---
title: "Signal timing: roads A 18 m and B 12 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-pedestrian-road-a-18m-b"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-pedestrian-road-a-18m-b
site: BCT Engineering
---

# Signal timing: roads A 18 m and B 12 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2081 Baisakh, 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-18m-b
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

An isolated signal with pedestrian indication is to be installed on a right angled intersection with road A, 18m wide and road B, 12m wide. The heaviest volume per hour for each lane of road A and B are 300 and 250, respectively. The approach speeds are 55 and 40 kmph for road A and B respectively. Design the timing of traffic and pedestrian signals.

## Short answer

Cycle time42 s; Road A vehicle green15 s + amber4 s, Road B20 s + amber3 s. Pedestrian crossing times (15 s for B,20 s for A) are accommodated during the opposing vehicle green.

## Answer

Isolated two-phase signal with pedestrian indications for road A (18 m) and road B (12 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 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 $I = t_r + \dfrac{v}{2a}$ 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:

$$
\begin{aligned} I_A &= 1 + \frac{55/3.6}{2(3)} = 3.55 \to 4\ \text{s}\\ I_B &= 1 + \frac{40/3.6}{2(3)} = 2.85 \to 3\ \text{s} \end{aligned}
$$

#### Step 2: Pedestrian crossing time

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

$$
\begin{aligned} t_{p,A} &= 5 + \frac{18}{1.2} = 20.0\ \text{s}\\ t_{p,B} &= 5 + \frac{12}{1.2} = 15.0\ \text{s} \end{aligned}
$$

Pedestrians crossing road A walk during Phase 2, so 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 15.0 s.

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

$$
\begin{aligned} y_A &= \frac{300}{1800} = 0.167, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.306\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.306} = 31.0\ \text{s} \end{aligned}
$$

$$
\begin{aligned} g_A &= \frac{0.167}{0.306}(20.0) = 10.9\ \text{s}, \quad G_A = g_A + 2 = 12.9\ \text{s}\\ g_B &= \frac{0.139}{0.306}(20.0) = 9.1\ \text{s}, \quad G_B = g_B + 2 = 11.1\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road A) | 12.9 | 15.0 | 15 |
| 2 (road B) | 11.1 | 20.0 | 20 |

Adopted cycle length $C = 15 + 4 + 20 + 3 = 42$ s.

#### Final timings

| Signal | Phase 1: road A | Phase 2: road B |
|---|---|---|
| Vehicle green | 15 s | 20 s |
| Vehicle amber | 4 s | 3 s |
| Vehicle red | 23 s | 19 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 15 s green); pedestrians crossing road A get WALK for 5 s then flashing for 15 s during Phase 2 (total 20 s of 20 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

#### Phase diagram

```text
Cycle = 42 s
Road A: [GREEN 15][A 4][  RED 23  ]
Road B: [ RED 19 ][GREEN 20][A 3]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 27]
Ped crossing A: [DON'T WALK 19][WALK+FLASH 20][DW]
```

## Check yourself

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

1. In the two-phase design, when do pedestrians crossing road A (18 m) get their WALK signal?
   A. During Phase 1, when road A traffic has green
   B. During Phase 2, when road B traffic has green
   C. During the amber of Phase 1
   D. During the all-red between phases
   Answer: B. Pedestrians cross a road while the cross road has green, so those crossing road A walk during Phase 2 (road B green). (Key point: Pedestrian phase assignment)

2. How is the minimum pedestrian green for crossing a road computed here?
   A. Width divided by walking speed only
   B. Initial walk time plus width divided by walking speed
   C. Initial walk time plus amber time
   D. Cycle length minus lost time
   Answer: B. Time to cross = initial walk (5 s) + width / walking speed (1.2 m/s), e.g. 5 + 18/1.2 = 20 s. (Key point: Pedestrian crossing time)

3. Amber time for road A at 55 km/h is found from I = t_r + v/(2a) with t_r = 1 s and a = 3 m/s^2. What result is adopted?
   A. 3 s
   B. 4 s
   C. 6 s
   D. 9 s
   Answer: B. I = 1 + (55/3.6)/(2*3) = 3.55 s, rounded up to 4 s within the IRC range of 2-4 s. (Key point: Amber time formula)

4. Webster's optimum cycle is C0 = (1.5L + 5)/(1 - Y). Here L = 11 s and Y = 0.306. What is C0 approximately?
   A. 21 s
   B. 31 s
   C. 42 s
   D. 65 s
   Answer: B. C0 = (1.5*11 + 5)/(1 - 0.306) = 21.5/0.694 = 31 s. (Key point: Webster cycle time)

5. Why is the adopted cycle 42 s rather than the Webster value of about 31 s?
   A. Pedestrian minimum greens (15 s and 20 s) exceed the traffic-based greens, so they govern
   B. Webster's formula always has to be doubled
   C. Amber times were added twice
   D. The saturation flow was too small
   Answer: A. Traffic needs only 12.9 s and 11.1 s of green, but pedestrians need 15 s and 20 s, so these are adopted: 15 + 4 + 20 + 3 = 42 s. (Key point: Governing green and final cycle)

## 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×

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