---
title: "Derivation of superelevation expression"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:super-derive"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/super-derive
site: BCT Engineering
---

# Derivation of superelevation expression

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2078 Baisakh, 4 marks; 2076 Baisakh; 2076 Bhadra; 2073 Bhadra, 8 marks; 2072 Ashwin; 2071 Bhadra; 2069 Bhadra (asked 7 times)
- Video: [Derivation of superelevation.part-1 (Highway)](https://www.youtube.com/watch?v=yrrdploBiis) — B-Civilian.
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/super-derive
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Define superelevation. Derive an expression for finding the superelevation required if the design coefficient of friction is 'f'.

## Short answer

Superelevation is the transverse slope raising the outer edge to counteract centrifugal force. Derived expression: e + f = V²/(127R), where e is superelevation, f is design friction coefficient (0.15), V is design speed in km/h, R is radius in meters. Maximum e is 7% (plain/rolling) or 10% (hills); if exceeded, adopt maximum and check friction. IRC uses e = V²/(225R) for 75% design speed.

## Answer

**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).

```text
              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 $f$.

## Check yourself

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

1. Superelevation is:
   A. Raising the outer edge above the inner edge on a curve to balance centrifugal force
   B. Crown of the road on a straight
   C. Widening on curves
   D. Slope of the shoulder
   Answer: A. It is the transverse slope on a curve, outer edge raised. (Key point: Definition)

2. The derived relation for e is:
   A. e = V^2 / 127R
   B. e + f = v^2 / (gR)
   C. e - f = v^2 / (gR)
   D. e f = v^2 / gR
   Answer: B. Force balance along the incline gives e + f = v^2/(gR) after neglecting f e v^2/gR. (Key point: Derived expression)

3. Which term is neglected in the derivation?
   A. e
   B. v^2/gR
   C. f e v^2/(gR), being very small
   D. f
   Answer: C. The product term is small and dropped. (Key point: Neglected term)

4. IRC/NRS design friction f and maximum e for plain terrain are:
   A. f = 0.35, e = 4%
   B. f = 0.15, e max = 10% in plains
   C. f = 0.5, e = 15%
   D. f = 0.15, e max = 7%
   Answer: D. f = 0.15, and maximum e is 7% in plain/rolling terrain, 10% in hills. (Key point: Value of f and e max)

## Related in Transportation Engineering I

- [Superelevation design for 150 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79j-super-150) — Transportation Engineering I, chapter 3
- [Design steps of superelevation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Asuper-design-steps) — Transportation Engineering I, chapter 3, asked 3×
- [Superelevation on 100 m radius curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-62j-super-100) — Transportation Engineering I, chapter 3
- [Extra widening: reasons and derivation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Aextra-widening-derive) — Transportation Engineering I, chapter 3, asked 8×
- [Raising outer edge for superelevation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-63k-super-raise) — Transportation Engineering I, chapter 3
- [Stopping sight distance and derivation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Assd-factors-derive) — Transportation Engineering I, chapter 3, asked 6×

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