ground; the distance from 0 to 100, one hundred yards on the ground,
etc.

If the above scale were applied to the road running from A to B in
Fig. 2, it would show that the length of the road is 675 yards.

[Illustration: Fig. 2]

=1863. Construction of Scales.= The following are the most usual
problems that arise in connection with the construction of scales:

1. Having given the R. F. on a map, to find how many miles on the
ground are represented by one inch on the map. Let us suppose that the
R. F. is 1/21120.

Solution

Now, as previously explained, 1/21120 simply means that one inch on
the map represents 21,120 inches on the ground. There are 63,360
inches in one mile. 21,120 goes into 63,360 three times--that is to
say, 21,120 is 1/3 of 63,360, and we, therefore, see from this that
one inch on the map represents 1/3 of a mile on the ground, and
consequently it would take three inches on the map to represent one
whole mile on the ground. So, we have this general rule: To find out
how many miles one inch on the map represents on the ground, divide
the denominator of the R. F. by 63,360.

2. Being given the R. F. to construct a graphical scale to read yards.
Let us assume that 1/21120 is the R. F. given--that is to say, one
inch on the map represents 21,120 inches on the ground, but, as there
are 36 inches in one yard, 21,120 inches = 21,120/36 yds. = 586.66
yds.--that is, one inch on the map represents 586.66 yds. on the
ground. Now, suppose about a 6-inch scale is desired. Since one inch
on the map = 586.66 yards on the ground, 6 inches (map) = 586.66 x 6 =
3,519.96 yards (ground). In order to get as nearly a 6-inch scale as
possible to represent even hundreds of yards, let us assume 3,500
yards to be the total number to be represented by the scale. The
question then resolves itself into this: How many inches on the map
are necessary to represent 3,500 yards on the ground. Since, as we
have seen, one inch (map) = 586.66 yards (ground), as many inches are
necessary to show 3,500 yards as 586.66 is contained in 3,500; or
3500/586.66 = 5.96 inches.

[Illustration: Fig. 3]

Now lay off with a scale of equal parts the distance A-I (Figure 3) =
5.96 inches (about 5 and 9-1/2 tenths), and divided it into 7 equal
parts by the construction shown in figure, as follows: Draw a line
A-H, making any convenient angle with A-I, and lay off 7 equal
convenient lengths (A-B, B-C, C-D, etc.), so as to bring H about
op