Gravesande, Willem Jakob, An essay on perspective, 1724

Bibliographic information

Author: Gravesande, Willem Jakob
Title: An essay on perspective
Year: 1724
Number of Pages: 205

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Document ID: MPIWG:248Y7C1N
Permanent URL: http://echo.mpiwg-berlin.mpg.de/MPIWG:248Y7C1N

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Copyright: Max Planck Institute for the History of Science (unless stated otherwise)
License: Internal use only, please contact library@mpiwg-berlin.mpg.de (unless stated otherwise)
Table of contents
1. Page: 0
2. South Librarp. Page: 2
3. AN ESSAY ON PERSPECTIVE. Page: 5
4. MAX--INSTITUT FOR WISSEESCHICHTE Bibliothek Page: 6
5. TO Mr. William Kent. Page: 7
6. The AuTHOR’S PREFACE. Page: 9
7. ERRATA. Page: 18
8. AN ESSAY ON PERSPECTIVE. CHAP. I. Definitions. Page: 19
9. CHAP. II. The Theory of Perſpective. Lemma. Page: 23
10. Theorem I. Page: 24
11. Corollary I. Page: 24
12. Corollary II. Page: 24
13. Corollary III. Page: 24
14. Theorem II. Page: 25
15. Corollary I. Page: 26
16. Corollary II. Page: 30
17. Theorem III. Page: 30
18. Theorem IV. Page: 30
19. Corollary I. Page: 31
20. Corollary II. Page: 31
21. Corollary III. Page: 32
22. Corollary IV. Page: 32
23. Theorem V. Page: 32
24. Theorem VI. Page: 33
25. Corollary. Page: 33
26. CHAP. III. Page: 33
27. Problem I. Page: 37
28. Operation. Page: 37
29. Demonstration. Page: 37
30. Remarks. Page: 41
31. Method II. Page: 42
32. Operation. Page: 42
33. Demonstration. Page: 42
34. Remarks. Page: 42
35. Method III. Page: 43
36. Operation. Page: 43
37. Demonstration. Page: 43
38. Remark. Page: 44
39. Method. IV. Page: 48
40. Operation, Without Compaſſes. Page: 48
41. Demonstration. Page: 48
42. Remarks. Page: 49
43. Method V. Page: 50
44. Operation, Without Compaſſes. Page: 50
45. Demonstration. Page: 50
46. Remark. Page: 50
47. Corollary. Page: 51
48. Method VI. Page: 51
49. Operation. Page: 51
50. Demonstration. Page: 55
51. Remarks. Page: 55
52. Corollary. Page: 55
53. Problem II. Page: 56
54. Remark. Page: 56
55. Problem III. Page: 60
56. Method. II. Page: 60
57. Problem IV. Page: 60
58. Example I. Page: 61
59. Example II. Page: 61
60. Remarks. Page: 62
61. Example III. 48. To throw a circle into Perſpective. Page: 62
62. Remarks. Page: 63
63. Prob. V. 50. To find the Repreſentation of a Point, elevated above the Geometrical Planc. Page: 70
64. Operation. Page: 70
65. Demonstration. Page: 70
66. Prob. VI. 52. To throm a Pyramid, or Cone, into Perſpective. Page: 71
67. 53. To determine the viſible Part of the Baſe of a Cone. Page: 72
68. Operation. Page: 72
69. Demonstration. Page: 73
70. Remarks. Page: 74
71. Problem VII. 55. To find the Perſpective of a Line, perpendicular to the Geometrical Plane. Page: 75
72. Operation. Page: 75
73. Demonstration. Page: 75
74. Method II. Page: 85
75. Demonstration. Page: 85
76. Method III. Page: 86
77. Operation, Without Compaſſes. Page: 87
78. Demonstration. Page: 87
79. Scholium. Page: 88
80. Corollary. Page: 88
81. Problem VIII. Page: 89
82. To do this another Way. Page: 90
83. Demonstration. Page: 90
84. Problem IX. Page: 94
85. Problem X. Page: 94
86. Demonstration. Page: 95
87. EG: EN:: GY: NM. Page: 96
88. Definition. Page: 97
89. Problem XI. Page: 97
90. Lemma. Page: 102
91. Demonstration. Page: 102
92. Remarks. Page: 108
93. Problem IX. Page: 109
94. Operation. Page: 109
95. Demonstration. Page: 113
96. Problem X. Page: 113
97. Operation. Page: 114
98. Demonstration. Page: 114
99. Remarks. Page: 114
100. Method II. 70. By the accidental Point of inclin’d Lines. Page: 115
101. Operation. Page: 115
102. Demonstration. Page: 115
103. Method. III. Page: 116
104. Operation. Page: 116
105. Method IV. Page: 116
106. Prob. XIV. Page: 120
107. Example I. Page: 120
108. Example II. Page: 121
109. Conclusion. Page: 121
110. CHAP. IV. Page: 125
111. Problem I. Page: 125
112. Example. Page: 126
113. Problem II. Page: 126
114. Operation. Page: 130
115. Demonstration. Page: 130
116. Proe. III. Page: 131
117. Operation. Page: 132
118. Demonstration. Page: 132
119. Prob. IV. Page: 133
120. CHAP. V. Page: 137
121. Problem I. Page: 137
122. Problem II. Page: 138
123. Operation. Page: 138
124. Demonstration. Page: 139
125. Remark. Page: 140
126. Problem III. Page: 140
127. Method II. Page: 144
128. Operation. Page: 144
129. Demonstration. Page: 144
130. Method III. Page: 145
131. Operation. Page: 145
132. Demonstration. Page: 145
133. Remark. Page: 146
134. Problem IV. Page: 146
135. Problem V. Page: 146
136. Operation. Page: 146
137. Demonstration. Page: 147
138. Problem VI. Page: 151
139. Method II. Page: 151
140. Operation. Page: 152
141. Demonstration. Page: 152
142. Method III. Page: 152
143. CHAP. VI. Page: 156
144. Prob. I. Page: 156
145. Prob. II. Page: 156
146. Demonstration. Page: 157
147. Corollary. Page: 161
148. Method II. Page: 161
149. Operation, Page: 161
150. Demonstration. Page: 161
151. Remarks. Page: 162
152. Method III. Page: 163
153. Operation. Page: 163
154. Demonstration. Page: 163
155. H I: T H:: a X: a T. Page: 164
156. Prob. III. Page: 164
157. Operation. Page: 164
158. Demonstration. Page: 164
159. Prob. IV. Page: 165
160. Method II. Page: 165
161. Operation. Page: 165
162. Method III. Page: 165
163. Remarks. Page: 166
164. CHAP. VII. Of Shadows. Page: 166
165. Of Solar Shadows. Problem I. Page: 167
166. Operation. Page: 168
167. Proe. II. Page: 168
168. Remarks. Page: 168
169. Problem III. Page: 172
170. Of the Shadows of a ſmall Light. Prob. IV. Page: 173
171. Problem V. Page: 173
172. Remarks. Page: 177
173. CHAP. VIII. Of mechanically ſhortning the Operations of Perſpective. 1. WHEN the perſpective Plane is ſup-pos’d perpendicular or upright. Problem I. Page: 177
174. Operation. Page: 178
175. Method II. Page: 178
176. Prob. II. Page: 182
177. Operation. Page: 182
178. Method II. Page: 183
179. Method III. Page: 183
180. The Demonſtration of the two laſt Ways. Page: 184
181. II. When the Perſpective Plane is inclined. Prob. III. Page: 185
182. Prob. IV. Page: 185
183. Remarks. Page: 185
184. III. When the Perſpective Plane is Parallel or Horizontal. Prob. V. Page: 189
185. Operation. Page: 189
186. Demonstration. Page: 190
187. Remarks. Page: 190
188. Prob. VI. Page: 190
189. Demonstration. Page: 191
190. Prob. VII. Page: 192
191. CHAP. IX. Page: 199
192. Prob. I. 122. To draw Vertical Dials. Page: 200
193. Demonstration. Page: 201
194. Remark. Page: 201
195. Prob. II. 123. To draw inclining Dials. Page: 201
196. The Uſe of the Camera Obscura in Deſigning. Advertisement. Page: 206
197. The Uſe of the Camera Obscura in Deſigning. Definition. Page: 208
198. Theorem I. Page: 208
199. Theorem II. Page: 209
200. The Deſcription of the Firſt Machine. Page: 209
201. Remarks. Page: 212
202. Uſe of the Machine. Problem I. Page: 212
203. Demonstration. Concerning the before-mention’d Inclination of the Mirrours. Page: 214
204. Prob. II. Page: 215
205. Prop. III. Page: 218
206. Problem IV. Page: 219
207. As the Machine’s Height above the Table, leſs the Glaſs’s focal Length, is to Page: 219
208. The Height of the Machine above the Table; So is The Glaſſes focal Length, to the Diſtance of the Figure from the Glaſs. Page: 220
209. 37 Remarks concerning the Repreſentation of Per-ſons Faces. Page: 220
210. A Deſcription of the Second Machine. Page: 222
211. The Uſe of this Machine. Page: 224
212. A Demonſtration of the Inclination of the Looking-Glaſs. Page: 225
213. FINIS. Page: 225
1
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211[Handwritten note 1] 1[Figure 1]
South Librarp.
Press mark, 181 & 28
Ent@in Gatalogue, ---
(1860)
3
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4
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5
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.
622[Handwritten note 2]
MAX--INSTITUT
FOR
WISSEESCHICHTE
Bibliothek
33[Handwritten note 3]
7
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
8 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
what
is here offer’d by
Your Humble Servant,
E. Stone.
9i
The AuTHOR’S
PREFACE
.
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
10iiThe PREFACE. Reaſons that induc’d me to publiſh the follow-
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.
11iiiThe PREFACE.
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’d.
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
12ivThe PREFACE. what Uſe ſuch an exceſſive Variety of Exam-
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
.
13vThe PREFACE.
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
14viThe PREFACE. ing purſued at first, tho’ it requires a little more
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:
15viiThe PREFACE. This I have done ſometimes; 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
16viiiThe PREFACE. ſtill difficulter: 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
.
17ixThe PREFACE.
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
iouſly conſider’d every Thing of his Profeſ-
ſion
, neceſſary to be known, among
18xThe PREFACE. Perſpective was not neglected. 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.
191
AN
ESSAY

ON

PERSPECTIVE
.
CHAP. I.
Definitions.
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.
202An ESSAY
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-
11Fig. 1. tween him and the Object he looks at, let us
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
.
213on PERSPECTIVE.
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
11Def. 1. an Object, (for theſe Three Words are ſynoni-
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
22Fig. 1. Plane C, is the Perſpective of the Object B;
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
33Def. 2. the Spectator [ſtands, or] is placed, as likewiſe the
Objects
that he views, is call’d the Geometrical Plane.
As A B C D.
44Fig. 2.
A Perſpective Plane [or Picture] is that which
55Def. 3. is placed between the Spectator and the Object, upon
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
224An ESSAY different Situations of the Perſpective Plane, [or
Picture
] which ought to be well obſerv’d.
The Interſection of the Perſpective Plane and the
11Def. 4. Geometrical Plane, is call’d the Baſe-Line:
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-
ſpect
to the Perſpective Plane, we ſuppoſe,
A Plane parallel to the Horizon, paſſing thro’ the
22Def. 5. Eye, and every way extending it ſelf;
and this is
call’d
the Horizontal Plane:
As OMVNL.
The Interſection of this Plane and the Perſpective
33Def. 6. Plane, is the Horizontal Line.
As M V N.
The Perpendicular drawn from the Eye to the Ho-
44Def. 7. rizontal Line, is the principal Ray.
As O V.
The Point V, wherein the ſaid Perpendicular meets
55Def. 8. the Horizontal Line, is the Point of Sight, or prin-
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
66Def. 9. the Geometrical Plane, is the Station-Point.
The Plane paſſing thro’ the aforeſaid Perpendicular,
77Def. 10. and the principal Ray, is call’d the Vertical Plane.
As SOLI.
235on PERSPECTIVE.
The Interſection V H of this Plane, and the Per-
11Def. 11. ſpective Plane, is the Vertical Line.
And S H I, the Interſection of it, and the Geo-
22Def. 12. metrical Plane, is the Station Line.
Points of Diſtance, are two Points in the Hori-
33Def. 13. zontal Line, each way diſtant from the Point of Sight
by
the length of the principal Ray;
as MN.
The Geometrical Line, is a Line, that paſſes
44Def. 14. through the Station Point, and is parallel to the baſe
Line
, as A B.
The Seat of an Object, is the Concurrence of Per-
55Def. 15. pendiculars let fall from every of its Points upon the
Geometrical
Plane, and the ſaid Plane.
The Direction of a Line inclined to the Geometri-
66Def. 16. cal Plane, is the Interſection of the ſaid Plane, and
another
Plane perpendicular thereto, paſſing through
the
ſaid inclined Line.
CHAP. II.
The Theory of Perſpective.
Lemma.
3. THE Perſpective, or Appearance of a
Right
Line, as A B, which being con-
tinued
, does not paſsthrough the Eye O, is like-
77Fig. 3. wiſe a right Line:
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.
246An ESSAY
Theorem I.
4. The Repreſentation of a Line Parallel to the
11Fig. 3. perſpective Plane, is parallel to the Line whereof it is
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.
Corollary I.
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.
Corollary II.
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.
Corollary III.
7. The Appearances of Lines parallel to the per-
ſpective
Plane, and equally inclin’d the ſame Way
257on PERSPECTIVE. the Geometrical Plane, make Angles with the baſe
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.
Theorem II.
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
11Fig. 4. prove, that its Repreſentation a b c d, is ſimilar
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 the 224. of which the two Figures conſiſt, are parallel be-
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
268An ESSAY Sides a d and d c of the Figure a b c d. 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.
Corollary I.
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
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. 1.
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309on PERSPECTIVE. right, is alſo perpendicular to the Geometrical
Plane
, and the third is then in the Direction of the
firſt
.
Corollary II.
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.
Theorem III.
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;
viz. As 119. Diſtance of the Eyes from this Plane, is to their
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.
Theorem IV.
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,
3110An ESSAY a right Line drawn from the Eye parallel to the pro-
poſed
Line, terminates.
The Line C D being continued, will meet
11Fig. 5. the perſpective Plane in the Point E.
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.
Corollary I.
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
22Def. 17. ſaid Parallels.
Corollary II.
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.
3211on PERSPECTIVE.
For the Horizontal Plane, is parallel to the
Geometrical
Plane.
Corollary III.
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.
Corollary IV.
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
, that the Appearances of the ſaid 1116. and parallel Lines concur in the principal
Point
.
Theorem V.
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.
3312An ESSAY
Theorem VI.
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
11Fig. 6. if B A be drawn in the Geometrical Plane, pa-
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, that 2213. need but demonſtrate, that O E is parallel to
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
.
Corollary.
20. If A B be made equal to A C, and E D to D O,
the
Appearance of B C will paſs thro’ the Point E,
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
3413on PERSPECTIVE. and the Objects ſtand; 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
3514An ESSAY ctive Plane; 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
3615on PERSPECTIVE. to the Eye. 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;
3716An ESSAY for then, what is on the Right, will appear on
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.
Problem I.
22. To find the Appearance of a Point, which is in
the
Geometrical Plane.
Let Z be the Geometrical Plane, I E the Baſe
11Fig. 7. Line, D V the Horizontal Line, V the Point
of
Sight, D one of the Points of Diſtance, and
A
the given Point.
Operation.
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.
Demonstration.
23. The Appearance of the Line A B, is a Part of the Line B V. Now, if we conceive a Line
2216. iſſuing from the Eye towards the Point D, and
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;
38
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396[Figure 6]Plate 2.
page
16.
Fig
. 5.
@ O H F c d E D C G
7[Figure 7]Fig. 6.E D O @ c F a b A C B G8[Figure 8]Fig. 7.D F H V C X a I B G E Z A
40
[Empty page]
4117on PERSPECTIVE. the Appearance of the Line A E, is a Part 1113. the Line E D. 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.
Remarks.
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
4218An ESSAY and drawing a Line from the Point I to the
other
Point of Diſtance;
which, by its Inter-
ſection
with E D, will give the Appearance of
the
Point A.
Method II.
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.
11Fig. 8.
Operation.
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
.
Demonstration.
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 be in the Lines B C, and E D;
2213. and therefore it will be in a, their Interſection.
Remarks.
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
4319on PERSPECTIVE. Sight to the Horizontal Line, equal in Length
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
.
Method III.
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
11Fig. 9. the Horizontal Line.
Operation.
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.
Demonstration.
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.
4420An ESSAY
Now, it is evident, that to prove the 1127. pearance of A is in the Line C H, we need but
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.
Remark.
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.
45
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469[Figure 9]Plate. 3.
page
20
Fig
. 8.
O Y D C X æ B E Z A
10[Figure 10]Fig. 9.O I Y H G D V X a B E F C Z L A
47
[Empty page]
4821on PERSPECTIVE.
Method. IV.
31. Draw the Line F O G through the Eye O,
parallel
to the Baſe Line, then aſſume F O in
11Fig. 10. this Line, equal to the Height of the Eye, and
O
G equal to the Length of the principal Ray.
A is the given Point.
Operation,
Without
Compaſſes.
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.
Demonstration.
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
the Appearance of A E, is a 2213 of the Line E D G.
It has alſo been 3327. that the Repreſentation of the Point A, is in the
Line
A O;
therefore i@ is in a the
4922An ESSAY of this laſt Line, and the Line E D G, which was
to
be demonſtrated.
Remarks.
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,
5023on PERSPECTIVE. Perpendiculars f P, O H, muſt be let fall from
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.
Method V.
35. When the Appearance of a Point is known,
Let A be a Point in the Geometrical Plane,
11Fig. 11. and a its Repreſentation in the perſpective Plane,
it
is requir’d to find the Appearance of the
Point
B.
Operation,
Without
Compaſſes.
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.
Demonstration.
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 2227. B O;
therefore it is in b the Interſection of the
Lines
E a, and B O.
Remark.
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
5124An ESSAY thod, unleſs by Means of the Point A, the Ap-
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
.
Corollary.
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.
Method VI.
39. The ſame things being given, as in the
11Fig. 12. ſecond Method, let F C be the Geometrical Line.
Operation.
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
52
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5311[Figure 11]Plate 4.
Page
24.
Fig
. 10.
f F O G g V D N L a P E H I M A
54
[Empty page]
5525on PERSPECTIVE. rence with the Geometrical Line; which would
come
to the ſame thing.
Demonstration.
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 1113: the Appearance of E A, is a Part of E D. And
after
the ſame Manner we prove, that the Re-
preſentation
of B A is a Part of B a.
Remarks.
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.
Corollary.
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
5626An ESSAY from the Eye O upon the Baſe Line, the Appear-
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.
Problem II.
42. To throw a Line in the Geometrical Plane
into
Perſpective.
It has been ſhewn , that to have the 1121. ſentation of a right Line, the Perſpective of
the
Extremes of the ſaid Line, need only be
found
;
and although it is difficult to find 2222. Repreſentation of two Points, nevertheleſs I
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
33Fig. 13. Line:
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
.
Remark.
If the Lines G O and C O cut E D very ob-
liquely
, and ſo their Interſection cannot be
57
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5812[Figure 12]Plate 5.
page
26.
Fig
. 11.
O Y b X a E Z A B
13[Figure 13]Fig. 12.M O Y F S C L D X a E B Z A
59
[Empty page]
6027on PERSPECTIVE. actly determined, this Method ought not then to
be
uſed.
Problem III.
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.
11Fig. 14. Now to find the Repreſentation of the Diviſions
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.
Method. II.
45. To find the Repreſentations of the Di-
viſions
of the Line G C, make choice of the
22Fig. 14. Point D at Pleaſure without this Line, and find
the Repreſentation d thereof;
then draw 3321. through the propoſed Diviſions to the Point D;
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.
Problem IV.
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 by any one of the Methods 4421. Problem I.
the fourth in general is the eaſieſt;
and may be firſt uſed in finding the 5522. ſentations of Points, or ſometimes of one
only
;
and then the fifth Method ſerves 6631.
6128An ESSAY ſinding the reſt. But yet the Work may be
ſhorten’d
by the two precedent Problems, as we
ſhall
ſhew in the following Examples.
Example I.
To throw a Pentagon having one Side parallel to
the
Baſe Line, into Perſpective.
Let A B C D E, be the propoſed Pentagon,
11Fig. 15. wherein draw the Line B D which will be paral-
lel
to A E, becauſe the Pentagon is a regular
one
.
Now find the Repreſentation of the ſaid 2242. Lines A E and B D, and you will have the Re-
preſentation
of four of the Corners of the Pen-
tagon
;
and to determine the Repreſentation of
the
fifth Corner, find the Appearance of 3343. Line drawn from C to E, which in this Example
is
parallel to A B, which is ſuppoſed parallel
to
the Baſe Line.
Example II.
To throw a Parallelogram, divided into ſeveral
other
Parallelograms into Perſpective.
Let A B C D be a Parallelogram, divided
44Fig. 17. into ſeveral other Parallelograms.
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
6229on PERSPECTIVE. drawn to the Point G, will give the Appearance
ſought
.
Remarks.
When this Method cannot be us’d, the Per-
ſpective
of the Diviſions dividing the Sides of
the
Parallelogram, mufl be found .
And we 1144. often oblig’d to have recourſe to this Expedient,
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.
Example III.
48. To throw a circle into Perſpective.
The Repreſentation of ſeveral Points of a
22Fig. 16. Circle, or any other Curve Line requir’d to be
thrown
into perſpective, muſt be found, 3321. may be well enough done, by drawing ſeveral
Chords
in the Circle, or Curve, parallel between
themſelves
, the Repreſentations of which muſt
be
found ;
then the Extremities of thoſe 4443.
6330An ESSAY preſentations being join’d, will give the Perſpe-
ctive
ſought.
The ſame may be done, in draw-
ing
the Chords thro’a Point, whoſe Repreſenta-
tion
is known.
Remarks.
49. Let G I be the Geometrical Line; and thro’
11Fig. 16. the Center P of the Circle, whoſe Perſpective is
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
.
64
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6514[Figure 14]Plate 6.
page
28.
Fig
. 13.
O D c b a g E G B A C
15[Figure 15]Fig. 14.O b 1 2 3 a c 1 2 3 g D A C 3 1 2 2 1 3 B G
66
[Empty page]
67
[Empty page]
6816[Figure 16]page 28.
Plate
. 7
Fig
. 16
Fig
. 15
O G F I V
l
d c e m n b a h
B
A H M N C E P D L
69
[Empty page]
7031on PERSPECTIVE.
Prob. V.
50. To find the Repreſentation of a Point, elevated
above
the Geometrical Planc.
Let G S be the Geometrical Line, and S the
11Fig. 18. Station Point:
Make S F, in the Geometrical
Line
, equal to the Height of the Eye;
and let
A
be the Seat of the given Line.
Operation.
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.
Demonstration.
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
22Fig. 19. mark’d with the ſame Letters in the precedent
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.
7132An ESSAY
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-
lar
, we have,
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.
Prob. VI.
52. To throm a Pyramid, or Cone, into Perſpective.
Now, to throw a Pyramid into perſpective,
11Fig. 20. the Appearance of its Baſe and Center muſt 2246. found :
After which, Lines muſt be drawn 3350. the Repreſentation of the Vertex, to the Ap-
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
44Fig. 21. Repreſentation of its Baſe and Vertex muſt 5546.
7233on PERSPECTIVE. firſt found ; and then if Lines be drawn 1147. the Repreſentation of the Vertex touching the
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.
Let the Circle L I F be the Baſe of a Cone
22Fig. 21. in the Geometrical Plane, and A the Center
thereof
.
Operation.
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)
7334An ESSAY the Circular Baſe of the Cone will be deter-
min’d
.
Demonstration.
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.
7435on PERSPECTIVE.
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.
Remarks.
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.
7536An ESSAY
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.
Problem VII.
55. To find the Perſpective of a Line, perpendicular
to
the Geometrical Plane.
It is requir’d to find the Appearance of a Line
11Fig. 22. equal to B C, and perpendicular to the Geome-
trical
Plane, in the Point A.
Operation.
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
a, the Repreſentation of the Point A;
2222. draw a H parallel to the Baſe Line, and aI per-
pendicular
thereto:
And if aI be made equal to
G
H, the ſaid a I will be the Perſpective
ſought
.
Demonstration.
56. The Appearance of the ſaid Line, is 336. pendicular to the Baſe Line, and equal to 4410. Perſpective of the Line A L, drawn from
76
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7717[Figure 17]page 36.
Plate
8
Fig
. 17
O G F c d b a A B D C
78
[Empty page]
79
[Empty page]
8018[Figure 18]Page 36.
Plate
9
Fig
. 18.
G F C S V I E B A
19[Figure 19]Fig. 19.O i M X L D @ b a20[Figure 20]Fig. 20.S x G n H S V D l R f Q m P t21[Figure 21]Fig. 21.I X f T L B N A C l M E F
81
[Empty page]
82
[Empty page]
8322[Figure 22]page 38
Plate
10.
Fig
. 22.
V F I N a G H M P D E B C L A
23[Figure 23]Fig. 23.O F I H a G D E B C L A M24[Figure 24]Fig. 24.@ o f X a e A
84
[Empty page]
8537on PERSPECTIVE. Point A parallel to the Baſe Line, and made e-
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 115, 16. the Perſpective of A L; 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.
Method II.
57. The ſame Things being given, as in the
precedent
Method, about the Point A, as a
22Fig. 23. Center, and with the Radius B C, deſcribe the
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.
Demonstration.
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-
33Fig. 24. preſents a Plane paſſing through the Eye and
the
Point A of the foregoing Figure, wherein
O
f here, repreſents O F there;
f e here, F
8638An ESSAY there; and finally e A here, E A in that Fi-
gure
.
This being ſuppoſed, o f is parallel to e 1127. and conſequently the Triangle o f a is ſimilar
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-
cedent
Figure, we ſhall have this,
O A : o a : : F D : F a.
Again, becauſe the Triangles O A L and O a G
are
ſimilar, we ſhall have
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-
ſider’d
, we ſhall have
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 2256. qual to the Repreſentation ſought.
Which was
to
be demonſtrated.
Method III.
58. Near one of the Sides of the perſpective
33Fig. 25. Plane, raiſe the Perpendicular C B to the Baſe
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
8739on PERSPECTIVE. the Station Point, and A the Point wherein the
Perpendicular
meets the Geometrical Plane.
Operation,
Without
Compaſſes.
Having firſt found the Perſpective a of 1131. Point A, draw the Line A S cutting the Baſe
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.
Demonstration.
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 its Repreſentation is 2241. dicular to the Baſe Line; therefore we are 3356. to prove, that a I is equal to a H.
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
ſimilar
, therefore
F G : F B : : G N : B L.
8840An ESSAY
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
give
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.
Scholium.
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.
Corollary.
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.
8941on PERSPECTIVE.
Problem VIII.
59. To throw a Priſm or Cylinder into Perſpective,
11Fig. 26. both of them being Perpendicular to the Geometri-
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
ſuch that they may repreſent 2255. lars to the Geometrical Plane, equal to the
Height
of the Priſm, and find the 3350. of the other Angular Points of the upper Sur-
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 the 4450. ance of ſeveral Points of the Periphery of its
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
9042An ESSAY B and C, which will be the two moſt extreme
ones
that can be ſeen.
To do this another Way.
61. If the upper Face of the Cylinder or Priſm
11Fig. 26,
27
.
be otherwiſe requir’d to be found, the ſame Things
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-
2246. wards P Q.
Finally we find the Perſpective of the Baſe of the Priſm or Cylinder, thus changed
in
Situation by uſing P Q for a Baſe Line, and
the
ſaid Perſpective is the Repreſentation of their
upper
Faces.
Demonstration.
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
91
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9225[Figure 25]page 42
Plate
11.
Fig
. 25.
S F V M I N P H a L D E G C A B
26[Figure 26]Fig. 26.
Fig
. 27.
S V P Q R n l g h G H B N I A C M L
93
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9443on PERSPECTIVE. Priſm is greater than the Height of the Eye, the
precedent
Method is the ſhorteſt.
Problem IX.
62. To throw a Concave Body into Perſpective.
11Fig. 28.
Having firſt ſound the Perſpective of the ſaid
Body
, afterwards find the Appearance of its
Cavity
, in conſidering the Cavity as a new
Body
.
Problem X.
63. To throw a Sphere into Perſpective.
22Fig. 29.
Let A be the Seat of the Centre of the Sphere;
then the Point I the Perſpective of the Centre
muſt
be found, and the Line IV drawn to 3350. Point of Sight V.
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
9544An ESSAY Line G E, in the Points p, from every of which
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.
Demonstration.
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
9645on PERSPECTIVE. and conſequently to V P. 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
11Fig. 30,
31
.
Figure 31 repreſent the Points denoted with the ſame
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
ſimilar
, whence
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
9746An ESSAY G m T, biſects the Axis G E: 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.
Definition.
The ſemicircular Part h m of a Column, en-
11Fig. 33. compaſſing the ſame like a Ring, is called the
Torus
.
Problem XI.
64. To throw the Torus of a Column into Per-
ſpective
.
Let B N C be the Baſe of the Column in the
22Fig. 32. Geometrical Plane;
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
33Fig. 33. draw the Line z 36, through the Center of the
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
98
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9927[Figure 27]page 46
Plate
12.
Fig
. 28.
28[Figure 28]Fig. 29.F S V q q q E L p p p I G H q D P n n n T R m m m C B Q A29[Figure 29]Fig. 30.O X E L N M G Z Y D30[Figure 30]Fig. 31.f 3 c l n m g 4
100
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10147on PERSPECTIVE. take the Part 2 s, equal to the Height of the Eye
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
11Fig. 32. the Column, therefore to avoid uſeleſs Operations,
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
10248An ESSAY might be done) the Point on the Semicircle
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.
Lemma.
65. If two Circles C D H E and D E F L cut
11Fig. 34. each other, thro’ whoſe Centers C and B the Line
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.
Demonstration.
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.
103
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10431[Figure 31]page 48.
Plate
13.
Fig
. 32.
V S R L P B D Q T M I F A E Y C G O H Z N
105
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10649on PERSPECTIVE.
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
into Perſpective upon it, (which 118. is alſo a Circle) the two Points of Interſection of this
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
22Fig. 35. the Column, A P a Perpendicular to the Baſe paſſing
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
10750An ESSAY Point R, wherein it is cut by the Line m R per-
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
({edx + ydx/x}) we ſhall have T D = y + dy - {ydx/x}
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 be divided 1165. Twice P T, and then half of P T muſt be taken there-
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
22Fig. 32,
33
.
be repreſented by e in this Computation;
as
10851on PERSPECTIVE. s i by x, and i h be y; it is manifeſt, that i 4 = a 5
being
Algebraially Expreſſed, will be {ydy/dx}
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.
Remarks.
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 1153.
10952An ESSAY viſible Portions of the ſaid Baſes. Note, This
Method
may be demonſtrated without Algebra,
but
it would be very long.
Problem IX.
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,
11Fig. 36. whoſe accidental Point is ſought;
and ECP, the
Angle
that the ſaid Lines make with the Geo-
metrical
Plane.
Operation.
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.
110
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11132[Figure 32]page 52.
Plate
. 14.
Fig
. 34
D C F G A B H L E
33[Figure 33]Fig. 33S X 8 1 h 6 g 3 z q 9 m 2 4 m a 5 Y34[Figure 34]Fig. 35O M D P T Q R m p q B A S C
112
[Empty page]
11353on PERSPECTIVE.
Demonstration.
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 the acciden- tal one ſought.
Note, This Demonſtration as well regards
1113, 14. inclin’d Lines entirely ſeparate from the Geo-
metrical
Plane, as thoſe that meet it in one of
their
Extremes only.
Problem X.
69. To find the Repreſentation of one or more
Lines
, inclin’d to the Geometrical Plane.
22Fig. 36.
Let A be a Point given in the Geometrical
Plane
;
whereon ſtands a Line, whoſe Length,
Direction
, and Angle of Inclination is known.
11454An ESSAY
Operation.
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 , the Perſpective of a Point 1150. above B, the Height of P E;
join the Points a
and
T by a right Line;
and the ſought Appear-
ance
will be had.
Demonstration.
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.
Remarks.
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 2268. rallel to the perſpective Plane. The Manner
11555on PERSPECTIVE. making theſe Abbreviations, will be laid down
in
the following Methods.
Method II.
70
. By the accidental Point of inclin’d Lines.
Thro’ F, the accidental Point of the inclin’d
11Fig. 36. parallel Lines, draw F H, parallel to the Baſe
Line
, and equal to F G.
And let A be the
Point
, wherein one of the inclin’d Lines meets
the
Geometrical Plane.
Operation.
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.
Demonstration.
By the Nature of the accidental Point, 2214. Perſpective ſought is a Part of the Line a F;
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 , 3356. L is the Perſpective of I; and conſequently,
L
H is the Appearance of a Line 44@@
11656An ESSAY through I, and the Extremity of the propoſed
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
that R Q muſt alfo have been taken 1119. to the one half, or third Part, &
c. of C E.
Method. III.
71. Forinclined Lines not meeting the Geometrical
Plane
.
Let A and B be the Seats of the Extremities of
22Fig. 37. the given Line.
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.
Operation.
Find the Point I , the Perſpective of a 3350. above the Point A, the Height of C N; 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.
Method IV.
72. For inclined Lines parallel to the perſpective
Plane
.
117
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11835[Figure 35]page 56.
Plate
. 15
F H O Z D G
36[Figure 36]Fig. 36T N M L a R Q E I A C P B37[Figure 37]Fig. 37F S V T I E M A N X P C B
119
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12057on PERSPECTIVE.
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.
Prob. XIV.
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. 1169. takes in all the Caſes. 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
.
Example I.
To throw ſeveral parallel Shores which ſtrengthen a
Wall
, into Perſpective.
I ſuppoſe here that the Baſes of theſe Shores,
22Fig. 38. which are the Places where they meet the Sur-
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 3368.
12158An ESSAY their accidental Point F, afterwards find the Re-
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.
Example II.
To throw ſeveral parallel Roofs of a Houſe into
Perſpective
.
Having found the accidental Points G and Q
11Fig. 39. of the ſaid Roofs, in the Repreſentation of
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.
Conclusion.
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
122
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12338[Figure 38]page 58.
Plate
. 16
Fig
. 39
Fig
. 38
F Q O p l r s 1 2 3 4 G
124
[Empty page]
12559on PERSPECTIVE. drawn upon the Geometrical Plane, and the
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.
Problem I.
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
12660An ESSAY tal Plane, or one of the Points of Diſtance on the
Horizontal
Line.
The Repreſentation of two Points of theſe
Figures
muſt be firſt found ;
and then by 1124. of theſe two Points the Appearances of others
may
be had .
2238.
Example.
Let A B C D E, be a Pentagon, whoſe Ap-
33Fig. 40. pearance is requir’d;
V the Point of Sight; and
V
F the ſixth Part of the Eye’s Diſtance from
the
perſpective Plane.
Now find b and e 4424. Appearance of B and E, by means of which,
the
Appearance of the Point A will be had .
5538. In like Manner, by means of the Repreſenta-
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 6655. inclined thereto , may be found by the 7769. of the precedent Chapter.
Problem II.
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
88Fig. 41. Line, and likewiſe draw the Line C E from the
ſame
Point in ſuch manner, that if it could
127
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12839[Figure 39]page 60.
Plate
. 17
F V
40[Figure 40]Fig. 40c θ b e a F G H I K L A B E C D
129
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13061on PERSPECTIVE. continued, it would cut the Horizontal Line in
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
.
Operation.
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
F G the Repreſentation of A B for 1142. ſecond Eye, which continue until it meets the
Line
C E in b, and in this Continuation aſſume
b
a equal to F G;
then a will be the Perſpective
ſought
.
Demonstration.
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 , the 2218. of A B muſt be a Part of F G continued, and
3312. alſo equal to F G:
And therefore becauſe the Perſpective of B is in the Line C E, a b is the
4416. Perſpective of A B;
and a , that of 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
.
13162An ESSAY
Proe. III.
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
11Fig. 42. the vertical Line;
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
13263on PERSPECTIVE. been drawn from the Baſe Line, equal to a fifth
Part
of the Eye’s Diſtance, and ſo on.
Now
A
is a Point whoſe Repreſentation is requir’d.
Operation.
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
.
Demonstration.
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
by Conſtruction, that the Line G F 1177. tinued, paſſes through the Eye O;
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
ſimilar
, we have
G M: G N: : G I: G C.
The Triangles G Z F and G R O being alſo
ſimilar
,
G Z: G R: : G F: G O.
13364An ESSAY
Whence
G I: G C: : G F: G O.
Again, becauſe the Triangles G I E and G C D
are
ſimilar, we have
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 , that the Perſpective 1113. E A, is a Part of E a D. 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
.
Prob. IV.
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,
22Fig. 43. equal in Length to the propoſed Perpendicular;
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
, in conſidering that D F and E 3356.
134
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13541[Figure 41]page 64
Plate
. 18.
d v
42[Figure 42]Fig. 41E b a G F H C B A D43[Figure 43]Fig. 42G Q A M I S E H T B L Z F P a X Y N C D R O
136
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13765on PERSPECTIVE. being produced, will meet each other in the
Horizontal
Line.
CHAP. V.
Of throwing Figures into Perſpective, when
the
Perſpective Plane is conſider’d as being
inclined
.
Problem I.
81. TO find the Perſpective of a Figure in the
11Fig. 44. Geometrical Plane.
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.
13866An ESSAY
Problem II.
82. To find the Appearance of a Point above the
Geometrical
Plane.
Let H C be the Baſe Line: And let T be the
11Fig. 45. accidental Point of the Lines perpendicular
to
the Geometrical Plane.
This Point will
be
in that Place of the Vertical Line, 2213. it is cut by the Prolongation of the Line mea-
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 .
3381. laſtly, let A be the Seat of the given Point.
Operation.
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
13967on PERSPECTIVE. Line T B X; 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.
Demonstration.
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 the Perſpective 1113. B A, is a Part of B X; 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 ;
2213, 14. therefore it is a Part of T X. 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 Part of C X;
and 3341. quently, the Appearance of L is in this Line.
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 446. ſpective is parallel to the Baſe Line. And
14068An ESSAY this Appearance paſſes thro’ that of the Point
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.
Remark.
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 to a 1181. pendicular, or upright one;
and the Repreſen-
tation
of a Point, above the Geometrical Plane,
(whoſe Seat is L, and Height M E) muſt be
found
.
2250.
Problem III.
84. To find the Repreſentation of a Line, perpendi-
cular
to the Geometrical Plane`.
The Appearance of the Extremity of the Per-
33Fig. 45. pendicular muſt be found , in conſidering 4482. ſaid Extremity as a Point above the Geometri-
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 5516. T X, will give the Appearance a of the Seat of
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
141
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14244[Figure 44]page 68
Plate
. 19
Fig
. 43
B D E a G H I C F L
45[Figure 45]Fig. 44O V X S H I T46[Figure 46]Fig. 45Q F V X S a H B C D E L M P T A
143
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14469on PERSPECTIVE. too obliquely, recourſe muſt be had to Problem I. 1181. to find the Appearance of a.
Method II.
85. A is the Foot of the Perpendicular: The
22Fig. 46. Triangle, E P M, is drawn as directed:
3382. T is the accidental Point of the Perpendiculars,
to
the Geometrical Plane.
Operation.
Thro’ the Point a, the Appearance of A,
draw
a Perpendicular to the Baſe Line;
which
make
equal in Repreſentation to the 4455. M E;
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.
Demonstration.
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 will 5516. thro’ the Point of Sight; and its Interſection
with
T a, will be the Extremity of the Repre-
ſentation
ſought.
But a I is the 6656. of the firſt Line, made equal to E M; and con-
ſequently
, V I is that of the ſecond.
Which was
to
be demonſtrated.
14570An ESSAY
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.
Method III.
86. Let T be the accidental Point of the Lines
11Fig. 47. perpendicular to the Geometrical Plane, thro’
which
Point draw a parallel to the Baſe Line,
in
which aſſume T R equal to O T of Fig.
44.
Operation.
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
.
Demonstration.
The Part a Q of the Line a H, is the 2256. pearance of a Line proceeding from A in the
Geometrical
Plane, and which is equal to the
propoſed
Line, and parallel to the Baſe Line;
and conſequently , the Line R Q paſſes 3320. the Perſpective of the Extremity of the propo-
ſed
Line:
And therefore X the Interſection of
R
Q and T a, is the Appearance of the ſaid
Extremity
.
14671on PERSPECTIVE.
Remark.
It is manifeſt , that T R may be 1119. equal to {1/2} or {1/3} & c. of what it is taken here,
provided
likewiſe that then D N be aſſumed
equal
to a correſpondent Part of the propoſed
Line
.
Problem IV.
87. To throw a Sphere into Perſpective.
The Method of ſolving this Problem before
laid
down , muſt be uſed here, but with 2263. Difference;
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.
Problem V.
88. To find the accidental Point of any Number
33Fig. 48. of Lines inclined to the Geometrical Plane.
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.
Operation.
Draw the Line O D, thro’ the Eye O parallel
to
A B, meeting the Horizontal Line in D, which
will
be the Accidental Point of the 4413, 14. of the given Line;
and thro’ the Station Point
14772An ESSAY draw S N perpendicular to the ſaid Line A B,
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.
Demonstration.
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 the 1113, 14. ctive Plane in the Accidental Point ſought.
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
148
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14947[Figure 47]page 72
Plate
. 20
Fig
. 46
V I X a E M P A T
48[Figure 48]Fig. 47V F X a Q G H D N A T R
150
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15173on PERSPECTIVE. Point in reſpect to each other, in the before ſup-
poſed
Plane.
Therefore the Line o F anſwers
@kewiſe
to the Line in the ſaid Plane imagined
to
be parallel to the propoſed Lines;
and con-
ſequently
the Point F, is that wherein the
ſaid
Parallel meets the Perſpective Plane;
and
therefore
it is the accidental Point ſought.
Note, Iſ the accidental Point T of Perpendi-
culars
to the Geometrical Plane be found, the
Operation
of this Problem may be ſhorten’d, in
drawing
the Line T D, which will neceſſarily
paſs
thro’ the Point N, and then the Point o will
be
found by the Interſection of the Arc O o,
and
a Semi-circle, whoſe Diameter is T D.
Problem VI.
89. To find the Perſpective of one or more Lines
inclin’d
to the Geometrical Plane.
Let A be the Foot of a Line inclin’d to the
11Fig. 48. Geometrical Plane, and a its Repreſentation.
Now determine, by Means of the Triangle C P E
according
to the Manner lay’d down for 2269. Perſpective Plane when ſuppoſed perpendicular,
the
Length A B of the Direction of the propoſed
Line
.
This being done, find the Point X the 3382. ſpective of a Point above the Geometrical Plane
by
the Length of P E;
and then a X will be the
Perſpective
ſought.
Method II.
90. To ſolve this Problem by the Accidental Points
of
inclined Lines, and their Directions.
Let AB be the Direction of an inclin’d Line; D
44Fig. 48. the Accidental Point of the Directions, &
F that
15274An ESSAY the Lines themſelves, and T the Accidental Point
oſ
Perpendiculars to the Geometrical Plane.
Operation.
Continue the Line A B until it meets the
Baſe
Line in G, and draw the Line G D, which
cut
in a and b by Lines drawn from A and B to
the
Eye.
Then draw the Lines a F and T b In-
terſecting
each other in X;
and a x will be the
Appearance
ſought.
Demonstration.
Becauſe a b is the Repreſentation of A 1143. the Appearance of the inclin’d Line is one Part
of
a F.
But the Extremity of the inclin’d Line
is
in a Perpendicular to the Geometrical Plane
raiſed
at the Point B;
therefore the Repreſen-
tation
of the ſaid Extremity is in T b, and con-
ſequently
x in the Interſection of this Line and
a
F.
Method III.
91. Draw F H thro’ the Accidental Point of
the
inclin’d Lines, parallel to the Baſe Line,
and
equal to o F in Fig.
48. then a is the Perpen-
22Fig. 49. dicular of the Foot of the inclin’d Line whoſe
Perſpective
a x may be found as directed n.
70.
153
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15449[Figure 49]Page 34.
Plate
. 21
Fig
. 48
F O D X S b a G N A E T B P C
50[Figure 50]Fig. 49H F O D G X a M N L R Q
155
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15675on PERSPECTIVE.
CHAP. VI.
Of throwing Figures into Perſpective, when
the
Perſpective Plane is conſider’d as being
parallel
to the Geometrical Plane.
Prob. I.
92. To find the Perſpective of a Figure, which
is
in the Geometrical Plane.
When the perſpective Plane is parallel to the
Horizon
, we commonly conſider it, as being
it
ſelf the Geometrical Plane, in which Caſe,
the
Problem is fully reſolved:
But when it
happens
, that another Geometrical Plane is
ſuppoſed
, either above or below the perſpective
Plane
, upon which the Figures upon the Geo-
metrical
Plane are requir’d to be drawn;
we
muſt
draw geometrically thereon, Figures ſimi-
lar
to thoſe in the Geometrical Plane;
ſo that
the
Lines on the perſpective Plane, be to their
correſpondent
ones on the Geometrical Plane,
as
the Eye’s Diſtance from the perſpective Plane,
is
to its Diſtance from the Geometrical Plane.
The Demonſtration of this is evident from
n
.
8, and 9.
Prob. II.
93. To find the Perſpective of a Line, perpen-
dicular
to the Geometrical Plane.
Draw the Line O S, in which aſſume O R e-
11Fig. 50. qual to the Eye’s Diſtance from the
15776An ESSAY Plane; and at the Points R and S, raiſe the in-
definite
perpendiculars R G and S M;
and aſſume
the
Point M at Pleaſure on S M;
from which
raiſe
the Perpendicular M N, equal to the given
Line
, and draw the Lines M O and N O, cutting
11Fig. 41. the Line R G in the Points E and G.
Then
having
drawn a Line at Pleaſure in the per-
ſpective
Plane through the Point T, which is
that
wherein a Perpendicular falling from the
Eye
on the perſpective Plane meets it, aſſume
T
H in the ſaid Line, equal to R E, and T I e-
qual
to R G;
draw the Lines Ta, Ha, through
the
Point a, the Perſpective of the Foot of the
given
Perpendicular, and through the Point I,
the
Line I X, parallel to Ha, and cutting Ta
in
X, then a X will be the Appearance ſought.
Demonstration.
It is manifeſt , that the Point T, is the 2213, 14. dental Point of Lines perpendicular to the Geo-
metrical
Plane;
and conſequently the Per-
ſpective
ſought is a Part of T a.
Moreover, it
is
manifeſt , that if the Feet and 334. of two equal right Lines, perpendicular to the
Geometrical
Plane be joyn’d by Lines, theſe
Lines
of Junction will have parallel Repreſen-
tations
;
becauſe they are parallel to each other,
as
likewiſe to the perſpective Plane.
And con-
ſequently
, ſince H I, by Conſtruction, is the
Perſpective
of a Line perpendicular to the
Geometrical
Plane, and equal to the given Line,
and
H a paſſes through the Appearances of the
Foot
of the ſaid Perpendicular, and the given
Perpendicular
;
I ſay, that X I, which is paral-
lel
to Ha, and paſſes through the Extremity of
the
Appeaarance H I, likewiſe paſſes through
the
Extremity of the given Line;
and
158
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15951[Figure 51]Page 36
Plate
22
Fig
. 50
O R E G N S M
52[Figure 52]Fig. 51I H T a X53[Figure 53]Fig. 52C D X I H G a F E L b T54[Figure 54]Fig. 53H I F T x d X L B C
160
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16177on PERSPECTIVE. fore the Point X is the Repreſentation of the
ſaid
Extremity.
Corollary.
94. It is manifeſt from hence, that when the
Perſpective
of a Line, perpendicular to the
Geometrical
Plane is once found, it is eaſy af-
terwards
to find the Repreſentations of any Per-
pendiculars
of the ſame Length as that.
Method II.
The Perſpective Plane being conſider’d as the Geo-
metrical
Plane.
95. Let T (as in Fig. 51.) be the accidental
11Fig. 53. Point of perpendicular Lines to the Geometri-
cal
Plane;
H I the Arc of a Circle, whoſe Cen-
ter
is T, and ſemidiameter the Eye’s Diſtance
from
the perſpective Plane:
Alſo let a be the
Point
where the Perpendicular, whoſe Appear-
ance
is ſought, meets the perſpective Plane, and
B
C the Length of this Perpendicular.
Operation,
About the Point a, as a Center, and with the
Semidiameter
B C, deſcribe the Circle L F, and
draw
the Line I L, or H F, touching each of the
Circles
H I, and F L;
and then a X or a x, is
the
Appearance ſought, viz.
a X, when the Per-
pendicular
is above that Surface of the per-
ſpective
Plane next to the Eye, and a x, when
the
Perpendicular is on the oppoſite Side.
Demonstration.
Draw the Radii a F, a L, T I, T H, to the
Points
of Contact F, L, I, and H.
Then
16278An ESSAY cauſe the Triangles THX and a F X are ſimilar,
TH
a F:
a F : : Ta: a X.
And becauſe the Triangles T I x and ax L, are
alſo
ſimilar, we have
TI
+ a L :
: a L: Ta : ax.
Now let PM NR be the perſpective Plane,
11Fig. 54. O the Eye, A Q the Perpendicular, whoſe
Perſpective
is requir’d, and O t a perpendicular
let
fall from the Eye upon the perſpective Plane,
and
ſo t will be the ſame, as the Point T in the
aforegoing
Figure, Now if the Lines O Q be
drawn
, it is manifeſt that A x, or A X, is the
Perſpective
of A Q, according as this Line is
above
or below the perſpective Plane in reſpect
to
the Eye.
Then becauſe the Triangles O t x
and
Q A x are ſimilar, we have
O
t A Q:
A Q : : t A: Ax.
And ſince the Triangles O t X and X A Q are
ſimilar
,
O
t + A Q:
A Q : : t A: A X.
Now Ot is equal to TH or TI of the afore-
going
Figure, and AQ to a F or a L of the
ſame
Figure;
as likewiſe At, Ta: Therefore
if
theſe two laſt Proportions be compared with
the
two precedent ones, we ſhall find A x = a X,
and
A X = a x;
which was to be demon-
ſtrated
.
Remarks.
96. When the two Circles interſect each other,
or
fall within one another, and ſo this Way be-
comes
uſeleſs;
a Line muſt be drawn at Pleaſure,
through
the Point T, equal to the Diſtance of
the
Eye from the perſpective Plane;
and then a
parallel
equal to the given Perpendicular muſt be
drawn
to the ſaid Line through the Point a, ei-
ther
towards L or F, according as the
16379on PERSPECTIVE. dicular is on one Side or other of the perſpective
Plane
with reſpect to the Eye.
And the Line
paſſing
through the Extremities of the ſaid Pa-
rallels
, will determine the Repreſentation ſought,
by
its interſecting the Line T a, as is evident by
what
is demonſtrated.
Method III.
97. To find the Repreſentation of ſeveral Perpen-
diculars
equal in Length to ſome one, whoſe Per-
ſpective
is already drawn.
Let H I, be the Perſpective of a Perpendicular
11Fig. 52. to the Geometrical or Perſpective Plane.
Now
about
the accidental Point T, as a Center, and
with
the Radius T H, deſcribe the circular Arc
H
G, whoſe Chord let be equal to H I, and draw
the
indefinite Line TGC, and let a and b, re-
preſent
the Feet of the Perpendiculars, whoſe
Repreſentations
are requir’d.
Operation.
Deſcribe about the Center T, the Arcs b F E,
and
a DC, paſſing through the Points a and b, and
draw
the Lines T b and T a;
in each of which
aſſume
b L equal to E F, and a X equal to C D;
and the ſought Repreſentations will be had.
Demonstration.
If H I, and a X repreſent Perpendiculars of
the
ſame Length, it follows from the Demon-
ſtration
of the precedent Method, that I H is to
H
T, and a X to a T as the Difference of the
ſaid
Perpendiculars, and Height of the Eye,
16480An ESSAY to the Length of the ſaid Perpendiculars: And
therefore
H I: T H:: a X: a T.
But in the Conſtruction of this Problem, be-
cauſe
the Triangles T C D and T H G, are
ſimilar
;
H G = H I: T H: : C D = a X: T D = a T;
and conſequently H I and a X, repreſent Perpen-
diculars
of the ſame Length.
Which was to be
demonſtrated
.
Prob. III.
98. To find the accidental Point of any Number
of
parallel Lines inclined to the Geometrical Plane.
Let a b be the Perſpective of the Direction
11Fig. 55. of one of the given Lines.
Operation.
Draw the Line F T L, parallel to a b, through
the
accidental Point T of the Lines perpendicu-
lar
to the Geometrical Plane;
and at the Point
T
, raiſe the Perpendicular T G, which make e-
qual
to the Diſtance of the Eye from the per-
ſpective
Plane;
then draw the Line G L, or
G
F, ſo that the Angle T L G, or T F G, be e-
qual
to the Angle of the Inclination of the given
Lines
;
and the Point L, will be the Accidental
Point
ſought, if the given Lines incline to-
wards
b;
but if they incline towards a, F will
be
the Accidental Point.
Demonstration.
It is manifeſt by Conſtruction, that if T G be
ſuppoſed
to be raiſed perpendicularly to the Geo-
metrical
Plane, G L or G F, will be parallel
16581on PERSPECTIVE. the given Lines; and conſequently L, or 1113, 14. will be the Accidental Point ſought.
Prob. IV.
99. To find the Repreſentation of one or more
Lines
inclined to the Geometrical Plane.
Let a b be the Perſpective of the Direction of
22Fig. 56. the given Line:
Now the Length of its Directi-
on
may be determin’d, by means of the Tri-
angle
ECP, according to the Directions of n.
69.
This being done, the Line b X muſt be drawn
through
the Point b, equal to E P, and this re-
preſents
a Perpendicular to the Geometrical
Plane
;
then a X being drawn, will be the Re-
preſentation
ſought.
Method II.
100. To ſolve the ſame Problem by means of the
accidental
Points F and T, the one being that of the
Lines
propoſed;
and the other, that of the Perpen-
33Fig. 56. diculars to the Geometrical Plane.
Operation.
Draw a Line from the Point F through the
Point
a, which interſect in the Point X, by ano-
ther
Line drawn from the Point T through b:
and then a X will be the Perſpective ſought.
Method III.
101. The ſame Things being given as in the
44Fig. 101. aforegoing Method, draw a I through the Point
a
, equal to E P, repreſenting a Line perpendicu-
lar
to the Geometrical Plane.
Then draw a pa-
rallel
to F T through the Point I, whoſe
16682An ESSAY ſection with F a, will determine a X the Repre-
ſentation
ſought.
Remarks.
Although we ſuppoſe the Eye in all the
Problems
in this Chapter to be above the per-
ſpective
Plane, yet it may likewiſe be under
the
ſaid Plane;
in which Cafe, the Geometrical
Plane
is ſuppoſed above the Objects, as we have
already
done on another Occaſion.
1179.
CHAP. VII.
Of Shadows.
FIRST we muſt obſerve, with thoſe who
have
already treated on this Subject, that
when
a luminous Body is equal to an opaque
Body
it enlightens, the Shadow of the ſaid Body
is
contain’d between parallel Lines, and con-
ſequently
, it is equal upon all parallel Lines
placed
at any Diſtance whatſoever beyond the
opaque
Body.
And when the luminous Body
is
leſſer than the opaque Body, the Shadow
thereof
, increaſes, and is infinitely augmented.
And on the contrary, when an opaque Body
is
leſs than the luminous Body, the Shadow there-
of
decreaſes and terminates in a Point.
Now becauſe the Sun is vaſtly greater than
any
of the Bodies on the Earth’s Surface it en-
lightens
, and is at ſo great a Diſtance therefrom,
therefore
its Rays may be conſider’d as being
parallel
;
and conſequently, the Bodies it ſhines
upon
as encloſed between parallels:
And this
is
the firſt Kind of Shadows I ſhall here explain;
16783on PERSPECTIVE. after which, I ſhall touch upon thoſe continually
increaſing
.
What I ſhall ſay on this Matter, is
ſufficient
for deſigning the Shadows of right-lin’d
Bodies
;
as to the Shadows of other Bodies, it
is
ſo difficult to determine them Geometrically
that
it is much better to examine thoſe which are
daily
obſerved, and ſo imitate them.
I ſhall not ſay any thing concerning Shadows
terminating
in a Point, becauſe their Variety
is
ſo great, that they cannot be geometrically de-
termin’d
.
Beſides, Painters ſcarcely ever ſup-
poſe
their perſpective Planes or Pictures en-
lightned
after this third Manner, unleſs only
when
they have a Mind to repreſent a Chamber,
wherein
the Light enters through the Windows;
but then the Number of Windows, their Situati-
on
, and the different Reflections that the Light
ſuffers
in the Chamber, produce ſo many dif-
ferent
Alterations, that a Painter had better
imitate
them, than have recourſe to Rules that
do
not take in all Caſes.
I ſhall likewiſe be ſi-
lent
as to the clair-obſcure, for a ſmall Atten-
tion
to daily Experience will better illuſtrate
this
Matter than a long Diſcourſe thereon.
Be-
ſides
, it is impoſſible to furniſh general Rules
on
this Subject;
and likewiſe the vaſt Number
of
Figures, will not permit us to ſeparately
examine
them;
add to all this, that a Painter
to
draw the clair-obſcure, he ought to have, not
only
regard to the Figures of Objects, but likwiſe
to
their Colour and Matter.
Of Solar Shadows.
Problem I.
103. To find the Perſpective of the Shadow of a
Point
above the Geometrical Plane, whoſe Height and
Seat
is given.
16884An ESSAY
Let Z be the Geometrical Plane; A the Seat
11Fig. 57. of the given Point;
and A B the Direction of
the
Sun’s Rays.
Operation.
Draw two right Lines, making a right Angle
with
each other;
in one of which, aſſume P E,
equal
to the Height of the given Point, above
the
Geometrical Plane:
Then draw the Line
E
C thro’ the Point E;
making an Angle with
C
P, equal to the Sun’s Altitude;
and make A B
equal
to C P.
Find the Appearance of the Point
B
;
and the Repreſentation ſought, will be
had
.
Note, This Operation, as likewiſe all the
others
in this Chapter, regard all the Situations
of
the perſpective Plane;
and is ſo evident, that
there
is no need of demonſtrating it.
Proe. II.
104. To find the Repreſentation of an elevated
Point
, whoſe Appearance, as alſo that of its Seat, is
given
, without uſing the Geometrical Plane.
Find F, the accidental Point of the 22Fig. 58. Rays, and D, that of their Direction: Then
3368, 88,
98
.
draw a Line from the Point D, through a, the
Perſpective
of the Seat of the given Point;
and
another
from F, through I, the Perſpective of
the
given Point:
And then b, the Interſection
of
the ſaid two Lines, will be the Point ſought;
as is manifeſt.
Remarks.
order to find the accidental Point of any
ber of inclin’d Lines, we have
169
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17055[Figure 55]page 64.
Plate
23.
Fig
. 54
O M P Q t A X x Q R N
56[Figure 56]Fig. 55G F b T L a57[Figure 57]Fig. 56I F a X b E T C P
171
[Empty page]
17285on PERSPECTIVE. one of the Directions drawn upon the 1168, 88. trical Plane; but it is ſufficient here, that the
Angle
the ſaid Directions make with the Baſe
Line
, be only known:
And ſo, as the Problem
expreſſes
it, the Geometrical Plane may be en-
tirely
laid aſide.
105. When the Perſpective Plane is parallel,
the
Sun’s Rays will have no accidental Point;
for their Repreſentations are then parallel; in
which
Caſe, one of the Parallels muſt be drawn
through
the Point a, inſtead of the Line D a.

Moreover
, when the Perſpective Plane is perpen-
dicular
, or inclin’d, and the Sun’s Rays are pa-
rallel
thereto;
a Line muſt be drawn through the
Point
a, parallel to the Baſe Line;
as likewiſe
another
Line through the Point I, parallel to the
Sun’s
Rays;
cutting the firſt Line in the Point
ſought
.
Problem III.
106. To find the Perſpective of the Shadow of an
elevated
Point, when there is ſome Body hindring its
falling
upon the Geometrical Plane.
The Perſpective of the Section of the Body
made
by a Plane paſſing through the given Point
perpendicular
to the Geometrical Plane, and pa-
rallel
to the Sun’s Rays, muſt be found:
And then
the
Interſection of the ſaid Perſpective, and a
Line
drawn from the Appearance of the given
Point
to the Repreſentation of its Shadow, is the
Repreſentation
ſought.
17386An ESSAY
Of the Shadows of a ſmall Light.
Prob. IV.
107. To find the Perſpective of the Shadow of a
Point
, whoſe Seat, and Height, above the Geometri-
cal
Plane is known.
Let Z be the Geometrical Plane, A the Seat
11Fig. 59. of the given Point;
and C that of the Light:
Draw the indefinite Line C A B; and about C,
as
a Center, with a Semidiameter equal to the
Height
of the Light above the Geometrical
Plane
, deſcribe the Arc F:
Again, about the
Point
A, with a Semidiameter equal to the
Height
of the given Point, deſcribe the Arc E.

This
being done, draw the Line E F touching
the
ſaid two Arcs, and cutting the Line C A in
B
.
Then if the Perſpective of B be found, the
Perſpective
of the Shadow of the Point will be
had
.
Problem V.
108. To find the Perſpective of the Shadow of an
elevated
Point, the Repreſentation of which, as alſo
of
its Seat being known, without uſing the Geometri-
cal
Plane.
The Operation of Prob. 3. for Solar 22104. dows muſt be uſed here, with this Difference;
that inſtead of the accidental Point of the Sun’s
Rays
, the Perſpective of the ſmall Sight muſt
be
uſed;
and in the Room of the accidental
Points
of the Directions of the ſaid Rays, the
Perſpective
of the Seat of the Light muſt be
aſſumed
.
174
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17558[Figure 58]page 66.
Plate
. 24.
Fig
. 57
E A Z C P B
59[Figure 59]Fig. 58F O D I a b60[Figure 60]Fig. 59F E Z C A B
176
[Empty page]
17787on PERSPECTIVE.
Remarks.
What hath been obſerved on the Solar 11105. dows, hath no Regard to thoſe mentioned here:
For it matters not in this Problem, whether the
perſpective
Plane be perpendicular, inclined, or
parallel
;
becauſe in theſe different Situations, the
two
Points that are uſed may always be found.
Moreover, it muſt be obſerved, that the third
Problem
takes in the Shadows of a ſmall 22106. as well as thoſe of the Sun;
but not without this
Difference
, that the Plane which in the third
Problem
is ſuppoſed parallel to the Sun’s Rays,
ought
here to be ſuppoſed to paſs through the
Light
whoſe Shadows are ſought.
CHAP. VIII.
Of mechanically ſhortning the Operations of
Perſpective
.
1. WHEN the perſpective Plane is ſup-
pos’d
perpendicular or upright.
Problem I.
109. To find the Repreſentations of Figures in
the
Geometrical Plane.
Let O be the Eye, R H the Baſe Line; F and
33Fig. 60 G the Points denoted with the ſame Letters in
Fig
.
10.
4431.
Then take a Ruler and faſten to the Point G,
ſo
that it may turn about the ſame in ſuch man.
17888An ESSAY on ner, that right Lines drawn along one of its Sides,
may
paſs through the Point G.
This being done,
faſten
one End of a Thread, put through the Eye
of
a Needle B, in the Point F;
and then put the
ſaid
Thread about a Pin faſten’d in the Point O;
ſo that when the Thread is uſing, it may be al-
ways
kept tight by means of a Plummet fix’d to
its
other End, and freely hanging under the Tube.

Note
, The aforeſaid Needle onght to be Braſs or
Silver
, ſharp at both Ends, and having its Eye
pretty
near one of them.
Operation.
Let A be a Point of the Figure to be thrown
into
Perſpective.
Now place that of the two
Points
of the Needle, which is nigheſt to the
Eye
thereof, upon the ſaid Point A, and move
the
Ruler G E, until it cuts the Thread A F, in
the
Point E, wherein the ſaid Thread cuts the
Baſe
Line;
and then the Point a, wherein the
Ruler
cuts the Thread A O, is the Point ſought,
and
it may be marked with that Point of the
Needle
fartheſt diſtant from its Eye, having firſt
preſſed
the Ruler down upon the Paper and Thread,
that
ſo the Plummet may not make the Thread
ſlip
.
And in this manner may any Number of
Points
be found.
This is demonſtrated in N. 32.
Sometimes it is more convenient to uſe the fol-
lowing
Method.
Method II.
110. Let O be the Eye, H E the Baſe Line,
11Fig. 61. and F I the Geometrical Line.
Take a Ruler
M
N, having two Threads equal in Length
faſten’d
to it, and about O as a Center, and
179
[Empty page]
18061[Figure 61]page 88.
Plate
. 25.
Fig
. 60
O G F f Z L R P D I T S M a Q E R H N A C B
181
[Empty page]
18289PERSPECTIVE. the Diſtance of the two Points whereat the
Thread
is faſten’d on the Ruler, deſcribe an
Arc
cutting the Geometrical Line in F;
then
faſten
the Extremity of one of the aforeſaid
Threads
in the ſaid Point F, and the Extremity
of
the other in the Point O:
Take moreover
another
Thread, put through the Eye of a
Needle
, as in the aforegoing Method;
one End
of
which, faſten in F, and afterwards put it a-
bout
a Pin placed in O.
Then the only Dif-
ference
between this Way and the precedent one,
is
, that the Ruler M N muſt be uſed, by always
keeping
the Threads M F, and N O tight, in-
ſtead
of one turning about a Point.
For the Demonſtration of this, vide n. 39.
Prob. II.
111. To find the Perſpective of one or more Lines
perpendicular
to the Geometrical Plane.
Take two Rulers L C and N Z, having the
11Fig. 60. four Ends of two Threads, or rather four Braſs
or
Steel Wires of equal Length fixed on them,
at
the Places L, I, N and M, ſo that L I be equal
to
M N.
Then fix one oſ theſe Rulers upon the
Edge
of the perſpective Plane, perpendicular to
the
Baſe Line.
Now take a Thread, put thro’
the
Eye of a Needle, hang a Plummet at one
End
, for the ſame Uſe as in Problem I , 22109. faſten the other End to the Slider or Curſor D,
freely
moveable on the Ruler L C;
put this
Thread
about a Pin, ſet up againſt the Ruler C L
inc
, ſo that C H be equal to the Height of the Eye.
Operation.
Let T be the Repreſentation of the Foot of a
Perpendicular
.
Move the Curſor D along C
18390An ESSAY until C D be equal to twice the Length of the
given
Perpendicular.
This being done, move
the
Needle along the Horizontal Line, ſuppoſe
to
R, until the Point of the Thread R S paſſes
through
the Point T;
then keeping the Thread
tight
in this Manner, if the Ruler M N be
mov’d
, till its Edge alſo paſſes through the
Point
T, and P is the Point wherein the Edge of
the
ſaid Ruler croſſes the Part of the Thread
R
D, the Line T P will be the Repreſentation
ſought
.
The Demonſtration of this is evident from
what
is ſaid in n.
59.
Method II.
112. When all the Perpendiculars have the ſame
Length
.
Let F G be parallel to the Baſe Line, and F O
11Fig. 60. equal to the Height of the Eye;
aſſume F f, e-
qual
to the Length of either of the given Per-
pendiculars
, and faſten the Thread fixed in F, in
the
Point f.
Then raiſe R S perpendicular to
the
Baſe Line, which make equal to F f, and
draw
S Q parallel to the Baſe Line.
This being
done
, tranſpoſe the Figures in the 2260. cal Plane, in ſuch Manner, that the Point R co-
incides
with S, and R H with S Q.
Then if S Q
be
taken for a Baſe Line, and the Appearances
of the Feet of the Perpendiculars be 33109. the Repreſentations of their Extremities will be
had
.
Method III.
113. For Perpendiculars of the ſame Length.
The Figures in the Geometrical Plane being
44Fig. 61. tranſpoſed in the Manner aforeſaid , aſſume T 55112.
18491on PERSPECTIVE. in the Perpendicular R S, continued equal to R S,
and
draw f i parallel to the Baſe Line, in which
take
the Point f in E I, the ſame as F in F I ;
11110. if the Threads which before were faſten’d in F,
being
placed in f, and by uſing them thus faſten’d,
as
likewiſe S Q for a Baſe Line, the Repreſen-
tations
of the Feet of the Perpendiculars be
found
, you will have the Repreſentations 22110. their Extremities.
The Demonſtration of the two laſt Ways.
114. If a Plane be imagined to paſs through
33Fig. 60,
and
61.
the Extremities of the equal Perpendiculars, it
will
be parallel to the Geometrical Plane, and
will
meet the perſpective Plane in S Q;
becauſe
R
S is equal to the ſaid Perpendiculars:
Moreover,
the
Extremities of theſe Perpendiculars form a
Figure
in this ſuppoſed Plane, ſimilar to that
which
their Feet form in the Geometrical Plane;
and the ſaid Figure hath the ſame Situation with
regard
to the Line Q S, as that on the Geome-
trical
Plane hath in reſpect of H R:
And conſe-
quently
, if the Figure in the Geometrical Plane
be
ſo raiſed up, that it hath the ſame Reſpect to
Q
S, as it had to H R, and if the Appearances
of
the Feet of the propoſed Perpendiculars be
found
, the Repreſentations of their Extremties
will
be had.
But the before ſuppoſed Tranſpo-
ſition
of the Figure in the Geometrical Plane,
gives
it the requiſite Situation with regard to
Q
S, and the Repreſentation of the Figure
conſider’d
in this new Geometrical Plane is
found
;
becauſe S Q, is taken for the Baſe 4432, 39. O f (Fig. 60.) equal to the Height of the Eye
above
this Plane, and fi (Fig.
61.) is the Geo-
metrical
Line in the ſaid Plane.
18592An ESSAY
II. When the Perſpective Plane is inclined.
Prob. III.
115. To find the Repreſentation of Figures which
are
in the Geometrical Plane.
The Operation laid down for the 11109,
110
.
Plane being perpendicular, may be uſed here,
becauſe
the inclined perſpective Plane may be
changed
into a perpendicular or upright one.
2281.
Prob. IV.
116. To find the Appearances of any Number of
Lines
of the ſame Length, which are perpendicular
to
the Geometrical Plane.
Raiſe a Perpendicular R C, in ſome Point on
33Fig. 62. the Baſe Line, in which aſſume R L equal in
Length
to the given Lines;
and draw the Line
L
P, through the Point L, ſo that the Angle
L
P R, be equal to the Angle of Inclination of
the
perſpective Plane;
then having made R S
equal
to P L, and S C equal to P R, draw the
Lines
S Q and C D, parallel to the Baſe Line.
This being done, raiſe up the Figures of the
Geometrical
Plane, ſo that the Point R coincides
with
c, and the Line R H, with C D;
then
proceed
as is directed for the 44112, 113. perſpective Plane, in uſing S Q for the Baſe
Line
.
This is demonſtrated in n. 114.
Remarks.
The Point C muſt be aſſumed below the Point
S
, when the perſpective Plane is inclined
186
[Empty page]
18762[Figure 62]Plate 26
Fig
. 61
O I F T N S Q S H E R M A
63[Figure 63]Fig. 62C D S Q L C D R P H
188
[Empty page]
18993on PERSPECTIVE. the Eye, and above it, when the ſaid Plane is in-
clined
towards the Objects.
Obſerve likewiſe,
that
F f of Fig.
60. muſt here be aſſumed e-
qual
to R S, and the Line Tt, Fig.
61. muſt
be
here a Part of the Line Z C continued, and
made
equal to R S.
III. When the Perſpective Plane is Parallel or
Horizontal
.
Prob. V.
117. To throw Figures which are in the Geometri-
cal
Plane into Perſpective.
Draw the Line C F at Pleaſure, in which aſ-
11Fig. 63. ſume the Point I, and make I H and I G, equal
to
the Eye’s Diſtance from the perſpective
Plane
:
Moreover, make I C, and I F equal to
the
Eye’s Diſtance from the Geometrical Plane,
or
at the leaſt, let I G and I H, be to I F and I C,
as
the Eye’s Diſtance from the perſpective Plane,
is
to its Diſtance from the Geometrical Plane.
This being done, raiſe two Perpendiculars to
the
Line C F, in the Points G and H, and take
two
Rulers, each of which has two equal Threads
ſo
faſten’d to them, that the Diſtance P Q, and
N
M be equal:
Then about F and C as Centers,
with
the Semidiameters M N or P Q, deſcribe
two
Arcs cutting the Perpendiculars raiſed at the
Points
G and H, in the Points C and D, and fix
the
Extremities of the two Threads of one Ruler,
in
the Points C and D, and the Threads of the
other
, in the Points F and E.
Operation.
Let Z be the Geometrical Plane, and A a
Point
of the given Figures.
Move the
19094An ESSAY Rulers, keeping all the Threads tight, ſo that
the
two Threads fix’d in the Points C and F,
croſs
each other in the Point A;
and then the
Point
a, wherein the two other Threads croſs
each
other, is the Perſpective ſought.
And in
this
Manner may the Repreſentations of any
Number
of Points be ſound.
Demonstration.
118. The Triangle D a E, is ſimilar to the
Triangle
C A F:
And becauſe all the Triangles
formed
for finding the Repreſentations of diffe-
rent
Points, have the ſame Baſes D E and C F,
which
are between themſelves, as the Eye’s Di-
ſtance
from the perſpective Plane, to its Diſtance
from
the Geometrical Plane;
whence their Ver-
tices
form ſimilar Figures, whoſe correſpondent
Lines
are in the ſame Proportion;
and which
conſequently
, are the Appearances ſought.
118, 9.
Remarks.
It will be convenient to have the two Threads
P
E, and M D of one Colour, and the two
Threads
Q F, and C N of another.
Prob. VI.
119. To find the Repreſentations of any Number
of
Lines, equal and perpendicular to the Geometri-
cal
Plane.
Let C D E F G I H, be the Points denoted
22Fig. 64. with the ſame Letters in the precedent Figure,
as
alſo let P Q and M N be the Rulers:
More-
over
, let B be the Point, wherein a
19195on PERSPECTIVE. lar drawn from the Eye to the Geometrical
Plane
, meets the ſaid Plane;
and T the Per-
ſpective
of that Point, found by the aforegoing
Problem
.
Now make F L and C R, equal to the
Length
of the given Lines;
and about the
Points
R and S, as Centers, with the Radius
M
N or P Q, deſcribe two Arcs cutting the
Perpendiculars
H D and G H, in the Points
X
and S:
Then fix the Extremities of the
Threads
, which before were faſten’d to the
Points
F and E, to the Points L and S;
and alſo
the
Extremities of thoſe two Threads, which
were
before faſten’d to the Points C and D, in
the
Points R and X:
Then moving the two
Rulers
, until the Threads S P and X M, croſs
each
other in the Point T, mark the Point O,
wherein
the two other Threads croſs one another,
through
which, and the Point B, draw the
indefinite
Line B O V:
this being done, tranſ-
poſe
the Lines of the Geometrical Plane, ſo that
the
Point B, coincides with the Point O, and
the
Line B O, with O V.
And by the precedent
Problem
, find the Appearances of the Feet of
the
Perpendiculars, and you will have the Re-
preſentations
of their Extremities.
Demonstration.
If a Plane be ſuppoſed to paſs through the
Extremities
of the Perpendiculars, it will be
parallel
to the Geometrical Plane, and conſe-
quently
, likewiſe to the perſpective Plane, be-
cauſe
all the Perpendiculars are ſuppoſed equal.
But the Figure formed by the Extremities of
the
Perpendiculars in the ſaid Plane, is ſimilar
and
equal to that form’d by their Feet in the
Geometrical
Plane:
and therefore, the
19296An ESSAY ſentation of the Figure which is in the ſecond
Plane
, is likewiſe ſimilar to the Figure which is
in
the Geometrical Plane, and the Lines com-
poſing
this Appearance to the correſpondent
Lines
in the ſecond Plane, as the Eye’s Diſtance
from
the perſpective Plane, is to its Diſtance
from
the before ſuppoſed Plane.
But by means
of
the Threads faſtned in the Manner aforeſaid,
we
find the Repreſentation of a Figure, whoſe
Lines
are in the aforenamed Proportion:
11118. Whence this Figure is the Perſpective ſought,
and
is ſituated as it ought to be, with regard to
the
Repreſentations of the Figures in the Geo-
metrical
Plane;
becauſe theſe Figures are ſo
ſlid
, that the Appearance of the Perpen-
dicular
in the Point B, is only a Point.
The
ſaid
Appearances are alſo in their proper Poſiti-
ons
, becauſe the Line B Q, is made to coincide
with
O N.
120. For Solar Shadows in all Situations of the Per-
ſpective
Plane.
Prob. VII.
6. To find the Repreſentations of the Shadows of
any
Number of Points, being at the ſame Height
above
the Geometrical Plane.
Find a Point in the Geometrical Plane, 22103. is the Shadow of one of the given Points: Tranſ-
poſe
the Figures in the Geometrical Plane, ſo that
the
Seat of the ſaid given Point coincides with
its
Shadow;
and the Line drawn through the
ſaid
Seat, and the Shadow of the given Point,
coincides
with its Prolongation.
Then if ac-
cording
to the Situation of the perſpective Plane,
the
Repreſentations of the Seats of the
193
[Empty page]
19464[Figure 64]page 96.
Plate
. 27
Fig
. 63
D E C F M H I G P A Q N
65[Figure 65]Fig. 64X S D E T C R L F H I G P M B O V Q N
195
[Empty page]
196
[Empty page]
19766[Figure 66]page 98.
Plate
. 28
Fig
. 65
L M F G D H C E I A B
67[Figure 67]Fig. 66A B VII VIII IV V H C VI VI P V VII IV S VIII E O I III II I XII XIX IX F D
198
[Empty page]
19997on PERSPECTIVE. Points be found , the Repreſentations of 11109,
115
, 117.
Shadows will be had.
CHAP. IX.
The Uſe of Perſpective in Dialling; ſhewing
how
to draw the Hour Lines upon any
Kind
of Plane, by means of an Horizon-
tal
Dial.
PERSPECTIVE is not only uſeſul in
drawing
, but likewiſe in other Parts of Ma-
thematicks
, and principally in Dialling:
For if
the
Extremity of the Style be conceived as the
Eye
, and the Sun’s Rays as viſual Rays, all
poſſible
Kinds of Dials may be drawn for the
ſame
Latitude, by Means of an Horizontal Dial,
as
we are now going to ſhew,
121. Let A B C D be an Horizontal Dial
made
for any given Latitude;
E F its Style, and
22Fig. 65. H I M L a Plane, upon which a Dial is to be
drawn
.
Now if this Plane be ſo ſituated, that
the
Extremity of its Style F G, coincides with
the
Extremity of the Style of the Horizontal
Dial
;
and if the Perſpective of one of the Hour
Lines
of the Horizontal Dial A B C D, be found
upon
the Plane H I M L, in conceiving the
Point
F as the Eye, it is evident , that 332. Shadow of the Point F, will fall upon the ſaid
Perſpective
, at the ſame Time that it falls
upon
the Hour Line, whereof it is the Per-
ſpective
;
and conſequently, the ſaid Shadow
will
ſhew the ſame Hour upon the Plane H I M L,
that
it ſhews upon the Horizontal Dial:
20098An ESSAY fore the before-mention’d Perſpective will be an
Hour-Line
of a Dial, drawn upon the Plane
H
L M I, and whoſe Style is G F.
The ſame may be demonſtrated of the Re-
preſentations
of other Hour-Lines, which form
a
Dial upon the Plane H L M I.
We now proceed to lay down the beſt way of
determining
the ſaid Repreſentations.
Prob. I.
122. To draw Vertical Dials.
Draw the Line E O, thro’ E, the Foot of the
11Fig. 66. Style of the Horizontal Dial A B F D, equal
to
the Length of the Style of the Dial to be
drawn
, and making an Angle with the Meri-
dian
C.
XII, equal to the Plane’s Declina-
tion
.
This Angle muſt be aſſum’d towards the
Point
D, when the Plane’s Declination is South-
Eaſt
, as here;
towards F, when the Declination
is
Weſtward;
towards A, when it is North-Eaſt-
wardly
;
and towards B, when it is North Weſt-
wardly
.
Now, thro’ the Extremity O of the ſaid Line,
draw
the Line I H, perpendicular thereto;
and
the
Line C P, thro’ the Center of the Dial, pa-
rallel
, and equal to E O;
thro’ whoſe Extre-
mity
, P, draw the Line P S, parallel to H I.
Then, to make the Dial, draw the Line h i,
22Fig. 67. in which prick down the Diviſions of the
Line
H I;
and in the Point o, (which is the
ſame
as O) raiſe the Perpendicular o p, equal to
the
Length of the Style of the Horizontal Dial
A
B D F.
This being done; draw a Parallel,
h
i, thro’ the Extremity of this Perpendicular;
on which, prick down the Diviſions of the
20199on PERSPECTIVE. P S, and making the Point p be the ſame as P.
then join each Diviſion of this Line, to the
correſpondent
Diviſion of the Line h i;
and the
Dial
ſought, will be drawn:
p being the Foot
of
the Style, and p s the Horizontal Line.
Demonstration.
The Baſe Line is h i; p s is the Horizontal
11Fig. 66,
67
.
Line;
p the Point of Sight; and E O, or C P,
of
Figure 66.
is the Length of the principal
Ray
.
Now, ſuppoſe the Plane p s h i, to be ſet per-
pendicularly
upon the Horizontal Dial, in ſuch
manner
, that the Line h i coincides with H I,
and
the Point o with O.
Suppoſe, moreover,
that
thro’ the Extremity of the Style, which we
conſider
as the Eye, Lines are drawn in the Ho-
rizontal
Plane, parallel to the Hour-Lines of
the
Dial;
it is evident, that theſe Lines will
meet
the Horizontal Line p s, in the Point al-
ready
prick’d down;
and conſequently , 2213. Appearances of the Hour-Lines, are the Lines
joining
the Diviſions of the Lines h i and p s.
Remark.
If the Line H I happens to meet the Meri-
dian
;
the common Method, by the Horizontal
Dial
, is eaſier than this.
Prob. II.
123. To draw inclining Dials.
Theſe Dials are drawn in the ſame manner as
Vertical
ones;
the following Preparations being
firſt
made.
202100An ESSAY
Draw the Line e c, equal to the Length of
11Fig. 68. the Style of the Horizontal Dial;
and, at its
two
Extremities raiſe the Perpendiculars e o
and
c p:
Then draw the Line c G, thro’ the
Point
c, equal in Length to the Style of the
Dial
to be drawn;
making an Angle with c e,
equal
to the Inclination of the Dial-Plane.
Af-
ter
which, draw the Line o G p, thro’ the Ex-
tremity
G of the ſaid Line, perpendicular there-
to
.
This Preparation being finiſh’d, we uſe the
Operations
of the precedent Problem, in making
E
O and C P, in the Horizontal Dial, equal to
e
o and c p of this Figure;
and o p, in the Dial
to
be drawn, equal to op of this Figure;
in
which
, the Point G is the Foot of the Style.
If it happens, in the aforeſaid Preparation,
22Fig. 69. that the Line p o cuts the Line e c;
E O muſt
not
then be aſſum’d in the ſame Line, in the
Horizontal
Dial, as otherwiſe it muſt have
been
;
but, in that Line, continued on the other
ſide
of the Foot of the Style.
The Demonſtration of this Problem, is the
ſame
as that of the precedent one;
in conſider-
ing
the Angle p o Q, equal to the Angle G c e,
which
is equal to the Inclination of the Plane.
We might have further ſhewn the Uſe of
Perſpective
, in facilitating the Operations of
Dialling
:
But this would be deviating from our
Subject
;
and ſo we ſhall content our ſelves with
this
Short Eſſay, touching the moſt common and
uſeful
Problem in Dialling.
203
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20468[Figure 68]page 100
Plate
. 29
Fig
. 67
5 6p 7 8 9 10 S V VI VII VIII IX X o XI ll l
69[Figure 69]Fig. 68c P G e o Q70[Figure 70]Fig. 69P c G o e Q
205
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206101on PERSPECTIVE. 71[Figure 71]
The Uſe of the Camera Obscura
in
Deſigning.
Advertisement.
EVery one knows how eaſy it is by one Convex Glaſs
only
, to repreſent outward Objects in any darken’d
Place
, according to their natural Appearances, where
the
Livelineſs of Colours, and the Diverſity of Mo-
tions
, are wonderfully pleaſant to behold;
and it is
ſo
eaſy to make this Invention uſeful in Deſigning,
that
our treating of it ſo fully as we have done, is
undoubtedly
ſomething neceſſary.
A few Hints and Obſervations ſeem to be ſufficient
for
putting a diligent Perſon in the way of contriving
himſelf
ſome Machine to perform the Uſes hereafter
mentioned
;
whence we might have let him bad the
Pleaſure
of the Invention himſelf, after it was made
eaſy
to him.
And this indeed is what I firſt reſolv’d
upon
;
but afterwards conſidering that in the Con-
ſtruction
of a Machine for facilitating the Buſineſs of
Deſigning
, ſeveral Things cannot be foreknown till
try’d
;
and that much Time may be ſpent in vain, and
ſeveral
Methods attempted, before one of theſe Ma-
chines
can be made ſimple and uſeful, as I have found
by
Experierce:
Therefore, that others may not be at
this
Trouble, I ſhall here lay down the Deſcription of
two
Machines (hoping it will not be unacceptable)
which
after ſeveral Alterations, in my Opinion, are
now
made convenient enough.
207102An ESSAY
The firſt of theſe Machines is undoubtedly much
preferable
to the other;
becauſe it is firmer, and
renders
the Work eaſier and exacter;
and Prints may
be
eaſier repreſented in it, than in the other.
Add to
all
this, that with a ſmall Alteration it may be made
capable
of a few Uſes, which are peculiar to
the
ſecond Machine;
yet ſince this latter one is
much
ſimpler, of a much leſs Price, and is eaſier to
be
carry’d from Place to Place, I thought it conveni-
ent
to lay down alſo its Deſcription in this ſmall
Tract
.
I ſhall not here take up the Reader’s Time in enu-
merating
the Advantages accruing to Painters from
Machines
of this Nature:
But only add, that they
are
of great Uſe for reducing or leſſening ſeveral ſe-
parate
Objects in the ſame Picture;
which therefore
may
be copy’d after Nature in the moſt perfect man-
ner
poſſible.
It is very difficult to give ſeveral Objects
their
true Bigneſs in a Picture, and to diſpoſe them
ſo
as to have the ſame Point of Sight;
but this is
done
extremely eaſy by means of Machines;
for the
Point
of Sight in them will be the ſame always, as
long
as the Convex Glaſs has the ſame Diſpoſition,
and
the Bigneſs of the Repreſentations of Objects in
the
Machines, do depend upon the Objects Diſtances
from
them.
This Invention, by Induſtry, may be certainly im-
prov’d
, and the following Obſervations will be not a
little
conducing thereto.
1. You muſt uſe but one
Convex
Glaſs;
for when there are two or more, the
true
Repreſentations of Objects will be loſt;
which is
an
Inconveniency one is alſo ſubject to, when a Con-
cave
Glaſs is any ways us’d in the Conſtruction of the
Machine
.
2. When more than two Mirrours or Look-
ing
glaſſes are us’d, the Rays, after having ſuffer’d a
triple
Reflection, are ſo weaken’d, that the Objects
will
not be well repreſented:
And even when but two
Mirrours
are us’d, they muſt be well poliſh’d 3.
208103on PERSPECTIVE. Mirrours muſt not be plac’d within the Machine; for
in
ſuch a cloſe Place, ones Breath will ſully them;
but
not
the Convex Glaſs, becauſe it is encloſed in a
Tube
.
72[Figure 72]
The Uſe of the Camera Obscura
in
Deſigning.
Definition.
1. ACamera Obſcura is any dark Place, in which
outward
Objects expoſed to Broad-Day-light,
are
repreſented upon Paper, or any other white
Body
.
The Way to repreſent Objects in the Camera
Obſcura
, is to make a ſinall Hole in that ſide
thereof
next to the Objects, and place a Convex-
Glaſs
therein, then if a Sheet of Paper be ex-
tended
in the Focus of the ſaid Glaſs, the Objects
will
appear inverted upon the Paper.
Theorem I.
2. The Camera Obſcura gives the true Repreſen-
tation
of Objects.
The Figures repreſented in the Camera Obſcura
are
form’d (as is demonſtrated in Dioptricks)
by
Rays which coming from all the Points of the
Objects
, paſs through the Centre of the Glaſs:
So that an Eye placed in the ſaid Centre, would
perceive
the Objects by the ſaid Rays, which
conſequently
by their Interſection with a Plane,
muſt
give the true Repreſentation of the Objects.

But
the Pyramid which the ſaid Rays forms
209104An ESSAY on out the Camera Obſcura, is ſimilar to that which
they
form, after having paſſed through the Glaſs:
Therefore the Rays which fall upon the Paper in
the
Camera Obſcura, likewiſe give the true Re-
preſentation
of the Objects thereon.
Which was
to
be demouſtrated.
Theſe Objects appear inverted, becauſe the
Rays
croſs each other in paſſing through the Glaſs;
thoſe coming from above going below, & c.
Theorem II.
3. The Reflection which the Rays of Light ſuffer
upon
a plain Mirrour or Speculum, before they fall
upon
a Convex Glaſs, no-wiſe deforms the Repreſen-
tation
of Objects.
This is evident: For the Speculum reflects the
Rays
in the ſame Order as it receives them.
Now to ſhew the Uſe that may be drawn from
the
Camera Obſcura in Deſigning, I ſhall here lay
down
the Deſcription and Uſe of Two Machines,
which
I uſe for this End.
The Deſcription of the Firſt Machine.
4. This Machine is ſomething in Figure of a
Chair
, (ſuch as People are carried in) the back Part
of
the Top is rounded, and its Foreſide P Q ſwells
out
in the middle:
Vide Fig. 70. which repreſents
11Fig. 70. the Machine, the Side whereof oppoſite to the
Door
, is ſuppoſed to be raiſed up, that ſo its In-
ſide
may be ſeen.
5. The Board A within-ſide, ſerves as a Ta-
ble
, and turns upon two Iron-pins, going into
the
Wood of the Fore-ſide of the Machine, and is
ſuſtain’d
by two ſmall Chains, that ſo the ſaid
Table
may be lifted up;
and therefore one
210105PERSPECTIVE. more conveniently go in at the Door in the Side
of
the Machine.
6. There are two Tin-Tubes bent at each End,
(one of which is repreſented in Fig.
76. becauſe
they
could not be ſhewn in the Figure of the
Machine
) placed in the Furniture, near the
Back-ſide
of the Machine, each having one End
without
the Machine.
Theſe Tubes ſerve to give
Air
to Perſons ſhut up in the Machine, yet ſo,
that
no Light may enter through them.
7. At the Places c, c, c, c, on the Out-ſide of
the
Back-part of the Machine, are four Iron-Sta-
ples
, in which ſlide two Wooden Rulers D E, D E,
about
3 Inches broad, having two other thin
Rulers
going through holes made at their Tops, at
the
Places D, D, to which thin Rulers the Board
F
is fixed;
and ſo by this Means the ſaid Board
may
be vertically moved backwards or for-
wards
.
8. On the Top of the Machine, there is a
Board
about 15 Inches long, and 9 Inches broad,
having
an hole PMOQ quite through it, about
9
or 10 Inches long, and four Inches broad.
9. Upon the aforeſaid Board are fix’d two Dove-
tail’d
Rulers, between which another Board ſlides
of
the ſame Length as that, and whoſe breadth is
about
6 Inches.
In the middle of this ſecond
Board
is a round Hole about three Inches Diame-
ter
, hollowed into a female Screw, in which is
fitted
a Cylinder about four Inches long, which
carries
the Convex Glaſs;
Of which more here-
after
.
10 The Figure X is a ſquare Box about 7 or
8
Inches broad, and 10 in height, which ſlides
upon
the plain Board mentioned in Numb.
8. the
Side
B ſerving as a Door, is next to the fore-ſide
of
the Machine, as it appears in the Figure;
211106An ESSAY the Back-ſide thereof hath in it a Square ope-
ning
N, each Side being about four Inches in
Length
, which may be ſhut by the little Board I
ſliding
between two Rulers.
11. Overthe ſaid Square opening, there is a Slit
parallel
to the Horizon, going along the whole
Breadth
of the Back-ſide of the Box;
through
which
Slit, a little Mirrour or Looking-Glaſs is to
be
put into the Box, whoſe two Sides may ſlide be-
tween
two Rulers ſo placed, that the poliſhed
Side
of the Glaſs being turn’d towards the Door
B
, may make an Angle of 112 {1/2} Degrees, with
the
Horizon.
Note, This Diſpoſition of the
Looking-glaſs
could not well be repreſented in
the
Figure.
12. The before-mention’d Mirrour hath a ſmall
Iron
Plate, on the Middle of that Side which is
without
the Box, (when the ſaid Looking-glaſs
is
in the Diſpoſition mention’d in N.
11.) being the
Baſe
of a Screw faſten’d to the Middle thereof, that
ſothe
Looking-glaſs may be fix’d upright (as appears
per
Figure) in any Place H upon the Top of the
Machine
, and vertically turn all ways;
and this
is
done by putting the Screw through a Hole
made
in the plain Board of N.
9. and through a
Slit
made for this End in the plain Board of N.
8.
and then fixing it with the Nut R. Now when
the
Mirrour is taken from this Situation, the ſaid
Slit
is ſhut by a ſmall Board ſliding between two
little
Rulers within the Machine.
As to the Slit
mention’d
N.
11. it is partly ſhut by the little
Board
I, when the Aperture N is open’d, and the
two
Ends thereof remaining open’d, are ſhut by
two
ſinall Rulers.
13. There are two Iron Staples on one Side of
the
Box, like thoſe which are on the 117. of the Machine;
in which a Ruler going ſeveral
Inches
out behind the Box X may ſlide,
212107on PERSPECTIVE. a Hole at its End, through which the above-men-
tion’d
Screw may paſs, and ſo the Mirrour H be
fixed
to any Inclination before the Aperture N.
14. Beſides the Mirrour H, there is another leſſer
one
, L, faſten’d near its Middle to a Ruler going
out
through the Middle of the Top of the Box.
This Ruler may Screw on, and ſerves to raiſe or
lower
the Mirrour faſten’d to it, ſo that it may be
fixed
to all Angles of Inclination.
Remarks.
If the Tubes mention’d in N. 6. be not thought
ſufficient
for giving Air to the Machine, a ſmall
Pair
of Bellows may be put under the Seat,
which
may be blown by one’s Foot.
And by
this
means the Air within the Machine may be
continually
remov’d;
for the Bellows driving the
Air
out of the Machine, obliges the external Air
to
enter through the Tubes therein.
Uſe of the Machine.
Problem I.
15. To repreſent Objects in their natural Diſpo-
ſition
.
When Objects are to be repreſented within the
11Fig. 70. Machine, we extend a Sheet of Paper upon the
Table
A;
or, which is better, we lay a Sheet of
Paper
upon another Board, ſo that it ſpreads be-
yond
the Edges of the Board;
then we ſqueeze
the
ſaid Paper and Board into a Frame, ſo that
it
be fixed therein by means of two Dove-tail’d
Rulers
.
This being done, we place a Convex
Glaſs
in the Cylinder C, which ſcrews into 229. Top of the Machine, having its Focal
213108An ESSAY nearly equal to the Height of the Top of the
Machine
above the Table:
Then we open the
Aperture
N at the Back-part of the Box upon the
Machine
;
and incline the Mirrour L, ſo as to make
an
Angle of 45 deg.
with the Horizon, when Ob-
jects
are to be repreſented for the perpendicular
Picture
.
Then, if the Mirrour H be taken away,
as
alſo the Board F, together with the two Ru-
lers
D E and D E, we ſhall perceive the Repre-
ſentation
upon the Sheet of Paper on the Table
A
, of all Objects, whoſe Rays falling upon the
Looking-glaſs
L, can be thereby reflected upon
the
Convex Glaſs;
which Convex Glaſs muſt be
rais’d
or lower’d, by means of the Screw about
the
Cylinder carrying it, until the ſaid Objects
appear
entirely diſtinct.
16. When the ſame Objects are requir’d to be
repreſented
for the inclin’d Picture, the Looking-
glaſs
L muſt have half the Inclination we would
give
to the Picture.
17. When the ſaid Objects are to be repreſent-
ed
upon the Picture being parallel, the Aperture
N
muſt be ſhut, and the Door B open’d;
then
the
Mirrour H muſt be rais’d to the Top of the
Box
, in putting it in a Situation parallel to the
Horizon
.
This Diſpoſition of the Machine may
ferve
when one is upon a Balcony, or ſome
other
high Place;
to deſign a Parterre under-
neath
.
18. If we have a mind to deſign a Statue
ſtanding
in a Place ſomething elevated, and it is
requir’d
to be ſo repreſented, as to be painted
againſt
a Cieling;
the Back-ſide of the Machine
muſt
be turned towards the Statue, and the Box
X
ſo turned, that the Door B may face the Sta-
tue
;
then, the Door being open’d, the Looking-
glaſs
L muſt be placed vertically, with its po-
liſh’d
Side towards the Statue;
and the Box
214109on PERSPECTIVE. ved backwards or forwards, or elſe the Looking-
glaſs
raiſed or lower’d, until the Rays proceed-
ing
from the Statue may be reflected by the Mir-
rour
upon the Convex Glaſs.
When theſe Alte-
rations
of the Box, or Mirrour, are not ſufficient to
throw
the Rays upon the Convex Glafs, the whole
Machine
muſt be removed backwards or for-
wards
.
Demonstration.
Concerning the before-mention’d Inclination of the
Mirrours
.
19. In order to demonſtrate, that the Mirrour
L
hath been conveniently inclin’d, we need on-
ly
prove, that the reflected Rays fall upon the
Table
A under the ſame Angle, as the direct
Rays
do upon a Plane, having the ſame Situation
as
one would give to the Picture.
Now let A B be a Ray falling from a Point of
11Fig. 71. ſome Object upon the Mirrour G H, and from
thence
is reflected in the Point a upon the Table
of
the Machine:
We are to demonſtrate, that if
the
Line D I be drawn, making an Angle with
FE
equal to the Inclination of the Picture;
that
is
, if the Angle DIE be the double of the 2215, 16. D F I;
I ſay, we are to demonſtrate, that the
Angle
B a f is equal to the Angle BCD.
The Angle DIE, by Conſtruction, is the double
of
the Angle DFI;
and conſequently this laſt Angle
is
equal to the Angle I D F;
and ſince the Angle
of
Incidence C B D is equal to the Angle of Re-
flection
a B F, the Triangle BCD is ſimilar to
the
Triangle F a B:
Whence it follows, that the
Angle
Ba F is equal to the Angle BCD.
Which
was
to be demonſtrated.
215110An ESSAY
20. Concerning what hath been ſaid of the
Picture
being parallel, it muſt be obſerv’d, that
in
the precedent Demonſtration, the Angle of
Inclination
of the Picture in this Demonſtration
is
meaſur’d next to the Objects;
and if this An-
gle
be diminiſh’d till it become nothing, we
ſhall
have a Picture parallel to the Horizon un-
derneath
the Eye.
But, by the Demonſtration,
the
Inclination of the Mirrour being half the In-
clination
of the Picture, it follows, that the In-
clination
of the Mirrour is alſo equal to nothing,
and
conſequently it ought to be likewiſe parallel
to
the Horizon.
In the ſame manner we de-
monſtrate
, that the Looking-glaſs muſt be verti-
cally
ſituated, when we conſider the Picture pa-
rallel
above the Eye:
For to give this Situation
to
the Picture, the Angle of the Inclination of
the
Picture meaſur’d next to the Objects, muſt
be
augmented till it be 180 Degrees, whoſe half
90
Degrees is conſequently the Inclination of the
Mirrour
.
Prob. II.
22. To repreſent Objects, ſo that what appears on
the
right Hand, ought to be on the left.
23. Having placed the Box X, in the Situation
11Fig. 70. as per Figure, the Door B muſt be opened, and
the
Aperture N ſhut;
then putting the Mirrour H
in
the Diſpoſition mentioned in Numb.
11. raiſe
up
the Mirrour L towards the Top of the Box;
and incline it towards the firſt Mirrour, in ſuch
manner
that it makes an Angle with the Hori-
zon
of 22 {1/2} Degrees;
that is, that the Top of the
Machine
, after a double Reflection, appears ver-
cal
in the Mirrour H.
24. Now if Objects are to be repreſented for
the
Picture inclin’d, the Mirrour L muſt make
216111on PERSPECTIVE. Angle with the Horizon, equal to half the
Inclination
of the Picture leſs {1/4} of a right Angle.
This Angle is found exactly enough for Practice,
by
inclining the Mirrour L, until the Repreſenta-
tion
of the Top of the Machine, after a double Re-
flection
, appears in the other Mirrour under an
Angle
with the Horizon, equal to the Inclination
one
would give the Picture.
Note, If the Incli-
nation
of the Picture be leſſer than {1/4} of 90 De-
grees
, the Looking-Glaſs L muſt not be inclin’d
towards
the other, as is directed, but 1123: contrary Way, in making the Angle of the In-
clination
of the Looking-Glaſs, equal to the
Difference
of the Inclination of the Picture, and
{1/4} of 90 Degrees.
25. When the Objects are to be repreſented
for
a parallel Picture, the Looking-Glaſs L muſt
be
placed in the Diſpoſition of Numb.
15. and
the
Looking-Glaſs H in that mentioned, Numb.
13. by inclining it towards the Horizon, under an
Angle
of 45 Degrees;
the poliſhed Side thereof
facing
downwards, when the Picture is ſuppoſed
underneath
the Eye;
and upwards, when it is
ſuppoſed
above the Eye.
26. This Diſpoſition of the Machine may be
likewiſe
uſeſul for inclin’d Pictures, making very
ſmall
Angles with the Horizon;
in which Caſe,
the
Inclination of one of the Looking-Glaſſes
muſt
be diminiſh’d, by half of the Inclination
of
the Picture.
27. A Demonſtration of the Inclination of the
Mirrours
.
We have mentioned, that for a 2222. lar Picture, one of the Mirrours muſt make an
Angle
of 112 {1/2} Degrees with the Horizon;
3311. the other, L, muſt be inclin’d towards the
217112An ESSAY and make an Angle of 22 {1/2} with the Horizon.
Let MN and GH be two Mirrours in the before-
11Fig. 72. mentioned Situation;
we are to demonſtrate,
that
if the Ray A B is parallel to the Horizon,
after
being reflected in B and C, it ought to fall
perpendicularly
upon the Machine.
The Angle
A
B N is 112 {1/2} Degrees;
and conſequently 2211. Angle A B M, and its equal, the Angle of Re-
flection
C B G, are each 67 {1/2} Degrees.
The An-
gle
B P Q, is the Complement of the Angle
N
B A, plus the Angle P Q B, which is 22 {1/2} 3323. grees;
whence the Angle B P Q is 45 Degrees.
Again, the Angle PC B is the Complement of
the
two Angles C B P and B PC to 180 Degrees;

and
conſequently it is 67 {1/2} Degrees, which is the
ſame
as its equal, the Angle Q C a of Reflection.

And
reaſoning after the ſame Manner, the An-
gle
C R Q of the Triangle R C Q, is a right one.

Which
was to be demonſtrated.
28. It is not abſolutely neceſſary to give the
Mirrours
the aforeſaid Inclinations;
for the An-
gle
A B N may be aſſumed at Pleaſure, from
which
muſt be taken an Angle of 135 Degrees,
to
have the Inclination of the Miror G H.
Ne-
vertheleſs
, the Angles we have determin’d, are
the
moſt advantagious for a perpendicular Pi-
cture
.
29. When a Picture is inclin’d, and makes
44Fig. 73. the Angle D I A with the Horizon, the Mirrour
M
N muſt keep its Situation, and the 5524. C Q R is equal to half the Angle D I A, leſs
{1/4} of a right Angle:
Then I ſay, the Angle
FaC
, or its equal C R Q, will be equal to the
Angle
B I D.
Now the Angle P B Q, is 112 {1/2}. 6611. Degrees; whence the Angle B P Q, which is the
Complement
of P B Q, and P Q B to two right
Angles
, is 90 Degrees, leſs the half of the 7727. gle D I A:
Wherefore becauſe N B C is 67 {1/2}
218113on PERSPECTIVE. Angle B C P, and its equal R C Q, is 22 {1/2} plus
{1/2} of D I A.
Now if the Angle R Q C be added
to
this Angle, their Sum will be equal to the An-
gle
D I A;
whence it follows, that the Angle
C
R Q is equal to D I R.
Which was to be demon-
ſtrated
.
30. If the Angle R B N be alter’d, and it be
called
a, the Angle D I A, b, and the right An-
gle
d;
then the Angle C Q R = d + {1/2} b a.
31. When a Picture is parallel, it appears ma-
11Fig. 74. nifeſt, the Mirrours GH and M N being each in-
clin’d
under an Angle of 45 Degrees, that a Ray,
which
is perpendicular to the Horizon, likewiſe
falls
, after a double Reflection, perpendicularly
upon
the Table A.
Prop. III.
32. To repreſent Objects which are roun’d about
the
Machine, and make them appear erect to the Per-
ſon
ſeated within the ſame.
The Back-ſide of the Machine muſt be turned
towards
the Sun, and the Objects behind the
ſame
repreſented by one Reflection only;
2215. their Appearance will always be clearer, altho
they
be in the Shade, than the Appearance of the
Objects
on the other Sides of the Machine, which
cannot
be perceiv’d unleſs by a double Refle-
ction
.
33. The Objects that are on the right and left
33Fig. 70. of the Machine, may be repreſented by means
of
the Mirrour H, ſituated as per Figure;
but 4412. ſaid Mirrour muſt be cover’d with a Paſtboard Caſe,
having
two Apertures therein;
the one next to
the
Objects, and the other next to the Aperture
N
, of the Box X.
The Reaſon of our uſing this
@recaution
is, becauſe when the Mirrour is not co-
ver’d
at all, it reflects the Rays of Light
219114An ESSAY Side-ways upon the Mirrour L, which being again
reflected
by the ſaid Mirrour L, and going through
the
Convex Glaſs, extremely weakens the Repre-
ſentation
.
34. The Objects before the Machine are re-
preſented
according to N.
22, and 28.
Problem IV.
35. To repreſent Pictures or Prints.
If we have a mind to repreſent Pictures and
11Fig. 70. Prints, they muſt be faſten’d againſt the Board
F
on that Side, regarding the Back of the Ma-
chine
, which muſt be ſo turned, that the Pictures
be
expoſed to the Sun.
Then they are repreſent-
ed
in this Situation as the other Objects, 2215. with this Difference, that the Convex Glaſs in the
Cylinder
C muſt be changed:
For if Prints are
requir’d
to have their true Bigneſs, the focal Di-
ſtance
of the Convex Glaſs muſt be equal to half
the
Height of the Machine above the Table;
that is, equal to half A C. Again, if the ſaid
Pictures
or Prints are requir’d to be repreſented
greater
than they really are, the focal Diſtance
of
the Convex Glaſs muſt ſtill be leſſer.
And
if
, on the contrary, they are to be repreſented
leſſer
than they really are, the focal Length of
the
Glaſs muſt be greater than the Length A C.

Moreover
, the proper Diſtance whereat the Pi-
ctures
or Prints muſt be placed, may be found
in
ſliding the Board F backwards or forwards,
until
they diſtinctly appear within the Machine.

This
Diſtance alſo may be determin’d by the
following
Proportion:
As the Machine’s Height above the Table, leſs the
Glaſs’s
focal Length,
is
to
220115on PERSPECTIVE.
The Height of the Machine above the Table;
So
is
The
Glaſſes focal Length,
to
the
Diſtance
of the Figure from the Glaſs.
Note, The ſaid Diſtance of the Convex Glaſs
from
the Figure, is meaſured by a Ray, pro-
ceeding
from the Figure parallel to the Horizon,
which
is perpendicularly reflected upon the Con-
vex
Glaſs, by the Mirrour.
Note, Moreover, that
when
we have a Mind to place the Figures out
beyond
the Back-ſide of the Machine, they muſt
be
faſtned againſt the Side F of the Board, which
muſt
be ſo turned, that the ſaid Side be next to
the
Aperture N.
37 Remarks concerning the Repreſentation of Per-
ſons
Faces.
It is certainly very curious and uſeful to de-
ſign
Perſons Faces to the Life;
which by this
Machine
, may be very well done in Miniature:
For if the Face of any known Perſon be thus re-
preſented
, by only looking at the Appearance,
we
may very readily know whoſe Face it is,
when
at the ſame time the Appearance of the
Whole
Perſon will not take up half an Inch
upon
the Paper on the Table:
But it is very dif-
ficult
to repreſent a Face diſtinctly as big as the
Life
;
for when we would repreſent a Face in its
natural
Bigneſs, ſuch a Convex Glaſs as is men-
tioned
in Numb.
35. muſt be uſed, and the Face
placed
where the Board F is .
But the 1135. Face which then appears diſtinct enough, that ſo
the
Perſon whereof it is the Repreſentation may
thereby
be known, hath not its Lineaments ſuffi-
ciently
enough repreſented, as to be followed by a
Painter
as they ought, in order to keep the true
221116An ESSAY on ſemblance. The Reaſon of which is, that the Li-
neaments
appear lively and diſtinct within the
Machine
, when the Re-union of the Rays pro-
ceeding
from a given Point in the Face, happens
exactly
upon the Paper in one Point only:
But
the
leaſt Diſtance that one Point is more than
another
from the Convex Glaſs, (when the Di-
ſtance
of the Face from the Glaſs is ſo ſmall, as
it
muſt be to repreſent it in its natural Bigneſs)
ſo
alters the Place of the ſaid Re-union, that for
different
Parts of the Face, thoſe Places of Re-
union
will differ about two Inches and a half.
Whence it is no wonder that all the Lineaments
be
not repreſented as could be wiſhed;
ſince in
all
Diſtances choſen, there will be always a great
many
Rays, whoſe Re-union will fall above an
Inch
beſides the Paper.
The Confuſion ariſing
from
this Diverſity, though not being very di-
ſtinguiſhable
by the Eye, yet is prejudicial, and
hinders
our getting the exact Reſemblance of the
Face
.
We have obſerved this, in order to give
an
exact Idea of the Goodneſs of this Machine,
in
equally ſhewing wherein it may be really
uſeful
, and wherein its apparent Uſefulneſs is
ſubject
to an Error rather diſcovered by Experi-
rience
than Reaſon.
38. We muſt not forget in all the precedent
Problems
, to examine the Aperture the Convex
Glaſs
ought to have;
for although we cannot re-
duce
this Aperture to a fixed Meaſure, yet it is
proper
to obſerve the following Remarks.
1. The
Convex
Glaſs may commonly have the ſame
Aperture
, as we would give a Perſpective Glaſs,
having
the ſaid Glaſs for its Object Glaſs.
2. When
Objects
are very much enlightned, the ſaid A-
perture
muſt be leſſened;
and contrariwiſe, when
they
are expoſed to a weaker Light, it muſt be
made
greater;
and when any Repreſentation
222117PERSPECTIVE. to be copyed, the Convex Glaſs muſt have the
leaſt
Aperture poſſible;
but yet with this Cau-
tion
, that the Light coming into the Machine,
muſt
not be too much extenuated.
From theſe
Obſervations
it is manifeſt, that we ought to be
provided
with ſeveral round Pieces of Tin or
thin
Braſs, having round Holes of different big-
neſſes
therein, in order to give a neceſſary A-
perture
to the Glaſs;
or Holes of different big-
neſſes
may be made in a long thin Piece of Braſs
which
may ſlide upon the Convex Glaſs:
Or
elſe
, we may uſe a round Plate, having Holes
of
different bigneſſes therein, which turning a-
bout
its Centre, may bring any deſired Hole for
the
Glaſſes Aperture.
A Deſcription of the Second Machine.
39. This Machine is a kind of Box, the Side
11Fig. 78. A C G B being open, whoſe breadth B D, and
height
AB, are equal, each being about 18 Inches:
Its greateſt width F B, is 10 Inches, and the Side
E
F is ſloping, ſo that A E is but about 6 Inches.
40. The Frame G ſlides at the Bottom of the
ſaid
Box, in which the Paper is faſtned.
2215.
41. There is a round Hole in the Middle of
the
Top of the Box, in which the Cylinder
carrying
the Convex Glaſs ſcrews.
339.
42. The two Sticks H I and L M, ſlide in four
little
Iron Staples fixed to the Inſide of the Top
of
the Box, like thoſe mentioned in Numb.
7.
Theſe Sticks come about two Feet without the
Box
, and the Diſtance of their Extremities I and
M
, is equal, or ſomething greater than the
Length
of the Box.
Their Uſe is to hang a black
Cloth
upon, which is faſtned to the Three Sides,
B
A, A C and C D, of the opening of the Box,
that
ſo the Box may be darkned, when
223118An ESSAY are to be repreſented upon the Paper in the Frame
G
.
43. There are two pieces of Wood, (one of
which
is repreſented in Fig.
77.) ſerving to
ſuſtain
the Box upon its Support or Foot.
One
of
theſe Pieces may be faſtned to one Side of the
Support
, and the other to the other Side thereof,
by
means of four Iron Pins, two of which go thro
the
Holes N and P in the Side of the Support,
and
the Holes T and V in the Piece R, and the
other
two in like manner, through Holes made
in
the other Side of the Support, and the other
Piece
, when we have Mind the Bottom of the
Box
ſhould be parallel to the Horizon;
but when
the
Box is to be a little inclin’d, that Pin going
through
the Hole P, muſt be put through the
Hole
O, in the Piece R.
Underſtand the ſame
of
the other Piece.
44. We are ſometimes obliged to ſet the Box
forwarder
on its Support;
and this is done, in
uſing
the Holes Q and S, inſtead of N and P.
It
is
likewiſe ſomething neceſſary to incline the
Box
a little backwards;
which may be done, by
putting
the Pin in S, into the Hole X, made in
a
Piece of Wood faſtned to the Back-ſide of the
Box
, and the correſpondent Pin on the other Side,
into
another Hole made on the other Side of the
ſaid
Piece.
45. The Box ſlides upon the Top of the
Machine
, and is like that already deſcribed, 1110, 11,
13
.
with this Difference only, that it is leſſer.
On
the
Top of the Box are two little Staples Z Z,
in
which a Ruler ſlides, having a Mirrour faſtned
to
it, in the Manner as is mentioned Numb.
13.
and ſo by this Means the ſaid Mirrour may be
put
in the ſame Situation, as that in the Figure
of
the firſt Machine it hath in H.
224119on PERSPECTIVE.
46. When we have a mind to remove this
Machine
from one place to another, we lay the
Box
B E C upon the croſs Pieces 2, 3, 4, 5, with
its
opening A B C upwards;
then we put the
little
Box Y, the Ruler and Mirrour (mentioned
Numb
.
13.) the black Cloth, and the two Sticks
M
L and I H, all into the ſaid great Box;
and
afterwards
partly cover it by the Frame G, 1140. which is ſuſtained by two very thin Rulers, and
then
by another little Board, when the Frame is
not
big enough.
The whole Machine thus taken
to
pieces, will take up no more room than the
Support
itſelf doth:
and ſo it is very eaſy to re-
move
from Place to Place.
Now when Objects
are
to be repreſented in this Machine, it muſt be
put
together again, as per Figure;
and the black
Cloth
, for a Perſon to put his Head under, hang-
ing
upon the Sticks, and faſtned to the Sides of
the
opening A B, A C, and C D.
The Uſe of this Machine.
47. The Uſe of this Second Machine is the
fame
as that of the Firſt;
but it ought to be ob-
ſerved
, that when we incline the Machine, 2243. Angle of Inclination of the Mirrour and Hori-
zon
muſt be made leſs, by half the Inclination of
the
Bottom of the Box;
and when the Machine
is
ſomewhat inclin’d backwards, the ſaid 3344. muſt be made greater by a like half.
You muſt
likewiſe
obſerve, that when Objects are to be
repreſented
for a perpendicular Picture, the Ma-
chine
muſt be placed according to the former
Part
of Numb.
44. Prints muſt be faſtned to a
Board
entirely ſeparated from the Machine, which
Board
muſt be ſet upon a Support, that may
conveniently
be moved backwards or forwards,
according
to Neceſſity.
225120An ESSAY, &c.
A Demonſtration of the Inclination of the Looking-
Glaſs
.
48. Let A B be a Ray, proceeding from ſome
11Fig, 75. Point of an Object.
We are to demonſtrate, 2219. if the Line D I hath the Inclination given to a
Picture
, and the Looking-Glaſs G H hath the In-
clination
we have preſcribed, that the Angle
B
a F, will be equal to the Angle B C D.
Now to
prove
this, draw the Line F I parallel to the Ho-
rizon
, then the two Angles I D F and D F I, of
the
Triangle I D F, are together equal to the
Angle
D I E;
but the Angle D F I, which is the
Inclination
of the Looking-Glaſs, is equal 3316. 47. half the Angle D I E, leſs half the Angle I F a;
and conſequently it is leſs than the Angle F D I,
by
the Quantity of the whole Angle I F a:

Therefore
if the Angle I F a be added to the An-
gle
D F I, we ſhall have the Angle D F a, equal
to
the Angle F D I:
Therefore the Angle
Fa
B will be likewiſe equal to the Angle BCD.
4419. Which was to he demonſtrated.
In reaſoning nearly after the ſame Manner,
we
demonſtrated what is mentioned 5547. ning the Inclination of the Mirrour, when the
Box
is inclin’d a little backwards.
FINIS.
73[Figure 73]
226
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22774[Figure 74]Page 120
Plate
. 30.
Fig. 70.
X I F B H D D P O M P R C C C C C E E Q
228
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229
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23075[Figure 75]Plate 31
page
120
Fig
. 71
D G C B A H F a I E
76[Figure 76]Fig. 72P G C H A N B R Q M a F77[Figure 77]Fig. 73P G C H D N B I A R Q M a F78[Figure 78]Fig. 74G N B C H M a A79[Figure 79]Fig. 75D G B C A H F I E a
231
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232
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23380[Figure 80]page 120
Plate
. 32.
Fig
. 76.
81[Figure 81]Fig. 77.R V T o82[Figure 82]Fig. 78.Z Z Y C M L I E A H D X G F B S Q P N 4 3 2
234
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