---
title: "Signal timing: roads P 13.5 m and Q 10.5 m"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:signal-pedestrian-road-p-13-5m"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-pedestrian-road-p-13-5m
site: BCT Engineering
---

# Signal timing: roads P 13.5 m and Q 10.5 m

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2080 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-p-13-5m
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Road P is 13.5 m wide and road Q is 10.5 m. An isolated signal with pedestrian indications are to be installed at right angled intersection. The peak volumes per hour for road P and road Q are 250 and 200 respectively. The approaching speeds for road P and Q are 60 kmph and 45 kmph. Design the traffic and pedestrian signal timings.

## Short answer

Amber is 4 s for both phases. Pedestrian minimum greens are 16.25 s for road P and 13.75 s for road Q. Webster traffic cycle is 30.7 s, but adopted cycle is 39 s with Phase 1 green 14 s, Phase 2 green 17 s, and 4 s amber each.

## Answer

Isolated two-phase signal with pedestrian indications for road P (13.5 m) and road Q (10.5 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 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:

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

#### Step 2: Pedestrian crossing time

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

$$
\begin{aligned} t_{p,P} &= 5 + \frac{13.5}{1.2} = 16.25\ \text{s}\\ t_{p,Q} &= 5 + \frac{10.5}{1.2} = 13.75\ \text{s} \end{aligned}
$$

Pedestrians crossing road P walk during Phase 2, so Phase 2 green must be at least 16.25 s. Pedestrians crossing road Q walk during Phase 1, so Phase 1 green must be at least 13.75 s.

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

$$
\begin{aligned} y_P &= \frac{250}{1800} = 0.139, \quad y_Q = \frac{200}{1800} = 0.111, \quad Y = 0.250\\ L &= (2+4) + (2+4) = 12\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(12)+5}{1-0.250} = 30.7\ \text{s} \end{aligned}
$$

$$
\begin{aligned} g_P &= \frac{0.139}{0.250}(18.7) = 10.4\ \text{s}, \quad G_P = g_P + 2 = 12.4\ \text{s}\\ g_Q &= \frac{0.111}{0.250}(18.7) = 8.3\ \text{s}, \quad G_Q = g_Q + 2 = 10.3\ \text{s} \end{aligned}
$$

#### Step 4: Check against pedestrian needs

| Phase | Green from traffic (s) | Pedestrian minimum (s) | Adopted green (s) |
|---|---|---|---|
| 1 (road P) | 12.4 | 13.75 | 14 |
| 2 (road Q) | 10.3 | 16.25 | 17 |

Adopted cycle length $C = 14 + 4 + 17 + 4 = 39$ s.

#### Final timings

| Signal | Phase 1: road P | Phase 2: road Q |
|---|---|---|
| Vehicle green | 14 s | 17 s |
| Vehicle amber | 4 s | 4 s |
| Vehicle red | 21 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 the 14 s green); pedestrians crossing road P get WALK for 5 s, then flashing for 12 s, during Phase 2 (total 17 s of the 17 s green, which covers the 16.25 s needed). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

#### Phase diagram

```text
Cycle = 39 s
Road P: [GREEN 14][A 4][  RED 21  ]
Road Q: [ RED 18 ][GREEN 17][A 4]
Ped crossing Q: [WALK+FLASH 14][DON'T WALK 25]
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. For road P at 60 km/h, I = 1 + (60/3.6)/(2 x 3) gives 3.78 s. What is adopted?
   A. 3 s
   B. 4 s
   C. 5 s
   D. 6 s
   Answer: B. The value is rounded up to 4 s, within the IRC:93 range of 2-4 s. (Key point: Amber time)

2. Pedestrians crossing road P (13.5 m) need a green of at least:
   A. 13.5 s
   B. 16.2 s
   C. 11.25 s
   D. 18.7 s
   Answer: B. t = 5 + 13.5/1.2 = 16.2 s, which falls in Phase 2 when road Q has green. (Key point: Pedestrian minimum green)

3. With Y = 0.250 and L = 12 s, Webster's cycle is about:
   A. 30.7 s
   B. 48 s
   C. 15 s
   D. 60 s
   Answer: A. C0 = (1.5 x 12 + 5)/(1 - 0.25) = 23/0.75 = 30.7 s. (Key point: Webster cycle and greens)

4. Why are the adopted greens 14 s and 17 s?
   A. Traffic greens (12.4 s and 10.3 s) are less than the pedestrian minimums (13.8 s and 16.2 s), so pedestrian needs govern
   B. Traffic needs the larger greens
   C. They are copied from road A
   D. Amber time is added to greens
   Answer: A. The larger of traffic-based and pedestrian-based green is adopted, then rounded up. (Key point: Pedestrian check governs)

5. What is the adopted cycle length?
   A. 14 + 17 = 31 s
   B. 14 + 4 + 17 + 4 = 39 s
   C. 30.7 s
   D. 12 + 4 + 10 + 4 = 30 s
   Answer: B. Cycle = green + amber for both phases: 14 + 4 + 17 + 4 = 39 s. (Key point: Final cycle)

## Related in Transportation Engineering II

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- [Signal timing: roads P 14 m and Q 10.5 m](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Asignal-pedestrian-road-p-14m-q) — Transportation Engineering II, chapter 1

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