Superelevation on 100 m radius curve

For a vehicle on a banked curve, the full balance of forces gives

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

with $V$ in km/h and $R$ in m. Here $\dfrac{V^2}{127R}=\dfrac{60^2}{127(100)}=0.2835$.

(i) Superelevation if full lateral friction is called into play

$$e=\frac{V^2}{127R}-f=0.2835-0.17=0.1135$$

e = 0.113 (about 11.3%). This is above the usual maximum of 7% (plain and rolling terrain, IRC/NRS) and 10% (hill roads, IRC), so it cannot be fully provided in practice. It means 60 km/h is too high for $R=100$ m with $f=0.17$ unless the maximum superelevation is accepted and speed or radius is changed.

(ii) Friction needed if no superelevation is provided ($e=0$)

$$f=\frac{V^2}{127R}=0.2835$$

f = 0.283. This is more than the permissible 0.15 to 0.17, so the vehicle would skid sideways. Hence superelevation is necessary.

Answer: (i) $e=0.113$; (ii) $f=0.283$.