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16-Civ-B10 Traffic Engineering · December 2013

Question 6 of 7: Volume and Travel Time by the Moving-Vehicle Method

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

Notes on this paper

Paper format. National Examinations, December 2013 — 98-Civ-B10 Traffic Engineering. Three-hour, open-book examination; any non-communicating calculator is permitted. Seven questions are printed and five complete solutions are required, all questions being of equal value. The printed grading scheme is Q1 (a) 6, (b) 6, (c) 8; Q2 (a) 10, (b) 10; Q3 (a) 10, (b) 10; Q4 20; Q5 20; Q6 (a)–(d) 5 each; Q7 (a)–(d) 5 each. The paper also states that if doubt exists as to the interpretation of a question the candidate should submit a clear statement of any assumptions made, and that any data required but not given can be assumed. All seven questions are worked below.

Reference texts.

Check: assumed reference-table values. Questions 2(a) and 2(b) are capacity problems whose input list — lane width, lateral obstruction, per-cent heavy vehicles, a specific grade, design speed and a target level of service — is exactly the argument list of the classical Highway Capacity Manual equations, but the paper does not reproduce the lookup tables. Consistent with the paper's own instruction that any data required but not given may be assumed, every table value used is stated explicitly at the point of use, drawn from one coherent edition family (HCM 1985/1994, ideal capacity 2,000 pc/h/ln for the freeway segment). Substituting another edition's tables rescales the final flow rate but changes neither the method nor the arithmetic chain; the sensitivity is discussed in the Question 2(a) concept note.

Question 6: Volume and Travel Time by the Moving-Vehicle Method (20 marks: (a)–(d) 5 each)

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. Averaged results of a two-direction moving-vehicle (Wardrop) survey over a test section.

QuantityNorth-bound runSouth-bound run
Average travel time of the test cartN = 3.00 mintS = 2.90 min
Opposing vehicles metMN = 80MS = 100
Vehicles overtaking the test carON = 2.00OS = 1.50
Vehicles passed by the test carPN = 1.00PS = 1.00

Find. The traffic volume in each direction and the average travel time of the traffic stream (as distinct from the test car) in each direction.

Approach. Apply Wardrop's two moving-observer relations. The vehicles counted while driving against a stream measure that stream's volume; the net overtaking count made while driving with it measures how much faster the stream is than the test car, and so corrects the test car's own travel time to the stream average.

  1. State the two moving-vehicle relations and fix the notation. For the direction of interest, with subscript \(a\) for the run against that direction and \(w\) for the run with it, $$q=\frac{M_{a}+O_{w}-P_{w}}{t_{a}+t_{w}},\qquad \bar{t}=t_{w}-\frac{O_{w}-P_{w}}{q}$$ The critical bookkeeping point is that \(M_{a}\) — the number of vehicles of the subject stream that the test car meets — is recorded on the opposite run. Vehicles travelling north are met while the test car drives south, so the north-bound volume takes the 100 counted on the south-bound trip, not the 80 counted on the north-bound trip.
  2. Part (a) — north-bound volume. Substituting \(M_{a}=M_{S}=100\) (met while driving south), \(O_{w}=O_{N}=2.00\), \(P_{w}=P_{N}=1.00\), and the combined run time \(t_{S}+t_{N}=2.90+3.00=5.90\) min, $$q_{N}=\frac{100+2.00-1.00}{5.90}=\frac{101}{5.90}=17.12\ \text{veh/min}$$ $$\boxed{q_{N}=17.12\ \text{veh/min}=1{,}027\ \text{veh/h}}$$
  3. Part (b) — south-bound volume. Now the roles reverse: south-bound vehicles are met while the test car drives north, so \(M_{a}=M_{N}=80\), with \(O_{w}=O_{S}=1.50\) and \(P_{w}=P_{S}=1.00\) from the south-bound run, over the same combined time of 5.90 min: $$q_{S}=\frac{80+1.50-1.00}{5.90}=\frac{80.5}{5.90}=13.64\ \text{veh/min}$$ $$\boxed{q_{S}=13.64\ \text{veh/min}=819\ \text{veh/h}}$$ The north-bound direction carries about 25 % more traffic, consistent with a morning peak on a radial route.
  4. Part (c) — average travel time of north-bound traffic. The test car was overtaken twice and overtook once on its north-bound run, a net one vehicle passing it, so the stream is on average slightly faster than the test car and its mean travel time must be slightly less than 3.00 min: $$\bar{t}_{N}=t_{N}-\frac{O_{N}-P_{N}}{q_{N}} =3.00-\frac{2.00-1.00}{17.12}=3.00-0.058$$ $$\boxed{\bar{t}_{N}=2.94\ \text{min}\ (\approx 2\ \text{min }57\ \text{s})}$$
  5. Part (d) — average travel time of south-bound traffic. By the same relation with the south-bound run's figures, $$\bar{t}_{S}=t_{S}-\frac{O_{S}-P_{S}}{q_{S}} =2.90-\frac{1.50-1.00}{13.64}=2.90-0.037$$ $$\boxed{\bar{t}_{S}=2.86\ \text{min}\ (\approx 2\ \text{min }52\ \text{s})}$$ Both corrections are small — under four seconds — which is the expected outcome of a properly conducted survey: the driver is instructed to "float" with the stream, so the net overtaking count should be near zero and the correction should be a refinement rather than a rescue. A large net count would indicate the test car was driven too slowly or too fast to represent the stream.
PartQuantityResult
(a)North-bound volume qN17.12 veh/min = 1,027 veh/h
(b)South-bound volume qS13.64 veh/min = 819 veh/h
(c)North-bound average travel time2.94 min (2 min 57 s)
(d)South-bound average travel time2.86 min (2 min 52 s)
—Combined two-way volume30.76 veh/min = 1,846 veh/h