0.5

1 cm

22

ENGINEER AND MACHINISTS’ DRAWING-BOOK.

(Fig. 72.)

(Fig. 73.)

Note.—When a series of parallels are required perpen-

dicular to a base-line, A B, they may be drawn by the

method of Prob. III.

through points in the

base-line, set off at the

required distances a-

part. This method is

convenient also where

a succession of parallels

are required to a given

line CD; for the per-

pendicular A B may be drawn to it, and any number of

parallels may be drawn upon the perpendicular.

Problem VIII.—To divide a straight line into a num-

ber of equal parts.

ls£ Method.—Suppose the line A B is required to be

divided into five parts.

From A and B, draw

two parallels AC, BD, \

on opposite sides. Set \

off any convenient dis- fW

tance four times (one a-^—-X——4--~-A-—-^b

less than the given \ \ \

number) from A on A \ \ ®

C, and from B on B D; \ *

join the first on A C 4

to the fourth on B D, p.W

and so on. The lines so drawn, divide A B as required.

2d Method.—Draw the line A C at an angle from A, set

off five equal parts, draw B 5, and draw parallels to it

from the other points of division

in A C. These parallels divide

A B as required.

3d Method.—When the length

of the given line, by a scale of

equal parts, is exactly divisible into the required number

of parts, set them off by the scale.

Note.—By a similar process a line may be divided into

a number of unequal parts: setting off divisions on A C,

proportional by a scale to the required divisions, and

drawing parallels to A B.

This problem is useful in the formation of drawing

scales, by setting off in the first place the total number of

feet, inches, and yards, required in the scale, and then

subdividing the line. This process, also, ensures greater

accuracy than if the unit of measure be taken in the

dividers, and simply set off along the line, as is frequently

done; for, by the latter process, any error of original

measurement is multiplied as the dividers proceed.

Problem IX.— Upon a straight line to draw an angle

equal to a given rectilineal angle.

(Fig. 75.)

(Fig. 7L)

Let A be the given angle, and F G the line. With any

radius, from the points A and F, describe arcs D E, I H,

(Fig. 76.)

cutting the sides of the angle A, and the line F G. Set

off the arc I H equal to D E, and draw F H. The angle

F is equal to A, as required.

To draw angles of 60° and 30°; from F, with any radius

F I (Fig. 76), describe an arc I H;

and from I, with the same radius, cut

the arc at H, and draw F H to form

the required angle I F H. Draw

the perpendicular H K to the base

line, to form the angle of 30°, F H K.

To draw an angle of 45°, set off the distance F I, draw

the perpendicular I H equal to I F,

and join H F to form the angle at F as

required.

By these constructions the triangu-

lar drawing instruments (Fig. 26) are

formed.

Problem X.—To bisect an angle (Fig. 78).

Let A C B be the angle; on the centre C, cut the sides

at A B. On A and B, as centres, describe arcs cutting at

D. Draw C D, dividing the angle into two equal parts.

Fig. 77.

(Fig. 78.)

(Fig. 79)

Problem XI.—To bisect the inclination of two lines,

of which the intersection is inaccessible (Fig. 79).

Upon the given lines C B, C H, at any points, draw

perpendiculars E F, G H, of equal length, and through

F and G draw parallels to the respective lines, cutting at

S; bisect the angle F S G so formed, by the line S D.

This line equally divides the inclination of the given fines.

SECTION II.

Problems on Straight Lines and Circles.

Problem XII.—Through two given points to describe

an arc of a circle with a given radius.

From the given points A and B

as centres, with the given radius in

the compasses, describe arcs cutting

at C, and from C, with the same

radius, describe an arc through A and

B, as required.

Problem XIII.—To find the centre of a circle, or of

an arc of a circle.

1st Method, for a circle (Fig. 81).—Draw the chord A B,

bisect it by the perpendicular C D, bounded both ways

by the circumference; and bisect C D for the centre G.

2d Method, for a circle or arc (Fig. 82).—Select three

ENGINEER AND MACHINISTS’ DRAWING-BOOK.

(Fig. 72.)

(Fig. 73.)

Note.—When a series of parallels are required perpen-

dicular to a base-line, A B, they may be drawn by the

method of Prob. III.

through points in the

base-line, set off at the

required distances a-

part. This method is

convenient also where

a succession of parallels

are required to a given

line CD; for the per-

pendicular A B may be drawn to it, and any number of

parallels may be drawn upon the perpendicular.

Problem VIII.—To divide a straight line into a num-

ber of equal parts.

ls£ Method.—Suppose the line A B is required to be

divided into five parts.

From A and B, draw

two parallels AC, BD, \

on opposite sides. Set \

off any convenient dis- fW

tance four times (one a-^—-X——4--~-A-—-^b

less than the given \ \ \

number) from A on A \ \ ®

C, and from B on B D; \ *

join the first on A C 4

to the fourth on B D, p.W

and so on. The lines so drawn, divide A B as required.

2d Method.—Draw the line A C at an angle from A, set

off five equal parts, draw B 5, and draw parallels to it

from the other points of division

in A C. These parallels divide

A B as required.

3d Method.—When the length

of the given line, by a scale of

equal parts, is exactly divisible into the required number

of parts, set them off by the scale.

Note.—By a similar process a line may be divided into

a number of unequal parts: setting off divisions on A C,

proportional by a scale to the required divisions, and

drawing parallels to A B.

This problem is useful in the formation of drawing

scales, by setting off in the first place the total number of

feet, inches, and yards, required in the scale, and then

subdividing the line. This process, also, ensures greater

accuracy than if the unit of measure be taken in the

dividers, and simply set off along the line, as is frequently

done; for, by the latter process, any error of original

measurement is multiplied as the dividers proceed.

Problem IX.— Upon a straight line to draw an angle

equal to a given rectilineal angle.

(Fig. 75.)

(Fig. 7L)

Let A be the given angle, and F G the line. With any

radius, from the points A and F, describe arcs D E, I H,

(Fig. 76.)

cutting the sides of the angle A, and the line F G. Set

off the arc I H equal to D E, and draw F H. The angle

F is equal to A, as required.

To draw angles of 60° and 30°; from F, with any radius

F I (Fig. 76), describe an arc I H;

and from I, with the same radius, cut

the arc at H, and draw F H to form

the required angle I F H. Draw

the perpendicular H K to the base

line, to form the angle of 30°, F H K.

To draw an angle of 45°, set off the distance F I, draw

the perpendicular I H equal to I F,

and join H F to form the angle at F as

required.

By these constructions the triangu-

lar drawing instruments (Fig. 26) are

formed.

Problem X.—To bisect an angle (Fig. 78).

Let A C B be the angle; on the centre C, cut the sides

at A B. On A and B, as centres, describe arcs cutting at

D. Draw C D, dividing the angle into two equal parts.

Fig. 77.

(Fig. 78.)

(Fig. 79)

Problem XI.—To bisect the inclination of two lines,

of which the intersection is inaccessible (Fig. 79).

Upon the given lines C B, C H, at any points, draw

perpendiculars E F, G H, of equal length, and through

F and G draw parallels to the respective lines, cutting at

S; bisect the angle F S G so formed, by the line S D.

This line equally divides the inclination of the given fines.

SECTION II.

Problems on Straight Lines and Circles.

Problem XII.—Through two given points to describe

an arc of a circle with a given radius.

From the given points A and B

as centres, with the given radius in

the compasses, describe arcs cutting

at C, and from C, with the same

radius, describe an arc through A and

B, as required.

Problem XIII.—To find the centre of a circle, or of

an arc of a circle.

1st Method, for a circle (Fig. 81).—Draw the chord A B,

bisect it by the perpendicular C D, bounded both ways

by the circumference; and bisect C D for the centre G.

2d Method, for a circle or arc (Fig. 82).—Select three