Question

There is a horizontal curve of radius $360m$ and length $180m$. Calculate the clearance required from the central line on the inner side of the curve, so as to provide an overtaking sight distance of $250m$.

• A. 19.85

Answer 1

The correct option is A 19.85

Given,
$R=360m$
$L_C=180m$
$S=250m$

Clearance required from the center line on the inner area of curve
$=M_1+M_2....\left(1\right)$

$\frac{\alpha }{2}=\frac{L_C}{R}\times \frac{180}{2\pi }$

$\frac{\alpha }{2}=\frac{180}{360}\times \frac{180}{2\pi }$

$\frac{\alpha }{2}=\frac{1}{2}\times \frac{180}{2\pi }={14.324}^{\circ }$

$M_1=R-R\cos \frac{\alpha }{2}$

$=360(1-\cos 14.324)$

$=11.191m$

$M_2=\left(\frac{S-L_C}{2}\right)\sin \frac{\alpha }{2}$

$=\left(\frac{250-180}{2}\right)\sin 14.324$

$=8.659m$

Hence,
Required clearance
$=11.191+8.659$
$=19.85m$