Derivation of superelevation expression

Superelevation (cant) is the transverse slope given to the carriageway at a horizontal curve by raising the outer edge above the inner edge, so that a component of the vehicle weight balances the centrifugal force.

Derivation

Consider a vehicle of weight $W$ moving at speed $v$ on a curve of radius $R$, on a pavement banked at angle $\theta$ ($\tan\theta = e$ = superelevation).

              P = Wv^2/gR  (centrifugal, horizontal)
        ----->  o
              /|\  W (down)
   F_A+F_B  / | \
   friction/  |   \  R_A + R_B (normal)
   ______/___\|/___\_____ banked pavement (angle theta)

Forces along the inclined plane (taking the vehicle as a point and the friction at both axles $F = f\,(R_A+R_B)$ where $R_A+R_B$ is the total normal reaction):

$$\begin{aligned} P\cos\theta &= W\sin\theta + F \\ R_A+R_B &= W\cos\theta + P\sin\theta \\ F &= f\,(R_A+R_B) \end{aligned}$$

So

$$P\cos\theta = W\sin\theta + f\,(W\cos\theta + P\sin\theta)$$

Dividing by $W\cos\theta$, using $P/W = v^2/(gR)$ and $\tan\theta = e$:

$$\frac{v^2}{gR} = e + f + f\,e\,\frac{v^2}{gR}$$

The term $f\,e\,v^2/(gR)$ is very small and neglected:

$$\boxed{\,e + f = \frac{v^2}{gR}\,}$$

With $V$ in km/h and $R$ in m (g = 9.81):

$$e + f = \frac{V^2}{127\,R}$$

where $e$ = rate of superelevation ($\tan\theta$) and $f$ = design coefficient of lateral friction (IRC/NRS take 0.15).

Use

• For given $V$, $R$ and $f = 0.15$: $e = \dfrac{V^2}{127R} - 0.15$. Maximum $e$ = 7% (plain/rolling, IRC) or 10% (hills); in snowy areas 7%. If $e$ computed exceeds the maximum, adopt the maximum and check $f$.
• IRC adopts $e = V^2/(225R)$, which is the superelevation for 75% of the design speed with zero friction; the check for full speed is then made with $f$.