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Signal timing: roads P 13.5 m and Q 10.5 m

Short answer

Amber is 4 s for both phases. Pedestrian minimum greens are 16.25 s for road P and 13.75 s for road Q. Webster traffic cycle is 30.7 s, but adopted cycle is 39 s with Phase 1 green 14 s, Phase 2 green 17 s, and 4 s amber each.

Road P is 13.5 m wide and road Q is 10.5 m. An isolated signal with pedestrian indications are to be installed at right angled intersection. The peak volumes per hour for road P and road Q are 250 and 200 respectively. The approaching speeds for road P and Q are 60 kmph and 45 kmph. Design the traffic and pedestrian signal timings.

Isolated two-phase signal with pedestrian indications for road P (13.5 m) and road Q (10.5 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 P moves, Phase 2 = road Q 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:

IP=1+60/3.62(3)=3.78→4 sIQ=1+45/3.62(3)=3.08→4 s\begin{aligned} I_P &= 1 + \frac{60/3.6}{2(3)} = 3.78 \to 4\ \text{s}\\ I_Q &= 1 + \frac{45/3.6}{2(3)} = 3.08 \to 4\ \text{s} \end{aligned}

Step 2: Pedestrian crossing time

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

tp,P=5+13.51.2=16.25 stp,Q=5+10.51.2=13.75 s\begin{aligned} t_{p,P} &= 5 + \frac{13.5}{1.2} = 16.25\ \text{s}\\ t_{p,Q} &= 5 + \frac{10.5}{1.2} = 13.75\ \text{s} \end{aligned}

Pedestrians crossing road P walk during Phase 2, so Phase 2 green must be at least 16.25 s. Pedestrians crossing road Q walk during Phase 1, so Phase 1 green must be at least 13.75 s.

Step 3: Webster cycle time from traffic

yP=2501800=0.139,yQ=2001800=0.111,Y=0.250L=(2+4)+(2+4)=12 sC0=1.5L+51−Y=1.5(12)+51−0.250=30.7 s\begin{aligned} y_P &= \frac{250}{1800} = 0.139, \quad y_Q = \frac{200}{1800} = 0.111, \quad Y = 0.250\\ L &= (2+4) + (2+4) = 12\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(12)+5}{1-0.250} = 30.7\ \text{s} \end{aligned} gP=0.1390.250(18.7)=10.4 s,GP=gP+2=12.4 sgQ=0.1110.250(18.7)=8.3 s,GQ=gQ+2=10.3 s\begin{aligned} g_P &= \frac{0.139}{0.250}(18.7) = 10.4\ \text{s}, \quad G_P = g_P + 2 = 12.4\ \text{s}\\ g_Q &= \frac{0.111}{0.250}(18.7) = 8.3\ \text{s}, \quad G_Q = g_Q + 2 = 10.3\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road P)12.413.7514
2 (road Q)10.316.2517

Adopted cycle length C=14+4+17+4=39C = 14 + 4 + 17 + 4 = 39 s.

Final timings

SignalPhase 1: road PPhase 2: road Q
Vehicle green14 s17 s
Vehicle amber4 s4 s
Vehicle red21 s18 s

Pedestrian signals: pedestrians crossing road Q get WALK for 5 s, then flashing (clearance) for 9 s, during Phase 1 (total 14 s of the 14 s green); pedestrians crossing road P get WALK for 5 s, then flashing for 12 s, during Phase 2 (total 17 s of the 17 s green, which covers the 16.25 s needed). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 39 s
Road P: [GREEN 14][A 4][  RED 21  ]
Road Q: [ RED 18 ][GREEN 17][A 4]
Ped crossing Q: [WALK+FLASH 14][DON'T WALK 25]
Ped crossing P: [DON'T WALK 18][WALK+FLASH 17][DW]

✓ Answer checked by Claude Opus 5.5 (high effort)

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