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

A falling grade running into a rising grade creates a valley (sag) curve, and for night driving the controlling sight distance is not obstruction-based but headlight-based: how far ahead the headlight beam, tilted slightly upward, actually lights up the road.

First combine the two grades into the total deviation angle $N$. Taking the descending grade as negative and the ascending grade as positive:

\[ N = \left|-\frac{1}{40}\right| + \left|\frac{1}{50}\right| = 0.025 + 0.02 = 0.045 \]

Next work out the reach term $h_1 + S\tan\alpha$, which combines the headlight's mounting height with the extra rise the beam picks up over the sight distance $S = 90$ m due to its $1.2^\circ$ upward tilt:

\[ \tan(1.2^\circ) \approx 0.02095, \qquad h_1 + S\tan\alpha = 0.75 + 90(0.02095) = 2.635 \text{ m} \]

There are two standard formulas depending on whether the curve is longer or shorter than the sight distance. Rather than guessing, test the short-curve formula ($S \geq L$) first, since a fairly flat pair of grades (1 in 40 and 1 in 50 are both gentle) usually gives a curve shorter than a 90 m sight distance:

\[ L = 2S - \frac{2(h_1+S\tan\alpha)}{N} = 2(90) - \frac{2(2.635)}{0.045} = 180 - 117.12 = 62.88 \text{ m} \]

Check the assumption: is $S \geq L$? Here $S = 90$ m and $L = 62.88$ m, so yes, $90 \geq 62.88$ holds, confirming this is the right formula (if it had failed, the other formula for $S < L$ would apply instead).

Let's summarize:

• The total grade change $N = 0.045$ combines the magnitudes of both grades since one is falling and the other rising.
• The reach term $h_1 + S\tan\alpha = 2.635$ m accounts for both the headlight's height and its upward tilt over the sight distance.
• Testing the $S \geq L$ formula gives a self-consistent answer of $L = 62.88$ m, confirming that branch applies.

So the design length of the vertical curve, rounded to the nearest integer, is $L \approx 63$ m.