0.5

1 cm

PERSPECTIVE.

99

the picture in the line A d perpendicular to its base. A d

produced would be the trace of a plane perpendicular to

the original plane, which would necessarily pass through

the point of sight. The line A o' will also contain the

projection of the vertical triangle D O' O, and the inter-

section of the triangle d AO with the picture will be the

line A d, which is the perspective of A I), and it will tend

to the point of sight 0.

The straight line B K, which is the base of the triangle

K B 0, is not in a plane perpendicular to the original

plane, but inclined to it; for KB is beyond the central

plane, while O is in that plane. Consequently the pro-

jection of the inclined plane in which this triangle is

situated will be the triangle K B 0 on the original plane,

and its vertical projection in the picture will be B o'; and,

as the intersection B k is part of B o', this perspective is

also directed towards the point of sight. It can be shown

that this also would be the case with all other lines perpen-

dicular to the picture; and, therefore, it can be concluded,

Rule III.—That the perspectives of all lines perpen-

dicular to the picture pass through the point of sight.

Let A B (Fig. 193) be a straight line, making with B C,

the base of the picture, an angle of 45°. The perspective

of the line will be a B, which, being produced, would

meet the horizon in the point d', which will also be

the point of convergence of the perspectives of all lines

parallel to A B. It is easy to perceive that the original

line A B is the base of a scalene triangle ABE,

formed by that line and the rays A E, B E, and which

triangle has its base in the original plane, and its summit

in the eye of the spectator. It will be inclined to both

planes of projection, and will cut the picture in the line

a' B. The vertical projection of this triangle in the pic-

ture will be the triangle C e B. Now we know that any

triangle may be regarded as the moiety of a quadrilateral

figure. Therefore, if through E we draw a line E F

parallel to B A, and another A F, parallel to the ray B E,

we shall obtain a quadrilateral figure A B E F, double the

first triangle ABE, divided into two equal parts by the

diagonal or ray A E. Since, then, the side E F is of the

height of the eye, it will necessarily meet the plane of

the picture at a point d' in the horizon, the line d' B will

be the intersection of the plane A B with the picture, so

the point d' will be the vanishing point of A B, and conse-

quently of all the original lines parallel to it.

Suppose perpendiculars let fall from E E upon the ori-

ginal plane, and we obtain the points ef as the projections

of E and F, draw e B, ef, f A, and we have the quadrila-

teral A B, ef, for the horizontal projection of the inclined

plane AE. The lines FE, fe, with the perpendiculars

E e, F/, form a rectangle eF, or/E, which passes through

the perpendiculars E e, F/. Consequently this plane is

vertical, and cuts the plane of the picture in the line d d'.

Observe that the lines E d', e d, being parallel to A B,

make with the picture an angle of 45°, and therefore

that E e is equal to e'd'. But E e is the principal ray,

or the distance of the eye from the picture, and therefore

d' may be regarded as the point of distance carried upon

the horizon from e to d'.

Rule IV.—Hence the perspectives of cdl original lines

making an angle of 45° with the picture, vanish in the

point of distance.

If the original line made a greater angle than 45° with

the picture, its vanishing point would be found between

the point of sight and the point of distance; and if a less

angle, its vanishing point would be beyond the point of

distance; and the general rule is thus expressed.

Rule V.—The perspective of an original line, making

any angle whatever with the picture, will have its va-

nishing point on the horizon at the intersection of the

picture, with a plane pamllel to the original line pass-

ing through the point of sight.

We have seen that the principal ray is necessarily

parallel to the lines which are perpendicular to the pic-

ture, and that its intersection with the picture, or the

point of sight, is the vanishing point for all such lines.

We have seen also that the vanishing point of lines mak-

ing an angle of 45° with the picture, is at the intersection

with the picture of a line drawn through the point of

sight parallel to the original line. And as in this, so in

the case of any line making any angle whatever with the

picture. Whence follows the general rule.

Rule VI.—That the vanishing point for horizontal

straight lines, forming any angle whatever with the pic-

ture, is at the intersection of a parallel to these lines,

drawn through the point of sight to the picture.

(Fig. 194.)

B C

To show this geometrically, let ABCDEFG (Fig. 194)

be the horizontal projection, or plane of an original ob-

99

the picture in the line A d perpendicular to its base. A d

produced would be the trace of a plane perpendicular to

the original plane, which would necessarily pass through

the point of sight. The line A o' will also contain the

projection of the vertical triangle D O' O, and the inter-

section of the triangle d AO with the picture will be the

line A d, which is the perspective of A I), and it will tend

to the point of sight 0.

The straight line B K, which is the base of the triangle

K B 0, is not in a plane perpendicular to the original

plane, but inclined to it; for KB is beyond the central

plane, while O is in that plane. Consequently the pro-

jection of the inclined plane in which this triangle is

situated will be the triangle K B 0 on the original plane,

and its vertical projection in the picture will be B o'; and,

as the intersection B k is part of B o', this perspective is

also directed towards the point of sight. It can be shown

that this also would be the case with all other lines perpen-

dicular to the picture; and, therefore, it can be concluded,

Rule III.—That the perspectives of all lines perpen-

dicular to the picture pass through the point of sight.

Let A B (Fig. 193) be a straight line, making with B C,

the base of the picture, an angle of 45°. The perspective

of the line will be a B, which, being produced, would

meet the horizon in the point d', which will also be

the point of convergence of the perspectives of all lines

parallel to A B. It is easy to perceive that the original

line A B is the base of a scalene triangle ABE,

formed by that line and the rays A E, B E, and which

triangle has its base in the original plane, and its summit

in the eye of the spectator. It will be inclined to both

planes of projection, and will cut the picture in the line

a' B. The vertical projection of this triangle in the pic-

ture will be the triangle C e B. Now we know that any

triangle may be regarded as the moiety of a quadrilateral

figure. Therefore, if through E we draw a line E F

parallel to B A, and another A F, parallel to the ray B E,

we shall obtain a quadrilateral figure A B E F, double the

first triangle ABE, divided into two equal parts by the

diagonal or ray A E. Since, then, the side E F is of the

height of the eye, it will necessarily meet the plane of

the picture at a point d' in the horizon, the line d' B will

be the intersection of the plane A B with the picture, so

the point d' will be the vanishing point of A B, and conse-

quently of all the original lines parallel to it.

Suppose perpendiculars let fall from E E upon the ori-

ginal plane, and we obtain the points ef as the projections

of E and F, draw e B, ef, f A, and we have the quadrila-

teral A B, ef, for the horizontal projection of the inclined

plane AE. The lines FE, fe, with the perpendiculars

E e, F/, form a rectangle eF, or/E, which passes through

the perpendiculars E e, F/. Consequently this plane is

vertical, and cuts the plane of the picture in the line d d'.

Observe that the lines E d', e d, being parallel to A B,

make with the picture an angle of 45°, and therefore

that E e is equal to e'd'. But E e is the principal ray,

or the distance of the eye from the picture, and therefore

d' may be regarded as the point of distance carried upon

the horizon from e to d'.

Rule IV.—Hence the perspectives of cdl original lines

making an angle of 45° with the picture, vanish in the

point of distance.

If the original line made a greater angle than 45° with

the picture, its vanishing point would be found between

the point of sight and the point of distance; and if a less

angle, its vanishing point would be beyond the point of

distance; and the general rule is thus expressed.

Rule V.—The perspective of an original line, making

any angle whatever with the picture, will have its va-

nishing point on the horizon at the intersection of the

picture, with a plane pamllel to the original line pass-

ing through the point of sight.

We have seen that the principal ray is necessarily

parallel to the lines which are perpendicular to the pic-

ture, and that its intersection with the picture, or the

point of sight, is the vanishing point for all such lines.

We have seen also that the vanishing point of lines mak-

ing an angle of 45° with the picture, is at the intersection

with the picture of a line drawn through the point of

sight parallel to the original line. And as in this, so in

the case of any line making any angle whatever with the

picture. Whence follows the general rule.

Rule VI.—That the vanishing point for horizontal

straight lines, forming any angle whatever with the pic-

ture, is at the intersection of a parallel to these lines,

drawn through the point of sight to the picture.

(Fig. 194.)

B C

To show this geometrically, let ABCDEFG (Fig. 194)

be the horizontal projection, or plane of an original ob-