ENGINEER AND MACHINISTS’ DRAWING-BOOK.
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.
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,
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
By these constructions the triangu-
lar drawing instruments (Fig. 26) are
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.
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.
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