Question

A falling gradient of 1 in 40 meets a rising gradient of 1 in 60. The length of valley curve to be provided for a head light sight distance of 120 m will be ______m.

• A. 103.2

Answer 1

The correct option is A 103.2
Based on head light sight distance length of valley curve,
$L_V=\frac{N{S}^2}{2h+2s\tan \alpha }$
$=\frac{N{S}^2}{1.5+0.035s}$
Assuming $L_V<SD$
$\therefore L_V=\frac{|-\frac{1}{40}-\frac{1}{60}|\times {120}^2}{1.5+0.035\times 120}$
$=105.26m\ngtr SD(=120m)$
Hence, the assumption is incorrect,
$\therefore For{L}_V<SD,$
$L_V=2s-\frac{1.5+0.035s}{N}$
$=2\times 120-\frac{1.5+0.035\times 120}{|-\frac{1}{40}-\frac{1}{60}|}$
$\therefore L_V=103.2m$