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16-Civ-B10 Traffic Engineering · December 2019

Question 1 of 7: Crest vertical curves and sight distance

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

Notes on this paper

Paper format. National Examinations, December 2019, 16-Civ-B10 Traffic Engineering. Three-hour duration, OPEN BOOK (any non-communicating calculator permitted). Seven questions on four pages, all of equal value at 20 marks each; the paper requires five solutions and marks only the first five as they appear in the answer book. Because the set is a study resource, all seven questions are solved here. The paper’s own Note 1 invites a clear statement of any assumptions made and Note 2 permits any required-but-not-given data to be assumed; every assumption used below is stated explicitly where it is introduced. Page 2 reproduces the AASHTO 2001 metric stopping-sight-distance table, whose footnote fixes the brake-reaction time at 2.5 s and the deceleration at 3.4 m/s2; Question 1 is built around it.

Reference texts.

Canadian context. These are Engineers Canada national examinations, so the answers are framed for Canadian practice: the TAC Geometric Design Guide for Canadian Roads and the MUTCDC are the governing documents and metric design controls are used throughout. Question 1 supplies the AASHTO 2001 metric table directly, so that table is used as printed — the TAC guide adopts the same 2.5 s / 3.4 m/s2 stopping model, so the two agree here. Lane widths, walking speed and crosswalk geometry follow TAC and MUTCDC practice.

Question 1: Crest vertical curves and sight distance (7 + 7 + 6 = 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. The AASHTO 2001 metric stopping-sight-distance table is printed on page 2 of the paper, and its footnote states the model behind it: a brake-reaction time of 2.5 s and a deceleration of 3.4 m/s2 (11.2 ft/s2). The two curve problems share the eye and object heights.

QuantitySymbolPart (a)Part (b)
Entering gradeg1+4.0 %+2.5 %
Departing gradeg2−2.0 %−2.5 %
Algebraic grade changeA6.0 %5.0 %
Design speedV113 km/hrequired
Curve lengthLrequired375 m
Driver eye heighth11050 mm1050 mm
Object heighth2300 mm300 mm
Brake-reaction timet2.5 s2.5 s
Decelerationa3.4 m/s23.4 m/s2

Find. (a) the minimum crest length that delivers the full stopping sight distance at 113 km/h; (b) the design speed a 375 m crest on a 5 % grade change can support; (c) a definition of sight distance and of two of its types.

Question 1(a): crest joining +4 per cent to -2 per cent at 113 km/h A = |g_2 − g_1| = 6 per cent  driver eye h_1 = 1050 mm  object h_2 = 300 mm vertical scale exaggerated; S < L, so both ends of the sight line lie on the curve PVI BVC EVC L = 614.27 m h_1 h_2 sight distance S = 225.00 m g_1 = +4 per cent g_2 = -2 per cent
Question 1(a): the +4 % / −2 % crest. The sight line from a 1050 mm eye to a 300 mm object lies wholly on the curve, so the S < L branch applies. Vertical scale exaggerated.

Approach. Compute the stopping sight distance from the same two-term model the printed table is built on, form the single crest constant that both curve-length branches share, solve the S < L branch, and confirm the branch by showing the S > L form is self-contradictory; part (b) inverts the same relation and reads the answer off the table.

  1. Part (a) — reproduce the printed table from its own model. The AASHTO metric stopping sight distance is the sum of a reaction term and a braking term, $$S = 0.278\,V t + \frac{0.039\,V^{2}}{a}$$ with $V$ in km/h, $t$ in s and $a$ in m/s\(^{2}\). Substituting the footnote values into the 20 km/h row gives \(0.278(20)(2.5) = 13.9\) m and \(0.039(20)^{2}/3.4 = 4.6\) m, i.e. exactly the printed entries, and the same is true of every one of the twelve metric rows. That single check validates the whole printed table before any of its numbers are used.
  2. Stopping sight distance at the stated design speed. At \(V = 113\) km/h, $$S = 0.278(113)(2.5) + \frac{0.039(113)^{2}}{3.4} = 78.54 + 146.46 = 225.00\ \text{m}$$ The speed is not tabulated, so this calculated value is used directly. It sits between the printed 110 km/h row (215.3 m calculated) and the 120 km/h row (248.6 m calculated), which is the sanity check that the arithmetic is right.
  3. Form the crest constant. For a crest curve the two standard branches are $$L = \frac{A S^{2}}{100\left(\sqrt{2h_{1}}+\sqrt{2h_{2}}\right)^{2}} \quad (S\lt L), \qquad L = 2S - \frac{200\left(\sqrt{h_{1}}+\sqrt{h_{2}}\right)^{2}}{A} \quad (S\gt L)$$ Because \(\left(\sqrt{2h_1}+\sqrt{2h_2}\right)^{2} = 2\left(\sqrt{h_1}+\sqrt{h_2}\right)^{2}\), the denominator of the first and the numerator of the second are the same quantity. With \(h_1 = 1.050\) m and \(h_2 = 0.300\) m, $$k = 100\left(\sqrt{2(1.050)}+\sqrt{2(0.300)}\right)^{2} = 494.50$$ so one number serves both branch tests and the commonest slip in this family — recomputing the second constant and getting it wrong — cannot occur.
  4. Solve on the S < L branch. Substituting \(A = 6\) %, \(S = 225.00\) m and \(k = 494.50\), $$L = \frac{A S^{2}}{k} = \frac{6(225.00)^{2}}{494.50} = \boxed{614.27\ \text{m}}$$ The assumption is self-consistent because \(S = 225.00 \lt L = 614.27\).
  5. Confirm the branch and the secondary criteria. Testing the other branch, \(L = 2(225.00) - 494.50/6 = 367.59\) m, which would require \(S \gt L\) and instead gives \(S = 225.00 \lt 367.59\): the S > L form is self-contradictory here, so the branch choice is proven rather than assumed. The rate of vertical curvature is \(K = L/A = 614.27/6 = 102.38\) m per per cent, and the comfort-and-appearance minimum \(0.6V = 0.6(113) = 67.8\) m is far below the sight-distance answer, so sight distance governs. Design length: 615 m after rounding up to the nearest 5 m.
Check: what the low object costs. The paper specifies a 300 mm object rather than the AASHTO standard 600 mm. Repeating the calculation with the standard 1080 mm / 600 mm heights gives \(k_{\text{std}} = 657.99\) and \(L = 461.64\) m, so the small-object criterion lengthens the curve by 33.06 %. That penalty is much larger than halving the object height suggests, because the constant depends on the square of the sum of the square roots. The US-Customary half of the printed table is an independent check on the speed: 113 km/h is 70.21 mph and the printed 70 mph row gives a calculated SSD of 727.6 ft = 221.8 m, within 1.4 % of the 225.0 m used above.
Question 1(b): the given 375 m crest joining +2.5 per cent to -2.5 per cent A = |g_2 − g_1| = 5 per cent  driver eye h_1 = 1050 mm  object h_2 = 300 mm vertical scale exaggerated; S < L, so both ends of the sight line lie on the curve PVI BVC EVC L = 375 m h_1 h_2 sight distance S = 192.58 m g_1 = +2.5 per cent g_2 = -2.5 per cent
Question 1(b): the given 375 m crest on a ±2.5 % grade change. The curve length is known and the sight distance it delivers is solved for, then matched against the printed table.
  1. Part (b) — invert the crest relation for the available sight distance. The curve length and the grade change are given, so the S < L branch is rearranged for S: $$S = \sqrt{\frac{L\,k}{A}} = \sqrt{\frac{375(494.50)}{5}} = \boxed{192.58\ \text{m}}$$ and \(S = 192.58 \lt L = 375\) confirms the branch.
  2. Read the design speed off the printed table. The design speed is the largest tabulated speed whose stopping sight distance the curve can actually deliver. The printed Design column requires 185 m at 100 km/h and 220 m at 110 km/h. Since \(185 \lt 192.58 \lt 220\), $$V_{\text{design}} = \boxed{100\ \text{km/h}}$$ The 110 km/h requirement is missed by \(220 - 192.58 = 27.42\) m, which is not a marginal shortfall that could be argued away. Using the Calculated column instead (184.2 m at 100 km/h, 215.3 m at 110 km/h) gives the same answer, so the reading does not depend on which column is adopted. Inverting the continuous SSD relation gives \(V = 102.77\) km/h, which confirms that the curve sits just above the 100 km/h step and nowhere near 110 km/h.
  3. Free cross-check against the published K table. The rate of vertical curvature is \(K = L/A = 375/5 = 75.0\). Recomputing the same curve at the AASHTO standard 1080 mm / 600 mm heights gives \(S_{\text{std}} = \sqrt{375(657.99)/5} = 222.15\) m, i.e. a 110 km/h curve, and \(S_{\text{std}}^{2}/k_{\text{std}} = 75.0\) against the published metric crest minimum of \(K = 74\) at 110 km/h. Landing on the published table to within one unit confirms the branch choice, the constant and the arithmetic in a single line — and it quantifies the cost of the 300 mm object as exactly one full design-speed increment, 110 km/h down to 100 km/h.

Part (c) — sight distance and its types. Sight distance is the length of roadway ahead that is continuously visible to a driver from a specified eye height to a specified target height. It is a geometric-design control rather than an operating characteristic: the designer chooses vertical and horizontal alignment, roadside clearances and cross-section so that the available sight distance everywhere equals or exceeds the distance a driver needs to complete a given manoeuvre safely at the design speed. Both curve problems above are that comparison in reverse — solving for the alignment that delivers a required sight distance.

The two types that matter most in traffic engineering are:

Stopping sight distance (SSD). The distance needed to perceive an unexpected object in the travelled way, react, and brake to a stop before reaching it. It is the sum of a perception-reaction distance and a braking distance, which is exactly the two-term model used in parts (a) and (b): \(S = 0.278Vt + 0.039V^{2}/a\). It is the minimum sight distance that must be available at every point on every highway, which is why it is the criterion that sizes crest curves, sets clearance to sight obstructions on the inside of horizontal curves, and fixes the advance placement of warning signs. On a grade the braking term becomes \(V^{2}/[254(f \pm G)]\), so downgrades increase the requirement.

Passing sight distance (PSD). The distance a driver on a two-lane, two-way highway needs to complete an overtaking manoeuvre in the opposing lane without conflicting with an oncoming vehicle: the sum of the initial manoeuvre distance, the distance travelled while occupying the opposing lane, a clearance to the opposing vehicle, and the distance the opposing vehicle covers during the manoeuvre. It is several times SSD at the same speed, so it is not a universal requirement — instead it determines the proportion of a two-lane highway that can be marked for passing, and every length that fails it is signed and marked as a no-passing zone.

Two further types are worth naming because they arise constantly in practice. Decision sight distance is the longer distance required where a driver must detect an unexpected or difficult-to-perceive information source, recognise it, select a speed and path, and complete the manoeuvre — interchange exits, lane drops and complex intersections. Intersection sight distance is the clear sight triangle a driver on a minor approach needs in order to judge an acceptable gap in the major-road stream.

Question 1 — final results
QuantityValue
(a) Stopping sight distance at 113 km/h225.00 m
(a) Crest constant k for 1050 mm / 300 mm494.50
(a) Minimum crest length614.27 m (design 615 m)
(a) Rate of vertical curvature K = L/A102.38 m per per cent
(a) Same curve at standard 1080/600 heights461.64 m (+33.06 %)
(b) Sight distance available on the 375 m crest192.58 m
(b) Design speed from the printed table100 km/h
(b) Shortfall against the 110 km/h requirement27.42 m
(c) Two types definedstopping sight distance; passing sight distance
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