---
title: "Signal design: roads X 15 m and Y 10.5 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-design-roads-x-15m-y"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-design-roads-x-15m-y
site: BCT Engineering
---

# Signal design: roads X 15 m and Y 10.5 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2082 Bhadra, 8 marks
- Video: [Design of Two Phase Traffic Signal Using Webster's Traffic Signal Design Method| Solved Problem](https://www.youtube.com/watch?v=HyIFzbsYXe0) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-design-roads-x-15m-y
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

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.

## 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.

## Answer

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 $I = t_r + \dfrac{v}{2a}$ with reaction time $t_r = 1$ s and comfortable deceleration $a = 3\ \text{m/s}^2$, kept within the IRC:93 range of 2-4 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:

$$
\begin{aligned} t_{p,X} &= 5 + \frac{15}{1.2} = 17.5\ \text{s}\\ t_{p,Y} &= 5 + \frac{10.5}{1.2} = 13.8\ \text{s} \end{aligned}
$$

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

$$
\begin{aligned} y_X &= \frac{250}{1800} = 0.139, \quad y_Y = \frac{200}{1800} = 0.111, \quad Y = 0.250\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.250} = 28.7\ \text{s} \end{aligned}
$$

$$
\begin{aligned} g_X &= \frac{0.139}{0.250}(17.7) = 9.8\ \text{s}, \quad G_X = g_X + 2 = 11.8\ \text{s}\\ g_Y &= \frac{0.111}{0.250}(17.7) = 7.9\ \text{s}, \quad G_Y = g_Y + 2 = 9.9\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road X) | 11.8 | 13.8 | 14 |
| 2 (road Y) | 9.9 | 17.5 | 18 |

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

#### Final timings

| Signal | Phase 1: road X | Phase 2: road Y |
|---|---|---|
| Vehicle green | 14 s | 18 s |
| Vehicle amber | 4 s | 3 s |
| Vehicle red | 21 s | 18 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

```text
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]
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the amber (change) time computed in this design?
   A. I = t_r + v/(2a) with t_r = 1 s and a = 3 m/s^2
   B. I = v/a only
   C. I = width/speed
   D. I = 2 x cycle time
   Answer: A. Amber = reaction time + v/(2a), kept within the IRC:93 range of 2-4 s; here 4 s for X and 3 s for Y. (Key point: Amber time formula)

2. How is the minimum pedestrian green computed?
   A. Width / walking speed only
   B. Initial walk time + width / walking speed
   C. Cycle time / 2
   D. Amber + red
   Answer: B. t = 5 + width/1.2, giving 17.5 s for the 15 m road and 13.8 s for the 10.5 m road. (Key point: Pedestrian crossing time)

3. When do pedestrians crossing road X walk, and what does this imply?
   A. During Phase 1, so Phase 1 green is at least 17.5 s
   B. During Phase 2 (when road Y has green), so Phase 2 green must be at least 17.5 s
   C. During amber of road X only
   D. Any time, with no effect on green
   Answer: B. Pedestrians cross while the cross road has green, so Phase 2 must last at least 17.5 s. (Key point: Phase of pedestrians crossing)

4. Traffic needs only 9.9 s green for Phase 2 but pedestrians need 17.5 s. What is adopted?
   A. 9.9 s, because traffic governs
   B. 18 s, because the pedestrian minimum governs
   C. 5 s
   D. 28.7 s
   Answer: B. The adopted green must satisfy pedestrians, so Phase 2 green is 18 s and cycle is 14+4+18+3 = 39 s. (Key point: Adopting pedestrian-controlled green)

## Related in Transportation Engineering II

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