16-Civ-A6 Highway Design, Construction, and Maintenance · May 2017

Question 3 of 7: Crest-curve collision — was the sight distance adequate?

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Crest-curve collision — was the sight distance adequate? (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A forensic reconstruction of a rear-end collision on a symmetrical crest curve, with the object height measured on the stalled vehicle itself rather than taken from a design table.

Find. (a) The sight distance the curve actually provides, compared with the distance an alert driver needs, so that the driver's claim can be accepted or rejected; (b) the other factors that plausibly contributed.

Approach. Compute the sight distance the geometry supplies from the crest-curve relation using the two real heights, compare it with the stopping sight distance required at 80 km/h, and then use the recorded impact speed to back out how much distance the driver actually consumed — which, set against the distance available, reveals the real cause.

1. Compute the available sight distance, assuming $S \lt L$. For a crest curve with the sight line entirely on the curve, the sheet gives $$S=\sqrt{\frac{200L}{A}}\left(\sqrt{H_{1}}+\sqrt{H_{2}}\right)=\sqrt{\frac{200(270)}{6}}\left(\sqrt{1.08}+\sqrt{1.10}\right)$$ $$S=94.868\,(1.0392+1.0488)=94.868(2.0880)=\boxed{198.09\ \text{m}}$$
2. Confirm the branch that was assumed. The result $S = 198.09\ \text{m}$ is less than $L = 270\ \text{m}$, so the $S \lt L$ formula was the correct one and no re-work on the $S \ge L$ branch is needed. Skipping this check is how candidates end up quoting a distance from the wrong equation.
3. Compute the stopping sight distance the design speed requires. The critical sight line spans the crest, where the mean grade is zero, so $G = 0$ and, with the design-table $f = 0.30$, $$\text{SSD}=\frac{22.222^{2}}{2(9.81)(0.30)}+22.222(2.5)=83.90+55.56=\boxed{139.45\ \text{m}}$$ Repeating this with the friction the test runs actually measured, $f = 0.32$, gives an even shorter $134.21\ \text{m}$.
4. Compare, and answer the claim. The curve supplies $198.09\ \text{m}$ against a requirement of at most $139.45\ \text{m}$ — a surplus of roughly $59\ \text{m}$, or 42 %. $$\boxed{\text{No; the sight distance was more than adequate.}}$$ The generosity has a specific cause: the object was a stalled car $1.10\ \text{m}$ tall, not the $0.60\ \text{m}$ design object, and a taller object is visible over a crest much sooner.
5. Reconstruct what the driver actually did. The distance consumed in slowing from 80 km/h to the recorded 5 km/h impact speed, on the measured friction, is $$d=\frac{V_{0}^{2}-V_{t}^{2}}{2g\,(f+G)}+V_{0}t_{pr}=\frac{493.83-1.93}{2(9.81)(0.32)}+55.56=78.35+55.56=133.90\ \text{m}$$ so a driver reacting within the normal $2.5\ \text{s}$ would have needed only $134\ \text{m}$ of the $198\ \text{m}$ he had.
6. Back out the reaction time the collision implies. The braking component alone accounts for $78.35\ \text{m}$ of the $198.09\ \text{m}$ available, so the time actually spent before the brakes took hold was $$t_{pr,\text{implied}}=\frac{198.09-78.35}{22.222}=\boxed{5.39\ \text{s}}$$ — more than twice the design value. The geometry was not the problem; the roughly three seconds of lost reaction time was.

(b) Other contributing factors. The reconstruction points first at driver state: an implied reaction of about $5.4\ \text{s}$ is characteristic of distraction (a mobile device, in-vehicle controls, a passenger), of fatigue or impairment, or of inattention on a familiar rural road where a stopped vehicle is not expected. A second family of factors concerns conspicuity of the hazard: a stalled vehicle with no hazard lights, no warning triangle and no high-visibility marking is a low-contrast object, and if it was stopped in the running lane rather than on a shoulder it also removed the escape path that would otherwise have allowed a steering avoidance instead of a stop. Third, the environment may have degraded either the sight line or the friction — low sun angle straight down a crest, headlight glare from opposing traffic at night, fog, rain, or a wet or contaminated surface. Note that the $0.32$ friction came from test runs made after the event; if the surface was wet at the time, the real value could easily have been $0.25$ or less, which lengthens the required distance to about $160\ \text{m}$ and narrows the margin considerably. Fourth, the vehicle itself may have contributed through worn brakes or tyres, a heavy load, or an absent or defective anti-lock system. Fifth, the driver may have been travelling faster than the 80 km/h assumed in the reconstruction; since braking distance scales with the square of speed, even 100 km/h would push the requirement past $195\ \text{m}$ and effectively consume the whole margin. Finally, operational factors matter: no advance warning of a stopped vehicle, no shoulder on which to stop, and no incident-response presence all reduce the chance that a slow-reacting driver recovers in time. A defensible investigation report would identify delayed perception–reaction as the proximate cause, list surface condition and actual travel speed as the two factors that most need independent corroboration, and record that the vertical alignment met the sight-distance standard for its design speed.