And,
M = ^(I-COs
1
X 2 A) (21)
M = 1200(1 - 0.9703) = 35.6 ft (10.85 m)
Lastly,
E = R tan^1 X 2 A tan^1 X 2 A (22)
E= 1200(0.2493)(0.1228) = 36.7 ft (11.19 m)
- Verify the results in step 1
Use the pythagorean theorem on triangle ADV. Or, AD = !/2(580.6) = 290.3 ft (88.48 m);
DV- 35.6 + 36.7 = 72.3 ft (22.04 m); then 290.3
2
- 72.3
2
= 89,500 ft
2
(8314.6 m
2
), to the
nearest hundred; 299.2
2
= 89,500 ft
2
(8314.6 m
2
); this is acceptable.
- Calculate the degree of curve D
Part b. In Fig. 13, let E represent a station along the curve. Angle VAE is termed the de-
flection angle oe of this station; it is equal to one-half the central angle AOE. In the field,
the curve is staked by setting up the transit at the PC and then locating each station by
means of its deflection angle and its chord distance from the preceding station.
Calculate the degree of curve D. This is the central angle formed by the radii to two
successive stations or, what is the same in this instance, the central angle subtended by a
chord of 100 ft (30.5 m). Then
sin
1
X 2 Z) - — (23)
R
So^1 X 2 D = arcsin 50/1200 = arcsin 0.04167;^1 X 2 Z) = 2°23.3'; Z) = 4°46.6'.
- Determine the station at the PT
Number of stations on the curve = 28°/4°46.6' = 5.862; station of PT = (82 + 30) +
(5+ 86.2)-88+ 16.2.
- Calculate the deflection angle of station 83 and the difference
between the deflection angles of station 88 and the PT
For simplicity, assume that central angles are directly proportional to their corresponding
chord lengths; the resulting error is negligible. Then S 83 = 0.70(2°23.3') = 1°40.3;; 6pT -
S 88 = 0.162(2°23.3') = 0°23.2'.
- Calculate the deflection angle of each station
Do this by adding^1 X 2 D to that of the preceding station. Record the results thus:
Station Deflection angle
82+30 O
83 1°40.3'
84 4°03,6'
85 6°26.9'
86 8°50.2'
87 11°13.5'
88 13°36.8'
88+16.2 140 OO'
