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| author | Klaus Thoden <kthoden@mpiwg-berlin.mpg.de> |
|---|---|
| date | Thu, 02 May 2013 11:29:00 +0200 |
| parents | 22d6a63640c6 |
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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:N1TU6UZF.xml</dcterms:identifier> <dcterms:creator identifier="GND:118939297">Gravesande, Willem Jakob</dcterms:creator> <dcterms:title xml:lang="en">An essay on perspective</dcterms:title> <dcterms:date xsi:type="dcterms:W3CDTF">1724</dcterms:date> <dcterms:language xsi:type="dcterms:ISO639-3">eng</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> </metadata> <text xml:lang="en" type="free"> <div xml:id="echoid-div1" type="section" level="1" n="1"><pb file="0001" n="1"/> <pb file="0002" n="2"/> <handwritten/> <figure> <image file="0002-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0002-01"/> </figure> </div> <div xml:id="echoid-div2" type="section" level="1" n="2"> <head xml:id="echoid-head1" xml:space="preserve">South L<unsure/>ibrarp.</head> <p> <s xml:id="echoid-s1" xml:space="preserve">Press mark, 181 & </s> <s xml:id="echoid-s2" xml:space="preserve">28</s> </p> <p> <s xml:id="echoid-s3" xml:space="preserve">Ent@in Gatalogue, ---</s> </p> <p> <s xml:id="echoid-s4" xml:space="preserve">(1860)</s> </p> <pb file="0003" n="3"/> <pb file="0004" n="4"/> <pb file="0005" n="5"/> </div> <div xml:id="echoid-div3" type="section" level="1" n="3"> <head xml:id="echoid-head2" xml:space="preserve">AN <lb/>ESSAY <lb/>ON <lb/>PERSPECTIVE.</head> <p> <s xml:id="echoid-s5" xml:space="preserve">Written in French by <lb/><emph style="sc">William</emph>-<emph style="sc">James ‘s</emph> <emph style="sc">Ggravesande</emph>, <lb/>Doctor of Laws and Philoſophy; </s> <s xml:id="echoid-s6" xml:space="preserve">Profeſſor <lb/>of Mathematicks and Aſtronomy at Leyden, <lb/>and Fellow of the Royal Society at London.</s> <s xml:id="echoid-s7" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s8" xml:space="preserve">And now Tranſlated into Engliſh.</s> <s xml:id="echoid-s9" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s10" xml:space="preserve">LONDON:</s> <s xml:id="echoid-s11" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s12" xml:space="preserve">Printed for J. </s> <s xml:id="echoid-s13" xml:space="preserve"><emph style="sc">Senex</emph>, in Fleetſtreet; </s> <s xml:id="echoid-s14" xml:space="preserve">W. </s> <s xml:id="echoid-s15" xml:space="preserve"><emph style="sc">Taylor</emph>, <lb/>in Pater-Noſter-Row; </s> <s xml:id="echoid-s16" xml:space="preserve">W. </s> <s xml:id="echoid-s17" xml:space="preserve">and J. </s> <s xml:id="echoid-s18" xml:space="preserve"><emph style="sc">Innys</emph>, in Ludgate-<lb/>ſtreet; </s> <s xml:id="echoid-s19" xml:space="preserve">J. </s> <s xml:id="echoid-s20" xml:space="preserve"><emph style="sc">Osborne</emph>, in Lombard-ſtreet; </s> <s xml:id="echoid-s21" xml:space="preserve">and E. </s> <s xml:id="echoid-s22" xml:space="preserve"><emph style="sc">Ssymon</emph>, <lb/>in Cornhill. </s> <s xml:id="echoid-s23" xml:space="preserve">M <emph style="sc">DCC XXIV</emph>.</s> <s xml:id="echoid-s24" xml:space="preserve"/> </p> <pb file="0006" n="6"/> <handwritten/> </div> <div xml:id="echoid-div4" type="section" level="1" n="4"> <head xml:id="echoid-head3" xml:space="preserve">MAX-<gap/>-INSTITUT <lb/>FOR WISSE<gap/>ESCHICHTE <lb/>Bibliothek</head> <handwritten/> <pb file="0007" n="7"/> </div> <div xml:id="echoid-div5" type="section" level="1" n="5"> <head xml:id="echoid-head4" style="it" xml:space="preserve">TO <lb/>Mr. William Kent.</head> <p> <s xml:id="echoid-s25" xml:space="preserve"><emph style="sp">SIR</emph>,</s> </p> <p> <s xml:id="echoid-s26" xml:space="preserve">IN Dedicating this <lb/>Tranſlation to you, <lb/>I have deſignedly <lb/>deviated from the general <lb/>Cuſtom obſerv’d by almoſt <lb/>all Dedicators, who make <lb/>choice of ſuch Patrons that <lb/>are Great and Rich, not at all <lb/>conſidering their Merit, or <lb/>whether they underſtand any <lb/>thing of what is offer’d to <lb/>them; </s> <s xml:id="echoid-s27" xml:space="preserve">ſince I have inſcrib’d <lb/>this Treatiſe of Perſpective to <lb/>one, whoſe daily Practice is <lb/>the very Art it ſelf, and whoſe <pb file="0008" n="8"/> Merit is undoubtedly excel-<lb/>lent, as evidently appears from <lb/>your own Works.</s> <s xml:id="echoid-s28" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s29" xml:space="preserve">I ſhall likewiſe be particu-<lb/>lar with regard to the Manner <lb/>of the Offering; </s> <s xml:id="echoid-s30" xml:space="preserve">being per-<lb/>ſuaded that Flattery, or even <lb/>due Praiſe, which are the com-<lb/>mon Topicks handled in De-<lb/>dications, muſt needs be offen-<lb/>ſive to an Ingenious Perſon; <lb/></s> <s xml:id="echoid-s31" xml:space="preserve">and ſo I ſhall be ſilent on theſe <lb/>Heads; </s> <s xml:id="echoid-s32" xml:space="preserve">and only crave your <lb/>Acceptance and Protection of <lb/>what is here offer’d by</s> </p> <p style="it"> <s xml:id="echoid-s33" xml:space="preserve">Your Humble Servant,</s> </p> <p> <s xml:id="echoid-s34" xml:space="preserve">E. </s> <s xml:id="echoid-s35" xml:space="preserve"><emph style="sc">Stone</emph>.</s> <s xml:id="echoid-s36" xml:space="preserve"/> </p> <pb o="i" file="0009" n="9"/> </div> <div xml:id="echoid-div6" type="section" level="1" n="6"> <head xml:id="echoid-head5" xml:space="preserve">The Au<emph style="sc">THOR’S</emph> <lb/>PREFACE.</head> <p style="it"> <s xml:id="echoid-s37" xml:space="preserve">THE Reader will wonder, per-<lb/>haps, to find me entring into <lb/>a Path, which ſeems to have <lb/>been too much trodden already; <lb/></s> <s xml:id="echoid-s38" xml:space="preserve">and eſteem as uſeleſs a New <lb/>Eſſay, on an Art, whoſe Sub-<lb/>ject (one would think) ſhould have been <lb/>long before this Time exhauſted; </s> <s xml:id="echoid-s39" xml:space="preserve">ſince there <lb/>have been ſo many Perſons, who have writ-<lb/>ten on the ſame.</s> <s xml:id="echoid-s40" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s41" xml:space="preserve">The Name of Perſpective now ſeems to <lb/>ſound unple aſant in the Ears of the Publick <lb/>Enemies of Repetition; </s> <s xml:id="echoid-s42" xml:space="preserve">and it may be look d <lb/>upon as a Piece of Inadvertency, to venture <lb/>to treat again on that ſame Subject. </s> <s xml:id="echoid-s43" xml:space="preserve">Yet, <lb/>notwithst anding this, I deſire the Render to <lb/>ſuſpend his Cenſure, until he h{as} heard the <pb o="ii" file="0010" n="10" rhead="The PREFACE."/> Reaſons that induc’d me to publiſh the follow-<lb/>ing Work.</s> <s xml:id="echoid-s44" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s45" xml:space="preserve">Having ſome Years ago buſied my ſelf in <lb/>drawing Figures by the common Methods, I <lb/>found out ſeveral Compendiums; </s> <s xml:id="echoid-s46" xml:space="preserve">which, by <lb/>diligent working, naturally enough fall in <lb/>one’s way, without being entirely beholden <lb/>to the Induſtry of others: </s> <s xml:id="echoid-s47" xml:space="preserve">And theſe firſt <lb/>Succeſſes made me hope for others more <lb/>conſiderable; </s> <s xml:id="echoid-s48" xml:space="preserve">and ſo I thought that a more <lb/>narrow Inſpection into the Theory of Per-<lb/>ſpective, might furniſh me with Rules more <lb/>general, for making the Practice thereof <lb/>eaſy.</s> <s xml:id="echoid-s49" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s50" xml:space="preserve">I then thought upon ſeveral Methods to this <lb/>Purpoſe; </s> <s xml:id="echoid-s51" xml:space="preserve">but, being ſuſpicious that they were <lb/>not ſo eaſy {as} they appear’d, I have try’d <lb/>their Goodneſs, by exactly applying them to <lb/>different Subjects; </s> <s xml:id="echoid-s52" xml:space="preserve">and have nicely examin’d <lb/>all the Caſes, and order’d it ſo {as} not to be de-<lb/>ceiv’d by certain Operations, which at firſt <lb/>ſeem eaſy, but, when put in Practice, are <lb/>quite otherwiſe. </s> <s xml:id="echoid-s53" xml:space="preserve">Moreover, at convenient <lb/>Times, I look’d over the beſt Part of the <lb/>Authors of this kind, (whoſe Number is <lb/>increas’d very much, without any manner of <lb/>Neceſſity) ſome of which being advantagi-<lb/>ouſly diſtinguiſh’d among the Crowd, have <lb/>been very uſeful to me: </s> <s xml:id="echoid-s54" xml:space="preserve">But I dare affirm, <lb/>there are but a very few that give a new <lb/>Turn to the practical Part of Perſpective.</s> <s xml:id="echoid-s55" xml:space="preserve"/> </p> <pb o="iii" file="0011" n="11" rhead="The PREFACE."/> <p style="it"> <s xml:id="echoid-s56" xml:space="preserve">Some content themſelves with the bare <lb/>Explication of the Theory, and have left to <lb/>the Reader the Trouble of applying the ſame <lb/>to Practice; </s> <s xml:id="echoid-s57" xml:space="preserve">or elſe have given only ſome of <lb/>the common Operations, and entertain us <lb/>with general Reflections on Painting; </s> <s xml:id="echoid-s58" xml:space="preserve">which <lb/>are indeed curio{us}, but foreign to my Pur-<lb/>poſe: </s> <s xml:id="echoid-s59" xml:space="preserve">For I intend not to make a Man a <lb/>Painter, but to render the Uſe and Exerciſe <lb/>of Perſpective eaſy to him.</s> <s xml:id="echoid-s60" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s61" xml:space="preserve">Other Authors, which (according to the <lb/>Bulk of their Works) might be thought to <lb/>have more carefully treated of the practical <lb/>Part of Perſpective, do indeed at firſt lay <lb/>down ſome general Rules, common to them <lb/>all; </s> <s xml:id="echoid-s62" xml:space="preserve">but are nothing the eaſier for having <lb/>paſs’d thro’ ſo many Hands; </s> <s xml:id="echoid-s63" xml:space="preserve">and that, in-<lb/>deed, becauſe they have not endeavour’d to <lb/>make them ſo. </s> <s xml:id="echoid-s64" xml:space="preserve">They thought that all Ob-<lb/>jects might be thrown into Perſpective by <lb/>theſe Rules, and therefore it would be uſe-<lb/>leſs to ſearch after others; </s> <s xml:id="echoid-s65" xml:space="preserve">and judg’d it <lb/>more neceſſary to ſhew Painters the Applica-<lb/>tion of them to an infinite Number of parti-<lb/>cular Examples; </s> <s xml:id="echoid-s66" xml:space="preserve">tho’ that Application, at <lb/>moſt, is but repeating over again the Uſe of <lb/>the Rules already preſcrib’<unsure/>d. </s> <s xml:id="echoid-s67" xml:space="preserve">But what Ad-<lb/>vantage can Painters gain from hence, if <lb/>they do not well underſtand general Opera-<lb/>tions? </s> <s xml:id="echoid-s68" xml:space="preserve">And if they do, I cannot conceive of <pb o="iv" file="0012" n="12" rhead="The PREFACE."/> what Uſe ſuch an exceſſive Variety of Exam-<lb/>ples will be to them.</s> <s xml:id="echoid-s69" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s70" xml:space="preserve">I believ’d then, that I might be able to <lb/>treat of this Art after another Manner: <lb/></s> <s xml:id="echoid-s71" xml:space="preserve">And altho’ I know my ſelf to be much infe-<lb/>rior to ſeveral of thoſe who have written on <lb/>this Subject; </s> <s xml:id="echoid-s72" xml:space="preserve">yet I am of Opinion, that if <lb/>Perſpective ſhould loſe any thing by me, on <lb/>account of my want of fudgment; </s> <s xml:id="echoid-s73" xml:space="preserve">yet that <lb/>may be regain’d, perhaps, (and with In-<lb/>tereſt too) by my great Diligence in this Bu-<lb/>ſineſs.</s> <s xml:id="echoid-s74" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s75" xml:space="preserve">I have conſider’d, moreover, that the te-<lb/>dious Particulars, inherent to the Subject on <lb/>which I have choſen to write, will always <lb/>hinder Genius’s capable of great Matters, <lb/>from undertaking a Subject ſo little worthy <lb/>their Endeavours, and ſo barren of great <lb/>Diſcoveries.</s> <s xml:id="echoid-s76" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s77" xml:space="preserve">Thus, hoping, on one hand, to give a new <lb/>Turn to the Practice of Perſpective, and <lb/>make it eaſier; </s> <s xml:id="echoid-s78" xml:space="preserve">and being perſuaded, on the <lb/>other, that more learned Perſons than my ſelf <lb/>will not take this Trouble upon them; </s> <s xml:id="echoid-s79" xml:space="preserve">I ven-<lb/>ture to publiſh this ſmall Work, and expoſe <lb/>it to the Taſte of the Learned World; </s> <s xml:id="echoid-s80" xml:space="preserve">from <lb/>whom I expect no other Praiſe, but what <lb/>may reaſonably be claim’d by an aſſiduo{us} Ap-<lb/>plication.</s> <s xml:id="echoid-s81" xml:space="preserve"/> </p> <pb o="v" file="0013" n="13" rhead="The PREFACE."/> <p style="it"> <s xml:id="echoid-s82" xml:space="preserve">The Practice of Perſpective may be made <lb/>eaſy, by the Three following Things in this <lb/>Treatiſe: </s> <s xml:id="echoid-s83" xml:space="preserve">Viz. </s> <s xml:id="echoid-s84" xml:space="preserve">1. </s> <s xml:id="echoid-s85" xml:space="preserve">In giving ſeveral new <lb/>and eaſier Ways (than thoſe commonly uſed) <lb/>of ſolving the moſt general Problems upon <lb/>which the whole Practice is founded: </s> <s xml:id="echoid-s86" xml:space="preserve">And the <lb/>Reaſon why we have laid down ſeveral So-<lb/>lutions, is, becauſe the ſame Way is not always <lb/>equally convenient in all Caſes; </s> <s xml:id="echoid-s87" xml:space="preserve">whence it is <lb/>neceſſary to have ſeveral, that ſo we may <lb/>chuſe one beſt ſuiting our Purpoſe. </s> <s xml:id="echoid-s88" xml:space="preserve">2. </s> <s xml:id="echoid-s89" xml:space="preserve">The <lb/>general Methods, which have been us’d hi-<lb/>therto, not being practicable on ſome particu-<lb/>lar Occaſions; </s> <s xml:id="echoid-s90" xml:space="preserve">to remedy this, we have ad-<lb/>ded others to them; </s> <s xml:id="echoid-s91" xml:space="preserve">which are indeed more <lb/>difficult, but (in ſome Caſes) there is an ab-<lb/>ſolute Neceſſity for them. </s> <s xml:id="echoid-s92" xml:space="preserve">3. </s> <s xml:id="echoid-s93" xml:space="preserve">When it is ve-<lb/>ry difficult to reſolve a particular Problem, <lb/>by means of general ones; </s> <s xml:id="echoid-s94" xml:space="preserve">then we have <lb/>thought it convenient to give a particular So-<lb/>lution thereof.</s> <s xml:id="echoid-s95" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s96" xml:space="preserve">By this means, the Study of Perſpective <lb/>becomes indeed more difficult; </s> <s xml:id="echoid-s97" xml:space="preserve">but the Diſ-<lb/>advantage is well recompenſed by the Faci-<lb/>cility of the Practice, which we have entire-<lb/>ly had in view. </s> <s xml:id="echoid-s98" xml:space="preserve">It is true, that a few ge-<lb/>neral Rules do not ſo much burthen the Me-<lb/>mory; </s> <s xml:id="echoid-s99" xml:space="preserve">but when one has ſeveral general <lb/>ones, and alſo particular ones, by them we <lb/>can abridge Matters. </s> <s xml:id="echoid-s100" xml:space="preserve">And this Method be- <pb o="vi" file="0014" n="14" rhead="The PREFACE."/> ing purſued at first, tho’ it requires a little more <lb/>Application, does afterwards ſave a great <lb/>many Hours Study, in an Art that always <lb/>appears difficult enough.</s> <s xml:id="echoid-s101" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s102" xml:space="preserve">A Painter, in a ſhort Time, may learn this <lb/>Work, and make the Rules thereof familiar to <lb/>him: </s> <s xml:id="echoid-s103" xml:space="preserve">And if this Study be repeated from <lb/>time to time, for a few Days, he will find <lb/>the Benefit thereof, in diminiſhing his Labour <lb/>and Trouble.</s> <s xml:id="echoid-s104" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s105" xml:space="preserve">But, that any one himſelf may ſee what <lb/>I promiſe in this Eſſay; </s> <s xml:id="echoid-s106" xml:space="preserve">take the following <lb/>ſhort Abſtract thereof. </s> <s xml:id="echoid-s107" xml:space="preserve">It is divided into <lb/>Nine Chapters: </s> <s xml:id="echoid-s108" xml:space="preserve">The Firſt, being as an In-<lb/>troduction to the reſt, ſhews the Uſefulneſs <lb/>of Perſpective, and gives you the Definiti-<lb/>ons of the Terms neceſſary for underſtanding <lb/>this Treatiſe.</s> <s xml:id="echoid-s109" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s110" xml:space="preserve">The whole Theory is contain’d in the Se-<lb/>cond Chapter: </s> <s xml:id="echoid-s111" xml:space="preserve">Where, what has been found <lb/>moſt uſeful in that Matter, is therein re-<lb/>duced to Three general Theorems; </s> <s xml:id="echoid-s112" xml:space="preserve">viz. </s> <s xml:id="echoid-s113" xml:space="preserve">the <lb/>firſt, ſecond, and fourth: </s> <s xml:id="echoid-s114" xml:space="preserve">All the reſt is de-<lb/>duced from them, by way of Corollary. </s> <s xml:id="echoid-s115" xml:space="preserve">To <lb/>theſe Theorems, already known, are added <lb/>ſome new ones, ſerving for the Demonſtra-<lb/>tion of ſome neceſſary Propoſitions. </s> <s xml:id="echoid-s116" xml:space="preserve">Perhaps <lb/>it might be wiſh’d, that I had ſhewn the Way <lb/>that led me to the Truths which I diſcover:</s> <s xml:id="echoid-s117" xml:space="preserve"> <pb o="vii" file="0015" n="15" rhead="The PREFACE."/> This I have done ſometimes; </s> <s xml:id="echoid-s118" xml:space="preserve">but it often <lb/>would have been very long and troubleſome. <lb/></s> <s xml:id="echoid-s119" xml:space="preserve">In Geometry, the eaſieſt and ſhorteſt Way, is <lb/>not always that which leads to Diſcoveries.</s> <s xml:id="echoid-s120" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s121" xml:space="preserve">In the following Chapter, the Practice of Per-<lb/>ſpective upon the perſpective Plane, or Picture, <lb/>conſider’d as upright, is explain’d: </s> <s xml:id="echoid-s122" xml:space="preserve">Wherein, <lb/>among the different Ways laid down for the <lb/>Solution of general Problems, you will find <lb/>ſome effected by a Ruler only; </s> <s xml:id="echoid-s123" xml:space="preserve">ſo that after <lb/>ſome Preparations, all Kinds of Objects may <lb/>be drawn without Compaſſes, and that eaſier <lb/>than by the common Operations. </s> <s xml:id="echoid-s124" xml:space="preserve">In that <lb/>Problem, to find the Appearance of a Point <lb/>out of the Geometrical Plane, it is commonly <lb/>conſider’d as the Extremity of a Perpendicu-<lb/>lar, whoſe Repreſentation muſt first be found, <lb/>before that of the Point can be had. </s> <s xml:id="echoid-s125" xml:space="preserve">But here <lb/>we avoid this round-about Way, and ſhew how <lb/>to find the Appearance of the Point given, <lb/>without being obliged to find the Perſpective <lb/>of its Seat.</s> <s xml:id="echoid-s126" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s127" xml:space="preserve">As to the Appearance of a Cone and Cy-<lb/>linder, we determine the viſible Portions of <lb/>the Baſe, and by this means avoid the uſe-<lb/>leſs Operations which the common Way is <lb/>ſubject to. </s> <s xml:id="echoid-s128" xml:space="preserve">It is very difficult, if not impoſ-<lb/>ſible, to throw a Sphere into Perſpective, by <lb/>means of general Problems; </s> <s xml:id="echoid-s129" xml:space="preserve">and in the Re-<lb/>preſentation of the Torus of a Column, it is <pb o="viii" file="0016" n="16" rhead="The PREFACE."/> ſtill difficulter: </s> <s xml:id="echoid-s130" xml:space="preserve">Whence we are obliged to <lb/>give particular Methods for the Reſolution of <lb/>theſe two Problems.</s> <s xml:id="echoid-s131" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s132" xml:space="preserve">The reſt of the Third Chapter is concern-<lb/>ing Inclin’d Lines, and how to find their Ap-<lb/>pearance by the Accidental Point.</s> <s xml:id="echoid-s133" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s134" xml:space="preserve">The Fourth Chapter ſhews the Manner of <lb/>working on a perſpective Plane, to be view’d <lb/>afar off, very obliquely, or which muſt ſtand <lb/>in an high Place. </s> <s xml:id="echoid-s135" xml:space="preserve">Theſe different Situations <lb/>require new Rules: </s> <s xml:id="echoid-s136" xml:space="preserve">For if the common Me-<lb/>thods were to be uſed here, the perſpective <lb/>Plane muſt be ſo large, as that it would be <lb/>impoſſible to work upon it.</s> <s xml:id="echoid-s137" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s138" xml:space="preserve">In the Two following Chapters, we treat <lb/>of the perſpective Plane, conſider’d as Hori-<lb/>zontal, or Inclin’d: </s> <s xml:id="echoid-s139" xml:space="preserve">Where there are laid <lb/>down ſeveral general Ways of working; <lb/></s> <s xml:id="echoid-s140" xml:space="preserve">which, together with thoſe of the foregoing <lb/>Chapters, will ſuffice (in my Opinion) for <lb/>throwing any Object whatſoever into Perſpe-<lb/>ctive, with Eaſe enough.</s> <s xml:id="echoid-s141" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s142" xml:space="preserve">In the Seventh Chapter, which treats of <lb/>Shadows, there is nothing particular, but <lb/>what may be ſeen elſewhere: </s> <s xml:id="echoid-s143" xml:space="preserve">But that lit-<lb/>tle we have ſaid concerning this Matter, is <lb/>enough for giving an Idea of them, which <lb/>the Reading of what goes before will make <lb/>eaſy.</s> <s xml:id="echoid-s144" xml:space="preserve"/> </p> <pb o="ix" file="0017" n="17" rhead="The PREFACE."/> <p style="it"> <s xml:id="echoid-s145" xml:space="preserve">In the Eighth Chapter, are laid down <lb/>ſome Mechanical ways for making the Uſe of <lb/>Perſpective eaſy, by means of Rulers and <lb/>Threads, (eaſily to be gotten by any body, <lb/>and not difficult to be put in practice) they <lb/>being eaſier to uſe than any of the Inſtru-<lb/>ments that have hitherto been invented for <lb/>this Purpoſe.</s> <s xml:id="echoid-s146" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s147" xml:space="preserve">The laſt Chapter ſhews the Uſefulneſs of <lb/>Perſpective in Dialling.</s> <s xml:id="echoid-s148" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s149" xml:space="preserve">Such is the Plan of this ſinall Work; <lb/></s> <s xml:id="echoid-s150" xml:space="preserve">wherein I have not ſo much endeavour’d to <lb/>advance Curioſities, as Things of real Uſe; </s> <s xml:id="echoid-s151" xml:space="preserve"><lb/>hoping that, without making a Shew of Skill <lb/>ill beſtow’d, I ſhall make my Book good enough, <lb/>if by its Uſe I make it neceſſary. </s> <s xml:id="echoid-s152" xml:space="preserve">For which <lb/>Reaſon, I have endeavour’d to lay down the <lb/>whole, ſo as to be underſtood by thoſe who <lb/>have only read the Elements of Euclid. </s> <s xml:id="echoid-s153" xml:space="preserve">And <lb/>tho’ I have deviated from this Rule in ſome <lb/>few Place; </s> <s xml:id="echoid-s154" xml:space="preserve">they are printed in Italick, that <lb/>ſo they may be paſs’d over without any Hin-<lb/>drance to the Learner.</s> <s xml:id="echoid-s155" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s156" xml:space="preserve">Here I muſt not forget to mention, that in <lb/>Reviſing this Eſſay, I had the Happineſs of <lb/>meeting with an able Painter; </s> <s xml:id="echoid-s157" xml:space="preserve">who has ſe-<lb/>r<unsure/>iouſly conſider’d every Thing of his Profeſ-<lb/>ſion, neceſſary to be known, among which, <pb o="x" file="0018" n="18" rhead="The PREFACE."/> Perſpective was not neglected. </s> <s xml:id="echoid-s158" xml:space="preserve">He has car-<lb/>ried the Matter farther than could have been <lb/>reaſonably expected from one ignorant of <lb/>Mathematicks; </s> <s xml:id="echoid-s159" xml:space="preserve">and I am indebted to him <lb/>for ſeveral Obſervations, which I my ſelf <lb/>ſhould perhaps have never thought on.</s> <s xml:id="echoid-s160" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div7" type="section" level="1" n="7"> <head xml:id="echoid-head6" xml:space="preserve">ERRATA.</head> <p> <s xml:id="echoid-s161" xml:space="preserve">Page 74. </s> <s xml:id="echoid-s162" xml:space="preserve">1. </s> <s xml:id="echoid-s163" xml:space="preserve">17. </s> <s xml:id="echoid-s164" xml:space="preserve">for x, r. </s> <s xml:id="echoid-s165" xml:space="preserve">in X.</s> <s xml:id="echoid-s166" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s167" xml:space="preserve">P. </s> <s xml:id="echoid-s168" xml:space="preserve">83. </s> <s xml:id="echoid-s169" xml:space="preserve">1. </s> <s xml:id="echoid-s170" xml:space="preserve">28. </s> <s xml:id="echoid-s171" xml:space="preserve">for that, r. </s> <s xml:id="echoid-s172" xml:space="preserve">that for.</s> <s xml:id="echoid-s173" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s174" xml:space="preserve">P. </s> <s xml:id="echoid-s175" xml:space="preserve">88. </s> <s xml:id="echoid-s176" xml:space="preserve">1. </s> <s xml:id="echoid-s177" xml:space="preserve">8. </s> <s xml:id="echoid-s178" xml:space="preserve">for Tube, r. </s> <s xml:id="echoid-s179" xml:space="preserve">Table.</s> <s xml:id="echoid-s180" xml:space="preserve"/> </p> <pb o="1" file="0019" n="19"/> </div> <div xml:id="echoid-div8" type="section" level="1" n="8"> <head xml:id="echoid-head7" xml:space="preserve">AN <lb/>ESSAY <lb/>ON <lb/>PERSPECTIVE.</head> <head xml:id="echoid-head8" xml:space="preserve">CHAP. I. <lb/><emph style="sc">Definitions</emph>.</head> <p> <s xml:id="echoid-s181" xml:space="preserve">1. </s> <s xml:id="echoid-s182" xml:space="preserve"><emph style="sp">PERSPECTIVE</emph> teaches us <lb/>the Manner of Delineating by <lb/>Mathematical Rules; </s> <s xml:id="echoid-s183" xml:space="preserve">that is, it <lb/>ſhews us how to draw geometri-<lb/>cally upon a Plane, the Repre-<lb/>ſentations of Objects according <lb/>to their Dimenſions and different Situations; </s> <s xml:id="echoid-s184" xml:space="preserve">in <lb/>ſuch manner, that the ſaid Repreſentations pro-<lb/>duce the ſame Effects upon our Eyes, as the Ob-<lb/>jects whereof they are the Pictures.</s> <s xml:id="echoid-s185" xml:space="preserve"/> </p> <pb o="2" file="0020" n="20" rhead="An ESSAY"/> <p> <s xml:id="echoid-s186" xml:space="preserve">In order to underſtand well how Mathema-<lb/>ticks may be apply’d to Drawing; </s> <s xml:id="echoid-s187" xml:space="preserve">let us ſup-<lb/>poſe a Man A, viewing an Object; </s> <s xml:id="echoid-s188" xml:space="preserve">and be-<lb/> <anchor type="note" xlink:label="note-0020-01a" xlink:href="note-0020-01"/> tween him and the Object he looks at, let us <lb/>imagine a tranſparent Plane C. </s> <s xml:id="echoid-s189" xml:space="preserve">Suppoſe more-<lb/>over, that Lines be drawn upon this Plane, as <lb/>in D, which cover the Bounds of the Object B <lb/>in reſpect of the Spectator A, and each Part that <lb/>he ſees thereof. </s> <s xml:id="echoid-s190" xml:space="preserve">Now, ſince all Objects are ſeen <lb/>by the Rays of Light coming from every of <lb/>their Points, and terminating at the Eye, and <lb/>not otherwiſe; </s> <s xml:id="echoid-s191" xml:space="preserve">and ſince that here all the Rays <lb/>proceeding from the Object B, likewiſe paſs <lb/>thro’ every Point of the Repreſentation D; </s> <s xml:id="echoid-s192" xml:space="preserve">it is <lb/>manifeſt, that this Repreſentation will have the <lb/>ſame Effect upon the Spectator’s Eye, as the <lb/>ſaid Object B hath. </s> <s xml:id="echoid-s193" xml:space="preserve">Now, by means of Geo-<lb/>metry, we can find the Points of the Figure D, <lb/>on the Plane C, placed in a given Situation, <lb/>thro’ which the Rays coming from the Object B <lb/>to the Eye of the Spectator A, do paſs; </s> <s xml:id="echoid-s194" xml:space="preserve">and <lb/>theſe Points are the Interſections of the Rays <lb/>and the Plane. </s> <s xml:id="echoid-s195" xml:space="preserve">Alſo, (as others have very well <lb/>obſerv’d) a Perſpective Plane, or Picture in Paint-<lb/>ing, may be conceiv’d as a Window, upon which <lb/>the Objects ſeen thro’ it are repreſented.</s> <s xml:id="echoid-s196" xml:space="preserve"/> </p> <div xml:id="echoid-div8" type="float" level="2" n="1"> <note position="left" xlink:label="note-0020-01" xlink:href="note-0020-01a" xml:space="preserve">Fig. 1.</note> </div> <p> <s xml:id="echoid-s197" xml:space="preserve">Now, without Mathematicks, this Repreſen-<lb/>tation cannot be well found: </s> <s xml:id="echoid-s198" xml:space="preserve">For when Objects <lb/>are drawn by only viewing, or looking at them; <lb/></s> <s xml:id="echoid-s199" xml:space="preserve">their true Repreſentations after this way, will <lb/>be very often miſs’d on; </s> <s xml:id="echoid-s200" xml:space="preserve">whereas, by Geome-<lb/>try, we can always obtain them.</s> <s xml:id="echoid-s201" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s202" xml:space="preserve">This Obſervation only, is ſufficient to eſta-<lb/>bliſh the Neceſſity of Perſpective: </s> <s xml:id="echoid-s203" xml:space="preserve">Tho’ there are <lb/>ſome Painters, who (according to the common <lb/>Maxim) affirm, That what they do not know <lb/>of this Art, is not worth the Pains of learn-<lb/>ing.</s> <s xml:id="echoid-s204" xml:space="preserve"/> </p> <pb o="3" file="0021" n="21" rhead="on PERSPECTIVE."/> <p> <s xml:id="echoid-s205" xml:space="preserve">Hitherto I have endeavour’d to give an Idea <lb/>of Perſpective in general: </s> <s xml:id="echoid-s206" xml:space="preserve">But there is yet ano-<lb/>ther particular Signification of this Word, which <lb/>it is neceſſary ſhould be explain’d, as well as <lb/>the other Terms of the Art, which are laid down <lb/>in the following Definitions; </s> <s xml:id="echoid-s207" xml:space="preserve">and which every <lb/>one, that intends to underſtand this Treatiſe, <lb/>ought to be well acquainted with.</s> <s xml:id="echoid-s208" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s209" xml:space="preserve">2. </s> <s xml:id="echoid-s210" xml:space="preserve">The Perſpective, Repreſentation, or Appearance of <lb/> <anchor type="note" xlink:label="note-0021-01a" xlink:href="note-0021-01"/> an Object, (for theſe Three Words are ſynoni-<lb/>mous) * is the Figure which the Rays, by which an <lb/>Object is perceiv’d, form in paſſing thro’ the tranſpa-<lb/>rent Plane: </s> <s xml:id="echoid-s211" xml:space="preserve">And the Perſpective of a Point, is the <lb/>Interſection of a Ray proceeding from that Point, <lb/>and the tranſparent Plane. </s> <s xml:id="echoid-s212" xml:space="preserve">Which Interſection is <lb/>a Point: </s> <s xml:id="echoid-s213" xml:space="preserve">As the Figure D in the tranſparent <lb/> <anchor type="note" xlink:label="note-0021-02a" xlink:href="note-0021-02"/> Plane C, is the Perſpective of the Object B; </s> <s xml:id="echoid-s214" xml:space="preserve">and <lb/>the Point e, in the ſame Plane, is the Perſpective <lb/>of the Point E, in that Object.</s> <s xml:id="echoid-s215" xml:space="preserve"/> </p> <div xml:id="echoid-div9" type="float" level="2" n="2"> <note position="right" xlink:label="note-0021-01" xlink:href="note-0021-01a" xml:space="preserve">Def. 1.</note> <note position="right" xlink:label="note-0021-02" xlink:href="note-0021-02a" xml:space="preserve">Fig. 1.</note> </div> <p style="it"> <s xml:id="echoid-s216" xml:space="preserve">The Plane parallel to the Horizon, upon which <lb/> <anchor type="note" xlink:label="note-0021-03a" xlink:href="note-0021-03"/> the Spectator [ſtands, or] is placed, as likewiſe the <lb/>Objects that he views, is call’d the Geometrical Plane. <lb/></s> <s xml:id="echoid-s217" xml:space="preserve">As A B C D.</s> <s xml:id="echoid-s218" xml:space="preserve"/> </p> <div xml:id="echoid-div10" type="float" level="2" n="3"> <note position="right" xlink:label="note-0021-03" xlink:href="note-0021-03a" xml:space="preserve">Def. 2.</note> </div> <note position="right" xml:space="preserve">Fig. 2.</note> <p> <s xml:id="echoid-s219" xml:space="preserve">A Perſpective Plane [or Picture] is that which <lb/> <anchor type="note" xlink:label="note-0021-05a" xlink:href="note-0021-05"/> is placed between the Spectator and the Object, upon <lb/>which the Objects are drawn: </s> <s xml:id="echoid-s220" xml:space="preserve">As F G R T. </s> <s xml:id="echoid-s221" xml:space="preserve">This <lb/>is commonly perpendicular to the Geometrical <lb/>Plane, and conſequently to the Horizon; </s> <s xml:id="echoid-s222" xml:space="preserve">becauſe <lb/>Pictures have generally this Situation: </s> <s xml:id="echoid-s223" xml:space="preserve">But yet <lb/>it may be ſometimes inclin’d, and even parallel <lb/>to the Geometrical Plane, according as one <lb/>would diſpoſe the Deſign, or Picture that we are <lb/>working. </s> <s xml:id="echoid-s224" xml:space="preserve">And for this Reaſon, in the follow-<lb/>ing Chapter, we have laid down General Theo-<lb/>rems, and their Corollaries, agreeing to all theſe <pb o="4" file="0022" n="22" rhead="An ESSAY"/> different Situations of the Perſpective Plane, [or <lb/>Picture] which ought to be well obſerv’d.</s> <s xml:id="echoid-s225" xml:space="preserve"/> </p> <div xml:id="echoid-div11" type="float" level="2" n="4"> <note position="right" xlink:label="note-0021-05" xlink:href="note-0021-05a" xml:space="preserve">Def. 3.</note> </div> <p style="it"> <s xml:id="echoid-s226" xml:space="preserve">The Interſection of the Perſpective Plane and the <lb/> <anchor type="note" xlink:label="note-0022-01a" xlink:href="note-0022-01"/> Geometrical Plane, is call’d the Baſe-Line: </s> <s xml:id="echoid-s227" xml:space="preserve">As <lb/>F G.</s> <s xml:id="echoid-s228" xml:space="preserve"/> </p> <div xml:id="echoid-div12" type="float" level="2" n="5"> <note position="left" xlink:label="note-0022-01" xlink:href="note-0022-01a" xml:space="preserve">Def. 4.</note> </div> <p> <s xml:id="echoid-s229" xml:space="preserve">The different Situation of the Eye, alters the <lb/>Repreſentation of Objects in the perſpective Plane; <lb/></s> <s xml:id="echoid-s230" xml:space="preserve">for the Rays proceeding from the Object, and <lb/>concurring in ſome other Point, will likewiſe fall <lb/>upon the Perſpective Plane in different Places. </s> <s xml:id="echoid-s231" xml:space="preserve">And <lb/>for determining this Situation of the Eye, in re-<lb/>ſpect to the Perſpective Plane, we ſuppoſe,</s> </p> <p style="it"> <s xml:id="echoid-s232" xml:space="preserve">A Plane parallel to the Horizon, paſſing thro’ the <lb/> <anchor type="note" xlink:label="note-0022-02a" xlink:href="note-0022-02"/> Eye, and every way extending it ſelf; </s> <s xml:id="echoid-s233" xml:space="preserve">and this is <lb/>call’d the Horizontal Plane: </s> <s xml:id="echoid-s234" xml:space="preserve">As OMVNL.</s> <s xml:id="echoid-s235" xml:space="preserve"/> </p> <div xml:id="echoid-div13" type="float" level="2" n="6"> <note position="left" xlink:label="note-0022-02" xlink:href="note-0022-02a" xml:space="preserve">Def. 5.</note> </div> <p style="it"> <s xml:id="echoid-s236" xml:space="preserve">The Interſection of this Plane and the Perſpective <lb/> <anchor type="note" xlink:label="note-0022-03a" xlink:href="note-0022-03"/> Plane, is the Horizontal Line. </s> <s xml:id="echoid-s237" xml:space="preserve">As M V N.</s> <s xml:id="echoid-s238" xml:space="preserve"/> </p> <div xml:id="echoid-div14" type="float" level="2" n="7"> <note position="left" xlink:label="note-0022-03" xlink:href="note-0022-03a" xml:space="preserve">Def. 6.</note> </div> <p style="it"> <s xml:id="echoid-s239" xml:space="preserve">The Perpendicular drawn from the Eye to the Ho-<lb/> <anchor type="note" xlink:label="note-0022-04a" xlink:href="note-0022-04"/> rizontal Line, is the principal Ray. </s> <s xml:id="echoid-s240" xml:space="preserve">As O V.</s> <s xml:id="echoid-s241" xml:space="preserve"/> </p> <div xml:id="echoid-div15" type="float" level="2" n="8"> <note position="left" xlink:label="note-0022-04" xlink:href="note-0022-04a" xml:space="preserve">Def. 7.</note> </div> <p style="it"> <s xml:id="echoid-s242" xml:space="preserve">The Point V, wherein the ſaid Perpendicular meets <lb/> <anchor type="note" xlink:label="note-0022-05a" xlink:href="note-0022-05"/> the Horizontal Line, is the Point of Sight, or prin-<lb/>cipal Point.</s> <s xml:id="echoid-s243" xml:space="preserve"/> </p> <div xml:id="echoid-div16" type="float" level="2" n="9"> <note position="left" xlink:label="note-0022-05" xlink:href="note-0022-05a" xml:space="preserve">Def. 8.</note> </div> <p> <s xml:id="echoid-s244" xml:space="preserve">Note, There is a Perpendicular, let fall from <lb/>the Eye upon the Geometrical Plane, meaſuring <lb/>the Height of the Eye.</s> <s xml:id="echoid-s245" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s246" xml:space="preserve">The Point S, wherein the ſaid Perpendicular meets <lb/> <anchor type="note" xlink:label="note-0022-06a" xlink:href="note-0022-06"/> the Geometrical Plane, is the Station-Point.</s> <s xml:id="echoid-s247" xml:space="preserve"/> </p> <div xml:id="echoid-div17" type="float" level="2" n="10"> <note position="left" xlink:label="note-0022-06" xlink:href="note-0022-06a" xml:space="preserve">Def. 9.</note> </div> <p style="it"> <s xml:id="echoid-s248" xml:space="preserve">The Plane paſſing thro’ the aforeſaid Perpendicular, <lb/> <anchor type="note" xlink:label="note-0022-07a" xlink:href="note-0022-07"/> and the principal Ray, is call’d the Vertical Plane. <lb/></s> <s xml:id="echoid-s249" xml:space="preserve">As SOLI.</s> <s xml:id="echoid-s250" xml:space="preserve"/> </p> <div xml:id="echoid-div18" type="float" level="2" n="11"> <note position="left" xlink:label="note-0022-07" xlink:href="note-0022-07a" xml:space="preserve">Def. 10.</note> </div> <pb o="5" file="0023" n="23" rhead="on PERSPECTIVE."/> <p style="it"> <s xml:id="echoid-s251" xml:space="preserve">The Interſection V H of this Plane, and the Per-<lb/> <anchor type="note" xlink:label="note-0023-01a" xlink:href="note-0023-01"/> ſpective Plane, is the Vertical Line.</s> <s xml:id="echoid-s252" xml:space="preserve"/> </p> <div xml:id="echoid-div19" type="float" level="2" n="12"> <note position="right" xlink:label="note-0023-01" xlink:href="note-0023-01a" xml:space="preserve">Def. 11.</note> </div> <p style="it"> <s xml:id="echoid-s253" xml:space="preserve">And S H I, the Interſection of it, and the Geo-<lb/> <anchor type="note" xlink:label="note-0023-02a" xlink:href="note-0023-02"/> metrical Plane, is the Station Line.</s> <s xml:id="echoid-s254" xml:space="preserve"/> </p> <div xml:id="echoid-div20" type="float" level="2" n="13"> <note position="right" xlink:label="note-0023-02" xlink:href="note-0023-02a" xml:space="preserve">Def. 12.</note> </div> <p style="it"> <s xml:id="echoid-s255" xml:space="preserve">Points of Diſtance, are two Points in the Hori-<lb/> <anchor type="note" xlink:label="note-0023-03a" xlink:href="note-0023-03"/> zontal Line, each way diſtant from the Point of Sight <lb/>by the length of the principal Ray; </s> <s xml:id="echoid-s256" xml:space="preserve">as MN.</s> <s xml:id="echoid-s257" xml:space="preserve"/> </p> <div xml:id="echoid-div21" type="float" level="2" n="14"> <note position="right" xlink:label="note-0023-03" xlink:href="note-0023-03a" xml:space="preserve">Def. 13.</note> </div> <p style="it"> <s xml:id="echoid-s258" xml:space="preserve">The Geometrical Line, is a Line, that paſſes <lb/> <anchor type="note" xlink:label="note-0023-04a" xlink:href="note-0023-04"/> through the Station Point, and is parallel to the baſe <lb/>Line, as A B.</s> <s xml:id="echoid-s259" xml:space="preserve"/> </p> <div xml:id="echoid-div22" type="float" level="2" n="15"> <note position="right" xlink:label="note-0023-04" xlink:href="note-0023-04a" xml:space="preserve">Def. 14.</note> </div> <p style="it"> <s xml:id="echoid-s260" xml:space="preserve">The Seat of an Object, is the Concurrence of Per-<lb/> <anchor type="note" xlink:label="note-0023-05a" xlink:href="note-0023-05"/> pendiculars let fall from every of its Points upon the <lb/>Geometrical Plane, and the ſaid Plane.</s> <s xml:id="echoid-s261" xml:space="preserve"/> </p> <div xml:id="echoid-div23" type="float" level="2" n="16"> <note position="right" xlink:label="note-0023-05" xlink:href="note-0023-05a" xml:space="preserve">Def. 15.</note> </div> <p style="it"> <s xml:id="echoid-s262" xml:space="preserve">The Direction of a Line inclined to the Geometri-<lb/> <anchor type="note" xlink:label="note-0023-06a" xlink:href="note-0023-06"/> cal Plane, is the Interſection of the ſaid Plane, and <lb/>another Plane perpendicular thereto, paſſing through <lb/>the ſaid inclined Line.</s> <s xml:id="echoid-s263" xml:space="preserve"/> </p> <div xml:id="echoid-div24" type="float" level="2" n="17"> <note position="right" xlink:label="note-0023-06" xlink:href="note-0023-06a" xml:space="preserve">Def. 16.</note> </div> </div> <div xml:id="echoid-div26" type="section" level="1" n="9"> <head xml:id="echoid-head9" xml:space="preserve">CHAP. II.</head> <head xml:id="echoid-head10" style="it" xml:space="preserve">The Theory of Perſpective.</head> <head xml:id="echoid-head11" xml:space="preserve"><emph style="sc">Lemma</emph>.</head> <p> <s xml:id="echoid-s264" xml:space="preserve">3. </s> <s xml:id="echoid-s265" xml:space="preserve">THE Perſpective, or Appearance of a <lb/>Right Line, as A B, which being con-<lb/>tinued, does not paſsthrough the Eye O, is like-<lb/> <anchor type="note" xlink:label="note-0023-07a" xlink:href="note-0023-07"/> wiſe a right Line: </s> <s xml:id="echoid-s266" xml:space="preserve">For the Rays, by which the <lb/>Line A B is perceived, form a Plane cutting <lb/>the perſpective Plane; </s> <s xml:id="echoid-s267" xml:space="preserve">and the common Section <lb/>of theſe two Planes is a right Line, as a b.</s> <s xml:id="echoid-s268" xml:space="preserve"/> </p> <div xml:id="echoid-div26" type="float" level="2" n="1"> <note position="right" xlink:label="note-0023-07" xlink:href="note-0023-07a" xml:space="preserve">Fig. 3.</note> </div> <pb o="6" file="0024" n="24" rhead="An ESSAY"/> </div> <div xml:id="echoid-div28" type="section" level="1" n="10"> <head xml:id="echoid-head12" xml:space="preserve"><emph style="sc">Theorem</emph> I.</head> <p style="it"> <s xml:id="echoid-s269" xml:space="preserve">4. </s> <s xml:id="echoid-s270" xml:space="preserve">The Repreſentation of a Line Parallel to the <lb/> <anchor type="note" xlink:label="note-0024-01a" xlink:href="note-0024-01"/> perſpective Plane, is parallel to the Line whereof it is <lb/>the Repreſentation.</s> <s xml:id="echoid-s271" xml:space="preserve"/> </p> <div xml:id="echoid-div28" type="float" level="2" n="1"> <note position="left" xlink:label="note-0024-01" xlink:href="note-0024-01a" xml:space="preserve">Fig. 3.</note> </div> <p> <s xml:id="echoid-s272" xml:space="preserve">Let A B be a Line Parallel to the perſpective <lb/>Plane; </s> <s xml:id="echoid-s273" xml:space="preserve">we are to prove that a b its Repreſentati-<lb/>on is Parallel thereto.</s> <s xml:id="echoid-s274" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s275" xml:space="preserve">Theſe two Lines A B and a b, will never <lb/>meet each other, becauſe a b is in the perſpective <lb/>Plane, and A B is ſuppoſed parallel to the ſaid <lb/>Plane. </s> <s xml:id="echoid-s276" xml:space="preserve">But they are alſo in one and the ſame <lb/>Plane, becauſe a b is the Interſection of the per-<lb/>ſpective Plane, and the Plane O A B, paſſing <lb/>through the Eye and the Line A B; </s> <s xml:id="echoid-s277" xml:space="preserve">and there-<lb/>fore they are parallel between themſelves: </s> <s xml:id="echoid-s278" xml:space="preserve">Which <lb/>was to be demonſtrated.</s> <s xml:id="echoid-s279" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div30" type="section" level="1" n="11"> <head xml:id="echoid-head13" xml:space="preserve"><emph style="sc">Corollary</emph> I.</head> <p style="it"> <s xml:id="echoid-s280" xml:space="preserve">5. </s> <s xml:id="echoid-s281" xml:space="preserve">The Appearance of a Line, parallel to the baſe <lb/>Line, is alſo parallel to the ſaid baſe Line.</s> <s xml:id="echoid-s282" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s283" xml:space="preserve">For the baſe Line, and the Repreſentation <lb/>being parallel to the ſame Line, are parallel to <lb/>one another.</s> <s xml:id="echoid-s284" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div31" type="section" level="1" n="12"> <head xml:id="echoid-head14" xml:space="preserve"><emph style="sc">Corollary</emph> II.</head> <p style="it"> <s xml:id="echoid-s285" xml:space="preserve">6. </s> <s xml:id="echoid-s286" xml:space="preserve">The Repreſentation of a Line parallel to the <lb/>vertical Line, is parallel to the ſaid vertical Line, and <lb/>conſequently perpendicular to the baſe Line. </s> <s xml:id="echoid-s287" xml:space="preserve">This <lb/>is demonſtrated as in the laſt Corollary.</s> <s xml:id="echoid-s288" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div32" type="section" level="1" n="13"> <head xml:id="echoid-head15" xml:space="preserve"><emph style="sc">Corollary</emph> III.</head> <p style="it"> <s xml:id="echoid-s289" xml:space="preserve">7. </s> <s xml:id="echoid-s290" xml:space="preserve">The Appearances of Lines parallel to the per-<lb/>ſpective Plane, and equally inclin’d the ſame Way upon <pb o="7" file="0025" n="25" rhead="on PERSPECTIVE."/> the Geometrical Plane, make Angles with the baſe <lb/>Line, equal to thoſe Angles that the Lines whereof <lb/>they are the Appearances, make with the Parallels <lb/>to the baſe Line, which cut them; </s> <s xml:id="echoid-s291" xml:space="preserve">and conſequently <lb/>the ſaid Appearances are parallel between them-<lb/>ſelves.</s> <s xml:id="echoid-s292" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s293" xml:space="preserve">This is evident, becauſe the Appearances of <lb/>Lines parallel to the baſe Line, are parallel to <lb/>the ſaid Line; </s> <s xml:id="echoid-s294" xml:space="preserve">and the Appearances of the in-<lb/>clined Lines are parallel to theſe Lines.</s> <s xml:id="echoid-s295" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div33" type="section" level="1" n="14"> <head xml:id="echoid-head16" xml:space="preserve"><emph style="sc">Theorem</emph> II.</head> <p style="it"> <s xml:id="echoid-s296" xml:space="preserve">8, 9. </s> <s xml:id="echoid-s297" xml:space="preserve">The Repreſentation of a Figure, parallel to <lb/>the perſpective Plane, is ſimilar to the ſaid Figure; </s> <s xml:id="echoid-s298" xml:space="preserve">and <lb/>the Sides of the ſaid Figure are to their Repreſen-<lb/>tations, as the Diſtance of the Eye from the Plane <lb/>of the Figure, to the Diſtance of the Eye from the <lb/>perſpective Plane.</s> <s xml:id="echoid-s299" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s300" xml:space="preserve">The given Figure is A B C D. </s> <s xml:id="echoid-s301" xml:space="preserve">We are firſt to <lb/> <anchor type="note" xlink:label="note-0025-01a" xlink:href="note-0025-01"/> prove, that its Repreſentation a b c d, is ſimilar <lb/>thereto; </s> <s xml:id="echoid-s302" xml:space="preserve">that is, that the correſponding Angles <lb/>of theſe two Figures A B C D, a b c d, are equal, <lb/>and their Sides proportional.</s> <s xml:id="echoid-s303" xml:space="preserve"/> </p> <div xml:id="echoid-div33" type="float" level="2" n="1"> <note position="right" xlink:label="note-0025-01" xlink:href="note-0025-01a" xml:space="preserve">Fig. 4.</note> </div> <p> <s xml:id="echoid-s304" xml:space="preserve">I. </s> <s xml:id="echoid-s305" xml:space="preserve">The Angles are equal, becauſe <anchor type="note" xlink:href="" symbol="*"/> the Lines <anchor type="note" xlink:label="note-0025-02a" xlink:href="note-0025-02"/> of which the two Figures conſiſt, are parallel be-<lb/>tween themſelves.</s> <s xml:id="echoid-s306" xml:space="preserve"/> </p> <div xml:id="echoid-div34" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0025-02" xlink:href="note-0025-02a" xml:space="preserve">4.</note> </div> <p> <s xml:id="echoid-s307" xml:space="preserve">II. </s> <s xml:id="echoid-s308" xml:space="preserve">In the ſimilar Triangles A D O, and a d o, <lb/>we have <lb/>A D: </s> <s xml:id="echoid-s309" xml:space="preserve">a d : </s> <s xml:id="echoid-s310" xml:space="preserve">: </s> <s xml:id="echoid-s311" xml:space="preserve">O D : </s> <s xml:id="echoid-s312" xml:space="preserve">O d.</s> <s xml:id="echoid-s313" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s314" xml:space="preserve">And in the ſimilar Triangles O D C, and O d c, <lb/>we have <lb/>D C : </s> <s xml:id="echoid-s315" xml:space="preserve">d c : </s> <s xml:id="echoid-s316" xml:space="preserve">: </s> <s xml:id="echoid-s317" xml:space="preserve">O D : </s> <s xml:id="echoid-s318" xml:space="preserve">O d. <lb/></s> <s xml:id="echoid-s319" xml:space="preserve">then <lb/>A D: </s> <s xml:id="echoid-s320" xml:space="preserve">a d : </s> <s xml:id="echoid-s321" xml:space="preserve">: </s> <s xml:id="echoid-s322" xml:space="preserve">D c : </s> <s xml:id="echoid-s323" xml:space="preserve">d c. </s> <s xml:id="echoid-s324" xml:space="preserve"><lb/>altern. </s> <s xml:id="echoid-s325" xml:space="preserve"><lb/>A D : </s> <s xml:id="echoid-s326" xml:space="preserve">D C : </s> <s xml:id="echoid-s327" xml:space="preserve">: </s> <s xml:id="echoid-s328" xml:space="preserve">a d : </s> <s xml:id="echoid-s329" xml:space="preserve">d c.</s> <s xml:id="echoid-s330" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s331" xml:space="preserve">And conſequently the Sides A D, and D C of <lb/>the Figure A B C D, are Proportional to the <pb o="8" file="0026" n="26" rhead="An ESSAY"/> Sides a d and d c of the Figure a b c d. </s> <s xml:id="echoid-s332" xml:space="preserve">The <lb/>ſame may be demonſtrated of the other Sides; <lb/></s> <s xml:id="echoid-s333" xml:space="preserve">and therefore the Figures are ſimilar.</s> <s xml:id="echoid-s334" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s335" xml:space="preserve">Now to prove the other Part of the Theorem: <lb/></s> <s xml:id="echoid-s336" xml:space="preserve">If a perpendicular be ſuppoſed to be let fall from <lb/>the Eye upon the Plane of the Figure, and con-<lb/>tinued as is neceſſary; </s> <s xml:id="echoid-s337" xml:space="preserve">it is evident, that O D, <lb/>will be to O d, as this Perpendicular, which <lb/>meaſures the Diſtance from the Eye to the Plane <lb/>of the Figure, is to the Diſtance of the Eye <lb/>from the Perſpective Plane, which is meaſur’d <lb/>by the Part of the perpendicular, contain’d be-<lb/>tween the Eye and the perſpective Plane. </s> <s xml:id="echoid-s338" xml:space="preserve">Now <lb/>this before was manifeſt; </s> <s xml:id="echoid-s339" xml:space="preserve">viz. </s> <s xml:id="echoid-s340" xml:space="preserve">that <lb/>O D : </s> <s xml:id="echoid-s341" xml:space="preserve">O d : </s> <s xml:id="echoid-s342" xml:space="preserve">: </s> <s xml:id="echoid-s343" xml:space="preserve">A D : </s> <s xml:id="echoid-s344" xml:space="preserve">a d :</s> <s xml:id="echoid-s345" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s346" xml:space="preserve">Whence there is the ſame Proportion between <lb/>A d one of the Sides of the Figure, and A D its <lb/>Appearance, as the Theorem expreſſes. </s> <s xml:id="echoid-s347" xml:space="preserve">The <lb/>ſame may be demonſtrated of the other Sides of <lb/>the Figure. </s> <s xml:id="echoid-s348" xml:space="preserve">Which was to be demonſtrated.</s> <s xml:id="echoid-s349" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div36" type="section" level="1" n="15"> <head xml:id="echoid-head17" xml:space="preserve"><emph style="sc">Corollary</emph> I.</head> <p style="it"> <s xml:id="echoid-s350" xml:space="preserve">10. </s> <s xml:id="echoid-s351" xml:space="preserve">If from a Point in the Geometrical Plane, three <lb/>right Lines proceed, which are equal between them-<lb/>ſelves, and parallel to the perſpective Plane; </s> <s xml:id="echoid-s352" xml:space="preserve">the firſt <lb/>of which is in the Geometrical Plane, the ſecond ele-<lb/>vated Perpendicular to the firſt, and the third in-<lb/>clined to it; </s> <s xml:id="echoid-s353" xml:space="preserve">the Appearances of theſe three right <lb/>Lines are equal.</s> <s xml:id="echoid-s354" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s355" xml:space="preserve">This will appear clear enough in conſidering <lb/>the Lines as a Figure parallel to the perſpective <lb/>Plane; </s> <s xml:id="echoid-s356" xml:space="preserve">and ſo conſequently they will have the <lb/>ſame Proportion as their Appearances.</s> <s xml:id="echoid-s357" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s358" xml:space="preserve">Note, The firſt of the aforeſaid Lines is always <lb/>parallel to the baſe Line; </s> <s xml:id="echoid-s359" xml:space="preserve">and the ſecond, when <lb/>the perſpective Plane is perpendicular or up- <pb file="0027" n="27"/> <pb file="0027a" n="28"/> <anchor type="figure" xlink:label="fig-0027a-01a" xlink:href="fig-0027a-01"/> <anchor type="figure" xlink:label="fig-0027a-02a" xlink:href="fig-0027a-02"/> <anchor type="figure" xlink:label="fig-0027a-03a" xlink:href="fig-0027a-03"/> <anchor type="figure" xlink:label="fig-0027a-04a" xlink:href="fig-0027a-04"/> <pb file="0028" n="29"/> <pb o="9" file="0029" n="30" rhead="on PERSPECTIVE."/> right, is alſo perpendicular to the Geometrical <lb/>Plane, and the third is then in the Direction of the <lb/>firſt.</s> <s xml:id="echoid-s360" xml:space="preserve"/> </p> <div xml:id="echoid-div36" type="float" level="2" n="1"> <figure xlink:label="fig-0027a-01" xlink:href="fig-0027a-01a"> <caption xml:id="echoid-caption1" style="it" xml:space="preserve">fronting page 8<lb/>Plate 1.<lb/>Fig. 1.</caption> <variables xml:id="echoid-variables1" xml:space="preserve">C A D B e E</variables> </figure> <figure xlink:label="fig-0027a-02" xlink:href="fig-0027a-02a"> <caption xml:id="echoid-caption2" xml:space="preserve">Fig. 2.</caption> <variables xml:id="echoid-variables2" xml:space="preserve">M T O V R L A N D F S I H B G C</variables> </figure> <figure xlink:label="fig-0027a-03" xlink:href="fig-0027a-03a"> <caption xml:id="echoid-caption3" style="it" xml:space="preserve">Fig. 3.</caption> <variables xml:id="echoid-variables3" xml:space="preserve">O b B F a A G</variables> </figure> <figure xlink:label="fig-0027a-04" xlink:href="fig-0027a-04a"> <caption xml:id="echoid-caption4" style="it" xml:space="preserve">Fig. 4.</caption> <variables xml:id="echoid-variables4" xml:space="preserve">c C b d F B D a A G</variables> </figure> </div> </div> <div xml:id="echoid-div38" type="section" level="1" n="16"> <head xml:id="echoid-head18" xml:space="preserve"><emph style="sc">Corollary</emph> II.</head> <p style="it"> <s xml:id="echoid-s361" xml:space="preserve">11. </s> <s xml:id="echoid-s362" xml:space="preserve">If tworight Lines, equal between themſelves, and <lb/>parallel to the perſpective Planes, be equally diſtant <lb/>from the perſpective Plane, their Appearances will be <lb/>equal.</s> <s xml:id="echoid-s363" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s364" xml:space="preserve">For, becauſe they are in a Plane, parallel to <lb/>the perſpective Plane, they will have the ſame <lb/>Proportion to each other, as their Repreſentations.</s> <s xml:id="echoid-s365" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div39" type="section" level="1" n="17"> <head xml:id="echoid-head19" xml:space="preserve"><emph style="sc">Theorem</emph> III.</head> <p style="it"> <s xml:id="echoid-s366" xml:space="preserve">12. </s> <s xml:id="echoid-s367" xml:space="preserve">If a Line parallel to the Perſpective Plane, be <lb/>view’d by two Eyes, both being in a Plane, parallel <lb/>to the perſpective Plane, the Repreſentations of the <lb/>ſaid Line will be equal.</s> <s xml:id="echoid-s368" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s369" xml:space="preserve">If we ſuppoſe a Plane, parallel to the Per-<lb/>ſpective Plane, to paſs through the propoſed <lb/>Line, this Proportion will be had; </s> <s xml:id="echoid-s370" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/> viz. </s> <s xml:id="echoid-s371" xml:space="preserve">As the <anchor type="note" xlink:label="note-0029-01a" xlink:href="note-0029-01"/> Diſtance of the Eyes from this Plane, is to their <lb/>Diſtance from the Perſpective Plane, ſo is the <lb/>given Line to the Repreſentation thereof. </s> <s xml:id="echoid-s372" xml:space="preserve">But <lb/>the three firſt Terms of this Proportion are <lb/>the ſame for each of the Eyes, which are <lb/>in one and the ſame Plane parallel to the Per-<lb/>ſpective Plane: </s> <s xml:id="echoid-s373" xml:space="preserve">Therefore, the fourth Term of <lb/>the Proportion will likewiſe be the ſame in both <lb/>Caſes: </s> <s xml:id="echoid-s374" xml:space="preserve">Which was to be demonſtrated.</s> <s xml:id="echoid-s375" xml:space="preserve"/> </p> <div xml:id="echoid-div39" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0029-01" xlink:href="note-0029-01a" xml:space="preserve">9.</note> </div> </div> <div xml:id="echoid-div41" type="section" level="1" n="18"> <head xml:id="echoid-head20" xml:space="preserve"><emph style="sc">Theorem</emph> IV.</head> <p style="it"> <s xml:id="echoid-s376" xml:space="preserve">13. </s> <s xml:id="echoid-s377" xml:space="preserve">If a right Line, being continued, meets the per-<lb/>ſpective Piane in one Point, the Appearance thereof <lb/>will be a Part of the Line drawn from the ſaid Point <lb/>in the perſpective Plane, to another Point, whereat <pb o="10" file="0030" n="31" rhead="An ESSAY"/> a right Line drawn from the Eye parallel to the pro-<lb/>poſed Line, terminates.</s> <s xml:id="echoid-s378" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s379" xml:space="preserve">The Line C D being continued, will meet<unsure/> <lb/> <anchor type="note" xlink:label="note-0030-01a" xlink:href="note-0030-01"/> the perſpective Plane in the Point E. </s> <s xml:id="echoid-s380" xml:space="preserve">We are <lb/>to prove, that its Appearance is a Part of the <lb/>Line E H, drawn from the Point E, to the <lb/>Point H, whereat the Line O H proceeding <lb/>from the Eye parallel to the given Line C D, <lb/>terminates.</s> <s xml:id="echoid-s381" xml:space="preserve"/> </p> <div xml:id="echoid-div41" type="float" level="2" n="1"> <note position="left" xlink:label="note-0030-01" xlink:href="note-0030-01a" xml:space="preserve">Fig. 5.</note> </div> <p> <s xml:id="echoid-s382" xml:space="preserve">The Interſection of the perſpective Plane, <lb/>and the Plane O D C, is the Repreſentation of the <lb/>given Line. </s> <s xml:id="echoid-s383" xml:space="preserve">Now the Plane O D C, is a Part of the <lb/>Plane paſſing through the parallels O H and EC.</s> <s xml:id="echoid-s384" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s385" xml:space="preserve">Therefore, this Repreſentation is a Part of the <lb/>Interſection of the laſt mentioned Plane, and <lb/>perſpective Plane; </s> <s xml:id="echoid-s386" xml:space="preserve">which Interſection is E H.</s> <s xml:id="echoid-s387" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div43" type="section" level="1" n="19"> <head xml:id="echoid-head21" xml:space="preserve"><emph style="sc">Corollary</emph> I.</head> <p style="it"> <s xml:id="echoid-s388" xml:space="preserve">14. </s> <s xml:id="echoid-s389" xml:space="preserve">All Lines parallel between themſelves, and be-<lb/>ing produced, do fall upon the perſpective Plane, have <lb/>Repreſentations, which being produced, will all con-<lb/>cur in one Point.</s> <s xml:id="echoid-s390" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s391" xml:space="preserve">This is evident, becauſe but only one Line <lb/>O H can be drawn from the Eye O, to the per-<lb/>ſpective Plane, parallel to the ſaid Parallels, and <lb/>becauſe all their Repreſentations are Parts of <lb/>Lines concurring in the Point H.</s> <s xml:id="echoid-s392" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s393" xml:space="preserve">And this Point is called the accidental Point of the <lb/> <anchor type="note" xlink:label="note-0030-02a" xlink:href="note-0030-02"/> ſaid Parallels.</s> <s xml:id="echoid-s394" xml:space="preserve"/> </p> <div xml:id="echoid-div43" type="float" level="2" n="1"> <note position="left" xlink:label="note-0030-02" xlink:href="note-0030-02a" xml:space="preserve">Def. 17.</note> </div> </div> <div xml:id="echoid-div45" type="section" level="1" n="20"> <head xml:id="echoid-head22" xml:space="preserve"><emph style="sc">Corollary</emph> II.</head> <p style="it"> <s xml:id="echoid-s395" xml:space="preserve">15. </s> <s xml:id="echoid-s396" xml:space="preserve">Two or more parallel Lines, which being produ-<lb/>ced, do fall on the perſpective Plane, parallel to the Geo-<lb/>metrical Plane, have their accidental Point in the <lb/>Horizontal Line.</s> <s xml:id="echoid-s397" xml:space="preserve"/> </p> <pb o="11" file="0031" n="32" rhead="on PERSPECTIVE."/> <p> <s xml:id="echoid-s398" xml:space="preserve">For the Horizontal Plane, is parallel to the <lb/>Geometrical Plane.</s> <s xml:id="echoid-s399" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div46" type="section" level="1" n="21"> <head xml:id="echoid-head23" xml:space="preserve"><emph style="sc">Corollary</emph> III.</head> <p style="it"> <s xml:id="echoid-s400" xml:space="preserve">16. </s> <s xml:id="echoid-s401" xml:space="preserve">The Repreſentations of all Lines parallel to <lb/>the ſtation Line, concur in the Point of Sight.</s> <s xml:id="echoid-s402" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s403" xml:space="preserve">This follows, becauſe the principal Ray is <lb/>parallel to the ſaid Lines.</s> <s xml:id="echoid-s404" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div47" type="section" level="1" n="22"> <head xml:id="echoid-head24" xml:space="preserve"><emph style="sc">Corollary</emph> IV.</head> <p style="it"> <s xml:id="echoid-s405" xml:space="preserve">17. </s> <s xml:id="echoid-s406" xml:space="preserve">Two or more equal Lines being perpendicular, <lb/>or equally inclined the ſame Way, to the ſame Line pa-<lb/>rallel to the Station Line, have their Repreſentations <lb/>concurring in the principal Point.</s> <s xml:id="echoid-s407" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s408" xml:space="preserve">Becauſe all theſe Lines are parallel and equal, <lb/>the Line paſſing through their Vertices, is pa-<lb/>rallel to that paſſing through their Baſes, and <lb/>this being parallel to the Station Line, it fol-<lb/>lows, <anchor type="note" xlink:href="" symbol="*"/> that the Appearances of the ſaid equal <anchor type="note" xlink:label="note-0031-01a" xlink:href="note-0031-01"/> and parallel Lines concur in the principal <lb/>Point.</s> <s xml:id="echoid-s409" xml:space="preserve"/> </p> <div xml:id="echoid-div47" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0031-01" xlink:href="note-0031-01a" xml:space="preserve">16.</note> </div> </div> <div xml:id="echoid-div49" type="section" level="1" n="23"> <head xml:id="echoid-head25" xml:space="preserve"><emph style="sc">Theorem</emph> V.</head> <p style="it"> <s xml:id="echoid-s410" xml:space="preserve">18. </s> <s xml:id="echoid-s411" xml:space="preserve">The Appearance of an indefinite Line does not <lb/>alter, when the Eye moves in a Line parallel to a <lb/>propoſed Line.</s> <s xml:id="echoid-s412" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s413" xml:space="preserve">The Repreſentation of this Line, is the In-<lb/>terſection of the perſpective Plane, and a Plane <lb/>paſſing through the Eye and the ſaid Line. </s> <s xml:id="echoid-s414" xml:space="preserve">Now <lb/>the Eye remains in the ſame Plane, when it <lb/>moves in a Line parallel to the propoſed Line; <lb/></s> <s xml:id="echoid-s415" xml:space="preserve">and conſequently the Appearance of this laſt <lb/>Line, will not be changed by that Motion.</s> <s xml:id="echoid-s416" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s417" xml:space="preserve">Note, This Demonſtration doth not extend <lb/>to any particular Part of the given Line, but on-<lb/>ly to the Line in general.</s> <s xml:id="echoid-s418" xml:space="preserve"/> </p> <pb o="12" file="0032" n="33" rhead="An ESSAY"/> </div> <div xml:id="echoid-div50" type="section" level="1" n="24"> <head xml:id="echoid-head26" xml:space="preserve"><emph style="sc">Theorem</emph> VI.</head> <p style="it"> <s xml:id="echoid-s419" xml:space="preserve">19. </s> <s xml:id="echoid-s420" xml:space="preserve">Let A C be a Line inclined to the Geometrical <lb/>Plane, and O D another Line drawn parallel to <lb/>A C, from the Eye to the perſpective Plane. </s> <s xml:id="echoid-s421" xml:space="preserve">Now <lb/> <anchor type="note" xlink:label="note-0032-01a" xlink:href="note-0032-01"/> if B A be drawn in the Geometrical Plane, pa-<lb/>rallel to the baſe Line, and likewiſe D E, in the <lb/>perſpective Plane, parallel to the ſaid Line, ſo that <lb/>B A be to A C, as E d to D O. </s> <s xml:id="echoid-s422" xml:space="preserve">I ſay, the Ap-<lb/>pearance of the Line B C, paſſing through the Point <lb/>B, and the Extremity of the Line A C, being con-<lb/>tinued, will meet the Point E.</s> <s xml:id="echoid-s423" xml:space="preserve"/> </p> <div xml:id="echoid-div50" type="float" level="2" n="1"> <note position="left" xlink:label="note-0032-01" xlink:href="note-0032-01a" xml:space="preserve">Fig. 6.</note> </div> <p> <s xml:id="echoid-s424" xml:space="preserve">Now to prove this; </s> <s xml:id="echoid-s425" xml:space="preserve">it is evident, <anchor type="note" xlink:href="" symbol="*"/> that we <anchor type="note" xlink:label="note-0032-02a" xlink:href="note-0032-02"/> need but demonſtrate, that O E is parallel to <lb/>B C: </s> <s xml:id="echoid-s426" xml:space="preserve">And this may be done in the following <lb/>Manner:</s> <s xml:id="echoid-s427" xml:space="preserve"/> </p> <div xml:id="echoid-div51" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0032-02" xlink:href="note-0032-02a" xml:space="preserve">13.</note> </div> <p> <s xml:id="echoid-s428" xml:space="preserve">A B is parallel to E D, and A C to O D; <lb/></s> <s xml:id="echoid-s429" xml:space="preserve">whence the Angle (E D O) of the Triangle <lb/>O E D, is equal to the Angle (B A C) of the <lb/>Triangle A C B: </s> <s xml:id="echoid-s430" xml:space="preserve">And ſo theſe two Triangles <lb/>are ſimilar; </s> <s xml:id="echoid-s431" xml:space="preserve">becauſe they have alſo their Sides <lb/>Proportional. </s> <s xml:id="echoid-s432" xml:space="preserve">But ſince theſe two ſimilar Tri-<lb/>angles, have two of their Sides parallel, the <lb/>third B C is alſo parallel to O E; </s> <s xml:id="echoid-s433" xml:space="preserve">which was to be <lb/>demonſtrated.</s> <s xml:id="echoid-s434" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div53" type="section" level="1" n="25"> <head xml:id="echoid-head27" xml:space="preserve"><emph style="sc">Corollary</emph>.</head> <p> <s xml:id="echoid-s435" xml:space="preserve">20. </s> <s xml:id="echoid-s436" xml:space="preserve">If A B be made equal to A C, and E D to D O, <lb/>the Appearance of B C will paſs thro’ the Point E,</s> </p> </div> <div xml:id="echoid-div54" type="section" level="1" n="26"> <head xml:id="echoid-head28" xml:space="preserve">CHAP. III.</head> <p style="it"> <s xml:id="echoid-s437" xml:space="preserve">The Practice of Perſpective upon the Per-<lb/>ſpective Plane, ſuppoſed to be perpendicu-<lb/>lar, or upright.</s> <s xml:id="echoid-s438" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s439" xml:space="preserve">IN order to give a diſtinct Idea of the Theory, I <lb/>have hitherto conſider’d the Geometrical Plane, <lb/>as it were the Ground upon which the Spectator <pb o="13" file="0033" n="34" rhead="on PERSPECTIVE."/> and the Objects ſtand; </s> <s xml:id="echoid-s440" xml:space="preserve">and the Perſpective <lb/>Plane, as a Window between the Spectator and <lb/>the Objects, in which the Objects are requir’d <lb/>to be repreſented. </s> <s xml:id="echoid-s441" xml:space="preserve">But, in Practice, this Matter <lb/>muſt be quite otherwiſe conceiv’d; </s> <s xml:id="echoid-s442" xml:space="preserve">which I <lb/>ſhall now endeavour to explain as clear as poſ-<lb/>ſible.</s> <s xml:id="echoid-s443" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s444" xml:space="preserve">Suppoſe then, that a Painter has a mind to <lb/>draw upon his Perſpective Plane, or Picture, <lb/>(whoſe Bigneſs is as he thinks fit) a Proſpect of <lb/>a Country, wherein are Trees, Houſes, Rivers, & </s> <s xml:id="echoid-s445" xml:space="preserve">c. <lb/></s> <s xml:id="echoid-s446" xml:space="preserve">Now, from what has been ſaid, this Country <lb/>will be his Geometrical Plane; </s> <s xml:id="echoid-s447" xml:space="preserve">and he ought to <lb/>conſider his Perſpective Plane as a Window, up-<lb/>on which the Points thro’ which the Rays com-<lb/>ing from all the Points of the Objects towards <lb/>the Eye, muſt be found. </s> <s xml:id="echoid-s448" xml:space="preserve">But theſe Interſections <lb/>of the Rays and the Window cannot be deter-<lb/>min’d, unleſs by Lines being drawn in the Geo-<lb/>metrical Plane to the Baſe Line.</s> <s xml:id="echoid-s449" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s450" xml:space="preserve">Now, it is impoſſible for Painters to draw Lines <lb/>of this Nature on the Ground; </s> <s xml:id="echoid-s451" xml:space="preserve">wherefore they <lb/>uſe another more convenient Geometrical Plane <lb/>thus. </s> <s xml:id="echoid-s452" xml:space="preserve">At the Foot of their Perſpective Plane, <lb/>they place a Plane, upon which are drawn in <lb/>Minature the Baſes of Houſes and Trees, which <lb/>are in the Country to be repreſented; </s> <s xml:id="echoid-s453" xml:space="preserve">and the <lb/>Seats of the Points which, in the Objects, are <lb/>elevated above the Country; </s> <s xml:id="echoid-s454" xml:space="preserve">always obſerving, <lb/>that there be the ſame Diſpoſition between the <lb/>Objects and their different Parts, upon this new <lb/>Geometrical Plane, as the Objects truly have in <lb/>the Country to be repreſented.</s> <s xml:id="echoid-s455" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s456" xml:space="preserve">Now, to determine the Magnitude of the <lb/>Space the Figures muſt take up upon this Geo-<lb/>metrical Plane, a Painter muſt firſt chuſe the <lb/>Diſpoſition of his Eye in reſpect to the Perſpe- <pb o="14" file="0034" n="35" rhead="An ESSAY"/> ctive Plane; </s> <s xml:id="echoid-s457" xml:space="preserve">and then (from the Station Point, <lb/>thro’ the Extremities of the Perſpective Plane) <lb/>he muſt draw right Lines; </s> <s xml:id="echoid-s458" xml:space="preserve">which will limit the <lb/>Space wherein the Figures muſt be placed; </s> <s xml:id="echoid-s459" xml:space="preserve">ſince <lb/>the Rays of Figures, without thoſe Lines coming <lb/>towards the Eye, will not paſs thro’ the Perſpe-<lb/>ctive Plane.</s> <s xml:id="echoid-s460" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s461" xml:space="preserve">21. </s> <s xml:id="echoid-s462" xml:space="preserve">The Figures being thus drawn on the Geo-<lb/>metrical Plane, the next Thing is to find their Ap-<lb/>pearance upon the Perſpective Plane. </s> <s xml:id="echoid-s463" xml:space="preserve">Now, <lb/>theſe Figures are made up of either ſtraight <lb/>Lines, or crooked ones. </s> <s xml:id="echoid-s464" xml:space="preserve">To find the Repreſen-<lb/>tation of a ſtraight Line, its Extremes need on-<lb/>ly be ſought: </s> <s xml:id="echoid-s465" xml:space="preserve">And to have the Appearance of a <lb/>crooked Line, ſeveral Points thereof need only <lb/>be found. </s> <s xml:id="echoid-s466" xml:space="preserve">Since all this is equally applicable <lb/>to Figures, as well in the Geometrical Plane, <lb/>as thoſe above it; </s> <s xml:id="echoid-s467" xml:space="preserve">it follows, that the whole Bu-<lb/>ſineſs of Perſpective conſiſts in only finding the <lb/>Repreſentation of a Point.</s> <s xml:id="echoid-s468" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s469" xml:space="preserve">And to find this Repreſentation in the follow-<lb/>ing Problems, we only uſe certain Lines drawn <lb/>in the Geometrical and Horizontal Planes; <lb/></s> <s xml:id="echoid-s470" xml:space="preserve">which, by their Interſection with the Baſe and <lb/>Horizontal Lines, ſhew the manner of drawing <lb/>new Lines upon the Perſpective Plane, which <lb/>determine the propos’d Appearances. </s> <s xml:id="echoid-s471" xml:space="preserve">Now, it <lb/>is plain, that in finding the ſaid Interſections, it <lb/>is not neceſſary to place the Perſpective Plane <lb/>perpendicular to the Geometrical and Horizontal <lb/>Planes; </s> <s xml:id="echoid-s472" xml:space="preserve">which would render the Work extream-<lb/>ly laborious: </s> <s xml:id="echoid-s473" xml:space="preserve">Whence the Perſpective and Hori-<lb/>zontal Planes may be conſider’d as lying upon <lb/>the Geometrical Plane, and ſo coinciding there-<lb/>with.</s> <s xml:id="echoid-s474" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s475" xml:space="preserve">The Perſpective Plane may lye upon the Geo-<lb/>metrical Plane two ways; </s> <s xml:id="echoid-s476" xml:space="preserve">viz. </s> <s xml:id="echoid-s477" xml:space="preserve">Either upon the <lb/>Face reſpecting the Objects, or upon that next <pb o="15" file="0035" n="36" rhead="on PERSPECTIVE."/> to the Eye. </s> <s xml:id="echoid-s478" xml:space="preserve">Now, as in the latter Situation, <lb/>Repreſentatious are drawn upon the Face of the <lb/>Perſpective Plane next to the Object, the Per-<lb/>ſpective Plane lying down upon its other Face; <lb/></s> <s xml:id="echoid-s479" xml:space="preserve">what ought to be on the Right Hand, appears on <lb/>the Left; </s> <s xml:id="echoid-s480" xml:space="preserve">and that on the Left, appears on the <lb/>Right; </s> <s xml:id="echoid-s481" xml:space="preserve">producing exactly the ſame Effect, as <lb/>looking thro’ the Back-ſide of a Paper, at a Pi-<lb/>cture drawn thereon.</s> <s xml:id="echoid-s482" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s483" xml:space="preserve">Yet, notwithſtanding this Deficiency, we pre-<lb/>fer the latter way of the Perſpective Plane’s ly-<lb/>ing down to the former, for the following Rea-<lb/>ſons.</s> <s xml:id="echoid-s484" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s485" xml:space="preserve">1. </s> <s xml:id="echoid-s486" xml:space="preserve">When the Perſpective Plane lies down in <lb/>the former manner, it lies upon the Part of the <lb/>Geometrical Plane wherein Figures have been <lb/>drawn; </s> <s xml:id="echoid-s487" xml:space="preserve">which, together with the new Lines that <lb/>muſt be drawn, cauſes a very great Confuſion, <lb/>and always obliges one to copy his Work. </s> <s xml:id="echoid-s488" xml:space="preserve">An <lb/>Inconveniency which the latter Method is ſel-<lb/>dom ſubject to.</s> <s xml:id="echoid-s489" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s490" xml:space="preserve">2. </s> <s xml:id="echoid-s491" xml:space="preserve">We work with much more Eaſe in the <lb/>manner I have choſen.</s> <s xml:id="echoid-s492" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s493" xml:space="preserve">Finally, The Default we have obſerv’d, may <lb/>ſeveral ways be remedied. </s> <s xml:id="echoid-s494" xml:space="preserve">For, in drawing up-<lb/>on the Geometrical Plane, we need but place <lb/>that on the Right Hand which we have a mind <lb/>ſhould appear on the Left; </s> <s xml:id="echoid-s495" xml:space="preserve">or if the Geometri-<lb/>cal Plane be drawn upon Paper, it may be oil’d, <lb/>or dipp’d in Varniſh, which will render it tran-<lb/>ſparent; </s> <s xml:id="echoid-s496" xml:space="preserve">and then the Back-ſide of the Paper <lb/>may be thrown into Perſpective.</s> <s xml:id="echoid-s497" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s498" xml:space="preserve">If all this be not found convenient, the ſaid <lb/>Default may be eaſily corrected geometrically, <lb/>in copying the Work after the Drawings are fi-<lb/>niſhed. </s> <s xml:id="echoid-s499" xml:space="preserve">And this may be yet eaſier done, if <lb/>the Figures are expoſed before a Looking-glaſs;</s> <s xml:id="echoid-s500" xml:space="preserve"> <pb o="16" file="0036" n="37" rhead="An ESSAY"/> for then, what is on the Right, will appear on <lb/>the Left.</s> <s xml:id="echoid-s501" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s502" xml:space="preserve">Therefore, I lay my Perſpective Plane upon <lb/>the Geometrical Plane; </s> <s xml:id="echoid-s503" xml:space="preserve">ſo that it be between <lb/>the Horizontal Plane, and the Figures requir’d <lb/>to be thrown into Perſpective.</s> <s xml:id="echoid-s504" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div55" type="section" level="1" n="27"> <head xml:id="echoid-head29" xml:space="preserve"><emph style="sc">Problem</emph> I.</head> <p style="it"> <s xml:id="echoid-s505" xml:space="preserve">22. </s> <s xml:id="echoid-s506" xml:space="preserve">To find the Appearance of a Point, which is in <lb/>the Geometrical Plane.</s> <s xml:id="echoid-s507" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s508" xml:space="preserve">Let Z be the Geometrical Plane, I E the Baſe <lb/> <anchor type="note" xlink:label="note-0036-01a" xlink:href="note-0036-01"/> Line, D V the Horizontal Line, V the Point <lb/>of Sight, D one of the Points of Diſtance, and <lb/>A the given Point.</s> <s xml:id="echoid-s509" xml:space="preserve"/> </p> <div xml:id="echoid-div55" type="float" level="2" n="1"> <note position="left" xlink:label="note-0036-01" xlink:href="note-0036-01a" xml:space="preserve">Fig. 7.</note> </div> </div> <div xml:id="echoid-div57" type="section" level="1" n="28"> <head xml:id="echoid-head30" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s510" xml:space="preserve">From the Point A, let fall the Perpendicular <lb/>A B upon the Baſe Line; </s> <s xml:id="echoid-s511" xml:space="preserve">and from the Point of <lb/>Concurrence B, draw the Line B V to the Point <lb/>of Sight; </s> <s xml:id="echoid-s512" xml:space="preserve">then aſſume B E in the Baſe Line <lb/>equal to B A, and from the Point E draw the <lb/>Line E D to the Point of Diſtance D: </s> <s xml:id="echoid-s513" xml:space="preserve">And the <lb/>Point (a), the Interſection of B V and E D, is <lb/>the Repreſentation ſought.</s> <s xml:id="echoid-s514" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div58" type="section" level="1" n="29"> <head xml:id="echoid-head31" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s515" xml:space="preserve">23. </s> <s xml:id="echoid-s516" xml:space="preserve">The Appearance of the Line A B, is <anchor type="note" xlink:href="" symbol="*"/> a Part of the Line B V. </s> <s xml:id="echoid-s517" xml:space="preserve">Now, if we conceive a Line <lb/> <anchor type="note" xlink:label="note-0036-02a" xlink:href="note-0036-02"/> iſſuing from the Eye towards the Point D, and <lb/>another from the Point A towards the Point E; <lb/></s> <s xml:id="echoid-s518" xml:space="preserve">theſe two Lines will be parallel, becauſe they <lb/>are in parallel Planes, aud each make half a <lb/>right Angle with the Perſpective Plane; </s> <s xml:id="echoid-s519" xml:space="preserve">whence <pb file="0037" n="38"/> <pb file="0037a" n="39"/> <anchor type="figure" xlink:label="fig-0037a-01a" xlink:href="fig-0037a-01"/> <anchor type="figure" xlink:label="fig-0037a-02a" xlink:href="fig-0037a-02"/> <anchor type="figure" xlink:label="fig-0037a-03a" xlink:href="fig-0037a-03"/> <pb file="0038" n="40"/> <pb o="17" file="0039" n="41" rhead="on PERSPECTIVE."/> the Appearance of the Line A E, is <anchor type="note" xlink:href="" symbol="*"/> a Part of <anchor type="note" xlink:label="note-0039-01a" xlink:href="note-0039-01"/> the Line E D. </s> <s xml:id="echoid-s520" xml:space="preserve">Now, ſince the Point A is in <lb/>the two Lines A B, A E; </s> <s xml:id="echoid-s521" xml:space="preserve">the Appearance of <lb/>the ſaid Point will likewiſe be in the Appear-<lb/>ances of the aforeſaid two Lines, and conſequent-<lb/>ly is in the Point a, the common Section of B V <lb/>and E D.</s> <s xml:id="echoid-s522" xml:space="preserve"/> </p> <div xml:id="echoid-div58" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0036-02" xlink:href="note-0036-02a" xml:space="preserve">16.</note> <figure xlink:label="fig-0037a-01" xlink:href="fig-0037a-01a"> <caption xml:id="echoid-caption5" style="it" xml:space="preserve">Plate 2.<lb/>page 16.<lb/>Fig. 5.</caption> <variables xml:id="echoid-variables5" xml:space="preserve">@ O H F c d E D C G</variables> </figure> <figure xlink:label="fig-0037a-02" xlink:href="fig-0037a-02a"> <caption xml:id="echoid-caption6" style="it" xml:space="preserve">Fig. 6.</caption> <variables xml:id="echoid-variables6" xml:space="preserve">E D O @ c F a b A C B G</variables> </figure> <figure xlink:label="fig-0037a-03" xlink:href="fig-0037a-03a"> <caption xml:id="echoid-caption7" style="it" xml:space="preserve">Fig. 7.</caption> <variables xml:id="echoid-variables7" xml:space="preserve">D F H V C X a I B G E Z A</variables> </figure> <note symbol="*" position="right" xlink:label="note-0039-01" xlink:href="note-0039-01a" xml:space="preserve">13.</note> </div> </div> <div xml:id="echoid-div60" type="section" level="1" n="30"> <head xml:id="echoid-head32" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s523" xml:space="preserve">24. </s> <s xml:id="echoid-s524" xml:space="preserve">If the Diſtance of the Eye be ſo great, that <lb/>one of the Points of Diſtance cannot be deno-<lb/>ted upon the horizontal Line; </s> <s xml:id="echoid-s525" xml:space="preserve">another Point, F, <lb/>muſt be uſed, diſtant from the Point of Sight <lb/>by about one third, or fourth Part of the Di-<lb/>ſtance of the Eye. </s> <s xml:id="echoid-s526" xml:space="preserve">But then, a correſpondent <lb/>Part of the Perpendicular A B muſt be likewiſe <lb/>taken, and laid off from B to G, in the Baſe <lb/>Line.</s> <s xml:id="echoid-s527" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s528" xml:space="preserve">25. </s> <s xml:id="echoid-s529" xml:space="preserve">And in this manner may the Repreſentation <lb/>of a very diſtant Point be found, if its Diſtance <lb/>from the Perſpective Plane be known, together <lb/>with the Place wherein a Perpendicular drawn <lb/>from that Point cuts the Baſe Line. </s> <s xml:id="echoid-s530" xml:space="preserve">For, having <lb/>firſt drawn a Line, as B V, from the ſaid Point <lb/>of Concurrence to the Point of Sight, then B E <lb/>muſt be aſſum’d in the Baſe Line; </s> <s xml:id="echoid-s531" xml:space="preserve">for Example, <lb/>equal to the tenth Part of the Diſtance of the <lb/>Point whoſe Repreſentation is ſought; </s> <s xml:id="echoid-s532" xml:space="preserve">and V H <lb/>in the Horizontal Line, likewiſe equal to the <lb/>tenth Part of the Eye’s Diſtance. </s> <s xml:id="echoid-s533" xml:space="preserve">Then C, the <lb/>Interſection of B V and E H, will be the Appear-<lb/>ance ſought.</s> <s xml:id="echoid-s534" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s535" xml:space="preserve">Note, By this Method may be found the Deep-<lb/>nings in Pictures.</s> <s xml:id="echoid-s536" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s537" xml:space="preserve">The Appearance of the Point A may yet be <lb/>otherwiſe found, without drawing the Line B V <lb/>from the Point A, in taking B I equal to B A, <pb o="18" file="0040" n="42" rhead="An ESSAY"/> and drawing a Line from the Point I to the <lb/>other Point of Diſtance; </s> <s xml:id="echoid-s538" xml:space="preserve">which, by its Inter-<lb/>ſection with E D, will give the Appearance of <lb/>the Point A.</s> <s xml:id="echoid-s539" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div61" type="section" level="1" n="31"> <head xml:id="echoid-head33" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p> <s xml:id="echoid-s540" xml:space="preserve">26. </s> <s xml:id="echoid-s541" xml:space="preserve">γ is the Horizontal Plane, X the Perſpe-<lb/>ctive Plane, Z the Geometrical Plane, O the Eye, <lb/>D C the Horizontal Line, B E the Baſe Line, and <lb/>A the given Point.</s> <s xml:id="echoid-s542" xml:space="preserve"/> </p> <note position="left" xml:space="preserve">Fig. 8.</note> </div> <div xml:id="echoid-div62" type="section" level="1" n="32"> <head xml:id="echoid-head34" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s543" xml:space="preserve">Draw a Line from the Point A, to the Eye O, cut-<lb/>ting the Baſe Line in the Point B, and the Horizon-<lb/>tal Line in the Point C: </s> <s xml:id="echoid-s544" xml:space="preserve">Then aſſume B E in the <lb/>Baſe Line, equal to B A; </s> <s xml:id="echoid-s545" xml:space="preserve">and C D in the Hori-<lb/>zontal Line, equal to C O; </s> <s xml:id="echoid-s546" xml:space="preserve">and join the Points <lb/>E and D, by a Line cutting the Line A O in <lb/>the Point a; </s> <s xml:id="echoid-s547" xml:space="preserve">which will be the Appearance <lb/>ſought.</s> <s xml:id="echoid-s548" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div63" type="section" level="1" n="33"> <head xml:id="echoid-head35" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s549" xml:space="preserve">27. </s> <s xml:id="echoid-s550" xml:space="preserve">The Triangle O D C in the Horizontal <lb/>Plane, is ſimilar to the Triangle A B E in the <lb/>Geometrical Plane; </s> <s xml:id="echoid-s551" xml:space="preserve">and conſequently A B is pa-<lb/>rallel to O C, and A E to O D. </s> <s xml:id="echoid-s552" xml:space="preserve">But the Appear-<lb/>ance of A muſt <anchor type="note" xlink:href="" symbol="*"/> be in the Lines B C, and E D;</s> <s xml:id="echoid-s553" xml:space="preserve"> <anchor type="note" xlink:label="note-0040-02a" xlink:href="note-0040-02"/> and therefore it will be in a, their Interſection.</s> <s xml:id="echoid-s554" xml:space="preserve"/> </p> <div xml:id="echoid-div63" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0040-02" xlink:href="note-0040-02a" xml:space="preserve"><unsure/>13.</note> </div> </div> <div xml:id="echoid-div65" type="section" level="1" n="34"> <head xml:id="echoid-head36" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s555" xml:space="preserve">28. </s> <s xml:id="echoid-s556" xml:space="preserve">If the Place wherein the Eye ought to be in <lb/>the Horizontal Plane be not known, but the Point <lb/>of Sight is; </s> <s xml:id="echoid-s557" xml:space="preserve">then, to find the Place of the Eye, <lb/>a Perpendicular muſt be rais’d from the Point of <pb o="19" file="0041" n="43" rhead="on PERSPECTIVE."/> Sight to the Horizontal Line, equal in Length <lb/>to the principal Ray; </s> <s xml:id="echoid-s558" xml:space="preserve">and the Extremity of this <lb/>Perpendicular will be the Point ſought.</s> <s xml:id="echoid-s559" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s560" xml:space="preserve">If nothing is determin’d, the Place of the Eye <lb/>may be taken at pleaſure in the Horizontal <lb/>Plane.</s> <s xml:id="echoid-s561" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div66" type="section" level="1" n="35"> <head xml:id="echoid-head37" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p> <s xml:id="echoid-s562" xml:space="preserve">29. </s> <s xml:id="echoid-s563" xml:space="preserve">The ſame Things being given as in the <lb/>precedent Method, about the Eye O, as a Cen-<lb/>ter; </s> <s xml:id="echoid-s564" xml:space="preserve">deſcribe the Arc of a Circle I H, touching <lb/> <anchor type="note" xlink:label="note-0041-01a" xlink:href="note-0041-01"/> the Horizontal Line.</s> <s xml:id="echoid-s565" xml:space="preserve"/> </p> <div xml:id="echoid-div66" type="float" level="2" n="1"> <note position="right" xlink:label="note-0041-01" xlink:href="note-0041-01a" xml:space="preserve">Fig. 9.</note> </div> </div> <div xml:id="echoid-div68" type="section" level="1" n="36"> <head xml:id="echoid-head38" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s566" xml:space="preserve">About the given Point A, as a Center, de-<lb/>ſcribe the Arc of a Circle L C, touching the Baſe <lb/>Line: </s> <s xml:id="echoid-s567" xml:space="preserve">Then draw two Lines, C H and L I, <lb/>touching the two Circles L C and H I; </s> <s xml:id="echoid-s568" xml:space="preserve">and the <lb/>Point a, the Interſection of the ſaid two Lines, <lb/>will be the Appearance ſought.</s> <s xml:id="echoid-s569" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div69" type="section" level="1" n="37"> <head xml:id="echoid-head39" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s570" xml:space="preserve">30. </s> <s xml:id="echoid-s571" xml:space="preserve">To demonſtrate this, draw the Line A B <lb/>perpendicular to the Baſe Line; </s> <s xml:id="echoid-s572" xml:space="preserve">O V perpendi-<lb/>cular to the Horizontal Line; </s> <s xml:id="echoid-s573" xml:space="preserve">and A C, O H <lb/>perpendicular to the Tangent H C. </s> <s xml:id="echoid-s574" xml:space="preserve">All theſe <lb/>Perpendiculars will cut the Lines to which they <lb/>are perpendicular, in the Points wherein theſe <lb/>laſt touch the Circle L B C, or H V I. </s> <s xml:id="echoid-s575" xml:space="preserve">Like-<lb/>wiſe draw the Line A E from the given Point <lb/>A, to the Point E, wherein the Line H C cuts <lb/>the Baſe Line. </s> <s xml:id="echoid-s576" xml:space="preserve">Finally, draw O D, from the <lb/>Eye O to the Point D, wherein the ſaid Line <lb/>HC cuts the Horizontal Line.</s> <s xml:id="echoid-s577" xml:space="preserve"/> </p> <pb o="20" file="0042" n="44" rhead="An ESSAY"/> <p> <s xml:id="echoid-s578" xml:space="preserve">Now, it is evident, <anchor type="note" xlink:href="" symbol="*"/> that to prove the Ap- <anchor type="note" xlink:label="note-0042-01a" xlink:href="note-0042-01"/> pearance of A is in the Line C H, we need but <lb/>demonſtrate that O D is parallel to A E; </s> <s xml:id="echoid-s579" xml:space="preserve">which <lb/>may be done thus:</s> <s xml:id="echoid-s580" xml:space="preserve"/> </p> <div xml:id="echoid-div69" type="float" level="2" n="1"> <note symbol="@" position="left" xlink:label="note-0042-01" xlink:href="note-0042-01a" xml:space="preserve">27.</note> </div> <p> <s xml:id="echoid-s581" xml:space="preserve">Becauſe the Triangles O G V, and A B F, are <lb/>ſimilar. <lb/></s> <s xml:id="echoid-s582" xml:space="preserve">A F: </s> <s xml:id="echoid-s583" xml:space="preserve">A B:</s> <s xml:id="echoid-s584" xml:space="preserve">: O G: </s> <s xml:id="echoid-s585" xml:space="preserve">O V: </s> <s xml:id="echoid-s586" xml:space="preserve"><lb/>altern. </s> <s xml:id="echoid-s587" xml:space="preserve"><lb/>A F: </s> <s xml:id="echoid-s588" xml:space="preserve">O G:</s> <s xml:id="echoid-s589" xml:space="preserve">: A B: </s> <s xml:id="echoid-s590" xml:space="preserve">O V: </s> <s xml:id="echoid-s591" xml:space="preserve"><lb/>Divid. </s> <s xml:id="echoid-s592" xml:space="preserve">and altern. </s> <s xml:id="echoid-s593" xml:space="preserve">the firſt <lb/>Proportion. </s> <s xml:id="echoid-s594" xml:space="preserve"><lb/>AF—AB (=CF): </s> <s xml:id="echoid-s595" xml:space="preserve">O G—O V=HG:</s> <s xml:id="echoid-s596" xml:space="preserve">:AB:</s> <s xml:id="echoid-s597" xml:space="preserve">OV.</s> <s xml:id="echoid-s598" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s599" xml:space="preserve">But becauſe the Triangles E C F, H G D are <lb/>ſimilar. <lb/></s> <s xml:id="echoid-s600" xml:space="preserve">C F: </s> <s xml:id="echoid-s601" xml:space="preserve">H G :</s> <s xml:id="echoid-s602" xml:space="preserve">: E F : </s> <s xml:id="echoid-s603" xml:space="preserve">G D.</s> <s xml:id="echoid-s604" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s605" xml:space="preserve">Now, by obſerving the two laſt Proportions of <lb/>the other two Triangles, <lb/>E F: </s> <s xml:id="echoid-s606" xml:space="preserve">G D:</s> <s xml:id="echoid-s607" xml:space="preserve">: A F: </s> <s xml:id="echoid-s608" xml:space="preserve">O G, <lb/>And the Angle A F E, being equal to the Angle <lb/>O G D, the Triangles A E F and O D G are <lb/>ſimilar; </s> <s xml:id="echoid-s609" xml:space="preserve">and therefore A E is parallel to O D: <lb/></s> <s xml:id="echoid-s610" xml:space="preserve">Which was to be demonſtrated.</s> <s xml:id="echoid-s611" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s612" xml:space="preserve">After the ſame manner we prove, that the <lb/>Appearance of the Point A is in the Line L I, <lb/>and conſequently is in the Interſection of this <lb/>Line and HC.</s> <s xml:id="echoid-s613" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div71" type="section" level="1" n="38"> <head xml:id="echoid-head40" xml:space="preserve"><emph style="sc">Remark</emph>.</head> <p> <s xml:id="echoid-s614" xml:space="preserve">Altho’ this Method appears more difficult than <lb/>the precedent one, as to the Geometrical Conſi-<lb/>deration thereof, yet the Operation is eaſier, if <lb/>the Points are not too far diſtant from the Baſe <lb/>Line: </s> <s xml:id="echoid-s615" xml:space="preserve">For Lines may well enough be drawn by <lb/>Gueſs, or Sight only, to touch Circles, and Cir-<lb/>cles to touch Lines.</s> <s xml:id="echoid-s616" xml:space="preserve"/> </p> <pb file="0043" n="45"/> <pb file="0043a" n="46"/> <figure> <caption xml:id="echoid-caption8" style="it" xml:space="preserve">Plate. 3.<lb/>page 20<lb/>Fig. 8.</caption> <variables xml:id="echoid-variables8" xml:space="preserve">O Y D C X æ B E Z A</variables> </figure> <figure> <caption xml:id="echoid-caption9" style="it" xml:space="preserve">Fig. 9.</caption> <variables xml:id="echoid-variables9" xml:space="preserve">O I Y H G D V X a B E F C Z L A</variables> </figure> <pb file="0044" n="47"/> <pb o="21" file="0045" n="48" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div72" type="section" level="1" n="39"> <head xml:id="echoid-head41" xml:space="preserve"><emph style="sc">Method</emph>. IV.</head> <p> <s xml:id="echoid-s617" xml:space="preserve">31. </s> <s xml:id="echoid-s618" xml:space="preserve">Draw the Line F O G through the Eye O, <lb/>parallel to the Baſe Line, then aſſume F O in <lb/> <anchor type="note" xlink:label="note-0045-01a" xlink:href="note-0045-01"/> this Line, equal to the Height of the Eye, and <lb/>O G equal to the Length of the principal Ray. <lb/></s> <s xml:id="echoid-s619" xml:space="preserve">A is the given Point.</s> <s xml:id="echoid-s620" xml:space="preserve"/> </p> <div xml:id="echoid-div72" type="float" level="2" n="1"> <note position="right" xlink:label="note-0045-01" xlink:href="note-0045-01a" xml:space="preserve">Fig. 10.</note> </div> </div> <div xml:id="echoid-div74" type="section" level="1" n="40"> <head xml:id="echoid-head42" xml:space="preserve"><emph style="sc">Operation</emph>, <lb/>Without Compaſſes.</head> <p> <s xml:id="echoid-s621" xml:space="preserve">From the given Point A, draw the Lines AO, <lb/>A F, to the Points O and F, and from the <lb/>Point E, wherein A F cuts the Baſe Line, draw <lb/>the Line E G to the Point G; </s> <s xml:id="echoid-s622" xml:space="preserve">then the Point a, <lb/>the Interſection of A O, and E G, is the Repre-<lb/>ſentation ſought.</s> <s xml:id="echoid-s623" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div75" type="section" level="1" n="41"> <head xml:id="echoid-head43" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s624" xml:space="preserve">Let fall the perpendicular G M from the Point <lb/>G, upon the Baſe Line, and through the Eye O, <lb/>draw the Line O D to the Point D, the Inter-<lb/>ſection of the Horizontal Line, and the Line G E.</s> <s xml:id="echoid-s625" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s626" xml:space="preserve">Then becauſe the Triangles G D L, G E M are <lb/>fimilar, <lb/>G D: </s> <s xml:id="echoid-s627" xml:space="preserve">G E:</s> <s xml:id="echoid-s628" xml:space="preserve">: G L: </s> <s xml:id="echoid-s629" xml:space="preserve">G M. <lb/></s> <s xml:id="echoid-s630" xml:space="preserve">But G O is equal to G L, and G F to L M. </s> <s xml:id="echoid-s631" xml:space="preserve"><lb/>whence <lb/>G D: </s> <s xml:id="echoid-s632" xml:space="preserve">G E:</s> <s xml:id="echoid-s633" xml:space="preserve">: G O: </s> <s xml:id="echoid-s634" xml:space="preserve">G F.</s> <s xml:id="echoid-s635" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s636" xml:space="preserve">And conſequenely the Triangles G O D: </s> <s xml:id="echoid-s637" xml:space="preserve">and <lb/>G F E, are ſimilar, and the Lines O D, and <lb/>A E F, are parallel between themſelves; </s> <s xml:id="echoid-s638" xml:space="preserve">and <lb/>therefore <anchor type="note" xlink:href="" symbol="*"/> the Appearance of A E, is a Part <anchor type="note" xlink:label="note-0045-02a" xlink:href="note-0045-02"/> of the Line E D G. </s> <s xml:id="echoid-s639" xml:space="preserve">It has alſo been prov’d<anchor type="note" xlink:href="" symbol="*"/>, <anchor type="note" xlink:label="note-0045-03a" xlink:href="note-0045-03"/> that the Repreſentation of the Point A, is in the <lb/>Line A O; </s> <s xml:id="echoid-s640" xml:space="preserve">therefore i@ is in a the Interſection <pb o="22" file="0046" n="49" rhead="An ESSAY"/> of this laſt Line, and the Line E D G, which was <lb/>to be demonſtrated.</s> <s xml:id="echoid-s641" xml:space="preserve"/> </p> <div xml:id="echoid-div75" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0045-02" xlink:href="note-0045-02a" xml:space="preserve">13</note> <note symbol="*" position="right" xlink:label="note-0045-03" xlink:href="note-0045-03a" xml:space="preserve">27.</note> </div> </div> <div xml:id="echoid-div77" type="section" level="1" n="42"> <head xml:id="echoid-head44" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s642" xml:space="preserve">33. </s> <s xml:id="echoid-s643" xml:space="preserve">By this Demonſtration it appears, that there <lb/>is no Neceſſity in taking G O exactly equal to <lb/>the Eye’s Diſtance, and O F equal to its Height: <lb/></s> <s xml:id="echoid-s644" xml:space="preserve">But it is ſufficient if they have the ſame Propor-<lb/>tion, as the aforeſaid Diſtance has to the Height. </s> <s xml:id="echoid-s645" xml:space="preserve"><lb/>Likewiſe there is no Neceſſity in aſſuming the <lb/>Points G and F, in a Line parallel to the Baſe <lb/>Line; </s> <s xml:id="echoid-s646" xml:space="preserve">for any other Line paſſing through the <lb/>Eye O may be uſed at Pleaſure. </s> <s xml:id="echoid-s647" xml:space="preserve">For Example, <lb/>let g O f be a Line any how drawn through the <lb/>Eye O, and take the Point g at Pleaſure in this <lb/>Line, through which draw alſo the Line g N I <lb/>at Pleaſure, cutting the Horizontal Line in N, <lb/>and the Baſe Line in I; </s> <s xml:id="echoid-s648" xml:space="preserve">and draw the Line O N, <lb/>and through the Point I, draw the Line I f pa-<lb/>rallelthereto, cutting the Line g O f in f.</s> <s xml:id="echoid-s649" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s650" xml:space="preserve">This being done, the Points g and f may be <lb/>uſed inſtead of G and F: </s> <s xml:id="echoid-s651" xml:space="preserve">for among all the <lb/>Lines that can be drawn (as G N I) it is mani-<lb/>ſeſt, that g N will always be to g I:</s> <s xml:id="echoid-s652" xml:space="preserve">: g O: </s> <s xml:id="echoid-s653" xml:space="preserve">g f, <lb/>which is ſufficient for the Demonſtration.</s> <s xml:id="echoid-s654" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s655" xml:space="preserve">If the Point f be firſt determin’d, the Point g <lb/>muſt be found by an Operation quite contrary <lb/>to that we have laid down.</s> <s xml:id="echoid-s656" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s657" xml:space="preserve">34. </s> <s xml:id="echoid-s658" xml:space="preserve">When nothing is determinate, we <lb/>may (a Baſe Line being firſt drawn) take at <lb/>Pleaſure, in another Line any how drawn, <lb/>the three Points g O f; </s> <s xml:id="echoid-s659" xml:space="preserve">ſo that in this Caſe, there <lb/>is no Manner of Neceſſity to uſe Compaſſes, in <lb/>throwing any Figure whatſoever, which is on <lb/>the Geometrical Plane, into Perſpective. </s> <s xml:id="echoid-s660" xml:space="preserve">But <lb/>if after having thus work’d, the Point of Sight, <lb/>Height and Diſtance of the Eye be requir’d, the <pb o="23" file="0047" n="50" rhead="on PERSPECTIVE."/> Perpendiculars f P, O H, muſt be let fall from <lb/>the Points f and O, on the Baſe Line, and the <lb/>Line P g drawn; </s> <s xml:id="echoid-s661" xml:space="preserve">then the Point V, wherein it <lb/>cuts the Perpendicular O H, is the Point of Sight <lb/>ſought, and the Parts O V, and V H determine <lb/>the Height and Diſtance of the Eye.</s> <s xml:id="echoid-s662" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div78" type="section" level="1" n="43"> <head xml:id="echoid-head45" xml:space="preserve"><emph style="sc">Method</emph> V.</head> <p style="it"> <s xml:id="echoid-s663" xml:space="preserve">35. </s> <s xml:id="echoid-s664" xml:space="preserve">When the Appearance of a Point is known,</s> </p> <p> <s xml:id="echoid-s665" xml:space="preserve">Let A be a Point in the Geometrical Plane, <lb/> <anchor type="note" xlink:label="note-0047-01a" xlink:href="note-0047-01"/> and a its Repreſentation in the perſpective Plane, <lb/>it is requir’d to find the Appearance of the <lb/>Point B.</s> <s xml:id="echoid-s666" xml:space="preserve"/> </p> <div xml:id="echoid-div78" type="float" level="2" n="1"> <note position="right" xlink:label="note-0047-01" xlink:href="note-0047-01a" xml:space="preserve">Fig. 11.</note> </div> </div> <div xml:id="echoid-div80" type="section" level="1" n="44"> <head xml:id="echoid-head46" xml:space="preserve"><emph style="sc">Operation</emph>, <lb/>Without Compaſſes.</head> <p> <s xml:id="echoid-s667" xml:space="preserve">Draw a Line from the Point B to the Eye O, <lb/>and another from the Point E, wherein the <lb/>ſaid Line continued, cuts the Baſe Line, to the <lb/>Point A; </s> <s xml:id="echoid-s668" xml:space="preserve">then draw the Line E a, and where <lb/>it cuts B O, is the Point b ſought.</s> <s xml:id="echoid-s669" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div81" type="section" level="1" n="45"> <head xml:id="echoid-head47" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s670" xml:space="preserve">The Point E is its own Repreſentation; </s> <s xml:id="echoid-s671" xml:space="preserve">and <lb/>becauſe the Point a is the Repreſentation of A, <lb/>the Line E a is that of E A. </s> <s xml:id="echoid-s672" xml:space="preserve">Now ſince the <lb/>Point B is in the Line E A, the Appearance of <lb/>this Point will be likewiſe in E a, as alſo <anchor type="note" xlink:href="" symbol="*"/> in <anchor type="note" xlink:label="note-0047-02a" xlink:href="note-0047-02"/> B O; </s> <s xml:id="echoid-s673" xml:space="preserve">therefore it is in b the Interſection of the <lb/>Lines E a, and B O.</s> <s xml:id="echoid-s674" xml:space="preserve"/> </p> <div xml:id="echoid-div81" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0047-02" xlink:href="note-0047-02a" xml:space="preserve">27.</note> </div> </div> <div xml:id="echoid-div83" type="section" level="1" n="46"> <head xml:id="echoid-head48" xml:space="preserve"><emph style="sc">Remark</emph>.</head> <p> <s xml:id="echoid-s675" xml:space="preserve">37. </s> <s xml:id="echoid-s676" xml:space="preserve">If the Point A be in the Line B O, or <lb/>the Line B A be parallel, or a very little inclined <lb/>to the Baſe Line, we cannot then uſe this Me- <pb o="24" file="0048" n="51" rhead="An ESSAY"/> thod, unleſs by Means of the Point A, the Ap-<lb/>pearance of ſome other Point taken at Pleaſure <lb/>upon the Geometrical Plane be found, by Means <lb/>of which, the Appearance of the Point B may <lb/>be afterwards gotten; </s> <s xml:id="echoid-s677" xml:space="preserve">but in theſe Caſes, the <lb/>ſhorteſt Way, is to uſe ſome one of the precedent <lb/>Methods.</s> <s xml:id="echoid-s678" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div84" type="section" level="1" n="47"> <head xml:id="echoid-head49" xml:space="preserve"><emph style="sc">Corollary</emph>.</head> <p> <s xml:id="echoid-s679" xml:space="preserve">38. </s> <s xml:id="echoid-s680" xml:space="preserve">It appears from this Method, that when <lb/>the Repreſentations of two Points are found, <lb/>the Appearance of any third Point whatſoever <lb/>may be had, without having any Regard to the <lb/>Situation of the Eye; </s> <s xml:id="echoid-s681" xml:space="preserve">becauſe two Lines as E a <lb/>may be drawn, whoſe Interſection will be the <lb/>Point ſought.</s> <s xml:id="echoid-s682" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div85" type="section" level="1" n="48"> <head xml:id="echoid-head50" xml:space="preserve"><emph style="sc">Method</emph> VI.</head> <p> <s xml:id="echoid-s683" xml:space="preserve">39. </s> <s xml:id="echoid-s684" xml:space="preserve">The ſame things being given, as in the <lb/> <anchor type="note" xlink:label="note-0048-01a" xlink:href="note-0048-01"/> ſecond Method, let F C be the Geometrical Line.</s> <s xml:id="echoid-s685" xml:space="preserve"/> </p> <div xml:id="echoid-div85" type="float" level="2" n="1"> <note position="left" xlink:label="note-0048-01" xlink:href="note-0048-01a" xml:space="preserve">Fig. 12.</note> </div> </div> <div xml:id="echoid-div87" type="section" level="1" n="49"> <head xml:id="echoid-head51" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s686" xml:space="preserve">Draw two Lines A F and A C from the given <lb/>Point A at Pleaſure, cutting the Baſe Line in the <lb/>Points E and B, and interſecting the Geometri-<lb/>cal Line in the Points F and C. </s> <s xml:id="echoid-s687" xml:space="preserve">From theſe <lb/>two laſt Points draw the Lines F O and C O to <lb/>the Eye; </s> <s xml:id="echoid-s688" xml:space="preserve">then draw E a through the Point E, <lb/>parallel to F O, and B a through the Point B, <lb/>parallel to C O, and the Point a the Interſection <lb/>of theſe two Lines will be that ſought.</s> <s xml:id="echoid-s689" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s690" xml:space="preserve">Note, We might firſt have drawn the Lines <lb/>O F and O C at Pleaſure, and then have drawn <lb/>the Lines AC and A F through their Concur- <pb file="0049" n="52"/> <pb file="0049a" n="53"/> <anchor type="figure" xlink:label="fig-0049a-01a" xlink:href="fig-0049a-01"/> <pb file="0050" n="54"/> <pb o="25" file="0051" n="55" rhead="on PERSPECTIVE."/> rence with the Geometrical Line; </s> <s xml:id="echoid-s691" xml:space="preserve">which would <lb/>come to the ſame thing.</s> <s xml:id="echoid-s692" xml:space="preserve"/> </p> <div xml:id="echoid-div87" type="float" level="2" n="1"> <figure xlink:label="fig-0049a-01" xlink:href="fig-0049a-01a"> <caption xml:id="echoid-caption10" xml:space="preserve">Plate 4.<lb/>Page 24.<lb/>Fig. 10.</caption> <variables xml:id="echoid-variables10" xml:space="preserve">f F O G g V D N L a P E H I M A</variables> </figure> </div> </div> <div xml:id="echoid-div89" type="section" level="1" n="50"> <head xml:id="echoid-head52" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s693" xml:space="preserve">Firſt continue the Line E a, until it meets <lb/>the Horizontal Line in D, and draw a Line <lb/>from D to the Eye, and another through the Eye <lb/>parallel to the Baſe Line.</s> <s xml:id="echoid-s694" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s695" xml:space="preserve">Then the Parallels O M and F C are at the <lb/>ſame Diſtance from each other, as L D is from <lb/>E B; </s> <s xml:id="echoid-s696" xml:space="preserve">whence it follows, that F O is equal to E D, <lb/>and therefore O D is parallel to A F. </s> <s xml:id="echoid-s697" xml:space="preserve">Whence <anchor type="note" xlink:href="" symbol="*"/> <anchor type="note" xlink:label="note-0051-01a" xlink:href="note-0051-01"/> the Appearance of E A, is a Part of E D. </s> <s xml:id="echoid-s698" xml:space="preserve">And <lb/>after the ſame Manner we prove, that the Re-<lb/>preſentation of B A is a Part of B a.</s> <s xml:id="echoid-s699" xml:space="preserve"/> </p> <div xml:id="echoid-div89" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0051-01" xlink:href="note-0051-01a" xml:space="preserve">13:</note> </div> </div> <div xml:id="echoid-div91" type="section" level="1" n="51"> <head xml:id="echoid-head53" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s700" xml:space="preserve">40. </s> <s xml:id="echoid-s701" xml:space="preserve">When there are no Lines drawn, and we <lb/>would uſe this Method, the Horizontal Line <lb/>may be laid aſide; </s> <s xml:id="echoid-s702" xml:space="preserve">and then having firſt drawn <lb/>the Geometrical Line, whoſe Diſtance from <lb/>the Baſe Line is equal to the Length of the Ray, <lb/>we aſſume the Diſtance from the Eye to the <lb/>Geometrical Line, equal to the Height of the <lb/>Eye.</s> <s xml:id="echoid-s703" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s704" xml:space="preserve">Although this Method appears uſeleſs, as being <lb/>more difficult than the precedent ones, yet in <lb/>the Eighth Chapter we have ſhewn the Uſe that <lb/>may be drawn from it.</s> <s xml:id="echoid-s705" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div92" type="section" level="1" n="52"> <head xml:id="echoid-head54" xml:space="preserve"><emph style="sc">Corollary</emph>.</head> <p> <s xml:id="echoid-s706" xml:space="preserve">41. </s> <s xml:id="echoid-s707" xml:space="preserve">It follows from this Demonſtration, that <lb/>the Appearances of Lines paſſing through the <lb/>Station Point, are all perpendicular to the Baſe <lb/>Line; </s> <s xml:id="echoid-s708" xml:space="preserve">for if the Perpendicular O S be let fall <pb o="26" file="0052" n="56" rhead="An ESSAY"/> from the Eye O upon the Baſe Line, the Appear-<lb/>ances of all Lines paſſing through S, will be <lb/>perpendicular to the Baſe Line; </s> <s xml:id="echoid-s709" xml:space="preserve">but the ſaid <lb/>Point S is the Station Point. </s> <s xml:id="echoid-s710" xml:space="preserve">Whence, &</s> <s xml:id="echoid-s711" xml:space="preserve">c.</s> <s xml:id="echoid-s712" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div93" type="section" level="1" n="53"> <head xml:id="echoid-head55" xml:space="preserve"><emph style="sc">Problem</emph> II.</head> <p style="it"> <s xml:id="echoid-s713" xml:space="preserve">42. </s> <s xml:id="echoid-s714" xml:space="preserve">To throw a Line in the Geometrical Plane <lb/>into Perſpective.</s> <s xml:id="echoid-s715" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s716" xml:space="preserve">It has been ſhewn <anchor type="note" xlink:href="" symbol="*"/>, that to have the Repre- <anchor type="note" xlink:label="note-0052-01a" xlink:href="note-0052-01"/> ſentation of a right Line, the Perſpective of <lb/>the Extremes of the ſaid Line, need only be <lb/>found; </s> <s xml:id="echoid-s717" xml:space="preserve">and although it is difficult to find <anchor type="note" xlink:href="" symbol="*"/> the <anchor type="note" xlink:label="note-0052-02a" xlink:href="note-0052-02"/> Repreſentation of two Points, nevertheleſs I <lb/>ſhall ſhew here how to find more eaſy the Re-<lb/>preſentation of a Line in ſome Caſes.</s> <s xml:id="echoid-s718" xml:space="preserve"/> </p> <div xml:id="echoid-div93" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0052-01" xlink:href="note-0052-01a" xml:space="preserve">21.</note> <note symbol="*" position="left" xlink:label="note-0052-02" xlink:href="note-0052-02a" xml:space="preserve">22.</note> </div> <p> <s xml:id="echoid-s719" xml:space="preserve">1. </s> <s xml:id="echoid-s720" xml:space="preserve">Let A B be a Line parallel to the Baſe <lb/> <anchor type="note" xlink:label="note-0052-03a" xlink:href="note-0052-03"/> Line: </s> <s xml:id="echoid-s721" xml:space="preserve">To draw the Repreſentation of which, <lb/>having firſt found the Point a the Appearance <lb/>of A, one of the Ends of the given Line; </s> <s xml:id="echoid-s722" xml:space="preserve">af-<lb/>terwards through that Appearance, draw a Pa-<lb/>rallel to the Baſe Line; </s> <s xml:id="echoid-s723" xml:space="preserve">then the Line B O, <lb/>drawn from B to the Eye O, will cut the ſaid <lb/>parallel in the Point b, and b a will be the Re-<lb/>preſentation ſought.</s> <s xml:id="echoid-s724" xml:space="preserve"/> </p> <div xml:id="echoid-div94" type="float" level="2" n="2"> <note position="left" xlink:label="note-0052-03" xlink:href="note-0052-03a" xml:space="preserve">Fig. 13.</note> </div> <p> <s xml:id="echoid-s725" xml:space="preserve">43. </s> <s xml:id="echoid-s726" xml:space="preserve">2. </s> <s xml:id="echoid-s727" xml:space="preserve">Let C G be a Line, which continued <lb/>out, cuts the baſe Line in E. </s> <s xml:id="echoid-s728" xml:space="preserve">Now to draw the <lb/>Appearance thereof; </s> <s xml:id="echoid-s729" xml:space="preserve">through the Eye O, draw a <lb/>Line parallel thereto, cutting the Horizontal <lb/>Line in D, and joyn the Points E and D, by <lb/>the Line E D, which cut in the Points c and g, <lb/>by Lines drawn from C and G to the Eye; </s> <s xml:id="echoid-s730" xml:space="preserve">then <lb/>the Part c g of the Line E D, is the Appearance <lb/>ſought.</s> <s xml:id="echoid-s731" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div96" type="section" level="1" n="54"> <head xml:id="echoid-head56" xml:space="preserve"><emph style="sc">Remark</emph>.</head> <p> <s xml:id="echoid-s732" xml:space="preserve">If the Lines G O and C O cut E D very ob-<lb/>liquely, and ſo their Interſection cannot be ex- <pb file="0053" n="57"/> <pb file="0053a" n="58"/> <anchor type="figure" xlink:label="fig-0053a-01a" xlink:href="fig-0053a-01"/> <anchor type="figure" xlink:label="fig-0053a-02a" xlink:href="fig-0053a-02"/> <pb file="0054" n="59"/> <pb o="27" file="0055" n="60" rhead="on PERSPECTIVE."/> actly determined, this Method ought not then to <lb/>be uſed.</s> <s xml:id="echoid-s733" xml:space="preserve"/> </p> <div xml:id="echoid-div96" type="float" level="2" n="1"> <figure xlink:label="fig-0053a-01" xlink:href="fig-0053a-01a"> <caption xml:id="echoid-caption11" style="it" xml:space="preserve">Plate 5.<lb/>page 26.<lb/>Fig. 11.</caption> <variables xml:id="echoid-variables11" xml:space="preserve">O Y b X a E Z A B</variables> </figure> <figure xlink:label="fig-0053a-02" xlink:href="fig-0053a-02a"> <caption xml:id="echoid-caption12" style="it" xml:space="preserve">Fig. 12.</caption> <variables xml:id="echoid-variables12" xml:space="preserve">M O Y F S C L D X a E B Z A</variables> </figure> </div> </div> <div xml:id="echoid-div98" type="section" level="1" n="55"> <head xml:id="echoid-head57" xml:space="preserve"><emph style="sc">Problem</emph> III.</head> <p style="it"> <s xml:id="echoid-s734" xml:space="preserve">44. </s> <s xml:id="echoid-s735" xml:space="preserve">To find the Appearance of the Diviſions of a <lb/>Line in the Geometrical Plane.</s> <s xml:id="echoid-s736" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s737" xml:space="preserve">Let A B be a Line, whoſe Appearance is ab. <lb/></s> <s xml:id="echoid-s738" xml:space="preserve"> <anchor type="note" xlink:label="note-0055-01a" xlink:href="note-0055-01"/> Now to find the Repreſentation of the Diviſions <lb/>of this Line, there muſt be Lines drawn from <lb/>the Diviſions of the Line to the Eye, whoſe <lb/>Interſections with a b will give the Points <lb/>ſought.</s> <s xml:id="echoid-s739" xml:space="preserve"/> </p> <div xml:id="echoid-div98" type="float" level="2" n="1"> <note position="right" xlink:label="note-0055-01" xlink:href="note-0055-01a" xml:space="preserve">Fig. 14.</note> </div> <p> <s xml:id="echoid-s740" xml:space="preserve">Note, When theſe Lines very obliquely cut <lb/>a b, the following Way ought to uſed.</s> <s xml:id="echoid-s741" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div100" type="section" level="1" n="56"> <head xml:id="echoid-head58" xml:space="preserve"><emph style="sc">Method</emph>. II.</head> <p> <s xml:id="echoid-s742" xml:space="preserve">45. </s> <s xml:id="echoid-s743" xml:space="preserve">To find the Repreſentations of the Di-<lb/>viſions of the Line G C, make choice of the <lb/> <anchor type="note" xlink:label="note-0055-02a" xlink:href="note-0055-02"/> Point D at Pleaſure without this Line, and find <lb/> <anchor type="note" xlink:href="" symbol="*"/> the Repreſentation d thereof; </s> <s xml:id="echoid-s744" xml:space="preserve">then draw Lines <anchor type="note" xlink:label="note-0055-03a" xlink:href="note-0055-03"/> through the propoſed Diviſions to the Point D; <lb/></s> <s xml:id="echoid-s745" xml:space="preserve">and from the Points wherein theſe Lines con-<lb/>tinued out cut the Baſe Line, draw other Lines <lb/>through the Repreſentation d, which will cut c g <lb/>the Repreſentation of C G in the Points ſought.</s> <s xml:id="echoid-s746" xml:space="preserve"/> </p> <div xml:id="echoid-div100" type="float" level="2" n="1"> <note position="right" xlink:label="note-0055-02" xlink:href="note-0055-02a" xml:space="preserve">Fig. 14.</note> <note symbol="*" position="right" xlink:label="note-0055-03" xlink:href="note-0055-03a" xml:space="preserve">21.</note> </div> </div> <div xml:id="echoid-div102" type="section" level="1" n="57"> <head xml:id="echoid-head59" xml:space="preserve"><emph style="sc">Problem</emph> IV.</head> <p style="it"> <s xml:id="echoid-s747" xml:space="preserve">46. </s> <s xml:id="echoid-s748" xml:space="preserve">To throw a Polygon, or any other regular <lb/>Figure on the Geometrical Plane into Perſpective.</s> <s xml:id="echoid-s749" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s750" xml:space="preserve">The Repreſentation of any Kinds of Figures <lb/>may be found <anchor type="note" xlink:href="" symbol="*"/> by any one of the Methods of <anchor type="note" xlink:label="note-0055-04a" xlink:href="note-0055-04"/> Problem I. </s> <s xml:id="echoid-s751" xml:space="preserve">the fourth in general is the eaſieſt; <lb/></s> <s xml:id="echoid-s752" xml:space="preserve">and may be firſt uſed in finding <anchor type="note" xlink:href="" symbol="*"/> the Repre- <anchor type="note" xlink:label="note-0055-05a" xlink:href="note-0055-05"/> ſentations of Points, or ſometimes of one <lb/>only; </s> <s xml:id="echoid-s753" xml:space="preserve">and then the fifth Method ſerves <anchor type="note" xlink:href="" symbol="*"/> for <anchor type="note" xlink:label="note-0055-06a" xlink:href="note-0055-06"/> <pb o="28" file="0056" n="61" rhead="An ESSAY"/> ſinding the reſt. </s> <s xml:id="echoid-s754" xml:space="preserve">But yet the Work may be <lb/>ſhorten’d by the two precedent Problems, as we <lb/>ſhall ſhew in the following Examples.</s> <s xml:id="echoid-s755" xml:space="preserve"/> </p> <div xml:id="echoid-div102" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0055-04" xlink:href="note-0055-04a" xml:space="preserve">21.</note> <note symbol="*" position="right" xlink:label="note-0055-05" xlink:href="note-0055-05a" xml:space="preserve">22.</note> <note symbol="*" position="right" xlink:label="note-0055-06" xlink:href="note-0055-06a" xml:space="preserve">31.</note> </div> </div> <div xml:id="echoid-div104" type="section" level="1" n="58"> <head xml:id="echoid-head60" xml:space="preserve"><emph style="sc">Example</emph> I.</head> <p style="it"> <s xml:id="echoid-s756" xml:space="preserve">To throw a Pentagon having one Side parallel to <lb/>the Baſe Line, into Perſpective.</s> <s xml:id="echoid-s757" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s758" xml:space="preserve">Let A B C D E, be the propoſed Pentagon, <lb/> <anchor type="note" xlink:label="note-0056-01a" xlink:href="note-0056-01"/> wherein draw the Line B D which will be paral-<lb/>lel to A E, becauſe the Pentagon is a regular <lb/>one.</s> <s xml:id="echoid-s759" xml:space="preserve"/> </p> <div xml:id="echoid-div104" type="float" level="2" n="1"> <note position="left" xlink:label="note-0056-01" xlink:href="note-0056-01a" xml:space="preserve">Fig. 15.</note> </div> <p> <s xml:id="echoid-s760" xml:space="preserve">Now find <anchor type="note" xlink:href="" symbol="*"/> the Repreſentation of the ſaid two <anchor type="note" xlink:label="note-0056-02a" xlink:href="note-0056-02"/> Lines A E and B D, and you will have the Re-<lb/>preſentation of four of the Corners of the Pen-<lb/>tagon; </s> <s xml:id="echoid-s761" xml:space="preserve">and to determine the Repreſentation of <lb/>the fifth Corner, find <anchor type="note" xlink:href="" symbol="*"/> the Appearance of a <anchor type="note" xlink:label="note-0056-03a" xlink:href="note-0056-03"/> Line drawn from C to E, which in this Example <lb/>is parallel to A B, which is ſuppoſed parallel <lb/>to the Baſe Line.</s> <s xml:id="echoid-s762" xml:space="preserve"/> </p> <div xml:id="echoid-div105" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0056-02" xlink:href="note-0056-02a" xml:space="preserve">42.</note> <note symbol="*" position="left" xlink:label="note-0056-03" xlink:href="note-0056-03a" xml:space="preserve">43.</note> </div> </div> <div xml:id="echoid-div107" type="section" level="1" n="59"> <head xml:id="echoid-head61" xml:space="preserve"><emph style="sc">Example</emph> II.</head> <p style="it"> <s xml:id="echoid-s763" xml:space="preserve">To throw a Parallelogram, divided into ſeveral <lb/>other Parallelograms into Perſpective.</s> <s xml:id="echoid-s764" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s765" xml:space="preserve">Let A B C D be a Parallelogram, divided <lb/> <anchor type="note" xlink:label="note-0056-04a" xlink:href="note-0056-04"/> into ſeveral other Parallelograms.</s> <s xml:id="echoid-s766" xml:space="preserve"/> </p> <div xml:id="echoid-div107" type="float" level="2" n="1"> <note position="left" xlink:label="note-0056-04" xlink:href="note-0056-04a" xml:space="preserve">Fig. 17.</note> </div> <p> <s xml:id="echoid-s767" xml:space="preserve">Draw the Line O G thro’ the Eye O parallel to <lb/>the Side A D, cutting the Horizontal Line in G; <lb/></s> <s xml:id="echoid-s768" xml:space="preserve">likewiſe draw O F parallel to the Side A B cutting <lb/>the Horizontal Line in F, and produce the Sides <lb/>of the Parallelogram, and the Lines dividing it, <lb/>to the Baſe Line; </s> <s xml:id="echoid-s769" xml:space="preserve">then from the Points wherein <lb/>A D and C B, and the Lines parallel thereto <lb/>cut the Baſe Line, draw Lines to the Point G. </s> <s xml:id="echoid-s770" xml:space="preserve"><lb/>Alſo from the Points wherein A B and C D and <lb/>their parallels cut the ſaid Line, draw Lines to <lb/>the Point F, whoſe Interſections with theſe <pb o="29" file="0057" n="62" rhead="on PERSPECTIVE."/> drawn to the Point G, will give the Appearance <lb/>ſought.</s> <s xml:id="echoid-s771" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div109" type="section" level="1" n="60"> <head xml:id="echoid-head62" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s772" xml:space="preserve">When this Method cannot be us’d, the Per-<lb/>ſpective of the Diviſions dividing the Sides of <lb/>the Parallelogram, mufl be found <anchor type="note" xlink:href="" symbol="*"/>. </s> <s xml:id="echoid-s773" xml:space="preserve">And we are <anchor type="note" xlink:label="note-0057-01a" xlink:href="note-0057-01"/> often oblig’d to have recourſe to this Expedient, <lb/>notwithſtanding the accidental Points, G and F, <lb/>being had. </s> <s xml:id="echoid-s774" xml:space="preserve">And this happens, when the Paralle-<lb/>logram is ſo far diſtant from the perſpective <lb/>Plane, that its Sides being produc’d, cannot meet <lb/>the Baſe Line.</s> <s xml:id="echoid-s775" xml:space="preserve"/> </p> <div xml:id="echoid-div109" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0057-01" xlink:href="note-0057-01a" xml:space="preserve">44.</note> </div> <p> <s xml:id="echoid-s776" xml:space="preserve">47. </s> <s xml:id="echoid-s777" xml:space="preserve">Note, moreover, that this one Example <lb/>is ſufficient to ſhew how to throw any Kinds of <lb/>Figures in the Geometrical Plane into Perſpe-<lb/>ctive. </s> <s xml:id="echoid-s778" xml:space="preserve">To effect which, we circumſcribe any <lb/>Parallelogram about the Figures, which we di-<lb/>vide into ſeveral others: </s> <s xml:id="echoid-s779" xml:space="preserve">Then we throw this <lb/>Parallelogram (thus divided) into Perſpective, <lb/>and transfer the given Figure therein, ſo that it <lb/>may have the ſame Situation with reſpect to <lb/>the little Parallelograms in the perſpective Plane, <lb/>as it had in regard to the ſmall Parallelograms <lb/>in the Geometrical Plane.</s> <s xml:id="echoid-s780" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div111" type="section" level="1" n="61"> <head xml:id="echoid-head63" xml:space="preserve"><emph style="sc">Example</emph> III.</head> <head xml:id="echoid-head64" style="it" xml:space="preserve">48. To throw a circle into Perſpective.</head> <p> <s xml:id="echoid-s781" xml:space="preserve">The Repreſentation of ſeveral Points of a <lb/> <anchor type="note" xlink:label="note-0057-02a" xlink:href="note-0057-02"/> Circle, or any other Curve Line requir’d to be <lb/>thrown into perſpective, muſt be <anchor type="note" xlink:href="" symbol="*"/> found, This <anchor type="note" xlink:label="note-0057-03a" xlink:href="note-0057-03"/> may be well enough done, by drawing ſeveral <lb/>Chords in the Circle, or Curve, parallel between <lb/>themſelves, the Repreſentations of which muſt <lb/>be found <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s782" xml:space="preserve">then the Extremities of thoſe Re- <anchor type="note" xlink:label="note-0057-04a" xlink:href="note-0057-04"/> <pb o="30" file="0058" n="63" rhead="An ESSAY"/> preſentations being join’d, will give the Perſpe-<lb/>ctive ſought. </s> <s xml:id="echoid-s783" xml:space="preserve">The ſame may be done, in draw-<lb/>ing the Chords thro’a Point, whoſe Repreſenta-<lb/>tion is known.</s> <s xml:id="echoid-s784" xml:space="preserve"/> </p> <div xml:id="echoid-div111" type="float" level="2" n="1"> <note position="right" xlink:label="note-0057-02" xlink:href="note-0057-02a" xml:space="preserve">Fig. 16.</note> <note symbol="*" position="right" xlink:label="note-0057-03" xlink:href="note-0057-03a" xml:space="preserve">21.</note> <note symbol="*" position="right" xlink:label="note-0057-04" xlink:href="note-0057-04a" xml:space="preserve">43.</note> </div> </div> <div xml:id="echoid-div113" type="section" level="1" n="62"> <head xml:id="echoid-head65" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p style="it"> <s xml:id="echoid-s785" xml:space="preserve">49. </s> <s xml:id="echoid-s786" xml:space="preserve">Let G I be the Geometrical Line; </s> <s xml:id="echoid-s787" xml:space="preserve">and thro’ <lb/> <anchor type="note" xlink:label="note-0058-01a" xlink:href="note-0058-01"/> the Center P of the Circle, whoſe Perſpective is <lb/>ſought, let fall the Perpendicular P F upon the ſaid <lb/>Line G I, which biſect in the Point R. </s> <s xml:id="echoid-s788" xml:space="preserve">About R, <lb/>as a Center, and with the Radius R P, deſcribe an <lb/>Arc of a Circle M P N, cutting the given Circle in <lb/>the Points M and N. </s> <s xml:id="echoid-s789" xml:space="preserve">Now, if the Perſpective of <lb/>L H and N M be found, the two Conjugate Dia-<lb/>meters of an Ellipſis, which is the Repreſentation of <lb/>the given Circle, will be bad. </s> <s xml:id="echoid-s790" xml:space="preserve">And, an Ellipſis may <lb/>be drawn by ſome one of the Methods laid down by <lb/>thoſe who have treated of Conick Sections.</s> <s xml:id="echoid-s791" xml:space="preserve"/> </p> <div xml:id="echoid-div113" type="float" level="2" n="1"> <note position="left" xlink:label="note-0058-01" xlink:href="note-0058-01a" xml:space="preserve">Fig. 16.</note> </div> <p style="it"> <s xml:id="echoid-s792" xml:space="preserve">I ſhall not ſpend time here in demonſtrating the <lb/>Truth of this. </s> <s xml:id="echoid-s793" xml:space="preserve">See Prop. </s> <s xml:id="echoid-s794" xml:space="preserve">10. </s> <s xml:id="echoid-s795" xml:space="preserve">lib. </s> <s xml:id="echoid-s796" xml:space="preserve">2. </s> <s xml:id="echoid-s797" xml:space="preserve">of the great <lb/>Latin Treatiſe of Conic Sections, written by M. </s> <s xml:id="echoid-s798" xml:space="preserve">de <lb/>la Hire; </s> <s xml:id="echoid-s799" xml:space="preserve">the Demonſtration of which may be here <lb/>apply’d. </s> <s xml:id="echoid-s800" xml:space="preserve">If we conſider, 1. </s> <s xml:id="echoid-s801" xml:space="preserve">That Lines drawn from <lb/>the Points M and N to the Point F, will touch the <lb/>Circle in the ſaid Points M and N. </s> <s xml:id="echoid-s802" xml:space="preserve">2. </s> <s xml:id="echoid-s803" xml:space="preserve">That the <lb/>viſual Rays, going from the Eye towards all the Parts <lb/>of the Circumference of the Circle, form a Cone, <lb/>3. </s> <s xml:id="echoid-s804" xml:space="preserve">That the Appearance of the Circle, is the Section <lb/>of a Cone, made by the perſpective Plane. </s> <s xml:id="echoid-s805" xml:space="preserve">Finally, <lb/>That the Line G I ought to be conceiv’d, as being <lb/>the Interſection of the Geometrical Plane, and a <lb/>Plane paſſing thro’ the Eye parallel to the perſpective <lb/>Plane.</s> <s xml:id="echoid-s806" xml:space="preserve"/> </p> <pb file="0059" n="64"/> <pb file="0059a" n="65"/> <figure> <caption xml:id="echoid-caption13" style="it" xml:space="preserve">Plate 6.<lb/>page 28.<lb/>Fig. 13.</caption> <variables xml:id="echoid-variables13" xml:space="preserve">O D c b a g E G B A C</variables> </figure> <figure> <caption xml:id="echoid-caption14" style="it" xml:space="preserve">Fig. 14.</caption> <variables xml:id="echoid-variables14" xml:space="preserve">O b 1 2 3 a c 1 2 3 g D A C 3 1 2 2 1 3 B G</variables> </figure> <pb file="0060" n="66"/> <pb file="0061" n="67"/> <pb file="0061a" n="68"/> <figure> <caption xml:id="echoid-caption15" style="it" xml:space="preserve">page 28.<lb/>Plate. 7<lb/>Fig. 16<lb/>Fig. 15</caption> <variables xml:id="echoid-variables15" xml:space="preserve">O G F I V<lb/>l d c e m n b a h <lb/>B A H M N C E P D L</variables> </figure> <pb file="0062" n="69"/> <pb o="31" file="0063" n="70" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div115" type="section" level="1" n="63"> <head xml:id="echoid-head66" xml:space="preserve"><emph style="sc">Prob</emph>. V.</head> <head xml:id="echoid-head67" style="it" xml:space="preserve">50. To find the Repreſentation of a Point, elevated <lb/>above the Geometrical Planc.</head> <p> <s xml:id="echoid-s807" xml:space="preserve">Let G S be the Geometrical Line, and S the <lb/> <anchor type="note" xlink:label="note-0063-01a" xlink:href="note-0063-01"/> Station Point: </s> <s xml:id="echoid-s808" xml:space="preserve">Make S F, in the Geometrical <lb/>Line, equal to the Height of the Eye; </s> <s xml:id="echoid-s809" xml:space="preserve">and let <lb/>A be the Seat of the given Line.</s> <s xml:id="echoid-s810" xml:space="preserve"/> </p> <div xml:id="echoid-div115" type="float" level="2" n="1"> <note position="right" xlink:label="note-0063-01" xlink:href="note-0063-01a" xml:space="preserve">Fig. 18.</note> </div> </div> <div xml:id="echoid-div117" type="section" level="1" n="64"> <head xml:id="echoid-head68" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s811" xml:space="preserve">Aſſume F C in the Geometrical Line, equal to <lb/>the Height of the Eye, above the Geometrical <lb/>Plane: </s> <s xml:id="echoid-s812" xml:space="preserve">Then draw Lines from the Point A to <lb/>the Points S and C, and on the Point B, the In-<lb/>terſection of the Line AS and the Baſe Line, <lb/>raiſe the Perpendicular BI to the Baſe Line, <lb/>equal to E B, plus FC; </s> <s xml:id="echoid-s813" xml:space="preserve">and the Point I will be <lb/>the Perſpective ſought.</s> <s xml:id="echoid-s814" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div118" type="section" level="1" n="65"> <head xml:id="echoid-head69" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s815" xml:space="preserve">51. </s> <s xml:id="echoid-s816" xml:space="preserve">Let us ſuppoſe a Plane to paſs thro’ the <lb/>given Point, and the Eye perpendicular to the <lb/>Geometrical Plane; </s> <s xml:id="echoid-s817" xml:space="preserve">then it is manifeſt, that the <lb/>Interſection of theſe two Planes is the Line <lb/>A B S, and the Interſection of the ſaid ſuppos’d <lb/>Plane and the perſpective Plane, is B I. </s> <s xml:id="echoid-s818" xml:space="preserve">Now, <lb/>let X be this ſuppos’d Plane; </s> <s xml:id="echoid-s819" xml:space="preserve">a, b, s, the Point <lb/> <anchor type="note" xlink:label="note-0063-02a" xlink:href="note-0063-02"/> mark’d with the ſame Letters in the precedent <lb/>Figure, bi the Interſection of this Plane and <lb/>the perſpective Plane; </s> <s xml:id="echoid-s820" xml:space="preserve">O the Eye, and D the <lb/>propos’d Point: </s> <s xml:id="echoid-s821" xml:space="preserve">We are to prove, that if O D <lb/>be drawn, the Line B I of the precedent Figure <lb/>will be equal to b i in this Figure.</s> <s xml:id="echoid-s822" xml:space="preserve"/> </p> <div xml:id="echoid-div118" type="float" level="2" n="1"> <note position="right" xlink:label="note-0063-02" xlink:href="note-0063-02a" xml:space="preserve">Fig. 19.</note> </div> <pb o="32" file="0064" n="71" rhead="An ESSAY"/> <p> <s xml:id="echoid-s823" xml:space="preserve">To demonſtrate which, draw the Line D L M <lb/>thro’ the Point D, parallel to a b s. </s> <s xml:id="echoid-s824" xml:space="preserve">Then, be-<lb/>cauſe the Triangles D M O and D L i are ſimi-<lb/>lar, we have,</s> </p> <p> <s xml:id="echoid-s825" xml:space="preserve">D M = as: </s> <s xml:id="echoid-s826" xml:space="preserve">D L = ab:</s> <s xml:id="echoid-s827" xml:space="preserve">: M O: </s> <s xml:id="echoid-s828" xml:space="preserve">L i. </s> <s xml:id="echoid-s829" xml:space="preserve">Again, <lb/>in the precedent Figure, the Triangles A S C and <lb/>A B E are ſimilar: </s> <s xml:id="echoid-s830" xml:space="preserve">Whence, <lb/>A S: </s> <s xml:id="echoid-s831" xml:space="preserve">A B:</s> <s xml:id="echoid-s832" xml:space="preserve">: C S: </s> <s xml:id="echoid-s833" xml:space="preserve">E B.</s> <s xml:id="echoid-s834" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s835" xml:space="preserve">The three firſt Terms of theſe two Progreſſions <lb/>are the ſame: </s> <s xml:id="echoid-s836" xml:space="preserve">For CS is equal to M O, ſince <lb/>they are each the Difference of the Height of the <lb/>Eye, and that of the given Point; </s> <s xml:id="echoid-s837" xml:space="preserve">and conſe-<lb/>quently, E B is equal to L i: </s> <s xml:id="echoid-s838" xml:space="preserve">But B I was made <lb/>equal to B E, pl{us} FC the Height of the given <lb/>Point above the Geometrical Plane; </s> <s xml:id="echoid-s839" xml:space="preserve">and b i is <lb/>equal to Li, pl{us} b L; </s> <s xml:id="echoid-s840" xml:space="preserve">which being equal to aD, <lb/>is likewiſe the Height of the given Point above <lb/>the Geometrical Plane; </s> <s xml:id="echoid-s841" xml:space="preserve">whence the Lines B I <lb/>and b i are equal. </s> <s xml:id="echoid-s842" xml:space="preserve">Which was to be demon-<lb/>ſtrated.</s> <s xml:id="echoid-s843" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s844" xml:space="preserve">Note, When the Height of the given Point is <lb/>greater than the Height of the Eye, E B muſt <lb/>be taken from that firſt Height, to have the <lb/>Magnitude of B I.</s> <s xml:id="echoid-s845" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div120" type="section" level="1" n="66"> <head xml:id="echoid-head70" xml:space="preserve"><emph style="sc">Prob</emph>. VI.</head> <head xml:id="echoid-head71" style="it" xml:space="preserve">52. To throm a Pyramid, or Cone, into Perſpective.</head> <p> <s xml:id="echoid-s846" xml:space="preserve">Now, to throw a Pyramid into perſpective, <lb/> <anchor type="note" xlink:label="note-0064-01a" xlink:href="note-0064-01"/> the Appearance of its Baſe <anchor type="note" xlink:href="" symbol="*"/> and Center muſt be <anchor type="note" xlink:label="note-0064-02a" xlink:href="note-0064-02"/> found <anchor type="note" xlink:href="" symbol="*"/>: </s> <s xml:id="echoid-s847" xml:space="preserve">After which, Lines muſt be drawn from <anchor type="note" xlink:label="note-0064-03a" xlink:href="note-0064-03"/> the Repreſentation of the Vertex, to the Ap-<lb/>pearance of thoſe Angles of the Baſe that are <lb/>viſible; </s> <s xml:id="echoid-s848" xml:space="preserve">and then the Perſpective ſought will be <lb/>had.</s> <s xml:id="echoid-s849" xml:space="preserve"/> </p> <div xml:id="echoid-div120" type="float" level="2" n="1"> <note position="left" xlink:label="note-0064-01" xlink:href="note-0064-01a" xml:space="preserve">Fig. 20.</note> <note symbol="*" position="left" xlink:label="note-0064-02" xlink:href="note-0064-02a" xml:space="preserve">46.</note> <note symbol="*" position="left" xlink:label="note-0064-03" xlink:href="note-0064-03a" xml:space="preserve">50.</note> </div> <p> <s xml:id="echoid-s850" xml:space="preserve">And to throw a Cone into perſpective, the <lb/> <anchor type="note" xlink:label="note-0064-04a" xlink:href="note-0064-04"/> Repreſentation of its Baſe <anchor type="note" xlink:href="" symbol="*"/> and Vertex muſt be <anchor type="note" xlink:label="note-0064-05a" xlink:href="note-0064-05"/> <pb o="33" file="0065" n="72" rhead="on PERSPECTIVE."/> firſt found <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s851" xml:space="preserve">and then if Lines be drawn from <anchor type="note" xlink:label="note-0065-01a" xlink:href="note-0065-01"/> the Repreſentation of the Vertex touching the <lb/>Repreſentation of the Baſe, the Repreſentation <lb/>of the Cone will be had.</s> <s xml:id="echoid-s852" xml:space="preserve"/> </p> <div xml:id="echoid-div121" type="float" level="2" n="2"> <note position="left" xlink:label="note-0064-04" xlink:href="note-0064-04a" xml:space="preserve">Fig. 21.</note> <note symbol="*" position="left" xlink:label="note-0064-05" xlink:href="note-0064-05a" xml:space="preserve">46.</note> <note symbol="*" position="right" xlink:label="note-0065-01" xlink:href="note-0065-01a" xml:space="preserve">47.</note> </div> <p> <s xml:id="echoid-s853" xml:space="preserve">But ſince, according to this Manner, we are <lb/>obliged to find the Perſpective of all the Baſe; <lb/></s> <s xml:id="echoid-s854" xml:space="preserve">whereas it often cannot be all ſeen; </s> <s xml:id="echoid-s855" xml:space="preserve">we may de-<lb/>termine, by the following Method, what Part <lb/>of the Baſe is viſible, and ſo only find the Re-<lb/>preſentation thereof. </s> <s xml:id="echoid-s856" xml:space="preserve">And then, to compleat <lb/>the Cone, we draw Lines from the Extremities <lb/>of the viſible Part of the Baſe, to the Repreſen-<lb/>tation of the Vertex.</s> <s xml:id="echoid-s857" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div123" type="section" level="1" n="67"> <head xml:id="echoid-head72" style="it" xml:space="preserve">53. To determine the viſible Part of the Baſe of <lb/>a Cone.</head> <p> <s xml:id="echoid-s858" xml:space="preserve">Let the Circle L I F be the Baſe of a Cone <lb/> <anchor type="note" xlink:label="note-0065-02a" xlink:href="note-0065-02"/> in the Geometrical Plane, and A the Center <lb/>thereof.</s> <s xml:id="echoid-s859" xml:space="preserve"/> </p> <div xml:id="echoid-div123" type="float" level="2" n="1"> <note position="right" xlink:label="note-0065-02" xlink:href="note-0065-02a" xml:space="preserve">Fig. 21.</note> </div> </div> <div xml:id="echoid-div125" type="section" level="1" n="68"> <head xml:id="echoid-head73" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s860" xml:space="preserve">Aſſume P Q ſomewhere in the Baſe Line, <lb/>equal to the Semidiameter of the Circle L F; <lb/></s> <s xml:id="echoid-s861" xml:space="preserve">and from the Point P, raiſe P D G perpendicu-<lb/>lar to the Baſe Line, meeting the Horizontal <lb/>Line in G; </s> <s xml:id="echoid-s862" xml:space="preserve">and in this Perpendicular, make <lb/>P D equal to the Height of the Cone; </s> <s xml:id="echoid-s863" xml:space="preserve">and draw <lb/>the Line Q D H, meeting the Horizontal Line <lb/>in H. </s> <s xml:id="echoid-s864" xml:space="preserve">Then, about the Point A as a Center, <lb/>and with the Radius G H, draw the Circle B C E; </s> <s xml:id="echoid-s865" xml:space="preserve"><lb/>and from the ſaid Point A, draw a Line to the <lb/>Station Point S: </s> <s xml:id="echoid-s866" xml:space="preserve">Biſect A S in R; </s> <s xml:id="echoid-s867" xml:space="preserve">and about <lb/>R, as a Center, with the Radius R A, deſcribe <lb/>the Circular Arc B A C, cutting the Circle BEC <lb/>in the Points B and C. </s> <s xml:id="echoid-s868" xml:space="preserve">Draw the Lines B A F, <lb/>and C A L; </s> <s xml:id="echoid-s869" xml:space="preserve">and the viſible Portion, (L I F) of <pb o="34" file="0066" n="73" rhead="An ESSAY"/> the Circular Baſe of the Cone will be deter-<lb/>min’d.</s> <s xml:id="echoid-s870" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div126" type="section" level="1" n="69"> <head xml:id="echoid-head74" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p style="it"> <s xml:id="echoid-s871" xml:space="preserve">To prove this, draw the Lines B C and L F, cut-<lb/>ting the Line A S in the Points N and M; </s> <s xml:id="echoid-s872" xml:space="preserve">and make <lb/>the Line G n equal to A N, and draw the Line <lb/>n D m. </s> <s xml:id="echoid-s873" xml:space="preserve">It is now manifeſt, that if the Cone be <lb/>continued out above its Vertex, (that is, if the oppo-<lb/>ſite Cone be form’d) it will cut the Horizontal Plane <lb/>in a Circle equal to B E C, whoſe Seat will be BEC: <lb/></s> <s xml:id="echoid-s874" xml:space="preserve">So that the Point S, in reſpect of B E C, is in the <lb/>ſame Situation as the Eye hath, with reſpect to the <lb/>Circle form’d in the Horizontal Plane, by the Conti-<lb/>nuation of the Cone. </s> <s xml:id="echoid-s875" xml:space="preserve">Whence it follows, that B C <lb/>is the Seat of the viſible Portion of that Circle. </s> <s xml:id="echoid-s876" xml:space="preserve">For, <lb/>by Conſtruction, B and C are the Points of Contact <lb/>of the Tangents to the Circle B E C, which paſs <lb/>thro’ the Point S; </s> <s xml:id="echoid-s877" xml:space="preserve">becauſe the Angle ABS, which <lb/>is in a Semicircle, is a right one.</s> <s xml:id="echoid-s878" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s879" xml:space="preserve">Now, if a Plane be conceiv’d, as paſſing thro’ ſome <lb/>Points in the Horizontal Plane, whoſe Seats are <lb/>B and C, and which cuts the two oppoſite Cones <lb/>thro’ their Vertex; </s> <s xml:id="echoid-s880" xml:space="preserve">it is evident, that this Plane <lb/>continued, will cut the Geometrical Plane in a Line <lb/>parallel to B N C; </s> <s xml:id="echoid-s881" xml:space="preserve">and that this Line upon the <lb/>ſaid Plane, will determine the viſible Part of the <lb/>Cone’s Baſe. </s> <s xml:id="echoid-s882" xml:space="preserve">So, ſince G n was made equal to <lb/>A N, we have only to prove, that P m is equal to <lb/>A M: </s> <s xml:id="echoid-s883" xml:space="preserve">For, it follows from thence, that L M F is <lb/>the Common Section of the Geometrical Plane, and <lb/>the Plane which we have here imagin’d.</s> <s xml:id="echoid-s884" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s885" xml:space="preserve">The Triangles D Q P and G H D are ſimilar, whence <lb/>D G: </s> <s xml:id="echoid-s886" xml:space="preserve">D P:</s> <s xml:id="echoid-s887" xml:space="preserve">: G H: </s> <s xml:id="echoid-s888" xml:space="preserve">P Q.</s> <s xml:id="echoid-s889" xml:space="preserve"/> </p> <pb o="35" file="0067" n="74" rhead="on PERSPECTIVE."/> <p style="it"> <s xml:id="echoid-s890" xml:space="preserve">And the Triangles D P m and D G n are ſimilar: <lb/></s> <s xml:id="echoid-s891" xml:space="preserve">Wherefore <lb/>D G: </s> <s xml:id="echoid-s892" xml:space="preserve">D P: </s> <s xml:id="echoid-s893" xml:space="preserve">G n:</s> <s xml:id="echoid-s894" xml:space="preserve">: P m. </s> <s xml:id="echoid-s895" xml:space="preserve"><lb/>And <lb/>G H: </s> <s xml:id="echoid-s896" xml:space="preserve">P Q:</s> <s xml:id="echoid-s897" xml:space="preserve">: G n: </s> <s xml:id="echoid-s898" xml:space="preserve">P m.</s> <s xml:id="echoid-s899" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s900" xml:space="preserve">The Triangles B A N and L A M are ſimilar: <lb/></s> <s xml:id="echoid-s901" xml:space="preserve">Therefore, <lb/>BA: </s> <s xml:id="echoid-s902" xml:space="preserve">AL:</s> <s xml:id="echoid-s903" xml:space="preserve">: AN: </s> <s xml:id="echoid-s904" xml:space="preserve">AM. </s> <s xml:id="echoid-s905" xml:space="preserve"><lb/>But the three firſt Terms of the two laſt Proportions, <lb/>are equal between themſelves; </s> <s xml:id="echoid-s906" xml:space="preserve">whence P m is alſo <lb/>equal to A M. </s> <s xml:id="echoid-s907" xml:space="preserve">Which was to be demonſtrated.</s> <s xml:id="echoid-s908" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div127" type="section" level="1" n="70"> <head xml:id="echoid-head75" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s909" xml:space="preserve">54. </s> <s xml:id="echoid-s910" xml:space="preserve">When the Height of the Cone is greater <lb/>than the Height of the Eye, the Points, G and <lb/>H, will fall below the Point D; </s> <s xml:id="echoid-s911" xml:space="preserve">in which Caſe, <lb/>the Lines A B and A C muſt be produc’d, till <lb/>they cut the Circle in the Points l and f, oppo-<lb/>ſite to L and F: </s> <s xml:id="echoid-s912" xml:space="preserve">Then lIf will be the viſible <lb/>Part of the Baſe.</s> <s xml:id="echoid-s913" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s914" xml:space="preserve">When the Cone is inclin’d, ſo that T (for Ex-<lb/>ample) is the Seat of its Vertex; </s> <s xml:id="echoid-s915" xml:space="preserve">AT muſt be <lb/>drawn: </s> <s xml:id="echoid-s916" xml:space="preserve">And then having aſſum’d P D equal to <lb/>the perpendicular Height of the Cone, and Pt <lb/>equal to A T; </s> <s xml:id="echoid-s917" xml:space="preserve">the Line t D x muſt be drawn; <lb/></s> <s xml:id="echoid-s918" xml:space="preserve">and the Part T X, taken in A T, equal to G x. </s> <s xml:id="echoid-s919" xml:space="preserve"><lb/>Alſo, X S muſt be drawn, and A s, equal and <lb/>parallel thereto.</s> <s xml:id="echoid-s920" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s921" xml:space="preserve">This being done; </s> <s xml:id="echoid-s922" xml:space="preserve">the ſame Method muſt be <lb/>apply’d here, that I have laid down for the up-<lb/>right Cone; </s> <s xml:id="echoid-s923" xml:space="preserve">with this Difference only, that the <lb/>Point s muſt be us’d inſtead of the Station Point <lb/>S. </s> <s xml:id="echoid-s924" xml:space="preserve">But when the Height of the Cone is greater <lb/>than the Height of the Eye, the Point X muſt <lb/>be aſſum’d in the Line T A, between the Points <lb/>T and A.</s> <s xml:id="echoid-s925" xml:space="preserve"/> </p> <pb o="36" file="0068" n="75" rhead="An ESSAY"/> <p style="it"> <s xml:id="echoid-s926" xml:space="preserve">The Reaſon of this is evident, from the Demon-<lb/>ſtration of the upright Cone: </s> <s xml:id="echoid-s927" xml:space="preserve">For, it is manifeſt, that <lb/>X is the Seat of the Center of the Circle, which the <lb/>Cone continued forms in the Horizontal Plane; </s> <s xml:id="echoid-s928" xml:space="preserve">and <lb/>conſequently, the Point s, in regard to the Circle BED, <lb/>is in the ſame Situation as the Eye is, in reſpect of <lb/>the Interſection of the continued Cone, and the Hori-<lb/>zontal Plane.</s> <s xml:id="echoid-s929" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s930" xml:space="preserve">Note, moreover, that a Cone can ſcarcely ever <lb/>be thrown into Perſpective, by the common Me-<lb/>thod, ſo exact as by this.</s> <s xml:id="echoid-s931" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div128" type="section" level="1" n="71"> <head xml:id="echoid-head76" xml:space="preserve"><emph style="sc">Problem</emph> VII.</head> <head xml:id="echoid-head77" style="it" xml:space="preserve">55. To find the Perſpective of a Line, perpendicular <lb/>to the Geometrical Plane.</head> <p> <s xml:id="echoid-s932" xml:space="preserve">It is requir’d to find the Appearance of a Line <lb/> <anchor type="note" xlink:label="note-0068-01a" xlink:href="note-0068-01"/> equal to B C, and perpendicular to the Geome-<lb/>trical Plane, in the Point A.</s> <s xml:id="echoid-s933" xml:space="preserve"/> </p> <div xml:id="echoid-div128" type="float" level="2" n="1"> <note position="left" xlink:label="note-0068-01" xlink:href="note-0068-01a" xml:space="preserve">Fig. 22.</note> </div> </div> <div xml:id="echoid-div130" type="section" level="1" n="72"> <head xml:id="echoid-head78" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s934" xml:space="preserve">Aſſume E D, any where in the Baſe Line, <lb/>equal to B C; </s> <s xml:id="echoid-s935" xml:space="preserve">and from the Points D and E, <lb/>draw D F and E F to ſome Point F, taken at <lb/>pleaſure in the Horizontal Line. </s> <s xml:id="echoid-s936" xml:space="preserve">Then having <lb/>found <anchor type="note" xlink:href="" symbol="*"/> a, the Repreſentation of the Point A;</s> <s xml:id="echoid-s937" xml:space="preserve"> <anchor type="note" xlink:label="note-0068-02a" xlink:href="note-0068-02"/> draw a H parallel to the Baſe Line, and aI per-<lb/>pendicular thereto: </s> <s xml:id="echoid-s938" xml:space="preserve">And if aI be made equal to <lb/>G H, the ſaid a I will be the Perſpective <lb/>ſought.</s> <s xml:id="echoid-s939" xml:space="preserve"/> </p> <div xml:id="echoid-div130" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0068-02" xlink:href="note-0068-02a" xml:space="preserve">22.</note> </div> </div> <div xml:id="echoid-div132" type="section" level="1" n="73"> <head xml:id="echoid-head79" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s940" xml:space="preserve">56. </s> <s xml:id="echoid-s941" xml:space="preserve">The Appearance of the ſaid Line, is <anchor type="note" xlink:href="" symbol="*"/> per- <anchor type="note" xlink:label="note-0068-03a" xlink:href="note-0068-03"/> pendicular to the Baſe Line, and equal <anchor type="note" xlink:href="" symbol="*"/> to the <anchor type="note" xlink:label="note-0068-04a" xlink:href="note-0068-04"/> Perſpective of the Line A L, drawn from the <pb file="0069" n="76"/> <pb file="0069a" n="77"/> <anchor type="figure" xlink:label="fig-0069a-01a" xlink:href="fig-0069a-01"/> <pb file="0070" n="78"/> <pb file="0071" n="79"/> <pb file="0071a" n="80"/> <anchor type="figure" xlink:label="fig-0071a-01a" xlink:href="fig-0071a-01"/> <anchor type="figure" xlink:label="fig-0071a-02a" xlink:href="fig-0071a-02"/> <anchor type="figure" xlink:label="fig-0071a-03a" xlink:href="fig-0071a-03"/> <anchor type="figure" xlink:label="fig-0071a-04a" xlink:href="fig-0071a-04"/> <pb file="0072" n="81"/> <pb file="0073" n="82"/> <pb file="0073a" n="83"/> <anchor type="figure" xlink:label="fig-0073a-01a" xlink:href="fig-0073a-01"/> <anchor type="figure" xlink:label="fig-0073a-02a" xlink:href="fig-0073a-02"/> <anchor type="figure" xlink:label="fig-0073a-03a" xlink:href="fig-0073a-03"/> <pb file="0074" n="84"/> <pb o="37" file="0075" n="85" rhead="on PERSPECTIVE."/> Point A parallel to the Baſe Line, and made e-<lb/>qual to B C. </s> <s xml:id="echoid-s942" xml:space="preserve">Now if from the Extremities of <lb/>the ſaid Line A L, Perpendiculars are let fall, <lb/>meeting the Baſe Line in the Points P and M, <lb/>and from theſe Points, Lines are drawn to the <lb/>Point of Sight V; </s> <s xml:id="echoid-s943" xml:space="preserve">then a N will likewiſe be <anchor type="note" xlink:href="" symbol="*"/> <anchor type="note" xlink:label="note-0075-01a" xlink:href="note-0075-01"/> the Perſpective of A L; </s> <s xml:id="echoid-s944" xml:space="preserve">and ſince P M is equal <lb/>to D E, a N will be likewiſe equal to G H, and <lb/>conſequently a N will be likewiſe equal to a I, <lb/>which is equal to G H.</s> <s xml:id="echoid-s945" xml:space="preserve"/> </p> <div xml:id="echoid-div132" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0068-03" xlink:href="note-0068-03a" xml:space="preserve">6.</note> <note symbol="*" position="left" xlink:label="note-0068-04" xlink:href="note-0068-04a" xml:space="preserve">10.</note> <figure xlink:label="fig-0069a-01" xlink:href="fig-0069a-01a"> <caption xml:id="echoid-caption16" style="it" xml:space="preserve">page 36.<lb/>Plate 8<lb/>Fig. 17</caption> <variables xml:id="echoid-variables16" xml:space="preserve">O G F c d b a A B D C</variables> </figure> <figure xlink:label="fig-0071a-01" xlink:href="fig-0071a-01a"> <caption xml:id="echoid-caption17" style="it" xml:space="preserve">Page 36.<lb/>Plate 9<lb/>Fig. 18.</caption> <variables xml:id="echoid-variables17" xml:space="preserve">G F C S V I E B A</variables> </figure> <figure xlink:label="fig-0071a-02" xlink:href="fig-0071a-02a"> <caption xml:id="echoid-caption18" style="it" xml:space="preserve">Fig. 19.</caption> <variables xml:id="echoid-variables18" xml:space="preserve">O i M X L D @ b a</variables> </figure> <figure xlink:label="fig-0071a-03" xlink:href="fig-0071a-03a"> <caption xml:id="echoid-caption19" style="it" xml:space="preserve">Fig. 20.</caption> <variables xml:id="echoid-variables19" xml:space="preserve">S x G n H S V D l R f Q m P t</variables> </figure> <figure xlink:label="fig-0071a-04" xlink:href="fig-0071a-04a"> <caption xml:id="echoid-caption20" style="it" xml:space="preserve">Fig. 21.</caption> <variables xml:id="echoid-variables20" xml:space="preserve">I X f T L B N A C l M E F</variables> </figure> <figure xlink:label="fig-0073a-01" xlink:href="fig-0073a-01a"> <caption xml:id="echoid-caption21" style="it" xml:space="preserve">page 38<lb/>Plate 10.<lb/>Fig. 22.</caption> <variables xml:id="echoid-variables21" xml:space="preserve">V F I N a G H M P D E B C L A</variables> </figure> <figure xlink:label="fig-0073a-02" xlink:href="fig-0073a-02a"> <caption xml:id="echoid-caption22" style="it" xml:space="preserve">Fig. 23.</caption> <variables xml:id="echoid-variables22" xml:space="preserve">O F I H a G D E B C L A M</variables> </figure> <figure xlink:label="fig-0073a-03" xlink:href="fig-0073a-03a"> <caption xml:id="echoid-caption23" style="it" xml:space="preserve">Fig. 24.</caption> <variables xml:id="echoid-variables23" xml:space="preserve">@ o f X a e A</variables> </figure> <note symbol="*" position="right" xlink:label="note-0075-01" xlink:href="note-0075-01a" xml:space="preserve">5, 16.</note> </div> </div> <div xml:id="echoid-div134" type="section" level="1" n="74"> <head xml:id="echoid-head80" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p> <s xml:id="echoid-s946" xml:space="preserve">57. </s> <s xml:id="echoid-s947" xml:space="preserve">The ſame Things being given, as in the <lb/>precedent Method, about the Point A, as a <lb/> <anchor type="note" xlink:label="note-0075-02a" xlink:href="note-0075-02"/> Center, and with the Radius B C, deſcribe the <lb/>Arc of a Circle L M, and draw the Line O L <lb/>from the Eye touching it; </s> <s xml:id="echoid-s948" xml:space="preserve">then about a, (which <lb/>is the Repreſentation of A) as a Center deſcribe <lb/>the Circular Arc G I touching the Line L O, <lb/>and cutting another Line drawn through a Per-<lb/>pendicular to the Baſe Line in the Point I: </s> <s xml:id="echoid-s949" xml:space="preserve">I <lb/>ſay the Point I is the Extremity of the Repre-<lb/>ſentation ſought.</s> <s xml:id="echoid-s950" xml:space="preserve"/> </p> <div xml:id="echoid-div134" type="float" level="2" n="1"> <note position="right" xlink:label="note-0075-02" xlink:href="note-0075-02a" xml:space="preserve">Fig. 23.</note> </div> </div> <div xml:id="echoid-div136" type="section" level="1" n="75"> <head xml:id="echoid-head81" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s951" xml:space="preserve">To prove this, let fall the Perpendiculars A L <lb/>and a G upon the Line O L, which will meet <lb/>the ſaid Line in the Points wherein it touches <lb/>the circular Arcs M L and G I.</s> <s xml:id="echoid-s952" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s953" xml:space="preserve">Alſo aſſume D E in the Baſe Line equal to B C <lb/>or A L, and draw the Line D F; </s> <s xml:id="echoid-s954" xml:space="preserve">then through a, <lb/>draw a H parallel to the Baſe Line.</s> <s xml:id="echoid-s955" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s956" xml:space="preserve">Now let us conſider the Figure X, which re-<lb/> <anchor type="note" xlink:label="note-0075-03a" xlink:href="note-0075-03"/> preſents a Plane paſſing through the Eye and <lb/>the Point A of the foregoing Figure, wherein <lb/>O f here, repreſents O F there; </s> <s xml:id="echoid-s957" xml:space="preserve">f e here, F E <pb o="38" file="0076" n="86" rhead="An ESSAY"/> there; </s> <s xml:id="echoid-s958" xml:space="preserve">and finally e A here, E A in that Fi-<lb/>gure.</s> <s xml:id="echoid-s959" xml:space="preserve"/> </p> <div xml:id="echoid-div136" type="float" level="2" n="1"> <note position="right" xlink:label="note-0075-03" xlink:href="note-0075-03a" xml:space="preserve">Fig. 24.</note> </div> <p> <s xml:id="echoid-s960" xml:space="preserve">This being ſuppoſed, o f is parallel <anchor type="note" xlink:href="" symbol="*"/> to e A, <anchor type="note" xlink:label="note-0076-01a" xlink:href="note-0076-01"/> and conſequently the Triangle o f a is ſimilar <lb/>to the Triangle a e A, and therefore we have this <lb/>Proportion.</s> <s xml:id="echoid-s961" xml:space="preserve"/> </p> <div xml:id="echoid-div137" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0076-01" xlink:href="note-0076-01a" xml:space="preserve">27.</note> </div> <p style="it"> <s xml:id="echoid-s962" xml:space="preserve">o f: </s> <s xml:id="echoid-s963" xml:space="preserve">f a:</s> <s xml:id="echoid-s964" xml:space="preserve">: A e: </s> <s xml:id="echoid-s965" xml:space="preserve">e o. <lb/></s> <s xml:id="echoid-s966" xml:space="preserve">Comp.</s> <s xml:id="echoid-s967" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s968" xml:space="preserve">o f + a f: </s> <s xml:id="echoid-s969" xml:space="preserve">f a:</s> <s xml:id="echoid-s970" xml:space="preserve">: A e + e a: </s> <s xml:id="echoid-s971" xml:space="preserve">e a. <lb/></s> <s xml:id="echoid-s972" xml:space="preserve">Altern.</s> <s xml:id="echoid-s973" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s974" xml:space="preserve">o f + a f: </s> <s xml:id="echoid-s975" xml:space="preserve">A e + e a:</s> <s xml:id="echoid-s976" xml:space="preserve">: f a: </s> <s xml:id="echoid-s977" xml:space="preserve">e a. <lb/></s> <s xml:id="echoid-s978" xml:space="preserve">Comp. </s> <s xml:id="echoid-s979" xml:space="preserve">and Perm.</s> <s xml:id="echoid-s980" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s981" xml:space="preserve">o f + f a + Ae + e a : </s> <s xml:id="echoid-s982" xml:space="preserve">o f + f a :</s> <s xml:id="echoid-s983" xml:space="preserve">: f a + e a : </s> <s xml:id="echoid-s984" xml:space="preserve">f a.</s> <s xml:id="echoid-s985" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s986" xml:space="preserve">This laſt Proportion being reduced to the pre-<lb/>cedent Figure, we ſhall have this,</s> </p> <p style="it"> <s xml:id="echoid-s987" xml:space="preserve">O A : </s> <s xml:id="echoid-s988" xml:space="preserve">o a : </s> <s xml:id="echoid-s989" xml:space="preserve">: </s> <s xml:id="echoid-s990" xml:space="preserve">F D : </s> <s xml:id="echoid-s991" xml:space="preserve">F a.</s> <s xml:id="echoid-s992" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s993" xml:space="preserve">Again, becauſe the Triangles O A L and O a G <lb/>are ſimilar, we ſhall have</s> </p> <p style="it"> <s xml:id="echoid-s994" xml:space="preserve">O A: </s> <s xml:id="echoid-s995" xml:space="preserve">O a : </s> <s xml:id="echoid-s996" xml:space="preserve">: </s> <s xml:id="echoid-s997" xml:space="preserve">A L : </s> <s xml:id="echoid-s998" xml:space="preserve">a G.</s> <s xml:id="echoid-s999" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1000" xml:space="preserve">And ſince the Triangle F E D and F a H are <lb/>ſimilar;</s> <s xml:id="echoid-s1001" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1002" xml:space="preserve">F E : </s> <s xml:id="echoid-s1003" xml:space="preserve">F a : </s> <s xml:id="echoid-s1004" xml:space="preserve">: </s> <s xml:id="echoid-s1005" xml:space="preserve">D E : </s> <s xml:id="echoid-s1006" xml:space="preserve">H a.</s> <s xml:id="echoid-s1007" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1008" xml:space="preserve">And ſo if theſe three laſt Propoſitions be con-<lb/>ſider’d, we ſhall have</s> </p> <p style="it"> <s xml:id="echoid-s1009" xml:space="preserve">A L : </s> <s xml:id="echoid-s1010" xml:space="preserve">a G : </s> <s xml:id="echoid-s1011" xml:space="preserve">: </s> <s xml:id="echoid-s1012" xml:space="preserve">D E : </s> <s xml:id="echoid-s1013" xml:space="preserve">H a.</s> <s xml:id="echoid-s1014" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1015" xml:space="preserve">But D E was made equal to A L, and there-<lb/>fore a G or a I is alſo equal to a H, which is <anchor type="note" xlink:href="" symbol="*"/> e- <anchor type="note" xlink:label="note-0076-02a" xlink:href="note-0076-02"/> qual to the Repreſentation ſought. </s> <s xml:id="echoid-s1016" xml:space="preserve">Which was <lb/>to be demonſtrated.</s> <s xml:id="echoid-s1017" xml:space="preserve"/> </p> <div xml:id="echoid-div138" type="float" level="2" n="3"> <note symbol="*" position="left" xlink:label="note-0076-02" xlink:href="note-0076-02a" xml:space="preserve">56.</note> </div> </div> <div xml:id="echoid-div140" type="section" level="1" n="76"> <head xml:id="echoid-head82" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p> <s xml:id="echoid-s1018" xml:space="preserve">58. </s> <s xml:id="echoid-s1019" xml:space="preserve">Near one of the Sides of the perſpective <lb/> <anchor type="note" xlink:label="note-0076-03a" xlink:href="note-0076-03"/> Plane, raiſe the Perpendicular C B to the Baſe <lb/>Line, equal to the Height of the Eye, in which <lb/>take B L equal in length to twice the Perpen-<lb/>dicular, whoſe Perſpective is requir’d. </s> <s xml:id="echoid-s1020" xml:space="preserve">Let S be <pb o="39" file="0077" n="87" rhead="on PERSPECTIVE."/> the Station Point, and A the Point wherein the <lb/>Perpendicular meets the Geometrical Plane.</s> <s xml:id="echoid-s1021" xml:space="preserve"/> </p> <div xml:id="echoid-div140" type="float" level="2" n="1"> <note position="left" xlink:label="note-0076-03" xlink:href="note-0076-03a" xml:space="preserve">Fig. 25.</note> </div> </div> <div xml:id="echoid-div142" type="section" level="1" n="77"> <head xml:id="echoid-head83" xml:space="preserve"><emph style="sc">Operation</emph>, <lb/>Without Compaſſes.</head> <p> <s xml:id="echoid-s1022" xml:space="preserve">Having firſt found <anchor type="note" xlink:href="" symbol="*"/> the Perſpective a of the <anchor type="note" xlink:label="note-0077-01a" xlink:href="note-0077-01"/> Point A, draw the Line A S cutting the Baſe <lb/>Line in E, through which Point E draw the <lb/>Line Ea; </s> <s xml:id="echoid-s1023" xml:space="preserve">then from the Point B draw a Line <lb/>B a to the Point a, cutting the Horizontal Line <lb/>in F. </s> <s xml:id="echoid-s1024" xml:space="preserve">Again through F draw a Line to the <lb/>Point L, cutting E a in I; </s> <s xml:id="echoid-s1025" xml:space="preserve">and a I is the Repre-<lb/>ſentation ſought.</s> <s xml:id="echoid-s1026" xml:space="preserve"/> </p> <div xml:id="echoid-div142" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0077-01" xlink:href="note-0077-01a" xml:space="preserve">31.</note> </div> </div> <div xml:id="echoid-div144" type="section" level="1" n="78"> <head xml:id="echoid-head84" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1027" xml:space="preserve">To prove this, let G N be a Perpendicular to <lb/>the Baſe Line drawn from the Point G, wherein <lb/>the ſaid Baſe Line is cut by the Line B F; </s> <s xml:id="echoid-s1028" xml:space="preserve">alſo <lb/>let G D be equal to the Perpendicular whoſe Ap-<lb/>pearance is ſought, and a H parallel to the Baſe <lb/>Line.</s> <s xml:id="echoid-s1029" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1030" xml:space="preserve">It is plain that the Perſpective of E A is <lb/>E a: </s> <s xml:id="echoid-s1031" xml:space="preserve">But E A paſſes through the Station Point; <lb/></s> <s xml:id="echoid-s1032" xml:space="preserve">and conſequently <anchor type="note" xlink:href="" symbol="*"/> its Repreſentation is perpen- <anchor type="note" xlink:label="note-0077-02a" xlink:href="note-0077-02"/> dicular to the Baſe Line; </s> <s xml:id="echoid-s1033" xml:space="preserve">therefore <anchor type="note" xlink:href="" symbol="*"/> we are only <anchor type="note" xlink:label="note-0077-03a" xlink:href="note-0077-03"/> to prove, that a I is equal to a H.</s> <s xml:id="echoid-s1034" xml:space="preserve"/> </p> <div xml:id="echoid-div144" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0077-02" xlink:href="note-0077-02a" xml:space="preserve">41.</note> <note symbol="*" position="right" xlink:label="note-0077-03" xlink:href="note-0077-03a" xml:space="preserve">56.</note> </div> <p> <s xml:id="echoid-s1035" xml:space="preserve">Now the Triangles B G C and B F M are ſimi-<lb/>lar; </s> <s xml:id="echoid-s1036" xml:space="preserve">and ſo</s> </p> <p style="it"> <s xml:id="echoid-s1037" xml:space="preserve">B C : </s> <s xml:id="echoid-s1038" xml:space="preserve">B M :</s> <s xml:id="echoid-s1039" xml:space="preserve">: B G: </s> <s xml:id="echoid-s1040" xml:space="preserve">B F.</s> <s xml:id="echoid-s1041" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1042" xml:space="preserve">But B M by Conſtruction is the double of B C; <lb/></s> <s xml:id="echoid-s1043" xml:space="preserve">whence B F is alſo the double of B G, which, <lb/>conſequently, is equal to G F.</s> <s xml:id="echoid-s1044" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1045" xml:space="preserve">Becauſe the Triangles F G N and F B L are <lb/>ſimilar, therefore</s> </p> <p style="it"> <s xml:id="echoid-s1046" xml:space="preserve">F G : </s> <s xml:id="echoid-s1047" xml:space="preserve">F B : </s> <s xml:id="echoid-s1048" xml:space="preserve">: </s> <s xml:id="echoid-s1049" xml:space="preserve">G N : </s> <s xml:id="echoid-s1050" xml:space="preserve">B L.</s> <s xml:id="echoid-s1051" xml:space="preserve"/> </p> <pb o="40" file="0078" n="88" rhead="An ESSAY"/> <p> <s xml:id="echoid-s1052" xml:space="preserve">Now we have proved, that F G is the half of <lb/>F B, therefore G N is likewiſe equal to the half <lb/>of B L, and conſequently equal to the Height <lb/>of the ſuppoſed Perpendicular.</s> <s xml:id="echoid-s1053" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1054" xml:space="preserve">Again, the ſimilar Triangles F G N and F a I <lb/>give</s> </p> <p style="it"> <s xml:id="echoid-s1055" xml:space="preserve">F G : </s> <s xml:id="echoid-s1056" xml:space="preserve">F a : </s> <s xml:id="echoid-s1057" xml:space="preserve">: </s> <s xml:id="echoid-s1058" xml:space="preserve">G N : </s> <s xml:id="echoid-s1059" xml:space="preserve">a I.</s> <s xml:id="echoid-s1060" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1061" xml:space="preserve">But F G : </s> <s xml:id="echoid-s1062" xml:space="preserve">F a : </s> <s xml:id="echoid-s1063" xml:space="preserve">: </s> <s xml:id="echoid-s1064" xml:space="preserve">G D : </s> <s xml:id="echoid-s1065" xml:space="preserve">a H; </s> <s xml:id="echoid-s1066" xml:space="preserve">becauſe the Tri-<lb/>angles F G D and F a H are ſimilar.</s> <s xml:id="echoid-s1067" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1068" xml:space="preserve">Whence</s> </p> <p style="it"> <s xml:id="echoid-s1069" xml:space="preserve">G N : </s> <s xml:id="echoid-s1070" xml:space="preserve">a I : </s> <s xml:id="echoid-s1071" xml:space="preserve">: </s> <s xml:id="echoid-s1072" xml:space="preserve">G D : </s> <s xml:id="echoid-s1073" xml:space="preserve">a H.</s> <s xml:id="echoid-s1074" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1075" xml:space="preserve">Now becauſe G N has been proved to be e-<lb/>qual to the Perpendicular, whoſe Perſpective is <lb/>requir’d and D G is ſuppoſed equal to that Per-<lb/>pendicular; </s> <s xml:id="echoid-s1076" xml:space="preserve">it follows, that G N and G D are <lb/>equal; </s> <s xml:id="echoid-s1077" xml:space="preserve">and therefore a I and a H are alſo equal. <lb/></s> <s xml:id="echoid-s1078" xml:space="preserve">Q E D.</s> <s xml:id="echoid-s1079" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div146" type="section" level="1" n="79"> <head xml:id="echoid-head85" xml:space="preserve"><emph style="sc">Scholium</emph>.</head> <p> <s xml:id="echoid-s1080" xml:space="preserve">I might have aſſumed C P equal to the Perpen-<lb/>dicular, and uſed the Points C and P inſtead of <lb/>B and L. </s> <s xml:id="echoid-s1081" xml:space="preserve">But uſing the ſaid Points B and L is <lb/>better: </s> <s xml:id="echoid-s1082" xml:space="preserve">For when the Points C and P are uſed, <lb/>the Horizontal Line muſt almoſt always be con-<lb/>tinued, that ſo a Line drawn through the Points <lb/>c and a may cut it; </s> <s xml:id="echoid-s1083" xml:space="preserve">moreover this Interſection <lb/>will ſometimes be at an infinite Diſtance; </s> <s xml:id="echoid-s1084" xml:space="preserve">where-<lb/>as in uſing the Point B, M N can never be <lb/>greater than thrice the Breadth of the Deſign to <lb/>be drawn.</s> <s xml:id="echoid-s1085" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div147" type="section" level="1" n="80"> <head xml:id="echoid-head86" xml:space="preserve"><emph style="sc">Corollary</emph>.</head> <p> <s xml:id="echoid-s1086" xml:space="preserve">The ſixth Problem may be ſolv’d by this; <lb/></s> <s xml:id="echoid-s1087" xml:space="preserve">for a Point elevated above the Geometrical <lb/>Plane, may be conceived as the Extremity of a <lb/>Perpendicular to the Geometrical Plane.</s> <s xml:id="echoid-s1088" xml:space="preserve"/> </p> <pb o="41" file="0079" n="89" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div148" type="section" level="1" n="81"> <head xml:id="echoid-head87" xml:space="preserve"><emph style="sc">Problem</emph> VIII.</head> <p style="it"> <s xml:id="echoid-s1089" xml:space="preserve">59. </s> <s xml:id="echoid-s1090" xml:space="preserve">To throw a Priſm or Cylinder into Perſpective, <lb/> <anchor type="note" xlink:label="note-0079-01a" xlink:href="note-0079-01"/> both of them being Perpendicular to the Geometri-<lb/>cal Plane.</s> <s xml:id="echoid-s1091" xml:space="preserve"/> </p> <div xml:id="echoid-div148" type="float" level="2" n="1"> <note position="right" xlink:label="note-0079-01" xlink:href="note-0079-01a" xml:space="preserve">Fig. 26.</note> </div> <p> <s xml:id="echoid-s1092" xml:space="preserve">Let G H I L M N be the Baſe of the Priſm <lb/>in the Geometrical Plane, and the viſible Part <lb/>thereof upon the perſpective Plane, let be n g h i; <lb/></s> <s xml:id="echoid-s1093" xml:space="preserve">then to compleat the Repreſentation of the <lb/>Priſm, draw Perpendiculars from the Points <lb/>n g h and i to the Baſe Line, whoſe Length let <lb/>be <anchor type="note" xlink:href="" symbol="*"/> ſuch that they may repreſent Perpendicu- <anchor type="note" xlink:label="note-0079-02a" xlink:href="note-0079-02"/> lars to the Geometrical Plane, equal to the <lb/>Height of the Priſm, and find <anchor type="note" xlink:href="" symbol="*"/> the Perſpective <anchor type="note" xlink:label="note-0079-03a" xlink:href="note-0079-03"/> of the other Angular Points of the upper Sur-<lb/>face of the Priſm, in confidering them as Points <lb/>elevated above the Geometrical Plane: </s> <s xml:id="echoid-s1094" xml:space="preserve">This <lb/>being done, if the Repreſentations of all the <lb/>ſaid Angular Points be joyn’d, the whole <lb/>Priſm will be thrown into Perſpective.</s> <s xml:id="echoid-s1095" xml:space="preserve"/> </p> <div xml:id="echoid-div149" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0079-02" xlink:href="note-0079-02a" xml:space="preserve">55.</note> <note symbol="*" position="right" xlink:label="note-0079-03" xlink:href="note-0079-03a" xml:space="preserve">50.</note> </div> <p> <s xml:id="echoid-s1096" xml:space="preserve">Now to throw a Cylinder into Perſpective, <lb/>the Repreſentation of its Baſe and upper Sur-<lb/>face muſt firſt be had, by finding <anchor type="note" xlink:href="" symbol="*"/> the Appear- <anchor type="note" xlink:label="note-0079-04a" xlink:href="note-0079-04"/> ance of ſeveral Points of the Periphery of its <lb/>upper Surface, and then two Perpendiculars <lb/>muſt be ſo drawn to the Baſe Line, that they <lb/>may touch the Appearances of the two circular <lb/>Euds of the Cylinder, and the Appearance of <lb/>the Cylinder will be had. </s> <s xml:id="echoid-s1097" xml:space="preserve">But to avoid uſeleſs <lb/>Operations, the viſible Part of the Baſe of the <lb/>Cylinder may be thus determin’d. </s> <s xml:id="echoid-s1098" xml:space="preserve">Draw the <lb/>Line A S from the Point A to the Station Point S, <lb/>then this Line muſt be biſected in the Point R, <lb/>about which, as a Centre, and with the Radius <lb/>R A, the Circular Arc B A C, muſt be deſcrib’d <lb/>cutting the Baſe of the Cylinder in the Points <pb o="42" file="0080" n="90" rhead="An ESSAY"/> B and C, which will be the two moſt extreme <lb/>ones that can be ſeen.</s> <s xml:id="echoid-s1099" xml:space="preserve"/> </p> <div xml:id="echoid-div150" type="float" level="2" n="3"> <note symbol="*" position="right" xlink:label="note-0079-04" xlink:href="note-0079-04a" xml:space="preserve">50.</note> </div> </div> <div xml:id="echoid-div152" type="section" level="1" n="82"> <head xml:id="echoid-head88" style="it" xml:space="preserve">To do this another Way.</head> <p> <s xml:id="echoid-s1100" xml:space="preserve">61. </s> <s xml:id="echoid-s1101" xml:space="preserve">If the upper Face of the Cylinder or Priſm <lb/> <anchor type="note" xlink:label="note-0080-01a" xlink:href="note-0080-01"/> be otherwiſe requir’d to be found, the ſame Things <lb/>being given as in the foregoing Method, we draw <lb/>the Line P Q in the perſpective Plane, parallel <lb/>to the Baſe Line, whoſe Diſtance therefrom we <lb/>make equal to the Height of the Priſm or Cy-<lb/>linder, whoſe Perſpective is requir’d. </s> <s xml:id="echoid-s1102" xml:space="preserve">Then we <lb/>change its Geometrical Plane, ſo that the Baſe <lb/>Line coincides with P Q, and that in this Tran-<lb/>ſpoſition a Perpendicular to the Baſe Line coin-<lb/>cides with this ſame Perpendicular continued to-<lb/> <anchor type="note" xlink:label="note-0080-02a" xlink:href="note-0080-02"/> wards P Q. </s> <s xml:id="echoid-s1103" xml:space="preserve">Finally we find <anchor type="note" xlink:href="" symbol="*"/> the Perſpective of the Baſe of the Priſm or Cylinder, thus changed <lb/>in Situation by uſing P Q for a Baſe Line, and <lb/>the ſaid Perſpective is the Repreſentation of their <lb/>upper Faces.</s> <s xml:id="echoid-s1104" xml:space="preserve"/> </p> <div xml:id="echoid-div152" type="float" level="2" n="1"> <note position="left" xlink:label="note-0080-01" xlink:href="note-0080-01a" xml:space="preserve">Fig. 26, <lb/>27.</note> <note symbol="*" position="left" xlink:label="note-0080-02" xlink:href="note-0080-02a" xml:space="preserve">46.</note> </div> </div> <div xml:id="echoid-div154" type="section" level="1" n="83"> <head xml:id="echoid-head89" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1105" xml:space="preserve">If we ſuppoſe the Plane of the upper Surface <lb/>of the Priſm to be continued, it will meet the <lb/>Perſpective Plane in P Q; </s> <s xml:id="echoid-s1106" xml:space="preserve">and the upper Face <lb/>in this Plane continued, will have the ſame <lb/>Situation in Reſpect to P Q, as the Baſe hath <lb/>on the Geometrical Plane with Regard to the <lb/>Baſe Line. </s> <s xml:id="echoid-s1107" xml:space="preserve">If then the ſaid continued Plane be <lb/>conceived to lye on the perſpective Plane, the <lb/>upper Faces of the Priſm or Cylinder, will be <lb/>as the Baſes changed in the Manner aforeſaid; <lb/></s> <s xml:id="echoid-s1108" xml:space="preserve">therefore the Appearance of the ſaid Baſes <lb/>changed, will be that of the upper Surfaces.</s> <s xml:id="echoid-s1109" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1110" xml:space="preserve">Note, By folding the Paper it is eaſy to <lb/>tranſpoſe Figures, and when the Height of the <pb file="0081" n="91"/> <pb file="0081a" n="92"/> <anchor type="figure" xlink:label="fig-0081a-01a" xlink:href="fig-0081a-01"/> <anchor type="figure" xlink:label="fig-0081a-02a" xlink:href="fig-0081a-02"/> <pb file="0082" n="93"/> <pb o="43" file="0083" n="94" rhead="on PERSPECTIVE."/> Priſm is greater than the Height of the Eye, the <lb/>precedent Method is the ſhorteſt.</s> <s xml:id="echoid-s1111" xml:space="preserve"/> </p> <div xml:id="echoid-div154" type="float" level="2" n="1"> <figure xlink:label="fig-0081a-01" xlink:href="fig-0081a-01a"> <caption xml:id="echoid-caption24" style="it" xml:space="preserve">page 42<lb/>Plate 11.<lb/>Fig. 25.</caption> <variables xml:id="echoid-variables24" xml:space="preserve">S F V M I N P H a L D E G C A B</variables> </figure> <figure xlink:label="fig-0081a-02" xlink:href="fig-0081a-02a"> <caption xml:id="echoid-caption25" style="it" xml:space="preserve">Fig. 26.<lb/>Fig. 27.</caption> <variables xml:id="echoid-variables25" xml:space="preserve">S V P Q R n l g h G H B N I A C M L</variables> </figure> </div> </div> <div xml:id="echoid-div156" type="section" level="1" n="84"> <head xml:id="echoid-head90" xml:space="preserve"><emph style="sc">Problem</emph> IX.</head> <p style="it"> <s xml:id="echoid-s1112" xml:space="preserve">62. </s> <s xml:id="echoid-s1113" xml:space="preserve">To throw a Concave Body into Perſpective.</s> <s xml:id="echoid-s1114" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Fig. 28.</note> <p> <s xml:id="echoid-s1115" xml:space="preserve">Having firſt ſound the Perſpective of the ſaid <lb/>Body, afterwards find the Appearance of its <lb/>Cavity, in conſidering the Cavity as a new <lb/>Body.</s> <s xml:id="echoid-s1116" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div157" type="section" level="1" n="85"> <head xml:id="echoid-head91" xml:space="preserve"><emph style="sc">Problem</emph> X.</head> <p style="it"> <s xml:id="echoid-s1117" xml:space="preserve">63. </s> <s xml:id="echoid-s1118" xml:space="preserve">To throw a Sphere into Perſpective.</s> <s xml:id="echoid-s1119" xml:space="preserve"/> </p> <note position="right" xml:space="preserve">Fig. 29.</note> <p> <s xml:id="echoid-s1120" xml:space="preserve">Let A be the Seat of the Centre of the Sphere; <lb/></s> <s xml:id="echoid-s1121" xml:space="preserve">then the Point I the Perſpective of the Centre <lb/>muſt be found, <anchor type="note" xlink:href="" symbol="*"/> and the Line IV drawn to the <anchor type="note" xlink:label="note-0083-03a" xlink:href="note-0083-03"/> Point of Sight V. </s> <s xml:id="echoid-s1122" xml:space="preserve">This being done, raiſe V F per-<lb/>pendicular to V I, which make equal to the <lb/>Diſtance from the Eye to the perſpective Plane; <lb/></s> <s xml:id="echoid-s1123" xml:space="preserve">and in this Perpendicular continued, take V P <lb/>equal to the Diſtance from the Centre of the <lb/>Sphere to the perſpective Plane. </s> <s xml:id="echoid-s1124" xml:space="preserve">Through the <lb/>Point P draw P Q parallel to V I cutting a Line <lb/>drawn from F through I, in Q; </s> <s xml:id="echoid-s1125" xml:space="preserve">and about Q as <lb/>a Centre, with the Semidiameter of the Sphere, <lb/>draw the Circle C B, to which from the Point F, <lb/>draw the Tangents F C and F B, cutting the <lb/>Line I V in the Points G and E. </s> <s xml:id="echoid-s1126" xml:space="preserve">On the Line <lb/>G E deſcribe the ſemicircle E D T G, wherein <lb/>draw the Line G D perpendicular to F I, which <lb/>biſect in H, and about H, as a Centre with the Ra-<lb/>dius H D, deſcribe the Arc of a Circle, L D R, cut-<lb/>ting the Line F I in the Points L and R. </s> <s xml:id="echoid-s1127" xml:space="preserve">Take <lb/>the Chord G T in the Semicircle E D T G equal <lb/>to R L, and deſcribe a Semicircle T m G upon <lb/>G T; </s> <s xml:id="echoid-s1128" xml:space="preserve">in which Semicircle draw ſeveral Lines, <lb/>as m n Perpendicular to G T; </s> <s xml:id="echoid-s1129" xml:space="preserve">and cutting the <pb o="44" file="0084" n="95" rhead="An ESSAY"/> Line G E, in the Points p, from every of which <lb/>raiſe Perpendiculars p q, each of which muſt <lb/>be continued on each Side the Line G E, equal <lb/>to m n the Part of the correſpondent Line p m. <lb/></s> <s xml:id="echoid-s1130" xml:space="preserve">Now if a great Number of the Points q be thus <lb/>found, and they are joyn’d by an even Hand, <lb/>you will have a Curve Line which will be the <lb/>Repreſentation ſought.</s> <s xml:id="echoid-s1131" xml:space="preserve"/> </p> <div xml:id="echoid-div157" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0083-03" xlink:href="note-0083-03a" xml:space="preserve">50.</note> </div> </div> <div xml:id="echoid-div159" type="section" level="1" n="86"> <head xml:id="echoid-head92" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p style="it"> <s xml:id="echoid-s1132" xml:space="preserve">The Rays by which we perceive a Sphere, do form <lb/>an upright Cone, whoſe Axis paſſes through the Cen-<lb/>ter of the Sphere, and whoſe Section made by the <lb/>Perſpective Plane, is the Repreſentation ſought: </s> <s xml:id="echoid-s1133" xml:space="preserve">from <lb/>whence it follows, that I is the Point in the Perſpe-<lb/>ctive Plane, through which the Cone’s Axis paſſes. <lb/></s> <s xml:id="echoid-s1134" xml:space="preserve">But when an upright Cone is ſo cut by a Plane, that <lb/>the Section is an Ellipſis, as in this Caſe, the tranſ-<lb/>verſe Diameter of this Ellipſis, will paſs through <lb/>the Point of Concurrence of the ſaid Plane, and <lb/>Axis of the Cone, and that Point wherein a Per-<lb/>pendicular drawn from the Vertex of the Cone, cuts <lb/>the ſaid Plane. </s> <s xml:id="echoid-s1135" xml:space="preserve">This will appear evident enough <lb/>to any one of but mean Knowledge in Conick Secti-<lb/>ons. </s> <s xml:id="echoid-s1136" xml:space="preserve">Therefore the tranſverſe Axis of the Ellipſis, <lb/>which is the Repreſentation of the Sphere, is ſome <lb/>Part of V I; </s> <s xml:id="echoid-s1137" xml:space="preserve">for the Eye is the Vertex of the Cone <lb/>formed by the viſual Rays of the Spbere.</s> <s xml:id="echoid-s1138" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1139" xml:space="preserve">Now let us conceive a Plane to paſs through the <lb/>Eye, and the Line I V; </s> <s xml:id="echoid-s1140" xml:space="preserve">this will paſs through the <lb/>Center of the Sphere: </s> <s xml:id="echoid-s1141" xml:space="preserve">And if a Perpendicular be <lb/>let fall from the Center upon the principal Ray con-<lb/>tinued, that Part of the ſaid Ray included between <lb/>the Point of Sight, and the Point wherein this Per-<lb/>pendicular falls, which is always parallel to the Per-<lb/>ſpective Plane, will be equal to the Diſtance from <lb/>the Center of the Sphere to the Perſpective Plane, <pb o="45" file="0085" n="96" rhead="on PERSPECTIVE."/> and conſequently to V P. </s> <s xml:id="echoid-s1142" xml:space="preserve">Therefore if the before-<lb/>mentioned Plane be ſuppoſed to revolve upon the Line <lb/>V I, as an Axis, until it coincides with the Per-<lb/>ſpective Plane, the Center of the Sphere will meet <lb/>the Perſpective Plane in Q, and the Eye in F; <lb/></s> <s xml:id="echoid-s1143" xml:space="preserve">whence the Part G E of the Line I V is the tranſ-<lb/>verſe Diameter of the Ellipſis.</s> <s xml:id="echoid-s1144" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1145" xml:space="preserve">Again let G D E in Figure 30, and g e f, in <lb/> <anchor type="note" xlink:label="note-0085-01a" xlink:href="note-0085-01"/> Figure 31 repreſent the Points denoted with the ſame <lb/>Letters in the foregoing Figure. </s> <s xml:id="echoid-s1146" xml:space="preserve">Now if the Cone, <lb/>whoſe Profile is denoted by the Lines f g and fe be ſup-<lb/>poſed to be compleated, and to be cut by a Plane paſ-<lb/>ſing through the Line g e perpendicular to the Plane <lb/>of the Figure; </s> <s xml:id="echoid-s1147" xml:space="preserve">we ſhall have an Ellipſis g 4 e 3 <lb/>ſimilar to that which is the ſought Repreſentation <lb/>of the Sphere. </s> <s xml:id="echoid-s1148" xml:space="preserve">Further if the ſaid Cone be conceived <lb/>to be cut by a Plane 14 m 3 parallel to its Baſe, <lb/>and biſecting g e in n, it is manifeſt, that 3 4, the <lb/>common Section of the Circle 14 m 3, and the Ellip-<lb/>ſis g 4 e 3, is the conjugate Axis of the Ellip-<lb/>ſis. </s> <s xml:id="echoid-s1149" xml:space="preserve">And therefore this conjugate Axis is equal <lb/>to the Line 3 4, Perpendicular in the Point n to the <lb/>Diameter 1 m of the Circle 14 m 3. </s> <s xml:id="echoid-s1150" xml:space="preserve">Now draw <lb/>the Lines E O and G Y in Figure 30, parallel to <lb/>L M, then the Triangles E G Y and E N M are <lb/>ſimilar, whence</s> </p> <div xml:id="echoid-div159" type="float" level="2" n="1"> <note position="right" xlink:label="note-0085-01" xlink:href="note-0085-01a" xml:space="preserve">Fig. 30, <lb/>31.</note> </div> </div> <div xml:id="echoid-div161" type="section" level="1" n="87"> <head xml:id="echoid-head93" xml:space="preserve">EG: EN:: GY: NM.</head> <p> <s xml:id="echoid-s1151" xml:space="preserve">But E G is twice E N; </s> <s xml:id="echoid-s1152" xml:space="preserve">wherefore G Y is alſo the <lb/>double of N M, and ſo N M equal to G Z. </s> <s xml:id="echoid-s1153" xml:space="preserve">After <lb/>the ſame manner we demonſtrate, that L N is equal <lb/>to X E; </s> <s xml:id="echoid-s1154" xml:space="preserve">whence it follows, that G D is equal to <lb/>L M, and is ſo cut in z as L M is in N; </s> <s xml:id="echoid-s1155" xml:space="preserve">and there-<lb/>fore R L or G T of Figure 29, is equal to 34 in <lb/>Figure 31; </s> <s xml:id="echoid-s1156" xml:space="preserve">and conſequently equal to the conjugate <lb/>Axis of the Ellipſis to be drawn. </s> <s xml:id="echoid-s1157" xml:space="preserve">On the other <lb/>Hand, it is manifeſt by Conſtruction, that ſome one <lb/>of the Perpendiculars m n, Figure 29, viz. </s> <s xml:id="echoid-s1158" xml:space="preserve">that <lb/>which paſſes through the Center of the ſemicircle <pb o="46" file="0086" n="97" rhead="An ESSAY"/> G m T, biſects the Axis G E: </s> <s xml:id="echoid-s1159" xml:space="preserve">For if a Line be <lb/>drawn from T to E, it will be perpendicular to G T, <lb/>and conſequently parallel to m n: </s> <s xml:id="echoid-s1160" xml:space="preserve">Whence the con-<lb/>jugate Axis of the Curve G q E, is equal to the <lb/>conjugate Axis of the Ellipſis to be drawn: </s> <s xml:id="echoid-s1161" xml:space="preserve">And <lb/>therefore we are only to prove, that the Curve paſ-<lb/>ſing through the Points q, is an Ellipſis. </s> <s xml:id="echoid-s1162" xml:space="preserve">Which may <lb/>be ſbewnthus.</s> <s xml:id="echoid-s1163" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1164" xml:space="preserve">The Parts G n of the Line G T, are Propor-<lb/>tional to the Parts G p of the Line G E: </s> <s xml:id="echoid-s1165" xml:space="preserve">Whence <lb/>the Rectangles under G p and p E, are Proportional <lb/>to the Rectangles under G n and n T; </s> <s xml:id="echoid-s1166" xml:space="preserve">but theſe laſt <lb/>Rectangles are equal to the Squares of the Ordinates <lb/>n m, which Squares are equal to the Squares of the <lb/>Ordinates p q; </s> <s xml:id="echoid-s1167" xml:space="preserve">therefore theſe laſt Squares are Pro-<lb/>portional to the Rectangles under G p and p E, which <lb/>is a Property of the Ellipſis.</s> <s xml:id="echoid-s1168" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div162" type="section" level="1" n="88"> <head xml:id="echoid-head94" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p> <s xml:id="echoid-s1169" xml:space="preserve">The ſemicircular Part h m of a Column, en-<lb/> <anchor type="note" xlink:label="note-0086-01a" xlink:href="note-0086-01"/> compaſſing the ſame like a Ring, is called the <lb/>Torus.</s> <s xml:id="echoid-s1170" xml:space="preserve"/> </p> <div xml:id="echoid-div162" type="float" level="2" n="1"> <note position="left" xlink:label="note-0086-01" xlink:href="note-0086-01a" xml:space="preserve">Fig. 33.</note> </div> </div> <div xml:id="echoid-div164" type="section" level="1" n="89"> <head xml:id="echoid-head95" xml:space="preserve"><emph style="sc">Problem</emph> XI.</head> <p style="it"> <s xml:id="echoid-s1171" xml:space="preserve">64. </s> <s xml:id="echoid-s1172" xml:space="preserve">To throw the Torus of a Column into Per-<lb/>ſpective.</s> <s xml:id="echoid-s1173" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1174" xml:space="preserve">Let B N C be the Baſe of the Column in the <lb/> <anchor type="note" xlink:label="note-0086-02a" xlink:href="note-0086-02"/> Geometrical Plane; </s> <s xml:id="echoid-s1175" xml:space="preserve">draw a Line from the Cen-<lb/>ter A to the Station Point S, which biſect in the <lb/>Point R, and deſcribe the Arc of a Circle B A C <lb/>about the Point R, as a Center with the Radius R A.</s> <s xml:id="echoid-s1176" xml:space="preserve"/> </p> <div xml:id="echoid-div164" type="float" level="2" n="1"> <note position="left" xlink:label="note-0086-02" xlink:href="note-0086-02a" xml:space="preserve">Fig. 32.</note> </div> <p> <s xml:id="echoid-s1177" xml:space="preserve">Let X be the Profile of the Column, in which <lb/> <anchor type="note" xlink:label="note-0086-03a" xlink:href="note-0086-03"/> draw the Line z 36, through the Center of the <lb/>ſemicircle h m, parallel to the Baſe of the Co-<lb/>lumn; </s> <s xml:id="echoid-s1178" xml:space="preserve">and in the Line s a, which goes through <lb/>the Center of the Column, parallel to its Sides, <pb file="0087" n="98"/> <pb file="0087a" n="99"/> <anchor type="figure" xlink:label="fig-0087a-01a" xlink:href="fig-0087a-01"/> <anchor type="figure" xlink:label="fig-0087a-02a" xlink:href="fig-0087a-02"/> <anchor type="figure" xlink:label="fig-0087a-03a" xlink:href="fig-0087a-03"/> <anchor type="figure" xlink:label="fig-0087a-04a" xlink:href="fig-0087a-04"/> <pb file="0088" n="100"/> <pb o="47" file="0089" n="101" rhead="on PERSPECTIVE."/> take the Part 2 s, equal to the Height of the Eye <lb/>above the Point 2, which is in the Baſe of the <lb/>Column; </s> <s xml:id="echoid-s1179" xml:space="preserve">likewiſe aſſume s a in the ſaid Line <lb/>equal to S A of the precedent Figure, and from the <lb/>Point a draw to a s the indefinite Perpendicular <lb/>a Y. </s> <s xml:id="echoid-s1180" xml:space="preserve">Theſe General Preparations being made, <lb/>take at Pleaſure the ſmall equal Parts 6 i and 69 <lb/>in the Line s a; </s> <s xml:id="echoid-s1181" xml:space="preserve">draw the Lines i h and 9 m Pa-<lb/>rallel to 63 z, and from the Point h draw the Line <lb/>h 3 4, thro’ the Center 3 of the Semicircle h m; <lb/></s> <s xml:id="echoid-s1182" xml:space="preserve">aſſume a 5 in a Y equal to i 4, and draw the Line <lb/>5 s cutting i b in g, and 9 m in q. </s> <s xml:id="echoid-s1183" xml:space="preserve">And, (in Figure <lb/>32.) </s> <s xml:id="echoid-s1184" xml:space="preserve">about the Point A, as a Center, with the <lb/>Radius i h or 9 m, which are equal, deſcribe the <lb/>Circle F L M H, cutting the Arc B A C in the <lb/>Points D and E; </s> <s xml:id="echoid-s1185" xml:space="preserve">then draw the Line D E cut-<lb/>ting the Line A S in I; </s> <s xml:id="echoid-s1186" xml:space="preserve">aſſume I G equal to i g, <lb/>and I Q equal to 9 q; </s> <s xml:id="echoid-s1187" xml:space="preserve">and thro’ the Points Q and <lb/>G, draw F H and L M, parallel to the Line E D, <lb/>cutting the Circle D M E F in the Points L, <lb/>M, F, and H. </s> <s xml:id="echoid-s1188" xml:space="preserve">Now if the Repreſentations of Four <lb/>Points, whereof L M F and H, are the Seats, and <lb/>the two firſt of which is equal to 29, and of the <lb/>two others 2 i, be found*; </s> <s xml:id="echoid-s1189" xml:space="preserve">the Repreſentation <lb/>of the ſaid four Points will be ſo many Points of <lb/>Appearance ſought. </s> <s xml:id="echoid-s1190" xml:space="preserve">And by drawing two other <lb/>Lines, as i h and 9 m, and proceeding as be-<lb/>fore, the Repreſentation of ſo many more Points <lb/>will be had.</s> <s xml:id="echoid-s1191" xml:space="preserve"/> </p> <div xml:id="echoid-div165" type="float" level="2" n="2"> <note position="left" xlink:label="note-0086-03" xlink:href="note-0086-03a" xml:space="preserve">Fig. 33.</note> <figure xlink:label="fig-0087a-01" xlink:href="fig-0087a-01a"> <caption xml:id="echoid-caption26" style="it" xml:space="preserve">page 46<lb/>Plate 12.<lb/>Fig. 28.</caption> </figure> <figure xlink:label="fig-0087a-02" xlink:href="fig-0087a-02a"> <caption xml:id="echoid-caption27" style="it" xml:space="preserve">Fig. 29.</caption> <variables xml:id="echoid-variables26" xml:space="preserve">F S V q q q E L p p p I G H q D P n n n T R m m m C B Q A</variables> </figure> <figure xlink:label="fig-0087a-03" xlink:href="fig-0087a-03a"> <caption xml:id="echoid-caption28" style="it" xml:space="preserve">Fig. 30.</caption> <variables xml:id="echoid-variables27" xml:space="preserve">O X E L N M G Z Y D</variables> </figure> <figure xlink:label="fig-0087a-04" xlink:href="fig-0087a-04a"> <caption xml:id="echoid-caption29" style="it" xml:space="preserve">Fig. 31.</caption> <variables xml:id="echoid-variables28" xml:space="preserve">f 3 c l n m g 4</variables> </figure> </div> <p> <s xml:id="echoid-s1192" xml:space="preserve">Note, Becauſe a part of the Torus is hid by <lb/> <anchor type="note" xlink:label="note-0089-01a" xlink:href="note-0089-01"/> the Column, therefore to avoid uſeleſs Operations, <lb/>a Circle muſt be deſcribed about the Center A, <lb/>with the Radius 36, cutting the Arc B C A in <lb/>the Points T and O, and the Lines STY and SOZ <lb/>muſt be drawn; </s> <s xml:id="echoid-s1193" xml:space="preserve">then all the Points as F and H, <lb/>falling between the Lines TY and O Z are uſeleſs, <lb/>and L and M not coming under this Obſervation <lb/>muſt only be uſed; </s> <s xml:id="echoid-s1194" xml:space="preserve">Note alſo, that there is no <lb/>neceſſity to determine Geometrically (which <pb o="48" file="0090" n="102" rhead="An ESSAY"/> might be done) the Point on the Semicircle <lb/>h z m, as far as the Parallels (as 9 m) are uſeful: <lb/></s> <s xml:id="echoid-s1195" xml:space="preserve">For when theſe Parallels are uſeleſs, the Point <lb/>q will fall beyond the Point m: </s> <s xml:id="echoid-s1196" xml:space="preserve">But then the <lb/>Perſpective of the Torus is entirely drawn alrea-<lb/>dy, if thoſe Parallels were firſt begun to be drawn <lb/>near to 6 3 z, and the others continually going <lb/>from it.</s> <s xml:id="echoid-s1197" xml:space="preserve"/> </p> <div xml:id="echoid-div166" type="float" level="2" n="3"> <note position="right" xlink:label="note-0089-01" xlink:href="note-0089-01a" xml:space="preserve">Fig. 32.</note> </div> <p> <s xml:id="echoid-s1198" xml:space="preserve">In order to demonſtrate this Problem, the fol-<lb/>lowing Lemma is neceſſary.</s> <s xml:id="echoid-s1199" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div168" type="section" level="1" n="90"> <head xml:id="echoid-head96" xml:space="preserve"><emph style="sc">Lemma</emph>.</head> <p> <s xml:id="echoid-s1200" xml:space="preserve">65. </s> <s xml:id="echoid-s1201" xml:space="preserve">If two Circles C D H E and D E F L cut <lb/> <anchor type="note" xlink:label="note-0090-01a" xlink:href="note-0090-01"/> each other, thro’ whoſe Centers C and B the Line <lb/>C L paſſes, and D E joyns their Interſections; <lb/></s> <s xml:id="echoid-s1202" xml:space="preserve">then, if the Radius A C or A H be called a, and <lb/>B F or BL, b, and the Diſtance A B between the <lb/>two Centers c, I ſay A G is equal to {bb—aa/ec}—{1/2}C.</s> <s xml:id="echoid-s1203" xml:space="preserve"/> </p> <div xml:id="echoid-div168" type="float" level="2" n="1"> <note position="left" xlink:label="note-0090-01" xlink:href="note-0090-01a" xml:space="preserve">Fig. 34.</note> </div> </div> <div xml:id="echoid-div170" type="section" level="1" n="91"> <head xml:id="echoid-head97" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p style="it"> <s xml:id="echoid-s1204" xml:space="preserve">Let us call A G, x, and G D or G E, y. <lb/></s> <s xml:id="echoid-s1205" xml:space="preserve">Then by the Property of the Circle, if y be conceiv’d <lb/>as an Ordinate of the Circle, C D H; </s> <s xml:id="echoid-s1206" xml:space="preserve">yy=aa—xx. </s> <s xml:id="echoid-s1207" xml:space="preserve"><lb/>And if it be likewiſe conſider’d as an Ordinate of <lb/>the Circle F D L, yy=bb—cc—2cx—xx: </s> <s xml:id="echoid-s1208" xml:space="preserve">Whence <lb/>aa—xx=bb—cc—2cx—xx, and ſo 2cx=bb <lb/>—aa—cc; </s> <s xml:id="echoid-s1209" xml:space="preserve">and dividing each Side of this laſt Equa-<lb/>tion by 2c, we have a x={bb—aa/2c}{1/2} c. </s> <s xml:id="echoid-s1210" xml:space="preserve">Which was <lb/>to be Demonſtrated.</s> <s xml:id="echoid-s1211" xml:space="preserve"/> </p> <pb file="0091" n="103"/> <pb file="0091a" n="104"/> <figure> <caption xml:id="echoid-caption30" style="it" xml:space="preserve">page 48.<lb/>Plate 13.<lb/>Fig. 32.</caption> <variables xml:id="echoid-variables29" xml:space="preserve">V S R L P B D Q T M I F A E Y C G O H Z N</variables> </figure> <pb file="0092" n="105"/> <pb o="49" file="0093" n="106" rhead="on PERSPECTIVE."/> <p> <s xml:id="echoid-s1212" xml:space="preserve">The Demonſtration of the <emph style="sc">Problem</emph>.</s> <s xml:id="echoid-s1213" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1214" xml:space="preserve">66. </s> <s xml:id="echoid-s1215" xml:space="preserve">The Torus of the Column muſt be conceiv’d as <lb/>made up of an Infinite Number of Circular Planes, <lb/>lying one upon another. </s> <s xml:id="echoid-s1216" xml:space="preserve">And it is evident that <lb/>the Reaſon why each of thoſe Circles cannot be wholly <lb/>ſeen, is becauſe that which is immediately under it <lb/>hides a Part thereof; </s> <s xml:id="echoid-s1217" xml:space="preserve">from whence it follows, that <lb/>if the Plane of one oſ theſe Circles be every way <lb/>continu’d, and the Circle immediately under it, be <lb/>thrown <anchor type="note" xlink:href="" symbol="*"/> into Perſpective upon it, (which Perſpective <anchor type="note" xlink:label="note-0093-01a" xlink:href="note-0093-01"/> is alſo a Circle) the two Points of Interſection of this <lb/>Repreſentation, and the Circle in the Plane, will deter-<lb/>mine the viſible Part of the ſaid Repreſentation; </s> <s xml:id="echoid-s1218" xml:space="preserve">and <lb/>conſequently if the Repreſentation of theſe two Points <lb/>of Interſection be found upon the Perſpective Plane, we <lb/>ſhall have two Points of the Perſpective of the Torus of <lb/>the propoſed Column. </s> <s xml:id="echoid-s1219" xml:space="preserve">This is what I have done in the <lb/>Solution of the Problem, as we ſhall now Analytically <lb/>demonſtrate.</s> <s xml:id="echoid-s1220" xml:space="preserve"/> </p> <div xml:id="echoid-div170" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0093-01" xlink:href="note-0093-01a" xml:space="preserve">8.</note> </div> <p style="it"> <s xml:id="echoid-s1221" xml:space="preserve">Let O be the Eye, A M a part of the Torus of <lb/> <anchor type="note" xlink:label="note-0093-02a" xlink:href="note-0093-02"/> the Column, A P a Perpendicular to the Baſe paſſing <lb/>through the Center of the Column, and A B a <lb/>Parallel to the Baſe, drawn thro’ the Center B of <lb/>the Semicircle Concavity of the Torus. </s> <s xml:id="echoid-s1222" xml:space="preserve">Let M P be <lb/>a Semidiameter of one of the Circles ſpoken of in the <lb/>the beginning of this Demonſtration. </s> <s xml:id="echoid-s1223" xml:space="preserve">Then if the <lb/>Line m p be drawn parallel and infinitely near C M P <lb/>and the Lines m O and p O are drawn cutting M P <lb/>in D and T, it is evident that D T, which is in the <lb/>Plane of the Circle paſſing thro’ M P, will be the <lb/>Semidiameter of the Perſpective of the Circle imme-<lb/>diately underneath.</s> <s xml:id="echoid-s1224" xml:space="preserve"/> </p> <div xml:id="echoid-div171" type="float" level="2" n="2"> <note position="right" xlink:label="note-0093-02" xlink:href="note-0093-02a" xml:space="preserve">Fig. 35.</note> </div> <p style="it"> <s xml:id="echoid-s1225" xml:space="preserve">Now let fall the Perpendicular O S from the Eye <lb/>to the Line A B, and continue the Lines M P and <lb/>m p, till they meet the ſaid Perpendicular in the Points <lb/>Q and q. </s> <s xml:id="echoid-s1226" xml:space="preserve">Moreover, continue the Line M P to the <pb o="50" file="0094" n="107" rhead="An ESSAY"/> Point R, wherein it is cut by the Line m R per-<lb/>pendicular to m p. </s> <s xml:id="echoid-s1227" xml:space="preserve">Aſſume A S = e, OQ = x, <lb/>and M P = y. </s> <s xml:id="echoid-s1228" xml:space="preserve">Then in the ſimilar Triangles O q m, <lb/>and m R D, we have, <lb/>O q (x): </s> <s xml:id="echoid-s1229" xml:space="preserve">q m (e + y):</s> <s xml:id="echoid-s1230" xml:space="preserve">: m R (d x) R D: </s> <s xml:id="echoid-s1231" xml:space="preserve">({edx + ydx/x}) <lb/>The ſimilar Triangles O p q and p T D, give, <lb/>oq (x): </s> <s xml:id="echoid-s1232" xml:space="preserve">q p (e):</s> <s xml:id="echoid-s1233" xml:space="preserve">: p P (d x): </s> <s xml:id="echoid-s1234" xml:space="preserve">P T ({edx/x}) <lb/>P R is equal to y + dy, and if P T ({edx/x}) be added <lb/>to it, and then from the Aggregate be taken R D <lb/>({edx + ydx/x}) we ſhall have T D = y + dy - {ydx/x}</s> </p> <p style="it"> <s xml:id="echoid-s1235" xml:space="preserve">Now to find the Points of Interſection of the two <lb/>Circles, whoſe Radii are T D and P M, and Centers <lb/>diſtant from each other, by the Space T P, the Square <lb/>of T D leſs the Square of P M muſt <anchor type="note" xlink:href="" symbol="*"/> be divided by <anchor type="note" xlink:label="note-0094-01a" xlink:href="note-0094-01"/> Twice P T, and then half of P T muſt be taken there-<lb/>from, which may be here neglected, becauſe it is infinite-<lb/>ly ſmall in compariſon of the reſt; </s> <s xml:id="echoid-s1236" xml:space="preserve">and we ſhall have <lb/>{xydy/dex} - {yy/e} for the Part of the Line PM included be-<lb/>tween P and the Point wherein this Line is cut by a <lb/>Line joyning the two Points of Interſection of the two <lb/>Circles.</s> <s xml:id="echoid-s1237" xml:space="preserve"/> </p> <div xml:id="echoid-div172" type="float" level="2" n="3"> <note symbol="*" position="left" xlink:label="note-0094-01" xlink:href="note-0094-01a" xml:space="preserve">65.</note> </div> <p style="it"> <s xml:id="echoid-s1238" xml:space="preserve">Now before what I have here demonſtrated be ap-<lb/>ply’d to the Problem, we muſt obſerve, that if from <lb/>the Point M, a Line be drawn thro’ the Center B, <lb/>the Triangles M P C and m R M will be ſimilar; <lb/></s> <s xml:id="echoid-s1239" xml:space="preserve">for the Angle m M P is the Exterior Angle of the <lb/>Triangle m R M, and the Angle m M C is a right <lb/>one. </s> <s xml:id="echoid-s1240" xml:space="preserve">And conſequently <lb/>m R (dx): </s> <s xml:id="echoid-s1241" xml:space="preserve">R M (d y):</s> <s xml:id="echoid-s1242" xml:space="preserve">: M P (y): </s> <s xml:id="echoid-s1243" xml:space="preserve">P C ({ydx/dx}).</s> <s xml:id="echoid-s1244" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1245" xml:space="preserve">Now if S A = s a in the 32d and 33d Figures <lb/> <anchor type="note" xlink:label="note-0094-02a" xlink:href="note-0094-02"/> be repreſented by e in this Computation; </s> <s xml:id="echoid-s1246" xml:space="preserve">as likewiſe <pb o="51" file="0095" n="108" rhead="on PERSPECTIVE."/> s i by x, and i h be y; </s> <s xml:id="echoid-s1247" xml:space="preserve">it is manifeſt, that i 4 = a 5 <lb/>being Algebraially Expreſſed, will be {ydy/dx}</s> </p> <div xml:id="echoid-div173" type="float" level="2" n="4"> <note position="left" xlink:label="note-0094-02" xlink:href="note-0094-02a" xml:space="preserve">Fig. 32, <lb/>33.</note> </div> <p style="it"> <s xml:id="echoid-s1248" xml:space="preserve">Again, the ſimilar Triangles, s a 5 and s i g give <lb/>s a (e): </s> <s xml:id="echoid-s1249" xml:space="preserve">a 5 ({ydy/dx}):</s> <s xml:id="echoid-s1250" xml:space="preserve">: s i (x): </s> <s xml:id="echoid-s1251" xml:space="preserve">i g ({xydx/edx}) Alſo by <lb/>the Conſtruction of Figure 32, <lb/>A S (e): </s> <s xml:id="echoid-s1252" xml:space="preserve">A P = i h (y):</s> <s xml:id="echoid-s1253" xml:space="preserve">: A P (y): </s> <s xml:id="echoid-s1254" xml:space="preserve">A I ({yy/e}); <lb/></s> <s xml:id="echoid-s1255" xml:space="preserve">Whence it follows, ſince I G = i g, that A G = <lb/>(I G - A I) = {xyd/cdx} - {yy/e}. </s> <s xml:id="echoid-s1256" xml:space="preserve">And conſequently, H <lb/>and F are the Seats of the two Points whoſe Perſpe-<lb/>ctive is required, and thoſe Points are both in a <lb/>Plane parallel to the Geometrical Plane, which is the <lb/>height of 21 above the Geometrical Plane.</s> <s xml:id="echoid-s1257" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1258" xml:space="preserve">If the precedent Calculation be apply’d to the Lower <lb/>Part of the Torus, the Expreſſion {xydy/edx} - {yy/e}, will <lb/>be chang’d into this, - {xydy/edx} - {yy/e;</s> <s xml:id="echoid-s1259" xml:space="preserve">} which ſhews that <lb/>theſe two Quantities muſt be aſſumed on the ſame Side <lb/>of A, viz. </s> <s xml:id="echoid-s1260" xml:space="preserve">towards S. </s> <s xml:id="echoid-s1261" xml:space="preserve">Moreover 9 q, inthe Line <lb/>9 m, is equal to {xydy/edx}; </s> <s xml:id="echoid-s1262" xml:space="preserve">for 98 ({ydy/e}) = i 4. <lb/></s> <s xml:id="echoid-s1263" xml:space="preserve">Which ſhews that M and L are alſo the Seats of two <lb/>Points whoſe Perſpective muſt be found, and which are <lb/>both in a Plane parallel to the Geometrical Plane, and <lb/>above it the Height of 29.</s> <s xml:id="echoid-s1264" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div175" type="section" level="1" n="92"> <head xml:id="echoid-head98" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s1265" xml:space="preserve">67. </s> <s xml:id="echoid-s1266" xml:space="preserve">This Problem may be likewiſe ſolved in <lb/>conſidering the Torus of a Column as made up of <lb/>an infinite Number of Baſes of Cones, whoſe Al-<lb/>titudes are determin’d by the concurrence of the <lb/>Tangents of the Semicircular Concavity of the <lb/>Axis of the Column; </s> <s xml:id="echoid-s1267" xml:space="preserve">and then determining <anchor type="note" xlink:href="" symbol="*"/> the <anchor type="note" xlink:label="note-0095-01a" xlink:href="note-0095-01"/> <pb o="52" file="0096" n="109" rhead="An ESSAY"/> viſible Portions of the ſaid Baſes. </s> <s xml:id="echoid-s1268" xml:space="preserve">Note, This <lb/>Method may be demonſtrated without Algebra, <lb/>but it would be very long.</s> <s xml:id="echoid-s1269" xml:space="preserve"/> </p> <div xml:id="echoid-div175" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0095-01" xlink:href="note-0095-01a" xml:space="preserve">53.</note> </div> </div> <div xml:id="echoid-div177" type="section" level="1" n="93"> <head xml:id="echoid-head99" xml:space="preserve"><emph style="sc">Problem</emph> IX.</head> <p style="it"> <s xml:id="echoid-s1270" xml:space="preserve">68. </s> <s xml:id="echoid-s1271" xml:space="preserve">To find the Accidental Point of ſeveral pa-<lb/>rallel Lines, which are inclin’d to the Geome-<lb/>trical Plane.</s> <s xml:id="echoid-s1272" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1273" xml:space="preserve">Let A B be the Direction of one of the Lines, <lb/> <anchor type="note" xlink:label="note-0096-01a" xlink:href="note-0096-01"/> whoſe accidental Point is ſought; </s> <s xml:id="echoid-s1274" xml:space="preserve">and ECP, the <lb/>Angle that the ſaid Lines make with the Geo-<lb/>metrical Plane.</s> <s xml:id="echoid-s1275" xml:space="preserve"/> </p> <div xml:id="echoid-div177" type="float" level="2" n="1"> <note position="left" xlink:label="note-0096-01" xlink:href="note-0096-01a" xml:space="preserve">Fig. 36.</note> </div> </div> <div xml:id="echoid-div179" type="section" level="1" n="94"> <head xml:id="echoid-head100" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1276" xml:space="preserve">Draw a Line, O D, thro’ the Eye O, parallel <lb/>to A B, and thro’ the Point D, wherein it cuts <lb/>the Horizontal Line, and which is the acciden-<lb/>tal Point of the Directions of the given Lines, <lb/>draw D F perpendicular to the ſaid Horizontal <lb/>Line; </s> <s xml:id="echoid-s1277" xml:space="preserve">in which aſſume D G, equal to DO. </s> <s xml:id="echoid-s1278" xml:space="preserve">Fi-<lb/>nally, thro’ the Point G, draw the Line G F, <lb/>making an Angle with the Horizontal Line, equal <lb/>to E C P; </s> <s xml:id="echoid-s1279" xml:space="preserve">and then the Point F, (the Interſection <lb/>of this Line) and the Perpendicular D F, is the <lb/>accidental Point ſought.</s> <s xml:id="echoid-s1280" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1281" xml:space="preserve">Note, When the Lines are inclin’d towards <lb/>the perſpective Plane, D F and G F muſt be <lb/>drawn below the Horizontal Line: </s> <s xml:id="echoid-s1282" xml:space="preserve">And, contra-<lb/>riwiſe, when the ſaid Lines are inclin’d towards <lb/>the oppoſite Part of the perſpective Plane, the <lb/>aforeſaid Lines muſt be drawn above the ſaid <lb/>Horizontal Line, as is done here.</s> <s xml:id="echoid-s1283" xml:space="preserve"/> </p> <pb file="0097" n="110"/> <pb file="0097a" n="111"/> <figure> <caption xml:id="echoid-caption31" style="it" xml:space="preserve">page 52.<lb/>Plate. 14.<lb/>Fig. 34</caption> <variables xml:id="echoid-variables30" xml:space="preserve">D C F G A B H L E</variables> </figure> <figure> <caption xml:id="echoid-caption32" style="it" xml:space="preserve">Fig. 33</caption> <variables xml:id="echoid-variables31" xml:space="preserve">S X 8 1 h 6 g 3 z q 9 m 2 4 m a 5 Y</variables> </figure> <figure> <caption xml:id="echoid-caption33" style="it" xml:space="preserve">Fig. 35</caption> <variables xml:id="echoid-variables32" xml:space="preserve">O M D P T Q R m p q B A S C</variables> </figure> <pb file="0098" n="112"/> <pb o="53" file="0099" n="113" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div180" type="section" level="1" n="95"> <head xml:id="echoid-head101" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1284" xml:space="preserve">If a Plane be conceiv’d to paſs thro’ the Eye, <lb/>perpendicular to the Geometrical Plane, and paral-<lb/>lel to the given Lines; </s> <s xml:id="echoid-s1285" xml:space="preserve">it is evident, that the ſaid <lb/>Plane will cut the Horizontal Plane in the Line <lb/>O D, and the perſpective Plane in D F. </s> <s xml:id="echoid-s1286" xml:space="preserve">It is, <lb/>moreover, manifeſt, that a Line drawn thro’ the <lb/>Eye, parallel to the given Line, is in the ſaid <lb/>Plane, and (with the Line O D) makes an An-<lb/>gle, equal to the Angle E C P, below the Hori-<lb/>zontal Plane, if the Lines be inclin’d towards <lb/>the perſpective Plane, and above it, if they in-<lb/>cline to the oppoſite ſide; </s> <s xml:id="echoid-s1287" xml:space="preserve">whence this laſt Line <lb/>makes a right-angled Triangle with O D and <lb/>D F, whoſe Angle at the Point O, is equal to <lb/>the Angle C E P. </s> <s xml:id="echoid-s1288" xml:space="preserve">But D G F is likewiſe a <lb/>right-angled Triangle, as having the Angle at the <lb/>Point G, equal to ECP; </s> <s xml:id="echoid-s1289" xml:space="preserve">therefore theſe two <lb/>Triangles are ſimilar. </s> <s xml:id="echoid-s1290" xml:space="preserve">And ſince the Side D G <lb/>is equal to the Side D O, the Triangles are alſo <lb/>equal: </s> <s xml:id="echoid-s1291" xml:space="preserve">Therefore the Line D F, being common <lb/>to theſe two Triangles; </s> <s xml:id="echoid-s1292" xml:space="preserve">the Point F, is the <lb/>Point wherein the Line, paſſing thro’ the Eye <lb/>parallel to the given Line, meets the Per-<lb/>ſpective Plane: </s> <s xml:id="echoid-s1293" xml:space="preserve">And this Point is <anchor type="note" xlink:href="" symbol="*"/> the acciden- tal one ſought.</s> <s xml:id="echoid-s1294" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1295" xml:space="preserve">Note, This Demonſtration as well regards <lb/> <anchor type="note" xlink:label="note-0099-01a" xlink:href="note-0099-01"/> inclin’d Lines entirely ſeparate from the Geo-<lb/>metrical Plane, as thoſe that meet it in one of <lb/>their Extremes only.</s> <s xml:id="echoid-s1296" xml:space="preserve"/> </p> <div xml:id="echoid-div180" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0099-01" xlink:href="note-0099-01a" xml:space="preserve">13, 14.</note> </div> </div> <div xml:id="echoid-div182" type="section" level="1" n="96"> <head xml:id="echoid-head102" xml:space="preserve"><emph style="sc">Problem</emph> X.</head> <p style="it"> <s xml:id="echoid-s1297" xml:space="preserve">69. </s> <s xml:id="echoid-s1298" xml:space="preserve">To find the Repreſentation of one or more <lb/>Lines, inclin’d to the Geometrical Plane. <lb/></s> <s xml:id="echoid-s1299" xml:space="preserve"> <anchor type="note" xlink:label="note-0099-02a" xlink:href="note-0099-02"/> </s> </p> <div xml:id="echoid-div182" type="float" level="2" n="1"> <note position="right" xlink:label="note-0099-02" xlink:href="note-0099-02a" xml:space="preserve">Fig. 36.</note> </div> <p> <s xml:id="echoid-s1300" xml:space="preserve">Let A be a Point given in the Geometrical <lb/>Plane; </s> <s xml:id="echoid-s1301" xml:space="preserve">whereon ſtands a Line, whoſe Length, <lb/>Direction, and Angle of Inclination is known.</s> <s xml:id="echoid-s1302" xml:space="preserve"/> </p> <pb o="54" file="0100" n="114" rhead="An ESSAY"/> </div> <div xml:id="echoid-div184" type="section" level="1" n="97"> <head xml:id="echoid-head103" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1303" xml:space="preserve">In ſome ſeparate Place, draw the Lines C E <lb/>and C P, making an Angle with each other <lb/>equal to the Angle of Inclination of the given <lb/>Line; </s> <s xml:id="echoid-s1304" xml:space="preserve">and in one of theſe Lines, aſſume C E <lb/>equal to the given Line, and let fall the Perpen-<lb/>dicular E P, from the Point E upon the other <lb/>Line. </s> <s xml:id="echoid-s1305" xml:space="preserve">Then aſſume A B, in the Direction of <lb/>the propos’d Line, equal to C P; </s> <s xml:id="echoid-s1306" xml:space="preserve">and after ha-<lb/>ving found a, the Perſpective of A, and the <lb/>Point T <anchor type="note" xlink:href="" symbol="*"/>, the Perſpective of a Point elevated <anchor type="note" xlink:label="note-0100-01a" xlink:href="note-0100-01"/> above B, the Height of P E; </s> <s xml:id="echoid-s1307" xml:space="preserve">join the Points a <lb/>and T by a right Line; </s> <s xml:id="echoid-s1308" xml:space="preserve">and the ſought Appear-<lb/>ance will be had.</s> <s xml:id="echoid-s1309" xml:space="preserve"/> </p> <div xml:id="echoid-div184" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0100-01" xlink:href="note-0100-01a" xml:space="preserve">50.</note> </div> </div> <div xml:id="echoid-div186" type="section" level="1" n="98"> <head xml:id="echoid-head104" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1310" xml:space="preserve">If from the Extremity of the inclin’d Line, a <lb/>Perpendicular be let fall upon the Geometrical <lb/>Plane, the ſaid Perpendicular will meet this <lb/>Plane in the Point B, and will be equal to P E; <lb/></s> <s xml:id="echoid-s1311" xml:space="preserve">as is evident by the Conſtruction of the Figure <lb/>C P E. </s> <s xml:id="echoid-s1312" xml:space="preserve">But the Point T is the Repreſentation <lb/>of the Extremity of this Perpendicular; </s> <s xml:id="echoid-s1313" xml:space="preserve">and <lb/>therefore it is alſo the Extremity of the inclin’d <lb/>Line. </s> <s xml:id="echoid-s1314" xml:space="preserve">Which was to be demonſtrated.</s> <s xml:id="echoid-s1315" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div187" type="section" level="1" n="99"> <head xml:id="echoid-head105" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s1316" xml:space="preserve">There are ſome Caſes of this Propoſition, that <lb/>may be ſhorten’d. </s> <s xml:id="echoid-s1317" xml:space="preserve">As, 1. </s> <s xml:id="echoid-s1318" xml:space="preserve">When there are ſeve-<lb/>ral Lines of this Kind parallel between them-<lb/>ſelves, and whoſe accidental Point can be <lb/>found <anchor type="note" xlink:href="" symbol="*"/>: </s> <s xml:id="echoid-s1319" xml:space="preserve">And, 2. </s> <s xml:id="echoid-s1320" xml:space="preserve">When an inclin’d Line is pa- <anchor type="note" xlink:label="note-0100-02a" xlink:href="note-0100-02"/> rallel to the perſpective Plane. </s> <s xml:id="echoid-s1321" xml:space="preserve">The Manner of <pb o="55" file="0101" n="115" rhead="on PERSPECTIVE."/> making theſe Abbreviations, will be laid down <lb/>in the following Methods.</s> <s xml:id="echoid-s1322" xml:space="preserve"/> </p> <div xml:id="echoid-div187" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0100-02" xlink:href="note-0100-02a" xml:space="preserve">68.</note> </div> </div> <div xml:id="echoid-div189" type="section" level="1" n="100"> <head xml:id="echoid-head106" xml:space="preserve"><emph style="sc">Method</emph> II. <lb/>70. By the accidental Point of inclin’d Lines.</head> <p> <s xml:id="echoid-s1323" xml:space="preserve">Thro’ F, the accidental Point of the inclin’d <lb/> <anchor type="note" xlink:label="note-0101-01a" xlink:href="note-0101-01"/> parallel Lines, draw F H, parallel to the Baſe <lb/>Line, and equal to F G. </s> <s xml:id="echoid-s1324" xml:space="preserve">And let A be the <lb/>Point, wherein one of the inclin’d Lines meets <lb/>the Geometrical Plane.</s> <s xml:id="echoid-s1325" xml:space="preserve"/> </p> <div xml:id="echoid-div189" type="float" level="2" n="1"> <note position="right" xlink:label="note-0101-01" xlink:href="note-0101-01a" xml:space="preserve">Fig. 36.</note> </div> </div> <div xml:id="echoid-div191" type="section" level="1" n="101"> <head xml:id="echoid-head107" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1326" xml:space="preserve">Aſſume R Q in the Baſe Line, equal to the <lb/>inclin’d Line; </s> <s xml:id="echoid-s1327" xml:space="preserve">and draw Lines from the Points <lb/>R and Q, to the Point Z, taken at pleaſure in <lb/>the Horizontal Plane.</s> <s xml:id="echoid-s1328" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1329" xml:space="preserve">Thro’ a, the Perſpective of A, draw a N pa-<lb/>rallel to the Baſe Line; </s> <s xml:id="echoid-s1330" xml:space="preserve">in which aſſume a L, <lb/>equal to M N; </s> <s xml:id="echoid-s1331" xml:space="preserve">and draw a Line from the Point <lb/>a, to the Point F; </s> <s xml:id="echoid-s1332" xml:space="preserve">and from the Point L, draw <lb/>another to the Point H. </s> <s xml:id="echoid-s1333" xml:space="preserve">Then a T will be the <lb/>Perſpective ſought.</s> <s xml:id="echoid-s1334" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div192" type="section" level="1" n="102"> <head xml:id="echoid-head108" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1335" xml:space="preserve">By <anchor type="note" xlink:href="" symbol="*"/> the Nature of the accidental Point, the <anchor type="note" xlink:label="note-0101-02a" xlink:href="note-0101-02"/> Perſpective ſought is a Part of the Line a F; <lb/></s> <s xml:id="echoid-s1336" xml:space="preserve">and therefore, we are only to demonſtrate, <lb/>that the Extremity of the faid Perpendicular is in <lb/>the Line L H. </s> <s xml:id="echoid-s1337" xml:space="preserve">Which may be thus done.</s> <s xml:id="echoid-s1338" xml:space="preserve"/> </p> <div xml:id="echoid-div192" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0101-02" xlink:href="note-0101-02a" xml:space="preserve">14.</note> </div> <p> <s xml:id="echoid-s1339" xml:space="preserve">Let us ſuppoſe a Line, A I, to paſs thro’ the <lb/>Point A, parallel to the Baſe Line, and equal to <lb/>the inclin’d Line. </s> <s xml:id="echoid-s1340" xml:space="preserve">It is then manifeſt <anchor type="note" xlink:href="" symbol="*"/>, that <anchor type="note" xlink:label="note-0101-03a" xlink:href="note-0101-03"/> L is the Perſpective of I; </s> <s xml:id="echoid-s1341" xml:space="preserve">and conſequently, <lb/>L H <anchor type="note" xlink:href="" symbol="*"/> is the Appearance of a Line paſſing <anchor type="note" xlink:label="note-0101-04a" xlink:href="note-0101-04"/> <pb o="56" file="0102" n="116" rhead="An ESSAY"/> through I, and the Extremity of the propoſed <lb/>Line; </s> <s xml:id="echoid-s1342" xml:space="preserve">and therefore the Perſpective of this Ex-<lb/>tremity is in the Line L H; </s> <s xml:id="echoid-s1343" xml:space="preserve">which was to be <lb/>demonſtrated.</s> <s xml:id="echoid-s1344" xml:space="preserve"/> </p> <div xml:id="echoid-div193" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0101-03" xlink:href="note-0101-03a" xml:space="preserve">56.</note> <note symbol="*" position="right" xlink:label="note-0101-04" xlink:href="note-0101-04a" xml:space="preserve">@@</note> </div> <p> <s xml:id="echoid-s1345" xml:space="preserve">Note, if F H had been aſſumed, the one half, <lb/>or one third, & </s> <s xml:id="echoid-s1346" xml:space="preserve">c. </s> <s xml:id="echoid-s1347" xml:space="preserve">of what it is; </s> <s xml:id="echoid-s1348" xml:space="preserve">then it is mani-<lb/>feſt <anchor type="note" xlink:href="" symbol="*"/> that R Q muſt alfo have been taken equal <anchor type="note" xlink:label="note-0102-01a" xlink:href="note-0102-01"/> to the one half, or third Part, & </s> <s xml:id="echoid-s1349" xml:space="preserve">c. </s> <s xml:id="echoid-s1350" xml:space="preserve">of C E.</s> <s xml:id="echoid-s1351" xml:space="preserve"/> </p> <div xml:id="echoid-div194" type="float" level="2" n="3"> <note symbol="*" position="left" xlink:label="note-0102-01" xlink:href="note-0102-01a" xml:space="preserve">19.</note> </div> </div> <div xml:id="echoid-div196" type="section" level="1" n="103"> <head xml:id="echoid-head109" xml:space="preserve"><emph style="sc">Method</emph>. III.</head> <p style="it"> <s xml:id="echoid-s1352" xml:space="preserve">71. </s> <s xml:id="echoid-s1353" xml:space="preserve">Forinclined Lines not meeting the Geometrical <lb/>Plane.</s> <s xml:id="echoid-s1354" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1355" xml:space="preserve">Let A and B be the Seats of the Extremities of <lb/> <anchor type="note" xlink:label="note-0102-02a" xlink:href="note-0102-02"/> the given Line. </s> <s xml:id="echoid-s1356" xml:space="preserve">Let X repreſent a Plane paſſing <lb/>through the given Line perpendicular to the <lb/>Geometrical Plane. </s> <s xml:id="echoid-s1357" xml:space="preserve">Likewiſe let M N in this <lb/>Plane, repreſent the Line whoſe Perſpective is re-<lb/>quir’d; </s> <s xml:id="echoid-s1358" xml:space="preserve">and let C N and P M be perpendicular to <lb/>the Geometrical Plane: </s> <s xml:id="echoid-s1359" xml:space="preserve">Whence P C repreſents <lb/>A B, and conſequently is equal thereto.</s> <s xml:id="echoid-s1360" xml:space="preserve"/> </p> <div xml:id="echoid-div196" type="float" level="2" n="1"> <note position="left" xlink:label="note-0102-02" xlink:href="note-0102-02a" xml:space="preserve">Fig. 37.</note> </div> </div> <div xml:id="echoid-div198" type="section" level="1" n="104"> <head xml:id="echoid-head110" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1361" xml:space="preserve">Find the Point I <anchor type="note" xlink:href="" symbol="*"/>, the Perſpective of a Point, <anchor type="note" xlink:label="note-0102-03a" xlink:href="note-0102-03"/> above the Point A, the Height of C N; </s> <s xml:id="echoid-s1362" xml:space="preserve">and <lb/>draw the Line B S, from the Point B, to the <lb/>Station Point I, cutting the Baſe Line in E; </s> <s xml:id="echoid-s1363" xml:space="preserve">and <lb/>from the Point I, draw a Line to the accidental <lb/>Point F; </s> <s xml:id="echoid-s1364" xml:space="preserve">which cut by a Perpendicular to the <lb/>Baſe Line, raiſed at the Point E; </s> <s xml:id="echoid-s1365" xml:space="preserve">and then I T <lb/>will be the Appearance ſought.</s> <s xml:id="echoid-s1366" xml:space="preserve"/> </p> <div xml:id="echoid-div198" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0102-03" xlink:href="note-0102-03a" xml:space="preserve">50.</note> </div> </div> <div xml:id="echoid-div200" type="section" level="1" n="105"> <head xml:id="echoid-head111" xml:space="preserve"><emph style="sc">Method</emph> IV.</head> <p style="it"> <s xml:id="echoid-s1367" xml:space="preserve">72. </s> <s xml:id="echoid-s1368" xml:space="preserve">For inclined Lines parallel to the perſpective <lb/>Plane.</s> <s xml:id="echoid-s1369" xml:space="preserve"/> </p> <pb file="0103" n="117"/> <pb file="0103a" n="118"/> <figure> <caption xml:id="echoid-caption34" style="it" xml:space="preserve">page 56.<lb/>Plate. 15</caption> <variables xml:id="echoid-variables33" xml:space="preserve">F H O Z D G</variables> </figure> <figure> <caption xml:id="echoid-caption35" style="it" xml:space="preserve">Fig. 36</caption> <variables xml:id="echoid-variables34" xml:space="preserve">T N M L a R Q E I A C P B</variables> </figure> <figure> <caption xml:id="echoid-caption36" xml:space="preserve">Fig. 37</caption> <variables xml:id="echoid-variables35" xml:space="preserve">F S V T I E M A N X P C B</variables> </figure> <pb file="0104" n="119"/> <pb o="57" file="0105" n="120" rhead="on PERSPECTIVE."/> <p> <s xml:id="echoid-s1370" xml:space="preserve">The Operation of Prob. </s> <s xml:id="echoid-s1371" xml:space="preserve">VII. </s> <s xml:id="echoid-s1372" xml:space="preserve">muſt be uſed <lb/>here, but with this Difference (ſee Fig. </s> <s xml:id="echoid-s1373" xml:space="preserve">of the <lb/>ſaid Prob.) </s> <s xml:id="echoid-s1374" xml:space="preserve">that whereas a I in the ſaid Problem <lb/>is perpendicular to the Baſe Line, here it muſt <lb/>make an Angle with the Baſe Line, equal to the <lb/>Angle of Inclination of the given Lines.</s> <s xml:id="echoid-s1375" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1376" xml:space="preserve">For the Demonſtration of this, ſee n. </s> <s xml:id="echoid-s1377" xml:space="preserve">7, and 10.</s> <s xml:id="echoid-s1378" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div201" type="section" level="1" n="106"> <head xml:id="echoid-head112" xml:space="preserve"><emph style="sc">Prob</emph>. XIV.</head> <p style="it"> <s xml:id="echoid-s1379" xml:space="preserve">73. </s> <s xml:id="echoid-s1380" xml:space="preserve">To throw a Body into Perſpective, having <lb/>ſome one or all of its Sides inclined to the Geometri-<lb/>cal Plane.</s> <s xml:id="echoid-s1381" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1382" xml:space="preserve">The Appearances of the Lines forming the <lb/>Angles of the propoſed Body muſt be found: </s> <s xml:id="echoid-s1383" xml:space="preserve">And <lb/>this may be eaſily done by Prob. </s> <s xml:id="echoid-s1384" xml:space="preserve">10. </s> <s xml:id="echoid-s1385" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/> which <anchor type="note" xlink:label="note-0105-01a" xlink:href="note-0105-01"/> takes in all the Caſes. </s> <s xml:id="echoid-s1386" xml:space="preserve">And in this Manner the <lb/>Appearance of a Pyramid, an inclined Priſm, <lb/>& </s> <s xml:id="echoid-s1387" xml:space="preserve">c. </s> <s xml:id="echoid-s1388" xml:space="preserve">may be found. </s> <s xml:id="echoid-s1389" xml:space="preserve">But nevertheleſs, it hap-<lb/>pens ſometimes, that the Operations of the pre-<lb/>cedent Problem may be abbreviated; </s> <s xml:id="echoid-s1390" xml:space="preserve">as when <lb/>the Extremity of ſeveral Lines are found in one <lb/>and the ſame Line, or when inclined Lines, that <lb/>have difficult accidental Points, interſect one <lb/>another, and ſo mutually determine each other. <lb/></s> <s xml:id="echoid-s1391" xml:space="preserve">This will appear manifeſt by the following Ex-<lb/>amples.</s> <s xml:id="echoid-s1392" xml:space="preserve"/> </p> <div xml:id="echoid-div201" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0105-01" xlink:href="note-0105-01a" xml:space="preserve">69.</note> </div> </div> <div xml:id="echoid-div203" type="section" level="1" n="107"> <head xml:id="echoid-head113" xml:space="preserve"><emph style="sc">Example</emph> I.</head> <p style="it"> <s xml:id="echoid-s1393" xml:space="preserve">To throw ſeveral parallel Shores which ſtrengthen a <lb/>Wall, into Perſpective.</s> <s xml:id="echoid-s1394" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1395" xml:space="preserve">I ſuppoſe here that the Baſes of theſe Shores, <lb/> <anchor type="note" xlink:label="note-0105-02a" xlink:href="note-0105-02"/> which are the Places where they meet the Sur-<lb/>face of the Ground, are all in a right Line, pa-<lb/>rallel to the Side of the Wall; </s> <s xml:id="echoid-s1396" xml:space="preserve">and then the <lb/>faid Shores may be thrown into Perſpective in <lb/>the following Manner: </s> <s xml:id="echoid-s1397" xml:space="preserve">Having firſt found <anchor type="note" xlink:href="" symbol="*"/> <anchor type="note" xlink:label="note-0105-03a" xlink:href="note-0105-03"/> <pb o="58" file="0106" n="121" rhead="An ESSAY"/> their accidental Point F, afterwards find the Re-<lb/>preſentation of their Baſes: </s> <s xml:id="echoid-s1398" xml:space="preserve">This being done, <lb/>denote the Appearances of the Lines wherein <lb/>the Shores meet the Wall, upon the Perſpective <lb/>of the Wall; </s> <s xml:id="echoid-s1399" xml:space="preserve">the Appearances here are the <lb/>Lines p t, r s, which repreſent Lines parallel to <lb/>the Geometrical Plane, from the Suppoſition, <lb/>that the Shores are parallel between themſelves, <lb/>and their Baſes equally diſtant from the Wall. <lb/></s> <s xml:id="echoid-s1400" xml:space="preserve">Finally, draw Lines from the Angles of the Re-<lb/>preſentations 1 2 3 4, to the Point F, which <lb/>will be terminated by their Interſections with <lb/>p t and r s, and will give the Appearances ſought, <lb/>as you ſee in the Figure.</s> <s xml:id="echoid-s1401" xml:space="preserve"/> </p> <div xml:id="echoid-div203" type="float" level="2" n="1"> <note position="right" xlink:label="note-0105-02" xlink:href="note-0105-02a" xml:space="preserve">Fig. 38.</note> <note symbol="*" position="right" xlink:label="note-0105-03" xlink:href="note-0105-03a" xml:space="preserve">68.</note> </div> </div> <div xml:id="echoid-div205" type="section" level="1" n="108"> <head xml:id="echoid-head114" xml:space="preserve"><emph style="sc">Example</emph> II.</head> <p style="it"> <s xml:id="echoid-s1402" xml:space="preserve">To throw ſeveral parallel Roofs of a Houſe into <lb/>Perſpective.</s> <s xml:id="echoid-s1403" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1404" xml:space="preserve">Having found the accidental Points G and Q <lb/> <anchor type="note" xlink:label="note-0106-01a" xlink:href="note-0106-01"/> of the ſaid Roofs, in the Repreſentation of <lb/>the Wall ſuſtaining them, denote the Points <lb/>a b c d, wherein the ſaid Roofs meet the Wall: <lb/></s> <s xml:id="echoid-s1405" xml:space="preserve">Then from the Point G draw Lines through the <lb/>Points a b c; </s> <s xml:id="echoid-s1406" xml:space="preserve">and from the Point Q others to <lb/>the Points b c d; </s> <s xml:id="echoid-s1407" xml:space="preserve">theſe Lines by their mutual <lb/>Interſection will determine each other, and give <lb/>the Repreſentations ſought.</s> <s xml:id="echoid-s1408" xml:space="preserve"/> </p> <div xml:id="echoid-div205" type="float" level="2" n="1"> <note position="left" xlink:label="note-0106-01" xlink:href="note-0106-01a" xml:space="preserve">Fig. 39.</note> </div> </div> <div xml:id="echoid-div207" type="section" level="1" n="109"> <head xml:id="echoid-head115" xml:space="preserve"><emph style="sc">Conclusion</emph>.</head> <p> <s xml:id="echoid-s1409" xml:space="preserve">74. </s> <s xml:id="echoid-s1410" xml:space="preserve">From what has been already ſaid, it will <lb/>not be difficult to throw any Objects whatſoever <lb/>into Perſpective. </s> <s xml:id="echoid-s1411" xml:space="preserve">But ſince it is very difficult, <lb/>and indeed impoſſible for a Painter to make a <lb/>Deſign entirely according to the Rules we have <lb/>preſcribed; </s> <s xml:id="echoid-s1412" xml:space="preserve">the Number of Points to be found <lb/>being almoſt infinite: </s> <s xml:id="echoid-s1413" xml:space="preserve">therefore the Figures <pb file="0107" n="122"/> <pb file="0107a" n="123"/> <anchor type="figure" xlink:label="fig-0107a-01a" xlink:href="fig-0107a-01"/> <pb file="0108" n="124"/> <pb o="59" file="0109" n="125" rhead="on PERSPECTIVE."/> drawn upon the Geometrical Plane, and the <lb/>principal Points of the Objects without the ſaid <lb/>Plane, need only be thrown into Perſpective. <lb/></s> <s xml:id="echoid-s1414" xml:space="preserve">Which being once obtained, he may make uſe <lb/>of theſe Appearances ſo found, as a Rule where-<lb/>by the reſt may be compleated by the Eye, with-<lb/>out running the Riſque of committing ſome <lb/>conſiderable Fault, which by this Means may be <lb/>avoided.</s> <s xml:id="echoid-s1415" xml:space="preserve"/> </p> <div xml:id="echoid-div207" type="float" level="2" n="1"> <figure xlink:label="fig-0107a-01" xlink:href="fig-0107a-01a"> <caption xml:id="echoid-caption37" style="it" xml:space="preserve">page 58.<lb/>Plate. 16<lb/>Fig. 39<lb/>Fig. 38</caption> <variables xml:id="echoid-variables36" xml:space="preserve">F Q O p l r s 1 2 3 4 G</variables> </figure> </div> </div> <div xml:id="echoid-div209" type="section" level="1" n="110"> <head xml:id="echoid-head116" xml:space="preserve">CHAP. IV.</head> <p style="it"> <s xml:id="echoid-s1416" xml:space="preserve">Of the Practice of Peſpective upon the Per-<lb/>ſpective Plane ſtill conſider’d as being upright.</s> <s xml:id="echoid-s1417" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1418" xml:space="preserve">IT often happens that Painters offend all <lb/>Rules of true Appearance when they paint <lb/>Pictures to ſtand aloft, to be ſeen Sideways, or at <lb/>a confiderable Diſtance. </s> <s xml:id="echoid-s1419" xml:space="preserve">Their Cuſtom is to <lb/>paint Pictures to be view’d, after the ſame Man-<lb/>ner as they themſelves look at them when they <lb/>are working; </s> <s xml:id="echoid-s1420" xml:space="preserve">whence in the following Caſes, <lb/>this Practice of theirs will be uſeleſs; </s> <s xml:id="echoid-s1421" xml:space="preserve">and ſo to <lb/>avoid enormous Faults, they are neceſſitated to <lb/>have recourſe to Perſpective But what has been <lb/>ſaid in the laſt Chapter, does not reach theſe <lb/>particular Caſes; </s> <s xml:id="echoid-s1422" xml:space="preserve">therefore we ſhall here add ſome <lb/>new Problems, which together with the former <lb/>ones, will take in all Caſes.</s> <s xml:id="echoid-s1423" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div210" type="section" level="1" n="111"> <head xml:id="echoid-head117" xml:space="preserve"><emph style="sc">Problem</emph> I.</head> <p style="it"> <s xml:id="echoid-s1424" xml:space="preserve">75. </s> <s xml:id="echoid-s1425" xml:space="preserve">To throw Figures which are in the Geometri-<lb/>cal Plane into Perſpective, when the Eye is at ſo great <lb/>a Diſtance that it cannot be denoted in the Horizon- <pb o="60" file="0110" n="126" rhead="An ESSAY"/> tal Plane, or one of the Points of Diſtance on the <lb/>Horizontal Line.</s> <s xml:id="echoid-s1426" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1427" xml:space="preserve">The Repreſentation of two Points of theſe <lb/>Figures muſt be firſt found <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s1428" xml:space="preserve">and then by Means <anchor type="note" xlink:label="note-0110-01a" xlink:href="note-0110-01"/> of theſe two Points the Appearances of others <lb/>may be had <anchor type="note" xlink:href="" symbol="*"/>.</s> <s xml:id="echoid-s1429" xml:space="preserve"/> </p> <div xml:id="echoid-div210" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0110-01" xlink:href="note-0110-01a" xml:space="preserve">24.</note> </div> <note symbol="*" position="left" xml:space="preserve">38.</note> </div> <div xml:id="echoid-div212" type="section" level="1" n="112"> <head xml:id="echoid-head118" xml:space="preserve"><emph style="sc">Example</emph>.</head> <p> <s xml:id="echoid-s1430" xml:space="preserve">Let A B C D E, be a Pentagon, whoſe Ap-<lb/> <anchor type="note" xlink:label="note-0110-03a" xlink:href="note-0110-03"/> pearance is requir’d; </s> <s xml:id="echoid-s1431" xml:space="preserve">V the Point of Sight; </s> <s xml:id="echoid-s1432" xml:space="preserve">and <lb/>V F the ſixth Part of the Eye’s Diſtance from <lb/>the perſpective Plane. </s> <s xml:id="echoid-s1433" xml:space="preserve">Now find <anchor type="note" xlink:href="" symbol="*"/> b and e the <anchor type="note" xlink:label="note-0110-04a" xlink:href="note-0110-04"/> Appearance of B and E, by means of which, <lb/>the Appearance of the Point A will be had <anchor type="note" xlink:href="" symbol="*"/>.</s> <s xml:id="echoid-s1434" xml:space="preserve"> <anchor type="note" xlink:label="note-0110-05a" xlink:href="note-0110-05"/> In like Manner, by means of the Repreſenta-<lb/>tion of A and E, will that of D be had; </s> <s xml:id="echoid-s1435" xml:space="preserve">and <lb/>by uſing B and A, the Perſpective of C may be <lb/>found.</s> <s xml:id="echoid-s1436" xml:space="preserve"/> </p> <div xml:id="echoid-div212" type="float" level="2" n="1"> <note position="left" xlink:label="note-0110-03" xlink:href="note-0110-03a" xml:space="preserve">Fig. 40.</note> <note symbol="*" position="left" xlink:label="note-0110-04" xlink:href="note-0110-04a" xml:space="preserve">24.</note> <note symbol="*" position="left" xlink:label="note-0110-05" xlink:href="note-0110-05a" xml:space="preserve">38.</note> </div> <p> <s xml:id="echoid-s1437" xml:space="preserve">76. </s> <s xml:id="echoid-s1438" xml:space="preserve">Note, the Perſpective of Lines perpendicu-<lb/>lar to the Geometrical Plane <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s1439" xml:space="preserve">as alſo of Lines <anchor type="note" xlink:label="note-0110-06a" xlink:href="note-0110-06"/> inclined thereto <anchor type="note" xlink:href="" symbol="*"/>, may be found by the Methods <anchor type="note" xlink:label="note-0110-07a" xlink:href="note-0110-07"/> of the precedent Chapter.</s> <s xml:id="echoid-s1440" xml:space="preserve"/> </p> <div xml:id="echoid-div213" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0110-06" xlink:href="note-0110-06a" xml:space="preserve">55.</note> <note symbol="*" position="left" xlink:label="note-0110-07" xlink:href="note-0110-07a" xml:space="preserve">69.</note> </div> </div> <div xml:id="echoid-div215" type="section" level="1" n="113"> <head xml:id="echoid-head119" xml:space="preserve"><emph style="sc">Problem</emph> II.</head> <p style="it"> <s xml:id="echoid-s1441" xml:space="preserve">77. </s> <s xml:id="echoid-s1442" xml:space="preserve">To throw Figures, which are in the Geometri-<lb/>cal Plane into Perſpective, when the Eye is ſo oblique <lb/>that it cannot be marked in the Horizontal Plane, <lb/>or the Point of Sight in the Horizontal Line.</s> <s xml:id="echoid-s1443" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1444" xml:space="preserve">We muſt proceed here according to the Di-<lb/>rections of the precedent Problem, after having <lb/>found the Perſpective of ſeveral Points of the <lb/>given Figures.</s> <s xml:id="echoid-s1445" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1446" xml:space="preserve">At any Point C, taken at Pleaſure in the Baſe <lb/>Line, draw the Perpendicular C D to the ſaid <lb/> <anchor type="note" xlink:label="note-0110-08a" xlink:href="note-0110-08"/> Line, and likewiſe draw the Line C E from the <lb/>ſame Point in ſuch manner, that if it could be <pb file="0111" n="127"/> <pb file="0111a" n="128"/> <anchor type="figure" xlink:label="fig-0111a-01a" xlink:href="fig-0111a-01"/> <anchor type="figure" xlink:label="fig-0111a-02a" xlink:href="fig-0111a-02"/> <pb file="0112" n="129"/> <pb o="61" file="0113" n="130" rhead="on PERSPECTIVE."/> continued, it would cut the Horizontal Line in <lb/>the Point of Sight.</s> <s xml:id="echoid-s1447" xml:space="preserve"/> </p> <div xml:id="echoid-div215" type="float" level="2" n="1"> <note position="left" xlink:label="note-0110-08" xlink:href="note-0110-08a" xml:space="preserve">Fig. 41.</note> <figure xlink:label="fig-0111a-01" xlink:href="fig-0111a-01a"> <caption xml:id="echoid-caption38" style="it" xml:space="preserve">page 60.<lb/>Plate. 17</caption> <variables xml:id="echoid-variables37" xml:space="preserve">F V</variables> </figure> <figure xlink:label="fig-0111a-02" xlink:href="fig-0111a-02a"> <caption xml:id="echoid-caption39" style="it" xml:space="preserve">Fig. 40</caption> <variables xml:id="echoid-variables38" xml:space="preserve">c θ b e a F G H I K L A B E C D</variables> </figure> </div> <p> <s xml:id="echoid-s1448" xml:space="preserve">This is done in aſſuming C H equal to {1/3}, or <lb/>{1/4} Part, &</s> <s xml:id="echoid-s1449" xml:space="preserve">c. </s> <s xml:id="echoid-s1450" xml:space="preserve">of the Diſtance from the Point C, <lb/>to the Foot of the vertical Line; </s> <s xml:id="echoid-s1451" xml:space="preserve">and in raiſing <lb/>the Perpendicular H E, in the Point H, equal <lb/>to {1/3} or {1/4} Part, &</s> <s xml:id="echoid-s1452" xml:space="preserve">c. </s> <s xml:id="echoid-s1453" xml:space="preserve">of the Height of the Eye. <lb/></s> <s xml:id="echoid-s1454" xml:space="preserve">Now A is a given Point, whoſe Appearance is <lb/>ſought.</s> <s xml:id="echoid-s1455" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div217" type="section" level="1" n="114"> <head xml:id="echoid-head120" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1456" xml:space="preserve">Draw a Parallel A B, through the Point A, <lb/>to the Baſe Line, meeting the Line C D in the <lb/>Point B, and let a ſecond Eye be ſuppoſed at <lb/>the ſame Height and Diſtance as the firſt; </s> <s xml:id="echoid-s1457" xml:space="preserve">then <lb/>find <anchor type="note" xlink:href="" symbol="*"/> F G the Repreſentation of A B for this <anchor type="note" xlink:label="note-0113-01a" xlink:href="note-0113-01"/> ſecond Eye, which continue until it meets the <lb/>Line C E in b, and in this Continuation aſſume <lb/>b a equal to F G; </s> <s xml:id="echoid-s1458" xml:space="preserve">then a will be the Perſpective <lb/>ſought.</s> <s xml:id="echoid-s1459" xml:space="preserve"/> </p> <div xml:id="echoid-div217" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0113-01" xlink:href="note-0113-01a" xml:space="preserve">42.</note> </div> </div> <div xml:id="echoid-div219" type="section" level="1" n="115"> <head xml:id="echoid-head121" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1460" xml:space="preserve">Becauſe the Height and Diſtance of the ſecond <lb/>Eye, is equal to the Height and Diſtance of the <lb/>firſt; </s> <s xml:id="echoid-s1461" xml:space="preserve">the ſaid two Eyes are both in one parallel <lb/>Line A B; </s> <s xml:id="echoid-s1462" xml:space="preserve">and conſequently <anchor type="note" xlink:href="" symbol="*"/>, the Perſpective <anchor type="note" xlink:label="note-0113-02a" xlink:href="note-0113-02"/> of A B muſt be a Part of F G continued, and <lb/> <anchor type="note" xlink:label="note-0113-03a" xlink:href="note-0113-03"/> alſo equal to F G: </s> <s xml:id="echoid-s1463" xml:space="preserve">And therefore becauſe <anchor type="note" xlink:href="" symbol="*"/> the Perſpective of B is in the Line C E, a b is the <lb/> <anchor type="note" xlink:label="note-0113-04a" xlink:href="note-0113-04"/> Perſpective of A B; </s> <s xml:id="echoid-s1464" xml:space="preserve">and a <anchor type="note" xlink:href="" symbol="*"/>, that of A. </s> <s xml:id="echoid-s1465" xml:space="preserve">Which was to be demonſtrated.</s> <s xml:id="echoid-s1466" xml:space="preserve"/> </p> <div xml:id="echoid-div219" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0113-02" xlink:href="note-0113-02a" xml:space="preserve">18.</note> <note symbol="*" position="right" xlink:label="note-0113-03" xlink:href="note-0113-03a" xml:space="preserve">12.</note> <note symbol="*" position="right" xlink:label="note-0113-04" xlink:href="note-0113-04a" xml:space="preserve">16.</note> </div> <p> <s xml:id="echoid-s1467" xml:space="preserve">78. </s> <s xml:id="echoid-s1468" xml:space="preserve">Note, as to Lines perpendicular, and inclined <lb/>to the Geometrical Plane, ſee n. </s> <s xml:id="echoid-s1469" xml:space="preserve">76. </s> <s xml:id="echoid-s1470" xml:space="preserve">This is <lb/>ſcarcely uſeful, unleſs for the Decorations of a <lb/>Theatre.</s> <s xml:id="echoid-s1471" xml:space="preserve"/> </p> <pb o="62" file="0114" n="131" rhead="An ESSAY"/> </div> <div xml:id="echoid-div221" type="section" level="1" n="116"> <head xml:id="echoid-head122" xml:space="preserve"><emph style="sc">Proe</emph>. III.</head> <p style="it"> <s xml:id="echoid-s1472" xml:space="preserve">79. </s> <s xml:id="echoid-s1473" xml:space="preserve">To find the Repreſentation of a Figure in the <lb/>Geometrical Plane, when the Perſpective Plane is <lb/>placed above the Eye.</s> <s xml:id="echoid-s1474" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1475" xml:space="preserve">When the perſpective Plane is ſituated above <lb/>the Eye, we ſuppoſe the Geometrical Plane to <lb/>paſs through the Top of the Perſpective Plane; <lb/></s> <s xml:id="echoid-s1476" xml:space="preserve">upon which Geometrical Plane are drawn the <lb/>Figures of Objects meeting it; </s> <s xml:id="echoid-s1477" xml:space="preserve">as alſo the Seats <lb/>of thoſe Objects that are underneath it, by <lb/>Means of Perpendiculars; </s> <s xml:id="echoid-s1478" xml:space="preserve">and the Height of the <lb/>Eye is here meaſur’d by a Perpendicular drawn <lb/>from the Eye to the Geometrical Plane; </s> <s xml:id="echoid-s1479" xml:space="preserve">whence <lb/>the perſpective Plane, elevated in reſpect to the <lb/>Eye, is the ſame thing, as an Eye elevated in <lb/>regard to the perſpective Plane.</s> <s xml:id="echoid-s1480" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1481" xml:space="preserve">Let I L be the Baſe Line, and H the Foot of <lb/> <anchor type="note" xlink:label="note-0114-01a" xlink:href="note-0114-01"/> the vertical Line; </s> <s xml:id="echoid-s1482" xml:space="preserve">then in the Baſe Line aſ-<lb/>ſume the Points I and L at Pleaſure, towards <lb/>the Sides of the perſpective Plane. </s> <s xml:id="echoid-s1483" xml:space="preserve">Make I S <lb/>equal to {1/3} or {1/4} Part of I H, and raiſe the Per-<lb/>pendicular S X, in the Point S, to the Baſe Line, <lb/>equal to a correſpondent Part of the Height and <lb/>Diſtance of the Eye taken together; </s> <s xml:id="echoid-s1484" xml:space="preserve">draw the <lb/>Line X I G, and moreover Y L Q, in aſſuming <lb/>L T equal to {1/3} or {1/4} & </s> <s xml:id="echoid-s1485" xml:space="preserve">c. </s> <s xml:id="echoid-s1486" xml:space="preserve">of L H. </s> <s xml:id="echoid-s1487" xml:space="preserve">Again draw <lb/>the Line G Q in the Geometrical Plane, pa-<lb/>rallel to the Baſe Line, and diſtant therefrom <lb/>(for Example) a third Part of the Height of <lb/>the Eye; </s> <s xml:id="echoid-s1488" xml:space="preserve">draw alſo F P in the perſpective Plane, <lb/>parallel to the Baſe Line, and diſtant therefrom, <lb/>a fourth Part of the Eye’s Diſtance; </s> <s xml:id="echoid-s1489" xml:space="preserve">theſe two <lb/>Lines will cut X I in G and F, and Y L in Q <lb/>and P. </s> <s xml:id="echoid-s1490" xml:space="preserve">Note, if the Diſtance of G Q from the <lb/>Baſe Line, had been aſſumed equal to a fourth <lb/>Part of the Eye’s Diſtance; </s> <s xml:id="echoid-s1491" xml:space="preserve">then F P muſt have <pb o="63" file="0115" n="132" rhead="on PERSPECTIVE."/> been drawn from the Baſe Line, equal to a fifth <lb/>Part of the Eye’s Diſtance, and ſo on. </s> <s xml:id="echoid-s1492" xml:space="preserve">Now <lb/>A is a Point whoſe Repreſentation is requir’d.</s> <s xml:id="echoid-s1493" xml:space="preserve"/> </p> <div xml:id="echoid-div221" type="float" level="2" n="1"> <note position="left" xlink:label="note-0114-01" xlink:href="note-0114-01a" xml:space="preserve">Fig. 42.</note> </div> </div> <div xml:id="echoid-div223" type="section" level="1" n="117"> <head xml:id="echoid-head123" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1494" xml:space="preserve">Draw the Lines A F and A P, from the Point <lb/>A to the Points F and P, cutting the Baſe Line <lb/>in the Points E and B; </s> <s xml:id="echoid-s1495" xml:space="preserve">then draw the Lines E G <lb/>and B Q, which continue till they interſect <lb/>each other in a, which is the Repreſentation <lb/>ſought.</s> <s xml:id="echoid-s1496" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div224" type="section" level="1" n="118"> <head xml:id="echoid-head124" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1497" xml:space="preserve">Let us ſuppoſe the perſpective Plane continu-<lb/>ed, C D the Horizontal Line, and O the Eye <lb/>denoted in the Horizontal Plane. </s> <s xml:id="echoid-s1498" xml:space="preserve">It is evi-<lb/>dent <anchor type="note" xlink:href="" symbol="*"/> by Conſtruction, that the Line G F con- <anchor type="note" xlink:label="note-0115-01a" xlink:href="note-0115-01"/> tinued, paſſes through the Eye O; </s> <s xml:id="echoid-s1499" xml:space="preserve">produce the <lb/>Line G S a, until it meets the Horizontal Line <lb/>in D, and draw the Line O D. </s> <s xml:id="echoid-s1500" xml:space="preserve">Let fall the <lb/>Perpendicular G N R, from the Point G upon <lb/>the Horizontal Line, which interſect in R, by <lb/>the Line O R, paſſing through the Eye parallel <lb/>to the Horizontal Line. </s> <s xml:id="echoid-s1501" xml:space="preserve">Now by Conſtruction, <lb/>G M is {1/3} of M N; </s> <s xml:id="echoid-s1502" xml:space="preserve">and conſequently it is {1/4} of <lb/>G N; </s> <s xml:id="echoid-s1503" xml:space="preserve">M Z is likewiſe {1/4} of N R: </s> <s xml:id="echoid-s1504" xml:space="preserve">Therefore</s> </p> <div xml:id="echoid-div224" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0115-01" xlink:href="note-0115-01a" xml:space="preserve">77.</note> </div> <p style="it"> <s xml:id="echoid-s1505" xml:space="preserve">G M: </s> <s xml:id="echoid-s1506" xml:space="preserve">M Z:</s> <s xml:id="echoid-s1507" xml:space="preserve">: G N: </s> <s xml:id="echoid-s1508" xml:space="preserve">N R.</s> <s xml:id="echoid-s1509" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1510" xml:space="preserve">Compon. </s> <s xml:id="echoid-s1511" xml:space="preserve">and Altern.</s> <s xml:id="echoid-s1512" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s1513" xml:space="preserve">G M: </s> <s xml:id="echoid-s1514" xml:space="preserve">G N:</s> <s xml:id="echoid-s1515" xml:space="preserve">: G M + M Z = G Z: </s> <s xml:id="echoid-s1516" xml:space="preserve">G N <lb/>+ N R = G R.</s> <s xml:id="echoid-s1517" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1518" xml:space="preserve">Becauſe the Triangles G M I and G N C are <lb/>ſimilar, we have</s> </p> <p style="it"> <s xml:id="echoid-s1519" xml:space="preserve">G M: </s> <s xml:id="echoid-s1520" xml:space="preserve">G N:</s> <s xml:id="echoid-s1521" xml:space="preserve">: G I: </s> <s xml:id="echoid-s1522" xml:space="preserve">G C.</s> <s xml:id="echoid-s1523" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1524" xml:space="preserve">The Triangles G Z F and G R O being alſo <lb/>ſimilar,</s> </p> <p style="it"> <s xml:id="echoid-s1525" xml:space="preserve">G Z: </s> <s xml:id="echoid-s1526" xml:space="preserve">G R:</s> <s xml:id="echoid-s1527" xml:space="preserve">: G F: </s> <s xml:id="echoid-s1528" xml:space="preserve">G O.</s> <s xml:id="echoid-s1529" xml:space="preserve"/> </p> <pb o="64" file="0116" n="133" rhead="An ESSAY"/> <p> <s xml:id="echoid-s1530" xml:space="preserve">Whence</s> </p> <p style="it"> <s xml:id="echoid-s1531" xml:space="preserve">G I: </s> <s xml:id="echoid-s1532" xml:space="preserve">G C:</s> <s xml:id="echoid-s1533" xml:space="preserve">: G F: </s> <s xml:id="echoid-s1534" xml:space="preserve">G O.</s> <s xml:id="echoid-s1535" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1536" xml:space="preserve">Again, becauſe the Triangles G I E and G C D <lb/>are ſimilar, we have</s> </p> <p style="it"> <s xml:id="echoid-s1537" xml:space="preserve">G I: </s> <s xml:id="echoid-s1538" xml:space="preserve">G C:</s> <s xml:id="echoid-s1539" xml:space="preserve">: G E: </s> <s xml:id="echoid-s1540" xml:space="preserve">G D.</s> <s xml:id="echoid-s1541" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1542" xml:space="preserve">And conſequently</s> </p> <p style="it"> <s xml:id="echoid-s1543" xml:space="preserve">G F: </s> <s xml:id="echoid-s1544" xml:space="preserve">G O:</s> <s xml:id="echoid-s1545" xml:space="preserve">: G E: </s> <s xml:id="echoid-s1546" xml:space="preserve">G D.</s> <s xml:id="echoid-s1547" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1548" xml:space="preserve">And ſo the Triangles G F E, and G O D are <lb/>ſimilar; </s> <s xml:id="echoid-s1549" xml:space="preserve">and the Line F E A is parallel to O D: <lb/></s> <s xml:id="echoid-s1550" xml:space="preserve">Whence it follows <anchor type="note" xlink:href="" symbol="*"/>, that the Perſpective of <anchor type="note" xlink:label="note-0116-01a" xlink:href="note-0116-01"/> E A, is a Part of E a D. </s> <s xml:id="echoid-s1551" xml:space="preserve">We demonſtrate in <lb/>the ſame Manner, that B a is the Perſpective <lb/>of B A, and ſo the Perſpective of the Point A, <lb/>the common Section of E A and B A, is a, the <lb/>Interſection of the Appearances of the ſaid two <lb/>Lines.</s> <s xml:id="echoid-s1552" xml:space="preserve"/> </p> <div xml:id="echoid-div225" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0116-01" xlink:href="note-0116-01a" xml:space="preserve">13.</note> </div> </div> <div xml:id="echoid-div227" type="section" level="1" n="119"> <head xml:id="echoid-head125" xml:space="preserve"><emph style="sc">Prob</emph>. IV.</head> <p style="it"> <s xml:id="echoid-s1553" xml:space="preserve">80. </s> <s xml:id="echoid-s1554" xml:space="preserve">To find the Repreſentation of a Line, per-<lb/>pendicular to the Geometrical Plane, when the per-<lb/>ſpective Plane is above the Eye.</s> <s xml:id="echoid-s1555" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1556" xml:space="preserve">In the Baſe Line B E, aſſume the Line E D, <lb/> <anchor type="note" xlink:label="note-0116-02a" xlink:href="note-0116-02"/> equal in Length to the propoſed Perpendicular; <lb/></s> <s xml:id="echoid-s1557" xml:space="preserve">and draw C L, parallel to the Baſe Line, and <lb/>diſtant therefrom (for Example) {1/4} of the Height <lb/>of the Eye; </s> <s xml:id="echoid-s1558" xml:space="preserve">make F L equal to {3/4} of D E, and <lb/>draw the Lines E L and D F. </s> <s xml:id="echoid-s1559" xml:space="preserve">Note, if the <lb/>Diſtance from C L to B E, had been aſſumed <lb/>equal to a fifth Part of the Height of the Eye, <lb/>F L muſt have been aſſumed equal to {4/5} Parts of <lb/>E D. </s> <s xml:id="echoid-s1560" xml:space="preserve">Now let a be the Perſpective of the Foot <lb/>of the propoſed Perpendicular; </s> <s xml:id="echoid-s1561" xml:space="preserve">through which <lb/>draw a H parallel to the Baſe Line, and a I per-<lb/>pendicular to the ſaid Line; </s> <s xml:id="echoid-s1562" xml:space="preserve">then make a I equal <lb/>to G H, and the propoſed Perſpective will be <lb/>had. </s> <s xml:id="echoid-s1563" xml:space="preserve">The Demonſtration of this Operation is <lb/>manifeſt <anchor type="note" xlink:href="" symbol="*"/>, in conſidering that D F and E L <anchor type="note" xlink:label="note-0116-03a" xlink:href="note-0116-03"/> <pb file="0117" n="134"/> <pb file="0117a" n="135"/> <anchor type="figure" xlink:label="fig-0117a-01a" xlink:href="fig-0117a-01"/> <anchor type="figure" xlink:label="fig-0117a-02a" xlink:href="fig-0117a-02"/> <anchor type="figure" xlink:label="fig-0117a-03a" xlink:href="fig-0117a-03"/> <pb file="0118" n="136"/> <pb o="65" file="0119" n="137" rhead="on PERSPECTIVE."/> being produced, will meet each other in the <lb/>Horizontal Line.</s> <s xml:id="echoid-s1564" xml:space="preserve"/> </p> <div xml:id="echoid-div227" type="float" level="2" n="1"> <note position="left" xlink:label="note-0116-02" xlink:href="note-0116-02a" xml:space="preserve">Fig. 43.</note> <note symbol="*" position="left" xlink:label="note-0116-03" xlink:href="note-0116-03a" xml:space="preserve">56.</note> <figure xlink:label="fig-0117a-01" xlink:href="fig-0117a-01a"> <caption xml:id="echoid-caption40" style="it" xml:space="preserve">page 64<lb/>Plate. 18.</caption> <variables xml:id="echoid-variables39" xml:space="preserve">d v</variables> </figure> <figure xlink:label="fig-0117a-02" xlink:href="fig-0117a-02a"> <caption xml:id="echoid-caption41" xml:space="preserve">Fig. 41</caption> <variables xml:id="echoid-variables40" xml:space="preserve">E b a G F H C B A D</variables> </figure> <figure xlink:label="fig-0117a-03" xlink:href="fig-0117a-03a"> <caption xml:id="echoid-caption42" xml:space="preserve">Fig. 42</caption> <variables xml:id="echoid-variables41" xml:space="preserve">G Q A M I S E H T B L Z F P a X Y N C D R O</variables> </figure> </div> </div> <div xml:id="echoid-div229" type="section" level="1" n="120"> <head xml:id="echoid-head126" xml:space="preserve">CHAP. V.</head> <p style="it"> <s xml:id="echoid-s1565" xml:space="preserve">Of throwing Figures into Perſpective, when <lb/>the Perſpective Plane is conſider’d as being <lb/>inclined.</s> <s xml:id="echoid-s1566" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div230" type="section" level="1" n="121"> <head xml:id="echoid-head127" xml:space="preserve"><emph style="sc">Problem</emph> I.</head> <p style="it"> <s xml:id="echoid-s1567" xml:space="preserve">81. </s> <s xml:id="echoid-s1568" xml:space="preserve">TO find the Perſpective of a Figure in the <lb/> <anchor type="note" xlink:label="note-0119-01a" xlink:href="note-0119-01"/> Geometrical Plane.</s> <s xml:id="echoid-s1569" xml:space="preserve"/> </p> <div xml:id="echoid-div230" type="float" level="2" n="1"> <note position="right" xlink:label="note-0119-01" xlink:href="note-0119-01a" xml:space="preserve">Fig. 44.</note> </div> <p> <s xml:id="echoid-s1570" xml:space="preserve">Let X be the Vertical Plane; </s> <s xml:id="echoid-s1571" xml:space="preserve">S I the Station <lb/>Line, S the Station Point, and H the Interſecti-<lb/>on of the Station Line and Baſe Line. </s> <s xml:id="echoid-s1572" xml:space="preserve">Now <lb/>draw the Vertical Line H V through the Point H, <lb/>making an Angle with S I, equal to the Angle <lb/>of Inclination of the perſpective Plane; </s> <s xml:id="echoid-s1573" xml:space="preserve">then <lb/>raiſe the Perpendicular I O to S I, in the Sta-<lb/>tion Point S, equal to the Height of the Eye; <lb/></s> <s xml:id="echoid-s1574" xml:space="preserve">and through the Extremity of the ſaid Perpen-<lb/>dicular, draw the principal Ray O V, paral-<lb/>lel to S I, and cutting H V in the Point of <lb/>Sight V.</s> <s xml:id="echoid-s1575" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1576" xml:space="preserve">Now it is evident, that O V determines the <lb/>Length of the principal Ray, and H V the Di-<lb/>ſtance from the Baſe Line to the Horizontal <lb/>Line; </s> <s xml:id="echoid-s1577" xml:space="preserve">and ſince the Demonſtration of the <lb/>Problems in the aforegoing Chapters regarding <lb/>the Geometrical Plane, have alſo Relation to <lb/>the perſpective Plane being inclined, the ſaid <lb/>Problems may be here uſed; </s> <s xml:id="echoid-s1578" xml:space="preserve">and conſequently, <lb/>this inclined perſpective Plane is reduced to a <lb/>Perpendicular one, view’d by an Eye, whoſe <lb/>Height is H V, and Diſtance O V.</s> <s xml:id="echoid-s1579" xml:space="preserve"/> </p> <pb o="66" file="0120" n="138" rhead="An ESSAY"/> </div> <div xml:id="echoid-div232" type="section" level="1" n="122"> <head xml:id="echoid-head128" xml:space="preserve"><emph style="sc">Problem</emph> II.</head> <p style="it"> <s xml:id="echoid-s1580" xml:space="preserve">82. </s> <s xml:id="echoid-s1581" xml:space="preserve">To find the Appearance of a Point above the <lb/>Geometrical Plane.</s> <s xml:id="echoid-s1582" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1583" xml:space="preserve">Let H C be the Baſe Line: </s> <s xml:id="echoid-s1584" xml:space="preserve">And let T be the <lb/> <anchor type="note" xlink:label="note-0120-01a" xlink:href="note-0120-01"/> accidental Point of the Lines perpendicular <lb/>to the Geometrical Plane. </s> <s xml:id="echoid-s1585" xml:space="preserve">This Point will <lb/>be <anchor type="note" xlink:href="" symbol="*"/> in that Place of the Vertical Line, wherein <anchor type="note" xlink:label="note-0120-02a" xlink:href="note-0120-02"/> it is cut by the Prolongation of the Line mea-<lb/>ſuring the Height of the Eye; </s> <s xml:id="echoid-s1586" xml:space="preserve">for this laſt <lb/>Line is parallel to the ſaid Perpendiculars. </s> <s xml:id="echoid-s1587" xml:space="preserve">And <lb/>ſo likewiſe the aforeſaid Point is the ſame as <lb/>the Point T of Fig. </s> <s xml:id="echoid-s1588" xml:space="preserve">44: </s> <s xml:id="echoid-s1589" xml:space="preserve">Let V be the Point of <lb/>Sight, S the Station Point, and Q the Station <lb/>Point of the upright perſpective Plane, to which <lb/>the inclined perſpective Plane is reduced <anchor type="note" xlink:href="" symbol="*"/>. </s> <s xml:id="echoid-s1590" xml:space="preserve">And <anchor type="note" xlink:label="note-0120-03a" xlink:href="note-0120-03"/> laſtly, let A be the Seat of the given Point.</s> <s xml:id="echoid-s1591" xml:space="preserve"/> </p> <div xml:id="echoid-div232" type="float" level="2" n="1"> <note position="left" xlink:label="note-0120-01" xlink:href="note-0120-01a" xml:space="preserve">Fig. 45.</note> <note symbol="*" position="left" xlink:label="note-0120-02" xlink:href="note-0120-02a" xml:space="preserve">13.</note> <note symbol="*" position="left" xlink:label="note-0120-03" xlink:href="note-0120-03a" xml:space="preserve">81.</note> </div> </div> <div xml:id="echoid-div234" type="section" level="1" n="123"> <head xml:id="echoid-head129" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1592" xml:space="preserve">Draw two Lines M P and P E ſeparately, <lb/>making a right Angle with each other; </s> <s xml:id="echoid-s1593" xml:space="preserve">in one <lb/>of which, aſſume P E, equal to the Height of <lb/>the given Point, whoſe Perſpective is ſought; <lb/></s> <s xml:id="echoid-s1594" xml:space="preserve">and draw the Line E M, making an Angle with <lb/>M P, equal to the Angle of Inclination of the <lb/>perſpective Plane. </s> <s xml:id="echoid-s1595" xml:space="preserve">Again let fall the Perpen-<lb/>dicular A D from the Point A to the Baſe <lb/>Line, in which aſſume A L equal to P M, to-<lb/>wards the Baſe Line, when the perſpective <lb/>Plane is inclined towards the Objects (as we <lb/>have here ſuppoſed) but on the other Side of A, <lb/>when the perſpective Plane inclines towards the <lb/>Eye. </s> <s xml:id="echoid-s1596" xml:space="preserve">Then from the Point A, draw a Line <lb/>to the Point S, cutting the Baſe Line in B, and <lb/>joyn the Points L and Q, by a Line cutting the <lb/>Baſe Line in C. </s> <s xml:id="echoid-s1597" xml:space="preserve">This being done, draw the <pb o="67" file="0121" n="139" rhead="on PERSPECTIVE."/> Line T B X; </s> <s xml:id="echoid-s1598" xml:space="preserve">which interſect in the Point X, <lb/>by a Perpendicular to the Baſe Line, in the <lb/>Point G; </s> <s xml:id="echoid-s1599" xml:space="preserve">and then the Point X is the Appear-<lb/>ance ſought.</s> <s xml:id="echoid-s1600" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div235" type="section" level="1" n="124"> <head xml:id="echoid-head130" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1601" xml:space="preserve">In Fig. </s> <s xml:id="echoid-s1602" xml:space="preserve">44. </s> <s xml:id="echoid-s1603" xml:space="preserve">where V, S, T, and H, repreſent <lb/>the ſame Points as thoſe that are denoted with <lb/>the ſame Letters in this Figure; </s> <s xml:id="echoid-s1604" xml:space="preserve">we have, <lb/></s> </p> <p> <s xml:id="echoid-s1605" xml:space="preserve">T H: </s> <s xml:id="echoid-s1606" xml:space="preserve">H S:</s> <s xml:id="echoid-s1607" xml:space="preserve">: T V: </s> <s xml:id="echoid-s1608" xml:space="preserve">V O.</s> <s xml:id="echoid-s1609" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1610" xml:space="preserve">Compon. </s> <s xml:id="echoid-s1611" xml:space="preserve">and altern.</s> <s xml:id="echoid-s1612" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1613" xml:space="preserve">T H: </s> <s xml:id="echoid-s1614" xml:space="preserve">T V:</s> <s xml:id="echoid-s1615" xml:space="preserve">: T H + H S: </s> <s xml:id="echoid-s1616" xml:space="preserve">T V + V O.</s> <s xml:id="echoid-s1617" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1618" xml:space="preserve">This being apply’d to Fig. </s> <s xml:id="echoid-s1619" xml:space="preserve">45. </s> <s xml:id="echoid-s1620" xml:space="preserve">and it will be, <lb/>T H: </s> <s xml:id="echoid-s1621" xml:space="preserve">T V:</s> <s xml:id="echoid-s1622" xml:space="preserve">: T S: </s> <s xml:id="echoid-s1623" xml:space="preserve">T V + V O.</s> <s xml:id="echoid-s1624" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1625" xml:space="preserve">If now T X be continued, till it cuts the Ho-<lb/>rizontal Line in F; </s> <s xml:id="echoid-s1626" xml:space="preserve">we ſhall have,</s> </p> <p> <s xml:id="echoid-s1627" xml:space="preserve">T H: </s> <s xml:id="echoid-s1628" xml:space="preserve">T V:</s> <s xml:id="echoid-s1629" xml:space="preserve">: T B: </s> <s xml:id="echoid-s1630" xml:space="preserve">T F.</s> <s xml:id="echoid-s1631" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1632" xml:space="preserve">And conſequently,</s> </p> <p> <s xml:id="echoid-s1633" xml:space="preserve">T B: </s> <s xml:id="echoid-s1634" xml:space="preserve">T F:</s> <s xml:id="echoid-s1635" xml:space="preserve">: T S: </s> <s xml:id="echoid-s1636" xml:space="preserve">T V + V O.</s> <s xml:id="echoid-s1637" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1638" xml:space="preserve">Whence it follows, that if a Line be drawn <lb/>ſrom the Eye, to the Point F, it will be paral-<lb/>lel to S B A. </s> <s xml:id="echoid-s1639" xml:space="preserve">Therefore <anchor type="note" xlink:href="" symbol="*"/> the Perſpective of <anchor type="note" xlink:label="note-0121-01a" xlink:href="note-0121-01"/> B A, is a Part of B X; </s> <s xml:id="echoid-s1640" xml:space="preserve">and ſo the Repreſenta-<lb/>tion of A is in the ſaid Line. </s> <s xml:id="echoid-s1641" xml:space="preserve">The Perſpective <lb/>of a Line perpendicular to the Geometrical <lb/>Plane, in the Point A, paſſes thro’ the Perſpe-<lb/>ctive of the Point A, and thro’ the Point T <anchor type="note" xlink:href="" symbol="*"/>;</s> <s xml:id="echoid-s1642" xml:space="preserve"> <anchor type="note" xlink:label="note-0121-02a" xlink:href="note-0121-02"/> therefore it is a Part of T X. </s> <s xml:id="echoid-s1643" xml:space="preserve">But the given Point <lb/>is in the aſoreſaid Perpendicular: </s> <s xml:id="echoid-s1644" xml:space="preserve">And ſo its Per-<lb/>ſpective is in T X.</s> <s xml:id="echoid-s1645" xml:space="preserve"/> </p> <div xml:id="echoid-div235" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0121-01" xlink:href="note-0121-01a" xml:space="preserve">13.</note> <note symbol="*" position="right" xlink:label="note-0121-02" xlink:href="note-0121-02a" xml:space="preserve">13, 14.</note> </div> <p> <s xml:id="echoid-s1646" xml:space="preserve">Again; </s> <s xml:id="echoid-s1647" xml:space="preserve">it is otherwiſe manifeſt, that the Per-<lb/>ſpective of C L, is a <anchor type="note" xlink:href="" symbol="*"/> Part of C X; </s> <s xml:id="echoid-s1648" xml:space="preserve">and conſe- <anchor type="note" xlink:label="note-0121-03a" xlink:href="note-0121-03"/> quently, the Appearance of L is in this Line. <lb/></s> <s xml:id="echoid-s1649" xml:space="preserve">Now, if a Line be ſuppos’d to be drawn from <lb/>the Point L, thro’ the propos’d Point, it will be <lb/>parallel to the Vertical Line; </s> <s xml:id="echoid-s1650" xml:space="preserve">and ſo its <anchor type="note" xlink:href="" symbol="*"/> Per- <anchor type="note" xlink:label="note-0121-04a" xlink:href="note-0121-04"/> ſpective is parallel to the Baſe Line. </s> <s xml:id="echoid-s1651" xml:space="preserve">And ſince <pb o="68" file="0122" n="140" rhead="An ESSAY"/> this Appearance paſſes thro’ that of the Point <lb/>L, it will be a Part of C X. </s> <s xml:id="echoid-s1652" xml:space="preserve">But becauſe that <lb/>Line, drawn from the Point L, paſſes thro’ the <lb/>propos’d Point; </s> <s xml:id="echoid-s1653" xml:space="preserve">the Repreſentation of the ſaid <lb/>Point is alſo in C X; </s> <s xml:id="echoid-s1654" xml:space="preserve">and ſo in X, the common <lb/>Interſection of C X, and T X.</s> <s xml:id="echoid-s1655" xml:space="preserve"/> </p> <div xml:id="echoid-div236" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0121-03" xlink:href="note-0121-03a" xml:space="preserve">41.</note> <note symbol="*" position="right" xlink:label="note-0121-04" xlink:href="note-0121-04a" xml:space="preserve">6.</note> </div> </div> <div xml:id="echoid-div238" type="section" level="1" n="125"> <head xml:id="echoid-head131" xml:space="preserve"><emph style="sc">Remark</emph>.</head> <p> <s xml:id="echoid-s1656" xml:space="preserve">83. </s> <s xml:id="echoid-s1657" xml:space="preserve">If the Point T ſhould be at too great a <lb/>Diſtance; </s> <s xml:id="echoid-s1658" xml:space="preserve">or if T B X, or C X, ſhould too ob-<lb/>liquely cut each other; </s> <s xml:id="echoid-s1659" xml:space="preserve">the perſpective Plane <lb/>muſt then be ſuppos’d to be reduc’d <anchor type="note" xlink:href="" symbol="*"/> to a per- <anchor type="note" xlink:label="note-0122-01a" xlink:href="note-0122-01"/> pendicular, or upright one; </s> <s xml:id="echoid-s1660" xml:space="preserve">and the Repreſen-<lb/>tation of a Point, above the Geometrical Plane, <lb/>(whoſe Seat is L, and Height M E) muſt be <lb/>found <anchor type="note" xlink:href="" symbol="*"/>.</s> <s xml:id="echoid-s1661" xml:space="preserve"/> </p> <div xml:id="echoid-div238" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0122-01" xlink:href="note-0122-01a" xml:space="preserve">81.</note> </div> <note symbol="*" position="left" xml:space="preserve">50.</note> </div> <div xml:id="echoid-div240" type="section" level="1" n="126"> <head xml:id="echoid-head132" xml:space="preserve"><emph style="sc">Problem</emph> III.</head> <p style="it"> <s xml:id="echoid-s1662" xml:space="preserve">84. </s> <s xml:id="echoid-s1663" xml:space="preserve">To find the Repreſentation of a Line, perpendi-<lb/>cular to the Geometrical Plane`.</s> <s xml:id="echoid-s1664" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1665" xml:space="preserve">The Appearance of the Extremity of the Per-<lb/> <anchor type="note" xlink:label="note-0122-03a" xlink:href="note-0122-03"/> pendicular muſt be found <anchor type="note" xlink:href="" symbol="*"/>, in conſidering the <anchor type="note" xlink:label="note-0122-04a" xlink:href="note-0122-04"/> ſaid Extremity as a Point above the Geometri-<lb/>cal Plane, by the Height of the propos’d Perpendi-<lb/>cular: </s> <s xml:id="echoid-s1666" xml:space="preserve">Then if a Line be drawn from the Point <lb/>D, to the Point of Sight; </s> <s xml:id="echoid-s1667" xml:space="preserve">its Interſection <anchor type="note" xlink:href="" symbol="*"/> with <anchor type="note" xlink:label="note-0122-05a" xlink:href="note-0122-05"/> T X, will give the Appearance a of the Seat of <lb/>the Perpendicular propos’d.</s> <s xml:id="echoid-s1668" xml:space="preserve"/> </p> <div xml:id="echoid-div240" type="float" level="2" n="1"> <note position="left" xlink:label="note-0122-03" xlink:href="note-0122-03a" xml:space="preserve">Fig. 45.</note> <note symbol="*" position="left" xlink:label="note-0122-04" xlink:href="note-0122-04a" xml:space="preserve">82.</note> <note symbol="*" position="left" xlink:label="note-0122-05" xlink:href="note-0122-05a" xml:space="preserve">16.</note> </div> <p> <s xml:id="echoid-s1669" xml:space="preserve">Note, When there is a Neceſſity of having re-<lb/>courſe to the Remarks of the foregoing Problem, <lb/>in order to find the Point X; </s> <s xml:id="echoid-s1670" xml:space="preserve">then the Point a <lb/>may be found, in drawing A S and D V, and <lb/>afterwards joining the Points B and X by a <lb/>Line. </s> <s xml:id="echoid-s1671" xml:space="preserve">And when B X and D V cut each other <pb file="0123" n="141"/> <pb file="0123a" n="142"/> <anchor type="figure" xlink:label="fig-0123a-01a" xlink:href="fig-0123a-01"/> <anchor type="figure" xlink:label="fig-0123a-02a" xlink:href="fig-0123a-02"/> <anchor type="figure" xlink:label="fig-0123a-03a" xlink:href="fig-0123a-03"/> <pb file="0124" n="143"/> <pb o="69" file="0125" n="144" rhead="on PERSPECTIVE."/> too obliquely, recourſe muſt be had to Problem I. </s> <s xml:id="echoid-s1672" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/>, <anchor type="note" xlink:label="note-0125-01a" xlink:href="note-0125-01"/> to find the Appearance of a.</s> <s xml:id="echoid-s1673" xml:space="preserve"/> </p> <div xml:id="echoid-div241" type="float" level="2" n="2"> <figure xlink:label="fig-0123a-01" xlink:href="fig-0123a-01a"> <caption xml:id="echoid-caption43" style="it" xml:space="preserve">page 68<lb/>Plate. 19<lb/>Fig. 43</caption> <variables xml:id="echoid-variables42" xml:space="preserve">B D E a G H I C F L</variables> </figure> <figure xlink:label="fig-0123a-02" xlink:href="fig-0123a-02a"> <caption xml:id="echoid-caption44" style="it" xml:space="preserve">Fig. 44</caption> <variables xml:id="echoid-variables43" xml:space="preserve">O V X S H I T</variables> </figure> <figure xlink:label="fig-0123a-03" xlink:href="fig-0123a-03a"> <caption xml:id="echoid-caption45" style="it" xml:space="preserve">Fig. 45</caption> <variables xml:id="echoid-variables44" xml:space="preserve">Q F V X S a H B C D E L M P T A</variables> </figure> <note symbol="*" position="right" xlink:label="note-0125-01" xlink:href="note-0125-01a" xml:space="preserve">81.</note> </div> </div> <div xml:id="echoid-div243" type="section" level="1" n="127"> <head xml:id="echoid-head133" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p> <s xml:id="echoid-s1674" xml:space="preserve">85. </s> <s xml:id="echoid-s1675" xml:space="preserve">A is the Foot of the Perpendicular: </s> <s xml:id="echoid-s1676" xml:space="preserve">The <lb/> <anchor type="note" xlink:label="note-0125-02a" xlink:href="note-0125-02"/> Triangle, E P M, is drawn <anchor type="note" xlink:href="" symbol="*"/> as directed: </s> <s xml:id="echoid-s1677" xml:space="preserve">And <anchor type="note" xlink:label="note-0125-03a" xlink:href="note-0125-03"/> T is the accidental Point of the Perpendiculars, <lb/>to the Geometrical Plane.</s> <s xml:id="echoid-s1678" xml:space="preserve"/> </p> <div xml:id="echoid-div243" type="float" level="2" n="1"> <note position="right" xlink:label="note-0125-02" xlink:href="note-0125-02a" xml:space="preserve">Fig. 46.</note> <note symbol="*" position="right" xlink:label="note-0125-03" xlink:href="note-0125-03a" xml:space="preserve">82.</note> </div> </div> <div xml:id="echoid-div245" type="section" level="1" n="128"> <head xml:id="echoid-head134" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1679" xml:space="preserve">Thro’ the Point a, the Appearance of A, <lb/>draw a Perpendicular to the Baſe Line; </s> <s xml:id="echoid-s1680" xml:space="preserve">which <lb/>make equal <anchor type="note" xlink:href="" symbol="*"/> in Repreſentation to the Line <anchor type="note" xlink:label="note-0125-04a" xlink:href="note-0125-04"/> M E; </s> <s xml:id="echoid-s1681" xml:space="preserve">in conſidering this laſt Line, as being <lb/>parallel to the Vertical Line. </s> <s xml:id="echoid-s1682" xml:space="preserve">Then, from the <lb/>Extremity I of this Perſpective, to the Point of <lb/>Sight V, draw a Line cutting the Line T a, in <lb/>the Point X; </s> <s xml:id="echoid-s1683" xml:space="preserve">which will be the Repreſentation <lb/>of the Extremity of the propos’d Line.</s> <s xml:id="echoid-s1684" xml:space="preserve"/> </p> <div xml:id="echoid-div245" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0125-04" xlink:href="note-0125-04a" xml:space="preserve">55.</note> </div> </div> <div xml:id="echoid-div247" type="section" level="1" n="129"> <head xml:id="echoid-head135" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1685" xml:space="preserve">Let us ſuppoſe a Line paſſing thro’ the Point <lb/>A, equal to M E, and parallel to the Verti-<lb/>cal Line. </s> <s xml:id="echoid-s1686" xml:space="preserve">Suppoſe, moreover, that another Line <lb/>is drawn thro’ the Extremity of this Line, and <lb/>that of the propos’d Perpendicular; </s> <s xml:id="echoid-s1687" xml:space="preserve">then this <lb/>laſt Line, by the Conſtruction of the Figure <lb/>M E P, will be parallel to the Station Line; <lb/></s> <s xml:id="echoid-s1688" xml:space="preserve">and conſequently, its Repreſentation <anchor type="note" xlink:href="" symbol="*"/> will paſs <anchor type="note" xlink:label="note-0125-05a" xlink:href="note-0125-05"/> thro’ the Point of Sight; </s> <s xml:id="echoid-s1689" xml:space="preserve">and its Interſection <lb/>with T a, will be the Extremity of the Repre-<lb/>ſentation ſought. </s> <s xml:id="echoid-s1690" xml:space="preserve">But a I is <anchor type="note" xlink:href="" symbol="*"/> the Perſpective <anchor type="note" xlink:label="note-0125-06a" xlink:href="note-0125-06"/> of the firſt Line, made equal to E M; </s> <s xml:id="echoid-s1691" xml:space="preserve">and con-<lb/>ſequently, V I is that of the ſecond. </s> <s xml:id="echoid-s1692" xml:space="preserve">Which was <lb/>to be demonſtrated.</s> <s xml:id="echoid-s1693" xml:space="preserve"/> </p> <div xml:id="echoid-div247" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0125-05" xlink:href="note-0125-05a" xml:space="preserve">16.</note> <note symbol="*" position="right" xlink:label="note-0125-06" xlink:href="note-0125-06a" xml:space="preserve">56.</note> </div> <pb o="70" file="0126" n="145" rhead="An ESSAY"/> <p> <s xml:id="echoid-s1694" xml:space="preserve">Note, When V I and T a cut each other ve-<lb/>ry obliquely, recourſe muſt be had to the Obſer-<lb/>vation at the End of the aforegoing Method, or <lb/>the following Way may be uſed.</s> <s xml:id="echoid-s1695" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div249" type="section" level="1" n="130"> <head xml:id="echoid-head136" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p> <s xml:id="echoid-s1696" xml:space="preserve">86. </s> <s xml:id="echoid-s1697" xml:space="preserve">Let T be the accidental Point of the Lines <lb/> <anchor type="note" xlink:label="note-0126-01a" xlink:href="note-0126-01"/> perpendicular to the Geometrical Plane, thro’ <lb/>which Point draw a parallel to the Baſe Line, <lb/>in which aſſume T R equal to O T of Fig. </s> <s xml:id="echoid-s1698" xml:space="preserve">44.</s> <s xml:id="echoid-s1699" xml:space="preserve"/> </p> <div xml:id="echoid-div249" type="float" level="2" n="1"> <note position="left" xlink:label="note-0126-01" xlink:href="note-0126-01a" xml:space="preserve">Fig. 47.</note> </div> </div> <div xml:id="echoid-div251" type="section" level="1" n="131"> <head xml:id="echoid-head137" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1700" xml:space="preserve">Aſſume D N ſomewhere in the Baſe Line, e-<lb/>qual to the propoſed Line, and draw the Lines <lb/>D F and N F to the Point F, taken at Pleaſure <lb/>in the Horizontal Line; </s> <s xml:id="echoid-s1701" xml:space="preserve">then through the Point <lb/>a, the Appearance of A, draw the parallel a H, <lb/>to the Baſe Line, in which aſſume a Q equal <lb/>to G H. </s> <s xml:id="echoid-s1702" xml:space="preserve">Then if the Lines T a, and R Q be <lb/>drawn, and continued, till they cut each other <lb/>in the Point X, a X will be the Appearance <lb/>ſought.</s> <s xml:id="echoid-s1703" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div252" type="section" level="1" n="132"> <head xml:id="echoid-head138" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1704" xml:space="preserve">The Part a Q of the Line a H, is <anchor type="note" xlink:href="" symbol="*"/> the Ap- <anchor type="note" xlink:label="note-0126-02a" xlink:href="note-0126-02"/> pearance of a Line proceeding from A in the <lb/>Geometrical Plane, and which is equal to the <lb/>propoſed Line, and parallel to the Baſe Line; <lb/></s> <s xml:id="echoid-s1705" xml:space="preserve">and conſequently <anchor type="note" xlink:href="" symbol="*"/>, the Line R Q paſſes through <anchor type="note" xlink:label="note-0126-03a" xlink:href="note-0126-03"/> the Perſpective of the Extremity of the propo-<lb/>ſed Line: </s> <s xml:id="echoid-s1706" xml:space="preserve">And therefore X the Interſection of <lb/>R Q and T a, is the Appearance of the ſaid <lb/>Extremity.</s> <s xml:id="echoid-s1707" xml:space="preserve"/> </p> <div xml:id="echoid-div252" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0126-02" xlink:href="note-0126-02a" xml:space="preserve">56.</note> <note symbol="*" position="left" xlink:label="note-0126-03" xlink:href="note-0126-03a" xml:space="preserve">20.</note> </div> <pb o="71" file="0127" n="146" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div254" type="section" level="1" n="133"> <head xml:id="echoid-head139" xml:space="preserve"><emph style="sc">Remark</emph>.</head> <p> <s xml:id="echoid-s1708" xml:space="preserve">It is manifeſt <anchor type="note" xlink:href="" symbol="*"/>, that T R may be aſſumed <anchor type="note" xlink:label="note-0127-01a" xlink:href="note-0127-01"/> equal to {1/2} or {1/3} & </s> <s xml:id="echoid-s1709" xml:space="preserve">c. </s> <s xml:id="echoid-s1710" xml:space="preserve">of what it is taken here, <lb/>provided likewiſe that then D N be aſſumed <lb/>equal to a correſpondent Part of the propoſed <lb/>Line.</s> <s xml:id="echoid-s1711" xml:space="preserve"/> </p> <div xml:id="echoid-div254" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0127-01" xlink:href="note-0127-01a" xml:space="preserve">19.</note> </div> </div> <div xml:id="echoid-div256" type="section" level="1" n="134"> <head xml:id="echoid-head140" xml:space="preserve"><emph style="sc">Problem</emph> IV.</head> <p> <s xml:id="echoid-s1712" xml:space="preserve">87. </s> <s xml:id="echoid-s1713" xml:space="preserve">To throw a Sphere into Perſpective.</s> <s xml:id="echoid-s1714" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1715" xml:space="preserve">The Method of ſolving this Problem before <lb/>laid down <anchor type="note" xlink:href="" symbol="*"/>, muſt be uſed here, but with this <anchor type="note" xlink:label="note-0127-02a" xlink:href="note-0127-02"/> Difference; </s> <s xml:id="echoid-s1716" xml:space="preserve">that inſtead of uſing the Point of <lb/>Sight, the Point wherein a Perpendicular drawn <lb/>from the Eye to the perſpective Plane, meets <lb/>the ſaid Plane, muſt be uſed. </s> <s xml:id="echoid-s1717" xml:space="preserve">And you muſt <lb/>obſerve, that this Perpendicular meaſures the <lb/>Eye’s Diſtance from the perſpective Plane.</s> <s xml:id="echoid-s1718" xml:space="preserve"/> </p> <div xml:id="echoid-div256" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0127-02" xlink:href="note-0127-02a" xml:space="preserve">63.</note> </div> </div> <div xml:id="echoid-div258" type="section" level="1" n="135"> <head xml:id="echoid-head141" xml:space="preserve"><emph style="sc">Problem</emph> V.</head> <p style="it"> <s xml:id="echoid-s1719" xml:space="preserve">88. </s> <s xml:id="echoid-s1720" xml:space="preserve">To find the accidental Point of any Number <lb/> <anchor type="note" xlink:label="note-0127-03a" xlink:href="note-0127-03"/> of Lines inclined to the Geometrical Plane.</s> <s xml:id="echoid-s1721" xml:space="preserve"/> </p> <div xml:id="echoid-div258" type="float" level="2" n="1"> <note position="right" xlink:label="note-0127-03" xlink:href="note-0127-03a" xml:space="preserve">Fig. 48.</note> </div> <p> <s xml:id="echoid-s1722" xml:space="preserve">Let A B be the Direction of one of the in-<lb/>clined Lines, O the Eye in the Horizontal Plane, <lb/>and S the Station Point.</s> <s xml:id="echoid-s1723" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div260" type="section" level="1" n="136"> <head xml:id="echoid-head142" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1724" xml:space="preserve">Draw the Line O D, thro’ the Eye O parallel <lb/>to A B, meeting the Horizontal Line in D, which <lb/>will be <anchor type="note" xlink:href="" symbol="*"/> the Accidental Point of the Directions <anchor type="note" xlink:label="note-0127-04a" xlink:href="note-0127-04"/> of the given Line; </s> <s xml:id="echoid-s1725" xml:space="preserve">and thro’ the Station Point S, <pb o="72" file="0128" n="147" rhead="An ESSAY"/> draw S N perpendicular to the ſaid Line A B, <lb/>cutting the Baſe Line in N, and draw the Line <lb/>N D. </s> <s xml:id="echoid-s1726" xml:space="preserve">Then about the Point D, as a Center, and <lb/>with the Radius D O, deſcribe the Circular Arc <lb/>O o: </s> <s xml:id="echoid-s1727" xml:space="preserve">And about N, as a Center, with the Radius <lb/>N S, draw the Circular Arc S s. </s> <s xml:id="echoid-s1728" xml:space="preserve">This being done, <lb/>draw the Line s o touching the ſaid two Arcs, <lb/>and the Line D o perpendicular to s o. </s> <s xml:id="echoid-s1729" xml:space="preserve">Then <lb/>draw o F, making an Angle with o D, equal to <lb/>the Angle of the Inclination of the Lines given, <lb/>and cutting N D continued in F: </s> <s xml:id="echoid-s1730" xml:space="preserve">Now I ſay F <lb/>is the Accidental Point ſought when the Lines <lb/>do not incline towards the Perſpective Plane: <lb/></s> <s xml:id="echoid-s1731" xml:space="preserve">But if they do, o F muſt be drawn below o D.</s> <s xml:id="echoid-s1732" xml:space="preserve"/> </p> <div xml:id="echoid-div260" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0127-04" xlink:href="note-0127-04a" xml:space="preserve">13, 14.</note> </div> </div> <div xml:id="echoid-div262" type="section" level="1" n="137"> <head xml:id="echoid-head143" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1733" xml:space="preserve">If a Plane be ſuppoſed to paſs thro’ the Eye <lb/>parallel to theinclin’d Lines; </s> <s xml:id="echoid-s1734" xml:space="preserve">the common Secti-<lb/>ons of this Plane, and the Horizontal and Geo-<lb/>metrical Planes, will be O D and S N. </s> <s xml:id="echoid-s1735" xml:space="preserve">It is <lb/>now manifeſt, that if a Line be drawn in the <lb/>ſaid Plane, below the Horizontal Plane, when <lb/>the Lines incline towards the perſpective Plane, <lb/>and above it when they incline the other way, <lb/>making an Angle with O D equal to the Angle <lb/>of Inclination of the propoſed Lines; </s> <s xml:id="echoid-s1736" xml:space="preserve">I ſay it is <lb/>manifeſt, that the ſaid Line will be parallel to <lb/>the propoſed Lines, and will meet <anchor type="note" xlink:href="" symbol="*"/> the perſpe- <anchor type="note" xlink:label="note-0128-01a" xlink:href="note-0128-01"/> ctive Plane in the Accidental Point ſought. </s> <s xml:id="echoid-s1737" xml:space="preserve">If <lb/>now the before ſuppoſed Plane be conceiv’d to <lb/>turn about the Line N D, the Eye, and the <lb/>Station Point in the ſaid Plain, will then meet <lb/>the perſpective Plane in the Points o and s; </s> <s xml:id="echoid-s1738" xml:space="preserve">for <lb/>the Lines D o and N s are equal to D O and N S, <lb/>and form right Angles with the Line s o joyning <lb/>their Extremities. </s> <s xml:id="echoid-s1739" xml:space="preserve">Now the two Points s and o <lb/>anſwer to the Situation of the Eye and Station <pb file="0129" n="148"/> <pb file="0129a" n="149"/> <anchor type="figure" xlink:label="fig-0129a-01a" xlink:href="fig-0129a-01"/> <anchor type="figure" xlink:label="fig-0129a-02a" xlink:href="fig-0129a-02"/> <pb file="0130" n="150"/> <pb o="73" file="0131" n="151" rhead="on PERSPECTIVE."/> Point in reſpect to each other, in the before ſup-<lb/>poſed Plane. </s> <s xml:id="echoid-s1740" xml:space="preserve">Therefore the Line o F anſwers <lb/>@kewiſe to the Line in the ſaid Plane imagined <lb/>to be parallel to the propoſed Lines; </s> <s xml:id="echoid-s1741" xml:space="preserve">and con-<lb/>ſequently the Point F, is that wherein the <lb/>ſaid Parallel meets the Perſpective Plane; </s> <s xml:id="echoid-s1742" xml:space="preserve">and <lb/>therefore it is the accidental Point ſought.</s> <s xml:id="echoid-s1743" xml:space="preserve"/> </p> <div xml:id="echoid-div262" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0128-01" xlink:href="note-0128-01a" xml:space="preserve">13, 14.</note> <figure xlink:label="fig-0129a-01" xlink:href="fig-0129a-01a"> <caption xml:id="echoid-caption46" style="it" xml:space="preserve">page 72<lb/>Plate. 20<lb/>Fig. 46</caption> <variables xml:id="echoid-variables45" xml:space="preserve">V I X a E M P A T</variables> </figure> <figure xlink:label="fig-0129a-02" xlink:href="fig-0129a-02a"> <caption xml:id="echoid-caption47" style="it" xml:space="preserve">Fig. 47</caption> <variables xml:id="echoid-variables46" xml:space="preserve">V F X a Q G H D N A T R</variables> </figure> </div> <p> <s xml:id="echoid-s1744" xml:space="preserve">Note, Iſ the accidental Point T of Perpendi-<lb/>culars to the Geometrical Plane be found, the <lb/>Operation of this Problem may be ſhorten’d, in <lb/>drawing the Line T D, which will neceſſarily <lb/>paſs thro’ the Point N, and then the Point o will <lb/>be found by the Interſection of the Arc O o, <lb/>and a Semi-circle, whoſe Diameter is T D.</s> <s xml:id="echoid-s1745" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div264" type="section" level="1" n="138"> <head xml:id="echoid-head144" xml:space="preserve"><emph style="sc">Problem</emph> VI.</head> <p style="it"> <s xml:id="echoid-s1746" xml:space="preserve">89. </s> <s xml:id="echoid-s1747" xml:space="preserve">To find the Perſpective of one or more Lines <lb/>inclin’d to the Geometrical Plane.</s> <s xml:id="echoid-s1748" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1749" xml:space="preserve">Let A be the Foot of a Line inclin’d to the <lb/> <anchor type="note" xlink:label="note-0131-01a" xlink:href="note-0131-01"/> Geometrical Plane, and a its Repreſentation. <lb/></s> <s xml:id="echoid-s1750" xml:space="preserve">Now determine, by Means of the Triangle C P E <lb/>according to the Manner lay’d down <anchor type="note" xlink:href="" symbol="*"/> for the <anchor type="note" xlink:label="note-0131-02a" xlink:href="note-0131-02"/> Perſpective Plane when ſuppoſed perpendicular, <lb/>the Length A B of the Direction of the propoſed <lb/>Line. </s> <s xml:id="echoid-s1751" xml:space="preserve">This being done, find the Point <anchor type="note" xlink:href="" symbol="*"/> X the Per- <anchor type="note" xlink:label="note-0131-03a" xlink:href="note-0131-03"/> ſpective of a Point above the Geometrical Plane <lb/>by the Length of P E; </s> <s xml:id="echoid-s1752" xml:space="preserve">and then a X will be the <lb/>Perſpective ſought.</s> <s xml:id="echoid-s1753" xml:space="preserve"/> </p> <div xml:id="echoid-div264" type="float" level="2" n="1"> <note position="right" xlink:label="note-0131-01" xlink:href="note-0131-01a" xml:space="preserve">Fig. 48.</note> <note symbol="*" position="right" xlink:label="note-0131-02" xlink:href="note-0131-02a" xml:space="preserve">69.</note> <note symbol="*" position="right" xlink:label="note-0131-03" xlink:href="note-0131-03a" xml:space="preserve">82.</note> </div> </div> <div xml:id="echoid-div266" type="section" level="1" n="139"> <head xml:id="echoid-head145" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p style="it"> <s xml:id="echoid-s1754" xml:space="preserve">90. </s> <s xml:id="echoid-s1755" xml:space="preserve">To ſolve this Problem by the Accidental Points <lb/>of inclined Lines, and their Directions.</s> <s xml:id="echoid-s1756" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1757" xml:space="preserve">Let AB be the Direction of an inclin’d Line; </s> <s xml:id="echoid-s1758" xml:space="preserve">D <lb/> <anchor type="note" xlink:label="note-0131-04a" xlink:href="note-0131-04"/> the Accidental Point of the Directions, & </s> <s xml:id="echoid-s1759" xml:space="preserve">F that of <pb o="74" file="0132" n="152" rhead="An ESSAY"/> the Lines themſelves, and T the Accidental Point <lb/>oſ Perpendiculars to the Geometrical Plane.</s> <s xml:id="echoid-s1760" xml:space="preserve"/> </p> <div xml:id="echoid-div266" type="float" level="2" n="1"> <note position="right" xlink:label="note-0131-04" xlink:href="note-0131-04a" xml:space="preserve">Fig. 48.</note> </div> </div> <div xml:id="echoid-div268" type="section" level="1" n="140"> <head xml:id="echoid-head146" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1761" xml:space="preserve">Continue the Line A B until it meets the <lb/>Baſe Line in G, and draw the Line G D, which <lb/>cut in a and b by Lines drawn from A and B to <lb/>the Eye. </s> <s xml:id="echoid-s1762" xml:space="preserve">Then draw the Lines a F and T b In-<lb/>terſecting each other in X; </s> <s xml:id="echoid-s1763" xml:space="preserve">and a x will be the <lb/>Appearance ſought.</s> <s xml:id="echoid-s1764" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div269" type="section" level="1" n="141"> <head xml:id="echoid-head147" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1765" xml:space="preserve">Becauſe a b is <anchor type="note" xlink:href="" symbol="*"/> the Repreſentation of A B, <anchor type="note" xlink:label="note-0132-01a" xlink:href="note-0132-01"/> the Appearance of the inclin’d Line is one Part <lb/>of a F. </s> <s xml:id="echoid-s1766" xml:space="preserve">But the Extremity of the inclin’d Line <lb/>is in a Perpendicular to the Geometrical Plane <lb/>raiſed at the Point B; </s> <s xml:id="echoid-s1767" xml:space="preserve">therefore the Repreſen-<lb/>tation of the ſaid Extremity is in T b, and con-<lb/>ſequently x in the Interſection of this Line and <lb/>a F.</s> <s xml:id="echoid-s1768" xml:space="preserve"/> </p> <div xml:id="echoid-div269" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0132-01" xlink:href="note-0132-01a" xml:space="preserve">43.</note> </div> </div> <div xml:id="echoid-div271" type="section" level="1" n="142"> <head xml:id="echoid-head148" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p> <s xml:id="echoid-s1769" xml:space="preserve">91. </s> <s xml:id="echoid-s1770" xml:space="preserve">Draw F H thro’ the Accidental Point of <lb/>the inclin’d Lines, parallel to the Baſe Line, <lb/>and equal to o F in Fig. </s> <s xml:id="echoid-s1771" xml:space="preserve">48. </s> <s xml:id="echoid-s1772" xml:space="preserve">then a is the Perpen-<lb/> <anchor type="note" xlink:label="note-0132-02a" xlink:href="note-0132-02"/> dicular of the Foot of the inclin’d Line whoſe <lb/>Perſpective a x may be found as directed n. </s> <s xml:id="echoid-s1773" xml:space="preserve">70.</s> <s xml:id="echoid-s1774" xml:space="preserve"/> </p> <div xml:id="echoid-div271" type="float" level="2" n="1"> <note position="left" xlink:label="note-0132-02" xlink:href="note-0132-02a" xml:space="preserve">Fig. 49.</note> </div> <pb file="0133" n="153"/> <pb file="0133a" n="154"/> <figure> <caption xml:id="echoid-caption48" style="it" xml:space="preserve">Page 34.<lb/>Plate. 21<lb/>Fig. 48</caption> <variables xml:id="echoid-variables47" xml:space="preserve">F O D X S b a G N A E T B P C</variables> </figure> <figure> <caption xml:id="echoid-caption49" style="it" xml:space="preserve">Fig. 49</caption> <variables xml:id="echoid-variables48" xml:space="preserve">H F O D G X a M N L R Q</variables> </figure> <pb file="0134" n="155"/> <pb o="75" file="0135" n="156" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div273" type="section" level="1" n="143"> <head xml:id="echoid-head149" xml:space="preserve">CHAP. VI.</head> <p style="it"> <s xml:id="echoid-s1775" xml:space="preserve">Of throwing Figures into Perſpective, when <lb/>the Perſpective Plane is conſider’d as being <lb/>parallel to the Geometrical Plane.</s> <s xml:id="echoid-s1776" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div274" type="section" level="1" n="144"> <head xml:id="echoid-head150" xml:space="preserve"><emph style="sc">Prob</emph>. I.</head> <p style="it"> <s xml:id="echoid-s1777" xml:space="preserve">92. </s> <s xml:id="echoid-s1778" xml:space="preserve">To find the Perſpective of a Figure, which <lb/>is in the Geometrical Plane.</s> <s xml:id="echoid-s1779" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1780" xml:space="preserve">When the perſpective Plane is parallel to the <lb/>Horizon, we commonly conſider it, as being <lb/>it ſelf the Geometrical Plane, in which Caſe, <lb/>the Problem is fully reſolved: </s> <s xml:id="echoid-s1781" xml:space="preserve">But when it <lb/>happens, that another Geometrical Plane is <lb/>ſuppoſed, either above or below the perſpective <lb/>Plane, upon which the Figures upon the Geo-<lb/>metrical Plane are requir’d to be drawn; </s> <s xml:id="echoid-s1782" xml:space="preserve">we <lb/>muſt draw geometrically thereon, Figures ſimi-<lb/>lar to thoſe in the Geometrical Plane; </s> <s xml:id="echoid-s1783" xml:space="preserve">ſo that <lb/>the Lines on the perſpective Plane, be to their <lb/>correſpondent ones on the Geometrical Plane, <lb/>as the Eye’s Diſtance from the perſpective Plane, <lb/>is to its Diſtance from the Geometrical Plane.</s> <s xml:id="echoid-s1784" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1785" xml:space="preserve">The Demonſtration of this is evident from <lb/>n. </s> <s xml:id="echoid-s1786" xml:space="preserve">8, and 9.</s> <s xml:id="echoid-s1787" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div275" type="section" level="1" n="145"> <head xml:id="echoid-head151" xml:space="preserve"><emph style="sc">Prob</emph>. II.</head> <p style="it"> <s xml:id="echoid-s1788" xml:space="preserve">93. </s> <s xml:id="echoid-s1789" xml:space="preserve">To find the Perſpective of a Line, perpen-<lb/>dicular to the Geometrical Plane.</s> <s xml:id="echoid-s1790" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1791" xml:space="preserve">Draw the Line O S, in which aſſume O R e-<lb/> <anchor type="note" xlink:label="note-0135-01a" xlink:href="note-0135-01"/> qual to the Eye’s Diſtance from the perſpective <pb o="76" file="0136" n="157" rhead="An ESSAY"/> Plane; </s> <s xml:id="echoid-s1792" xml:space="preserve">and at the Points R and S, raiſe the in-<lb/>definite perpendiculars R G and S M; </s> <s xml:id="echoid-s1793" xml:space="preserve">and aſſume <lb/>the Point M at Pleaſure on S M; </s> <s xml:id="echoid-s1794" xml:space="preserve">from which <lb/>raiſe the Perpendicular M N, equal to the given <lb/>Line, and draw the Lines M O and N O, cutting <lb/> <anchor type="note" xlink:label="note-0136-01a" xlink:href="note-0136-01"/> the Line R G in the Points E and G. </s> <s xml:id="echoid-s1795" xml:space="preserve">Then <lb/>having drawn a Line at Pleaſure in the per-<lb/>ſpective Plane through the Point T, which is <lb/>that wherein a Perpendicular falling from the <lb/>Eye on the perſpective Plane meets it, aſſume <lb/>T H in the ſaid Line, equal to R E, and T I e-<lb/>qual to R G; </s> <s xml:id="echoid-s1796" xml:space="preserve">draw the Lines Ta, Ha, through <lb/>the Point a, the Perſpective of the Foot of the <lb/>given Perpendicular, and through the Point I, <lb/>the Line I X, parallel to Ha, and cutting Ta <lb/>in X, then a X will be the Appearance ſought.</s> <s xml:id="echoid-s1797" xml:space="preserve"/> </p> <div xml:id="echoid-div275" type="float" level="2" n="1"> <note position="right" xlink:label="note-0135-01" xlink:href="note-0135-01a" xml:space="preserve">Fig. 50.</note> <note position="left" xlink:label="note-0136-01" xlink:href="note-0136-01a" xml:space="preserve">Fig. 41.</note> </div> </div> <div xml:id="echoid-div277" type="section" level="1" n="146"> <head xml:id="echoid-head152" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1798" xml:space="preserve">It is manifeſt <anchor type="note" xlink:href="" symbol="*"/>, that the Point T, is the acci- <anchor type="note" xlink:label="note-0136-02a" xlink:href="note-0136-02"/> dental Point of Lines perpendicular to the Geo-<lb/>metrical Plane; </s> <s xml:id="echoid-s1799" xml:space="preserve">and conſequently the Per-<lb/>ſpective ſought is a Part of T a. </s> <s xml:id="echoid-s1800" xml:space="preserve">Moreover, it <lb/>is manifeſt <anchor type="note" xlink:href="" symbol="*"/>, that if the Feet and Extremities <anchor type="note" xlink:label="note-0136-03a" xlink:href="note-0136-03"/> of two equal right Lines, perpendicular to the <lb/>Geometrical Plane be joyn’d by Lines, theſe <lb/>Lines of Junction will have parallel Repreſen-<lb/>tations; </s> <s xml:id="echoid-s1801" xml:space="preserve">becauſe they are parallel to each other, <lb/>as likewiſe to the perſpective Plane. </s> <s xml:id="echoid-s1802" xml:space="preserve">And con-<lb/>ſequently, ſince H I, by Conſtruction, is the <lb/>Perſpective of a Line perpendicular to the <lb/>Geometrical Plane, and equal to the given Line, <lb/>and H a paſſes through the Appearances of the <lb/>Foot of the ſaid Perpendicular, and the given <lb/>Perpendicular; </s> <s xml:id="echoid-s1803" xml:space="preserve">I ſay, that X I, which is paral-<lb/>lel to Ha, and paſſes through the Extremity of <lb/>the Appeaarance H I, likewiſe paſſes through <lb/>the Extremity of the given Line; </s> <s xml:id="echoid-s1804" xml:space="preserve">and there- <pb file="0137" n="158"/> <pb file="0137a" n="159"/> <anchor type="figure" xlink:label="fig-0137a-01a" xlink:href="fig-0137a-01"/> <anchor type="figure" xlink:label="fig-0137a-02a" xlink:href="fig-0137a-02"/> <anchor type="figure" xlink:label="fig-0137a-03a" xlink:href="fig-0137a-03"/> <anchor type="figure" xlink:label="fig-0137a-04a" xlink:href="fig-0137a-04"/> <pb file="0138" n="160"/> <pb o="77" file="0139" n="161" rhead="on PERSPECTIVE."/> fore the Point X is the Repreſentation of the <lb/>ſaid Extremity.</s> <s xml:id="echoid-s1805" xml:space="preserve"/> </p> <div xml:id="echoid-div277" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0136-02" xlink:href="note-0136-02a" xml:space="preserve">13, 14.</note> <note symbol="*" position="left" xlink:label="note-0136-03" xlink:href="note-0136-03a" xml:space="preserve">4.</note> <figure xlink:label="fig-0137a-01" xlink:href="fig-0137a-01a"> <caption xml:id="echoid-caption50" style="it" xml:space="preserve">Page 36<lb/>Plate 22<lb/>Fig. 50</caption> <variables xml:id="echoid-variables49" xml:space="preserve">O R E G N S M</variables> </figure> <figure xlink:label="fig-0137a-02" xlink:href="fig-0137a-02a"> <caption xml:id="echoid-caption51" style="it" xml:space="preserve">Fig. 51</caption> <variables xml:id="echoid-variables50" xml:space="preserve">I H T a X</variables> </figure> <figure xlink:label="fig-0137a-03" xlink:href="fig-0137a-03a"> <caption xml:id="echoid-caption52" style="it" xml:space="preserve">Fig. 52</caption> <variables xml:id="echoid-variables51" xml:space="preserve">C D X I H G a F E L b T</variables> </figure> <figure xlink:label="fig-0137a-04" xlink:href="fig-0137a-04a"> <caption xml:id="echoid-caption53" style="it" xml:space="preserve">Fig. 53</caption> <variables xml:id="echoid-variables52" xml:space="preserve">H I F T x d X L B C</variables> </figure> </div> </div> <div xml:id="echoid-div279" type="section" level="1" n="147"> <head xml:id="echoid-head153" xml:space="preserve"><emph style="sc">Corollary</emph>.</head> <p> <s xml:id="echoid-s1806" xml:space="preserve">94. </s> <s xml:id="echoid-s1807" xml:space="preserve">It is manifeſt from hence, that when the <lb/>Perſpective of a Line, perpendicular to the <lb/>Geometrical Plane is once found, it is eaſy af-<lb/>terwards to find the Repreſentations of any Per-<lb/>pendiculars of the ſame Length as that.</s> <s xml:id="echoid-s1808" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div280" type="section" level="1" n="148"> <head xml:id="echoid-head154" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p style="it"> <s xml:id="echoid-s1809" xml:space="preserve">The Perſpective Plane being conſider’d as the Geo-<lb/>metrical Plane.</s> <s xml:id="echoid-s1810" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1811" xml:space="preserve">95. </s> <s xml:id="echoid-s1812" xml:space="preserve">Let T (as in Fig. </s> <s xml:id="echoid-s1813" xml:space="preserve">51.) </s> <s xml:id="echoid-s1814" xml:space="preserve">be the accidental <lb/> <anchor type="note" xlink:label="note-0139-01a" xlink:href="note-0139-01"/> Point of perpendicular Lines to the Geometri-<lb/>cal Plane; </s> <s xml:id="echoid-s1815" xml:space="preserve">H I the Arc of a Circle, whoſe Cen-<lb/>ter is T, and ſemidiameter the Eye’s Diſtance <lb/>from the perſpective Plane: </s> <s xml:id="echoid-s1816" xml:space="preserve">Alſo let a be the <lb/>Point where the Perpendicular, whoſe Appear-<lb/>ance is ſought, meets the perſpective Plane, and <lb/>B C the Length of this Perpendicular.</s> <s xml:id="echoid-s1817" xml:space="preserve"/> </p> <div xml:id="echoid-div280" type="float" level="2" n="1"> <note position="right" xlink:label="note-0139-01" xlink:href="note-0139-01a" xml:space="preserve">Fig. 53.</note> </div> </div> <div xml:id="echoid-div282" type="section" level="1" n="149"> <head xml:id="echoid-head155" xml:space="preserve"><emph style="sc">Operation</emph>,</head> <p> <s xml:id="echoid-s1818" xml:space="preserve">About the Point a, as a Center, and with the <lb/>Semidiameter B C, deſcribe the Circle L F, and <lb/>draw the Line I L, or H F, touching each of the <lb/>Circles H I, and F L; </s> <s xml:id="echoid-s1819" xml:space="preserve">and then a X or a x, is <lb/>the Appearance ſought, viz. </s> <s xml:id="echoid-s1820" xml:space="preserve">a X, when the Per-<lb/>pendicular is above that Surface of the per-<lb/>ſpective Plane next to the Eye, and a x, when <lb/>the Perpendicular is on the oppoſite Side.</s> <s xml:id="echoid-s1821" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div283" type="section" level="1" n="150"> <head xml:id="echoid-head156" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1822" xml:space="preserve">Draw the Radii a F, a L, T I, T H, to the <lb/>Points of Contact F, L, I, and H. </s> <s xml:id="echoid-s1823" xml:space="preserve">Then be- <pb o="78" file="0140" n="162" rhead="An ESSAY"/> cauſe the Triangles THX and a F X are ſimilar, <lb/>TH — a F: </s> <s xml:id="echoid-s1824" xml:space="preserve">a F : </s> <s xml:id="echoid-s1825" xml:space="preserve">: </s> <s xml:id="echoid-s1826" xml:space="preserve">Ta: </s> <s xml:id="echoid-s1827" xml:space="preserve">a X.</s> <s xml:id="echoid-s1828" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1829" xml:space="preserve">And becauſe the Triangles T I x and ax L, are <lb/>alſo ſimilar, we have <lb/>TI + a L : </s> <s xml:id="echoid-s1830" xml:space="preserve">: </s> <s xml:id="echoid-s1831" xml:space="preserve">a L: </s> <s xml:id="echoid-s1832" xml:space="preserve">Ta : </s> <s xml:id="echoid-s1833" xml:space="preserve">ax.</s> <s xml:id="echoid-s1834" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1835" xml:space="preserve">Now let PM NR be the perſpective Plane, <lb/> <anchor type="note" xlink:label="note-0140-01a" xlink:href="note-0140-01"/> O the Eye, A Q the Perpendicular, whoſe <lb/>Perſpective is requir’d, and O t a perpendicular <lb/>let fall from the Eye upon the perſpective Plane, <lb/>and ſo t will be the ſame, as the Point T in the <lb/>aforegoing Figure, Now if the Lines O Q be <lb/>drawn, it is manifeſt that A x, or A X, is the <lb/>Perſpective of A Q, according as this Line is <lb/>above or below the perſpective Plane in reſpect <lb/>to the Eye. </s> <s xml:id="echoid-s1836" xml:space="preserve">Then becauſe the Triangles O t x <lb/>and Q A x are ſimilar, we have <lb/>O t — A Q: </s> <s xml:id="echoid-s1837" xml:space="preserve">A Q :</s> <s xml:id="echoid-s1838" xml:space="preserve">: t A: </s> <s xml:id="echoid-s1839" xml:space="preserve">Ax.</s> <s xml:id="echoid-s1840" xml:space="preserve"/> </p> <div xml:id="echoid-div283" type="float" level="2" n="1"> <note position="left" xlink:label="note-0140-01" xlink:href="note-0140-01a" xml:space="preserve">Fig. 54.</note> </div> <p> <s xml:id="echoid-s1841" xml:space="preserve">And ſince the Triangles O t X and X A Q are <lb/>ſimilar, <lb/>O t + A Q: </s> <s xml:id="echoid-s1842" xml:space="preserve">A Q :</s> <s xml:id="echoid-s1843" xml:space="preserve">: t A: </s> <s xml:id="echoid-s1844" xml:space="preserve">A X.</s> <s xml:id="echoid-s1845" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1846" xml:space="preserve">Now Ot is equal to TH or TI of the afore-<lb/>going Figure, and AQ to a F or a L of the <lb/>ſame Figure; </s> <s xml:id="echoid-s1847" xml:space="preserve">as likewiſe At, Ta: </s> <s xml:id="echoid-s1848" xml:space="preserve">Therefore <lb/>if theſe two laſt Proportions be compared with <lb/>the two precedent ones, we ſhall find A x = a X, <lb/>and A X = a x; </s> <s xml:id="echoid-s1849" xml:space="preserve">which was to be demon-<lb/>ſtrated.</s> <s xml:id="echoid-s1850" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div285" type="section" level="1" n="151"> <head xml:id="echoid-head157" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s1851" xml:space="preserve">96. </s> <s xml:id="echoid-s1852" xml:space="preserve">When the two Circles interſect each other, <lb/>or fall within one another, and ſo this Way be-<lb/>comes uſeleſs; </s> <s xml:id="echoid-s1853" xml:space="preserve">a Line muſt be drawn at Pleaſure, <lb/>through the Point T, equal to the Diſtance of <lb/>the Eye from the perſpective Plane; </s> <s xml:id="echoid-s1854" xml:space="preserve">and then a <lb/>parallel equal to the given Perpendicular muſt be <lb/>drawn to the ſaid Line through the Point a, ei-<lb/>ther towards L or F, according as the Perpen- <pb o="79" file="0141" n="163" rhead="on PERSPECTIVE."/> dicular is on one Side or other of the perſpective <lb/>Plane with reſpect to the Eye. </s> <s xml:id="echoid-s1855" xml:space="preserve">And the Line <lb/>paſſing through the Extremities of the ſaid Pa-<lb/>rallels, will determine the Repreſentation ſought, <lb/>by its interſecting the Line T a, as is evident by <lb/>what is demonſtrated.</s> <s xml:id="echoid-s1856" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div286" type="section" level="1" n="152"> <head xml:id="echoid-head158" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p style="it"> <s xml:id="echoid-s1857" xml:space="preserve">97. </s> <s xml:id="echoid-s1858" xml:space="preserve">To find the Repreſentation of ſeveral Perpen-<lb/>diculars equal in Length to ſome one, whoſe Per-<lb/>ſpective is already drawn.</s> <s xml:id="echoid-s1859" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1860" xml:space="preserve">Let H I, be the Perſpective of a Perpendicular <lb/> <anchor type="note" xlink:label="note-0141-01a" xlink:href="note-0141-01"/> to the Geometrical or Perſpective Plane. </s> <s xml:id="echoid-s1861" xml:space="preserve">Now <lb/>about the accidental Point T, as a Center, and <lb/>with the Radius T H, deſcribe the circular Arc <lb/>H G, whoſe Chord let be equal to H I, and draw <lb/>the indefinite Line TGC, and let a and b, re-<lb/>preſent the Feet of the Perpendiculars, whoſe <lb/>Repreſentations are requir’d.</s> <s xml:id="echoid-s1862" xml:space="preserve"/> </p> <div xml:id="echoid-div286" type="float" level="2" n="1"> <note position="right" xlink:label="note-0141-01" xlink:href="note-0141-01a" xml:space="preserve">Fig. 52.</note> </div> </div> <div xml:id="echoid-div288" type="section" level="1" n="153"> <head xml:id="echoid-head159" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1863" xml:space="preserve">Deſcribe about the Center T, the Arcs b F E, <lb/>and a DC, paſſing through the Points a and b, and <lb/>draw the Lines T b and T a; </s> <s xml:id="echoid-s1864" xml:space="preserve">in each of which <lb/>aſſume b L equal to E F, and a X equal to C D; <lb/></s> <s xml:id="echoid-s1865" xml:space="preserve">and the ſought Repreſentations will be had.</s> <s xml:id="echoid-s1866" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div289" type="section" level="1" n="154"> <head xml:id="echoid-head160" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1867" xml:space="preserve">If H I, and a X repreſent Perpendiculars of <lb/>the ſame Length, it follows from the Demon-<lb/>ſtration of the precedent Method, that I H is to <lb/>H T, and a X to a T as the Difference of the <lb/>ſaid Perpendiculars, and Height of the Eye, is <pb o="80" file="0142" n="164" rhead="An ESSAY"/> to the Length of the ſaid Perpendiculars: </s> <s xml:id="echoid-s1868" xml:space="preserve">And <lb/>therefore</s> </p> </div> <div xml:id="echoid-div290" type="section" level="1" n="155"> <head xml:id="echoid-head161" xml:space="preserve">H I: T H:: a X: a T.</head> <p> <s xml:id="echoid-s1869" xml:space="preserve">But in the Conſtruction of this Problem, be-<lb/>cauſe the Triangles T C D and T H G, are <lb/>ſimilar;</s> <s xml:id="echoid-s1870" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1871" xml:space="preserve">H G = H I: </s> <s xml:id="echoid-s1872" xml:space="preserve">T H:</s> <s xml:id="echoid-s1873" xml:space="preserve">: C D = a X: </s> <s xml:id="echoid-s1874" xml:space="preserve">T D = a T; <lb/></s> <s xml:id="echoid-s1875" xml:space="preserve">and conſequently H I and a X, repreſent Perpen-<lb/>diculars of the ſame Length. </s> <s xml:id="echoid-s1876" xml:space="preserve">Which was to be <lb/>demonſtrated.</s> <s xml:id="echoid-s1877" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div291" type="section" level="1" n="156"> <head xml:id="echoid-head162" xml:space="preserve"><emph style="sc">Prob</emph>. III.</head> <p style="it"> <s xml:id="echoid-s1878" xml:space="preserve">98. </s> <s xml:id="echoid-s1879" xml:space="preserve">To find the accidental Point of any Number <lb/>of parallel Lines inclined to the Geometrical Plane.</s> <s xml:id="echoid-s1880" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1881" xml:space="preserve">Let a b be the Perſpective of the Direction <lb/> <anchor type="note" xlink:label="note-0142-01a" xlink:href="note-0142-01"/> of one of the given Lines.</s> <s xml:id="echoid-s1882" xml:space="preserve"/> </p> <div xml:id="echoid-div291" type="float" level="2" n="1"> <note position="left" xlink:label="note-0142-01" xlink:href="note-0142-01a" xml:space="preserve">Fig. 55.</note> </div> </div> <div xml:id="echoid-div293" type="section" level="1" n="157"> <head xml:id="echoid-head163" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1883" xml:space="preserve">Draw the Line F T L, parallel to a b, through <lb/>the accidental Point T of the Lines perpendicu-<lb/>lar to the Geometrical Plane; </s> <s xml:id="echoid-s1884" xml:space="preserve">and at the Point <lb/>T, raiſe the Perpendicular T G, which make e-<lb/>qual to the Diſtance of the Eye from the per-<lb/>ſpective Plane; </s> <s xml:id="echoid-s1885" xml:space="preserve">then draw the Line G L, or <lb/>G F, ſo that the Angle T L G, or T F G, be e-<lb/>qual to the Angle of the Inclination of the given <lb/>Lines; </s> <s xml:id="echoid-s1886" xml:space="preserve">and the Point L, will be the Accidental <lb/>Point ſought, if the given Lines incline to-<lb/>wards b; </s> <s xml:id="echoid-s1887" xml:space="preserve">but if they incline towards a, F will <lb/>be the Accidental Point.</s> <s xml:id="echoid-s1888" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div294" type="section" level="1" n="158"> <head xml:id="echoid-head164" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s1889" xml:space="preserve">It is manifeſt by Conſtruction, that if T G be <lb/>ſuppoſed to be raiſed perpendicularly to the Geo-<lb/>metrical Plane, G L or G F, will be parallel to <pb o="81" file="0143" n="165" rhead="on PERSPECTIVE."/> the given Lines; </s> <s xml:id="echoid-s1890" xml:space="preserve">and conſequently <anchor type="note" xlink:href="" symbol="*"/> L, or F, <anchor type="note" xlink:label="note-0143-01a" xlink:href="note-0143-01"/> will be the Accidental Point ſought.</s> <s xml:id="echoid-s1891" xml:space="preserve"/> </p> <div xml:id="echoid-div294" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0143-01" xlink:href="note-0143-01a" xml:space="preserve">13, 14.</note> </div> </div> <div xml:id="echoid-div296" type="section" level="1" n="159"> <head xml:id="echoid-head165" xml:space="preserve"><emph style="sc">Prob</emph>. IV.</head> <p style="it"> <s xml:id="echoid-s1892" xml:space="preserve">99. </s> <s xml:id="echoid-s1893" xml:space="preserve">To find the Repreſentation of one or more <lb/>Lines inclined to the Geometrical Plane.</s> <s xml:id="echoid-s1894" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1895" xml:space="preserve">Let a b be the Perſpective of the Direction of <lb/> <anchor type="note" xlink:label="note-0143-02a" xlink:href="note-0143-02"/> the given Line: </s> <s xml:id="echoid-s1896" xml:space="preserve">Now the Length of its Directi-<lb/>on may be determin’d, by means of the Tri-<lb/>angle ECP, according to the Directions of n. </s> <s xml:id="echoid-s1897" xml:space="preserve">69. <lb/></s> <s xml:id="echoid-s1898" xml:space="preserve">This being done, the Line b X muſt be drawn <lb/>through the Point b, equal to E P, and this re-<lb/>preſents a Perpendicular to the Geometrical <lb/>Plane; </s> <s xml:id="echoid-s1899" xml:space="preserve">then a X being drawn, will be the Re-<lb/>preſentation ſought.</s> <s xml:id="echoid-s1900" xml:space="preserve"/> </p> <div xml:id="echoid-div296" type="float" level="2" n="1"> <note position="right" xlink:label="note-0143-02" xlink:href="note-0143-02a" xml:space="preserve">Fig. 56.</note> </div> </div> <div xml:id="echoid-div298" type="section" level="1" n="160"> <head xml:id="echoid-head166" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p style="it"> <s xml:id="echoid-s1901" xml:space="preserve">100. </s> <s xml:id="echoid-s1902" xml:space="preserve">To ſolve the ſame Problem by means of the <lb/>accidental Points F and T, the one being that of the <lb/>Lines propoſed; </s> <s xml:id="echoid-s1903" xml:space="preserve">and the other, that of the Perpen-<lb/> <anchor type="note" xlink:label="note-0143-03a" xlink:href="note-0143-03"/> diculars to the Geometrical Plane.</s> <s xml:id="echoid-s1904" xml:space="preserve"/> </p> <div xml:id="echoid-div298" type="float" level="2" n="1"> <note position="right" xlink:label="note-0143-03" xlink:href="note-0143-03a" xml:space="preserve">Fig. 56.</note> </div> </div> <div xml:id="echoid-div300" type="section" level="1" n="161"> <head xml:id="echoid-head167" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1905" xml:space="preserve">Draw a Line from the Point F through the <lb/>Point a, which interſect in the Point X, by ano-<lb/>ther Line drawn from the Point T through b: <lb/></s> <s xml:id="echoid-s1906" xml:space="preserve">and then a X will be the Perſpective ſought.</s> <s xml:id="echoid-s1907" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div301" type="section" level="1" n="162"> <head xml:id="echoid-head168" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p> <s xml:id="echoid-s1908" xml:space="preserve">101. </s> <s xml:id="echoid-s1909" xml:space="preserve">The ſame Things being given as in the <lb/> <anchor type="note" xlink:label="note-0143-04a" xlink:href="note-0143-04"/> aforegoing Method, draw a I through the Point <lb/>a, equal to E P, repreſenting a Line perpendicu-<lb/>lar to the Geometrical Plane. </s> <s xml:id="echoid-s1910" xml:space="preserve">Then draw a pa-<lb/>rallel to F T through the Point I, whoſe Inter- <pb o="82" file="0144" n="166" rhead="An ESSAY"/> ſection with F a, will determine a X the Repre-<lb/>ſentation ſought.</s> <s xml:id="echoid-s1911" xml:space="preserve"/> </p> <div xml:id="echoid-div301" type="float" level="2" n="1"> <note position="right" xlink:label="note-0143-04" xlink:href="note-0143-04a" xml:space="preserve">Fig. 101.</note> </div> </div> <div xml:id="echoid-div303" type="section" level="1" n="163"> <head xml:id="echoid-head169" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s1912" xml:space="preserve">Although we ſuppoſe the Eye in all the <lb/>Problems in this Chapter to be above the per-<lb/>ſpective Plane, yet it may likewiſe be under <lb/>the ſaid Plane; </s> <s xml:id="echoid-s1913" xml:space="preserve">in which Cafe, the Geometrical <lb/>Plane is ſuppoſed above the Objects, as we have <lb/>already done <anchor type="note" xlink:href="" symbol="*"/> on another Occaſion.</s> <s xml:id="echoid-s1914" xml:space="preserve"/> </p> <note symbol="*" position="left" xml:space="preserve">79.</note> </div> <div xml:id="echoid-div304" type="section" level="1" n="164"> <head xml:id="echoid-head170" xml:space="preserve">CHAP. VII.</head> <head xml:id="echoid-head171" style="it" xml:space="preserve">Of Shadows.</head> <p> <s xml:id="echoid-s1915" xml:space="preserve">FIRST we muſt obſerve, with thoſe who <lb/>have already treated on this Subject, that <lb/>when a luminous Body is equal to an opaque <lb/>Body it enlightens, the Shadow of the ſaid Body <lb/>is contain’d between parallel Lines, and con-<lb/>ſequently, it is equal upon all parallel Lines <lb/>placed at any Diſtance whatſoever beyond the <lb/>opaque Body. </s> <s xml:id="echoid-s1916" xml:space="preserve">And when the luminous Body <lb/>is leſſer than the opaque Body, the Shadow <lb/>thereof, increaſes, and is infinitely augmented. <lb/></s> <s xml:id="echoid-s1917" xml:space="preserve">And on the contrary, when an opaque Body <lb/>is leſs than the luminous Body, the Shadow there-<lb/>of decreaſes and terminates in a Point.</s> <s xml:id="echoid-s1918" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1919" xml:space="preserve">Now becauſe the Sun is vaſtly greater than <lb/>any of the Bodies on the Earth’s Surface it en-<lb/>lightens, and is at ſo great a Diſtance therefrom, <lb/>therefore its Rays may be conſider’d as being <lb/>parallel; </s> <s xml:id="echoid-s1920" xml:space="preserve">and conſequently, the Bodies it ſhines <lb/>upon as encloſed between parallels: </s> <s xml:id="echoid-s1921" xml:space="preserve">And this <lb/>is the firſt Kind of Shadows I ſhall here explain;</s> <s xml:id="echoid-s1922" xml:space="preserve"> <pb o="83" file="0145" n="167" rhead="on PERSPECTIVE."/> after which, I ſhall touch upon thoſe continually <lb/>increaſing. </s> <s xml:id="echoid-s1923" xml:space="preserve">What I ſhall ſay on this Matter, is <lb/>ſufficient for deſigning the Shadows of right-lin’d <lb/>Bodies; </s> <s xml:id="echoid-s1924" xml:space="preserve">as to the Shadows of other Bodies, it <lb/>is ſo difficult to determine them Geometrically <lb/>that it is much better to examine thoſe which are <lb/>daily obſerved, and ſo imitate them.</s> <s xml:id="echoid-s1925" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1926" xml:space="preserve">I ſhall not ſay any thing concerning Shadows <lb/>terminating in a Point, becauſe their Variety <lb/>is ſo great, that they cannot be geometrically de-<lb/>termin’d. </s> <s xml:id="echoid-s1927" xml:space="preserve">Beſides, Painters ſcarcely ever ſup-<lb/>poſe their perſpective Planes or Pictures en-<lb/>lightned after this third Manner, unleſs only <lb/>when they have a Mind to repreſent a Chamber, <lb/>wherein the Light enters through the Windows; <lb/></s> <s xml:id="echoid-s1928" xml:space="preserve">but then the Number of Windows, their Situati-<lb/>on, and the different Reflections that the Light <lb/>ſuffers in the Chamber, produce ſo many dif-<lb/>ferent Alterations, that a Painter had better <lb/>imitate them, than have recourſe to Rules that <lb/>do not take in all Caſes. </s> <s xml:id="echoid-s1929" xml:space="preserve">I ſhall likewiſe be ſi-<lb/>lent as to the clair-obſcure, for a ſmall Atten-<lb/>tion to daily Experience will better illuſtrate <lb/>this Matter than a long Diſcourſe thereon. </s> <s xml:id="echoid-s1930" xml:space="preserve">Be-<lb/>ſides, it is impoſſible to furniſh general Rules <lb/>on this Subject; </s> <s xml:id="echoid-s1931" xml:space="preserve">and likewiſe the vaſt Number <lb/>of Figures, will not permit us to ſeparately <lb/>examine them; </s> <s xml:id="echoid-s1932" xml:space="preserve">add to all this, that a Painter <lb/>to draw the clair-obſcure, he ought to have, not <lb/>only regard to the Figures of Objects, but likwiſe <lb/>to their Colour and Matter.</s> <s xml:id="echoid-s1933" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div305" type="section" level="1" n="165"> <head xml:id="echoid-head172" style="it" xml:space="preserve">Of Solar Shadows.</head> <head xml:id="echoid-head173" xml:space="preserve"><emph style="sc">Problem</emph> I.</head> <p style="it"> <s xml:id="echoid-s1934" xml:space="preserve">103. </s> <s xml:id="echoid-s1935" xml:space="preserve">To find the Perſpective of the Shadow of a <lb/>Point above the Geometrical Plane, whoſe Height and <lb/>Seat is given.</s> <s xml:id="echoid-s1936" xml:space="preserve"/> </p> <pb o="84" file="0146" n="168" rhead="An ESSAY"/> <p> <s xml:id="echoid-s1937" xml:space="preserve">Let Z be the Geometrical Plane; </s> <s xml:id="echoid-s1938" xml:space="preserve">A the Seat <lb/> <anchor type="note" xlink:label="note-0146-01a" xlink:href="note-0146-01"/> of the given Point; </s> <s xml:id="echoid-s1939" xml:space="preserve">and A B the Direction of <lb/>the Sun’s Rays.</s> <s xml:id="echoid-s1940" xml:space="preserve"/> </p> <div xml:id="echoid-div305" type="float" level="2" n="1"> <note position="left" xlink:label="note-0146-01" xlink:href="note-0146-01a" xml:space="preserve">Fig. 57.</note> </div> </div> <div xml:id="echoid-div307" type="section" level="1" n="166"> <head xml:id="echoid-head174" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s1941" xml:space="preserve">Draw two right Lines, making a right Angle <lb/>with each other; </s> <s xml:id="echoid-s1942" xml:space="preserve">in one of which, aſſume P E, <lb/>equal to the Height of the given Point, above <lb/>the Geometrical Plane: </s> <s xml:id="echoid-s1943" xml:space="preserve">Then draw the Line <lb/>E C thro’ the Point E; </s> <s xml:id="echoid-s1944" xml:space="preserve">making an Angle with <lb/>C P, equal to the Sun’s Altitude; </s> <s xml:id="echoid-s1945" xml:space="preserve">and make A B <lb/>equal to C P. </s> <s xml:id="echoid-s1946" xml:space="preserve">Find the Appearance of the Point <lb/>B; </s> <s xml:id="echoid-s1947" xml:space="preserve">and the Repreſentation ſought, will be <lb/>had.</s> <s xml:id="echoid-s1948" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1949" xml:space="preserve">Note, This Operation, as likewiſe all the <lb/>others in this Chapter, regard all the Situations <lb/>of the perſpective Plane; </s> <s xml:id="echoid-s1950" xml:space="preserve">and is ſo evident, that <lb/>there is no need of demonſtrating it.</s> <s xml:id="echoid-s1951" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div308" type="section" level="1" n="167"> <head xml:id="echoid-head175" xml:space="preserve"><emph style="sc">Proe</emph>. II.</head> <p style="it"> <s xml:id="echoid-s1952" xml:space="preserve">104. </s> <s xml:id="echoid-s1953" xml:space="preserve">To find the Repreſentation of an elevated <lb/>Point, whoſe Appearance, as alſo that of its Seat, is <lb/>given, without uſing the Geometrical Plane.</s> <s xml:id="echoid-s1954" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1955" xml:space="preserve">Find <anchor type="note" xlink:href="" symbol="*"/> F, the accidental Point of the Sun’s <anchor type="note" xlink:label="note-0146-02a" xlink:href="note-0146-02"/> Rays, and D, that of their Direction: </s> <s xml:id="echoid-s1956" xml:space="preserve">Then <lb/> <anchor type="note" xlink:label="note-0146-03a" xlink:href="note-0146-03"/> draw a Line from the Point D, through a, the <lb/>Perſpective of the Seat of the given Point; </s> <s xml:id="echoid-s1957" xml:space="preserve">and <lb/>another from F, through I, the Perſpective of <lb/>the given Point: </s> <s xml:id="echoid-s1958" xml:space="preserve">And then b, the Interſection <lb/>of the ſaid two Lines, will be the Point ſought; <lb/></s> <s xml:id="echoid-s1959" xml:space="preserve">as is manifeſt.</s> <s xml:id="echoid-s1960" xml:space="preserve"/> </p> <div xml:id="echoid-div308" type="float" level="2" n="1"> <note position="left" xlink:label="note-0146-02" xlink:href="note-0146-02a" xml:space="preserve">Fig. 58.</note> <note symbol="*" position="left" xlink:label="note-0146-03" xlink:href="note-0146-03a" xml:space="preserve">68, 88, <lb/>98.</note> </div> </div> <div xml:id="echoid-div310" type="section" level="1" n="168"> <head xml:id="echoid-head176" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s1961" xml:space="preserve"><gap/>order to find the accidental Point of any <lb/><gap/>ber of inclin’d Lines, we have ſuppos’d <pb file="0147" n="169"/> <pb file="0147a" n="170"/> <anchor type="figure" xlink:label="fig-0147a-01a" xlink:href="fig-0147a-01"/> <anchor type="figure" xlink:label="fig-0147a-02a" xlink:href="fig-0147a-02"/> <anchor type="figure" xlink:label="fig-0147a-03a" xlink:href="fig-0147a-03"/> <pb file="0148" n="171"/> <pb o="85" file="0149" n="172" rhead="on PERSPECTIVE."/> <anchor type="note" xlink:href="" symbol="*"/> one of the Directions drawn upon the Geome- <anchor type="note" xlink:label="note-0149-01a" xlink:href="note-0149-01"/> trical Plane; </s> <s xml:id="echoid-s1962" xml:space="preserve">but it is ſufficient here, that the <lb/>Angle the ſaid Directions make with the Baſe <lb/>Line, be only known: </s> <s xml:id="echoid-s1963" xml:space="preserve">And ſo, as the Problem <lb/>expreſſes it, the Geometrical Plane may be en-<lb/>tirely laid aſide.</s> <s xml:id="echoid-s1964" xml:space="preserve"/> </p> <div xml:id="echoid-div310" type="float" level="2" n="1"> <figure xlink:label="fig-0147a-01" xlink:href="fig-0147a-01a"> <caption xml:id="echoid-caption54" style="it" xml:space="preserve">page 64.<lb/>Plate 23.<lb/>Fig. 54</caption> <variables xml:id="echoid-variables53" xml:space="preserve">O M P Q t A X x Q R N</variables> </figure> <figure xlink:label="fig-0147a-02" xlink:href="fig-0147a-02a"> <caption xml:id="echoid-caption55" style="it" xml:space="preserve">Fig. 55</caption> <variables xml:id="echoid-variables54" xml:space="preserve">G F b T L a</variables> </figure> <figure xlink:label="fig-0147a-03" xlink:href="fig-0147a-03a"> <caption xml:id="echoid-caption56" style="it" xml:space="preserve">Fig. 56</caption> <variables xml:id="echoid-variables55" xml:space="preserve">I F a X b E T C P</variables> </figure> <note symbol="*" position="right" xlink:label="note-0149-01" xlink:href="note-0149-01a" xml:space="preserve">68, 88.</note> </div> <p> <s xml:id="echoid-s1965" xml:space="preserve">105. </s> <s xml:id="echoid-s1966" xml:space="preserve">When the Perſpective Plane is parallel, <lb/>the Sun’s Rays will have no accidental Point; <lb/></s> <s xml:id="echoid-s1967" xml:space="preserve">for their Repreſentations are then parallel; </s> <s xml:id="echoid-s1968" xml:space="preserve">in <lb/>which Caſe, one of the Parallels muſt be drawn <lb/>through the Point a, inſtead of the Line D a. </s> <s xml:id="echoid-s1969" xml:space="preserve"><lb/>Moreover, when the Perſpective Plane is perpen-<lb/>dicular, or inclin’d, and the Sun’s Rays are pa-<lb/>rallel thereto; </s> <s xml:id="echoid-s1970" xml:space="preserve">a Line muſt be drawn through the <lb/>Point a, parallel to the Baſe Line; </s> <s xml:id="echoid-s1971" xml:space="preserve">as likewiſe <lb/>another Line through the Point I, parallel to the <lb/>Sun’s Rays; </s> <s xml:id="echoid-s1972" xml:space="preserve">cutting the firſt Line in the Point <lb/>ſought.</s> <s xml:id="echoid-s1973" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div312" type="section" level="1" n="169"> <head xml:id="echoid-head177" xml:space="preserve"><emph style="sc">Problem</emph> III.</head> <p style="it"> <s xml:id="echoid-s1974" xml:space="preserve">106. </s> <s xml:id="echoid-s1975" xml:space="preserve">To find the Perſpective of the Shadow of an <lb/>elevated Point, when there is ſome Body hindring its <lb/>falling upon the Geometrical Plane.</s> <s xml:id="echoid-s1976" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1977" xml:space="preserve">The Perſpective of the Section of the Body <lb/>made by a Plane paſſing through the given Point <lb/>perpendicular to the Geometrical Plane, and pa-<lb/>rallel to the Sun’s Rays, muſt be found: </s> <s xml:id="echoid-s1978" xml:space="preserve">And then <lb/>the Interſection of the ſaid Perſpective, and a <lb/>Line drawn from the Appearance of the given <lb/>Point to the Repreſentation of its Shadow, is the <lb/>Repreſentation ſought.</s> <s xml:id="echoid-s1979" xml:space="preserve"/> </p> <pb o="86" file="0150" n="173" rhead="An ESSAY"/> </div> <div xml:id="echoid-div313" type="section" level="1" n="170"> <head xml:id="echoid-head178" style="it" xml:space="preserve">Of the Shadows of a ſmall Light.</head> <head xml:id="echoid-head179" xml:space="preserve"><emph style="sc">Prob</emph>. IV.</head> <p style="it"> <s xml:id="echoid-s1980" xml:space="preserve">107. </s> <s xml:id="echoid-s1981" xml:space="preserve">To find the Perſpective of the Shadow of a <lb/>Point, whoſe Seat, and Height, above the Geometri-<lb/>cal Plane is known.</s> <s xml:id="echoid-s1982" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1983" xml:space="preserve">Let Z be the Geometrical Plane, A the Seat <lb/> <anchor type="note" xlink:label="note-0150-01a" xlink:href="note-0150-01"/> of the given Point; </s> <s xml:id="echoid-s1984" xml:space="preserve">and C that of the Light: <lb/></s> <s xml:id="echoid-s1985" xml:space="preserve">Draw the indefinite Line C A B; </s> <s xml:id="echoid-s1986" xml:space="preserve">and about C, <lb/>as a Center, with a Semidiameter equal to the <lb/>Height of the Light above the Geometrical <lb/>Plane, deſcribe the Arc F: </s> <s xml:id="echoid-s1987" xml:space="preserve">Again, about the <lb/>Point A, with a Semidiameter equal to the <lb/>Height of the given Point, deſcribe the Arc E. </s> <s xml:id="echoid-s1988" xml:space="preserve"><lb/>This being done, draw the Line E F touching <lb/>the ſaid two Arcs, and cutting the Line C A in <lb/>B. </s> <s xml:id="echoid-s1989" xml:space="preserve">Then if the Perſpective of B be found, the <lb/>Perſpective of the Shadow of the Point will be <lb/>had.</s> <s xml:id="echoid-s1990" xml:space="preserve"/> </p> <div xml:id="echoid-div313" type="float" level="2" n="1"> <note position="left" xlink:label="note-0150-01" xlink:href="note-0150-01a" xml:space="preserve">Fig. 59.</note> </div> </div> <div xml:id="echoid-div315" type="section" level="1" n="171"> <head xml:id="echoid-head180" xml:space="preserve"><emph style="sc">Problem</emph> V.</head> <p style="it"> <s xml:id="echoid-s1991" xml:space="preserve">108. </s> <s xml:id="echoid-s1992" xml:space="preserve">To find the Perſpective of the Shadow of an <lb/>elevated Point, the Repreſentation of which, as alſo <lb/>of its Seat being known, without uſing the Geometri-<lb/>cal Plane.</s> <s xml:id="echoid-s1993" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s1994" xml:space="preserve">The Operation of Prob. </s> <s xml:id="echoid-s1995" xml:space="preserve">3. </s> <s xml:id="echoid-s1996" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/> for Solar Sha- <anchor type="note" xlink:label="note-0150-02a" xlink:href="note-0150-02"/> dows muſt be uſed here, with this Difference; <lb/></s> <s xml:id="echoid-s1997" xml:space="preserve">that inſtead of the accidental Point of the Sun’s <lb/>Rays, the Perſpective of the ſmall Sight muſt <lb/>be uſed; </s> <s xml:id="echoid-s1998" xml:space="preserve">and in the Room of the accidental <lb/>Points of the Directions of the ſaid Rays, the <lb/>Perſpective of the Seat of the Light muſt be <lb/>aſſumed.</s> <s xml:id="echoid-s1999" xml:space="preserve"/> </p> <div xml:id="echoid-div315" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0150-02" xlink:href="note-0150-02a" xml:space="preserve">104.</note> </div> <pb file="0151" n="174"/> <pb file="0151a" n="175"/> <figure> <caption xml:id="echoid-caption57" style="it" xml:space="preserve">page 66.<lb/>Plate. 24.<lb/>Fig. 57</caption> <variables xml:id="echoid-variables56" xml:space="preserve">E A Z C P B</variables> </figure> <figure> <caption xml:id="echoid-caption58" style="it" xml:space="preserve">Fig. 58</caption> <variables xml:id="echoid-variables57" xml:space="preserve">F O D I a b</variables> </figure> <figure> <caption xml:id="echoid-caption59" style="it" xml:space="preserve">Fig. 59</caption> <variables xml:id="echoid-variables58" xml:space="preserve">F E Z C A B</variables> </figure> <pb file="0152" n="176"/> <pb o="87" file="0153" n="177" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div317" type="section" level="1" n="172"> <head xml:id="echoid-head181" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s2000" xml:space="preserve">What hath been obſerved <anchor type="note" xlink:href="" symbol="*"/> on the Solar Sha- <anchor type="note" xlink:label="note-0153-01a" xlink:href="note-0153-01"/> dows, hath no Regard to thoſe mentioned here: <lb/></s> <s xml:id="echoid-s2001" xml:space="preserve">For it matters not in this Problem, whether the <lb/>perſpective Plane be perpendicular, inclined, or <lb/>parallel; </s> <s xml:id="echoid-s2002" xml:space="preserve">becauſe in theſe different Situations, the <lb/>two Points that are uſed may always be found.</s> <s xml:id="echoid-s2003" xml:space="preserve"/> </p> <div xml:id="echoid-div317" type="float" level="2" n="1"> <note position="right" xlink:label="note-0153-01" xlink:href="note-0153-01a" xml:space="preserve">105.</note> </div> <p> <s xml:id="echoid-s2004" xml:space="preserve">Moreover, it muſt be obſerved, that the third <lb/>Problem <anchor type="note" xlink:href="" symbol="*"/> takes in the Shadows of a ſmall Sight, <anchor type="note" xlink:label="note-0153-02a" xlink:href="note-0153-02"/> as well as thoſe of the Sun; </s> <s xml:id="echoid-s2005" xml:space="preserve">but not without this <lb/>Difference, that the Plane which in the third <lb/>Problem is ſuppoſed parallel to the Sun’s Rays, <lb/>ought here to be ſuppoſed to paſs through the <lb/>Light whoſe Shadows are ſought.</s> <s xml:id="echoid-s2006" xml:space="preserve"/> </p> <div xml:id="echoid-div318" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0153-02" xlink:href="note-0153-02a" xml:space="preserve">106.</note> </div> </div> <div xml:id="echoid-div320" type="section" level="1" n="173"> <head xml:id="echoid-head182" xml:space="preserve">CHAP. VIII.</head> <head xml:id="echoid-head183" style="it" xml:space="preserve">Of mechanically ſhortning the Operations of <lb/>Perſpective.</head> <head xml:id="echoid-head184" xml:space="preserve">1. WHEN the perſpective Plane is ſup-<lb/>pos’d perpendicular or upright.</head> <head xml:id="echoid-head185" xml:space="preserve"><emph style="sc">Problem</emph> I.</head> <p style="it"> <s xml:id="echoid-s2007" xml:space="preserve">109. </s> <s xml:id="echoid-s2008" xml:space="preserve">To find the Repreſentations of Figures in <lb/>the Geometrical Plane.</s> <s xml:id="echoid-s2009" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2010" xml:space="preserve">Let O be the Eye, R H the Baſe Line; </s> <s xml:id="echoid-s2011" xml:space="preserve">F and <lb/> <anchor type="note" xlink:label="note-0153-03a" xlink:href="note-0153-03"/> G the Points denoted with the ſame Letters in <lb/>Fig. </s> <s xml:id="echoid-s2012" xml:space="preserve">10. </s> <s xml:id="echoid-s2013" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/></s> </p> <div xml:id="echoid-div320" type="float" level="2" n="1"> <note position="right" xlink:label="note-0153-03" xlink:href="note-0153-03a" xml:space="preserve">Fig. 60</note> </div> <note symbol="*" position="right" xml:space="preserve">31.</note> <p> <s xml:id="echoid-s2014" xml:space="preserve">Then take a Ruler and faſten to the Point G, <lb/>ſo that it may turn about the ſame in ſuch man.</s> <s xml:id="echoid-s2015" xml:space="preserve"> <pb o="88" file="0154" n="178" rhead="An ESSAY on"/> ner, that right Lines drawn along one of its Sides, <lb/>may paſs through the Point G. </s> <s xml:id="echoid-s2016" xml:space="preserve">This being done, <lb/>faſten one End of a Thread, put through the Eye <lb/>of a Needle B, in the Point F; </s> <s xml:id="echoid-s2017" xml:space="preserve">and then put the <lb/>ſaid Thread about a Pin faſten’d in the Point O; <lb/></s> <s xml:id="echoid-s2018" xml:space="preserve">ſo that when the Thread is uſing, it may be al-<lb/>ways kept tight by means of a Plummet fix’d to <lb/>its other End, and freely hanging under the Tube. </s> <s xml:id="echoid-s2019" xml:space="preserve"><lb/>Note, The aforeſaid Needle onght to be Braſs or <lb/>Silver, ſharp at both Ends, and having its Eye <lb/>pretty near one of them.</s> <s xml:id="echoid-s2020" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div322" type="section" level="1" n="174"> <head xml:id="echoid-head186" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s2021" xml:space="preserve">Let A be a Point of the Figure to be thrown <lb/>into Perſpective. </s> <s xml:id="echoid-s2022" xml:space="preserve">Now place that of the two <lb/>Points of the Needle, which is nigheſt to the <lb/>Eye thereof, upon the ſaid Point A, and move <lb/>the Ruler G E, until it cuts the Thread A F, in <lb/>the Point E, wherein the ſaid Thread cuts the <lb/>Baſe Line; </s> <s xml:id="echoid-s2023" xml:space="preserve">and then the Point a, wherein the <lb/>Ruler cuts the Thread A O, is the Point ſought, <lb/>and it may be marked with that Point of the <lb/>Needle fartheſt diſtant from its Eye, having firſt <lb/>preſſed the Ruler down upon the Paper and Thread, <lb/>that ſo the Plummet may not make the Thread <lb/>ſlip. </s> <s xml:id="echoid-s2024" xml:space="preserve">And in this manner may any Number of <lb/>Points be found. </s> <s xml:id="echoid-s2025" xml:space="preserve">This is demonſtrated in N. </s> <s xml:id="echoid-s2026" xml:space="preserve">32. <lb/></s> <s xml:id="echoid-s2027" xml:space="preserve">Sometimes it is more convenient to uſe the fol-<lb/>lowing Method.</s> <s xml:id="echoid-s2028" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div323" type="section" level="1" n="175"> <head xml:id="echoid-head187" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p> <s xml:id="echoid-s2029" xml:space="preserve">110. </s> <s xml:id="echoid-s2030" xml:space="preserve">Let O be the Eye, H E the Baſe Line, <lb/> <anchor type="note" xlink:label="note-0154-01a" xlink:href="note-0154-01"/> and F I the Geometrical Line. </s> <s xml:id="echoid-s2031" xml:space="preserve">Take a Ruler <lb/>M N, having two Threads equal in Length <lb/>faſten’d to it, and about O as a Center, and with <pb file="0155" n="179"/> <pb file="0155a" n="180"/> <anchor type="figure" xlink:label="fig-0155a-01a" xlink:href="fig-0155a-01"/> <pb file="0156" n="181"/> <pb o="89" file="0157" n="182" rhead="PERSPECTIVE."/> the Diſtance of the two Points whereat the <lb/>Thread is faſten’d on the Ruler, deſcribe an <lb/>Arc cutting the Geometrical Line in F; </s> <s xml:id="echoid-s2032" xml:space="preserve">then <lb/>faſten the Extremity of one of the aforeſaid <lb/>Threads in the ſaid Point F, and the Extremity <lb/>of the other in the Point O: </s> <s xml:id="echoid-s2033" xml:space="preserve">Take moreover <lb/>another Thread, put through the Eye of a <lb/>Needle, as in the aforegoing Method; </s> <s xml:id="echoid-s2034" xml:space="preserve">one End <lb/>of which, faſten in F, and afterwards put it a-<lb/>bout a Pin placed in O. </s> <s xml:id="echoid-s2035" xml:space="preserve">Then the only Dif-<lb/>ference between this Way and the precedent one, <lb/>is, that the Ruler M N muſt be uſed, by always <lb/>keeping the Threads M F, and N O tight, in-<lb/>ſtead of one turning about a Point.</s> <s xml:id="echoid-s2036" xml:space="preserve"/> </p> <div xml:id="echoid-div323" type="float" level="2" n="1"> <note position="left" xlink:label="note-0154-01" xlink:href="note-0154-01a" xml:space="preserve">Fig. 61.</note> <figure xlink:label="fig-0155a-01" xlink:href="fig-0155a-01a"> <caption xml:id="echoid-caption60" style="it" xml:space="preserve">page 88.<lb/>Plate. 25.<lb/>Fig. 60</caption> <variables xml:id="echoid-variables59" xml:space="preserve">O G F f Z L R P D I T S M a Q E R H N A C B</variables> </figure> </div> <p> <s xml:id="echoid-s2037" xml:space="preserve">For the Demonſtration of this, vide n. </s> <s xml:id="echoid-s2038" xml:space="preserve">39.</s> <s xml:id="echoid-s2039" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div325" type="section" level="1" n="176"> <head xml:id="echoid-head188" xml:space="preserve"><emph style="sc">Prob</emph>. II.</head> <p style="it"> <s xml:id="echoid-s2040" xml:space="preserve">111. </s> <s xml:id="echoid-s2041" xml:space="preserve">To find the Perſpective of one or more Lines <lb/>perpendicular to the Geometrical Plane.</s> <s xml:id="echoid-s2042" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2043" xml:space="preserve">Take two Rulers L C and N Z, having the <lb/> <anchor type="note" xlink:label="note-0157-01a" xlink:href="note-0157-01"/> four Ends of two Threads, or rather four Braſs <lb/>or Steel Wires of equal Length fixed on them, <lb/>at the Places L, I, N and M, ſo that L I be equal <lb/>to M N. </s> <s xml:id="echoid-s2044" xml:space="preserve">Then fix one oſ theſe Rulers upon the <lb/>Edge of the perſpective Plane, perpendicular to <lb/>the Baſe Line. </s> <s xml:id="echoid-s2045" xml:space="preserve">Now take a Thread, put thro’ <lb/>the Eye of a Needle, hang a Plummet at one <lb/>End, for the ſame Uſe as in Problem I <anchor type="note" xlink:href="" symbol="*"/>, and <anchor type="note" xlink:label="note-0157-02a" xlink:href="note-0157-02"/> faſten the other End to the Slider or Curſor D, <lb/>freely moveable on the Ruler L C; </s> <s xml:id="echoid-s2046" xml:space="preserve">put this <lb/>Thread about a Pin, ſet up againſt the Ruler C L <lb/>inc, ſo that C H be equal to the Height of the Eye.</s> <s xml:id="echoid-s2047" xml:space="preserve"/> </p> <div xml:id="echoid-div325" type="float" level="2" n="1"> <note position="right" xlink:label="note-0157-01" xlink:href="note-0157-01a" xml:space="preserve">Fig. 60.</note> <note symbol="*" position="right" xlink:label="note-0157-02" xlink:href="note-0157-02a" xml:space="preserve">109.</note> </div> </div> <div xml:id="echoid-div327" type="section" level="1" n="177"> <head xml:id="echoid-head189" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s2048" xml:space="preserve">Let T be the Repreſentation of the Foot of a <lb/>Perpendicular. </s> <s xml:id="echoid-s2049" xml:space="preserve">Move the Curſor D along C L, <pb o="90" file="0158" n="183" rhead="An ESSAY"/> until C D be equal to twice the Length of the <lb/>given Perpendicular. </s> <s xml:id="echoid-s2050" xml:space="preserve">This being done, move <lb/>the Needle along the Horizontal Line, ſuppoſe <lb/>to R, until the Point of the Thread R S paſſes <lb/>through the Point T; </s> <s xml:id="echoid-s2051" xml:space="preserve">then keeping the Thread <lb/>tight in this Manner, if the Ruler M N be <lb/>mov’d, till its Edge alſo paſſes through the <lb/>Point T, and P is the Point wherein the Edge of <lb/>the ſaid Ruler croſſes the Part of the Thread <lb/>R D, the Line T P will be the Repreſentation <lb/>ſought.</s> <s xml:id="echoid-s2052" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2053" xml:space="preserve">The Demonſtration of this is evident from <lb/>what is ſaid in n. </s> <s xml:id="echoid-s2054" xml:space="preserve">59.</s> <s xml:id="echoid-s2055" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div328" type="section" level="1" n="178"> <head xml:id="echoid-head190" xml:space="preserve"><emph style="sc">Method</emph> II.</head> <p style="it"> <s xml:id="echoid-s2056" xml:space="preserve">112. </s> <s xml:id="echoid-s2057" xml:space="preserve">When all the Perpendiculars have the ſame <lb/>Length.</s> <s xml:id="echoid-s2058" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2059" xml:space="preserve">Let F G be parallel to the Baſe Line, and F O <lb/> <anchor type="note" xlink:label="note-0158-01a" xlink:href="note-0158-01"/> equal to the Height of the Eye; </s> <s xml:id="echoid-s2060" xml:space="preserve">aſſume F f, e-<lb/>qual to the Length of either of the given Per-<lb/>pendiculars, and faſten the Thread fixed in F, in <lb/>the Point f. </s> <s xml:id="echoid-s2061" xml:space="preserve">Then raiſe R S perpendicular to <lb/>the Baſe Line, which make equal to F f, and <lb/>draw S Q parallel to the Baſe Line. </s> <s xml:id="echoid-s2062" xml:space="preserve">This being <lb/>done, tranſpoſe <anchor type="note" xlink:href="" symbol="*"/> the Figures in the Geometri- <anchor type="note" xlink:label="note-0158-02a" xlink:href="note-0158-02"/> cal Plane, in ſuch Manner, that the Point R co-<lb/>incides with S, and R H with S Q. </s> <s xml:id="echoid-s2063" xml:space="preserve">Then if S Q <lb/>be taken for a Baſe Line, and the Appearances <lb/> <anchor type="note" xlink:href="" symbol="*"/> of the Feet of the Perpendiculars be found, <anchor type="note" xlink:label="note-0158-03a" xlink:href="note-0158-03"/> the Repreſentations of their Extremities will be <lb/>had.</s> <s xml:id="echoid-s2064" xml:space="preserve"/> </p> <div xml:id="echoid-div328" type="float" level="2" n="1"> <note position="left" xlink:label="note-0158-01" xlink:href="note-0158-01a" xml:space="preserve">Fig. 60.</note> <note symbol="*" position="left" xlink:label="note-0158-02" xlink:href="note-0158-02a" xml:space="preserve">60.</note> <note symbol="*" position="left" xlink:label="note-0158-03" xlink:href="note-0158-03a" xml:space="preserve">109.</note> </div> </div> <div xml:id="echoid-div330" type="section" level="1" n="179"> <head xml:id="echoid-head191" xml:space="preserve"><emph style="sc">Method</emph> III.</head> <p style="it"> <s xml:id="echoid-s2065" xml:space="preserve">113. </s> <s xml:id="echoid-s2066" xml:space="preserve">For Perpendiculars of the ſame Length.</s> <s xml:id="echoid-s2067" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2068" xml:space="preserve">The Figures in the Geometrical Plane being <lb/> <anchor type="note" xlink:label="note-0158-04a" xlink:href="note-0158-04"/> tranſpoſed in the Manner aforeſaid <anchor type="note" xlink:href="" symbol="*"/>, aſſume T t, <anchor type="note" xlink:label="note-0158-05a" xlink:href="note-0158-05"/> <pb o="91" file="0159" n="184" rhead="on PERSPECTIVE."/> in the Perpendicular R S, continued equal to R S, <lb/>and draw f i parallel to the Baſe Line, in which <lb/>take the Point f in E I, the ſame as F in F I <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s2069" xml:space="preserve">then <anchor type="note" xlink:label="note-0159-01a" xlink:href="note-0159-01"/> if the Threads which before were faſten’d in F, <lb/>being placed in f, and by uſing them thus faſten’d, <lb/>as likewiſe S Q for a Baſe Line, the Repreſen-<lb/>tations of the Feet of the Perpendiculars be <lb/>found <anchor type="note" xlink:href="" symbol="*"/>, you will have the Repreſentations of <anchor type="note" xlink:label="note-0159-02a" xlink:href="note-0159-02"/> their Extremities.</s> <s xml:id="echoid-s2070" xml:space="preserve"/> </p> <div xml:id="echoid-div330" type="float" level="2" n="1"> <note position="left" xlink:label="note-0158-04" xlink:href="note-0158-04a" xml:space="preserve">Fig. 61.</note> <note symbol="*" position="left" xlink:label="note-0158-05" xlink:href="note-0158-05a" xml:space="preserve">112.</note> <note symbol="*" position="right" xlink:label="note-0159-01" xlink:href="note-0159-01a" xml:space="preserve">110.</note> <note symbol="*" position="right" xlink:label="note-0159-02" xlink:href="note-0159-02a" xml:space="preserve">110.</note> </div> </div> <div xml:id="echoid-div332" type="section" level="1" n="180"> <head xml:id="echoid-head192" style="it" xml:space="preserve">The Demonſtration of the two laſt Ways.</head> <p> <s xml:id="echoid-s2071" xml:space="preserve">114. </s> <s xml:id="echoid-s2072" xml:space="preserve">If a Plane be imagined to paſs through <lb/> <anchor type="note" xlink:label="note-0159-03a" xlink:href="note-0159-03"/> the Extremities of the equal Perpendiculars, it <lb/>will be parallel to the Geometrical Plane, and <lb/>will meet the perſpective Plane in S Q; </s> <s xml:id="echoid-s2073" xml:space="preserve">becauſe <lb/>R S is equal to the ſaid Perpendiculars: </s> <s xml:id="echoid-s2074" xml:space="preserve">Moreover, <lb/>the Extremities of theſe Perpendiculars form a <lb/>Figure in this ſuppoſed Plane, ſimilar to that <lb/>which their Feet form in the Geometrical Plane; <lb/></s> <s xml:id="echoid-s2075" xml:space="preserve">and the ſaid Figure hath the ſame Situation with <lb/>regard to the Line Q S, as that on the Geome-<lb/>trical Plane hath in reſpect of H R: </s> <s xml:id="echoid-s2076" xml:space="preserve">And conſe-<lb/>quently, if the Figure in the Geometrical Plane <lb/>be ſo raiſed up, that it hath the ſame Reſpect to <lb/>Q S, as it had to H R, and if the Appearances <lb/>of the Feet of the propoſed Perpendiculars be <lb/>found, the Repreſentations of their Extremties <lb/>will be had. </s> <s xml:id="echoid-s2077" xml:space="preserve">But the before ſuppoſed Tranſpo-<lb/>ſition of the Figure in the Geometrical Plane, <lb/>gives it the requiſite Situation with regard to <lb/>Q S, and the Repreſentation of the Figure <lb/>conſider’d in this new Geometrical Plane is <lb/>found <anchor type="note" xlink:href="" symbol="*"/>; </s> <s xml:id="echoid-s2078" xml:space="preserve">becauſe S Q, is taken for the Baſe Line, <anchor type="note" xlink:label="note-0159-04a" xlink:href="note-0159-04"/> O f (Fig. </s> <s xml:id="echoid-s2079" xml:space="preserve">60.) </s> <s xml:id="echoid-s2080" xml:space="preserve">equal to the Height of the Eye <lb/>above this Plane, and fi (Fig. </s> <s xml:id="echoid-s2081" xml:space="preserve">61.) </s> <s xml:id="echoid-s2082" xml:space="preserve">is the Geo-<lb/>metrical Line in the ſaid Plane.</s> <s xml:id="echoid-s2083" xml:space="preserve"/> </p> <div xml:id="echoid-div332" type="float" level="2" n="1"> <note position="right" xlink:label="note-0159-03" xlink:href="note-0159-03a" xml:space="preserve">Fig. 60, <lb/>and 61.</note> <note symbol="*" position="right" xlink:label="note-0159-04" xlink:href="note-0159-04a" xml:space="preserve">32, 39.</note> </div> <pb o="92" file="0160" n="185" rhead="An ESSAY"/> </div> <div xml:id="echoid-div334" type="section" level="1" n="181"> <head xml:id="echoid-head193" style="it" xml:space="preserve">II. When the Perſpective Plane is inclined.</head> <head xml:id="echoid-head194" xml:space="preserve"><emph style="sc">Prob</emph>. III.</head> <p style="it"> <s xml:id="echoid-s2084" xml:space="preserve">115. </s> <s xml:id="echoid-s2085" xml:space="preserve">To find the Repreſentation of Figures which <lb/>are in the Geometrical Plane.</s> <s xml:id="echoid-s2086" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2087" xml:space="preserve">The Operation laid down <anchor type="note" xlink:href="" symbol="*"/> for the perſpective <anchor type="note" xlink:label="note-0160-01a" xlink:href="note-0160-01"/> Plane being perpendicular, may be uſed here, <lb/>becauſe the inclined perſpective Plane may be <lb/>changed <anchor type="note" xlink:href="" symbol="*"/> into a perpendicular or upright one.</s> <s xml:id="echoid-s2088" xml:space="preserve"/> </p> <div xml:id="echoid-div334" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0160-01" xlink:href="note-0160-01a" xml:space="preserve">109, <lb/>110.</note> </div> <note symbol="*" position="left" xml:space="preserve">81.</note> </div> <div xml:id="echoid-div336" type="section" level="1" n="182"> <head xml:id="echoid-head195" xml:space="preserve"><emph style="sc">Prob</emph>. IV.</head> <p style="it"> <s xml:id="echoid-s2089" xml:space="preserve">116. </s> <s xml:id="echoid-s2090" xml:space="preserve">To find the Appearances of any Number of <lb/>Lines of the ſame Length, which are perpendicular <lb/>to the Geometrical Plane.</s> <s xml:id="echoid-s2091" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2092" xml:space="preserve">Raiſe a Perpendicular R C, in ſome Point on <lb/> <anchor type="note" xlink:label="note-0160-03a" xlink:href="note-0160-03"/> the Baſe Line, in which aſſume R L equal in <lb/>Length to the given Lines; </s> <s xml:id="echoid-s2093" xml:space="preserve">and draw the Line <lb/>L P, through the Point L, ſo that the Angle <lb/>L P R, be equal to the Angle of Inclination of <lb/>the perſpective Plane; </s> <s xml:id="echoid-s2094" xml:space="preserve">then having made R S <lb/>equal to P L, and S C equal to P R, draw the <lb/>Lines S Q and C D, parallel to the Baſe Line. <lb/></s> <s xml:id="echoid-s2095" xml:space="preserve">This being done, raiſe up the Figures of the <lb/>Geometrical Plane, ſo that the Point R coincides <lb/>with c, and the Line R H, with C D; </s> <s xml:id="echoid-s2096" xml:space="preserve">then <lb/>proceed as is directed <anchor type="note" xlink:href="" symbol="*"/> for the perpendicular <anchor type="note" xlink:label="note-0160-04a" xlink:href="note-0160-04"/> perſpective Plane, in uſing S Q for the Baſe <lb/>Line.</s> <s xml:id="echoid-s2097" xml:space="preserve"/> </p> <div xml:id="echoid-div336" type="float" level="2" n="1"> <note position="left" xlink:label="note-0160-03" xlink:href="note-0160-03a" xml:space="preserve">Fig. 62.</note> <note symbol="*" position="left" xlink:label="note-0160-04" xlink:href="note-0160-04a" xml:space="preserve">112, 113.</note> </div> <p> <s xml:id="echoid-s2098" xml:space="preserve">This is demonſtrated in n. </s> <s xml:id="echoid-s2099" xml:space="preserve">114.</s> <s xml:id="echoid-s2100" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div338" type="section" level="1" n="183"> <head xml:id="echoid-head196" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s2101" xml:space="preserve">The Point C muſt be aſſumed below the Point <lb/>S, when the perſpective Plane is inclined towards <pb file="0161" n="186"/> <pb file="0161a" n="187"/> <anchor type="figure" xlink:label="fig-0161a-01a" xlink:href="fig-0161a-01"/> <anchor type="figure" xlink:label="fig-0161a-02a" xlink:href="fig-0161a-02"/> <pb file="0162" n="188"/> <pb o="93" file="0163" n="189" rhead="on PERSPECTIVE."/> the Eye, and above it, when the ſaid Plane is in-<lb/>clined towards the Objects. </s> <s xml:id="echoid-s2102" xml:space="preserve">Obſerve likewiſe, <lb/>that F f of Fig. </s> <s xml:id="echoid-s2103" xml:space="preserve">60. </s> <s xml:id="echoid-s2104" xml:space="preserve">muſt here be aſſumed e-<lb/>qual to R S, and the Line Tt, Fig. </s> <s xml:id="echoid-s2105" xml:space="preserve">61. </s> <s xml:id="echoid-s2106" xml:space="preserve">muſt <lb/>be here a Part of the Line Z C continued, and <lb/>made equal to R S.</s> <s xml:id="echoid-s2107" xml:space="preserve"/> </p> <div xml:id="echoid-div338" type="float" level="2" n="1"> <figure xlink:label="fig-0161a-01" xlink:href="fig-0161a-01a"> <caption xml:id="echoid-caption61" style="it" xml:space="preserve">Plate 26<lb/>Fig. 61</caption> <variables xml:id="echoid-variables60" xml:space="preserve">O I F T N S Q S H E R M A</variables> </figure> <figure xlink:label="fig-0161a-02" xlink:href="fig-0161a-02a"> <caption xml:id="echoid-caption62" style="it" xml:space="preserve">Fig. 62</caption> <variables xml:id="echoid-variables61" xml:space="preserve">C D S Q L C D R P H</variables> </figure> </div> </div> <div xml:id="echoid-div340" type="section" level="1" n="184"> <head xml:id="echoid-head197" style="it" xml:space="preserve">III. When the Perſpective Plane is Parallel or <lb/>Horizontal.</head> <head xml:id="echoid-head198" xml:space="preserve"><emph style="sc">Prob</emph>. V.</head> <p style="it"> <s xml:id="echoid-s2108" xml:space="preserve">117. </s> <s xml:id="echoid-s2109" xml:space="preserve">To throw Figures which are in the Geometri-<lb/>cal Plane into Perſpective.</s> <s xml:id="echoid-s2110" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2111" xml:space="preserve">Draw the Line C F at Pleaſure, in which aſ-<lb/> <anchor type="note" xlink:label="note-0163-01a" xlink:href="note-0163-01"/> ſume the Point I, and make I H and I G, equal <lb/>to the Eye’s Diſtance from the perſpective <lb/>Plane: </s> <s xml:id="echoid-s2112" xml:space="preserve">Moreover, make I C, and I F equal to <lb/>the Eye’s Diſtance from the Geometrical Plane, <lb/>or at the leaſt, let I G and I H, be to I F and I C, <lb/>as the Eye’s Diſtance from the perſpective Plane, <lb/>is to its Diſtance from the Geometrical Plane. <lb/></s> <s xml:id="echoid-s2113" xml:space="preserve">This being done, raiſe two Perpendiculars to <lb/>the Line C F, in the Points G and H, and take <lb/>two Rulers, each of which has two equal Threads <lb/>ſo faſten’d to them, that the Diſtance P Q, and <lb/>N M be equal: </s> <s xml:id="echoid-s2114" xml:space="preserve">Then about F and C as Centers, <lb/>with the Semidiameters M N or P Q, deſcribe <lb/>two Arcs cutting the Perpendiculars raiſed at the <lb/>Points G and H, in the Points C and D, and fix <lb/>the Extremities of the two Threads of one Ruler, <lb/>in the Points C and D, and the Threads of the <lb/>other, in the Points F and E.</s> <s xml:id="echoid-s2115" xml:space="preserve"/> </p> <div xml:id="echoid-div340" type="float" level="2" n="1"> <note position="right" xlink:label="note-0163-01" xlink:href="note-0163-01a" xml:space="preserve">Fig. 63.</note> </div> </div> <div xml:id="echoid-div342" type="section" level="1" n="185"> <head xml:id="echoid-head199" xml:space="preserve"><emph style="sc">Operation</emph>.</head> <p> <s xml:id="echoid-s2116" xml:space="preserve">Let Z be the Geometrical Plane, and A a <lb/>Point of the given Figures. </s> <s xml:id="echoid-s2117" xml:space="preserve">Move the two <pb o="94" file="0164" n="190" rhead="An ESSAY"/> Rulers, keeping all the Threads tight, ſo that <lb/>the two Threads fix’d in the Points C and F, <lb/>croſs each other in the Point A; </s> <s xml:id="echoid-s2118" xml:space="preserve">and then the <lb/>Point a, wherein the two other Threads croſs <lb/>each other, is the Perſpective ſought. </s> <s xml:id="echoid-s2119" xml:space="preserve">And in <lb/>this Manner may the Repreſentations of any <lb/>Number of Points be ſound.</s> <s xml:id="echoid-s2120" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div343" type="section" level="1" n="186"> <head xml:id="echoid-head200" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s2121" xml:space="preserve">118. </s> <s xml:id="echoid-s2122" xml:space="preserve">The Triangle D a E, is ſimilar to the <lb/>Triangle C A F: </s> <s xml:id="echoid-s2123" xml:space="preserve">And becauſe all the Triangles <lb/>formed for finding the Repreſentations of diffe-<lb/>rent Points, have the ſame Baſes D E and C F, <lb/>which are between themſelves, as the Eye’s Di-<lb/>ſtance from the perſpective Plane, to its Diſtance <lb/>from the Geometrical Plane; </s> <s xml:id="echoid-s2124" xml:space="preserve">whence their Ver-<lb/>tices form ſimilar Figures, whoſe correſpondent <lb/>Lines are in the ſame Proportion; </s> <s xml:id="echoid-s2125" xml:space="preserve">and which <lb/>conſequently <anchor type="note" xlink:href="" symbol="*"/>, are the Appearances ſought.</s> <s xml:id="echoid-s2126" xml:space="preserve"/> </p> <note symbol="*" position="left" xml:space="preserve">8, 9.</note> </div> <div xml:id="echoid-div344" type="section" level="1" n="187"> <head xml:id="echoid-head201" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s2127" xml:space="preserve">It will be convenient to have the two Threads <lb/>P E, and M D of one Colour, and the two <lb/>Threads Q F, and C N of another.</s> <s xml:id="echoid-s2128" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div345" type="section" level="1" n="188"> <head xml:id="echoid-head202" xml:space="preserve"><emph style="sc">Prob</emph>. VI.</head> <p style="it"> <s xml:id="echoid-s2129" xml:space="preserve">119. </s> <s xml:id="echoid-s2130" xml:space="preserve">To find the Repreſentations of any Number <lb/>of Lines, equal and perpendicular to the Geometri-<lb/>cal Plane.</s> <s xml:id="echoid-s2131" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2132" xml:space="preserve">Let C D E F G I H, be the Points denoted <lb/> <anchor type="note" xlink:label="note-0164-02a" xlink:href="note-0164-02"/> with the ſame Letters in the precedent Figure, <lb/>as alſo let P Q and M N be the Rulers: </s> <s xml:id="echoid-s2133" xml:space="preserve">More-<lb/>over, let B be the Point, wherein a Perpendicu- <pb o="95" file="0165" n="191" rhead="on PERSPECTIVE."/> lar drawn from the Eye to the Geometrical <lb/>Plane, meets the ſaid Plane; </s> <s xml:id="echoid-s2134" xml:space="preserve">and T the Per-<lb/>ſpective of that Point, found by the aforegoing <lb/>Problem. </s> <s xml:id="echoid-s2135" xml:space="preserve">Now make F L and C R, equal to the <lb/>Length of the given Lines; </s> <s xml:id="echoid-s2136" xml:space="preserve">and about the <lb/>Points R and S, as Centers, with the Radius <lb/>M N or P Q, deſcribe two Arcs cutting the <lb/>Perpendiculars H D and G H, in the Points <lb/>X and S: </s> <s xml:id="echoid-s2137" xml:space="preserve">Then fix the Extremities of the <lb/>Threads, which before were faſten’d to the <lb/>Points F and E, to the Points L and S; </s> <s xml:id="echoid-s2138" xml:space="preserve">and alſo <lb/>the Extremities of thoſe two Threads, which <lb/>were before faſten’d to the Points C and D, in <lb/>the Points R and X: </s> <s xml:id="echoid-s2139" xml:space="preserve">Then moving the two <lb/>Rulers, until the Threads S P and X M, croſs <lb/>each other in the Point T, mark the Point O, <lb/>wherein the two other Threads croſs one another, <lb/>through which, and the Point B, draw the <lb/>indefinite Line B O V: </s> <s xml:id="echoid-s2140" xml:space="preserve">this being done, tranſ-<lb/>poſe the Lines of the Geometrical Plane, ſo that <lb/>the Point B, coincides with the Point O, and <lb/>the Line B O, with O V. </s> <s xml:id="echoid-s2141" xml:space="preserve">And by the precedent <lb/>Problem, find the Appearances of the Feet of <lb/>the Perpendiculars, and you will have the Re-<lb/>preſentations of their Extremities.</s> <s xml:id="echoid-s2142" xml:space="preserve"/> </p> <div xml:id="echoid-div345" type="float" level="2" n="1"> <note position="left" xlink:label="note-0164-02" xlink:href="note-0164-02a" xml:space="preserve">Fig. 64.</note> </div> </div> <div xml:id="echoid-div347" type="section" level="1" n="189"> <head xml:id="echoid-head203" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s2143" xml:space="preserve">If a Plane be ſuppoſed to paſs through the <lb/>Extremities of the Perpendiculars, it will be <lb/>parallel to the Geometrical Plane, and conſe-<lb/>quently, likewiſe to the perſpective Plane, be-<lb/>cauſe all the Perpendiculars are ſuppoſed equal. <lb/></s> <s xml:id="echoid-s2144" xml:space="preserve">But the Figure formed by the Extremities of <lb/>the Perpendiculars in the ſaid Plane, is ſimilar <lb/>and equal to that form’d by their Feet in the <lb/>Geometrical Plane: </s> <s xml:id="echoid-s2145" xml:space="preserve">and therefore, the Repre- <pb o="96" file="0166" n="192" rhead="An ESSAY"/> ſentation of the Figure which is in the ſecond <lb/>Plane, is likewiſe ſimilar to the Figure which is <lb/>in the Geometrical Plane, and the Lines com-<lb/>poſing this Appearance to the correſpondent <lb/>Lines in the ſecond Plane, as the Eye’s Diſtance <lb/>from the perſpective Plane, is to its Diſtance <lb/>from the before ſuppoſed Plane. </s> <s xml:id="echoid-s2146" xml:space="preserve">But by means <lb/>of the Threads faſtned in the Manner aforeſaid, <lb/>we find the Repreſentation of a Figure, whoſe <lb/>Lines are <anchor type="note" xlink:href="" symbol="*"/> in the aforenamed Proportion:</s> <s xml:id="echoid-s2147" xml:space="preserve"> <anchor type="note" xlink:label="note-0166-01a" xlink:href="note-0166-01"/> Whence this Figure is the Perſpective ſought, <lb/>and is ſituated as it ought to be, with regard to <lb/>the Repreſentations of the Figures in the Geo-<lb/>metrical Plane; </s> <s xml:id="echoid-s2148" xml:space="preserve">becauſe theſe Figures are ſo <lb/>ſlid, that the Appearance of the Perpen-<lb/>dicular in the Point B, is only a Point. </s> <s xml:id="echoid-s2149" xml:space="preserve">The <lb/>ſaid Appearances are alſo in their proper Poſiti-<lb/>ons, becauſe the Line B Q, is made to coincide <lb/>with O N.</s> <s xml:id="echoid-s2150" xml:space="preserve"/> </p> <div xml:id="echoid-div347" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0166-01" xlink:href="note-0166-01a" xml:space="preserve">118.</note> </div> <p style="it"> <s xml:id="echoid-s2151" xml:space="preserve">120. </s> <s xml:id="echoid-s2152" xml:space="preserve">For Solar Shadows in all Situations of the Per-<lb/>ſpective Plane.</s> <s xml:id="echoid-s2153" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div349" type="section" level="1" n="190"> <head xml:id="echoid-head204" xml:space="preserve"><emph style="sc">Prob</emph>. VII.</head> <p style="it"> <s xml:id="echoid-s2154" xml:space="preserve">6. </s> <s xml:id="echoid-s2155" xml:space="preserve">To find the Repreſentations of the Shadows of <lb/>any Number of Points, being at the ſame Height <lb/>above the Geometrical Plane.</s> <s xml:id="echoid-s2156" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2157" xml:space="preserve">Find <anchor type="note" xlink:href="" symbol="*"/> a Point in the Geometrical Plane, which <anchor type="note" xlink:label="note-0166-02a" xlink:href="note-0166-02"/> is the Shadow of one of the given Points: </s> <s xml:id="echoid-s2158" xml:space="preserve">Tranſ-<lb/>poſe the Figures in the Geometrical Plane, ſo that <lb/>the Seat of the ſaid given Point coincides with <lb/>its Shadow; </s> <s xml:id="echoid-s2159" xml:space="preserve">and the Line drawn through the <lb/>ſaid Seat, and the Shadow of the given Point, <lb/>coincides with its Prolongation. </s> <s xml:id="echoid-s2160" xml:space="preserve">Then if ac-<lb/>cording to the Situation of the perſpective Plane, <lb/>the Repreſentations of the Seats of the given <pb file="0167" n="193"/> <pb file="0167a" n="194"/> <anchor type="figure" xlink:label="fig-0167a-01a" xlink:href="fig-0167a-01"/> <anchor type="figure" xlink:label="fig-0167a-02a" xlink:href="fig-0167a-02"/> <pb file="0168" n="195"/> <pb file="0169" n="196"/> <pb file="0169a" n="197"/> <anchor type="figure" xlink:label="fig-0169a-01a" xlink:href="fig-0169a-01"/> <anchor type="figure" xlink:label="fig-0169a-02a" xlink:href="fig-0169a-02"/> <pb file="0170" n="198"/> <pb o="97" file="0171" n="199" rhead="on PERSPECTIVE."/> Points be found <anchor type="note" xlink:href="" symbol="*"/>, the Repreſentations of their <anchor type="note" xlink:label="note-0171-01a" xlink:href="note-0171-01"/> Shadows will be had.</s> <s xml:id="echoid-s2161" xml:space="preserve"/> </p> <div xml:id="echoid-div349" type="float" level="2" n="1"> <note symbol="*" position="left" xlink:label="note-0166-02" xlink:href="note-0166-02a" xml:space="preserve">103.</note> <figure xlink:label="fig-0167a-01" xlink:href="fig-0167a-01a"> <caption xml:id="echoid-caption63" style="it" xml:space="preserve">page 96.<lb/>Plate. 27<lb/>Fig. 63</caption> <variables xml:id="echoid-variables62" xml:space="preserve">D E C F M H I G P A Q N</variables> </figure> <figure xlink:label="fig-0167a-02" xlink:href="fig-0167a-02a"> <caption xml:id="echoid-caption64" style="it" xml:space="preserve">Fig. 64</caption> <variables xml:id="echoid-variables63" xml:space="preserve">X S D E T C R L F H I G P M B O V Q N</variables> </figure> <figure xlink:label="fig-0169a-01" xlink:href="fig-0169a-01a"> <caption xml:id="echoid-caption65" style="it" xml:space="preserve">page 98.<lb/>Plate. 28<lb/>Fig. 65</caption> <variables xml:id="echoid-variables64" xml:space="preserve">L M F G D H C E I A B</variables> </figure> <figure xlink:label="fig-0169a-02" xlink:href="fig-0169a-02a"> <caption xml:id="echoid-caption66" style="it" xml:space="preserve">Fig. 66</caption> <variables xml:id="echoid-variables65" xml:space="preserve">A B VII VIII IV V H C VI VI P V VII IV S VIII E O I III II I XII XIX IX F D</variables> </figure> <note symbol="*" position="right" xlink:label="note-0171-01" xlink:href="note-0171-01a" xml:space="preserve">109, <lb/>115, 117.</note> </div> </div> <div xml:id="echoid-div351" type="section" level="1" n="191"> <head xml:id="echoid-head205" xml:space="preserve">CHAP. IX.</head> <p style="it"> <s xml:id="echoid-s2162" xml:space="preserve">The Uſe of Perſpective in Dialling; </s> <s xml:id="echoid-s2163" xml:space="preserve">ſhewing <lb/>how to draw the Hour Lines upon any <lb/>Kind of Plane, by means of an Horizon-<lb/>tal Dial.</s> <s xml:id="echoid-s2164" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2165" xml:space="preserve">PERSPECTIVE is not only uſeſul in <lb/>drawing, but likewiſe in other Parts of Ma-<lb/>thematicks, and principally in Dialling: </s> <s xml:id="echoid-s2166" xml:space="preserve">For if <lb/>the Extremity of the Style be conceived as the <lb/>Eye, and the Sun’s Rays as viſual Rays, all <lb/>poſſible Kinds of Dials may be drawn for the <lb/>ſame Latitude, by Means of an Horizontal Dial, <lb/>as we are now going to ſhew,</s> </p> <p> <s xml:id="echoid-s2167" xml:space="preserve">121. </s> <s xml:id="echoid-s2168" xml:space="preserve">Let A B C D be an Horizontal Dial <lb/>made for any given Latitude; </s> <s xml:id="echoid-s2169" xml:space="preserve">E F its Style, and <lb/> <anchor type="note" xlink:label="note-0171-02a" xlink:href="note-0171-02"/> H I M L a Plane, upon which a Dial is to be <lb/>drawn. </s> <s xml:id="echoid-s2170" xml:space="preserve">Now if this Plane be ſo ſituated, that <lb/>the Extremity of its Style F G, coincides with <lb/>the Extremity of the Style of the Horizontal <lb/>Dial; </s> <s xml:id="echoid-s2171" xml:space="preserve">and if the Perſpective of one of the Hour <lb/>Lines of the Horizontal Dial A B C D, be found <lb/>upon the Plane H I M L, in conceiving the <lb/>Point F as the Eye, it is evident <anchor type="note" xlink:href="" symbol="*"/>, that the <anchor type="note" xlink:label="note-0171-03a" xlink:href="note-0171-03"/> Shadow of the Point F, will fall upon the ſaid <lb/>Perſpective, at the ſame Time that it falls <lb/>upon the Hour Line, whereof it is the Per-<lb/>ſpective; </s> <s xml:id="echoid-s2172" xml:space="preserve">and conſequently, the ſaid Shadow <lb/>will ſhew the ſame Hour upon the Plane H I M L, <lb/>that it ſhews upon the Horizontal Dial: </s> <s xml:id="echoid-s2173" xml:space="preserve">There- <pb o="98" file="0172" n="200" rhead="An ESSAY"/> fore the before-mention’d Perſpective will be an <lb/>Hour-Line of a Dial, drawn upon the Plane <lb/>H L M I, and whoſe Style is G F.</s> <s xml:id="echoid-s2174" xml:space="preserve"/> </p> <div xml:id="echoid-div351" type="float" level="2" n="1"> <note position="right" xlink:label="note-0171-02" xlink:href="note-0171-02a" xml:space="preserve">Fig. 65.</note> <note symbol="*" position="right" xlink:label="note-0171-03" xlink:href="note-0171-03a" xml:space="preserve">2.</note> </div> <p> <s xml:id="echoid-s2175" xml:space="preserve">The ſame may be demonſtrated of the Re-<lb/>preſentations of other Hour-Lines, which form <lb/>a Dial upon the Plane H L M I.</s> <s xml:id="echoid-s2176" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2177" xml:space="preserve">We now proceed to lay down the beſt way of <lb/>determining the ſaid Repreſentations.</s> <s xml:id="echoid-s2178" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div353" type="section" level="1" n="192"> <head xml:id="echoid-head206" xml:space="preserve"><emph style="sc">Prob</emph>. I.</head> <head xml:id="echoid-head207" style="it" xml:space="preserve">122. To draw Vertical Dials.</head> <p> <s xml:id="echoid-s2179" xml:space="preserve">Draw the Line E O, thro’ E, the Foot of the <lb/> <anchor type="note" xlink:label="note-0172-01a" xlink:href="note-0172-01"/> Style of the Horizontal Dial A B F D, equal <lb/>to the Length of the Style of the Dial to be <lb/>drawn, and making an Angle with the Meri-<lb/>dian C. </s> <s xml:id="echoid-s2180" xml:space="preserve">XII, equal to the Plane’s Declina-<lb/>tion.</s> <s xml:id="echoid-s2181" xml:space="preserve"/> </p> <div xml:id="echoid-div353" type="float" level="2" n="1"> <note position="left" xlink:label="note-0172-01" xlink:href="note-0172-01a" xml:space="preserve">Fig. 66.</note> </div> <p> <s xml:id="echoid-s2182" xml:space="preserve">This Angle muſt be aſſum’d towards the <lb/>Point D, when the Plane’s Declination is South-<lb/>Eaſt, as here; </s> <s xml:id="echoid-s2183" xml:space="preserve">towards F, when the Declination <lb/>is Weſtward; </s> <s xml:id="echoid-s2184" xml:space="preserve">towards A, when it is North-Eaſt-<lb/>wardly; </s> <s xml:id="echoid-s2185" xml:space="preserve">and towards B, when it is North Weſt-<lb/>wardly.</s> <s xml:id="echoid-s2186" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2187" xml:space="preserve">Now, thro’ the Extremity O of the ſaid Line, <lb/>draw the Line I H, perpendicular thereto; </s> <s xml:id="echoid-s2188" xml:space="preserve">and <lb/>the Line C P, thro’ the Center of the Dial, pa-<lb/>rallel, and equal to E O; </s> <s xml:id="echoid-s2189" xml:space="preserve">thro’ whoſe Extre-<lb/>mity, P, draw the Line P S, parallel to H I.</s> <s xml:id="echoid-s2190" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2191" xml:space="preserve">Then, to make the Dial, draw the Line h i, <lb/> <anchor type="note" xlink:label="note-0172-02a" xlink:href="note-0172-02"/> in which prick down the Diviſions of the <lb/>Line H I; </s> <s xml:id="echoid-s2192" xml:space="preserve">and in the Point o, (which is the <lb/>ſame as O) raiſe the Perpendicular o p, equal to <lb/>the Length of the Style of the Horizontal Dial <lb/>A B D F. </s> <s xml:id="echoid-s2193" xml:space="preserve">This being done; </s> <s xml:id="echoid-s2194" xml:space="preserve">draw a Parallel, <lb/>h i, thro’ the Extremity of this Perpendicular; <lb/></s> <s xml:id="echoid-s2195" xml:space="preserve">on which, prick down the Diviſions of the Line <pb o="99" file="0173" n="201" rhead="on PERSPECTIVE."/> P S, and making the Point p be the ſame as P. <lb/></s> <s xml:id="echoid-s2196" xml:space="preserve">then join each Diviſion of this Line, to the <lb/>correſpondent Diviſion of the Line h i; </s> <s xml:id="echoid-s2197" xml:space="preserve">and the <lb/>Dial ſought, will be drawn: </s> <s xml:id="echoid-s2198" xml:space="preserve">p being the Foot <lb/>of the Style, and p s the Horizontal Line.</s> <s xml:id="echoid-s2199" xml:space="preserve"/> </p> <div xml:id="echoid-div354" type="float" level="2" n="2"> <note position="left" xlink:label="note-0172-02" xlink:href="note-0172-02a" xml:space="preserve">Fig. 67.</note> </div> </div> <div xml:id="echoid-div356" type="section" level="1" n="193"> <head xml:id="echoid-head208" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <p> <s xml:id="echoid-s2200" xml:space="preserve">The Baſe Line is h i; </s> <s xml:id="echoid-s2201" xml:space="preserve">p s is the Horizontal <lb/> <anchor type="note" xlink:label="note-0173-01a" xlink:href="note-0173-01"/> Line; </s> <s xml:id="echoid-s2202" xml:space="preserve">p the Point of Sight; </s> <s xml:id="echoid-s2203" xml:space="preserve">and E O, or C P, <lb/>of Figure 66. </s> <s xml:id="echoid-s2204" xml:space="preserve">is the Length of the principal <lb/>Ray.</s> <s xml:id="echoid-s2205" xml:space="preserve"/> </p> <div xml:id="echoid-div356" type="float" level="2" n="1"> <note position="right" xlink:label="note-0173-01" xlink:href="note-0173-01a" xml:space="preserve">Fig. 66, <lb/>67.</note> </div> <p> <s xml:id="echoid-s2206" xml:space="preserve">Now, ſuppoſe the Plane p s h i, to be ſet per-<lb/>pendicularly upon the Horizontal Dial, in ſuch <lb/>manner, that the Line h i coincides with H I, <lb/>and the Point o with O. </s> <s xml:id="echoid-s2207" xml:space="preserve">Suppoſe, moreover, <lb/>that thro’ the Extremity of the Style, which we <lb/>conſider as the Eye, Lines are drawn in the Ho-<lb/>rizontal Plane, parallel to the Hour-Lines of <lb/>the Dial; </s> <s xml:id="echoid-s2208" xml:space="preserve">it is evident, that theſe Lines will <lb/>meet the Horizontal Line p s, in the Point al-<lb/>ready prick’d down; </s> <s xml:id="echoid-s2209" xml:space="preserve">and conſequently <anchor type="note" xlink:href="" symbol="*"/>, the <anchor type="note" xlink:label="note-0173-02a" xlink:href="note-0173-02"/> Appearances of the Hour-Lines, are the Lines <lb/>joining the Diviſions of the Lines h i and p s.</s> <s xml:id="echoid-s2210" xml:space="preserve"/> </p> <div xml:id="echoid-div357" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0173-02" xlink:href="note-0173-02a" xml:space="preserve">13.</note> </div> </div> <div xml:id="echoid-div359" type="section" level="1" n="194"> <head xml:id="echoid-head209" xml:space="preserve"><emph style="sc">Remark</emph>.</head> <p> <s xml:id="echoid-s2211" xml:space="preserve">If the Line H I happens to meet the Meri-<lb/>dian; </s> <s xml:id="echoid-s2212" xml:space="preserve">the common Method, by the Horizontal <lb/>Dial, is eaſier than this.</s> <s xml:id="echoid-s2213" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div360" type="section" level="1" n="195"> <head xml:id="echoid-head210" xml:space="preserve"><emph style="sc">Prob</emph>. II.</head> <head xml:id="echoid-head211" style="it" xml:space="preserve">123. To draw inclining Dials.</head> <p> <s xml:id="echoid-s2214" xml:space="preserve">Theſe Dials are drawn in the ſame manner as <lb/>Vertical ones; </s> <s xml:id="echoid-s2215" xml:space="preserve">the following Preparations being <lb/>firſt made.</s> <s xml:id="echoid-s2216" xml:space="preserve"/> </p> <pb o="100" file="0174" n="202" rhead="An ESSAY"/> <p> <s xml:id="echoid-s2217" xml:space="preserve">Draw the Line e c, equal to the Length of <lb/> <anchor type="note" xlink:label="note-0174-01a" xlink:href="note-0174-01"/> the Style of the Horizontal Dial; </s> <s xml:id="echoid-s2218" xml:space="preserve">and, at its <lb/>two Extremities raiſe the Perpendiculars e o <lb/>and c p: </s> <s xml:id="echoid-s2219" xml:space="preserve">Then draw the Line c G, thro’ the <lb/>Point c, equal in Length to the Style of the <lb/>Dial to be drawn; </s> <s xml:id="echoid-s2220" xml:space="preserve">making an Angle with c e, <lb/>equal to the Inclination of the Dial-Plane. </s> <s xml:id="echoid-s2221" xml:space="preserve">Af-<lb/>ter which, draw the Line o G p, thro’ the Ex-<lb/>tremity G of the ſaid Line, perpendicular there-<lb/>to. </s> <s xml:id="echoid-s2222" xml:space="preserve">This Preparation being finiſh’d, we uſe the <lb/>Operations of the precedent Problem, in making <lb/>E O and C P, in the Horizontal Dial, equal to <lb/>e o and c p of this Figure; </s> <s xml:id="echoid-s2223" xml:space="preserve">and o p, in the Dial <lb/>to be drawn, equal to op of this Figure; </s> <s xml:id="echoid-s2224" xml:space="preserve">in <lb/>which, the Point G is the Foot of the Style.</s> <s xml:id="echoid-s2225" xml:space="preserve"/> </p> <div xml:id="echoid-div360" type="float" level="2" n="1"> <note position="left" xlink:label="note-0174-01" xlink:href="note-0174-01a" xml:space="preserve">Fig. 68.</note> </div> <p> <s xml:id="echoid-s2226" xml:space="preserve">If it happens, in the aforeſaid Preparation, <lb/> <anchor type="note" xlink:label="note-0174-02a" xlink:href="note-0174-02"/> that the Line p o cuts the Line e c; </s> <s xml:id="echoid-s2227" xml:space="preserve">E O muſt <lb/>not then be aſſum’d in the ſame Line, in the <lb/>Horizontal Dial, as otherwiſe it muſt have <lb/>been; </s> <s xml:id="echoid-s2228" xml:space="preserve">but, in that Line, continued on the other <lb/>ſide of the Foot of the Style.</s> <s xml:id="echoid-s2229" xml:space="preserve"/> </p> <div xml:id="echoid-div361" type="float" level="2" n="2"> <note position="left" xlink:label="note-0174-02" xlink:href="note-0174-02a" xml:space="preserve">Fig. 69.</note> </div> <p> <s xml:id="echoid-s2230" xml:space="preserve">The Demonſtration of this Problem, is the <lb/>ſame as that of the precedent one; </s> <s xml:id="echoid-s2231" xml:space="preserve">in conſider-<lb/>ing the Angle p o Q, equal to the Angle G c e, <lb/>which is equal to the Inclination of the Plane.</s> <s xml:id="echoid-s2232" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2233" xml:space="preserve">We might have further ſhewn the Uſe of <lb/>Perſpective, in facilitating the Operations of <lb/>Dialling: </s> <s xml:id="echoid-s2234" xml:space="preserve">But this would be deviating from our <lb/>Subject; </s> <s xml:id="echoid-s2235" xml:space="preserve">and ſo we ſhall content our ſelves with <lb/>this Short Eſſay, touching the moſt common and <lb/>uſeful Problem in Dialling.</s> <s xml:id="echoid-s2236" xml:space="preserve"/> </p> <pb file="0175" n="203"/> <pb file="0175a" n="204"/> <figure> <caption xml:id="echoid-caption67" style="it" xml:space="preserve">page 100<lb/>Plate. 29<lb/>Fig. 67</caption> <variables xml:id="echoid-variables66" xml:space="preserve">5 6p 7 8 9 10 S V VI VII VIII IX X o XI ll l</variables> </figure> <figure> <caption xml:id="echoid-caption68" style="it" xml:space="preserve">Fig. 68</caption> <variables xml:id="echoid-variables67" xml:space="preserve">c P G e o Q</variables> </figure> <figure> <caption xml:id="echoid-caption69" style="it" xml:space="preserve">Fig. 69</caption> <variables xml:id="echoid-variables68" xml:space="preserve">P c G o e Q</variables> </figure> <pb file="0176" n="205"/> <pb o="101" file="0177" n="206" rhead="on PERSPECTIVE."/> <figure> <image file="0177-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0177-01"/> </figure> </div> <div xml:id="echoid-div363" type="section" level="1" n="196"> <head xml:id="echoid-head212" style="it" xml:space="preserve">The Uſe of the <emph style="sc">Camera</emph> <emph style="sc">Obscura</emph> <lb/>in Deſigning.</head> <head xml:id="echoid-head213" xml:space="preserve"><emph style="sc">Advertisement</emph>.</head> <p style="it"> <s xml:id="echoid-s2237" xml:space="preserve">EVery one knows how eaſy it is by one Convex Glaſs <lb/>only, to repreſent outward Objects in any darken’d <lb/>Place, according to their natural Appearances, where <lb/>the Livelineſs of Colours, and the Diverſity of Mo-<lb/>tions, are wonderfully pleaſant to behold; </s> <s xml:id="echoid-s2238" xml:space="preserve">and it is <lb/>ſo eaſy to make this Invention uſeful in Deſigning, <lb/>that our treating of it ſo fully as we have done, is <lb/>undoubtedly ſomething neceſſary.</s> <s xml:id="echoid-s2239" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2240" xml:space="preserve">A few Hints and Obſervations ſeem to be ſufficient <lb/>for putting a diligent Perſon in the way of contriving <lb/>himſelf ſome Machine to perform the Uſes hereafter <lb/>mentioned; </s> <s xml:id="echoid-s2241" xml:space="preserve">whence we might have let him bad the <lb/>Pleaſure of the Invention himſelf, after it was made <lb/>eaſy to him. </s> <s xml:id="echoid-s2242" xml:space="preserve">And this indeed is what I firſt reſolv’d <lb/>upon; </s> <s xml:id="echoid-s2243" xml:space="preserve">but afterwards conſidering that in the Con-<lb/>ſtruction of a Machine for facilitating the Buſineſs of <lb/>Deſigning, ſeveral Things cannot be foreknown till <lb/>try’d; </s> <s xml:id="echoid-s2244" xml:space="preserve">and that much Time may be ſpent in vain, and <lb/>ſeveral Methods attempted, before one of theſe Ma-<lb/>chines can be made ſimple and uſeful, as I have found <lb/>by Experierce: </s> <s xml:id="echoid-s2245" xml:space="preserve">Therefore, that others may not be at <lb/>this Trouble, I ſhall here lay down the Deſcription of <lb/>two Machines (hoping it will not be unacceptable) <lb/>which after ſeveral Alterations, in my Opinion, are <lb/>now made convenient enough.</s> <s xml:id="echoid-s2246" xml:space="preserve"/> </p> <pb o="102" file="0178" n="207" rhead="An ESSAY"/> <p style="it"> <s xml:id="echoid-s2247" xml:space="preserve">The firſt of theſe Machines is undoubtedly much <lb/>preferable to the other; </s> <s xml:id="echoid-s2248" xml:space="preserve">becauſe it is firmer, and <lb/>renders the Work eaſier and exacter; </s> <s xml:id="echoid-s2249" xml:space="preserve">and Prints may <lb/>be eaſier repreſented in it, than in the other. </s> <s xml:id="echoid-s2250" xml:space="preserve">Add to <lb/>all this, that with a ſmall Alteration it may be made <lb/>capable of a few Uſes, which are peculiar to <lb/>the ſecond Machine; </s> <s xml:id="echoid-s2251" xml:space="preserve">yet ſince this latter one is <lb/>much ſimpler, of a much leſs Price, and is eaſier to <lb/>be carry’d from Place to Place, I thought it conveni-<lb/>ent to lay down alſo its Deſcription in this ſmall <lb/>Tract.</s> <s xml:id="echoid-s2252" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2253" xml:space="preserve">I ſhall not here take up the Reader’s Time in enu-<lb/>merating the Advantages accruing to Painters from <lb/>Machines of this Nature: </s> <s xml:id="echoid-s2254" xml:space="preserve">But only add, that they <lb/>are of great Uſe for reducing or leſſening ſeveral ſe-<lb/>parate Objects in the ſame Picture; </s> <s xml:id="echoid-s2255" xml:space="preserve">which therefore <lb/>may be copy’d after Nature in the moſt perfect man-<lb/>ner poſſible. </s> <s xml:id="echoid-s2256" xml:space="preserve">It is very difficult to give ſeveral Objects <lb/>their true Bigneſs in a Picture, and to diſpoſe them <lb/>ſo as to have the ſame Point of Sight; </s> <s xml:id="echoid-s2257" xml:space="preserve">but this is <lb/>done extremely eaſy by means of Machines; </s> <s xml:id="echoid-s2258" xml:space="preserve">for the <lb/>Point of Sight in them will be the ſame always, as <lb/>long as the Convex Glaſs has the ſame Diſpoſition, <lb/>and the Bigneſs of the Repreſentations of Objects in <lb/>the Machines, do depend upon the Objects Diſtances <lb/>from them.</s> <s xml:id="echoid-s2259" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2260" xml:space="preserve">This Invention, by Induſtry, may be certainly im-<lb/>prov’d, and the following Obſervations will be not a <lb/>little conducing thereto. </s> <s xml:id="echoid-s2261" xml:space="preserve">1. </s> <s xml:id="echoid-s2262" xml:space="preserve">You muſt uſe but one <lb/>Convex Glaſs; </s> <s xml:id="echoid-s2263" xml:space="preserve">for when there are two or more, the <lb/>true Repreſentations of Objects will be loſt; </s> <s xml:id="echoid-s2264" xml:space="preserve">which is <lb/>an Inconveniency one is alſo ſubject to, when a Con-<lb/>cave Glaſs is any ways us’d in the Conſtruction of the <lb/>Machine. </s> <s xml:id="echoid-s2265" xml:space="preserve">2. </s> <s xml:id="echoid-s2266" xml:space="preserve">When more than two Mirrours or Look-<lb/>ing glaſſes are us’d, the Rays, after having ſuffer’d a <lb/>triple Reflection, are ſo weaken’d, that the Objects <lb/>will not be well repreſented: </s> <s xml:id="echoid-s2267" xml:space="preserve">And even when but two <lb/>Mirrours are us’d, they muſt be well poliſh’d 3. </s> <s xml:id="echoid-s2268" xml:space="preserve">The <pb o="103" file="0179" n="208" rhead="on PERSPECTIVE."/> Mirrours muſt not be plac’d within the Machine; </s> <s xml:id="echoid-s2269" xml:space="preserve">for <lb/>in ſuch a cloſe Place, ones Breath will ſully them; </s> <s xml:id="echoid-s2270" xml:space="preserve">but <lb/>not the Convex Glaſs, becauſe it is encloſed in a <lb/>Tube.</s> <s xml:id="echoid-s2271" xml:space="preserve"/> </p> <figure> <image file="0179-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0179-01"/> </figure> </div> <div xml:id="echoid-div364" type="section" level="1" n="197"> <head xml:id="echoid-head214" style="it" xml:space="preserve">The Uſe of the <emph style="sc">Camera</emph> <emph style="sc">Obscura</emph> <lb/>in Deſigning.</head> <head xml:id="echoid-head215" xml:space="preserve"><emph style="sc">Definition</emph>.</head> <p style="it"> <s xml:id="echoid-s2272" xml:space="preserve">1. </s> <s xml:id="echoid-s2273" xml:space="preserve">ACamera Obſcura is any dark Place, in which <lb/>outward Objects expoſed to Broad-Day-light, <lb/>are repreſented upon Paper, or any other white <lb/>Body.</s> <s xml:id="echoid-s2274" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2275" xml:space="preserve">The Way to repreſent Objects in the Camera <lb/>Obſcura, is to make a ſinall Hole in that ſide <lb/>thereof next to the Objects, and place a Convex-<lb/>Glaſs therein, then if a Sheet of Paper be ex-<lb/>tended in the Focus of the ſaid Glaſs, the Objects <lb/>will appear inverted upon the Paper.</s> <s xml:id="echoid-s2276" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div365" type="section" level="1" n="198"> <head xml:id="echoid-head216" xml:space="preserve"><emph style="sc">Theorem</emph> I.</head> <p style="it"> <s xml:id="echoid-s2277" xml:space="preserve">2. </s> <s xml:id="echoid-s2278" xml:space="preserve">The Camera Obſcura gives the true Repreſen-<lb/>tation of Objects.</s> <s xml:id="echoid-s2279" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2280" xml:space="preserve">The Figures repreſented in the Camera Obſcura <lb/>are form’d (as is demonſtrated in Dioptricks) <lb/>by Rays which coming from all the Points of the <lb/>Objects, paſs through the Centre of the Glaſs: <lb/></s> <s xml:id="echoid-s2281" xml:space="preserve">So that an Eye placed in the ſaid Centre, would <lb/>perceive the Objects by the ſaid Rays, which <lb/>conſequently by their Interſection with a Plane, <lb/>muſt give the true Repreſentation of the Objects. </s> <s xml:id="echoid-s2282" xml:space="preserve"><lb/>But the Pyramid which the ſaid Rays forms with- <pb o="104" file="0180" n="209" rhead="An ESSAY on"/> out the Camera Obſcura, is ſimilar to that which <lb/>they form, after having paſſed through the Glaſs: <lb/></s> <s xml:id="echoid-s2283" xml:space="preserve">Therefore the Rays which fall upon the Paper in <lb/>the Camera Obſcura, likewiſe give the true Re-<lb/>preſentation of the Objects thereon. </s> <s xml:id="echoid-s2284" xml:space="preserve">Which was <lb/>to be demouſtrated.</s> <s xml:id="echoid-s2285" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2286" xml:space="preserve">Theſe Objects appear inverted, becauſe the <lb/>Rays croſs each other in paſſing through the Glaſs; <lb/></s> <s xml:id="echoid-s2287" xml:space="preserve">thoſe coming from above going below, &</s> <s xml:id="echoid-s2288" xml:space="preserve">c.</s> <s xml:id="echoid-s2289" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div366" type="section" level="1" n="199"> <head xml:id="echoid-head217" xml:space="preserve"><emph style="sc">Theorem</emph> II.</head> <p style="it"> <s xml:id="echoid-s2290" xml:space="preserve">3. </s> <s xml:id="echoid-s2291" xml:space="preserve">The Reflection which the Rays of Light ſuffer <lb/>upon a plain Mirrour or Speculum, before they fall <lb/>upon a Convex Glaſs, no-wiſe deforms the Repreſen-<lb/>tation of Objects.</s> <s xml:id="echoid-s2292" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2293" xml:space="preserve">This is evident: </s> <s xml:id="echoid-s2294" xml:space="preserve">For the Speculum reflects the <lb/>Rays in the ſame Order as it receives them.</s> <s xml:id="echoid-s2295" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2296" xml:space="preserve">Now to ſhew the Uſe that may be drawn from <lb/>the Camera Obſcura in Deſigning, I ſhall here lay <lb/>down the Deſcription and Uſe of Two Machines, <lb/>which I uſe for this End.</s> <s xml:id="echoid-s2297" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div367" type="section" level="1" n="200"> <head xml:id="echoid-head218" style="it" xml:space="preserve">The Deſcription of the Firſt Machine.</head> <p> <s xml:id="echoid-s2298" xml:space="preserve">4. </s> <s xml:id="echoid-s2299" xml:space="preserve">This Machine is ſomething in Figure of a <lb/>Chair, (ſuch as People are carried in) the back Part <lb/>of the Top is rounded, and its Foreſide P Q ſwells <lb/>out in the middle: </s> <s xml:id="echoid-s2300" xml:space="preserve">Vide Fig. </s> <s xml:id="echoid-s2301" xml:space="preserve">70. </s> <s xml:id="echoid-s2302" xml:space="preserve">which repreſents <lb/> <anchor type="note" xlink:label="note-0180-01a" xlink:href="note-0180-01"/> the Machine, the Side whereof oppoſite to the <lb/>Door, is ſuppoſed to be raiſed up, that ſo its In-<lb/>ſide may be ſeen.</s> <s xml:id="echoid-s2303" xml:space="preserve"/> </p> <div xml:id="echoid-div367" type="float" level="2" n="1"> <note position="left" xlink:label="note-0180-01" xlink:href="note-0180-01a" xml:space="preserve">Fig. 70.</note> </div> <p> <s xml:id="echoid-s2304" xml:space="preserve">5. </s> <s xml:id="echoid-s2305" xml:space="preserve">The Board A within-ſide, ſerves as a Ta-<lb/>ble, and turns upon two Iron-pins, going into <lb/>the Wood of the Fore-ſide of the Machine, and is <lb/>ſuſtain’d by two ſmall Chains, that ſo the ſaid <lb/>Table may be lifted up; </s> <s xml:id="echoid-s2306" xml:space="preserve">and therefore one may <pb o="105" file="0181" n="210" rhead="PERSPECTIVE."/> more conveniently go in at the Door in the Side <lb/>of the Machine.</s> <s xml:id="echoid-s2307" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2308" xml:space="preserve">6. </s> <s xml:id="echoid-s2309" xml:space="preserve">There are two Tin-Tubes bent at each End, <lb/>(one of which is repreſented in Fig. </s> <s xml:id="echoid-s2310" xml:space="preserve">76. </s> <s xml:id="echoid-s2311" xml:space="preserve">becauſe <lb/>they could not be ſhewn in the Figure of the <lb/>Machine) placed in the Furniture, near the <lb/>Back-ſide of the Machine, each having one End <lb/>without the Machine. </s> <s xml:id="echoid-s2312" xml:space="preserve">Theſe Tubes ſerve to give <lb/>Air to Perſons ſhut up in the Machine, yet ſo, <lb/>that no Light may enter through them.</s> <s xml:id="echoid-s2313" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2314" xml:space="preserve">7. </s> <s xml:id="echoid-s2315" xml:space="preserve">At the Places c, c, c, c, on the Out-ſide of <lb/>the Back-part of the Machine, are four Iron-Sta-<lb/>ples, in which ſlide two Wooden Rulers D E, D E, <lb/>about 3 Inches broad, having two other thin <lb/>Rulers going through holes made at their Tops, at <lb/>the Places D, D, to which thin Rulers the Board <lb/>F is fixed; </s> <s xml:id="echoid-s2316" xml:space="preserve">and ſo by this Means the ſaid Board <lb/>may be vertically moved backwards or for-<lb/>wards.</s> <s xml:id="echoid-s2317" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2318" xml:space="preserve">8. </s> <s xml:id="echoid-s2319" xml:space="preserve">On the Top of the Machine, there is a <lb/>Board about 15 Inches long, and 9 Inches broad, <lb/>having an hole PMOQ quite through it, about <lb/>9 or 10 Inches long, and four Inches broad.</s> <s xml:id="echoid-s2320" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2321" xml:space="preserve">9. </s> <s xml:id="echoid-s2322" xml:space="preserve">Upon the aforeſaid Board are fix’d two Dove-<lb/>tail’d Rulers, between which another Board ſlides <lb/>of the ſame Length as that, and whoſe breadth is <lb/>about 6 Inches. </s> <s xml:id="echoid-s2323" xml:space="preserve">In the middle of this ſecond <lb/>Board is a round Hole about three Inches Diame-<lb/>ter, hollowed into a female Screw, in which is <lb/>fitted a Cylinder about four Inches long, which <lb/>carries the Convex Glaſs; </s> <s xml:id="echoid-s2324" xml:space="preserve">Of which more here-<lb/>after.</s> <s xml:id="echoid-s2325" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2326" xml:space="preserve">10 The Figure X is a ſquare Box about 7 or <lb/>8 Inches broad, and 10 in height, which ſlides <lb/>upon the plain Board mentioned in Numb. </s> <s xml:id="echoid-s2327" xml:space="preserve">8. </s> <s xml:id="echoid-s2328" xml:space="preserve">the <lb/>Side B ſerving as a Door, is next to the fore-ſide <lb/>of the Machine, as it appears in the Figure; </s> <s xml:id="echoid-s2329" xml:space="preserve">and <pb o="106" file="0182" n="211" rhead="An ESSAY"/> the Back-ſide thereof hath in it a Square ope-<lb/>ning N, each Side being about four Inches in <lb/>Length, which may be ſhut by the little Board I <lb/>ſliding between two Rulers.</s> <s xml:id="echoid-s2330" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2331" xml:space="preserve">11. </s> <s xml:id="echoid-s2332" xml:space="preserve">Overthe ſaid Square opening, there is a Slit <lb/>parallel to the Horizon, going along the whole <lb/>Breadth of the Back-ſide of the Box; </s> <s xml:id="echoid-s2333" xml:space="preserve">through <lb/>which Slit, a little Mirrour or Looking-Glaſs is to <lb/>be put into the Box, whoſe two Sides may ſlide be-<lb/>tween two Rulers ſo placed, that the poliſhed <lb/>Side of the Glaſs being turn’d towards the Door <lb/>B, may make an Angle of 112 {1/2} Degrees, with <lb/>the Horizon. </s> <s xml:id="echoid-s2334" xml:space="preserve">Note, This Diſpoſition of the <lb/>Looking-glaſs could not well be repreſented in <lb/>the Figure.</s> <s xml:id="echoid-s2335" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2336" xml:space="preserve">12. </s> <s xml:id="echoid-s2337" xml:space="preserve">The before-mention’d Mirrour hath a ſmall <lb/>Iron Plate, on the Middle of that Side which is <lb/>without the Box, (when the ſaid Looking-glaſs <lb/>is in the Diſpoſition mention’d in N. </s> <s xml:id="echoid-s2338" xml:space="preserve">11.) </s> <s xml:id="echoid-s2339" xml:space="preserve">being the <lb/>Baſe of a Screw faſten’d to the Middle thereof, that <lb/>ſothe Looking-glaſs may be fix’d upright (as appears <lb/>per Figure) in any Place H upon the Top of the <lb/>Machine, and vertically turn all ways; </s> <s xml:id="echoid-s2340" xml:space="preserve">and this <lb/>is done by putting the Screw through a Hole <lb/>made in the plain Board of N. </s> <s xml:id="echoid-s2341" xml:space="preserve">9. </s> <s xml:id="echoid-s2342" xml:space="preserve">and through a <lb/>Slit made for this End in the plain Board of N. </s> <s xml:id="echoid-s2343" xml:space="preserve">8. <lb/></s> <s xml:id="echoid-s2344" xml:space="preserve">and then fixing it with the Nut R. </s> <s xml:id="echoid-s2345" xml:space="preserve">Now when <lb/>the Mirrour is taken from this Situation, the ſaid <lb/>Slit is ſhut by a ſmall Board ſliding between two <lb/>little Rulers within the Machine. </s> <s xml:id="echoid-s2346" xml:space="preserve">As to the Slit <lb/>mention’d N. </s> <s xml:id="echoid-s2347" xml:space="preserve">11. </s> <s xml:id="echoid-s2348" xml:space="preserve">it is partly ſhut by the little <lb/>Board I, when the Aperture N is open’d, and the <lb/>two Ends thereof remaining open’d, are ſhut by <lb/>two ſinall Rulers.</s> <s xml:id="echoid-s2349" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2350" xml:space="preserve">13. </s> <s xml:id="echoid-s2351" xml:space="preserve">There are two Iron Staples on one Side of <lb/>the Box, like thoſe which are <anchor type="note" xlink:href="" symbol="*"/> on the Back-ſide <anchor type="note" xlink:label="note-0182-01a" xlink:href="note-0182-01"/> of the Machine; </s> <s xml:id="echoid-s2352" xml:space="preserve">in which a Ruler going ſeveral <lb/>Inches out behind the Box X may ſlide, having <pb o="107" file="0183" n="212" rhead="on PERSPECTIVE."/> a Hole at its End, through which the above-men-<lb/>tion’d Screw may paſs, and ſo the Mirrour H be <lb/>fixed to any Inclination before the Aperture N.</s> <s xml:id="echoid-s2353" xml:space="preserve"/> </p> <div xml:id="echoid-div368" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0182-01" xlink:href="note-0182-01a" xml:space="preserve">7.</note> </div> <p> <s xml:id="echoid-s2354" xml:space="preserve">14. </s> <s xml:id="echoid-s2355" xml:space="preserve">Beſides the Mirrour H, there is another leſſer <lb/>one, L, faſten’d near its Middle to a Ruler going <lb/>out through the Middle of the Top of the Box. <lb/></s> <s xml:id="echoid-s2356" xml:space="preserve">This Ruler may Screw on, and ſerves to raiſe or <lb/>lower the Mirrour faſten’d to it, ſo that it may be <lb/>fixed to all Angles of Inclination.</s> <s xml:id="echoid-s2357" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div370" type="section" level="1" n="201"> <head xml:id="echoid-head219" xml:space="preserve"><emph style="sc">Remarks</emph>.</head> <p> <s xml:id="echoid-s2358" xml:space="preserve">If the Tubes mention’d in N. </s> <s xml:id="echoid-s2359" xml:space="preserve">6. </s> <s xml:id="echoid-s2360" xml:space="preserve">be not thought <lb/>ſufficient for giving Air to the Machine, a ſmall <lb/>Pair of Bellows may be put under the Seat, <lb/>which may be blown by one’s Foot. </s> <s xml:id="echoid-s2361" xml:space="preserve">And by <lb/>this means the Air within the Machine may be <lb/>continually remov’d; </s> <s xml:id="echoid-s2362" xml:space="preserve">for the Bellows driving the <lb/>Air out of the Machine, obliges the external Air <lb/>to enter through the Tubes therein.</s> <s xml:id="echoid-s2363" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div371" type="section" level="1" n="202"> <head xml:id="echoid-head220" style="it" xml:space="preserve">Uſe of the Machine.</head> <head xml:id="echoid-head221" xml:space="preserve"><emph style="sc">Problem</emph> I.</head> <p style="it"> <s xml:id="echoid-s2364" xml:space="preserve">15. </s> <s xml:id="echoid-s2365" xml:space="preserve">To repreſent Objects in their natural Diſpo-<lb/>ſition.</s> <s xml:id="echoid-s2366" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2367" xml:space="preserve">When Objects are to be repreſented within the <lb/> <anchor type="note" xlink:label="note-0183-01a" xlink:href="note-0183-01"/> Machine, we extend a Sheet of Paper upon the <lb/>Table A; </s> <s xml:id="echoid-s2368" xml:space="preserve">or, which is better, we lay a Sheet of <lb/>Paper upon another Board, ſo that it ſpreads be-<lb/>yond the Edges of the Board; </s> <s xml:id="echoid-s2369" xml:space="preserve">then we ſqueeze <lb/>the ſaid Paper and Board into a Frame, ſo that <lb/>it be fixed therein by means of two Dove-tail’d <lb/>Rulers. </s> <s xml:id="echoid-s2370" xml:space="preserve">This being done, we place a Convex <lb/>Glaſs in the Cylinder C, <anchor type="note" xlink:href="" symbol="*"/> which ſcrews into the <anchor type="note" xlink:label="note-0183-02a" xlink:href="note-0183-02"/> Top of the Machine, having its Focal Length <pb o="108" file="0184" n="213" rhead="An ESSAY"/> nearly equal to the Height of the Top of the <lb/>Machine above the Table: </s> <s xml:id="echoid-s2371" xml:space="preserve">Then we open the <lb/>Aperture N at the Back-part of the Box upon the <lb/>Machine; </s> <s xml:id="echoid-s2372" xml:space="preserve">and incline the Mirrour L, ſo as to make <lb/>an Angle of 45 deg. </s> <s xml:id="echoid-s2373" xml:space="preserve">with the Horizon, when Ob-<lb/>jects are to be repreſented for the perpendicular <lb/>Picture. </s> <s xml:id="echoid-s2374" xml:space="preserve">Then, if the Mirrour H be taken away, <lb/>as alſo the Board F, together with the two Ru-<lb/>lers D E and D E, we ſhall perceive the Repre-<lb/>ſentation upon the Sheet of Paper on the Table <lb/>A, of all Objects, whoſe Rays falling upon the <lb/>Looking-glaſs L, can be thereby reflected upon <lb/>the Convex Glaſs; </s> <s xml:id="echoid-s2375" xml:space="preserve">which Convex Glaſs muſt be <lb/>rais’d or lower’d, by means of the Screw about <lb/>the Cylinder carrying it, until the ſaid Objects <lb/>appear entirely diſtinct.</s> <s xml:id="echoid-s2376" xml:space="preserve"/> </p> <div xml:id="echoid-div371" type="float" level="2" n="1"> <note position="right" xlink:label="note-0183-01" xlink:href="note-0183-01a" xml:space="preserve">Fig. 70.</note> <note symbol="*" position="right" xlink:label="note-0183-02" xlink:href="note-0183-02a" xml:space="preserve">9.</note> </div> <p> <s xml:id="echoid-s2377" xml:space="preserve">16. </s> <s xml:id="echoid-s2378" xml:space="preserve">When the ſame Objects are requir’d to be <lb/>repreſented for the inclin’d Picture, the Looking-<lb/>glaſs L muſt have half the Inclination we would <lb/>give to the Picture.</s> <s xml:id="echoid-s2379" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2380" xml:space="preserve">17. </s> <s xml:id="echoid-s2381" xml:space="preserve">When the ſaid Objects are to be repreſent-<lb/>ed upon the Picture being parallel, the Aperture <lb/>N muſt be ſhut, and the Door B open’d; </s> <s xml:id="echoid-s2382" xml:space="preserve">then <lb/>the Mirrour H muſt be rais’d to the Top of the <lb/>Box, in putting it in a Situation parallel to the <lb/>Horizon. </s> <s xml:id="echoid-s2383" xml:space="preserve">This Diſpoſition of the Machine may <lb/>ferve when one is upon a Balcony, or ſome <lb/>other high Place; </s> <s xml:id="echoid-s2384" xml:space="preserve">to deſign a Parterre under-<lb/>neath.</s> <s xml:id="echoid-s2385" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2386" xml:space="preserve">18. </s> <s xml:id="echoid-s2387" xml:space="preserve">If we have a mind to deſign a Statue <lb/>ſtanding in a Place ſomething elevated, and it is <lb/>requir’d to be ſo repreſented, as to be painted <lb/>againſt a Cieling; </s> <s xml:id="echoid-s2388" xml:space="preserve">the Back-ſide of the Machine <lb/>muſt be turned towards the Statue, and the Box <lb/>X ſo turned, that the Door B may face the Sta-<lb/>tue; </s> <s xml:id="echoid-s2389" xml:space="preserve">then, the Door being open’d, the Looking-<lb/>glaſs L muſt be placed vertically, with its po-<lb/>liſh’d Side towards the Statue; </s> <s xml:id="echoid-s2390" xml:space="preserve">and the Box mo- <pb o="109" file="0185" n="214" rhead="on PERSPECTIVE."/> ved backwards or forwards, or elſe the Looking-<lb/>glaſs raiſed or lower’d, until the Rays proceed-<lb/>ing from the Statue may be reflected by the Mir-<lb/>rour upon the Convex Glaſs. </s> <s xml:id="echoid-s2391" xml:space="preserve">When theſe Alte-<lb/>rations of the Box, or Mirrour, are not ſufficient to <lb/>throw the Rays upon the Convex Glafs, the whole <lb/>Machine muſt be removed backwards or for-<lb/>wards.</s> <s xml:id="echoid-s2392" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div373" type="section" level="1" n="203"> <head xml:id="echoid-head222" xml:space="preserve"><emph style="sc">Demonstration</emph>.</head> <head xml:id="echoid-head223" style="it" xml:space="preserve">Concerning the before-mention’d Inclination of the <lb/>Mirrours.</head> <p> <s xml:id="echoid-s2393" xml:space="preserve">19. </s> <s xml:id="echoid-s2394" xml:space="preserve">In order to demonſtrate, that the Mirrour <lb/>L hath been conveniently inclin’d, we need on-<lb/>ly prove, that the reflected Rays fall upon the <lb/>Table A under the ſame Angle, as the direct <lb/>Rays do upon a Plane, having the ſame Situation <lb/>as one would give to the Picture.</s> <s xml:id="echoid-s2395" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2396" xml:space="preserve">Now let A B be a Ray falling from a Point of <lb/> <anchor type="note" xlink:label="note-0185-01a" xlink:href="note-0185-01"/> ſome Object upon the Mirrour G H, and from <lb/>thence is reflected in the Point a upon the Table <lb/>of the Machine: </s> <s xml:id="echoid-s2397" xml:space="preserve">We are to demonſtrate, that if <lb/>the Line D I be drawn, making an Angle with <lb/>FE equal to the Inclination of the Picture; </s> <s xml:id="echoid-s2398" xml:space="preserve">that <lb/>is, <anchor type="note" xlink:href="" symbol="*"/> if the Angle DIE be the double of the Angle <anchor type="note" xlink:label="note-0185-02a" xlink:href="note-0185-02"/> D F I; </s> <s xml:id="echoid-s2399" xml:space="preserve">I ſay, we are to demonſtrate, that the <lb/>Angle B a f is equal to the Angle BCD.</s> <s xml:id="echoid-s2400" xml:space="preserve"/> </p> <div xml:id="echoid-div373" type="float" level="2" n="1"> <note position="right" xlink:label="note-0185-01" xlink:href="note-0185-01a" xml:space="preserve">Fig. 71.</note> <note symbol="*" position="right" xlink:label="note-0185-02" xlink:href="note-0185-02a" xml:space="preserve">15, 16.</note> </div> <p> <s xml:id="echoid-s2401" xml:space="preserve">The Angle DIE, by Conſtruction, is the double <lb/>of the Angle DFI; </s> <s xml:id="echoid-s2402" xml:space="preserve">and conſequently this laſt Angle <lb/>is equal to the Angle I D F; </s> <s xml:id="echoid-s2403" xml:space="preserve">and ſince the Angle <lb/>of Incidence C B D is equal to the Angle of Re-<lb/>flection a B F, the Triangle BCD is ſimilar to <lb/>the Triangle F a B: </s> <s xml:id="echoid-s2404" xml:space="preserve">Whence it follows, that the <lb/>Angle Ba F is equal to the Angle BCD. </s> <s xml:id="echoid-s2405" xml:space="preserve">Which <lb/>was to be demonſtrated.</s> <s xml:id="echoid-s2406" xml:space="preserve"/> </p> <pb o="110" file="0186" n="215" rhead="An ESSAY"/> <p> <s xml:id="echoid-s2407" xml:space="preserve">20. </s> <s xml:id="echoid-s2408" xml:space="preserve">Concerning what hath been ſaid of the <lb/>Picture being parallel, it muſt be obſerv’d, that <lb/>in the precedent Demonſtration, the Angle of <lb/>Inclination of the Picture in this Demonſtration <lb/>is meaſur’d next to the Objects; </s> <s xml:id="echoid-s2409" xml:space="preserve">and if this An-<lb/>gle be diminiſh’d till it become nothing, we <lb/>ſhall have a Picture parallel to the Horizon un-<lb/>derneath the Eye. </s> <s xml:id="echoid-s2410" xml:space="preserve">But, by the Demonſtration, <lb/>the Inclination of the Mirrour being half the In-<lb/>clination of the Picture, it follows, that the In-<lb/>clination of the Mirrour is alſo equal to nothing, <lb/>and conſequently it ought to be likewiſe parallel <lb/>to the Horizon. </s> <s xml:id="echoid-s2411" xml:space="preserve">In the ſame manner we de-<lb/>monſtrate, that the Looking-glaſs muſt be verti-<lb/>cally ſituated, when we conſider the Picture pa-<lb/>rallel above the Eye: </s> <s xml:id="echoid-s2412" xml:space="preserve">For to give this Situation <lb/>to the Picture, the Angle of the Inclination of <lb/>the Picture meaſur’d next to the Objects, muſt <lb/>be augmented till it be 180 Degrees, whoſe half <lb/>90 Degrees is conſequently the Inclination of the <lb/>Mirrour.</s> <s xml:id="echoid-s2413" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div375" type="section" level="1" n="204"> <head xml:id="echoid-head224" xml:space="preserve"><emph style="sc">Prob</emph>. II.</head> <p style="it"> <s xml:id="echoid-s2414" xml:space="preserve">22. </s> <s xml:id="echoid-s2415" xml:space="preserve">To repreſent Objects, ſo that what appears on <lb/>the right Hand, ought to be on the left.</s> <s xml:id="echoid-s2416" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2417" xml:space="preserve">23. </s> <s xml:id="echoid-s2418" xml:space="preserve">Having placed the Box X, in the Situation <lb/> <anchor type="note" xlink:label="note-0186-01a" xlink:href="note-0186-01"/> as per Figure, the Door B muſt be opened, and <lb/>the Aperture N ſhut; </s> <s xml:id="echoid-s2419" xml:space="preserve">then putting the Mirrour H <lb/>in the Diſpoſition mentioned in Numb. </s> <s xml:id="echoid-s2420" xml:space="preserve">11. </s> <s xml:id="echoid-s2421" xml:space="preserve">raiſe <lb/>up the Mirrour L towards the Top of the Box; <lb/></s> <s xml:id="echoid-s2422" xml:space="preserve">and incline it towards the firſt Mirrour, in ſuch <lb/>manner that it makes an Angle with the Hori-<lb/>zon of 22 {1/2} Degrees; </s> <s xml:id="echoid-s2423" xml:space="preserve">that is, that the Top of the <lb/>Machine, after a double Reflection, appears ver-<lb/>cal in the Mirrour H.</s> <s xml:id="echoid-s2424" xml:space="preserve"/> </p> <div xml:id="echoid-div375" type="float" level="2" n="1"> <note position="left" xlink:label="note-0186-01" xlink:href="note-0186-01a" xml:space="preserve">Fig. 70.</note> </div> <p> <s xml:id="echoid-s2425" xml:space="preserve">24. </s> <s xml:id="echoid-s2426" xml:space="preserve">Now if Objects are to be repreſented for <lb/>the Picture inclin’d, the Mirrour L muſt make an <pb o="111" file="0187" n="216" rhead="on PERSPECTIVE."/> Angle with the Horizon, equal to half the <lb/>Inclination of the Picture leſs {1/4} of a right Angle. <lb/></s> <s xml:id="echoid-s2427" xml:space="preserve">This Angle is found exactly enough for Practice, <lb/>by inclining the Mirrour L, until the Repreſenta-<lb/>tion of the Top of the Machine, after a double Re-<lb/>flection, appears in the other Mirrour under an <lb/>Angle with the Horizon, equal to the Inclination <lb/>one would give the Picture. </s> <s xml:id="echoid-s2428" xml:space="preserve">Note, If the Incli-<lb/>nation of the Picture be leſſer than {1/4} of 90 De-<lb/>grees, the Looking-Glaſs L muſt not be inclin’d <lb/>towards the other, as is directed, <anchor type="note" xlink:href="" symbol="*"/> but the <anchor type="note" xlink:label="note-0187-01a" xlink:href="note-0187-01"/> contrary Way, in making the Angle of the In-<lb/>clination of the Looking-Glaſs, equal to the <lb/>Difference of the Inclination of the Picture, and <lb/>{1/4} of 90 Degrees.</s> <s xml:id="echoid-s2429" xml:space="preserve"/> </p> <div xml:id="echoid-div376" type="float" level="2" n="2"> <note symbol="*" position="right" xlink:label="note-0187-01" xlink:href="note-0187-01a" xml:space="preserve">23:<unsure/></note> </div> <p> <s xml:id="echoid-s2430" xml:space="preserve">25. </s> <s xml:id="echoid-s2431" xml:space="preserve">When the Objects are to be repreſented <lb/>for a parallel Picture, the Looking-Glaſs L muſt <lb/>be placed in the Diſpoſition of Numb. </s> <s xml:id="echoid-s2432" xml:space="preserve">15. </s> <s xml:id="echoid-s2433" xml:space="preserve">and <lb/>the Looking-Glaſs H in that mentioned, Numb. <lb/></s> <s xml:id="echoid-s2434" xml:space="preserve">13. </s> <s xml:id="echoid-s2435" xml:space="preserve">by inclining it towards the Horizon, under an <lb/>Angle of 45 Degrees; </s> <s xml:id="echoid-s2436" xml:space="preserve">the poliſhed Side thereof <lb/>facing downwards, when the Picture is ſuppoſed <lb/>underneath the Eye; </s> <s xml:id="echoid-s2437" xml:space="preserve">and upwards, when it is <lb/>ſuppoſed above the Eye.</s> <s xml:id="echoid-s2438" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2439" xml:space="preserve">26. </s> <s xml:id="echoid-s2440" xml:space="preserve">This Diſpoſition of the Machine may be <lb/>likewiſe uſeſul for inclin’d Pictures, making very <lb/>ſmall Angles with the Horizon; </s> <s xml:id="echoid-s2441" xml:space="preserve">in which Caſe, <lb/>the Inclination of one of the Looking-Glaſſes <lb/>muſt be diminiſh’d, by half of the Inclination <lb/>of the Picture.</s> <s xml:id="echoid-s2442" xml:space="preserve"/> </p> <p style="it"> <s xml:id="echoid-s2443" xml:space="preserve">27. </s> <s xml:id="echoid-s2444" xml:space="preserve">A Demonſtration of the Inclination of the <lb/>Mirrours.</s> <s xml:id="echoid-s2445" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2446" xml:space="preserve">We have mentioned, <anchor type="note" xlink:href="" symbol="*"/> that for a perpendicu- <anchor type="note" xlink:label="note-0187-02a" xlink:href="note-0187-02"/> lar Picture, one of the Mirrours muſt make an <lb/>Angle <anchor type="note" xlink:href="" symbol="*"/> of 112 {1/2} Degrees with the Horizon; </s> <s xml:id="echoid-s2447" xml:space="preserve">and <anchor type="note" xlink:label="note-0187-03a" xlink:href="note-0187-03"/> the other, L, muſt be inclin’d towards the firſt, <pb o="112" file="0188" n="217" rhead="An ESSAY"/> and make an Angle of 22 {1/2} with the Horizon. <lb/></s> <s xml:id="echoid-s2448" xml:space="preserve">Let MN and GH be two Mirrours in the before-<lb/> <anchor type="note" xlink:label="note-0188-01a" xlink:href="note-0188-01"/> mentioned Situation; </s> <s xml:id="echoid-s2449" xml:space="preserve">we are to demonſtrate, <lb/>that if the Ray A B is parallel to the Horizon, <lb/>after being reflected in B and C, it ought to fall <lb/>perpendicularly upon the Machine. </s> <s xml:id="echoid-s2450" xml:space="preserve">The Angle <lb/>A B N is <anchor type="note" xlink:href="" symbol="*"/> 112 {1/2} Degrees; </s> <s xml:id="echoid-s2451" xml:space="preserve">and conſequently the <anchor type="note" xlink:label="note-0188-02a" xlink:href="note-0188-02"/> Angle A B M, and its equal, the Angle of Re-<lb/>flection C B G, are each 67 {1/2} Degrees. </s> <s xml:id="echoid-s2452" xml:space="preserve">The An-<lb/>gle B P Q, is the Complement of the Angle <lb/>N B A, plus the Angle P Q B, which is <anchor type="note" xlink:href="" symbol="*"/> 22 {1/2} De- <anchor type="note" xlink:label="note-0188-03a" xlink:href="note-0188-03"/> grees; </s> <s xml:id="echoid-s2453" xml:space="preserve">whence the Angle B P Q is 45 Degrees. <lb/></s> <s xml:id="echoid-s2454" xml:space="preserve">Again, the Angle PC B is the Complement of <lb/>the two Angles C B P and B PC to 180 Degrees; </s> <s xml:id="echoid-s2455" xml:space="preserve"><lb/>and conſequently it is 67 {1/2} Degrees, which is the <lb/>ſame as its equal, the Angle Q C a of Reflection. </s> <s xml:id="echoid-s2456" xml:space="preserve"><lb/>And reaſoning after the ſame Manner, the An-<lb/>gle C R Q of the Triangle R C Q, is a right one. </s> <s xml:id="echoid-s2457" xml:space="preserve"><lb/>Which was to be demonſtrated.</s> <s xml:id="echoid-s2458" xml:space="preserve"/> </p> <div xml:id="echoid-div377" type="float" level="2" n="3"> <note symbol="*" position="right" xlink:label="note-0187-02" xlink:href="note-0187-02a" xml:space="preserve">22.</note> <note symbol="*" position="right" xlink:label="note-0187-03" xlink:href="note-0187-03a" xml:space="preserve">11.</note> <note position="left" xlink:label="note-0188-01" xlink:href="note-0188-01a" xml:space="preserve">Fig. 72.</note> <note symbol="*" position="left" xlink:label="note-0188-02" xlink:href="note-0188-02a" xml:space="preserve">11.</note> <note symbol="*" position="left" xlink:label="note-0188-03" xlink:href="note-0188-03a" xml:space="preserve">23.</note> </div> <p> <s xml:id="echoid-s2459" xml:space="preserve">28. </s> <s xml:id="echoid-s2460" xml:space="preserve">It is not abſolutely neceſſary to give the <lb/>Mirrours the aforeſaid Inclinations; </s> <s xml:id="echoid-s2461" xml:space="preserve">for the An-<lb/>gle A B N may be aſſumed at Pleaſure, from <lb/>which muſt be taken an Angle of 135 Degrees, <lb/>to have the Inclination of the Miror G H. </s> <s xml:id="echoid-s2462" xml:space="preserve">Ne-<lb/>vertheleſs, the Angles we have determin’d, are <lb/>the moſt advantagious for a perpendicular Pi-<lb/>cture.</s> <s xml:id="echoid-s2463" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2464" xml:space="preserve">29. </s> <s xml:id="echoid-s2465" xml:space="preserve">When a Picture is inclin’d, and makes <lb/> <anchor type="note" xlink:label="note-0188-04a" xlink:href="note-0188-04"/> the Angle D I A with the Horizon, the Mirrour <lb/>M N muſt <anchor type="note" xlink:href="" symbol="*"/> keep its Situation, and the Angle <anchor type="note" xlink:label="note-0188-05a" xlink:href="note-0188-05"/> C Q R is equal to half the Angle D I A, leſs <lb/>{1/4} of a right Angle: </s> <s xml:id="echoid-s2466" xml:space="preserve">Then I ſay, the Angle <lb/>FaC, or its equal C R Q, will be equal to the <lb/>Angle B I D. </s> <s xml:id="echoid-s2467" xml:space="preserve">Now the Angle P B Q, is <anchor type="note" xlink:href="" symbol="*"/> 112 {1/2}.</s> <s xml:id="echoid-s2468" xml:space="preserve"> <anchor type="note" xlink:label="note-0188-06a" xlink:href="note-0188-06"/> Degrees; </s> <s xml:id="echoid-s2469" xml:space="preserve">whence the Angle B P Q, which is the <lb/>Complement of P B Q, and P Q B to two right <lb/>Angles, is <anchor type="note" xlink:href="" symbol="*"/> 90 Degrees, leſs the half of the An- <anchor type="note" xlink:label="note-0188-07a" xlink:href="note-0188-07"/> gle D I A: </s> <s xml:id="echoid-s2470" xml:space="preserve">Wherefore becauſe N B C is 67 {1/2} the <pb o="113" file="0189" n="218" rhead="on PERSPECTIVE."/> Angle B C P, and its equal R C Q, is 22 {1/2} plus <lb/>{1/2} of D I A. </s> <s xml:id="echoid-s2471" xml:space="preserve">Now if the Angle R Q C be added <lb/>to this Angle, their Sum will be equal to the An-<lb/>gle D I A; </s> <s xml:id="echoid-s2472" xml:space="preserve">whence it follows, that the Angle <lb/>C R Q is equal to D I R. </s> <s xml:id="echoid-s2473" xml:space="preserve">Which was to be demon-<lb/>ſtrated.</s> <s xml:id="echoid-s2474" xml:space="preserve"/> </p> <div xml:id="echoid-div378" type="float" level="2" n="4"> <note position="left" xlink:label="note-0188-04" xlink:href="note-0188-04a" xml:space="preserve">Fig. 73.</note> <note symbol="*" position="left" xlink:label="note-0188-05" xlink:href="note-0188-05a" xml:space="preserve">24.</note> <note symbol="*" position="left" xlink:label="note-0188-06" xlink:href="note-0188-06a" xml:space="preserve">11.</note> <note symbol="*" position="left" xlink:label="note-0188-07" xlink:href="note-0188-07a" xml:space="preserve">27.</note> </div> <p> <s xml:id="echoid-s2475" xml:space="preserve">30. </s> <s xml:id="echoid-s2476" xml:space="preserve">If the Angle R B N be alter’d, and it be <lb/>called a, the Angle D I A, b, and the right An-<lb/>gle d; </s> <s xml:id="echoid-s2477" xml:space="preserve">then the Angle C Q R = d + {1/2} b — a.</s> <s xml:id="echoid-s2478" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2479" xml:space="preserve">31. </s> <s xml:id="echoid-s2480" xml:space="preserve">When a Picture is parallel, it appears ma-<lb/> <anchor type="note" xlink:label="note-0189-01a" xlink:href="note-0189-01"/> nifeſt, the Mirrours GH and M N being each in-<lb/>clin’d under an Angle of 45 Degrees, that a Ray, <lb/>which is perpendicular to the Horizon, likewiſe <lb/>falls, after a double Reflection, perpendicularly <lb/>upon the Table A.</s> <s xml:id="echoid-s2481" xml:space="preserve"/> </p> <div xml:id="echoid-div379" type="float" level="2" n="5"> <note position="right" xlink:label="note-0189-01" xlink:href="note-0189-01a" xml:space="preserve">Fig. 74.</note> </div> </div> <div xml:id="echoid-div381" type="section" level="1" n="205"> <head xml:id="echoid-head225" xml:space="preserve"><emph style="sc">Prop</emph>. III.</head> <p style="it"> <s xml:id="echoid-s2482" xml:space="preserve">32. </s> <s xml:id="echoid-s2483" xml:space="preserve">To repreſent Objects which are roun’d about <lb/>the Machine, and make them appear erect to the Per-<lb/>ſon ſeated within the ſame.</s> <s xml:id="echoid-s2484" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2485" xml:space="preserve">The Back-ſide of the Machine muſt be turned <lb/>towards the Sun, and the Objects behind the <lb/>ſame repreſented <anchor type="note" xlink:href="" symbol="*"/> by one Reflection only; </s> <s xml:id="echoid-s2486" xml:space="preserve">then <anchor type="note" xlink:label="note-0189-02a" xlink:href="note-0189-02"/> their Appearance will always be clearer, altho <lb/>they be in the Shade, than the Appearance of the <lb/>Objects on the other Sides of the Machine, which <lb/>cannot be perceiv’d unleſs by a double Refle-<lb/>ction.</s> <s xml:id="echoid-s2487" xml:space="preserve"/> </p> <div xml:id="echoid-div381" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0189-02" xlink:href="note-0189-02a" xml:space="preserve">15.</note> </div> <p> <s xml:id="echoid-s2488" xml:space="preserve">33. </s> <s xml:id="echoid-s2489" xml:space="preserve">The Objects that are on the right and left <lb/> <anchor type="note" xlink:label="note-0189-03a" xlink:href="note-0189-03"/> of the Machine, may be repreſented by means <lb/>of the Mirrour H, ſituated <anchor type="note" xlink:href="" symbol="*"/> as per Figure; </s> <s xml:id="echoid-s2490" xml:space="preserve">but the <anchor type="note" xlink:label="note-0189-04a" xlink:href="note-0189-04"/> ſaid Mirrour muſt be cover’d with a Paſtboard Caſe, <lb/>having two Apertures therein; </s> <s xml:id="echoid-s2491" xml:space="preserve">the one next to <lb/>the Objects, and the other next to the Aperture <lb/>N, of the Box X. </s> <s xml:id="echoid-s2492" xml:space="preserve">The Reaſon of our uſing this <lb/>@recaution is, becauſe when the Mirrour is not co-<lb/>ver’d at all, it reflects the Rays of Light coming <pb o="114" file="0190" n="219" rhead="An ESSAY"/> Side-ways upon the Mirrour L, which being again <lb/>reflected by the ſaid Mirrour L, and going through <lb/>the Convex Glaſs, extremely weakens the Repre-<lb/>ſentation.</s> <s xml:id="echoid-s2493" xml:space="preserve"/> </p> <div xml:id="echoid-div382" type="float" level="2" n="2"> <note position="right" xlink:label="note-0189-03" xlink:href="note-0189-03a" xml:space="preserve">Fig. 70.</note> <note symbol="*" position="right" xlink:label="note-0189-04" xlink:href="note-0189-04a" xml:space="preserve">12.</note> </div> <p> <s xml:id="echoid-s2494" xml:space="preserve">34. </s> <s xml:id="echoid-s2495" xml:space="preserve">The Objects before the Machine are re-<lb/>preſented according to N. </s> <s xml:id="echoid-s2496" xml:space="preserve">22, and 28.</s> <s xml:id="echoid-s2497" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div384" type="section" level="1" n="206"> <head xml:id="echoid-head226" xml:space="preserve"><emph style="sc">Problem</emph> IV.</head> <p style="it"> <s xml:id="echoid-s2498" xml:space="preserve">35. </s> <s xml:id="echoid-s2499" xml:space="preserve">To repreſent Pictures or Prints.</s> <s xml:id="echoid-s2500" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2501" xml:space="preserve">If we have a mind to repreſent Pictures and <lb/> <anchor type="note" xlink:label="note-0190-01a" xlink:href="note-0190-01"/> Prints, they muſt be faſten’d againſt the Board <lb/>F on that Side, regarding the Back of the Ma-<lb/>chine, which muſt be ſo turned, that the Pictures <lb/>be expoſed to the Sun. </s> <s xml:id="echoid-s2502" xml:space="preserve">Then they are repreſent-<lb/>ed in this Situation as <anchor type="note" xlink:href="" symbol="*"/> the other Objects, but <anchor type="note" xlink:label="note-0190-02a" xlink:href="note-0190-02"/> with this Difference, that the Convex Glaſs in the <lb/>Cylinder C muſt be changed: </s> <s xml:id="echoid-s2503" xml:space="preserve">For if Prints are <lb/>requir’d to have their true Bigneſs, the focal Di-<lb/>ſtance of the Convex Glaſs muſt be equal to half <lb/>the Height of the Machine above the Table; <lb/></s> <s xml:id="echoid-s2504" xml:space="preserve">that is, equal to half A C. </s> <s xml:id="echoid-s2505" xml:space="preserve">Again, if the ſaid <lb/>Pictures or Prints are requir’d to be repreſented <lb/>greater than they really are, the focal Diſtance <lb/>of the Convex Glaſs muſt ſtill be leſſer. </s> <s xml:id="echoid-s2506" xml:space="preserve">And <lb/>if, on the contrary, they are to be repreſented <lb/>leſſer than they really are, the focal Length of <lb/>the Glaſs muſt be greater than the Length A C. </s> <s xml:id="echoid-s2507" xml:space="preserve"><lb/>Moreover, the proper Diſtance whereat the Pi-<lb/>ctures or Prints muſt be placed, may be found <lb/>in ſliding the Board F backwards or forwards, <lb/>until they diſtinctly appear within the Machine. </s> <s xml:id="echoid-s2508" xml:space="preserve"><lb/>This Diſtance alſo may be determin’d by the <lb/>following Proportion:</s> <s xml:id="echoid-s2509" xml:space="preserve"/> </p> <div xml:id="echoid-div384" type="float" level="2" n="1"> <note position="left" xlink:label="note-0190-01" xlink:href="note-0190-01a" xml:space="preserve">Fig. 70.</note> <note symbol="*" position="left" xlink:label="note-0190-02" xlink:href="note-0190-02a" xml:space="preserve">15.</note> </div> </div> <div xml:id="echoid-div386" type="section" level="1" n="207"> <head xml:id="echoid-head227" style="it" xml:space="preserve">As the Machine’s Height above the Table, leſs the <lb/>Glaſs’s focal Length, <lb/>is to</head> <pb o="115" file="0191" n="220" rhead="on PERSPECTIVE."/> </div> <div xml:id="echoid-div387" type="section" level="1" n="208"> <head xml:id="echoid-head228" style="it" xml:space="preserve">The Height of the Machine above the Table; <lb/>So is <lb/>The Glaſſes focal Length, <lb/>to the <lb/>Diſtance of the Figure from the Glaſs.</head> <p> <s xml:id="echoid-s2510" xml:space="preserve">Note, The ſaid Diſtance of the Convex Glaſs <lb/>from the Figure, is meaſured by a Ray, pro-<lb/>ceeding from the Figure parallel to the Horizon, <lb/>which is perpendicularly reflected upon the Con-<lb/>vex Glaſs, by the Mirrour. </s> <s xml:id="echoid-s2511" xml:space="preserve">Note, Moreover, that <lb/>when we have a Mind to place the Figures out <lb/>beyond the Back-ſide of the Machine, they muſt <lb/>be faſtned againſt the Side F of the Board, which <lb/>muſt be ſo turned, that the ſaid Side be next to <lb/>the Aperture N.</s> <s xml:id="echoid-s2512" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div388" type="section" level="1" n="209"> <head xml:id="echoid-head229" style="it" xml:space="preserve">37 Remarks concerning the Repreſentation of Per-<lb/>ſons Faces.</head> <p> <s xml:id="echoid-s2513" xml:space="preserve">It is certainly very curious and uſeful to de-<lb/>ſign Perſons Faces to the Life; </s> <s xml:id="echoid-s2514" xml:space="preserve">which by this <lb/>Machine, may be very well done in Miniature: <lb/></s> <s xml:id="echoid-s2515" xml:space="preserve">For if the Face of any known Perſon be thus re-<lb/>preſented, by only looking at the Appearance, <lb/>we may very readily know whoſe Face it is, <lb/>when at the ſame time the Appearance of the <lb/>Whole Perſon will not take up half an Inch <lb/>upon the Paper on the Table: </s> <s xml:id="echoid-s2516" xml:space="preserve">But it is very dif-<lb/>ficult to repreſent a Face diſtinctly as big as the <lb/>Life; </s> <s xml:id="echoid-s2517" xml:space="preserve">for when we would repreſent a Face in its <lb/>natural Bigneſs, ſuch a Convex Glaſs as is men-<lb/>tioned in Numb. </s> <s xml:id="echoid-s2518" xml:space="preserve">35. </s> <s xml:id="echoid-s2519" xml:space="preserve">muſt be uſed, and the Face <lb/>placed where the Board F is <anchor type="note" xlink:href="" symbol="*"/>. </s> <s xml:id="echoid-s2520" xml:space="preserve">But the ſaid <anchor type="note" xlink:label="note-0191-01a" xlink:href="note-0191-01"/> Face which then appears diſtinct enough, that ſo <lb/>the Perſon whereof it is the Repreſentation may <lb/>thereby be known, hath not its Lineaments ſuffi-<lb/>ciently enough repreſented, as to be followed by a <lb/>Painter as they ought, in order to keep the true Re- <pb o="116" file="0192" n="221" rhead="An ESSAY on"/> ſemblance. </s> <s xml:id="echoid-s2521" xml:space="preserve">The Reaſon of which is, that the Li-<lb/>neaments appear lively and diſtinct within the <lb/>Machine, when the Re-union of the Rays pro-<lb/>ceeding from a given Point in the Face, happens <lb/>exactly upon the Paper in one Point only: </s> <s xml:id="echoid-s2522" xml:space="preserve">But <lb/>the leaſt Diſtance that one Point is more than <lb/>another from the Convex Glaſs, (when the Di-<lb/>ſtance of the Face from the Glaſs is ſo ſmall, as <lb/>it muſt be to repreſent it in its natural Bigneſs) <lb/>ſo alters the Place of the ſaid Re-union, that for <lb/>different Parts of the Face, thoſe Places of Re-<lb/>union will differ about two Inches and a half. <lb/></s> <s xml:id="echoid-s2523" xml:space="preserve">Whence it is no wonder that all the Lineaments <lb/>be not repreſented as could be wiſhed; </s> <s xml:id="echoid-s2524" xml:space="preserve">ſince in <lb/>all Diſtances choſen, there will be always a great <lb/>many Rays, whoſe Re-union will fall above an <lb/>Inch beſides the Paper. </s> <s xml:id="echoid-s2525" xml:space="preserve">The Confuſion ariſing <lb/>from this Diverſity, though not being very di-<lb/>ſtinguiſhable by the Eye, yet is prejudicial, and <lb/>hinders our getting the exact Reſemblance of the <lb/>Face. </s> <s xml:id="echoid-s2526" xml:space="preserve">We have obſerved this, in order to give <lb/>an exact Idea of the Goodneſs of this Machine, <lb/>in equally ſhewing wherein it may be really <lb/>uſeful, and wherein its apparent Uſefulneſs is <lb/>ſubject to an Error rather diſcovered by Experi-<lb/>rience than Reaſon.</s> <s xml:id="echoid-s2527" xml:space="preserve"/> </p> <div xml:id="echoid-div388" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0191-01" xlink:href="note-0191-01a" xml:space="preserve">35.</note> </div> <p> <s xml:id="echoid-s2528" xml:space="preserve">38. </s> <s xml:id="echoid-s2529" xml:space="preserve">We muſt not forget in all the precedent <lb/>Problems, to examine the Aperture the Convex <lb/>Glaſs ought to have; </s> <s xml:id="echoid-s2530" xml:space="preserve">for although we cannot re-<lb/>duce this Aperture to a fixed Meaſure, yet it is <lb/>proper to obſerve the following Remarks. </s> <s xml:id="echoid-s2531" xml:space="preserve">1. </s> <s xml:id="echoid-s2532" xml:space="preserve">The <lb/>Convex Glaſs may commonly have the ſame <lb/>Aperture, as we would give a Perſpective Glaſs, <lb/>having the ſaid Glaſs for its Object Glaſs. </s> <s xml:id="echoid-s2533" xml:space="preserve">2. </s> <s xml:id="echoid-s2534" xml:space="preserve">When <lb/>Objects are very much enlightned, the ſaid A-<lb/>perture muſt be leſſened; </s> <s xml:id="echoid-s2535" xml:space="preserve">and contrariwiſe, when <lb/>they are expoſed to a weaker Light, it muſt be <lb/>made greater; </s> <s xml:id="echoid-s2536" xml:space="preserve">and when any Repreſentation is <pb o="117" file="0193" n="222" rhead="PERSPECTIVE."/> to be copyed, the Convex Glaſs muſt have the <lb/>leaſt Aperture poſſible; </s> <s xml:id="echoid-s2537" xml:space="preserve">but yet with this Cau-<lb/>tion, that the Light coming into the Machine, <lb/>muſt not be too much extenuated. </s> <s xml:id="echoid-s2538" xml:space="preserve">From theſe <lb/>Obſervations it is manifeſt, that we ought to be <lb/>provided with ſeveral round Pieces of Tin or <lb/>thin Braſs, having round Holes of different big-<lb/>neſſes therein, in order to give a neceſſary A-<lb/>perture to the Glaſs; </s> <s xml:id="echoid-s2539" xml:space="preserve">or Holes of different big-<lb/>neſſes may be made in a long thin Piece of Braſs <lb/>which may ſlide upon the Convex Glaſs: </s> <s xml:id="echoid-s2540" xml:space="preserve">Or <lb/>elſe, we may uſe a round Plate, having Holes <lb/>of different bigneſſes therein, which turning a-<lb/>bout its Centre, may bring any deſired Hole for <lb/>the Glaſſes Aperture.</s> <s xml:id="echoid-s2541" xml:space="preserve"/> </p> </div> <div xml:id="echoid-div390" type="section" level="1" n="210"> <head xml:id="echoid-head230" style="it" xml:space="preserve">A Deſcription of the Second Machine.</head> <p> <s xml:id="echoid-s2542" xml:space="preserve">39. </s> <s xml:id="echoid-s2543" xml:space="preserve">This Machine is a kind of Box, the Side <lb/> <anchor type="note" xlink:label="note-0193-01a" xlink:href="note-0193-01"/> A C G B being open, whoſe breadth B D, and <lb/>height AB, are equal, each being about 18 Inches: <lb/></s> <s xml:id="echoid-s2544" xml:space="preserve">Its greateſt width F B, is 10 Inches, and the Side <lb/>E F is ſloping, ſo that A E is but about 6 Inches.</s> <s xml:id="echoid-s2545" xml:space="preserve"/> </p> <div xml:id="echoid-div390" type="float" level="2" n="1"> <note position="right" xlink:label="note-0193-01" xlink:href="note-0193-01a" xml:space="preserve">Fig. 78.</note> </div> <p> <s xml:id="echoid-s2546" xml:space="preserve">40. </s> <s xml:id="echoid-s2547" xml:space="preserve">The Frame G ſlides at the Bottom of the <lb/>ſaid Box, in which the Paper is <anchor type="note" xlink:href="" symbol="*"/> faſtned.</s> <s xml:id="echoid-s2548" xml:space="preserve"/> </p> <note symbol="*" position="right" xml:space="preserve">15.</note> <p> <s xml:id="echoid-s2549" xml:space="preserve">41. </s> <s xml:id="echoid-s2550" xml:space="preserve">There is a round Hole in the Middle of <lb/>the Top of the Box, in which the Cylinder <lb/>carrying the Convex Glaſs ſcrews.</s> <s xml:id="echoid-s2551" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/></s> </p> <note symbol="*" position="right" xml:space="preserve">9.</note> <p> <s xml:id="echoid-s2552" xml:space="preserve">42. </s> <s xml:id="echoid-s2553" xml:space="preserve">The two Sticks H I and L M, ſlide in four <lb/>little Iron Staples fixed to the Inſide of the Top <lb/>of the Box, like thoſe mentioned in Numb. </s> <s xml:id="echoid-s2554" xml:space="preserve">7. <lb/></s> <s xml:id="echoid-s2555" xml:space="preserve">Theſe Sticks come about two Feet without the <lb/>Box, and the Diſtance of their Extremities I and <lb/>M, is equal, or ſomething greater than the <lb/>Length of the Box. </s> <s xml:id="echoid-s2556" xml:space="preserve">Their Uſe is to hang a black <lb/>Cloth upon, which is faſtned to the Three Sides, <lb/>B A, A C and C D, of the opening of the Box, <lb/>that ſo the Box may be darkned, when Objects <pb o="118" file="0194" n="223" rhead="An ESSAY"/> are to be repreſented upon the Paper in the Frame <lb/>G.</s> <s xml:id="echoid-s2557" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2558" xml:space="preserve">43. </s> <s xml:id="echoid-s2559" xml:space="preserve">There are two pieces of Wood, (one of <lb/>which is repreſented in Fig. </s> <s xml:id="echoid-s2560" xml:space="preserve">77.) </s> <s xml:id="echoid-s2561" xml:space="preserve">ſerving to <lb/>ſuſtain the Box upon its Support or Foot. </s> <s xml:id="echoid-s2562" xml:space="preserve">One <lb/>of theſe Pieces may be faſtned to one Side of the <lb/>Support, and the other to the other Side thereof, <lb/>by means of four Iron Pins, two of which go thro <lb/>the Holes N and P in the Side of the Support, <lb/>and the Holes T and V in the Piece R, and the <lb/>other two in like manner, through Holes made <lb/>in the other Side of the Support, and the other <lb/>Piece, when we have Mind the Bottom of the <lb/>Box ſhould be parallel to the Horizon; </s> <s xml:id="echoid-s2563" xml:space="preserve">but when <lb/>the Box is to be a little inclin’d, that Pin going <lb/>through the Hole P, muſt be put through the <lb/>Hole O, in the Piece R. </s> <s xml:id="echoid-s2564" xml:space="preserve">Underſtand the ſame <lb/>of the other Piece.</s> <s xml:id="echoid-s2565" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2566" xml:space="preserve">44. </s> <s xml:id="echoid-s2567" xml:space="preserve">We are ſometimes obliged to ſet the Box <lb/>forwarder on its Support; </s> <s xml:id="echoid-s2568" xml:space="preserve">and this is done, in <lb/>uſing the Holes Q and S, inſtead of N and P. </s> <s xml:id="echoid-s2569" xml:space="preserve">It <lb/>is likewiſe ſomething neceſſary to incline the <lb/>Box a little backwards; </s> <s xml:id="echoid-s2570" xml:space="preserve">which may be done, by <lb/>putting the Pin in S, into the Hole X, made in <lb/>a Piece of Wood faſtned to the Back-ſide of the <lb/>Box, and the correſpondent Pin on the other Side, <lb/>into another Hole made on the other Side of the <lb/>ſaid Piece.</s> <s xml:id="echoid-s2571" xml:space="preserve"/> </p> <p> <s xml:id="echoid-s2572" xml:space="preserve">45. </s> <s xml:id="echoid-s2573" xml:space="preserve">The Box ♈ ſlides upon the Top of the <lb/>Machine, and is like that already deſcribed, <anchor type="note" xlink:href="" symbol="*"/> but <anchor type="note" xlink:label="note-0194-01a" xlink:href="note-0194-01"/> with this Difference only, that it is leſſer. </s> <s xml:id="echoid-s2574" xml:space="preserve">On <lb/>the Top of the Box are two little Staples Z Z, <lb/>in which a Ruler ſlides, having a Mirrour faſtned <lb/>to it, in the Manner as is mentioned Numb. </s> <s xml:id="echoid-s2575" xml:space="preserve">13. <lb/></s> <s xml:id="echoid-s2576" xml:space="preserve">and ſo by this Means the ſaid Mirrour may be <lb/>put in the ſame Situation, as that in the Figure <lb/>of the firſt Machine it hath in H.</s> <s xml:id="echoid-s2577" xml:space="preserve"/> </p> <div xml:id="echoid-div391" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0194-01" xlink:href="note-0194-01a" xml:space="preserve">10, 11, <lb/>13.</note> </div> <pb o="119" file="0195" n="224" rhead="on PERSPECTIVE."/> <p> <s xml:id="echoid-s2578" xml:space="preserve">46. </s> <s xml:id="echoid-s2579" xml:space="preserve">When we have a mind to remove this <lb/>Machine from one place to another, we lay the <lb/>Box B E C upon the croſs Pieces 2, 3, 4, 5, with <lb/>its opening A B C upwards; </s> <s xml:id="echoid-s2580" xml:space="preserve">then we put the <lb/>little Box Y, the Ruler and Mirrour (mentioned <lb/>Numb. </s> <s xml:id="echoid-s2581" xml:space="preserve">13.) </s> <s xml:id="echoid-s2582" xml:space="preserve">the black Cloth, and the two Sticks <lb/>M L and I H, all into the ſaid great Box; </s> <s xml:id="echoid-s2583" xml:space="preserve">and <lb/>afterwards partly cover it by the Frame G, <anchor type="note" xlink:href="" symbol="*"/> <anchor type="note" xlink:label="note-0195-01a" xlink:href="note-0195-01"/> which is ſuſtained by two very thin Rulers, and <lb/>then by another little Board, when the Frame is <lb/>not big enough. </s> <s xml:id="echoid-s2584" xml:space="preserve">The whole Machine thus taken <lb/>to pieces, will take up no more room than the <lb/>Support itſelf doth: </s> <s xml:id="echoid-s2585" xml:space="preserve">and ſo it is very eaſy to re-<lb/>move from Place to Place. </s> <s xml:id="echoid-s2586" xml:space="preserve">Now when Objects <lb/>are to be repreſented in this Machine, it muſt be <lb/>put together again, as per Figure; </s> <s xml:id="echoid-s2587" xml:space="preserve">and the black <lb/>Cloth, for a Perſon to put his Head under, hang-<lb/>ing upon the Sticks, and faſtned to the Sides of <lb/>the opening A B, A C, and C D.</s> <s xml:id="echoid-s2588" xml:space="preserve"/> </p> <div xml:id="echoid-div392" type="float" level="2" n="3"> <note symbol="*" position="right" xlink:label="note-0195-01" xlink:href="note-0195-01a" xml:space="preserve">40.</note> </div> </div> <div xml:id="echoid-div394" type="section" level="1" n="211"> <head xml:id="echoid-head231" style="it" xml:space="preserve">The Uſe of this Machine.</head> <p> <s xml:id="echoid-s2589" xml:space="preserve">47. </s> <s xml:id="echoid-s2590" xml:space="preserve">The Uſe of this Second Machine is the <lb/>fame as that of the Firſt; </s> <s xml:id="echoid-s2591" xml:space="preserve">but it ought to be ob-<lb/>ſerved, that when we incline <anchor type="note" xlink:href="" symbol="*"/> the Machine, the <anchor type="note" xlink:label="note-0195-02a" xlink:href="note-0195-02"/> Angle of Inclination of the Mirrour and Hori-<lb/>zon muſt be made leſs, by half the Inclination of <lb/>the Bottom of the Box; </s> <s xml:id="echoid-s2592" xml:space="preserve">and when the Machine <lb/>is ſomewhat inclin’d backwards, <anchor type="note" xlink:href="" symbol="*"/> the ſaid Angle <anchor type="note" xlink:label="note-0195-03a" xlink:href="note-0195-03"/> muſt be made greater by a like half. </s> <s xml:id="echoid-s2593" xml:space="preserve">You muſt <lb/>likewiſe obſerve, that when Objects are to be <lb/>repreſented for a perpendicular Picture, the Ma-<lb/>chine muſt be placed according to the former <lb/>Part of Numb. </s> <s xml:id="echoid-s2594" xml:space="preserve">44. </s> <s xml:id="echoid-s2595" xml:space="preserve">Prints muſt be faſtned to a <lb/>Board entirely ſeparated from the Machine, which <lb/>Board muſt be ſet upon a Support, that may <lb/>conveniently be moved backwards or forwards, <lb/>according to Neceſſity.</s> <s xml:id="echoid-s2596" xml:space="preserve"/> </p> <div xml:id="echoid-div394" type="float" level="2" n="1"> <note symbol="*" position="right" xlink:label="note-0195-02" xlink:href="note-0195-02a" xml:space="preserve">43.</note> <note symbol="*" position="right" xlink:label="note-0195-03" xlink:href="note-0195-03a" xml:space="preserve">44.</note> </div> <pb o="120" file="0196" n="225" rhead="An ESSAY, &c."/> </div> <div xml:id="echoid-div396" type="section" level="1" n="212"> <head xml:id="echoid-head232" style="it" xml:space="preserve">A Demonſtration of the Inclination of the Looking-<lb/>Glaſs.</head> <p> <s xml:id="echoid-s2597" xml:space="preserve">48. </s> <s xml:id="echoid-s2598" xml:space="preserve">Let A B be a Ray, proceeding from ſome <lb/> <anchor type="note" xlink:label="note-0196-01a" xlink:href="note-0196-01"/> Point of an Object. </s> <s xml:id="echoid-s2599" xml:space="preserve">We are to demonſtrate, <anchor type="note" xlink:href="" symbol="*"/> <anchor type="note" xlink:label="note-0196-02a" xlink:href="note-0196-02"/> if the Line D I hath the Inclination given to a <lb/>Picture, and the Looking-Glaſs G H hath the In-<lb/>clination we have preſcribed, that the Angle <lb/>B a F, will be equal to the Angle B C D. </s> <s xml:id="echoid-s2600" xml:space="preserve">Now to <lb/>prove this, draw the Line F I parallel to the Ho-<lb/>rizon, then the two Angles I D F and D F I, of <lb/>the Triangle I D F, are together equal to the <lb/>Angle D I E; </s> <s xml:id="echoid-s2601" xml:space="preserve">but the Angle D F I, which is the <lb/>Inclination of the Looking-Glaſs, is equal <anchor type="note" xlink:href="" symbol="*"/> to <anchor type="note" xlink:label="note-0196-03a" xlink:href="note-0196-03"/> half the Angle D I E, leſs half the Angle I F a; <lb/></s> <s xml:id="echoid-s2602" xml:space="preserve">and conſequently it is leſs than the Angle F D I, <lb/>by the Quantity of the whole Angle I F a: </s> <s xml:id="echoid-s2603" xml:space="preserve"><lb/>Therefore if the Angle I F a be added to the An-<lb/>gle D F I, we ſhall have the Angle D F a, equal <lb/>to the Angle F D I: </s> <s xml:id="echoid-s2604" xml:space="preserve">Therefore the Angle <lb/>Fa B will be <anchor type="note" xlink:href="" symbol="*"/> likewiſe equal to the Angle BCD.</s> <s xml:id="echoid-s2605" xml:space="preserve"> <anchor type="note" xlink:label="note-0196-04a" xlink:href="note-0196-04"/> Which was to he demonſtrated.</s> <s xml:id="echoid-s2606" xml:space="preserve"/> </p> <div xml:id="echoid-div396" type="float" level="2" n="1"> <note position="left" xlink:label="note-0196-01" xlink:href="note-0196-01a" xml:space="preserve">Fig, 75.</note> <note symbol="*" position="left" xlink:label="note-0196-02" xlink:href="note-0196-02a" xml:space="preserve">19.</note> <note symbol="*" position="left" xlink:label="note-0196-03" xlink:href="note-0196-03a" xml:space="preserve">16. 47.</note> <note symbol="*" position="left" xlink:label="note-0196-04" xlink:href="note-0196-04a" xml:space="preserve">19.</note> </div> <p> <s xml:id="echoid-s2607" xml:space="preserve">In reaſoning nearly after the ſame Manner, <lb/>we demonſtrated what is mentioned <anchor type="note" xlink:href="" symbol="*"/> concer- <anchor type="note" xlink:label="note-0196-05a" xlink:href="note-0196-05"/> ning the Inclination of the Mirrour, when the <lb/>Box is inclin’d a little backwards.</s> <s xml:id="echoid-s2608" xml:space="preserve"/> </p> <div xml:id="echoid-div397" type="float" level="2" n="2"> <note symbol="*" position="left" xlink:label="note-0196-05" xlink:href="note-0196-05a" xml:space="preserve">47.</note> </div> </div> <div xml:id="echoid-div399" type="section" level="1" n="213"> <head xml:id="echoid-head233" xml:space="preserve">FINIS.</head> <figure> <image file="0196-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0196-01"/> </figure> <pb file="0197" n="226"/> <pb file="0197a" n="227"/> <figure> <caption xml:id="echoid-caption70" style="it" xml:space="preserve">Page 120<lb/>Plate. 30.<lb/><emph style="sc">Fig</emph>. 70.</caption> <variables xml:id="echoid-variables69" xml:space="preserve">X I F B H D D P O M P R C C C C C E E Q</variables> </figure> <pb file="0198" n="228"/> <pb file="0199" n="229"/> <pb file="0199a" n="230"/> <figure> <caption xml:id="echoid-caption71" style="it" xml:space="preserve">Plate 31<lb/>page 120<lb/>Fig. 71</caption> <variables xml:id="echoid-variables70" xml:space="preserve">D G C B A H F a I E</variables> </figure> <figure> <caption xml:id="echoid-caption72" style="it" xml:space="preserve">Fig. 72</caption> <variables xml:id="echoid-variables71" xml:space="preserve">P G C H A N B R Q M a F</variables> </figure> <figure> <caption xml:id="echoid-caption73" style="it" xml:space="preserve">Fig. 73</caption> <variables xml:id="echoid-variables72" xml:space="preserve">P G C H D N B I A R Q M a F</variables> </figure> <figure> <caption xml:id="echoid-caption74" style="it" xml:space="preserve">Fig. 74</caption> <variables xml:id="echoid-variables73" xml:space="preserve">G N B C H M a A</variables> </figure> <figure> <caption xml:id="echoid-caption75" style="it" xml:space="preserve">Fig. 75</caption> <variables xml:id="echoid-variables74" xml:space="preserve">D G B C A H F I E a</variables> </figure> <pb file="0200" n="231"/> <pb file="0201" n="232"/> <pb file="0201a" n="233"/> <figure> <caption xml:id="echoid-caption76" style="it" xml:space="preserve">page 120<lb/>Plate. 32.<lb/>Fig. 76.</caption> </figure> <figure> <caption xml:id="echoid-caption77" style="it" xml:space="preserve">Fig. 77.</caption> <variables xml:id="echoid-variables75" xml:space="preserve">R V T o</variables> </figure> <figure> <caption xml:id="echoid-caption78" style="it" xml:space="preserve">Fig. 78.</caption> <variables xml:id="echoid-variables76" xml:space="preserve">Z Z Y C M L I E A H D X G F B S Q P N 4 3 2</variables> </figure> <pb file="0202" n="234"/> <pb file="0203" n="235"/> <pb file="0204" n="236"/> <pb file="0205" n="237"/> </div></text> </echo>