grades is to pass through station 42 + 60 of elevation 213.70 ft (65.136 m). Compute the
required length of curve.
Calculation Procedure:
- Compute the tangent offsets; establish the horizontal location
of P in terms of L
Refer to Fig. 22, where P denotes the specified point. The given data enable computation
of the tangent offsets CP and DP, thus giving a relationship between the horizontal dis-
tances from A to P and from P to B. Since the distance from the centerline of curve to P is
known, the length of curve may readily be found.
Computing the tangent offsets gives CP = V- Gah and DP = V- Gbh\ but CPIDP =
(L/2 + h)
2
/(L/2 - h)
2
= (L + 2H)
2
I(L - 2h)
2
; therefore,
L + 2h _/v-Gah\m
L-2h ~(v-Gbh)
(45)
- Substitute numerical values and solve for L
Thus, G 0 = -1.6 percent; Gb = +3.8 percent; h = 60 ft (18.3 m); v = 3.70 ft (1.128 m); then
(L + UQ)I(L - 120) = [(3.7 x 0.016 x 60)7(3.7 - 0.038 x 60)]1/2 - 1.81; so L = 416 ft
(126.8 m).
- Verify the solution
There are many ways of verifying the solution. The simplest way is to compare the offsets
of P and B from a tangent through A. By Eq. 44b, offset of B from tangent through A =
208(0.016 + 0.038) = 11.232 ft (3.4235 m). From the preceding calculations, offset of P
FIGURE 22
