---
title: "Signal timing: roads A 14.4 m and B 12 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-pedestrian-road-a-14-4m"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-pedestrian-road-a-14-4m
site: BCT Engineering
---

# Signal timing: roads A 14.4 m and B 12 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2078 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-pedestrian-road-a-14-4m
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

At right angled crossing of road A and road B isolated signal with pedestrain indicators is to be installed. The road A is 14.4m wide and road B is 12m wide and the peak hour volumes are 280 and 230 PCU per hour and the approach speeds are 50 and 35 kmph respectively. Design the vehicular and pedestrian signal timing.

## Short answer

Use Webster method with amber times 4 s for road A and 3 s for road B. Pedestrian minimums are 15 s for road B and 17 s for road A, giving adopted greens 15 s and 17 s and a 39 s cycle.

## Answer

Isolated two-phase signal with pedestrian indicators for road A (14.4 m) and road B (12 m). Flows of 280 and 230 PCU/h are taken as the heaviest flow per lane.

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:

$$
\begin{aligned} I_A &= 1 + \frac{50/3.6}{2(3)} = 3.31 \to 4\ \text{s}\\ I_B &= 1 + \frac{35/3.6}{2(3)} = 2.62 \to 3\ \text{s} \end{aligned}
$$

#### Step 2: Pedestrian crossing time

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

$$
\begin{aligned} t_{p,A} &= 5 + \frac{14.4}{1.2} = 17.0\ \text{s}\\ t_{p,B} &= 5 + \frac{12}{1.2} = 15.0\ \text{s} \end{aligned}
$$

Pedestrians crossing road A walk during Phase 2, so Phase 2 green must be at least 17.0 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

$$
\begin{aligned} y_A &= \frac{280}{1800} = 0.156, \quad y_B = \frac{230}{1800} = 0.128, \quad Y = 0.283\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.283} = 30.0\ \text{s} \end{aligned}
$$

$$
\begin{aligned} g_A &= \frac{0.156}{0.283}(19.0) = 10.4\ \text{s}, \quad G_A = g_A + 2 = 12.4\ \text{s}\\ g_B &= \frac{0.128}{0.283}(19.0) = 8.6\ \text{s}, \quad G_B = g_B + 2 = 10.6\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road A) | 12.4 | 15.0 | 15 |
| 2 (road B) | 10.6 | 17.0 | 17 |

Adopted cycle length $C = 15 + 4 + 17 + 3 = 39$ s.

#### Final timings

| Signal | Phase 1: road A | Phase 2: road B |
|---|---|---|
| Vehicle green | 15 s | 17 s |
| Vehicle amber | 4 s | 3 s |
| Vehicle red | 20 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 12 s during Phase 2 (total 17 s of 17 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

#### Phase diagram

```text
Cycle = 39 s
Road A: [GREEN 15][A 4][  RED 20  ]
Road B: [ RED 19 ][GREEN 17][A 3]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 24]
Ped crossing A: [DON'T WALK 19][WALK+FLASH 17][DW]
```

## Check yourself

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

1. Pedestrians crossing road A (14.4 m) need at least:
   A. 12 s
   B. 14.4 s
   C. 17 s
   D. 20 s
   Answer: C. t = 5 + 14.4/1.2 = 17 s, provided during Phase 2. (Key point: Pedestrian minimum green)

2. Amber times adopted for road A (50 km/h) and road B (35 km/h) are:
   A. 3 s and 3 s
   B. 4 s and 3 s
   C. 4 s and 4 s
   D. 5 s and 2 s
   Answer: B. I_A = 3.31 rounded to 4 s; I_B = 2.62 rounded to 3 s. (Key point: Amber times)

3. How is y calculated for road A?
   A. Flow divided by saturation flow, 280/1800 = 0.156
   B. Saturation flow divided by flow
   C. Flow times 1800
   D. Flow minus 1800
   Answer: A. y = q/s, taking the heaviest flow per lane and s = 1800 PCU/h. (Key point: Flow ratio)

4. With L = 11 s and Y = 0.283, C0 is about:
   A. 30 s
   B. 39 s
   C. 15 s
   D. 60 s
   Answer: A. C0 = (1.5 x 11 + 5)/(1 - 0.283) = 21.5/0.717 = 30 s. (Key point: Webster cycle)

5. Why is the final cycle 39 s?
   A. Adopted greens 15 s and 17 s are set by pedestrian needs, plus ambers 4 and 3 s
   B. Webster value is rounded up to 39
   C. Reds are added to the cycle
   D. Greens are 12.4 and 10.6 s
   Answer: A. Pedestrian minimums (15 s, 17 s) exceed traffic greens, so C = 15 + 4 + 17 + 3 = 39 s. (Key point: Final timings)

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