Question

A valley curve of length 240 m is to be formed where a descending gradient of 1 in 20 meets an ascending gradient of 1 in 40 . The distance of the lowest point on the valley curve from its first tangent point will be

Answer 1

The distance is given by
$x=L\sqrt{\frac{N_1}{2N}}$
$where,L=Lengthofvalleycurve=240m,N_1=1in20$
$N=N_1+N_2=\frac{1}{20}+\frac{1}{40}=0.075$
$\Rightarrow x=240\sqrt{\frac{1/20}{2\times 0.075}}=138.56\simeq 139m$