16-Civ-B7 Transportation Planning and Engineering · December 2015

Question 1 of 8: Crest Curve Sight Distance and an Unsuperelevated Curve

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

Notes on this paper

Paper format. 98-Civ-B7 Highway Engineering, National Examinations, December 2015. Three hours, open book, any non-communicating calculator. Eight questions of equal value (20 marks each); five solutions constitute a complete paper and only the first five in the answer book are marked. Note 1 invites the candidate to state any assumption made about an ambiguous input, and Note 2 permits any datum that is required but not given to be assumed. All eight questions are solved here, because the set is a study resource rather than a timed attempt.

Reference texts. Garber & Hoel, Traffic and Highway Engineering, 5th ed. (geometric design, sight distance, pavement design); Transportation Association of Canada, Geometric Design Guide for Canadian Roads (superelevation and spiral tables — the paper's Table 2.1.2.5 is TAC page 2.1.2.12); AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book) for runoff distribution and relative-gradient limits; AASHTO, Guide for Design of Pavement Structures (1993) for the flexible pavement equation and layer/drainage coefficients; Asphalt Institute MS-2, Asphalt Mix Design Methods and Mamlouk & Zaniewski, Materials for Civil and Construction Engineers, for mixture volumetrics and binder grading.

Check: assumptions carried through this paper. Under the paper's own Note 2 the following values are assumed and stated where used: the AASHTO maximum relative gradient (0.50 % at 80 km/h) and the 70 % / 30 % split of superelevation runoff either side of the PC for two lanes rotated (Question 2); a truck factor of 0.52 for all trucks on a rural Interstate and a lane-distribution factor of 0.70 for three lanes in one direction (Question 6); and a downhill 2 % ramp grade in Question 4, since the freeway is elevated above the local street. Each is flagged again at the point of use with the sensitivity of the answer to it.

Question 1: Crest Curve Sight Distance and an Unsuperelevated Curve (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.

Part A — is 600 m of crest curve enough at 110 km/h?

Given. A crest vertical curve with grades $g_1 = +1.0\%$ and $g_2 = -1.0\%$, length $L = 600\ \text{m}$; driver eye height $h_1 = 1.00\ \text{m}$, object height $h_2 = 0.50\ \text{m}$; perception-reaction time $t = 2.5\ \text{s}$; longitudinal friction $f = 0.10$; existing posted design speed 100 km/h, proposed 110 km/h.

Find. The stopping sight distance the 600 m curve actually provides, the stopping sight distance a driver needs at 110 km/h, and therefore whether the residents' objection is technically sound.

Crest curve at Dead Man's Hill. The sight line from a 1.00 m eye height to a 0.50 m object grazes the crest; the geometry gives the available sight distance S from L and the algebraic grade difference A.

Approach. Compute the required stopping sight distance from the AASHTO braking model at each speed, compute the sight distance the existing crest geometry supplies from the crest-curve relation quoted on the examination paper, and compare the two.

Required stopping sight distance at the proposed speed. The metric AASHTO model splits the distance into a reaction leg and a braking leg, $$SSD = 0.278\,V t + \frac{V^{2}}{254\,(f \pm G)}$$ with $V$ in km/h. On the crest the average grade is nil, so $G = 0$. At $V = 110$ km/h, $$SSD_{110} = 0.278(110)(2.5) + \frac{110^{2}}{254(0.10)} = 76.45 + 476.38 = \boxed{552.83\ \text{m}}$$ The braking leg dominates because the pavement friction is only 0.10 — a wet or iced surface. Repeating the same arithmetic at the present 100 km/h gives $SSD_{100} = 69.50 + 393.70 = 463.20$ m.
Sight distance the existing curve supplies. The paper prints the standard crest relation, which for the case $S < L$ is $$L = \frac{A S^{2}}{200\left(\sqrt{h_1}+\sqrt{h_2}\right)^{2}}$$ Here $A = |g_2 - g_1| = 2.0\ \%$ and the height constant is $$200\left(\sqrt{1.00}+\sqrt{0.50}\right)^{2} = 200(1.70711)^{2} = 582.84$$ Solving the relation for $S$ with $L = 600$ m, $$S = \sqrt{\frac{L \cdot 582.84}{A}} = \sqrt{\frac{600(582.84)}{2.0}} = \boxed{418.15\ \text{m}}$$ and since 418.15 m is indeed less than the 600 m curve length, the $S < L$ branch was the right one to use.
Compare supply with demand. The curve offers 418 m of sight distance. A driver at 110 km/h needs 553 m — a shortfall of 134.7 m, or 24 % of the required distance. The politician's "600 metres is a long way" confuses the length of the curve with the distance a driver can see; on a crest those are not the same quantity, and the sight distance is always the smaller of the two.
Curve length that would be required. Rearranging the same relation for the proposed speed, $$L_{req} = \frac{A\,SSD_{110}^{2}}{582.84} = \frac{2.0(552.83)^{2}}{582.84} = \boxed{1048.7\ \text{m}}$$ so the crest would have to be lengthened from 600 m to about 1050 m, equivalent to raising the rate of vertical curvature from $K = L/A = 300$ to $K = 524$.
The curve is already deficient at the present speed. Repeating step 4 at 100 km/h gives $L_{req} = 2.0(463.20)^{2}/582.84 = 736.2$ m, still well above the 600 m built. Inverting the sight-distance model for the 418.15 m actually available, $$0.695\,V + \frac{V^{2}}{25.4} = 418.15 \quad\Longrightarrow\quad \boxed{V = 94.6\ \text{km/h}}$$ The clean absence of accidents to date reflects light traffic and drivers behaving below the posted speed, not an adequate design.

The residents are making a valid engineering point, and a stronger one than they realise: the curve does not satisfy stopping sight distance at 110 km/h, and it does not satisfy it at the present 100 km/h either. The defensible recommendation is to reconstruct the crest to $K \ge 524$ before any speed increase, or, if reconstruction is not funded, to post an advisory speed of 90 km/h through the curve and to treat the surface so that the friction supply rises above 0.10.

Part B — proving the unsuperelevated curve is infeasible

Given. Two-lane road, lane width 4.0 m, radius $R = 500\ \text{m}$, design speed $V = 80\ \text{km/h}$, superelevation $e = 0$, available side friction $f = 0.05$, vehicles of mass 900 kg with a high centre of gravity.

Find. A quantitative demonstration that the curve cannot be driven at its posted speed, and the remedies that would make it safe.

Approach. Apply the point-mass curve equation to find the side friction the curve demands, compare it with the friction the pavement can supply, and express the same deficiency three further ways — as a minimum radius, as a safe speed, and as the superelevation that would be needed.

Side friction demanded. Equilibrium of a vehicle on a curve gives the familiar point-mass relation $$e + f = \frac{V^{2}}{127\,R}$$ with $V$ in km/h and $R$ in metres. With no superelevation the pavement must supply all of it: $$f_{req} = \frac{80^{2}}{127(500)} - 0 = \frac{6400}{63\,500} = \boxed{0.101}$$ The measured supply is 0.05. The curve therefore asks the tyres for twice the friction the surface can deliver, and the vehicle must slide outward — the demonstration the county engineer needs.
The same result as a minimum radius. Solving the relation for $R$ at the posted speed and the available friction, $$R_{min} = \frac{V^{2}}{127(e+f)} = \frac{6400}{127(0.05)} = \boxed{1008\ \text{m}}$$ The curve as built is less than half the radius that 80 km/h requires on this surface.
The same result as a safe speed. Holding the geometry and the friction fixed, $$V_{safe} = \sqrt{127\,R\,(e+f)} = \sqrt{127(500)(0.05)} = \boxed{56.3\ \text{km/h}}$$ so the curve is a 55 km/h curve wearing an 80 km/h sign.
The superelevation that would be needed. Retaining $f = 0.05$, $$e_{req} = \frac{V^{2}}{127R} - f = 0.101 - 0.05 = 0.051$$ A 5.1 % cross slope is buildable on a rural road with a 6 % maximum, so rebuilding the curve with superelevation is the direct fix. If instead the surface were restored to a normal design side-friction factor of $f = 0.14$ at 80 km/h and given $e = 0.04$, the required radius falls to $R = 6400/[127(0.18)] = 280$ m, comfortably inside the existing 500 m — which shows that the friction supply, not the curvature, is the governing defect.
Why the 900 kg mass and the high centre of gravity do not change the conclusion. Mass cancels from the equilibrium equation, so the sliding threshold is identical for a 900 kg car and a loaded truck; quoting the mass is a distractor. Rollover is governed instead by the static stability factor $SSF = t/(2h)$, the ratio of track width to twice the centre-of-gravity height, and a lateral acceleration of $$a_{lat} = \frac{v^{2}}{R} = \frac{(22.22)^{2}}{500} = 0.99\ \text{m/s}^{2} = 0.101\,g$$ is far below the 0.8–1.2 g that even a top-heavy passenger vehicle needs to roll. The tyres therefore break away first: the vehicle slides off the curve, and rollover follows only if the sliding vehicle trips on a curb, a soft shoulder or a ditch. The correct message to the politicians is that the curve fails by skidding, and that the "two-wheeling" the local drivers enjoy is the last warning before departure from the road.

Question 1 — final results
Quantity	Value	Comment
Stopping sight distance required at 110 km/h	552.83 m	0.10 friction dominates
Stopping sight distance required at 100 km/h	463.20 m	present design speed
Sight distance supplied by the 600 m crest	418.15 m	$S < L$ case
Curve length required at 110 km/h	1048.7 m ($K = 524$)	versus 600 m built
Highest speed the existing crest serves	94.6 km/h	already below the posted 100
Side friction demanded by the 500 m curve	0.101	supply is 0.05
Radius required at 80 km/h with $f=0.05$	1008 m	versus 500 m built
Safe speed on the curve as built	56.3 km/h	or superelevate to 5.1 %

Topic. Stopping sight distance on crest vertical curves, and the point-mass equilibrium of a vehicle on a horizontal curve.

Key relations. $SSD = 0.278Vt + V^{2}/[254(f \pm G)]$; crest curve with $S < L$: $L = A S^{2}/[200(\sqrt{h_1}+\sqrt{h_2})^{2}]$, and with $S > L$: $L = 2S - 200(\sqrt{h_1}+\sqrt{h_2})^{2}/A$; horizontal curve: $e + f = V^{2}/(127R)$; rate of vertical curvature $K = L/A$.

Why this works. A crest hides an object because the pavement itself is the obstruction; the sight line is a chord that grazes the parabola, so the available sight distance grows with the square root of the curve length and falls with the grade difference. Both sight-distance and curve equations are independent of vehicle mass, because the gravitational term that resists sliding and the inertial term that drives it scale together — which is why an argument built on friction and geometry is unanswerable, while one built on vehicle weight is not. The two parts of this question are the vertical and the horizontal expressions of the same idea: a design speed is a promise about stopping and about cornering, and geometry has to keep both promises.

Common pitfall. Assuming that a 600 m curve gives 600 m of sight distance — the length of the curve and the sight distance across it are different quantities, and mixing them is exactly the politician's error. The second pitfall is choosing the wrong branch of the crest equation: always compute $S$ on the assumed branch and then check the assumption, as in step 2.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §3.3 (sight distance) and §3.5 (vertical and horizontal curve design); AASHTO Green Book, Chapter 3.

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