0.5

1 cm

104

ENGINEER AND MACHINIST’S DRAWING-BOOK.

and set along it three equal divisions of the compasses,

opened at pleasure in 1 2 3, through the last division, and

(Fig. 208.)

O

through m draw a line to the horizon, intersecting it in 0,

which will be the vanishing point of all parallels to 3 O,

and, of course, of the lines 2 0, 1 O, and by drawing these

the perspective line m n is divided into three equal parts.

When the given line is inclined both to the picture and

the ground plane, as A B (Fig. 209), it is necessary to find

r, for example, be the point on the original plane at which

the feet of a figure are supposed to be situated, and it is

required to know its height, and suppose AB the real

height of the figure, then through r draw a horizontal

line cutting the parallels O e, b e, in s and t, and t s will be

the height sought, which has to be carried vertically from

r to H. Or otherwise, from the point of intersection t

raise a perpendicular, which gives the point h, and t h

will equally be the height sought.

Problem: IY.—To draw the perspective of a pavement

of squares, two of the sides of the squares being parallel

to the base of the picture.

The squares having two of their sides parallel to the

base of the picture, their other two sides will have for

their vanishing point the point of sight, and the point of

distance will be the vanishing point of the diagonals.

No plan is required. Set off on the base of the picture

(Fig. 209.)

(Fig. 211.)

the divisions, first on its horizontal projection Ah, and from

them to draw vertical lines which will cut the perspective

line in the points required.

Problem III.—Through a given point in a picture to

draw a line parallel to the base or side of the picture, and

perspectively equal to another given line A B.

Let a (Fig. 210) be the given point. From it draw an

indefinite line parallel to the base of the picture, or to its

side, and then to determine on either of these lines a

length perspectively equal to the given line, draw through a

any line at pleasure, cutting the base of the picture, say at

O, and the horizon at e. Set off from O a length, O b, equal

to the given line, and draw b e, which will cut the indefi-

nite horizontal line in g, and a g will be equal to A B.

In the same way proceed to obtain the length of the line

a f parallel to the side of the picture.

. (Fig. 210.)

at 1 2 3, &c. (Fig. 211), divisions equal to the sides of the

square, draw through these to the point of sight O, lines

which are the perspectives of the sides of the squares,

perpendicular to the picture, and intersect these by a

diagonal drawn from any point, as 1 to D, the point of

distance, and through the points of intersection draw lines

parallel to the base of the picture. The squares can be

extended so as to fill the picture by extending the base line

and setting off more divisions. But, if there is not room

to do this, any of its parallels, as A B, may be produced,

and the perspective lengths of the side of the squares set

out on it, and again as at E F if necessary.

Problem Y.—To draw a pavement of squares in per-

spective when the sides are inclined at angle of 45° to

the base of the picture.

In this case let a b (Fig. 212) be the side of the square,

(Fig. 212.)

It is easy by this means to find the perspective height

of a figure at any distance from the base of the picture. Let

and its diagonal will consequently be a 1. Lay off along

the base of the picture the divisions 12 3, &c., each equal to

the length of the diagonal. Now, as the sides of the square

make angles of 45° with the picture, the distance points

will be their vanishing points, and nothing more is re-

quired to be done than to draw from 12 3, and to D D the

lines 1 D, 2 D. If it is required to fill the space from a or b

ENGINEER AND MACHINIST’S DRAWING-BOOK.

and set along it three equal divisions of the compasses,

opened at pleasure in 1 2 3, through the last division, and

(Fig. 208.)

O

through m draw a line to the horizon, intersecting it in 0,

which will be the vanishing point of all parallels to 3 O,

and, of course, of the lines 2 0, 1 O, and by drawing these

the perspective line m n is divided into three equal parts.

When the given line is inclined both to the picture and

the ground plane, as A B (Fig. 209), it is necessary to find

r, for example, be the point on the original plane at which

the feet of a figure are supposed to be situated, and it is

required to know its height, and suppose AB the real

height of the figure, then through r draw a horizontal

line cutting the parallels O e, b e, in s and t, and t s will be

the height sought, which has to be carried vertically from

r to H. Or otherwise, from the point of intersection t

raise a perpendicular, which gives the point h, and t h

will equally be the height sought.

Problem: IY.—To draw the perspective of a pavement

of squares, two of the sides of the squares being parallel

to the base of the picture.

The squares having two of their sides parallel to the

base of the picture, their other two sides will have for

their vanishing point the point of sight, and the point of

distance will be the vanishing point of the diagonals.

No plan is required. Set off on the base of the picture

(Fig. 209.)

(Fig. 211.)

the divisions, first on its horizontal projection Ah, and from

them to draw vertical lines which will cut the perspective

line in the points required.

Problem III.—Through a given point in a picture to

draw a line parallel to the base or side of the picture, and

perspectively equal to another given line A B.

Let a (Fig. 210) be the given point. From it draw an

indefinite line parallel to the base of the picture, or to its

side, and then to determine on either of these lines a

length perspectively equal to the given line, draw through a

any line at pleasure, cutting the base of the picture, say at

O, and the horizon at e. Set off from O a length, O b, equal

to the given line, and draw b e, which will cut the indefi-

nite horizontal line in g, and a g will be equal to A B.

In the same way proceed to obtain the length of the line

a f parallel to the side of the picture.

. (Fig. 210.)

at 1 2 3, &c. (Fig. 211), divisions equal to the sides of the

square, draw through these to the point of sight O, lines

which are the perspectives of the sides of the squares,

perpendicular to the picture, and intersect these by a

diagonal drawn from any point, as 1 to D, the point of

distance, and through the points of intersection draw lines

parallel to the base of the picture. The squares can be

extended so as to fill the picture by extending the base line

and setting off more divisions. But, if there is not room

to do this, any of its parallels, as A B, may be produced,

and the perspective lengths of the side of the squares set

out on it, and again as at E F if necessary.

Problem Y.—To draw a pavement of squares in per-

spective when the sides are inclined at angle of 45° to

the base of the picture.

In this case let a b (Fig. 212) be the side of the square,

(Fig. 212.)

It is easy by this means to find the perspective height

of a figure at any distance from the base of the picture. Let

and its diagonal will consequently be a 1. Lay off along

the base of the picture the divisions 12 3, &c., each equal to

the length of the diagonal. Now, as the sides of the square

make angles of 45° with the picture, the distance points

will be their vanishing points, and nothing more is re-

quired to be done than to draw from 12 3, and to D D the

lines 1 D, 2 D. If it is required to fill the space from a or b