Question 2 of 7: Webster Signal Design for a Four-Leg Intersection

Question 2: 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.

Saturation flows, per lane: through 3075, through-right 2250, left 1995, left-through 2015, left-through-right 2500 vphpl. Lost time per phase from acceleration and deceleration \(\ell = 3.5\) s; all-red \(AR = 1.5\) s per phase.

Find. An appropriate phasing plan (number and arrangement of phases, with the reason it is appropriate), the optimum cycle length by Webster's method checked against the pedestrian minimum-green requirement, and the resulting phase lengths — presented with a layout of the intersection geometry and of the phasing.

Approach. Convert each movement to a peak-15-minute design flow with the PHF, assign lanes and balance the through demand between them so both lanes of an approach carry the same flow ratio, take the critical flow ratio in each phase, price the four-phase alternative to show it is infeasible, then compute Webster's optimum cycle and — because the table supplies conflicting pedestrian volumes — solve the HCM pedestrian minimum-green requirement as an inequality in \(C\) and adopt the larger of the two.

1. Part (a of the design) — Establish the geometry from the stated approach widths. The "(Width)" entry is the full curb-to-curb width shared by both directions, and the standard urban lane is 3.5 m. A 20 m carriageway takes four 3.5 m lanes (14.0 m) plus a 6.0 m median — six lanes would need 21.0 m and will not fit — so the north and south approaches carry two lanes each. A 15 m carriageway takes four 3.5 m lanes plus a 1.0 m centre island, so the east and west approaches also carry two lanes each. Reading the width as one direction would put the West through movement at over 950/3 ≈ 300 veh/h per lane on a phantom third lane and is the classic error in this family.
2. Convert the hourly volumes to design flow rates. Webster's method is applied to the flow rate within the peak 15 minutes, obtained by dividing by the peak-hour factor: $$v=\frac{V}{\mathrm{PHF}}$$ For the West approach, \(v_L = 300/0.85 = 352.94\), \(v_T = 950/0.85 = 1117.65\) and \(v_R = 300/0.85 = 352.94\) veh/h; the other approaches follow identically.
3. Assign lanes and balance the through demand. With two lanes and three movements on each approach and permitted left turns, the natural assignment is an inner left-through lane and a kerb through-right lane; the given saturation-flow table supplies exactly those two lane types (2015 and 2250 vphpl), which confirms the intended arrangement. Let \(x\) be the share of the through demand that uses the inner lane. The lanes are balanced when both carry the same flow ratio: $$\frac{v_L+x}{s_{LT}}=\frac{v_T-x+v_R}{s_{TR}}\qquad\Longrightarrow\qquad x=\frac{s_{LT}\,(v_T+v_R)-s_{TR}\,v_L}{s_{LT}+s_{TR}}$$ For the West approach, \(x = [2015(1117.65+352.94)-2250(352.94)]/(2015+2250) = 508.59\) veh/h, so the inner lane carries \(352.94+508.59 = 861.53\) and the kerb lane \(962.00\) veh/h.
4. Compute the flow ratio of each approach. With the lanes balanced, the approach flow ratio is simply the common lane ratio \(y = v_{\text{lane}}/s_{\text{lane}}\): $$y_W=\frac{861.53}{2015}=0.42756$$ Repeating for the other three approaches gives \(y_N = 0.30343\), \(y_S = 0.26205\) and \(y_E = 0.24826\).
5. Price the four-phase (protected-left) option before choosing a plan. "Determine an appropriate phasing system" asks for the justification, not merely the plan, so the protected-left alternative must be tested. Giving each approach an exclusive left bay on a two-lane approach forces the entire through-plus-right demand onto one through-right lane. The four critical ratios then become 0.11794 (N–S lefts), 0.47059 (N–S through-right), 0.17691 (E–W lefts) and 0.65359 (E–W through-right), so $$Y_{4\text{-phase}}=1.41904 \;>\; 1$$ which is infeasible at any cycle length — the demand exceeds the intersection's capacity to serve it in four separate phases. A four-phase plan would also cost a second pedestrian minimum green on each street, which Step 8 shows is already the binding constraint. Two phases with permitted left turns are therefore forced, on both counts.
6. Take the critical flow ratio in each phase and sum them. Each phase serves one street in both directions, so the phase is governed by its worse approach: $$Y=\max(y_N,y_S)+\max(y_E,y_W)=0.30343+0.42756=\boxed{0.73098}$$ Note that North governs the N–S phase although South carries a similar volume, and West governs the E–W phase decisively.
7. Compute the total lost time and Webster's optimum cycle. Each of the two phases loses 3.5 s to acceleration and deceleration and carries a 1.5 s all-red, so $$L=n(\ell+AR)=2(3.5+1.5)=10.0\ \text{s}$$ $$C_o=\frac{1.5L+5}{1-Y}=\frac{1.5(10.0)+5}{1-0.73098}=\frac{20.0}{0.26902}=\boxed{74.35\ \text{s}}$$
8. Check the pedestrian requirement — and solve it as an inequality, not as a one-shot test. The table supplies conflicting pedestrian volumes of 750–800 ped/h on every leg, which is heavy, so the HCM minimum green must be satisfied by each phase: $$G_p=3.2+\frac{L_c}{S_p}+2.7\,\frac{N_{\text{ped}}}{W_E},\qquad N_{\text{ped}}=\frac{v_{\text{ped}}\,C}{3600}$$ The crosswalk a phase must clear is the one whose pedestrians walk parallel to that phase's vehicles: the N–S phase releases pedestrians across the east and west legs, which span the 15 m E–W carriageway, so \(L_c = 15\) m; the E–W phase clears the north and south legs, \(L_c = 20\) m. Both take the governing 800 ped/h. Because \(N_{\text{ped}}\) is itself proportional to \(C\), testing at \(C_o\) alone always passes and always under-designs; the constraint must be written per phase and solved for \(C\): $$\frac{y_i}{Y}\,(C-L)\;\ge\;3.2+\frac{L_c}{S_p}+\frac{2.7}{W_E}\cdot\frac{v_{\text{ped}}C}{3600}$$ With \(S_p = 1.2\) m/s and an assumed \(W_E = 4.0\) m (see the callout below), the N–S phase requires \(0.41509(C-10)\ge 15.70+0.15C\), i.e. \(C \ge 74.88\) s, and the E–W phase requires \(0.58491(C-10)\ge 19.87+0.15C\), i.e. \(C \ge 59.13\) s.
9. Adopt the cycle length. The design must satisfy both controls, so $$C=\max\bigl(C_o,\;C_{\text{ped}}\bigr)=\max(74.35,\ 74.88)=74.88\ \text{s}\;\longrightarrow\;\boxed{C=75\ \text{s}}$$ rounded up to the nearest 5 s as is standard. The pedestrians govern, but only just — 0.5 s separates the two bounds, which is why the inequality had to be solved rather than spot-checked.
10. Split the green in proportion to the critical flow ratios. The total effective green available is \(C-L = 75-10 = 65.0\) s, apportioned as $$g_i=\frac{y_i}{Y}\,(C-L)$$ $$g_{NS}=\frac{0.30343}{0.73098}(65.0)=26.98\ \text{s},\qquad g_{EW}=\frac{0.42756}{0.73098}(65.0)=38.02\ \text{s}$$ Confirming the pedestrian check at the adopted cycle: \(G_{p,NS} = 3.2+12.50+2.7(16.67)/4.0 = 26.95\) s ≤ 26.98 s, and \(G_{p,EW} = 3.2+16.67+11.25 = 31.12\) s ≤ 38.02 s. Both are satisfied, the N–S phase with 0.03 s to spare, which is the signature of a pedestrian-governed design.
11. Convert effective green to displayed intervals. With \(g_i = G_i + A_i - \ell_i\) and an assumed amber of \(A = 3.5\) s equal to the stated lost time, the displayed green equals the effective green numerically and each phase occupies \(g_i + A + AR = g_i + 5.0\) s. The cycle closes exactly: \((26.98+5.0)+(38.02+5.0) = 75.0\) s.
12. Verify the design with the degree of saturation. The critical degree of saturation is $$x_{\text{crit}}=\frac{Y\,C}{C-L}=\frac{0.73098(75)}{65}=\boxed{0.843}$$ and it must reappear as \(q/c\) on both critical approaches, which tests the lane assignment, the flow ratios and the split in one line: North, \(861\!\rightarrow\!611.41/(2015\times 26.98/75) = 0.843\); West, \(861.53/(2015\times 38.02/75) = 0.843\). Both agree exactly. At 0.84 the intersection is busy but below the 0.90 practical limit, so the design stands.