0.5

1 cm

PERSPECTIVE.

105

draw a 0. 6 0, which will he the perspective direction of

the diagonals of the square, and their intersections with

the perspectives of the-sides produced will give their points.

Problem VI.—To draw the perspective of a pavement of

squares, the sides of which form any angle with the picture.

Let AB {Fig. 218) be the horizontal projection of the

base of the picture, cutting the squares (of which only

two rows are here shown) at any angle.

{Fig. 213.)

Transfer to the vertical projection a 1 of the base of the

picture the divisions 1, 2, 3, 4, 5, 6, 7.* Through the seat

of the eye 0 draw Op parallel to the sides of the squares,

and p will be the vanishing point for them. Transfer the

distance of p from the central plane op to the horizon at

o' p, and draw Ip’ 2pi, &c., which gives us the perspec-

tives of the sides of the squares parallel to m n, and we

have now to find those which are perpendicular to them.

This we might do by finding their vanishing points; but

this mode would be inconvenient, as it would extend the

vauishing point so far beyond the limits of the paper,

and we shall operate by the diagonals. Draw, therefore,

through 0 the seat of the eye, the lines 0 s, 0 t, parallel

to the diagonals, to meet the picture-line produced in s t,

and transfer os, ot, to the horizon-line in o' s', o' t'. Then

transfer the points 4 or v, where the diagonal meets the

picture-line A B, to the vertical projection of the picture-

line a b, and from these points draw fines to the vanishing

points, as shown in the figure.

Problem VII.—To draw a hexagonal pavement in per-

spective, when one of the sides of the hexagon is parallel

to the base of the picture.

Let A B {Fig. 214) be the given hexagon. Set off along

the base of the picture, divisions equal to the side of the

hexagon; then draw its diagonals, which will divide it

into six equilateral triangles; and find the vanishing

points of the diagonals. This may be done in the follow-

* This is most conveniently done by transferring them first to

the edge of a strip of paper, from which they can be transferred

to the picture-line.

ing simple manner. Let O be the intersection of the

central plane with the horizon, and 0 C be equal to the

{Fig. 214.)

distance of the spectator; then from C draw the fines

C D, C D, parallel to the diagonals of the hexagon, and

their intersection with the horizon in the points D D will

be the vanishing points of the diagonals. The remainder

of the operation requires no description.

Problem VIII.—To draw the perspective of a circle.

Let A B a b {Fig. 215) be the projections of the picture,

c c those of the eye, o the point of distance, I J K L the

given circle.

(Fig. 215.)

The most expeditious method of operating is to cir-

cumscribe the circle by a square. The circle touches the

square at four points; and if the diagonals of the square

are drawn, they intersect the circle at four other points,

which gives eight points, the perspective of which are

easily found.

Thus, then, we draw the perspective of the square and

of its diameters and diagonals, and then project on the

base of the picture ab the points IJKL in K'and L

and K" and L", and from these points draw the fines K" c,

L"c, which cut the diagonals in ij hi, and through these and

the four others EFGH the circle is to be traced by hand.

As the circle is a figure which has very frequently to

be drawn in perspective, we shall consider it under ano-

ther aspect.

Let there be any number of points taken in the circum-

ference of the original circle {same Figure), and suppose

fines drawn to them from the eye c, as the tangents M c,

N c. Now it is evident that the collection of all these

fines forms the projection of a scalene cone, having its base

p

105

draw a 0. 6 0, which will he the perspective direction of

the diagonals of the square, and their intersections with

the perspectives of the-sides produced will give their points.

Problem VI.—To draw the perspective of a pavement of

squares, the sides of which form any angle with the picture.

Let AB {Fig. 218) be the horizontal projection of the

base of the picture, cutting the squares (of which only

two rows are here shown) at any angle.

{Fig. 213.)

Transfer to the vertical projection a 1 of the base of the

picture the divisions 1, 2, 3, 4, 5, 6, 7.* Through the seat

of the eye 0 draw Op parallel to the sides of the squares,

and p will be the vanishing point for them. Transfer the

distance of p from the central plane op to the horizon at

o' p, and draw Ip’ 2pi, &c., which gives us the perspec-

tives of the sides of the squares parallel to m n, and we

have now to find those which are perpendicular to them.

This we might do by finding their vanishing points; but

this mode would be inconvenient, as it would extend the

vauishing point so far beyond the limits of the paper,

and we shall operate by the diagonals. Draw, therefore,

through 0 the seat of the eye, the lines 0 s, 0 t, parallel

to the diagonals, to meet the picture-line produced in s t,

and transfer os, ot, to the horizon-line in o' s', o' t'. Then

transfer the points 4 or v, where the diagonal meets the

picture-line A B, to the vertical projection of the picture-

line a b, and from these points draw fines to the vanishing

points, as shown in the figure.

Problem VII.—To draw a hexagonal pavement in per-

spective, when one of the sides of the hexagon is parallel

to the base of the picture.

Let A B {Fig. 214) be the given hexagon. Set off along

the base of the picture, divisions equal to the side of the

hexagon; then draw its diagonals, which will divide it

into six equilateral triangles; and find the vanishing

points of the diagonals. This may be done in the follow-

* This is most conveniently done by transferring them first to

the edge of a strip of paper, from which they can be transferred

to the picture-line.

ing simple manner. Let O be the intersection of the

central plane with the horizon, and 0 C be equal to the

{Fig. 214.)

distance of the spectator; then from C draw the fines

C D, C D, parallel to the diagonals of the hexagon, and

their intersection with the horizon in the points D D will

be the vanishing points of the diagonals. The remainder

of the operation requires no description.

Problem VIII.—To draw the perspective of a circle.

Let A B a b {Fig. 215) be the projections of the picture,

c c those of the eye, o the point of distance, I J K L the

given circle.

(Fig. 215.)

The most expeditious method of operating is to cir-

cumscribe the circle by a square. The circle touches the

square at four points; and if the diagonals of the square

are drawn, they intersect the circle at four other points,

which gives eight points, the perspective of which are

easily found.

Thus, then, we draw the perspective of the square and

of its diameters and diagonals, and then project on the

base of the picture ab the points IJKL in K'and L

and K" and L", and from these points draw the fines K" c,

L"c, which cut the diagonals in ij hi, and through these and

the four others EFGH the circle is to be traced by hand.

As the circle is a figure which has very frequently to

be drawn in perspective, we shall consider it under ano-

ther aspect.

Let there be any number of points taken in the circum-

ference of the original circle {same Figure), and suppose

fines drawn to them from the eye c, as the tangents M c,

N c. Now it is evident that the collection of all these

fines forms the projection of a scalene cone, having its base

p