Question 3 of 7: Webster Signal Design for a Four-Leg Intersection

Question 3: 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. Peak-hour turning volumes, conflicting pedestrian volumes and a peak-hour factor for each of the four approaches, together with the roadway width of each street and a schedule of saturation flow rates by lane type.

Find. A defensible phasing plan, the Webster optimum cycle length, the green split, and a capacity check confirming the design serves the demand.

Approach. Convert the hourly volumes to peak flow rates, assign lanes to the available width, choose the phasing that the geometry can actually support, form the critical flow ratios, obtain Webster’s optimum cycle, then test that cycle against the pedestrian crossing-time requirement and lengthen it if the pedestrians govern.

1. Convert peak-hour volumes to design flow rates. The flow rate within the peak fifteen minutes is the hourly volume divided by the peak-hour factor, $v = V/\text{PHF}$. For the north approach $$v_N = \frac{200 + 630 + 210}{0.95} = \frac{1{,}040}{0.95} = 1{,}094.7\ \text{veh/h}$$ and repeating for the other three approaches gives 919.0 (south), 1,423.2 (east) and 1,311.6 (west) veh/h.
2. Assign lanes and total the approach saturation flow. With two lanes on each north–south approach the only workable marking is a left-through lane beside a through-right lane, so $$s_{NS} = 1{,}800 + 2{,}100 = 3{,}900\ \text{veh/h of green}$$ and with three lanes on each east–west approach a left-through, a through and a through-right lane give $$s_{EW} = 1{,}800 + 2{,}400 + 2{,}100 = 6{,}300\ \text{veh/h of green}$$ Note that the available width will not accommodate exclusive left-turn bays on the north–south street without removing a through lane.
3. Select the phasing system. A two-phase plan is adopted — north–south, then east–west, each phase carrying all movements of its street with permitted left turns. Two independent arguments force it. First, the geometry: laying out a four-phase plan with protected lefts requires an exclusive left lane on every approach, which on the north–south street leaves the entire through-plus-right demand of 884 veh/h on a single 2,100 vphpl lane and drives the sum of critical flow ratios to 0.996 — an infeasible design. Second, the pedestrians: with 1,200 to 1,350 pedestrians per hour on every leg, each phase must be long enough to walk a pedestrian across, so every phase added costs roughly half a minute of cycle. Two phases is the maximum this site can afford.
4. Form the critical flow ratios. With one lane group per approach, $y = v/s$:

   Flow ratios by approach

   Approach	$v$ (veh/h)	$s$ (veh/h green)	$y = v/s$
   North	1,094.7	3,900	0.2807
   South	919.0	3,900	0.2356
   East	1,423.2	6,300	0.2259
   West	1,311.6	6,300	0.2082

   The critical (larger) ratio of each phase governs, so$$Y = y_{NS} + y_{EW} = 0.2807 + 0.2259 = \boxed{0.5066}$$
5. Compute the change intervals and the lost time. The yellow interval follows the standard clearance relation with zero grade, $$\tau = t + \frac{u}{2a} = 1.0 + \frac{13.89}{2(3.4)} = 3.04\ \text{s} \rightarrow 3.0\ \text{s}$$ The all-red must let a vehicle that has just entered clear the stop-bar set-back, the near crosswalk, the full cross-street carriageway, the far crosswalk and its own length. For the north–south phase, which crosses the 21 m street, $$d_{NS} = 1.0 + 3.0 + 21.0 + 3.0 + 6.0 = 34.0\ \text{m}, \qquad R_{NS} = \frac{34.0}{13.89} = 2.45 \rightarrow 2.5\ \text{s}$$ and for the east–west phase, crossing the 17 m street, $d_{EW} = 30.0$ m and $R_{EW} = 2.16 \rightarrow 2.5$ s. Each intergreen is therefore $I = 3.0 + 2.5 = 5.5$ s and the total lost time is $L = 2(5.5) = 11.0$ s.
6. Webster’s optimum cycle for the vehicles alone. Substituting into Webster’s delay-minimising expression, $$C_o = \frac{1.5L + 5}{1 - Y} = \frac{1.5(11.0) + 5}{1 - 0.5066} = \frac{21.5}{0.4934} = 43.6\ \text{s}$$ so on vehicular grounds a cycle of about 45 s would minimise delay. That cycle cannot be used, because it will not walk a pedestrian across either street.
7. Impose the pedestrian crossing requirement. Pedestrians crossing the east and west legs walk parallel to the north–south traffic and are served in the north–south phase; they cross the 21 m east–west carriageway, and the heavier of the two legs carries 1,350 ped/h. Pedestrians crossing the north and south legs are served in the east–west phase, cross 17 m, and again the heavier leg carries 1,350 ped/h. The HCM crossing-time model for a crosswalk wider than 3.0 m is $$G_p = 3.2 + \frac{L_c}{S_p} + 2.7\,\frac{N_{ped}}{W_E}$$ with $N_{ped} = 1{,}350\,C/3{,}600$ pedestrians arriving per cycle, $S_p = 1.2$ m/s and $W_E = 4.0$ m. Because $N_{ped}$ grows with the cycle while the green available grows only as $(C - L)$ times the phase’s share of $Y$, the requirement must be solved for $C$.
8. Solve for the shortest cycle that satisfies both phases. The phase shares are $y_{NS}/Y = 0.5541$ and $y_{EW}/Y = 0.4459$. Setting green available equal to green required, $$0.4459\,(C - 11.0) \;\ge\; 3.2 + \frac{17.0}{1.2} + \frac{2.7(1{,}350\,C/3{,}600)}{4.0}$$ which reduces to $0.1928\,C \ge 22.27$, so $C \ge 115.5$ s for the east–west phase; the corresponding north–south condition returns the milder $C \ge 89.0$ s. Rounding up to a practical five-second increment,$$\boxed{C = 120\ \text{s}}$$
9. Split the green by the Webster rule. The effective green available is $C - L = 120 - 11 = 109$ s, apportioned in proportion to the critical flow ratios: $$g_{NS} = \frac{0.2807}{0.5066}(109) = 60.4\ \text{s}, \qquad g_{EW} = \frac{0.2259}{0.5066}(109) = 48.6\ \text{s}$$ Because no start-up lost time was quoted, the effective and displayed greens coincide, so the timing plan is a 60 s north–south green, a 5.5 s change interval, a 49 s east–west green and a second 5.5 s change interval — which sums to exactly 120 s.
10. Check the pedestrians against the greens actually provided. At $C = 120$ s each crosswalk receives $N_{ped} = 1{,}350(120)/3{,}600 = 45$ pedestrians per cycle, so $$G_{p,NS} = 3.2 + \frac{21.0}{1.2} + \frac{2.7(45)}{4.0} = 51.1\ \text{s} \le 60\ \text{s} \quad \checkmark$$ $$G_{p,EW} = 3.2 + \frac{17.0}{1.2} + \frac{2.7(45)}{4.0} = 47.7\ \text{s} \le 49\ \text{s} \quad \checkmark$$ Both crossings fit, the east–west phase with only 1.3 s to spare, which confirms that this leg is the binding constraint on the whole design.
11. Check capacity. Each approach receives $c = s\,g/C$, and the degree of saturation is $X = v/c$: north $1{,}950$ veh/h and $X = 0.56$; south $1{,}950$ veh/h and $X = 0.47$; east $2{,}572$ veh/h and $X = 0.55$; west $2{,}572$ veh/h and $X = 0.51$. Every movement is comfortably below the 0.85 practical limit, so the pedestrian-driven cycle delivers ample vehicular capacity as a by-product.

The design is worth reading back as a story. The vehicles want a short cycle — 45 s would minimise their delay — but 1,350 pedestrians per hour on every leg of a 21 m street cannot be walked across in the green that a 45 s cycle provides. Stretching the cycle to serve them delivers roughly twice the vehicle capacity actually required, which is the classic signature of a pedestrian-controlled downtown signal: the intersection is not short of green, it is short of crossing time. If the 120 s cycle proves unacceptable in operation, the productive levers are a wider crosswalk or a pedestrian refuge in the median that lets the 21 m crossing be made in two stages — not more green.