Question

A vertical curve is formed by a descending gradient of 1 in 40 meeting an ascending gradient of 1 in 50. Consider the following:

Stopping Sight Distance (SSD) = 90 m
Height of headlight of a vehicle above the road surface = 0.75 m
Headlight beam angle with respect to the longitudinal axis of the vehicle = \(1.2^\circ\)

Based on the sight distance criteria, the design length (in m) of the vertical curve is (rounded off to the nearest integer).

Hint:

This is a valley curve; find \(N = |g_1| + |g_2|\), then test the headlight sight distance formulas for \(S < L\) and \(S \geq L\) to see which is self-consistent.

Answer 1

Correct Answer: 63

Step 1: Identify the curve type and the two grades.
A descending gradient meeting an ascending gradient forms a valley (sag) curve. Taking the falling grade as negative and the rising grade as positive:
\[ g_1 = -\frac{1}{40} = -0.025, \qquad g_2 = +\frac{1}{50} = +0.02 \]
The total change in grade (deviation angle) is
\[ N = g_2 - g_1 = 0.02 - (-0.025) = 0.045 \]

Step 2: Recall the headlight sight distance formula for a valley curve.
At night, the limiting sight distance on a valley curve is set by how far the headlight beam reaches ahead. Let \(h_1\) be the headlight height and \(\alpha\) the upward beam divergence angle. Two cases arise depending on whether the sight distance \(S\) is less than or greater than the curve length \(L\):
\[ \text{if } S < L: \quad L = \frac{NS^2}{2(h_1 + S\tan\alpha)} \]
\[ \text{if } S \geq L: \quad L = 2S - \frac{2(h_1 + S\tan\alpha)}{N} \]

Step 3: Compute the common term \(h_1 + S\tan\alpha\).
\(\tan(1.2^\circ) \approx 0.02095\), so
\[ h_1 + S\tan\alpha = 0.75 + 90(0.02095) = 0.75 + 1.885 = 2.635 \]

Step 4: Try the \(S < L\) case first and check consistency.
\[ L = \frac{NS^2}{2(h_1+S\tan\alpha)} = \frac{0.045 \times 90^2}{2(2.635)} = \frac{364.5}{5.270} = 69.17 \text{ m} \]
This gives \(L = 69.17\) m, which is less than \(S = 90\) m, contradicting the assumption \(S < L\). So this case is not valid; the other case must be used.

Step 5: Use the \(S \geq L\) case.
\[ L = 2S - \frac{2(h_1+S\tan\alpha)}{N} = 2(90) - \frac{2(2.635)}{0.045} = 180 - 117.12 \]
\[ L = 62.88 \text{ m} \]
Here \(S = 90\) m \(\geq L = 62.88\) m, consistent with this case's own assumption, confirming it is the correct branch.

Final Answer:
Rounded to the nearest integer,
\[ \boxed{L \approx 63 \text{ m}} \]