Question

A falling gradient of 1 in 25 meets a rising gradient of 1 in 30. The length of the vertical curve to fulfil the sight distance criteria if stopping sight distance is 130 m in 'm' is_____

• A. 204.84

Answer 1

The correct option is A 204.84
Deviation angle $=\left(\frac{-1}{25}\right)-\left(\frac{1}{30}\right)=\left(\frac{-11}{150}\right)$
Hence valley curve is to be designed
Assuming length of valley curve (L) > SSD
$L=\frac{N{S}^2}{1.5+0.035S}=\frac{\left(\frac{11}{150}\right)\times {130}^2}{1.5+(0.035\times 130)}=204.84m>130m$
Hence assumption is correct. Length of valley curve $=204.84m$