Question

An ascending gradient of 1 in 100 meets a descending gradient of 1 in 50. The length of summit curve required to provide over turning sight distance of 500 m will be:
1. 938.25 m
2. 781.25 m
3. 470.25 m
4. 170.25 m

Answer 1

Correct Answer - Option 2 : 781.25 m

Concept:

Length of summit curve is given by,

If we assume L > OSD

\({\rm{L}} = \frac{{{\rm{N}}{{\rm{S}}^2}}}{{2{{\left[ {\sqrt {{{\rm{h}}_1}} + \sqrt {{{\rm{h}}_2}} } \right]}^2}}}\)

If we assume L < OSD

\({\rm{L}} = 2{\rm{S}} - \frac{{2{{\left[ {\sqrt {{{\rm{h}}_1}} + \sqrt {{{\rm{h}}_2}} } \right]}^2}}}{{\rm{N}}}\)

Where, N = Deviation angle & S = Stoping distance

Calculation:

Given: Upward gradient = 1 in 100, Downward gradient = 1 in 50, Stopping sight distance = 500 m, and \({\rm{N}} = \frac{1}{{100}} - \left( { - \frac{1}{{50}}} \right) = 0.03\)

For overturning sight distance, h₁ = 1.2 m & h₂ = 1.2 m

Case 1: If L > OSD

\({\rm{L}} = \frac{{0.03 \times {{500}^2}}}{{2{{\left[ {\sqrt {1.2} + \sqrt {1.2} } \right]}^2}}} = 781.25{\rm{\;m}}\)

Case 2: If L < OSD

\({\rm{L}} = 2 \times 500 - \frac{{2{{\left[ {\sqrt {1.2} + \sqrt {1.2} } \right]}^2}}}{{0.03}} = 680{\rm{\;m}}\)

Thus the assumption is not correct.

∴ Length of summit curve is 781.25 m.