Even though several of the foregoing equations are actually approximations, their use is
valid when the value of D 0 is relatively small.
Calculating the basic values yields A = 52°48'; Ls = 350 ft (106.7 m); D 0 = 6°; Os =
LA/200 - 350(6)7200 = 10.5° = 10°30', or 0, = 10.5(0.017453) = 0.18326 rad; Ac =
52°48' - 2(10°30') = 31°48'; D 0 = 6(0.017453) = 0.10472 rad; R 0 = 100/DC = 954.93 ft
(291.063 m); xc = 350(1 - 0.18326
2
/10) = 348.83 ft (106.323 m); yc = 350(0.18326/3 -
0.18326^2 /42) = 21.33 ft (6.501 m); k =348.83 -954.93 sin 10°30' = 174.80 ft (53.279m);
p = 21.33 - 954.93(1 - cos 10°30') = 5.34 ft (1.628 m).
- Locate the TS and SC
Thus, T 5 = (954.93 + 5.34) tan 26°24' + 174.80 = 651.47; station of TS = (34 + 93.81)
-(6 + 51.47) = 28 + 42.34; station of SC = (28 + 42.34) + (3 + 50.00) -31+ 92.34. - Calculate the deflection angles
Thus, S 5 = 10°30V3 = 3°30' - 3.5°. Apply Eq. 32 to find the deflection angles at the inter-
mediate stations. For example, for point 7, 8 = 0.7^2 (3.5°) - 1.715° = 1°42.9'.
Record the results in Table 1. The chord lengths between successive stations differ
from the corresponding arc lengths by negligible amounts, and therefore each chord
length may be taken as 35.00 ft (1066.8 cm). - Compute the LJ 1 57, and E 3
Thus, LT = 348.83 - 21.33 cot 10°30' = 233.75 ft (7124.7 cm); ST = 21.33 esc 10°30' =
117.04 ft (3567.4 cm); E 3 = (954.93 + 5.34)(sec 26°24' - 1) + 5.34 = 117.14 ft (3570.4
cm). - Verify the last three calculations by substituting
in the following test equation
Thus
ST + RC 0 tan C_ V 2 Ac T = Ss = - LT E £ £V 3 - R 0 (sec C^1 M (^0) J - 1) ,
cos^1 /2&C cos V^A 0 sin (^6) S
TABLE 1. Deflection Angles on Approach Spiral
Point Station Deflection angle
TS 28 + 42.34 O
1 77.34 0°02.1'
2 29+12.34 0°08.4'
3 47.34 0°18.9'
4 82.34 0°33.6'
5 30+17.34 0°52.5'
6 52.34 1°15.6'
7 87.34 1°42.9'
8 31+22.34 2°14.4'
9 57.34 2°50.1'
SC 92.34 3°30.0'