Practice Question: Max Elevation on a Vertical Curve

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Solution

Answer: C) 102.8 ft

Elevation at PVC:
\[Y_{PVC}=98 ft\]
Length of Curve:
\[L=750 ft\]
Grade of Back Tangent:
\[g_1=3.0 \%\]
Grade of Forward Tangent:
\[g_2=-4.0 \%\]
Horizontal Distance to Max Elevation:
\[x_m= \frac{g_1 L}{g_1-g_2 }\]
\[x_m= \frac{3 \% \times 750 ft}{3 \%-(-4 \%)}=321.43 ft\]
Curve Max Elevation:
\[EL_{max}=Y_{PVC}+g_1 x_m+x_m^2 (\frac{g_2-g_1}{2L})\]
\[EL_{max}=98 ft+3.0 \% \times 321.43 ft + (321.43 ft)^2 (\frac{-4 \%-3 \%}{2*750 ft})\]
\[EL_{max}=102.82 ft\]
 

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