Question

A sharp horizontal curve of radius 100m, the gradient is 4.5%. The percentage reduction in gradient to compensate the loss of traction force due to curvature is

• A. 3.75
• B. 0.5
• C. 1.87
• D. 0.75

Answer 1

The correct option is B 0.5

Grade compensation= min $[\frac{(30+R)}{R},\frac{75}{R}]$ such that compensated gradient should not be less than 4%
$=min(\frac{130}{100},\frac{75}{100})=0.75\%$
Gradient provided after grade compensation
$=4.5-0.75=3.75\%<4\%$
Hence grade compensation provided is $4.5-4=0.5\%$