Given a sag vertical curve connecting a –1.5% grade with a +2.5% grade on a rural arterial highway, use the rate of vertical curvature and a design speed of 70 mi/h to compute the elevation of the curve at 100 ft stations if the grades intersect at station (475+00) at an elevation of 300.00 ft. Identify the station and elevation of the low point.

Question

What will be an ideal response?

Answer 1

From Table 15.5, K = 181

L = KA = 181 (1.5 – (–2.5)) = 724 ft

Station of BVC = (475+00) – (724 ft)/2 = 471+38.00

Station of EVC = (475+00) + (724 ft)/2 = 478+62.00

Elevation of BVC = 300 + (0.015)(724/2) = 305.43 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 adding the elevation on the leading

tangent to 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 low point can be found as:

xlow = LG1 / (G1 – G2) = (724)(1.5)/(1.5 – (–2.5)) = 271.50 ft

The station of the low point is (471+38) + (2+71.50) = 474+09.50

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

point can be found as:

ylow = LG12 / (200(G1 – G2)) = (724)(1.5)2 / (200)(1.5 – (–2.5)) = 2.04 ft

Therefore, the elevation of the low point is 305.43 – 2.04 = 303.39 ft