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An equal-tangent sag vertical curve is designed for 45 mi/h. The low point is 237 ft from the PVC at station 112 37 and the final offset at the PVT is 19.355 ft. If the PVC is at station 110 00, what is
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An equal-tangent sag vertical curve is designed for 45 mi/h. The low point is 237 ft from the PVC at station 112 37 and the final offset at the PVT is 19.355 ft. If the PVC is at station 110 00, what is the elevation difference between the PVT and a point on the curve at station 111 00

User Kdh
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1 Answer

3 votes

Answer:

18.722 ft

Step-by-step explanation:

The elevation difference between the PVT and a point on the curve at station

111 + 00

attached below is a detailed solution to the problem

Δelevation ( elevation difference )

= Yt - Y

= 19.355 - 0.632 = 18.722 ft

An equal-tangent sag vertical curve is designed for 45 mi/h. The low point is 237 ft-example-1
User ArendE
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