A descending gradient of 1 in 20 meets an ascending gradient of 1 in 30. What will be length(in m) of a valley curve to fulfill comfort criterion for a design speed of 80 kmph.
[Assume allowable rate of change of centrifugal acceleration 0.6 $m / s e c^{3}$ and answer correct upto one decimal place].
78.2

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Solution

The correct option is A 78.2
As per comfort criterion,
$L = 2 [\frac{N v^{3}}{c}]^{\frac{1}{2}}$
$N = \frac{- 1}{20} - \frac{1}{30} = - \frac{1}{12}$
$∵ L = 2 [\frac{\frac{1}{12} \times (0.278 \times 80)^{3}}{0.6}]^{\frac{1}{2}}$
=78.2 m

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A descending gradient of 1 in 20 meets an ascending gradient of 1 in 30. What will be length(in m) of a valley curve to fulfill comfort criterion for a design speed of 80 kmph. [Assume allowable rate of change of centrifugal acceleration 0.6 m/sec3 and answer correct upto one decimal place].78.2

The correct option is A 78.2As per comfort criterion, L=2[Nv3c]12 N=−120−130=−112 ∵L=2[112×(0.278×80)30.6]12 =78.2 m

The correct option is A 78.2
As per comfort criterion,
$L = 2 [\frac{N v^{3}}{c}]^{\frac{1}{2}}$
$N = \frac{- 1}{20} - \frac{1}{30} = - \frac{1}{12}$
$∵ L = 2 [\frac{\frac{1}{12} \times (0.278 \times 80)^{3}}{0.6}]^{\frac{1}{2}}$
=78.2 m