Skip to main content

Signal timing: roads A 15 m and B 7 m

Short answer

Using Webster's method with amber 4 s and pedestrian walking speed 1.0 m/s, the adopted cycle is 44 s. Road A vehicle green 16 s, amber 4 s; road B green 20 s, amber 4 s. Pedestrian crossing road B gets 16 s, road A gets 20 s.

An isolated signal with pedestrian indicators to be installed on a right angled intersection with road A of 15 m and road B of 7 m respectively. The heaviest volume per hour for each lane of road A and road B are 350 and 250. The approaching speeds for road A and road B are 60 kmph and 45 kmph. Design the traffic and pedestrian signal timings. (Take amber period for yellow light signal as 2 to 4 secs and pedestrian walking speed of 1.0 m/sec)

Isolated two-phase signal with pedestrian indicators for road A (15 m) and road B (7 m); the pedestrian walking speed is 1.0 m/s and the amber is taken within 2-4 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 and initial walk (pedestrian start-up) time 5 s; the walking speed of 1.0 m/s is given.

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 given range of 2-4 s:

IA=1+60/3.62(3)=3.78→4 sIB=1+45/3.62(3)=3.08→4 s (rounded up)\begin{aligned} I_A &= 1 + \frac{60/3.6}{2(3)} = 3.78 \to 4\ \text{s}\\ I_B &= 1 + \frac{45/3.6}{2(3)} = 3.08 \to 4\ \text{s (rounded up)} \end{aligned}

Step 2: Pedestrian crossing time

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

tp,A=5+151.0=20.0 stp,B=5+71.0=12.0 s\begin{aligned} t_{p,A} &= 5 + \frac{15}{1.0} = 20.0\ \text{s}\\ t_{p,B} &= 5 + \frac{7}{1.0} = 12.0\ \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 12.0 s.

Step 3: Webster cycle time from traffic

yA=3501800=0.194,yB=2501800=0.139,Y=0.333L=(2+4)+(2+4)=12 sC0=1.5L+51−Y=1.5(12)+51−0.333=34.5 s\begin{aligned} y_A &= \frac{350}{1800} = 0.194, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.333\\ L &= (2+4) + (2+4) = 12\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(12)+5}{1-0.333} = 34.5\ \text{s} \end{aligned} gA=0.1940.333(22.5)=13.1 s,GA=gA+2=15.1 sgB=0.1390.333(22.5)=9.4 s,GB=gB+2=11.4 s\begin{aligned} g_A &= \frac{0.194}{0.333}(22.5) = 13.1\ \text{s}, \quad G_A = g_A + 2 = 15.1\ \text{s}\\ g_B &= \frac{0.139}{0.333}(22.5) = 9.4\ \text{s}, \quad G_B = g_B + 2 = 11.4\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road A)15.112.016
2 (road B)11.420.020

Adopted cycle length C=16+4+20+4=44C = 16 + 4 + 20 + 4 = 44 s.

Final timings

SignalPhase 1: road APhase 2: road B
Vehicle green16 s20 s
Vehicle amber4 s4 s
Vehicle red24 s20 s

Pedestrian signals: pedestrians crossing road B get WALK for 5 s then flashing (clearance) for 7 s during Phase 1 (total 12 s of the 16 s green); pedestrians crossing road A get WALK for 5 s then flashing for 15 s during Phase 2 (total 20 s of the 20 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 44 s
Road A: [GREEN 16][A 4][  RED 24  ]
Road B: [ RED 20 ][GREEN 20][A 4]
Ped crossing B: [WALK+FLASH 16][DON'T WALK 28]
Ped crossing A: [DON'T WALK 20][WALK+FLASH 20][DW]

✓ Answer checked by Claude Opus 5.5 (high effort)

All Transportation Engineering II chapter 1 questions →