Question 3 of 7: Webster signal design for a four-approach intersection

Question 3 — Webster signal design for a four-approach 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.

Approach. Establish the lane geometry implied by the stated approach widths, convert the peak-hour volumes to peak-flow rates with the PHF, assign movements to lanes and balance the through movement so both lanes on an approach carry the same degree of saturation, take the critical flow ratio on each street, apply Webster's optimum-cycle formula, then test the resulting green times against the pedestrian minimum-green requirement and adopt whichever control governs.

1. Interpret the approach widths and fix the lane geometry. The widths in the volume table are curb-to-curb widths shared by both directions of travel. With 3.5 m lanes,
   $$18\ \text{m} = 4 \times 3.5 + 4.0\ \text{m (median)}, \qquad 20\ \text{m} = 4 \times 3.5 + 6.0\ \text{m (median)}$$
   so every approach has two lanes. Reading 18 m as a one-direction width would imply five lanes per approach at about 250 veh/h/lane, which is not a credible loading for a peak-hour intersection carrying 1200–1500 veh/h per approach. Since 20 m will not take six 3.5 m lanes, the East and West approaches also have two lanes each despite being wider. With three movements sharing two lanes the natural assignment is lane 1 = left + through ($s = 1900$ vphpl) and lane 2 = through + right ($s = 2080$ vphpl), which is the geometry drawn in Figure 3.1(i).
2. Convert peak-hour volumes to peak 15-minute flow rates. Design must be for the peak rate within the hour, so every movement is divided by the PHF:
   $$v = \frac{V}{\text{PHF}} = \frac{V}{0.95}$$

   Approach	Left (v)	Through (v)	Right (v)	Total (v)
   North	247.4	774.7	228.4	1250.5
   South	215.8	688.4	327.4	1231.6
   East	231.6	815.8	360.0	1407.4
   West	231.6	894.7	369.5	1495.8

3. Balance the through movement across the two lanes and obtain each approach flow ratio. Drivers distribute themselves so that the two lanes on an approach reach the same degree of saturation. Letting $x$ be the through volume choosing lane 1,
   $$\frac{v_L + x}{1900} = \frac{(v_T - x) + v_R}{2080} \;\Longrightarrow\; x = \frac{1900\,(v_T + v_R) - 2080\,v_L}{1900 + 2080}$$
   and the common value of that ratio is the approach flow ratio $y = v/s$. For the North approach,
   $$x_N = \frac{1900(774.7 + 228.4) - 2080(247.4)}{3980} = 349.6\ \text{veh}\,\text{h}^{-1}$$
   $$y_N = \frac{247.4 + 349.6}{1900} = \frac{597.0}{1900} = 0.3142$$
   and the check on lane 2 returns the same number, $653.5/2080 = 0.3142$. Repeating for the other three approaches:

   Approach	Lane 1 flow (L + part T)	Lane 2 flow (rest of T + R)	Flow ratio $y$
   North	597.0	653.5	0.3142
   South	587.9	643.6	0.3094
   East	671.9	735.5	0.3536
   West	714.1	781.7	0.3758

   Note that the flow ratio, not the volume, decides which movement is critical: West is critical on the E–W street and North on the N–S street even though South carries more right turns than North.
4. Determine an appropriate phasing system — and price the alternative before rejecting it. The question asks for a phasing system and its justification, so the four-phase option (exclusive protected left turns on each street) must be tested rather than dismissed. With only two lanes per approach, an exclusive left bay forces the entire through-plus-right demand onto one through-right lane, and the critical flow ratios become
   $$y_{\text{crit,NS}} = \frac{1015.8}{2080} = 0.488, \qquad y_{\text{crit,EW}} = \frac{1264.2}{2080} = 0.608$$
   $$Y_{4\text{-phase}} = 0.488 + 0.608 = \boxed{1.10 \gt 1.0 \;\Rightarrow\; \text{infeasible}}$$
   A four-phase plan cannot be timed at any cycle length. Two further arguments point the same way: each additional phase costs another 5.0 s of lost time and another full pedestrian minimum green, and the pedestrian volumes here (985–1345 per hour on every leg) are already the binding constraint. A two-phase plan with permitted left turns is therefore adopted — Phase A serving North and South, Phase B serving East and West, as drawn in Figure 3.1(ii).
5. Sum the critical flow ratios and the lost time. The critical approach in each phase governs:
   $$Y = y_A + y_B = \max(y_N, y_S) + \max(y_E, y_W) = 0.3142 + 0.3758 = 0.6900$$
   Lost time is 3.5 s of acceleration/deceleration plus a 1.5 s all-red at each of the two phase changes:
   $$L = n(\ell + AR) = 2\,(3.5 + 1.5) = 10.0\ \text{s}$$
6. Compute Webster's optimum cycle for the vehicles. Webster's minimum-delay cycle is
   $$C_o = \frac{1.5L + 5}{1 - Y} = \frac{1.5(10.0) + 5}{1 - 0.6900} = \frac{20.0}{0.3100} = 64.5\ \text{s}$$
   On vehicular grounds alone a cycle of about 65 s would minimise delay. This value must not be adopted before the pedestrians are checked, because the table's "conflicting pedestrian volumes" row signals a pedestrian-controlled design.
7. Write the pedestrian minimum-green requirement for each phase. The HCM pedestrian green-time requirement for a crosswalk wider than 3.0 m is
   $$G_p = 3.2 + \frac{L_c}{S_p} + 2.7\,\frac{N_{ped}}{W_E}, \qquad N_{ped} = \frac{v_{ped}\,C}{3600}$$
   with $S_p = 1.2$ m/s. The crosswalks a phase must clear are those on which pedestrians walk parallel to that phase's vehicles: the N–S phase releases pedestrians across the East and West legs, and those legs span the E–W carriageway, so $L_c = 20$ m with $v_{ped} = \max(1200, 1345) = 1345$/h. Symmetrically the E–W phase clears the North and South legs, $L_c = 18$ m and $v_{ped} = \max(1000, 985) = 1000$/h. Taking $W_E = 4.0$ m (see the assumption callout),
   $$G_{p,A} = 3.2 + \frac{20}{1.2} + 2.7\,\frac{1345\,C/3600}{4.0} = 19.87 + 0.2522\,C$$
   $$G_{p,B} = 3.2 + \frac{18}{1.2} + 2.7\,\frac{1000\,C/3600}{4.0} = 18.20 + 0.1875\,C$$
   Because $N_{ped}$ itself grows with $C$, this is an inequality in $C$, not a one-shot check at $C_o$.
8. Solve for the cycle length that satisfies the pedestrians. Webster splits the available effective green in proportion to the critical flow ratios,
   $$g_i = \frac{y_i}{Y}\,(C - L), \qquad \frac{y_A}{Y} = 0.4554, \quad \frac{y_B}{Y} = 0.5446$$
   so the requirement $g_i \ge G_{p,i}$ becomes, for Phase A,
   $$0.4554\,(C - 10) \ge 19.87 + 0.2522\,C \;\Longrightarrow\; 0.2032\,C \ge 24.42 \;\Longrightarrow\; C \ge 120.2\ \text{s}$$
   and for Phase B,
   $$0.5446\,(C - 10) \ge 18.20 + 0.1875\,C \;\Longrightarrow\; 0.3572\,C \ge 23.65 \;\Longrightarrow\; C \ge 66.2\ \text{s}$$
   Phase A governs. Rounding up to the nearest 5 s as is normal controller practice,
   $$C = \boxed{125\ \text{s}}$$
   which is nearly double the vehicular optimum — the design is pedestrian-controlled.
9. Distribute the green and convert to displayed timings. With $C = 125$ s the effective green available is $C - L = 115$ s, split as
   $$g_A = 0.4554 \times 115 = \boxed{52.4\ \text{s}}, \qquad g_B = 0.5446 \times 115 = \boxed{62.6\ \text{s}}$$
   The displayed green adds back the 3.5 s of lost time absorbed at each phase change, $G_i = g_i + \ell$:
   $$G_A = 55.9\ \text{s}, \qquad G_B = 66.1\ \text{s}$$
   and the cycle closes exactly: $55.9 + 1.5 + 66.1 + 1.5 = 125.0$ s.
10. Confirm the pedestrian and capacity checks at the adopted timing. Substituting $C = 125$ s into the two requirements,
    $$G_{p,A} = 19.87 + 0.2522(125) = 51.4\ \text{s} \le 52.4\ \text{s} \quad\checkmark$$
    $$G_{p,B} = 18.20 + 0.1875(125) = 41.6\ \text{s} \le 62.6\ \text{s} \quad\checkmark$$
    The critical degree of saturation is
    $$x_{crit} = \frac{Y\,C}{C - L} = \frac{0.6900 \times 125}{115} = 0.750$$
    and because the green is split in proportion to the flow ratios this same value must reappear as $q/c$ on both critical approaches, which validates the lane assignment, the flow ratios and the split in one line:
    $$\text{North lane 1: } \frac{597.0}{1900(52.4/125)} = 0.750, \qquad \text{West lane 1: } \frac{714.1}{1900(62.6/125)} = 0.750 \;\checkmark$$
    A critical degree of saturation of 0.75 is comfortable — the low value is itself the signature of a pedestrian-controlled design, in which the vehicles receive far more green than their demand requires.