Superelevation & Side Friction

\frac{e + f}{1 - e f} \approx e + f = \frac{V ^{2}}{g R} = \frac{V ^{2}}{15 R} (V mph, R ft)

A vehicle on a curve needs centripetal acceleration V²/R supplied by superelevation e (decimal) and side friction f. AASHTO’s 15 in the US formula is g in mph-ft units with the small ef term neglected. Maximum e is often 0.06–0.12 depending on climate and urban/rural.

$e$: Superelevation rate (decimal)
$f$: Side friction factor
$R$ (ft or m): Radius of curve

Step-by-step solved example

V = 45 mph, e = 0.06, f = 0.15. Minimum R?

R = V² / [g(e+f)]. PC to PT along the circular arc.

1. AASHTO US
R = V² / [15(e+f)] = 2025 / [15×0.21] = 2025 / 3.15 = 643 ft.
$R_{m i n} = 643 ft$

Answer: 643 ft

10 practice questions

0/10 correct

1.If e + f is fixed, R_min varies with

2.Degree of curve D (arc definition) relates to R by

3.Length of circular curve L =

4.Tangent length T of a circular curve is

5.Spiral transitions are used to

6.Metric: e + f = V² / (127 R) with V in

7.Increasing superelevation allows

8.Normal crown is typically about

9.V = 30 mph, e+f = 0.20. R_min ≈

10.Reverse curves without a tangent are generally