Question

A rising gradient of 1 in 50 meets a falling gradient of 1 in 30. Which one of the following is the length of vertical curve if the stopping sight distance is 120 m?

• A. 174 m
• B. 140 m
• C. 158 m
• D. 120 m

Answer 1

The correct option is A 174 m

Deviation angle,

$N=\frac{1}{50}+\frac{1}{30}=\frac{4}{75}$

Assuming L > S

$L=\frac{N{S}^2}{4.4}=\frac{4}{75}\times \frac{{120}^2}{4.4}$

= 174.5 m