Question

A descending gradient of 1 in 20 meets an ascending gradient of 1 in 30. What will be length(in m) of a valley curve to fulfill comfort criterion for a design speed of 80 kmph.
[Assume allowable rate of change of centrifugal acceleration 0.6 $m/se{c}^3$ and answer correct upto one decimal place].

Answer 1

As per comfort criterion,
$L=2[\frac{N{v}^3}{c}{]}^{\frac{1}{2}}$
$N=\frac{-1}{20}-\frac{1}{30}=-\frac{1}{12}$
$\because L=2[\frac{\frac{1}{12}\times (0.278\times 80{)}^3}{0.6}{]}^{\frac{1}{2}}$
=78.2 m