An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at station 240 + 00 and elevation 122 ft. The PVT is at station 242 + 30 and elevation 127.75 ft. What is the station and elevation of the lowest point on the curve?

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

is given as

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A is given as

With initial grade, the elevation of PVC is

The station is given as

Low point is given as

The station of low point is given as

The elevation is given as

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

answer: report them

explanation: