Question

There is a Horizontal curve having a radius of 500 m and a length of 250 m on a National highway. The distance between the centerline of the road and the inner lane is 1.8 m. Consider OSD as 300 m. The setback distance is

• A. 22.62
• B. 21.62
• C. 23.62
• D. 20.62

Answer 1

The correct option is C 23.62


$R=500m,d=1.8m$

$l=250m;OSD=S=300m;$

As i < s

$l=(R-d)$

$250=(500-1.8)$

$\alpha =0.5018\times \frac{180}{\pi }={28.75}^{\circ }$

$\frac{\alpha }{2}={14.38}^{\circ }$

For $i<s$

Set back distance $=R-(R-d)\cos \frac{\alpha }{2}+\frac{S-l}{2}\sin \frac{\alpha }{2}$

$=500-498.2\cos \left(14.38\right)+\frac{300-250}{2}\sin \left(14.38\right)$

$=500-482.59+6.208=23.619\approx 23.62m$