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Superelevation Calculator — PE Civil Transportation Worked Example
On a horizontal curve, the demand for lateral friction and banking is set by the point-mass relation e + f = V²/(15R). Enter the design speed, curve radius and the side friction factor supplied in the problem, and solve for the rate of superelevation the design must provide.
For study and checking only. This calculator runs entirely in your browser — nothing is sent anywhere. It is meant to verify your hand calculations while you study; on the exam you will work by hand with the NCEES reference handbook.
Quick answer: the point-mass relation
\[e + f = \frac{V^{2}}{15\,R} \qquad\Longrightarrow\qquad e = \frac{V^{2}}{15\,R} - f\]
The left-hand side is the total lateral demand; the design splits it between banking (e, the superelevation rate) and tyre–pavement friction (f). A common exam distractor reports the total demand as if it were e alone — always subtract f.
How to use it — and how to sanity-check the answer
Use the radius to the vehicle path (centreline radius is the usual exam convention) and the friction factor given in the stem — f depends on speed and is supplied, not chosen. Sanity-check: e should come out as a small decimal (a few percent). If e is negative, friction alone carries the demand and the required superelevation is effectively zero; if e exceeds the maximum rate in the problem’s criteria, the radius or speed has to change.
Worked example (reproduces in the tool above)
Worked example Required superelevation
Given: A curve of radius \(R = 1{,}200\) ft is designed for \(V = 55\) mph with a given side friction factor \(f = 0.13\).
Solution:
- Total demand: \(e + f = V^{2}/(15R) = 55^{2}/(15 \times 1{,}200) = 3{,}025/18{,}000 = 0.1681\).
- Superelevation: \(e = 0.1681 - 0.13 = 0.0381 \approx 0.038\).
Answer: \(e \approx 3.8\%\).