Question 2 of 7: Webster signal design for the four-leg intersection

Question 2 — Webster signal design for the 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.

Find. An appropriate phasing system with its justification; the intersection geometry and lane assignment used; the Webster optimum cycle; the governing cycle length after the pedestrian check; and the green and all-red intervals for each phase.

Approach. Convert the volumes to design flow rates with the peak-hour factor, assign lanes and balance the through demand between them so both lanes of an approach carry the same degree of saturation, price the protected (four-phase) option and reject it, then run Webster's optimum-cycle formula on the two-phase plan and test the result against the HCM pedestrian crossing requirement, which at these pedestrian volumes is what actually governs.

1. Convert peak-hour volumes to design flow rates. The peak-hour factor converts an hourly volume to the flow rate sustained during the peak 15 minutes, which is the rate the signal must serve:
   $$q=\frac{V}{\text{PHF}}$$
   With PHF = 0.95 throughout, for example the north left turn becomes $300/0.95=315.8$ veh/h and the west through movement $850/0.95=894.7$ veh/h. The full set is

   Design flow rate (veh/h)	Left	Through	Right	Approach total
   North	315.8	749.5	255.8	1321.1
   South	211.6	631.6	415.8	1258.9
   East	315.8	647.4	268.4	1231.6
   West	211.6	894.7	315.8	1422.1

2. Price the protected four-phase option before choosing a plan. The question asks for an appropriate phasing system, so the protected alternative must be evaluated rather than dismissed. With only two lanes per approach, an exclusive left-turn lane leaves the entire through-plus-right demand on a single through-right lane. Taking the critical movement in each of the four phases,
   $$y_{\text{NS,L}}=\frac{315.8}{1835}=0.172,\qquad y_{\text{NS,T}}=\frac{1047.4}{2010}=0.521$$
   $$y_{\text{EW,L}}=\frac{315.8}{1835}=0.172,\qquad y_{\text{EW,T}}=\frac{1210.5}{2010}=0.602$$
   which sum to
   $$Y_{4\text{-phase}}=0.172+0.521+0.172+0.602=1.468$$
   A value of $Y$ above unity means the intersection cannot be timed at all under that plan — no cycle length exists. The protected plan is therefore infeasible on geometry alone, and a wider approach would be needed to consider it. A second, independent argument points the same way: each extra phase costs roughly one pedestrian crossing interval, and at 1050 to 1250 ped/h those intervals are large.
3. Adopt a two-phase plan with permitted left turns, and assign the lanes. Phase A serves north and south, phase B serves east and west, and in each case the left turns are permitted, filtering through gaps in the opposing through stream. Each approach then has an inner left-plus-through lane ($s_1=1800$) and an outer through-plus-right lane ($s_2=2010$).
4. Balance the through demand between the two lanes. Drivers distribute themselves so that neither lane is systematically worse, which means equal degrees of saturation. Letting $x$ be the share of the through demand that uses the inner lane,
   $$\frac{v_L+x}{s_1}=\frac{(v_T-x)+v_R}{s_2}\quad\Longrightarrow\quad x=\frac{s_1\left(v_T+v_R\right)-s_2\,v_L}{s_1+s_2}$$
   For the north approach,
   $$x_N=\frac{1800\left(749.5+255.8\right)-2010\left(315.8\right)}{1800+2010}=308.3\ \text{veh/h}$$
   so the inner lane carries $315.8+308.3=624.1$ veh/h and the outer lane $1005.3-308.3=697.0$ veh/h. Both give the same flow ratio, $624.1/1800=697.0/2010=0.3467$, which is the check that the split was done correctly.
5. Collect the flow ratio of every approach and pick the two critical movements. Repeating the balance on the other three approaches:

   Approach	Through split $x$ (veh/h)	Inner lane flow (veh/h)	Flow ratio $y$	Critical?
   North	308.3	624.1	0.3467	yes, phase A
   South	383.2	594.8	0.3305	no
   East	266.1	581.9	0.3233	no
   West	460.3	671.9	0.3733	yes, phase B

   Note that the flow ratio, not the volume, decides which movement is critical: the west approach carries the largest volume and is critical for phase B, but for phase A it is the north approach at 1321 veh/h rather than the south approach that governs, because north has the heavier left turn feeding the lower-saturation inner lane. The sum over phases is
   $$Y=y_{\text{NS}}+y_{\text{EW}}=0.3467+0.3733=\boxed{0.7200}$$
6. Total the lost time. No start-up lost time is given separately from the 3.5 s acceleration-and-deceleration allowance, and no amber interval is quoted, so the whole intergreen is treated as lost:
   $$L=n\left(l+R\right)=2\left(3.5+1.5\right)=10.0\ \text{s}$$
7. Compute Webster's optimum cycle for the vehicles. Webster's delay-minimising cycle is
   $$C_o=\frac{1.5L+5}{1-Y}=\frac{1.5(10.0)+5}{1-0.7200}=\frac{20.0}{0.2800}=71.4\ \text{s}$$
   If vehicles were the only consideration, a 70 s or 75 s cycle would be adopted here. They are not: the conflicting pedestrian volumes of 1050 to 1250 ped/h are exceptionally high and must be checked before any cycle is accepted.
8. Set up the pedestrian requirement as an inequality in the cycle length. The HCM pedestrian crossing time for a crosswalk wider than 3.0 m is
   $$G_p=3.2+\frac{L_c}{S_p}+2.7\,\frac{N_{ped}}{W_E},\qquad N_{ped}=\frac{v_{ped}\,C}{3600}$$
   Because $N_{ped}$ itself grows with $C$, evaluating this once at $C_o$ is not a valid test — the demand moves with the cycle. Writing the green available to a phase as $g_i=(y_i/Y)(C-L)$, the requirement becomes a linear inequality in $C$:
   $$\frac{y_i}{Y}\left(C-L\right)\ \ge\ 3.2+\frac{L_c}{S_p}+\frac{2.7\,v_{ped}}{3600\,W_E}\,C$$
9. Identify which crosswalk each phase must clear. Pedestrians walk during the phase whose vehicles move parallel to them. Phase A moves north–south traffic, so it serves the pedestrians on the east and west legs, who cross the 15 m E–W carriageway; the governing volume is $\max(1250,1050)=1250$ ped/h. Phase B serves the north and south legs, crossing the 12 m N–S carriageway at $\max(1150,1150)=1150$ ped/h. Getting this pairing the wrong way round is the classic error in this question, because it swaps a 15 m crosswalk against a 12 m one.
10. Solve the inequality for each phase. For phase A, with $L_c=15$ m, $v_{ped}=1250$ ped/h, $y_i/Y=0.3467/0.7200=0.4815$:
    $$0.4815\left(C-10\right)\ \ge\ 3.2+\frac{15}{1.2}+\frac{2.7(1250)}{3600(4.0)}\,C=15.70+0.23438\,C$$
    $$\left(0.4815-0.23438\right)C\ \ge\ 15.70+4.815\quad\Longrightarrow\quad C\ \ge\ 83.0\ \text{s}$$
    and for phase B, with $L_c=12$ m, $v_{ped}=1150$ ped/h, $y_i/Y=0.5185$:
    $$\left(0.5185-0.21563\right)C\ \ge\ 3.2+\frac{12}{1.2}+5.185\quad\Longrightarrow\quad C\ \ge\ 60.7\ \text{s}$$
    Phase A governs. The pedestrian requirement of 83.0 s exceeds Webster's 71.4 s, so this intersection is pedestrian-controlled, and rounding up to the next 5 s increment gives
    $$\boxed{C=85\ \text{s}}$$
11. Split the green in proportion to the flow ratios. The effective green available is $C-L=85-10=75$ s, apportioned as
    $$g_i=\frac{y_i}{Y}\left(C-L\right)$$
    $$g_{\text{NS}}=0.4815(75)=36.1\ \text{s},\qquad g_{\text{EW}}=0.5185(75)=38.9\ \text{s}$$
    and the two sum to 75.0 s exactly, as they must. With no amber given, the displayed green recovers the 3.5 s of acceleration-and-deceleration lost time, so
    $$G_{\text{NS}}=36.1+3.5=39.6\ \text{s},\qquad G_{\text{EW}}=38.9+3.5=42.4\ \text{s}$$
    and the cycle closes: $39.6+1.5+42.4+1.5=85.0$ s.
12. Confirm the pedestrian intervals are actually delivered. Evaluating $G_p$ at the adopted 85 s cycle,
    $$G_{p,\text{A}}=3.2+12.50+\frac{2.7\left(1250\times85/3600\right)}{4.0}=35.6\ \text{s}\ \le\ 36.1\ \text{s}$$
    $$G_{p,\text{B}}=3.2+10.00+\frac{2.7\left(1150\times85/3600\right)}{4.0}=31.5\ \text{s}\ \le\ 38.9\ \text{s}$$
    Both pass, phase A with only 0.5 s to spare, which confirms that phase A's crosswalk is what set the cycle.
13. Check the degree of saturation on both critical lanes. Because the green was split in proportion to the flow ratios, every critical movement must end up at the same degree of saturation,
    $$x_{crit}=\frac{Y\,C}{C-L}=\frac{0.7200(85)}{75}=0.816$$
    Verifying it directly, the capacities of the two critical lanes are $c=s\,g/C$:
    $$c_{\text{N}}=1800\times\frac{36.1}{85}=764.7\ \text{veh/h},\qquad c_{\text{W}}=1800\times\frac{38.9}{85}=823.3\ \text{veh/h}$$
    giving $624.1/764.7=0.816$ and $671.9/823.3=0.816$. The two agreeing validates the lane assignment, the flow ratios and the green split in a single line. A degree of saturation of 0.82 is a satisfactory design value, and Webster's delay formula gives about 28 s per vehicle on the critical north lane and 26 s on the critical west lane — level of service C.

Results.