Question 5 of 6: Signalised Intersection — Demand, Saturation Flow, Clearance Intervals and Flow Ratio

Question 5: Signalised Intersection — Demand, Saturation Flow, Clearance Intervals and Flow Ratio (5 + 5 + 5 + 5 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. A four-legged intersection with two 3.75 m lanes per direction on each street, 3 m crosswalks with no pedestrian refuge, and stop lines set back 1 m from the crosswalks. Demand, occupancy and timing data are as tabulated.

Find. (a) approach demand in pcu/h and in persons/h; (b) the heavy-vehicle-adjusted saturation flow for the NB and SB approaches; (c) the all-red interval, the intergreen period and the intersection lost time; and (d) the intersection flow ratio.

Approach. Convert the mixed traffic to passenger-car units and to person flows for part (a); apply the heavy-vehicle saturation-flow adjustment factor for part (b); build the clearance distance from the intersection geometry and divide by the clearing speed for part (c); and sum the critical-movement flow ratios of the two phases for part (d).

1. Part (a) — declare the passenger-car equivalent, then convert. No bus equivalent is given, so under the paper's NOTE 2 take the standard level-terrain value $E_B = 2.0$ pcu per bus. Each approach flow in passenger-car units is $$q_{\text{pcu}} = (\text{cars}) + E_B(\text{buses})$$ so $q_{\text{NB}} = 900 + 2.0(25) = 950$ pcu/h and $q_{\text{SB}} = 800 + 2.0(22) = 844$ pcu/h, while the two car-only approaches are unchanged at 600 and 800 pcu/h. The intersection total is $$\sum q_{\text{pcu}} = 950 + 844 + 600 + 800 = \boxed{3194\ \text{pcu/h}}$$
2. Part (a), continued — apply occupancy to vehicles, never to pcu. Person flow is a property of real vehicles, so the occupancies multiply the raw car and bus counts and not their pcu equivalents: $$P_{\text{NB}} = 900(2.0) + 25(25) = 1800 + 625 = 2425\ \text{persons/h}$$ $$P_{\text{SB}} = 800(2.0) + 22(15) = 1600 + 330 = 1930\ \text{persons/h}$$ with $P_{\text{EB}} = 600(2.0) = 1200$ and $P_{\text{WB}} = 800(2.0) = 1600$ persons/h. The intersection therefore moves $$\sum P = 2425 + 1930 + 1200 + 1600 = \boxed{7155\ \text{persons/h}}$$ The comparison is instructive: buses are 2.7 % of the north-bound vehicles but carry 25.8 % of its people, which is the standard argument for transit priority at such a site.
3. Part (b) — adjust the saturation flow for heavy vehicles. The heavy-vehicle factor converts a saturation flow expressed in passenger cars into one expressed in mixed vehicles: $$f_{HV} = \frac{1}{1 + P_{HV}\left(E_B - 1\right)}$$ For the north-bound approach $P_{HV} = 25/925 = 0.02703$, so $f_{HV} = 1/1.02703 = 0.9737$; for the south-bound approach $P_{HV} = 22/822 = 0.02677$ and $f_{HV} = 0.9739$. Applying them to the basic saturation flow gives, per lane, $$s_{\text{NB}} = 1900(0.9737) = 1850\ \text{veh/h/lane}, \qquad s_{\text{SB}} = 1900(0.9739) = 1850\ \text{veh/h/lane}$$ and, since each approach has two lanes, $\boxed{3700\ \text{veh/h}}$ north-bound and $3701\ \text{veh/h}$ south-bound for the approach as a whole. The near-identity of the two approaches is a coincidence of their similar bus percentages, not an error.
4. Confirm the adjustment factor is exactly the unit conversion. Because $f_{HV} = 1/[1 + P_{HV}(E_B-1)]$ is algebraically the ratio of vehicles to passenger-car units on the approach, the adjusted saturation flow must satisfy $$s = s_0\,\frac{\text{vehicles}}{\text{pcu}} = 1900 \times \frac{925}{950} = 1850\ \text{veh/h/lane}$$ which reproduces the step above. This identity is the cheapest available check on both parts (b) and (d).
5. Part (c) — build the clearance distance from the geometry. A vehicle that crosses the stop line at the end of amber must clear the whole conflict area before the cross street is released. The path is the 1 m stop-line set-back, the 3 m near crosswalk, the full cross-street carriageway (two lanes each way at 3.75 m, so $2 \times 3.75 \times 2 = 15.0$ m), the 3 m far crosswalk, and finally the vehicle's own 6 m length so that its rear is clear: $$d_{\text{clear}} = 1.0 + 3.0 + 15.0 + 3.0 + 6.0 = 28.0\ \text{m}$$ With no pedestrian refuge, both crosswalks must be crossed in one movement, so neither may be omitted.
6. Convert the clearance distance to the all-red interval. At the stated clearing speed of 30 km/h, that is $30/3.6 = 8.333$ m/s, so $$R = \frac{d_{\text{clear}}}{v} = \frac{28.0}{8.333} = 3.36\ \text{s} \;\rightarrow\; \boxed{R = 3\ \text{s}}$$ rounded to the nearest second as the question directs. The intergreen period is the amber plus the all-red: $$I = A + R = 3.0 + 3 = 6.0\ \text{s per phase}$$
7. Obtain the intersection lost time. A four-legged intersection with these movements is signalised in two phases (north–south, then east–west). No start-up lost time is given, so the whole intergreen of each phase is taken as lost: $$L = \sum_{i=1}^{n} I_i = n\,(A + R) = 2(6.0) = \boxed{12.0\ \text{s per cycle}}$$
8. Part (d) — identify the critical movement in each phase. The flow ratio of a movement is its demand divided by the saturation capacity of the lanes serving it. Two lanes at 1900 pc/h each give 3800 pcu/h per approach, and the governing approach in each phase is the heavier one — north-bound (950 pcu/h) against south-bound (844), and west-bound (800 pcu/h) against east-bound (600): $$y_{\text{NS}} = \frac{950}{3800} = 0.2500, \qquad y_{\text{EW}} = \frac{800}{3800} = 0.2105$$
9. Sum the critical flow ratios. The intersection flow ratio is the sum over phases of the critical movement ratios: $$Y = \sum y_{ci} = 0.2500 + 0.2105 = \boxed{0.461}$$ Working in vehicles instead of pcu gives the same figure — $925/3700 = 0.2500$ north-bound — because the heavy-vehicle factor is exactly the pcu conversion, as established in step 4. With $Y = 0.461$ well below 1.0 the intersection is not saturated, and Webster's optimum cycle would be $$C_o = \frac{1.5L + 5}{1 - Y} = \frac{1.5(12) + 5}{1 - 0.461} = \frac{23}{0.539} = 42.6\ \text{s}$$ which in practice would be raised to a 45–50 s cycle so the pedestrian minimum green on the 15 m crossings can be met.