Condition for skidding and overturning

Consider a vehicle of weight $W$ moving at speed $v$ on a horizontal curve of radius $R$ on a pavement with superelevation $e$ (neglect the small term $f\,e$). Wheel track $= b$, height of centre of gravity above the road $= h$, coefficient of lateral friction $= f$. Centrifugal force $P = \dfrac{Wv^2}{gR}$.

        P ---> o  (c.g. at height h)
               |
   <- b/2 ->   |
   inner wheel   outer wheel (overturning about this)

1. Condition for lateral skidding

Skidding occurs when the component of the centrifugal force along the pavement exceeds the sum of the friction and the weight component down the slope:

$$\begin{aligned} P\cos\theta &\ge W\sin\theta + F \\ \frac{v^2}{gR} &\ge e + f \end{aligned}$$

So the maximum safe speed against skidding is $v = \sqrt{gR(e+f)}$, or $V = \sqrt{127R(e+f)}$ km/h. For a flat road ($e=0$): $\dfrac{v^2}{gR} \ge f$.

2. Condition for overturning

Take moments about the contact point of the outer wheel. The overturning moment $= P\,h$ and the restoring moment $= W\,b/2$ (flat road). Overturning starts when the reaction at the inner wheel becomes zero:

$$P\,h \ge W\frac{b}{2} \ \Rightarrow\ \frac{v^2}{gR} \ge \frac{b}{2h}$$

With superelevation, $\dfrac{v^2}{gR} \ge \dfrac{b}{2h} + e$ (approximately).

Result

• Safe against skidding if $\dfrac{v^2}{gR} < e + f$.
• Safe against overturning if $\dfrac{v^2}{gR} < \dfrac{b}{2h} + e$.
• For usual vehicles $b/2h$ is about 0.5 or more (well above 1 for cars), which is much greater than $f$ (0.15), so a vehicle will skid before it overturns.