---
title: "Signal timing: roads A 15 m and B 7 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-pedestrian-road-a-15m-7m"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-pedestrian-road-a-15m-7m
site: BCT Engineering
---

# Signal timing: roads A 15 m and B 7 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2070 Chaitra (old course), 10 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-7m
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

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)

## 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.

## Answer

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

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

#### Step 2: Pedestrian crossing time

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

$$
\begin{aligned} t_{p,A} &= 5 + \frac{15}{1.0} = 20.0\ \text{s}\\ t_{p,B} &= 5 + \frac{7}{1.0} = 12.0\ \text{s} \end{aligned}
$$

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

$$
\begin{aligned} y_A &= \frac{350}{1800} = 0.194, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.333\\ L &= (2+4) + (2+4) = 12\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(12)+5}{1-0.333} = 34.5\ \text{s} \end{aligned}
$$

$$
\begin{aligned} g_A &= \frac{0.194}{0.333}(22.5) = 13.1\ \text{s}, \quad G_A = g_A + 2 = 15.1\ \text{s}\\ g_B &= \frac{0.139}{0.333}(22.5) = 9.4\ \text{s}, \quad G_B = g_B + 2 = 11.4\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road A) | 15.1 | 12.0 | 16 |
| 2 (road B) | 11.4 | 20.0 | 20 |

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

#### Final timings

| Signal | Phase 1: road A | Phase 2: road B |
|---|---|---|
| Vehicle green | 16 s | 20 s |
| Vehicle amber | 4 s | 4 s |
| Vehicle red | 24 s | 20 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

```text
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]
```

## Check yourself

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

1. Using I = t_r + v/(2a) with t_r = 1 s, a = 3 m/s^2, what amber is adopted for road B at 45 km/h?
   A. 3 s
   B. 4 s (3.08 s rounded up)
   C. 2 s
   D. 5 s
   Answer: B. I = 1 + 12.5/6 = 3.08 s, which is rounded up to 4 s within the 2-4 s range. (Key point: Amber time)

2. With walking speed 1.0 m/s and 5 s initial walk, how long is needed to cross road A (15 m)?
   A. 15 s
   B. 12 s
   C. 18 s
   D. 20 s
   Answer: D. t = 5 + 15/1.0 = 20 s, so Phase 2 green must be at least 20 s. (Key point: Pedestrian crossing time)

3. With L = 12 s and Y = 0.333, what is C0?
   A. 34.5 s
   B. 18.0 s
   C. 44.0 s
   D. 22.5 s
   Answer: A. C0 = (1.5x12 + 5)/(1 - 0.333) = 23/0.667 = 34.5 s. (Key point: Webster cycle)

4. Why is Phase 2 green 20 s (not 11.4 s) and what is the cycle?
   A. Pedestrian need of 20 s governs; cycle = 16+4+20+4 = 44 s
   B. Traffic need governs; cycle = 34.5 s
   C. Amber governs; cycle = 36 s
   D. Pedestrian need governs; cycle = 40 s
   Answer: A. The pedestrian minimum is larger than the traffic green, so it is adopted, giving a 44 s cycle. (Key point: Adopted greens and cycle)

## Related in Transportation Engineering II

- [Signal timing: roads A 15 m and B 12 m](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-pedestrian-road-a-15m-b) — Transportation Engineering II, chapter 1, asked 2×
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