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Signal timing: roads P 14 m and Q 10.5 m

Short answer

Using Webster's method, amber 4 s (P) and 3 s (Q), pedestrian walking speed 1.2 m/s, and 1800 PCU/h saturation flow, the adopted cycle is 38 s. Road P vehicle green 14 s, amber 4 s; road Q green 17 s, amber 3 s. Pedestrian crossing Q gets 14 s, crossing P gets 17 s.

Design the timing of traffic and pedestrian signals of an isolated signal to be installed at a right angle intersection when road P and Q cross. The data available are:

Road PRoad Q
i) Width14.0010.50
ii) Peak hour traffic volume, Vehicle/hour/lane210120
iii) Approach speed, Kmph5035

Isolated two-phase signal with pedestrian indications for road P (14 m wide) and road Q (10.5 m wide).

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+50/3.62(3)=3.31→4 sIQ=1+35/3.62(3)=2.62→3 s\begin{aligned} I_P &= 1 + \frac{50/3.6}{2(3)} = 3.31 \to 4\ \text{s}\\ I_Q &= 1 + \frac{35/3.6}{2(3)} = 2.62 \to 3\ \text{s} \end{aligned}

Step 2: Pedestrian crossing time

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

tp,P=5+141.2=16.7 stp,Q=5+10.51.2=13.8 s\begin{aligned} t_{p,P} &= 5 + \frac{14}{1.2} = 16.7\ \text{s}\\ t_{p,Q} &= 5 + \frac{10.5}{1.2} = 13.8\ \text{s} \end{aligned}

Pedestrians crossing road P walk during Phase 2, so Phase 2 green must be at least 16.7 s. Pedestrians crossing road Q walk during Phase 1, so Phase 1 green must be at least 13.8 s.

Step 3: Webster cycle time from traffic

yP=2101800=0.117,yQ=1201800=0.067,Y=0.183L=(2+4)+(2+3)=11 sC0=1.5L+51−Y=1.5(11)+51−0.183=26.3 s\begin{aligned} y_P &= \frac{210}{1800} = 0.117, \quad y_Q = \frac{120}{1800} = 0.067, \quad Y = 0.183\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.183} = 26.3\ \text{s} \end{aligned} gP=0.1170.183(15.3)=9.8 s,GP=gP+2=11.8 sgQ=0.0670.183(15.3)=5.6 s,GQ=gQ+2=7.6 s\begin{aligned} g_P &= \frac{0.117}{0.183}(15.3) = 9.8\ \text{s}, \quad G_P = g_P + 2 = 11.8\ \text{s}\\ g_Q &= \frac{0.067}{0.183}(15.3) = 5.6\ \text{s}, \quad G_Q = g_Q + 2 = 7.6\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road P)11.813.814
2 (road Q)7.616.717

Adopted cycle length C=14+4+17+3=38C = 14 + 4 + 17 + 3 = 38 s.

Final timings

SignalPhase 1: road PPhase 2: road Q
Vehicle green14 s17 s
Vehicle amber4 s3 s
Vehicle red20 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 14 s green); pedestrians crossing road P get WALK for 5 s then flashing for 12 s during Phase 2 (total 17 s of 17 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 38 s
Road P: [GREEN 14][A 4][  RED 20  ]
Road Q: [ RED 18 ][GREEN 17][A 3]
Ped crossing Q: [WALK+FLASH 14][DON'T WALK 24]
Ped crossing P: [DON'T WALK 18][WALK+FLASH 17][DW]

✓ Answer checked by Claude Opus 5.5 (high effort)

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