---
title: "Signal timing: roads A 15 m and B 12 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-pedestrian-road-a-15m-b"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-pedestrian-road-a-15m-b
site: BCT Engineering
---

# Signal timing: roads A 15 m and B 12 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2071 Chaitra, 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-pedestrian-road-a-15m-b
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

An isolated signal with pedestrians indication is to be installed on a right angled intersection with road A 15 m wide and road B 12 m wide. The heaviest volume per hour for each lane of road A and road B are 300 and 250 respectively. The amber times for roads A and B are 3 and 2 seconds respectively. Design the timings of traffic and pedestrian signal.

## Short answer

Pedestrian crossing times govern: minimum green for road A (crossing road B) is 15 s, for road B (crossing road A) 18 s. Adopted cycle 38 s: road A green 15 s, amber 3 s; road B green 18 s, amber 2 s. Pedestrian signals show 5 s WALK then flashing clearance for the respective crossing duration.

## Answer

Isolated two-phase signal with pedestrian indications for road A (15 m) and road B (12 m). Approach speeds are not needed because the ambers are given (3 s and 2 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, pedestrian walking speed 1.2 m/s and initial walk (pedestrian start-up) time 5 s.

#### Step 1: Amber (change) time

Amber times are given: road A = 3 s, road B = 2 s.

#### Step 2: Pedestrian crossing time

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

$$
\begin{aligned} t_{p,A} &= 5 + \frac{15}{1.2} = 17.5\ \text{s}\\ t_{p,B} &= 5 + \frac{12}{1.2} = 15.0\ \text{s} \end{aligned}
$$

Pedestrians crossing road A walk during Phase 2, so the Phase 2 green must be at least 17.5 s. Pedestrians crossing road B walk during Phase 1, so the Phase 1 green must be at least 15.0 s.

#### Step 3: Webster cycle time from traffic

$$
\begin{aligned} y_A &= \frac{300}{1800} = 0.167, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.306\\ L &= (2+3) + (2+2) = 9\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(9)+5}{1-0.306} = 26.6\ \text{s} \end{aligned}
$$

With $C_0 - L = 17.6$ s of effective green:

$$
\begin{aligned} g_A &= \frac{0.167}{0.306}(17.6) = 9.6\ \text{s}, \quad G_A = g_A + 2 = 11.6\ \text{s}\\ g_B &= \frac{0.139}{0.306}(17.6) = 8.0\ \text{s}, \quad G_B = g_B + 2 = 10.0\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road A) | 11.6 | 15.0 | 15 |
| 2 (road B) | 10.0 | 17.5 | 18 |

The pedestrian requirement governs. Adopted cycle length $C = 15 + 3 + 18 + 2 = 38$ s.

#### Final timings

| Signal | Phase 1: road A | Phase 2: road B |
|---|---|---|
| Vehicle green | 15 s | 18 s |
| Vehicle amber | 3 s | 2 s |
| Vehicle red | 20 s | 18 s |

Pedestrian signals: pedestrians crossing road B get WALK for 5 s, then flashing (clearance) for 10 s during Phase 1 (total 15 s of the 15 s green); pedestrians crossing road A get WALK for 5 s, then flashing for 13 s during Phase 2 (total 18 s of the 18 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

#### Phase diagram

```text
Cycle = 38 s
Road A: [GREEN 15][A 3][  RED 20  ]
Road B: [ RED 18 ][GREEN 18][A 2]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 23]
Ped crossing A: [DON'T WALK 18][WALK+FLASH 18][DW 2]
```

## Check yourself

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

1. With 5 s initial walk and 1.2 m/s walking speed, what is the crossing time for the 15 m road A?
   A. 12.5 s
   B. 17.5 s
   C. 15.0 s
   D. 20.0 s
   Answer: B. t = 5 + 15/1.2 = 17.5 s. (Key point: Pedestrian crossing time)

2. Pedestrians crossing road A walk during which phase, and so which green must be at least 17.5 s?
   A. Phase 1 (road A green)
   B. Phase 2 (road B green)
   C. Both phases equally
   D. The amber of road A
   Answer: B. Pedestrians cross a road while the cross road has green, so road A pedestrians use Phase 2. (Key point: Which phase serves which pedestrians)

3. With lost time L = 9 s and Y = 0.306, what is Webster's cycle C0 = (1.5L + 5)/(1 - Y)?
   A. 26.6 s
   B. 13.5 s
   C. 38.0 s
   D. 18.5 s
   Answer: A. C0 = (1.5x9 + 5)/(1 - 0.306) = 18.5/0.694 = 26.6 s. (Key point: Webster optimum cycle)

4. Traffic needs greens of 11.6 s and 10.0 s, but pedestrians need 15 s and 17.5 s. What cycle is adopted?
   A. 26.6 s
   B. 33 s
   C. 38 s
   D. 43 s
   Answer: C. Pedestrian minimums govern: 15 + 3 + 18 + 2 = 38 s. (Key point: Adopted cycle length)

## Close variants set in other exams

- [Signal timing: roads H 12 m and F 9.6 m](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-pedestrian-road-h-12m-f) — Transportation Engineering II, chapter 1, asked 3×

## Related in Transportation Engineering II

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