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Signal design: roads X 15 m and Y 10.5 m

Short answer

Use Webster two-phase with amber 4 s for X and 3 s for Y. Pedestrian crossing needs 17.5 s for X and 13.8 s for Y, so adopt greens 14 s and 18 s; cycle length 39 s. Phase diagram shows green/amber/red and pedestrian walk/flash intervals.

Design an isolated traffic signal at an intersection of road X (15 m width), and road Y (10.5 m width). The highest observed traffic volume per lane is 250 vehicles per hour on road X and 200 vehicles per hour on road Y. The approach speeds are 50 km/h for road X and 35 km/h for road Y. Design the traffic and pedestrian signal timings for this intersection. Also, prepare a phase diagram showing both vehicular and pedestrian signal timing.

Design the vehicular and pedestrian timings of an isolated two-phase signal for roads X (15 m) and Y (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 X moves, Phase 2 = road Y 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:

IX=1+50/3.62(3)=3.31→4 sIY=1+35/3.62(3)=2.62→3 s\begin{aligned} I_X &= 1 + \frac{50/3.6}{2(3)} = 3.31 \to 4\ \text{s}\\ I_Y &= 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,X=5+151.2=17.5 stp,Y=5+10.51.2=13.8 s\begin{aligned} t_{p,X} &= 5 + \frac{15}{1.2} = 17.5\ \text{s}\\ t_{p,Y} &= 5 + \frac{10.5}{1.2} = 13.8\ \text{s} \end{aligned}

Pedestrians crossing road X walk during Phase 2, so Phase 2 green must be at least 17.5 s. Pedestrians crossing road Y walk during Phase 1, so Phase 1 green must be at least 13.8 s.

Step 3: Webster cycle time from traffic

yX=2501800=0.139,yY=2001800=0.111,Y=0.250L=(2+4)+(2+3)=11 sC0=1.5L+51−Y=1.5(11)+51−0.250=28.7 s\begin{aligned} y_X &= \frac{250}{1800} = 0.139, \quad y_Y = \frac{200}{1800} = 0.111, \quad Y = 0.250\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.250} = 28.7\ \text{s} \end{aligned} gX=0.1390.250(17.7)=9.8 s,GX=gX+2=11.8 sgY=0.1110.250(17.7)=7.9 s,GY=gY+2=9.9 s\begin{aligned} g_X &= \frac{0.139}{0.250}(17.7) = 9.8\ \text{s}, \quad G_X = g_X + 2 = 11.8\ \text{s}\\ g_Y &= \frac{0.111}{0.250}(17.7) = 7.9\ \text{s}, \quad G_Y = g_Y + 2 = 9.9\ \text{s} \end{aligned}

Step 4: Check against pedestrian needs

PhaseGreen from traffic (s)Pedestrian minimum (s)Adopted green (s)
1 (road X)11.813.814
2 (road Y)9.917.518

Adopted cycle length C=14+4+18+3=39C = 14 + 4 + 18 + 3 = 39 s.

Final timings

SignalPhase 1: road XPhase 2: road Y
Vehicle green14 s18 s
Vehicle amber4 s3 s
Vehicle red21 s18 s

Pedestrian signals: pedestrians crossing road Y 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 X get WALK for 5 s, then flashing for 13 s, during Phase 2 (total 18 s of the 18 s green, which covers the 17.5 s needed). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 39 s
Road X: [GREEN 14][A 4][  RED 21  ]
Road Y: [ RED 18 ][GREEN 18][A 3]
Ped crossing Y: [WALK+FLASH 14][DON'T WALK 25]
Ped crossing X: [DON'T WALK 18][WALK+FLASH 18][DW]

✓ Answer checked by Claude Opus 5.5 (high effort)

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