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PRACTICAL TREATISE

ON

PERSPECTIVE AND ISOMETRICAL

DRAWING.

EDITED BY

WM. E. WORTHEN.

’y' A“ .\
‘_,\
, -: f’v 'y" "1

NEW YORK :
I). APPLETON & COMPANY,
443 & 445 BROADWAY.
1862.

 

ENTERED; ucoordmg to Act bf Congas, inflf‘eyeu- 1857, by
D. APPLETORfiQQuPANY,
In the Clerk’s Oflice of the District Court of the United States for the Southern District of
New York.

 

 

\

 

PERSPECTIVE DRAWING. 355

 

PERSPECTIVE DRAWING.

are

, .. science of Perspective' is the representation by geometrical rules, upon
" _, 2:1)1ane surface, of objects as they appear to the eye, from any point of

   

hew.
All the points of the surface of a body
are Visible by means of luminous rays
' proceeding from these points to the eye.
1 Thus, let the line AB (fig. 1) be placed
before the eye, C, the lines drawn from
the different points 1, 2, 3, 4, &c., repre-
sent the Visual rays emanating from each

 

of these points. It is easy to understand

 

that, if in the place of a line a plane or -
curved surface is substituted, the result will be a cone of rays.

Let A B (fig. 2) be
a straight line, and let
the globe of the eye be

TI,

/

represented by a cir-
cle, and its pupil by.
the point C. The ray

 

emanating from A, en- 21

 

tering through C, Will
proceed to the retina

L
Lbi \
g 2
Q
”adage

 

 

of the eye, and be de— A' A

picted at at. And as

it follows that all the

points of A B will send /

rays, entering the eye Fig. 2.

through 0, the whole image of A B will be depicted on the retina of the
25

1H 10!?)

 

386 PERSPECTIVE DRAWING.

eye in a curved line a 3 I). Conceive the line AB moved to a greater
distance from the eye, and placed at A’ B’, then the optic angle will be
reduced, and the image a’ 3 b’ will be less than before ; and as our visual
sensations are in proportion to the magnitude of the image painted on the
retina, it may be concluded that the more distant an object is from the eye,
the smaller the angle under which it is seen becomes, and consequently
the farther the same object is removed from the eye the less it appears.

Observation has rendered it evident, that the greatest angle under
- which one or more objects can be distinctly seen, is one of 90°. If be-
tween the object and the eye there be interposed a transparent plane (such
as one of glass m 72,), the intersection of this plane with the visual rays are
termed perspectives of the points from which the 'ays emanate. Thus a
is the perspective of A, I) of B, and so on of all the intermediate points;
but, as two points determine the length of a straight line, it follovs that
a I) is the perspective of A B, and a/ I)’ the perspective of A" B”.

It is evident from the figure that objects appear more or less great ac—
cording to the angle under which they are viewed; and further, that ob-
jects of unequal size may appear equal if seen under the same angle.
For draw f g, and its perspective will be found to be the same as that of
A’ B’.

It follows also, that a line near the eye may be viewed under an angle
much greater than a line of greater dimensions but more distant, and
hence a little object may appear to be much greater than a similar object
of larger dimensions. Since, therefore, unequally sized objects may ap~
pear equal in size, and equally sized objects unequal, and since objects are
not seen as they‘are in effect, but as they appear under certain conditions,
perspective may be defined to be a science which affords the means of rep-
resenting, on any surface whatever, objects such as they appear when seen
from a given point of view. It is divided into two branches, the one
called linear perspective, occupying itself with the delineation of the con-
tours of bodies, the other called aerial perspective, with the gradations of
colors produced by distance. It is the former of these only, that is pro-
posed here to be discussed.

The perspective of objects, then, is obtained by the intersection of the
rays which emanate from them to the eye, by a plane or other surface
(which is called the picture), situated between the eye and the objects.

From the explanation and definition just given, it is easy to conceive
that linear perspective is in reality the problem of constructing the section,
by a surface of some kind, of a pyramid of ass of which the summit. and
the base are given. The eye is the summit, the base may be regarded as

PERSPECTIVE DRAWING. 3 87

the whole visible extent of the object or objects to be represented, and the
intersecting surface is the picture.

A good idea of this will be obtained by supposing the picture to be a
transparent plane, through which the object may be Viewed, and on which

it may be depicted.
In addition to the vertical and horizontal planes with which we are fa-

miliar in the operations of projection, several auxiliary planes are em-
ployed in perspective, and particularly the four following:

N

0

 

Fig. 3.

1. The horizontal plane AB (fig. 3), on which the spectator and the
objects viewed are supposed to stand, for convenience supposed perfectly
level, is termed the ground plane.

2. The plane M N, which has been considered as a transparent plane
placed in front of the spectator, on which the objects are delineated, is
called the plane of projection or the plane of the picture. The intersec-
tion M M of the first and second planes is called the line of projection, the
ground, or base line of tire picture. I

3. The plane E F passing horizontally through the eye of the spectator,
and cutting the plane of the picture at right angles, is called the horizontal
plane, and its intersection at D D with the plane of the picture is called the
liorizon line, the horizon of the picture, or simply the liorizon.

4. The plane S T passing vertically through the eye of the spectator,
and cutting each of the other planes at a right angle, is called the central
plane.

Point of view, or point of sight, is the point where the eye is supposed
to be placed to View the object, as at C, and is the vertex of the
optic cone. Its projection on the ground plane S is termed the station
point.

 

388 ' PERSPECTIVE DRAWING.

The projection of any point on the ground plane is called the seat of
that point.

Centre of view (commonly, though erroneously, called the point of
sight), is the point V where the central vertical line intersects the horizon
line ; a line drawn from this point to the eye would be in every way per-
pendicular to the plane of the picture.

Points of distance, are points on the horizontal line, as remote from
the centre of View as the eye.

Vanishing points, are points in a picture to which all lines converge
that in the original object are parallel to each other.

Parallel perspective—An object is said to be seen in parallel perspec-
tive when one of its sides is parallel to the plane of the picture.

Angular perspective—An object is said to be seen in angular perspec-
tive when none of its sides are parallel to the picture.

To find the perspective of points, as the points 772, 8, (fig. 4) in the ground

 

 

 

 

plane, the same letters designating similar planes and points as in fig.
3. From the point. m draw a line to the point of sight 0, and also to the
station point S, at the intersection of the line 722 S with the base line M S’,
erect a perpendicular cutting the line m C, the intersection 112’ will be the
perspective projection of the point 712, on the plane of the picture M V.
The point 8 being in the central plane, its projection must be in the in-
tersection of that plane by the plane of the picture, as the point 8' the
intersection of the central vertical line by the line a C‘. The point ,1,
being both in the central and horizontal plane, its projection in the plane
of the picture must be in the intersection of all three planes, or at the

DRAWING IN PERSPECTIVE. 389

point of view V. The point h being in the horizontal plane, its projec-
tion must be in the intersection of this plane with the plane of the picture,
or the intersection h’ of the horizon line by the line h C. The points h
and m being in the same vertical line, the points h’ and m’ must also be in
the same vertical line in the plane of the picture, and the position of 72/
might be determined by the intersection of h C by the perpendicular to
the base line at its intersection by m S.

Connect the points h v s m, and also their projected perspective points
h’ V 8' 9n’, and we find that when an original line is parallel or perpendic-
ular to the base of the picture, the perspective of that line will also be par-
allel or perpendicular to it.

Fig 5. Draw the diagonals M s and m S’, project as in preceding fig-
ure the points m and 8 into the plane of the picture, draw M m, M S’ and
S’ m’; now since m and M are the extremities of a line perpendicular to

w .Z—erwe- Ct

’Lg/ D

 

 

 

the plane of the picture, the line m’ M must be the projection of this line
on the plane of the pieture, and if this line be extended it will pass through
V, which may be demonstrated of all lines perpendicular to the plane of
the picture; hence the perspective direction of lines perpendicular to the
picture is to the centre of view.

If the line 777 ’S’ be extended, it will pass through the point D, and if
M s be extended it will pass through a point in the line of the horizon at
a distance from V equal to V D ; by construction D V has been made equal
to V C, and as this demonstration is applicable to other similar lines, and
since M msS’ is a square; hence the perspective direction of all lines,

e. as:

390 DRAWING IN PERSPECTIVE.

making an angle of 45° with the plane of the picture, is towards the point
of distance. .

Having thus illustrated the rules of parallel perspective, we now pro-
ceed to apply them to the drawing of a square and cube, Plate I. The
same letters are employed in similar position as in preceding figures.

It is necessary to premise that the student should draw these examples
at least three times the size of those in the plate.

Let A and 13 (fig 1,) represent the plan, or situation upon the ground,
of two squares, of which a perspective representation is required. First
draw the line M M, which represents the base line of the picture; make' S
the station point or place of the observer, and draw lines or rays from all
visible angles of the squares, to S; then draw the lines S M. parallel to the
diagonal lines of the squares. Now draw M’ M’ parallel to M represent-
ing the base line of the picture in elevation; then draw 8’ V, the vertical
line immediately opposite the eye; let the distance, 8’ V, be the height of
the eye from the ground, and draw D D the horizontal line;V being the
centre of View; let fall perpendicular lines from the angles a and b of the

plan of the square A, and also from the point e, where the ray from the _ .

angle einterseets the base line, M M, and from a’ and h’, where a a’ and h b'
intersect the base or ground line M’ M’, draw lines to the centre of view,
V; and e’ where the perpendicular line from c intersects the line h'V,“will
give the apparent or perspective width of the side be; from e’ draw a
line parallel to a’ Z)’, and the perspective representation of the near-
est square A, is complete. In order to prove the accuracy of this per-
formance, it is necessary to try if the diagonal lines, a’ c’, and 6’] ’, incline
respectively to the points of distance, D D, on the horizontal line: if so, it
is correct. The square B is drawn in precisely the same manner, and will
be easily understood by observing the example.

The plans of the two cubes 0 and D, are the same as the plans of the
squares A and B. As neither of these cubes appears to touch the plane
of the picture M M, it will be necessary to imagine the sides 1 g, and h h,
to be continued until they do so ; now draw down perpendicular lines from
where the continuations of these sides intersect the base line, and set off
on them from the line M’ M’, the height of the cube, as 1—2 which
is the same as the width, and complete the square shown by the dotted
lines: from all four angles of this square draw lines to the centre of view ‘
-—this will give the representation of four lines at right angles with the
picture carried on as far as it would be possible to see them; then it only
remains to cut off the required perspective widths of the cubes by the per-
pendicular lines from the intersection of the visual rays with the plane of

 

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PERSPECTIVE DRAWING. 391

the picture: the completion of this problem will be very easy, if the
drawing of the squares is well understood.

In such simple objects as these it will not be necessary to draw a plan ;
when one side is parallel to the picture, and dimensions are known. In
fig. 2, the same objects as those in fig. 1 are drawn without a plan thus :—

Draw the ground line M M, then the vertical line S’ V, and the horizon-
tal line D D, at the height of the eye; making D D the same distance on
each side of V, that the eye is from the transparent plane; for drawing
the squares mark off from S’ to 6’, on the ground line, the distance that
the square is on one side of the Observer; let 6’ a’ be the length of one
side of the square; from b’ and a’ draw lines to V, which represent the
sides of the square carried on indefinitely; to cut off the required per—
spective width of the side 6’ 6’ of the square, lay off the width, a” 1/, from
b' to 1), then draw from 19 to D on the left, and the point (5’ where the line
D p intersects ‘b’ V, will give the apparent width required; then draw f 6’
parallel to a’b’, and the square is complete: this may be proved in the
same way as in fig. 1. The further square may be obtained in a similar
manner, setting off the distance between the squares from 29 to g, and the
width of the square beyond that, and drawing lines to D as before:
some of the lines in this plate are not continued to the ground line,
in order to avoid confusion. Proceed with the cubes by the same rule.
Let 1, 2, 3, 4, be the size of one side of the cube if continued until touch-
ing the picture ; from these points draw rays to V: from 3 to t set off the
distance the cube is from the picture, and from t to 7‘, the Width of the
cube; draw from these points to D on the right, and their intersection of
the line 3 V in m, 0, will give the perspective width and position of that
side of the cube : draw lines perpendicular to the ground line from m and
0, and lines parallel to 4—2 from the angles of the cube, Z’,g’, m ,' then
draw the side 7; 72/, and the cube is complete. The operation of drawing
the other cube is similar, and easy to be understood.

From the drawing of a square in parallel perspective, we deduce rules
for the construction of a scale in perspective. Let D M M D, (fig 6,) be
the plane of the picture, the same letters of reference being used as in
preceding figures. From S’ lay off the distance 0 8’ equal to some unit
of measure, as may be most convenient; from 0 draw the diagonal to D
the point of distance; now draw 1 1’ parallel to the ground line M M,
again draw from 1’ the diagonal 1’ D, and lay off the parallel 2 2’, pro-
ceed in the same way with the diagonal 2D and the parallel 3 3/, and
extend the construction as far as may be necessary. It is evident 0 S’ 1 1/,
1’ 1 2 2’, 2’ 2 3 3’ are the perspective projectors of equal squares, and

392 _ PERSPECTIVE DRAWING.

therefore 0 S’, 1 1’, 2 2’ 3 3/, etc., and S’ 1, 1 2, 2 3, etc., are equal to each
other, and that if 0 S’ is set off to represent” any unit of measure, as one
foot, one yard, or ten feet, &c., each of these lines represents the same dis-

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—-l 11]

Fig. 6.

tance, the one being measures parallel to the'base line, the others perpen-
dicular to it. In making a perspective drawing a scale thus drawn will be
found very convenient ; but as in the centre of the picture it might inter-
fere with the construction lines of the object to be put in perspective, it is
better that the scale be transferred to the side of the picture a M 0, the di-
agonals to be laid off to a point to the right of D equal to the point of
distance.

The scales thus projected are for lines in the base or ground plane; for
lines perpendicular to thisplane the following construction is to be adopt-
ed; upon any point of the base line removed from S’, as a for instance,
erect a perpendicular, ad ,' on this line, lay off as many of the units 0 S’
as may be necessary; in this example three have been laid ofi“, that is, a d
:3 0 S’. From a and d draw lines to the centre of View, and extend the
parallels 1 1’, 2 2’, 3 3 / ; at the intersection of these lines with a V erect
perpendiculars. The portions comprehended between the lines a V and (Z
V will be the perspective representations of the line a (I, in planes at dis-
tances of 1, 2, 3, 0 S’ from the base line, and as b, c, d are laid off at inter-
vals equal to 0 S’, by drawing the lines 0 V and b V six equal squares are
constructed, of which the sides correspond to the unit of measure, 0 8’.

To determine t/ee Perspective Position of any/point in the Ground Plane.
Thus (fig. 7), to determine the position of the point p, which in plane
would be six feet distant from the plane of the picture, M M, and ten feet
from the central plane, to the left.

Lay off from S’, to the left, the distance a 8’, equal to six feet on the

PERSPECTIVE DRAWING. 393

scale adopted; draw the diagonal to the point of distance D, on the right;
at its intersection a’ with the vertical line V S’, draw a parallel to M M
the base line; lay off from S’, S’ 6 equal to ten feet, draw I) V; the inter-
section of this line I” with the parallel previously drawn, will be the posi~
tion of the point required.

 

 

 

 

 

 

 

 

 

 

D - 7’ 17
{1/ '
. /v~~
‘ / I! S V
M I) / a // 1.1
J
111 a' is" ' ill
Fig.7.

By a similar construction the position of any point in the ground plan
may be determined. It is not necessary that the distances should be ex—
pressed numerically; they may be shown on the plan and thence be trans-
ferred to the base line, and thrown into perspective by the diagonals and
parallels. As the intersections of the various lines of the outlines of ob—
jects are points, by projecting perspectively these points, and afterwards
connecting by lines, the perspective of any plane surface, on the ground
plane, may be shown.

If the point 1) were not in the ground plane, but in a position directly
above that already assumed, that is, the distances from the plane of the pic-
ture, and the central plane being the same, but its distances above the
ground plane were, say, five feet; then at b erect a perpendicular, and lay
off I; 5’ equal to five feet, connect 6’ V, at 1) erect another perpendicular,
and its intersection 19’ with the line (2’ V will be the position of the point
required.

Or the plane of the point [9’ might be assumed as the position of the
ground plane, M’ M’ becoming the base line, and laying off from S”, S” a”
and S"b’——equal respectively to six and ten feet; drawing the diagonal
a” I) and b’ V and the parallel as before, the point 19’ will be determined.

To draw an Octagon in Parallel Perspectloe.—Let A (fig. 8) represent
the plan of an octagon. Draw M M, S’ V, and D D, as before ; from the
points M, a, b, c, draw rays to V. Set off on M M from c to the right the
distances 06, 0d, cf, from which draw diagonals to D on the left, and
at their intersection with the ray 0 V', draw parallels e’g’, cl’ lt’, k’ l’, to
the base line; these points will correspond to the angles on the-plan. Now

394 PERSPECTIVE DRAWING.

connect the angles on the perspective View, in the proper succession, and
the perspective projection is complete.

It will be observed, that in this construction the plan has been placed
forward of the plane of the picture, contrary to the position it should oc-
cupy, which should be the same relative position back of this plane; but
it will be found much simpler in construction than if it were placed as in
Plate 1-,, and the points were all projected to the base line ; it is, of course,
equally correct in its perspective projection.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

{3 _____

To draw a Circle in Parallel Perspective.—-—Let 0, (fig. 8) represent the
plan of a circle, round which let the square a, e c m be described. two of
its sides being parallel to the base line M M; draw diagonals across the
square, and where these intersect the circumference of the circle draw the
lines l2}: 'and (l 9 parallel to the base line, and the lines o n and 1) g at right
angles thereto. Draw also the lines fl and clay at right angles to each
other through the centre of the circle, project the points a, 0, l, 1), m, to
the base and draw rays to V; set off from a! to the left the distances a" a,
a’l), a’c, a’d, a’e, and draw diagonals to the point of distance D on the
right; at their intersection with the line a’ V draw horizontal lines, or
parallels to the base, and there will be projected in perspective the square
accm, with'all the lines of parallels and perpendiculars; connect the
intersections corresponding to the points 0, 72, f, g, It, Ir, l, r, and we

PERSPECTIVE DRAWING. 395

have the perspective projection of the required circle, which will be an
ellipse.

To erect upon the octagonal base A an octagonal pillar or tower. This
construction resolves itself into simply constructing another octagon on an
upper plane, and connecting the visible angles by perpendiculars, or per-
pendiculars may be erected at the points M, a, b, 0, and the heights of the
tower laid 05 upon them, and from these extremities rays drawn to the
centre of View; the intersection of these rays by perpendiculars from the
angles of the octagon beneath will determine the projection of the upper
surface of I the pillar; represent in full lines all visible outlines, and the
projection is complete. A

In the same manner a pillar may be erected on the circular base. If
the pillars be inclined, the first method of projecting the upper outline on
a plane assumed at the height of the pillar, must be adopted.

T 0 draw a Pyramid in Parallel Perspective—Let A (fig. 9) be the
plan of a pyramid, the diagonal lines represent the angles, and their in-
tersection the vertex; project the plan as in previous examples of squares.
Draw diagonal lines from M to b, and a to 0, their intersection gives the
perspective centre of the square; upon this point raise a perpendicular
line which is the axis of the pyramid; draw a perpendicular line 6 f, in

 

 

 

 

 

 

 

 

 

 

 

 

F ig. 9.

the centre of the line M a, upon which set up the height of the pyramid
r: f ; from f draw a line to V,'and its intersection of the axis of the pyramid
at cl will give the perspective height; complete the figure by drawing
lines from (Z, the apex, to M, a, b, the three visible angles. The other
two pyramids are drawn in a similar manner, by setting their distances
from the plane of the picture off from a, on the ground line to the right,
and drawing diagonals to the point of distance on the left.

3‘; .

396 PERSPECTIVE DRAWING.

T 0 draw a Gone in Parallel Perspectives—Let B (fig. 9) represent the I
plan of a cone, apply the same lines of construction as to 0 (fig. 8); and
draw the perspective View of a circle, upon the perspective centre of
which draw a perpendicular line, a, l) ,' on the centre of the line cl 6, raise
a perpendicular, upon which set up the height of the cone, from the
ground line to c ,' from c draw a ray to V, and the point where this line
intersects the axis of the cone a l), in I), will give the perspective height
of the axis; from I) draw lines to f and g, and the figure is complete.

To draw the prism, C, which consists of two triangular ends and three
rectangular sides, place the length of the side a M upon the ground line,
and draw lines to V; mark off the width of one end from a to the left upon
the ground line, and draw to the point of distance on the right, which
gives the perspective width, a cl ,' find the perspective centre f of the
side in the same way, and from f and (Z draw horizontal lines until they
intersect the line from M V; upon f and g draw perpendiculars; set up
the height a b, of the end of the prism, and from I) draw a line to V, and
the point where it intersects f0, in c, will give the perspective height of
the end of the figure ; from 0 draw 08, parallel to a M, from c draw 0d,
and c a, and the visible end is complete ; the other end is dotted in to show
the process only.

E is the representation of a cylinder, with one end towards the spec-
tator; its projection will be easily understood by examination.

To draw a Square and Cube in Angular Perspective. Plate II. Let
A (fig. 1) be the plan of the square, and B the plan of the cube, M M the
base or ground line, and S the station point. Draw M' M’, and D D' par-
allel to M M, the one being the ground line and the other the horizon of
the plane of the picture; project the point (I on M M to (l', on M’ M’. It
has been shown in parallel perspective that the vanishing points of diago-
nals of squares lie in the points of distance; if through the station point
S, in any of the preceding figures, lines be drawn parallel to the diagonals,
they will intersect the base lines at distances from the central plane equal
to the points of distance. In like manner to find the vanishing points of
lines in the ground planes, or in planes parallel to the ground plane. in-
clined to the plane of the picture, through the station point S draw lines par-
allel to the inclined lines, and project their intersection with the base line
to the horizon of the picture; thus, in the present example draw S M, S M
parallel to a (l, c 72, and to (l c, la y ,' project their intersections M. M, with
the base line to D, D', the horizon of the picture, and I), D'. will be the
vanishing points of all lines parallel to)a d and d 0. Draw (l, D and (l' D',
the perspective projection of (l a will lie in the former of these lines and

PLATE II.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PERSPECTIVE DRAWING. 397

do in the latter. To determine the perspective position of the points a
and c, or the length of these lines, draw the rays a S and c S, project their
intersection with the base M M, upon the lines d' D and d’ D’, and their in-
tersections a’, 0' will be the perspective projection of the points a and c.
To complete the projection of the square, draw the lines a’ D’ and c' D,
their intersection will be the perspective projection of the point 6, and the
square is complete. To prove the construction, draw the ray 6 S and pro-
ject its intersection with the base M M, and if the construction be correct
it will fall upon the point 6’.

As the cube is placed at some distance from the plane of the picture,
it will be necessary to continue either 6 It or g 71, or both, till they intersect
the base line M M at n and m ,' drop perpendiculars or project these points
upon M M' at n' and m’ ,' on these perpendiculars set up the height of the
cube m’ o and 72/ s, draw the lines m’ D’, 0’ D’ and n’ D, s D ; connect the
intersections I2" and It” ,' draw the rays S e and S g, and project their inter-
sections with M M, to g’ 6'; draw the lines 6”D’ and g" D; if the construction
be correct, the projection of the intersection of the ray Sf with the base
will fall upon f’, and of the ray S IL will fall upon It” and 70’.

T 0 Solve the Same Problem by a Difermt Construction—Let A and
B, (fig. 1,) be as before the plans of the square and of the cube; to pro-
ject them perspectively on the plane of the picture M D D’ M, (fig. 2).

From the point M and M, (fig. 1,) set off distances equal, to M S, M S,
to p and 19’ ,' project these points upon D D/ fig. 2, the point 29, fig 2, will be
that from which any number of parts may be laid off on lines vanishing in
D’; the point 1; will be the corresponding point for lines vanishing in D.
These points may be called the points of division. In parallel perspec~
tive the points of distance were the points of division, the one for the
other. To illustrate their application in the present example, project the
point d, (fig. 1,) to d’ (fig. 2,) draw 62’ D and d’ D’, from d’ on either lay
off a distance d’ 2', (Z’ L" equal to the side of the square a (l. N OW since 10
is the division point of lines vanishing in D from i, draw the line 2'19, and
its intersection with d/ D cuts off a line d'a’ equal perspectively to the
line (1’ 2' or a (I measured on the base line. Again since 19’ is the division
point of lines vanishing in D/, the line kp’ cuts off on d’ D’, a line d” 0’ equal
perspectively to the line d’ In, or a (1 measured on the base: having a’ d’ c,
the square is completed by drawing the lines c’b’ towards D, and co’b/
towards D'.

To construct the cube, project the point m, (fig. 1,) to m’, (fig. 2) ; lay
off on the perpendicular forming the projection, the height m’ 0’ of the
cube; draw the lines 727/ D’ and 0’ D’. Lay oflP the distance m’ 7' equal to

398 PERSPECTIVE DRAWING.

m It, (fig. 1,) and draw the line r 19’, its intersection with m’ D’ will cut of?
m’ li’, equal to m it, (fig. 1,) and establish the angle lb of the cube. From
7* lay off rs, equalto li 9, (fig. 1,) draw sp’, and its intersection with m’ D’
establishes the angle 9’. From li’ draw a line vanishing in D. Through
li’ extend a line p li’ to t, from t lay off to the left ta, equal to the side of
the cube lb 6 ; draw a p, and its intersection with the line li’ D, establishes
a third point e of the cube. Upon these points li’ g’ c’ erect perpendicu-
lars; those upon li’ and 9’ will, by their intersection with o’ D, determine
k” 9”. Draw lb” D’, its intersection with the perpendicular at 0 determines
6”. Draw g” D’ and e” D to their intersection, and the cube is complete.

To Draw the Perspective Projection of an Octagonal Pillar in Angu-
lar Perspcctioa—Plate III. Let A, (fig. 1,) be the plan of the pillar. En-
close it by a square. Let M M be the base line, and S the station point ; de-
termine the position of the vanishing points for the sides of the square as in
Plate IL, and project the square upon the plane of the picture M D D’ M’
by either of the methods already explained. These lines of construction are
omitted, as on the necessarily small diagrams they would confuse the stu-
dent; but in drawing these examples to the scale recommended, they might
be retained. From the angles of the octagon visible to the spectator
draw rays to the station point S project their intersection with the base
line M M, to the perspective square, (fig. 2,) which will thus determine on
the sides of the square the positions of the points a’, l;’, o’, (l’, 6’, correspond-
ing to the visible angles of the octagon ; connect these points by lines. To
construct the pillar upon this base, upon the perpendicular let fall from
the corner f of the square upon M M’ at f set oil the height of the pillar;
from this point f draw lines to the vanishing points D, D’, and construct
three sides of an upper square similar to the lower one. The lines of this
square will determine the length of the sides of the tower, which are the
perpendiculars let fall upon a’ l)’ c' c" c’.

To Construct a Circular Pillar in Angular Pcrspccz‘ivc.—Plate III.
Let B, (fig. 1,) be the plan of the base ; enclose it with a square whose sides
are parallel respectively to S M and S M; project this square upon the plane
of the picture, (fig. 2,) divide the plan into four equal squares by lines par-
allel to the sides ; draw rays through the points lo and i, and project their
intersection with M M upon the perspective square. From the points la" and
i’ thus formed, draw lines to vanishing points D’ and D, and the persPec-
tive square is divided similarly to the original, and there are four points of
the circle established: through these draw the perspective of the. circle.
By the division of the base into smaller squares more points of the curve
might be determined, but for the present purpose they are unnecessary.

 

 

 

Fig. 1.

 

 

 

 

 

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PERSPECTIVE DRAWING. _ 399

To determine the outline of the pillar, draw from S rays tangent to the
sides of the plan at k and i, the perpendiculars let fall from their intersec.
tion with M M will be the outline of, the cylinder. To cut them off to the
proper height, and to determine the top of the cylinder, upon the perpen-
dicular let fall upon i, set 013? the height of the cylinder l’ l”, and upon this
plane project the square as before, and draw in through the points thus
determined the outline of the curve. As a still further elucidation of the
principle of projection, an enlarged cap is represented on the pillar, of
which the circumscribing circle (fig. 1,) is the plan. In this by extending
the central lines of the square, both in plan and perspective, we are en-
abled to project readily eight points in the larger circle through which
the curve may be drawn.

T 0 Draw an Octagonal Pyramid in Angular Pcrspcctine.—I’latc III.
Let f, (fig. 1,) be the base of the pyramid; project upon the plane of the
picture, (fig. 3,) the visible angles of the base, as in the case of the pillar.
Through the centre of the plan draw a line parallel to one of the sides and
intersecting M M at m ,' from this point let fall a perpendicular to m’ on
M M’, (fig. 3,); on this perpendicular set off the height of the pyramid m'
0 from m’ and draw lines to D’. From the centre of the plan draw a ray
to S, and project its intersection with M M, upon the line 0 D’, its intersec-
tion 0' with this line will be the apex of the pyramid: from this point
draw lines to the angles of the base already projected, and the pyramid is
complete.

To Draw a Cone in Angular Perspectiva—Plate III. Let the inner
circle B, (fig. 1,) be the base of the cone project its visible outline to fig.
3, as in case of the cylinder. To determine its height extend one of the
diameters of the plan to the base line at p ; from this point let fall a per-
pendicular to j)’ on M M’, and set off upon it 19’ g, the height of the cone;
from j)’ and g draw lines to the vanishing point D’. From the centre of
the plan, (fig. 1,) draw rays to S, and project its intersection with M M,
upon 7" on the line g D’, and 7" will be the apex of the cone: connect the
apex with the extremities of the perspective of the base, and the projec-
tion of the cone is complete.

To Draw the Elevation (If a Building in Angular Perspectice.—Plate
IV. For example, take the school—house, Plate XXIX. of architecture.
Plot so much of the plan of the building at it as may be seen from the po-
sition of the spectator at S. Draw a base line, and through the station
point draw parallels to the sides of the building cutting the base as at M
M : draw M M’ for a base, and D D’ for the horizontal line of the picture.
Project M and M to D and D’, for the vanishing points, the one of the

400 PERSPECTIVE DRAWING.

lines parallel to a c, the other to a 6 ,' extend a c, a 6; project d. e, to
d’, e’, and on d’ (i set off the height of the eaves (l’ o, and of the ridge (1
n ,' from (Z’, 0 and n draw lines to D’, and from 0 to D, draw rays from c
and I) to S’ , and project their intersection with the base to the vanishing
lines just drawn. To find the perspective of the ridge draw a ray from
the centre of a I), and project its intersection with the base to r on the line
n D’, the point is the apex of the gable, the line 7° D will be the perspective
of the ridge; to determine its length erect a perpendicular at the intersec-
tion of tD’ and s D, draw the sloping lines of the roof, and the outline of
the building is complete. The filling in of the details will be readily under-
stood ; it will only be necessary to keep in mind, that all lines parallel to
a 5 must meet in D’, those to a 0 in D: all measures laid off on any lines
of the plan must be connected with the point of sight S, and their inter-
sections with the base projected. All vertical heights must be laid off on
the line 03’ d, and referred to the proper position by lines to D or D’, as the
case may be.

As an example of the other method of constructing this same problem,
let the scholar lay off to the double of the present scale the plane of the
picture M D D’ M’, and the division points 10’ and 1), and without drawing
plan or elevation take the dimensions from Plate XXIX. of architecture.

To Draw an Arched Bridge in Angular Pcrqwcz‘ilat—Plate V. Let
A and 13, (fig. 1,) be the plans of the piers; on the line a It. one of the sides
of the bridge, lay down the curve of the arch as it would appear in eleva-
tion, in this example an ellipse. Divide the width of the arch as at b. c.
(I. af. 9. 12., carry up lines perpendicular to I) it until they intersect the
curve of the arch, and through these points, draw lines parallel to b it as 1:.
1.722. ,' let 0 r be the height of the parapet of the bridge above the spring of
the arch. Through the station point draw lines parallel to the side a It
and end a a of the bridge, till they intersect the assumed base line M )1:
project these intersections to the horizon line of the picture for the vanish-
ing points D, D’ of perspective lines parallel to a It and (1/ a. Let- fall a
perpendicular from a to a’, and on this perpendicular set (at from a' the
heights 8 l‘, 8 7, S m, and 82‘,“ from a' and I" draw lines to D and D', and from
the points 222’, Z’, Z" to D'. Draw rays from the points (I. 1). (1d. (hf. g. b.
to the station point S, and project their intersection with the base lines to
the perspective line (1’ D’ as in previous examples: the intersection of the
lines I" D’, Z" D', m’ D’ by the perpendiculars thus projected, will establish
the points of the curve of the arch on the side nearest. the spectator. To
determine the position of the opposite side of the arch, from a”, the per-
spective width of the bridge, draw a” D', and from 727 draw lines to D;

i

400

PLATE IV.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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PERSPECTIVE DRA‘VING. 4:0 1

the line h’p’ will be the perspective width of the pier ; draw 76’ D 3 and from
k", 76" D; from g" the intersection of the curve of the arch by the perpen—
dicular to g’, draw 9” D, the intersection with 7v” D' will be one point in
the curve of the arch on the opposite side of the bridge: in the same way,
from any point in the nearer arc draw lines to D, and the intersection
with lines in the same planes on the opposite side of the bridge, will fur-
nish points for the further arch: all below the first only will be visible to
the spectator.

To Draw in Parallel Perspective tire Interior of .a Emma—Plate V.
We propose to construct this by scale without laying down the plan.
Draw the horizon line D V D', and the base M M’, making D and D’ the‘
point of distance. Let the room be 20 feet wide, 14 feet high, and 12 feet
deep; on the base M M’, lay off the rectangle of the section in our figure
on a scale of 8 feet to the inch, 20 feet x 14 feet. From the four corners
draw lines to the centre of view V; from S’ lay off to the right or left on
M M' 12 feet, and through this point draw lines to D’ or D as the case may
be: through the point of intersection a’ of this line with S’ V draw a line
parallel to M M’ ; at the intersections of this line with M V and M’ V erect
a perpendicular, cutting .the vanishing lines of the upper angle of the
room at d- and e ,' connect (2 e and the perspective of the room is complete.
To draw the aperture for a door or window on the side, measure off from
S' the distance of the near side from the plane of the picture, and in addi-
tion thereto the width of the aperture ; from these two points draw lines to
the proper point of distance, and at their intersection with S’ V, draw par-
allels to M M’, cutting the lower angles of the room, and erect perpendic-
ulars, the height of which will be determined by a line drawn from f, the
height of the window above the floor measured on M D. Should the win- .
(low be recessed, the farther jalnb will be visible ; extend the farther par-
allel to M M’, and cut it by a line gV; M g being the depth of the recess,
the rest of the construction may be easily understood bypinspection‘of the
figure. At the extremity of the apartment a door is represented half open,
hence as the plane of the door is at right angles to the plane of the picture,
the top and bottom lines will meet in the point of View ; if the door were
open at an angle of 45°, these lines would meet in the points of distance;
if at any other angle, the vanishing points would have to be determined ,
by constructing a plan, drawing a line parallel to the side of the door.
through the station point, and projecting it upon the horizon line. The
chair in the middle of the room is placed diagonally, and the table parallel
to the plane of the picture; their projection is simple.

T 0' Draw in I ‘erspegtgz'ee a Flay/at of Stairs.—I’late VI. Lay off the

402 PERSPECTIVE DRAWING.

base line, horizon, centre of View, and point of distance of the picture,
construct the solid at b o d, efg A, containing the stairs, and in the required
position in the plane of the picture, divide the rise a 0 into equal parts ac-
cording to the number of stairs, four for instance; divide perspectively the
line a, 6 into the‘ same number of parts; at the points of division of this
latter erect perpendiculars, and through the former draw lines to the cen-
tre of view; one will form the rise and the other the tread of the steps.
Fr om the top of the first step to the top of the upper continue a line a d,
till it meets the perpendicular S’ V p1010nged 1n '1); this line W ill be the
inclination or pitch of the stair; it through the top of the step at the other
-extre1nity a similar line be drawn, it will meet the central perpendicular
at the same point 7), and Will define the length of the lines of nosing of
the steps, and the other lines may be completed. As the pitch lines of both
sides of the stairs meet the central vertical in the same point, in like man-
1101 ’U will be the vanishing point of all lines having a simila1 inclination
to the plane of the pictu1e. The p1ojection of the other flight of stairs
will be easily unde1stood from the lines of construction perpendicular to
the base line or pa1 allel the1et1, 1y 111g in planes.

To Find the Prflectzon of Olyects m Me U ate; .—Plate VI. Let B be
a cube suspended above the water; we find the reflection of the point a,
but letting fall a perpendicular from it, and setting off the distance a’ to
below the plane of the water equal to the line a to above this line ; the line
zef’ will also be equal to the line 'wf,‘ find in the same way the points 3/
and 6’, through these points construct perspectively a cube in this lower
plane, and we have the reflection of the cube above.

To find the reflection of the square pillar D removed from the shore :
suppose the plane ot: the water extended beneath the pillar, and proceed
as in the previous example. i

It will be observed that those lines of an object which meet in the cen-
tre of View V, in the original; their corresponding reflected lines will con—
verge to the same point. If the originals converge ~to the points of distance,
the reflected ones will do the same. To find the reflection of any inclined
line, find the reflection of the rectangle of which it is the diagonal, if the
plane of the rectangle is perpendicular to the plane of the picture; if the
line is inclined in both directions enclose it in a parallelopided and project
the reflection of the solid.

To find 25/16 Perspective Projection of Shadows. Plate VII. Let the

construction points and lines of the picture be plotted. Let A be the per-

 

spective projection of a cube placed against another block, of which the
face is parallel to the plane of the picture: to find the shadow upon the

402

1.

PLATE

 

 

 

EH11 in. é.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PERSPECTIVE DRAWING. 403

block and upon the ground plane, {Supposing the light to come into the
picture from the upper left-hand corner and at an angle of 45°. Since the
angle of light is the diagonal of a cube, construct another cube similar to
A, and adjacent to the face cl 0 g ,' draw the diagonal b 76, it will be the
direction of the rays of light, and It will be the shadow of b ,' connect f It:
and c k, f I: must be the shadow of the line 6/, and 0 l: of 6 c ,' the one
upon the horizontal plane and the other in a vertical one: the former will
have its direction, being a diagonal, toward the point of distance D’, the
other being a diagonal in a plane, parallel to that of the picture, will be
always projected upon this plane in a parallel direction.

Let B be a cube similar to A; to find its projection upon a horizontal
plane, the shadow of the point 5’ may be determined as in the preceding
example, but the shadow of the point 0’, instead of falling upon a plane
parallel to the picture, falls upon a horizontal one; its position must be de-
termined as we did before by 6. Construct the cube and draw the diag-
onal c’ l ,' in the same way determine the point 777/ the shadow of d’ ,' con-
meet 0 lc’ lm n, and we have the shadow of the cube in perspective on a
horizontal plane.

On examination of these projected shadows, it will he found that as the
rays of light fall in a parallel direction to the diagonal of the cube, the
vanishing point of these rays will be in one point V’ on the line D’ M’
prolonged, at a distance below D’ equal VD’; and since the shadows of
vertical lines upon a horizontal plane are always directed towards the point
of sight, the extent of the shadow of a vertical line may be determined
by the intersection of the shadow of the ground point of the line by the
line of light, from the other extremity. Thus, the point 76, cube A, is the
intersection off D’ by I) V’; the points 16’, l, m are the intersections of c D’,
o D’, 71’ D’ by 6’ V’, c’ V'by cl’ V.’ Similarly on planes parallel to that of the
picture, k, cube A is intersection of the diagonal C- ]v, by the ray of light 6 V’.

Applying this rule to the frame 0, from 7’, 8,19, draw lines to D’ from 7",
s’,p’, draw rays to V'; their intersections define the outline of the shadow of
the post. To draw the shadow of the projection, the shadow upon the post
from 25 will follow the direction of the diagonal 0 In. Project u and 7) upon
the ground plane at u' and v’ ,' from t” u’ 7/ and I) draw lines to D'; from t,
u, v, w and {I} draw rays to V’, and the intersection of these lines with their
corresponding lines from their bases will give the outline required; as '0 and w
are on the same perpendicular, their rays will intersect the same line 7/ V’.

“Tith reference to the intensity of “ shade and shadow ” and the neces-
sary manipulation to produce the required effect, the reader is referred to
the article on this subject.

40$ PERSPECTIVE DRAWING.

In treating of Perspective it has been considered not in an artistic
point, as enabling a person to draw from nature, but rather as a useful art to
assist the architect or engineer to complete his designs, by exhibiting them
in aview such as they would have to the eye of a spectator when con-
structed. In our examples, owing to size of the page, we have been limit-
ed in the scale of the figures, and in the distance of the point of view, or
distance of the eye from the plane of the picture, and as it was unimportant
to the mathematical demonstration, few of the figures extend above the
line of the horizon. In these particular points it is unnecessary that the
examples should be copied. The most agreeable perspective representa-
tions are generally considered to be produced by fixing the angle of vision
M S M’, at from 45 to 50°, and the distance of the horizon above the
ground line at about one-third the height of the picture.

Linear perspective is more adapted to the representation of edifices,
bridges, interiors, &c., than to that of machinery; it belongs, therefore,
rather to the architect than to the engineer or the mechanic; for the pur-
poses of the latter we would recommend Isometrical Perspective, uniting
accuracy of measures with graphic perspective representation.

ISOMETRIOAL DRAWING. 405

ISOMETRICAL DRAWING.

PROFESSOR FARISH, of Cambridge, has given the term Isometrical Per-
spective to a particular projection which represents a cube, as in fig. 1.
The words imply that the measure of the representations of the lines
forming the sides'of each face are equal.

The principle of isometric representation con-
sists in selecting for the plane of the projection,
one equally inclined to three principal axes, at
right angles to each other, so that all straight lines
coincident with or parallel to these axes, are drawn
in projection to the same scale. The axes are
called isometric axes, and all lines parallel to them

 

are called isometric lines. The planes containing
the isometric axes are isometric planes; the point “3- 1‘

in the object projected, assumed as the origin of the axes, is called the reg-
ulating point.

To draw the isometrical projection of a cube, (fig. 2,) draw the hori-
zontal line A B indefinitely ; at the point D erect the perpendicular D 0,
equal to one side of the cube required; through D draw the line D 7) and
Df to the right and left, makinr ng B and I) D A each equal an angle of
. 3°O . Consequently the angles F Df and F D?) are each equal to 60°.
Make D b and D f each equal to the side of the cube, and at I) and f erect
perpendiculars, making bar, and f 6 each equal to the side of the cube;
connect F a and F e and draw 6 9 parallel to aF, and a 9 parallel to F e,
and we obtain the projection of the cube.

If from the point F, with a radius F D, a circle be described, and com-
mencing at the point D radii be laid off around the circumference, forming
a regular inscribed hexagon, and the points D a e be connected with the

 

406 ISOMETRICAL DRAWING.

centre of the circle F, we have an isometrical representation of a cube.
The point D is called the regulating point.

If a cube be projected according to the principles of isometrical per-
spective, in a similar manner as we have constructed one according to the
rules of linear perspective, the length of the isometrical lines would be to
the original lines as .8164 to 1, but since the value of isometrical perspec-
tive as a practical art lies in the applicability of common and known
scales to the isometric lines, in our constructions we have not thought it
necessary to exemplify the principles of the projection, but have drawn our
figures Without any reference to what would be the comparative size of the
original and of the projection, transferring measures directly from plans and
elevations in orthographic projections, to those in isometry. It will be ob-
served that the isometric scale adopted applies only to isometric lines, as
F D, F a and F e or lines parallel thereto; the diagonals which are abso-
lutely equal to each other, and longer than the sides of the cube, are the
one less,the other greater ; the minor axis being unity, the isometrical lines
and the major axis are to each other as, 1. 4/2. 4/3.

Understanding the isometrical projection of a cube, any surface or
solid may be similarly constructed, since it is easy to suppose acube suffi-
ciently large to contain within it the whole of the model intended to be
represented, and as hereafter will be farther illustrated, the position of any
point on or within the cube, the direction of any line or the inclination of
any plane to which it may be cut, can be easily ascertained and repre-
sented.

In figs. 1 and 2 one face of the
cube appears horizontal, and the
other two faces appear vertical. If

now the figures be inverted, that

 

which before appeared to be the top
of the object, will 110w appear to
be its under side.

‘ The angle of the cube formed
by the three radii meeting in the
centre of the hexagon, may be
made to appear either an internal or

external angle; in the one case the

 

 

faces representing the interior, and

Fir. 2.

in the other the exterior of a cube.

’31.
{132

Cl)
”03

5+.

, 5, illustrate the application of isometrical drawing to simple

ISOMETRICAL DRAWING. , 407

combinations of the cube and parallelopipedon. The mode of construction

 

   
  

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will“

    

Figs. 8, 4, 5.

of these figures will be easily understood by inspection, as they contain no
lines except isometrical ones.
To draw Angles to the Boundary Lines of am Isometrical Cube.

   

4U 30 20 10
Fig. 6, 7.

Draw a square 0 (fig. 6,) whose sides are equal to those of the isome—
trical cube A, and from any of its angles describe a quadrant, which di-
vide into 90°, and draw radii through the divisions meeting the sides of
the square. These will then form a scale to be applied to the faces of the
cube; thus on D E, or any other, by making the same divisions along their
respective edges.

As the figure has twelve isometrical sides, and the scale of tangents

~$08 ISOMETRICAL DRAWING.

may be applied two ways to each, it can be applied therefore twenty-four
ways in all. We thus have a simple means of drawing, on the isometrical
faces of the cube, lines, forming any angles with their boundaries.

Figs. 1, 2, 3, 4:, 5, 6, Plate 1., show the section of a cube by single
planes, at various inclinations to the faces of the cubes. Figs. 7, 8 are the
same cube, but turned round, with pieces cut out .of it. Fig. 9 is a cube
cut by two planes forming the projection of a roof. Fig. 10 is a cube with
all of the angles cut off by planes, so as to leave each face an octagon.
Fig. 11, represents the angles cut off by planes perpendicular to the base
of the cube, forming thereby a regular octagonal cylinder. By drawing
lines from each of the angles of an octagonal base to the centre point of
the upper face of the cube, we have the isometrical representation of an
octagonal prism.

As the lines of construction have all been retained in these figures,
they will be easily understood and copied, and are suflicient illustrations
of the method of representing any solid by enclosing it in a cube.

we have now to consider the application of this species of projection
to curved lines.

 

 

 

 

 

Let A B (fig. 8,) be the side of a cube with a circle inscribed; and
suppose all the faces of the cube to have similarly inscribed circles. Draw
the diagonals A B, C D, and at their intersection with the circumference
lines parallel to A C, B D. New draw the isometrical projection of the
cube, ttig. 9,) and lay out on the several faces the diagonals and the par-
allels; the projection of the circle will be an ellipse, of which the diago-
nals being the axes, their extremities are defined by their intersections f 6,

PLATE I.

 

 

 

 

 

 

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ISOMETRICAL DRAWING. 409
t
e 5, a 2, I) 1, d 3, c 4, by the parallels; having thus the major and minor
axis, construct the ellipse by the trammel, or since the curve is tangent at
the centre of the sides, we have eight points in the curve; it may be put
in by sweeps or by the hand.

 

 

To Divide Me Circumference qf a Circle—First method. On the
centre of the line A B erect a perpendicular O D, making it equal to O A
or C B; then from D, with any radius, describe an arc and divide it in
the ratio required, and draw the divisions radii from D meeting A B;
then from the isometric centre of the circle draw radii from the divisions
on A B, cutting the circumference in the points required.

Second method. On the major axis of the ellipse describe a semi-cir-
cle, and divide it in the manner required. Through the points of division
draw lines perpendicular to A E, which will divide the circumference of
the ellipse in the same ratio. On the right hand of the figure both
methods are shown in combination, and the intersection of the lines give
the points in the ellipse.

The drawing of the mitre-wheel, (fig. 1, Plate II.) is given as an illus-
tration and example for practice.

Fig 2, Plate 11., represents a pillow block.

Plate III. represents various combinations of framing met with in ar-
chitectural and mechanical constructions. ‘

Plate IV. represents a culvert such as were constructed beneath the
Croton Aqueduct.

The Plates and figures thus represented show the applicability of this

410 ISOMETRICAL DRAWING.

method of projection to various examples. Many others might be intro-
duced, but the principles of this perspective are so easy and intelligible,
that to multiply examples would be entirely unnecessary.

Isometrical projection is especially valuable to the mechanical draughts-
man, embracing as it does the applicability of a scale With pictorial rep-
resentation. For drawings for the Patent Office it is especially desirable,
yet when pictorial representation alone is intended, it is not as truthful as

a drawing in linear perspective. Its office is rather practical than orna-
mental.

 

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