Practice Question: Maximum Elevation of a Symmetrical Vertical Curve

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Solution

Answer: C) +231 ft

Refer to PE Civil Reference handbook Section 5.3.1 for symmetrical vertical curves:
 

Horizontal Distance to Maximum Elevation:
\[X_{m} = g_{1} \times L/(g_{1}-g_{2}) = 0.04 \times 2500/(0.04+0.025) = 1538 \ ft\]Maximum Elevation:
\[Y_{max} = Y_{PVC} + g_{1} X_{m} + aX_{m}^2 \\\]Parabola Constant, a:
\[a= (g_{2}-g_{1}) / 2L = (-0.025-0.04)/(2\times2500) = -1.3\times10^{-5}\\\]
\[Y_{max} = 200 + 0.04 \times1538 -1.3\times10^{-5} \times 1538^2 =230.8\ ft\\\]

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