la propor-<lb/>tione de’cubi, eſprimono anche quella delle grandezze de<emph style="sub">2</emph> <lb/>ſolidi metallici, e per conſeguenza reciprocamente preſi an-<lb/>che la proportione delle grauità ſpecifiche.</s> <s xml:id="echoid-s2858" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2859" xml:space="preserve">Quindi è, che ſaputoſi il peſo d’vna palla di ferro, che por-<lb/>ta vn cannone, ſi potrà facilmente ſapere, quante libre porti <lb/>di palla di pietra; </s> <s xml:id="echoid-s2860" xml:space="preserve">poiche trouata la proportione delle graui-<lb/>tà ſpecifiche, come 3 à 1, ſe la palla di ferro è di libre 60, <lb/>quella di pietra vguale è libre 20.</s> <s xml:id="echoid-s2861" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2862" xml:space="preserve">E quì ſi può auuertire la diuerſa forma, con cui ſi può in <lb/>diſſegno eſprimere la proportione delle grauità di due corpi; <lb/></s> <s xml:id="echoid-s2863" xml:space="preserve">perche ſe ſi vuol’ eſprimere con sfere, ò con cubi, baſterà <lb/>prendere gl’interualli della linea metallica, e ſopra quelli, co-<lb/> <anchor type="figure" xlink:label="fig-0167-01a" xlink:href="fig-0167-01"/> me ſopra diametri, ò ſemidiametri deſcri-<lb/>uere le sfere, ò come ſopra lati deſcriuer i <lb/>cubi, ò altri ſolidi ſimili, poiche recipro-<lb/>camente preſi eſprimeranno la proportio-<lb/>ne delle grauità ſpecifiche. </s> <s xml:id="echoid-s2864" xml:space="preserve">Così per eſ-<lb/>primere la proportione dell<unsure/>’ oro al ferro, <lb/>nella linea metallica all’interuallo dell’oro <lb/>prendo qualunque ſemidiametro, e de-<lb/>ſcriuo la sfera A; </s> <s xml:id="echoid-s2865" xml:space="preserve">e ritenuta la ſteſſa aper-<lb/>tura dello Stromento, prendo l’interuallo <lb/>del ferro, e queſto mi ſerue di ſemidiame-<lb/>tro per la sfera B, & </s> <s xml:id="echoid-s2866" xml:space="preserve">in tal maniera la pro-<lb/>portione della grauità dell’oro alla grauità del ferro, è quella <lb/>della sfera B alla sfera A. </s> <s xml:id="echoid-s2867" xml:space="preserve">Mà ſe ſi vorrà con linee eſprimere <lb/>la ſteſſa proportione, non baſterà deſcriuere due linee, che <pb o="154" file="0168" n="170" rhead="CAPO V."/> ſiano gl’interualli dell’oro, e del ferro nella linea metallica; <lb/></s> <s xml:id="echoid-s2868" xml:space="preserve">mà ò conuiene continuar la proportio ne di dette linee ſin al-<lb/>la quarta proportionale, e come la proportione della prima <lb/>alla quarta è la proportione della grandezza de’ peſi vguali <lb/>di oro, e di ferro, così la proportione della quarta alla prima <lb/>è la proportione della grauità ſpecifica dell’ oro alla grauità <lb/>del ferro; </s> <s xml:id="echoid-s2869" xml:space="preserve">ò traportati queſti interualli alla linea cubica, ve-<lb/>dendo, che l’interuallo del ferro poſto al 50. </s> <s xml:id="echoid-s2870" xml:space="preserve">50, l’interuallo <lb/>dell’ oro cade nel 21. </s> <s xml:id="echoid-s2871" xml:space="preserve">21, conuiene nella linea Aritmetica <lb/>prendere due interualli nella proportione di 50 à 21, e ſiano <lb/>le linee R, S, onde l’oro al ferro di mole vguale è in grauità, <lb/>come R ad S.</s> <s xml:id="echoid-s2872" xml:space="preserve"/> </p> <div xml:id="echoid-div88" type="float" level="2" n="1"> <figure xlink:label="fig-0167-01" xlink:href="fig-0167-01a"> <image file="0167-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0167-01"/> </figure> </div> </div> <div xml:id="echoid-div90" type="section" level="1" n="50"> <head xml:id="echoid-head88" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head89" style="it" xml:space="preserve">Dato vn corpo, la cui grandezz<unsure/>a, e grauità ſiano note, come ſi <lb/>poſſa trouarne vn’altro d’altra materia, che in grauità <lb/>habbia la proportione data.</head> <p> <s xml:id="echoid-s2873" xml:space="preserve">PErche in queſta queſtione ſi ſuppone nota la grauità, ela <lb/>grandezza del corpo, poco importa, che detto corpo <lb/>ſia regolare, eſſendo che ſi può operare, come ſe ſi haueſ@e <lb/>vna sfera di peſo vguale, mentre non ſi cerca im mediatamen-<lb/>te la proportione, ne ſa ſimilitudine della grandezza, mà <lb/>de’peſi.</s> <s xml:id="echoid-s2874" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2875" xml:space="preserve">Sia per eſſempio vn pezzo di marmo di peſo 40 libre, e ſi <lb/>voglia hauer’vna palla, ò vn cubo di piombo vguale di peſo <lb/>al marmo. </s> <s xml:id="echoid-s2876" xml:space="preserve">Conuien dunque trouar, ò il diametro d’vna sfe-<lb/>ra, ò il lato d’vn cubo di marmo vguale alla grauità del pez-<lb/>zo di marmo dato. </s> <s xml:id="echoid-s2877" xml:space="preserve">Sia per eſſempio conoſciuto il lato d’vn <pb o="155" file="0169" n="171" rhead="Linea Metallica"/> cubo di marmo, che peſi due libre, e ſia la linea M: </s> <s xml:id="echoid-s2878" xml:space="preserve">que ſta <lb/> <anchor type="figure" xlink:label="fig-0169-01a" xlink:href="fig-0169-01"/> nella linea cubica s’ applichi <lb/>all’interuallo 2. </s> <s xml:id="echoid-s2879" xml:space="preserve">2, & </s> <s xml:id="echoid-s2880" xml:space="preserve">all’inter-<lb/>uallo 40. </s> <s xml:id="echoid-s2881" xml:space="preserve">40, s’haurà la linea <lb/>N lato d’ vn cubo di marmo <lb/>di libre 40 vguale al pezzo <lb/>dato. </s> <s xml:id="echoid-s2882" xml:space="preserve">Si porti dunque la li-<lb/>nea N nella linea metallica all’interuallo del marmo MM, e <lb/>nella ſteſſa linea metallica ritenuta l’apertura dello Stromen-<lb/>to, l’interuallo del piombo PP, darà la linea P lato d’vn cubo <lb/>di piombo di libre 40.</s> <s xml:id="echoid-s2883" xml:space="preserve"/> </p> <div xml:id="echoid-div90" type="float" level="2" n="1"> <figure xlink:label="fig-0169-01" xlink:href="fig-0169-01a"> <image file="0169-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0169-01"/> </figure> </div> <p> <s xml:id="echoid-s2884" xml:space="preserve">Mà ſe ſi cercaſſe vn cubo di piombo, ch’in vna ſtadiera <lb/>equilibraſſe vn’altro peſo maggiore, è manifeſto dalle ragio-<lb/>ni ſtatiche, che li peſi deuono hauere la proportione recipro-<lb/>ca delle lunghezze de bracci della ſtadiera, pigliandoli dal <lb/>punto, da cui ella ſtà ſoſpeſa; </s> <s xml:id="echoid-s2885" xml:space="preserve">e perciò al peſo dato conuien <lb/>trouar v’altro peſo della ſteſſa materia, che ſia minore nella <lb/>proportione de’bracci della ſtadiera; </s> <s xml:id="echoid-s2886" xml:space="preserve">& </s> <s xml:id="echoid-s2887" xml:space="preserve">hauuto il lato cubico, <lb/>ò diametro sferico di tal peſo minore applicato alla linea <lb/>metallica, ſubito ſi trouerà il lato, ò il diametro del cubo, ò <lb/>della sfera dell’altra materia, che ſi cerca. </s> <s xml:id="echoid-s2888" xml:space="preserve">Così ſia la ſtadie-<lb/>ra AB ſoſtenuta nel punto C, ſi che il braccio CB ſia noue <lb/>volte maggiore del braccio CA, e dall’ eſtremità A debba <lb/>ſoſpenderſi vn peſo di 450 libre di ſtagno; </s> <s xml:id="echoid-s2889" xml:space="preserve">dunque eſſendo <lb/>BC à CA, come 9 à 1, il peſo che in A è 450 libre, vien equi-<lb/>librato in B da libre 50. </s> <s xml:id="echoid-s2890" xml:space="preserve">Ora facciamo, che ſia noto il dia-<lb/>metro di vna palla di ſtagno di lib. </s> <s xml:id="echoid-s2891" xml:space="preserve">3, s’appli chi tal diametro <lb/>nella linea cubica all’interuallo 3. </s> <s xml:id="echoid-s2892" xml:space="preserve">3, e l’interuallo 50. </s> <s xml:id="echoid-s2893" xml:space="preserve">50, da-<lb/>rà il diametro d’vna palla di ſtagno di lib. </s> <s xml:id="echoid-s2894" xml:space="preserve">50. </s> <s xml:id="echoid-s2895" xml:space="preserve">Queſto dia-<lb/>metro trouato ſi porti nella linea metallica all’ interuallo SS <pb o="156" file="0170" n="172" rhead="CAPO V."/> dello ſtagno, poiche l’interuallo PP del piombo darà il dia-<lb/>metro d’vna palla di piombo dilibre 50, che poſta in B, equi-<unsure/> <lb/>librerà le libre 450 di ſtagno poſte in A.</s> <s xml:id="echoid-s2896" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2897" xml:space="preserve">Quì però deue intenderſi la ſtadiera equilibrata da ſe me-<lb/>deſima, perche altrimenti nelle ſtadiere communi non riuſci-<lb/>rebbe aggiuſtato il peſo, a cagione che il braccio lungo del-<lb/>la ſtadiera hà li ſuoi momenti di grauità.</s> <s xml:id="echoid-s2898" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2899" xml:space="preserve">Auuertaſi in queſte operationi riuſcir aſſai commodo <lb/>prendere le sfere; </s> <s xml:id="echoid-s2900" xml:space="preserve">perche quando foſſero grandi aſſai, ſi può <lb/>operare col ſemidiametro più toſto, che col diametro, e s’hà <lb/>l’apertura del Compaſſo per deſcriuer la sfera; </s> <s xml:id="echoid-s2901" xml:space="preserve">ma ſe ſi pren-<lb/>deſ@e la metà dellato cubico, conuerria pigliar il cubo otto <lb/>volte minore del peſo dato, e ſi trouerebbe il lato d’vn cubo <lb/>otto volte minoré del douere: </s> <s xml:id="echoid-s2902" xml:space="preserve">onde finita l’operatione, ſaria <lb/>di meſtieri raddoppiar il lato trouato.</s> <s xml:id="echoid-s2903" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2904" xml:space="preserve">In oltre ſi deue auuertire da chi non foſſe tanto prattico <lb/>della Geometria, che quando ſi tratta ſolamente d’eſprimere <lb/>la proportione, tanto è trouar li diametri delle sfere, quanto <lb/>ilati de’cubi; </s> <s xml:id="echoid-s2905" xml:space="preserve">perche le sfere eſſendo tra di ſe nella triplicata <lb/>proportione de’loro diametri, hanno la proportione de’cubi <lb/>de gli ſteſſi diametri; </s> <s xml:id="echoid-s2906" xml:space="preserve">Mà ſe ſi trattaſſe d’eſprimere le gran-<lb/>dezze, non è l’iſteſſo prender le sfere, & </s> <s xml:id="echoid-s2907" xml:space="preserve">i cubi, come è ma-<lb/>nifeſto; </s> <s xml:id="echoid-s2908" xml:space="preserve">poiche la sfera circoſcritta dal cilindro è à queſto co-<lb/>me 2 a 3, & </s> <s xml:id="echoid-s2909" xml:space="preserve">il cilindro cir@oſcritto dal cubo è nella propor-<lb/>tione del circolo al quadrato d@l diametro, cioè come 11 a <lb/>14: </s> <s xml:id="echoid-s2910" xml:space="preserve">onde ne viene, che queſti tre corpi sfera, cilindro, e cu-<lb/>bo, à quali ſerue l’iſteſſa linea di diametro alli rotondi, e di la-<lb/>to al cubo, ſono nella proportione di 22. </s> <s xml:id="echoid-s2911" xml:space="preserve">33. </s> <s xml:id="echoid-s2912" xml:space="preserve">42, e così il <lb/>cubo alla sfera è come 21 à 11; </s> <s xml:id="echoid-s2913" xml:space="preserve">dal che appariſce quanto <lb/>enorme sbaglio faria chi in ciò operaſſe ſenza la douuta rifleſ-<lb/>ſione.</s> <s xml:id="echoid-s2914" xml:space="preserve"/> </p> <pb o="157" file="0171" n="173" rhead="Linea Metallica"/> <p> <s xml:id="echoid-s2915" xml:space="preserve">Dal che così di paſſaggio poſſiamo raccogliere, come ſi <lb/>poſſa trasformar vn cubo in vna sfera, & </s> <s xml:id="echoid-s2916" xml:space="preserve">al contrario. </s> <s xml:id="echoid-s2917" xml:space="preserve">Perche <lb/>ſe ſarà dato il lato d’vn cubo, è manifeſto, ehe di quali parti <lb/>quel cubo è 21, la sfera che habbia diametro vguale ſarà ſolo <lb/>11: </s> <s xml:id="echoid-s2918" xml:space="preserve">pongaſi dunque quel lato del cubo dato nella linea cubi-<lb/>ca, come ſe foſſe diametro d’vna sfera all’interuallo II. </s> <s xml:id="echoid-s2919" xml:space="preserve">II, e <lb/>preſo l’interuallo 21.</s> <s xml:id="echoid-s2920" xml:space="preserve">21, queſto ſarà il diametro della sfera, <lb/>la quale eſſendo alla sfera del primo diametro, come 21 à 11, <lb/>vien ad eſſer vgual al cubo dato, perla 7 del lib. </s> <s xml:id="echoid-s2921" xml:space="preserve">5. </s> <s xml:id="echoid-s2922" xml:space="preserve">E ſe la <lb/>sfera s’haurà à cangiar in cubo, pongaſi il diametto di detta <lb/>sfera come latò d’vn cubo all’interuallo 21. </s> <s xml:id="echoid-s2923" xml:space="preserve">21, e preſo l’in-<lb/>teruallo 11. </s> <s xml:id="echoid-s2924" xml:space="preserve">11, ſarà lato d’vn cubo, che ſarà al cubo del pri-<lb/>mo lato, come 11 à 21, e perciò vguale alla sfera del primo <lb/>diametro preſo, come lato di cubo.</s> <s xml:id="echoid-s2925" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2926" xml:space="preserve">Fatta poi queſta trasformatione di sfera in cubo vguale <lb/>della ſteſſa materia, ſarà facile, per quel ches’è detto con la <lb/>linea metallica trouar la sfera, c̀’lcubo vguale di peſo, che <lb/>ſia d’altra materia.</s> <s xml:id="echoid-s2927" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2928" xml:space="preserve">L’iſteſſa forma d’operare ſi terrà nella trasformatione di <lb/>sfera, ò cubo in cilindro, hauendo riſguardo alla propor-<lb/>tione delle loro grandezze; </s> <s xml:id="echoid-s2929" xml:space="preserve">e ſeruendoſi della linea Cubica, <lb/>Geometrica, e poi della linea Metallica per la diuerſità della <lb/>materia in ordine al peſo. </s> <s xml:id="echoid-s2930" xml:space="preserve">Così eſſendo data la sfera S d’ar-<lb/>gento, e ſi voglia vn cilindro d’oro vguale di peſo, il cilindro <lb/>quadrato CE, che hà per baſe il circolo maſſimo della sfera, <lb/>e per altezza il diametro della ſteſſa sfera, è ſeſquialtero alla <lb/>sfera: </s> <s xml:id="echoid-s2931" xml:space="preserve">dunque trouandoſi con la linea Geometrica il diame-<lb/>tro d’vn circolo ſubſeſquialtero, e ſia CF, il cilindro CG d’al-<lb/>tezza vguale al diametro della sfera ſarà vguale alla ſteſſa sfe-<lb/>ra, poiche anch’egliè ſubſel<unsure/>quia<unsure/>ltero del cilindro CE, ha- <pb o="158" file="0172" n="174" rhead="CAPOV."/> <anchor type="figure" xlink:label="fig-0172-01a" xlink:href="fig-0172-01"/> uendo la proportione delle baſi, per <lb/>la 11 del lib. </s> <s xml:id="echoid-s2932" xml:space="preserve">12. </s> <s xml:id="echoid-s2933" xml:space="preserve">Dunque il cilindro <lb/>CG d’argento è vguate alla sfera S <lb/>d’argento. </s> <s xml:id="echoid-s2934" xml:space="preserve">Or volendoſi vn cilindro <lb/>quadrato, che fia vguale al cilindo <lb/>CG, e per confeguenza alla sfera da-<lb/>ta S, tra il diametro della baſe CF, e <lb/>l’altezza FG ſitroui la ſeconda delle <lb/>quattro continuatamente proportio-<lb/>nali, per la Queſt. </s> <s xml:id="echoid-s2935" xml:space="preserve">1. </s> <s xml:id="echoid-s2936" xml:space="preserve">del Cap. </s> <s xml:id="echoid-s2937" xml:space="preserve">4. </s> <s xml:id="echoid-s2938" xml:space="preserve">col <lb/>mezzo della linea cubica, e ſia CO, <lb/>diametro della baſe del cilindro, à cui <lb/>eſſendo vguale l’altezza OL, ſarà il ci-<lb/>lindro CL quadrato vguale al cilindro <lb/>CG, cioè alla sfera; </s> <s xml:id="echoid-s2939" xml:space="preserve">eſſendo che le ba-<lb/>ſi, e l’altezze di queſti due cilindri ſo-<lb/>no reciproche, come s’è dimoſtrato <lb/>nella Queſt. </s> <s xml:id="echoid-s2940" xml:space="preserve">6. </s> <s xml:id="echoid-s2941" xml:space="preserve">del Cap. </s> <s xml:id="echoid-s2942" xml:space="preserve">4. </s> <s xml:id="echoid-s2943" xml:space="preserve">perche per <lb/>la coſtruttione il circolo del diametro <lb/>CF al circolo del diametro CO è co-<lb/>mela prima alla terza proportionale, <lb/>tra le quali la linea CO è la ſeconda. <lb/></s> <s xml:id="echoid-s2944" xml:space="preserve">Or cſſendo come la prima alla terza, così la ſeconda alla <lb/>quarta, cioè CO, ouero OL vguale altezza, all’ altezza FG, <lb/>ſi rende manifeſto, che ſi reciprocano le baſi, e l’altezze. </s> <s xml:id="echoid-s2945" xml:space="preserve">Tra-<lb/>portato dunque CO nella linea metallica all’interuallo AA <lb/>dell’argento, prendaſi l’interuallo OO dell’oro, e ſia la linea <lb/>IM diametro della baſe, & </s> <s xml:id="echoid-s2946" xml:space="preserve">MK altezza vguale: </s> <s xml:id="echoid-s2947" xml:space="preserve">onde il cilin-<lb/>dro d’oro IK eſſendo ſimile al cilindro CL d’argento, & </s> <s xml:id="echoid-s2948" xml:space="preserve">eſ-<lb/>ſendo per la coſtruttione dello ſtromento nella proportione <pb o="159" file="0173" n="175" rhead="Linea Metallica"/> reciproca delle grauità ſpecifiche, ſaranno detti due cilindri <lb/>equiponderanti, e perciò il cilindro d’oro IK ſarà di peſo <lb/>vguale alla sfera S d’argento.</s> <s xml:id="echoid-s2949" xml:space="preserve"/> </p> <div xml:id="echoid-div91" type="float" level="2" n="2"> <figure xlink:label="fig-0172-01" xlink:href="fig-0172-01a"> <image file="0172-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0172-01"/> </figure> </div> </div> <div xml:id="echoid-div93" type="section" level="1" n="51"> <head xml:id="echoid-head90" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head91" style="it" xml:space="preserve">Come ſi poſſa trouare la grandezza di qualſiuoglia peſo, <lb/>conoſcendone vn’altro d’alira materia.</head> <p> <s xml:id="echoid-s2950" xml:space="preserve">DAlle coſe dette ſin’ora è manifeſto, che ſapendoſi la, <lb/>grandezza d’vn peſo in materia determinata di quel-<lb/>le, che ſono nella linea metallica ſubito ſi troua la grandezza <lb/>del corpo d’vgual peſo in figura ſimile, e di materia diuerſa. <lb/></s> <s xml:id="echoid-s2951" xml:space="preserve">Poſcia con la linea cubica ſi troua la grandezza del peſo, che <lb/>ſi cerca. </s> <s xml:id="echoid-s2952" xml:space="preserve">Per cagione d’eſſempio ſi cerca di far’ vn vaſo di <lb/>capacità cubica in modo, che capiſca libre 3200 d’argento <lb/>viuo: </s> <s xml:id="echoid-s2953" xml:space="preserve">è noto il diametro d’vna palla diferro di 3 libre. </s> <s xml:id="echoid-s2954" xml:space="preserve">Per-<lb/>che ſi cerca illato cubico del vaſo, ſi riduca la grandezza del-<lb/>la palla ad vn cubo vguale, trouando il lato del cubo di ferro <lb/>di 3 libre, come s’è detto nella Queſt. </s> <s xml:id="echoid-s2955" xml:space="preserve">precedente: </s> <s xml:id="echoid-s2956" xml:space="preserve">e queſto <lb/>lato cubico nella linea metallica s’appl<unsure/>ichi all’interuallo del <lb/>ferro FF, perche l’interuallo del mercurio MM darà il lato di <lb/>vn cubo d’argento viuo di 3 libre. </s> <s xml:id="echoid-s2957" xml:space="preserve">Queſto lato trouato s’ap-<lb/>plichi nella linea cubica all’interuallo 3. </s> <s xml:id="echoid-s2958" xml:space="preserve">3, e l’interuall 0 50. </s> <s xml:id="echoid-s2959" xml:space="preserve"><lb/>50, darà illato d’vn cubo di 50 libre d’argento viuo. </s> <s xml:id="echoid-s2960" xml:space="preserve">Dun que <lb/>queſto lato quadruplicato darà il lato d’vn cubo 64 volte <lb/>maggiore del cubo di libre 50, cioè del cubo di lib. </s> <s xml:id="echoid-s2961" xml:space="preserve">3200 <lb/>d’argento viuo, come ſi cercaua.</s> <s xml:id="echoid-s2962" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2963" xml:space="preserve">Quando il numero, che denomina il peſo è grande aſſai, <lb/>per trouar preſto vn lato, che con replicarlo alcune volte dia <pb o="160" file="0174" n="176" rhead="CAPOV."/> il lato, che ſi cerca, prendaſi vn numero cubo, che lo miſuri <lb/>per vn’altro num. </s> <s xml:id="echoid-s2964" xml:space="preserve">minore del 50 (poſto che la linea cubica, <lb/>dello ſtromento non ecceda li 50) ò di qualſiuoglia altro, che <lb/>ſia il maſſimo de’numeri notati nella linea cubica. </s> <s xml:id="echoid-s2965" xml:space="preserve">Così per <lb/>trouar’il diametro d’vna sfera di marmo, che peſi libre 4000, <lb/>ſe prendeſſi il cubo di 4, cioè 64, verrebbe il quotiente 62 {1/2} <lb/>maggiore del 50, che è il maſſimo delli notati nella linea cu-<lb/>bica; </s> <s xml:id="echoid-s2966" xml:space="preserve">perciò preſo il cubo di 5, cioè 125, e per 125 diuiſo il <lb/>4000, viene il quotiente 32. </s> <s xml:id="echoid-s2967" xml:space="preserve">Et in tal maniera operando, <lb/>come prima, cioè trouato il diametro della sfera di marmo <lb/>di lib. </s> <s xml:id="echoid-s2968" xml:space="preserve">3 vguale alla sfera di ferro conoſciuta, & </s> <s xml:id="echoid-s2969" xml:space="preserve">applicato <lb/>nella linea cubica tal diametro all’interuallo 3. </s> <s xml:id="echoid-s2970" xml:space="preserve">3, prendaſi <lb/>l’interuallo 32. </s> <s xml:id="echoid-s2971" xml:space="preserve">32; </s> <s xml:id="echoid-s2972" xml:space="preserve">e perche il 4000 fù diuiſo per il cubo di <lb/>5. </s> <s xml:id="echoid-s2973" xml:space="preserve">per queſto quell’interuallo 32. </s> <s xml:id="echoid-s2974" xml:space="preserve">32 deue replicarſi cinque <lb/>volte, e quello ſarà il diametro d’vna palla di marmo di <lb/>4000 libre.</s> <s xml:id="echoid-s2975" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div94" type="section" level="1" n="52"> <head xml:id="echoid-head92" xml:space="preserve">CAPO VI.</head> <head xml:id="echoid-head93" style="it" xml:space="preserve">In qual maniera s’habbiano à notare nello Stromento li Gradi <lb/>del Circolo: & vſo di tal linea.</head> <p> <s xml:id="echoid-s2976" xml:space="preserve">PEr la neceſſità, che s’hà molte volte di diſſegnar’ alcune <lb/>piante di campi, e coſe ſimili, ò per l’vſo della Gnomo-<lb/>nica, conuien fare angoli di miſure determinate in gradi, i <lb/>quali ſono quelle 360 parti, in cui s’intende diuiſa la circon-<lb/>ferenza di ciaſcun circolo, come è noto. </s> <s xml:id="echoid-s2977" xml:space="preserve">A queſto fine mol-<lb/>ti hanno deſcritta vna quarta parte dicerchio diuiſa ne’ſuoi <lb/>gradi, e dalla circonferenza vltima tirate per ciaſcun grado <lb/>linee rette al centro, vengono à diuidere ſimilmente altri <pb o="161" file="0175" n="177" rhead="Gradi del Circolo"/> archi più piccoli deſcritti dal medeſimo centro, per poterſi <lb/>ſeruire ora di queſto, ora di quell’ arco di maggior, ò minor <lb/>diſtanza dal centro, conforme al biſogno occorrente. </s> <s xml:id="echoid-s2978" xml:space="preserve">Mà di <lb/>quanta imperfettione ciò ſia, è manifeſto, per la confuſione, <lb/>che ſaria, ſe foſſero molti gli archi deſcritti l’vno vicino all’al-<lb/>tro, e per la difficoltà, che tutte le linee ſiano giuſtiſſimamen-<lb/>te tirate; </s> <s xml:id="echoid-s2979" xml:space="preserve">oltre che coll’auuicinarſi tra di loro, quanto più s’ac-<lb/>coſtano al centro, vengon’ à far confuſione, eſpeſſo non ſal-<lb/>uano l’vguaglianza della diuiſione. </s> <s xml:id="echoid-s2980" xml:space="preserve">Perciò ſi sfuggono tutti <lb/>queſti inconuenienti nello Stromento di Proportione, il qua-<lb/>le ſerue per diuider tutti li circoli poſſibili, li cui ſemidia me-<lb/>tri puonno capire tra la minima, e la maſſima dilatatione <lb/>dello ſtromento nel luogo, doue s’applica il ſemidiametro, <lb/>come ſi dirà.</s> <s xml:id="echoid-s2981" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2982" xml:space="preserve">Tirandoſi dunque nello ſtromento vna linea retta, è certo, <lb/>che queſta non và diuiſa in parti vguali, come vna linea cir-<lb/>colare è diuiſa in parti vguali, che ſi chiamano Grandi; </s> <s xml:id="echoid-s2983" xml:space="preserve">poi-<lb/>che in tal linea reta dello ſtromento ſi ſegnano non gl’archi, <lb/>mà le corde ſottendenti à gl’archi, e con eſſe s’opera nel mo-<lb/>do, che ſi ſpiegarà à baſſo. </s> <s xml:id="echoid-s2984" xml:space="preserve">E che tali corde de gl’archi, che <lb/>creſcono vgualmente in numero di grandi, non creſcono <lb/>anch’eſſe vgualmente, è manifeſto dalla dottrina de’Seni, che <lb/>quì ſi ſuppone. </s> <s xml:id="echoid-s2985" xml:space="preserve">Onde grauemente errarebbe l’ Artefice, <lb/>che vna tal linea tirata nello ſtromento per vn quadrante di <lb/>cerchio, voleſſe diuider’in 90 parti vguali; </s> <s xml:id="echoid-s2986" xml:space="preserve">perche così fa-<lb/>cendo, queſta linea non ſaria punto differente dalla ſinea <lb/>Aritmetica, di cuis’è parlato nel Capo 2. </s> <s xml:id="echoid-s2987" xml:space="preserve">E così eſſendoci <lb/>oſferto vno Stromento di Proportione, ſe applicati due com-<lb/>paſſi à due numeri nella linea Aritmetica, quelle due diſtan-<lb/>ze vengono ad applicarſi à due numeri ſimili nella linea de’ <pb o="162" file="0176" n="178" rhead="CAPO VI."/> gradi, ò del quadrante del cerchio, ſarà ſegno euidente non <lb/>eſſerſi fatta tal linea dall’ Artefice ſecondole regole debite, e <lb/>lo ſtromento è inutile.</s> <s xml:id="echoid-s2988" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2989" xml:space="preserve">Ora douendoſi notare nello ſtromeuto le corde de gl’ar-<lb/>chi, ſi puonno notare, ò quelle di tutto vn ſemicircolo, ò ſol <lb/>quelle d’vn quadrante; </s> <s xml:id="echoid-s2990" xml:space="preserve">e torna più à conto notar ſol queſte <lb/>del quadrante, perche in tal modo rieſcono le diuiſioni della <lb/>linea più diſtinte, e notabili, e per altro queſte baſtano per <lb/>qualſiuoglia arco anche maggiore. </s> <s xml:id="echoid-s2991" xml:space="preserve">Se pur non foſſe così lun-<lb/>go lo ſtromento, che riuſciſſe commodo il notarui tutto vn <lb/>ſemicircolo. </s> <s xml:id="echoid-s2992" xml:space="preserve">Perciò qui parleremo ſolo della diuiſione per il <lb/>quadrante, perche da ciò ſarà manifeſto, quanto s’habbia à <lb/>fare volendoſi fare per il ſemicircolo.</s> <s xml:id="echoid-s2993" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2994" xml:space="preserve">Per tanto voltato lo ſtromento dall’altra faccia oppoſta <lb/>alla ſegnata già per linee rette ſenza relatione al circolo, ſi ti-<lb/>rino dal centro nell’vno, e nell’altro braccio due linee rette <lb/>vguali, cialcuna delle quali ſi ſuppone eſſer corda dell’arco <lb/>di 90 gradi. </s> <s xml:id="echoid-s2995" xml:space="preserve">Conuien dunque trouare, qual ſia il ſemidia-<lb/>metro d’vn circolo, la di cui quarta parte habbia per corda <lb/>la linea data. </s> <s xml:id="echoid-s2996" xml:space="preserve">Il che ſi fà in tal maniera. </s> <s xml:id="echoid-s2997" xml:space="preserve">Suppongaſi, che la <lb/> <anchor type="figure" xlink:label="fig-0176-01a" xlink:href="fig-0176-01"/> linea retta tirata nello ſtromento <lb/>ſia la A B corda dell’ arco di gradi <lb/>90, ecerchiſi il ſemidiametro, cioè <lb/>la corda di gr. </s> <s xml:id="echoid-s2998" xml:space="preserve">60. </s> <s xml:id="echoid-s2999" xml:space="preserve">Diuidaſi vgual-<lb/>mente la AB in C, e ſi alzi la per-<lb/>pendicolare CD vguale alla CB, e <lb/>per il punto D ſi tiri la retta BD, à <lb/>cui prendaſi vguale BE, & </s> <s xml:id="echoid-s3000" xml:space="preserve">il punto <lb/>Eè il termine della corda di gr. </s> <s xml:id="echoid-s3001" xml:space="preserve">60 nel cerchio, di cui la AB <lb/>è corda di gr. </s> <s xml:id="echoid-s3002" xml:space="preserve">90. </s> <s xml:id="echoid-s3003" xml:space="preserve">Perche ſe ſi tira la retta DA, li due triangoli <pb o="163" file="0177" n="179" rhead="Gradi del Circolo"/> ACD, BCD hanno per la coſtruttione vguali i lati CA, CB, <lb/>c la CD è commune, e gl’angoli al punto C ſono fatti vguali <lb/>dalla perpendicolare CD, dunque, per la 4 del lib. </s> <s xml:id="echoid-s3004" xml:space="preserve">1, le baſi <lb/>DB, DA ſono vguali, e gl’angoli vguali. </s> <s xml:id="echoid-s3005" xml:space="preserve">E perche per la co-<lb/>ſtruttione ambidue ſono iſoſceli, eſſendo le tre line AC, CD, <lb/>CB vguali, gl’angoli CDB, CDA ſono ſemiretti, per la 5, e <lb/>32 del lib. </s> <s xml:id="echoid-s3006" xml:space="preserve">1. </s> <s xml:id="echoid-s3007" xml:space="preserve">e così tutto l’angolo ADB è retto: </s> <s xml:id="echoid-s3008" xml:space="preserve">Onde eſſendo <lb/>ſimili li triangoli BCD, BDA, come CB ſemidiametro à BD <lb/>corda di gr. </s> <s xml:id="echoid-s3009" xml:space="preserve">90. </s> <s xml:id="echoid-s3010" xml:space="preserve">così anche BD ſemidia metro, cioè BE, à BA <lb/>corda di gradi 90. </s> <s xml:id="echoid-s3011" xml:space="preserve">E per prouare ſe habbi operato giuſta-<lb/>mente, prolonghiſi la BD in F, tanto che BF ſia vguale alla <lb/>BA, e fatto centro in E all’interuallo EB, ſi deſcriua l’arco <lb/>BF, eſe paſſerà preciſa mente per il punto F, ſarà ſegno, che <lb/>s’operò giuſtamente: </s> <s xml:id="echoid-s3012" xml:space="preserve">Perche dal centro C deſcritto il qua-<lb/>drante BD, ſono due circoli, che ſi toccano interior mente <lb/>nel punto B, e così la retta BDF tagliando dell’vno, e dell’al-<lb/>tro archi ſimili (come ſi può facilmente raccogliere dalla 20, <lb/>ò anche dalla 32 del lib. </s> <s xml:id="echoid-s3013" xml:space="preserve">3.) </s> <s xml:id="echoid-s3014" xml:space="preserve">fà che tanto l’arco BF, quanto <lb/>l’arco BD ſiano di gr. </s> <s xml:id="echoid-s3015" xml:space="preserve">90. </s> <s xml:id="echoid-s3016" xml:space="preserve">Similmente ſi prouerà con alzare <lb/>dal punto E vna perpendicolare, e perciò parallela alla CD, <lb/>la quale cadendo nel punto F, ſarà indicio, che s’oprò giu-<lb/>ſtamente. </s> <s xml:id="echoid-s3017" xml:space="preserve">Perche eſſendo ſimili li triango li BCD, BEF, co-<lb/>me BD à BC, così BF, cioè BA à BE, per la 4 del lib. </s> <s xml:id="echoid-s3018" xml:space="preserve">6. </s> <s xml:id="echoid-s3019" xml:space="preserve">Ne <lb/>ſono inutili queſte proue, perche conuien’operare con eſſat-<lb/>tezza nel for mare lo ſtro mento.</s> <s xml:id="echoid-s3020" xml:space="preserve"/> </p> <div xml:id="echoid-div94" type="float" level="2" n="1"> <figure xlink:label="fig-0176-01" xlink:href="fig-0176-01a"> <image file="0176-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0176-01"/> </figure> </div> <p> <s xml:id="echoid-s3021" xml:space="preserve">Sia dunque ſo pra vna laſtra piana di rame, ò altra materia <lb/>piana conſiſtente, la linea RS longhezza della linea, che può <lb/>tirarſi nel lato dello ſtromento, e conforme al modo detto ſia <lb/>R C la corda di gr. </s> <s xml:id="echoid-s3022" xml:space="preserve">60. </s> <s xml:id="echoid-s3023" xml:space="preserve">Perciò all’interuallo CR fatto centro <lb/>in C, ſi deſcriua vn àrco, & </s> <s xml:id="echoid-s3024" xml:space="preserve">applicata l’apertura del Compaſ- <pb o="164" file="0178" n="180" rhead="CAPO VI."/> ſo dal punto R, ſi taglia l’arco nel punto 60. </s> <s xml:id="echoid-s3025" xml:space="preserve">Queſt’arco R 60 <lb/>diuiſo per metà, per la 30 del lib. </s> <s xml:id="echoid-s3026" xml:space="preserve">3. </s> <s xml:id="echoid-s3027" xml:space="preserve">darà il punto 30; </s> <s xml:id="echoid-s3028" xml:space="preserve">onde <lb/>la diſtanza di R 30 replicata dal punto 60, darà 60. </s> <s xml:id="echoid-s3029" xml:space="preserve">90, e così <lb/>R 90 ſarà il quadrante del cerchio, e ſi ſarà operato giuſta-<lb/>mente, ſe l’apertura R 90 comprenderà preciſa mente la li-<lb/>nea R S. </s> <s xml:id="echoid-s3030" xml:space="preserve">Così le ſolite ſubdiuiſioni daranno tutti li 90 gradi <lb/>del quadrante, quali conuien notare con grandiſſima eſatez-<lb/>za, quanto ſarà pòſſibile; </s> <s xml:id="echoid-s3031" xml:space="preserve">poiche diuiſo R 30 per metà, darà <lb/>R 15; </s> <s xml:id="echoid-s3032" xml:space="preserve">e diuiſo R 30 in tre parti vguali, darà R 10; </s> <s xml:id="echoid-s3033" xml:space="preserve">le quali <lb/>parti R 10, & </s> <s xml:id="echoid-s3034" xml:space="preserve">R 15 replicate, daranno la diuiſione di tutte le <lb/>decine per metà. </s> <s xml:id="echoid-s3035" xml:space="preserve">Sì che ſol reſta diuidere R 5 in cinque gradi <lb/>vguali: </s> <s xml:id="echoid-s3036" xml:space="preserve">il che forſi non riuſcirebbe così aggiuſtato, ſe ſi ten-<lb/>taſſe immediatamente replicando cinque volte la piccola <lb/>apertura del Compaſſo; </s> <s xml:id="echoid-s3037" xml:space="preserve">perciò prendo vn’ interuallo mag-<lb/>giore, e lo diuido con ogni diligenza in cinque parti vguali, e <lb/>ſia R 45, poiche la ſ<unsure/>ua quinta parte RI contine 9 gradi;</s> <s xml:id="echoid-s3038" xml:space="preserve">e così <lb/>queſt’a pertura replicata, caderà in O, E, V, cioè ne’gradi 18, <lb/>27, 36, e così di mano in mano. </s> <s xml:id="echoid-s3039" xml:space="preserve">Applicata poi queſta ſteſ-<lb/>ſa apertura alli punti già notati, e replicata conuenientemen-<lb/>te, verranno ad eſſer ſegnati tuttili gradi.</s> <s xml:id="echoid-s3040" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3041" xml:space="preserve">Che ſe più toſto voleſſimo prendere vn’interuallo minore, <lb/>e replicarlo più ſpeſſo (il che forſi non riuſcirà tanto accura-<lb/>to, poiche quanto più ſi replica il Compaſſo, la punta tanto <lb/>più ſpatio rubba) ſi può diuidere R 30 in cinque parti vgua-<lb/>li, ciaſcuna delle quali contiene 6 gradi, e replicato quell’in-<lb/>teruallo conuenientemente al modo detto, cominciando or <lb/>da vno, or da vn’altro de’punti già ſegnati, verranno ad eſſer <lb/>notati tutti li gradi.</s> <s xml:id="echoid-s3042" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3043" xml:space="preserve">Fatta queſta diuiſione del quadrantc ne’ſuoi gradi, ſi pren-<lb/>dano dal punto R gl’interualli à ciaſcun grado, e ſi notino <pb file="0179" n="181"/> <pb file="0179a" n="182" rhead="Capo VI."/> <anchor type="figure" xlink:label="fig-0179a-01a" xlink:href="fig-0179a-01"/> <pb file="0180" n="183"/> <pb o="165" file="0181" n="184" rhead="Gradi del Circolo"/> nella linea RS, e queſte ſono le corde di ciaſcuno di quegl’ar-<lb/>chi, che deuono notarſi nello ſtromento: </s> <s xml:id="echoid-s3044" xml:space="preserve">e perciò tali diuiſio-<lb/>ni deuono trasferirſi nelle linee AC, AQ dello ſtromento. <lb/></s> <s xml:id="echoid-s3045" xml:space="preserve">Se bene io conſegliarei più toſto prendere nell’arco R 90 <lb/>immediatamente le corde di ciaſcun’arco, e traſportarle sù <lb/>lo ſtromento; </s> <s xml:id="echoid-s3046" xml:space="preserve">poiche così pare l’operatione ſia per riuſcire <lb/>più eſatta.</s> <s xml:id="echoid-s3047" xml:space="preserve"/> </p> <div xml:id="echoid-div95" type="float" level="2" n="2"> <figure xlink:label="fig-0179a-01" xlink:href="fig-0179a-01a"> <image file="0179a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0179a-01"/> </figure> </div> <p> <s xml:id="echoid-s3048" xml:space="preserve">Da queſta coſtruttione, e dalle ragioni di ſopra più volte <lb/>addotte, ſi rende manifeſto, che eſſendo lilati AC, AQ diuiſi <lb/>nella proportione ditutte le corde de gl’archi del quadrante, <lb/>il cui ſemidiametro è A 60, data quaſſiuoglia apertura dello <lb/>ſtromento, l’interuallo 60. </s> <s xml:id="echoid-s3049" xml:space="preserve">60 ſarà la quantità del ſemidia. <lb/></s> <s xml:id="echoid-s3050" xml:space="preserve">metro del circolo, e tutti gl’altri interualli daranno le corde <lb/>de gl’archi corriſpondenti di detto circolo.</s> <s xml:id="echoid-s3051" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div97" type="section" level="1" n="53"> <head xml:id="echoid-head94" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head95" style="it" xml:space="preserve">Come ſi poſſa deſcriuer’ vn’angolo di quantità determinata.</head> <p> <s xml:id="echoid-s3052" xml:space="preserve">GIà ſi sà, che la quantità de gl’angoli ſi denomina dalla <lb/>moltitudine de’gradi del circolo, che habbia il centro <lb/>nel punto, doue s’vniſcono le due linee, che fanno l’angolo; <lb/></s> <s xml:id="echoid-s3053" xml:space="preserve">e la quantità de’gradi della circonferenza compreſa tra dette <lb/>due linee denomina l’angolo di tanti, ò tanti gradi. </s> <s xml:id="echoid-s3054" xml:space="preserve">Onde ne <lb/>viene, che douendoſi deſcriuer’vn’angolo, dall’eſtremo d’vna <lb/>linea data, come da centro à qualunque interuallo, ſi deſcri-<lb/>ue occultamente vn’arco minore della ſemicirconferenza, <lb/>più, ò meno, ſecondo che l’angolo deu’eſſer maggior, ò mi-<lb/>nore; </s> <s xml:id="echoid-s3055" xml:space="preserve">poiche dal punto, doue la data linea taglia la detta cir-<lb/>conferenza, prendendoſi l’arco della determinata quantità, <pb o="166" file="0182" n="185" rhead="CAPO VI."/> ſi trouerà il punto, per il quale dalcentro tirata vna linea <lb/>farà l’angolo cercato.</s> <s xml:id="echoid-s3056" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3057" xml:space="preserve">Debbaſi per cagione d’eſſempio deſcriuere l’Angolo del <lb/> <anchor type="figure" xlink:label="fig-0182-01a" xlink:href="fig-0182-01"/> centro d’vna Fortezza rego-<lb/>lare di cinque baloardi; </s> <s xml:id="echoid-s3058" xml:space="preserve">il <lb/>qual’è digr. </s> <s xml:id="echoid-s3059" xml:space="preserve">72. </s> <s xml:id="echoid-s3060" xml:space="preserve">Sia la linea <lb/>CL, che partendo dal centro <lb/>della fortezza, ſia inſieme <lb/>ſemidiametro del circolo, in <lb/>cui ſi deſcriue il Poligono in-<lb/>teriore. </s> <s xml:id="echoid-s3061" xml:space="preserve">Dal punto C, come <lb/>centro all’interuallo CL ſi deſcriua l’arco LM. </s> <s xml:id="echoid-s3062" xml:space="preserve">Poſcia nello <lb/>Stromento s’applichi la linea CL all’interuallo de’gradi 60. <lb/></s> <s xml:id="echoid-s3063" xml:space="preserve">60: </s> <s xml:id="echoid-s3064" xml:space="preserve">& </s> <s xml:id="echoid-s3065" xml:space="preserve">in quella apertura dello Stromento prendaſi l’interual-<lb/>1072. </s> <s xml:id="echoid-s3066" xml:space="preserve">72; </s> <s xml:id="echoid-s3067" xml:space="preserve">e queſto applicato all’arco deſcritto, ſarà LN. </s> <s xml:id="echoid-s3068" xml:space="preserve">Dun-<lb/>que dal punto C al punto N tirata la CN, ſarà LCN l’angolo <lb/>del centro d’vn Pentagono regolare, cioè digradi 72.</s> <s xml:id="echoid-s3069" xml:space="preserve"/> </p> <div xml:id="echoid-div97" type="float" level="2" n="1"> <figure xlink:label="fig-0182-01" xlink:href="fig-0182-01a"> <image file="0182-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0182-01"/> </figure> </div> <p> <s xml:id="echoid-s3070" xml:space="preserve">Mà ſe ſi voleſſe deſcriuere l’angolo del medeſimo Penta-<lb/>gono ſenza ſaperſi il centro della figura, per deſcriuerui vn <lb/>Baloardo, baſterà leuare l’angolo del centro, che è gr. </s> <s xml:id="echoid-s3071" xml:space="preserve">72 da <lb/> <anchor type="figure" xlink:label="fig-0182-02a" xlink:href="fig-0182-02"/> due Retti, cioè da 180, ereſtano <lb/>gr. </s> <s xml:id="echoid-s3072" xml:space="preserve">108. </s> <s xml:id="echoid-s3073" xml:space="preserve">Sia dunque la linea BA, <lb/>& </s> <s xml:id="echoid-s3074" xml:space="preserve">il punto A, doue deu’eſſer l’an-<lb/>golo, ſia centro dell’arco BO (pre-<lb/>ſo l’interuallo AB, ò tutto, come <lb/>in quefta figura, ò ſol parte d’vna <lb/>linea maggiore, ſe AB foſſe aſſai <lb/>più lunga) da cui ſi deuono pren-<lb/>dere gr. </s> <s xml:id="echoid-s3075" xml:space="preserve">108. </s> <s xml:id="echoid-s3076" xml:space="preserve">Nello Stromento <lb/>s’applica AB all’interuallo de’gr.</s> <s xml:id="echoid-s3077" xml:space="preserve"> <pb o="167" file="0183" n="186" rhead="Gradi del Circolo"/> 60. </s> <s xml:id="echoid-s3078" xml:space="preserve">60; </s> <s xml:id="echoid-s3079" xml:space="preserve">e perche non vi ſon notati ſe non i gradi del quadran-<lb/>te, e queſto angolo è aſſai maggiore, perciò con la ſteſſa <lb/>a pertura del Compaſſo prendo primieramente BC, che è <lb/>gradi 60; </s> <s xml:id="echoid-s3080" xml:space="preserve">e perche il reſiduo ſin alli 108, ſono gradi 48, pren-<lb/>do l’interuallo 48. </s> <s xml:id="echoid-s3081" xml:space="preserve">48, e lo trasferiſco in CD; </s> <s xml:id="echoid-s3082" xml:space="preserve">onde vien ad <lb/>eſſerel’arco BD gr. </s> <s xml:id="echoid-s3083" xml:space="preserve">108 e tirara la linea AD darà l’angolo del <lb/>pentagono BAD.</s> <s xml:id="echoid-s3084" xml:space="preserve"/> </p> <div xml:id="echoid-div98" type="float" level="2" n="2"> <figure xlink:label="fig-0182-02" xlink:href="fig-0182-02a"> <image file="0182-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0182-02"/> </figure> </div> <p> <s xml:id="echoid-s3085" xml:space="preserve">Ora ſe ſopra l’angolo BAD del pentagono voleſſimo de-<lb/>ſcriuere il baloardo col ſuo angolo proportionato, primiera-<lb/>mente ſi diuide l’angolo BAD per metà, onde eſſendo BD gr. <lb/></s> <s xml:id="echoid-s3086" xml:space="preserve">108, prendaſi nello Stromento l’interuallo 54. </s> <s xml:id="echoid-s3087" xml:space="preserve">54, e ſarà BE: </s> <s xml:id="echoid-s3088" xml:space="preserve"><lb/>e così applicata la riga alli punti AE, ſi tiri la Capitale I. </s> <s xml:id="echoid-s3089" xml:space="preserve">A, <lb/>che prolongata taglia per mezzo l’angolo del Poligono, e <lb/>giungerebbe ſin al centro. </s> <s xml:id="echoid-s3090" xml:space="preserve">Suppongaſi che in L debba eſſer <lb/>la punta del Baloardo. </s> <s xml:id="echoid-s3091" xml:space="preserve">E perche alla forma aſſai commune, <lb/>e pratticata ſi fà l’angolo del Baloardo, che ſia due terzi dell’ <lb/>angolo del Poligono, eſſendo queſto gr. </s> <s xml:id="echoid-s3092" xml:space="preserve">108, quello ſarà gr. </s> <s xml:id="echoid-s3093" xml:space="preserve"><lb/>72, & </s> <s xml:id="echoid-s3094" xml:space="preserve">il ſemiangolo del Baloardo gr. </s> <s xml:id="echoid-s3095" xml:space="preserve">36. </s> <s xml:id="echoid-s3096" xml:space="preserve">Fatto dunque centro <lb/>in L à qualunque interuallo, per eſſempio LM, ſi deſcriua vn <lb/>arco di quà, e di là; </s> <s xml:id="echoid-s3097" xml:space="preserve">& </s> <s xml:id="echoid-s3098" xml:space="preserve">applicata nello Stromento la linea LM <lb/>all’interuallo 60. </s> <s xml:id="echoid-s3099" xml:space="preserve">60, prendaſi l’interuallo 36. </s> <s xml:id="echoid-s3100" xml:space="preserve">36, & </s> <s xml:id="echoid-s3101" xml:space="preserve">applica-<lb/>to nell’arco deſcritto, dal punto M ſi prenda vguale MN, & </s> <s xml:id="echoid-s3102" xml:space="preserve"><lb/>MO: </s> <s xml:id="echoid-s3103" xml:space="preserve">e tirate le linee LN, LO, ſarà l’angolo del Baloardo <lb/>NLO di gr. </s> <s xml:id="echoid-s3104" xml:space="preserve">72, come ſi richiedeua.</s> <s xml:id="echoid-s3105" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3106" xml:space="preserve">Che ſe occorreſſe deſcriuer vn’angolo, che oltre li gradi <lb/>haueſſe anco li minuti, conuien auuertire, ſe la figura da de-<lb/>ſcriuerſi è grande, ò pur piccola; </s> <s xml:id="echoid-s3107" xml:space="preserve">perche nelle piccole vua <lb/>cotal differenza di minuti non è notabile: </s> <s xml:id="echoid-s3108" xml:space="preserve">onde ſe li minuti <lb/>ſono aſſai meno di 30, ſi puonno laſciare, ſe paſſano notabil-<lb/>mente li 30, ſi puonno prendere per vn grado di più; </s> <s xml:id="echoid-s3109" xml:space="preserve">così in <pb o="168" file="0184" n="187" rhead="CAPO VI."/> vece digr. </s> <s xml:id="echoid-s3110" xml:space="preserve">10. </s> <s xml:id="echoid-s3111" xml:space="preserve">m. </s> <s xml:id="echoid-s3112" xml:space="preserve">12. </s> <s xml:id="echoid-s3113" xml:space="preserve">baſta prendere nello Stromento l’inter-<lb/>uallo 10. </s> <s xml:id="echoid-s3114" xml:space="preserve">10: </s> <s xml:id="echoid-s3115" xml:space="preserve">& </s> <s xml:id="echoid-s3116" xml:space="preserve">in vece digr. </s> <s xml:id="echoid-s3117" xml:space="preserve">10. </s> <s xml:id="echoid-s3118" xml:space="preserve">m. </s> <s xml:id="echoid-s3119" xml:space="preserve">49. </s> <s xml:id="echoid-s3120" xml:space="preserve">ſi può prendere nello <lb/>Stromento l’interuallo 11. </s> <s xml:id="echoid-s3121" xml:space="preserve">11. </s> <s xml:id="echoid-s3122" xml:space="preserve">Che ſe li minuti aggionti alli <lb/>gradi s’auuicinano più, ò meno alli 30, ſi puonno pigliare <lb/>nello Stromento li due numeri vicini, cioè il minore in vn <lb/>braccio, & </s> <s xml:id="echoid-s3123" xml:space="preserve">il maggiore nell’altro braccìo dello Stromento; <lb/></s> <s xml:id="echoid-s3124" xml:space="preserve">così per gr. </s> <s xml:id="echoid-s3125" xml:space="preserve">10. </s> <s xml:id="echoid-s3126" xml:space="preserve">m. </s> <s xml:id="echoid-s3127" xml:space="preserve">28, ouero per gr. </s> <s xml:id="echoid-s3128" xml:space="preserve">10. </s> <s xml:id="echoid-s3129" xml:space="preserve">m. </s> <s xml:id="echoid-s3130" xml:space="preserve">36. </s> <s xml:id="echoid-s3131" xml:space="preserve">ſi può prende-<lb/>re nello Stromento l’interuallo 10. </s> <s xml:id="echoid-s3132" xml:space="preserve">11, & </s> <s xml:id="echoid-s3133" xml:space="preserve">ſarà proſſimamen-<lb/>te ciò che ſi deſidera. </s> <s xml:id="echoid-s3134" xml:space="preserve">Ma ſela figura foſſe notabilmente <lb/>grande, in tal caſo conuerrà deſcriuer vn arco con vna grand’ <lb/>apertura di Compaſſo, ſiche il ſemidiametro ſia grande da <lb/>applicarſi all’interuallo 60. </s> <s xml:id="echoid-s3135" xml:space="preserve">60, dipoi ſi prenda nell’ arco de-<lb/>ſcritto il numero de’gradi intieri, e poi il numero d’vn grado <lb/>di più, e quella differenza à occhio ſi può diuidere ſecondo il <lb/>numero de’minuti aggionti; </s> <s xml:id="echoid-s3136" xml:space="preserve">così per l’angolo digr. </s> <s xml:id="echoid-s3137" xml:space="preserve">10. </s> <s xml:id="echoid-s3138" xml:space="preserve">m. </s> <s xml:id="echoid-s3139" xml:space="preserve">12, <lb/>prendo prima l’interuallo 10. </s> <s xml:id="echoid-s3140" xml:space="preserve">10, e poil’interuallo 11. </s> <s xml:id="echoid-s3141" xml:space="preserve">11, e <lb/>ſegnati nell’arco deſcritto, piglio à occhio la quinta parte <lb/>della differenza tra queſti due ſegni, che corriſponde alli mi-<lb/>nuti 12; </s> <s xml:id="echoid-s3142" xml:space="preserve">e tirata la linea darà l’angolo deſiderato.</s> <s xml:id="echoid-s3143" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div100" type="section" level="1" n="54"> <head xml:id="echoid-head96" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head97" style="it" xml:space="preserve">Come ſi eonoſca la grandezza, e quantità d’vn’angolo dato.</head> <p> <s xml:id="echoid-s3144" xml:space="preserve">DA ciò, che s’è detto nella precedente Queſtione è coſa <lb/>faciliſſima, ſe ſarà dato vn’angolo, conoſcere deter-<lb/>minatamente in gradi, quanta ſia la ſua grandezza, fatto cen-<lb/>tro nel punto, oue le due linee ſi toccano, & </s> <s xml:id="echoid-s3145" xml:space="preserve">à qualunquein-<lb/>teruallo deſcritto vn arco, che tagli amendue quelle linee, <lb/>perche applicata la larghezza del Compaſſo, alla cui apertura <pb o="169" file="0185" n="188" rhead="Gradi del Circolo"/> ſi deſcriſſe l’arco alli punti 60. </s> <s xml:id="echoid-s3146" xml:space="preserve">60, dello Stromento poſcia <lb/>co’l Compaſſo preſa la grandezza dell’arco deſcritto com-<lb/>preſo tra le due linee date, s’applichi allo Stromento, & </s> <s xml:id="echoid-s3147" xml:space="preserve">ap-<lb/>parirà di quanti gradi ſia l’angolo dato. </s> <s xml:id="echoid-s3148" xml:space="preserve">Così le due linee RS, <lb/> <anchor type="figure" xlink:label="fig-0185-01a" xlink:href="fig-0185-01"/> RT fanno l’angolo SRT, la cui quantità ſi <lb/>deſidera conoſcere. </s> <s xml:id="echoid-s3149" xml:space="preserve">Dal punto R all’inter-<lb/>uallo RA deſcriuo l’arco AB occulto (ouero <lb/>per più facilità ſegno le due linee ne’punti A, <lb/>e B ſenza deſcriuere l’arco) e l’apertura del <lb/>Compaſſo RA applico all’interuallo 60. </s> <s xml:id="echoid-s3150" xml:space="preserve">60 <lb/>nello Stromento. </s> <s xml:id="echoid-s3151" xml:space="preserve">Dipoi prendo col Com-<lb/>paſſo la diſtanza AB, & </s> <s xml:id="echoid-s3152" xml:space="preserve">applicata allo Stro-<lb/>mento ritenuto nella ſteſſa apertura, trouo, <lb/>che caſca all’interuallo 25<emph style="sub">3</emph><unsure/>. </s> <s xml:id="echoid-s3153" xml:space="preserve">25 {1/3}, e così dico l’angolo SRT <lb/>eſſere digr. </s> <s xml:id="echoid-s3154" xml:space="preserve">25. </s> <s xml:id="echoid-s3155" xml:space="preserve">m. </s> <s xml:id="echoid-s3156" xml:space="preserve">20.</s> <s xml:id="echoid-s3157" xml:space="preserve"/> </p> <div xml:id="echoid-div100" type="float" level="2" n="1"> <figure xlink:label="fig-0185-01" xlink:href="fig-0185-01a"> <image file="0185-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0185-01"/> </figure> </div> <p> <s xml:id="echoid-s3158" xml:space="preserve">Similmente ſe ſarà tirata la linea TS, e fatto il triangolo, <lb/>conoſcerò, quanto ſia l’ang. </s> <s xml:id="echoid-s3159" xml:space="preserve">S, ſe alla lunghezza ST prendeiò <lb/>vguale SC, & </s> <s xml:id="echoid-s3160" xml:space="preserve">applicata queſta lunghezza ST alli punti 60. <lb/></s> <s xml:id="echoid-s3161" xml:space="preserve">60 dello Stromento, prenderò col Compaſſo la diſtanza TC, <lb/>e ritenuta la ſteſſa apertura dello Stromento, trouando, che <lb/>la diſtanza TC s’applica giuſtamente nello Stromento all’in-<lb/>teruallo 90. </s> <s xml:id="echoid-s3162" xml:space="preserve">90, dico che l’angolo Sè retto, e perciò l’angolo <lb/>T è il complemento dell’angolo R, e per conſeguenza è di <lb/>gr. </s> <s xml:id="echoid-s3163" xml:space="preserve">64. </s> <s xml:id="echoid-s3164" xml:space="preserve">m. </s> <s xml:id="echoid-s3165" xml:space="preserve">40.</s> <s xml:id="echoid-s3166" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3167" xml:space="preserve">Di qui è manifeſto il modo di cauare dall’ombra d’vn cor-<lb/>po, la cui altezza è conoſciuta, quanta ſia l’altezza del Sole <lb/>ſopra l’Orizonte. </s> <s xml:id="echoid-s3168" xml:space="preserve">Sia dunque l’altezza perpendicolare d’vn <lb/>baſtone piedi 6, e miſurando la longhezza dell’ombra, trouo <lb/>che è piedi 2. </s> <s xml:id="echoid-s3169" xml:space="preserve">oncie 10 {1/2}. </s> <s xml:id="echoid-s3170" xml:space="preserve">Si che queſte due miſure ſono oncie <lb/>72, & </s> <s xml:id="echoid-s3171" xml:space="preserve">oncie 34 {1/2}. </s> <s xml:id="echoid-s3172" xml:space="preserve">Dunque alargatolo Stromento à mio pia- <pb o="170" file="0186" n="189" rhead="CAPO VI."/> cere, prendo nella linea Aritmetica l’interuallo 72. </s> <s xml:id="echoid-s3173" xml:space="preserve">72, & </s> <s xml:id="echoid-s3174" xml:space="preserve">in <lb/>vn piano delcriuo à tal’interuallo vguale la linea RS: </s> <s xml:id="echoid-s3175" xml:space="preserve">e poi <lb/>preſo l’interuallo 34 {1/2}. </s> <s xml:id="echoid-s3176" xml:space="preserve">34 {1/2}, glideſcriuo vguale la linea ST, <lb/>che cade perpendicolarmente in S. </s> <s xml:id="echoid-s3177" xml:space="preserve">Quindi tirata la linea RT <lb/>moſtrarà il raggio del ſole, come RS rappreſenta l’altezza <lb/>del baſtone, & </s> <s xml:id="echoid-s3178" xml:space="preserve">ST la longhezza dell’ ombra. </s> <s xml:id="echoid-s3179" xml:space="preserve">Cerco dunque <lb/>nel modo detto di ſopra la quantità dell’angolo T, e queſta è <lb/>l’altezza del Sole ſopra l’Orizonte.</s> <s xml:id="echoid-s3180" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3181" xml:space="preserve">Di queſto modo potranno ſeruirſi iPittori, per non far <lb/>l’ombre de’corpi, ò troppo corte, òtroppo lunghe, quando <lb/>la coſa dipinta rappreſenta vn fatto operato in ora determi-<lb/>nata del giorno in vn luogo determinato; </s> <s xml:id="echoid-s3182" xml:space="preserve">perche per eſſem-<lb/>pio ſe ſi dourà dipinger il Miracolo di S. </s> <s xml:id="echoid-s3183" xml:space="preserve">Pietro, quando riſa-<lb/>nò lo ſtorpiato alla Porta ſpecioſa del Tempio di Gierufa-<lb/>lemme, biſogna auuertire di non far l’ombre delle fabriche in <lb/>modo, che non corriſpondano con le altezze, all’hora nona, <lb/>cioè tre ore doppo mezzo dì (parlando dell’ ore diſuguali) <lb/>circa il fine di Maggio in Gierufalemme. </s> <s xml:id="echoid-s3184" xml:space="preserve">Che ſe bene nonè <lb/>neceſſaria in ciò vna certa preciſione Mattematica per l’vſo <lb/>de’ Pittori, ad ogni modo ſi può errare aſſai in ciò, e moſtra. <lb/></s> <s xml:id="echoid-s3185" xml:space="preserve">re d’hauer fatto l’ombre, & </s> <s xml:id="echoid-s3186" xml:space="preserve">il ſito del Sole à caſo.</s> <s xml:id="echoid-s3187" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3188" xml:space="preserve">Mà ſe l’angolo dato foſſe così grande, che deſcritto l’arco, <lb/>non ſi poteſſe nello Stromento trouare la ſua quantità, ſi po-<lb/>trà prender in due volte: </s> <s xml:id="echoid-s3189" xml:space="preserve">Come nella figura della queſtione <lb/>precedente l’angolo BAD è tale, che aperto lo Stromento <lb/>all’interuallo AB applicato alli punti 60. </s> <s xml:id="echoid-s3190" xml:space="preserve">60, la diſtanza <lb/>BD non capiſce nello Stromento, perciò preſo ad arbitrio <lb/>vn’interuallo, pereſſempio 80. </s> <s xml:id="echoid-s3191" xml:space="preserve">80, & </s> <s xml:id="echoid-s3192" xml:space="preserve">applicato all’arco de-<lb/>ſcritto BD, ſaranno BI gr. </s> <s xml:id="echoid-s3193" xml:space="preserve">80; </s> <s xml:id="echoid-s3194" xml:space="preserve">il reſto dell’arco ID applico al-<lb/>lo Stromento, ecade nell’interuallo 28. </s> <s xml:id="echoid-s3195" xml:space="preserve">28; </s> <s xml:id="echoid-s3196" xml:space="preserve">onde alli gradi <pb o="171" file="0187" n="190" rhead="Gradi del Circolo"/> 80. </s> <s xml:id="echoid-s3197" xml:space="preserve">aggiontigradi 28, tutto l’arco BD, e per conſeguenzala <lb/>quantità dell’angolo dato BAD, ègr. </s> <s xml:id="echoid-s3198" xml:space="preserve">108.</s> <s xml:id="echoid-s3199" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div102" type="section" level="1" n="55"> <head xml:id="echoid-head98" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head99" style="it" xml:space="preserve">come con lo Stromento ſi poſa pratticare tutta la Trigonometria <lb/>ſenza Tauole.</head> <p> <s xml:id="echoid-s3200" xml:space="preserve">SE Bene di queſto ſi parlò qualche coſa nel cap. </s> <s xml:id="echoid-s3201" xml:space="preserve">2. </s> <s xml:id="echoid-s3202" xml:space="preserve">Queſt. <lb/></s> <s xml:id="echoid-s3203" xml:space="preserve">6, ad ogni modo ſarà meglio più vniuerſalmente ſpie-<lb/>gare quì l’vſo dello Stromento nella ſolutione prattica de’ <lb/>triangoli, e ſeruirà per quelli che non ſi curano di tanta pre-<lb/>ciſione, quanta oprando co’numeri ſi troua coforme alle re-<lb/>gole della Trigonometria.</s> <s xml:id="echoid-s3204" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3205" xml:space="preserve">E quì ſuppongo ciò che è noto, che delle ſei parti, cioè di <lb/>tre lati, etre angoli, che ſono in vn triangolo, conuien ſaper-<lb/>ne tre, per conoſcere l’altre tre. </s> <s xml:id="echoid-s3206" xml:space="preserve">Se ſono datitutti tre gl’an-<lb/>goli, non ſi può conoſcere, quanta ſia la longhezza de’lati, <lb/>ma ſolo la proportione, che li lati hanno tra di loro, eſſendo-<lb/>che li triangoli equiangoli, eſimili tra di loro, hanno ben ſi i <lb/>lati proportionali, ma non vguali. </s> <s xml:id="echoid-s3207" xml:space="preserve">Onde ſe ſaranno dati tre <lb/> <anchor type="figure" xlink:label="fig-0187-01a" xlink:href="fig-0187-01"/> angoli d’vn triangolo, facciaſi qualunque <lb/>triangolo con detti tre angoli, enella linea <lb/>Aritmet. </s> <s xml:id="echoid-s3208" xml:space="preserve">applicato vno de’lati all’interuallo, <lb/>che più piacerà, ſi troueranno gl’altri, e ſarà <lb/>manifeſta la lor proportione. </s> <s xml:id="echoid-s3209" xml:space="preserve">Siano litte <lb/>angoli dati gr. </s> <s xml:id="echoid-s3210" xml:space="preserve">25. </s> <s xml:id="echoid-s3211" xml:space="preserve">m. </s> <s xml:id="echoid-s3212" xml:space="preserve">20, gr. </s> <s xml:id="echoid-s3213" xml:space="preserve">19. </s> <s xml:id="echoid-s3214" xml:space="preserve">m. </s> <s xml:id="echoid-s3215" xml:space="preserve">40, gradi <lb/>135. </s> <s xml:id="echoid-s3216" xml:space="preserve">Sopra la linea RT, faccio l’angolo <lb/>TRC gr. </s> <s xml:id="echoid-s3217" xml:space="preserve">25. </s> <s xml:id="echoid-s3218" xml:space="preserve">m. </s> <s xml:id="echoid-s3219" xml:space="preserve">20, el’angolo RTC digradi <lb/>19. </s> <s xml:id="echoid-s3220" xml:space="preserve">m. </s> <s xml:id="echoid-s3221" xml:space="preserve">40, ecosì rieſce il terzo angolo TCR <pb o="172" file="0188" n="191" rhead="CAPO VI."/> gradi 135. </s> <s xml:id="echoid-s3222" xml:space="preserve">Ora applico la linea RT nella linea Aritmetica <lb/>all’interuallo 80. </s> <s xml:id="echoid-s3223" xml:space="preserve">80, eritenuta quell’apertura dello Stromen-<lb/>to, veggo che il lato RC cade all’interuallo 38. </s> <s xml:id="echoid-s3224" xml:space="preserve">38, & </s> <s xml:id="echoid-s3225" xml:space="preserve">il lato <lb/>CT cade all’interuallo 48. </s> <s xml:id="echoid-s3226" xml:space="preserve">48, dal che cauo la proportione <lb/>de’tre lati eſſere 160, 76, 96.</s> <s xml:id="echoid-s3227" xml:space="preserve"/> </p> <div xml:id="echoid-div102" type="float" level="2" n="1"> <figure xlink:label="fig-0187-01" xlink:href="fig-0187-01a"> <image file="0187-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0187-01"/> </figure> </div> <p> <s xml:id="echoid-s3228" xml:space="preserve">Mà ſe ſaranno dati li tre lati d’vn triangolo, ſi troueranno <lb/>li tre angoli, prendendo nella linea Aritmetica tre interualli <lb/>nella proportione de’lati dati; </s> <s xml:id="echoid-s3229" xml:space="preserve">e formatone vn triangolo, ſi <lb/>cerchi la quantità di due angoli nel modo detto nella Que-<lb/>ſtione precedente, perche il terzo angolo ſarà noto, eſſendo <lb/>il complemento ſin a’ gradi 180. </s> <s xml:id="echoid-s3230" xml:space="preserve">Così date le diſtanze di tre <lb/>luoghi di paſſi 160. </s> <s xml:id="echoid-s3231" xml:space="preserve">76. </s> <s xml:id="echoid-s3232" xml:space="preserve">96, prendo nella linea Aritmetica <lb/>gl’interualli della metà di detti num. </s> <s xml:id="echoid-s3233" xml:space="preserve">cioè 80. </s> <s xml:id="echoid-s3234" xml:space="preserve">38. </s> <s xml:id="echoid-s3235" xml:space="preserve">48, e forma-<lb/>to il triangolo TCR, cerco come ſopra s’è detto gl’angoli R, <lb/>& </s> <s xml:id="echoid-s3236" xml:space="preserve">T, e così ſi fà noto anche il terzo angolo.</s> <s xml:id="echoid-s3237" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3238" xml:space="preserve">Mà ſe non foſſero date le miſure delli trè lati, eſo<unsure/>lamente <lb/>foſſe propoſ<unsure/>to vn triangolo, dicui ſi deſidera ſapere gli ango-<lb/>li:</s> <s xml:id="echoid-s3239" xml:space="preserve">circa il Triangolo ſi deſcriua il circolo per la 5. </s> <s xml:id="echoid-s3240" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3241" xml:space="preserve">4. <lb/></s> <s xml:id="echoid-s3242" xml:space="preserve">(cioè ſi troui il centro, e da quel punto ſin all’eſtremità d’vno <lb/>de gliangoli ſi prenda la diſtanza, che è il Raggio del circo-<lb/>lo) & </s> <s xml:id="echoid-s3243" xml:space="preserve">ilſemidiametro di tal circolo portato tra li punti 60. </s> <s xml:id="echoid-s3244" xml:space="preserve"><lb/>60, veggaſi à qual interuallo capiſca ciaſcuno de’lati dati; </s> <s xml:id="echoid-s3245" xml:space="preserve"><lb/>poiche il numero corriſpondente nello Stromento, darà il <lb/>doppio dell’angolo oppoſto allato applicato: </s> <s xml:id="echoid-s3246" xml:space="preserve">eſſendoche tal <lb/>lato è Corda dell’ arco notato, & </s> <s xml:id="echoid-s3247" xml:space="preserve">è ſottenſa all’angolo fatto <lb/>nelcentro, che è doppio dell’angolo alla circonferenza, qual <lb/>è l’angolo cercato oppoſto al lato dato.</s> <s xml:id="echoid-s3248" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3249" xml:space="preserve">Quando li dati ſono miſti d’angoli, elati, ò ſono due an-<lb/>goli, & </s> <s xml:id="echoid-s3250" xml:space="preserve">vn lato, ò due lati, & </s> <s xml:id="echoid-s3251" xml:space="preserve">vn angolo: </s> <s xml:id="echoid-s3252" xml:space="preserve">e queſto in due ma-<lb/>niere, poiche è il lato adiacente alli due angoli dati, ouero <pb o="173" file="0189" n="192" rhead="Gradi del Circolo"/> oppoſto ad vn di loro; </s> <s xml:id="echoid-s3253" xml:space="preserve">e ſimilmente ò è l’angolo compreſo <lb/>dalli due lati dati, ouero oppoſto ad vno di detti lati.</s> <s xml:id="echoid-s3254" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3255" xml:space="preserve">Sia dato vn lato, e gl’angoli adiacenti; </s> <s xml:id="echoid-s3256" xml:space="preserve">ſia AB parte delſa <lb/>riua d’vn fiume, conoſciuta in miſura di piedi 90; </s> <s xml:id="echoid-s3257" xml:space="preserve">e ſi deſideri <lb/>ſapere la diſtanza AC, che trauerſa il fiume. </s> <s xml:id="echoid-s3258" xml:space="preserve">Sia oſſeruato in <lb/>A l’angolo CAB, di gradi 78, & </s> <s xml:id="echoid-s3259" xml:space="preserve">in B l’angolo ABC di gradi <lb/>35; </s> <s xml:id="echoid-s3260" xml:space="preserve">deſcriuo nell’eſtremità della linea AB li due angoli con-<lb/>forme alle ſopradette miſure oſſeruate, cioè ABC gr. </s> <s xml:id="echoid-s3261" xml:space="preserve">35, e <lb/>BAC gr. </s> <s xml:id="echoid-s3262" xml:space="preserve">78; </s> <s xml:id="echoid-s3263" xml:space="preserve">onde le linee BC, AC ſi rincontrano in C. </s> <s xml:id="echoid-s3264" xml:space="preserve">Ap-<lb/>plicata dunque la linea AB sù la linea Aritmetica alli punti <lb/>90. </s> <s xml:id="echoid-s3265" xml:space="preserve">90, trouo, che AC cade nell’interuallo 56. </s> <s xml:id="echoid-s3266" xml:space="preserve">56, dal che cõ-<lb/> <anchor type="figure" xlink:label="fig-0189-01a" xlink:href="fig-0189-01"/> chiudo, che la diſtanza dal <lb/>punto A al punto C, che tra-<lb/>uerſa il fiume è di piedi 56: <lb/></s> <s xml:id="echoid-s3267" xml:space="preserve">e così la diſtanza BC è di <lb/>piedi 95 {1/2}.</s> <s xml:id="echoid-s3268" xml:space="preserve"/> </p> <div xml:id="echoid-div103" type="float" level="2" n="2"> <figure xlink:label="fig-0189-01" xlink:href="fig-0189-01a"> <image file="0189-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0189-01"/> </figure> </div> <p> <s xml:id="echoid-s3269" xml:space="preserve">Mà ſe foſſe dato illato A <lb/>B con l’angolo B adiacente, <lb/>e l’angolo C oppoſto, ſarà <lb/>anche noto il terzo angolo <lb/>A, che è complemento alli <lb/>due retti; </s> <s xml:id="echoid-s3270" xml:space="preserve">e così ſi deſcriuerà la figura, come ſe foſſe dato il <lb/>lato con li due angoli B, & </s> <s xml:id="echoid-s3271" xml:space="preserve">A adiacenti, e s’operarà, come <lb/>poco fà ſi diceua.</s> <s xml:id="echoid-s3272" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3273" xml:space="preserve">Ora ſian dati due lati con l’angolo compreſo: </s> <s xml:id="echoid-s3274" xml:space="preserve">deſcriuaſi <lb/>l’angolo dato, come s’è detto nella prima Queſtione, e ſi <lb/>prenda la lunghezza de’lati proportionata à ilati dati; </s> <s xml:id="echoid-s3275" xml:space="preserve">poile <lb/>eſtremità de’lati ſi congiungano, e s’haurà il triangolo, in cui <lb/>ſi conoſceranno l’altre parti, come ſopra. </s> <s xml:id="echoid-s3276" xml:space="preserve">Sia nella figura <lb/>antecedente, dato l’angolo compreſo dalli lati dati di gr. </s> <s xml:id="echoid-s3277" xml:space="preserve">25.</s> <s xml:id="echoid-s3278" xml:space="preserve"> <pb o="174" file="0190" n="193" rhead="CAPO VI."/> 20. </s> <s xml:id="echoid-s3279" xml:space="preserve">& </s> <s xml:id="echoid-s3280" xml:space="preserve">il lato RT ſia paſſi 92, & </s> <s xml:id="echoid-s3281" xml:space="preserve">RS paſſi 83; </s> <s xml:id="echoid-s3282" xml:space="preserve">& </s> <s xml:id="echoid-s3283" xml:space="preserve">appunto con <lb/>tal proportione ſiano le linee RT, RS: </s> <s xml:id="echoid-s3284" xml:space="preserve">tiro la linea TS; </s> <s xml:id="echoid-s3285" xml:space="preserve">& </s> <s xml:id="echoid-s3286" xml:space="preserve">ap-<lb/>plicata RT nella linea Aritmetica all’interuallo 92. </s> <s xml:id="echoid-s3287" xml:space="preserve">92, tro-<lb/>uo che TS cadendo nell’interuallo 40. </s> <s xml:id="echoid-s3288" xml:space="preserve">40, moſtra che la di-<lb/>ſtanza di S da T è di paſſi 40. </s> <s xml:id="echoid-s3289" xml:space="preserve">Così cercando nel modo ſpie-<lb/>gato nella 2. </s> <s xml:id="echoid-s3290" xml:space="preserve">Queſtione, ſi trouerà l’angolo S retto, e l’altro <lb/>reſta noto, per eſſer il complemento delli due conoſciuti ſin’à <lb/>gradi 180.</s> <s xml:id="echoid-s3291" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3292" xml:space="preserve">Siano finalmente dati due lati, & </s> <s xml:id="echoid-s3293" xml:space="preserve">vn angolo oppoſto ad <lb/>vno diloro. </s> <s xml:id="echoid-s3294" xml:space="preserve">In queſto caſo conuien oſſeruare ſe l’angolo da-<lb/>to è oppoſto allato maggiore, ò pur al minore de’dati; </s> <s xml:id="echoid-s3295" xml:space="preserve">per-<lb/>che ſe è oppoſto al lato maggiore, non v’è biſogno d’altra <lb/>precognitione; </s> <s xml:id="echoid-s3296" xml:space="preserve">mà ſe foſſe oppoſto allato minore, allhora <lb/>può darſi caſo, in cui ſia neceſſario ſaperela ſpecie dell’ango-<lb/>lo oppoſto allato maggiore, cioè ſe ſia ottuſo, ò pur acuto. <lb/></s> <s xml:id="echoid-s3297" xml:space="preserve">ll che ſi vedrà chiaramente dalla prattica, che quì ſoggionge-<lb/>rò. </s> <s xml:id="echoid-s3298" xml:space="preserve">Sia dato vn’angolo di gr. </s> <s xml:id="echoid-s3299" xml:space="preserve">67. </s> <s xml:id="echoid-s3300" xml:space="preserve">oppoſto ad vn lato di piedi <lb/>90, & </s> <s xml:id="echoid-s3301" xml:space="preserve">adiacente ad vn lato di piedi 56. </s> <s xml:id="echoid-s3302" xml:space="preserve">Tiro la linea CA di <lb/>piedi 56, e faccio l’angolo C di gr. </s> <s xml:id="echoid-s3303" xml:space="preserve">67. </s> <s xml:id="echoid-s3304" xml:space="preserve">tirando la CB indefi-<lb/>nita. </s> <s xml:id="echoid-s3305" xml:space="preserve">Poi nella linea Aritmetica poſto il lato CA all’inter-<lb/>uallo 56. </s> <s xml:id="echoid-s3306" xml:space="preserve">56, prendo l’interuallo 90. </s> <s xml:id="echoid-s3307" xml:space="preserve">90, e dal punto A, come <lb/>da centro deſcriuo con quell’apertura di Compaſſo vn’arco, <lb/>che taglia l’indefinita CB nel punto B: </s> <s xml:id="echoid-s3308" xml:space="preserve">e così tirata la retta <lb/>AB, ſarà l’altro lato de’dati oppoſto all’angolo dato: </s> <s xml:id="echoid-s3309" xml:space="preserve">onde <lb/>ſarà conſtituito tutto il triangolo ABC, e nel modo detto ſi <lb/>conoſceranno l’altre parti incognite. </s> <s xml:id="echoid-s3310" xml:space="preserve">Ora perche la linea <lb/>AB è maggiore, che AC, è manifeſto chel’arco occulto de-<lb/>ſcritto non taglia l’indefinita CB, ſe non nel pnnto B da que-<lb/>ſta parte oppoſta all’angolo dato: </s> <s xml:id="echoid-s3311" xml:space="preserve">e così il lato dato non può <lb/>hauer altra poſitura che AB.</s> <s xml:id="echoid-s3312" xml:space="preserve"/> </p> <pb o="175" file="0191" n="194" rhead="Gradi del Circolo"/> <p> <s xml:id="echoid-s3313" xml:space="preserve">Mà ſe dato l’iſteſſo angolo C gr. </s> <s xml:id="echoid-s3314" xml:space="preserve">67. </s> <s xml:id="echoid-s3315" xml:space="preserve">il lato adiacente foſſe <lb/>70 piedi, cioè C D, & </s> <s xml:id="echoid-s3316" xml:space="preserve">il lato oppoſto foſſe piedi 65, applicata <lb/>C D nella linea Aritmetica all’interuallo 70. </s> <s xml:id="echoid-s3317" xml:space="preserve">70, e preſa la di-<lb/>ſtanza 65. </s> <s xml:id="echoid-s3318" xml:space="preserve">65, deſcritto dal centro D vn’arco, che tocchi l’in-<lb/>definita C B nel punto E, tirata la linea D E, è manifeſto, che <lb/>l’angolo D E C è retto, ne altra può eſſere la poſitione del lato <lb/>oppoſto di piedi 65.</s> <s xml:id="echoid-s3319" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3320" xml:space="preserve">Che ſe finalmente dati gl’iſteſſi lati di piedi 90, e piedi 56, <lb/>ſia dato l’angolo B digr. </s> <s xml:id="echoid-s3321" xml:space="preserve">35. </s> <s xml:id="echoid-s3322" xml:space="preserve">oppoſto allato minore, preſa <lb/>A C di tali parti 56, delle quali A B è 90, e dal punto A deſcrit-<lb/>to vn’arco, ſi vede, che tagſia l’indefinita B C in due punti C, <lb/>& </s> <s xml:id="echoid-s3323" xml:space="preserve">l, e così non ſappiamo ſe dobbiamo più toſto ſeruirci della <lb/>A C, ò pure della A I, ſe non ſi sà, ſe l’angolo oppoſto al lato <lb/>maggiore dato A B, ſia acuto, come A C B, ò pur ottuſo, co-<lb/>me A I B.</s> <s xml:id="echoid-s3324" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div105" type="section" level="1" n="56"> <head xml:id="echoid-head100" style="it" xml:space="preserve">QVESTIONE QVARTA.</head> <head xml:id="echoid-head101" style="it" xml:space="preserve">Trouar in numeri la proportione di due rette con l’ aiuto <lb/>delle T auole de’ Seni.</head> <p> <s xml:id="echoid-s3325" xml:space="preserve">COn tutto, che nell’ vſo della linea Aritmetica dello <lb/>Stromento ſi ſia moſtrato, come poſſa trouarſi la pro-<lb/>portione di due linee date, ad ogni modo chi deſideraſſe <lb/>auuicinarſi anche più alla preciſione, & </s> <s xml:id="echoid-s3326" xml:space="preserve">eſprimerla con nu-<lb/>meri maggiori, potria ſeruirſi di queſta linea de’ gradi, doue <lb/>ſono notate le corde de gl’archi del Quadrante: </s> <s xml:id="echoid-s3327" xml:space="preserve">le quali cor-<lb/>de ſono il doppio del ſeno della metà dell’arco: </s> <s xml:id="echoid-s3328" xml:space="preserve">cosila metà <lb/>della corda di gradi 74, è il ſeno di gradi 37.</s> <s xml:id="echoid-s3329" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3330" xml:space="preserve">Date dunque due linee, la maggiote s’applichi in queſta <pb o="176" file="0192" n="195" rhead="CAPO VI."/> linea de’gradi all’interuallo 60. </s> <s xml:id="echoid-s3331" xml:space="preserve">60, e s’intenderà diuiſa in tan-<lb/>te particelle, di quante è il raggio delle Tauole de’ Seni, poi <lb/>la linea minore delle date ſi vegga à qual interuallo preciſa-<lb/>mente cade nella ſteſſa linea de’ gradi dello Stromento, e <lb/>prendaſi la metà di detti gradi, il cui ſeno trouato nelle tauo-<lb/>le ſi raddoppia, e ſi hà il numero corriſpondente alle particel-<lb/>le contenute nella linea minore data: </s> <s xml:id="echoid-s3332" xml:space="preserve">Come ſe delle due li-<lb/>nee R T, R S, nella figura dell’ antecedente queſtione 3. </s> <s xml:id="echoid-s3333" xml:space="preserve">pag. <lb/></s> <s xml:id="echoid-s3334" xml:space="preserve">171. </s> <s xml:id="echoid-s3335" xml:space="preserve">io cerco la proportione, applico la maggiore R T nella <lb/>linea de’gradi all’interuallo 60. </s> <s xml:id="echoid-s3336" xml:space="preserve">60; </s> <s xml:id="echoid-s3337" xml:space="preserve">poi veggendo, che la mi-<lb/>nore R S cade nell’interuallo di gr. </s> <s xml:id="echoid-s3338" xml:space="preserve">53 {1/2}, cerco nelle tauole <lb/>il leno di gr. </s> <s xml:id="echoid-s3339" xml:space="preserve">26. </s> <s xml:id="echoid-s3340" xml:space="preserve">m. </s> <s xml:id="echoid-s3341" xml:space="preserve">45. </s> <s xml:id="echoid-s3342" xml:space="preserve">(che è la metà di detti gr. </s> <s xml:id="echoid-s3343" xml:space="preserve">53 {1/2}) e rad-<lb/>doppiato il numero di queſto ſeno trouato, haurò il numero <lb/>deſle particelle corriſpondenti alla linea R S, dando alla R T <lb/>il numero del raggio delle tauole.</s> <s xml:id="echoid-s3344" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3345" xml:space="preserve">Che ſe le due linee date non foſſero con notabil ecceſſo <lb/>differenti, potria la minore applicarſi all’interuallo 60. </s> <s xml:id="echoid-s3346" xml:space="preserve">60, <lb/>e poi vedere doue capiſca la maggiore, e cercare come pri-<lb/>ma il ſeno della metà de’gradi, e raddoppiarlo; </s> <s xml:id="echoid-s3347" xml:space="preserve">e queſte ſaran-<lb/>no le particelle della linea maggiore, poſta la minore col nu-<lb/>mero del raggio.</s> <s xml:id="echoid-s3348" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3349" xml:space="preserve">Mà ſe dato il numero del raggio alla minore, ſa linea mag-<lb/>giore foſſe così grande, che eccedeſſe l’ interuallo 90. </s> <s xml:id="echoid-s3350" xml:space="preserve">90. <lb/></s> <s xml:id="echoid-s3351" xml:space="preserve">(come nella ſteſſa figura applicata T S all’interuallo 60. </s> <s xml:id="echoid-s3352" xml:space="preserve">60, e <lb/>cercandoſi il numero delle particelle di T R) prendaſi l’inter-<lb/>uallo 90. </s> <s xml:id="echoid-s3353" xml:space="preserve">90; </s> <s xml:id="echoid-s3354" xml:space="preserve">e leuiſi dalla linea maggiore, quante volte ſi <lb/>può, e quante volte s’è preſo, tante volte ſi pigli iſ doppio del <lb/>ſeno di gr. </s> <s xml:id="echoid-s3355" xml:space="preserve">45, e ſia T E vna volta il doppio del ſeno di gradi <lb/>45. </s> <s xml:id="echoid-s3356" xml:space="preserve">Dipoi il reſtante della linea, cioè E R s’applichi nello <lb/>Stromento alla linea de’ gradi, e cadendo nell’interuallo 54.</s> <s xml:id="echoid-s3357" xml:space="preserve"> <pb o="177" file="0193" n="196" rhead="Gradi del Circolo"/> 54, prendaſi il ſeno di gr. </s> <s xml:id="echoid-s3358" xml:space="preserve">27, e ſi raddoppij, e queſto s’ag-<lb/>giunga al doppio deſ ſeno di gr. </s> <s xml:id="echoid-s3359" xml:space="preserve">45 già preſo, e così s’haurà <lb/>il numero delle particelle della linea T R corriſpondenti alle <lb/>parti del raggio aſſegnate alla linea minore T S.</s> <s xml:id="echoid-s3360" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div106" type="section" level="1" n="57"> <head xml:id="echoid-head102" style="it" xml:space="preserve">QVESTIONE QVINTA.</head> <head xml:id="echoid-head103" style="it" xml:space="preserve">Trouar in piccolinumeri iſeni de’ gradi del quadrante.</head> <p> <s xml:id="echoid-s3361" xml:space="preserve">ALcuna volta conuien operare ſenza hauer le tauole de’ <lb/>Seni, e pur ſi vuole riſoluer il triangolo non così me-<lb/>canicamente, come s’è detto nella Queſt. </s> <s xml:id="echoid-s3362" xml:space="preserve">3. </s> <s xml:id="echoid-s3363" xml:space="preserve">di queſto Capo; <lb/></s> <s xml:id="echoid-s3364" xml:space="preserve">& </s> <s xml:id="echoid-s3365" xml:space="preserve">in tal caſo potiamo ſeruirci dello Stromento per trouar i <lb/>Seni de gl’angoli. </s> <s xml:id="echoid-s3366" xml:space="preserve">E perche nello Stromento ſono ſegnate le <lb/>corde de gl’archi, già ſi vede, che volendo il ſeno d’vn’agolo, <lb/>conuien prendere la corda d’vn arco doppio; </s> <s xml:id="echoid-s3367" xml:space="preserve">così per trouar <lb/>il ſeno dell’ angolo di gr. </s> <s xml:id="echoid-s3368" xml:space="preserve">37, ſi deue prendere la corda dell’ <lb/>arco di gr. </s> <s xml:id="echoid-s3369" xml:space="preserve">74.</s> <s xml:id="echoid-s3370" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3371" xml:space="preserve">Primieramente dunque allargato ad arbitrio lo Stromen-<lb/>to, con vn Compaſſo prendo l’interuallo 60. </s> <s xml:id="echoid-s3372" xml:space="preserve">60 nella linea <lb/>de’ gradi, e queſto è il raggio. </s> <s xml:id="echoid-s3373" xml:space="preserve">Poi ritenuta la ſteſſa apertura <lb/>dello Stromento, con vn’altro Compaſſo prendo l’interuallo <lb/>dell’arco doppio dell’angolo, il cui ſeno ſi deſidera, e volen-<lb/>doſi il ſeno di gr. </s> <s xml:id="echoid-s3374" xml:space="preserve">37, prendo l’interuallo 74. </s> <s xml:id="echoid-s3375" xml:space="preserve">74. </s> <s xml:id="echoid-s3376" xml:space="preserve">Fatto que-<lb/>ſto, ritenuta l’apertura de’due Compaſſi, applico nella linea <lb/>Aritmetica l’apertura del Compaſſo, che dà il raggio alli pun-<lb/>ti 50. </s> <s xml:id="echoid-s3377" xml:space="preserve">50 (intendendoſi ciaſcuno diuiſo in due, onde è come <lb/>ſe il raggio foſle 100) e l’altro Compaſſo con la ſua apertura <lb/>applico nella ſteſſa linea Aritmetica, e cade nelli punti 60. <lb/></s> <s xml:id="echoid-s3378" xml:space="preserve">60; </s> <s xml:id="echoid-s3379" xml:space="preserve">il che moſtra, che la corda di gr. </s> <s xml:id="echoid-s3380" xml:space="preserve">74 è di parti 120 di quel- <pb o="178" file="0194" n="197" rhead="CAPO VI."/> le, delle quali il raggio è 100; </s> <s xml:id="echoid-s3381" xml:space="preserve">e per conſegûenza il ſeno di <lb/>gr. </s> <s xml:id="echoid-s3382" xml:space="preserve">37. </s> <s xml:id="echoid-s3383" xml:space="preserve">è particelle 60. </s> <s xml:id="echoid-s3384" xml:space="preserve">L’iſteſſa forma ſi tiene per trouare <lb/>qualſiuoglia altro ſeno.</s> <s xml:id="echoid-s3385" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3386" xml:space="preserve">Quì perc̀ conuien’ oſſeruare, che eſſendo nello Stromento <lb/>fatta la diuiſione delle corde ſolo per il quadrante, non ſi po-<lb/>trà trouar’ il ſeno, ſe non di gr. </s> <s xml:id="echoid-s3387" xml:space="preserve">45. </s> <s xml:id="echoid-s3388" xml:space="preserve">nel modo detto; </s> <s xml:id="echoid-s3389" xml:space="preserve">doue che <lb/>ſe nello Stromento foſſero le corde per tutto il ſemicircolo, <lb/>come ſi può fare nelli Stromenti, che ſono aſſai lunghi, con <lb/>queſto metodo ſi trouerebbono li ſeni di tutti i gradi del qua-<lb/>drante. </s> <s xml:id="echoid-s3390" xml:space="preserve">Ma non hauendoſi ſe non le corde del quadrante <lb/>nello Stromento, in occaſione, che il doppio dell’angolo, il <lb/>cui ſeno ſi cerca, eccedeſſe li gr. </s> <s xml:id="echoid-s3391" xml:space="preserve">90, cerchiſi il ſeno del com-<lb/>plemento dell’angolo dato, e queſto moltiplicato in ſe ſteſſo, <lb/>ſi caui dal 10000 quadrato del raggio; </s> <s xml:id="echoid-s3392" xml:space="preserve">poiche il reſtante è il <lb/>quadrato del ſeno, che ſi cerca. </s> <s xml:id="echoid-s3393" xml:space="preserve">Per eſſempio, deſidero il ſe-<lb/>no di gr. </s> <s xml:id="echoid-s3394" xml:space="preserve">50: </s> <s xml:id="echoid-s3395" xml:space="preserve">queſt’arco raddoppiato è gr. </s> <s xml:id="echoid-s3396" xml:space="preserve">100, i quali non ſo-<lb/>no nello Stromento. </s> <s xml:id="echoid-s3397" xml:space="preserve">Cerco dunque nel modo detto di ſopra <lb/>il ſeno del complemento, cioè di gr. </s> <s xml:id="echoid-s3398" xml:space="preserve">40, prendendo la corda <lb/>di gr. </s> <s xml:id="echoid-s3399" xml:space="preserve">80. </s> <s xml:id="echoid-s3400" xml:space="preserve">la quale trouo di particelle 129; </s> <s xml:id="echoid-s3401" xml:space="preserve">onde il ſeno di gr. <lb/></s> <s xml:id="echoid-s3402" xml:space="preserve">40 è 64 {1/2}: </s> <s xml:id="echoid-s3403" xml:space="preserve">il cui quadrato 4160, leuato dal 10000 quadra-<lb/>to del raggio 100, laſcia 5840, la cui radice quadrata 76 è il <lb/>ſenocercato di gr. </s> <s xml:id="echoid-s3404" xml:space="preserve">50, le quali coſe ſon manifeſte, per la dot-<lb/>trina de’ſeni, eſſendo che il quadrato del raggio è vguale alli <lb/>quadrati de’ſeni di due angoli, che inſieme fanno gr. </s> <s xml:id="echoid-s3405" xml:space="preserve">90.</s> <s xml:id="echoid-s3406" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3407" xml:space="preserve">Aggiongaſi quì, che moſte volte potrà oprarſi con la cor-<lb/>da dell’arco doppio così bene, come col ſeno dell’angolo da-<lb/>to, poiche hanno tra diloro la ſteſſa proportione le parti, & </s> <s xml:id="echoid-s3408" xml:space="preserve">i <lb/>moltiplici: </s> <s xml:id="echoid-s3409" xml:space="preserve">ne meno ſarà neceſſario prendere il raggio, ma <lb/>baſterà nella linea de’gradi prendere le corde de gl’archi dop-<lb/>pij, e poi trasferitele à gl’interualli della linea Aritmetica, ſi <pb o="179" file="0195" n="198" rhead="Gradi del Circolo"/> conoſcerà la loro proportione, e s’operarà, come ſe s’haueſ-<lb/> <anchor type="figure" xlink:label="fig-0195-01a" xlink:href="fig-0195-01"/> ſero li ſeni de gl’angoli. </s> <s xml:id="echoid-s3410" xml:space="preserve">Sia <lb/>per eſſempio il triangolo <lb/>AIB, di cui ſono dati gl’an-<lb/>goli IAB gr. </s> <s xml:id="echoid-s3411" xml:space="preserve">32, IBA gr. </s> <s xml:id="echoid-s3412" xml:space="preserve">35, <lb/>& </s> <s xml:id="echoid-s3413" xml:space="preserve">il lato A I piedi 56: </s> <s xml:id="echoid-s3414" xml:space="preserve">cer-<lb/>chiſi la quantità del lato I B. <lb/></s> <s xml:id="echoid-s3415" xml:space="preserve">Ora perche i lati, & </s> <s xml:id="echoid-s3416" xml:space="preserve">i ſeni de <lb/>gl’angoli oppoſti ſono pro-<lb/>portionali, e le corde de gl’-<lb/>archi doppij ſono propor-<lb/>tionali alli ſeni delle loro metà, anche i lati del triangolo, e <lb/>le corde de gl’archi doppij de gl’angoli dati, ſono tra di loro <lb/>proportionali. </s> <s xml:id="echoid-s3417" xml:space="preserve">Prendo dunque nella linea de’ gradi le corde <lb/>de gl’archi 70, e 64, e traportata nella linea Aritmetica la <lb/>corda di gr. </s> <s xml:id="echoid-s3418" xml:space="preserve">70 all’interuallo 100. </s> <s xml:id="echoid-s3419" xml:space="preserve">100, trouo, che la corda <lb/>di gr. </s> <s xml:id="echoid-s3420" xml:space="preserve">64 cade all’interuallo 91 {1/2}, 91 {1/2}. </s> <s xml:id="echoid-s3421" xml:space="preserve">Dunque oprando, <lb/>come ſe queſti foſſero li ſeni de gl’angoli dati, dico, come <lb/>100 à 91 {1/2}, eosì A I piedi 56 à I B piedi 51 {1/48}.</s> <s xml:id="echoid-s3422" xml:space="preserve"/> </p> <div xml:id="echoid-div106" type="float" level="2" n="1"> <figure xlink:label="fig-0195-01" xlink:href="fig-0195-01a"> <image file="0195-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0195-01"/> </figure> </div> </div> <div xml:id="echoid-div108" type="section" level="1" n="58"> <head xml:id="echoid-head104" style="it" xml:space="preserve">QVESTIONE SESTA.</head> <head xml:id="echoid-head105" style="it" xml:space="preserve">Data vna linea corda d’ vn arco di determniata quantità, <lb/>come ſi iroui il ſuo circolo.</head> <p> <s xml:id="echoid-s3423" xml:space="preserve">SIa dato vn triangolo ABC, e ſia il lato A B oppoſto ad <lb/>ad vn’angolo di gr. </s> <s xml:id="echoid-s3424" xml:space="preserve">42, e voglia deſctiuerſi vn circolo <lb/>intorno ad vn taltriangolo. </s> <s xml:id="echoid-s3425" xml:space="preserve">E dunque manifeſto, che la da-<lb/>ta linea del triangolo inſcritto nel circolo è corda d’vn’arco <lb/>doppio dell’angolo oppoſto, che è angolo alla circonferen- <pb o="180" file="0196" n="199" rhead="CAPO VI."/> za di cuiè doppio l’angolo al centro, per la 20, del libro 3. <lb/></s> <s xml:id="echoid-s3426" xml:space="preserve"> <anchor type="figure" xlink:label="fig-0196-01a" xlink:href="fig-0196-01"/> Dunque la data linea A B applico nella <lb/>linea de’gradi dello Stromento all’ inter-<lb/>uallo 84 84, eritenuta quell’ apertura di <lb/>Stromento, prendo l’interuallo 60. </s> <s xml:id="echoid-s3427" xml:space="preserve">60; </s> <s xml:id="echoid-s3428" xml:space="preserve">e <lb/>queſto è il ſemidiametro del circolo, in <lb/>cui il triangolo dato ſi deſcriue. </s> <s xml:id="echoid-s3429" xml:space="preserve">Per tan-<lb/>to con quell’ apertura di Compaſſo dalli <lb/>punti A, & </s> <s xml:id="echoid-s3430" xml:space="preserve">B deſcriuo due archi occulti, <lb/>che ſi tagliano in D, onde è il ſemidiame-<lb/>tro A D, & </s> <s xml:id="echoid-s3431" xml:space="preserve">èil punto D centro del circo-<lb/>lo circoſcritto al dato triangolo.</s> <s xml:id="echoid-s3432" xml:space="preserve"/> </p> <div xml:id="echoid-div108" type="float" level="2" n="1"> <figure xlink:label="fig-0196-01" xlink:href="fig-0196-01a"> <image file="0196-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0196-01"/> </figure> </div> <p> <s xml:id="echoid-s3433" xml:space="preserve">E così generalmente data vna linea, che ſia corda d’vn’ <lb/>arco, quella s’applichi al numero de’gradi di detto arco; </s> <s xml:id="echoid-s3434" xml:space="preserve">poi <lb/>ritenuta quell’a pertura di Stromento, ſi prenda l’interuallo <lb/>60. </s> <s xml:id="echoid-s3435" xml:space="preserve">60, e queſta ſarà la quantità del ſemidiametro del circolo, <lb/>in cui la data linea è corda dell’arco determinato.</s> <s xml:id="echoid-s3436" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3437" xml:space="preserve">Che ſe la linea data ſoffe corda d’vn’arco maggiore del <lb/>quadrante, alſhora queſta ſi diuide per mezzo con vna linea <lb/>perpendicolare indefinita: </s> <s xml:id="echoid-s3438" xml:space="preserve">poiad vn’eſtremità di detta linea <lb/>ſi faccia vn’angolo, che ſia la metà del reſiduo ſin’ al ſemicir-<lb/>colo, cioè ſin a gradi 180; </s> <s xml:id="echoid-s3439" xml:space="preserve">poiche doue ſarà tagliata la per-<lb/>pendicolare indefinita, iuiſaràil centro del circolo, che ſi de-<lb/>ſidera. </s> <s xml:id="echoid-s3440" xml:space="preserve">Così ſia la linea MN corda digr. </s> <s xml:id="echoid-s3441" xml:space="preserve">136, la quale non è <lb/>nello Stromento, in cui ſolo ſon’i gradi del quadrante. </s> <s xml:id="echoid-s3442" xml:space="preserve">Que-<lb/>ſta ſi diuida per mezzo in P, e ſia la perpendicolar indefinita <lb/>PK. </s> <s xml:id="echoid-s3443" xml:space="preserve">Or il reſiduo da 136 ſin à 180 è 44, la cui metà è gradi <lb/>22. </s> <s xml:id="echoid-s3444" xml:space="preserve">Facciaſi dunque nell’eſtremità M l’angolo PMO, come <lb/>s’è detto nella prima Queſtione, digr. </s> <s xml:id="echoid-s3445" xml:space="preserve">22: </s> <s xml:id="echoid-s3446" xml:space="preserve">e la linea MO ſarà <lb/>il ſemidiametro del Circolo, il cui centro è il punto O, & </s> <s xml:id="echoid-s3447" xml:space="preserve">in <pb o="181" file="0197" n="200" rhead="Gradi del Circolo"/> cui la linea MN è corda di gr. </s> <s xml:id="echoid-s3448" xml:space="preserve">136. </s> <s xml:id="echoid-s3449" xml:space="preserve">Il che è manifeſto, per-<lb/>che ſe ſi tira la linea ON, li due triangoli OPM, OPN rettan-<lb/>goli in P hanno il lato OP commune, elilati PM, PN vguali <lb/>per la coſtruttione, dunque per la 4 del lib. </s> <s xml:id="echoid-s3450" xml:space="preserve">1. </s> <s xml:id="echoid-s3451" xml:space="preserve">gl’angoli POM, <lb/>PON ſono vguali: </s> <s xml:id="echoid-s3452" xml:space="preserve">l’angolo POM è complemento dell’an-<lb/>golo OMP digr. </s> <s xml:id="echoid-s3453" xml:space="preserve">22, dunque POM è gr 68. </s> <s xml:id="echoid-s3454" xml:space="preserve">e per conſeguen-<lb/>za anche PON è gr. </s> <s xml:id="echoid-s3455" xml:space="preserve">68; </s> <s xml:id="echoid-s3456" xml:space="preserve">ondetutto l’angolo MON, cioè l’ar-<lb/>co di cui MN è corda, è di gr. </s> <s xml:id="echoid-s3457" xml:space="preserve">136.</s> <s xml:id="echoid-s3458" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div110" type="section" level="1" n="59"> <head xml:id="echoid-head106" xml:space="preserve">QVESTIONE SETTIMA.</head> <head xml:id="echoid-head107" style="it" xml:space="preserve">Come ſi poſſa prendere qualſiuoglia parte determinata del circolo, <lb/>e deſcriuere qualſiuoglia figura regolare.</head> <p> <s xml:id="echoid-s3459" xml:space="preserve">SE il circolo è dato, e ſi deſidera vna ſua parte aliquota, <lb/>diuidaſi il numero de’ gradi 360 per il denominatore <lb/>della parte che ſi deſidera, & </s> <s xml:id="echoid-s3460" xml:space="preserve">il quotiente ſarà il numero de’ <lb/>gradi, la corda de’quali applicata al circolo prenderà la parte <lb/>cercata. </s> <s xml:id="echoid-s3461" xml:space="preserve">Il che ſi fà applicando prima il ſemidiametro del <lb/>circolo dato all’interuallo 60. </s> <s xml:id="echoid-s3462" xml:space="preserve">60 nella linea de’ gradi nello <lb/> <anchor type="figure" xlink:label="fig-0197-01a" xlink:href="fig-0197-01"/> Stromento: </s> <s xml:id="echoid-s3463" xml:space="preserve">e poi prendendo l’in-<lb/>teruallo corriſpondente al nume-<lb/>ro de’ gradi trouati nel quotiente <lb/>della diuiſione.</s> <s xml:id="echoid-s3464" xml:space="preserve"/> </p> <div xml:id="echoid-div110" type="float" level="2" n="1"> <figure xlink:label="fig-0197-01" xlink:href="fig-0197-01a"> <image file="0197-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0197-01"/> </figure> </div> <p> <s xml:id="echoid-s3465" xml:space="preserve">Sia dato il circolo, il cui ſemi-<lb/>diametro BC; </s> <s xml:id="echoid-s3466" xml:space="preserve">e ſi cerchi l’ottaua, <lb/>parte: </s> <s xml:id="echoid-s3467" xml:space="preserve">Diuido 360 per 8, evien <lb/>il quotiente 45. </s> <s xml:id="echoid-s3468" xml:space="preserve">Applico dunque <lb/>nello Stromento nella linea de’ <lb/>gradi all’interuallo 60. </s> <s xml:id="echoid-s3469" xml:space="preserve">60 la linea <pb o="182" file="0198" n="201" rhead="CAPO VI."/> BC; </s> <s xml:id="echoid-s3470" xml:space="preserve">e ritenuta quell’apertura, prendo l’interuallo 45. </s> <s xml:id="echoid-s3471" xml:space="preserve">45, e <lb/>queſto applicato al circolo dato in CD, queſta è l’ottaua <lb/>patte di detto circolo; </s> <s xml:id="echoid-s3472" xml:space="preserve">e così replicata diuiderà il circolo in ot-<lb/>to parti vguali; </s> <s xml:id="echoid-s3473" xml:space="preserve">e le linee tirate alli punti di dette diuiſioni de-<lb/>ſcriueranno vn’ottangolo regolare. </s> <s xml:id="echoid-s3474" xml:space="preserve">Così per deſcriuere vna <lb/>figura di noue lati vguali, diuido 360 per 9, & </s> <s xml:id="echoid-s3475" xml:space="preserve">il quotiente <lb/>40 moſtra, che deuo prendere la corda digr. </s> <s xml:id="echoid-s3476" xml:space="preserve">40. </s> <s xml:id="echoid-s3477" xml:space="preserve">& </s> <s xml:id="echoid-s3478" xml:space="preserve">oprare <lb/>come ſopra, e ſarà CE la nona parte del circolo.</s> <s xml:id="echoid-s3479" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3480" xml:space="preserve">Mà ſe la parte del circolo cercata non foſſe aliquota, fac-<lb/>ciaſi come il denominatore al numeratore della parte cerca-<lb/>ta, così gr. </s> <s xml:id="echoid-s3481" xml:space="preserve">360. </s> <s xml:id="echoid-s3482" xml:space="preserve">ad vn’altro numero, e verrà il numero de’ <lb/>gradicompetenti alla parte, che ſi deſidera. </s> <s xml:id="echoid-s3483" xml:space="preserve">Così deſideran-<lb/>doſi hauere d’vn circolo vn’arco, che ſia {5/9}, facciaſi come 9 à 5, <lb/>così 360 à 200. </s> <s xml:id="echoid-s3484" xml:space="preserve">Dunque deuono pigliarſi dal circolo dato <lb/>gradi 200; </s> <s xml:id="echoid-s3485" xml:space="preserve">i quali ſe bene non ſi puonno pigliare nello Stro-<lb/>mento tutti inſieme, ad ogni modo ſi puonno pigliar per par-<lb/>ti; </s> <s xml:id="echoid-s3486" xml:space="preserve">onde eſſendo più del ſemicircolo, prolongato il ſemidia-<lb/>metro CB in F, ſarà CEDF gr. </s> <s xml:id="echoid-s3487" xml:space="preserve">180; </s> <s xml:id="echoid-s3488" xml:space="preserve">e rimanendo gradi 20 <lb/>fin’à 200, prendo gr. </s> <s xml:id="echoid-s3489" xml:space="preserve">20 nello Stromento allargato in 60. </s> <s xml:id="echoid-s3490" xml:space="preserve">60, <lb/>all’interuallo di BC, e ſono FG; </s> <s xml:id="echoid-s3491" xml:space="preserve">e così tutto l’arco CDGè {5/9} del <lb/>circolo, cioè gr. </s> <s xml:id="echoid-s3492" xml:space="preserve">200. </s> <s xml:id="echoid-s3493" xml:space="preserve">In ſomigliante maniera, per prender la <lb/>terza parte del circolo, che è gr. </s> <s xml:id="echoid-s3494" xml:space="preserve">120, ſi prendono due volte <lb/>60, ò qualſiuoglia aſtri due numeri, che aggiunti inſieme fac-<lb/>ciano la ſteſſa ſomma digr. </s> <s xml:id="echoid-s3495" xml:space="preserve">120.</s> <s xml:id="echoid-s3496" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3497" xml:space="preserve">Che ſe foſſe data vna linea, e conueniſſe farne vn poligo-<lb/>no regolare, diuidanſi gr. </s> <s xml:id="echoid-s3498" xml:space="preserve">360 per il denominatore del poli-<lb/>gono; </s> <s xml:id="echoid-s3499" xml:space="preserve">alli gradi del quotiente s’applichi nello ſtromento la <lb/>linea data, e ritenuta quell’ apertura dello Stromento, pren-<lb/>daſi l’interuallo 60. </s> <s xml:id="echoid-s3500" xml:space="preserve">60, e ſarà quello il ſemidiametro del cir-<lb/>colo, a cui applicata la linea data, ſarà il lato del poligono, e <pb o="183" file="0199" n="202" rhead="Gradidel Circolo"/> replicata formarà il detto poligono cercato. </s> <s xml:id="echoid-s3501" xml:space="preserve">Sia data la linea <lb/>KL, e ſi deſideri vn pentagono regolare, di cui ella ſia lato. <lb/></s> <s xml:id="echoid-s3502" xml:space="preserve">Diuido 360 per 5 denominatore del poligono, & </s> <s xml:id="echoid-s3503" xml:space="preserve">è il quo-<lb/>tiente 72: </s> <s xml:id="echoid-s3504" xml:space="preserve">perciò cerco il circolo, in cui KL ſia corda di gradi <lb/>72 nel modo detto nella precedente Queſtione: </s> <s xml:id="echoid-s3505" xml:space="preserve">il che faccio, <lb/>applicando la linea KL all’interuallo 72. </s> <s xml:id="echoid-s3506" xml:space="preserve">72 nella linea de’ <lb/>gradi; </s> <s xml:id="echoid-s3507" xml:space="preserve">e poi preſo l’interuallo 60. </s> <s xml:id="echoid-s3508" xml:space="preserve">60, trouo eſſer’vguale alla <lb/>linea BC; </s> <s xml:id="echoid-s3509" xml:space="preserve">e di queſta ſeruendomi, come di ſemidia metro, de-<lb/>ſcriuo il circolo CDG, à cui applicata, e replicata la linea <lb/>KL, formarà il pentagono.</s> <s xml:id="echoid-s3510" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div112" type="section" level="1" n="60"> <head xml:id="echoid-head108" xml:space="preserve">QVESTIONE OTTAVA.</head> <head xml:id="echoid-head109" style="it" xml:space="preserve">Dato il diametro d’vna sfera, come ſi troui la ſuperficie sferica, ela <lb/>ſolidita di qualſiuoglia ſegmento di detta sfera, conoſciuto <lb/>nella quantità de’ gradi d’vn circolo maſsimo perpen-<lb/>dicolare al piano della baſe di detto <lb/>ſegmento.</head> <p> <s xml:id="echoid-s3511" xml:space="preserve">SI come nel circolo altra coſa è il ſegmento, & </s> <s xml:id="echoid-s3512" xml:space="preserve">altra il ſet-<lb/>tore, poiche ſegmento è quello, che da vna linea retta, <lb/>e parte della circonferenza ſi comprende, e ſettore è quello, <lb/>che vien compreſo da due linee rette vſcite dal centro, e dalla <lb/>circonferenza, che da dette linee rette vien’intercetta: </s> <s xml:id="echoid-s3513" xml:space="preserve">Così <lb/>parimente nella sfera ſegmento è quella parte ſolida, che ſi <lb/>comprende da vn piano, che taglia la sfera, e dalla ſuperficie <lb/>sferica: </s> <s xml:id="echoid-s3514" xml:space="preserve">doue che il ſettore è compreſo da vna ſuperficie <lb/>conica, la cui cima è nel centro della sfera, e della ſuperficie <lb/>sferica, che vientagliata dalla detta ſuperficie conica. </s> <s xml:id="echoid-s3515" xml:space="preserve">Quindi <lb/>ciò che ſi comprende dal piano CTRH, e dalla ſuperficie sfe- <pb o="184" file="0200" n="203" rhead="CAPO VI."/> rica CAR, ouero dalla ſuperficie sferica CBR, è ſegmento <lb/> <anchor type="figure" xlink:label="fig-0200-01a" xlink:href="fig-0200-01"/> della sfera: </s> <s xml:id="echoid-s3516" xml:space="preserve">mà il ſolido compreſo dal-<lb/>la ſuperficie conica CSR, e dalla ſuper-<lb/>ficie sferica CAR, è ſettore della sfera.</s> <s xml:id="echoid-s3517" xml:space="preserve"/> </p> <div xml:id="echoid-div112" type="float" level="2" n="1"> <figure xlink:label="fig-0200-01" xlink:href="fig-0200-01a"> <image file="0200-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0200-01"/> </figure> </div> <p> <s xml:id="echoid-s3518" xml:space="preserve">Or per trouare la ſuperficie di tutta <lb/>la sfera data, baſta prendere per ſemi-<lb/>dia metro d’vn circolo tutto il diame-<lb/>tro della sfera, poiche quel circolo ſarà <lb/>vguale alla ſuperficie della sfera;</s> <s xml:id="echoid-s3519" xml:space="preserve">eſſendo <lb/>che la ſuperficie di qualſiuoglia sfera, <lb/>come dimoſtra Archimede lib. </s> <s xml:id="echoid-s3520" xml:space="preserve">1. </s> <s xml:id="echoid-s3521" xml:space="preserve">de <lb/>Sphoer. </s> <s xml:id="echoid-s3522" xml:space="preserve">& </s> <s xml:id="echoid-s3523" xml:space="preserve">Cylindro, prop. </s> <s xml:id="echoid-s3524" xml:space="preserve">30, è qua-<lb/>drupla del circolo maſſimo di detta sfe-<lb/>ra; </s> <s xml:id="echoid-s3525" xml:space="preserve">& </s> <s xml:id="echoid-s3526" xml:space="preserve">il circolo, il cui diametro è dop-<lb/>pio del diametro dell’ifteſſo circolo maſ-<lb/>ſimo, è quadruplo di detto circolo, per <lb/>la 2. </s> <s xml:id="echoid-s3527" xml:space="preserve">dellib. </s> <s xml:id="echoid-s3528" xml:space="preserve">12, e perciò il circolo, il cui <lb/>raggio è vguale al diametro della sfera, <lb/>è vguale alla ſuperficie di tutta la sfera, <lb/>per la 7. </s> <s xml:id="echoid-s3529" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3530" xml:space="preserve">5. </s> <s xml:id="echoid-s3531" xml:space="preserve">E perche il circolo è <lb/>vguale al triangolo, li di cui lati poſti ad <lb/>angolo retto, ſono il raggio, e la circon-<lb/>ferenza (come nel lib. </s> <s xml:id="echoid-s3532" xml:space="preserve">de dimenſ. </s> <s xml:id="echoid-s3533" xml:space="preserve">circ. <lb/></s> <s xml:id="echoid-s3534" xml:space="preserve">moſtra Archimede) e perciò al paralle-<lb/>logrammorettangolo fatto dal raggio, e dalla ſemicirconfe-<lb/>renza; </s> <s xml:id="echoid-s3535" xml:space="preserve">perla 41 del lib. </s> <s xml:id="echoid-s3536" xml:space="preserve">1. </s> <s xml:id="echoid-s3537" xml:space="preserve">d’Euclide; </s> <s xml:id="echoid-s3538" xml:space="preserve">ne ſeguita, che il ret-<lb/>tangolo fatto da tutto il diametro, etutta la circonferenza <lb/>ſarà quadruplo del circolo. </s> <s xml:id="echoid-s3539" xml:space="preserve">Dunque dato il diametro della <lb/>sfera, ſi conoſce la circonferenza, la quale è al diametro proſ-<lb/>ſimamente come 355 à 113; </s> <s xml:id="echoid-s3540" xml:space="preserve">e moltiplicato il diametro per <pb o="185" file="0201" n="204" rhead="Gradi del Circolo"/> la circonferenza del circolo maſſimo, s’haurà tutta la ſuperfi-<lb/>cie delſa sfera. </s> <s xml:id="echoid-s3541" xml:space="preserve">In queſta maniera facilmente troueremo tut-<lb/>ta la ſuperſicie della terra, il di cui giro nel libro, che intitolai, <lb/>Terra Machinis mota diſſert. </s> <s xml:id="echoid-s3542" xml:space="preserve">2. </s> <s xml:id="echoid-s3543" xml:space="preserve">n. </s> <s xml:id="echoid-s3544" xml:space="preserve">22. </s> <s xml:id="echoid-s3545" xml:space="preserve">moſtrai molto proba-<lb/>bilmente eſſere di paſſi romani antichi 30598162. </s> <s xml:id="echoid-s3546" xml:space="preserve">ſe queſto <lb/>giro moltiplicato per 113, diuideremo il prodotto per 355, <lb/>poiche verrà il diametro della terra di paſſi romani antichi <lb/>9739696. </s> <s xml:id="echoid-s3547" xml:space="preserve">moltiplicato dunque il giro per il diametro, ſi tro-<lb/>uerà la ſuperficie di tutta la terra eſſere di paſſi romani antichi <lb/>quadrati 298016796038752, cioè miglia quadrate <lb/>298016796, e paſſi quadrati 38752.</s> <s xml:id="echoid-s3548" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3549" xml:space="preserve">Mà per trouare la ſuperficie d’vn ſegmento di sfera, ſe ſi <lb/>cerca la ſola ſuperficie sferica conoſciuta ne’gradi del circolo <lb/>maſſimo perpendicolare alla baſe di detto ſegmento, pren-<lb/>daſi la metà del numero di detti gradi, & </s> <s xml:id="echoid-s3550" xml:space="preserve">applicato nelle linee <lb/>de’gradi neli<unsure/>o Stromento il ſemidiametro della sfera, il qual <lb/>è anche ſemidiametro del circolo maſſimo, all’interuallo de’ <lb/>gradi 60. </s> <s xml:id="echoid-s3551" xml:space="preserve">60, prendaſi l’interuallo della metà di detti gradi, e <lb/>queſto ſarà il ſemidiametro del circolo vguale alla ſuperficie <lb/>sferica cercata di detto ſegmento. </s> <s xml:id="echoid-s3552" xml:space="preserve">Mà ſe ſi prenderà l’inter-<lb/>uallo del numero intiero de’gradi dati, queſto ſarà tutto il dia-<lb/>metro del circolo, che è la baſe del ſegmento. </s> <s xml:id="echoid-s3553" xml:space="preserve">Il<unsure/> che è mani-<lb/>feſto nella ſteſſa figura, in cui al piano CHRT è perpendico-<lb/>lare, il circolo maſſimo BCAR, & </s> <s xml:id="echoid-s3554" xml:space="preserve">il punto A è l’apice del <lb/>ſegmento C A R, come il punto B è l’apice del ſegmento <lb/>C B R: </s> <s xml:id="echoid-s3555" xml:space="preserve">dunque per la prop. </s> <s xml:id="echoid-s3556" xml:space="preserve">36. </s> <s xml:id="echoid-s3557" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3558" xml:space="preserve">1. </s> <s xml:id="echoid-s3559" xml:space="preserve">de Sphœra, & </s> <s xml:id="echoid-s3560" xml:space="preserve">Cylind. <lb/></s> <s xml:id="echoid-s3561" xml:space="preserve">d’Archimede, la linea A C è raggio del circolo vguale alla ſu-<lb/>perficie sferica C A R, e per la prop. </s> <s xml:id="echoid-s3562" xml:space="preserve">37. </s> <s xml:id="echoid-s3563" xml:space="preserve">la linea BC è raggio <lb/>del circolo vguale alla ſuperficie sferica CBR. </s> <s xml:id="echoid-s3564" xml:space="preserve">Ora tanto la <lb/>linea A C, quanto la B C, ſottendono la metà de’gradi del cir- <pb o="186" file="0202" n="205" rhead="CAPO VI."/> colo maſſimo, che paſſa per detti ſegmenti. </s> <s xml:id="echoid-s3565" xml:space="preserve">Doue che la <lb/>CR, che ſottende tutto l’arco di detto circolo maſſimo, è il <lb/>diametro del circolo, che è baſe delli ſegmenti.</s> <s xml:id="echoid-s3566" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3567" xml:space="preserve">E ſe vorremo trouar in numeri la ſuperficie sferica ſudetta, <lb/>cerchiamo per eſſempio nella terra, quanta ſia la ſuperficie, <lb/>compreſa dal circolo polare, e ſia il polo A, nel meridiano <lb/>BRAC ſia AC gr. </s> <s xml:id="echoid-s3568" xml:space="preserve">23 {1/2}. </s> <s xml:id="echoid-s3569" xml:space="preserve">Apro lo Stromento ad arbitrio, e <lb/>con vn Compaſſo preſo l’interuallo de’gradi 60. </s> <s xml:id="echoid-s3570" xml:space="preserve">60, con vn <lb/>altro Compaſſo prendo l’interuallo 23 {1/2}. </s> <s xml:id="echoid-s3571" xml:space="preserve">23 {1/2}. </s> <s xml:id="echoid-s3572" xml:space="preserve">Dipoi appli-<lb/>cato l’vno, el’altro Compaſſo nella linea Aritmetica, il primo <lb/>all’interuallo 100. </s> <s xml:id="echoid-s3573" xml:space="preserve">100, el’altro doue s’addata, trouo, che di <lb/>quali parti il ſemidiametro è 100, & </s> <s xml:id="echoid-s3574" xml:space="preserve">il diametro è 200, di ta-<lb/>li quaſi 41 è AC ſottendente gr. </s> <s xml:id="echoid-s3575" xml:space="preserve">23 {1/2}. </s> <s xml:id="echoid-s3576" xml:space="preserve">Dunque come 200 à <lb/>41, così il diametro della terra di paſſi 9739696, alla ſotten-<lb/>dente di gr. </s> <s xml:id="echoid-s3577" xml:space="preserve">23 {1/2}, cioè paſſi 1996637. </s> <s xml:id="echoid-s3578" xml:space="preserve">ſemidiametro del cir-<lb/>colo vguale alla ſuperficie sferica CAR compreſa dal circolo <lb/>Polare. </s> <s xml:id="echoid-s3579" xml:space="preserve">Facciaſi per tanto come 113 à 355, così il ſemidia-<lb/>metro 1996637 alla ſemicirconferenza di detto circolo, che <lb/>è paſſi 6272620; </s> <s xml:id="echoid-s3580" xml:space="preserve">e moltiplicato il ſemidiametro per la ſemi-<lb/>circonferenza ſarà tutta l’area del circolo paſſi quadrati <lb/>12524145178940, e così la ſuperficie sferica compreſa nel <lb/>circolo polare è miglia quadrate 12524145, e paſſi quadra-<lb/>ti 178940.</s> <s xml:id="echoid-s3581" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3582" xml:space="preserve">Trouata queſta ſuperficie sferica, ſitrouarà la ſolidità del <lb/>ſettore SRAC, poiche queſta è vguale al cono, la cui baſe è <lb/>vguale alla ſuperficie sferica, CAR, è l’altezza vguale al rag-<lb/>gio della sfera AS, come inſegna Archimede lib. </s> <s xml:id="echoid-s3583" xml:space="preserve">1. </s> <s xml:id="echoid-s3584" xml:space="preserve">de Sphęr. <lb/></s> <s xml:id="echoid-s3585" xml:space="preserve">& </s> <s xml:id="echoid-s3586" xml:space="preserve">Cylind. </s> <s xml:id="echoid-s3587" xml:space="preserve">prop. </s> <s xml:id="echoid-s3588" xml:space="preserve">38. </s> <s xml:id="echoid-s3589" xml:space="preserve">Dunque moltiplicata la baſe perla terza <lb/>parte dell’altezza, s’haurà la ſolidità del cono vguale al ſetto-<lb/>r<unsure/>e. </s> <s xml:id="echoid-s3590" xml:space="preserve">Si che la terza parte del raggio del globo della terra, eſ- <pb o="187" file="0203" n="206" rhead="Gradi del Circolo"/> ſendo paſſi 1623282 moltiplicata per la ſuperficie sferica <lb/>trouata 12524145178940, dà la ſolidità di tutto il ſettore, <lb/>migſia cubiche 20330219434. </s> <s xml:id="echoid-s3591" xml:space="preserve">e paſſi ſolidi 360081080.</s> <s xml:id="echoid-s3592" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3593" xml:space="preserve">Finalmente per hauere la ſolidità del ſolo ſegmento CRA, <lb/>ſi cerchi la ſolidità del cono CSR, trouando la ſubtenſa di tut-<lb/>to l’arco CAR, che è gradi 47. </s> <s xml:id="echoid-s3594" xml:space="preserve">il che ſi fà applicando il ſemi-<lb/>diametro della sfera alli gr. </s> <s xml:id="echoid-s3595" xml:space="preserve">60. </s> <s xml:id="echoid-s3596" xml:space="preserve">60, e poi preſo l’interuallo <lb/>47. </s> <s xml:id="echoid-s3597" xml:space="preserve">47, e nella linea Aritmetica applicato il raggio della <lb/>sfera al 100. </s> <s xml:id="echoid-s3598" xml:space="preserve">100, la ſubtenſa di gr. </s> <s xml:id="echoid-s3599" xml:space="preserve">47, cioè CR è quaſi 80; <lb/></s> <s xml:id="echoid-s3600" xml:space="preserve">e queſta come diametro darà la grandezza del circolo CT <lb/>RH; </s> <s xml:id="echoid-s3601" xml:space="preserve">e la SI ſeno del complemento della metà de’gradi dati, <lb/>ſarà l’altezza del cono, la terza parte dunque di tal altezza <lb/>moltiplicando la grandezza del circolo baſe del cono, dà la di <lb/>lui ſohdità; </s> <s xml:id="echoid-s3602" xml:space="preserve">la quale leuata dalla ſolidità del ſettore, laſcierà la <lb/>ſolidità cercata del ſegmento CRA.</s> <s xml:id="echoid-s3603" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3604" xml:space="preserve">Vn’altra maniera vi ſarà per trouar la ſuperficie sferica di <lb/>qualſiuoglia ſegmento, e delle zone, ſe faremo rifleſſione, che <lb/>Archimede al manifeſto 9. </s> <s xml:id="echoid-s3605" xml:space="preserve">doppo la prop. </s> <s xml:id="echoid-s3606" xml:space="preserve">31. </s> <s xml:id="echoid-s3607" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3608" xml:space="preserve">1. </s> <s xml:id="echoid-s3609" xml:space="preserve">de <lb/>Sphœra, & </s> <s xml:id="echoid-s3610" xml:space="preserve">Cylindro, moſtra, che la ſuperficie del cilindro <lb/>con le baſi è ſelquialtera alla ſuperficie della sfera, il cui maſ-<lb/>ſimo circolo è vguale alla baſe di detto cilindro circoſcritto à <lb/>detta sfera: </s> <s xml:id="echoid-s3611" xml:space="preserve">onde neſegue, che detratte le baſi, reſta la ſuper-<lb/>ficie cilindrica vguale alla ſuperficie sferica. </s> <s xml:id="echoid-s3612" xml:space="preserve">Ora ſia alla sfe-<lb/>ra BRAC circoſcritto il cilindro IK, e con li piani OF, ZP pa-<lb/>ralleli ſia tagliata la sfera, & </s> <s xml:id="echoid-s3613" xml:space="preserve">il cilindro. </s> <s xml:id="echoid-s3614" xml:space="preserve">Come di ſopra ſi è <lb/>detto, il circolo, di cui ſia raggio la linea AC, è vguale alla <lb/>ſuperficie sferica CAR. </s> <s xml:id="echoid-s3615" xml:space="preserve">Ma per la prop. </s> <s xml:id="echoid-s3616" xml:space="preserve">13. </s> <s xml:id="echoid-s3617" xml:space="preserve">dello ſteſſo lib. <lb/></s> <s xml:id="echoid-s3618" xml:space="preserve">d’Archimede, la linea media proportionale trà il lato, & </s> <s xml:id="echoid-s3619" xml:space="preserve">il <lb/>diametro della baſe del cilindro retto, è raggio d’vn circolo <lb/>vguale alla ſuperficie cilindrica; </s> <s xml:id="echoid-s3620" xml:space="preserve">dunque ſela ſteſla CA è me- <pb o="188" file="0204" n="207" rhead="CAPO VI."/> dia proportionale tra il lato del cilindro KF, & </s> <s xml:id="echoid-s3621" xml:space="preserve">il diametro <lb/> <anchor type="figure" xlink:label="fig-0204-01a" xlink:href="fig-0204-01"/> della baſe OF, ſarà la ſuperficie cilindri-<lb/>ca KO vguale alla ſuperficie sferica d’al-<lb/>tezza vguale CAR. </s> <s xml:id="echoid-s3622" xml:space="preserve">E che CA ſia media <lb/>proportionale trà KF, & </s> <s xml:id="echoid-s3623" xml:space="preserve">OF, così è ma-<lb/>nifeſto. </s> <s xml:id="echoid-s3624" xml:space="preserve">OF è vguale ad IM, cioè à KM, <lb/>cioè ad AB diametro del circolo, e tirata <lb/>la BC, l’angolo BCA nel ſemicircolo è <lb/>retto; </s> <s xml:id="echoid-s3625" xml:space="preserve">e la CH è perpendicolare alla <lb/>baſe BA, dunque, per l’8. </s> <s xml:id="echoid-s3626" xml:space="preserve">del 6. </s> <s xml:id="echoid-s3627" xml:space="preserve">CA è <lb/>media tra BA, & </s> <s xml:id="echoid-s3628" xml:space="preserve">AH, cioè tra OF, <lb/>e KF.</s> <s xml:id="echoid-s3629" xml:space="preserve"/> </p> <div xml:id="echoid-div113" type="float" level="2" n="2"> <figure xlink:label="fig-0204-01" xlink:href="fig-0204-01a"> <image file="0204-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0204-01"/> </figure> </div> <p> <s xml:id="echoid-s3630" xml:space="preserve">Nella ſteſſa maniera ſi moſtra, che la <lb/>ſuperficie cilindrica KZ è vguale al cir-<lb/>colo, di cui è raggio l’AD; </s> <s xml:id="echoid-s3631" xml:space="preserve">& </s> <s xml:id="echoid-s3632" xml:space="preserve">all’iſteſſo <lb/>circolo è vguale la ſuperficie sferica <lb/>D A E. </s> <s xml:id="echoid-s3633" xml:space="preserve">Dunque leuata la cilindrica <lb/>K O, e la sferica CAR vguali, rimane la <lb/>cilindrica FZ vguale alla zona della sfe-<lb/>rica D C R E.</s> <s xml:id="echoid-s3634" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3635" xml:space="preserve">Sì che ſe la ſuperficie sferica è di ſeg-<lb/>mento, trouiſi il ſeno verſo della metà <lb/>de’gradi dati, cioè AH, e queſto ſi mol-<lb/>tiplichi per il giro del circolo maſſimo <lb/>della sfera: </s> <s xml:id="echoid-s3636" xml:space="preserve">e ſe la ſuperficie sferica è d’vna zona, prendaſi la <lb/>differenza de’ſeni verſi de’ due gradi eſtremi della larghezza <lb/>di detta zona, cioè HV, e ſi moltiplichi per l’iſteſſo giro del <lb/>circolo maſſimo della sfera, e s’haurà la ſuperficie, così sfe-<lb/>rica CRED, come cilindrica FZ corriſpondente. </s> <s xml:id="echoid-s3637" xml:space="preserve">Mà ſe<unsure/> <lb/>nelle linee Geometriche applicarai le due linee AC; </s> <s xml:id="echoid-s3638" xml:space="preserve">AD, e per <pb o="189" file="0205" n="208" rhead="Gradi del Circolo"/> la Queſt. </s> <s xml:id="echoid-s3639" xml:space="preserve">6. </s> <s xml:id="echoid-s3640" xml:space="preserve">del Capo 3. </s> <s xml:id="echoid-s3641" xml:space="preserve">trouerai il raggio del circolo vguale <lb/>alla differenza de’circoli di dette due linee AC, AD, haurai il <lb/>circolo vguale alla zona C R E D.</s> <s xml:id="echoid-s3642" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div115" type="section" level="1" n="61"> <head xml:id="echoid-head110" xml:space="preserve">QVESTIONE NONA.</head> <head xml:id="echoid-head111" style="it" xml:space="preserve">Data in gradi la circonferenza d’vn ſegmento di circolo, come <lb/>ſi troui l’area di detto ſegmento.</head> <p> <s xml:id="echoid-s3643" xml:space="preserve">ESſendo che per l’vltima del 6. </s> <s xml:id="echoid-s3644" xml:space="preserve">d’Euclide li ſettori del cir-<lb/>colo hanno tra di ſe la proportione de gl’archi, da’ quali <lb/>ſono compreſi, il ſettore à tutto il circolo hà la proportione <lb/>del ſuo arco à tutta la circonferenza. </s> <s xml:id="echoid-s3645" xml:space="preserve">Si che nella figura 24, <lb/>ſe ſarà dato il circolo BR AC, & </s> <s xml:id="echoid-s3646" xml:space="preserve">il ſegmento di circolo CRA, <lb/>tirate dal centro le linee SC, SR, il ſettore SCAR à tutto il cir-<lb/>colo, hà la proportione, che hà l’arco CAR à tutta la circon-<lb/>ferenza. </s> <s xml:id="echoid-s3647" xml:space="preserve">Quindi è, che conoſciuti li gradi dell’arco del ſeg-<lb/>mento, ſe ſi fà come gr. </s> <s xml:id="echoid-s3648" xml:space="preserve">360, alli gradi conoſciuti del ſegmen-<lb/>to, così l’area di tutto il circolo ad altro, verrà ad hauerſi l’a-<lb/>rea del ſettore SCAR: </s> <s xml:id="echoid-s3649" xml:space="preserve">E ſe da queſto ſi leua il triangolo CSR <lb/>(il quale ſi troua moltiplicando CI ſeno della metà de’gradi <lb/>conoſciuti del ſegmento, per SI ſeno del complemento di <lb/>di detta metà) rimane l’area del ſegmento CRA.</s> <s xml:id="echoid-s3650" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3651" xml:space="preserve">Dunque applicato il raggio del circolo dato all’interuallo <lb/>de’gradi 60. </s> <s xml:id="echoid-s3652" xml:space="preserve">60. </s> <s xml:id="echoid-s3653" xml:space="preserve">prendaſi l’interuallo congruente alli gradi <lb/>dati del ſegmento: </s> <s xml:id="echoid-s3654" xml:space="preserve">ouero ſe ſolo ſoſſe dato il ſegmento, per la <lb/>Queſt. </s> <s xml:id="echoid-s3655" xml:space="preserve">6. </s> <s xml:id="echoid-s3656" xml:space="preserve">di queſto Capo, ſi troui il raggio del ſuo circolo. </s> <s xml:id="echoid-s3657" xml:space="preserve">Et <lb/>applicati queſti due interualli (cioè il raggio del circolo, e la <lb/>corda del ſegmento) nelle linee Aritmetiche ſi troui la lor <lb/>proportione, e della CR già conoſciuta in numeri ſi prenda <pb o="190" file="0206" n="209" rhead="CAPO VI."/> la metà CI. </s> <s xml:id="echoid-s3658" xml:space="preserve">Quindi per la Queſt. </s> <s xml:id="echoid-s3659" xml:space="preserve">5. </s> <s xml:id="echoid-s3660" xml:space="preserve">ſi troui il ſeno del <unsure/>com-<lb/>plemento della metà de’gradi dati, cioè la SI, e queſto molti-<lb/>plicato per CI darà la quantità del triangoſo da leuarſi dal <lb/>ſettore, acciò reſti l’area del ſegmento.</s> <s xml:id="echoid-s3661" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3662" xml:space="preserve">Sia dato il ſegmento, il cui arco ſia di gr. </s> <s xml:id="echoid-s3663" xml:space="preserve">47. </s> <s xml:id="echoid-s3664" xml:space="preserve">Se iſ diametro <lb/>è 100000, e la circonferenza 314159, l’area del circolo fat-<lb/>ta dalla metà del diametro, e dalla metà della circonferenza <lb/>è di particelle quadrate 7853975000. </s> <s xml:id="echoid-s3665" xml:space="preserve">Dunque come gr. <lb/></s> <s xml:id="echoid-s3666" xml:space="preserve">360 à gr. </s> <s xml:id="echoid-s3667" xml:space="preserve">47, così 7853975000 all’area del ſettore di gr. </s> <s xml:id="echoid-s3668" xml:space="preserve">47, <lb/>cioè à 1025380069. </s> <s xml:id="echoid-s3669" xml:space="preserve">Quindi aperto lo Stromento, e preſi <lb/>gl’interualli 47. </s> <s xml:id="echoid-s3670" xml:space="preserve">47, e 60. </s> <s xml:id="echoid-s3671" xml:space="preserve">60, trouo che di quali parti 50 è il <lb/>raggio di tali quaſi 40 è la ſubtenſa di gr. </s> <s xml:id="echoid-s3672" xml:space="preserve">47. </s> <s xml:id="echoid-s3673" xml:space="preserve">dunque la metà <lb/>è parti quaſi 20. </s> <s xml:id="echoid-s3674" xml:space="preserve">E perche la metà de’gr. </s> <s xml:id="echoid-s3675" xml:space="preserve">47 è 23 {1/2}, il cui <lb/>complemento è gr. </s> <s xml:id="echoid-s3676" xml:space="preserve">66{@/2}<unsure/> trouo con aprire di nuouo lo Stro-<lb/>mento, come prima, che il ſeno di gr. </s> <s xml:id="echoid-s3677" xml:space="preserve">66{@/2}<unsure/> è di parti 45, del-<lb/>le quali il raggio è 50. </s> <s xml:id="echoid-s3678" xml:space="preserve">Ora perche il diametro ſi poſe 100000 <lb/>il raggio non è 50; </s> <s xml:id="echoid-s3679" xml:space="preserve">ma 50000, e così alli numeri trouati con <lb/>lo Stromento aggiongo trè zeri; </s> <s xml:id="echoid-s3680" xml:space="preserve">onde moltiplco 20000 per <lb/>45000, e ſi produce l’area nel triangolo 900000000, che <lb/>leuata dal ſettore trouato 1025380069 laſcia per area del <lb/>ſegmento dato 125380069.</s> <s xml:id="echoid-s3681" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3682" xml:space="preserve">Di quì ſi vede ciò, che debba farſi, quando il ſegmento <lb/>dato è maggiore del ſemicircolo, come il ſegmento CRB: <lb/></s> <s xml:id="echoid-s3683" xml:space="preserve">poiche operandoſi, come prima, ſi troua da principio tutto <lb/>il ſettore SCBR: </s> <s xml:id="echoid-s3684" xml:space="preserve">e poi trouata l’area del triangolo CSR, que-<lb/>ſta non ſi leua dal ſettore trouato; </s> <s xml:id="echoid-s3685" xml:space="preserve">mà ſe gl’aggionge per ha-<lb/>uer tutto il ſegmento CRB.</s> <s xml:id="echoid-s3686" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3687" xml:space="preserve">E ſe ſarà vna parte di circolo compreſa da due linee’ paral-<lb/>lele, trouiſi la quantità de’due ſegmenti, che eſſe fanno, e la <lb/>differenza di detti ſeg menti, è l’area dello ſpatio compre- <pb o="191" file="0207" n="210" rhead="Linea de’ Poligoni"/> ſo dalle due linee parallele, e da gl’archi trà eſſe intercetti, <lb/>come è manifeſto.</s> <s xml:id="echoid-s3688" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div116" type="section" level="1" n="62"> <head xml:id="echoid-head112" xml:space="preserve">CAPO VII.</head> <head xml:id="echoid-head113" style="it" xml:space="preserve">Come nello Stromeni<unsure/>o s’ habbiano à ſegnare ilati delle figure <lb/>regolari; vſo di queſta linea de’ Poligoni.</head> <p> <s xml:id="echoid-s3689" xml:space="preserve">DA quello, che s’è detto nella Queſt. </s> <s xml:id="echoid-s3690" xml:space="preserve">7. </s> <s xml:id="echoid-s3691" xml:space="preserve">del Capo pre-<lb/>cedente, doue habbia mo inſegnato il modo di troua-<lb/>uare il lato di qualſiuoglia figura regolare, non pare neceſſa-<lb/>rio deſcriuere nello Stromento i lati delle figure iegolari, che <lb/>puonno deſcriuerſi nello ſteſſo circolo, ad ogni modo per la <lb/>breuità dell’operare, ſarà vtile porre nello Stromento queſta <lb/>linea de’Poligoni.</s> <s xml:id="echoid-s3692" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3693" xml:space="preserve">Tirate dunque ne’ <unsure/>lati dello Stromento le due linee AR, <lb/>AT, acciò rieſcano più diſtinte le diuiſioni, prendaſi tutta <lb/>la linea A R, per il lato del triangolo equilatero, che può de-<lb/>ſcriuerſi nel circolo: </s> <s xml:id="echoid-s3694" xml:space="preserve">poiche come queſta figura è la minore <lb/>di tutte quelle, che nello ſteſſo circolo puonno deſcriuerſi, ſe <lb/>ſi conſidera l’area, e capacità ſua, così il ſuo lato è il maggio-<lb/>re di tutti. </s> <s xml:id="echoid-s3695" xml:space="preserve">Ora poſta la detta linea AR, per lato del trian-<lb/>golo, è manifeſto, ch’ella è corda della terza parte del circo-<lb/>lo, cioè di gr. </s> <s xml:id="echoid-s3696" xml:space="preserve">120. </s> <s xml:id="echoid-s3697" xml:space="preserve">Conuien dunque trouar il ſemidiametro <lb/>del ſuo circolo: </s> <s xml:id="echoid-s3698" xml:space="preserve">il quale ſe non ſi troua nel modo detto nella <lb/>Queſtione 6. </s> <s xml:id="echoid-s3699" xml:space="preserve">del Capo precedente, può trouarſi nel modo <lb/>ſeguente.</s> <s xml:id="echoid-s3700" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3701" xml:space="preserve">Sia la linea A B lato del triangolo, e corda di gr. </s> <s xml:id="echoid-s3702" xml:space="preserve">120; </s> <s xml:id="echoid-s3703" xml:space="preserve">dun-<lb/>que dal centro del circolo tirati li ſemidia metri, faranno gli <lb/>angoli alla baſe vguali di gr. </s> <s xml:id="echoid-s3704" xml:space="preserve">30 per ciaſcuno. </s> <s xml:id="echoid-s3705" xml:space="preserve">E per far ciò, <pb o="192" file="0208" n="211" rhead="CAPO VII."/> prendo nell’eſtreml<unsure/>tà della data linea due parti vguali tra di <lb/>loro BC, AD, & </s> <s xml:id="echoid-s3706" xml:space="preserve">allo ſteſſo interuallo dalli punti B, & </s> <s xml:id="echoid-s3707" xml:space="preserve">C de-<lb/>ſcriuo due archi occuſti, che ſi ſegano in E; </s> <s xml:id="echoid-s3708" xml:space="preserve">e ſimilmente dal-<lb/>li punti C, & </s> <s xml:id="echoid-s3709" xml:space="preserve">E deſcriuo due altri archi occulti, che ſi taglia-<lb/>no in F. </s> <s xml:id="echoid-s3710" xml:space="preserve">Nella ſteſſa maniera opero dalli punti A, & </s> <s xml:id="echoid-s3711" xml:space="preserve">D allo <lb/>ſteſſo interuallo deſcriuendo due archi, che ſi tagliano in G: </s> <s xml:id="echoid-s3712" xml:space="preserve">e <lb/>dalli punti G, & </s> <s xml:id="echoid-s3713" xml:space="preserve">D due altri, che ſi ſegano in H. </s> <s xml:id="echoid-s3714" xml:space="preserve">Poſcia dal <lb/>punto B per F, & </s> <s xml:id="echoid-s3715" xml:space="preserve">dal punto A per H, tiro due linee, che ſi <lb/>incontrano in I, e dico, che I è il centro del circolo, e l’ango-<lb/> <anchor type="figure" xlink:label="fig-0208-01a" xlink:href="fig-0208-01"/> lo AIB, è di gr. </s> <s xml:id="echoid-s3716" xml:space="preserve">120. </s> <s xml:id="echoid-s3717" xml:space="preserve">eſſendo, che li due angoli ABI, BAI ſo-<lb/>no ciaſcuno di gradi 30. </s> <s xml:id="echoid-s3718" xml:space="preserve">Il che così ſi rende manifeſto. </s> <s xml:id="echoid-s3719" xml:space="preserve">Ti-<lb/>rinſi le linee AG, GD, DH, HG, e perche per la coſtruttione <lb/>gl archi occulti tutti ſono ſtati deſcritti allo ſteſſo interuallo, <lb/>li due triangoli ADG, DHG ſono equilateri, e tra di loro <lb/>vguali; </s> <s xml:id="echoid-s3720" xml:space="preserve">dunquel’angolo DAG è di gradi 60, come anche <pb file="0209" n="212"/> <pb o="192" file="0209a" n="213" rhead="Capo VII."/> <anchor type="figure" xlink:label="fig-0209a-01a" xlink:href="fig-0209a-01"/> <pb file="0210" n="214"/> <pb o="193" file="0211" n="215" rhead="Linea de’ Poligoni"/> tutti gl’altri. </s> <s xml:id="echoid-s3721" xml:space="preserve">Or eſſendo ne’ triangoli ADH, AGH li due <lb/>lati AD, DH vguali alli due lati AG, GH, ela baſe AH com-<lb/>mune, per l’8. </s> <s xml:id="echoid-s3722" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3723" xml:space="preserve">1. </s> <s xml:id="echoid-s3724" xml:space="preserve">gl’angoli DAH, GAH ſono vguali; <lb/></s> <s xml:id="echoid-s3725" xml:space="preserve">dunque l’angolo DAH è gr. </s> <s xml:id="echoid-s3726" xml:space="preserve">30. </s> <s xml:id="echoid-s3727" xml:space="preserve">E la ſteſſa forma di dimo-<lb/>ſtrare ſaria per prouare, che CBF ſia digr. </s> <s xml:id="echoid-s3728" xml:space="preserve">30. </s> <s xml:id="echoid-s3729" xml:space="preserve">Dunque eſ-<lb/>ſendo vguali li due angoli BAI, ABI, anche i ſ<unsure/>ati IA, IB ſono <lb/>vguali: </s> <s xml:id="echoid-s3730" xml:space="preserve">Dunque fatto centro in l all’interuallo IB ſi deſcriua il <lb/>circolo, e l’arco oppoſto all’ angolo AIB ſarà gr. </s> <s xml:id="echoid-s3731" xml:space="preserve">120; </s> <s xml:id="echoid-s3732" xml:space="preserve">il che ſi <lb/>renderà manife<unsure/>ſto ſe dal punto A applicato il ſemidiametro <lb/>alla circonferenza diuiderà in L preciſamente per metà, in <lb/>modo, che AL; </s> <s xml:id="echoid-s3733" xml:space="preserve">LB ſiano vguali, e prolongata la AI in K, ſi <lb/>che ſia diametro del circolo, riuſcirà parimenti BK vguale à <lb/>BL, & </s> <s xml:id="echoid-s3734" xml:space="preserve">LA.</s> <s xml:id="echoid-s3735" xml:space="preserve"/> </p> <div xml:id="echoid-div116" type="float" level="2" n="1"> <figure xlink:label="fig-0208-01" xlink:href="fig-0208-01a"> <image file="0208-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0208-01"/> </figure> <figure xlink:label="fig-0209a-01" xlink:href="fig-0209a-01a"> <image file="0209a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0209a-01"/> </figure> </div> <p> <s xml:id="echoid-s3736" xml:space="preserve">Trouato illato dell’eſſagono, che è la corda dell’arco AL, <lb/>la quale nella linea AB traportata è A 6, ſi cerca il lato del <lb/>quadrato nello ſteſſo circolo: </s> <s xml:id="echoid-s3737" xml:space="preserve">il che ſi fà diuidendo per mezzo <lb/>l’arco LB, ouero dal centro I, tirando vna perpendicolare al <lb/>diametto AK, e cade in M, ſi che AM traportata nella linea <lb/>data AB, ſia A 4 lato del quadrato.</s> <s xml:id="echoid-s3738" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3739" xml:space="preserve">Per hauer il lato del pentagono, diuidaſi, come inſegna <lb/>Ptolomeo nel lib. </s> <s xml:id="echoid-s3740" xml:space="preserve">1. </s> <s xml:id="echoid-s3741" xml:space="preserve">dell’ Almageſto, per mezzo il ſemidia-<lb/>metro IK, nel punto N, e dal punto N all’interuallo NM, ſi <lb/>deſcriua vn’arco occulto, che taglia il diametro in O; </s> <s xml:id="echoid-s3742" xml:space="preserve">poiche <lb/>dal punto O, tirata la linea OM, queſta è illato del pentago-<lb/>no da applicarſi all’arco AP, e nella linea A B ſarà A 5. </s> <s xml:id="echoid-s3743" xml:space="preserve">E per <lb/>conſeguenza OI<unsure/> è il lato della figura di dieci angoli applicata <lb/>all’arco A Q, e nella linea Ab ſarà A 10.</s> <s xml:id="echoid-s3744" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3745" xml:space="preserve">Per illato della figura di ſette lati non v’è forma propria-<lb/>mente Geometrica; </s> <s xml:id="echoid-s3746" xml:space="preserve">ma tentando ſi può trouare, ò la ſettin a <lb/>parte di tutto il circolo, e queſt’ arco darà la corda, che ſarà <pb o="194" file="0212" n="216" rhead="CAPO VII."/> lato dell’eptagono, ouero la ſettima parte del ſemicircolo, e <lb/>due di queſte ſaranno la ſettima ditutto il circolo.</s> <s xml:id="echoid-s3747" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3748" xml:space="preserve">Or hauendo gl’archi, che ſonola 4. </s> <s xml:id="echoid-s3749" xml:space="preserve">5. </s> <s xml:id="echoid-s3750" xml:space="preserve">6. </s> <s xml:id="echoid-s3751" xml:space="preserve">7. </s> <s xml:id="echoid-s3752" xml:space="preserve">10. </s> <s xml:id="echoid-s3753" xml:space="preserve">parte del <lb/>circolo, diuidendoli per mezzo, e ſubdiuidendoli hauremo la <lb/>8. </s> <s xml:id="echoid-s3754" xml:space="preserve">16. </s> <s xml:id="echoid-s3755" xml:space="preserve">12. </s> <s xml:id="echoid-s3756" xml:space="preserve">14. </s> <s xml:id="echoid-s3757" xml:space="preserve">20. </s> <s xml:id="echoid-s3758" xml:space="preserve">parte del circolo con la ſua corda da ſe-<lb/>gnarſi nella linea A B. </s> <s xml:id="echoid-s3759" xml:space="preserve">Pertrouare la 9 parte, ſi può diuider <lb/>in 3 parti l’ arco ALB, e la terza parte ſia A R, quale perciò <lb/>ſarà la 9 di tutto il circolo. </s> <s xml:id="echoid-s3760" xml:space="preserve">E queſta diuiſa per mezzo da-<lb/>rà la 18.</s> <s xml:id="echoid-s3761" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3762" xml:space="preserve">Mà per la decimaquinta parte, ſi prenderà l’ arco A P, che <lb/>è la quinta, e l’ arco A B, che è la terza parte del circolo, e la <lb/>loro differenza PB diuiſa per mezzo s’applichi all’arco A S, <lb/>che queſta ſarà la 15 parte di tutto il circolo, come conſta <lb/>dalla 16. </s> <s xml:id="echoid-s3763" xml:space="preserve">dellib. </s> <s xml:id="echoid-s3764" xml:space="preserve">4.</s> <s xml:id="echoid-s3765" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3766" xml:space="preserve">Si che non reſtano, che la 11. </s> <s xml:id="echoid-s3767" xml:space="preserve">13. </s> <s xml:id="echoid-s3768" xml:space="preserve">17. </s> <s xml:id="echoid-s3769" xml:space="preserve">19. </s> <s xml:id="echoid-s3770" xml:space="preserve">parte del circo-<lb/>lo, la quale non ſi troua, che mecanicamente tentando con la <lb/>replicatione del Compaſſo. </s> <s xml:id="echoid-s3771" xml:space="preserve">ll che ſe bene è di qualche noia <lb/>nella fabrica dello Stromento, ad ogni modo apporta poi fa-<lb/>cilità per ſempre nell’ altre occaſioni: </s> <s xml:id="echoid-s3772" xml:space="preserve">e la prattica di tal di-<lb/>uiſione non rieſce tanto ſcommoda, quando il circolo è così <lb/>grande, che la corda della terza parte ſia vguale alla linea <lb/>dello Stromento, e di tal grandezza deue intenderſi la linea <lb/>A B della preſente figura, ſe bene s’è fatta quì aſſai più <lb/>piccola.</s> <s xml:id="echoid-s3773" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3774" xml:space="preserve">Che ſe bene quando lo Stromento è aſſai lungo, vi ſi puon-<lb/>no commodamente notare li lati delle figure anche di più an-<lb/>goli, nulladimeno ne’ mediocri baſterà ſin alla figura di 20 <lb/>angoli, come s’è fatto nella figura 27.</s> <s xml:id="echoid-s3775" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3776" xml:space="preserve">Mà ſe queſtaforma d’oprare ſin’ ora accennata, non pia-<lb/>ceſſe come troppo operoſa, potremo hauere l’iſteſſo intento <pb o="195" file="0213" n="217" rhead="Linea de’Poligoni"/> con l’ aiuto della tauola de’ ſeni, e della linea aritmetica dello <lb/>Stromento; </s> <s xml:id="echoid-s3777" xml:space="preserve">eſſendo che in tal modo hauremo, quanto baſte-<lb/>rà, per le operationi Fiſiche. </s> <s xml:id="echoid-s3778" xml:space="preserve">Ora primieramente diuidaſi il <lb/>circolo, cioè gr. </s> <s xml:id="echoid-s3779" xml:space="preserve">360. </s> <s xml:id="echoid-s3780" xml:space="preserve">per il numero de’lati della figura, e <lb/>s’haurà ſa quantità de’ gradi, che toccano à ciaſcun lato. </s> <s xml:id="echoid-s3781" xml:space="preserve">Di-<lb/>poi queſto numero de’ gradi trouati diuidaſi per metà, e di <lb/>queſta metà ſi cerchi il ſeno nelle tauole, come ſi vede fatto <lb/>nella ſeguente tauoletta, in cui nella prima colonna ſonoi <lb/>numeri de’lati delle figure regolari; </s> <s xml:id="echoid-s3782" xml:space="preserve">nella ſeconda ſono i gra-<lb/>di de gl’archi, che toccano à ciaſcun lato di ciaſcuna figura <lb/> <anchor type="note" xlink:label="note-0213-01a" xlink:href="note-0213-01"/> <anchor type="note" xlink:label="note-0213-02a" xlink:href="note-0213-02"/> nella terza la metà di detti gradi, e nella quarta il ſeno di cia-<lb/>ſcuna. </s> <s xml:id="echoid-s3783" xml:space="preserve">Ciò fatto tiriſi ſopra vn piano vna linea retta vgua- <pb o="196" file="0214" n="218" rhead="CAPO VII."/> le alla linea AR, ouero AT dello Stromento nella figura 27, <lb/>e preſa col Compaſſo la lunghezza di tal linea, s’applichi nel-<lb/>la linea Aritmetica dello Stromento all’ interuallo 86 {1/2}, 86 {1/2}, <lb/>poiche douendo quella eſſer corda digr. </s> <s xml:id="echoid-s3784" xml:space="preserve">120, il ſeno di gradi <lb/>60 è 866. </s> <s xml:id="echoid-s3785" xml:space="preserve">E ritenuto lo Stromento in quell’ apertura, pren-<lb/>daſi il ſeno 707, all’interuallo 70 {1/2}. </s> <s xml:id="echoid-s3786" xml:space="preserve">70 {1/2} per il lato del qua-<lb/>drato, e queſto ſi ſegni nella linea tirata, che rappreſenta la <lb/>linea dello Stromento AR. </s> <s xml:id="echoid-s3787" xml:space="preserve">E così di mano in mano confor-<lb/>me alla quantità de’ſeni notati: </s> <s xml:id="echoid-s3788" xml:space="preserve">perche ſe bene queſti ſono ſe-<lb/>ni della metà de gl’archi, ſono metà delle corde, e queſte han-<lb/>no tra loro la medeſima proportione, che detti ſeni.</s> <s xml:id="echoid-s3789" xml:space="preserve"/> </p> <div xml:id="echoid-div117" type="float" level="2" n="2"> <note position="right" xlink:label="note-0213-01" xlink:href="note-0213-01a" xml:space="preserve"> <lb/>Proportione de’lati de’Poligoni deſcritti nello ſteſſo circolo, enumero\\de’ gradi, che prende ciaſcun lato di dette figure. <lb/>Fig. # # Arco # # Metà # Seno <lb/>1 # G. # M. # G. # M. <lb/>2 # # # <lb/>3 # 120 # # 60 # # 866 <lb/>4 # 90 # # 45 # # 707 <lb/>5 # 72 # # 36 # # 587 <lb/>6 # 60 # # 30 # # 500 <lb/>7 # 51 # 25 # 25 # 42 # 433 <lb/>8 # 45 # # 22 # 30 # 382 <lb/>9 # 40 # # 20 # # 342 <lb/>10 # 36 # # 18 # # 309 <lb/></note> <note position="right" xlink:label="note-0213-02" xlink:href="note-0213-02a" xml:space="preserve"> <lb/>Fig. # # Arco # # Metà # Seno <lb/>11 # 32 # 43 # 16 # 21 # 281 <lb/>12 # 30 # # 15 # # 258 <lb/>13 # 27 # 41 # 13 # 50 # 239 <lb/>14 # 25 # 42 # 12 # 51 # 222 <lb/>15 # 24 # # 12 # # 204 <lb/>16 # 22 # 30 # 11 # 15 # 195 <lb/>17 # 21 # 10 # 10 # 35 # 183 <lb/>18 # 20 # # 10 # # 173 <lb/>19 # 18 # 54 # 9 # 27 # 164 <lb/>20 # 18 # # 9 # # 156 <lb/></note> </div> <p> <s xml:id="echoid-s3790" xml:space="preserve">Finita, che ſia nella linea tirata queſta diuiſione, ſi traporta <lb/>sù le linee AR, AT dello Stromento, il quale hauendo le li-<lb/>nee laterali diuiſe nella proportione de’ lati delle figure rego-<lb/>laririſpetto al medeſimo circolo, in cui capifcano, è manife-<lb/>ſto, che anche gl’interualli hauranno ſimile proportione, co-<lb/>me più volte s’è dimoſtrato.</s> <s xml:id="echoid-s3791" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div119" type="section" level="1" n="63"> <head xml:id="echoid-head114" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head115" style="it" xml:space="preserve">Come data vna linea ſi poſſa farne vna figura Regolare, qual più <lb/>piace, ò deſcriuere l’ angolo d’vna figura Regolare, di quelle, <lb/>che ſon ſegnate nello Stromento.</head> <p> <s xml:id="echoid-s3792" xml:space="preserve">SIa data vna linea AB nella figura 35, e di eſſa voglia farſi <lb/>vna figura di cinque lati vguali. </s> <s xml:id="echoid-s3793" xml:space="preserve">Queſta s’applichi nella <lb/>linea de’poligoni AR, AT dello Stromento, all’ interuallo 5. <lb/></s> <s xml:id="echoid-s3794" xml:space="preserve">5: </s> <s xml:id="echoid-s3795" xml:space="preserve">e perche il lato dell’aſſagono è vguale al ſemidiametro del <lb/>circolo, in cui hà da formarſi il cercato pentagono, ritenuta <lb/>quell’apertura dello Stromento, prendaſi l’interuallo 6. </s> <s xml:id="echoid-s3796" xml:space="preserve">6, e <pb o="197" file="0215" n="219" rhead="Linea de’Poligoni"/> con tal’interuallo dall’ eſtremità A, & </s> <s xml:id="echoid-s3797" xml:space="preserve">B della linea data ſi de-<lb/> <anchor type="figure" xlink:label="fig-0215-01a" xlink:href="fig-0215-01"/> ſcriuano due archetti, che ſi tagliano <lb/>in C, e con quello ſteſſo interuallo dal <lb/>centro C ſi deſcriua il circolo ABD-<lb/>EF, nel quale replicata la linea A B, <lb/>s’haurà il pentagono cercato.</s> <s xml:id="echoid-s3798" xml:space="preserve"/> </p> <div xml:id="echoid-div119" type="float" level="2" n="1"> <figure xlink:label="fig-0215-01" xlink:href="fig-0215-01a"> <image file="0215-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0215-01"/> </figure> </div> <p> <s xml:id="echoid-s3799" xml:space="preserve">Che ſeſolo ſicercaſſe di far’vn’ an-<lb/>golo del Pentagono all’ eſtremità A <lb/>della linea data, trouato come prima <lb/>il centro C, baſterà deſcriuere occul-<lb/>tamente l’arco A F, & </s> <s xml:id="echoid-s3800" xml:space="preserve">ad eſſo applicare la linea A B, ſiche ſra <lb/>la retta A F, e ſarà fatto l’angolo BAF del pentagono. </s> <s xml:id="echoid-s3801" xml:space="preserve">Il che <lb/>è vn pran compendio d’operare per chi hà da far in grande il <lb/>diſſegno d’vna fortezza regolare.</s> <s xml:id="echoid-s3802" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3803" xml:space="preserve">Quindi è, che ſe la linea data foſſe molto grande, in modo, <lb/>che non ſi poteſſe prender tutta col Compaſſo, ò non capiſſe <lb/>nell’interuallo dello Stromento, baſterà ſolo pigliarne vna <lb/>parte nell’eſtremità, qualunque ella ſia ad arbitrio, ò ſia ali-<lb/>quota, ò nò, e con quella far l angolo deſiderato del poligo-<lb/>no, nel modo che s’è detto: </s> <s xml:id="echoid-s3804" xml:space="preserve">perche allongata poi queſta linea <lb/>tirata per far l’angolo, ſinche ſia tanto quanto la prima, fatto <lb/>nella ſua eſtremità vn angolo vguale al già trouato, e co-<lb/>sì di mano in mano verrà à compirſi la figura bramata. </s> <s xml:id="echoid-s3805" xml:space="preserve">Co-<lb/>me per eſſempio, ſe c’imaginiamo la linea A B prolongata <lb/>alla lunghezza di quattro palmi, queſta non può tutta capire <lb/>nello Stromento: </s> <s xml:id="echoid-s3806" xml:space="preserve">perciò ne prendo ſolo la parte A B, e come <lb/>ſe con quella ſola doueſſi operare, quella applico nello Stro-<lb/>mento, & </s> <s xml:id="echoid-s3807" xml:space="preserve">opero come s’è detto: </s> <s xml:id="echoid-s3808" xml:space="preserve">poiche prolongata poi A F <lb/>tanto ch’anch’ella ſia di quattro palmi, nella ſua eſtremità fac-<lb/>cio vn’altr’angolo vguale all’ angolo BAF, e così di mano in <lb/>mano ſin che ſia compita la figura.</s> <s xml:id="echoid-s3809" xml:space="preserve"/> </p> <pb o="198" file="0216" n="220" rhead="CAPO VII."/> </div> <div xml:id="echoid-div121" type="section" level="1" n="64"> <head xml:id="echoid-head116" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head117" style="it" xml:space="preserve">Data vna figura regolare, come ſe le poſſa circoſcriuere, <lb/>ò inſcriuer’ vn circolo.</head> <p> <s xml:id="echoid-s3810" xml:space="preserve">PEr la circoſerittione del circolo non ſi richiede più che <lb/>trouar’il centro della figura regolare data: </s> <s xml:id="echoid-s3811" xml:space="preserve">la quale ſe <lb/>hà numero pari di ſati, come 6, 8, &</s> <s xml:id="echoid-s3812" xml:space="preserve">c. </s> <s xml:id="echoid-s3813" xml:space="preserve">baſta dalli due angoli <lb/>oppoſti tirar’vna diagonale, e da altri due angoli oppoſti vn’ <lb/>altra diagonale, la quale diuiderà per mezzo la prima, & </s> <s xml:id="echoid-s3814" xml:space="preserve">il <lb/>punto dell’interſettione è il centro della figura; </s> <s xml:id="echoid-s3815" xml:space="preserve">e l’interuallo <lb/>dal detto punto ſin’ad vno de gl’angoli è il ſemidia metro del <lb/>circolo, che ſi circoſcriue alla figura.</s> <s xml:id="echoid-s3816" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3817" xml:space="preserve">Mà ſe la data figura è di numero diſuguale dilati, conuien’ <lb/>applicar’ il lato di detta figura nella linea de’poligoni nello <lb/>Stromento all’interuallo corriſpondente alla figura (così ſe <lb/>è vn pentagono s’applica all’interuallo 5.</s> <s xml:id="echoid-s3818" xml:space="preserve">5.) </s> <s xml:id="echoid-s3819" xml:space="preserve">e poi preſo l’in-<lb/>teruallo 6. </s> <s xml:id="echoid-s3820" xml:space="preserve">6, deſcriuere, come nella Queſtione precedente, <lb/>due archi occulti, che ſi tagliano in C; </s> <s xml:id="echoid-s3821" xml:space="preserve">e queſto è il centro <lb/>della figura, & </s> <s xml:id="echoid-s3822" xml:space="preserve">all’interuallo CA ſe le circoſcriue il circolo <lb/>ABDF.</s> <s xml:id="echoid-s3823" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3824" xml:space="preserve">Per iſcriuere poi il circolo, baſta, trouato come prima il <lb/>centro della data figura, diuider per mezzo vno de’lati, come <lb/>AB in H, e dal centro C all’interuallo CH deſcriuer’ il circo-<lb/>lo HIKLM, il quale ſarà inſcritto alla detta figura, poiche <lb/>tutti i lati di eſſa lo toccano; </s> <s xml:id="echoid-s3825" xml:space="preserve">come facilmente ſi può dimo-<lb/>ſtrare dalle coſe, che dice Euclide nellib. </s> <s xml:id="echoid-s3826" xml:space="preserve">4. </s> <s xml:id="echoid-s3827" xml:space="preserve">in ſomigliante <lb/>propoſito.</s> <s xml:id="echoid-s3828" xml:space="preserve"/> </p> <pb o="199" file="0217" n="221" rhead="Linea de’Poligoni"/> </div> <div xml:id="echoid-div122" type="section" level="1" n="65"> <head xml:id="echoid-head118" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head119" style="it" xml:space="preserve">Dato vn’arco, come ſi poſſa facil mente trouare in eſſo la quantità <lb/>d’vn’ grado, & altre partidel circolo non ſegnate <lb/>nella linea de’ poligoni.</head> <p> <s xml:id="echoid-s3829" xml:space="preserve">SE bene queſto problema facilmente ſi mette in prattica <lb/>con la linea de’gradi dello Stromento, nondimeno con-<lb/>uien pratticarlo con queſta linea de’ poligoni, perche queſta <lb/>prattica darà lume per varie diuiſioni aſſai minute anche di <lb/>lin ee rette.</s> <s xml:id="echoid-s3830" xml:space="preserve"/> </p> <figure> <image file="0217-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0217-01"/> </figure> <p> <s xml:id="echoid-s3831" xml:space="preserve">Sia dato l’arco AB, di cui <lb/>ſi deſidera ſapere, quanto <lb/>ſia grande la quantità d’vn <lb/>grado. </s> <s xml:id="echoid-s3832" xml:space="preserve">Cerchiſi, per la 25. <lb/></s> <s xml:id="echoid-s3833" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3834" xml:space="preserve">3. </s> <s xml:id="echoid-s3835" xml:space="preserve">il centro di tal’ar-<lb/>co; </s> <s xml:id="echoid-s3836" xml:space="preserve">il che breuemente ſi fà <lb/>prendendo ad arbitrio AC, <lb/>e dalli punti A, & </s> <s xml:id="echoid-s3837" xml:space="preserve">C deſcrit-<lb/>ti occultamente à qualſiuo-<lb/>glia interuallo due archi, <lb/>che ſi tagliano in E, & </s> <s xml:id="echoid-s3838" xml:space="preserve">F, per <lb/>li punti E, & </s> <s xml:id="echoid-s3839" xml:space="preserve">F ſi tiri vna li-<lb/>nea retta indefinita, e lo ſteſ-<lb/>ſo facciaſi prendendo ad ar-<lb/>bitrio BD, e per li punti del-<lb/>l’interſettioni de gl’archi occulti G, & </s> <s xml:id="echoid-s3840" xml:space="preserve">H ſimilmente ſi tiri vna <lb/>linea retta indefinita; </s> <s xml:id="echoid-s3841" xml:space="preserve">la quale taglierà la prima nel punto I; </s> <s xml:id="echoid-s3842" xml:space="preserve"><lb/>e queſto è il centro del circolo, di cui l’arco dato AB è parte.</s> <s xml:id="echoid-s3843" xml:space="preserve"> <pb o="200" file="0218" n="222" rhead="CAPO VII."/> Preſo dunque il ſemidiametro di tal circolo, cioè l’interuallo <lb/>IA, ouero IB, l’applico nella linea de’poligoni alli punti 6.</s> <s xml:id="echoid-s3844" xml:space="preserve">6, <lb/>e ritengo queſta apertura dello Stromento.</s> <s xml:id="echoid-s3845" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3846" xml:space="preserve">Ora quì conuiene far rifieſſione à ciò, che oſſeruò Euclide <lb/>nell’vltima propoſitione del libro 4. </s> <s xml:id="echoid-s3847" xml:space="preserve">doue inſegnò à deſcri-<lb/>uere la figura di quindici lati, col beneficio de’lati del trian-<lb/>golo, e del pentagono: </s> <s xml:id="echoid-s3848" xml:space="preserve">& </s> <s xml:id="echoid-s3849" xml:space="preserve">è, che moltiplicando inſieme li de-<lb/>nominatori di due figure regolari, cioè i numeri de’loro lati, <lb/>ſi hà il denominatore d’vn’altra nuoua figura; </s> <s xml:id="echoid-s3850" xml:space="preserve">e la differenza <lb/>de gl’archi corriſpondenti al lato di dette due figure contiene <lb/>tante parti di queſta nuoua figura, quanta è la differenza de’ <lb/>numeri de’lati di quelle figure. </s> <s xml:id="echoid-s3851" xml:space="preserve">Così il triangolo hà trè lati, <lb/>il pentagono cinque, moltiplico 3, per 5, & </s> <s xml:id="echoid-s3852" xml:space="preserve">hò 15; </s> <s xml:id="echoid-s3853" xml:space="preserve">e perche <lb/>la differenza di 3 à 5 è 2, perciò dall’ iſteſſo punto del circolo <lb/>applicato il lato del triangolo, & </s> <s xml:id="echoid-s3854" xml:space="preserve">il lato del pentagono, la dif-<lb/>ferenza de gl’archi corriſpondenti à queſti lati contiene due <lb/>parti delle quindici del circolo. </s> <s xml:id="echoid-s3855" xml:space="preserve">E ſe la differenza del nume-<lb/>ro de lati delle figure ſia l’vnità, applicati i loro latial circolo, <lb/>reſtarà la differenza de gl’archi la parte competente alla nuo-<lb/>ua figura: </s> <s xml:id="echoid-s3856" xml:space="preserve">Così applicato il lato del quadrato, e del pentago-<lb/>no, la differenza è la venteſima parte del circoſ<unsure/>o, perche 4 <lb/>moltiplicato per 5, fà 20. </s> <s xml:id="echoid-s3857" xml:space="preserve">Il che è manifeſto, perche delle <lb/>20 parti vn quarto ne leua 5, e delle ſteſle 20 vn quinto ne <lb/>leua quattro; </s> <s xml:id="echoid-s3858" xml:space="preserve">dunque la differenza d’vn quarto, e d’vn quin-<lb/>to è vna venteſima.</s> <s xml:id="echoid-s3859" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3860" xml:space="preserve">Suppoſta queſta dottrina veriſſima, e chiariſſima, hauendo <lb/>noi nella linea de’poligoni illato della figura di 20, & </s> <s xml:id="echoid-s3861" xml:space="preserve">il lato <lb/>della fig. </s> <s xml:id="echoid-s3862" xml:space="preserve">di 18 lati, moltiplicando 20 per 18, habbiamo 360, <lb/>che è il numero de’gradi ditutto il circolo; </s> <s xml:id="echoid-s3863" xml:space="preserve">e perche la diffe-<lb/>renzatra 20, e 18 è 2, perciò preſo nello Stromento nella li- <pb o="201" file="0219" n="223" rhead="Linea de’Poligoni"/> nea de’poligoni l’interuallo 18. </s> <s xml:id="echoid-s3864" xml:space="preserve">18, l’applico all’arco dato, & </s> <s xml:id="echoid-s3865" xml:space="preserve"><lb/>è A K: </s> <s xml:id="echoid-s3866" xml:space="preserve">di poi preſo l’interuallo 20. </s> <s xml:id="echoid-s3867" xml:space="preserve">20, l’applico nello ſteſſo <lb/>arco dal punto K, & </s> <s xml:id="echoid-s3868" xml:space="preserve">è K L; </s> <s xml:id="echoid-s3869" xml:space="preserve">onde reſta A L due trecenſeſſante-<lb/>ſime di circolo, e ſe A L ſi diuiderà per mezzo, hauremo il gra-<lb/>do del circolo.</s> <s xml:id="echoid-s3870" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3871" xml:space="preserve">Che ſe prendeſſimo l’interuallo, che diuide il circolo in 20, <lb/>e quello, che lo diuide in 19 parti, la differenza loro ſarà{1/380} del <lb/>circolo, così per diuider il circolo in 63 parti, prendo due nu-<lb/>meri, che moltiplicati facciano 63, e queſti ſono 7, e 9, la <lb/>differenza de’quali è 2. </s> <s xml:id="echoid-s3872" xml:space="preserve">Dunque applicato al circolo il lato <lb/>della figura di ſette, e quello di noue lati, la differenza ſarà <lb/>{2/63} del circolo, e diuiſa per mezzo, darà l’arco, la cui corda è <lb/>lato della figura di 63 lati.</s> <s xml:id="echoid-s3873" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3874" xml:space="preserve">Di quì ſi vede, che hauendo noi nella linea de’poligonii <lb/>lati di diciotto figure, combinandole à due à due, ſi ponno fa-<lb/>re 162 combinationi, e trouar’i lati di altre 162 figure, oltre <lb/>le notate nello Stromento. </s> <s xml:id="echoid-s3875" xml:space="preserve">Mà perche alcune differenze <lb/>comprenderebbono numero diſuguale di parti, ſaria aſſai dif-<lb/>ficile il trouarle, perciò meglio è ſeruirſi ſolo di quelli, che <lb/>hanno ne’numeri la differenza, che è numero pari, e riceue <lb/>ſubdiuiſione. </s> <s xml:id="echoid-s3876" xml:space="preserve">Come per eſſempio, ſe prendiamo il lato di <lb/>20, e queilo di 13, la differenza ſarà {9/26<unsure/>0} del circolo; </s> <s xml:id="echoid-s3877" xml:space="preserve">e troppo <lb/>difficile riuſcirebbe diuidere in ſette parti quella particella, <lb/>che è la differenza de gl’archi: </s> <s xml:id="echoid-s3878" xml:space="preserve">ſe pur non s’adopraſſe ne gli <lb/>archi l’induſtria, che nelle linee rette habbiamo moſtrata nel <lb/>Cap.</s> <s xml:id="echoid-s3879" xml:space="preserve">2. </s> <s xml:id="echoid-s3880" xml:space="preserve">eſpreſſa doue vna venteſima ſi diuiſe in cinque parti. <lb/></s> <s xml:id="echoid-s3881" xml:space="preserve">Mà ſe prendiamo il lato di 11, e quello di 19, la difſerenza <lb/>ſarà{8/209} del circolo; </s> <s xml:id="echoid-s3882" xml:space="preserve">la qual differenza diuiſa, e due altre volte <lb/>ſubdiuiſa, finalmente reſta {1/209} del circolo.</s> <s xml:id="echoid-s3883" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3884" xml:space="preserve">Da queſte coſe quì dette ſi raccoglie vn modo faciliſſi mo <pb o="202" file="0220" n="224" rhead="CAPO VII."/> per pigliar in vna retta linea data vna particella, che per altro <lb/>ſaria difficile à trouare, quando il numero delle parti è nume-<lb/>ro compoſto: </s> <s xml:id="echoid-s3885" xml:space="preserve">cioè trouando due numeri differenti tra di loro <lb/>ſolamente per l’vnità, ouero per il binario, ò quaternario, i <lb/>quali inſieme moltiplicati, facciano il numero, che denomi-<lb/>na le parti.</s> <s xml:id="echoid-s3886" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3887" xml:space="preserve">Per eſſempio voglio vna ſettanteſima ſeconda della linea <lb/> <anchor type="figure" xlink:label="fig-0220-01a" xlink:href="fig-0220-01"/> retta MN. </s> <s xml:id="echoid-s3888" xml:space="preserve">Veggo, che il 72 ſi fà dalla molti-<lb/>plicatione di 8 per 9, onde cauo, che la diffe-<lb/>renza dell’ottaua, e della nona parte di detta <lb/>linea MN è la ſettanteſima ſeconda cercata. <lb/></s> <s xml:id="echoid-s3889" xml:space="preserve">Applico dunque nella linea Aritmetica dello <lb/>Stromento la linea M N al interuallo 80. </s> <s xml:id="echoid-s3890" xml:space="preserve">80, <lb/>perche all’ interuallo 10. </s> <s xml:id="echoid-s3891" xml:space="preserve">10, haurò l’ ottaua <lb/>parte, che ſarà ML. </s> <s xml:id="echoid-s3892" xml:space="preserve">Dipoi l’iſteſſa MN appli-<lb/>co all’interuallo 90. </s> <s xml:id="echoid-s3893" xml:space="preserve">90, & </s> <s xml:id="echoid-s3894" xml:space="preserve">all’interuallo 10. </s> <s xml:id="echoid-s3895" xml:space="preserve"><lb/>10, haurò la nona parte, la quale ſarà LI, e la-<lb/>ſcierà la differenza IM {1/72} di tutta la linea; </s> <s xml:id="echoid-s3896" xml:space="preserve">per-<lb/>che delle 72 particelle vn’ ottauo ne contiene <lb/>9, & </s> <s xml:id="echoid-s3897" xml:space="preserve">vn nono ne contiene 8, dunque la diffe-<lb/>renza d’vn’ottauo, e d’vn nono è {1/72}.</s> <s xml:id="echoid-s3898" xml:space="preserve"/> </p> <div xml:id="echoid-div122" type="float" level="2" n="1"> <figure xlink:label="fig-0220-01" xlink:href="fig-0220-01a"> <image file="0220-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0220-01"/> </figure> </div> <p> <s xml:id="echoid-s3899" xml:space="preserve">E' vero, che ſi può fare più breuemente, e <lb/>ſarà maniera commune anche quando la parte <lb/>è denominata da vn numero primo; </s> <s xml:id="echoid-s3900" xml:space="preserve">cioè ſi <lb/>metta la linea data all’interuallo della deno-<lb/>minatione delle parti, & </s> <s xml:id="echoid-s3901" xml:space="preserve">all’ apertura medeſi-<lb/>ma ſi prenda l’interuallo proſſi mamente mi-<lb/>nore, poiche leuato queſto dalla linea data, il <lb/>rimanente ſarà la parte cercata. </s> <s xml:id="echoid-s3902" xml:space="preserve">Così poſta <lb/>la MN all’ interuallo 72. </s> <s xml:id="echoid-s3903" xml:space="preserve">72, prendaſi l’inter- <pb o="203" file="0221" n="225" rhead="Linea de’Poligoni"/> uallo 71. </s> <s xml:id="echoid-s3904" xml:space="preserve">71, e ſarà NI; </s> <s xml:id="echoid-s3905" xml:space="preserve">dunque IM è vna ſettanteſima ſecon-<lb/>da, come ſi cercaua. </s> <s xml:id="echoid-s3906" xml:space="preserve">E di queſta maniera conuerà operare, <lb/>quando il numero della parte cercata cadeſſe nelli punti vi-<lb/>cini al centro dello Stromento, che per il gruppo dello ſteſſo <lb/>Stromento, non viſi puonno prendere: </s> <s xml:id="echoid-s3907" xml:space="preserve">onde conuiene pren-<lb/>dere l’interuallo, che porta la differenza tra il Numeratore, <lb/>& </s> <s xml:id="echoid-s3908" xml:space="preserve">il Denominatore della parte cercata. </s> <s xml:id="echoid-s3909" xml:space="preserve">Così ſe voleſti @<unsure/> del-<lb/>la MN, veggo che la differenza tra il 3, e 72 è 69; </s> <s xml:id="echoid-s3910" xml:space="preserve">perciò po-<lb/>ſta la MN alli punti 72 72, prendo 69. </s> <s xml:id="echoid-s3911" xml:space="preserve">69, e leuato dalla <lb/>MN queſt’interuallo, il reſiduo ſarebbe {3/72}.</s> <s xml:id="echoid-s3912" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3913" xml:space="preserve">Che ſe la linea data foſſe piccola aſſai, come ML, e ſi vedeſ-<lb/>ſe diuidere in parti 9; </s> <s xml:id="echoid-s3914" xml:space="preserve">perche ſaria ſcommodo l’applicarla <lb/>allo Stromento, prolongo la linea ML tanto, che la replico <lb/>otto volte ſin ad N: </s> <s xml:id="echoid-s3915" xml:space="preserve">dipoi applicata la MN all’ottuplo di par-<lb/>ti 9, cioè al 72, prendo poi 71. </s> <s xml:id="echoid-s3916" xml:space="preserve">71, e ſarà NI, onde reſtando <lb/>IM {1/72} di MN, ſarà per conſequenza {1/9} di ML: </s> <s xml:id="echoid-s3917" xml:space="preserve">e così potrà, ſe <lb/>ſi vorà, continuar ladiuiſione di ML in tutte le ſue none par-<lb/>ti prendendoſi 70. </s> <s xml:id="echoid-s3918" xml:space="preserve">70, e traportandolo dal punto N verſo M, <lb/>che laſciarà {2/72}, cioè {2/9} di ML, &</s> <s xml:id="echoid-s3919" xml:space="preserve">c.</s> <s xml:id="echoid-s3920" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div124" type="section" level="1" n="66"> <head xml:id="echoid-head120" xml:space="preserve">QVESTIONE QVARTA.</head> <head xml:id="echoid-head121" style="it" xml:space="preserve">Come ſi conoſca la proportione de’lati delli poligoni deſcritti nello <lb/>ſteſſo circolo; e poi anche la proportione delli ſteſsi poligoni.</head> <p> <s xml:id="echoid-s3921" xml:space="preserve">DAlla tauoletta poſta in queſto Capo è manifeſta la pro-<lb/>portione de’lati de’poligoni; </s> <s xml:id="echoid-s3922" xml:space="preserve">mà non ſi può ſempre <lb/>hauere queſta tauoletta alla mano, come s’hà lo Stromento. <lb/></s> <s xml:id="echoid-s3923" xml:space="preserve">Per conoſcer dunque la proportione di detti lati conuiene <lb/>vedere, ſe ſi vogſiono con relatione alſemidiametro, ò ſolo <pb o="204" file="0222" n="226" rhead="CAPO VII."/> tra di loro. </s> <s xml:id="echoid-s3924" xml:space="preserve">Per eſſempio voglio ſapere, che proportione <lb/>habbia il lato del pentagono al lato del decagono. </s> <s xml:id="echoid-s3925" xml:space="preserve">Poſſo <lb/>conſiderarli aſſolutamente tra di loro ſenza riguardo del lato <lb/>dell’eſſagono, che è vgual al ſemidia metro; </s> <s xml:id="echoid-s3926" xml:space="preserve">ouero determina-<lb/>ta la quantità delle particelle del ſemidiametro, conſiderare <lb/>quante di quelle particelle contenga ciaſcuno di detti lati. <lb/></s> <s xml:id="echoid-s3927" xml:space="preserve">Nel primo caſo con due Compaſſi prendo gl’interualli 5. </s> <s xml:id="echoid-s3928" xml:space="preserve">5, <lb/>e 10. </s> <s xml:id="echoid-s3929" xml:space="preserve">10, nella linea de’poligoni. </s> <s xml:id="echoid-s3930" xml:space="preserve">Dipoi nella linea Aritme-<lb/>tica applico il lato del pentagono all’interuallo 100. </s> <s xml:id="echoid-s3931" xml:space="preserve">100, e <lb/>trouando, che il lato del decagono cade nell’interuallo 52. </s> <s xml:id="echoid-s3932" xml:space="preserve"><lb/>53, dico, che la loro proportione è come di 100 à 52 {1/2}. </s> <s xml:id="echoid-s3933" xml:space="preserve">Mà <lb/>volendoſi la loro proportione in riguardo del lato dell’ eſſa-<lb/>gono, conuiene prendere trè miſure, cioè oltre li due detti in-<lb/>terualli pigliar’anche quello di 6. </s> <s xml:id="echoid-s3934" xml:space="preserve">6, e queſto nella linea Arit-<lb/>metica porre all’interuallo 100. </s> <s xml:id="echoid-s3935" xml:space="preserve">100, e così troueraſſi la pro-<lb/>portione del lato del pentagono à quello del decagono, come <lb/>58 {1/2} à quaſi 31.</s> <s xml:id="echoid-s3936" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3937" xml:space="preserve">Trouata la proportione de’lati di due figure, in riguardo <lb/>al lato dell’eſſagono poſto come 100, ſi trouerà la proportio-<lb/>ne di dette figure, cercando l’area d’vno de’triangoli di cia-<lb/>ſcuna<unsure/>, e poi moltiplicando queſt’area, per il numero de’lati di <lb/>ciaſcuna. </s> <s xml:id="echoid-s3938" xml:space="preserve">L’area poi di ciaſcun trian-<lb/>golo ſi troua con la moltiplicatione <lb/> <anchor type="figure" xlink:label="fig-0222-01a" xlink:href="fig-0222-01"/> della metà del lato per la perpendico-<lb/>lare, che in eſſo cade dal centro; </s> <s xml:id="echoid-s3939" xml:space="preserve">cioè <lb/>moltiplicando AH per CH, come ſi <lb/>caua dalla 42. </s> <s xml:id="echoid-s3940" xml:space="preserve">dellib. </s> <s xml:id="echoid-s3941" xml:space="preserve">1. </s> <s xml:id="echoid-s3942" xml:space="preserve">Si troua poi <lb/>la grandezza della perpendicolare <lb/>CH, ò con lo Stromento applicando <lb/>CA ſemidiametro nella linea Aritme- <pb o="205" file="0223" n="227" rhead="Linea de’Poligoni"/> tica all’interuallo 100. </s> <s xml:id="echoid-s3943" xml:space="preserve">100, ò dal quadrato della CA 100, <lb/>cauando il quadrato della metà del lato conoſciuto. </s> <s xml:id="echoid-s3944" xml:space="preserve">Eſſen-<lb/>do dunque il lato del pentagono in riguardo del ſemidiame-<lb/>tro del circolo, à cui è inſcritto, come 58 {1/2}, la ſua metà è 29 {1/4}, <lb/>il cui quadrato è 855 {9/16}, il quale ſottratto dal quadrato del <lb/>ſemidiametro, reſta il quadrato della CH, e la radice 95 {1/2} in <lb/>circa è la quantità della perpendicolare CH. </s> <s xml:id="echoid-s3945" xml:space="preserve">Moltiplicato <lb/>dunque CH 95 {1/2} per HA 29 {1/4}, l’area d’vn triangolo quinta <lb/>parte del pentagono è 2793 {1/2}, e queſta moltiplicata per 5. <lb/></s> <s xml:id="echoid-s3946" xml:space="preserve">numero de’lati per conſeguenza de’triangoli del pentagono, <lb/>ſarà tutta l’area del pentagono 13976. </s> <s xml:id="echoid-s3947" xml:space="preserve">Il che pure ſi ſaria <lb/>trouato, ſe preſa la metà del giro del pentagono (che è 292 {1/2}) <lb/>cioè 146 {1/4} ſi foſle moltiplicata per la perpendicolare 95 {1/2} @ <lb/>poiche ſaria venuta l’area del pentagono allo ſteſſo modo <lb/>13967.</s> <s xml:id="echoid-s3948" xml:space="preserve"/> </p> <div xml:id="echoid-div124" type="float" level="2" n="1"> <figure xlink:label="fig-0222-01" xlink:href="fig-0222-01a"> <image file="0222-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0222-01"/> </figure> </div> <p> <s xml:id="echoid-s3949" xml:space="preserve">Ora per trouar l’area del decagono, il cui lato è quaſi 31, <lb/>& </s> <s xml:id="echoid-s3950" xml:space="preserve">il mezzo giro 155, in circa, trouola perpendicolare cauan-<lb/>do dal quadrato del ſemidiametro, cioè da 10000, il quadra-<lb/>to della metà del lato 15 {1/2}, cioè 240, e reſtano 9760 qua-<lb/>drato della perpendicolare, quale perciò è 98<unsure/>. </s> <s xml:id="echoid-s3951" xml:space="preserve">Moltiplica-<lb/>to dunque 155 per 98 {3/4}, ſi produce l’area del decagono <lb/>15306. </s> <s xml:id="echoid-s3952" xml:space="preserve">Dal che conchiudo, che il pentagono, & </s> <s xml:id="echoid-s3953" xml:space="preserve">il deca-<lb/>gono deſcritti nello ſteſſo circolo ſono come 13967, e <lb/>15306, & </s> <s xml:id="echoid-s3954" xml:space="preserve">in minori termini, poiche li numeri non ſon tanto <lb/>preciſi, come 14 à 15. </s> <s xml:id="echoid-s3955" xml:space="preserve">E nella ſteſſa forma ſi procederà nel-<lb/>la comparatione dell’altre figure, doue ſi vedrà, che quanto <lb/>minore è il lato, tanto più và creſcendo l’area.</s> <s xml:id="echoid-s3956" xml:space="preserve"/> </p> <pb o="206" file="0224" n="228" rhead="CAPO VII."/> </div> <div xml:id="echoid-div126" type="section" level="1" n="67"> <head xml:id="echoid-head122" xml:space="preserve">QVESTIONE QVINTA.</head> <head xml:id="echoid-head123" xml:space="preserve">Dato vn poligono regolare, trouarne vn’altro à lui vguale.</head> <p> <s xml:id="echoid-s3957" xml:space="preserve">SE ſarà data vna figura regolare, & </s> <s xml:id="echoid-s3958" xml:space="preserve">vn’altra diuerſa ſe ne <lb/>deſideri à lei vguale, primieramente per la Queſtione <lb/>antecedente ſi troui la proportione di tali figure nello ſteſſo <lb/>circolo, come ſe ſia dato vn pentagono, e ſi voglia vn deca-<lb/>gono à lui vguale, ſi troua, che il pentagono al decagono nel-<lb/>lo ſteſſo circolo è come 14 à 15. </s> <s xml:id="echoid-s3959" xml:space="preserve">Dipoi il lato della data figu-<lb/>ra s’applichi nelle linee de’poligoni all’interuallo conuenien-<lb/>te, come nel caſo noſtro all<unsure/> interuallo 5.</s> <s xml:id="echoid-s3960" xml:space="preserve">5, e ſi prenda l’inter-<lb/>uallo della ſpecie della figura, che ſi cerca, come quì è il de-<lb/>cagono, e ſarà 10. </s> <s xml:id="echoid-s3961" xml:space="preserve">10. </s> <s xml:id="echoid-s3962" xml:space="preserve">Finalmente perche il decagono è co-<lb/>me 15, al pentagono, che è come 14; </s> <s xml:id="echoid-s3963" xml:space="preserve">nelle linee Geometri-<lb/>che all’interuallo 15. </s> <s xml:id="echoid-s3964" xml:space="preserve">15, applico queſto lato trouato del de-<lb/>cagono; </s> <s xml:id="echoid-s3965" xml:space="preserve">e preſo l’internallo 14. </s> <s xml:id="echoid-s3966" xml:space="preserve">14, ſarà illato d’vn decago-<lb/>no, che è al decagono inſcritto nello ſteſſo circolo col penta-<lb/>gono dato, come 14 à 15, cioè come il pentagono dato al <lb/>decagono nello ſteſſo circolo: </s> <s xml:id="echoid-s3967" xml:space="preserve">Dunque queſt’ vltimo inter-<lb/>uallo preſo è il lato del decagono vguale al dato pentagono; <lb/></s> <s xml:id="echoid-s3968" xml:space="preserve">poiche così il decagono di queſto lato, come il pentagono <lb/>dato hanno la ſteſſa proportione di 14 à 15 al decagono nel-<lb/>lo ſteſſo circolo con la figura data, per la 7 del 5.</s> <s xml:id="echoid-s3969" xml:space="preserve"/> </p> <pb o="207" file="0225" n="229" rhead="Trasformatoria de’ Piani"/> </div> <div xml:id="echoid-div127" type="section" level="1" n="68"> <head xml:id="echoid-head124" xml:space="preserve">CAPO VIII.</head> <head xml:id="echoid-head125" style="it" xml:space="preserve">In qual maniera s’<unsure/> habbia à ſegnare nello Stromento la linea <lb/>d’vgualianza trà piani regolari diſſomiglianu: <lb/>& vſo di queſta linea trasformatoria.</head> <p> <s xml:id="echoid-s3970" xml:space="preserve">COnuien talhora cangiar’vna figura piana in vn’altra di <lb/>ſpecie differente, e ſe bene di ciò s’e parlato nel Capo <lb/>antecedente alla Queſt. </s> <s xml:id="echoid-s3971" xml:space="preserve">1. </s> <s xml:id="echoid-s3972" xml:space="preserve">nientedimeno per farlo più preſto, <lb/>e con facilità, ſi può nel noſtro Stromento ſegnar<unsure/> il lato di <lb/>ciaſcuna figura. </s> <s xml:id="echoid-s3973" xml:space="preserve">E perche le figure Irregolari non hanno al-<lb/>cuna determinatione, potendo eſſer molto varia la loro irre-<lb/>golarità, perciò ſolamente ſi conſiderano le reg<unsure/>lari, poiche <lb/>conoſciuto vn lato, tutti gl’altri ſon noti, eſſendo tra di ſe <lb/>vguali.</s> <s xml:id="echoid-s3974" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3975" xml:space="preserve">Primieramente fà di meſtieri conoſcere la proportione <lb/>de’lati delle figure diſſomiglianti, ma ſecondo l’area, ò ſuper-<lb/>ficie tra diſe vguali. </s> <s xml:id="echoid-s3976" xml:space="preserve">E perche tutte le figure regolari puon-<lb/>no concepirſi, come deſcritte nel circolo; </s> <s xml:id="echoid-s3977" xml:space="preserve">dal cui centro tira-<lb/>te à ciaſcun’ angolo linee rette, l’area ſi diuide in tantitrian-<lb/>goli vguali, quanti ſono i lati di ciaſcuna di dette figure, per-<lb/>ciò baſterà trouar la baſe d’vno di detti triangoli. </s> <s xml:id="echoid-s3978" xml:space="preserve">Onde no-<lb/>ta, che ſia l’area d’vna figura, queſta ſi diuiderà in tante parti, <lb/>quanti ſono i lati della figura, che ſi deſidera, e queſto quo-<lb/>tiente ſarà l’area del triangolo, che è tal parte di detta figu-<lb/>ra. </s> <s xml:id="echoid-s3979" xml:space="preserve">Del qual triangolo iſoſcele eſſendo conoſciuta l’area, e <lb/>la proportione de’lati (poiche per il Capo antecedente ſi co-<lb/>noſce la proportione dellato della figura al ſemidiametro del <lb/>circolo, in cui è deſcritta, ò almeno ſi può cauare dalle rauole <lb/>de’ſeni) ſi troua la grandezza della baſe.</s> <s xml:id="echoid-s3980" xml:space="preserve"/> </p> <pb o="208" file="0226" n="230" rhead="CAPO VIII."/> <p> <s xml:id="echoid-s3981" xml:space="preserve">Dunque ſuppoſto il lato del triangolo equilatero eſſer <lb/>1000, trouo la ſua area nel modo commune à tutti li trian-<lb/>goli, cioè dalla metà del giro di tutto il triangolo ſottraendo <lb/>ciaſcuno de’lati, e moltiplicate inſieme le trè differenze, e <lb/>queſto prodotto moltiplicato per la detta metà del giro, cauo <lb/>la radice quadrata, che ſarà l’area cercata. </s> <s xml:id="echoid-s3982" xml:space="preserve">Perciò eſſendo <lb/>vn lato 1000, tutto il giro è 3000, e la metà 1500; </s> <s xml:id="echoid-s3983" xml:space="preserve">dunque <lb/>le trè differenze ſono 500, 500, 500, le quali moltiplicate <lb/>inſieme, fanno 125000000, e queſto prodotto moltiplicato <lb/>per 1500 metà del giro del triangolo, dà 187500000000; <lb/></s> <s xml:id="echoid-s3984" xml:space="preserve">la cui radice quadrata è 433012 area del dato triangolo <lb/>equilatero.</s> <s xml:id="echoid-s3985" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s3986" xml:space="preserve">Ora volendoſi illato d’vn quadrato vguale al dato trian-<lb/>golo, prendo la quarta parte dell’area trouata del triangolo, <lb/>& </s> <s xml:id="echoid-s3987" xml:space="preserve">è 108253, e queſta è l’area del triangolo, che è la quarta <lb/>parte del quadrato vguà<unsure/>le al dato triangolo. </s> <s xml:id="echoid-s3988" xml:space="preserve">Et in queſto <lb/>piccolo triangolo, quarta parte del quadrato li lati poſti, co-<lb/>me 1000, la baſe è 1414 ℞ 2000000. </s> <s xml:id="echoid-s3989" xml:space="preserve">Dunque perche li <lb/>triangoli ſimili ſono nella proportione duplicata de’lati, cioè <lb/>le lor’aree ſono come li quadrati de’lati homologi, per la <lb/>19. </s> <s xml:id="echoid-s3990" xml:space="preserve">del lib. </s> <s xml:id="echoid-s3991" xml:space="preserve">6, trouata l’area corriſpondente à queſti trè lati <lb/>ne’termini della proportione conoſciuta, ſe ſi farà come l’area <lb/>trouata all’area conoſciuta 108253, così il quadrato della <lb/>baſe 1414 ad vn’altro verrà il quadrato della baſe, che ſi cer-<lb/>ca. </s> <s xml:id="echoid-s3992" xml:space="preserve">Quindiè, che data la proportione de’lati del triangolo <lb/>1000, 1000, 1414, ſi troua l’area 499999: </s> <s xml:id="echoid-s3993" xml:space="preserve">ecosì come <lb/>queſta à 108253, così il quadrato della baſe, che è 2000000 <lb/>(ouero 1999396 ſe ſi prende per baſe 1414 preciſamente) <lb/>à 433012, quadrato della vera baſe, che ſi cerca; </s> <s xml:id="echoid-s3994" xml:space="preserve">quale per-<lb/>ciò ſarà 658 +, etale ſarà illato del quadrato vguale al da-<lb/>to triangolo.</s> <s xml:id="echoid-s3995" xml:space="preserve"/> </p> <pb o="209" file="0227" n="231" rhead="Trasformatoria de’ Piani"/> <p> <s xml:id="echoid-s3996" xml:space="preserve">Con l’iſteſſo metodo ſi trouano i lati del pentagono, eſſa-<lb/>gono, & </s> <s xml:id="echoid-s3997" xml:space="preserve">altri vguali al dato triangolo, cioè prendendo per il <lb/>pentagono la quinta parte dell’area del triangolo equilatero <lb/>poſto, per l’Eptagono la ſettima parte, &</s> <s xml:id="echoid-s3998" xml:space="preserve">c. </s> <s xml:id="echoid-s3999" xml:space="preserve">E poi conoſciu-<lb/>ta la proportione del lato di ciaſcuna figura al ſemidiametro <lb/>del circolo, in cui ella può deſcriuerſi, ſi troua l’area di queſto <lb/>triangolo iſoſcele; </s> <s xml:id="echoid-s4000" xml:space="preserve">e finalmente facendoſi, come la quinta, ò <lb/>ſettima, &</s> <s xml:id="echoid-s4001" xml:space="preserve">c. </s> <s xml:id="echoid-s4002" xml:space="preserve">parte del triangolo equilatero poſto, à queſt’area <lb/>vltimamente trouata, così il quadrato del lato del pentago-<lb/>no, ò eptagono, &</s> <s xml:id="echoid-s4003" xml:space="preserve">c. </s> <s xml:id="echoid-s4004" xml:space="preserve">al quadrato del lato vero cercato; </s> <s xml:id="echoid-s4005" xml:space="preserve">onde <lb/>la radice di queſt’vltimo quadrato ſarà il lato, che ſi cerca: </s> <s xml:id="echoid-s4006" xml:space="preserve">e <lb/>così ſi ſono trouati ilati d’aſcune figure regolari, come nell’ <lb/>anneſſa Tauoletta ſi troua notato. </s> <s xml:id="echoid-s4007" xml:space="preserve">Econ queſta proportione <lb/> <anchor type="note" xlink:label="note-0227-01a" xlink:href="note-0227-01"/> ſi diuidono le linee AN, AV nella fig. </s> <s xml:id="echoid-s4008" xml:space="preserve">dello Stromento pag.</s> <s xml:id="echoid-s4009" xml:space="preserve"><unsure/> <lb/>164. </s> <s xml:id="echoid-s4010" xml:space="preserve">pigſiando tutta la AN per 1000 lato del triangolo, il <lb/>quale ſi ſegna con la nota Δ per contradiſtinguerlo dal 3, che <lb/>ſi ſegna nell’altra linea, in cui ſono le parti del circolo, e chia-<lb/>miamo linea de’poligoni. </s> <s xml:id="echoid-s4011" xml:space="preserve">Così per il pentagono ſi prende <lb/>A 5 di pati 502-- di quelle, delle quali tutta la AN è 1000; <lb/></s> <s xml:id="echoid-s4012" xml:space="preserve">e neſlo ſteſſo modo dell’altre tutte.</s> <s xml:id="echoid-s4013" xml:space="preserve"/> </p> <div xml:id="echoid-div127" type="float" level="2" n="1"> <note position="right" xlink:label="note-0227-01" xlink:href="note-0227-01a" xml:space="preserve"> <lb/># # # # Lati di figure regolari tra di loro vguali. <lb/>Triangolo # 1000. # Ottangolo # 299+ <lb/>Circolo # 742+ # Nonangolo # 264+ <lb/>Quadrato # 658+ # Decangolo # 237+ <lb/>Pentagono # 502 --- # Vndecangolo # 214+ <lb/>Eſſagono # 408+ # Dodecangolo # 197 ---<lb/>Eptagono # 342 ---<lb/></note> </div> <pb o="210" file="0228" n="232" rhead="CAPO VIII."/> <p> <s xml:id="echoid-s4014" xml:space="preserve">Col medeſimo metodo approuarei, che nella ſteſſa <lb/>linea ſi ſegnaſf<unsure/>e il Diametro del circolo vguale all’iſteſſo <lb/>triangolo, la cui area è di parti 433012 quadrate. </s> <s xml:id="echoid-s4015" xml:space="preserve">Perche <lb/>il circolo è vguale al triangolo rettangolo fatto dal ſemidia-<lb/>metro, edalla circonferenza, e perciò vguale al Rettangolo <lb/>ſotto il ſemidiametro, ela ſemicirconſerenza, onde queſti la-<lb/>ti hanno la proportione medeſima del diametro alla circon-<lb/>ferenza, cioè di 113 à 355; </s> <s xml:id="echoid-s4016" xml:space="preserve">perciò moltiplicato 355 per <lb/>113 l’area del circolo ſarà 40115. </s> <s xml:id="echoid-s4017" xml:space="preserve">Siche habbiamo due aree <lb/>di circoli, vna di 40115, l’altra di 433012; </s> <s xml:id="echoid-s4018" xml:space="preserve">e perche ſono i <lb/>circoli comei quadrati del diametro, prendaſi il quadrato del <lb/>diametro 226, cioè 51076, e facciaſi, come il circolo 40115 <lb/>al circolo 433012, così il quadrato 51076 al quadro 5513-<lb/>28: </s> <s xml:id="echoid-s4019" xml:space="preserve">la cui radice quadrata 742 + è la quantità del diame-<lb/>tro del circolo, che dourà prenderſi dal punto A, e verrà à <lb/>cadere tra’l quadrato, & </s> <s xml:id="echoid-s4020" xml:space="preserve">il Triangolo, e ſi potrà ſegnare ò <lb/>con la figura circolare Θ, ouero con le lettere Dia; </s> <s xml:id="echoid-s4021" xml:space="preserve">acciò s’in-<lb/>tenda quello eſſer il diametro del circolo, la cui area è di parti <lb/>433012, vguale al Triangolo equilatero, li cui lati f<unsure/>ono <lb/>vguali alla linea AN di parti 1000. </s> <s xml:id="echoid-s4022" xml:space="preserve">Così con vna tal diui-<lb/>ſione ſegnata per il circolo, ſi potrà immediatamente quadra-<lb/>re il circolo, eſſendoui il quadrato vguale al dato Triangolo, <lb/>al qual è vguale il Circolo del diametro notato.</s> <s xml:id="echoid-s4023" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4024" xml:space="preserve">Quindi è manifeſto, che dato qualunque lato di triangolo, <lb/>à cui ſi deſidera altra figura regolare vguale, gl’interualli dell’ <lb/>apertura dello Stromento ſaranno nella ſteſſa proportione, in <lb/>cui ſono diuiſi i lati dello ſteſſo Stromento, come più volte di <lb/>ſopra s’è detto.</s> <s xml:id="echoid-s4025" xml:space="preserve"/> </p> <pb o="211" file="0229" n="233" rhead="Trasformatoria de’ Piani"/> </div> <div xml:id="echoid-div129" type="section" level="1" n="69"> <head xml:id="echoid-head126" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head127" style="it" xml:space="preserve">Data vna figura regolare, trasformarla in vn’altra vguale <lb/>dipiù, ò meno lati.</head> <p> <s xml:id="echoid-s4026" xml:space="preserve">HAbbiaſi per cagione d’eſſempio vna laſtra d’argento <lb/>quadrata, e vogliaſi farne vn’altra d’vgual groſlezza, <lb/>mà di figura eſſagona, ſi cerca la grandezza del lato dell’eſſa-<lb/>gona. </s> <s xml:id="echoid-s4027" xml:space="preserve">Nella linea trasformatoria, ò d’vguaglianza, comun-<lb/>que chiamar la vogliamo, s’applichi all’interuallo del qua-<lb/>drato il lato dato; </s> <s xml:id="echoid-s4028" xml:space="preserve">e ritenuta quell’<unsure/> apertura, prendaſi nella <lb/>ſteſſa linea l’interuallo 6. </s> <s xml:id="echoid-s4029" xml:space="preserve">6, e queſto riuſcirà il lato cercato <lb/>dell’eſſagono.</s> <s xml:id="echoid-s4030" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4031" xml:space="preserve">Mà ſe foſſe la laſtra così grande, che non capiſce il lato del <lb/>quadrato ne gl’interualli dello Stromento, e ſi voleſſe ſape-<lb/>re in numeri di quanti deti ſarà la lunghezza del lato trouato <lb/>dell’eſſagono, così può operarſi. </s> <s xml:id="echoid-s4032" xml:space="preserve">Allargato lo Stromento à <lb/>qualſiuoglia apertura, prendaſi con due Compaſſi gl’inter-<lb/>ualli corriſpondenti al quadrato, & </s> <s xml:id="echoid-s4033" xml:space="preserve">all’eſſagono nella linea <lb/>trasformatoria. </s> <s xml:id="echoid-s4034" xml:space="preserve">Dipoi nella linea Aritmetica ſi vegga con <lb/>l’applicatione de’due Compaſſi, che proportione habbiano <lb/>tra diloro que’ due lati; </s> <s xml:id="echoid-s4035" xml:space="preserve">e trouando che il lato del quadrato <lb/>à quello dell’eſſagono vguale è come 100 à 62, con la rego-<lb/>la del trè dico, ſe 100 danno 62, illato d’vna laſtra quadrata <lb/>di deti 20, mi darà in vna laſtra vguale eſſagona, il lato di <lb/>deti 12 {2/3}.</s> <s xml:id="echoid-s4036" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4037" xml:space="preserve">Che ſe non ſi poteſſe prendere preciſamente in denomina-<lb/>tione di miſura conoſciuta di palmi, deti, &</s> <s xml:id="echoid-s4038" xml:space="preserve">c. </s> <s xml:id="echoid-s4039" xml:space="preserve">il lato del qua-<lb/>drato, e nondimeno foſſe aſſai grande, prendo la metà, ò al- <pb o="212" file="0230" n="234" rhead="CAPO VIII."/> tra parte aliquota di detto lato, e l’applico all’interuallo del <lb/>quadrato nella linea trasformatoria, e poi prendo il lato del-<lb/>la figura, che ſi deſidera, nell’interuallo della ſteſſa linea tra-<lb/>sformatoria; </s> <s xml:id="echoid-s4040" xml:space="preserve">perche moltiplicando queſta tante volte, in <lb/>quante parti fù diuiſo l’altro lato della figura data, s’haurà il <lb/>lato cercato. </s> <s xml:id="echoid-s4041" xml:space="preserve">La ragione di ciò è manifeſta; </s> <s xml:id="echoid-s4042" xml:space="preserve">perche i lati del-<lb/>le figure ſimili ſono nella proportione ſubduplicata nelle ſteſ-<lb/>ſe figure, dunque preſa la metà del lato dato, queſta è lato <lb/>d’vn quadrato ſubquadruplo del primo: </s> <s xml:id="echoid-s4043" xml:space="preserve">Dunque illato del-<lb/>l’altra figura trouato (eſſendo al quadrato di quella metà <lb/>vguale l’eſſagono di queſto lato trouato) è lato d’vn’eſſagono <lb/>ſubquadruplo al dato quadrato. </s> <s xml:id="echoid-s4044" xml:space="preserve">Ora raddoppiato il lato <lb/>trouato ſarà lato d’vn’altro eſſagono quadruplo di queſto; <lb/></s> <s xml:id="echoid-s4045" xml:space="preserve">Dunque l’eſſagono della linea doppia del lato trouato è <lb/>vguale al quadrato dato.</s> <s xml:id="echoid-s4046" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div130" type="section" level="1" n="70"> <head xml:id="echoid-head128" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head129" style="it" xml:space="preserve">Data vna figura regolare trouarne vn’altra regolare diuerſa, à cui <lb/>habbia la data Proportione.</head> <p> <s xml:id="echoid-s4047" xml:space="preserve">QVeſta operatione è facile adoprandoſi la linea trasfor-<lb/>matoria, e la linea Geometrica: </s> <s xml:id="echoid-s4048" xml:space="preserve">poiche prima nelia <lb/>trasformatoria ſi troua l’vguale, poi nella Geometrica ſi tro-<lb/>ua quella, che hà la data proportione. </s> <s xml:id="echoid-s4049" xml:space="preserve">Sia dato vn triango-<lb/>lo, e ſi deſidera vn’ottangolo, che contenga tre volte, e mez-<lb/>za detto triangolo, cioè che ſia al triangolo, come 7 à 2. </s> <s xml:id="echoid-s4050" xml:space="preserve">Pon-<lb/>go dunque nella linea trasformatoria il lato dato del triango-<lb/>lo all’interuallo proprio: </s> <s xml:id="echoid-s4051" xml:space="preserve">quindi prendo nella ſteſſa linea l’in-<lb/>teruallo 8. </s> <s xml:id="echoid-s4052" xml:space="preserve">8, e queſto è l’ottangolo vguale al triangolo dato.</s> <s xml:id="echoid-s4053" xml:space="preserve"> <pb o="213" file="0231" n="235" rhead="Trasformatoria de’ Piani"/> Conuien dunque trouare vn’ottangolo, che à queſto ſteſſo <lb/>ottangolo ſia come 7 à 2: </s> <s xml:id="echoid-s4054" xml:space="preserve">perciò il lato trouato dell’ottan-<lb/>golo vguale applico nella linea Geometrica all’interuallo 2. <lb/></s> <s xml:id="echoid-s4055" xml:space="preserve">2: </s> <s xml:id="echoid-s4056" xml:space="preserve">e preſo nella ſteſſa linea Geometrica l’interuallo7.</s> <s xml:id="echoid-s4057" xml:space="preserve">7, que-<lb/>ſto ſarà il lato dell’ ottangolo, che è come 7, in riguardo del <lb/>primo ottangolo, cioè del triangolo dato, che è come 2.</s> <s xml:id="echoid-s4058" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4059" xml:space="preserve">Che ſe deſideri conoſcer in numeri illato di queſto ottan-<lb/>golo, che è al triangolo dato, come 7 à 2: </s> <s xml:id="echoid-s4060" xml:space="preserve">ſi troua con l’ap-<lb/>plicatione de’lati del triangolo, & </s> <s xml:id="echoid-s4061" xml:space="preserve">ottangolo vguali nella <lb/>linea Aritmetica, che ſono come 100 à quaſi 30: </s> <s xml:id="echoid-s4062" xml:space="preserve">dipoii lati <lb/>de gl’ottangoli, che ſono come 2 à 7, applicati ſimilmente al-<lb/>la linea Aritmetica, trouo che ſono come 30 à 56, onde <lb/>raccolgo, cheil lato del triangolo dato allato d’vn’ottango-<lb/>lo, che lo contiene trè volte, e mezza è come 100 à 56.</s> <s xml:id="echoid-s4063" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div131" type="section" level="1" n="71"> <head xml:id="echoid-head130" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head131" xml:space="preserve">Date due figure regolari diuerſe, conoſcere, che proportione <lb/>habbiano tra di loro.</head> <p> <s xml:id="echoid-s4064" xml:space="preserve">SIano date due figure diuerſe regolari, per eſſempio vn <lb/>pentagono, & </s> <s xml:id="echoid-s4065" xml:space="preserve">vn triangolo: </s> <s xml:id="echoid-s4066" xml:space="preserve">applico nella linea trasfor-<lb/>matoria il lato della figura, che hà meno angoli, cioèil lato <lb/>del triangolo, & </s> <s xml:id="echoid-s4067" xml:space="preserve">à queſta appertura all’interuallo 5. </s> <s xml:id="echoid-s4068" xml:space="preserve">5. </s> <s xml:id="echoid-s4069" xml:space="preserve">nella <lb/>ſteſſa trasformatoria prendo il lato del pentagono vg uale. <lb/></s> <s xml:id="echoid-s4070" xml:space="preserve">Poſcia queſto lato d’vn pentagono vguale al triangolo dato, <lb/>& </s> <s xml:id="echoid-s4071" xml:space="preserve">il lato del pentagono dato, applico nella linea Geometri-<lb/>ca, come ſi diſſe nel Capo 3. </s> <s xml:id="echoid-s4072" xml:space="preserve">Queſt. </s> <s xml:id="echoid-s4073" xml:space="preserve">4. </s> <s xml:id="echoid-s4074" xml:space="preserve">e così trouata la pro-<lb/>portione de’pentagoni di queſti due lati, ſi fà manifefta la <lb/>proportione del pentagono, etriangolo dati.</s> <s xml:id="echoid-s4075" xml:space="preserve"/> </p> <pb o="214" file="0232" n="236" rhead="CAPO VIII."/> <p> <s xml:id="echoid-s4076" xml:space="preserve">La ragione di queſta operatione è manifeſta dalle coſe più <lb/>volte dette, e dalla coſtruttione dello Stromento nella diui-<lb/>ſione di queſte linee, delle quali ci ſeruiamo.</s> <s xml:id="echoid-s4077" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div132" type="section" level="1" n="72"> <head xml:id="echoid-head132" xml:space="preserve">QVESTIONE QVART A.</head> <head xml:id="echoid-head133" xml:space="preserve">Data l’area d’vn poligono regolare, trouar il ſuo lato.</head> <p> <s xml:id="echoid-s4078" xml:space="preserve">ESſendoche ogni area s’intende compoſta di quadretti di <lb/>determinata miſura, data l’area, deue eſler dato il lato <lb/>di ciaſcun quadretto. </s> <s xml:id="echoid-s4079" xml:space="preserve">Ora ſuppongaſi data l’area d’vn pen-<lb/>tagono di 400 palmi quadrati, e cerchiſi quanto grande ſia il <lb/>lato del detto pentagono. </s> <s xml:id="echoid-s4080" xml:space="preserve">Trouifi il lato d’vn quadrato di <lb/>400 palmi, cauando dal dato numero la radice quadrata, che <lb/>è 20, & </s> <s xml:id="echoid-s4081" xml:space="preserve">in vn piano ſi deſcriua vna linea, che ſi ſupponga di <lb/>20 particelle, ciaſcuna delle quali ſe ben piccola rappreſenti <lb/>vn palmo. </s> <s xml:id="echoid-s4082" xml:space="preserve">Queſta linea s’applichi nella linea trasformato-<lb/>ria all’interuallo proprio del quadrato, & </s> <s xml:id="echoid-s4083" xml:space="preserve">à quella apertura <lb/>dello Stromento ſi prenda l’interuallo 5. </s> <s xml:id="echoid-s4084" xml:space="preserve">5, del pentagono. </s> <s xml:id="echoid-s4085" xml:space="preserve">Il <lb/>che fatto, queſti due interualli del quadrato, e del pentago-<lb/>no s’applichino nella linea Aritmetica, e ſi trouerà, che ſe il <lb/>lato del quadrato 400, è 20, il lato del pentagono di 400 pal-<lb/>miè 15 {1/4}.</s> <s xml:id="echoid-s4086" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4087" xml:space="preserve">Si che data qualſiuoglia area ſi caua la radice quadrata; </s> <s xml:id="echoid-s4088" xml:space="preserve">e <lb/>poſta vna linea di tante miſure s’applica nella trasformatoria <lb/>all’interuallo del quadrato; </s> <s xml:id="echoid-s4089" xml:space="preserve">poiche l’interuallo corriſponden-<lb/>te alla denominatione del poligono dato, ſarà il lato della fi-<lb/>gura, la cui area è vguale al quadrato della linea ſuppoſta, <lb/>cloè all’area data.</s> <s xml:id="echoid-s4090" xml:space="preserve"/> </p> <pb o="215" file="0233" n="237" rhead="Trasformatoria de’ Piani"/> </div> <div xml:id="echoid-div133" type="section" level="1" n="73"> <head xml:id="echoid-head134" xml:space="preserve">QVESTIONE QVINT A.</head> <head xml:id="echoid-head135" xml:space="preserve">Dati due poligoni regolari diuerſi vguali, trouare la porportione <lb/>de’ circoli, ne’ quali eſsi ſt deſcriuono.</head> <p> <s xml:id="echoid-s4091" xml:space="preserve">E’Manifeſto, che li poligoni vguali diuerſi non ſi puonno <lb/>deſcriuere nello ſteſſo circolo; </s> <s xml:id="echoid-s4092" xml:space="preserve">dunque il poligono di <lb/>più lati ſi deſcriue in vn circolo minore, che quello di meno <lb/>lati, ma vguale d’area. </s> <s xml:id="echoid-s4093" xml:space="preserve">Cerchiſi dunque la proportione de’ <lb/>circoli.</s> <s xml:id="echoid-s4094" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4095" xml:space="preserve">Il che ſi fà trouando la proportione de’ ſemidiametri. </s> <s xml:id="echoid-s4096" xml:space="preserve">E ſia <lb/>per eſſe mpio vn triangolo, & </s> <s xml:id="echoid-s4097" xml:space="preserve">vn’eptagono vguali.</s> <s xml:id="echoid-s4098" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4099" xml:space="preserve">Primieramente applico nella linea de’poligoni il lato del <lb/>triangolo all’interuallo 3.</s> <s xml:id="echoid-s4100" xml:space="preserve">3, e prendo l’interuallo 6.</s> <s xml:id="echoid-s4101" xml:space="preserve">6, e que-<lb/>ſto è il ſemidiametro del circolo, in cui ſi deſcriue il dato <lb/>triangolo. </s> <s xml:id="echoid-s4102" xml:space="preserve">Similmente nella ſteſſa linea de’poligoni applico <lb/>il lato dell’eptagono all’interuallo 7. </s> <s xml:id="echoid-s4103" xml:space="preserve">7, e con quell’ apertura <lb/>prendo l’interuallo 6. </s> <s xml:id="echoid-s4104" xml:space="preserve">6, il quale ſfarà il femidiametro del cir-<lb/>colo, in cui ſi deſcriue il dato eptagono. </s> <s xml:id="echoid-s4105" xml:space="preserve">Preſi dipoi queſti <lb/>due ſemidiametri, s’applicano nella linea Geometrica, & </s> <s xml:id="echoid-s4106" xml:space="preserve">in <lb/>quella ſi troua la proportione de’circoli, come s’è detto nella <lb/>Queſt. </s> <s xml:id="echoid-s4107" xml:space="preserve">4. </s> <s xml:id="echoid-s4108" xml:space="preserve">del Cap. </s> <s xml:id="echoid-s4109" xml:space="preserve">3.</s> <s xml:id="echoid-s4110" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div134" type="section" level="1" n="74"> <head xml:id="echoid-head136" xml:space="preserve">QVESTIONE SESTA.</head> <head xml:id="echoid-head137" xml:space="preserve">Data vna figura regolare far’vn circolo à lei vguale, e dato <lb/>vn circolo far vn quadrato vguale.</head> <p> <s xml:id="echoid-s4111" xml:space="preserve">SE non foſſe nella linea ſegnato anche il diametro del cir-<lb/>colo vguale à ciaſcuna delle figure notate nella linea <pb o="216" file="0234" n="238" rhead="CAPO VIII."/> trasformatoria; </s> <s xml:id="echoid-s4112" xml:space="preserve">è facile il trouarſi in queſto modo. </s> <s xml:id="echoid-s4113" xml:space="preserve">Data la <lb/>figura, ſi trasformi in quadrato: </s> <s xml:id="echoid-s4114" xml:space="preserve">il lato di queſto quadrato <lb/>nella linea Geometrica s’applichi all’interuallo 11. </s> <s xml:id="echoid-s4115" xml:space="preserve">11; </s> <s xml:id="echoid-s4116" xml:space="preserve">pren-<lb/>daſi nella ſteſſa linea Geometrica l’interuallo 14. </s> <s xml:id="echoid-s4117" xml:space="preserve">14, e queſto <lb/>è il diametro del circolo, che ſicerca; </s> <s xml:id="echoid-s4118" xml:space="preserve">la ragione è manifeſta, <lb/>perche per le coſe dimoſtrate da Archim. </s> <s xml:id="echoid-s4119" xml:space="preserve">il quadrato del dia-<lb/>metro è al circolo, come 14, à 11; </s> <s xml:id="echoid-s4120" xml:space="preserve">il quadrato di queſt’ vltima <lb/>linea è al quadrato poſto all’interuallo 11. </s> <s xml:id="echoid-s4121" xml:space="preserve">11, cioè al poligo-<lb/>no dato, come 14 à 11, dunque il dato poligono, & </s> <s xml:id="echoid-s4122" xml:space="preserve">iſ circo-<lb/>lo del diametro vltimamente trouato ſono tra di ſe vguali per <lb/>la 7. </s> <s xml:id="echoid-s4123" xml:space="preserve">del 5.</s> <s xml:id="echoid-s4124" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4125" xml:space="preserve">Quindi dato vn circolo, ſarà faciliſſimo il quadrarlo: </s> <s xml:id="echoid-s4126" xml:space="preserve">per-<lb/>che applicato il diametro dato alli punti 14. </s> <s xml:id="echoid-s4127" xml:space="preserve">14: </s> <s xml:id="echoid-s4128" xml:space="preserve">prendaſi l’in-<lb/>teruallo 11. </s> <s xml:id="echoid-s4129" xml:space="preserve">11, e queſta linea darà vn quadrato vguale al cir-<lb/>colo dato; </s> <s xml:id="echoid-s4130" xml:space="preserve">eſſendoche il circolo al quadrato del ſuo diametro <lb/>è come 11 à 14.</s> <s xml:id="echoid-s4131" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div135" type="section" level="1" n="75"> <head xml:id="echoid-head138" xml:space="preserve">QVESTIONE SETTIMA.</head> <head xml:id="echoid-head139" xml:space="preserve">Date due figure regolari diſsimili, e diſuguali, farne vna vguale <lb/>à tutte due, e diſſomigliante.</head> <p> <s xml:id="echoid-s4132" xml:space="preserve">QVeſta operatione ſi fà con ridurre le due diſſimili à <lb/>ſomiglianza, e poi vnirle in vna ſimile, e finalmente <lb/>trouare vna diſſimile. </s> <s xml:id="echoid-s4133" xml:space="preserve">Sia dato vn pentagono, & </s> <s xml:id="echoid-s4134" xml:space="preserve"><lb/>vn quadrato diſuguali, e ſi voglia far vn triangolo vguale alla <lb/>ſomma del pentagono, e del quadrato. </s> <s xml:id="echoid-s4135" xml:space="preserve">Prima riducaſi il pen-<lb/>tagono in quadrato, in queſto modo. </s> <s xml:id="echoid-s4136" xml:space="preserve">Nella linea trasforma-<lb/>toria s’applichi il lato del pentagono dato aſ<unsure/>l’interuallo 5. <lb/></s> <s xml:id="echoid-s4137" xml:space="preserve">5, e poi prendaſi l’interualſ<unsure/>o de’quadrati, *** *** che ſarà il la- <pb o="217" file="0235" n="239" rhead="Trasformatoria de’ Piani"/> to del quadrato, vguale al dato pentagono. </s> <s xml:id="echoid-s4138" xml:space="preserve">Di poihauen-<lb/>doſi già queſto lato d’vn quadrato, & </s> <s xml:id="echoid-s4139" xml:space="preserve">il lato del quadrato da-<lb/>to, s’applichino tutti due nelle linee Geometriche, per tro-<lb/>uar la lor proportione, e ſi faccia vn quadrato vguale à tutti <lb/>due, come s’è detto nel Cap. </s> <s xml:id="echoid-s4140" xml:space="preserve">3. </s> <s xml:id="echoid-s4141" xml:space="preserve">Queſt. </s> <s xml:id="echoid-s4142" xml:space="preserve">5. </s> <s xml:id="echoid-s4143" xml:space="preserve">e ſarà queſto qua-<lb/>drato vguale al pentagono, & </s> <s xml:id="echoid-s4144" xml:space="preserve">al quadrato dati. </s> <s xml:id="echoid-s4145" xml:space="preserve">Finalmente <lb/>il lato di queſto quadrato nelle linee trasformatorie s’appli-<lb/>chi all’interuallo proprio de’quadrati, e con quella apertura <lb/>s’haura all’interuallo ΔΔ proprio de’triangoli il lato deltrian-<lb/>golo vguale al dato quadrato, e per conſeguenza alle due fi-<lb/>gure date diſſimili, e diſeguali.</s> <s xml:id="echoid-s4146" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4147" xml:space="preserve">E ſe foſſero molte le figure date da vnirſi, ſi continui l’o-<lb/>peratione nello ſteſſo modo; </s> <s xml:id="echoid-s4148" xml:space="preserve">come ſe oltre iſ pentagono, e <lb/>quadrato dati vi foſſe anche vn triangolo, e poitutti inſieme <lb/>haueſſero à far’ vn’ottangolo; </s> <s xml:id="echoid-s4149" xml:space="preserve">trouato il triangolo vguale al <lb/>pentagono, & </s> <s xml:id="echoid-s4150" xml:space="preserve">al quadrato dati, così il lato di queſto, come <lb/>del dato triangolo s’applichino nelle linee Geometriche, e ſi <lb/>troui vn triangolo eguale à tutti due; </s> <s xml:id="echoid-s4151" xml:space="preserve">e finalmente il lato di <lb/>tal triangolo vguale à tutte trè le figure date s’applichi nelle <lb/>linee trasformatorie all’interuallo del triangolo, poiche rite-<lb/>nuta quell’ a pertura di Stromento, l’interuallo 8. </s> <s xml:id="echoid-s4152" xml:space="preserve">8, darà il <lb/>lato dell’ottangolo vguale alle trè figure date.</s> <s xml:id="echoid-s4153" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div136" type="section" level="1" n="76"> <head xml:id="echoid-head140" xml:space="preserve">QVESTIONE OTTAVA.</head> <head xml:id="echoid-head141" xml:space="preserve">Dati due poligoni regolari diſsimili, e diſuguali, trouar’ vn’ altra <lb/>figura diſsimile, che ſia vguale alla loro differenza.</head> <p> <s xml:id="echoid-s4154" xml:space="preserve">SIa dato nello ſteſſo circolo vn triangolo, & </s> <s xml:id="echoid-s4155" xml:space="preserve">vn quadrato, <lb/>li quali neceſſariamente ſono diſuguali, e ſi voglia far <pb o="218" file="0236" n="240" rhead="CAPO VIII."/> vn’eſſagono vguale alla differenza tra il triangolo, e quadra-<lb/>to dati. </s> <s xml:id="echoid-s4156" xml:space="preserve">Nelle linee trasformatorie applicato il lato del trian-<lb/>golo dato, ſi troui il lato d’vn quadrato à lui vguale; </s> <s xml:id="echoid-s4157" xml:space="preserve">Dipoi <lb/>queſto lato trouato, & </s> <s xml:id="echoid-s4158" xml:space="preserve">il lato dato del quadrato, s’applichi-<lb/>no nelle linee Geometriche, e trouata la loro proportione ſi <lb/>troui il lato del quadrato vguale alla ſoro differenza, per quel <lb/>che s’è detto nel Cap. </s> <s xml:id="echoid-s4159" xml:space="preserve">3. </s> <s xml:id="echoid-s4160" xml:space="preserve">Queſt. </s> <s xml:id="echoid-s4161" xml:space="preserve">6. </s> <s xml:id="echoid-s4162" xml:space="preserve">Finalmente queſto lato del <lb/>quadrato vltimamente trouato s’applichi nelle linee trasfor-<lb/>matorie all’interuallo de’quadrati, poiche nelle ſteſſe linee <lb/>l’interuallo 6. </s> <s xml:id="echoid-s4163" xml:space="preserve">6, darà il lato dell’eſſagono vguale à quel qua-<lb/>drato, che è la differenza de’ due quadrati applicati, cioè del <lb/>triangolo, e del quadrato dati.</s> <s xml:id="echoid-s4164" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4165" xml:space="preserve">In tutte queſte operationi ſe le linee, che ſono lati delle <lb/>figure date, foſſero troppo grandi, ſi prendano le parti ali-<lb/>quote, ricordandoſi poi di moltiplicare l’vltima linea troua-<lb/>ta ſecondo la denominatione della parte aliquota preſa; </s> <s xml:id="echoid-s4166" xml:space="preserve">co-<lb/>me ſe ſi preſe il terzo della linea, quella trouata ſarà ſola-<lb/>mente il terzo di quella, che ſi cerca, e così dourà triplicarſi: <lb/></s> <s xml:id="echoid-s4167" xml:space="preserve">ſe ſi preſe il quarto, queſta dourà quadruplicarſi, e così <lb/>dell’altre.</s> <s xml:id="echoid-s4168" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div137" type="section" level="1" n="77"> <head xml:id="echoid-head142" style="it" xml:space="preserve">CAPO IX.</head> <head xml:id="echoid-head143" style="it" xml:space="preserve">In qual maniera habbia à ſegnarſi la linea de’ corpi regolari, <lb/>& vſo di queſta linea.</head> <p> <s xml:id="echoid-s4169" xml:space="preserve">COrpi regolari ſi chiamano quelli, che hanno le loro ſu-<lb/>perficie piane, dalle quali ſono compreſi, ſimili, & </s> <s xml:id="echoid-s4170" xml:space="preserve"><lb/>vguali: </s> <s xml:id="echoid-s4171" xml:space="preserve">E perche ogni angolo ſolido è fatto almeno da trè <lb/>ſuperficie, ne può eſſere ſe non minore di quattro angoli ret- <pb o="219" file="0237" n="241" rhead="Corpi Regolari"/> ti, perciò niun corpo regolare può hauere l’angolo ſolido fat-<lb/>to, ò da ſei triangoli equilateri, ò da quattro quadrati, perche <lb/>queſti inſieme fanno quattro angoli retti, e non ſaria ango-<lb/>lo, mà vn piano: </s> <s xml:id="echoid-s4172" xml:space="preserve">quattro pentagoni vguali farebbono più <lb/>di quartro retti; </s> <s xml:id="echoid-s4173" xml:space="preserve">tre eſſagoni fariano giuſtamente quattro ret-<lb/>ti, e tre eptagoni ò di più lati fariano più di quattro retti; </s> <s xml:id="echoid-s4174" xml:space="preserve">on-<lb/>de conſta, che l<unsure/>’angolo ſolido non può eſſer fatto, che ò da <lb/>tre, quattro, e cinque triangoli equilateri, ò datre quadrati, <lb/>ò da tre pentagoni equilateri; </s> <s xml:id="echoid-s4175" xml:space="preserve">e per conſequenza ſolo cinque <lb/>corpi regolari ſono poſſ@bili. </s> <s xml:id="echoid-s4176" xml:space="preserve">Ora ſe di trè triangoli equila-<lb/>teri ſi faccia vn’angolo ſolido, tutto il corpo haurà quattro <lb/>faccie, e ſi chiama retraedro, che vuol dire di quattro faccie, <lb/>ouero piramide; </s> <s xml:id="echoid-s4177" xml:space="preserve">ſe ſi faccia vn’angolo ſolido di quattro trian-<lb/>goli equilateri ſi forma l’octaedro, cioè d’otto faccie; </s> <s xml:id="echoid-s4178" xml:space="preserve">ſe di <lb/>cinque triangoli equilateri, ſi formi l’angolo ſolido, ne viene <lb/>l<unsure/>’icoſaedro di venti faccie. </s> <s xml:id="echoid-s4179" xml:space="preserve">Dipoi l’angolo ſolido ſi fà di trè <lb/>quadrati, e ſe ne forma il cubo, ouero exaedro di ſei faccie: </s> <s xml:id="echoid-s4180" xml:space="preserve">e <lb/>finalmente di tre pentagoni equilateri ſi fà l’angolo ſolido <lb/>del dodecaedro di dodici faccie.</s> <s xml:id="echoid-s4181" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4182" xml:space="preserve">Per trouar dunque i lati di queſti cinque corpi regolari <lb/>contenuti in vna medeſima sfera, ci ſeruiremo del modo da-<lb/>to da Euclide nell’vltima propoſitione del lib. </s> <s xml:id="echoid-s4183" xml:space="preserve">13. </s> <s xml:id="echoid-s4184" xml:space="preserve">Si tiri nel-<lb/>l<unsure/>o Stromento la linea, che deue à queſto effetto ſeruire, e ſia <lb/>l<unsure/>a linea AP, ouero AM. </s> <s xml:id="echoid-s4185" xml:space="preserve">A queſta linea ſe ne tiri in vn pia-<lb/>no vna vguale, e ſia la linea AB, la quale diuidaſi in modo, <lb/>che BC ſia la metà, BD la terza parte, BE la quinta parte. <lb/></s> <s xml:id="echoid-s4186" xml:space="preserve">E dal centro C ſi deſcriua il ſemicircolo AFB. </s> <s xml:id="echoid-s4187" xml:space="preserve">S’alzino poi <lb/>le perpendicolari CF, DG, EH, e ſi tirino le linee AF, che <lb/>è lato dell’octaedro, AG, che è lato della piramide, ouero <lb/>tetraedro BG, che è lato del cubo. </s> <s xml:id="echoid-s4188" xml:space="preserve">E queſta linea BG ſi ta- <pb o="220" file="0238" n="242" rhead="CAPO IX."/> gli nell’eſtrema, e media ragione, cioè in modo, che il qua-<lb/>drato del ſegmento mag-<lb/>giore ſia vgual’al rettango-<lb/> <anchor type="figure" xlink:label="fig-0238-01a" xlink:href="fig-0238-01"/> lo fatto da tutta, e dal ſeg-<lb/>mento minore, come s’in-<lb/>ſegna nella 30 del libro 6, <lb/>ouero nell’ 11. </s> <s xml:id="echoid-s4189" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4190" xml:space="preserve">2; </s> <s xml:id="echoid-s4191" xml:space="preserve">e <lb/>ſia il ſegmento maggiore <lb/>BK, che è lato del dodecae-<lb/>dro. </s> <s xml:id="echoid-s4192" xml:space="preserve">Finalmente della linea BH, come <lb/> <anchor type="figure" xlink:label="fig-0238-02a" xlink:href="fig-0238-02"/> di ſemidiametro ſi formi il ſemicircolo <lb/>BOL; </s> <s xml:id="echoid-s4193" xml:space="preserve">diuidaſi l’arco per metà in O, & </s> <s xml:id="echoid-s4194" xml:space="preserve">il <lb/>ſemidiametro HL per metà in N: </s> <s xml:id="echoid-s4195" xml:space="preserve">pren-<lb/>daſi l’interuallo NO; </s> <s xml:id="echoid-s4196" xml:space="preserve">& </s> <s xml:id="echoid-s4197" xml:space="preserve">à queſto ſia vgua-<lb/>le NI: </s> <s xml:id="echoid-s4198" xml:space="preserve">e così ſarà HI lato del decagono, <lb/>& </s> <s xml:id="echoid-s4199" xml:space="preserve">IO lato del pentagono; </s> <s xml:id="echoid-s4200" xml:space="preserve">e ſi trasferi-<lb/>ſcano nell’altra figura in modo, che BI ſia <lb/>vguale à IO, & </s> <s xml:id="echoid-s4201" xml:space="preserve">IH ſia il lato del decago-<lb/>no nel circolo BOL, ſarà dunque BI lato <lb/>dell’ icoſaedro.</s> <s xml:id="echoid-s4202" xml:space="preserve"/> </p> <div xml:id="echoid-div137" type="float" level="2" n="1"> <figure xlink:label="fig-0238-01" xlink:href="fig-0238-01a"> <image file="0238-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0238-01"/> </figure> <figure xlink:label="fig-0238-02" xlink:href="fig-0238-02a"> <image file="0238-02" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0238-02"/> </figure> </div> <p> <s xml:id="echoid-s4203" xml:space="preserve">Trouate queſte miſure, ſitrasferiſcono ſopra lo Stromen-<lb/>to, in cui AP è diametro della sfera, A4 vguale ad AG, A8 <lb/>vguale ad AF, A 6 vguale à BG, A 20 vguale à BI, A 12 <lb/>vguale à BK; </s> <s xml:id="echoid-s4204" xml:space="preserve">& </s> <s xml:id="echoid-s4205" xml:space="preserve">in tal maniera ſono ſegnati i lati de’corpi re-<lb/>golari, che puonno deſcriuerſi nella ſteſſa sfera.</s> <s xml:id="echoid-s4206" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4207" xml:space="preserve">E perche ſe bene tutte queſte linee ſono tra di loro incom-<lb/>menſurabili di longhezza, nondimeno li lati del tetraedro, <lb/>octaedro, e cubo ſono col diametro della sfera commenſura-<lb/>bili di potenza (gl’altri due lati del dodecaedro, & </s> <s xml:id="echoid-s4208" xml:space="preserve">icoſaedro <lb/>ſon’affatto irrationali) e ſono iloro quadrati in queſta pro- <pb o="221" file="0239" n="243" rhead="Corpi Regolari"/> portione, cioè del diametro della sf<unsure/>era, come 6, del lato del-<lb/>la piramide, come 4, del lato dell’octaedro, come 3, del la-<lb/>to del cubo, come 2, come ſi vede appreſſo il<unsure/> Clauio nella <lb/>dimoſtratione della ſudetta prop. </s> <s xml:id="echoid-s4209" xml:space="preserve">vlt. </s> <s xml:id="echoid-s4210" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4211" xml:space="preserve">13. </s> <s xml:id="echoid-s4212" xml:space="preserve">perciò ſi <lb/>potrà prouare con la linea Geometrica dello Stromento, ſe <lb/>tali lati da noi trouati nel primo modo applicati in eſſa corri-<lb/>ſpondano giuſtamente alli numeri di 6. </s> <s xml:id="echoid-s4213" xml:space="preserve">4. </s> <s xml:id="echoid-s4214" xml:space="preserve">3. </s> <s xml:id="echoid-s4215" xml:space="preserve">2. </s> <s xml:id="echoid-s4216" xml:space="preserve">acciò ſiamo <lb/>ſicuri, che l’operatione fù giuſta.</s> <s xml:id="echoid-s4217" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4218" xml:space="preserve">Quindi ſi potrà in numeri determinare la quantità di que-<lb/>ſte linee in proportione al diametro della sfera, quale mettia-<lb/>mo eſſere di particelle 2000. </s> <s xml:id="echoid-s4219" xml:space="preserve">Dunque il ſuo quadrato <lb/>4000000, che è al quadrato del lato della Piramide come <lb/>6 à 4, darà 2666666 quadrato, la cui radice 1633-<unsure/> è il la-<lb/>to della Piramide. </s> <s xml:id="echoid-s4220" xml:space="preserve">Similmente come 6 à 3, così il quadrato <lb/>4000000 al quadrato 2000000, la cui radice 1414 + è il <lb/>lato dell’octaedro. </s> <s xml:id="echoid-s4221" xml:space="preserve">E come 6 à 2, così il quadrato 4000000 <lb/>al quadrato 1333333, la cui radice 1154 + è il lato del <lb/>Cubo.</s> <s xml:id="echoid-s4222" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4223" xml:space="preserve">Mà per ilati delli altri due corpi regolari ſi richiede mag-<lb/>gior induſtria, poiche il lato del Cubo 1154 deue diuiderſi <lb/>nella media, & </s> <s xml:id="echoid-s4224" xml:space="preserve">eſtrema ragione, cioè come 1000 à 618. </s> <s xml:id="echoid-s4225" xml:space="preserve">proſ. <lb/></s> <s xml:id="echoid-s4226" xml:space="preserve">ſimamente, & </s> <s xml:id="echoid-s4227" xml:space="preserve">il ſegmento maggiore 713 ſarà il lato del do-<lb/>decaedro, come ſi hà dal primo corollario della prop. </s> <s xml:id="echoid-s4228" xml:space="preserve">17. </s> <s xml:id="echoid-s4229" xml:space="preserve">del <lb/>lib. </s> <s xml:id="echoid-s4230" xml:space="preserve">13. </s> <s xml:id="echoid-s4231" xml:space="preserve">d’Euclide. </s> <s xml:id="echoid-s4232" xml:space="preserve">E per trouar il lato dell’Icoſaedro, primie-<lb/>ramente deue trouarſi il raggio di quel circolo, che compren. </s> <s xml:id="echoid-s4233" xml:space="preserve"><lb/>de lecinque baſi delli cinque triangoli, che coſtituiſcono l’an-<lb/>golo ſolido di queſto corpo: </s> <s xml:id="echoid-s4234" xml:space="preserve">Ora per il primo corollario del-<lb/>la prop 16. </s> <s xml:id="echoid-s4235" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4236" xml:space="preserve">13. </s> <s xml:id="echoid-s4237" xml:space="preserve">il quadrato di quel raggio è la quinta <lb/>parte del quadrato del diametro della sfera; </s> <s xml:id="echoid-s4238" xml:space="preserve">onde ſarà <lb/>800000 il quadrato, e la ſua radice 894 + è il raggio di det- <pb o="222" file="0240" n="244" rhead="CAPO IX."/> to circolo. </s> <s xml:id="echoid-s4239" xml:space="preserve">Dipoi eſſendo noto queſto circolo, deue trou<unsure/>arſi <lb/>il lato del Pentagono compreſ<unsure/>o in queſto circolo; </s> <s xml:id="echoid-s4240" xml:space="preserve">poiche <lb/>queſto è il ſato cercato dell’ Icoſaedro, eſſendo baſe d’vno <lb/>delli cinque triangoli equilateri, che fanno l’angolo ſolido. <lb/></s> <s xml:id="echoid-s4241" xml:space="preserve">Per trouar queſto lato del Pentagono (il cui quadrato per la <lb/>10 del 13. </s> <s xml:id="echoid-s4242" xml:space="preserve">è vguale alli quadrati del Raggio, e del Decago-<lb/>no nell’ iſteſſo circolo) biſogna trouar il lato del Decagono <lb/>poſto il Raggio 894, cioè tagliar il Raggio nella eſtrema, e <lb/>media ragione, eſſendoche il ſegmento maggiore è il lato del <lb/>Decagono per il corollario della 9. </s> <s xml:id="echoid-s4243" xml:space="preserve">del 13. </s> <s xml:id="echoid-s4244" xml:space="preserve">Quindi ſarà il la-<lb/>to del Decagono 552: </s> <s xml:id="echoid-s4245" xml:space="preserve">il cui quadrato 304704 aggionto al <lb/>quadrato del Raggio, che è 800000 dà 1104704 quadrato <lb/>del lato del Pentagono; </s> <s xml:id="echoid-s4246" xml:space="preserve">e perciò ſarà la ſua radice 1051 il la-<lb/>to cercato dell’Icoſaedro.</s> <s xml:id="echoid-s4247" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"> <lb/># # Diuiſioni della linea per i corpi regolari inſcritti \\ nella medeſima sfera. <lb/>Diametro della sfera. # 2000 <lb/>Piramide. # 1633---<lb/>Octaedro. # 1414+ <lb/>Cubo. # 1154+ <lb/>Icoſaedro. # 1051 <lb/>Dodecaedro. # 713+ <lb/></note> <pb o="223" file="0241" n="245" rhead="Corpi Regolari"/> </div> <div xml:id="echoid-div139" type="section" level="1" n="78"> <head xml:id="echoid-head144" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head145" style="it" xml:space="preserve">Conoſciuto il diametro d’vna sfera, come ſi poſſa formar’ vn cubo, <lb/>ò altro ſolidoregolare, che capiſca in eſſa.</head> <p> <s xml:id="echoid-s4248" xml:space="preserve">QVelli, che ſi dilettano dentro sfere di vetro formare di <lb/>piccole regolette teſſute inſieme varie figure, come ſe <lb/>foſſero linee, hauranno l’vſo di queſto problema. <lb/></s> <s xml:id="echoid-s4249" xml:space="preserve">Il diametro della sfera dato s’applichi all’ interuallo vltimo <lb/>della linea de’ corpi regolari; </s> <s xml:id="echoid-s4250" xml:space="preserve">e di poi preſo l’interuallo del <lb/>cubo, ſe ſi deſidera formare vn cubo, ò di qualunque altro ſo-<lb/>lido, che vogli<unsure/>a formarſi, cioè l’interuallo 6. </s> <s xml:id="echoid-s4251" xml:space="preserve">6, in quella ſteſ-<lb/>ſa linea, e s’haurà il lato del cubo. </s> <s xml:id="echoid-s4252" xml:space="preserve">Se ſi voleſſe formar’ vna <lb/>piramide, prendaſi l’interuallo 4.</s> <s xml:id="echoid-s4253" xml:space="preserve">4, in quella linea de’cor-<lb/>piregolari.</s> <s xml:id="echoid-s4254" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div140" type="section" level="1" n="79"> <head xml:id="echoid-head146" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head147" style="it" xml:space="preserve">Data vna piramide trouar la sfera, che contenga vn’ altra <lb/>piramide in data proportione.</head> <p> <s xml:id="echoid-s4255" xml:space="preserve">SIa data vna piramide, e ſi deſideri vna sfera, che conten-<lb/>ga vna piramide, che alſa data ſia come 9, à 8. </s> <s xml:id="echoid-s4256" xml:space="preserve">Trouiſi <lb/>illato della piramide, che ſia come 9 à 8, riſpetto della pira-<lb/>mide data: </s> <s xml:id="echoid-s4257" xml:space="preserve">e perche i ſolidi ſimili ſono nella triplicata pro-<lb/>portione de’lati Homologi, cioè, come i cubi de’lati, illato <lb/>della piramide data s’applichi nella linea cubica dello Stro-<lb/>mento all’interuallo 8. </s> <s xml:id="echoid-s4258" xml:space="preserve">8; </s> <s xml:id="echoid-s4259" xml:space="preserve">e preſo l’interuallo 9.</s> <s xml:id="echoid-s4260" xml:space="preserve">9, ſarà lato <lb/>della piramide, che alla prima ſarà come 9 à 8. </s> <s xml:id="echoid-s4261" xml:space="preserve">Quelſto lato <pb o="224" file="0242" n="246" rhead="CAPO IX."/> trouato s’applichi nella linea de’corpi regolari all’interuallo <lb/>4.</s> <s xml:id="echoid-s4262" xml:space="preserve">4, proprio del tetraedro, el’interuallo eſtremo darà il dia-<lb/>metro della sfera, che contiene vna piramide, che è ſeſquiot-<lb/>taua della piramide data.</s> <s xml:id="echoid-s4263" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div141" type="section" level="1" n="80"> <head xml:id="echoid-head148" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head149" style="it" xml:space="preserve">Dato il diametro della sfera trouar la proportione de’corpi <lb/>regolari inſcritti.</head> <p> <s xml:id="echoid-s4264" xml:space="preserve">SIa data vna sfera, il<unsure/> cui diametro è noto, eſi cerchi la <lb/>proportione di detta sfera à ciaſcuno de’corpi regolari <lb/>inſcritti. </s> <s xml:id="echoid-s4265" xml:space="preserve">Ogni sfera è vguale al cono, la cui baſe è vguale <lb/>alla ſuperficie sferica, e l’altezza vguale al raggio, come di-<lb/>moſtra Archimede nel li<unsure/>b. </s> <s xml:id="echoid-s4266" xml:space="preserve">1. </s> <s xml:id="echoid-s4267" xml:space="preserve">de Sphęr. </s> <s xml:id="echoid-s4268" xml:space="preserve">Cyl. </s> <s xml:id="echoid-s4269" xml:space="preserve">dunque dato il <lb/>diametro ſi troua la circonferenza del maſſimo circolo, e que-<lb/>ſta moltiplicata per il ſudetto diametro dà la ſuperficie sferi-<lb/>ca, baſe del cono, e queſta poi moltiplicara per la terza par-<lb/>te del raggio, cioè il ſeſto del diametro dà la ſolidità del cono <lb/>vguale alla sfera; </s> <s xml:id="echoid-s4270" xml:space="preserve">perche ſe la baſe ſi moltiplicaſſe per tuttta <lb/>l’altezza, ſaria la ſolidità del cilindro di baſe, & </s> <s xml:id="echoid-s4271" xml:space="preserve">altezza vgua-<lb/>le; </s> <s xml:id="echoid-s4272" xml:space="preserve">dunque eſſendo il cono la terza parte di tal cilindro, perla <lb/>10. </s> <s xml:id="echoid-s4273" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4274" xml:space="preserve">12. </s> <s xml:id="echoid-s4275" xml:space="preserve">è manifeſto, cheſi deue moltipliar ſolo per la <lb/>terza parte dell’altezza. </s> <s xml:id="echoid-s4276" xml:space="preserve">Per trouar poila ſolidità d’vn corpo <lb/>regolare inſcritto; </s> <s xml:id="echoid-s4277" xml:space="preserve">Primo, ſi troua il lato di detto corpo, ap-<lb/>plicando il dia metro della sfera all’eſtremità della linea de’ <lb/>corpi regolari, econ vn’altro Compaſſo ſi prenda l’interual-<lb/>lo competente al corpo, che ſi cerca: </s> <s xml:id="echoid-s4278" xml:space="preserve">e queſti due interualli <lb/>applicati nella linea Aritmetica, danno in numeri homologi <lb/>al diametro della sfera, illato del corpo, per eſſempio dell’ <pb o="225" file="0243" n="247" rhead="Corpi Regolari"/> icoſaedro, che conſta di 20 faccie triangolari equilatere. </s> <s xml:id="echoid-s4279" xml:space="preserve">Se-<lb/>condo trouato il lato del triangolo equilatero ſi cerchi la ſua <lb/>area, trouando ſ<unsure/>a perpendicolare, che da vn’angolo cade nel <lb/>mezzo del lato oppoſto: </s> <s xml:id="echoid-s4280" xml:space="preserve">il che ſi fà nella linea Geometrica, <lb/>applicando il lato del triangolo, ela metà di detto lato, à due <lb/>numeri, de’quali neceſſariamente vno è quadruplo dell’altro, <lb/>per eſſempio 48, e 12, e preſa la differenza 36 piglio l’inter-<lb/>uallo 36. </s> <s xml:id="echoid-s4281" xml:space="preserve">36, & </s> <s xml:id="echoid-s4282" xml:space="preserve">applico nella linea Aritmetica il lato del <lb/>triangolo al ſuo numero competente trouato nella prima <lb/>operatione, e poi veggo qual interuallo comprenda quella <lb/>diſtanza vltimamente preſa, che è illato d’vn quadrato, a cui <lb/>il quadrato del lato del triangolo è come 4 à 3, e queſto mol-<lb/>tiplicato per la metà del lato del triangolo dà l’area del trian-<lb/>golo. </s> <s xml:id="echoid-s4283" xml:space="preserve">Terzo, perche il corpo iſcritto nella sfera è vguale à <lb/>tante piramidi, che hanno la cima nel centro della sfera tra <lb/>di ſ<unsure/>oro vguali, per hauer le baſi, e gl’aſſi vguali, conuien tro-<lb/>uare la perpendicolare, che dal centro della sfera cade nel <lb/>piano del triangolo. </s> <s xml:id="echoid-s4284" xml:space="preserve">Ora ſe il piano del triangolo s’intenda <lb/>prolongato per ogni parte, taglia la sfera, e fà vn circolo, in <lb/>cui è iſcritto detto triangolo. </s> <s xml:id="echoid-s4285" xml:space="preserve">Prendaſi dunque il lato del <lb/>triangolo, e nella linea de’poligoni s’applichi all’interuallo <lb/>proprio del triangolo, econ vn’altro compaſſo ſi prenda il <lb/>raggio del ſuo circolo, cioè il lato dell’eſſagono: </s> <s xml:id="echoid-s4286" xml:space="preserve">e nella linea <lb/>Aritmetica applicato il lato del triangolo al numero, che gli <lb/>compete già trouato, veggaſi à qual numero cada il raggio <lb/>del circolo. </s> <s xml:id="echoid-s4287" xml:space="preserve">Cadendo dunque dal centro della sfera la per-<lb/>pendicolare nel centro di tal circolo, è noto il raggio del cir-<lb/>colo, & </s> <s xml:id="echoid-s4288" xml:space="preserve">è noto il raggio della sfera oppoſto all’angolo retto, <lb/>dunque applicati queſti due raggi alla linea Geometrica, ſi <lb/>troua la proportione de’l<unsure/>oro quadrati, & </s> <s xml:id="echoid-s4289" xml:space="preserve">alla differenza di <pb o="226" file="0244" n="248" rhead="CAPO IX."/> tali quadrati applicato il Compaſſo, ſi troui poi nella linea <lb/>Aritmetica la ſua quantità in parti homologhe al raggio della <lb/>sfera, e per conſeguenza al lato del corpo, che ſicerca. <lb/></s> <s xml:id="echoid-s4290" xml:space="preserve">E queſta è l’altezza della piramide triangolare. </s> <s xml:id="echoid-s4291" xml:space="preserve">Quarto, per-<lb/>che la piramide per la 7. </s> <s xml:id="echoid-s4292" xml:space="preserve">del 12 è la terza parte del priſma, <lb/>che hà l’iſteſſa baſe, e la iſteſſa altezza, ſi moltiplichi l’area <lb/>trouata del triangolo per la terza parte di queſta altezza tro-<lb/>uata, e ſarà la ſolidità della piramide. </s> <s xml:id="echoid-s4293" xml:space="preserve">Finalmente queſta ſo-<lb/>lidità trouata ſi moltiplichi per il numero delle faccie del cor-<lb/>po regolare, che ſi cerca, e s’haurà tutta la ſolidità di detto <lb/>corpo; </s> <s xml:id="echoid-s4294" xml:space="preserve">e per conſeguenza la proportione, che hà alla sfera.</s> <s xml:id="echoid-s4295" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4296" xml:space="preserve">Ciò che s’è detto de’corpi, le cui faccie ſono triangolari, ſi <lb/>deue proportionata mente intendere del dodecaedro, le cui <lb/>faccie ſono pentagone: </s> <s xml:id="echoid-s4297" xml:space="preserve">perche trouato il lato del dodecae-<lb/>dro, che è il lato del pentagono, ſi troua il raggio del circolo, <lb/>in cui capiſce detto pentagono, e diuiſo per metà il lato del <lb/>pentagono in eſſo cade ſa perpendicolare dal centro, la qua-<lb/>le può il quadrato, che è differenza trà il quadrato del rag-<lb/>gio trouato del circolo, & </s> <s xml:id="echoid-s4298" xml:space="preserve">il quadrato della metà del lato del <lb/>pentagono: </s> <s xml:id="echoid-s4299" xml:space="preserve">e cosi<unsure/>4; </s> <s xml:id="echoid-s4300" xml:space="preserve">ſi troua l’area d’vno de’cinque triangoli <lb/>iſoſceli, ne’quali ſi diuide il pentagono; </s> <s xml:id="echoid-s4301" xml:space="preserve">onde ſi vien à cono-<lb/>ſcerel’area di detto pentagono. </s> <s xml:id="echoid-s4302" xml:space="preserve">Poi dal quadrato del raggio <lb/>della sfera leuato il quadrato del raggio di detto circolo, re-<lb/>ſta il quadrato della linea, che dal centro della sfera cade <lb/>perpendicolarmente nel piano pentagonico, & </s> <s xml:id="echoid-s4303" xml:space="preserve">è l’altezza <lb/>della piramide, che è la duodecima parte dell’octaedro: </s> <s xml:id="echoid-s4304" xml:space="preserve">co-<lb/>me è manifeſto.</s> <s xml:id="echoid-s4305" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4306" xml:space="preserve">Quanto poi al cubo è manifeſto, ch’egli è alla sfera dello <lb/>ſteſſo diametro con il ſato del cubo, come 21 à 11, come s’oſ-<lb/>ſeruò nel Cap. </s> <s xml:id="echoid-s4307" xml:space="preserve">5. </s> <s xml:id="echoid-s4308" xml:space="preserve">queſt. </s> <s xml:id="echoid-s4309" xml:space="preserve">2. </s> <s xml:id="echoid-s4310" xml:space="preserve">Mà il cubo inſcritto nella sfera è <pb o="227" file="0245" n="249" rhead="Corpi Regolari"/> tale, che il ſuo lato è di potenza ſubtripla alla potenza del <lb/>diametro della sfera, per la 15. </s> <s xml:id="echoid-s4311" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4312" xml:space="preserve">13. </s> <s xml:id="echoid-s4313" xml:space="preserve">Dunque prendaſi <lb/>la terza parte del quadrato del diametro della sfera, e di <lb/>queſta prendaſi la radice quadrata: </s> <s xml:id="echoid-s4314" xml:space="preserve">la quale moltiplicata nel <lb/>ſuo quadrato darà la ſolidità del cubo inſcritto. </s> <s xml:id="echoid-s4315" xml:space="preserve">Cosi<unsure/>4; </s> <s xml:id="echoid-s4316" xml:space="preserve">poſto <lb/>il diametro della sfera eſſer 2000, il ſuo quadrato è 4000000 <lb/>di cui la terza parte è 1333333 {1/2}; </s> <s xml:id="echoid-s4317" xml:space="preserve">e la radice quaſi 1154 {1/2} è <lb/>lato del cubo, che moſtiplicato per il ſuo quadrato, dà la ſoſi-<lb/>dità 1537999990, doue che il cubo circoſcritto vien’ad eſ-<lb/>ſere 8000000000.</s> <s xml:id="echoid-s4318" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div142" type="section" level="1" n="81"> <head xml:id="echoid-head150" xml:space="preserve">QVESTIONE QVART A.</head> <head xml:id="echoid-head151" xml:space="preserve">Data vna sfera trouar i lati de’corpi or dinati circoſcritti.</head> <p> <s xml:id="echoid-s4319" xml:space="preserve">LI corpicircoſcritti alla sfera hanno i loro piani, che toc-<lb/>cano la sfera; </s> <s xml:id="echoid-s4320" xml:space="preserve">e perciò l’altezza delle piramidi, che han-<lb/>no per bale tali piani, è vguale al raggio della sfera data. </s> <s xml:id="echoid-s4321" xml:space="preserve">Ora <lb/>perche il corpo inſcritto, & </s> <s xml:id="echoid-s4322" xml:space="preserve">il circoſcritto ſono ſimili, hanno <lb/>anche ilati homologi, e li piani ſono ſimili: </s> <s xml:id="echoid-s4323" xml:space="preserve">e per conſeguen-<lb/>za le pitamidi, nelle quali ſi riſoluono, hauendo trà di loro la <lb/>proportione de’ſuoi tutti, per la 15. </s> <s xml:id="echoid-s4324" xml:space="preserve">del 5. </s> <s xml:id="echoid-s4325" xml:space="preserve">hanno la propor-<lb/>tione triplicata de’lati homologi. </s> <s xml:id="echoid-s4326" xml:space="preserve">Mà perche le piramidi <lb/>hanno le baſi ſimili, queſte baſi hanno la proportione dupli-<lb/>cata de’lati homologi; </s> <s xml:id="echoid-s4327" xml:space="preserve">e perche le piramidi hanno trà diſe <lb/>la proportione compoſta della proportione delle baſi, e del-<lb/>le altezze, eſſendo le baſi nella duplicata proportione de’lati, <lb/>ſeguita, che le altezze habbiano la ſteſſa proportione de’lati. <lb/></s> <s xml:id="echoid-s4328" xml:space="preserve">Ora eſſendo data la sfera, & </s> <s xml:id="echoid-s4329" xml:space="preserve">il ſuo raggio, habbiamo l’altez-<lb/>za del<unsure/>la piramide maggiore, che è parte del corpo circoſcrit- <pb o="228" file="0246" n="250" rhead="CAPO IX."/> to. </s> <s xml:id="echoid-s4330" xml:space="preserve">Nello Stromento data la sfera habbiamo il lato del cor-<lb/>poinſcritto. </s> <s xml:id="echoid-s4331" xml:space="preserve">dunque nel modo detto nella Queſtione pre-<lb/>cedente, ſi troui la perpendicolare, che dal centro della sfera <lb/>cade ſul piano del corpo inſcritto. </s> <s xml:id="echoid-s4332" xml:space="preserve">E poi facciaſi, come la <lb/>perpendicolare trouata, allato del corpo inſcritto, così il ſe. <lb/></s> <s xml:id="echoid-s4333" xml:space="preserve">midiametro della sfera al lato del corpo circoſcritto, che ſi <lb/>cerca.</s> <s xml:id="echoid-s4334" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4335" xml:space="preserve">Di quì è manifeſto, che hauendo ſe piramidi ſudette la <lb/>proportione triplicata de’lati delle baſi, cioè la triplicata del-<lb/>l’altezze, anche il corpo inſcritto, & </s> <s xml:id="echoid-s4336" xml:space="preserve">il circoſcritto hanno la, <lb/>proportione triplicata della perpendicolare dal centro della <lb/>sfera sù la faccia del corpo inſcritto, al ſemidiametro della <lb/>ſteſſa sfera; </s> <s xml:id="echoid-s4337" xml:space="preserve">e così conoſciuta detta perpendicolare, & </s> <s xml:id="echoid-s4338" xml:space="preserve">il rag-<lb/>gio della sfera, e preſi i loro cubi, queſti daranno la propor-<lb/>tione delcorpo inſcritto, al circoſcritto, nella ſteſſa sfera.</s> <s xml:id="echoid-s4339" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div143" type="section" level="1" n="82"> <head xml:id="echoid-head152" xml:space="preserve">QVESTIONE QVINT A.</head> <head xml:id="echoid-head153" xml:space="preserve">Come dato vn corpo regolare ſi trasformi in vn’altro, <lb/>che gli ſia vguale.</head> <p> <s xml:id="echoid-s4340" xml:space="preserve">SIa dato vn’icoſaedro, e ſi voglia far’vna piramide à lui <lb/>vguale. </s> <s xml:id="echoid-s4341" xml:space="preserve">Come s’è detto nella Queſt. </s> <s xml:id="echoid-s4342" xml:space="preserve">3. </s> <s xml:id="echoid-s4343" xml:space="preserve">ſi troui la pro-<lb/>portione dell’icoſaedro, e della piramide inſcritti nella ſteſſa <lb/>sfera. </s> <s xml:id="echoid-s4344" xml:space="preserve">Dipoinella linea delli corpi regolari applicato il lato <lb/>dato dell’icoſaedro all’interuallo 20. </s> <s xml:id="echoid-s4345" xml:space="preserve">20, ſi prenda il lato del-<lb/>la piramide nella ſteſſa sfera all’interuallo 4.</s> <s xml:id="echoid-s4346" xml:space="preserve">4. </s> <s xml:id="echoid-s4347" xml:space="preserve">E finalmente <lb/>nelle linee cubiche s’applichi queſto lato della piramide all’ <lb/>nteruallo d’vn numero, à cui ſia vn’altro numero di dette <lb/>lineenella proportione, che ſi trouò eſſere l’icoſaedro alla <pb o="229" file="0247" n="251" rhead="Corpi Regolari"/> piramide; </s> <s xml:id="echoid-s4348" xml:space="preserve">perche l’interuallo di quell’ altro numero darà il <lb/>lato della piramide, che alla piramide inſcritta nella ſteſſa <lb/>sfera con l’icoſaedro hà la proportione, che l’iſteſſo icoſaedro <lb/>hà alla piramide ſeco inſe<unsure/>titta; </s> <s xml:id="echoid-s4349" xml:space="preserve">Dunque per la 7. </s> <s xml:id="echoid-s4350" xml:space="preserve">del 5. </s> <s xml:id="echoid-s4351" xml:space="preserve">la pi-<lb/>ramide di queſt’ vltimo lato trouato è vgualle all’icoſaedro <lb/>dato.</s> <s xml:id="echoid-s4352" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4353" xml:space="preserve">Da ciò, che quì ſi è detto, potranno ad imitatione della <lb/>linea Trasformatoria de’Poligoni trouarſi ilati di tutti i cin-<lb/>que corpi regolari, & </s> <s xml:id="echoid-s4354" xml:space="preserve">il diametro della sfera, i quali corpi <lb/>ſiano tra di ſe vguali; </s> <s xml:id="echoid-s4355" xml:space="preserve">onde ſi potriano ſegnare nella ſteſſa <lb/>linea de’corpi regolari, mà tirata (non così à trauerſo, come <lb/>per più diſtintione ſi è fatto nella figura poſta alla pag. </s> <s xml:id="echoid-s4356" xml:space="preserve">164.) <lb/></s> <s xml:id="echoid-s4357" xml:space="preserve">per il lungo de’lati dello Stromento come l’altre linee, acciò <lb/>così rimanendo le diſtanze delle miſure notate alquanto <lb/>maggiori, vi ſi poſſano con diſtintione ſegnar i punti, che cor-<lb/>riſpondono alli lati de’corpi, che ſi vguagliano. </s> <s xml:id="echoid-s4358" xml:space="preserve">Nel che ſi <lb/>deuono auuertire due coſe: </s> <s xml:id="echoid-s4359" xml:space="preserve">la prima è, che queſti punti no-<lb/>tati per l’vguaglianza ſudetta non ſi notino con inumeri, co-<lb/>me ſi ſon notati li corpi inſcritti nella ſteſſa sfera, mà con la <lb/>lettera capitale de’loro nomi; </s> <s xml:id="echoid-s4360" xml:space="preserve">cioè il Dodecaedro col D, <lb/>l’Icoſaedro con l’I, il Cubo col C, la Sfera con S, l’Ottaedro <lb/>con l’O, e la Piramide con P. </s> <s xml:id="echoid-s4361" xml:space="preserve">La ſeconda è, che creſcendo <lb/>ilati con l’ordine, con cui quì ſi ſono annouerati, conuien. </s> <s xml:id="echoid-s4362" xml:space="preserve"><lb/>auuertire, che il maggior lato di tutti è quello della Pirami-<lb/>de, ò Tetraedro: </s> <s xml:id="echoid-s4363" xml:space="preserve">e così queſto deue metterſi nel fine della <lb/>linea, ò più à baſſo, ò alquanto più ſopra del punto, doue è <lb/>notato il diametro della sfera per li corpi inſcritti: </s> <s xml:id="echoid-s4364" xml:space="preserve">altrimen-<lb/>ti ſe à ciò non ſi haueſſe il douuto riguardo, correrebbe peri-<lb/>colo, che non vi foſſe luogo per il lato della Piramide, che <lb/>douria eſſere più l<unsure/>ungo ditutta la linea tirata ſul lato dello <pb o="230" file="0248" n="252" rhead="CAPO IX."/> Stromento. </s> <s xml:id="echoid-s4365" xml:space="preserve">Perciò auuertaſi di metter il diametro della Sfe-<lb/>ra notato con la lettera S, come ſi è detto, circa li trè quinti <lb/>di tutta la linea AP, ouero altra più lunga tirata ſul lato dello <lb/>Stromento; </s> <s xml:id="echoid-s4366" xml:space="preserve">perche in tal modo viſarà luogo per il lato della <lb/>Piramide: </s> <s xml:id="echoid-s4367" xml:space="preserve">eſſendo, che li lati de’corpi vguagliati ſono proſ-<lb/>ſimamente nella proportione, che quì metto per facilità de <lb/>gli artefici, che voleſſero valerſi delli numeri per far la ſudet-<lb/>ta diuiſione, per trasformar vn corpo in vn’altro vguale.</s> <s xml:id="echoid-s4368" xml:space="preserve"/> </p> <note position="right" xml:space="preserve"> <lb/># # Lati de’ corpi vguagliati. <lb/>Piramide # 100. <lb/>Octaedro # 63---<lb/>Sfera # 61---<lb/>Cubo # 49. <lb/>Icoſaedro # 37. <lb/>Dodecaedro # 24+ <lb/></note> <pb o="231" file="0249" n="253" rhead="Quadratrice de’Segmenti del Circolo"/> </div> <div xml:id="echoid-div144" type="section" level="1" n="83"> <head xml:id="echoid-head154" xml:space="preserve">CAPO X.</head> <head xml:id="echoid-head155" xml:space="preserve">Come ſi poſſa diuidere vna linea, che ſerua per quadrare <lb/>tutti i Segmenti del Circolo, e figure inſcritte: <lb/>& vſo diqueſta linea Quadratrice.</head> <p> <s xml:id="echoid-s4369" xml:space="preserve">ESſendoſi queſto opuſcolo ſtampato alcuni anni ſono, ec-<lb/>co mi capitan’in mano le Operationi del Compaſſo <lb/>Geometrico del Galilei; </s> <s xml:id="echoid-s4370" xml:space="preserve">& </s> <s xml:id="echoid-s4371" xml:space="preserve">all’Operat. </s> <s xml:id="echoid-s4372" xml:space="preserve">31. </s> <s xml:id="echoid-s4373" xml:space="preserve">trouo vſarſi da <lb/>lui certe linee, che chiama Aggiunte, e ſeruono à riquadrare <lb/>i Segmenti del Circolo, e per conſeguenza anche le figure <lb/>inſcritte al Circolo benche Trapezie, cioè à ritrouar vna li-<lb/>nea, che fatta lato d’vn quadrato, darà vn’area vguale al pro-<lb/>poſto Segmento, ouero alla figura rettilinea, ò miſta, che ſia <lb/>di linee rette, e di curue circolari. </s> <s xml:id="echoid-s4374" xml:space="preserve">Mi pare vtile queſta linea, <lb/>perciò in queſta ſeconda impreſſione aggiongo quì la ſua de-<lb/>ſcrittione, & </s> <s xml:id="echoid-s4375" xml:space="preserve">vſo, à fine che chi haueſſe alcuno Stromento for-<lb/>mato à ſo miglianza di quello del Galilei, ſappia valerſene, <lb/>& </s> <s xml:id="echoid-s4376" xml:space="preserve">intenda come ſia fatta la diuiſione dital linea, la quale <lb/>io chiamo Quadratrice; </s> <s xml:id="echoid-s4377" xml:space="preserve">eſſendo che dà li lati de’quadrati <lb/>vguali alli Segmenti di circolo propoſti.</s> <s xml:id="echoid-s4378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4379" xml:space="preserve">Primieramente è neceſſario determinare la lunghezza del-<lb/>la linea da tirarſi ſul lato dello Stromento; </s> <s xml:id="echoid-s4380" xml:space="preserve">e queſto ſi farà tro-<lb/>uando la linea, ilcui quadrato ſia vguale al ſemicircolo, che ſi <lb/>ſuppone eſſer il maggior delli ſegmenti, che ſi notano nella <lb/>linea. </s> <s xml:id="echoid-s4381" xml:space="preserve">L’area dunque del ſemicircolo è vguale al rettangolo <lb/>fatto dal Raggio, e dalla quarta parte della circonferenza: <lb/></s> <s xml:id="echoid-s4382" xml:space="preserve">perciò inteſo iſ diametro eſſere 200000, la circonferenza è <lb/>628318; </s> <s xml:id="echoid-s4383" xml:space="preserve">ela quarta parte 157079 moltiplicata per il Rag- <pb o="232" file="0250" n="254" rhead="CAPO X."/> gio 100000. </s> <s xml:id="echoid-s4384" xml:space="preserve">darà l’area 15707900000: </s> <s xml:id="echoid-s4385" xml:space="preserve">ſ<unsure/>a radice quadra-<lb/>ta di queſto numero è 125331 di quelle parti, delle quali il <lb/>Raggio è 100000. </s> <s xml:id="echoid-s4386" xml:space="preserve">Dal che ſi vede, che tutta la linea tirata <lb/>dal centro deue in maniera diuiderſi, che <lb/>delle cinque parti di tutta; </s> <s xml:id="echoid-s4387" xml:space="preserve">le quattro par-<lb/> <anchor type="figure" xlink:label="fig-0250-01a" xlink:href="fig-0250-01"/> ti cominciando dal centro ſi diano al Rag-<lb/>gio, e tutta ſarà il lato del quadrato vgua-<lb/>le al ſemicircolo: </s> <s xml:id="echoid-s4388" xml:space="preserve">Perciò prendaſi A *** <lb/>100, & </s> <s xml:id="echoid-s4389" xml:space="preserve">A ◻ 125 {1/3} perche poi sì come <lb/>l’interuallo *** *** ſarà il Raggio, così l’in-<lb/>teruallo ◻ ◻ ſarà il lato del quadrato vgua-<lb/>le al ſemicircolo di quel Raggio.</s> <s xml:id="echoid-s4390" xml:space="preserve"/> </p> <div xml:id="echoid-div144" type="float" level="2" n="1"> <figure xlink:label="fig-0250-01" xlink:href="fig-0250-01a"> <image file="0250-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0250-01"/> </figure> </div> <p> <s xml:id="echoid-s4391" xml:space="preserve">Fatto queſto, ſi deue determinare in <lb/>quante parti vguali ſi vuole diuidere l’al-<lb/>tezza del ſemicircolo, la qual è vguale <lb/>al Raggio, per hauer con ciò le diuerſe <lb/>altezze di varj ſegmenti. </s> <s xml:id="echoid-s4392" xml:space="preserve">Eſſendoche l’i-<lb/>ſteſſa linea A ***, che ſi è poſta raggio <lb/>d’vn ſemicircolo, può in vn’altro circolo <lb/>maggiore eſſere la metà della corda d’vn’ <lb/>arco minore del ſemicircolo, e perciò l’al-<lb/>tezza del ſegmento ſarà minore di A ◻. <lb/></s> <s xml:id="echoid-s4393" xml:space="preserve">Il Galilei la diuiſe in 20 parti vguali, onde <lb/>non ne ſegnò ſe non 18, perche l’vltime <lb/>due cadeuano nel gruppo dello Stromen-<lb/>to. </s> <s xml:id="echoid-s4394" xml:space="preserve">Veroè, che ſe la linea foſſe aſſai lun-<lb/>ga, ſi potria la parte A *** diuidere in <lb/>maggior numero di parti; </s> <s xml:id="echoid-s4395" xml:space="preserve">mà auuertaſi, <lb/>che poſſano eſſer i punti ſenza confuſio-<lb/>ne. </s> <s xml:id="echoid-s4396" xml:space="preserve">Qui<unsure/> per chiarezza maggiore ſi è fat- <pb o="233" file="0251" n="255" rhead="Quadratrice de’Segmenti del Circolo"/> ta la diuiſione in 20 parti, e dal modo, che in queſte ſi ado-<lb/>prarà, ſarà manifeſto ciò, che douria pratticarſi in qualunque <lb/>altra diuiſione. </s> <s xml:id="echoid-s4397" xml:space="preserve">Solo auuertaſi, che il ſegno ***, e linumeri <lb/>ſi mettono dalla parte di fuori della linea, perche nell’iſteſſa <lb/>linea ſi deuono far le altre diuiſioni, che ſeruano per ilati de’ <lb/>quadrati corriſpondenti, & </s> <s xml:id="echoid-s4398" xml:space="preserve">inumeri ſi metteranno dalla <lb/>parte di dentro.</s> <s xml:id="echoid-s4399" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4400" xml:space="preserve">Ora per intender il <lb/> <anchor type="figure" xlink:label="fig-0251-01a" xlink:href="fig-0251-01"/> modo da tenerſi in tro-<lb/>uare l’ aree di ciaſcun <lb/>ſegmento, la metà del-<lb/>la cui corda ſia vguale <lb/>al Raggio A *** dello <lb/>Stromento, e le altez-<lb/>ze ſiano differenti ciaſ-<lb/>cuna per vna ò più ven-<lb/>teſime parti del Rag-<lb/>gio di che manchino; <lb/></s> <s xml:id="echoid-s4401" xml:space="preserve">Conſideriſi la preſente <lb/>figura, nella quale CD <lb/>è corda del ſegmento <lb/>C O D A, e la medeſima linea era diametro del circolo mi-<lb/>nore già ſtatuito, e così la meta della corda ſudetta AD è <lb/>100000. </s> <s xml:id="echoid-s4402" xml:space="preserve">Sia altezza del ſegmento la perpendicolare OA, <lb/>la quale s’intenda prolongata ſin alla circonferenza in B; </s> <s xml:id="echoid-s4403" xml:space="preserve">e <lb/>per<unsure/>ciò OB è di<unsure/>ametro del circolo, eſſendo che paſſa per il <lb/>centro, come quella, che taglia CD per mezzo ad angoli ret-<lb/>ti; </s> <s xml:id="echoid-s4404" xml:space="preserve">come ſi caua dalla terza del libro terzo. </s> <s xml:id="echoid-s4405" xml:space="preserve">Dun<unsure/>que DA è <lb/>media proportionale tra OA, & </s> <s xml:id="echoid-s4406" xml:space="preserve">AB per la 13. </s> <s xml:id="echoid-s4407" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4408" xml:space="preserve">ſeſto, <lb/>e così eſſendo nota la prima OA altezza del ſegmento, e la <pb o="234" file="0252" n="256" rhead="CAPO X."/> ſeconda, AD metà della corda, ſi verrà in cognitione della <lb/>terza AB. </s> <s xml:id="echoid-s4409" xml:space="preserve">Sia dunque OA 19 di quelle parti delle quali ne <lb/>ſono 20 in AD: </s> <s xml:id="echoid-s4410" xml:space="preserve">ſi che diuiſo il quadrato di AD 100000. <lb/></s> <s xml:id="echoid-s4411" xml:space="preserve">00000. </s> <s xml:id="echoid-s4412" xml:space="preserve">per OA 95000, il quotiente darà AB 105263; </s> <s xml:id="echoid-s4413" xml:space="preserve">à <lb/>cui aggionto AO 95000, tutto il diametro OB è noto <lb/>200263; </s> <s xml:id="echoid-s4414" xml:space="preserve">e queſto diuiſo per mezzo dà il Raggio OI 100131 <lb/>dal qual Raggio leuata l’altezza del ſegmento OA 95000, <lb/>rimane AI 5131 altezza perpendicolare del triangolo CID, <lb/>che dourà leuarſi dal ſettore I C O D, per hauere la quantità <lb/>del ſegmẽto dato. </s> <s xml:id="echoid-s4415" xml:space="preserve">Dũque il triangolo CID ſarà 513100000, <lb/>vguale al rettangolo fatto dal perpendicolo IA, e da A D me-<lb/>tà della baſe CD.</s> <s xml:id="echoid-s4416" xml:space="preserve"/> </p> <div xml:id="echoid-div145" type="float" level="2" n="2"> <figure xlink:label="fig-0251-01" xlink:href="fig-0251-01a"> <image file="0251-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0251-01"/> </figure> </div> <p> <s xml:id="echoid-s4417" xml:space="preserve">Ora perche il Settore ſi fà dal Raggio, e dalla metà dell’ <lb/>arco, perciò conuien inueſtigare la metà dell’arco COD, cioè <lb/>l’arco O D, che è miſura dell’angolo OID. </s> <s xml:id="echoid-s4418" xml:space="preserve">Mà perche nel <lb/>triangolo rettangolo D A I è noto il lato D A 100000, & </s> <s xml:id="echoid-s4419" xml:space="preserve">il <lb/>lato AI 5131, prendaſi queſto numero come Tangente dell’ <lb/>Angolo A D I, e nella tauola delle Tangenti ſi troua corri-<lb/>ſpondere à gr. </s> <s xml:id="echoid-s4420" xml:space="preserve">2. </s> <s xml:id="echoid-s4421" xml:space="preserve">5' 6 {1/4}; </s> <s xml:id="echoid-s4422" xml:space="preserve">perilche ſi notifica il ſuo complemen-<lb/>to quantità dell’angolo DIA, e dell’arco OD gr. </s> <s xml:id="echoid-s4423" xml:space="preserve">87. </s> <s xml:id="echoid-s4424" xml:space="preserve">3’{3/4}.</s> <s xml:id="echoid-s4425" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4426" xml:space="preserve">Notificata la quantità dell’arco OD in gradi, reſta ridurla <lb/>à parti ſimili alle particelle del ſuo Raggio OI. </s> <s xml:id="echoid-s4427" xml:space="preserve">E perche in <lb/>ogni circolo la proportione del Raggio alla ſemicirconferen-<lb/>za è come 100000 à 314159, facciaſi il terzo termine dell’ <lb/>analogia il Raggio OI già trouato 100131, e ſarà il quarto <lb/>termine 314570 ſemicirconferenza del circolo, di cui è Rag-<lb/>gio OI. </s> <s xml:id="echoid-s4428" xml:space="preserve">Il che fatto inſtituiſcaſi queſta ſeconda analogia: </s> <s xml:id="echoid-s4429" xml:space="preserve">ſe <lb/>gr. </s> <s xml:id="echoid-s4430" xml:space="preserve">180 danno particelle 314570, che daranno gr. </s> <s xml:id="echoid-s4431" xml:space="preserve">87. </s> <s xml:id="echoid-s4432" xml:space="preserve">3' {3/4}? <lb/></s> <s xml:id="echoid-s4433" xml:space="preserve">e trouaremmo particelle 152151, che ſono l’arco O D. </s> <s xml:id="echoid-s4434" xml:space="preserve">Mol-<lb/>tiplichiſi queſt’arco OD trouato per il Raggio IO, e ſarà tut- <pb o="235" file="0253" n="257" rhead="Quadr atrice de’ Segmenti del Circole"/> ta la quantità del Settore ICOD 15235031781: </s> <s xml:id="echoid-s4435" xml:space="preserve">dal Settore <lb/>ſi leua l’area del triangolo CID 513100000, & </s> <s xml:id="echoid-s4436" xml:space="preserve">il reſiduo <lb/>14721931781 èla quantità cercata del ſegmento dato CO <lb/>DA. </s> <s xml:id="echoid-s4437" xml:space="preserve">Queſto numero ſi accorci delle due vltime figure 81, e <lb/>dal reſto ſi caui la Radice quadrata 121.</s> <s xml:id="echoid-s4438" xml:space="preserve">33. </s> <s xml:id="echoid-s4439" xml:space="preserve">nella quale le <lb/>due vltime figure 33 ſi ſon ſeparate con vn punto, per ſigni-<lb/>ficare, che di quali 100 partiè la metà della corda del ſeg-<lb/>mento dato, di tali 121, e di più 33 centeſime, cioè{1/3} deue <lb/>eſſere la linea, il cui quadrato ſia vguale al dato ſegmento. <lb/></s> <s xml:id="echoid-s4440" xml:space="preserve">E così di tal lunghezza è A 1 de’numeri interiori in propor-<lb/>tione di A *** come 100.</s> <s xml:id="echoid-s4441" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4442" xml:space="preserve">Con queſto metodo ſi trouano le altre linee quadratrici de’ <lb/>ſegmenti, che hanno minor altezza: </s> <s xml:id="echoid-s4443" xml:space="preserve">e così nell’ anneſſa Ta-<lb/>uoletta nella prima colonna ſi mettono per ordine li ſegmen-<lb/>ti, come ſon notate le ſue altezze nella linea dello ſtromento <lb/>cominciando dalli più alti, e così il primo hà pet altezza {19/20<unsure/>}, <lb/>il ſecondo ne hà 18 venteſime, e così per ordine, come di-<lb/>moſtra la ſeconda colonna. </s> <s xml:id="echoid-s4444" xml:space="preserve">Il reſtante è chiaro dal titolo di <lb/>ciaſcuna colonna. </s> <s xml:id="echoid-s4445" xml:space="preserve">E finaſmente l’vltima colonna contiene <lb/>le Radici abbreuiate del quadrato vguale all’ area del ſeg-<lb/>mento, poiche queſte ſon quelle, ehe deuono notarſi nella <lb/>linea Quadratrice dello Stromento; </s> <s xml:id="echoid-s4446" xml:space="preserve">e le due vltime figure <lb/>ſeparate col punto, dinotano le parti centeſime d’vn’ intiero; <lb/></s> <s xml:id="echoid-s4447" xml:space="preserve">acciò ſi vegga quel che ſi deue aggiongere all’ intieri: </s> <s xml:id="echoid-s4448" xml:space="preserve">così al <lb/>numero 6 interiore deue eſſere A6 parti 100.</s> <s xml:id="echoid-s4449" xml:space="preserve">95, cioè pochiſ-<lb/>ſimo meno di parti 101 delle quali A *** è 100.</s> <s xml:id="echoid-s4450" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4451" xml:space="preserve">Dalla conſtruttione di queſta linea Quadratrice ſi rende <lb/>manifeſto il ſuo vſo: </s> <s xml:id="echoid-s4452" xml:space="preserve">eſſendoche A *** e la metà della corda <lb/>d’vn legmento: </s> <s xml:id="echoid-s4453" xml:space="preserve">A 3, per eſſem pio, de’numeri eſteriori è l’al-<lb/>tezza del ſegmento, & </s> <s xml:id="echoid-s4454" xml:space="preserve">A 3 de’numeri ntieriori è la linea, che <pb file="0254" n="258" rhead="Quadratrice de’S"/> <anchor type="note" xlink:label="note-0254-01a" xlink:href="note-0254-01"/> <pb file="0255" n="259" rhead="enti del Circolo."/> <anchor type="note" xlink:label="note-0255-01a" xlink:href="note-0255-01"/> <pb o="236" file="0256" n="260" rhead="CAPO X."/> dà vn quadrato vguale à quel ſegmento. </s> <s xml:id="echoid-s4455" xml:space="preserve">Dunque dato qua-<lb/>lunque ſegmento di circolo, la metà della ſua corda ſi appli-<lb/>chi all’interuallo *** ***: </s> <s xml:id="echoid-s4456" xml:space="preserve">poi ritenuta l’apertura medeſima <lb/>dello Stromento ſi veda à che interuallo delli numeri eſteriori <lb/>capiſca l’altezza data del ſegmento, e ſia per eſſempio alli <lb/>punti 3. </s> <s xml:id="echoid-s4457" xml:space="preserve">3. </s> <s xml:id="echoid-s4458" xml:space="preserve">eſteriori; </s> <s xml:id="echoid-s4459" xml:space="preserve">perciò prendendoſi l’interuallo 3. </s> <s xml:id="echoid-s4460" xml:space="preserve">3. <lb/></s> <s xml:id="echoid-s4461" xml:space="preserve">delli numeri interiori ſi haurà la linea, che dà il quadrato <lb/>vguale al dato ſegmento.</s> <s xml:id="echoid-s4462" xml:space="preserve"/> </p> <div xml:id="echoid-div146" type="float" level="2" n="3"> <note position="right" xlink:label="note-0254-01" xlink:href="note-0254-01a" xml:space="preserve"> <lb/>Ordine de’ Segmenti # Altezza de’Segmenti # Metà dell’ Angolo \\ del \\ Settore # # # In parti de<gap/> <lb/># # Gr. M. # Perpen- \\ dico. del \\ Triang. # Raggio \\ del \\ Circolo # <gap/> <lb/>1 # 19 # 87 # 3 {3/4} # 5131 # 100131 # 1<gap/> <lb/>2 # 18 # 83 # 58 {1/2} # 10555 # 100555 # 1<gap/> <lb/>3 # 17 # 80 # 43 {3/4} # 16323 # 101323 # 1<gap/> <lb/>4 # 16 # 77 # 19 {1/6} # 22500 # 102500 # 1<gap/> <lb/>5 # 15 # 73 # 44 {2/5} # 29166 # 104166 # 1<gap/> <lb/>6 # 14 # 69 # 59. # 36428 # 106428 # 1<gap/> <lb/>7 # 13 # 66 # 2 {6/7} # 44423 # 109423 # 1<gap/> <lb/>8 # 12 # 61 # 55 {2/3} # 53333 # 113333 # 1<gap/> <lb/>9 # 11 # 57 # 37 {1/3} # 63409 # 118409 # 1<gap/> <lb/>10 # 10 # 53 # 7 {4/5} # 75000 # 125000 # 1<gap/> <lb/>11 # 9 # 48 # 27 {1/3} # 88611 # 133611 # 1<gap/> <lb/>12 # 8 # 43 # 36 {1/6} # 105000 # 145000 # 1<gap/> <lb/>13 # 7 # 38 # 34 {5/6} # 125357 # 160357 # 1<gap/> <lb/>14 # 6 # 33 # 24 # 151666 # 181666 # 1<gap/> <lb/>15 # 5 # 28 # 4 {1/3} # 187500 # 212500 # 1<gap/> <lb/>16 # 4 # 22 # 37 {1/6} # 240000 # 260000 # 1<gap/> <lb/>17 # 3 # 17 # 3 {2/3} # 325833 # 340833 # 1<gap/> <lb/>18 # 2 # 11 # 26 # 495000 # 505000 # 1<gap/> <lb/></note> <note position="right" xlink:label="note-0255-01" xlink:href="note-0255-01a" xml:space="preserve"> <lb/># # # # <gap/>ali la metà della corda è 100000. <lb/><gap/> # Area \\ del \\ Settore # Area \\ del \\ Segmento # Radici quara- \\ te abbreuiare. <lb/><gap/>1 # 15235031781 # 14721931781 # 121.33 <lb/><gap/>7 # 14819494235 # 13763994235 # 117.32 <lb/><gap/>2 # 14465074126 # 12832774126 # 113.28 <lb/><gap/>9 # 14177697500 # 11927697500 # 109.21 <lb/><gap/>2 # 13964702292 # 11048102292 # 105.10 <lb/><gap/>8 # 13835427144 # 10192627144 # 10095 <lb/><gap/>7 # 13802288951 # 9359988951 # 96.74 <lb/><gap/>5 # 13882499169 # 8549199169 # 92.46 <lb/><gap/>3 # 14100498947 # 7759598947 # 88.08 <lb/><gap/>1 # 14488875000 # 6988875000 # 83.59 <lb/><gap/>5 # 15097374945 # 6236274945 # 78.97 <lb/><gap/>6 # 16000170000 # 5500170000 # 74.16 <lb/><gap/>7 # 17314867789 # 4779167789 # 69.13 <lb/><gap/>0 # 19238429400 # 4071829400 # 63.81 <lb/><gap/>4 # 22124225000 # 3374225000 # 58.08 <lb/><gap/>3 # 26687180000 # 2687180000 # 51.83 <lb/><gap/>0 # 34591141170 # 2007841170 # 44.80 <lb/><gap/>7 # 50892385000 # 1392385000 # 37.31 <lb/></note> </div> </div> <div xml:id="echoid-div148" type="section" level="1" n="84"> <head xml:id="echoid-head156" xml:space="preserve">QVESTIONE PRIMA.</head> <head xml:id="echoid-head157" style="it" xml:space="preserve">Se due Circoli diſuguali ſi tagliano, come ſi troui la quantità <lb/>dell’area, in cui communicano, e la lunula che reſta.</head> <p> <s xml:id="echoid-s4463" xml:space="preserve">HAbbiaſi riceuuta ſopra vna carta la ſpecie optica dell’ <lb/>Ecliſſe del Sole, e ſia ADB il termine dell’oſcuratione, <lb/>e vogliaſi ſapere, quanta ſia la parte del diſco Solare oſcura-<lb/>ta, e coperta dalla luna. </s> <s xml:id="echoid-s4464" xml:space="preserve">Tiriſi alli punti A & </s> <s xml:id="echoid-s4465" xml:space="preserve">B, doue le cir-<lb/>conferenze ſi tagliano, la corda A B, e queſta diuiſa per mez-<lb/>zo in F ſia tagliata dalla perpendicolare DC: </s> <s xml:id="echoid-s4466" xml:space="preserve">Quindi la metà <lb/>della corda A B, cioè F B, ſi applichi nelle linee Quadratrici <lb/>all’interuallo *** ***, poi preſa l’altezza <lb/> <anchor type="figure" xlink:label="fig-0256-01a" xlink:href="fig-0256-01"/> FD veggaſi à quall’interuallo de’nume-<lb/>ri eſteriori<unsure/> ella capiſca; </s> <s xml:id="echoid-s4467" xml:space="preserve">& </s> <s xml:id="echoid-s4468" xml:space="preserve">alli numeri <lb/>interiori corriſpondenti ſi haurà la li-<lb/>nea del quadrato vguale al ſegmento <lb/>A D B F.</s> <s xml:id="echoid-s4469" xml:space="preserve"/> </p> <div xml:id="echoid-div148" type="float" level="2" n="1"> <figure xlink:label="fig-0256-01" xlink:href="fig-0256-01a"> <image file="0256-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0256-01"/> </figure> </div> <p> <s xml:id="echoid-s4470" xml:space="preserve">Similmente preſa la altezza FC, & </s> <s xml:id="echoid-s4471" xml:space="preserve"><lb/>applicata alli numeri eſteriori, doue <lb/>capiſce, ſi vedrà qual interuallo debba <pb o="237" file="0257" n="261" rhead="Quadratrice de’Segmenti del Circolo"/> pigliarſi de’numeri interiori per hauer la linea del quadrato <lb/>vguale al ſegmento ACBF. </s> <s xml:id="echoid-s4472" xml:space="preserve">Hauute queſte due linee de’ <lb/>quadrati vguali alli due ſegmenti, conforme alla Queſt 5. </s> <s xml:id="echoid-s4473" xml:space="preserve">del <lb/>capo 3. </s> <s xml:id="echoid-s4474" xml:space="preserve">ſi trouarà il lato d’vn quadrato vguale à tutti due li ſu-<lb/>detti quadrati, cioè à tutta la parte oſcurata ADBCA. </s> <s xml:id="echoid-s4475" xml:space="preserve">E que-<lb/>ſto, che ſi è detto dell’Ecliſſe del Sole, deue intenderſi anche <lb/>di quello della Luna, che cade nel cono ombroſo della Terra, <lb/>come è manifeſto.</s> <s xml:id="echoid-s4476" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4477" xml:space="preserve">Et acciò qualche principiante non ſtimaſſe difficile l’haue-<lb/>re queſte linee, cioè la corda AB, e le altezze FD, FC, à ca-<lb/>gione del moto, che fà la ſpecie optica del Sole, ò della Luna <lb/>ſopra il piano, doue ſi riceue; </s> <s xml:id="echoid-s4478" xml:space="preserve">ſappia che baſta notare con vn <lb/>punto li due termini A e B, che ſon manifeſti, e ſubito ad ar-<lb/>bitrio notare vn punto, per eſſempio 1 nel giro dell’ombra, <lb/>& </s> <s xml:id="echoid-s4479" xml:space="preserve">vn’altro punto arbitrario nelgiro dell’imagine lucida, per <lb/>eſſempio S. </s> <s xml:id="echoid-s4480" xml:space="preserve">Poiche hauuti queſti punti ſarà facile con ſuo <lb/>agio finire l’imagine circolare, e trouare i centri delli due cir-<lb/>coli; </s> <s xml:id="echoid-s4481" xml:space="preserve">eſſendo che perla 25. </s> <s xml:id="echoid-s4482" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4483" xml:space="preserve">3. </s> <s xml:id="echoid-s4484" xml:space="preserve">e la quinta del lib. </s> <s xml:id="echoid-s4485" xml:space="preserve">4. </s> <s xml:id="echoid-s4486" xml:space="preserve">per <lb/>li tre punti S A B ſi tira il circolo, il dicui centro ſi troua O, e <lb/>perli trè punti A I B ſimilmente ſi tira il circolo, il dicui cen-<lb/>tro ſitroua V. </s> <s xml:id="echoid-s4487" xml:space="preserve">E di queſta maniera ſarà facile trouare il dia-<lb/>metro del circolo, da cui ſi deue cauare la parte oſcurata <lb/>ADBCA.</s> <s xml:id="echoid-s4488" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4489" xml:space="preserve">Per vedere quanta ſia la parte oſcurata di tutto il diſco lu-<lb/>minoſo, prendaſi il diametro del diſco luminoſo, e nelle linee <lb/>Geometriche ſi applichi all’interuallo 14. </s> <s xml:id="echoid-s4490" xml:space="preserve">14, e ritenuta quel-<lb/>l’apertura dello Stromento prendaſi l’interuallo 11. </s> <s xml:id="echoid-s4491" xml:space="preserve">11. </s> <s xml:id="echoid-s4492" xml:space="preserve">poi-<lb/>che queſto è il lato del quadrato vguale à tutto il circolo, il <lb/>cui diametro ſi è preſo. </s> <s xml:id="echoid-s4493" xml:space="preserve">Di poi ritenuta pure l’iſteſſa apertu-<lb/>ra, nelle medeſime linee ſi vegga, doue capiſca la linea tro- <pb o="238" file="0258" n="262" rhead="CAPO X."/> uata lato del quadrato vguale alla parte oſcurat<unsure/>a ADBCA, <lb/>& </s> <s xml:id="echoid-s4494" xml:space="preserve">il numero corriſpondente à queſto interuallo paragonato <lb/>con 11, moſtrarà la proportione di detta parte oſcurata al <lb/>circolo intiero: </s> <s xml:id="echoid-s4495" xml:space="preserve">onde la differenza ſarà la quantità della par-<lb/>te ancora luminoſa: </s> <s xml:id="echoid-s4496" xml:space="preserve">e così ſarà quadrata anche la lunula <lb/>ASBDA.</s> <s xml:id="echoid-s4497" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4498" xml:space="preserve">Di quì ſi vede, che ſia meglio compire tutto il cerchio <lb/>quando ſia data vna ſunula, in cui tirata la corda, che vniſca <lb/>le punte eſtreme, e queſta diuiſa per mezzo da vna perpen-<lb/>dicolare, veniſſe l’altezza maggiore della metà della ſudetta <lb/>corda; </s> <s xml:id="echoid-s4499" xml:space="preserve">perche ſaria ſegno, che il ſegmento ſia maggiore del <lb/>ſemicircolo: </s> <s xml:id="echoid-s4500" xml:space="preserve">come ſe la ſunula data foſſe AGBDA, trouiſi il <lb/>centro O del circolo eſteriore, e ſi compiſca il circolo con <lb/>l’aggionta dell’arco ACB: </s> <s xml:id="echoid-s4501" xml:space="preserve">poiche trouata, come ſopra, la <lb/>quantità della parte ADBCA, e leuata, come ſi è detto dal <lb/>circolo intiero, rimarrà la cercata quantità della lunula <lb/>AGBDA.</s> <s xml:id="echoid-s4502" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4503" xml:space="preserve">Mà ſe l’altezza della perpendicolare, che cade in mezzo <lb/>della corda, che vniſce le punte eſtreme della Lunula data, <lb/>ſarà minore della metà di detta corda, ſarà ſegno, ch’il ſeg-<lb/>mento è minore del ſemicircolo: </s> <s xml:id="echoid-s4504" xml:space="preserve">tale ſarebbe la lunula SGE <lb/>LS. </s> <s xml:id="echoid-s4505" xml:space="preserve">Tirata la corda SE, diuidaſi per mezzo in H dalla per-<lb/>pendicolare GH; </s> <s xml:id="echoid-s4506" xml:space="preserve">così ſi hanno due ſegmenti ſull’iſteſſa corda, <lb/>l’altezza del minore è H L, quella del maggiore è H G. </s> <s xml:id="echoid-s4507" xml:space="preserve">Dun-<lb/>que applicata HE all’interuallo *** ***, conforme alle due al-<lb/>tezze HG, HL ſi trouino le linee de’quadrati vguali alli ſeg-<lb/>menti predetti: </s> <s xml:id="echoid-s4508" xml:space="preserve">Quindi per la Queſt. </s> <s xml:id="echoid-s4509" xml:space="preserve">6. </s> <s xml:id="echoid-s4510" xml:space="preserve">del capo 3. </s> <s xml:id="echoid-s4511" xml:space="preserve">nelle linee <lb/>Geometriche ſi troui la differenza di queſti quadrati, e la li-<lb/>nea, il cui quadrato è vguale à tal differenza, darà il quadra-<lb/>to vguale alla lunula SGELS.</s> <s xml:id="echoid-s4512" xml:space="preserve"/> </p> <pb o="239" file="0259" n="263" rhead="Quadratrice de’Segmenti del Circolo"/> </div> <div xml:id="echoid-div150" type="section" level="1" n="85"> <head xml:id="echoid-head158" xml:space="preserve">QVESTIONE SECONDA.</head> <head xml:id="echoid-head159" style="it" xml:space="preserve">Dato vn trapeZio in vn Circolo, e ſegmento di circolo, <lb/>trouare la ſua quantità.</head> <p> <s xml:id="echoid-s4513" xml:space="preserve">NOn tuttili trapezj ſon tali, che poſſa loro circoſcriuerſi <lb/>vn circolo; </s> <s xml:id="echoid-s4514" xml:space="preserve">perche i quadrilateri deſcritti in vn circo-<lb/>lo hanno gli angoli oppoſti vguali à due retti per la 22. </s> <s xml:id="echoid-s4515" xml:space="preserve">del <lb/>lib.</s> <s xml:id="echoid-s4516" xml:space="preserve">3. </s> <s xml:id="echoid-s4517" xml:space="preserve">Onde à queſti ſoli è riſtretta la preſente Queſtione. </s> <s xml:id="echoid-s4518" xml:space="preserve">Sia <lb/>dato il Trapezio ABCD nel ſegmento circolare AOD. </s> <s xml:id="echoid-s4519" xml:space="preserve">Pri-<lb/>mieramente diuidaſi in mezzo nel <lb/> <anchor type="figure" xlink:label="fig-0259-01a" xlink:href="fig-0259-01"/> punto Elacorda AD, & </s> <s xml:id="echoid-s4520" xml:space="preserve">alzata la <lb/>perpendicolare EO, cerchiſi nel <lb/>modo detto in queſto Capo la li-<lb/>nea, che dà il quadrato vguale al <lb/>ſegmento AODEA. </s> <s xml:id="echoid-s4521" xml:space="preserve">Dipoi ciaſ-<lb/>cheduno de gl’altri lati del Trape-<lb/>zio, i quali ſono corde di particolari ſegmenti, ſimilmente ſi <lb/>diuidano per mezzo, e ſi habbiano dalle perpendicolari le <lb/>altezze delli ſegmenti. </s> <s xml:id="echoid-s4522" xml:space="preserve">E con quelle corde, & </s> <s xml:id="echoid-s4523" xml:space="preserve">altezze nel <lb/>modo predetto ſi trouino i quadrati vguali à ciaſcun delli trè <lb/>ſegmenti. </s> <s xml:id="echoid-s4524" xml:space="preserve">Queſti trè quadrati minori ſi vniſcano in vn ſol <lb/>quadrato, per la Queſt. </s> <s xml:id="echoid-s4525" xml:space="preserve">5. </s> <s xml:id="echoid-s4526" xml:space="preserve">del capo 3. </s> <s xml:id="echoid-s4527" xml:space="preserve">e queſto quadrato ſi le-<lb/>ui dal quadrato vguale à tutto il ſegmento AODEA, per la <lb/>Queſt. </s> <s xml:id="echoid-s4528" xml:space="preserve">6. </s> <s xml:id="echoid-s4529" xml:space="preserve">del capo 3. </s> <s xml:id="echoid-s4530" xml:space="preserve">& </s> <s xml:id="echoid-s4531" xml:space="preserve">il quadrato vgualle alla differenza, che <lb/>rimane è la quantità del Trapezio propoſto.</s> <s xml:id="echoid-s4532" xml:space="preserve"/> </p> <div xml:id="echoid-div150" type="float" level="2" n="1"> <figure xlink:label="fig-0259-01" xlink:href="fig-0259-01a"> <image file="0259-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0259-01"/> </figure> </div> <p> <s xml:id="echoid-s4533" xml:space="preserve">Queſto, che ſi è detto del modo di trouare l’area de’Trape-<lb/>zj inſcritti nel circolo, deue intenderſi dell’ altre figure m ol-<lb/>tilatere, ò ſiano dilati vguali, ò diſuguali, trouando le linee <pb o="240" file="0260" n="264" rhead="CAPO X."/> de’ quadrati vguali alli particolari ſegmenti, e queſti quadra-<lb/>ti vniti leuandoli dal quadrato vguale à tutto il ſegmento, che <lb/>capiſce tutta la figura; </s> <s xml:id="echoid-s4534" xml:space="preserve">poiche la differenza che reſta è la cer-<lb/>cata quantità della figura propoſta.</s> <s xml:id="echoid-s4535" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div152" type="section" level="1" n="86"> <head xml:id="echoid-head160" xml:space="preserve">QVESTIONE TERZA.</head> <head xml:id="echoid-head161" style="it" xml:space="preserve">Dato vn ſegmento di circolo, ò troppo grande, ò troppo piccolo, come <lb/>ſi debba operare per trouar la linea, che dia il quadr ato <lb/>vguale al ſegmento.</head> <p> <s xml:id="echoid-s4536" xml:space="preserve">ALle volte occorre, che ſia propoſto vn ſegmento con <lb/>la corda, ò con l’altezza così piccola, ò così grande, <lb/>che non ſi poſſano commodamente applicare à gl’interualli <lb/>della linea quadratrice, perciò ſarà neceſſario nelle troppo <lb/>piccole valerſi delle moltiplici, e nelle troppo grandi ſeruirſi <lb/>d’vna parte aliquota; </s> <s xml:id="echoid-s4537" xml:space="preserve">perche poi la linea trouata nella ſteſſa <lb/>proportione ſi ſminuiſce, con cui l’altre ſi accrebero, ò ſi ac-<lb/>creſce, ſe l’altre furono ſminuite. </s> <s xml:id="echoid-s4538" xml:space="preserve">Così ſe le miſure del ſeg-<lb/>mento furono raddoppiate, ſi toglie la metà della linea tro-<lb/>uata; </s> <s xml:id="echoid-s4539" xml:space="preserve">ſe quelle furono dimezzate, queſta ſi raddoppia.</s> <s xml:id="echoid-s4540" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4541" xml:space="preserve">Mà può accadere, che ſe bene la metà della corda commo-<lb/>damente capiſce nell’interuallo *** ***, l’altezza del ſegmento <lb/>ſia minore di quelle, che corriſpondono à gl’interualli de’pun-<lb/>ti notati eſteriormente, il che occorrerà ogni volta, che la <lb/>proportione dell’altezza alla metà della corda ſarà minore <lb/>d’vna decima parte di detta metà; </s> <s xml:id="echoid-s4542" xml:space="preserve">poiche ſolamente vi ſono <lb/>ſegnate 18 venteſime di tutta la A ***. </s> <s xml:id="echoid-s4543" xml:space="preserve">Et in tal caſo non va-<lb/>lerebbe raddoppiar, ò triplicare la mezzacorda, e l’altezza; <lb/></s> <s xml:id="echoid-s4544" xml:space="preserve">perche rimanendo ſempre la medeſima proportione, non ſi <pb o="241" file="0261" n="265" rhead="Quadratrice de’Segmenti del Circolo"/> potria trouar ſegnato alcun punto, che deſſe interuallo ſoffi-<lb/>ciente all’iutento. </s> <s xml:id="echoid-s4545" xml:space="preserve">Perciò ſi vede, che in quante più parti <lb/>vguali ſi potrà commodamente diuidere la corda propoſta <lb/>A *** nello ftromento, tanto maggiore ſarà ilſuo vſo, eſſendo <lb/>che più dirado occorrerà hauere vn ſegmento, la cui altezza <lb/>ſia molto minore; </s> <s xml:id="echoid-s4546" xml:space="preserve">eſe il gruppo dello ſtromento impediſce <lb/>ilſ<unsure/>uogo per li punti 19. </s> <s xml:id="echoid-s4547" xml:space="preserve">19, forſi non impedirà per li punti <lb/>37 37; </s> <s xml:id="echoid-s4548" xml:space="preserve">ſe tutta la linea fofſe diuiſa in 40 parti vguali. </s> <s xml:id="echoid-s4549" xml:space="preserve">Oltre <lb/>di che queſte minori diuiſioni daranno più eſattamente le al-<lb/>tre altezze de’ſegmenti.</s> <s xml:id="echoid-s4550" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4551" xml:space="preserve">In caſo però che ſi faceſſero queſte più minute diuiſioni, <lb/>deue auuertirſi, che caderanno alle volte i punti delli numeri <lb/>eſteriori, e delli interiori, così vicini, che ſi dubitarà, à quali <lb/>numeri eſſi appartengono. </s> <s xml:id="echoid-s4552" xml:space="preserve">Perciò io conſigliarei, che alla <lb/>linea Quadratrice ſi tiraſſe parallela dalla parte difuori vn’al-<lb/>tra linea vicina, alla quale dalli punti delle parti vguali ſi tiraſ-<lb/>ſero lineette, poiche tali punti, da quali vſciſſero tali lineette <lb/>traſuerſali, ſi riconoſcerebbero per appartenenti alli numeri <lb/>eſteriori; </s> <s xml:id="echoid-s4553" xml:space="preserve">e così alli numeri interiori apparterebbono gli al-<lb/>tri punti, dalli quali non vſciſſero ſimili lineette, e ſi togliereb-<lb/>be il pericolo di prender vn punto per vn’altro vicino.</s> <s xml:id="echoid-s4554" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4555" xml:space="preserve">Quando dunque l’altezza del ſegmento è minore della de-<lb/>cima parte della metà della corda, trouiſi la loro proportio<unsure/>-<lb/>ne, come ſi diſſe alla Queſt. </s> <s xml:id="echoid-s4556" xml:space="preserve">5. </s> <s xml:id="echoid-s4557" xml:space="preserve">del capo 2, e ſtatuita la mezza <lb/>corda come 100000, ſi faccia l’altezza data del ſegmento à <lb/>queſto numero nella proportione trouata: </s> <s xml:id="echoid-s4558" xml:space="preserve">così trouata la <lb/>proportione della mezza corda all’altezza eſſere di 12 à 1, <lb/>diuidaſi 100000 per 12, eſarà l’altezza 8333. </s> <s xml:id="echoid-s4559" xml:space="preserve">dipoi con <lb/>queſta miſura ſi operi nella maniera adoperata in queſto ca-<lb/>po per trouare le quantità de’lati del quadrato da notarſi sù <pb o="242" file="0262" n="266" rhead="CAPO X."/> lo ſtromento (il che quìnon fà biſogno di replicare) e cosi <lb/>ſi haurà cognitione di quel piccoſ<unsure/>o ſegmento.</s> <s xml:id="echoid-s4560" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div153" type="section" level="1" n="87"> <head xml:id="echoid-head162" style="it" xml:space="preserve">QVESTIONE QVART A.</head> <head xml:id="echoid-head163" xml:space="preserve">Data vna portione di Circolo trouare la ſua grandezza <lb/>in miſura determinata.</head> <p> <s xml:id="echoid-s4561" xml:space="preserve">SOno alle voſte date alcune portioni circolari, che non <lb/>ſono deſcritte in carta da poterſene traportare le linee <lb/>con il Compaſſo; </s> <s xml:id="echoid-s4562" xml:space="preserve">perciò date le loro miſure, ſi trouano linee <lb/>nella ſteſſa proportione, e con quelle ſi opera sù lo Stromen-<lb/>to nel modo detto. </s> <s xml:id="echoid-s4563" xml:space="preserve">Sia, per cagione d’eſempio, data nella <lb/>parte ſuperiore d’vna porta, che tondeggia, vna portione <lb/>circolare, e ſi vuol ſapere di quante braccie, ouer oncie, qua-<lb/>drate ſia quello ſpatio.</s> <s xml:id="echoid-s4564" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4565" xml:space="preserve">Prendaſi la miſura della larghezza, che ſia braccia 5, e del-<lb/>l’altezza, che ſia braccia vno, & </s> <s xml:id="echoid-s4566" xml:space="preserve">oncie noue: </s> <s xml:id="echoid-s4567" xml:space="preserve">la metà della <lb/>corda è braccia 2 {1/2}, cioè oncie 30, e l’altezza è oncie 21. <lb/></s> <s xml:id="echoid-s4568" xml:space="preserve">Nelle linee Aritmetiche con due Compaſſi prendanſi due <lb/>interualli, che habbiano la ſteſſa proportione di 30 à 21; </s> <s xml:id="echoid-s4569" xml:space="preserve">e <lb/>ſiano 100. </s> <s xml:id="echoid-s4570" xml:space="preserve">100, e 70. </s> <s xml:id="echoid-s4571" xml:space="preserve">70. </s> <s xml:id="echoid-s4572" xml:space="preserve">le quali lunghezze quanto ſi pren-<lb/>deranno maggiori, tanto più eſatta riuſcirà l’operatione. </s> <s xml:id="echoid-s4573" xml:space="preserve">La <lb/>lunghezza, che rappreſenta la metà della corda del ſegmen-<lb/>to circolare, ſi applichi nelle Quadratrici all’interuallo *** ***, <lb/>e l’altra che rappreſenta l’altezza, ſi applichi alli punti de’nu-<lb/>meri eſteriori doue capiſce, e ſarà all’interuallo 6. </s> <s xml:id="echoid-s4574" xml:space="preserve">6. </s> <s xml:id="echoid-s4575" xml:space="preserve">Perciò <lb/>ritenuta l’apertura ſteſſa dello Stromento, con queſto mede-<lb/>fimo Compaſſo allargato ſi prenda nelli punti de’numeri in-<lb/>teriori l’interuallo 6. </s> <s xml:id="echoid-s4576" xml:space="preserve">6. </s> <s xml:id="echoid-s4577" xml:space="preserve">Poſcia ritornando alle linee Aritme- <pb o="243" file="0263" n="267" rhead="Quadratrice de’ſegmenti del Circolo"/> ti<unsure/>che, di nuouo ſiapplichi il primo Compaſſo all’interuallo <lb/>100. </s> <s xml:id="echoid-s4578" xml:space="preserve">100, eveggaſi doue darà l’apertura di queſto ſecondo <lb/>Compaſſo, che ſarà alquanto maggiore; </s> <s xml:id="echoid-s4579" xml:space="preserve">e ſi trouarà eſſere <lb/>101, ſe il primo Compaſſo ſi applicarà alli punti 50. </s> <s xml:id="echoid-s4580" xml:space="preserve">50, per-<lb/>che il ſecondo caderà nel 50 {1/2}. </s> <s xml:id="echoid-s4581" xml:space="preserve">50 {1/2}. </s> <s xml:id="echoid-s4582" xml:space="preserve">Ora dicaſi, ſe la mezza <lb/>corda 100 dà la linea 101, il cui quadrato è vguale al ſegmen-<lb/>to, vna linea di oncie 30 darà vna linea di oncie 30 {3/10}; </s> <s xml:id="echoid-s4583" xml:space="preserve">il cui <lb/>quadrato {91809/100} ſarà l’area di detta portione circolare data, <lb/>cioè oncie quadrate 918: </s> <s xml:id="echoid-s4584" xml:space="preserve">e perche ogni braccio quadro con-<lb/>tiene oncie 144, la ſua area ſarà braccia 6, oncie 54, cioè <lb/>braccia 6 {3/8} di miſura piana.</s> <s xml:id="echoid-s4585" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4586" xml:space="preserve">Mà ſe miſurando il ſegmento propoſto, ſi trouaſſe l’altez-<lb/>za eſſere maggiore della metà della larghezza, ſaria ſegno, <lb/>che quel ſegmento foſſe maggiore del ſemicircolo: </s> <s xml:id="echoid-s4587" xml:space="preserve">& </s> <s xml:id="echoid-s4588" xml:space="preserve">in tal <lb/>caſo conuerrebbe trouare l’altezza dell’altro ſegmento mi-<lb/>nore, e con quella ſi operarebbe nel modo ſodetto, trouan-<lb/>do la quantità di quel ſegmento minore; </s> <s xml:id="echoid-s4589" xml:space="preserve">e queſta leuata dal-<lb/>la quantità di tutto il circolo, il reſiduo darebbe la grandez-<lb/>za del propoſto ſegmento. </s> <s xml:id="echoid-s4590" xml:space="preserve">Per trouar dunque l’altezza del <lb/>ſegmento minore, facciaſi come l’aſ<unsure/>tezza data D C alla C B <lb/>metà della data larghezza, così C B à C E: </s> <s xml:id="echoid-s4591" xml:space="preserve">e queſta terza pro-<lb/>portionale, trouata per la Queſt. </s> <s xml:id="echoid-s4592" xml:space="preserve">7. <lb/></s> <s xml:id="echoid-s4593" xml:space="preserve"> <anchor type="figure" xlink:label="fig-0263-01a" xlink:href="fig-0263-01"/> del capo 3. </s> <s xml:id="echoid-s4594" xml:space="preserve">è il reſiduo del diametro <lb/>del Circolo, altezza del ſegmento <lb/>minore. </s> <s xml:id="echoid-s4595" xml:space="preserve">Siche applicata C B all’in-<lb/>teruallo *** ***, e C E all’interuallo <lb/>de’numeri eſteriori doue capiſce, ſi <lb/>haurà dall’interuallo de’numeri inte-<lb/>riori corriſpondentila linea del qua-<lb/>drato vguale al ſegmento minore.</s> <s xml:id="echoid-s4596" xml:space="preserve"> <pb o="244" file="0264" n="268" rhead="CAPO X."/> Or eſſendo già noto il diametro del circolo, ſi troui la linea <lb/>del quadrato à lui vguale, per quello che ſi è detto nel capo 8. <lb/></s> <s xml:id="echoid-s4597" xml:space="preserve">e dal quadrato vguale al circolo ſi leui il quadrato vguale al <lb/>ſegmento minore, come per la Queſt. </s> <s xml:id="echoid-s4598" xml:space="preserve">6. </s> <s xml:id="echoid-s4599" xml:space="preserve">del capo 3. </s> <s xml:id="echoid-s4600" xml:space="preserve">& </s> <s xml:id="echoid-s4601" xml:space="preserve">ilre-<lb/>ſiduo ſarà la cercata quantità del ſegmento maggiore pro-<lb/>poſto.</s> <s xml:id="echoid-s4602" xml:space="preserve"/> </p> <div xml:id="echoid-div153" type="float" level="2" n="1"> <figure xlink:label="fig-0263-01" xlink:href="fig-0263-01a"> <image file="0263-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0263-01"/> </figure> </div> </div> <div xml:id="echoid-div155" type="section" level="1" n="88"> <head xml:id="echoid-head164" style="it" xml:space="preserve">QVESTIONE QVINT A.</head> <head xml:id="echoid-head165" xml:space="preserve">Dato vn Segmento di Circolo, trouare la proportione, cheil <lb/>Segmento hàad vn dato Triangolo, che in eſſo capiſce.</head> <p> <s xml:id="echoid-s4603" xml:space="preserve">SIa dato il Segmento di circolo C O D B C, in cui il maſſi-<lb/>mo triangolo è quello, la cui altezza è la medeſima <lb/>con l’altezza <lb/>del Segmẽto, <lb/> <anchor type="figure" xlink:label="fig-0264-01a" xlink:href="fig-0264-01"/> cioè la perpẽ-<lb/>dicolare, che <lb/>cade nel mez-<lb/>zo della corda <lb/>C D, cioè BO. <lb/></s> <s xml:id="echoid-s4604" xml:space="preserve">Ora ſia dato il Triangolo C A D, di cui ſi voglia ſapere, che <lb/>parte ſia del ſegmento dato. </s> <s xml:id="echoid-s4605" xml:space="preserve">Compiſcaſi il maſſimo Trian-<lb/>golo COD, il quale eſſendo sù la medeſima baſe CD, hà al <lb/>Triangolo CAD la proportione delli perpendicoli, cioè di <lb/>OB ad AE.</s> <s xml:id="echoid-s4606" xml:space="preserve"/> </p> <div xml:id="echoid-div155" type="float" level="2" n="1"> <figure xlink:label="fig-0264-01" xlink:href="fig-0264-01a"> <image file="0264-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0264-01"/> </figure> </div> <p> <s xml:id="echoid-s4607" xml:space="preserve">Primieramente eſſendo larea del maſſimo triangolo vgua-<lb/>le al rettangolo fatto da OB, e BC, trouiſi tra queſte due linee <lb/>la media proportionale, eſia H, per la Queſt. </s> <s xml:id="echoid-s4608" xml:space="preserve">8. </s> <s xml:id="echoid-s4609" xml:space="preserve">del capo 3. <lb/></s> <s xml:id="echoid-s4610" xml:space="preserve">& </s> <s xml:id="echoid-s4611" xml:space="preserve">il quadrato diqueſta linea H ſarà vguale al detto Triango-<lb/>lo maſſimo COD, perla 17. </s> <s xml:id="echoid-s4612" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4613" xml:space="preserve">6.</s> <s xml:id="echoid-s4614" xml:space="preserve"/> </p> <pb o="245" file="0265" n="269" rhead="Quadratrice de’Segmenti del Circolo"/> <p> <s xml:id="echoid-s4615" xml:space="preserve">Dipoi nelle linee Quadratrici di queſto capo ſi applichi <lb/>BC metà della corda alli punti *** ***, e l’altezza BO ſi troui <lb/>ne gl’interualli de’numeri eſteriori, poiche all’interuallo de’ <lb/>numeri interiori corriſpondenti ſi haurà la linea I, che dà il <lb/>quadrato vguale al ſegmento dato. </s> <s xml:id="echoid-s4616" xml:space="preserve">Si che il dato ſegmento <lb/>di circolo al Triangolo maſſimo che capiſce, hà la proportio-<lb/>ne del quadrato di I al quadrato di H, cioè la duplicara pro-<lb/>portione di queſta ſeconda linea I trouata, à quella H, che in <lb/>primo luogo ſi trouò. </s> <s xml:id="echoid-s4617" xml:space="preserve">Dunque cerchiſi, per la Queſt. </s> <s xml:id="echoid-s4618" xml:space="preserve">7. </s> <s xml:id="echoid-s4619" xml:space="preserve">del <lb/>capo 3, à queſte due la terza proportionale K; </s> <s xml:id="echoid-s4620" xml:space="preserve">& </s> <s xml:id="echoid-s4621" xml:space="preserve">il ſegmen-<lb/>to al Triangolo maſſimo hà la proportione della linea I alla <lb/>linea K.</s> <s xml:id="echoid-s4622" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4623" xml:space="preserve">Finalmente per la Queſt. </s> <s xml:id="echoid-s4624" xml:space="preserve">3. </s> <s xml:id="echoid-s4625" xml:space="preserve">del capo 2. </s> <s xml:id="echoid-s4626" xml:space="preserve">ſi faccia come B O <lb/>ad EA, così K ad L: </s> <s xml:id="echoid-s4627" xml:space="preserve">onde neſiegue, perl’ 11. </s> <s xml:id="echoid-s4628" xml:space="preserve">del lib. </s> <s xml:id="echoid-s4629" xml:space="preserve">5, che <lb/>il triangolo C O D al triangolo C A D ſia come K ad L. </s> <s xml:id="echoid-s4630" xml:space="preserve">Dun-<lb/>que il ſegmento del circolo al Triangolo C O D è come la <lb/>linea I alla linea K; </s> <s xml:id="echoid-s4631" xml:space="preserve">& </s> <s xml:id="echoid-s4632" xml:space="preserve">il Triangolo C O D al Triangolo C A D <lb/>è come la linea K alla linea L: </s> <s xml:id="echoid-s4633" xml:space="preserve">dunque perla 22. </s> <s xml:id="echoid-s4634" xml:space="preserve">del libro 5. <lb/></s> <s xml:id="echoid-s4635" xml:space="preserve">ſaràil dato ſegmento del circolo al triangolo dato C A D in-<lb/>chiuſo, come la linea I alla linea L. </s> <s xml:id="echoid-s4636" xml:space="preserve">Perciò volendoſi ſaper in <lb/>numerila proportione, ſi portino le dette due linee I, & </s> <s xml:id="echoid-s4637" xml:space="preserve">L <lb/>sù le linee Aritmetiche; </s> <s xml:id="echoid-s4638" xml:space="preserve">e gl’interualli, ne’quali capiranne, <lb/>daranno i numeri, che eſprimono la cercata proportione del <lb/>ſegmento al triangolo dato in eſſo.</s> <s xml:id="echoid-s4639" xml:space="preserve"/> </p> <pb o="246" file="0266" n="270" rhead="CAPO VLTIMO."/> </div> <div xml:id="echoid-div157" type="section" level="1" n="89"> <head xml:id="echoid-head166" xml:space="preserve">Come ſi poſſano con gran facilità fabricare molti Compaſsi <lb/>di proportione altri grandi, altri piccoli.</head> <p> <s xml:id="echoid-s4640" xml:space="preserve">DAlle coſe dette in tutto queſto Trattato della diligen-<lb/>za, con cui deuono farſi le diuiſioni delle linee de-<lb/>ſcritte (alcune delle quali non ſi può negare, che ricercano <lb/>molto particolar’ attentione, acciò ſiano diuiſe accuratamen-<lb/>te) potrà per auuentura ſpauentarſi qualche. </s> <s xml:id="echoid-s4641" xml:space="preserve">Artefice, te-<lb/>mendo, che rieſca la fattura così lunga, e trauaglioſa, che <lb/>douendoſi condegnamente ricompenſare, venga à riuſcire <lb/>tanto cara, che trouandoſi pochicompratori, venga à trarne <lb/>poco guadagno. </s> <s xml:id="echoid-s4642" xml:space="preserve">Per facilità dunque de gl’Artefici, a’ quali <lb/>non baſta hauerne fatto vno, ò anche d’altri, i quali voleſ-<lb/>ſero con poca fatica diuidere le linee tirate nel ſuo Compaſſo <lb/>di proportione, ſoggiongo per fine di queſto Trattato que-<lb/>ſto Capo, iſ quale in ſoſtanza non è altro, che la prattica di <lb/>quanto diſopra s’è detto.</s> <s xml:id="echoid-s4643" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4644" xml:space="preserve">Proueggaſi dunque l’Artefice d’vn Compaſſo di propor-<lb/>tione con le regole aſſai lunghe, ſopra delle quali ſiano tira-<lb/>te dal centro varie linee rette nell’vna, e nell’ altra faccia, e <lb/>queſte linee diuida nella maniera, che habbiamo moſtrato, <lb/>ne ſtimi alcuna diligenza ſuperflua, ne perduto il rempo, che <lb/>v’impiegarà, à fine, che le diuiſioni ſiano accuratiſſime; </s> <s xml:id="echoid-s4645" xml:space="preserve">per-<lb/>che fatta vna volta queſta fatica, non haurà più à replicarla, <lb/>e gli ſeruirà per tutta la ſua vita, e de’ ſuoi figliuoli, perche <lb/>queſto Compaſſo di proportione dourà ritener appreſſo di <lb/>ſe, e non venderlo, per non neceſſitarſi ad vna nuoua fatica.</s> <s xml:id="echoid-s4646" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4647" xml:space="preserve">Occorrendo poi far vn’ aſtro Stromento vguale, ò più <pb o="247" file="0267" n="271" rhead="Fabricar Compaſsi"/> grande, ò più piccolo del ſuo gia fatto, qual però ſi ſuppone <lb/>de’ più lunghi, che ſogliano communemente farſi, ſi tirino <lb/>dal centro le linee, che poiſi vogliono diuidere; </s> <s xml:id="echoid-s4648" xml:space="preserve">e fatto que-<lb/>ſto, la lunghezza di ciaſcuna linea pongaſi nell eſtremo in-<lb/>teruallo della linea ſimile dello Stromento già perfettionato: <lb/></s> <s xml:id="echoid-s4649" xml:space="preserve">poiche ritenuta quell’a pertura dello Stromento, baſterà tra-<lb/>portare ciaſcun’ interuallo ſopra la linea, che ſi vuol diuidere; </s> <s xml:id="echoid-s4650" xml:space="preserve"><lb/>& </s> <s xml:id="echoid-s4651" xml:space="preserve">in tal maniera queſta ſarà diuiſa nella ſteſſa proportione, <lb/>che la linea dello Stromento maggiore. </s> <s xml:id="echoid-s4652" xml:space="preserve">Così volendo ſe-<lb/>gnare la linea metallica, per eſſempio, prendo la diſtanza dal <lb/>centro dello Stromento, ſin’all’ eſtremità della linea da diui-<lb/>derſi, & </s> <s xml:id="echoid-s4653" xml:space="preserve">alargo lo Stromento già fatto, in modo, che tutta <lb/>quella linea capiſca nell’vltimo interuallo della linea metal-<lb/>lica PP, doue è ſegnata la pietra. </s> <s xml:id="echoid-s4654" xml:space="preserve">Dipoi prendo l’interuallo <lb/>MM per il marmo, e queſta longhezza traporto dal centro <lb/>ſopra la linea che ſi diuide, nell’vno, e nell’altro braccio, e ſi <lb/>ſegnarà il punto per il marmo. </s> <s xml:id="echoid-s4655" xml:space="preserve">E così ſuſſeguentemente ne <lb/>gl’altri punti CC, SS, &</s> <s xml:id="echoid-s4656" xml:space="preserve">c. </s> <s xml:id="echoid-s4657" xml:space="preserve">onde ſarà diuiſa la linea Metallica <lb/>nel nuouo Stromento, ſecondo la proportione, con cui fù <lb/>diuiſa quella del primo Stromento: </s> <s xml:id="echoid-s4658" xml:space="preserve">l’iſteſſo s’intende di qual-<lb/>ſiuoglia altra linea da diuiderſi. </s> <s xml:id="echoid-s4659" xml:space="preserve">Nel che ſi vede quanto gran <lb/>compendio di fatica ſia queſto.</s> <s xml:id="echoid-s4660" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4661" xml:space="preserve">Di quì ſi vede, che ſe vn’amico habbia vn Compaſſo di <lb/>proportione, diligentemente fatto da buon’artefice, ciaſcuno <lb/>potrà con gran facilità farſene vno da ſe, cauando da quello <lb/>le diuiſioni nel modo, che s’è detto douer fare l’Artefice. </s> <s xml:id="echoid-s4662" xml:space="preserve">On-<lb/>de con molto poca ſpeſa può eſlere prouiſto d’vn buono <lb/>Stromento.</s> <s xml:id="echoid-s4663" xml:space="preserve"/> </p> <pb o="248" file="0268" n="272" rhead="Conchiuſione."/> <p> <s xml:id="echoid-s4664" xml:space="preserve">EQueſte coſe baſtino per la ſpiegatione della Fabrica, & </s> <s xml:id="echoid-s4665" xml:space="preserve"><lb/>Vſo del Compaſlo di proportione, dalle quali ciaſcu-<lb/>no potrà andar inuentando altre operationi. </s> <s xml:id="echoid-s4666" xml:space="preserve">Sì come anche <lb/>puonno deſcriuerſi altre linee, nelle quali ſiano altre propor-<lb/>tioni, ſecondo il piacere di ciaſcuno: </s> <s xml:id="echoid-s4667" xml:space="preserve">come ſarebbe vna linea <lb/>delle fortificationi, nella quale ſi ſegnaſſe la proportione delle <lb/>parti di eſſa, cioè la capitale, & </s> <s xml:id="echoid-s4668" xml:space="preserve">il fianco del baloardo in cia-<lb/>ſcuna fortezza di più angoli, ſupponendoſi la mezzagola, & </s> <s xml:id="echoid-s4669" xml:space="preserve"><lb/>il fianco vguali al ſeſto di tutto il lato del poligono: </s> <s xml:id="echoid-s4670" xml:space="preserve">& </s> <s xml:id="echoid-s4671" xml:space="preserve">io per <lb/> <anchor type="figure" xlink:label="fig-0268-01a" xlink:href="fig-0268-01"/> sfuggire la confuſione, tal linea ſegnarei, co-<lb/>me nella preſente figura, pigliando per eſ-<lb/>ſempio A4 per la capitale in vna fortezza <lb/>di 4 baloardi, e perciò notarei al punto 4 <lb/>anche la lettera C, per denotare, che è ſa ca-<lb/>pitale, e poi il fianco del baloardo di tal for-<lb/>tezza notarei AF. </s> <s xml:id="echoid-s4672" xml:space="preserve">Dal che ne verrebbe, che <lb/>data vna fortezza di 4 baloardi da deſcri-<lb/>uerſi, tagliato per mezzo l’angolo con vna <lb/>capitale indefinita, ſi prenderebbe il ſeſto <lb/>del lato del poligono fortificabile, e queſto <lb/>applicato all’interuallo FF, che è tra il 4, & </s> <s xml:id="echoid-s4673" xml:space="preserve"><lb/>ilcentro A, l’interuallo CC, che è di rimpetto al 4, daria la <lb/>quantità della capitale determinata. </s> <s xml:id="echoid-s4674" xml:space="preserve">Per la fortezza poi di <lb/>cinque baloardi hauuta ſi la proportione della capitale, e del <lb/>fianco per mezzo del calcolo, prenderei dal centro A tal di-<lb/>ſtanza per A 5, la quale foſſe la capitale del baloardo dital <lb/>fortezza, che prendendoſi il fianco proportionato AF, cadeſ-<lb/>ſe tra il punto ſegnato 5, & </s> <s xml:id="echoid-s4675" xml:space="preserve">il ſegnato 4; </s> <s xml:id="echoid-s4676" xml:space="preserve">perche in tal modo <pb o="249" file="0269" n="273" rhead="Conchiuſione."/> queſte lettere CF, ſignificarebbono la capitale, & </s> <s xml:id="echoid-s4677" xml:space="preserve">il fianco <lb/>del baloardo di fortezza di cinque baſtioni. </s> <s xml:id="echoid-s4678" xml:space="preserve">L’iſteſſo dico in <lb/>or dine ad altri punti per fortezza di più baloardi. </s> <s xml:id="echoid-s4679" xml:space="preserve">A me poi <lb/>pìace più ſegnar il fianco, e la capitale, perche con queſteſi <lb/>può anche operare per la fortificatione irregolare, quanto lo <lb/>permetterà la ſteſſa irregolat<unsure/>ità.</s> <s xml:id="echoid-s4680" xml:space="preserve"/> </p> <div xml:id="echoid-div157" type="float" level="2" n="1"> <figure xlink:label="fig-0268-01" xlink:href="fig-0268-01a"> <image file="0268-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0268-01"/> </figure> </div> <p> <s xml:id="echoid-s4681" xml:space="preserve">Ciò che per modo d’eſſe m pio s’è detto della linea delle <lb/>fortificationi, con notare queſte due ſole diuiſioni, s’intenda <lb/>anche, ò notando altre proportioni d’altre linee appartenen-<lb/>ti alla fortificatione, ò pur anche altre linee d’altre coſe, e <lb/>proportioni, ſecondo il piacere di ciaſcuno. </s> <s xml:id="echoid-s4682" xml:space="preserve">Così perche <lb/>ſpeſſo può venir’ occaſione di tagliar’ vna linea nella media, <lb/>& </s> <s xml:id="echoid-s4683" xml:space="preserve">eſtrema ragione, potrebbeſi nello Stromento tirar’ vna li-<lb/>nea nell’vno, e nell’ altro braccio, la quale à queſt’ effetto ſer-<lb/>uiſſe, tagliandola con queſta proportione, poiche qualſiuo-<lb/>glia linea data applicata all’eſtremo interuallo, ſaria tagliata <lb/>ſimilmente, prendendo l’interuallo de’ punti, ne’ quali le li-<lb/>nee laterali furono così diuiſe. </s> <s xml:id="echoid-s4684" xml:space="preserve">Se bene ſe non hai tal linea <lb/>preciſamente diuiſa nello Stromento, baſterà, che applica-<lb/>ta tutta la linea all’interuallo 100. </s> <s xml:id="echoid-s4685" xml:space="preserve">100, prendi l’interuallo <lb/>38. </s> <s xml:id="echoid-s4686" xml:space="preserve">38, e con queſto diuidaſi la linea data; </s> <s xml:id="echoid-s4687" xml:space="preserve">perche il ſegmen-<lb/>to maggiore 62. </s> <s xml:id="echoid-s4688" xml:space="preserve">hà per ſuo quadrato 3844. </s> <s xml:id="echoid-s4689" xml:space="preserve">poco maggio-<lb/>re del rettangolo fatto da tutta 100, e dal minor ſegmento <lb/>38, cioè poco maggiore di 3800, come richiede cotal ſet-<lb/>tione. </s> <s xml:id="echoid-s4690" xml:space="preserve">Se tutta la linea foſſe 1000, le parti ſariano 618, e <lb/>382, & </s> <s xml:id="echoid-s4691" xml:space="preserve">il quadrato del maggior ſegmento è 381924 poco <lb/>minore del rettangolo 382000.</s> <s xml:id="echoid-s4692" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s4693" xml:space="preserve">Mà ciò ſi fà con preciſione maggiore ſe la data linea ſi ap-<lb/>plichi nelle linee che moftranole corde de gliarchi, all’inter-<lb/>uallo 60. </s> <s xml:id="echoid-s4694" xml:space="preserve">60; </s> <s xml:id="echoid-s4695" xml:space="preserve">poiprendaſil’interuallo 36. </s> <s xml:id="echoid-s4696" xml:space="preserve">36, che queſto da- <pb o="250" file="0270" n="274" rhead="Concbiuſione."/> rà il ſeg mento maggiore; </s> <s xml:id="echoid-s4697" xml:space="preserve">eſſendo che il primo interuallo è la-<lb/>to dell’ Eſſagono, il ſecondo è lato del Decagono deſcritti <lb/>nell’iſteſſo cerchio; </s> <s xml:id="echoid-s4698" xml:space="preserve">e dalla Prop.</s> <s xml:id="echoid-s4699" xml:space="preserve">9. </s> <s xml:id="echoid-s4700" xml:space="preserve">dellib.</s> <s xml:id="echoid-s4701" xml:space="preserve">13.</s> <s xml:id="echoid-s4702" xml:space="preserve">d’Euclide ſi hà <lb/>il Corollario, che tagliato illato dell’Eſſagono nella media, <lb/>& </s> <s xml:id="echoid-s4703" xml:space="preserve">eſtrema ragione, il ſeg mento maggiore è il lato del Deca-<lb/>gono. </s> <s xml:id="echoid-s4704" xml:space="preserve">Che ſe ſi voleſſe, che la data linea foſſe l’vno de<unsure/>’ſeg-<lb/>menti, e biſognaſſe farui vn’aggionta, ſi che tutta foſſe taglia-<lb/>ta nella media, & </s> <s xml:id="echoid-s4705" xml:space="preserve">eſtrema ragione, ſarà pronto il modo per <lb/>la ſteſſa Prop. </s> <s xml:id="echoid-s4706" xml:space="preserve">9. </s> <s xml:id="echoid-s4707" xml:space="preserve">dellib. </s> <s xml:id="echoid-s4708" xml:space="preserve">13. </s> <s xml:id="echoid-s4709" xml:space="preserve">S’ella è il ſegmento maggiore, ſi <lb/>applichi al 60.</s> <s xml:id="echoid-s4710" xml:space="preserve">60, e preſo l’interuallo 36.</s> <s xml:id="echoid-s4711" xml:space="preserve">36. </s> <s xml:id="echoid-s4712" xml:space="preserve">gli ſi aggionga: <lb/></s> <s xml:id="echoid-s4713" xml:space="preserve">per il contrario, ſe la linea data è il ſegmento minore, ſi ap-<lb/>plichi al 36.</s> <s xml:id="echoid-s4714" xml:space="preserve">36, egli ſi aggiongerà l’interuallo 60.</s> <s xml:id="echoid-s4715" xml:space="preserve">60; </s> <s xml:id="echoid-s4716" xml:space="preserve">che <lb/>così tutta la compoſta ſarà, qualeſiricerca.</s> <s xml:id="echoid-s4717" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div159" type="section" level="1" n="90"> <head xml:id="echoid-head167" xml:space="preserve">IL FINE.</head> <pb file="0271" n="275"/> <pb file="0272" n="276"/> <pb file="0273" n="277"/> <pb file="0274" n="278"/> </div></text> </echo>