---
title: "Rotary capacity with weaving section numerical"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:rotary-capacity-weaving-numerical"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/rotary-capacity-weaving-numerical
site: BCT Engineering
---

# Rotary capacity with weaving section numerical

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2076 Chaitra, 4 marks
- Video: [Weaving ratio in rotary intersection | traffic engineering | Numerical Example](https://www.youtube.com/watch?v=bJRnRAlklpE) — Civil Engineering & Stuff
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/rotary-capacity-weaving-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

Calculate the capacity of rotary with entry and exit width of 8m, the width of non-weaving section is 9m, width of rotary is 12m, length of weaving section is 60m. The ratio of weaving to total traffic in the weaving section is 0.7.

## Short answer

Use the IRC rotary capacity formula with weaving width 12 m, average entry width 8.5 m, weaving length 60 m, and weaving proportion 0.7. The capacity is 3667 PCU/h.

## Answer

Capacity of the weaving section of a rotary (IRC:65, based on the TRL/Wardrop formula):

$$
Q_p = \frac{280\,w\left(1+\dfrac{e}{w}\right)\left(1-\dfrac{p}{3}\right)}{1+\dfrac{w}{l}}\ \text{PCU/h}
$$

**Data:** weaving width $w = 12$ m (width of the rotary), entry width $e_1 = 8$ m, non-weaving width $e_2 = 9$ m, so $e = (e_1+e_2)/2 = 8.5$ m; $l = 60$ m; $p = 0.7$.

**Check of limits:** $e/w = 0.708$ (0.4 to 1.0 OK), $w/l = 0.20$ (0.12 to 0.4 OK), $p = 0.7$ (0.4 to 1.0 OK).

$$
\begin{aligned} Q_p &= \frac{280(12)\left(1+\dfrac{8.5}{12}\right)\left(1-\dfrac{0.7}{3}\right)}{1+\dfrac{12}{60}}\\ &= \frac{3360(1.7083)(0.7667)}{1.2} = 3667\ \text{PCU/h} \end{aligned}
$$

**Answer: capacity of the rotary = 3667 PCU/h.**

## Check yourself

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

1. In the problem, which value is used as the weaving width w?
   A. 8 m
   B. 9 m
   C. 12 m (width of the rotary)
   D. 60 m
   Answer: C. The rotary width of 12 m is the weaving width w. (Key point: Weaving width)

2. How is e calculated here?
   A. (8 + 9)/2 = 8.5 m
   B. 8 + 9 = 17 m
   C. 9 - 8 = 1 m
   D. 12/2 = 6 m
   Answer: A. e = (e1 + e2)/2 with e1 = 8 m and e2 = 9 m gives 8.5 m. (Key point: Mean width e)

3. What is w/l and is it within the allowed range 0.12 to 0.4?
   A. 0.20, within range
   B. 5.0, out of range
   C. 0.71, out of range
   D. 0.08, out of range
   Answer: A. w/l = 12/60 = 0.20, which lies within 0.12-0.4. (Key point: Validity check)

4. What is the capacity of the weaving section?
   A. About 1273 PCU/h
   B. About 2400 PCU/h
   C. About 3667 PCU/h
   D. About 4500 PCU/h
   Answer: C. Qp = 280(12)(1.7083)(0.7667)/1.2 = about 3667 PCU/h. (Key point: Capacity result)

## Related in Transportation Engineering II

- [Traffic capacity types; rotary intersection sketch](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Atraffic-capacity-types-rotary-sketch) — Transportation Engineering II, chapter 1
- [Dowel bar length and spacing numerical](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Adowel-bar-design-expansion-joint-25mm) — Transportation Engineering II, chapter 2
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