Question 1 of 7: Signalised Four-Legged Intersection — Arrival Flow, Saturation Flow, Clearance Intervals and Flow Ratio

Question 1: Signalised Four-Legged Intersection — Arrival Flow, Saturation Flow, Clearance Intervals and Flow Ratio (4 × 5 = 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.

Find. The approach demands expressed both in passenger-car units and in persons carried; the heavy-vehicle-adjusted saturation flow on the two north–south approaches; the all-red, intergreen and lost time implied by the intersection geometry; and the intersection flow ratio that any subsequent cycle-length calculation would rest on.

Approach. Convert buses to passenger-car units with an assumed equivalent, apply occupancy to vehicles (never to pcu) to get person flow, invert the same equivalent to obtain the heavy-vehicle saturation-flow factor, walk the physical clearance path across the intersection to size the all-red, and finally take the critical-movement flow ratio in each phase.

1. Part (a) — convert each approach to passenger-car units. A bus occupies more road space and discharges more slowly than a car, so it is counted as several cars. On level urban approaches the standard equivalent is $$E_B = 2.0 \text{ pcu per bus}$$ so that each approach flow becomes $q_{\text{pcu}} = (\text{cars}) + E_B(\text{buses})$. Substituting the four approach counts, $$\begin{aligned} q_{NB} &= 700 + 14(2.0) = 728 \text{ pcu/h}\\ q_{SB} &= 600 + 10(2.0) = 620 \text{ pcu/h}\\ q_{EB} &= 550 + 0 = 550 \text{ pcu/h}\\ q_{WB} &= 650 + 0 = 650 \text{ pcu/h} \end{aligned}$$ Summing over all four legs gives the total vehicular arrival flow at the intersection, $$\boxed{\;\sum q = 728 + 620 + 550 + 650 = 2548 \text{ pcu/h}\;}$$
2. Convert the same demand into persons carried per hour. Occupancy is a property of a real vehicle, not of a passenger-car unit, so the occupancies multiply the original counts. Approach by approach, $$\begin{aligned} P_{NB} &= 700(2.0) + 14(25) = 1400 + 350 = 1750 \text{ persons/h}\\ P_{SB} &= 600(2.0) + 10(15) = 1200 + 150 = 1350 \text{ persons/h}\\ P_{EB} &= 550(2.0) = 1100 \text{ persons/h}\\ P_{WB} &= 650(2.0) = 1300 \text{ persons/h} \end{aligned}$$ The intersection therefore serves $$\boxed{\;\sum P = 1750 + 1350 + 1100 + 1300 = 5500 \text{ persons/h in vehicles}\;}$$ Four crosswalks at 120 pedestrians per hour add a further $4(120) = 480$ people crossing on foot, so 5980 people in total pass through the intersection each hour. Notice that the north-bound leg carries fewer vehicles than it does people relative to the west-bound leg: 14 well-loaded buses move 350 people, which is more than the 300 people carried by the extra 150 cars on the west-bound approach. That contrast is the whole argument for transit priority, and it is invisible in a pcu-only count.
3. Part (b) — form the heavy-vehicle adjustment factor. The basic saturation flow of 1800 pc/h/lane is quoted for a stream of passenger cars only. A stream containing a proportion $P_{HV}$ of heavy vehicles, each worth $E_B$ cars, discharges fewer vehicles per hour in the ratio $$f_{HV} = \frac{1}{1 + P_{HV}\,(E_B - 1)}$$ The proportions are taken on the true vehicle counts, not the pcu counts: $$P_{HV,NB} = \frac{14}{700+14} = \frac{14}{714} = 0.01961, \qquad P_{HV,SB} = \frac{10}{600+10} = \frac{10}{610} = 0.01639$$
4. Apply the factor to obtain the adjusted saturation flow in veh/h. Substituting each proportion with $E_B = 2.0$, $$f_{HV,NB} = \frac{1}{1+0.01961(1.0)} = 0.9808, \qquad f_{HV,SB} = \frac{1}{1+0.01639(1.0)} = 0.9839$$ and multiplying by the basic rate, $$\begin{aligned} s_{NB} &= 1800(0.9808) = 1765.4 \text{ veh/h per lane}\\ s_{SB} &= 1800(0.9839) = 1771.0 \text{ veh/h per lane} \end{aligned}$$ Because each approach has two lanes, the approach saturation flows are $$\boxed{\;s_{NB} = 3530.8 \text{ veh/h}, \qquad s_{SB} = 3541.9 \text{ veh/h}\;}$$ The reduction is small — under 2 % — precisely because the bus share is small; on a busy transit corridor with 10 % buses the same formula would take almost 10 % off the saturation flow.
5. Part (c) — trace the physical clearance path to size the all-red. The all-red interval exists so that a vehicle which has legally crossed the stop line on amber can leave the conflict area completely before the cross street receives green. The distance it must cover is the sum of every element between the stop line and the far side of the far crosswalk, plus its own length, because the rear bumper must clear too: $$d = \underbrace{1.0}_{\text{set-back}} + \underbrace{3.0}_{\text{near crosswalk}} + \underbrace{2(3.75)(2)}_{\text{cross-street carriageway}} + \underbrace{3.0}_{\text{far crosswalk}} + \underbrace{6.0}_{\text{car length}}$$ The cross-street carriageway is two lanes of 3.75 m in each of two directions, so 15.0 m, and $$d = 1.0 + 3.0 + 15.0 + 3.0 + 6.0 = 28.0 \text{ m}$$ Stopping the count at the far kerb, as is often done, would leave out the far crosswalk and the car length and under-size the interval by 9 m.
6. Convert the clearance distance to an all-red time and build the intergreen. At the stated clearing speed, $$v = 30 \text{ km/h} = \frac{30}{3.6} = 8.333 \text{ m/s}, \qquad R = \frac{d}{v} = \frac{28.0}{8.333} = 3.36 \text{ s}$$ Rounded to the nearest second as the question instructs, $R = 3\text{ s}$. The intergreen is the amber plus the all-red, $$\boxed{\;R = 3 \text{ s}, \qquad I = A + R = 3.0 + 3 = 6.0 \text{ s}\;}$$
7. Accumulate the lost time over the whole cycle. No start-up lost time is quoted, so the whole intergreen of each phase is treated as time during which no useful flow crosses the stop line. With two phases — one serving the north–south pair, one the east–west pair — $$L = n\,(A+R) = 2(6.0) = \boxed{\;12.0 \text{ s per cycle}\;}$$ At a 75 s cycle that is 16 % of the cycle spent clearing rather than serving, which is why adding phases is expensive: a four-phase version of this same intersection would lose 24 s per cycle and roughly a third of its capacity.
8. Part (d) — select the critical movement in each phase and sum the flow ratios. The flow ratio of a movement is its demand divided by its saturation flow, both per lane: $$y = \frac{q}{s}$$ Working in passenger-car units against the basic rate of 1800 pc/h/lane, $$\begin{aligned} y_{NB} &= \frac{728/2}{1800} = \frac{364}{1800} = 0.2022 \qquad y_{SB} = \frac{620/2}{1800} = \frac{310}{1800} = 0.1722\\ y_{WB} &= \frac{650/2}{1800} = \frac{325}{1800} = 0.1806 \qquad y_{EB} = \frac{550/2}{1800} = \frac{275}{1800} = 0.1528 \end{aligned}$$ Each phase is governed by its heaviest movement, so the north-bound approach governs the north–south phase and the west-bound approach governs the east–west phase: $$Y = y_{NB} + y_{WB} = 0.2022 + 0.1806 = \boxed{\;Y = 0.383\;}$$
9. Cross-check the flow ratio in vehicle units. The same ratio must appear if the calculation is done entirely in vehicles against the adjusted saturation flow, because $f_{HV}$ is the vehicle-to-pcu conversion. For the north-bound approach, $$\frac{q_{veh}/2}{s_{NB}} = \frac{714/2}{1765.4} = \frac{357}{1765.4} = 0.2022$$ which reproduces $y_{NB}$ exactly. That agreement validates the assumed $E_B = 2.0$, the $f_{HV}$ arithmetic and the choice of critical movements in one line. With $Y = 0.383$ and $L = 12$ s the intersection is comfortably under-saturated: at a 75 s cycle the effective green fraction is $(75-12)/75 = 0.84$, well above the 0.383 the demand requires, so the reserve capacity is large.