A +2% grade intersects with a –1% grade at station (535+24.25) at an elevation of 300 ft. If the design speed is 65 mi/h, determine:

Question

(a) the minimum length of vertical curve using the rate of vertical curvature
Then, using the length found in part (a),
(b) the stations and elevations of the BVC and EVC
(c) the elevation of each 100-ft station
(d) the station and elevation of the highpoint

What will be an ideal response?

Answer 1

From Table 15.4, K = 193

L = KA = 193 (2 – (–1)) = 579 ft

Station of BVC = (535+24.25) – (579 ft)/2 = 532+34.75

Station of EVC = (535+24.25) + (579 ft)/2 = 538+13.75

Elevation of BVC = 300 – (0.02)(579/2) = 294.21 ft

Elevation at any station on the leading tangent can be found in a similar manner.

The elevation on the curve can be found by subtracting the elevation on the

leading tangent by the offset, which can be found using Equation 15.12,



Using this procedure, the following table, which tabulates the elevation at 100 ft

stations on the curve, can be generated.



The distance from the BVC to the high point can be found as:

xhigh = LG1 / (G1 – G2) = (579)(2)/(2 – (–1)) = 386 ft

The station of the high point is (532+34.75) + (386 ft) = 536+20.75

The difference between the elevation of the BVC and the elevation of the high

point can be found as:

yhigh = LG12 / (200(G1 – G2)) = (579)(2)2 / (200)(2 – (–1)) = 3.86 ft

Therefore, the elevation of the high point is 294.21 + 3.86 = 298.07 ft