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16-Civ-B10 Traffic Engineering · May 2018

Question 3 of 7: Moving-vehicle method: volumes and travel times

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

Notes on this paper

Paper format. National Examinations, May 2018 — 16-Civ-B10 Traffic Engineering, 3-hour duration, OPEN BOOK (any non-communicating calculator permitted). Seven questions, all of equal value (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 4 reproduces the AASHTO 2001 metric stopping-sight-distance table, which Question 5 is built around.

Reference texts.



Question 3 — Moving-vehicle method: volumes and travel times (a) to (e), 4 marks each — 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. Eight test-vehicle runs in each direction over the same section, each recording the travel time and three counts. The counts are reproduced below exactly as tabulated.

RunSouthboundNorthbound
 t (min)MetOvertook testOvertakent (min)MetOvertook testOvertaken
14.55162324.0313932
24.20155434.2716323
34.72178234.4116213
44.42157433.8914142
55.21174124.1913521
65.26166333.7115732
74.92187324.4416622
84.75160424.1515633

Find. (a) the mean of every recorded quantity for each direction; (b) and (c) the northbound and southbound volumes in veh/h; (d) and (e) the average travel time of the traffic stream in each direction, in minutes.

Southbound lane (test-car run S: t = 4.754 min) Northbound lane (test-car run N: t = 4.136 min) TEST TEST MET running south: M_s = 167.375 northbound vehicles O_s = 3.00 overtake the test car  |  P_s = 2.50 are overtaken by it On the northbound run: M_n = 152.375  O_n = 2.50  P_n = 2.25 Direction-crossed: the NORTHBOUND volume uses M_s — the vehicles met on the run made AGAINST it.
Figure 3.1 — The moving-vehicle (Wardrop) method. On each run the observer records the travel time, the vehicles met in the opposing lane, the vehicles that overtake the test car and the vehicles it overtakes. The volume of a direction is built from the count of vehicles MET on the run made against it.

Approach. Average the four recorded quantities over the eight runs in each direction, then apply Wardrop's two relations. The single point on which this question is won or lost is that the relations are direction-crossed: the volume of one direction uses the vehicles-met count from the run made in the opposite direction.

  1. Part (a) — average every recorded quantity for the southbound runs. Summing the eight southbound records and dividing by eight:
    $$\bar t_s=\frac{4.55+4.20+4.72+4.42+5.21+5.26+4.92+4.75}{8}=\frac{38.03}{8}=4.754\ \text{min}$$
    $$\bar M_s=\frac{1339}{8}=167.375,\qquad \bar O_s=\frac{24}{8}=3.000,\qquad \bar P_s=\frac{20}{8}=2.500$$
    where $M$ is the number met in the opposing direction, $O$ the number that overtook the test vehicle and $P$ the number it overtook.
  2. Part (a), continued — the same averages for the northbound runs.
    $$\bar t_n=\frac{33.09}{8}=4.136\ \text{min},\qquad \bar M_n=\frac{1219}{8}=152.375$$
    $$\bar O_n=\frac{20}{8}=2.500,\qquad \bar P_n=\frac{18}{8}=2.250$$
    Collecting part (a):
    Average valueSouthbound runsNorthbound runs
    Travel time of the test vehicle, $\bar t$ (min)4.7544.136
    Vehicles met in the opposing direction, $\bar M$167.375152.375
    Vehicles that overtook the test vehicle, $\bar O$3.0002.500
    Vehicles overtaken by the test vehicle, $\bar P$2.5002.250
    Net overtaking, $\bar O-\bar P$+0.500+0.250
    That both net-overtaking values are small and positive is the physical sanity check: the test car was driven at close to the average speed of the stream, slightly slower, which is exactly how a floating-car run should be conducted.
  3. Set up Wardrop's relations and note which run supplies which count. For the direction of interest, taken here as northbound,
    $$q_N=\frac{M_s+O_n-P_n}{t_s+t_n}$$
    The numerator mixes runs deliberately. $M_s$ is the count of vehicles met while the test car ran southbound — and every vehicle it met head-on was a northbound vehicle, so $M_s$ is a sample of the northbound stream. The overtaking terms come from the northbound run, because they correct for the test car having travelled slower or faster than the northbound traffic it was embedded in. Using $M_n$ here instead of $M_s$ is the classic error in this question and produces a plausible answer that nothing else in the data contradicts.
  4. Part (b) — northbound volume. The denominator is the sum of the two run times, 4.754 + 4.136 = 8.890 min, so
    $$q_N=\frac{167.375+2.500-2.250}{8.890}=\frac{167.625}{8.890}=18.855\ \text{veh/min}$$
    $$q_N=18.855\times60=\boxed{1131\ \text{veh/h}}$$
  5. Part (c) — southbound volume. Crossing the directions the other way, the southbound volume takes the vehicles met on the northbound run:
    $$q_S=\frac{M_n+O_s-P_s}{t_s+t_n}=\frac{152.375+3.000-2.500}{8.890}=17.196\ \text{veh/min}$$
    $$q_S=17.196\times60=\boxed{1032\ \text{veh/h}}$$
    The two-way volume is therefore about 2163 veh/h, with roughly a 52:48 directional split favouring northbound — consistent with the northbound runs being the faster ones only if the section is not near capacity, which the travel times support.
  6. Part (d) — average travel time of the northbound stream. The test vehicle's own run time is not the stream's average travel time; it must be corrected for the net overtaking, which measures how far the test car departed from the average speed:
    $$\bar t_N=t_n-\frac{O_n-P_n}{q_N}=4.136-\frac{2.500-2.250}{18.855}=4.136-0.0133$$
    $$\bar t_N=\boxed{4.123\ \text{min}}$$
    The correction is subtracted because more vehicles overtook the test car than it overtook, so the stream was travelling slightly faster than the test car and its average travel time is slightly shorter.
  7. Part (e) — average travel time of the southbound stream. The same correction with the southbound quantities:
    $$\bar t_S=t_s-\frac{O_s-P_s}{q_S}=4.754-\frac{3.000-2.500}{17.196}=4.754-0.0291$$
    $$\bar t_S=\boxed{4.725\ \text{min}}$$
    Southbound traffic takes about 0.60 min (36 s) longer over the section than northbound traffic, a 15 per cent difference, which together with the higher northbound volume is the signature of a peak-direction flow: the heavier direction here is also the faster one, so the section is operating below capacity and the southbound delay must come from something other than its own volume — roadside friction, signal progression favouring northbound, or an upstream bottleneck.

Check: the direction-crossing is worth one explicit check. Had $M_n$ been used in place of $M_s$ in part (b), the northbound volume would have come out as 1030 veh/h — a 9 per cent error, and one that looks entirely reasonable because it happens to be close to the true southbound volume. The mistake is only detectable by reasoning about what the test car actually saw, which is why the derivation above names the physical meaning of each count before substituting.

Results.

PartQuantityValue
(a)Southbound averages: $\bar t_s$, $\bar M_s$, $\bar O_s$, $\bar P_s$4.754 min, 167.375, 3.000, 2.500
(a)Northbound averages: $\bar t_n$, $\bar M_n$, $\bar O_n$, $\bar P_n$4.136 min, 152.375, 2.500, 2.250
(b)Northbound traffic volume1131 veh/h (18.86 veh/min)
(c)Southbound traffic volume1032 veh/h (17.20 veh/min)
(d)Average travel time, northbound traffic4.123 min
(e)Average travel time, southbound traffic4.725 min
—Two-way volume and directional split2163 veh/h, 52 per cent northbound