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16-Civ-B10 Traffic Engineering · December 2013

Question 2 of 7: Freeway Service Flow Rate and Signalised-Intersection Capacity

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 2: Freeway Service Flow Rate and Signalised-Intersection Capacity (20 marks: (a) 10, (b) 10)

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. Two independent capacity calculations, one for a freeway mainline segment and one for a signalised approach.

Quantity(a) Freeway segment(b) Signalised approach
Lanes in the direction analysed2 (four-lane freeway)2 approach lanes
Lane width3.5 m3.5 m
Lateral obstruction1.0 m, one side only—
Grade5 % up, 1.5 km long3 % down
Heavy vehicles10 % trucks and buses, 3 % RVs8 % heavy trucks
Design speed100 km/h—
Area type—Central Business District
Parking / bus stop / turns—none / none / through only
Signal timing—C = 75 s, g/C = 0.50
TargetLevel of service CCapacity

Find. (a) the service flow rate the segment can carry while still operating at level of service C, in vehicles per hour in the direction analysed; and (b) the capacity of the signalised approach in vehicles per hour.

Approach. Both parts follow the same logic: start from an ideal flow rate per lane, multiply by the number of lanes, then apply multiplicative adjustment factors for every way in which the real facility falls short of ideal — geometry, heavy vehicles, grade, and for the signal the fraction of each cycle that is actually green.

  1. Part (a) — state the governing service-flow equation. For a basic freeway segment the service flow rate at level of service \(i\) is $$SF_{i}=MSF_{i}\times N\times f_{w}\times f_{HV}\times f_{p}$$ where \(MSF_{i}\) is the maximum service flow rate per lane at that level of service, \(N\) the number of lanes in one direction, \(f_{w}\) the lane-width and lateral-clearance factor, \(f_{HV}\) the heavy-vehicle factor and \(f_{p}\) the driver-population factor.
  2. Read the maximum service flow rate for level of service C. For a 100 km/h design speed on a four-lane freeway the assumed table values are an ideal capacity of 2,000 pc/h/ln and a volume-to-capacity ratio of 0.70 at level of service C, so $$MSF_{C}=\left(\frac{v}{c}\right)_{C}\times c_{j}=0.70\times 2000 =1400\ \text{pc/h/ln}$$ This is a rate in passenger cars per hour per lane under ideal geometry; the factors that follow convert it into mixed vehicles on the real cross-section.
  3. Evaluate the geometric factor. The lanes are 3.5 m rather than the ideal 3.6 m, and a lateral obstruction stands 1.0 m from the travelled pavement on one side rather than the ideal 1.8 m. Interpolating the lane-width and lateral-clearance table for a four-lane freeway with an obstruction on one side gives $$f_{w}=0.97$$ a 3 % penalty. Physically, narrow lanes and a close obstruction make drivers shy away from the edge and increase their following distances.
  4. Evaluate the heavy-vehicle factor on the specific grade. A sustained 5 % grade 1.5 km long is long enough that trucks reach crawl speed, so specific-grade passenger-car equivalents apply rather than the much smaller general-terrain values. Taking \(E_{T}=10\) for trucks and buses and \(E_{R}=4\) for recreational vehicles on this grade and length, $$f_{HV}=\frac{1}{1+P_{T}(E_{T}-1)+P_{R}(E_{R}-1)} =\frac{1}{1+0.10(9)+0.03(3)}=\frac{1}{1.99}$$ $$f_{HV}=0.5025$$ The denominator of 1.99 says that the 13 % of the traffic stream that is heavy occupies almost as much road space as the whole passenger-car stream — the single dominant effect in this problem.
  5. Assemble the service flow rate. Taking \(f_{p}=1.00\) for a familiar, largely commuting driver population and substituting all four factors, $$SF_{C}=1400\times 2\times 0.97\times 0.5025\times 1.00$$ $$\boxed{SF_{C}\approx 1{,}365\ \text{veh/h in the direction analysed}}$$ which is about 682 veh/h per lane. Set against an ideal capacity of 2,000 pc/h/ln, the grade and the heavy vehicles between them have cost this segment roughly two thirds of its nominal ability: a two-lane direction that would carry 2,800 pc/h at level of service C on level terrain manages fewer than 1,400 real vehicles per hour on this climb. That is the engineering message, and it is the standard justification for a climbing lane.
  6. Part (b) — state the capacity equation for a signalised approach. Capacity is the saturation flow rate reduced by the fraction of the cycle during which the approach actually has green: $$c=s\times\frac{g}{C},\qquad s=s_{o}\,N\,f_{w}\,f_{HV}\,f_{g}\,f_{p}\,f_{bb}\,f_{a}\,f_{RT}\,f_{LT}$$ where \(s_{o}\) is the ideal saturation flow per lane and the remaining factors adjust for width, heavy vehicles, grade, parking, bus blockage, area type and turning movements.
  7. Evaluate the adjustment factors one by one. Taking \(s_{o}=1900\) pc/h/ln of green: the width factor for a 3.5 m lane is $$f_{w}=1+\frac{W-3.6}{9}=1+\frac{3.5-3.6}{9}=0.9889$$ the heavy-vehicle factor with 8 % trucks at \(E_{T}=2.0\) is $$f_{HV}=\frac{100}{100+\%HV\,(E_{T}-1)}=\frac{100}{100+8(1)}=0.9259$$ and the grade factor for a 3 % down grade is $$f_{g}=1-\frac{\%G}{200}=1-\frac{-3}{200}=1.015$$ The down grade is a small bonus, because vehicles accelerate away from the stop line more briskly. There is no parking and no bus stop, and the movement is through only, so \(f_{p}=f_{bb}=f_{RT}=f_{LT}=1.00\); the Central Business District location carries the area-type penalty \(f_{a}=0.90\) for narrow lanes, pedestrian friction and general activity.
  8. Compute the saturation flow rate. Multiplying the ideal rate by the two lanes and by every factor in turn, $$s=1900\times 2\times 0.9889\times 0.9259\times 1.015\times 0.90 =3{,}178\ \text{veh/h of green}$$ i.e. about 1,589 veh/h of green per lane, or a saturation headway close to 2.3 seconds per vehicle per lane.
  9. Reduce to capacity over the whole cycle. With a green-to-cycle ratio of 0.50 the approach is discharging for only half of each 75-second cycle, so $$c=s\times\frac{g}{C}=3{,}178\times 0.50$$ $$\boxed{c\approx 1{,}589\ \text{veh/h}}$$ Note that the cycle length itself does not enter the capacity — only the ratio \(g/C\) does. The 75-second cycle matters instead for the quantum of discharge: there are 48 cycles per hour, each with 37.5 s of effective green, so capacity corresponds to about 33 vehicles per cycle across the two lanes, or roughly 17 per lane per cycle. A queue longer than about 17 vehicles per lane will therefore not clear in one green, which is the practical check an engineer applies next.
PartQuantityValue
(a)MSF at level of service C1,400 pc/h/ln
(a)Lane-width / clearance factor fw0.97
(a)Heavy-vehicle factor fHV0.5025
(a)Service flow rate SFC≈ 1,365 veh/h (682 veh/h per lane)
(b)Width / heavy-vehicle / grade factors0.9889 / 0.9259 / 1.015
(b)Area-type factor fa (CBD)0.90
(b)Saturation flow rate s3,178 veh/h of green
(b)Capacity c = s (g/C)≈ 1,589 veh/h (≈ 33 veh per cycle)