Question

What will be the non-passing sight distance on a highway for a design speed of 100 kmph when its ascending gradient is 2%? Assume coefficient of friction as 0.7 and brake efficiency as 50%.
1. 176 m
2. 200 m
3. 150 m
4. 185 m

Answer 1

Correct Answer - Option 1 : 176 m

Concept:

Crossing sight distance = Stopping sight distance for rising vehicle + Stopping sight distance for approaching vehicle

Stopping sight distance \({\rm{(SSD)}} = {\rm{V}} \times {\rm{t}} + \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}\left( {{\rm{\eta }} \times {\rm{f}} \pm {\rm{N}}} \right)}}\)

Where,

v = Design speed (98th percentile speed), t = reaction time of the driver, η = efficiency of the brake,

f = friction coefficient on road, N = gradient of road,

‘+’ → for raising gradient, ‘-' → for falling gradient.

Calculation:

V = 100 kmph = 27.78 m/s

n = ± 0.02

F = 0.7

η_brake = 50%

As the brake efficiency is 50%, consider f_m  = 0.5 × 0.7 = 0.35

\(SSD = V \times t + \frac{{{V^2}}}{{2g\left( {{F_m} + n} \right)}}\)

\(= 27.78 \times 2.5 + \frac{{{{\left( {27.78} \right)}^2}}}{{2 \times 9.81\;\left( {0.35 + 0.02} \right)}}\)

= 69.44 + 106.30

= 175.75 m ∼ 176 m