\fdO'^ - Old Forge Coal Mines

RAILROAD LOCATION. 277
4.0827 ft., which we add to 128.6979 ft., the elevation of
grade at Sta. 28 + 70, giving 132.7806 ft. for the elevation
of grade at Sea. 31 + 80. Here we commence an 8° curve
for 450 ft. =4.5 stations. The compensation in grade for
an 8° curve is .03 ft. X 8 = 0.24 ft. per station. Hence, the
grade for that curve is 1.317 ft. - 0.24 ft. = + 1.077 ft. per
station, and the total rise on the 8° curve is 1.077 ft. x 4.5 =
4.84G5 ft., which we add to 132.7806, the elevation of
grade at Sta. 31 + 80, giving 137.6271 ft. for the elevation
of grade at Sta. 36 + 30, the P. T. of the 8° curve. The
line between Sta. 36 + 30 and Sta. 40 is a tangent, and has
a grade of 4-1.317 ft. per station. The distance between
these stations is 370 ft. = 3.7 stations, and the total rise is
1.317 X 3.7 = 4.8729 ft., which, added to 137.6271 ft., the
elevation of grade at Sta. 36 + 30, gives 142.5 ft. for the
elevation of grade at Sta. 40.
(805)
See Art. 1439.
(806) In this question, ^=+1.0 ft., ^'z=— 0.8 ft.
and « = 3. Substituting these values in formula lOl,
d= ^'~ ^ (see Art. 1440), we have a = -— ~ ~ '
'
) =
4« 12
1 8
-^=0.15 ft. The successive grades or additions for the
6 stations of the vertical curve are the following: g — a,
g — da, g — 5a, g — 7a, g — S)a, g — \\a. Substituting known
values of g and a, we have for the successive grades :
Heights of Curve
Above Starting
Stations. Point.
1. g- rt = 1.0 ft. -0.15 ft. = 0.85 ft 0.85 ft.
2. g- 3rt = 1.0 ft. —0.45 ft; = 0.55 ft. ... .1.40 ft.
3. g- 5rt= 1.0ft. -0.75 ft. = 0.25 ft 1.65 ft.
4. g- 7a = 1.0 ft. -1.05 ft. = -0.05 ft 1.60 ft.
5. ^ - 9