Question

A descending gradient of 4 % meets an ascending grade of 1 in 40 where a valley curve of length 200 must to be formed. What will be the distance of the lowest point on the valley curve from its first tangent point?
1. 100 m
2. 111 m
3. 125 m
4. 118 m

Answer 1

Correct Answer - Option 2 : 111 m

Concept:

The lowest point on the valley curve:

The lowest point on the valley curve is to be located for providing the cross drainage facility. The lowest

point on the valley curve will be on the bisector of the angle between the grades if the gradients on

either side are equal. When the gradients are not equal, the lowest point lies on the side of the flatter

grade, and this point is from the tangent point of the first gradient, n₁ at a distance, X_o given by:

\({X_0} = L \times {\left( {\frac{{{n_1}}}{{2N}}} \right)^{1/2}}\)

Where X₀ = for distance from the end of the first tangent point t₀ the lowest point in meters.

n₁ = first gradient.

N = |n₁ – n₂|, L = length of valley curve

Calculation:

Descending gradient = 4%

\({n_1} = \frac{{ - 4}}{{100}}\)

Ascending gradient = 1 in 40

\({n_2} = \frac{1}{{40}}\)

\(V = \left[ {\frac{{ - 4}}{{100}} - \frac{1}{{40}}} \right] = 0.065\)

L = 200 m

\(x = 200 \times {\left( {\frac{{0.04}}{{2 \times 0.065}}} \right)^{1/2}}\)

\(= 110.94\;m \sim 111\;m\)