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Calculate the stopping sight distance on a highway at a descending gradient of 2% for a design. Speed of 80kmph.Assume other data as per IRC standards if required
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Descending gradient = 2%

V = 80 kmph = 22.22 m/sec

$SSD = Vt + \frac{V^2}{2g(f - \frac{\pi}{100})}....\textit{-ve for descending gradient}\\ = 22.22\times2.5 + \frac{22.22^2}{2\times9.81(0.35 - \frac{0.2}{100})}\\ = 127.86 m \approx128 m$

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