Question 4 of 5: Collision reconstruction on a crest vertical curve

Question 4: Collision reconstruction on a crest vertical curve (20 marks: a 8, b 8, c 4)

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 crest vertical curve joining a +3 % grade to a −2 % grade over 350 m on a rural highway posted at 90 km/h, with a driver eye height of 1.38 m and a stalled-vehicle (object) height of 1.10 m.

Find. (a) whether the sight distance was in fact deficient at 90 km/h; (b) the sight distance the geometry actually provides; (c) other contributing factors, one of them quantified.

Approach. Invert the crest-curve sight-distance formula with the real object height to get the sight distance the geometry supplies, compare it with the stopping sight distance required at the posted speed, then work backwards from the collision to see what would have had to be true for the crash to happen.

1. Characterise the curve. The algebraic difference of grades and the rate of vertical curvature are $$A = |G_2 - G_1| = |-2 - 3| = 5\,\%, \qquad K = \frac{L}{A} = \frac{350}{5} = 70\ \text{m per \% of grade}$$ The curve is a crest, so the controlling sight obstruction is the pavement surface itself, and both the driver eye height and the object height enter the calculation.
2. (b) Invert the crest-curve formula using the measured object height. The appendix gives $$L = \frac{S^{2}A}{200\left(\sqrt{H}+\sqrt{h_1}\right)^{2}} \quad \Rightarrow \quad S = \sqrt{\frac{200\,L\left(\sqrt{H}+\sqrt{h_1}\right)^{2}}{A}}$$ With $H = 1.38\ \text{m}$ and $h_1 = 1.10\ \text{m}$, $\left(\sqrt{1.38}+\sqrt{1.10}\right)^{2} = (1.1747+1.0488)^{2} = 4.944$, so $$S = \sqrt{\frac{200 \times 350 \times 4.944}{5}} = \sqrt{69\,218} = \boxed{263.1\ \text{m}}$$ The result satisfies $S \lt L$, so the branch used is the correct one. This is the single most important step in the question: a standard design chart would have used the 0.60 m design object height, giving only 208 m. The obstruction here was a whole stalled car standing 1.10 m tall, and the higher the object the sooner it appears over a crest.
3. (a) Compute the stopping sight distance actually required at the posted speed. The two test runs bracket the posted speed, so interpolate linearly: $$f_{90} = \tfrac{1}{2}(0.32 + 0.29) = 0.305$$ which agrees with the 0.31 the appendix friction table lists at 90 km/h. Braking takes place on the descending 2 % leg, which is the unfavourable direction, so with $v = 90/3.6 = 25\ \text{m/s}$ and $t_{pr} = 2.5\ \text{s}$, $$SSD = v\,t_{pr} + \frac{v^{2}}{2g(f+G)} = 25(2.5) + \frac{25^{2}}{2(9.81)(0.305-0.02)} = 62.5 + 111.8 = \boxed{174.3\ \text{m}}$$ On the level the requirement would be 166.9 m, so the downgrade adds only about 7 m.
4. Deliver the verdict on the claim. $$S_{\text{available}} = 263.1\ \text{m} \quad \text{versus} \quad SSD_{\text{required}} = 174.3\ \text{m}$$ The curve provides 88.8 m — more than half as much again — beyond what a driver travelling at the posted 90 km/h needs. The driver’s claim is refuted. The geometry of this crest was not deficient; something about the driving was.
5. Quantify how much speed the geometry would in fact have tolerated. Solving $v\,t_{pr} + v^{2}/[2g(f(V)-0.02)] = 263.1\ \text{m}$ with the friction taken from the appendix table at each trial speed gives $$\boxed{V_{\max} \approx 111\ \text{km/h}}$$ So the available sight distance would have allowed a stop from about 111 km/h. The driver could have been 20 km/h over the limit and still have stopped short of the stalled vehicle.
6. (c) Back-calculate what the collision implies about the driver, at the stated speed. If the speed really was 90 km/h and the friction really was 0.305, then for the vehicle to consume all 263.1 m the perception–reaction time must have been $$t_{pr} = \frac{S - d_b}{v} = \frac{263.1 - 111.8}{25} = \boxed{6.05\ \text{s}}$$ against the 2.5 s design value and the roughly 1.5 s an alert driver actually needs. A 6-second delay is not a normal reaction; it is the signature of inattention, distraction, drowsiness or impairment.
7. Test the competing explanation — a surface that could not deliver the tested friction. Alternatively, hold $t_{pr}$ at 2.5 s and ask what friction the collision implies: $$f = \frac{v^{2}}{2g\left(S - v\,t_{pr}\right)} + G = \frac{625}{2(9.81)(263.1-62.5)} + 0.02 = \boxed{0.179}$$ That is 41 % below the tested dry-surface value and squarely in the range measured on wet asphalt or on worn tyres. Either explanation is arithmetically sufficient, and the two are not exclusive.
8. Assemble the list of contributing factors. The calculations above narrow the field to a short list, of which the wet-surface branch is the one quantified: (i) surface condition — rain, frost or contamination reducing the available friction to about 0.18, which is exactly what the collision implies; (ii) tyre condition — worn or under-inflated tyres produce the same reduction; (iii) driver attention — a perception–reaction time of 6 s, indicating distraction, fatigue or impairment; (iv) actual speed — the claim “very close to the posted speed” is unverified, and the available sight distance only becomes marginal above about 111 km/h; (v) conspicuity of the stalled vehicle — darkness, no hazard lights or no advance warning device, which delays recognition even when the sight line is geometrically clear; and (vi) the absence of a shoulder wide enough to remove the stalled vehicle from the travelled way, which is the only one of the six that is a highway-design deficiency rather than a driving one.