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Signal timing: roads A 18 m and B 12 m

Short answer

Cycle time42 s; Road A vehicle green15 s + amber4 s, Road B20 s + amber3 s. Pedestrian crossing times (15 s for B,20 s for A) are accommodated during the opposing vehicle green.

An isolated signal with pedestrian indication is to be installed on a right angled intersection with road A, 18m wide and road B, 12m wide. The heaviest volume per hour for each lane of road A and B are 300 and 250, respectively. The approach speeds are 55 and 40 kmph for road A and B respectively. Design the timing of traffic and pedestrian signals.

Isolated two-phase signal with pedestrian indications for road A (18 m) and road B (12 m).

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 I=tr+v2aI = t_r + \dfrac{v}{2a} with reaction time tr=1t_r = 1 s and comfortable deceleration a=3 m/s2a = 3\ \text{m/s}^2, kept within the IRC:93 range of 2-4 s:

IA=1+55/3.62(3)=3.55→4 sIB=1+40/3.62(3)=2.85→3 s\begin{aligned} I_A &= 1 + \frac{55/3.6}{2(3)} = 3.55 \to 4\ \text{s}\\ I_B &= 1 + \frac{40/3.6}{2(3)} = 2.85 \to 3\ \text{s} \end{aligned}

Step 2: Pedestrian crossing time

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

tp,A=5+181.2=20.0 stp,B=5+121.2=15.0 s\begin{aligned} t_{p,A} &= 5 + \frac{18}{1.2} = 20.0\ \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 Phase 2 green must be at least 20.0 s. Pedestrians crossing road B walk during Phase 1, so 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+4)+(2+3)=11 sC0=1.5L+51−Y=1.5(11)+51−0.306=31.0 s\begin{aligned} y_A &= \frac{300}{1800} = 0.167, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.306\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.306} = 31.0\ \text{s} \end{aligned} gA=0.1670.306(20.0)=10.9 s,GA=gA+2=12.9 sgB=0.1390.306(20.0)=9.1 s,GB=gB+2=11.1 s\begin{aligned} g_A &= \frac{0.167}{0.306}(20.0) = 10.9\ \text{s}, \quad G_A = g_A + 2 = 12.9\ \text{s}\\ g_B &= \frac{0.139}{0.306}(20.0) = 9.1\ \text{s}, \quad G_B = g_B + 2 = 11.1\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road A)12.915.015
2 (road B)11.120.020

Adopted cycle length C=15+4+20+3=42C = 15 + 4 + 20 + 3 = 42 s.

Final timings

SignalPhase 1: road APhase 2: road B
Vehicle green15 s20 s
Vehicle amber4 s3 s
Vehicle red23 s19 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 15 s green); pedestrians crossing road A get WALK for 5 s then flashing for 15 s during Phase 2 (total 20 s of 20 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 42 s
Road A: [GREEN 15][A 4][  RED 23  ]
Road B: [ RED 19 ][GREEN 20][A 3]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 27]
Ped crossing A: [DON'T WALK 19][WALK+FLASH 20][DW]

✓ Answer checked by Claude Opus 5.5 (high effort)

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