Question 6 of 7: Flows, Saturation Flow and Clearance at an Isolated Intersection

Question 6: Flows, Saturation Flow and Clearance at an Isolated Intersection (20 marks — 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. A four-leg intersection, one lane per approach, no turning movements, so each approach is a single lane group and the two phases are simply north–south and east–west.

Geometry and control: lane width 4.0 m; crosswalks 3.0 m wide on all four legs; stop line set back 1.0 m from the crosswalk; car occupancy 2.0 persons; basic saturation flow 2200 pcu/h/lane; two phases; amber 3.0 s; clearing speed 36 km/h; passenger-car length 6.0 m; level grades.

Find. (a) approach flows in veh/h, pcu/h and persons/h; (b) the saturation flow adjusted for buses, expressed in veh/h; (c) the all-red interval, intergreen period and intersection lost time; (d) the intersection flow ratio.

Approach. Convert the bus and car counts into the three units in turn, using the assumed bus equivalent \(E_B = 2.0\); obtain the heavy-vehicle factor from the same equivalent and apply it to the basic saturation flow; build the all-red from the physical distance a vehicle must travel to clear the conflict area; and finally take the critical flow ratio of each phase and add them.

1. Part (a) — vehicle flows. The vehicle flow is simply the sum of cars and buses, since a bus is one vehicle: \[ q_{\text{veh}} = q_{\text{car}} + q_{\text{bus}} \] giving NB \(1000+17 = 1017\), SB \(600+10 = 610\), EB \(400\) and WB \(800\) veh/h, a total of 2827 veh/h entering the intersection.
2. Convert to passenger-car units. A bus occupies more road space and discharges more slowly than a car; with the assumed equivalent \(E_B = 2.0\), \[ q_{\text{pcu}} = q_{\text{car}} + E_B\,q_{\text{bus}} \] so NB \(= 1000 + 2(17) = 1034\), SB \(= 600 + 2(10) = 620\), EB \(= 400\) and WB \(= 800\) pcu/h, totalling 2854 pcu/h.
3. Convert to persons per hour. Occupancy applies to vehicles, never to passenger-car units, and the bus occupancy differs by direction: \[ q_{\text{per}} = 2.0\,q_{\text{car}} + O_{\text{bus}}\,q_{\text{bus}} \] NB \(= 2(1000) + 30(17) = 2000 + 510 = 2510\); SB \(= 2(600) + 15(10) = 1200 + 150 = 1350\); EB \(= 2(400) = 800\); WB \(= 2(800) = 1600\) persons/h, totalling 6260 persons/h. The NB approach is the busiest in every unit, and it is dramatically the busiest in person terms: seventeen buses an hour move 510 people, the equivalent of 255 cars.
4. Part (b) — heavy-vehicle adjustment factor. The proportion of heavy vehicles on each approach is \(P_{HV} = q_{\text{bus}}/q_{\text{veh}}\), and \[ f_{HV} = \frac{1}{1 + P_{HV}\,(E_B - 1)} \] For NB, \(P_{HV} = 17/1017 = 0.01672\) so \(f_{HV} = 1/1.01672 = 0.9836\); for SB, \(P_{HV} = 10/610 = 0.01639\) so \(f_{HV} = 0.9839\). The EB and WB approaches carry no buses, so \(f_{HV} = 1.000\).
5. Apply the factor to obtain the saturation flow in veh/h. The basic value 2200 is in pcu/h; multiplying by \(f_{HV}\) converts it into the number of mixed vehicles the lane can discharge per hour of green: \[ s_{\text{veh}} = s_0\,f_{HV} \] \[ s_{NB} = 2200(0.9836) = \boxed{2164\ \text{veh/h}} \qquad s_{SB} = 2200(0.9839) = 2165\ \text{veh/h} \] with \(s_{EB} = s_{WB} = 2200\) veh/h. The cross-check is that \(f_{HV}\) is precisely the vehicle-to-pcu conversion, so \(s_0\,q_{\text{veh}}/q_{\text{pcu}}\) must reproduce the same numbers: \(2200(1017/1034) = 2163.8\) and \(2200(610/620) = 2164.5\), which it does. The adjustment is small here because buses are less than 2 % of the traffic.
6. Part (c) — distance a vehicle must travel to clear the intersection. The all-red interval must let a vehicle that entered on amber pass completely beyond the conflict area. Starting from the stop line, that distance is the set-back, the near crosswalk, the full cross-street carriageway (two 4 m lanes), the far crosswalk, and finally the length of the vehicle itself: \[ d = 1.0 + 3.0 + (2 \times 4.0) + 3.0 + 6.0 = 21.0\ \text{m} \] The geometry is identical for both phases because all four approaches share the same lane and crosswalk widths.
7. All-red interval and intergreen period. At the clearing speed \(v = 36\ \text{km/h} = 10.0\ \text{m/s}\), \[ R = \frac{d}{v} = \frac{21.0}{10.0} = \boxed{2.1\ \text{s}} \] and the intergreen — the whole change-and-clearance period between one phase’s green and the next — is \[ I = A + R = 3.0 + 2.1 = \boxed{5.1\ \text{s}} \]
8. Intersection lost time. No start-up lost time is quoted, so the standard assumption is that the whole intergreen of each phase is lost. With two phases, \[ L = n\,(A + R) = 2 \times 5.1 = \boxed{10.2\ \text{s}} \] Only \(C - L\) of any cycle is therefore available as effective green — 88.7 % of a 90 s cycle, but only 79.6 % of a 50 s cycle, which is why very short cycles are inefficient.
9. Part (d) — flow ratios and the critical movements. The flow ratio of a movement is its demand divided by its saturation flow, both in pcu: \[ y = \frac{q_{\text{pcu}}}{s_0} \] \[ y_{NB} = \frac{1034}{2200} = 0.470 \qquad y_{SB} = \frac{620}{2200} = 0.282 \] \[ y_{EB} = \frac{400}{2200} = 0.182 \qquad y_{WB} = \frac{800}{2200} = 0.364 \] Each phase is governed by its heaviest movement, so the critical movements are NB in the north–south phase and WB in the east–west phase.
10. Intersection flow ratio. Summing the critical flow ratios, \[ Y = y_{NB} + y_{WB} = 0.470 + 0.364 = \boxed{0.834} \] Since \(Y < 1\) the intersection is theoretically able to carry the demand, but at 0.834 it is heavily loaded — a long cycle will be required, and there is little reserve. The result can be confirmed in vehicle units, \(1017/2163.8 + 800/2200 = 0.470 + 0.364 = 0.834\), which is identical because \(f_{HV}\) is exactly the pcu conversion; agreement here validates the assumed \(E_B = 2.0\), the heavy-vehicle arithmetic and the choice of critical movements in one line.