---
title: "Signal timing: roads P 14 m and Q 10.5 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-pedestrian-road-p-14m-q"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-pedestrian-road-p-14m-q
site: BCT Engineering
---

# Signal timing: roads P 14 m and Q 10.5 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2068 Baisakh, 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-p-14m-q
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Design the timing of traffic and pedestrian signals of an isolated signal to be installed at a right angle intersection when road P and Q cross. The data available are:

| | Road P | Road Q |
|---|---|---|
| i) Width | 14.00 | 10.50 |
| ii) Peak hour traffic volume, Vehicle/hour/lane | 210 | 120 |
| iii) Approach speed, Kmph | 50 | 35 |

## Short answer

Using Webster's method, amber 4 s (P) and 3 s (Q), pedestrian walking speed 1.2 m/s, and 1800 PCU/h saturation flow, the adopted cycle is 38 s. Road P vehicle green 14 s, amber 4 s; road Q green 17 s, amber 3 s. Pedestrian crossing Q gets 14 s, crossing P gets 17 s.

## Answer

Isolated two-phase signal with pedestrian indications for road P (14 m wide) and road Q (10.5 m wide).

Design basis (IRC:93-1985, *Guidelines on Design and Installation of Road Traffic Signals*, with Webster's method): two-phase signal, Phase 1 = road P moves, Phase 2 = road Q 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:

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

#### Step 2: Pedestrian crossing time

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

$$
\begin{aligned} t_{p,P} &= 5 + \frac{14}{1.2} = 16.7\ \text{s}\\ t_{p,Q} &= 5 + \frac{10.5}{1.2} = 13.8\ \text{s} \end{aligned}
$$

Pedestrians crossing road P walk during Phase 2, so Phase 2 green must be at least 16.7 s. Pedestrians crossing road Q walk during Phase 1, so Phase 1 green must be at least 13.8 s.

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

$$
\begin{aligned} y_P &= \frac{210}{1800} = 0.117, \quad y_Q = \frac{120}{1800} = 0.067, \quad Y = 0.183\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.183} = 26.3\ \text{s} \end{aligned}
$$

$$
\begin{aligned} g_P &= \frac{0.117}{0.183}(15.3) = 9.8\ \text{s}, \quad G_P = g_P + 2 = 11.8\ \text{s}\\ g_Q &= \frac{0.067}{0.183}(15.3) = 5.6\ \text{s}, \quad G_Q = g_Q + 2 = 7.6\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road P) | 11.8 | 13.8 | 14 |
| 2 (road Q) | 7.6 | 16.7 | 17 |

Adopted cycle length $C = 14 + 4 + 17 + 3 = 38$ s.

#### Final timings

| Signal | Phase 1: road P | Phase 2: road Q |
|---|---|---|
| Vehicle green | 14 s | 17 s |
| Vehicle amber | 4 s | 3 s |
| Vehicle red | 20 s | 18 s |

Pedestrian signals: pedestrians crossing road Q get WALK for 5 s then flashing (clearance) for 9 s during Phase 1 (total 14 s of 14 s green); pedestrians crossing road P 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 = 38 s
Road P: [GREEN 14][A 4][  RED 20  ]
Road Q: [ RED 18 ][GREEN 17][A 3]
Ped crossing Q: [WALK+FLASH 14][DON'T WALK 24]
Ped crossing P: [DON'T WALK 18][WALK+FLASH 17][DW]
```

## Check yourself

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

1. What amber is adopted for road Q (35 km/h)?
   A. 2 s
   B. 2.62 s kept as is
   C. 3 s (2.62 s rounded up)
   D. 4 s
   Answer: C. I = 1 + 9.72/6 = 2.62 s, rounded up to 3 s. (Key point: Amber time)

2. Pedestrians crossing road P need 5 + 14/1.2 = 16.7 s. In which phase must this fit?
   A. Phase 1 (road P green)
   B. Phase 2 (road Q green)
   C. Road P amber
   D. Either phase
   Answer: B. They cross while road Q has green, so Phase 2 must be at least 16.7 s. (Key point: Pedestrian minimum for road P)

3. Why is Phase 2 green adopted as 17 s when traffic needs only 7.6 s?
   A. Webster's formula gives 17 s
   B. Pedestrian crossing time governs
   C. Amber sets it
   D. Road Q has more traffic
   Answer: B. The pedestrian minimum (16.7 s) is larger than the traffic green, so it is adopted. (Key point: Governing criterion)

4. What is the adopted cycle with greens 14 s and 17 s and ambers 4 s and 3 s?
   A. 26.3 s
   B. 34 s
   C. 38 s
   D. 41 s
   Answer: C. C = 14 + 4 + 17 + 3 = 38 s. (Key point: Cycle length)

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