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Ent@in Gatalogue, ---
(1860)
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.
SIR,
IN Dedicating this
Tranſlation to you,
I have deſignedly
deviated from the general
Cuſtom obſerv’d by almoſt
all Dedicators, who make
choice of ſuch Patrons that
are Great and Rich, not at all
conſidering their Merit, or
whether they underſtand any
thing of what is offer’d to
them; ſince I have inſcrib’d
this Treatiſe of Perſpective to
one, whoſe daily Practice is
the very Art it ſelf, and whoſe
Merit is undoubtedly excel-
lent, as evidently appears from
your own Works.
I ſhall likewiſe be particu-
lar with regard to the Manner
of the Offering; being per-
ſuaded that Flattery, or even
due Praiſe, which are the com-
mon Topicks handled in De-
dications, muſt needs be offen-
ſive to an Ingenious Perſon; and ſo I ſhall be ſilent on theſe
Heads; and only crave your
Acceptance and Protection of
what is here offer’d by
Your Humble Servant,
E.
Stone.
THE Reader will wonder, per-
haps, to find me entring into
a Path, which ſeems to have
been too much trodden already; and eſteem as uſeleſs a New
Eſſay, on an Art, whoſe Sub-
ject (one would think) ſhould have been
long before this Time exhauſted; ſince there
have been ſo many Perſons, who have writ-
ten on the ſame.
The Name of Perſpective now ſeems to
ſound unple aſant in the Ears of the Publick
Enemies of Repetition; and it may be look d
upon as a Piece of Inadvertency, to venture
to treat again on that ſame Subject. Yet,
notwithst anding this, I deſire the Render to
ſuſpend his Cenſure, until he h{as} heard the
ing Work.
Having ſome Years ago buſied my ſelf in
drawing Figures by the common Methods, I
found out ſeveral Compendiums; which, by
diligent working, naturally enough fall in
one’s way, without being entirely beholden
to the Induſtry of others: And theſe firſt
Succeſſes made me hope for others more
conſiderable; and ſo I thought that a more
narrow Inſpection into the Theory of Per-
ſpective, might furniſh me with Rules more
general, for making the Practice thereof
eaſy.
I then thought upon ſeveral Methods to this
Purpoſe; but, being ſuſpicious that they were
not ſo eaſy {as} they appear’d, I have try’d
their Goodneſs, by exactly applying them to
different Subjects; and have nicely examin’d
all the Caſes, and order’d it ſo {as} not to be de-
ceiv’d by certain Operations, which at firſt
ſeem eaſy, but, when put in Practice, are
quite otherwiſe. Moreover, at convenient
Times, I look’d over the beſt Part of the
Authors of this kind, (whoſe Number is
increas’d very much, without any manner of
Neceſſity) ſome of which being advantagi-
ouſly diſtinguiſh’d among the Crowd, have
been very uſeful to me: But I dare affirm,
there are but a very few that give a new
Turn to the practical Part of Perſpective.
Some content themſelves with the bare
Explication of the Theory, and have left to
the Reader the Trouble of applying the ſame
to Practice; or elſe have given only ſome of
the common Operations, and entertain us
with general Reflections on Painting; which
are indeed curio{us}, but foreign to my Pur-
poſe: For I intend not to make a Man a
Painter, but to render the Uſe and Exerciſe
of Perſpective eaſy to him.
Other Authors, which (according to the
Bulk of their Works) might be thought to
have more carefully treated of the practical
Part of Perſpective, do indeed at firſt lay
down ſome general Rules, common to them
all; but are nothing the eaſier for having
paſs’d thro’ ſo many Hands; and that, in-
deed, becauſe they have not endeavour’d to
make them ſo. They thought that all Ob-
jects might be thrown into Perſpective by
theſe Rules, and therefore it would be uſe-
leſs to ſearch after others; and judg’d it
more neceſſary to ſhew Painters the Applica-
tion of them to an infinite Number of parti-
cular Examples; tho’ that Application, at
moſt, is but repeating over again the Uſe of
the Rules already preſcrib’But what Ad-
vantage can Painters gain from hence, if
they do not well underſtand general Opera-
tions? And if they do, I cannot conceive of
ples will be to them.
I believ’d then, that I might be able to
treat of this Art after another Manner: And altho’ I know my ſelf to be much infe-
rior to ſeveral of thoſe who have written on
this Subject; yet I am of Opinion, that if
Perſpective ſhould loſe any thing by me, on
account of my want of fudgment; yet that
may be regain’d, perhaps, (and with In-
tereſt too) by my great Diligence in this Bu-
ſineſs.
I have conſider’d, moreover, that the te-
dious Particulars, inherent to the Subject on
which I have choſen to write, will always
hinder Genius’s capable of great Matters,
from undertaking a Subject ſo little worthy
their Endeavours, and ſo barren of great
Diſcoveries.
Thus, hoping, on one hand, to give a new
Turn to the Practice of Perſpective, and
make it eaſier; and being perſuaded, on the
other, that more learned Perſons than my ſelf
will not take this Trouble upon them; I ven-
ture to publiſh this ſmall Work, and expoſe
it to the Taſte of the Learned World; from
whom I expect no other Praiſe, but what
may reaſonably be claim’d by an aſſiduo{us} Ap-
plication.
The Practice of Perſpective may be made
eaſy, by the Three following Things in this
Treatiſe: Viz.
1.
In giving ſeveral new
and eaſier Ways (than thoſe commonly uſed)
of ſolving the moſt general Problems upon
which the whole Practice is founded: And the
Reaſon why we have laid down ſeveral So-
lutions, is, becauſe the ſame Way is not always
equally convenient in all Caſes; whence it is
neceſſary to have ſeveral, that ſo we may
chuſe one beſt ſuiting our Purpoſe. 2.
The
general Methods, which have been us’d hi-
therto, not being practicable on ſome particu-
lar Occaſions; to remedy this, we have ad-
ded others to them; which are indeed more
difficult, but (in ſome Caſes) there is an ab-
ſolute Neceſſity for them. 3.
When it is ve-
ry difficult to reſolve a particular Problem,
by means of general ones; then we have
thought it convenient to give a particular So-
lution thereof.
By this means, the Study of Perſpective
becomes indeed more difficult; but the Diſ-
advantage is well recompenſed by the Faci-
cility of the Practice, which we have entire-
ly had in view. It is true, that a few ge-
neral Rules do not ſo much burthen the Me-
mory; but when one has ſeveral general
ones, and alſo particular ones, by them we
can abridge Matters. And this Method be-
Application, does afterwards ſave a great
many Hours Study, in an Art that always
appears difficult enough.
A Painter, in a ſhort Time, may learn this
Work, and make the Rules thereof familiar to
him: And if this Study be repeated from
time to time, for a few Days, he will find
the Benefit thereof, in diminiſhing his Labour
and Trouble.
But, that any one himſelf may ſee what
I promiſe in this Eſſay; take the following
ſhort Abſtract thereof. It is divided into
Nine Chapters: The Firſt, being as an In-
troduction to the reſt, ſhews the Uſefulneſs
of Perſpective, and gives you the Definiti-
ons of the Terms neceſſary for underſtanding
this Treatiſe.
The whole Theory is contain’d in the Se-
cond Chapter: Where, what has been found
moſt uſeful in that Matter, is therein re-
duced to Three general Theorems; viz.
the
firſt, ſecond, and fourth: All the reſt is de-
duced from them, by way of Corollary. To
theſe Theorems, already known, are added
ſome new ones, ſerving for the Demonſtra-
tion of ſome neceſſary Propoſitions. Perhaps
it might be wiſh’d, that I had ſhewn the Way
that led me to the Truths which I diſcover:
but it often
would have been very long and troubleſome. In Geometry, the eaſieſt and ſhorteſt Way, is
not always that which leads to Diſcoveries.
In the following Chapter, the Practice of Per-
ſpective upon the perſpective Plane, or Picture,
conſider’d as upright, is explain’d: Wherein,
among the different Ways laid down for the
Solution of general Problems, you will find
ſome effected by a Ruler only; ſo that after
ſome Preparations, all Kinds of Objects may
be drawn without Compaſſes, and that eaſier
than by the common Operations. In that
Problem, to find the Appearance of a Point
out of the Geometrical Plane, it is commonly
conſider’d as the Extremity of a Perpendicu-
lar, whoſe Repreſentation muſt first be found,
before that of the Point can be had. But here
we avoid this round-about Way, and ſhew how
to find the Appearance of the Point given,
without being obliged to find the Perſpective
of its Seat.
As to the Appearance of a Cone and Cy-
linder, we determine the viſible Portions of
the Baſe, and by this means avoid the uſe-
leſs Operations which the common Way is
ſubject to. It is very difficult, if not impoſ-
ſible, to throw a Sphere into Perſpective, by
means of general Problems; and in the Re-
preſentation of the Torus of a Column, it is
Whence we are obliged to
give particular Methods for the Reſolution of
theſe two Problems.
The reſt of the Third Chapter is concern-
ing Inclin’d Lines, and how to find their Ap-
pearance by the Accidental Point.
The Fourth Chapter ſhews the Manner of
working on a perſpective Plane, to be view’d
afar off, very obliquely, or which muſt ſtand
in an high Place. Theſe different Situations
require new Rules: For if the common Me-
thods were to be uſed here, the perſpective
Plane muſt be ſo large, as that it would be
impoſſible to work upon it.
In the Two following Chapters, we treat
of the perſpective Plane, conſider’d as Hori-
zontal, or Inclin’d: Where there are laid
down ſeveral general Ways of working; which, together with thoſe of the foregoing
Chapters, will ſuffice (in my Opinion) for
throwing any Object whatſoever into Perſpe-
ctive, with Eaſe enough.
In the Seventh Chapter, which treats of
Shadows, there is nothing particular, but
what may be ſeen elſewhere: But that lit-
tle we have ſaid concerning this Matter, is
enough for giving an Idea of them, which
the Reading of what goes before will make
eaſy.
In the Eighth Chapter, are laid down
ſome Mechanical ways for making the Uſe of
Perſpective eaſy, by means of Rulers and
Threads, (eaſily to be gotten by any body,
and not difficult to be put in practice) they
being eaſier to uſe than any of the Inſtru-
ments that have hitherto been invented for
this Purpoſe.
The laſt Chapter ſhews the Uſefulneſs of
Perſpective in Dialling.
Such is the Plan of this ſinall Work;
wherein I have not ſo much endeavour’d to
advance Curioſities, as Things of real Uſe;
hoping that, without making a Shew of Skill
ill beſtow’d, I ſhall make my Book good enough,
if by its Uſe I make it neceſſary. For which
Reaſon, I have endeavour’d to lay down the
whole, ſo as to be underſtood by thoſe who
have only read the Elements of Euclid. And
tho’ I have deviated from this Rule in ſome
few Place; they are printed in Italick, that
ſo they may be paſs’d over without any Hin-
drance to the Learner.
Here I muſt not forget to mention, that in
Reviſing this Eſſay, I had the Happineſs of
meeting with an able Painter; who has ſe-
r
ſion, neceſſary to be known, among which,
He has car-
ried the Matter farther than could have been
reaſonably expected from one ignorant of
Mathematicks; and I am indebted to him
for ſeveral Obſervations, which I my ſelf
ſhould perhaps have never thought on.
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.
1.
PERSPECTIVE teaches us
the Manner of Delineating by
Mathematical Rules; that is, it
ſhews us how to draw geometri-
cally upon a Plane, the Repre-
ſentations of Objects according
to their Dimenſions and different Situations; in
ſuch manner, that the ſaid Repreſentations pro-
duce the ſame Effects upon our Eyes, as the Ob-
jects whereof they are the Pictures.
In order to underſtand well how Mathema-
ticks may be apply’d to Drawing; let us ſup-
poſe a Man A, viewing an Object; and be-
imagine a tranſparent Plane C. Suppoſe more-
over, that Lines be drawn upon this Plane, as
in D, which cover the Bounds of the Object B
in reſpect of the Spectator A, and each Part that
he ſees thereof. Now, ſince all Objects are ſeen
by the Rays of Light coming from every of
their Points, and terminating at the Eye, and
not otherwiſe; and ſince that here all the Rays
proceeding from the Object B, likewiſe paſs
thro’ every Point of the Repreſentation D; it is
manifeſt, that this Repreſentation will have the
ſame Effect upon the Spectator’s Eye, as the
ſaid Object B hath. Now, by means of Geo-
metry, we can find the Points of the Figure D,
on the Plane C, placed in a given Situation,
thro’ which the Rays coming from the Object B
to the Eye of the Spectator A, do paſs; and
theſe Points are the Interſections of the Rays
and the Plane. Alſo, (as others have very well
obſerv’d) a Perſpective Plane, or Picture in Paint-
ing, may be conceiv’d as a Window, upon which
the Objects ſeen thro’ it are repreſented.
Now, without Mathematicks, this Repreſen-
tation cannot be well found: For when Objects
are drawn by only viewing, or looking at them; their true Repreſentations after this way, will
be very often miſs’d on; whereas, by Geome-
try, we can always obtain them.
This Obſervation only, is ſufficient to eſta-
bliſh the Neceſſity of Perſpective: Tho’ there are
ſome Painters, who (according to the common
Maxim) affirm, That what they do not know
of this Art, is not worth the Pains of learn-
ing.
Hitherto I have endeavour’d to give an Idea
of Perſpective in general: But there is yet ano-
ther particular Signification of this Word, which
it is neceſſary ſhould be explain’d, as well as
the other Terms of the Art, which are laid down
in the following Definitions; and which every
one, that intends to underſtand this Treatiſe,
ought to be well acquainted with.
2.
The Perſpective, Repreſentation, or Appearance of
mous) * is the Figure which the Rays, by which an
Object is perceiv’d, form in paſſing thro’ the tranſpa-
rent Plane: And the Perſpective of a Point, is the
Interſection of a Ray proceeding from that Point,
and the tranſparent Plane. Which Interſection is
a Point: As the Figure D in the tranſparent
and
the Point e, in the ſame Plane, is the Perſpective
of the Point E, in that Object.
The Plane parallel to the Horizon, upon which
Objects that he views, is call’d the Geometrical Plane. As A B C D.
A Perſpective Plane [or Picture] is that which
which the Objects are drawn: As F G R T.
This
is commonly perpendicular to the Geometrical
Plane, and conſequently to the Horizon; becauſe
Pictures have generally this Situation: But yet
it may be ſometimes inclin’d, and even parallel
to the Geometrical Plane, according as one
would diſpoſe the Deſign, or Picture that we are
working. And for this Reaſon, in the follow-
ing Chapter, we have laid down General Theo-
rems, and their Corollaries, agreeing to all theſe
Picture] which ought to be well obſerv’d.
The Interſection of the Perſpective Plane and the
As
F G.
The different Situation of the Eye, alters the
Repreſentation of Objects in the perſpective Plane; for the Rays proceeding from the Object, and
concurring in ſome other Point, will likewiſe fall
upon the Perſpective Plane in different Places. And
for determining this Situation of the Eye, in re-
ſpect to the Perſpective Plane, we ſuppoſe,
A Plane parallel to the Horizon, paſſing thro’ the
and this is
call’d the Horizontal Plane: As OMVNL.
The Interſection of this Plane and the Perſpective
As M V N.
The Perpendicular drawn from the Eye to the Ho-
As O V.
The Point V, wherein the ſaid Perpendicular meets
cipal Point.
Note, There is a Perpendicular, let fall from
the Eye upon the Geometrical Plane, meaſuring
the Height of the Eye.
The Point S, wherein the ſaid Perpendicular meets
The Plane paſſing thro’ the aforeſaid Perpendicular,
As SOLI.
The Interſection V H of this Plane, and the Per-
And S H I, the Interſection of it, and the Geo-
Points of Diſtance, are two Points in the Hori-
by the length of the principal Ray; as MN.
The Geometrical Line, is a Line, that paſſes
Line, as A B.
The Seat of an Object, is the Concurrence of Per-
Geometrical Plane, and the ſaid Plane.
The Direction of a Line inclined to the Geometri-
another Plane perpendicular thereto, paſſing through
the ſaid inclined Line.
3.
THE Perſpective, or Appearance of a
Right Line, as A B, which being con-
tinued, does not paſsthrough the Eye O, is like-
For the Rays, by which the
Line A B is perceived, form a Plane cutting
the perſpective Plane; and the common Section
of theſe two Planes is a right Line, as a b.
4.
The Repreſentation of a Line Parallel to the
the Repreſentation.
Let A B be a Line Parallel to the perſpective
Plane; we are to prove that a b its Repreſentati-
on is Parallel thereto.
Theſe two Lines A B and a b, will never
meet each other, becauſe a b is in the perſpective
Plane, and A B is ſuppoſed parallel to the ſaid
Plane. But they are alſo in one and the ſame
Plane, becauſe a b is the Interſection of the per-
ſpective Plane, and the Plane O A B, paſſing
through the Eye and the Line A B; and there-
fore they are parallel between themſelves: Which
was to be demonſtrated.
5.
The Appearance of a Line, parallel to the baſe
Line, is alſo parallel to the ſaid baſe Line.
For the baſe Line, and the Repreſentation
being parallel to the ſame Line, are parallel to
one another.
6.
The Repreſentation of a Line parallel to the
vertical Line, is parallel to the ſaid vertical Line, and
conſequently perpendicular to the baſe Line. This
is demonſtrated as in the laſt Corollary.
7.
The Appearances of Lines parallel to the per-
ſpective Plane, and equally inclin’d the ſame Way upon
Line, equal to thoſe Angles that the Lines whereof
they are the Appearances, make with the Parallels
to the baſe Line, which cut them; and conſequently
the ſaid Appearances are parallel between them-
ſelves.
This is evident, becauſe the Appearances of
Lines parallel to the baſe Line, are parallel to
the ſaid Line; and the Appearances of the in-
clined Lines are parallel to theſe Lines.
8, 9.
The Repreſentation of a Figure, parallel to
the perſpective Plane, is ſimilar to the ſaid Figure; and
the Sides of the ſaid Figure are to their Repreſen-
tations, as the Diſtance of the Eye from the Plane
of the Figure, to the Diſtance of the Eye from the
perſpective Plane.
The given Figure is A B C D.
We are firſt to
thereto; that is, that the correſponding Angles
of theſe two Figures A B C D, a b c d, are equal,
and their Sides proportional.
I.
The Angles are equal, becauſe
tween themſelves.
II.
In the ſimilar Triangles A D O, and a d o,
we have
A D: a d :
:
O D :
O d.
And in the ſimilar Triangles O D C, and O d c,
we have
D C : d c :
:
O D :
O d.
then
A D: a d :
:
D c :
d c.
altern.
A D : D C :
:
a d :
d c.
And conſequently the Sides A D, and D C of
the Figure A B C D, are Proportional to the
The
ſame may be demonſtrated of the other Sides; and therefore the Figures are ſimilar.
Now to prove the other Part of the Theorem:
If a perpendicular be ſuppoſed to be let fall from
the Eye upon the Plane of the Figure, and con-
tinued as is neceſſary; it is evident, that O D,
will be to O d, as this Perpendicular, which
meaſures the Diſtance from the Eye to the Plane
of the Figure, is to the Diſtance of the Eye
from the Perſpective Plane, which is meaſur’d
by the Part of the perpendicular, contain’d be-
tween the Eye and the perſpective Plane. Now
this before was manifeſt; viz.
that
O D : O d :
:
A D :
a d :
Whence there is the ſame Proportion between
A d one of the Sides of the Figure, and A D its
Appearance, as the Theorem expreſſes. The
ſame may be demonſtrated of the other Sides of
the Figure. Which was to be demonſtrated.
10.
If from a Point in the Geometrical Plane, three
right Lines proceed, which are equal between them-
ſelves, and parallel to the perſpective Plane; the firſt
of which is in the Geometrical Plane, the ſecond ele-
vated Perpendicular to the firſt, and the third in-
clined to it; the Appearances of theſe three right
Lines are equal.
This will appear clear enough in conſidering
the Lines as a Figure parallel to the perſpective
Plane; and ſo conſequently they will have the
ſame Proportion as their Appearances.
Note, The firſt of the aforeſaid Lines is always
parallel to the baſe Line; and the ſecond, when
the perſpective Plane is perpendicular or up-
Plane, and the third is then in the Direction of the
firſt.
11.
If tworight Lines, equal between themſelves, and
parallel to the perſpective Planes, be equally diſtant
from the perſpective Plane, their Appearances will be
equal.
For, becauſe they are in a Plane, parallel to
the perſpective Plane, they will have the ſame
Proportion to each other, as their Repreſentations.
12.
If a Line parallel to the Perſpective Plane, be
view’d by two Eyes, both being in a Plane, parallel
to the perſpective Plane, the Repreſentations of the
ſaid Line will be equal.
If we ſuppoſe a Plane, parallel to the Per-
ſpective Plane, to paſs through the propoſed
Line, this Proportion will be had; As the
Diſtance from the Perſpective Plane, ſo is the
given Line to the Repreſentation thereof. But
the three firſt Terms of this Proportion are
the ſame for each of the Eyes, which are
in one and the ſame Plane parallel to the Per-
ſpective Plane: Therefore, the fourth Term of
the Proportion will likewiſe be the ſame in both
Caſes: Which was to be demonſtrated.
13.
If a right Line, being continued, meets the per-
ſpective Piane in one Point, the Appearance thereof
will be a Part of the Line drawn from the ſaid Point
in the perſpective Plane, to another Point, whereat
poſed Line, terminates.
The Line C D being continued, will meet
We are
to prove, that its Appearance is a Part of the
Line E H, drawn from the Point E, to the
Point H, whereat the Line O H proceeding
from the Eye parallel to the given Line C D,
terminates.
The Interſection of the perſpective Plane,
and the Plane O D C, is the Repreſentation of the
given Line. Now the Plane O D C, is a Part of the
Plane paſſing through the parallels O H and EC.
Therefore, this Repreſentation is a Part of the
Interſection of the laſt mentioned Plane, and
perſpective Plane; which Interſection is E H.
14.
All Lines parallel between themſelves, and be-
ing produced, do fall upon the perſpective Plane, have
Repreſentations, which being produced, will all con-
cur in one Point.
This is evident, becauſe but only one Line
O H can be drawn from the Eye O, to the per-
ſpective Plane, parallel to the ſaid Parallels, and
becauſe all their Repreſentations are Parts of
Lines concurring in the Point H.
And this Point is called the accidental Point of the
15.
Two or more parallel Lines, which being produ-
ced, do fall on the perſpective Plane, parallel to the Geo-
metrical Plane, have their accidental Point in the
Horizontal Line.
For the Horizontal Plane, is parallel to the
Geometrical Plane.
16.
The Repreſentations of all Lines parallel to
the ſtation Line, concur in the Point of Sight.
This follows, becauſe the principal Ray is
parallel to the ſaid Lines.
17.
Two or more equal Lines being perpendicular,
or equally inclined the ſame Way, to the ſame Line pa-
rallel to the Station Line, have their Repreſentations
concurring in the principal Point.
Becauſe all theſe Lines are parallel and equal,
the Line paſſing through their Vertices, is pa-
rallel to that paſſing through their Baſes, and
this being parallel to the Station Line, it fol-
lows,
Point.
18.
The Appearance of an indefinite Line does not
alter, when the Eye moves in a Line parallel to a
propoſed Line.
The Repreſentation of this Line, is the In-
terſection of the perſpective Plane, and a Plane
paſſing through the Eye and the ſaid Line. Now
the Eye remains in the ſame Plane, when it
moves in a Line parallel to the propoſed Line; and conſequently the Appearance of this laſt
Line, will not be changed by that Motion.
Note, This Demonſtration doth not extend
to any particular Part of the given Line, but on-
ly to the Line in general.
19.
Let A C be a Line inclined to the Geometrical
Plane, and O D another Line drawn parallel to
A C, from the Eye to the perſpective Plane. Now
rallel to the baſe Line, and likewiſe D E, in the
perſpective Plane, parallel to the ſaid Line, ſo that
B A be to A C, as E d to D O. I ſay, the Ap-
pearance of the Line B C, paſſing through the Point
B, and the Extremity of the Line A C, being con-
tinued, will meet the Point E.
Now to prove this;
it is evident,
B C: And this may be done in the following
Manner:
A B is parallel to E D, and A C to O D;
whence the Angle (E D O) of the Triangle
O E D, is equal to the Angle (B A C) of the
Triangle A C B: And ſo theſe two Triangles
are ſimilar; becauſe they have alſo their Sides
Proportional. But ſince theſe two ſimilar Tri-
angles, have two of their Sides parallel, the
third B C is alſo parallel to O E; which was to be
demonſtrated.
20.
If A B be made equal to A C, and E D to D O,
the Appearance of B C will paſs thro’ the Point E,
The Practice of Perſpective upon the Per-
ſpective Plane, ſuppoſed to be perpendicu-
lar, or upright.
IN order to give a diſtinct Idea of the Theory, I
have hitherto conſider’d the Geometrical Plane,
as it were the Ground upon which the Spectator
and the Perſpective
Plane, as a Window between the Spectator and
the Objects, in which the Objects are requir’d
to be repreſented. But, in Practice, this Matter
muſt be quite otherwiſe conceiv’d; which I
ſhall now endeavour to explain as clear as poſ-
ſible.
Suppoſe then, that a Painter has a mind to
draw upon his Perſpective Plane, or Picture,
(whoſe Bigneſs is as he thinks fit) a Proſpect of
a Country, wherein are Trees, Houſes, Rivers, & c.
Now, from what has been ſaid, this Country
will be his Geometrical Plane; and he ought to
conſider his Perſpective Plane as a Window, up-
on which the Points thro’ which the Rays com-
ing from all the Points of the Objects towards
the Eye, muſt be found. But theſe Interſections
of the Rays and the Window cannot be deter-
min’d, unleſs by Lines being drawn in the Geo-
metrical Plane to the Baſe Line.
Now, it is impoſſible for Painters to draw Lines
of this Nature on the Ground; wherefore they
uſe another more convenient Geometrical Plane
thus. At the Foot of their Perſpective Plane,
they place a Plane, upon which are drawn in
Minature the Baſes of Houſes and Trees, which
are in the Country to be repreſented; and the
Seats of the Points which, in the Objects, are
elevated above the Country; always obſerving,
that there be the ſame Diſpoſition between the
Objects and their different Parts, upon this new
Geometrical Plane, as the Objects truly have in
the Country to be repreſented.
Now, to determine the Magnitude of the
Space the Figures muſt take up upon this Geo-
metrical Plane, a Painter muſt firſt chuſe the
Diſpoſition of his Eye in reſpect to the Perſpe-
and then (from the Station Point,
thro’ the Extremities of the Perſpective Plane)
he muſt draw right Lines; which will limit the
Space wherein the Figures muſt be placed; ſince
the Rays of Figures, without thoſe Lines coming
towards the Eye, will not paſs thro’ the Perſpe-
ctive Plane.
21.
The Figures being thus drawn on the Geo-
metrical Plane, the next Thing is to find their Ap-
pearance upon the Perſpective Plane. Now,
theſe Figures are made up of either ſtraight
Lines, or crooked ones. To find the Repreſen-
tation of a ſtraight Line, its Extremes need on-
ly be ſought: And to have the Appearance of a
crooked Line, ſeveral Points thereof need only
be found. Since all this is equally applicable
to Figures, as well in the Geometrical Plane,
as thoſe above it; it follows, that the whole Bu-
ſineſs of Perſpective conſiſts in only finding the
Repreſentation of a Point.
And to find this Repreſentation in the follow-
ing Problems, we only uſe certain Lines drawn
in the Geometrical and Horizontal Planes; which, by their Interſection with the Baſe and
Horizontal Lines, ſhew the manner of drawing
new Lines upon the Perſpective Plane, which
determine the propos’d Appearances. Now, it
is plain, that in finding the ſaid Interſections, it
is not neceſſary to place the Perſpective Plane
perpendicular to the Geometrical and Horizontal
Planes; which would render the Work extream-
ly laborious: Whence the Perſpective and Hori-
zontal Planes may be conſider’d as lying upon
the Geometrical Plane, and ſo coinciding there-
with.
The Perſpective Plane may lye upon the Geo-
metrical Plane two ways; viz.
Either upon the
Face reſpecting the Objects, or upon that next
Now, as in the latter Situation,
Repreſentatious are drawn upon the Face of the
Perſpective Plane next to the Object, the Per-
ſpective Plane lying down upon its other Face; what ought to be on the Right Hand, appears on
the Left; and that on the Left, appears on the
Right; producing exactly the ſame Effect, as
looking thro’ the Back-ſide of a Paper, at a Pi-
cture drawn thereon.
Yet, notwithſtanding this Deficiency, we pre-
fer the latter way of the Perſpective Plane’s ly-
ing down to the former, for the following Rea-
ſons.
1.
When the Perſpective Plane lies down in
the former manner, it lies upon the Part of the
Geometrical Plane wherein Figures have been
drawn; which, together with the new Lines that
muſt be drawn, cauſes a very great Confuſion,
and always obliges one to copy his Work. An
Inconveniency which the latter Method is ſel-
dom ſubject to.
2.
We work with much more Eaſe in the
manner I have choſen.
Finally, The Default we have obſerv’d, may
ſeveral ways be remedied. For, in drawing up-
on the Geometrical Plane, we need but place
that on the Right Hand which we have a mind
ſhould appear on the Left; or if the Geometri-
cal Plane be drawn upon Paper, it may be oil’d,
or dipp’d in Varniſh, which will render it tran-
ſparent; and then the Back-ſide of the Paper
may be thrown into Perſpective.
If all this be not found convenient, the ſaid
Default may be eaſily corrected geometrically,
in copying the Work after the Drawings are fi-
niſhed. And this may be yet eaſier done, if
the Figures are expoſed before a Looking-glaſs;
the Left.
Therefore, I lay my Perſpective Plane upon
the Geometrical Plane; ſo that it be between
the Horizontal Plane, and the Figures requir’d
to be thrown into Perſpective.
22.
To find the Appearance of a Point, which is in
the Geometrical Plane.
Let Z be the Geometrical Plane, I E the Baſe
of Sight, D one of the Points of Diſtance, and
A the given Point.
From the Point A, let fall the Perpendicular
A B upon the Baſe Line; and from the Point of
Concurrence B, draw the Line B V to the Point
of Sight; then aſſume B E in the Baſe Line
equal to B A, and from the Point E draw the
Line E D to the Point of Diſtance D: And the
Point (a), the Interſection of B V and E D, is
the Repreſentation ſought.
23.
The Appearance of the Line A B, is
Now, if we conceive a Line
another from the Point A towards the Point E; theſe two Lines will be parallel, becauſe they
are in parallel Planes, aud each make half a
right Angle with the Perſpective Plane; whence
Now, ſince the Point A is in
the two Lines A B, A E; the Appearance of
the ſaid Point will likewiſe be in the Appear-
ances of the aforeſaid two Lines, and conſequent-
ly is in the Point a, the common Section of B V
and E D.
24.
If the Diſtance of the Eye be ſo great, that
one of the Points of Diſtance cannot be deno-
ted upon the horizontal Line; another Point, F,
muſt be uſed, diſtant from the Point of Sight
by about one third, or fourth Part of the Di-
ſtance of the Eye. But then, a correſpondent
Part of the Perpendicular A B muſt be likewiſe
taken, and laid off from B to G, in the Baſe
Line.
25.
And in this manner may the Repreſentation
of a very diſtant Point be found, if its Diſtance
from the Perſpective Plane be known, together
with the Place wherein a Perpendicular drawn
from that Point cuts the Baſe Line. For, having
firſt drawn a Line, as B V, from the ſaid Point
of Concurrence to the Point of Sight, then B E
muſt be aſſum’d in the Baſe Line; for Example,
equal to the tenth Part of the Diſtance of the
Point whoſe Repreſentation is ſought; and V H
in the Horizontal Line, likewiſe equal to the
tenth Part of the Eye’s Diſtance. Then C, the
Interſection of B V and E H, will be the Appear-
ance ſought.
Note, By this Method may be found the Deep-
nings in Pictures.
The Appearance of the Point A may yet be
otherwiſe found, without drawing the Line B V
from the Point A, in taking B I equal to B A,
other Point of Diſtance; which, by its Inter-
ſection with E D, will give the Appearance of
the Point A.
26.
γ is the Horizontal Plane, X the Perſpe-
ctive Plane, Z the Geometrical Plane, O the Eye,
D C the Horizontal Line, B E the Baſe Line, and
A the given Point.
Draw a Line from the Point A, to the Eye O, cut-
ting the Baſe Line in the Point B, and the Horizon-
tal Line in the Point C: Then aſſume B E in the
Baſe Line, equal to B A; and C D in the Hori-
zontal Line, equal to C O; and join the Points
E and D, by a Line cutting the Line A O in
the Point a; which will be the Appearance
ſought.
27.
The Triangle O D C in the Horizontal
Plane, is ſimilar to the Triangle A B E in the
Geometrical Plane; and conſequently A B is pa-
rallel to O C, and A E to O D. But the Appear-
ance of A muſt
28.
If the Place wherein the Eye ought to be in
the Horizontal Plane be not known, but the Point
of Sight is; then, to find the Place of the Eye,
a Perpendicular muſt be rais’d from the Point of
to the principal Ray; and the Extremity of this
Perpendicular will be the Point ſought.
If nothing is determin’d, the Place of the Eye
may be taken at pleaſure in the Horizontal
Plane.
29.
The ſame Things being given as in the
precedent Method, about the Eye O, as a Cen-
ter; deſcribe the Arc of a Circle I H, touching
About the given Point A, as a Center, de-
ſcribe the Arc of a Circle L C, touching the Baſe
Line: Then draw two Lines, C H and L I,
touching the two Circles L C and H I; and the
Point a, the Interſection of the ſaid two Lines,
will be the Appearance ſought.
30.
To demonſtrate this, draw the Line A B
perpendicular to the Baſe Line; O V perpendi-
cular to the Horizontal Line; and A C, O H
perpendicular to the Tangent H C. All theſe
Perpendiculars will cut the Lines to which they
are perpendicular, in the Points wherein theſe
laſt touch the Circle L B C, or H V I. Like-
wiſe draw the Line A E from the given Point
A, to the Point E, wherein the Line H C cuts
the Baſe Line. Finally, draw O D, from the
Eye O to the Point D, wherein the ſaid Line
HC cuts the Horizontal Line.
Now, it is evident,
demonſtrate that O D is parallel to A E; which
may be done thus:
Becauſe the Triangles O G V, and A B F, are
ſimilar. A F:
A B:
: O G:
O V:
altern.
A F: O G:
: A B:
O V:
Divid. and altern.
the firſt
Proportion.
AF—AB (=CF): O G—O V=HG:
:AB:
OV.
But becauſe the Triangles E C F, H G D are
ſimilar. C F:
H G :
: E F :
G D.
Now, by obſerving the two laſt Proportions of
the other two Triangles,
E F: G D:
: A F:
O G,
And the Angle A F E, being equal to the Angle
O G D, the Triangles A E F and O D G are
ſimilar; and therefore A E is parallel to O D:
Which was to be demonſtrated.
After the ſame manner we prove, that the
Appearance of the Point A is in the Line L I,
and conſequently is in the Interſection of this
Line and HC.
Altho’ this Method appears more difficult than
the precedent one, as to the Geometrical Conſi-
deration thereof, yet the Operation is eaſier, if
the Points are not too far diſtant from the Baſe
Line: For Lines may well enough be drawn by
Gueſs, or Sight only, to touch Circles, and Cir-
cles to touch Lines.
31.
Draw the Line F O G through the Eye O,
parallel to the Baſe Line, then aſſume F O in
O G equal to the Length of the principal Ray. A is the given Point.
From the given Point A, draw the Lines AO,
A F, to the Points O and F, and from the
Point E, wherein A F cuts the Baſe Line, draw
the Line E G to the Point G; then the Point a,
the Interſection of A O, and E G, is the Repre-
ſentation ſought.
Let fall the perpendicular G M from the Point
G, upon the Baſe Line, and through the Eye O,
draw the Line O D to the Point D, the Inter-
ſection of the Horizontal Line, and the Line G E.
Then becauſe the Triangles G D L, G E M are
fimilar,
G D: G E:
: G L:
G M.
But G O is equal to G L, and G F to L M.
whence
G D: G E:
: G O:
G F.
And conſequenely the Triangles G O D:
and
G F E, are ſimilar, and the Lines O D, and
A E F, are parallel between themſelves; and
therefore It has alſo been prov’d
Line A O; therefore i@ is in a the Interſection
to be demonſtrated.
33.
By this Demonſtration it appears, that there
is no Neceſſity in taking G O exactly equal to
the Eye’s Diſtance, and O F equal to its Height: But it is ſufficient if they have the ſame Propor-
tion, as the aforeſaid Diſtance has to the Height.
Likewiſe there is no Neceſſity in aſſuming the
Points G and F, in a Line parallel to the Baſe
Line; for any other Line paſſing through the
Eye O may be uſed at Pleaſure. For Example,
let g O f be a Line any how drawn through the
Eye O, and take the Point g at Pleaſure in this
Line, through which draw alſo the Line g N I
at Pleaſure, cutting the Horizontal Line in N,
and the Baſe Line in I; and draw the Line O N,
and through the Point I, draw the Line I f pa-
rallelthereto, cutting the Line g O f in f.
This being done, the Points g and f may be
uſed inſtead of G and F: for among all the
Lines that can be drawn (as G N I) it is mani-
ſeſt, that g N will always be to g I:: g O:
g f,
which is ſufficient for the Demonſtration.
If the Point f be firſt determin’d, the Point g
muſt be found by an Operation quite contrary
to that we have laid down.
34.
When nothing is determinate, we
may (a Baſe Line being firſt drawn) take at
Pleaſure, in another Line any how drawn,
the three Points g O f; ſo that in this Caſe, there
is no Manner of Neceſſity to uſe Compaſſes, in
throwing any Figure whatſoever, which is on
the Geometrical Plane, into Perſpective. But
if after having thus work’d, the Point of Sight,
Height and Diſtance of the Eye be requir’d, the
the Points f and O, on the Baſe Line, and the
Line P g drawn; then the Point V, wherein it
cuts the Perpendicular O H, is the Point of Sight
ſought, and the Parts O V, and V H determine
the Height and Diſtance of the Eye.
35.
When the Appearance of a Point is known,
Let A be a Point in the Geometrical Plane,
it is requir’d to find the Appearance of the
Point B.
Draw a Line from the Point B to the Eye O,
and another from the Point E, wherein the
ſaid Line continued, cuts the Baſe Line, to the
Point A; then draw the Line E a, and where
it cuts B O, is the Point b ſought.
The Point E is its own Repreſentation;
and
becauſe the Point a is the Repreſentation of A,
the Line E a is that of E A. Now ſince the
Point B is in the Line E A, the Appearance of
this Point will be likewiſe in E a, as alſo therefore it is in b the Interſection of the
Lines E a, and B O.
37.
If the Point A be in the Line B O, or
the Line B A be parallel, or a very little inclined
to the Baſe Line, we cannot then uſe this Me-
pearance of ſome other Point taken at Pleaſure
upon the Geometrical Plane be found, by Means
of which, the Appearance of the Point B may
be afterwards gotten; but in theſe Caſes, the
ſhorteſt Way, is to uſe ſome one of the precedent
Methods.
38.
It appears from this Method, that when
the Repreſentations of two Points are found,
the Appearance of any third Point whatſoever
may be had, without having any Regard to the
Situation of the Eye; becauſe two Lines as E a
may be drawn, whoſe Interſection will be the
Point ſought.
39.
The ſame things being given, as in the
Draw two Lines A F and A C from the given
Point A at Pleaſure, cutting the Baſe Line in the
Points E and B, and interſecting the Geometri-
cal Line in the Points F and C. From theſe
two laſt Points draw the Lines F O and C O to
the Eye; then draw E a through the Point E,
parallel to F O, and B a through the Point B,
parallel to C O, and the Point a the Interſection
of theſe two Lines will be that ſought.
Note, We might firſt have drawn the Lines
O F and O C at Pleaſure, and then have drawn
the Lines AC and A F through their Concur-
which would
come to the ſame thing.
Firſt continue the Line E a, until it meets
the Horizontal Line in D, and draw a Line
from D to the Eye, and another through the Eye
parallel to the Baſe Line.
Then the Parallels O M and F C are at the
ſame Diſtance from each other, as L D is from
E B; whence it follows, that F O is equal to E D,
and therefore O D is parallel to A F. Whence
And
after the ſame Manner we prove, that the Re-
preſentation of B A is a Part of B a.
40.
When there are no Lines drawn, and we
would uſe this Method, the Horizontal Line
may be laid aſide; and then having firſt drawn
the Geometrical Line, whoſe Diſtance from
the Baſe Line is equal to the Length of the Ray,
we aſſume the Diſtance from the Eye to the
Geometrical Line, equal to the Height of the
Eye.
Although this Method appears uſeleſs, as being
more difficult than the precedent ones, yet in
the Eighth Chapter we have ſhewn the Uſe that
may be drawn from it.
41.
It follows from this Demonſtration, that
the Appearances of Lines paſſing through the
Station Point, are all perpendicular to the Baſe
Line; for if the Perpendicular O S be let fall
ances of all Lines paſſing through S, will be
perpendicular to the Baſe Line; but the ſaid
Point S is the Station Point. Whence, &
c.
42.
To throw a Line in the Geometrical Plane
into Perſpective.
It has been ſhewn
the Extremes of the ſaid Line, need only be
found; and although it is difficult to find
ſhall ſhew here how to find more eaſy the Re-
preſentation of a Line in ſome Caſes.
1.
Let A B be a Line parallel to the Baſe
To draw the Repreſentation of which,
having firſt found the Point a the Appearance
of A, one of the Ends of the given Line; af-
terwards through that Appearance, draw a Pa-
rallel to the Baſe Line; then the Line B O,
drawn from B to the Eye O, will cut the ſaid
parallel in the Point b, and b a will be the Re-
preſentation ſought.
43.
2.
Let C G be a Line, which continued
out, cuts the baſe Line in E. Now to draw the
Appearance thereof; through the Eye O, draw a
Line parallel thereto, cutting the Horizontal
Line in D, and joyn the Points E and D, by
the Line E D, which cut in the Points c and g,
by Lines drawn from C and G to the Eye; then
the Part c g of the Line E D, is the Appearance
ſought.
If the Lines G O and C O cut E D very ob-
liquely, and ſo their Interſection cannot be ex-
be uſed.
44.
To find the Appearance of the Diviſions of a
Line in the Geometrical Plane.
Let A B be a Line, whoſe Appearance is ab.
of this Line, there muſt be Lines drawn from
the Diviſions of the Line to the Eye, whoſe
Interſections with a b will give the Points
ſought.
Note, When theſe Lines very obliquely cut
a b, the following Way ought to uſed.
45.
To find the Repreſentations of the Di-
viſions of the Line G C, make choice of the
then draw Lines
and from the Points wherein theſe Lines con-
tinued out cut the Baſe Line, draw other Lines
through the Repreſentation d, which will cut c g
the Repreſentation of C G in the Points ſought.
46.
To throw a Polygon, or any other regular
Figure on the Geometrical Plane into Perſpective.
The Repreſentation of any Kinds of Figures
may be found the fourth in general is the eaſieſt;
and may be firſt uſed in finding
only; and then the fifth Method ſerves
But yet the Work may be
ſhorten’d by the two precedent Problems, as we
ſhall ſhew in the following Examples.
To throw a Pentagon having one Side parallel to
the Baſe Line, into Perſpective.
Let A B C D E, be the propoſed Pentagon,
lel to A E, becauſe the Pentagon is a regular
one.
Now find
preſentation of four of the Corners of the Pen-
tagon; and to determine the Repreſentation of
the fifth Corner, find
is parallel to A B, which is ſuppoſed parallel
to the Baſe Line.
To throw a Parallelogram, divided into ſeveral
other Parallelograms into Perſpective.
Let A B C D be a Parallelogram, divided
Draw the Line O G thro’ the Eye O parallel to
the Side A D, cutting the Horizontal Line in G; likewiſe draw O F parallel to the Side A B cutting
the Horizontal Line in F, and produce the Sides
of the Parallelogram, and the Lines dividing it,
to the Baſe Line; then from the Points wherein
A D and C B, and the Lines parallel thereto
cut the Baſe Line, draw Lines to the Point G.
Alſo from the Points wherein A B and C D and
their parallels cut the ſaid Line, draw Lines to
the Point F, whoſe Interſections with theſe
ſought.
When this Method cannot be us’d, the Per-
ſpective of the Diviſions dividing the Sides of
the Parallelogram, mufl be found And we are
notwithſtanding the accidental Points, G and F,
being had. And this happens, when the Paralle-
logram is ſo far diſtant from the perſpective
Plane, that its Sides being produc’d, cannot meet
the Baſe Line.
47.
Note, moreover, that this one Example
is ſufficient to ſhew how to throw any Kinds of
Figures in the Geometrical Plane into Perſpe-
ctive. To effect which, we circumſcribe any
Parallelogram about the Figures, which we di-
vide into ſeveral others: Then we throw this
Parallelogram (thus divided) into Perſpective,
and transfer the given Figure therein, ſo that it
may have the ſame Situation with reſpect to
the little Parallelograms in the perſpective Plane,
as it had in regard to the ſmall Parallelograms
in the Geometrical Plane.
The Repreſentation of ſeveral Points of a
thrown into perſpective, muſt be
Chords in the Circle, or Curve, parallel between
themſelves, the Repreſentations of which muſt
be found then the Extremities of thoſe Re-
ctive ſought. The ſame may be done, in draw-
ing the Chords thro’a Point, whoſe Repreſenta-
tion is known.
49.
Let G I be the Geometrical Line;
and thro’
ſought, let fall the Perpendicular P F upon the ſaid
Line G I, which biſect in the Point R. About R,
as a Center, and with the Radius R P, deſcribe an
Arc of a Circle M P N, cutting the given Circle in
the Points M and N. Now, if the Perſpective of
L H and N M be found, the two Conjugate Dia-
meters of an Ellipſis, which is the Repreſentation of
the given Circle, will be bad. And, an Ellipſis may
be drawn by ſome one of the Methods laid down by
thoſe who have treated of Conick Sections.
I ſhall not ſpend time here in demonſtrating the
Truth of this. See Prop.
10.
lib.
2.
of the great
Latin Treatiſe of Conic Sections, written by M. de
la Hire; the Demonſtration of which may be here
apply’d. If we conſider, 1.
That Lines drawn from
the Points M and N to the Point F, will touch the
Circle in the ſaid Points M and N. 2.
That the
viſual Rays, going from the Eye towards all the Parts
of the Circumference of the Circle, form a Cone,
3. That the Appearance of the Circle, is the Section
of a Cone, made by the perſpective Plane. Finally,
That the Line G I ought to be conceiv’d, as being
the Interſection of the Geometrical Plane, and a
Plane paſſing thro’ the Eye parallel to the perſpective
Plane.
Let G S be the Geometrical Line, and S the
Make S F, in the Geometrical
Line, equal to the Height of the Eye; and let
A be the Seat of the given Line.
Aſſume F C in the Geometrical Line, equal to
the Height of the Eye, above the Geometrical
Plane: Then draw Lines from the Point A to
the Points S and C, and on the Point B, the In-
terſection of the Line AS and the Baſe Line,
raiſe the Perpendicular BI to the Baſe Line,
equal to E B, plus FC; and the Point I will be
the Perſpective ſought.
51.
Let us ſuppoſe a Plane to paſs thro’ the
given Point, and the Eye perpendicular to the
Geometrical Plane; then it is manifeſt, that the
Interſection of theſe two Planes is the Line
A B S, and the Interſection of the ſaid ſuppos’d
Plane and the perſpective Plane, is B I. Now,
let X be this ſuppos’d Plane; a, b, s, the Point
Figure, bi the Interſection of this Plane and
the perſpective Plane; O the Eye, and D the
propos’d Point: We are to prove, that if O D
be drawn, the Line B I of the precedent Figure
will be equal to b i in this Figure.
To demonſtrate which, draw the Line D L M
thro’ the Point D, parallel to a b s. Then, be-
cauſe the Triangles D M O and D L i are ſimi-
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.
Now, to throw a Pyramid into perſpective,
After which, Lines muſt be drawn from
pearance of thoſe Angles of the Baſe that are
viſible; and then the Perſpective ſought will be
had.
And to throw a Cone into perſpective, the
and then if Lines be drawn from
Repreſentation of the Baſe, the Repreſentation
of the Cone will be had.
But ſince, according to this Manner, we are
obliged to find the Perſpective of all the Baſe; whereas it often cannot be all ſeen;
we may de-
termine, by the following Method, what Part
of the Baſe is viſible, and ſo only find the Re-
preſentation thereof. And then, to compleat
the Cone, we draw Lines from the Extremities
of the viſible Part of the Baſe, to the Repreſen-
tation of the Vertex.
Let the Circle L I F be the Baſe of a Cone
thereof.
Aſſume P Q ſomewhere in the Baſe Line,
equal to the Semidiameter of the Circle L F; and from the Point P, raiſe P D G perpendicu-
lar to the Baſe Line, meeting the Horizontal
Line in G; and in this Perpendicular, make
P D equal to the Height of the Cone; and draw
the Line Q D H, meeting the Horizontal Line
in H. Then, about the Point A as a Center,
and with the Radius G H, draw the Circle B C E;
and from the ſaid Point A, draw a Line to the
Station Point S: Biſect A S in R;
and about
R, as a Center, with the Radius R A, deſcribe
the Circular Arc B A C, cutting the Circle BEC
in the Points B and C. Draw the Lines B A F,
and C A L; and the viſible Portion, (L I F) of
min’d.
To prove this, draw the Lines B C and L F, cut-
ting the Line A S in the Points N and M; and make
the Line G n equal to A N, and draw the Line
n D m. It is now manifeſt, that if the Cone be
continued out above its Vertex, (that is, if the oppo-
ſite Cone be form’d) it will cut the Horizontal Plane
in a Circle equal to B E C, whoſe Seat will be BEC: So that the Point S, in reſpect of B E C, is in the
ſame Situation as the Eye hath, with reſpect to the
Circle form’d in the Horizontal Plane, by the Conti-
nuation of the Cone. Whence it follows, that B C
is the Seat of the viſible Portion of that Circle. For,
by Conſtruction, B and C are the Points of Contact
of the Tangents to the Circle B E C, which paſs
thro’ the Point S; becauſe the Angle ABS, which
is in a Semicircle, is a right one.
Now, if a Plane be conceiv’d, as paſſing thro’ ſome
Points in the Horizontal Plane, whoſe Seats are
B and C, and which cuts the two oppoſite Cones
thro’ their Vertex; it is evident, that this Plane
continued, will cut the Geometrical Plane in a Line
parallel to B N C; and that this Line upon the
ſaid Plane, will determine the viſible Part of the
Cone’s Baſe. So, ſince G n was made equal to
A N, we have only to prove, that P m is equal to
A M: For, it follows from thence, that L M F is
the Common Section of the Geometrical Plane, and
the Plane which we have here imagin’d.
The Triangles D Q P and G H D are ſimilar, whence
D G: D P:
: G H:
P Q.
And the Triangles D P m and D G n are ſimilar:
Wherefore
D G: D P:
G n:
: P m.
And
G H: P Q:
: G n:
P m.
The Triangles B A N and L A M are ſimilar:
Therefore,
BA: AL:
: AN:
AM.
But the three firſt Terms of the two laſt Proportions,
are equal between themſelves; whence P m is alſo
equal to A M. Which was to be demonſtrated.
54.
When the Height of the Cone is greater
than the Height of the Eye, the Points, G and
H, will fall below the Point D; in which Caſe,
the Lines A B and A C muſt be produc’d, till
they cut the Circle in the Points l and f, oppo-
ſite to L and F: Then lIf will be the viſible
Part of the Baſe.
When the Cone is inclin’d, ſo that T (for Ex-
ample) is the Seat of its Vertex; AT muſt be
drawn: And then having aſſum’d P D equal to
the perpendicular Height of the Cone, and Pt
equal to A T; the Line t D x muſt be drawn;
and the Part T X, taken in A T, equal to G x.
Alſo, X S muſt be drawn, and A s, equal and
parallel thereto.
This being done;
the ſame Method muſt be
apply’d here, that I have laid down for the up-
right Cone; with this Difference only, that the
Point s muſt be us’d inſtead of the Station Point
S. But when the Height of the Cone is greater
than the Height of the Eye, the Point X muſt
be aſſum’d in the Line T A, between the Points
T and A.
The Reaſon of this is evident, from the Demon-
ſtration of the upright Cone: For, it is manifeſt, that
X is the Seat of the Center of the Circle, which the
Cone continued forms in the Horizontal Plane; and
conſequently, the Point s, in regard to the Circle BED,
is in the ſame Situation as the Eye is, in reſpect of
the Interſection of the continued Cone, and the Hori-
zontal Plane.
Note, moreover, that a Cone can ſcarcely ever
be thrown into Perſpective, by the common Me-
thod, ſo exact as by this.
It is requir’d to find the Appearance of a Line
trical Plane, in the Point A.
Aſſume E D, any where in the Baſe Line,
equal to B C; and from the Points D and E,
draw D F and E F to ſome Point F, taken at
pleaſure in the Horizontal Line. Then having
found
pendicular thereto: And if aI be made equal to
G H, the ſaid a I will be the Perſpective
ſought.
56.
The Appearance of the ſaid Line, is
qual to B C. Now if from the Extremities of
the ſaid Line A L, Perpendiculars are let fall,
meeting the Baſe Line in the Points P and M,
and from theſe Points, Lines are drawn to the
Point of Sight V; then a N will likewiſe be
and ſince P M is equal
to D E, a N will be likewiſe equal to G H, and
conſequently a N will be likewiſe equal to a I,
which is equal to G H.
57.
The ſame Things being given, as in the
precedent Method, about the Point A, as a
Arc of a Circle L M, and draw the Line O L
from the Eye touching it; then about a, (which
is the Repreſentation of A) as a Center deſcribe
the Circular Arc G I touching the Line L O,
and cutting another Line drawn through a Per-
pendicular to the Baſe Line in the Point I: I
ſay the Point I is the Extremity of the Repre-
ſentation ſought.
To prove this, let fall the Perpendiculars A L
and a G upon the Line O L, which will meet
the ſaid Line in the Points wherein it touches
the circular Arcs M L and G I.
Alſo aſſume D E in the Baſe Line equal to B C
or A L, and draw the Line D F; then through a,
draw a H parallel to the Baſe Line.
Now let us conſider the Figure X, which re-
the Point A of the foregoing Figure, wherein
O f here, repreſents O F there; f e here, F E
and finally e A here, E A in that Fi-
gure.
This being ſuppoſed, o f is parallel
to the Triangle a e A, and therefore we have this
Proportion.
o f:
f a:
: A e:
e o.
Comp.
o f + a f:
f a:
: A e + e a:
e a.
Altern.
o f + a f:
A e + e a:
: f a:
e a.
Comp.
and Perm.
o f + f a + Ae + e a :
o f + f a :
: f a + e a :
f a.
This laſt Proportion being reduced to the pre-
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 Which was
to be demonſtrated.
58.
Near one of the Sides of the perſpective
Line, equal to the Height of the Eye, in which
take B L equal in length to twice the Perpen-
dicular, whoſe Perſpective is requir’d. Let S be
Perpendicular meets the Geometrical Plane.
Having firſt found
Line in E, through which Point E draw the
Line Ea; then from the Point B draw a Line
B a to the Point a, cutting the Horizontal Line
in F. Again through F draw a Line to the
Point L, cutting E a in I; and a I is the Repre-
ſentation ſought.
To prove this, let G N be a Perpendicular to
the Baſe Line drawn from the Point G, wherein
the ſaid Baſe Line is cut by the Line B F; alſo
let G D be equal to the Perpendicular whoſe Ap-
pearance is ſought, and a H parallel to the Baſe
Line.
It is plain that the Perſpective of E A is
E a: But E A paſſes through the Station Point;
and conſequently
therefore
Now the Triangles B G C and B F M are ſimi-
lar; and ſo
B C :
B M :
: B G:
B F.
But B M by Conſtruction is the double of B C;
whence B F is alſo the double of B G, which,
conſequently, is equal to G F.
Becauſe the Triangles F G N and F B L are
ſimilar, therefore
F G :
F B :
:
G N :
B L.
Now we have proved, that F G is the half of
F B, therefore G N is likewiſe equal to the half
of B L, and conſequently equal to the Height
of the ſuppoſed Perpendicular.
Again, the ſimilar Triangles F G N and F a I
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.
I might have aſſumed C P equal to the Perpen-
dicular, and uſed the Points C and P inſtead of
B and L. But uſing the ſaid Points B and L is
better: For when the Points C and P are uſed,
the Horizontal Line muſt almoſt always be con-
tinued, that ſo a Line drawn through the Points
c and a may cut it; moreover this Interſection
will ſometimes be at an infinite Diſtance; where-
as in uſing the Point B, M N can never be
greater than thrice the Breadth of the Deſign to
be drawn.
The ſixth Problem may be ſolv’d by this;
for a Point elevated above the Geometrical
Plane, may be conceived as the Extremity of a
Perpendicular to the Geometrical Plane.
59.
To throw a Priſm or Cylinder into Perſpective,
cal Plane.
Let G H I L M N be the Baſe of the Priſm
in the Geometrical Plane, and the viſible Part
thereof upon the perſpective Plane, let be n g h i; then to compleat the Repreſentation of the
Priſm, draw Perpendiculars from the Points
n g h and i to the Baſe Line, whoſe Length let
be
Height of the Priſm, and find
face of the Priſm, in confidering them as Points
elevated above the Geometrical Plane: This
being done, if the Repreſentations of all the
ſaid Angular Points be joyn’d, the whole
Priſm will be thrown into Perſpective.
Now to throw a Cylinder into Perſpective,
the Repreſentation of its Baſe and upper Sur-
face muſt firſt be had, by finding
upper Surface, and then two Perpendiculars
muſt be ſo drawn to the Baſe Line, that they
may touch the Appearances of the two circular
Euds of the Cylinder, and the Appearance of
the Cylinder will be had. But to avoid uſeleſs
Operations, the viſible Part of the Baſe of the
Cylinder may be thus determin’d. Draw the
Line A S from the Point A to the Station Point S,
then this Line muſt be biſected in the Point R,
about which, as a Centre, and with the Radius
R A, the Circular Arc B A C, muſt be deſcrib’d
cutting the Baſe of the Cylinder in the Points
ones that can be ſeen.
61.
If the upper Face of the Cylinder or Priſm
being given as in the foregoing Method, we draw
the Line P Q in the perſpective Plane, parallel
to the Baſe Line, whoſe Diſtance therefrom we
make equal to the Height of the Priſm or Cy-
linder, whoſe Perſpective is requir’d. Then we
change its Geometrical Plane, ſo that the Baſe
Line coincides with P Q, and that in this Tran-
ſpoſition a Perpendicular to the Baſe Line coin-
cides with this ſame Perpendicular continued to-
Finally we find
in Situation by uſing P Q for a Baſe Line, and
the ſaid Perſpective is the Repreſentation of their
upper Faces.
If we ſuppoſe the Plane of the upper Surface
of the Priſm to be continued, it will meet the
Perſpective Plane in P Q; and the upper Face
in this Plane continued, will have the ſame
Situation in Reſpect to P Q, as the Baſe hath
on the Geometrical Plane with Regard to the
Baſe Line. If then the ſaid continued Plane be
conceived to lye on the perſpective Plane, the
upper Faces of the Priſm or Cylinder, will be
as the Baſes changed in the Manner aforeſaid; therefore the Appearance of the ſaid Baſes
changed, will be that of the upper Surfaces.
Note, By folding the Paper it is eaſy to
tranſpoſe Figures, and when the Height of the
precedent Method is the ſhorteſt.
62.
To throw a Concave Body into Perſpective.
Having firſt ſound the Perſpective of the ſaid
Body, afterwards find the Appearance of its
Cavity, in conſidering the Cavity as a new
Body.
63.
To throw a Sphere into Perſpective.
Let A be the Seat of the Centre of the Sphere;
then the Point I the Perſpective of the Centre
muſt be found, This being done, raiſe V F per-
pendicular to V I, which make equal to the
Diſtance from the Eye to the perſpective Plane; and in this Perpendicular continued, take V P
equal to the Diſtance from the Centre of the
Sphere to the perſpective Plane. Through the
Point P draw P Q parallel to V I cutting a Line
drawn from F through I, in Q; and about Q as
a Centre, with the Semidiameter of the Sphere,
draw the Circle C B, to which from the Point F,
draw the Tangents F C and F B, cutting the
Line I V in the Points G and E. On the Line
G E deſcribe the ſemicircle E D T G, wherein
draw the Line G D perpendicular to F I, which
biſect in H, and about H, as a Centre with the Ra-
dius H D, deſcribe the Arc of a Circle, L D R, cut-
ting the Line F I in the Points L and R. Take
the Chord G T in the Semicircle E D T G equal
to R L, and deſcribe a Semicircle T m G upon
G T; in which Semicircle draw ſeveral Lines,
as m n Perpendicular to G T; and cutting the
raiſe Perpendiculars p q, each of which muſt
be continued on each Side the Line G E, equal
to m n the Part of the correſpondent Line p m. Now if a great Number of the Points q be thus
found, and they are joyn’d by an even Hand,
you will have a Curve Line which will be the
Repreſentation ſought.
The Rays by which we perceive a Sphere, do form
an upright Cone, whoſe Axis paſſes through the Cen-
ter of the Sphere, and whoſe Section made by the
Perſpective Plane, is the Repreſentation ſought: from
whence it follows, that I is the Point in the Perſpe-
ctive Plane, through which the Cone’s Axis paſſes. But when an upright Cone is ſo cut by a Plane, that
the Section is an Ellipſis, as in this Caſe, the tranſ-
verſe Diameter of this Ellipſis, will paſs through
the Point of Concurrence of the ſaid Plane, and
Axis of the Cone, and that Point wherein a Per-
pendicular drawn from the Vertex of the Cone, cuts
the ſaid Plane. This will appear evident enough
to any one of but mean Knowledge in Conick Secti-
ons. Therefore the tranſverſe Axis of the Ellipſis,
which is the Repreſentation of the Sphere, is ſome
Part of V I; for the Eye is the Vertex of the Cone
formed by the viſual Rays of the Spbere.
Now let us conceive a Plane to paſs through the
Eye, and the Line I V; this will paſs through the
Center of the Sphere: And if a Perpendicular be
let fall from the Center upon the principal Ray con-
tinued, that Part of the ſaid Ray included between
the Point of Sight, and the Point wherein this Per-
pendicular falls, which is always parallel to the Per-
ſpective Plane, will be equal to the Diſtance from
the Center of the Sphere to the Perſpective Plane,
Therefore if the before-
mentioned Plane be ſuppoſed to revolve upon the Line
V I, as an Axis, until it coincides with the Per-
ſpective Plane, the Center of the Sphere will meet
the Perſpective Plane in Q, and the Eye in F; whence the Part G E of the Line I V is the tranſ-
verſe Diameter of the Ellipſis.
Again let G D E in Figure 30, and g e f, in
Letters in the foregoing Figure. Now if the Cone,
whoſe Profile is denoted by the Lines f g and fe be ſup-
poſed to be compleated, and to be cut by a Plane paſ-
ſing through the Line g e perpendicular to the Plane
of the Figure; we ſhall have an Ellipſis g 4 e 3
ſimilar to that which is the ſought Repreſentation
of the Sphere. Further if the ſaid Cone be conceived
to be cut by a Plane 14 m 3 parallel to its Baſe,
and biſecting g e in n, it is manifeſt, that 3 4, the
common Section of the Circle 14 m 3, and the Ellip-
ſis g 4 e 3, is the conjugate Axis of the Ellip-
ſis. And therefore this conjugate Axis is equal
to the Line 3 4, Perpendicular in the Point n to the
Diameter 1 m of the Circle 14 m 3. Now draw
the Lines E O and G Y in Figure 30, parallel to
L M, then the Triangles E G Y and E N M are
ſimilar, whence
But E G is twice E N;
wherefore G Y is alſo the
double of N M, and ſo N M equal to G Z. After
the ſame manner we demonſtrate, that L N is equal
to X E; whence it follows, that G D is equal to
L M, and is ſo cut in z as L M is in N; and there-
fore R L or G T of Figure 29, is equal to 34 in
Figure 31; and conſequently equal to the conjugate
Axis of the Ellipſis to be drawn. On the other
Hand, it is manifeſt by Conſtruction, that ſome one
of the Perpendiculars m n, Figure 29, viz. that
which paſſes through the Center of the ſemicircle
For if a Line be
drawn from T to E, it will be perpendicular to G T,
and conſequently parallel to m n: Whence the con-
jugate Axis of the Curve G q E, is equal to the
conjugate Axis of the Ellipſis to be drawn: And
therefore we are only to prove, that the Curve paſ-
ſing through the Points q, is an Ellipſis. Which may
be ſbewnthus.
The Parts G n of the Line G T, are Propor-
tional to the Parts G p of the Line G E: Whence
the Rectangles under G p and p E, are Proportional
to the Rectangles under G n and n T; but theſe laſt
Rectangles are equal to the Squares of the Ordinates
n m, which Squares are equal to the Squares of the
Ordinates p q; therefore theſe laſt Squares are Pro-
portional to the Rectangles under G p and p E, which
is a Property of the Ellipſis.
The ſemicircular Part h m of a Column, en-
Torus.
64.
To throw the Torus of a Column into Per-
ſpective.
Let B N C be the Baſe of the Column in the
draw a Line from the Cen-
ter A to the Station Point S, which biſect in the
Point R, and deſcribe the Arc of a Circle B A C
about the Point R, as a Center with the Radius R A.
Let X be the Profile of the Column, in which
ſemicircle h m, parallel to the Baſe of the Co-
lumn; and in the Line s a, which goes through
the Center of the Column, parallel to its Sides,
above the Point 2, which is in the Baſe of the
Column; likewiſe aſſume s a in the ſaid Line
equal to S A of the precedent Figure, and from the
Point a draw to a s the indefinite Perpendicular
a Y. Theſe General Preparations being made,
take at Pleaſure the ſmall equal Parts 6 i and 69
in the Line s a; draw the Lines i h and 9 m Pa-
rallel to 63 z, and from the Point h draw the Line
h 3 4, thro’ the Center 3 of the Semicircle h m; aſſume a 5 in a Y equal to i 4, and draw the Line
5 s cutting i b in g, and 9 m in q. And, (in Figure
32.) about the Point A, as a Center, with the
Radius i h or 9 m, which are equal, deſcribe the
Circle F L M H, cutting the Arc B A C in the
Points D and E; then draw the Line D E cut-
ting the Line A S in I; aſſume I G equal to i g,
and I Q equal to 9 q; and thro’ the Points Q and
G, draw F H and L M, parallel to the Line E D,
cutting the Circle D M E F in the Points L,
M, F, and H. Now if the Repreſentations of Four
Points, whereof L M F and H, are the Seats, and
the two firſt of which is equal to 29, and of the
two others 2 i, be found*; the Repreſentation
of the ſaid four Points will be ſo many Points of
Appearance ſought. And by drawing two other
Lines, as i h and 9 m, and proceeding as be-
fore, the Repreſentation of ſo many more Points
will be had.
Note, Becauſe a part of the Torus is hid by
a Circle muſt be deſcribed about the Center A,
with the Radius 36, cutting the Arc B C A in
the Points T and O, and the Lines STY and SOZ
muſt be drawn; then all the Points as F and H,
falling between the Lines TY and O Z are uſeleſs,
and L and M not coming under this Obſervation
muſt only be uſed; Note alſo, that there is no
neceſſity to determine Geometrically (which
h z m, as far as the Parallels (as 9 m) are uſeful: For when theſe Parallels are uſeleſs, the Point
q will fall beyond the Point m: But then the
Perſpective of the Torus is entirely drawn alrea-
dy, if thoſe Parallels were firſt begun to be drawn
near to 6 3 z, and the others continually going
from it.
In order to demonſtrate this Problem, the fol-
lowing Lemma is neceſſary.
65.
If two Circles C D H E and D E F L cut
C L paſſes, and D E joyns their Interſections; then, if the Radius A C or A H be called a, and
B F or BL, b, and the Diſtance A B between the
two Centers c, I ſay A G is equal to {bb—aa/ec}—{1/2}C.
Let us call A G, x, and G D or G E, y.
Then by the Property of the Circle, if y be conceiv’d
as an Ordinate of the Circle, C D H; yy=aa—xx.
And if it be likewiſe conſider’d as an Ordinate of
the Circle F D L, yy=bb—cc—2cx—xx: Whence
aa—xx=bb—cc—2cx—xx, and ſo 2cx=bb
—aa—cc; and dividing each Side of this laſt Equa-
tion by 2c, we have a x={bb—aa/2c}{1/2} c. Which was
to be Demonſtrated.
The Demonſtration of the
Problem.
66.
The Torus of the Column muſt be conceiv’d as
made up of an Infinite Number of Circular Planes,
lying one upon another. And it is evident that
the Reaſon why each of thoſe Circles cannot be wholly
ſeen, is becauſe that which is immediately under it
hides a Part thereof; from whence it follows, that
if the Plane of one oſ theſe Circles be every way
continu’d, and the Circle immediately under it, be
thrown
Repreſentation, and the Circle in the Plane, will deter-
mine the viſible Part of the ſaid Repreſentation; and
conſequently if the Repreſentation of theſe two Points
of Interſection be found upon the Perſpective Plane, we
ſhall have two Points of the Perſpective of the Torus of
the propoſed Column. This is what I have done in the
Solution of the Problem, as we ſhall now Analytically
demonſtrate.
Let O be the Eye, A M a part of the Torus of
through the Center of the Column, and A B a
Parallel to the Baſe, drawn thro’ the Center B of
the Semicircle Concavity of the Torus. Let M P be
a Semidiameter of one of the Circles ſpoken of in the
the beginning of this Demonſtration. Then if the
Line m p be drawn parallel and infinitely near C M P
and the Lines m O and p O are drawn cutting M P
in D and T, it is evident that D T, which is in the
Plane of the Circle paſſing thro’ M P, will be the
Semidiameter of the Perſpective of the Circle imme-
diately underneath.
Now let fall the Perpendicular O S from the Eye
to the Line A B, and continue the Lines M P and
m p, till they meet the ſaid Perpendicular in the Points
Q and q. Moreover, continue the Line M P to the
pendicular to m p. Aſſume A S = e, OQ = x,
and M P = y. Then in the ſimilar Triangles O q m,
and m R D, we have,
O q (x): q m (e + y):
: m R (d x) R D:
({edx + ydx/x})
The ſimilar Triangles O p q and p T D, give,
oq (x): q p (e):
: p P (d x):
P T ({edx/x})
P R is equal to y + dy, and if P T ({edx/x}) be added
to it, and then from the Aggregate be taken R D
({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
from, which may be here neglected, becauſe it is infinite-
ly ſmall in compariſon of the reſt; and we ſhall have
{xydy/dex} - {yy/e} for the Part of the Line PM included be-
tween P and the Point wherein this Line is cut by a
Line joyning the two Points of Interſection of the two
Circles.
Now before what I have here demonſtrated be ap-
ply’d to the Problem, we muſt obſerve, that if from
the Point M, a Line be drawn thro’ the Center B,
the Triangles M P C and m R M will be ſimilar; for the Angle m M P is the Exterior Angle of the
Triangle m R M, and the Angle m M C is a right
one. And conſequently
m R (dx): R M (d y):
: M P (y):
P C ({ydx/dx}).
Now if S A = s a in the 32d and 33d Figures
as likewiſe
it is manifeſt, that i 4 = a 5
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.
67.
This Problem may be likewiſe ſolved in
conſidering the Torus of a Column as made up of
an infinite Number of Baſes of Cones, whoſe Al-
titudes are determin’d by the concurrence of the
Tangents of the Semicircular Concavity of the
Axis of the Column; and then determining
Note, This
Method may be demonſtrated without Algebra,
but it would be very long.
68.
To find the Accidental Point of ſeveral pa-
rallel Lines, which are inclin’d to the Geome-
trical Plane.
Let A B be the Direction of one of the Lines,
and ECP, the
Angle that the ſaid Lines make with the Geo-
metrical Plane.
Draw a Line, O D, thro’ the Eye O, parallel
to A B, and thro’ the Point D, wherein it cuts
the Horizontal Line, and which is the acciden-
tal Point of the Directions of the given Lines,
draw D F perpendicular to the ſaid Horizontal
Line; in which aſſume D G, equal to DO.
Fi-
nally, thro’ the Point G, draw the Line G F,
making an Angle with the Horizontal Line, equal
to E C P; and then the Point F, (the Interſection
of this Line) and the Perpendicular D F, is the
accidental Point ſought.
Note, When the Lines are inclin’d towards
the perſpective Plane, D F and G F muſt be
drawn below the Horizontal Line: And, contra-
riwiſe, when the ſaid Lines are inclin’d towards
the oppoſite Part of the perſpective Plane, the
aforeſaid Lines muſt be drawn above the ſaid
Horizontal Line, as is done here.
If a Plane be conceiv’d to paſs thro’ the Eye,
perpendicular to the Geometrical Plane, and paral-
lel to the given Lines; it is evident, that the ſaid
Plane will cut the Horizontal Plane in the Line
O D, and the perſpective Plane in D F. It is,
moreover, manifeſt, that a Line drawn thro’ the
Eye, parallel to the given Line, is in the ſaid
Plane, and (with the Line O D) makes an An-
gle, equal to the Angle E C P, below the Hori-
zontal Plane, if the Lines be inclin’d towards
the perſpective Plane, and above it, if they in-
cline to the oppoſite ſide; whence this laſt Line
makes a right-angled Triangle with O D and
D F, whoſe Angle at the Point O, is equal to
the Angle C E P. But D G F is likewiſe a
right-angled Triangle, as having the Angle at the
Point G, equal to ECP; therefore theſe two
Triangles are ſimilar. And ſince the Side D G
is equal to the Side D O, the Triangles are alſo
equal: Therefore the Line D F, being common
to theſe two Triangles; the Point F, is the
Point wherein the Line, paſſing thro’ the Eye
parallel to the given Line, meets the Per-
ſpective Plane: And this Point is
Note, This Demonſtration as well regards
metrical Plane, as thoſe that meet it in one of
their Extremes only.
69.
To find the Repreſentation of one or more
Lines, inclin’d to the Geometrical Plane.
Let A be a Point given in the Geometrical
Plane; whereon ſtands a Line, whoſe Length,
Direction, and Angle of Inclination is known.
In ſome ſeparate Place, draw the Lines C E
and C P, making an Angle with each other
equal to the Angle of Inclination of the given
Line; and in one of theſe Lines, aſſume C E
equal to the given Line, and let fall the Perpen-
dicular E P, from the Point E upon the other
Line. Then aſſume A B, in the Direction of
the propos’d Line, equal to C P; and after ha-
ving found a, the Perſpective of A, and the
Point T join the Points a
and T by a right Line; and the ſought Appear-
ance will be had.
If from the Extremity of the inclin’d Line, a
Perpendicular be let fall upon the Geometrical
Plane, the ſaid Perpendicular will meet this
Plane in the Point B, and will be equal to P E; as is evident by the Conſtruction of the Figure
C P E. But the Point T is the Repreſentation
of the Extremity of this Perpendicular; and
therefore it is alſo the Extremity of the inclin’d
Line. Which was to be demonſtrated.
There are ſome Caſes of this Propoſition, that
may be ſhorten’d. As, 1.
When there are ſeve-
ral Lines of this Kind parallel between them-
ſelves, and whoſe accidental Point can be
found And, 2.
When an inclin’d Line is pa-
The Manner of
in the following Methods.
Thro’ F, the accidental Point of the inclin’d
Line, and equal to F G. And let A be the
Point, wherein one of the inclin’d Lines meets
the Geometrical Plane.
Aſſume R Q in the Baſe Line, equal to the
inclin’d Line; and draw Lines from the Points
R and Q, to the Point Z, taken at pleaſure in
the Horizontal Plane.
Thro’ a, the Perſpective of A, draw a N pa-
rallel to the Baſe Line; in which aſſume a L,
equal to M N; and draw a Line from the Point
a, to the Point F; and from the Point L, draw
another to the Point H. Then a T will be the
Perſpective ſought.
By
and therefore, we are only to demonſtrate,
that the Extremity of the faid Perpendicular is in
the Line L H. Which may be thus done.
Let us ſuppoſe a Line, A I, to paſs thro’ the
Point A, parallel to the Baſe Line, and equal to
the inclin’d Line. It is then manifeſt
and conſequently,
L H
Line; and therefore the Perſpective of this Ex-
tremity is in the Line L H; which was to be
demonſtrated.
Note, if F H had been aſſumed, the one half,
or one third, & c.
of what it is;
then it is mani-
feſt c.
of C E.
71.
Forinclined Lines not meeting the Geometrical
Plane.
Let A and B be the Seats of the Extremities of
Let X repreſent a Plane paſſing
through the given Line perpendicular to the
Geometrical Plane. Likewiſe let M N in this
Plane, repreſent the Line whoſe Perſpective is re-
quir’d; and let C N and P M be perpendicular to
the Geometrical Plane: Whence P C repreſents
A B, and conſequently is equal thereto.
Find the Point I
and
draw the Line B S, from the Point B, to the
Station Point I, cutting the Baſe Line in E; and
from the Point I, draw a Line to the accidental
Point F; which cut by a Perpendicular to the
Baſe Line, raiſed at the Point E; and then I T
will be the Appearance ſought.
72.
For inclined Lines parallel to the perſpective
Plane.
The Operation of Prob.
VII.
muſt be uſed
here, but with this Difference (ſee Fig. of the
ſaid Prob.) that whereas a I in the ſaid Problem
is perpendicular to the Baſe Line, here it muſt
make an Angle with the Baſe Line, equal to the
Angle of Inclination of the given Lines.
For the Demonſtration of this, ſee n.
7, and 10.
73.
To throw a Body into Perſpective, having
ſome one or all of its Sides inclined to the Geometri-
cal Plane.
The Appearances of the Lines forming the
Angles of the propoſed Body muſt be found: And
this may be eaſily done by Prob. 10.
And in this Manner the
Appearance of a Pyramid, an inclined Priſm,
& c.
may be found.
But nevertheleſs, it hap-
pens ſometimes, that the Operations of the pre-
cedent Problem may be abbreviated; as when
the Extremity of ſeveral Lines are found in one
and the ſame Line, or when inclined Lines, that
have difficult accidental Points, interſect one
another, and ſo mutually determine each other. This will appear manifeſt by the following Ex-
amples.
To throw ſeveral parallel Shores which ſtrengthen a
Wall, into Perſpective.
I ſuppoſe here that the Baſes of theſe Shores,
face of the Ground, are all in a right Line, pa-
rallel to the Side of the Wall; and then the
faid Shores may be thrown into Perſpective in
the following Manner: Having firſt found
preſentation of their Baſes: This being done,
denote the Appearances of the Lines wherein
the Shores meet the Wall, upon the Perſpective
of the Wall; the Appearances here are the
Lines p t, r s, which repreſent Lines parallel to
the Geometrical Plane, from the Suppoſition,
that the Shores are parallel between themſelves,
and their Baſes equally diſtant from the Wall. Finally, draw Lines from the Angles of the Re-
preſentations 1 2 3 4, to the Point F, which
will be terminated by their Interſections with
p t and r s, and will give the Appearances ſought,
as you ſee in the Figure.
To throw ſeveral parallel Roofs of a Houſe into
Perſpective.
Having found the accidental Points G and Q
the Wall ſuſtaining them, denote the Points
a b c d, wherein the ſaid Roofs meet the Wall: Then from the Point G draw Lines through the
Points a b c; and from the Point Q others to
the Points b c d; theſe Lines by their mutual
Interſection will determine each other, and give
the Repreſentations ſought.
74.
From what has been already ſaid, it will
not be difficult to throw any Objects whatſoever
into Perſpective. But ſince it is very difficult,
and indeed impoſſible for a Painter to make a
Deſign entirely according to the Rules we have
preſcribed; the Number of Points to be found
being almoſt infinite: therefore the Figures
principal Points of the Objects without the ſaid
Plane, need only be thrown into Perſpective. Which being once obtained, he may make uſe
of theſe Appearances ſo found, as a Rule where-
by the reſt may be compleated by the Eye, with-
out running the Riſque of committing ſome
conſiderable Fault, which by this Means may be
avoided.
Of the Practice of Peſpective upon the Per-
ſpective Plane ſtill conſider’d as being upright.
IT often happens that Painters offend all
Rules of true Appearance when they paint
Pictures to ſtand aloft, to be ſeen Sideways, or at
a confiderable Diſtance. Their Cuſtom is to
paint Pictures to be view’d, after the ſame Man-
ner as they themſelves look at them when they
are working; whence in the following Caſes,
this Practice of theirs will be uſeleſs; and ſo to
avoid enormous Faults, they are neceſſitated to
have recourſe to Perſpective But what has been
ſaid in the laſt Chapter, does not reach theſe
particular Caſes; therefore we ſhall here add ſome
new Problems, which together with the former
ones, will take in all Caſes.
75.
To throw Figures which are in the Geometri-
cal Plane into Perſpective, when the Eye is at ſo great
a Diſtance that it cannot be denoted in the Horizon-
Horizontal Line.
The Repreſentation of two Points of theſe
Figures muſt be firſt found and then by Means
may be had
Let A B C D E, be a Pentagon, whoſe Ap-
V the Point of Sight;
and
V F the ſixth Part of the Eye’s Diſtance from
the perſpective Plane. Now find
the Appearance of the Point A will be had
tion of A and E, will that of D be had; and
by uſing B and A, the Perſpective of C may be
found.
76.
Note, the Perſpective of Lines perpendicu-
lar to the Geometrical Plane as alſo of Lines
77.
To throw Figures, which are in the Geometri-
cal Plane into Perſpective, when the Eye is ſo oblique
that it cannot be marked in the Horizontal Plane,
or the Point of Sight in the Horizontal Line.
We muſt proceed here according to the Di-
rections of the precedent Problem, after having
found the Perſpective of ſeveral Points of the
given Figures.
At any Point C, taken at Pleaſure in the Baſe
Line, draw the Perpendicular C D to the ſaid
ſame Point in ſuch manner, that if it could be
the Point of Sight.
This is done in aſſuming C H equal to {1/3}, or
{1/4} Part, &c.
of the Diſtance from the Point C,
to the Foot of the vertical Line; and in raiſing
the Perpendicular H E, in the Point H, equal
to {1/3} or {1/4} Part, &c.
of the Height of the Eye.
Now A is a given Point, whoſe Appearance is
ſought.
Draw a Parallel A B, through the Point A,
to the Baſe Line, meeting the Line C D in the
Point B, and let a ſecond Eye be ſuppoſed at
the ſame Height and Diſtance as the firſt; then
find
Line C E in b, and in this Continuation aſſume
b a equal to F G; then a will be the Perſpective
ſought.
Becauſe the Height and Diſtance of the ſecond
Eye, is equal to the Height and Diſtance of the
firſt; the ſaid two Eyes are both in one parallel
Line A B; and conſequently
And therefore becauſe
and a
Which
was to be demonſtrated.
78.
Note, as to Lines perpendicular, and inclined
to the Geometrical Plane, ſee n. 76.
This is
ſcarcely uſeful, unleſs for the Decorations of a
Theatre.
79.
To find the Repreſentation of a Figure in the
Geometrical Plane, when the Perſpective Plane is
placed above the Eye.
When the perſpective Plane is ſituated above
the Eye, we ſuppoſe the Geometrical Plane to
paſs through the Top of the Perſpective Plane; upon which Geometrical Plane are drawn the
Figures of Objects meeting it; as alſo the Seats
of thoſe Objects that are underneath it, by
Means of Perpendiculars; and the Height of the
Eye is here meaſur’d by a Perpendicular drawn
from the Eye to the Geometrical Plane; whence
the perſpective Plane, elevated in reſpect to the
Eye, is the ſame thing, as an Eye elevated in
regard to the perſpective Plane.
Let I L be the Baſe Line, and H the Foot of
then in the Baſe Line aſ-
ſume the Points I and L at Pleaſure, towards
the Sides of the perſpective Plane. Make I S
equal to {1/3} or {1/4} Part of I H, and raiſe the Per-
pendicular S X, in the Point S, to the Baſe Line,
equal to a correſpondent Part of the Height and
Diſtance of the Eye taken together; draw the
Line X I G, and moreover Y L Q, in aſſuming
L T equal to {1/3} or {1/4} & c.
of L H.
Again draw
the Line G Q in the Geometrical Plane, pa-
rallel to the Baſe Line, and diſtant therefrom
(for Example) a third Part of the Height of
the Eye; draw alſo F P in the perſpective Plane,
parallel to the Baſe Line, and diſtant therefrom,
a fourth Part of the Eye’s Diſtance; theſe two
Lines will cut X I in G and F, and Y L in Q
and P. Note, if the Diſtance of G Q from the
Baſe Line, had been aſſumed equal to a fourth
Part of the Eye’s Diſtance; then F P muſt have
Part of the Eye’s Diſtance, and ſo on. Now
A is a Point whoſe Repreſentation is requir’d.
Draw the Lines A F and A P, from the Point
A to the Points F and P, cutting the Baſe Line
in the Points E and B; then draw the Lines E G
and B Q, which continue till they interſect
each other in a, which is the Repreſentation
ſought.
Let us ſuppoſe the perſpective Plane continu-
ed, C D the Horizontal Line, and O the Eye
denoted in the Horizontal Plane. It is evi-
dent produce the
Line G S a, until it meets the Horizontal Line
in D, and draw the Line O D. Let fall the
Perpendicular G N R, from the Point G upon
the Horizontal Line, which interſect in R, by
the Line O R, paſſing through the Eye parallel
to the Horizontal Line. Now by Conſtruction,
G M is {1/3} of M N; and conſequently it is {1/4} of
G N; M Z is likewiſe {1/4} of N R:
Therefore
G M:
M Z:
: G N:
N R.
Compon.
and Altern.
G M:
G N:
: G M + M Z = G Z:
G N
+ N R = G R.
Becauſe the Triangles G M I and G N C are
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.
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
We demonſtrate in
the ſame Manner, that B a is the Perſpective
of B A, and ſo the Perſpective of the Point A,
the common Section of E A and B A, is a, the
Interſection of the Appearances of the ſaid two
Lines.
80.
To find the Repreſentation of a Line, per-
pendicular to the Geometrical Plane, when the per-
ſpective Plane is above the Eye.
In the Baſe Line B E, aſſume the Line E D,
and draw C L, parallel to the Baſe Line, and
diſtant therefrom (for Example) {1/4} of the Height
of the Eye; make F L equal to {3/4} of D E, and
draw the Lines E L and D F. Note, if the
Diſtance from C L to B E, had been aſſumed
equal to a fifth Part of the Height of the Eye,
F L muſt have been aſſumed equal to {4/5} Parts of
E D. Now let a be the Perſpective of the Foot
of the propoſed Perpendicular; through which
draw a H parallel to the Baſe Line, and a I per-
pendicular to the ſaid Line; then make a I equal
to G H, and the propoſed Perſpective will be
had. The Demonſtration of this Operation is
manifeſt
Horizontal Line.
Of throwing Figures into Perſpective, when
the Perſpective Plane is conſider’d as being
inclined.
81.
TO find the Perſpective of a Figure in the
Let X be the Vertical Plane;
S I the Station
Line, S the Station Point, and H the Interſecti-
on of the Station Line and Baſe Line. Now
draw the Vertical Line H V through the Point H,
making an Angle with S I, equal to the Angle
of Inclination of the perſpective Plane; then
raiſe the Perpendicular I O to S I, in the Sta-
tion Point S, equal to the Height of the Eye; and through the Extremity of the ſaid Perpen-
dicular, draw the principal Ray O V, paral-
lel to S I, and cutting H V in the Point of
Sight V.
Now it is evident, that O V determines the
Length of the principal Ray, and H V the Di-
ſtance from the Baſe Line to the Horizontal
Line; and ſince the Demonſtration of the
Problems in the aforegoing Chapters regarding
the Geometrical Plane, have alſo Relation to
the perſpective Plane being inclined, the ſaid
Problems may be here uſed; and conſequently,
this inclined perſpective Plane is reduced to a
Perpendicular one, view’d by an Eye, whoſe
Height is H V, and Diſtance O V.
82.
To find the Appearance of a Point above the
Geometrical Plane.
Let H C be the Baſe Line:
And let T be the
to the Geometrical Plane. This Point will
be
ſuring the Height of the Eye; for this laſt
Line is parallel to the ſaid Perpendiculars. And
ſo likewiſe the aforeſaid Point is the ſame as
the Point T of Fig. 44:
Let V be the Point of
Sight, S the Station Point, and Q the Station
Point of the upright perſpective Plane, to which
the inclined perſpective Plane is reduced And
Draw two Lines M P and P E ſeparately,
making a right Angle with each other; in one
of which, aſſume P E, equal to the Height of
the given Point, whoſe Perſpective is ſought; and draw the Line E M, making an Angle with
M P, equal to the Angle of Inclination of the
perſpective Plane. Again let fall the Perpen-
dicular A D from the Point A to the Baſe
Line, in which aſſume A L equal to P M, to-
wards the Baſe Line, when the perſpective
Plane is inclined towards the Objects (as we
have here ſuppoſed) but on the other Side of A,
when the perſpective Plane inclines towards the
Eye. Then from the Point A, draw a Line
to the Point S, cutting the Baſe Line in B, and
joyn the Points L and Q, by a Line cutting the
Baſe Line in C. This being done, draw the
which interſect in the Point X,
by a Perpendicular to the Baſe Line, in the
Point G; and then the Point X is the Appear-
ance ſought.
In Fig.
44.
where V, S, T, and H, repreſent
the ſame Points as thoſe that are denoted with
the ſame Letters in this Figure; we have,
T H:
H S:
: T V:
V O.
Compon.
and altern.
T H:
T V:
: T H + H S:
T V + V O.
This being apply’d to Fig.
45.
and it will be,
T H: T V:
: T S:
T V + V O.
If now T X be continued, till it cuts the Ho-
rizontal Line in F; we ſhall have,
T H:
T V:
: T B:
T F.
And conſequently,
T B:
T F:
: T S:
T V + V O.
Whence it follows, that if a Line be drawn
ſrom the Eye, to the Point F, it will be paral-
lel to S B A. Therefore
and ſo the Repreſenta-
tion of A is in the ſaid Line. The Perſpective
of a Line perpendicular to the Geometrical
Plane, in the Point A, paſſes thro’ the Perſpe-
ctive of the Point A, and thro’ the Point T
But the given Point
is in the aſoreſaid Perpendicular: And ſo its Per-
ſpective is in T X.
Again;
it is otherwiſe manifeſt, that the Per-
ſpective of C L, is a and conſe-
Now, if a Line be ſuppos’d to be drawn from
the Point L, thro’ the propos’d Point, it will be
parallel to the Vertical Line; and ſo its
And ſince
L, it will be a Part of C X. But becauſe that
Line, drawn from the Point L, paſſes thro’ the
propos’d Point; the Repreſentation of the ſaid
Point is alſo in C X; and ſo in X, the common
Interſection of C X, and T X.
83.
If the Point T ſhould be at too great a
Diſtance; or if T B X, or C X, ſhould too ob-
liquely cut each other; the perſpective Plane
muſt then be ſuppos’d to be reduc’d and the Repreſen-
tation of a Point, above the Geometrical Plane,
(whoſe Seat is L, and Height M E) muſt be
found
84.
To find the Repreſentation of a Line, perpendi-
cular to the Geometrical Plane`.
The Appearance of the Extremity of the Per-
cal Plane, by the Height of the propos’d Perpendi-
cular: Then if a Line be drawn from the Point
D, to the Point of Sight; its Interſection
the Perpendicular propos’d.
Note, When there is a Neceſſity of having re-
courſe to the Remarks of the foregoing Problem,
in order to find the Point X; then the Point a
may be found, in drawing A S and D V, and
afterwards joining the Points B and X by a
Line. And when B X and D V cut each other
85.
A is the Foot of the Perpendicular:
The
And
to the Geometrical Plane.
Thro’ the Point a, the Appearance of A,
draw a Perpendicular to the Baſe Line; which
make equal in conſidering this laſt Line, as being
parallel to the Vertical Line. Then, from the
Extremity I of this Perſpective, to the Point of
Sight V, draw a Line cutting the Line T a, in
the Point X; which will be the Repreſentation
of the Extremity of the propos’d Line.
Let us ſuppoſe a Line paſſing thro’ the Point
A, equal to M E, and parallel to the Verti-
cal Line. Suppoſe, moreover, that another Line
is drawn thro’ the Extremity of this Line, and
that of the propos’d Perpendicular; then this
laſt Line, by the Conſtruction of the Figure
M E P, will be parallel to the Station Line; and conſequently, its Repreſentation
and its Interſection
with T a, will be the Extremity of the Repre-
ſentation ſought. But a I is
and con-
ſequently, V I is that of the ſecond. Which was
to be demonſtrated.
Note, When V I and T a cut each other ve-
ry obliquely, recourſe muſt be had to the Obſer-
vation at the End of the aforegoing Method, or
the following Way may be uſed.
86.
Let T be the accidental Point of the Lines
which Point draw a parallel to the Baſe Line,
in which aſſume T R equal to O T of Fig. 44.
Aſſume D N ſomewhere in the Baſe Line, e-
qual to the propoſed Line, and draw the Lines
D F and N F to the Point F, taken at Pleaſure
in the Horizontal Line; then through the Point
a, the Appearance of A, draw the parallel a H,
to the Baſe Line, in which aſſume a Q equal
to G H. Then if the Lines T a, and R Q be
drawn, and continued, till they cut each other
in the Point X, a X will be the Appearance
ſought.
The Part a Q of the Line a H, is
Geometrical Plane, and which is equal to the
propoſed Line, and parallel to the Baſe Line; and conſequently
ſed Line: And therefore X the Interſection of
R Q and T a, is the Appearance of the ſaid
Extremity.
It is manifeſt
c.
of what it is taken here,
provided likewiſe that then D N be aſſumed
equal to a correſpondent Part of the propoſed
Line.
87.
To throw a Sphere into Perſpective.
The Method of ſolving this Problem before
laid down that inſtead of uſing the Point of
Sight, the Point wherein a Perpendicular drawn
from the Eye to the perſpective Plane, meets
the ſaid Plane, muſt be uſed. And you muſt
obſerve, that this Perpendicular meaſures the
Eye’s Diſtance from the perſpective Plane.
88.
To find the accidental Point of any Number
Let A B be the Direction of one of the in-
clined Lines, O the Eye in the Horizontal Plane,
and S the Station Point.
Draw the Line O D, thro’ the Eye O parallel
to A B, meeting the Horizontal Line in D, which
will be and thro’ the Station Point S,
cutting the Baſe Line in N, and draw the Line
N D. Then about the Point D, as a Center, and
with the Radius D O, deſcribe the Circular Arc
O o: And about N, as a Center, with the Radius
N S, draw the Circular Arc S s. This being done,
draw the Line s o touching the ſaid two Arcs,
and the Line D o perpendicular to s o. Then
draw o F, making an Angle with o D, equal to
the Angle of the Inclination of the Lines given,
and cutting N D continued in F: Now I ſay F
is the Accidental Point ſought when the Lines
do not incline towards the Perſpective Plane: But if they do, o F muſt be drawn below o D.
If a Plane be ſuppoſed to paſs thro’ the Eye
parallel to theinclin’d Lines; the common Secti-
ons of this Plane, and the Horizontal and Geo-
metrical Planes, will be O D and S N. It is
now manifeſt, that if a Line be drawn in the
ſaid Plane, below the Horizontal Plane, when
the Lines incline towards the perſpective Plane,
and above it when they incline the other way,
making an Angle with O D equal to the Angle
of Inclination of the propoſed Lines; I ſay it is
manifeſt, that the ſaid Line will be parallel to
the propoſed Lines, and will meet If
now the before ſuppoſed Plane be conceiv’d to
turn about the Line N D, the Eye, and the
Station Point in the ſaid Plain, will then meet
the perſpective Plane in the Points o and s; for
the Lines D o and N s are equal to D O and N S,
and form right Angles with the Line s o joyning
their Extremities. Now the two Points s and o
anſwer to the Situation of the Eye and Station
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.
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
Now determine, by Means of the Triangle C P E
according to the Manner lay’d down
the Length A B of the Direction of the propoſed
Line. This being done, find the Point
by the Length of P E; and then a X will be the
Perſpective ſought.
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
F that of
oſ Perpendiculars to the Geometrical Plane.
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.
Becauſe a b is
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.
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-
Perſpective a x may be found as directed n. 70.
Of throwing Figures into Perſpective, when
the Perſpective Plane is conſider’d as being
parallel to the Geometrical Plane.
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.
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-
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
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.
It is manifeſt
metrical Plane; and conſequently the Per-
ſpective ſought is a Part of T a. Moreover, it
is manifeſt
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 there-
ſaid Extremity.
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.
The Perſpective Plane being conſider’d as the Geo-
metrical Plane.
95.
Let T (as in Fig.
51.)
be the accidental
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.
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.
Draw the Radii a F, a L, T I, T H, to the
Points of Contact F, L, I, and H. Then be-
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,
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.
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 Perpen-
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.
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
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.
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.
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, is
And
therefore
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.
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
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.
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 to
and conſequently
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
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.
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-
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.
101.
The ſame Things being given as in the
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 Inter-
ſentation ſought.
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
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;
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.
103.
To find the Perſpective of the Shadow of a
Point above the Geometrical Plane, whoſe Height and
Seat is given.
Let Z be the Geometrical Plane;
A the Seat
and A B the Direction of
the Sun’s Rays.
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.
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
Then
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.
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.
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.
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
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.
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.
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.
What hath been obſerved
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 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.
109.
To find the Repreſentations of Figures in
the Geometrical Plane.
Let O be the Eye, R H the Baſe Line;
F and
Fig. 10.
Then take a Ruler and faſten to the Point G,
ſo that it may turn about the ſame in ſuch man.
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.
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.
110.
Let O be the Eye, H E the Baſe Line,
Take a Ruler
M N, having two Threads equal in Length
faſten’d to it, and about O as a Center, and with
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.
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
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
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.
Let T be the Repreſentation of the Foot of a
Perpendicular. Move the Curſor D along C L,
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.
112.
When all the Perpendiculars have the ſame
Length.
Let F G be parallel to the Baſe Line, and F O
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
incides with S, and R H with S Q. Then if S Q
be taken for a Baſe Line, and the Appearances
had.
113.
For Perpendiculars of the ſame Length.
The Figures in the Geometrical Plane being
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 then
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
114.
If a Plane be imagined to paſs through
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 Line,
60.)
equal to the Height of the Eye
above this Plane, and fi (Fig. 61.)
is the Geo-
metrical Line in the ſaid Plane.
115.
To find the Repreſentation of Figures which
are in the Geometrical Plane.
The Operation laid down
becauſe the inclined perſpective Plane may be
changed
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
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
Line.
This is demonſtrated in n.
114.
The Point C muſt be aſſumed below the Point
S, when the perſpective Plane is inclined towards
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.
117.
To throw Figures which are in the Geometri-
cal Plane into Perſpective.
Draw the Line C F at Pleaſure, in which aſ-
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.
Let Z be the Geometrical Plane, and A a
Point of the given Figures. Move the two
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.
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
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.
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
as alſo let P Q and M N be the Rulers: More-
over, let B be the Point, wherein a Perpendicu-
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.
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 Repre-
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
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.
6.
To find the Repreſentations of the Shadows of
any Number of Points, being at the ſame Height
above the Geometrical Plane.
Find
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 given
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
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
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: There-
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.
Draw the Line E O, thro’ E, the Foot of the
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,
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 Line
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.
The Baſe Line is h i;
p s is the Horizontal
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
joining the Diviſions of the Lines h i and p s.
If the Line H I happens to meet the Meri-
dian; the common Method, by the Horizontal
Dial, is eaſier than this.
Theſe Dials are drawn in the ſame manner as
Vertical ones; the following Preparations being
firſt made.
Draw the Line e c, equal to the Length of
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,
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.
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.
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. The
for
in ſuch a cloſe Place, ones Breath will ſully them; but
not the Convex Glaſs, becauſe it is encloſed in a
Tube.
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.
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 with-
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.
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.
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
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 may
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; and
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 in which a Ruler going ſeveral
Inches out behind the Box X may ſlide, having
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.
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.
15.
To repreſent Objects in their natural Diſpo-
ſition.
When Objects are to be repreſented within 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,
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 mo-
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.
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
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, 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.
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.
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
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 an
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,
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,
Angle and
Let MN and GH be two Mirrours in the before-
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 and conſequently the
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 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
M N muſt
{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
whence the Angle B P Q, which is the
Complement of P B Q, and P Q B to two right
Angles, is Wherefore becauſe N B C is 67 {1/2} the
{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-
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.
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 then
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
of the Mirrour H, ſituated but the
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 coming
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.
35.
To repreſent Pictures or Prints.
If we have a mind to repreſent Pictures and
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
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:
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.
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 ſaid
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 Re-
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 is
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.
39.
This Machine is a kind of Box, the Side
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
41.
There is a round Hole in the Middle of
the Top of the Box, in which the Cylinder
carrying the Convex Glaſs ſcrews.
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 Objects
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, 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.
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,
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.
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
zon muſt be made leſs, by half the Inclination of
the Bottom of the Box; and when the Machine
is ſomewhat inclin’d backwards, 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.
48.
Let A B be a Ray, proceeding from ſome
We are to demonſtrate,
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 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
In reaſoning nearly after the ſame Manner,
we demonſtrated what is mentioned
Box is inclin’d a little backwards.