Question

The rate of superelevation for a horizontal curve of radius 500 m in a national highway for a design speed of 100 kmph is

• A. 0.07
• B. 0.70
• C. 0.04
• D. 0.063

Answer 1

The correct option is A 0.07

$e=\frac{V^2}{225R}=\frac{(100{)}^2}{225\times 500}$
$=0.088$

Now limit e = 0.07

$e+f=\frac{V^2}{127R}$
$\Rightarrow f=\frac{{100}^2}{127\times 500}=0.07$
$=0.087\left(OK\right)<0.15$

Provide, e = 0.07