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Signal timing: roads H 12 m and F 9.6 m

Short answer

Using Webster’s method with given ambers, pedestrian crossing times (H: 15 s, F: 13 s), and saturation flow 1800 PCU/h/lane, cycle length 45 s. Road H: green 21 s, amber 5 s; Road F: green 15 s, amber 4 s. Pedestrian WALK 5 s plus clearance for each phase.

An isolated signal with pedestrians indication is to be installed on a right angled intersection with road H of 12 m wide and road F of 9.6 m wide. The heaviest volume per hour for each lane of H and F are 475 and 325 respectively. The approach speeds are 60 and 45 kmph for road H and road F respectively. Design the timings of traffic and pedestrian signals. Assume amber times for road H and road F as 5 and 4 secs respectively.

Isolated two-phase signal with pedestrian indications for road H (12 m) and road F (9.6 m). The ambers are given (5 s and 4 s), so the approach speeds are not needed for them.

Design basis (IRC:93-1985, Guidelines on Design and Installation of Road Traffic Signals, with Webster's method): two-phase signal, Phase 1 = road H moves, Phase 2 = road F 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 H = 5 s, road F = 4 s.

Step 2: Pedestrian crossing time

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

tp,H=5+121.2=15.0 stp,F=5+9.61.2=13.0 s\begin{aligned} t_{p,H} &= 5 + \frac{12}{1.2} = 15.0\ \text{s}\\ t_{p,F} &= 5 + \frac{9.6}{1.2} = 13.0\ \text{s} \end{aligned}

Pedestrians crossing road H walk during Phase 2, so Phase 2 green must be at least 15.0 s. Pedestrians crossing road F walk during Phase 1, so Phase 1 green must be at least 13.0 s.

Step 3: Webster cycle time from traffic

yH=4751800=0.264,yF=3251800=0.181,Y=0.444L=(2+5)+(2+4)=13 sC0=1.5L+51−Y=1.5(13)+51−0.444=44.1 s\begin{aligned} y_H &= \frac{475}{1800} = 0.264, \quad y_F = \frac{325}{1800} = 0.181, \quad Y = 0.444\\ L &= (2+5) + (2+4) = 13\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(13)+5}{1-0.444} = 44.1\ \text{s} \end{aligned} gH=0.2640.444(31.1)=18.5 s,GH=gH+2=20.5 sgF=0.1810.444(31.1)=12.6 s,GF=gF+2=14.6 s\begin{aligned} g_H &= \frac{0.264}{0.444}(31.1) = 18.5\ \text{s}, \quad G_H = g_H + 2 = 20.5\ \text{s}\\ g_F &= \frac{0.181}{0.444}(31.1) = 12.6\ \text{s}, \quad G_F = g_F + 2 = 14.6\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road H)20.513.021
2 (road F)14.615.015

Adopted cycle length C=21+5+15+4=45C = 21 + 5 + 15 + 4 = 45 s.

Final timings

SignalPhase 1: road HPhase 2: road F
Vehicle green21 s15 s
Vehicle amber5 s4 s
Vehicle red19 s26 s

Pedestrian signals: pedestrians crossing road F get WALK for 5 s then flashing (clearance) for 8 s during Phase 1 (total 13 s of 21 s green); pedestrians crossing road H get WALK for 5 s then flashing for 10 s during Phase 2 (total 15 s of 15 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 45 s
Road H: [GREEN 21][A 5][  RED 19  ]
Road F: [ RED 26 ][GREEN 15][A 4]
Ped crossing F: [WALK+FLASH 21][DON'T WALK 24]
Ped crossing H: [DON'T WALK 26][WALK+FLASH 15][DW]

✓ Answer checked by Claude Opus 5.5 (high effort)

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