Question

An ascending gradient of 1 in 40 meets a descending gradient of 1 in 60. What will be the length of summit curve for a stopping sight distance of 100 m?

• A. 80 m
• B. 100 m
• C. 120 m
• D. 95 m

Answer 1

The correct option is D 95 m

$N=\frac{1}{40}-\left(\frac{-1}{60}\right)=\frac{1}{24}$
S = 100 m

Assuming , L > S
$\therefore L=\frac{N{S}^2}{4.4}=\frac{1}{24}\times \frac{(100{)}^2}{4.4}=94.7m$
Our assumption is wrong

For L < S
$\therefore L=2S-\frac{4.4}{N}$
$=2\times 100-\frac{4.4}{\left(1/24\right)}=94.4m$
$\therefore L=95m$