Question 6 of 7: Webster Signal Design for a Four-Leg Intersection

Question 6: Webster Signal Design for a Four-Leg Intersection (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. A four-leg intersection of an 18 m north–south street with a 20 m east–west street, the turning volumes, pedestrian volumes and peak hour factors tabulated above, and a menu of saturation flows by lane type. Not given, and therefore assumed below: the lane widths and lane arrangement, the crosswalk width, the pedestrian walking speed, the amber interval and the clearance speed.

Find. A defensible intersection geometry, a phasing system with its justification, and the Webster phase lengths — together with the degrees of saturation the design delivers.

Approach. Fix the geometry from the given widths, convert volumes to flow rates through the PHF, assign each approach's demand to its lanes so the lane flow ratios balance, take the critical flow ratio in each phase, price the four-phase alternative to justify the phasing choice, compute the lost time from a clearance chain, then take the cycle as the larger of Webster's optimum and the pedestrian minimum before splitting the green.

1. Fix the geometry from the given widths. The widths in the volume table are curb-to-curb widths of the whole street, shared by both directions — reading 18 m as one direction would imply five lanes carrying only about 220 veh/h each, which is absurd. With 3.5 m lanes, 18 m accommodates four lanes (14.0 m) plus a 4.0 m median, and 20 m accommodates the same four 3.5 m lanes plus a 6.0 m median. So every approach has two lanes: an inner (median) lane taking lefts and throughs, and a curb lane taking throughs and rights.
2. Convert volumes to design flow rates. Capacity analysis works at the peak rate of flow, so every movement is divided by the peak hour factor, $v = V/\mathrm{PHF} = V/0.95$. The west approach, for example, becomes 221.1 left, 770.5 through and 561.1 right, a total of 1552.6 veh/h — the heaviest approach at the intersection. The corresponding totals are 1113.7 N, 891.6 S, 1407.4 E and 1552.6 W.
3. Assign demand to lanes and take the lane flow ratio. With two lanes and permitted lefts, the through demand splits between them; the correct split is the one that equalises the two lane flow ratios, since the larger of the two governs the approach. Writing $a$ for the through volume in the median lane, $$\frac{v_L + a}{1850} = \frac{v_R + v_T - a}{2150}$$ On the west approach this gives $a = 497.0$ veh/h and a balanced lane flow ratio of $$y_W = \frac{221.1 + 497.0}{1850} = \boxed{0.3882}$$ The same construction gives $y_N = 0.2784$, $y_S = 0.2229$ and $y_E = 0.3518$.
4. Critical flow ratios and the sum $Y$. Each phase is governed by its heaviest approach, so the north–south phase takes $y_N = 0.2784$ and the east–west phase takes $y_W = 0.3882$, giving $$Y = 0.2784 + 0.3882 = \boxed{0.6666}$$ comfortably below 1, so the intersection can be timed. Note that the flow ratio, not the volume, decides which street gets the larger green: the east approach carries 1407 veh/h against the north's 1114 and still yields to the west.
5. Choose the phasing system, and say why. Price the four-phase alternative first. Exclusive left-turn bays on a two-lane approach would strand the whole through-plus-right demand in one curb lane, giving through flow ratios of 0.4088 N, 0.3682 S, 0.5346 E and 0.6193 W and left-turn ratios of 0.1423 N, 0.0606 S, 0.1563 E and 0.1339 W. The four critical values sum to $$Y_4 = 0.6193 + 0.4088 + 0.1563 + 0.1423 = 1.327 \gt 1$$ which cannot be timed at any cycle length. Two phases with permitted left turns are therefore forced by the geometry — and independently by the pedestrians, since every additional phase costs roughly a full pedestrian green. The question asks for “an appropriate phasing system”; this calculation is the justification, not merely the choice.
6. Intergreen and lost time from a clearance chain. A vehicle clearing the north–south phase must cross the set-back to the near crosswalk, the near crosswalk, the full east–west carriageway, the far crosswalk and its own length. Assuming a 1.0 m set-back, 4.0 m crosswalks, a 6.0 m vehicle and a 50 km/h clearance speed, $$d = 1.0 + 4.0 + 20.0 + 4.0 + 6.0 = 35.0 \text{ m} \qquad R = \frac{35.0}{50/3.6} = 2.52 \text{ s}$$ and the east–west phase needs 33.0 m, or 2.38 s. Rounding both to $R = 2.5$ s and taking an amber of $A = 3.0$ s gives an intergreen of 5.5 s per phase. With no start-up lost time given, the whole intergreen is taken as lost: $$L = n(A + R) = 2(5.5) = \boxed{11.0 \text{ s}}$$
7. Webster's vehicle optimum — and why it is not the answer. The cycle that minimises vehicular delay is $$C_o = \frac{1.5L + 5}{1 - Y} = \frac{1.5(11) + 5}{1 - 0.6666} = 64.5 \text{ s}$$ But this intersection carries 1100–1225 ped/h on every leg, and the pedestrian green requirement grows with the cycle, because the number of pedestrians arriving per cycle does. Evaluating the pedestrian check once at $C_o$ would always pass and always under-design; the constraint has to be solved as an inequality in $C$.
8. Solve the pedestrian minimum as a fixed point. The HCM pedestrian green, for a crosswalk wider than 3.0 m, is $$G_p = 3.2 + \frac{L_c}{S_p} + 2.7\,\frac{N_{ped}}{W_E}, \qquad N_{ped} = \frac{v_{ped}\,C}{3600}$$ Pedestrians crossing the east and west legs walk parallel to the north–south vehicles, so they are served by the north–south green and they span the 20 m east–west carriageway; those crossing the north and south legs span 18 m during the east–west green. Both directions are governed by 1225 ped/h. Taking $S_p = 1.2$ m/s and $W_E = 4.0$ m, $$G_{p,NS} = 19.87 + 0.2297\,C \qquad G_{p,EW} = 18.20 + 0.2297\,C$$
9. Impose the constraint on each phase. Splitting the green in proportion to the critical flow ratios gives $g_i = (y_i/Y)(C - L)$, that is 0.4177 and 0.5823 of the effective green. The north–south phase therefore requires $$0.4177(C - 11) \ge 19.87 + 0.2297\,C \qquad\Longrightarrow\qquad C \ge 130.1 \text{ s}$$ while the east–west phase needs only $C \ge 69.8$ s. The pedestrians on the lighter vehicular phase govern, because that phase gets the smaller share of the green while facing the wider crossing. Taking the largest of the three requirements and rounding up to a 5 s multiple, $$C = \boxed{135 \text{ s}}$$
10. Phase lengths, and the pedestrian check. The effective green available is $C - L = 135 - 11 = 124$ s, split as $$g_{NS} = 0.4177(124) = \boxed{51.8 \text{ s}} \qquad g_{EW} = 0.5823(124) = \boxed{72.2 \text{ s}}$$ and $51.8 + 72.2 + 11.0 = 135$ s closes the cycle exactly. Checking the pedestrians at the adopted cycle, 1225 ped/h delivers $1225(135)/3600 = 45.9$ pedestrians per cycle, so $G_{p,NS} = 50.9$ s against 51.8 s available and $G_{p,EW} = 49.2$ s against 72.2 s available. Both clear, the north–south phase only just — which is the signature of a pedestrian-controlled design.
11. Confirm the design with the degrees of saturation. Because the green was split in proportion to the flow ratios, the two critical movements must share one degree of saturation, $$x_{crit} = \frac{Y\,C}{C - L} = \frac{0.6666(135)}{124} = 0.726$$ and computing $q/c$ separately for the north and west approaches reproduces 0.726 on both, which validates the flow ratios, the lane assignment and the green split in a single line. The non-critical approaches sit lower, at $x_S = 0.581$ and $x_E = 0.658$. Approach capacities are $(1850 + 2150)(g/C)$, that is 1535 veh/h on each of north and south and 2139 veh/h on each of east and west, and Webster's uniform delay term on the critical east–west approach is 23.9 s per vehicle.