The rate of equilibrium super-elevation on a road is

1. Directly proportional to the square of vehicle velocity

2. Inversely proportional to the radius of the horizontal curve

3. Directly proportional to the square of the radius of the horizontal curve

Which of the above statements are correct?

This question was previously asked in

ESE Civil 2018 Official Paper

Option 1 : 1 and 2 only

CT 1: विलोम शब्द

4735

10 Questions
30 Marks
10 Mins

__Concept:__

**General equation of superelevation** is given by,

\(e + f = \frac{{{V^2}}}{{g\;R}}\)

Where,

e = rate of superelevation = tan θ

R = radius of the horizontal curve in m

f = coefficient of lateral friction = 0.15

**Equilibrium superelevation:**

It is that superelevation which when provided imposes **equal pressure on both outsides and inside** of tyres of the vehicle, i.e **f = 0**

**The rate of equilibrium superelevation on a road is given by**

\(e = \frac{{{V^2}}}{{g\;R}}\)

__Explanation:__

The rate of equilibrium superelevation on a road is given by

\(e = \frac{{{V^2}}}{{g\;R}}\)

Where,

e α V^{2} i.e The rate of equilibrium superelevation on a road is directly proportional to **the square of the velocity.**

e α 1/R i.e The rate of equilibrium superelevation on a road is inversely proportional to the **radius of the horizontal curve.**

Hence only statement 1 and 2 are **correct**