An equal-tangent sag vertical curve connects a 1% and 3% initial and final grades, respectively, and is designed for 70 mph. The highest point on the curve is at elevation 822 ft. If the PVC is at station 110 00, what is the elevation of the curve at station 112+12?

Explanation:

An equal-tangent crest vertical curve is designed for 70 mi/h. The high point is at elevation 822 ft. The initial grade is +1% and the final grade is +3%. If the PVC is at station 110+00, What is the elevation of the curve at 112+62?

Solution:

Determine the design length of the vertical curve based on the stopping sight distance (SSD):

L = KA

L = K X [G1 – G2]

K = 247

L = 247 * [ 1- 3] = 494

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