Skip to main content

Signal timing: roads A 15 m and B 12 m

Short answer

Pedestrian crossing times govern: minimum green for road A (crossing road B) is 15 s, for road B (crossing road A) 18 s. Adopted cycle 38 s: road A green 15 s, amber 3 s; road B green 18 s, amber 2 s. Pedestrian signals show 5 s WALK then flashing clearance for the respective crossing duration.

An isolated signal with pedestrians indication is to be installed on a right angled intersection with road A 15 m wide and road B 12 m wide. The heaviest volume per hour for each lane of road A and road B are 300 and 250 respectively. The amber times for roads A and B are 3 and 2 seconds respectively. Design the timings of traffic and pedestrian signal.

Isolated two-phase signal with pedestrian indications for road A (15 m) and road B (12 m). Approach speeds are not needed because the ambers are given (3 s and 2 s).

Design basis (IRC:93-1985, Guidelines on Design and Installation of Road Traffic Signals, with Webster's method): two-phase signal, Phase 1 = road A moves, Phase 2 = road B moves. Pedestrians cross a road while the traffic on the cross road has green. Assumed (not given): saturation flow 1800 PCU/h per lane, start-up lost time 2 s per phase, pedestrian walking speed 1.2 m/s and initial walk (pedestrian start-up) time 5 s.

Step 1: Amber (change) time

Amber times are given: road A = 3 s, road B = 2 s.

Step 2: Pedestrian crossing time

Time to cross = initial walk + width / walking speed:

tp,A=5+151.2=17.5 stp,B=5+121.2=15.0 s\begin{aligned} t_{p,A} &= 5 + \frac{15}{1.2} = 17.5\ \text{s}\\ t_{p,B} &= 5 + \frac{12}{1.2} = 15.0\ \text{s} \end{aligned}

Pedestrians crossing road A walk during Phase 2, so the Phase 2 green must be at least 17.5 s. Pedestrians crossing road B walk during Phase 1, so the Phase 1 green must be at least 15.0 s.

Step 3: Webster cycle time from traffic

yA=3001800=0.167,yB=2501800=0.139,Y=0.306L=(2+3)+(2+2)=9 sC0=1.5L+51−Y=1.5(9)+51−0.306=26.6 s\begin{aligned} y_A &= \frac{300}{1800} = 0.167, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.306\\ L &= (2+3) + (2+2) = 9\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(9)+5}{1-0.306} = 26.6\ \text{s} \end{aligned}

With C0−L=17.6C_0 - L = 17.6 s of effective green:

gA=0.1670.306(17.6)=9.6 s,GA=gA+2=11.6 sgB=0.1390.306(17.6)=8.0 s,GB=gB+2=10.0 s\begin{aligned} g_A &= \frac{0.167}{0.306}(17.6) = 9.6\ \text{s}, \quad G_A = g_A + 2 = 11.6\ \text{s}\\ g_B &= \frac{0.139}{0.306}(17.6) = 8.0\ \text{s}, \quad G_B = g_B + 2 = 10.0\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road A)11.615.015
2 (road B)10.017.518

The pedestrian requirement governs. Adopted cycle length C=15+3+18+2=38C = 15 + 3 + 18 + 2 = 38 s.

Final timings

SignalPhase 1: road APhase 2: road B
Vehicle green15 s18 s
Vehicle amber3 s2 s
Vehicle red20 s18 s

Pedestrian signals: pedestrians crossing road B get WALK for 5 s, then flashing (clearance) for 10 s during Phase 1 (total 15 s of the 15 s green); pedestrians crossing road A get WALK for 5 s, then flashing for 13 s during Phase 2 (total 18 s of the 18 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 38 s
Road A: [GREEN 15][A 3][  RED 20  ]
Road B: [ RED 18 ][GREEN 18][A 2]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 23]
Ped crossing A: [DON'T WALK 18][WALK+FLASH 18][DW 2]

✓ Answer checked by Claude Opus 5.5 (high effort)

All Transportation Engineering II chapter 1 questions →