Crest-curve sight distance and accident causation: 16-Civ-A6 Highway Design, Construction, and Maintenance, May 2018, Q1

Question 1: Crest-curve sight distance and accident causation (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 symmetrical parabolic crest curve joining a rising grade to a falling grade, with the geometry, the two sight heights and the measured friction all taken from the police investigation:

Data recovered from the investigation
Quantity	Symbol	Value
Approach grade	$G_{1}$	$+2\ \%$
Departure grade	$G_{2}$	$-3\ \%$
Curve length	$L$	$300\ \text{m}$
Driver eye height	$h_{1}$	$1.08\ \text{m}$
Height of the stalled vehicle	$h_{2}$	$1.10\ \text{m}$
Travelling speed	$V_{0}$	$90\ \text{km/h}$
Impact speed	$V_{t}$	$10\ \text{km/h}$
Posted design speed	$V_{d}$	$80\ \text{km/h}$
Longitudinal friction (test runs)	$f$	$0.36$

Find. Whether the sight distance the curve actually provides was less than the stopping sight distance the driver needed, and — if it was not — what the collision then tells us about the other contributing factors.

Figure 1.1 — the +2 % / −3 % crest curve. The limiting sight line runs from the 1.08 m eye height to the 1.10 m target and grazes the pavement at the apex.

Approach. Invert the crest-curve sight-distance relation printed on appendix page 8 to obtain the sight distance the geometry supplies, compare it with the stopping sight distance demanded at 90 km/h, and then use the 10 km/h impact speed to back-calculate the perception–reaction time the driver actually used.

Algebraic difference of grades. The rate of change of grade over the curve is set by$$A=\left|G_{2}-G_{1}\right|=\left|-3-(+2)\right|=5\ \%$$so the rate of vertical curvature is $K=L/A=300/5=60\ \text{m per }\%$.
Sight distance the curve provides. For $S\lt L$ the appendix gives$$L=\dfrac{S^{2}A}{200\left(\sqrt{h_{1}}+\sqrt{h_{2}}\right)^{2}}\qquad\Longrightarrow\qquad S=\sqrt{\dfrac{200\,L\left(\sqrt{h_{1}}+\sqrt{h_{2}}\right)^{2}}{A}}$$Substituting the two heights, $\left(\sqrt{1.08}+\sqrt{1.10}\right)^{2}=(1.0392+1.0488)^{2}=4.3599$, so$$S=\sqrt{\dfrac{200\times 300\times 4.3599}{5}}=\boxed{228.7\ \text{m}}$$The assumption is self-consistent because $228.7\ \text{m}\lt 300\ \text{m}=L$.
Sight distance the driver needed. With the appendix braking relation and a 2.5 s perception–reaction time, at $V_{0}=90\ \text{km/h}$ ($v_{0}=25.0\ \text{m/s}$),$$\text{SSD}=v_{0}t_{pr}+\dfrac{v_{0}^{2}}{2g(f+G)}=25.0(2.5)+\dfrac{25.0^{2}}{2(9.81)(0.36)}=62.5+88.5=\boxed{151.0\ \text{m}}$$Taking the crest as effectively level is the correct reading here: the limiting sight line straddles the apex, where the instantaneous grade passes through zero.
Check the least favourable grade and the design speed. Even if the whole braking manoeuvre is charged to the steepest downgrade limb, $G=-0.03$ gives $\text{SSD}=62.5+25.0^{2}/[2(9.81)(0.33)]=159.0\ \text{m}$, and at the posted design speed of 80 km/h the demand falls to $\text{SSD}=55.6+69.9=125.5\ \text{m}$. Both remain far below the 228.7 m available.
Verdict on the claim. The margin is large in every reading:$$S_{\text{available}}=228.7\ \text{m}\ \gt\ \text{SSD}_{90}=151.0\ \text{m}\quad(\text{a surplus of }77.7\ \text{m},\ 51\ \%)$$so the curve supplied more than enough sight distance even for the 90 km/h travel speed, which is 10 km/h above the design speed. I would not agree with the driver.
What the impact speed implies. Braking from 90 km/h to the 10 km/h impact speed consumes$$d_{b}=\dfrac{v_{0}^{2}-v_{t}^{2}}{2g f}=\dfrac{25.0^{2}-2.778^{2}}{2(9.81)(0.36)}=87.4\ \text{m}$$If the driver first saw the obstruction at the limit of sight, the remaining 228.7 − 87.4 = 141.3 m was covered before the brakes took hold, which corresponds to$$t_{pr,\text{implied}}=\dfrac{141.3}{25.0}=\boxed{5.65\ \text{s}}$$more than twice the 2.5 s design value. The sight distance was not the problem; the response to it was.
Independent check on the geometry itself. Against the standard AASHTO heights (1.08 m eye, 0.60 m object) the same curve yields$$S=\sqrt{200\,K\left(\sqrt{1.08}+\sqrt{0.60}\right)^{2}}=\sqrt{200(60)(3.2900)}=198.7\ \text{m}$$against a requirement of $200.9\ \text{m}$ at 100 km/h ($f=0.30$). The alignment is therefore adequate for roughly 100 km/h, comfortably above its own 80 km/h design speed.

Question 1 — results
Quantity	Value	Comment
Algebraic grade difference $A$	5 %	$K=60$ m per %
Sight distance available	228.7 m	$S\lt L$ branch confirmed
SSD required at 90 km/h	151.0 m	2.5 s reaction, $f=0.36$
SSD required at 90 km/h on −3 %	159.0 m	least favourable reading
SSD required at the 80 km/h design speed	125.5 m	design condition
Implied perception–reaction time	5.65 s	versus 2.5 s design
Speed the geometry supports (0.60 m object)	about 100 km/h	$S=198.7$ m available
Answer to (a)	Claim rejected	sight distance exceeded the requirement by 51 %

(b) Other factors that contributed to the collision (6 marks)

Because the geometry is exonerated, the causes must lie with the driver, the stalled vehicle and the operating environment. The single largest contributor is the travel speed: 90 km/h on a highway designed for 80 km/h lengthens the braking distance by about 27 % and shortens every decision window in proportion. The second is the driver’s reaction. A perception–reaction time of 5.65 s is characteristic of distraction, fatigue, impairment or an unexpected stimulus that the driver took time to recognise as a hazard rather than as roadside furniture — a stationary vehicle on a crest is a classic “looked-but-failed-to-see” target because it presents no relative motion.

The stalled vehicle itself is the third factor. If it was left in the travelled lane without hazard lights, reflective triangles or a flare, and with no shoulder wide enough to move it clear, it converted a routine breakdown into an unlit fixed object at the point of least conspicuity. A related deficiency is the absence of an emergency shoulder or refuge over the crest.

Fourth, the pavement and vehicle condition deserve scrutiny. A measured longitudinal friction of 0.36 is close to the 0.31–0.35 band the appendix table assigns to 80–90 km/h, so the surface was not grossly polished, but wet or contaminated pavement, worn tyres or a degraded brake system would each erode the reserve. Finally, operational and traffic-control shortcomings complete the picture: no advance warning of a crest with restricted forward visibility, no delineation or chevrons, no lighting, and no incident-detection or rapid-response arrangement to remove disabled vehicles. A recommendation set follows directly — speed enforcement, an advance “limited sight distance” warning, shoulder widening over the crest, and a motorist-assistance patrol.

Topic. Stopping sight distance on crest vertical curves, and its use in collision reconstruction.

Key relations. Crest-curve sight distance $L=S^{2}A/[200(\sqrt{h_{1}}+\sqrt{h_{2}})^{2}]$ for $S\lt L$ and $L=2S-200(\sqrt{h_{1}}+\sqrt{h_{2}})^{2}/A$ for $S\ge L$; rate of vertical curvature $K=L/A$; stopping sight distance $\text{SSD}=v t_{pr}+v^{2}/[2g(f\pm G)]$.

Why this works. On a crest the pavement itself is the obstruction: the limiting line of sight is the chord that just grazes the parabola, running from the driver’s eye to the top of the target. Because the parabola’s offset from its tangent grows with the square of the distance from the tangent point, the geometry collapses into the single relation above, in which the two heights enter only through the combination $(\sqrt{h_{1}}+\sqrt{h_{2}})^{2}$. That combination is why the object height matters so much: raising the target from the standard 0.60 m to the 1.10 m roof line of a real car increases the available sight distance by about 15 %, and using the packaged $K=\text{SSD}^{2}/658$ — which silently assumes a 0.60 m object — would have understated the sight distance and wrongly supported the driver’s claim. Once the available distance is known, the impact speed closes the system: the energy equation fixes the braking distance, and the remainder of the sight distance must have been consumed by reaction, so the reaction time becomes a measured quantity rather than an assumption.

Common pitfall. Two recur. Students compare the available sight distance with the SSD at the design speed and conclude the curve is fine without ever testing the actual 90 km/h travel speed; and they use the 0.60 m standard object height even though the question supplies a measured 1.10 m, which changes the answer by 15 % in exactly the direction that matters.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §3.4 (stopping sight distance) and §15.6 (vertical curves); AASHTO Green Book, 7th ed., §3.2 and §3.4.

Question 1 of 7: Crest-curve sight distance and accident causation

Question 1: Crest-curve sight distance and accident causation (20 marks)

(b) Other factors that contributed to the collision (6 marks)