---
title: "Condition for skidding and overturning"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:skidding-overturning-derive"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/skidding-overturning-derive
site: BCT Engineering
---

# Condition for skidding and overturning

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2081 Ashwin, 5 marks; 2075 Bhadra (asked 2 times)
- Video: [Why Vehicles Skid & Overturn on Curves? | Overturning and Skidding |Horizontal Curves | Abrar Khan](https://www.youtube.com/watch?v=Au6Z0uWlTH0) — CIVILIAN WAYS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/skidding-overturning-derive
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Derive the condition of lateral skidding and overturning of a vehicle moving at a horizontal curve section.

## Short answer

For a horizontal curve, lateral skidding occurs when v^2/(gR) >= e+f, so safe speed is sqrt(gR(e+f)) or sqrt(127R(e+f)) km/h. Overturning occurs when v^2/(gR) >= b/(2h)+e. Since b/(2h) usually exceeds f, a vehicle skids before overturning.

## Answer

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 $P = \dfrac{Wv^2}{gR}$.

```text
        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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With superelevation e and friction f, the vehicle is on the point of skidding when v^2/(gR) equals
   A. e - f
   B. b/(2h)
   C. f only
   D. e + f
   Answer: D. Balancing P cos(theta) against W sin(theta) + F gives v^2/(gR) = e + f. (Key point: Skidding condition)

2. The maximum safe speed against skidding (km/h) is
   A. sqrt(127 R (e+f))
   B. sqrt(127 R / (e+f))
   C. sqrt(R (e+f)/127)
   D. 127 R (e+f)
   Answer: A. From v^2 = gR(e+f), V = sqrt(127 R (e+f)) in km/h. (Key point: Maximum safe speed)

3. In deriving overturning, moments are taken about
   A. The inner wheel contact point
   B. The outer wheel contact point
   C. The centre of gravity
   D. The road centre line
   Answer: B. Overturning moment P h and restoring moment W b/2 are taken about the outer wheel; overturning begins when the inner wheel reaction is zero. (Key point: Overturning moments)

4. Why does a vehicle usually skid before it overturns?
   A. Because f (about 0.15) is less than b/2h (about 0.45-0.6)
   B. Because b/2h is less than f
   C. Because superelevation is zero
   D. Because the radius is always large
   Answer: A. The skidding limit (f) is reached at a lower centrifugal ratio than the overturning limit b/2h. (Key point: Which happens first)

## Related in Transportation Engineering I

- [Effect of centrifugal force on horizontal curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Acentrifugal-effect) — Transportation Engineering I, chapter 3, asked 2×
- [Vertical curve with horizontal curve, 45 degrees](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-77c-vc-horizontal) — Transportation Engineering I, chapter 3
- [Composite curve, 65 kmph, centrifugal ratio 1/5](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-71bh-composite-65) — Transportation Engineering I, chapter 3
- [Transition curve 200 m radius, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-tc-200) — Transportation Engineering I, chapter 3
- [Transition curve for 230 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-81c-tc-230) — Transportation Engineering I, chapter 3
- [Transition curve for 300 m radius, 7 m pavement](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-64p-tc-300) — Transportation Engineering I, chapter 3

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