Question

A descending gradient of 4% meets an ascending grade of 1 in 40 where a valley curve of length 200 m is to be formed. What will be the distance of the lowest point on the valley curve from its first tangent point?

Answer 1

$x=L{\left(\frac{n_1}{2N}\right)}^{1/2}$

x = Horizontal distance from end of first tangent point to the lowest point in metres.

$n_1=Naturaltangentoffirsttangent$

$N=n_1+n_2$

$=|-\frac{4}{10}-\frac{1}{40}|=0.065$

$x=200\left(\frac{0.04}{2\times 0.065}\right)61/2$

$=110.94\simeq 111m$