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~~Press mark, 181 & ~~
~~28~~

~~Ent@in Gatalogue, ---~~

~~(1860)~~

AN ESSAY ON PERSPECTIVE.
~~
~~~~
~~~~
~~~~
~~

~~Written in French by ~~

William-James ‘s Ggravesande, Doctor of Laws and Philoſophy;

~~Profeſſor ~~
of Mathematicks and Aſtronomy at Leyden,

and Fellow of the Royal Society at London.

~~
~~

~~And now Tranſlated into Engliſh.~~
~~
~~

~~LONDON:~~
~~
~~

~~Printed for J. ~~
Senex, in Fleetſtreet; ~~W. ~~
Taylor, in Pater-Noſter-Row;

~~W. ~~
~~and J. ~~
Innys, in Ludgate-ſtreet;

~~J. ~~
Osborne, in Lombard-ſtreet; ~~and E. ~~
Ssymon, in Cornhill.

~~M ~~
DCC XXIV.~~
~~

MAX--INSTITUT FOR WISSEESCHICHTE Bibliothek

TO Mr. William Kent.
~~
~~~~
~~

SIR,

~~IN Dedicating this ~~
Tranſlation to you,

I have deſignedly

deviated from the general

Cuſtom obſerv’d by almoſt

all Dedicators, who make

choice of ſuch Patrons that

are Great and Rich, not at all

conſidering their Merit, or

whether they underſtand any

thing of what is offer’d to

them;

~~ſince I have inſcrib’d ~~
this Treatiſe of Perſpective to

one, whoſe daily Practice is

the very Art it ſelf, and whoſe

~~
Merit is undoubtedly excel-~~
lent, as evidently appears from

your own Works.

~~
~~

~~I ſhall likewiſe be particu-~~
lar with regard to the Manner

of the Offering;

~~being per-~~
ſuaded that Flattery, or even

due Praiſe, which are the com-

mon Topicks handled in De-

dications, muſt needs be offen-

ſive to an Ingenious Perſon;

~~and ſo I ſhall be ſilent on theſe ~~
Heads;

~~and only crave your ~~
Acceptance and Protection of

~~Your Humble Servant,~~

~~E. ~~
Stone.~~
~~

The AuTHOR’S PREFACE.
~~
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~~
The PREFACE.
~~
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~~~~
~~~~
~~~~
~~
The PREFACE.
~~
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~~~~
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~~~~
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~~
The PREFACE.
~~
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~~

~~THE Reader will wonder, per-~~
haps, to find me entring into

a Path, which ſeems to have

been too much trodden already;

~~and eſteem as uſeleſs a New ~~
Eſſay, on an Art, whoſe Sub-

ject (one would think) ſhould have been

long before this Time exhauſted;

~~ſince there ~~
have been ſo many Perſons, who have writ-

ten on the ſame.

~~
~~

~~The Name of Perſpective now ſeems to ~~
ſound unple aſant in the Ears of the Publick

Enemies of Repetition;

~~and it may be look d ~~
upon as a Piece of Inadvertency, to venture

to treat again on that ſame Subject.

~~Yet, ~~
notwithst anding this, I deſire the Render to

ſuſpend his Cenſure, until he h{as} heard the

ing Work.

~~
~~

~~Having ſome Years ago buſied my ſelf in ~~
drawing Figures by the common Methods, I

found out ſeveral Compendiums;

~~which, by ~~
diligent working, naturally enough fall in

one’s way, without being entirely beholden

to the Induſtry of others:

~~And theſe firſt ~~
Succeſſes made me hope for others more

conſiderable;

~~and ſo I thought that a more ~~
narrow Inſpection into the Theory of Per-

ſpective, might furniſh me with Rules more

general, for making the Practice thereof

eaſy.

~~
~~

~~I then thought upon ſeveral Methods to this ~~
Purpoſe;

~~but, being ſuſpicious that they were ~~
not ſo eaſy {as} they appear’d, I have try’d

their Goodneſs, by exactly applying them to

different Subjects;

~~and have nicely examin’d ~~
all the Caſes, and order’d it ſo {as} not to be de-

ceiv’d by certain Operations, which at firſt

ſeem eaſy, but, when put in Practice, are

quite otherwiſe.

~~Moreover, at convenient ~~
Times, I look’d over the beſt Part of the

Authors of this kind, (whoſe Number is

increas’d very much, without any manner of

Neceſſity) ſome of which being advantagi-

ouſly diſtinguiſh’d among the Crowd, have

been very uſeful to me:

~~But I dare affirm, ~~
there are but a very few that give a new

Turn to the practical Part of Perſpective.

~~
~~

~~Some content themſelves with the bare ~~
Explication of the Theory, and have left to

the Reader the Trouble of applying the ſame

to Practice;

~~or elſe have given only ſome of ~~
the common Operations, and entertain us

with general Reflections on Painting;

~~which ~~
are indeed curio{us}, but foreign to my Pur-

poſe:

~~For I intend not to make a Man a ~~
Painter, but to render the Uſe and Exerciſe

of Perſpective eaſy to him.

~~
~~

~~Other Authors, which (according to the ~~
Bulk of their Works) might be thought to

have more carefully treated of the practical

Part of Perſpective, do indeed at firſt lay

down ſome general Rules, common to them

all;

~~but are nothing the eaſier for having ~~
paſs’d thro’ ſo many Hands;

~~and that, in-~~
deed, becauſe they have not endeavour’d to

make them ſo.

~~They thought that all Ob-~~
jects might be thrown into Perſpective by

theſe Rules, and therefore it would be uſe-

leſs to ſearch after others;

~~and judg’d it ~~
more neceſſary to ſhew Painters the Applica-

tion of them to an infinite Number of parti-

cular Examples;

~~tho’ that Application, at ~~
moſt, is but repeating over again the Uſe of

the Rules already preſcrib’

~~But what Ad-~~
vantage can Painters gain from hence, if

they do not well underſtand general Opera-

tions?

~~And if they do, I cannot conceive of
~~
ples will be to them.

~~
~~

~~I believ’d then, that I might be able to ~~
treat of this Art after another Manner:

~~And altho’ I know my ſelf to be much infe-~~
rior to ſeveral of thoſe who have written on

this Subject;

~~yet I am of Opinion, that if ~~
Perſpective ſhould loſe any thing by me, on

account of my want of fudgment;

~~yet that ~~
may be regain’d, perhaps, (and with In-

tereſt too) by my great Diligence in this Bu-

ſineſs.

~~
~~

~~I have conſider’d, moreover, that the te-~~
dious Particulars, inherent to the Subject on

which I have choſen to write, will always

hinder Genius’s capable of great Matters,

from undertaking a Subject ſo little worthy

their Endeavours, and ſo barren of great

Diſcoveries.

~~
~~

~~Thus, hoping, on one hand, to give a new ~~
Turn to the Practice of Perſpective, and

make it eaſier;

~~and being perſuaded, on the ~~
other, that more learned Perſons than my ſelf

will not take this Trouble upon them;

~~I ven-~~
ture to publiſh this ſmall Work, and expoſe

it to the Taſte of the Learned World;

~~from ~~
whom I expect no other Praiſe, but what

may reaſonably be claim’d by an aſſiduo{us} Ap-

plication.

~~
~~

~~The Practice of Perſpective may be made ~~
eaſy, by the Three following Things in this

Treatiſe:

~~Viz. ~~
~~1. ~~
~~In giving ſeveral new ~~
and eaſier Ways (than thoſe commonly uſed)

of ſolving the moſt general Problems upon

which the whole Practice is founded:

~~And the ~~
Reaſon why we have laid down ſeveral So-

lutions, is, becauſe the ſame Way is not always

equally convenient in all Caſes;

~~whence it is ~~
neceſſary to have ſeveral, that ſo we may

chuſe one beſt ſuiting our Purpoſe.

~~2. ~~
~~The ~~
general Methods, which have been us’d hi-

therto, not being practicable on ſome particu-

lar Occaſions;

~~to remedy this, we have ad-~~
ded others to them;

~~which are indeed more ~~
difficult, but (in ſome Caſes) there is an ab-

ſolute Neceſſity for them.

~~3. ~~
~~When it is ve-~~
ry difficult to reſolve a particular Problem,

by means of general ones;

~~then we have ~~
thought it convenient to give a particular So-

lution thereof.

~~
~~

~~By this means, the Study of Perſpective ~~
becomes indeed more difficult;

~~but the Diſ-~~
advantage is well recompenſed by the Faci-

cility of the Practice, which we have entire-

ly had in view.

~~It is true, that a few ge-~~
neral Rules do not ſo much burthen the Me-

mory;

~~but when one has ſeveral general ~~
ones, and alſo particular ones, by them we

can abridge Matters.

~~And this Method be-
~~
Application, does afterwards ſave a great

many Hours Study, in an Art that always

appears difficult enough.

~~
~~

~~A Painter, in a ſhort Time, may learn this ~~
Work, and make the Rules thereof familiar to

him:

~~And if this Study be repeated from ~~
time to time, for a few Days, he will find

the Benefit thereof, in diminiſhing his Labour

and Trouble.

~~
~~

~~But, that any one himſelf may ſee what ~~
I promiſe in this Eſſay;

~~take the following ~~
ſhort Abſtract thereof.

~~It is divided into ~~
Nine Chapters:

~~The Firſt, being as an In-~~
troduction to the reſt, ſhews the Uſefulneſs

of Perſpective, and gives you the Definiti-

ons of the Terms neceſſary for underſtanding

this Treatiſe.

~~
~~

~~The whole Theory is contain’d in the Se-~~
cond Chapter:

~~Where, what has been found ~~
moſt uſeful in that Matter, is therein re-

duced to Three general Theorems;

~~viz. ~~
~~the ~~
firſt, ſecond, and fourth:

~~All the reſt is de-~~
duced from them, by way of Corollary.

~~To ~~
theſe Theorems, already known, are added

ſome new ones, ſerving for the Demonſtra-

tion of ſome neceſſary Propoſitions.

~~Perhaps ~~
it might be wiſh’d, that I had ſhewn the Way

that led me to the Truths which I diſcover:

~~
~~
~~but it often ~~
would have been very long and troubleſome.

~~In Geometry, the eaſieſt and ſhorteſt Way, is ~~
not always that which leads to Diſcoveries.

~~
~~

~~In the following Chapter, the Practice of Per-~~
ſpective upon the perſpective Plane, or Picture,

conſider’d as upright, is explain’d:

~~Wherein, ~~
among the different Ways laid down for the

Solution of general Problems, you will find

ſome effected by a Ruler only;

~~ſo that after ~~
ſome Preparations, all Kinds of Objects may

be drawn without Compaſſes, and that eaſier

than by the common Operations.

~~In that ~~
Problem, to find the Appearance of a Point

out of the Geometrical Plane, it is commonly

conſider’d as the Extremity of a Perpendicu-

lar, whoſe Repreſentation muſt first be found,

before that of the Point can be had.

~~But here ~~
we avoid this round-about Way, and ſhew how

to find the Appearance of the Point given,

without being obliged to find the Perſpective

of its Seat.

~~
~~

~~As to the Appearance of a Cone and Cy-~~
linder, we determine the viſible Portions of

the Baſe, and by this means avoid the uſe-

leſs Operations which the common Way is

ſubject to.

~~It is very difficult, if not impoſ-~~
ſible, to throw a Sphere into Perſpective, by

means of general Problems;

~~and in the Re-~~
preſentation of the Torus of a Column, it is

~~Whence we are obliged to ~~
give particular Methods for the Reſolution of

theſe two Problems.

~~
~~

~~The reſt of the Third Chapter is concern-~~
ing Inclin’d Lines, and how to find their Ap-

pearance by the Accidental Point.

~~
~~

~~The Fourth Chapter ſhews the Manner of ~~
working on a perſpective Plane, to be view’d

afar off, very obliquely, or which muſt ſtand

in an high Place.

~~Theſe different Situations ~~
require new Rules:

~~For if the common Me-~~
thods were to be uſed here, the perſpective

Plane muſt be ſo large, as that it would be

impoſſible to work upon it.

~~
~~

~~In the Two following Chapters, we treat ~~
of the perſpective Plane, conſider’d as Hori-

zontal, or Inclin’d:

~~Where there are laid ~~
down ſeveral general Ways of working;

~~which, together with thoſe of the foregoing ~~
Chapters, will ſuffice (in my Opinion) for

throwing any Object whatſoever into Perſpe-

ctive, with Eaſe enough.

~~
~~

~~In the Seventh Chapter, which treats of ~~
Shadows, there is nothing particular, but

what may be ſeen elſewhere:

~~But that lit-~~
tle we have ſaid concerning this Matter, is

enough for giving an Idea of them, which

the Reading of what goes before will make

eaſy.

~~
~~

~~In the Eighth Chapter, are laid down ~~
ſome Mechanical ways for making the Uſe of

Perſpective eaſy, by means of Rulers and

Threads, (eaſily to be gotten by any body,

and not difficult to be put in practice) they

being eaſier to uſe than any of the Inſtru-

ments that have hitherto been invented for

this Purpoſe.

~~
~~

~~The laſt Chapter ſhews the Uſefulneſs of ~~
Perſpective in Dialling.

~~
~~

~~Such is the Plan of this ſinall Work; ~~
~~wherein I have not ſo much endeavour’d to ~~
advance Curioſities, as Things of real Uſe;

hoping that, without making a Shew of Skill

ill beſtow’d, I ſhall make my Book good enough,

if by its Uſe I make it neceſſary.

~~For which ~~
Reaſon, I have endeavour’d to lay down the

whole, ſo as to be underſtood by thoſe who

have only read the Elements of Euclid.

~~And ~~
tho’ I have deviated from this Rule in ſome

few Place;

~~they are printed in Italick, that ~~
ſo they may be paſs’d over without any Hin-

drance to the Learner.

~~
~~

~~Here I muſt not forget to mention, that in ~~
Reviſing this Eſſay, I had the Happineſs of

meeting with an able Painter;

~~who has ſe-~~
r

ſion, neceſſary to be known, among which,

~~He has car-~~
ried the Matter farther than could have been

reaſonably expected from one ignorant of

Mathematicks;

~~and I am indebted to him ~~
for ſeveral Obſervations, which I my ſelf

ſhould perhaps have never thought on.

~~
~~

ERRATA.
~~
~~~~
~~~~
~~

~~Page 74. ~~
~~1. ~~
~~17. ~~
~~for x, r. ~~
~~in X.~~
~~
~~

~~P. ~~
~~83. ~~
~~1. ~~
~~28. ~~
~~for that, r. ~~
~~that for.~~
~~
~~

~~P. ~~
~~88. ~~
~~1. ~~
~~8. ~~
~~for Tube, r. ~~
~~Table.~~
~~
~~

AN ESSAY ON PERSPECTIVE.
CHAP. I. Definitions.
~~
~~
An ESSAY
~~
~~
Fig. 1.
~~
~~~~
~~
on PERSPECTIVE.
~~
~~~~
~~
Def. 1.
Fig. 1.
~~
~~
Def. 2.
Fig. 2.
~~
~~
Def. 3.
~~
~~
Def. 4.
~~
~~
Def. 5.
~~
~~
Def. 6.
~~
~~
Def. 7.
~~
~~
Def. 8.
~~
~~~~
~~
Def. 9.
~~
~~
Def. 10.
on PERSPECTIVE.
~~
~~
Def. 11.
~~
~~
Def. 12.
~~
~~
Def. 13.
~~
~~
Def. 14.
~~
~~
Def. 15.
~~
~~
Def. 16.

~~1. ~~
PERSPECTIVE teaches us the Manner of Delineating by

Mathematical Rules;

~~that is, it ~~
ſhews us how to draw geometri-

cally upon a Plane, the Repre-

ſentations of Objects according

to their Dimenſions and different Situations;

~~in ~~
ſuch manner, that the ſaid Repreſentations pro-

duce the ſame Effects upon our Eyes, as the Ob-

jects whereof they are the Pictures.

~~
~~

~~In order to underſtand well how Mathema-~~
ticks may be apply’d to Drawing;

~~let us ſup-~~
poſe a Man A, viewing an Object;

~~and be-~~

imagine a tranſparent Plane C.

~~Suppoſe more-~~
over, that Lines be drawn upon this Plane, as

in D, which cover the Bounds of the Object B

in reſpect of the Spectator A, and each Part that

he ſees thereof.

~~Now, ſince all Objects are ſeen ~~
by the Rays of Light coming from every of

their Points, and terminating at the Eye, and

not otherwiſe;

~~and ſince that here all the Rays ~~
proceeding from the Object B, likewiſe paſs

thro’ every Point of the Repreſentation D;

~~it is ~~
manifeſt, that this Repreſentation will have the

ſame Effect upon the Spectator’s Eye, as the

ſaid Object B hath.

~~Now, by means of Geo-~~
metry, we can find the Points of the Figure D,

on the Plane C, placed in a given Situation,

thro’ which the Rays coming from the Object B

to the Eye of the Spectator A, do paſs;

~~and ~~
theſe Points are the Interſections of the Rays

and the Plane.

~~Alſo, (as others have very well ~~
obſerv’d) a Perſpective Plane, or Picture in Paint-

ing, may be conceiv’d as a Window, upon which

the Objects ſeen thro’ it are repreſented.

~~
~~

~~Now, without Mathematicks, this Repreſen-~~
tation cannot be well found:

~~For when Objects ~~
are drawn by only viewing, or looking at them;

~~their true Repreſentations after this way, will ~~
be very often miſs’d on;

~~whereas, by Geome-~~
try, we can always obtain them.

~~
~~

~~This Obſervation only, is ſufficient to eſta-~~
bliſh the Neceſſity of Perſpective:

~~Tho’ there are ~~
ſome Painters, who (according to the common

Maxim) affirm, That what they do not know

of this Art, is not worth the Pains of learn-

ing.

~~
~~

~~Hitherto I have endeavour’d to give an Idea ~~
of Perſpective in general:

~~But there is yet ano-~~
ther particular Signification of this Word, which

it is neceſſary ſhould be explain’d, as well as

the other Terms of the Art, which are laid down

in the following Definitions;

~~and which every ~~
one, that intends to underſtand this Treatiſe,

ought to be well acquainted with.

~~
~~

~~2. ~~
~~The Perſpective, Repreſentation, or Appearance of ~~

mous) * is the Figure which the Rays, by which an

Object is perceiv’d, form in paſſing thro’ the tranſpa-

rent Plane:

~~And the Perſpective of a Point, is the ~~
Interſection of a Ray proceeding from that Point,

and the tranſparent Plane.

~~Which Interſection is ~~
a Point:

~~As the Figure D in the tranſparent ~~

~~and ~~
the Point e, in the ſame Plane, is the Perſpective

of the Point E, in that Object.

~~
~~

~~The Plane parallel to the Horizon, upon which ~~

Objects that he views, is call’d the Geometrical Plane.

~~As A B C D.~~
~~
~~

~~A Perſpective Plane [or Picture] is that which ~~

which the Objects are drawn:

~~As F G R T. ~~
~~This ~~
is commonly perpendicular to the Geometrical

Plane, and conſequently to the Horizon;

~~becauſe ~~
Pictures have generally this Situation:

~~But yet ~~
it may be ſometimes inclin’d, and even parallel

to the Geometrical Plane, according as one

would diſpoſe the Deſign, or Picture that we are

working.

~~And for this Reaſon, in the follow-~~
ing Chapter, we have laid down General Theo-

rems, and their Corollaries, agreeing to all theſe

Picture] which ought to be well obſerv’d.

~~
~~

~~The Interſection of the Perſpective Plane and the ~~

~~As ~~
F G.

~~
~~

~~The different Situation of the Eye, alters the ~~
Repreſentation of Objects in the perſpective Plane;

~~for the Rays proceeding from the Object, and ~~
concurring in ſome other Point, will likewiſe fall

upon the Perſpective Plane in different Places.

~~And ~~
for determining this Situation of the Eye, in re-

~~A Plane parallel to the Horizon, paſſing thro’ the ~~

~~and this is ~~
call’d the Horizontal Plane:

~~As OMVNL.~~
~~
~~

~~The Interſection of this Plane and the Perſpective ~~

~~As M V N.~~
~~
~~

~~The Perpendicular drawn from the Eye to the Ho-~~

~~As O V.~~
~~
~~

~~The Point V, wherein the ſaid Perpendicular meets ~~

cipal Point.

~~
~~

~~Note, There is a Perpendicular, let fall from ~~
the Eye upon the Geometrical Plane, meaſuring

the Height of the Eye.

~~
~~

~~The Point S, wherein the ſaid Perpendicular meets ~~

~~
~~

~~The Plane paſſing thro’ the aforeſaid Perpendicular, ~~

~~As SOLI.~~
~~
~~

~~The Interſection V H of this Plane, and the Per-~~

~~
~~

~~And S H I, the Interſection of it, and the Geo-~~

~~
~~

~~Points of Diſtance, are two Points in the Hori-~~

by the length of the principal Ray;

~~as MN.~~
~~
~~

~~The Geometrical Line, is a Line, that paſſes ~~

Line, as A B.

~~
~~

~~The Seat of an Object, is the Concurrence of Per-~~

Geometrical Plane, and the ſaid Plane.

~~
~~

~~The Direction of a Line inclined to the Geometri-~~

another Plane perpendicular thereto, paſſing through

the ſaid inclined Line.

~~
~~

CHAP. II.
The Theory of Perſpective.
Lemma.
~~
~~
Fig. 3.
An ESSAY

~~3. ~~
~~THE Perſpective, or Appearance of a ~~
Right Line, as A B, which being con-

tinued, does not paſsthrough the Eye O, is like-

~~For the Rays, by which the ~~
Line A B is perceived, form a Plane cutting

the perſpective Plane;

~~and the common Section ~~
of theſe two Planes is a right Line, as a b.

~~
~~

~~4. ~~
~~The Repreſentation of a Line Parallel to the ~~

the Repreſentation.

~~
~~

~~Let A B be a Line Parallel to the perſpective ~~
Plane;

~~we are to prove that a b its Repreſentati-~~
on is Parallel thereto.

~~
~~

~~Theſe two Lines A B and a b, will never ~~
meet each other, becauſe a b is in the perſpective

Plane, and A B is ſuppoſed parallel to the ſaid

Plane.

~~But they are alſo in one and the ſame ~~
Plane, becauſe a b is the Interſection of the per-

ſpective Plane, and the Plane O A B, paſſing

through the Eye and the Line A B;

~~and there-~~
fore they are parallel between themſelves:

~~Which ~~
was to be demonſtrated.

~~
~~

~~5. ~~
~~The Appearance of a Line, parallel to the baſe ~~
Line, is alſo parallel to the ſaid baſe Line.

~~
~~

~~For the baſe Line, and the Repreſentation ~~
being parallel to the ſame Line, are parallel to

one another.

~~
~~

~~6. ~~
~~The Repreſentation of a Line parallel to the ~~
vertical Line, is parallel to the ſaid vertical Line, and

conſequently perpendicular to the baſe Line.

~~This ~~
is demonſtrated as in the laſt Corollary.

~~
~~

~~7. ~~
~~The Appearances of Lines parallel to the per-~~
ſpective Plane, and equally inclin’d the ſame Way upon

Line, equal to thoſe Angles that the Lines whereof

they are the Appearances, make with the Parallels

to the baſe Line, which cut them;

~~and conſequently ~~
the ſaid Appearances are parallel between them-

ſelves.

~~
~~

~~This is evident, becauſe the Appearances of ~~
Lines parallel to the baſe Line, are parallel to

the ſaid Line;

~~and the Appearances of the in-~~
clined Lines are parallel to theſe Lines.

~~
~~

~~8, 9. ~~
~~The Repreſentation of a Figure, parallel to ~~
the perſpective Plane, is ſimilar to the ſaid Figure;

~~and ~~
the Sides of the ſaid Figure are to their Repreſen-

tations, as the Diſtance of the Eye from the Plane

of the Figure, to the Diſtance of the Eye from the

perſpective Plane.

~~
~~

~~The given Figure is A B C D. ~~
~~We are firſt to ~~

thereto;

~~that is, that the correſponding Angles ~~
of theſe two Figures A B C D, a b c d, are equal,

and their Sides proportional.

~~
~~

~~I. ~~
~~The Angles are equal, becauſe ~~
tween themſelves.

~~
~~

~~II. ~~
~~In the ſimilar Triangles A D O, and a d o, ~~
we have

A D:

~~a d : ~~
~~: ~~
~~O D : ~~
~~O d.~~
~~
~~

~~And in the ſimilar Triangles O D C, and O d c, ~~
we have

D C :

~~d c : ~~
~~: ~~
~~O D : ~~
~~O d. ~~
~~then ~~
A D:

~~a d : ~~
~~: ~~
~~D c : ~~
~~d c. ~~
altern.

A D :

~~D C : ~~
~~: ~~
~~a d : ~~
~~d c.~~
~~
~~

~~And conſequently the Sides A D, and D C of ~~
the Figure A B C D, are Proportional to the

~~The ~~
ſame may be demonſtrated of the other Sides;

~~and therefore the Figures are ſimilar.~~
~~
~~

~~Now to prove the other Part of the Theorem: ~~
~~If a perpendicular be ſuppoſed to be let fall from ~~
the Eye upon the Plane of the Figure, and con-

tinued as is neceſſary;

~~it is evident, that O D, ~~
will be to O d, as this Perpendicular, which

meaſures the Diſtance from the Eye to the Plane

of the Figure, is to the Diſtance of the Eye

from the Perſpective Plane, which is meaſur’d

by the Part of the perpendicular, contain’d be-

tween the Eye and the perſpective Plane.

~~Now ~~
this before was manifeſt;

~~viz. ~~
~~that ~~
O D :

~~O d : ~~
~~: ~~
~~A D : ~~
~~a d :~~
~~
~~

~~Whence there is the ſame Proportion between ~~
A d one of the Sides of the Figure, and A D its

Appearance, as the Theorem expreſſes.

~~The ~~
ſame may be demonſtrated of the other Sides of

the Figure.

~~Which was to be demonſtrated.~~
~~
~~

~~10. ~~
~~If from a Point in the Geometrical Plane, three ~~
right Lines proceed, which are equal between them-

ſelves, and parallel to the perſpective Plane;

~~the firſt ~~
of which is in the Geometrical Plane, the ſecond ele-

vated Perpendicular to the firſt, and the third in-

clined to it;

~~the Appearances of theſe three right ~~
Lines are equal.

~~
~~

~~This will appear clear enough in conſidering ~~
the Lines as a Figure parallel to the perſpective

Plane;

~~and ſo conſequently they will have the ~~
ſame Proportion as their Appearances.

~~
~~

~~Note, The firſt of the aforeſaid Lines is always ~~
parallel to the baſe Line;

~~and the ſecond, when ~~
the perſpective Plane is perpendicular or up-

~~
~~
~~
~~
~~
~~
Plane, and the third is then in the Direction of the

firſt.

~~
~~

~~11. ~~
~~If tworight Lines, equal between themſelves, and ~~
parallel to the perſpective Planes, be equally diſtant

from the perſpective Plane, their Appearances will be

equal.

~~
~~

~~For, becauſe they are in a Plane, parallel to ~~
the perſpective Plane, they will have the ſame

Proportion to each other, as their Repreſentations.

~~
~~

~~12. ~~
~~If a Line parallel to the Perſpective Plane, be ~~
view’d by two Eyes, both being in a Plane, parallel

to the perſpective Plane, the Repreſentations of the

ſaid Line will be equal.

~~
~~

~~If we ſuppoſe a Plane, parallel to the Per-~~
ſpective Plane, to paſs through the propoſed

Line, this Proportion will be had;

~~As the
~~
Diſtance from the Perſpective Plane, ſo is the

given Line to the Repreſentation thereof.

~~But ~~
the three firſt Terms of this Proportion are

the ſame for each of the Eyes, which are

in one and the ſame Plane parallel to the Per-

ſpective Plane:

~~Therefore, the fourth Term of ~~
the Proportion will likewiſe be the ſame in both

Caſes:

~~Which was to be demonſtrated.~~
~~
~~

~~13. ~~
~~If a right Line, being continued, meets the per-~~
ſpective Piane in one Point, the Appearance thereof

will be a Part of the Line drawn from the ſaid Point

in the perſpective Plane, to another Point, whereat

poſed Line, terminates.

~~
~~

~~The Line C D being continued, will meet~~

~~We are ~~
to prove, that its Appearance is a Part of the

Line E H, drawn from the Point E, to the

Point H, whereat the Line O H proceeding

from the Eye parallel to the given Line C D,

terminates.

~~
~~

~~The Interſection of the perſpective Plane, ~~
and the Plane O D C, is the Repreſentation of the

given Line.

~~Now the Plane O D C, is a Part of the ~~
Plane paſſing through the parallels O H and EC.

~~
~~

~~Therefore, this Repreſentation is a Part of the ~~
Interſection of the laſt mentioned Plane, and

perſpective Plane;

~~which Interſection is E H.~~
~~
~~

~~14. ~~
~~All Lines parallel between themſelves, and be-~~
ing produced, do fall upon the perſpective Plane, have

Repreſentations, which being produced, will all con-

cur in one Point.

~~
~~

~~This is evident, becauſe but only one Line ~~
O H can be drawn from the Eye O, to the per-

ſpective Plane, parallel to the ſaid Parallels, and

becauſe all their Repreſentations are Parts of

Lines concurring in the Point H.

~~
~~

~~And this Point is called the accidental Point of the ~~

~~
~~

~~15. ~~
~~Two or more parallel Lines, which being produ-~~
ced, do fall on the perſpective Plane, parallel to the Geo-

metrical Plane, have their accidental Point in the

Horizontal Line.

~~
~~

~~For the Horizontal Plane, is parallel to the ~~
Geometrical Plane.

~~
~~

~~16. ~~
~~The Repreſentations of all Lines parallel to ~~
the ſtation Line, concur in the Point of Sight.

~~
~~

~~This follows, becauſe the principal Ray is ~~
parallel to the ſaid Lines.

~~
~~

~~17. ~~
~~Two or more equal Lines being perpendicular, ~~
or equally inclined the ſame Way, to the ſame Line pa-

rallel to the Station Line, have their Repreſentations

concurring in the principal Point.

~~
~~

~~Becauſe all theſe Lines are parallel and equal, ~~
the Line paſſing through their Vertices, is pa-

rallel to that paſſing through their Baſes, and

this being parallel to the Station Line, it fol-

lows,

Point.

~~
~~

~~18. ~~
~~The Appearance of an indefinite Line does not ~~
alter, when the Eye moves in a Line parallel to a

propoſed Line.

~~
~~

~~The Repreſentation of this Line, is the In-~~
terſection of the perſpective Plane, and a Plane

paſſing through the Eye and the ſaid Line.

~~Now ~~
the Eye remains in the ſame Plane, when it

moves in a Line parallel to the propoſed Line;

~~and conſequently the Appearance of this laſt ~~
Line, will not be changed by that Motion.

~~
~~

~~Note, This Demonſtration doth not extend ~~
to any particular Part of the given Line, but on-

ly to the Line in general.

~~
~~

~~19. ~~
~~Let A C be a Line inclined to the Geometrical ~~
Plane, and O D another Line drawn parallel to

A C, from the Eye to the perſpective Plane.

~~Now ~~

rallel to the baſe Line, and likewiſe D E, in the

perſpective Plane, parallel to the ſaid Line, ſo that

B A be to A C, as E d to D O.

~~I ſay, the Ap-~~
pearance of the Line B C, paſſing through the Point

B, and the Extremity of the Line A C, being con-

tinued, will meet the Point E.

~~
~~

~~Now to prove this; ~~
~~it is evident, ~~
B C:

~~And this may be done in the following ~~
Manner:

~~
~~

~~A B is parallel to E D, and A C to O D; ~~
~~whence the Angle (E D O) of the Triangle ~~
O E D, is equal to the Angle (B A C) of the

Triangle A C B:

~~And ſo theſe two Triangles ~~
are ſimilar;

~~becauſe they have alſo their Sides ~~
Proportional.

~~But ſince theſe two ſimilar Tri-~~
angles, have two of their Sides parallel, the

third B C is alſo parallel to O E;

~~which was to be ~~
demonſtrated.

~~
~~

~~20. ~~
~~If A B be made equal to A C, and E D to D O, ~~

CHAP. III.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~The Practice of Perſpective upon the Per-~~
ſpective Plane, ſuppoſed to be perpendicu-

lar, or upright.

~~
~~

~~IN order to give a diſtinct Idea of the Theory, I ~~
have hitherto conſider’d the Geometrical Plane,

as it were the Ground upon which the Spectator

~~and the Perſpective ~~
Plane, as a Window between the Spectator and

the Objects, in which the Objects are requir’d

to be repreſented.

~~But, in Practice, this Matter ~~
muſt be quite otherwiſe conceiv’d;

~~which I ~~
ſhall now endeavour to explain as clear as poſ-

ſible.

~~
~~

~~Suppoſe then, that a Painter has a mind to ~~
draw upon his Perſpective Plane, or Picture,

(whoſe Bigneſs is as he thinks fit) a Proſpect of

a Country, wherein are Trees, Houſes, Rivers, &

~~c. ~~
~~Now, from what has been ſaid, this Country ~~
will be his Geometrical Plane;

~~and he ought to ~~
conſider his Perſpective Plane as a Window, up-

on which the Points thro’ which the Rays com-

ing from all the Points of the Objects towards

the Eye, muſt be found.

~~But theſe Interſections ~~
of the Rays and the Window cannot be deter-

min’d, unleſs by Lines being drawn in the Geo-

metrical Plane to the Baſe Line.

~~
~~

~~Now, it is impoſſible for Painters to draw Lines ~~
of this Nature on the Ground;

~~wherefore they ~~
uſe another more convenient Geometrical Plane

thus.

~~At the Foot of their Perſpective Plane, ~~
they place a Plane, upon which are drawn in

Minature the Baſes of Houſes and Trees, which

are in the Country to be repreſented;

~~and the ~~
Seats of the Points which, in the Objects, are

elevated above the Country;

~~always obſerving, ~~
that there be the ſame Diſpoſition between the

Objects and their different Parts, upon this new

Geometrical Plane, as the Objects truly have in

the Country to be repreſented.

~~
~~

~~Now, to determine the Magnitude of the ~~
Space the Figures muſt take up upon this Geo-

metrical Plane, a Painter muſt firſt chuſe the

Diſpoſition of his Eye in reſpect to the Perſpe-

~~and then (from the Station Point, ~~
thro’ the Extremities of the Perſpective Plane)

he muſt draw right Lines;

~~which will limit the ~~
Space wherein the Figures muſt be placed;

~~ſince ~~
the Rays of Figures, without thoſe Lines coming

towards the Eye, will not paſs thro’ the Perſpe-

ctive Plane.

~~
~~

~~21. ~~
~~The Figures being thus drawn on the Geo-~~
metrical Plane, the next Thing is to find their Ap-

pearance upon the Perſpective Plane.

~~Now, ~~
theſe Figures are made up of either ſtraight

Lines, or crooked ones.

~~To find the Repreſen-~~
tation of a ſtraight Line, its Extremes need on-

ly be ſought:

~~And to have the Appearance of a ~~
crooked Line, ſeveral Points thereof need only

be found.

~~Since all this is equally applicable ~~
to Figures, as well in the Geometrical Plane,

as thoſe above it;

~~it follows, that the whole Bu-~~
ſineſs of Perſpective conſiſts in only finding the

Repreſentation of a Point.

~~
~~

~~And to find this Repreſentation in the follow-~~
ing Problems, we only uſe certain Lines drawn

in the Geometrical and Horizontal Planes;

~~which, by their Interſection with the Baſe and ~~
Horizontal Lines, ſhew the manner of drawing

new Lines upon the Perſpective Plane, which

determine the propos’d Appearances.

~~Now, it ~~
is plain, that in finding the ſaid Interſections, it

is not neceſſary to place the Perſpective Plane

perpendicular to the Geometrical and Horizontal

Planes;

~~which would render the Work extream-~~
ly laborious:

~~Whence the Perſpective and Hori-~~
zontal Planes may be conſider’d as lying upon

the Geometrical Plane, and ſo coinciding there-

with.

~~
~~

~~The Perſpective Plane may lye upon the Geo-~~
metrical Plane two ways;

~~viz. ~~
~~Either upon the ~~
Face reſpecting the Objects, or upon that next

~~Now, as in the latter Situation, ~~
Repreſentatious are drawn upon the Face of the

Perſpective Plane next to the Object, the Per-

ſpective Plane lying down upon its other Face;

~~what ought to be on the Right Hand, appears on ~~
the Left;

~~and that on the Left, appears on the ~~
Right;

~~producing exactly the ſame Effect, as ~~
looking thro’ the Back-ſide of a Paper, at a Pi-

cture drawn thereon.

~~
~~

~~Yet, notwithſtanding this Deficiency, we pre-~~
fer the latter way of the Perſpective Plane’s ly-

ing down to the former, for the following Rea-

ſons.

~~
~~

~~1. ~~
~~When the Perſpective Plane lies down in ~~
the former manner, it lies upon the Part of the

Geometrical Plane wherein Figures have been

drawn;

~~which, together with the new Lines that ~~
muſt be drawn, cauſes a very great Confuſion,

and always obliges one to copy his Work.

~~An ~~
Inconveniency which the latter Method is ſel-

dom ſubject to.

~~
~~

~~2. ~~
~~We work with much more Eaſe in the ~~
manner I have choſen.

~~
~~

~~Finally, The Default we have obſerv’d, may ~~
ſeveral ways be remedied.

~~For, in drawing up-~~
on the Geometrical Plane, we need but place

that on the Right Hand which we have a mind

ſhould appear on the Left;

~~or if the Geometri-~~
cal Plane be drawn upon Paper, it may be oil’d,

or dipp’d in Varniſh, which will render it tran-

ſparent;

~~and then the Back-ſide of the Paper ~~
may be thrown into Perſpective.

~~
~~

~~If all this be not found convenient, the ſaid ~~
Default may be eaſily corrected geometrically,

in copying the Work after the Drawings are fi-

niſhed.

~~And this may be yet eaſier done, if ~~
the Figures are expoſed before a Looking-glaſs;

~~
~~
the Left.

~~
~~

~~Therefore, I lay my Perſpective Plane upon ~~
the Geometrical Plane;

~~ſo that it be between ~~
the Horizontal Plane, and the Figures requir’d

to be thrown into Perſpective.

~~
~~

~~22. ~~
~~To find the Appearance of a Point, which is in ~~
the Geometrical Plane.

~~
~~

~~Let Z be the Geometrical Plane, I E the Baſe ~~

of Sight, D one of the Points of Diſtance, and

A the given Point.

~~
~~

~~From the Point A, let fall the Perpendicular ~~
A B upon the Baſe Line;

~~and from the Point of ~~
Concurrence B, draw the Line B V to the Point

of Sight;

~~then aſſume B E in the Baſe Line ~~
equal to B A, and from the Point E draw the

Line E D to the Point of Diſtance D:

~~And the ~~
Point (a), the Interſection of B V and E D, is

the Repreſentation ſought.

~~
~~

~~23. ~~
~~The Appearance of the Line A B, is ~~
~~Now, if we conceive a Line ~~

another from the Point A towards the Point E;

~~theſe two Lines will be parallel, becauſe they ~~
are in parallel Planes, aud each make half a

right Angle with the Perſpective Plane;

~~whence
~~
~~
~~
~~
~~
~~
~~
~~Now, ſince the Point A is in ~~
the two Lines A B, A E;

~~the Appearance of ~~
the ſaid Point will likewiſe be in the Appear-

ances of the aforeſaid two Lines, and conſequent-

ly is in the Point a, the common Section of B V

and E D.

~~
~~

~~24. ~~
~~If the Diſtance of the Eye be ſo great, that ~~
one of the Points of Diſtance cannot be deno-

ted upon the horizontal Line;

~~another Point, F, ~~
muſt be uſed, diſtant from the Point of Sight

by about one third, or fourth Part of the Di-

ſtance of the Eye.

~~But then, a correſpondent ~~
Part of the Perpendicular A B muſt be likewiſe

taken, and laid off from B to G, in the Baſe

Line.

~~
~~

~~25. ~~
~~And in this manner may the Repreſentation ~~
of a very diſtant Point be found, if its Diſtance

from the Perſpective Plane be known, together

with the Place wherein a Perpendicular drawn

from that Point cuts the Baſe Line.

~~For, having ~~
firſt drawn a Line, as B V, from the ſaid Point

of Concurrence to the Point of Sight, then B E

muſt be aſſum’d in the Baſe Line;

~~for Example, ~~
equal to the tenth Part of the Diſtance of the

Point whoſe Repreſentation is ſought;

~~and V H ~~
in the Horizontal Line, likewiſe equal to the

tenth Part of the Eye’s Diſtance.

~~Then C, the ~~
Interſection of B V and E H, will be the Appear-

ance ſought.

~~
~~

~~Note, By this Method may be found the Deep-~~
nings in Pictures.

~~
~~

~~The Appearance of the Point A may yet be ~~
otherwiſe found, without drawing the Line B V

from the Point A, in taking B I equal to B A,

other Point of Diſtance;

~~which, by its Inter-~~
ſection with E D, will give the Appearance of

the Point A.

~~
~~

~~26. ~~
~~γ is the Horizontal Plane, X the Perſpe-~~
ctive Plane, Z the Geometrical Plane, O the Eye,

D C the Horizontal Line, B E the Baſe Line, and

A the given Point.

~~
~~

~~Draw a Line from the Point A, to the Eye O, cut-~~
ting the Baſe Line in the Point B, and the Horizon-

tal Line in the Point C:

~~Then aſſume B E in the ~~
Baſe Line, equal to B A;

~~and C D in the Hori-~~
zontal Line, equal to C O;

~~and join the Points ~~
E and D, by a Line cutting the Line A O in

the Point a;

~~which will be the Appearance ~~
ſought.

~~
~~

~~27. ~~
~~The Triangle O D C in the Horizontal ~~
Plane, is ſimilar to the Triangle A B E in the

Geometrical Plane;

~~and conſequently A B is pa-~~
rallel to O C, and A E to O D.

~~But the Appear-~~
ance of A muſt

~~
~~
~~
~~

~~28. ~~
~~If the Place wherein the Eye ought to be in ~~
the Horizontal Plane be not known, but the Point

of Sight is;

~~then, to find the Place of the Eye, ~~
a Perpendicular muſt be rais’d from the Point of

to the principal Ray;

~~and the Extremity of this ~~
Perpendicular will be the Point ſought.

~~
~~

~~If nothing is determin’d, the Place of the Eye ~~
may be taken at pleaſure in the Horizontal

Plane.

~~
~~

~~29. ~~
~~The ſame Things being given as in the ~~
precedent Method, about the Eye O, as a Cen-

ter;

~~deſcribe the Arc of a Circle I H, touching ~~

~~
~~

~~About the given Point A, as a Center, de-~~
ſcribe the Arc of a Circle L C, touching the Baſe

Line:

~~Then draw two Lines, C H and L I, ~~
touching the two Circles L C and H I;

~~and the ~~
Point a, the Interſection of the ſaid two Lines,

will be the Appearance ſought.

~~
~~

~~30. ~~
~~To demonſtrate this, draw the Line A B ~~
perpendicular to the Baſe Line;

~~O V perpendi-~~
cular to the Horizontal Line;

~~and A C, O H ~~
perpendicular to the Tangent H C.

~~All theſe ~~
Perpendiculars will cut the Lines to which they

are perpendicular, in the Points wherein theſe

laſt touch the Circle L B C, or H V I.

~~Like-~~
wiſe draw the Line A E from the given Point

A, to the Point E, wherein the Line H C cuts

the Baſe Line.

~~Finally, draw O D, from the ~~
Eye O to the Point D, wherein the ſaid Line

HC cuts the Horizontal Line.

~~
~~

~~Now, it is evident, ~~
demonſtrate that O D is parallel to A E;

~~which ~~
may be done thus:

~~
~~

~~Becauſe the Triangles O G V, and A B F, are ~~
ſimilar.

~~A F: ~~
~~A B:~~
~~: O G: ~~
~~O V: ~~
altern.

A F:

~~O G:~~
~~: A B: ~~
~~O V: ~~
Divid.

~~and altern. ~~
~~the firſt ~~
Proportion.

AF—AB (=CF):

~~O G—O V=HG:~~
~~:AB:~~
~~OV.~~
~~
~~

~~But becauſe the Triangles E C F, H G D are ~~
ſimilar.

~~C F: ~~
~~H G :~~
~~: E F : ~~
~~G D.~~
~~
~~

~~Now, by obſerving the two laſt Proportions of ~~
the other two Triangles,

E F:

~~G D:~~
~~: A F: ~~
~~O G, ~~
And the Angle A F E, being equal to the Angle

O G D, the Triangles A E F and O D G are

ſimilar;

~~and therefore A E is parallel to O D: ~~
~~Which was to be demonſtrated.~~
~~
~~

~~After the ſame manner we prove, that the ~~
Appearance of the Point A is in the Line L I,

and conſequently is in the Interſection of this

Line and HC.

~~
~~

~~Altho’ this Method appears more difficult than ~~
the precedent one, as to the Geometrical Conſi-

deration thereof, yet the Operation is eaſier, if

the Points are not too far diſtant from the Baſe

Line:

~~For Lines may well enough be drawn by ~~
Gueſs, or Sight only, to touch Circles, and Cir-

cles to touch Lines.

~~
~~

~~31. ~~
~~Draw the Line F O G through the Eye O, ~~
parallel to the Baſe Line, then aſſume F O in

O G equal to the Length of the principal Ray.

~~A is the given Point.~~
~~
~~

~~From the given Point A, draw the Lines AO, ~~
A F, to the Points O and F, and from the

Point E, wherein A F cuts the Baſe Line, draw

the Line E G to the Point G;

~~then the Point a, ~~
the Interſection of A O, and E G, is the Repre-

ſentation ſought.

~~
~~

~~Let fall the perpendicular G M from the Point ~~
G, upon the Baſe Line, and through the Eye O,

draw the Line O D to the Point D, the Inter-

ſection of the Horizontal Line, and the Line G E.

~~
~~

~~Then becauſe the Triangles G D L, G E M are ~~
fimilar,

G D:

~~G E:~~
~~: G L: ~~
~~G M. ~~
~~But G O is equal to G L, and G F to L M. ~~
whence

G D:

~~G E:~~
~~: G O: ~~
~~G F.~~
~~
~~

~~And conſequenely the Triangles G O D: ~~
~~and ~~
G F E, are ſimilar, and the Lines O D, and

A E F, are parallel between themſelves;

~~and ~~
therefore

~~It has alſo been prov’d~~
Line A O;

~~therefore i@ is in a the Interſection
~~
to be demonſtrated.

~~
~~

~~33. ~~
~~By this Demonſtration it appears, that there ~~
is no Neceſſity in taking G O exactly equal to

the Eye’s Diſtance, and O F equal to its Height:

~~But it is ſufficient if they have the ſame Propor-~~
tion, as the aforeſaid Diſtance has to the Height.

Likewiſe there is no Neceſſity in aſſuming the

Points G and F, in a Line parallel to the Baſe

Line;

~~for any other Line paſſing through the ~~
Eye O may be uſed at Pleaſure.

~~For Example, ~~
let g O f be a Line any how drawn through the

Eye O, and take the Point g at Pleaſure in this

Line, through which draw alſo the Line g N I

at Pleaſure, cutting the Horizontal Line in N,

and the Baſe Line in I;

~~and draw the Line O N, ~~
and through the Point I, draw the Line I f pa-

rallelthereto, cutting the Line g O f in f.

~~
~~

~~This being done, the Points g and f may be ~~
uſed inſtead of G and F:

~~for among all the ~~
Lines that can be drawn (as G N I) it is mani-

ſeſt, that g N will always be to g I:

~~: g O: ~~
~~g f, ~~
which is ſufficient for the Demonſtration.

~~
~~

~~If the Point f be firſt determin’d, the Point g ~~
muſt be found by an Operation quite contrary

to that we have laid down.

~~
~~

~~34. ~~
~~When nothing is determinate, we ~~
may (a Baſe Line being firſt drawn) take at

Pleaſure, in another Line any how drawn,

the three Points g O f;

~~ſo that in this Caſe, there ~~
is no Manner of Neceſſity to uſe Compaſſes, in

throwing any Figure whatſoever, which is on

the Geometrical Plane, into Perſpective.

~~But ~~
if after having thus work’d, the Point of Sight,

Height and Diſtance of the Eye be requir’d, the

the Points f and O, on the Baſe Line, and the

Line P g drawn;

~~then the Point V, wherein it ~~
cuts the Perpendicular O H, is the Point of Sight

ſought, and the Parts O V, and V H determine

the Height and Diſtance of the Eye.

~~
~~

~~35. ~~
~~When the Appearance of a Point is known,~~

~~Let A be a Point in the Geometrical Plane, ~~

it is requir’d to find the Appearance of the

Point B.

~~
~~

~~Draw a Line from the Point B to the Eye O, ~~
and another from the Point E, wherein the

ſaid Line continued, cuts the Baſe Line, to the

Point A;

~~then draw the Line E a, and where ~~
it cuts B O, is the Point b ſought.

~~
~~

~~The Point E is its own Repreſentation; ~~
~~and ~~
becauſe the Point a is the Repreſentation of A,

the Line E a is that of E A.

~~Now ſince the ~~
Point B is in the Line E A, the Appearance of

this Point will be likewiſe in E a, as alſo

~~therefore it is in b the Interſection of the ~~
Lines E a, and B O.

~~
~~

~~37. ~~
~~If the Point A be in the Line B O, or ~~
the Line B A be parallel, or a very little inclined

to the Baſe Line, we cannot then uſe this Me-

pearance of ſome other Point taken at Pleaſure

upon the Geometrical Plane be found, by Means

of which, the Appearance of the Point B may

be afterwards gotten;

~~but in theſe Caſes, the ~~
ſhorteſt Way, is to uſe ſome one of the precedent

Methods.

~~
~~

~~38. ~~
~~It appears from this Method, that when ~~
the Repreſentations of two Points are found,

the Appearance of any third Point whatſoever

may be had, without having any Regard to the

Situation of the Eye;

~~becauſe two Lines as E a ~~
may be drawn, whoſe Interſection will be the

Point ſought.

~~
~~

~~39. ~~
~~The ſame things being given, as in the ~~

~~
~~

~~Draw two Lines A F and A C from the given ~~
Point A at Pleaſure, cutting the Baſe Line in the

Points E and B, and interſecting the Geometri-

cal Line in the Points F and C.

~~From theſe ~~
two laſt Points draw the Lines F O and C O to

the Eye;

~~then draw E a through the Point E, ~~
parallel to F O, and B a through the Point B,

parallel to C O, and the Point a the Interſection

of theſe two Lines will be that ſought.

~~
~~

~~Note, We might firſt have drawn the Lines ~~
O F and O C at Pleaſure, and then have drawn

the Lines AC and A F through their Concur-

~~
~~
~~
~~
~~
~~
~~which would ~~
come to the ſame thing.

~~
~~

~~Firſt continue the Line E a, until it meets ~~
the Horizontal Line in D, and draw a Line

from D to the Eye, and another through the Eye

parallel to the Baſe Line.

~~
~~

~~Then the Parallels O M and F C are at the ~~
ſame Diſtance from each other, as L D is from

E B;

~~whence it follows, that F O is equal to E D, ~~
and therefore O D is parallel to A F.

~~Whence ~~
~~And ~~
after the ſame Manner we prove, that the Re-

preſentation of B A is a Part of B a.

~~
~~

~~40. ~~
~~When there are no Lines drawn, and we ~~
would uſe this Method, the Horizontal Line

may be laid aſide;

~~and then having firſt drawn ~~
the Geometrical Line, whoſe Diſtance from

the Baſe Line is equal to the Length of the Ray,

we aſſume the Diſtance from the Eye to the

Geometrical Line, equal to the Height of the

Eye.

~~
~~

~~Although this Method appears uſeleſs, as being ~~
more difficult than the precedent ones, yet in

the Eighth Chapter we have ſhewn the Uſe that

may be drawn from it.

~~
~~

~~41. ~~
~~It follows from this Demonſtration, that ~~
the Appearances of Lines paſſing through the

Station Point, are all perpendicular to the Baſe

Line;

~~for if the Perpendicular O S be let fall
~~
ances of all Lines paſſing through S, will be

perpendicular to the Baſe Line;

~~but the ſaid ~~
Point S is the Station Point.

~~Whence, &~~
~~c.~~
~~
~~

~~42. ~~
~~To throw a Line in the Geometrical Plane ~~
into Perſpective.

~~
~~

~~It has been ſhewn ~~
the Extremes of the ſaid Line, need only be

found;

~~and although it is difficult to find ~~
ſhall ſhew here how to find more eaſy the Re-

preſentation of a Line in ſome Caſes.

~~
~~

~~1. ~~
~~Let A B be a Line parallel to the Baſe ~~

~~To draw the Repreſentation of which, ~~
having firſt found the Point a the Appearance

of A, one of the Ends of the given Line;

~~af-~~
terwards through that Appearance, draw a Pa-

rallel to the Baſe Line;

~~then the Line B O, ~~
drawn from B to the Eye O, will cut the ſaid

parallel in the Point b, and b a will be the Re-

preſentation ſought.

~~
~~

~~43. ~~
~~2. ~~
~~Let C G be a Line, which continued ~~
out, cuts the baſe Line in E.

~~Now to draw the ~~
Appearance thereof;

~~through the Eye O, draw a ~~
Line parallel thereto, cutting the Horizontal

Line in D, and joyn the Points E and D, by

the Line E D, which cut in the Points c and g,

by Lines drawn from C and G to the Eye;

~~then ~~
the Part c g of the Line E D, is the Appearance

ſought.

~~
~~

~~If the Lines G O and C O cut E D very ob-~~
liquely, and ſo their Interſection cannot be ex-

~~
~~
~~
~~
~~
~~
be uſed.

~~
~~

~~44. ~~
~~To find the Appearance of the Diviſions of a ~~
Line in the Geometrical Plane.

~~
~~

~~Let A B be a Line, whoſe Appearance is ab. ~~
~~
~~
of this Line, there muſt be Lines drawn from

the Diviſions of the Line to the Eye, whoſe

Interſections with a b will give the Points

ſought.

~~
~~

~~Note, When theſe Lines very obliquely cut ~~
a b, the following Way ought to uſed.

~~
~~

~~45. ~~
~~To find the Repreſentations of the Di-~~
viſions of the Line G C, make choice of the

~~then draw Lines
~~
~~and from the Points wherein theſe Lines con-~~
tinued out cut the Baſe Line, draw other Lines

through the Repreſentation d, which will cut c g

the Repreſentation of C G in the Points ſought.

~~
~~

~~46. ~~
~~To throw a Polygon, or any other regular ~~
Figure on the Geometrical Plane into Perſpective.

~~
~~

~~The Repreſentation of any Kinds of Figures ~~
may be found

~~the fourth in general is the eaſieſt; ~~
~~and may be firſt uſed in finding ~~
only;

~~and then the fifth Method ſerves ~~
~~But yet the Work may be ~~
ſhorten’d by the two precedent Problems, as we

ſhall ſhew in the following Examples.

~~
~~

~~To throw a Pentagon having one Side parallel to ~~
the Baſe Line, into Perſpective.

~~
~~

~~Let A B C D E, be the propoſed Pentagon, ~~

lel to A E, becauſe the Pentagon is a regular

one.

~~
~~

~~Now find ~~
preſentation of four of the Corners of the Pen-

tagon;

~~and to determine the Repreſentation of ~~
the fifth Corner, find

is parallel to A B, which is ſuppoſed parallel

to the Baſe Line.

~~
~~

~~To throw a Parallelogram, divided into ſeveral ~~
other Parallelograms into Perſpective.

~~
~~

~~Let A B C D be a Parallelogram, divided ~~

~~
~~

~~Draw the Line O G thro’ the Eye O parallel to ~~
the Side A D, cutting the Horizontal Line in G;

~~likewiſe draw O F parallel to the Side A B cutting ~~
the Horizontal Line in F, and produce the Sides

of the Parallelogram, and the Lines dividing it,

to the Baſe Line;

~~then from the Points wherein ~~
A D and C B, and the Lines parallel thereto

cut the Baſe Line, draw Lines to the Point G.

Alſo from the Points wherein A B and C D and

their parallels cut the ſaid Line, draw Lines to

the Point F, whoſe Interſections with theſe

ſought.

~~
~~

~~When this Method cannot be us’d, the Per-~~
ſpective of the Diviſions dividing the Sides of

the Parallelogram, mufl be found

~~And we are
~~
notwithſtanding the accidental Points, G and F,

being had.

~~And this happens, when the Paralle-~~
logram is ſo far diſtant from the perſpective

Plane, that its Sides being produc’d, cannot meet

the Baſe Line.

~~
~~

~~47. ~~
~~Note, moreover, that this one Example ~~
is ſufficient to ſhew how to throw any Kinds of

Figures in the Geometrical Plane into Perſpe-

ctive.

~~To effect which, we circumſcribe any ~~
Parallelogram about the Figures, which we di-

vide into ſeveral others:

~~Then we throw this ~~
Parallelogram (thus divided) into Perſpective,

and transfer the given Figure therein, ſo that it

may have the ſame Situation with reſpect to

the little Parallelograms in the perſpective Plane,

as it had in regard to the ſmall Parallelograms

in the Geometrical Plane.

~~
~~

~~The Repreſentation of ſeveral Points of a ~~

thrown into perſpective, muſt be

Chords in the Circle, or Curve, parallel between

themſelves, the Repreſentations of which muſt

be found

~~then the Extremities of thoſe Re-
~~
ctive ſought.

~~The ſame may be done, in draw-~~
ing the Chords thro’a Point, whoſe Repreſenta-

tion is known.

~~
~~

~~49. ~~
~~Let G I be the Geometrical Line; ~~
~~and thro’ ~~

ſought, let fall the Perpendicular P F upon the ſaid

Line G I, which biſect in the Point R.

~~About R, ~~
as a Center, and with the Radius R P, deſcribe an

Arc of a Circle M P N, cutting the given Circle in

the Points M and N.

~~Now, if the Perſpective of ~~
L H and N M be found, the two Conjugate Dia-

meters of an Ellipſis, which is the Repreſentation of

the given Circle, will be bad.

~~And, an Ellipſis may ~~
be drawn by ſome one of the Methods laid down by

thoſe who have treated of Conick Sections.

~~
~~

~~I ſhall not ſpend time here in demonſtrating the ~~
Truth of this.

~~See Prop. ~~
~~10. ~~
~~lib. ~~
~~2. ~~
~~of the great ~~
Latin Treatiſe of Conic Sections, written by M.

~~de ~~
la Hire;

~~the Demonſtration of which may be here ~~
apply’d.

~~If we conſider, 1. ~~
~~That Lines drawn from ~~
the Points M and N to the Point F, will touch the

Circle in the ſaid Points M and N.

~~2. ~~
~~That the ~~
viſual Rays, going from the Eye towards all the Parts

of the Circumference of the Circle, form a Cone,

3.

~~That the Appearance of the Circle, is the Section ~~
of a Cone, made by the perſpective Plane.

~~Finally, ~~
That the Line G I ought to be conceiv’d, as being

the Interſection of the Geometrical Plane, and a

Plane paſſing thro’ the Eye parallel to the perſpective

Plane.

~~
~~

~~Let G S be the Geometrical Line, and S the ~~

~~Make S F, in the Geometrical ~~
Line, equal to the Height of the Eye;

~~and let ~~
A be the Seat of the given Line.

~~
~~

~~Aſſume F C in the Geometrical Line, equal to ~~
the Height of the Eye, above the Geometrical

Plane:

~~Then draw Lines from the Point A to ~~
the Points S and C, and on the Point B, the In-

terſection of the Line AS and the Baſe Line,

raiſe the Perpendicular BI to the Baſe Line,

equal to E B, plus FC;

~~and the Point I will be ~~
the Perſpective ſought.

~~
~~

~~51. ~~
~~Let us ſuppoſe a Plane to paſs thro’ the ~~
given Point, and the Eye perpendicular to the

Geometrical Plane;

~~then it is manifeſt, that the ~~
Interſection of theſe two Planes is the Line

A B S, and the Interſection of the ſaid ſuppos’d

Plane and the perſpective Plane, is B I.

~~Now, ~~
let X be this ſuppos’d Plane;

~~a, b, s, the Point ~~

Figure, bi the Interſection of this Plane and

the perſpective Plane;

~~O the Eye, and D the ~~
propos’d Point:

~~We are to prove, that if O D ~~
be drawn, the Line B I of the precedent Figure

will be equal to b i in this Figure.

~~
~~

~~To demonſtrate which, draw the Line D L M ~~
thro’ the Point D, parallel to a b s.

~~Then, be-~~
cauſe the Triangles D M O and D L i are ſimi-

~~D M = as: ~~
~~D L = ab:~~
~~: M O: ~~
~~L i. ~~
~~Again, ~~
in the precedent Figure, the Triangles A S C and

A B E are ſimilar:

~~Whence, ~~
A S:

~~A B:~~
~~: C S: ~~
~~E B.~~
~~
~~

~~The three firſt Terms of theſe two Progreſſions ~~
are the ſame:

~~For CS is equal to M O, ſince ~~
they are each the Difference of the Height of the

Eye, and that of the given Point;

~~and conſe-~~
quently, E B is equal to L i:

~~But B I was made ~~
equal to B E, pl{us} FC the Height of the given

Point above the Geometrical Plane;

~~and b i is ~~
equal to Li, pl{us} b L;

~~which being equal to aD, ~~
is likewiſe the Height of the given Point above

the Geometrical Plane;

~~whence the Lines B I ~~
and b i are equal.

~~Which was to be demon-~~
ſtrated.

~~
~~

~~Note, When the Height of the given Point is ~~
greater than the Height of the Eye, E B muſt

be taken from that firſt Height, to have the

Magnitude of B I.

~~
~~

~~Now, to throw a Pyramid into perſpective, ~~

~~After which, Lines muſt be drawn from
~~
pearance of thoſe Angles of the Baſe that are

viſible;

~~and then the Perſpective ſought will be ~~
had.

~~
~~

~~And to throw a Cone into perſpective, the ~~

~~and then if Lines be drawn from
~~
Repreſentation of the Baſe, the Repreſentation

of the Cone will be had.

~~
~~

~~But ſince, according to this Manner, we are ~~
obliged to find the Perſpective of all the Baſe;

~~whereas it often cannot be all ſeen; ~~
~~we may de-~~
termine, by the following Method, what Part

of the Baſe is viſible, and ſo only find the Re-

preſentation thereof.

~~And then, to compleat ~~
the Cone, we draw Lines from the Extremities

of the viſible Part of the Baſe, to the Repreſen-

tation of the Vertex.

~~
~~

53. To determine the viſible Part of the Baſe of a Cone.
~~
~~
Fig. 21.

~~Let the Circle L I F be the Baſe of a Cone ~~

thereof.

~~
~~

~~Aſſume P Q ſomewhere in the Baſe Line, ~~
equal to the Semidiameter of the Circle L F;

~~and from the Point P, raiſe P D G perpendicu-~~
lar to the Baſe Line, meeting the Horizontal

Line in G;

~~and in this Perpendicular, make ~~
P D equal to the Height of the Cone;

~~and draw ~~
the Line Q D H, meeting the Horizontal Line

in H.

~~Then, about the Point A as a Center, ~~
and with the Radius G H, draw the Circle B C E;

and from the ſaid Point A, draw a Line to the

Station Point S:

~~Biſect A S in R; ~~
~~and about ~~
R, as a Center, with the Radius R A, deſcribe

the Circular Arc B A C, cutting the Circle BEC

in the Points B and C.

~~Draw the Lines B A F, ~~
and C A L;

~~and the viſible Portion, (L I F) of
~~
min’d.

~~
~~

~~To prove this, draw the Lines B C and L F, cut-~~
ting the Line A S in the Points N and M;

~~and make ~~
the Line G n equal to A N, and draw the Line

n D m.

~~It is now manifeſt, that if the Cone be ~~
continued out above its Vertex, (that is, if the oppo-

ſite Cone be form’d) it will cut the Horizontal Plane

in a Circle equal to B E C, whoſe Seat will be BEC:

~~So that the Point S, in reſpect of B E C, is in the ~~
ſame Situation as the Eye hath, with reſpect to the

Circle form’d in the Horizontal Plane, by the Conti-

nuation of the Cone.

~~Whence it follows, that B C ~~
is the Seat of the viſible Portion of that Circle.

~~For, ~~
by Conſtruction, B and C are the Points of Contact

of the Tangents to the Circle B E C, which paſs

thro’ the Point S;

~~becauſe the Angle ABS, which ~~
is in a Semicircle, is a right one.

~~
~~

~~Now, if a Plane be conceiv’d, as paſſing thro’ ſome ~~
Points in the Horizontal Plane, whoſe Seats are

B and C, and which cuts the two oppoſite Cones

thro’ their Vertex;

~~it is evident, that this Plane ~~
continued, will cut the Geometrical Plane in a Line

parallel to B N C;

~~and that this Line upon the ~~
ſaid Plane, will determine the viſible Part of the

Cone’s Baſe.

~~So, ſince G n was made equal to ~~
A N, we have only to prove, that P m is equal to

A M:

~~For, it follows from thence, that L M F is ~~
the Common Section of the Geometrical Plane, and

the Plane which we have here imagin’d.

~~
~~

~~The Triangles D Q P and G H D are ſimilar, whence ~~
D G:

~~D P:~~
~~: G H: ~~
~~P Q.~~
~~
~~

~~And the Triangles D P m and D G n are ſimilar: ~~
~~Wherefore ~~
D G:

~~D P: ~~
~~G n:~~
~~: P m. ~~
And

G H:

~~P Q:~~
~~: G n: ~~
~~P m.~~
~~
~~

~~The Triangles B A N and L A M are ſimilar: ~~
~~Therefore, ~~
BA:

~~AL:~~
~~: AN: ~~
~~AM. ~~
But the three firſt Terms of the two laſt Proportions,

are equal between themſelves;

~~whence P m is alſo ~~
equal to A M.

~~Which was to be demonſtrated.~~
~~
~~

~~54. ~~
~~When the Height of the Cone is greater ~~
than the Height of the Eye, the Points, G and

H, will fall below the Point D;

~~in which Caſe, ~~
the Lines A B and A C muſt be produc’d, till

they cut the Circle in the Points l and f, oppo-

ſite to L and F:

~~Then lIf will be the viſible ~~
Part of the Baſe.

~~
~~

~~When the Cone is inclin’d, ſo that T (for Ex-~~
ample) is the Seat of its Vertex;

~~AT muſt be ~~
drawn:

~~And then having aſſum’d P D equal to ~~
the perpendicular Height of the Cone, and Pt

equal to A T;

~~the Line t D x muſt be drawn; ~~
~~and the Part T X, taken in A T, equal to G x. ~~
Alſo, X S muſt be drawn, and A s, equal and

parallel thereto.

~~
~~

~~This being done; ~~
~~the ſame Method muſt be ~~
apply’d here, that I have laid down for the up-

right Cone;

~~with this Difference only, that the ~~
Point s muſt be us’d inſtead of the Station Point

S.

~~But when the Height of the Cone is greater ~~
than the Height of the Eye, the Point X muſt

be aſſum’d in the Line T A, between the Points

T and A.

~~
~~

~~The Reaſon of this is evident, from the Demon-~~
ſtration of the upright Cone:

~~For, it is manifeſt, that ~~
X is the Seat of the Center of the Circle, which the

Cone continued forms in the Horizontal Plane;

~~and ~~
conſequently, the Point s, in regard to the Circle BED,

is in the ſame Situation as the Eye is, in reſpect of

the Interſection of the continued Cone, and the Hori-

zontal Plane.

~~
~~

~~Note, moreover, that a Cone can ſcarcely ever ~~
be thrown into Perſpective, by the common Me-

thod, ſo exact as by this.

~~
~~

~~It is requir’d to find the Appearance of a Line ~~

trical Plane, in the Point A.

~~
~~

~~Aſſume E D, any where in the Baſe Line, ~~
equal to B C;

~~and from the Points D and E, ~~
draw D F and E F to ſome Point F, taken at

pleaſure in the Horizontal Line.

~~Then having ~~
found

~~
~~
pendicular thereto:

~~And if aI be made equal to ~~
G H, the ſaid a I will be the Perſpective

ſought.

~~
~~

~~56. ~~
~~The Appearance of the ſaid Line, is ~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
~~
qual to B C.

~~Now if from the Extremities of ~~
the ſaid Line A L, Perpendiculars are let fall,

meeting the Baſe Line in the Points P and M,

and from theſe Points, Lines are drawn to the

Point of Sight V;

~~then a N will likewiſe be ~~
~~and ſince P M is equal ~~
to D E, a N will be likewiſe equal to G H, and

conſequently a N will be likewiſe equal to a I,

which is equal to G H.

~~
~~

~~57. ~~
~~The ſame Things being given, as in the ~~
precedent Method, about the Point A, as a

Arc of a Circle L M, and draw the Line O L

from the Eye touching it;

~~then about a, (which ~~
is the Repreſentation of A) as a Center deſcribe

the Circular Arc G I touching the Line L O,

and cutting another Line drawn through a Per-

pendicular to the Baſe Line in the Point I:

~~I ~~
ſay the Point I is the Extremity of the Repre-

ſentation ſought.

~~
~~

~~To prove this, let fall the Perpendiculars A L ~~
and a G upon the Line O L, which will meet

the ſaid Line in the Points wherein it touches

the circular Arcs M L and G I.

~~
~~

~~Alſo aſſume D E in the Baſe Line equal to B C ~~
or A L, and draw the Line D F;

~~then through a, ~~
draw a H parallel to the Baſe Line.

~~
~~

~~Now let us conſider the Figure X, which re-~~

the Point A of the foregoing Figure, wherein

O f here, repreſents O F there;

~~f e here, F E
~~
~~and finally e A here, E A in that Fi-~~
gure.

~~
~~

~~This being ſuppoſed, o f is parallel ~~
to the Triangle a e A, and therefore we have this

Proportion.

~~
~~

~~o f: ~~
~~f a:~~
~~: A e: ~~
~~e o. ~~
~~Comp.~~
~~
~~

~~o f + a f: ~~
~~f a:~~
~~: A e + e a: ~~
~~e a. ~~
~~Altern.~~
~~
~~

~~o f + a f: ~~
~~A e + e a:~~
~~: f a: ~~
~~e a. ~~
~~Comp. ~~
~~and Perm.~~
~~
~~

~~o f + f a + Ae + e a : ~~
~~o f + f a :~~
~~: f a + e a : ~~
~~f a.~~
~~
~~

~~This laſt Proportion being reduced to the pre-~~

~~O A : ~~
~~o a : ~~
~~: ~~
~~F D : ~~
~~F a.~~
~~
~~

~~Again, becauſe the Triangles O A L and O a G ~~

~~O A: ~~
~~O a : ~~
~~: ~~
~~A L : ~~
~~a G.~~
~~
~~

~~And ſince the Triangle F E D and F a H are ~~
ſimilar;

~~
~~

~~F E : ~~
~~F a : ~~
~~: ~~
~~D E : ~~
~~H a.~~
~~
~~

~~And ſo if theſe three laſt Propoſitions be con-~~

~~A L : ~~
~~a G : ~~
~~: ~~
~~D E : ~~
~~H a.~~
~~
~~

~~But D E was made equal to A L, and there-~~
fore a G or a I is alſo equal to a H, which is

~~Which was ~~
to be demonſtrated.

~~
~~

~~58. ~~
~~Near one of the Sides of the perſpective ~~

Line, equal to the Height of the Eye, in which

take B L equal in length to twice the Perpen-

dicular, whoſe Perſpective is requir’d.

~~Let S be
~~
Perpendicular meets the Geometrical Plane.

~~
~~

~~Having firſt found ~~
Line in E, through which Point E draw the

Line Ea;

~~then from the Point B draw a Line ~~
B a to the Point a, cutting the Horizontal Line

in F.

~~Again through F draw a Line to the ~~
Point L, cutting E a in I;

~~and a I is the Repre-~~
ſentation ſought.

~~
~~

~~To prove this, let G N be a Perpendicular to ~~
the Baſe Line drawn from the Point G, wherein

the ſaid Baſe Line is cut by the Line B F;

~~alſo ~~
let G D be equal to the Perpendicular whoſe Ap-

pearance is ſought, and a H parallel to the Baſe

Line.

~~
~~

~~It is plain that the Perſpective of E A is ~~
E a:

~~But E A paſſes through the Station Point; ~~
~~and conſequently ~~
~~therefore ~~
~~
~~

~~Now the Triangles B G C and B F M are ſimi-~~
lar;

~~and ſo~~

~~B C : ~~
~~B M :~~
~~: B G: ~~
~~B F.~~
~~
~~

~~But B M by Conſtruction is the double of B C; ~~
~~whence B F is alſo the double of B G, which, ~~
conſequently, is equal to G F.

~~
~~

~~Becauſe the Triangles F G N and F B L are ~~

~~F G : ~~
~~F B : ~~
~~: ~~
~~G N : ~~
~~B L.~~
~~
~~

~~Now we have proved, that F G is the half of ~~
F B, therefore G N is likewiſe equal to the half

of B L, and conſequently equal to the Height

of the ſuppoſed Perpendicular.

~~
~~

~~Again, the ſimilar Triangles F G N and F a I ~~

~~F G : ~~
~~F a : ~~
~~: ~~
~~G N : ~~
~~a I.~~
~~
~~

~~But F G : ~~
~~F a : ~~
~~: ~~
~~G D : ~~
~~a H; ~~
~~becauſe the Tri-~~
angles F G D and F a H are ſimilar.

~~
~~

~~Whence~~

~~G N : ~~
~~a I : ~~
~~: ~~
~~G D : ~~
~~a H.~~
~~
~~

~~Now becauſe G N has been proved to be e-~~
qual to the Perpendicular, whoſe Perſpective is

requir’d and D G is ſuppoſed equal to that Per-

pendicular;

~~it follows, that G N and G D are ~~
equal;

~~and therefore a I and a H are alſo equal. ~~
~~Q E D.~~
~~
~~

~~I might have aſſumed C P equal to the Perpen-~~
dicular, and uſed the Points C and P inſtead of

B and L.

~~But uſing the ſaid Points B and L is ~~
better:

~~For when the Points C and P are uſed, ~~
the Horizontal Line muſt almoſt always be con-

tinued, that ſo a Line drawn through the Points

c and a may cut it;

~~moreover this Interſection ~~
will ſometimes be at an infinite Diſtance;

~~where-~~
as in uſing the Point B, M N can never be

greater than thrice the Breadth of the Deſign to

be drawn.

~~
~~

~~The ſixth Problem may be ſolv’d by this; ~~
~~for a Point elevated above the Geometrical ~~
Plane, may be conceived as the Extremity of a

Perpendicular to the Geometrical Plane.

~~
~~

~~59. ~~
~~To throw a Priſm or Cylinder into Perſpective, ~~

cal Plane.

~~
~~

~~Let G H I L M N be the Baſe of the Priſm ~~
in the Geometrical Plane, and the viſible Part

thereof upon the perſpective Plane, let be n g h i;

~~then to compleat the Repreſentation of the ~~
Priſm, draw Perpendiculars from the Points

n g h and i to the Baſe Line, whoſe Length let

be

Height of the Priſm, and find

face of the Priſm, in confidering them as Points

elevated above the Geometrical Plane:

~~This ~~
being done, if the Repreſentations of all the

ſaid Angular Points be joyn’d, the whole

Priſm will be thrown into Perſpective.

~~
~~

~~Now to throw a Cylinder into Perſpective, ~~
the Repreſentation of its Baſe and upper Sur-

face muſt firſt be had, by finding

upper Surface, and then two Perpendiculars

muſt be ſo drawn to the Baſe Line, that they

may touch the Appearances of the two circular

Euds of the Cylinder, and the Appearance of

the Cylinder will be had.

~~But to avoid uſeleſs ~~
Operations, the viſible Part of the Baſe of the

Cylinder may be thus determin’d.

~~Draw the ~~
Line A S from the Point A to the Station Point S,

then this Line muſt be biſected in the Point R,

about which, as a Centre, and with the Radius

R A, the Circular Arc B A C, muſt be deſcrib’d

cutting the Baſe of the Cylinder in the Points

ones that can be ſeen.

~~
~~

To do this another Way.
~~
~~
Fig. 26, 27.

46.

~~61. ~~
~~If the upper Face of the Cylinder or Priſm ~~

being given as in the foregoing Method, we draw

the Line P Q in the perſpective Plane, parallel

to the Baſe Line, whoſe Diſtance therefrom we

make equal to the Height of the Priſm or Cy-

linder, whoſe Perſpective is requir’d.

~~Then we ~~
change its Geometrical Plane, ſo that the Baſe

Line coincides with P Q, and that in this Tran-

ſpoſition a Perpendicular to the Baſe Line coin-

cides with this ſame Perpendicular continued to-

~~Finally we find ~~
in Situation by uſing P Q for a Baſe Line, and

the ſaid Perſpective is the Repreſentation of their

upper Faces.

~~
~~

~~If we ſuppoſe the Plane of the upper Surface ~~
of the Priſm to be continued, it will meet the

Perſpective Plane in P Q;

~~and the upper Face ~~
in this Plane continued, will have the ſame

Situation in Reſpect to P Q, as the Baſe hath

on the Geometrical Plane with Regard to the

Baſe Line.

~~If then the ſaid continued Plane be ~~
conceived to lye on the perſpective Plane, the

upper Faces of the Priſm or Cylinder, will be

as the Baſes changed in the Manner aforeſaid;

~~therefore the Appearance of the ſaid Baſes ~~
changed, will be that of the upper Surfaces.

~~
~~

~~Note, By folding the Paper it is eaſy to ~~
tranſpoſe Figures, and when the Height of the

~~
~~
~~
~~
~~
~~
precedent Method is the ſhorteſt.

~~
~~

~~62. ~~
~~To throw a Concave Body into Perſpective.~~
~~
~~

~~Having firſt ſound the Perſpective of the ſaid ~~
Body, afterwards find the Appearance of its

Cavity, in conſidering the Cavity as a new

Body.

~~
~~

~~63. ~~
~~To throw a Sphere into Perſpective.~~
~~
~~

~~Let A be the Seat of the Centre of the Sphere; ~~
~~then the Point I the Perſpective of the Centre ~~
muſt be found,

~~This being done, raiſe V F per-~~
pendicular to V I, which make equal to the

Diſtance from the Eye to the perſpective Plane;

~~and in this Perpendicular continued, take V P ~~
equal to the Diſtance from the Centre of the

Sphere to the perſpective Plane.

~~Through the ~~
Point P draw P Q parallel to V I cutting a Line

drawn from F through I, in Q;

~~and about Q as ~~
a Centre, with the Semidiameter of the Sphere,

draw the Circle C B, to which from the Point F,

draw the Tangents F C and F B, cutting the

Line I V in the Points G and E.

~~On the Line ~~
G E deſcribe the ſemicircle E D T G, wherein

draw the Line G D perpendicular to F I, which

biſect in H, and about H, as a Centre with the Ra-

dius H D, deſcribe the Arc of a Circle, L D R, cut-

ting the Line F I in the Points L and R.

~~Take ~~
the Chord G T in the Semicircle E D T G equal

to R L, and deſcribe a Semicircle T m G upon

G T;

~~in which Semicircle draw ſeveral Lines, ~~
as m n Perpendicular to G T;

~~and cutting the
~~
raiſe Perpendiculars p q, each of which muſt

be continued on each Side the Line G E, equal

to m n the Part of the correſpondent Line p m.

~~Now if a great Number of the Points q be thus ~~
found, and they are joyn’d by an even Hand,

you will have a Curve Line which will be the

Repreſentation ſought.

~~
~~

~~The Rays by which we perceive a Sphere, do form ~~
an upright Cone, whoſe Axis paſſes through the Cen-

ter of the Sphere, and whoſe Section made by the

Perſpective Plane, is the Repreſentation ſought:

~~from ~~
whence it follows, that I is the Point in the Perſpe-

ctive Plane, through which the Cone’s Axis paſſes.

~~But when an upright Cone is ſo cut by a Plane, that ~~
the Section is an Ellipſis, as in this Caſe, the tranſ-

verſe Diameter of this Ellipſis, will paſs through

the Point of Concurrence of the ſaid Plane, and

Axis of the Cone, and that Point wherein a Per-

pendicular drawn from the Vertex of the Cone, cuts

the ſaid Plane.

~~This will appear evident enough ~~
to any one of but mean Knowledge in Conick Secti-

ons.

~~Therefore the tranſverſe Axis of the Ellipſis, ~~
which is the Repreſentation of the Sphere, is ſome

Part of V I;

~~for the Eye is the Vertex of the Cone ~~
formed by the viſual Rays of the Spbere.

~~
~~

~~Now let us conceive a Plane to paſs through the ~~
Eye, and the Line I V;

~~this will paſs through the ~~
Center of the Sphere:

~~And if a Perpendicular be ~~
let fall from the Center upon the principal Ray con-

tinued, that Part of the ſaid Ray included between

the Point of Sight, and the Point wherein this Per-

pendicular falls, which is always parallel to the Per-

ſpective Plane, will be equal to the Diſtance from

the Center of the Sphere to the Perſpective Plane,

~~Therefore if the before-~~
mentioned Plane be ſuppoſed to revolve upon the Line

V I, as an Axis, until it coincides with the Per-

ſpective Plane, the Center of the Sphere will meet

the Perſpective Plane in Q, and the Eye in F;

~~whence the Part G E of the Line I V is the tranſ-~~
verſe Diameter of the Ellipſis.

~~
~~

~~Again let G D E in Figure 30, and g e f, in ~~

Letters in the foregoing Figure.

~~Now if the Cone, ~~
whoſe Profile is denoted by the Lines f g and fe be ſup-

poſed to be compleated, and to be cut by a Plane paſ-

ſing through the Line g e perpendicular to the Plane

of the Figure;

~~we ſhall have an Ellipſis g 4 e 3 ~~
ſimilar to that which is the ſought Repreſentation

of the Sphere.

~~Further if the ſaid Cone be conceived ~~
to be cut by a Plane 14 m 3 parallel to its Baſe,

and biſecting g e in n, it is manifeſt, that 3 4, the

common Section of the Circle 14 m 3, and the Ellip-

ſis g 4 e 3, is the conjugate Axis of the Ellip-

ſis.

~~And therefore this conjugate Axis is equal ~~
to the Line 3 4, Perpendicular in the Point n to the

Diameter 1 m of the Circle 14 m 3.

~~Now draw ~~
the Lines E O and G Y in Figure 30, parallel to

L M, then the Triangles E G Y and E N M are

EG: EN:: GY: NM.
~~
~~~~
~~

~~But E G is twice E N; ~~
~~wherefore G Y is alſo the ~~
double of N M, and ſo N M equal to G Z.

~~After ~~
the ſame manner we demonſtrate, that L N is equal

to X E;

~~whence it follows, that G D is equal to ~~
L M, and is ſo cut in z as L M is in N;

~~and there-~~
fore R L or G T of Figure 29, is equal to 34 in

Figure 31;

~~and conſequently equal to the conjugate ~~
Axis of the Ellipſis to be drawn.

~~On the other ~~
Hand, it is manifeſt by Conſtruction, that ſome one

of the Perpendiculars m n, Figure 29, viz.

~~that ~~
which paſſes through the Center of the ſemicircle

~~For if a Line be ~~
drawn from T to E, it will be perpendicular to G T,

and conſequently parallel to m n:

~~Whence the con-~~
jugate Axis of the Curve G q E, is equal to the

conjugate Axis of the Ellipſis to be drawn:

~~And ~~
therefore we are only to prove, that the Curve paſ-

ſing through the Points q, is an Ellipſis.

~~Which may ~~
be ſbewnthus.

~~
~~

~~The Parts G n of the Line G T, are Propor-~~
tional to the Parts G p of the Line G E:

~~Whence ~~
the Rectangles under G p and p E, are Proportional

to the Rectangles under G n and n T;

~~but theſe laſt ~~
Rectangles are equal to the Squares of the Ordinates

n m, which Squares are equal to the Squares of the

Ordinates p q;

~~therefore theſe laſt Squares are Pro-~~
portional to the Rectangles under G p and p E, which

is a Property of the Ellipſis.

~~
~~

~~The ſemicircular Part h m of a Column, en-~~

Torus.

~~
~~

~~64. ~~
~~To throw the Torus of a Column into Per-~~
ſpective.

~~
~~

~~Let B N C be the Baſe of the Column in the ~~

~~draw a Line from the Cen-~~
ter A to the Station Point S, which biſect in the

Point R, and deſcribe the Arc of a Circle B A C

about the Point R, as a Center with the Radius R A.

~~
~~

~~Let X be the Profile of the Column, in which ~~

ſemicircle h m, parallel to the Baſe of the Co-

lumn;

~~and in the Line s a, which goes through ~~
the Center of the Column, parallel to its Sides,

~~
~~
above the Point 2, which is in the Baſe of the

Column;

~~likewiſe aſſume s a in the ſaid Line ~~
equal to S A of the precedent Figure, and from the

Point a draw to a s the indefinite Perpendicular

a Y.

~~Theſe General Preparations being made, ~~
take at Pleaſure the ſmall equal Parts 6 i and 69

in the Line s a;

~~draw the Lines i h and 9 m Pa-~~
rallel to 63 z, and from the Point h draw the Line

h 3 4, thro’ the Center 3 of the Semicircle h m;

~~aſſume a 5 in a Y equal to i 4, and draw the Line ~~
5 s cutting i b in g, and 9 m in q.

~~And, (in Figure ~~
32.)

~~about the Point A, as a Center, with the ~~
Radius i h or 9 m, which are equal, deſcribe the

Circle F L M H, cutting the Arc B A C in the

Points D and E;

~~then draw the Line D E cut-~~
ting the Line A S in I;

~~aſſume I G equal to i g, ~~
and I Q equal to 9 q;

~~and thro’ the Points Q and ~~
G, draw F H and L M, parallel to the Line E D,

cutting the Circle D M E F in the Points L,

M, F, and H.

~~Now if the Repreſentations of Four ~~
Points, whereof L M F and H, are the Seats, and

the two firſt of which is equal to 29, and of the

two others 2 i, be found*;

~~the Repreſentation ~~
of the ſaid four Points will be ſo many Points of

Appearance ſought.

~~And by drawing two other ~~
Lines, as i h and 9 m, and proceeding as be-

fore, the Repreſentation of ſo many more Points

will be had.

~~
~~

~~Note, Becauſe a part of the Torus is hid by ~~

a Circle muſt be deſcribed about the Center A,

with the Radius 36, cutting the Arc B C A in

the Points T and O, and the Lines STY and SOZ

muſt be drawn;

~~then all the Points as F and H, ~~
falling between the Lines TY and O Z are uſeleſs,

and L and M not coming under this Obſervation

muſt only be uſed;

~~Note alſo, that there is no ~~
neceſſity to determine Geometrically (which

h z m, as far as the Parallels (as 9 m) are uſeful:

~~For when theſe Parallels are uſeleſs, the Point ~~
q will fall beyond the Point m:

~~But then the ~~
Perſpective of the Torus is entirely drawn alrea-

dy, if thoſe Parallels were firſt begun to be drawn

near to 6 3 z, and the others continually going

from it.

~~
~~

~~In order to demonſtrate this Problem, the fol-~~
lowing Lemma is neceſſary.

~~
~~

~~65. ~~
~~If two Circles C D H E and D E F L cut ~~

C L paſſes, and D E joyns their Interſections;

~~then, if the Radius A C or A H be called a, and ~~
B F or BL, b, and the Diſtance A B between the

two Centers c, I ſay A G is equal to {bb—aa/ec}—{1/2}C.

~~
~~

~~Let us call A G, x, and G D or G E, y. ~~
~~Then by the Property of the Circle, if y be conceiv’d ~~
as an Ordinate of the Circle, C D H;

~~yy=aa—xx. ~~
And if it be likewiſe conſider’d as an Ordinate of

the Circle F D L, yy=bb—cc—2cx—xx:

~~Whence ~~
aa—xx=bb—cc—2cx—xx, and ſo 2cx=bb

—aa—cc;

~~and dividing each Side of this laſt Equa-~~
tion by 2c, we have a x={bb—aa/2c}{1/2} c.

~~Which was ~~
to be Demonſtrated.

~~
~~

~~The Demonſtration of the ~~
Problem.~~
~~

~~66. ~~
~~The Torus of the Column muſt be conceiv’d as ~~
made up of an Infinite Number of Circular Planes,

lying one upon another.

~~And it is evident that ~~
the Reaſon why each of thoſe Circles cannot be wholly

ſeen, is becauſe that which is immediately under it

hides a Part thereof;

~~from whence it follows, that ~~
if the Plane of one oſ theſe Circles be every way

continu’d, and the Circle immediately under it, be

thrown

Repreſentation, and the Circle in the Plane, will deter-

mine the viſible Part of the ſaid Repreſentation;

~~and ~~
conſequently if the Repreſentation of theſe two Points

of Interſection be found upon the Perſpective Plane, we

ſhall have two Points of the Perſpective of the Torus of

the propoſed Column.

~~This is what I have done in the ~~
Solution of the Problem, as we ſhall now Analytically

demonſtrate.

~~
~~

~~Let O be the Eye, A M a part of the Torus of ~~

through the Center of the Column, and A B a

Parallel to the Baſe, drawn thro’ the Center B of

the Semicircle Concavity of the Torus.

~~Let M P be ~~
a Semidiameter of one of the Circles ſpoken of in the

the beginning of this Demonſtration.

~~Then if the ~~
Line m p be drawn parallel and infinitely near C M P

and the Lines m O and p O are drawn cutting M P

in D and T, it is evident that D T, which is in the

Plane of the Circle paſſing thro’ M P, will be the

Semidiameter of the Perſpective of the Circle imme-

diately underneath.

~~
~~

~~Now let fall the Perpendicular O S from the Eye ~~
to the Line A B, and continue the Lines M P and

m p, till they meet the ſaid Perpendicular in the Points

Q and q.

~~Moreover, continue the Line M P to the
~~
pendicular to m p.

~~Aſſume A S = e, OQ = x, ~~
and M P = y.

~~Then in the ſimilar Triangles O q m, ~~
and m R D, we have,

O q (x):

~~q m (e + y):~~
~~: m R (d x) R D: ~~
~~({edx + ydx/x}) ~~
The ſimilar Triangles O p q and p T D, give,

oq (x):

~~q p (e):~~
~~: p P (d x): ~~
~~P T ({edx/x}) ~~
P R is equal to y + dy, and if P T ({edx/x}) be added

to it, and then from the Aggregate be taken R D

~~Now to find the Points of Interſection of the two ~~
Circles, whoſe Radii are T D and P M, and Centers

diſtant from each other, by the Space T P, the Square

of T D leſs the Square of P M muſt

from, which may be here neglected, becauſe it is infinite-

ly ſmall in compariſon of the reſt;

~~and we ſhall have ~~
{xydy/dex} - {yy/e} for the Part of the Line PM included be-

tween P and the Point wherein this Line is cut by a

Line joyning the two Points of Interſection of the two

Circles.

~~
~~

~~Now before what I have here demonſtrated be ap-~~
ply’d to the Problem, we muſt obſerve, that if from

the Point M, a Line be drawn thro’ the Center B,

the Triangles M P C and m R M will be ſimilar;

~~for the Angle m M P is the Exterior Angle of the ~~
Triangle m R M, and the Angle m M C is a right

one.

~~And conſequently ~~
m R (dx):

~~R M (d y):~~
~~: M P (y): ~~
~~P C ({ydx/dx}).~~
~~
~~

~~Now if S A = s a in the 32d and 33d Figures ~~

~~as likewiſe
~~
~~it is manifeſt, that i 4 = a 5 ~~

~~Again, the ſimilar Triangles, s a 5 and s i g give ~~
s a (e):

~~a 5 ({ydy/dx}):~~
~~: s i (x): ~~
~~i g ({xydx/edx}) Alſo by ~~
the Conſtruction of Figure 32,

A S (e):

~~A P = i h (y):~~
~~: A P (y): ~~
~~A I ({yy/e}); ~~
~~Whence it follows, ſince I G = i g, that A G = ~~
(I G - A I) = {xyd/cdx} - {yy/e}.

~~And conſequently, H ~~
and F are the Seats of the two Points whoſe Perſpe-

ctive is required, and thoſe Points are both in a

Plane parallel to the Geometrical Plane, which is the

height of 21 above the Geometrical Plane.

~~
~~

~~If the precedent Calculation be apply’d to the Lower ~~
Part of the Torus, the Expreſſion {xydy/edx} - {yy/e}, will

be chang’d into this, - {xydy/edx} - {yy/e;

~~} which ſhews that ~~
theſe two Quantities muſt be aſſumed on the ſame Side

of A, viz.

~~towards S. ~~
~~Moreover 9 q, inthe Line ~~
9 m, is equal to {xydy/edx};

~~for 98 ({ydy/e}) = i 4. ~~
~~Which ſhews that M and L are alſo the Seats of two ~~
Points whoſe Perſpective muſt be found, and which are

both in a Plane parallel to the Geometrical Plane, and

above it the Height of 29.

~~
~~

~~67. ~~
~~This Problem may be likewiſe ſolved in ~~
conſidering the Torus of a Column as made up of

an infinite Number of Baſes of Cones, whoſe Al-

titudes are determin’d by the concurrence of the

Tangents of the Semicircular Concavity of the

Axis of the Column;

~~and then determining ~~
~~Note, This ~~
Method may be demonſtrated without Algebra,

but it would be very long.

~~
~~

~~68. ~~
~~To find the Accidental Point of ſeveral pa-~~
rallel Lines, which are inclin’d to the Geome-

trical Plane.

~~
~~

~~Let A B be the Direction of one of the Lines, ~~

~~and ECP, the ~~
Angle that the ſaid Lines make with the Geo-

metrical Plane.

~~
~~

~~Draw a Line, O D, thro’ the Eye O, parallel ~~
to A B, and thro’ the Point D, wherein it cuts

the Horizontal Line, and which is the acciden-

tal Point of the Directions of the given Lines,

draw D F perpendicular to the ſaid Horizontal

Line;

~~in which aſſume D G, equal to DO. ~~
~~Fi-~~
nally, thro’ the Point G, draw the Line G F,

making an Angle with the Horizontal Line, equal

to E C P;

~~and then the Point F, (the Interſection ~~
of this Line) and the Perpendicular D F, is the

accidental Point ſought.

~~
~~

~~Note, When the Lines are inclin’d towards ~~
the perſpective Plane, D F and G F muſt be

drawn below the Horizontal Line:

~~And, contra-~~
riwiſe, when the ſaid Lines are inclin’d towards

the oppoſite Part of the perſpective Plane, the

aforeſaid Lines muſt be drawn above the ſaid

Horizontal Line, as is done here.

~~
~~

~~If a Plane be conceiv’d to paſs thro’ the Eye, ~~
perpendicular to the Geometrical Plane, and paral-

lel to the given Lines;

~~it is evident, that the ſaid ~~
Plane will cut the Horizontal Plane in the Line

O D, and the perſpective Plane in D F.

~~It is, ~~
moreover, manifeſt, that a Line drawn thro’ the

Eye, parallel to the given Line, is in the ſaid

Plane, and (with the Line O D) makes an An-

gle, equal to the Angle E C P, below the Hori-

zontal Plane, if the Lines be inclin’d towards

the perſpective Plane, and above it, if they in-

cline to the oppoſite ſide;

~~whence this laſt Line ~~
makes a right-angled Triangle with O D and

D F, whoſe Angle at the Point O, is equal to

the Angle C E P.

~~But D G F is likewiſe a ~~
right-angled Triangle, as having the Angle at the

Point G, equal to ECP;

~~therefore theſe two ~~
Triangles are ſimilar.

~~And ſince the Side D G ~~
is equal to the Side D O, the Triangles are alſo

equal:

~~Therefore the Line D F, being common ~~
to theſe two Triangles;

~~the Point F, is the ~~
Point wherein the Line, paſſing thro’ the Eye

parallel to the given Line, meets the Per-

ſpective Plane:

~~And this Point is ~~
~~
~~

~~Note, This Demonſtration as well regards ~~

metrical Plane, as thoſe that meet it in one of

their Extremes only.

~~
~~

~~69. ~~
~~To find the Repreſentation of one or more ~~
Lines, inclin’d to the Geometrical Plane.

~~
~~

~~Let A be a Point given in the Geometrical ~~
Plane;

~~whereon ſtands a Line, whoſe Length, ~~
Direction, and Angle of Inclination is known.

~~
~~

~~In ſome ſeparate Place, draw the Lines C E ~~
and C P, making an Angle with each other

equal to the Angle of Inclination of the given

Line;

~~and in one of theſe Lines, aſſume C E ~~
equal to the given Line, and let fall the Perpen-

dicular E P, from the Point E upon the other

Line.

~~Then aſſume A B, in the Direction of ~~
the propos’d Line, equal to C P;

~~and after ha-~~
ving found a, the Perſpective of A, and the

Point T

~~join the Points a ~~
and T by a right Line;

~~and the ſought Appear-~~
ance will be had.

~~
~~

~~If from the Extremity of the inclin’d Line, a ~~
Perpendicular be let fall upon the Geometrical

Plane, the ſaid Perpendicular will meet this

Plane in the Point B, and will be equal to P E;

~~as is evident by the Conſtruction of the Figure ~~
C P E.

~~But the Point T is the Repreſentation ~~
of the Extremity of this Perpendicular;

~~and ~~
therefore it is alſo the Extremity of the inclin’d

Line.

~~Which was to be demonſtrated.~~
~~
~~

~~There are ſome Caſes of this Propoſition, that ~~
may be ſhorten’d.

~~As, 1. ~~
~~When there are ſeve-~~
ral Lines of this Kind parallel between them-

ſelves, and whoſe accidental Point can be

found

~~And, 2. ~~
~~When an inclin’d Line is pa-
~~
~~The Manner of
~~
in the following Methods.

~~
~~

~~Thro’ F, the accidental Point of the inclin’d ~~

Line, and equal to F G.

~~And let A be the ~~
Point, wherein one of the inclin’d Lines meets

the Geometrical Plane.

~~
~~

~~Aſſume R Q in the Baſe Line, equal to the ~~
inclin’d Line;

~~and draw Lines from the Points ~~
R and Q, to the Point Z, taken at pleaſure in

the Horizontal Plane.

~~
~~

~~Thro’ a, the Perſpective of A, draw a N pa-~~
rallel to the Baſe Line;

~~in which aſſume a L, ~~
equal to M N;

~~and draw a Line from the Point ~~
a, to the Point F;

~~and from the Point L, draw ~~
another to the Point H.

~~Then a T will be the ~~
Perſpective ſought.

~~
~~

~~By ~~
~~and therefore, we are only to demonſtrate, ~~
that the Extremity of the faid Perpendicular is in

the Line L H.

~~Which may be thus done.~~
~~
~~

~~Let us ſuppoſe a Line, A I, to paſs thro’ the ~~
Point A, parallel to the Baſe Line, and equal to

the inclin’d Line.

~~It is then manifeſt ~~
~~and conſequently, ~~
L H

Line;

~~and therefore the Perſpective of this Ex-~~
tremity is in the Line L H;

~~which was to be ~~
demonſtrated.

~~
~~

~~Note, if F H had been aſſumed, the one half, ~~
or one third, &

~~c. ~~
~~of what it is; ~~
~~then it is mani-~~
feſt

~~c. ~~
~~of C E.~~
~~
~~

~~71. ~~
~~Forinclined Lines not meeting the Geometrical ~~
Plane.

~~
~~

~~Let A and B be the Seats of the Extremities of ~~

~~Let X repreſent a Plane paſſing ~~
through the given Line perpendicular to the

Geometrical Plane.

~~Likewiſe let M N in this ~~
Plane, repreſent the Line whoſe Perſpective is re-

quir’d;

~~and let C N and P M be perpendicular to ~~
the Geometrical Plane:

~~Whence P C repreſents ~~
A B, and conſequently is equal thereto.

~~
~~

~~Find the Point I ~~
~~and ~~
draw the Line B S, from the Point B, to the

Station Point I, cutting the Baſe Line in E;

~~and ~~
from the Point I, draw a Line to the accidental

Point F;

~~which cut by a Perpendicular to the ~~
Baſe Line, raiſed at the Point E;

~~and then I T ~~
will be the Appearance ſought.

~~
~~

~~72. ~~
~~For inclined Lines parallel to the perſpective ~~
Plane.

~~
~~

~~The Operation of Prob. ~~
~~VII. ~~
~~muſt be uſed ~~
here, but with this Difference (ſee Fig.

~~of the ~~
ſaid Prob.)

~~that whereas a I in the ſaid Problem ~~
is perpendicular to the Baſe Line, here it muſt

make an Angle with the Baſe Line, equal to the

Angle of Inclination of the given Lines.

~~
~~

~~For the Demonſtration of this, ſee n. ~~
~~7, and 10.~~
~~
~~

~~73. ~~
~~To throw a Body into Perſpective, having ~~
ſome one or all of its Sides inclined to the Geometri-

cal Plane.

~~
~~

~~The Appearances of the Lines forming the ~~
Angles of the propoſed Body muſt be found:

~~And ~~
this may be eaſily done by Prob.

~~10. ~~
~~And in this Manner the ~~
Appearance of a Pyramid, an inclined Priſm,

&

~~c. ~~
~~may be found. ~~
~~But nevertheleſs, it hap-~~
pens ſometimes, that the Operations of the pre-

cedent Problem may be abbreviated;

~~as when ~~
the Extremity of ſeveral Lines are found in one

and the ſame Line, or when inclined Lines, that

have difficult accidental Points, interſect one

another, and ſo mutually determine each other.

~~This will appear manifeſt by the following Ex-~~
amples.

~~
~~

~~To throw ſeveral parallel Shores which ſtrengthen a ~~
Wall, into Perſpective.

~~
~~

~~I ſuppoſe here that the Baſes of theſe Shores, ~~

face of the Ground, are all in a right Line, pa-

rallel to the Side of the Wall;

~~and then the ~~
faid Shores may be thrown into Perſpective in

the following Manner:

~~Having firſt found ~~
preſentation of their Baſes:

~~This being done, ~~
denote the Appearances of the Lines wherein

the Shores meet the Wall, upon the Perſpective

of the Wall;

~~the Appearances here are the ~~
Lines p t, r s, which repreſent Lines parallel to

the Geometrical Plane, from the Suppoſition,

that the Shores are parallel between themſelves,

and their Baſes equally diſtant from the Wall.

~~Finally, draw Lines from the Angles of the Re-~~
preſentations 1 2 3 4, to the Point F, which

will be terminated by their Interſections with

p t and r s, and will give the Appearances ſought,

as you ſee in the Figure.

~~
~~

~~To throw ſeveral parallel Roofs of a Houſe into ~~
Perſpective.

~~
~~

~~Having found the accidental Points G and Q ~~

the Wall ſuſtaining them, denote the Points

a b c d, wherein the ſaid Roofs meet the Wall:

~~Then from the Point G draw Lines through the ~~
Points a b c;

~~and from the Point Q others to ~~
the Points b c d;

~~theſe Lines by their mutual ~~
Interſection will determine each other, and give

the Repreſentations ſought.

~~
~~

~~74. ~~
~~From what has been already ſaid, it will ~~
not be difficult to throw any Objects whatſoever

into Perſpective.

~~But ſince it is very difficult, ~~
and indeed impoſſible for a Painter to make a

Deſign entirely according to the Rules we have

preſcribed;

~~the Number of Points to be found ~~
being almoſt infinite:

~~therefore the Figures
~~
~~
~~
~~
~~
~~
~~
principal Points of the Objects without the ſaid

Plane, need only be thrown into Perſpective.

~~Which being once obtained, he may make uſe ~~
of theſe Appearances ſo found, as a Rule where-

by the reſt may be compleated by the Eye, with-

out running the Riſque of committing ſome

conſiderable Fault, which by this Means may be

avoided.

~~
~~

CHAP. IV.
~~
~~~~
~~

~~Of the Practice of Peſpective upon the Per-~~
ſpective Plane ſtill conſider’d as being upright.

~~
~~

~~IT often happens that Painters offend all ~~
Rules of true Appearance when they paint

Pictures to ſtand aloft, to be ſeen Sideways, or at

a confiderable Diſtance.

~~Their Cuſtom is to ~~
paint Pictures to be view’d, after the ſame Man-

ner as they themſelves look at them when they

are working;

~~whence in the following Caſes, ~~
this Practice of theirs will be uſeleſs;

~~and ſo to ~~
avoid enormous Faults, they are neceſſitated to

have recourſe to Perſpective But what has been

ſaid in the laſt Chapter, does not reach theſe

particular Caſes;

~~therefore we ſhall here add ſome ~~
new Problems, which together with the former

ones, will take in all Caſes.

~~
~~

~~75. ~~
~~To throw Figures which are in the Geometri-~~
cal Plane into Perſpective, when the Eye is at ſo great

a Diſtance that it cannot be denoted in the Horizon-

Horizontal Line.

~~
~~

~~The Repreſentation of two Points of theſe ~~
Figures muſt be firſt found

~~and then by Means
~~
may be had

~~
~~

~~Let A B C D E, be a Pentagon, whoſe Ap-~~

~~V the Point of Sight; ~~
~~and ~~
V F the ſixth Part of the Eye’s Diſtance from

the perſpective Plane.

~~Now find ~~
the Appearance of the Point A will be had

~~
~~
tion of A and E, will that of D be had;

~~and ~~
by uſing B and A, the Perſpective of C may be

found.

~~
~~

~~76. ~~
~~Note, the Perſpective of Lines perpendicu-~~
lar to the Geometrical Plane

~~as alſo of Lines
~~
~~
~~

~~77. ~~
~~To throw Figures, which are in the Geometri-~~
cal Plane into Perſpective, when the Eye is ſo oblique

that it cannot be marked in the Horizontal Plane,

or the Point of Sight in the Horizontal Line.

~~
~~

~~We muſt proceed here according to the Di-~~
rections of the precedent Problem, after having

found the Perſpective of ſeveral Points of the

given Figures.

~~
~~

~~At any Point C, taken at Pleaſure in the Baſe ~~
Line, draw the Perpendicular C D to the ſaid

ſame Point in ſuch manner, that if it could be

~~
~~
the Point of Sight.

~~
~~

~~This is done in aſſuming C H equal to {1/3}, or ~~
{1/4} Part, &

~~c. ~~
~~of the Diſtance from the Point C, ~~
to the Foot of the vertical Line;

~~and in raiſing ~~
the Perpendicular H E, in the Point H, equal

to {1/3} or {1/4} Part, &

~~c. ~~
~~of the Height of the Eye. ~~
~~Now A is a given Point, whoſe Appearance is ~~
ſought.

~~
~~

~~Draw a Parallel A B, through the Point A, ~~
to the Baſe Line, meeting the Line C D in the

Point B, and let a ſecond Eye be ſuppoſed at

the ſame Height and Diſtance as the firſt;

~~then ~~
find

Line C E in b, and in this Continuation aſſume

b a equal to F G;

~~then a will be the Perſpective ~~
ſought.

~~
~~

~~Becauſe the Height and Diſtance of the ſecond ~~
Eye, is equal to the Height and Diſtance of the

firſt;

~~the ſaid two Eyes are both in one parallel ~~
Line A B;

~~and conſequently ~~

~~And therefore becauſe ~~

~~and a ~~
~~Which
was to be demonſtrated.~~
~~
~~

~~78. ~~
~~Note, as to Lines perpendicular, and inclined ~~
to the Geometrical Plane, ſee n.

~~76. ~~
~~This is ~~
ſcarcely uſeful, unleſs for the Decorations of a

Theatre.

~~
~~

~~79. ~~
~~To find the Repreſentation of a Figure in the ~~
Geometrical Plane, when the Perſpective Plane is

placed above the Eye.

~~
~~

~~When the perſpective Plane is ſituated above ~~
the Eye, we ſuppoſe the Geometrical Plane to

paſs through the Top of the Perſpective Plane;

~~upon which Geometrical Plane are drawn the ~~
Figures of Objects meeting it;

~~as alſo the Seats ~~
of thoſe Objects that are underneath it, by

Means of Perpendiculars;

~~and the Height of the ~~
Eye is here meaſur’d by a Perpendicular drawn

from the Eye to the Geometrical Plane;

~~whence ~~
the perſpective Plane, elevated in reſpect to the

Eye, is the ſame thing, as an Eye elevated in

regard to the perſpective Plane.

~~
~~

~~Let I L be the Baſe Line, and H the Foot of ~~

~~then in the Baſe Line aſ-~~
ſume the Points I and L at Pleaſure, towards

the Sides of the perſpective Plane.

~~Make I S ~~
equal to {1/3} or {1/4} Part of I H, and raiſe the Per-

pendicular S X, in the Point S, to the Baſe Line,

equal to a correſpondent Part of the Height and

Diſtance of the Eye taken together;

~~draw the ~~
Line X I G, and moreover Y L Q, in aſſuming

L T equal to {1/3} or {1/4} &

~~c. ~~
~~of L H. ~~
~~Again draw ~~
the Line G Q in the Geometrical Plane, pa-

rallel to the Baſe Line, and diſtant therefrom

(for Example) a third Part of the Height of

the Eye;

~~draw alſo F P in the perſpective Plane, ~~
parallel to the Baſe Line, and diſtant therefrom,

a fourth Part of the Eye’s Diſtance;

~~theſe two ~~
Lines will cut X I in G and F, and Y L in Q

and P.

~~Note, if the Diſtance of G Q from the ~~
Baſe Line, had been aſſumed equal to a fourth

Part of the Eye’s Diſtance;

~~then F P muſt have
~~
Part of the Eye’s Diſtance, and ſo on.

~~Now ~~
A is a Point whoſe Repreſentation is requir’d.

~~
~~

~~Draw the Lines A F and A P, from the Point ~~
A to the Points F and P, cutting the Baſe Line

in the Points E and B;

~~then draw the Lines E G ~~
and B Q, which continue till they interſect

each other in a, which is the Repreſentation

ſought.

~~
~~

~~Let us ſuppoſe the perſpective Plane continu-~~
ed, C D the Horizontal Line, and O the Eye

denoted in the Horizontal Plane.

~~It is evi-~~
dent

~~produce the ~~
Line G S a, until it meets the Horizontal Line

in D, and draw the Line O D.

~~Let fall the ~~
Perpendicular G N R, from the Point G upon

the Horizontal Line, which interſect in R, by

the Line O R, paſſing through the Eye parallel

to the Horizontal Line.

~~Now by Conſtruction, ~~
G M is {1/3} of M N;

~~and conſequently it is {1/4} of ~~
G N;

~~M Z is likewiſe {1/4} of N R: ~~
~~Therefore~~

~~G M: ~~
~~M Z:~~
~~: G N: ~~
~~N R.~~
~~
~~

~~Compon. ~~
~~and Altern.~~
~~
~~

~~G M: ~~
~~G N:~~
~~: G M + M Z = G Z: ~~
~~G N ~~
+ N R = G R.

~~
~~

~~Becauſe the Triangles G M I and G N C are ~~

~~G M: ~~
~~G N:~~
~~: G I: ~~
~~G C.~~
~~
~~

~~The Triangles G Z F and G R O being alſo ~~

~~G Z: ~~
~~G R:~~
~~: G F: ~~
~~G O.~~
~~
~~

~~Whence~~

~~G I: ~~
~~G C:~~
~~: G F: ~~
~~G O.~~
~~
~~

~~Again, becauſe the Triangles G I E and G C D ~~

~~G I: ~~
~~G C:~~
~~: G E: ~~
~~G D.~~
~~
~~

~~And conſequently~~

~~G F: ~~
~~G O:~~
~~: G E: ~~
~~G D.~~
~~
~~

~~And ſo the Triangles G F E, and G O D are ~~
ſimilar;

~~and the Line F E A is parallel to O D: ~~
~~Whence it follows ~~
~~We demonſtrate in ~~
the ſame Manner, that B a is the Perſpective

of B A, and ſo the Perſpective of the Point A,

the common Section of E A and B A, is a, the

Interſection of the Appearances of the ſaid two

Lines.

~~
~~

~~80. ~~
~~To find the Repreſentation of a Line, per-~~
pendicular to the Geometrical Plane, when the per-

ſpective Plane is above the Eye.

~~
~~

~~In the Baſe Line B E, aſſume the Line E D, ~~

~~and draw C L, parallel to the Baſe Line, and ~~
diſtant therefrom (for Example) {1/4} of the Height

of the Eye;

~~make F L equal to {3/4} of D E, and ~~
draw the Lines E L and D F.

~~Note, if the ~~
Diſtance from C L to B E, had been aſſumed

equal to a fifth Part of the Height of the Eye,

F L muſt have been aſſumed equal to {4/5} Parts of

E D.

~~Now let a be the Perſpective of the Foot ~~
of the propoſed Perpendicular;

~~through which ~~
draw a H parallel to the Baſe Line, and a I per-

pendicular to the ſaid Line;

~~then make a I equal ~~
to G H, and the propoſed Perſpective will be

had.

~~The Demonſtration of this Operation is ~~
manifeſt

~~
~~
~~
~~
~~
~~
Horizontal Line.

~~
~~

CHAP. V.
~~
~~

~~Of throwing Figures into Perſpective, when ~~
the Perſpective Plane is conſider’d as being

inclined.

~~
~~

~~81. ~~
~~TO find the Perſpective of a Figure in the ~~

~~
~~

~~Let X be the Vertical Plane; ~~
~~S I the Station ~~
Line, S the Station Point, and H the Interſecti-

on of the Station Line and Baſe Line.

~~Now ~~
draw the Vertical Line H V through the Point H,

making an Angle with S I, equal to the Angle

of Inclination of the perſpective Plane;

~~then ~~
raiſe the Perpendicular I O to S I, in the Sta-

tion Point S, equal to the Height of the Eye;

~~and through the Extremity of the ſaid Perpen-~~
dicular, draw the principal Ray O V, paral-

lel to S I, and cutting H V in the Point of

Sight V.

~~
~~

~~Now it is evident, that O V determines the ~~
Length of the principal Ray, and H V the Di-

ſtance from the Baſe Line to the Horizontal

Line;

~~and ſince the Demonſtration of the ~~
Problems in the aforegoing Chapters regarding

the Geometrical Plane, have alſo Relation to

the perſpective Plane being inclined, the ſaid

Problems may be here uſed;

~~and conſequently, ~~
this inclined perſpective Plane is reduced to a

Perpendicular one, view’d by an Eye, whoſe

Height is H V, and Diſtance O V.

~~
~~

~~82. ~~
~~To find the Appearance of a Point above the ~~
Geometrical Plane.

~~
~~

~~Let H C be the Baſe Line: ~~
~~And let T be the ~~

to the Geometrical Plane.

~~This Point will ~~
be

ſuring the Height of the Eye;

~~for this laſt ~~
Line is parallel to the ſaid Perpendiculars.

~~And ~~
ſo likewiſe the aforeſaid Point is the ſame as

the Point T of Fig.

~~44: ~~
~~Let V be the Point of ~~
Sight, S the Station Point, and Q the Station

Point of the upright perſpective Plane, to which

the inclined perſpective Plane is reduced

~~And
~~
~~
~~

~~Draw two Lines M P and P E ſeparately, ~~
making a right Angle with each other;

~~in one ~~
of which, aſſume P E, equal to the Height of

the given Point, whoſe Perſpective is ſought;

~~and draw the Line E M, making an Angle with ~~
M P, equal to the Angle of Inclination of the

perſpective Plane.

~~Again let fall the Perpen-~~
dicular A D from the Point A to the Baſe

Line, in which aſſume A L equal to P M, to-

wards the Baſe Line, when the perſpective

Plane is inclined towards the Objects (as we

have here ſuppoſed) but on the other Side of A,

when the perſpective Plane inclines towards the

Eye.

~~Then from the Point A, draw a Line ~~
to the Point S, cutting the Baſe Line in B, and

joyn the Points L and Q, by a Line cutting the

Baſe Line in C.

~~This being done, draw the
~~
~~which interſect in the Point X, ~~
by a Perpendicular to the Baſe Line, in the

Point G;

~~and then the Point X is the Appear-~~
ance ſought.

~~
~~

~~In Fig. ~~
~~44. ~~
~~where V, S, T, and H, repreſent ~~
the ſame Points as thoſe that are denoted with

the ſame Letters in this Figure;

~~we have, ~~

~~T H: ~~
~~H S:~~
~~: T V: ~~
~~V O.~~
~~
~~

~~Compon. ~~
~~and altern.~~
~~
~~

~~T H: ~~
~~T V:~~
~~: T H + H S: ~~
~~T V + V O.~~
~~
~~

~~This being apply’d to Fig. ~~
~~45. ~~
~~and it will be, ~~
T H:

~~T V:~~
~~: T S: ~~
~~T V + V O.~~
~~
~~

~~If now T X be continued, till it cuts the Ho-~~
rizontal Line in F;

~~we ſhall have,~~

~~T H: ~~
~~T V:~~
~~: T B: ~~
~~T F.~~
~~
~~

~~And conſequently,~~

~~T B: ~~
~~T F:~~
~~: T S: ~~
~~T V + V O.~~
~~
~~

~~Whence it follows, that if a Line be drawn ~~
ſrom the Eye, to the Point F, it will be paral-

lel to S B A.

~~Therefore ~~
~~and ſo the Repreſenta-~~
tion of A is in the ſaid Line.

~~The Perſpective ~~
of a Line perpendicular to the Geometrical

Plane, in the Point A, paſſes thro’ the Perſpe-

ctive of the Point A, and thro’ the Point T

~~
~~
~~But the given Point ~~
is in the aſoreſaid Perpendicular:

~~And ſo its Per-~~
ſpective is in T X.

~~
~~

~~Again; ~~
~~it is otherwiſe manifeſt, that the Per-~~
ſpective of C L, is a

~~and conſe-
~~
~~Now, if a Line be ſuppos’d to be drawn from ~~
the Point L, thro’ the propos’d Point, it will be

parallel to the Vertical Line;

~~and ſo its ~~
~~And ſince
~~
L, it will be a Part of C X.

~~But becauſe that ~~
Line, drawn from the Point L, paſſes thro’ the

propos’d Point;

~~the Repreſentation of the ſaid ~~
Point is alſo in C X;

~~and ſo in X, the common ~~
Interſection of C X, and T X.

~~
~~

~~83. ~~
~~If the Point T ſhould be at too great a ~~
Diſtance;

~~or if T B X, or C X, ſhould too ob-~~
liquely cut each other;

~~the perſpective Plane ~~
muſt then be ſuppos’d to be reduc’d

~~and the Repreſen-~~
tation of a Point, above the Geometrical Plane,

(whoſe Seat is L, and Height M E) muſt be

found

~~
~~

~~84. ~~
~~To find the Repreſentation of a Line, perpendi-~~
cular to the Geometrical Plane`.

~~
~~

~~The Appearance of the Extremity of the Per-~~

cal Plane, by the Height of the propos’d Perpendi-

cular:

~~Then if a Line be drawn from the Point ~~
D, to the Point of Sight;

~~its Interſection ~~
the Perpendicular propos’d.

~~
~~

~~Note, When there is a Neceſſity of having re-~~
courſe to the Remarks of the foregoing Problem,

in order to find the Point X;

~~then the Point a ~~
may be found, in drawing A S and D V, and

afterwards joining the Points B and X by a

Line.

~~And when B X and D V cut each other
~~
~~
~~
~~
~~
~~
~~
~~
~~

~~85. ~~
~~A is the Foot of the Perpendicular: ~~
~~The ~~

~~And
~~
to the Geometrical Plane.

~~
~~

~~Thro’ the Point a, the Appearance of A, ~~
draw a Perpendicular to the Baſe Line;

~~which ~~
make equal

~~in conſidering this laſt Line, as being ~~
parallel to the Vertical Line.

~~Then, from the ~~
Extremity I of this Perſpective, to the Point of

Sight V, draw a Line cutting the Line T a, in

the Point X;

~~which will be the Repreſentation ~~
of the Extremity of the propos’d Line.

~~
~~

~~Let us ſuppoſe a Line paſſing thro’ the Point ~~
A, equal to M E, and parallel to the Verti-

cal Line.

~~Suppoſe, moreover, that another Line ~~
is drawn thro’ the Extremity of this Line, and

that of the propos’d Perpendicular;

~~then this ~~
laſt Line, by the Conſtruction of the Figure

M E P, will be parallel to the Station Line;

~~and conſequently, its Repreſentation ~~
~~and its Interſection ~~
with T a, will be the Extremity of the Repre-

ſentation ſought.

~~But a I is ~~
~~and con-~~
ſequently, V I is that of the ſecond.

~~Which was ~~
to be demonſtrated.

~~
~~

~~Note, When V I and T a cut each other ve-~~
ry obliquely, recourſe muſt be had to the Obſer-

vation at the End of the aforegoing Method, or

the following Way may be uſed.

~~
~~

~~86. ~~
~~Let T be the accidental Point of the Lines ~~

which Point draw a parallel to the Baſe Line,

in which aſſume T R equal to O T of Fig.

~~44.~~
~~
~~

~~Aſſume D N ſomewhere in the Baſe Line, e-~~
qual to the propoſed Line, and draw the Lines

D F and N F to the Point F, taken at Pleaſure

in the Horizontal Line;

~~then through the Point ~~
a, the Appearance of A, draw the parallel a H,

to the Baſe Line, in which aſſume a Q equal

to G H.

~~Then if the Lines T a, and R Q be ~~
drawn, and continued, till they cut each other

in the Point X, a X will be the Appearance

ſought.

~~
~~

~~The Part a Q of the Line a H, is ~~
Geometrical Plane, and which is equal to the

propoſed Line, and parallel to the Baſe Line;

~~and conſequently ~~
ſed Line:

~~And therefore X the Interſection of ~~
R Q and T a, is the Appearance of the ſaid

Extremity.

~~
~~

~~It is manifeſt ~~
~~c. ~~
~~of what it is taken here, ~~
provided likewiſe that then D N be aſſumed

equal to a correſpondent Part of the propoſed

Line.

~~
~~

~~87. ~~
~~To throw a Sphere into Perſpective.~~
~~
~~

~~The Method of ſolving this Problem before ~~
laid down

~~that inſtead of uſing the Point of ~~
Sight, the Point wherein a Perpendicular drawn

from the Eye to the perſpective Plane, meets

the ſaid Plane, muſt be uſed.

~~And you muſt ~~
obſerve, that this Perpendicular meaſures the

Eye’s Diſtance from the perſpective Plane.

~~
~~

~~88. ~~
~~To find the accidental Point of any Number ~~

~~
~~

~~Let A B be the Direction of one of the in-~~
clined Lines, O the Eye in the Horizontal Plane,

and S the Station Point.

~~
~~

~~Draw the Line O D, thro’ the Eye O parallel ~~
to A B, meeting the Horizontal Line in D, which

will be

~~and thro’ the Station Point S,
~~
cutting the Baſe Line in N, and draw the Line

N D.

~~Then about the Point D, as a Center, and ~~
with the Radius D O, deſcribe the Circular Arc

O o:

~~And about N, as a Center, with the Radius ~~
N S, draw the Circular Arc S s.

~~This being done, ~~
draw the Line s o touching the ſaid two Arcs,

and the Line D o perpendicular to s o.

~~Then ~~
draw o F, making an Angle with o D, equal to

the Angle of the Inclination of the Lines given,

and cutting N D continued in F:

~~Now I ſay F ~~
is the Accidental Point ſought when the Lines

do not incline towards the Perſpective Plane:

~~But if they do, o F muſt be drawn below o D.~~
~~
~~

~~If a Plane be ſuppoſed to paſs thro’ the Eye ~~
parallel to theinclin’d Lines;

~~the common Secti-~~
ons of this Plane, and the Horizontal and Geo-

metrical Planes, will be O D and S N.

~~It is ~~
now manifeſt, that if a Line be drawn in the

ſaid Plane, below the Horizontal Plane, when

the Lines incline towards the perſpective Plane,

and above it when they incline the other way,

making an Angle with O D equal to the Angle

of Inclination of the propoſed Lines;

~~I ſay it is ~~
manifeſt, that the ſaid Line will be parallel to

the propoſed Lines, and will meet

~~If ~~
now the before ſuppoſed Plane be conceiv’d to

turn about the Line N D, the Eye, and the

Station Point in the ſaid Plain, will then meet

the perſpective Plane in the Points o and s;

~~for ~~
the Lines D o and N s are equal to D O and N S,

and form right Angles with the Line s o joyning

their Extremities.

~~Now the two Points s and o ~~
anſwer to the Situation of the Eye and Station

~~
~~