Question

A falling gradient of $1$ in $20$ meets a rising gradient of $1$ in $40$ on a highway having design speed of $80km/hr$. If safe stopping sight distance is $120m$ then length of valley curve is _____ $m$

Answer 1

Design speed = $80km/hr$

$N=|n_1=n_2|=|\frac{1}{40}-(-\frac{1}{20})|=0.075$

Stopping sight distance $=120m$

Length of valley curve by sight distance criterion:

Assuming

(L > SSD)

$L_V\frac{N{S}^2}{1.5+0.035S}$

$=\frac{0.075\times {120}^2}{1.5+0.035\times 120}=189.47m>SSD$
OK
Form comfort criterion,
= $L_V=2{\left[\frac{N{V}^3}{c}\right]}^{1/2}=0.38(N{V}^3{)}^{1/2}$

$=0.38[0.075\times 80{]}^{1/2}$

$=74.46m$

$therefore,$ Length of valley curve to be provided $=189.47m$. [Maximum of the above two criteria]