You have been instructed to design a crest vertical curve that will connect a highway segment with a 3% grade to an adjoining segment with a -1% grade. Assume that the minimum stopping sight distance for the highway is 540 feet. If the elevation of the VPC is 1500 ft, what will the elevation of the curve be at L/2?

[Solution Shown Below]

Solution

The first step in the analysis is to find the length of the crest vertical curve. The grade changes from 3% to -1%, which is a change of -4% or A = |-4%|.  In addition, for the stopping sight distance h1 = 3.5 ft and h2 = 0.5 ft. Since we know S = 540 ft, we can go ahead and solve for the length of the crest vertical curve.

If S > L then (invalid because L > S)

If S < L then

Where:
L = Length of the crest vertical curve (ft)
S = Sight distance, 540 ft
A = The change in grades, |-4%|
h1 = Height of the driver's eyes above the ground, 3.5 ft
h2 = Height of the object above the roadway, 0.5 ft

The curve length calculated from the 'S < L' equation was 877.5 feet, which is greater than the sight distance of 540 feet. To find the elevation of the curve at a horizontal distance of L/2 from the VPC, we need to use the equation below.

Y = VPCy + B*x + (A*x2)/(200*L)

Where:
Y = Elevation of the curve at a distance x from the VPC (ft)
VPCy = Elevation of the VPC, 1500 ft
B = Slope of the approaching roadway, or the roadway that intersects the VPC, 0.03
A = The change in grade between the disjointed segments, -4 (From 3% to -1% would be a change of -4%)
x = L/2 = 877.5 ft /2 = 438.75 ft