---
title: "Superelevation design for 150 m radius"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-79j-super-150"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-79j-super-150
site: BCT Engineering
---

# Superelevation design for 150 m radius

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2079 Jestha, 8 marks
- Video: [Super Elevation calculation](https://www.youtube.com/watch?v=HS4EP8JSliY) — Engr Suliman Khan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-79j-super-150
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

The design speed is 60 kmph and the maximum permissible value of super elevation and coefficient of lateral friction are 0.07 and 0.15. Highway engineer is planning to design super elevation for a curve of radius 150 m. (i) Find out the equilibrium super elevation. ii) Design the super elevation for this curve. (iii) if the maximum super elevation is not to be exceeded, calculate the maximum allowable speed on this curve. (iv) What would be the ruling radius required if speed of 60 kmph is to be maintained.

## Short answer

(i) Equilibrium e=0.189. (ii) Design e=0.07 (f=0.119). (iii) Maximum allowable speed = 64.7 km/h. (iv) Ruling radius for 60 km/h ≈129 m.

## Answer

#### Given
$V=60$ km/h, $R=150$ m, $e_{max}=0.07$, $f_{max}=0.15$ (IRC 73 / NRS 2070 approach).

#### (i) Equilibrium superelevation
Equilibrium superelevation is the value for which the centrifugal force is balanced entirely by the superelevation (no side friction, $f=0$):

$$
e=\frac{V^2}{127R}=\frac{60^2}{127(150)}=0.189\ (=18.9\%)
$$

#### (ii) Design superelevation (IRC method)
1. Superelevation for 75% of design speed, no friction: $e=\dfrac{V^2}{225R}=\dfrac{3600}{225(150)}=0.107$.
2. This exceeds $e_{max}=0.07$, so provide $e=e_{max}=\mathbf{0.07}$ (7%).
3. Check friction needed at full design speed:

$$
f=\frac{V^2}{127R}-e=0.189-0.07=0.119
$$

$f=0.119<0.15$, so the design is safe.

#### (iii) Maximum allowable speed if $e$ is not to exceed 0.07
With $e=0.07$ and $f=0.15$:

$$
V=\sqrt{127R(e+f)}=\sqrt{127(150)(0.22)}=64.7\ \text{km/h}
$$

Allowable speed is about **65 km/h**.

#### (iv) Ruling radius for 60 km/h

$$
R=\frac{V^2}{127(e+f)}=\frac{60^2}{127(0.07+0.15)}=128.8\ \text{m}
$$

**Answer:** (i) 0.189; (ii) 0.07 (friction required 0.119); (iii) 64.7 km/h; (iv) about 129 m.

## Check yourself

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

1. What is the equilibrium superelevation at 60 km/h and R = 150 m?
   A. 0.107
   B. 0.189
   C. 0.07
   D. 0.119
   Answer: B. Equilibrium means f = 0: e = V^2/(127R) = 3600/19050 = 0.189. (Key point: Equilibrium superelevation)

2. In the IRC method, superelevation is first computed for
   A. full design speed with friction
   B. 75% of design speed with no friction, e = V^2/225R
   C. double the design speed
   D. zero speed
   Answer: B. e = V^2/(225R) = 0.107, which exceeds 0.07, so e_max = 0.07 is provided. (Key point: IRC design method)

3. After providing e = 0.07, the friction needed at full speed is
   A. 0.15
   B. 0.119, which is less than 0.15, so safe
   C. 0.189
   D. 0.07
   Answer: B. f = V^2/127R - e = 0.189 - 0.07 = 0.119 < 0.15. (Key point: Friction check)

4. With e = 0.07 and f = 0.15 on R = 150 m, the maximum speed is about
   A. 60 km/h
   B. 64.7 km/h
   C. 75 km/h
   D. 50 km/h
   Answer: B. V = sqrt(127 x 150 x 0.22) = 64.7 km/h. (Key point: Maximum allowable speed)

## Related in Transportation Engineering I

- [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
- [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 220 m radius, 65 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-68bh-tc-65) — Transportation Engineering I, chapter 3
- [Vertical curve +2% and +5.5%, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79c-vc-2-55) — Transportation Engineering I, chapter 3
- [Transition curve 400 m radius, 85 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-tc-400-85) — Transportation Engineering I, chapter 3, asked 2×
- [Transition curve in hilly region, 60 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65c-tc-hilly) — Transportation Engineering I, chapter 3

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