Question 7 of 7: Critical Lanes, Left-Turn Adjustment and Green Splits

Question 7: Critical Lanes, Left-Turn Adjustment and Green Splits (20 marks — 4+2+4+2+8)

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. Four independent sub-problems on the same theme: (a) four lane flows and saturation flows in one phase; (b) a three-lane approach with 18 % permitted left turns, a through-vehicle equivalent of 7.25 and a base saturation flow of 2150 vphgpl; (c) and (d) build on (b), with an effective green of 50 s in an 80 s cycle; (e) a total available green of 80 s to be split at four intersections whose phase flow rates are tabulated above, all with the same saturation flow.

Find. (a) the critical flow ratio and critical lane; (b) the left-turn adjustment factor; (c) the approach saturation flow rate and saturation headway; (d) the approach capacity; (e) the green time for each phase at each of the four intersections.

Approach. Compute a flow ratio per lane and take the maximum for (a); apply the passenger-car-equivalent form of the left-turn adjustment for (b) and propagate it through saturation flow, headway and capacity for (c) and (d); split the green in the ratio of the phase flow rates for (e), which is legitimate only because the saturation flows are equal.

1. Part (a) — a flow ratio for each lane. The flow ratio of a lane is its design flow divided by its own saturation flow, $y = q/s$: $$y_{NB\text{-}LS} = \frac{550}{1350} = 0.4074 \qquad y_{NB\text{-}RS} = \frac{450}{1650} = 0.2727$$ $$y_{SB\text{-}LS} = \frac{400}{1200} = 0.3333 \qquad y_{SB\text{-}RS} = \frac{745}{1450} = 0.5138$$
2. Part (a) — identify the critical lane. A phase must be timed for its worst lane, because green that satisfies the heaviest lane satisfies all the others. The largest ratio is $$y_{crit} = \boxed{0.514 \text{ on lane SB - R,S}}$$ so the southbound right-and-through lane is the critical lane. It happens to carry the largest flow as well, but that is coincidence: NB - L,S carries 550 pcu/h against SB - L,S's 400 and yet has a higher ratio than NB - R,S, which carries less traffic on a much better lane. It is the ratio, never the volume, that decides.
3. Part (b) — left-turn adjustment factor. With permitted lefts, each left-turning vehicle consumes the green of $E_L$ through vehicles, so a lane group carrying a proportion $P_{LT}$ of lefts discharges $1 + P_{LT}(E_L - 1)$ through-equivalents per vehicle. The adjustment factor is the reciprocal: $$f_{LT} = \frac{1}{1 + P_{LT}(E_L - 1)} = \frac{1}{1 + 0.18(7.25 - 1)} = \frac{1}{1 + 1.125} = \boxed{0.471}$$ An $E_L$ of 7.25 is very high — it implies a heavy opposing through movement with few usable gaps — and it is why 18 % of lefts destroys more than half the approach's capacity.
4. Part (c) — approach saturation flow rate. Applying the factor to the base flow of all three lanes, $$s = s_0\,N\,f_{LT} = 2150 \times 3 \times 0.4706 = \boxed{3035 \text{ veh/h}}$$ against an unadjusted $2150 \times 3 = 6450$ veh/h. The permitted lefts cost 3415 veh/h of green capacity, more than a whole lane and a half.
5. Part (c) — saturation headway. The saturation headway is the reciprocal of the saturation flow expressed per vehicle: $$h_s = \frac{3600}{s} = \frac{3600}{3035} = \boxed{1.19 \text{ s}}$$ for the approach as a whole. Per lane it is $3600/(2150 \times 0.4706) = 3.56$ s, against a base of $3600/2150 = 1.67$ s — the ratio of the two, 2.13, is exactly $1/f_{LT}$, which is the cleanest check on part (b).
6. Part (d) — approach capacity. Capacity is the saturation flow prorated by the fraction of the cycle that is effective green: $$c = s\,\frac{g}{C} = 3035 \times \frac{50}{80} = 3035 \times 0.625 = \boxed{1897 \text{ veh/h}}$$ Only 62.5 per cent of the cycle is usable green, so the approach delivers about 632 veh/h per lane in practice.
7. Part (e) — split the green in the ratio of the phase flows. When both phases share the same saturation flow, the flow ratios are proportional to the flow rates themselves, so equalising the degrees of saturation reduces to a simple proportional split: $$g_1 = G\,\frac{q_1}{q_1 + q_2} \qquad g_2 = G\,\frac{q_2}{q_1 + q_2}$$ with $G = 80$ s. Intersection 1, with 350 and 185 pcu/h, gives $g_1 = 80(350/535) = 52.3$ s and $g_2 = 80(185/535) = 27.7$ s.
8. Part (e) — the four intersections. Repeating the same split: $$\text{2: } 59.1 / 20.9 \text{ s} \qquad \text{3: } 63.6 / 16.4 \text{ s} \qquad \text{4: } 42.2 / 37.8 \text{ s}$$ each pair summing to 80 s. The spread is instructive: intersection 3, with a 3.9-to-1 demand ratio, gives phase 2 barely 16 s — likely below the pedestrian and minimum-green requirements, so in practice its split would be overridden by a minimum green rather than by the flow ratio.