\fdO'^ - Old Forge Coal Mines

248 SURVEYING.
^^^ = 1.875'. The deflection angle for a chord of 72.7 ft:
is, therefore, 1.875' X 72.7 = 136.31' = 2° 10.31'.
(677) We find the tangent deflection by applying
tan def = —75. (See Art. 1 255.) c = 50.
formula 93, .
50^ = 2,500. The radius R of 5° 30' curve = 1,042.14 ft.
(See table of Radii and Deflections.) Substituting these
2 500
values in formula 93, we have tan def. = ^ '
, ^^ = 1.199
2,084.28
ft. Ans.
(678) The formula for chord deflections is d= -^. (See
Art. 1255, formula 92.) ^ = 35.2. 35.2' := 1,239.04. The
radius 7? of a 4° 15' curve is 1,348.45 ft. Substituting these
1 239 04
values in formula 92, we have d= '
^,' , ^ = .919 ft. Ans.
1,348.45
(679) The formula for finding
the radius ^ is 7? =
-.^,. (See Art. 1 249.) The degree of curve is 3° 10'. D,
sm V / o
3° 10'
the deflection angle, is -—— = 1° 35'; sin 1° 35' = .02763.
Substituting the value of sin D in the formula, we have R =
50
—-7^—-; whence, R = 1,809.63 ft. Ans.
The answer given with the question, viz., 1,809.57 ft.,
agrees with the radius given in the table of Radii and
Deflections, which was probably calculated with sine given to
eight places instead of five places, as in the above calculation,
which accounts for the discrepancy in results.
(680) In Fig. 68, let A B and A C represent the given
lines, and BC the amount of their divergence, viz., 18.22
B ft. The lines will form a
A ^.^---— '^ J^f^-^^ triangle ABC, of which
the angle A =. V. Draw
Fig. 68. a perpendicular from A to
D, the middle point of the base. The perpendicular will