--- title: "Design factors of rotary intersection" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:rotary-design-factors" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/rotary-design-factors site: BCT Engineering --- # Design factors of rotary intersection Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2078 Bhadra, 8 marks - Video: [Lec-21 Guidelines for Rotary Intersection 1| Traffic Engineering and Management | Civil Engineering](https://www.youtube.com/watch?v=Mco1FgsYkac) — Civil Engineering Department_LJIET - 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-design-factors - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question Explain the design factors to be considered in rotary intersection design. ## Short answer Key factors are design speed, central island radius, entry and exit curve radii, weaving length and width, carriageway width, entry and non-weaving widths, central island shape, proportion of weaving traffic, superelevation, sight distance, and channelising islands. ## Answer **Design elements/factors for a rotary (IRC:65-1976, Recommended Practice for Traffic Rotaries)** 1. **Design speed:** 30 km/h for urban and 40 km/h for rural rotaries (IRC:65). 2. **Radius of the central island and entry/exit curves:** the central island radius is about 1.33 times the entry curve radius (IRC:65); the entry curve radius is about 20 m (urban) to 25 m (rural); exit curves are about 1.5 to 2 times the entry curve radius, to speed up exits. 3. **Weaving length and width:** the weaving length between the entry and exit of consecutive arms should be at least 4 times the weaving width (IRC:65), and is checked so that $w/l$ lies between 0.12 and 0.4 for the capacity formula. 4. **Width of the rotary carriageway:** the width of the weaving section is the average of the entry width and the non-weaving width plus 3.5 m ($w = (e_1 + e_2)/2 + 3.5$ m); minimum 7 m for a rotary on two-lane approaches. 5. **Entry and exit widths:** the entry width $e_1$ and non-weaving width $e_2$ are provided to suit the approach carriageways, keeping $e/w$ between 0.4 and 1.0. 6. **Shape of the central island:** circular (preferred), elliptical, tangent or turbine shapes to suit the angles of the approach roads. 7. **Traffic volume and proportion of weaving traffic:** the capacity formula depends on this proportion $p$ (0.4-1.0). 8. **Super-elevation and cross slope:** the carriageway is given a reverse camber of about 1.5 to 2 percent (cross fall outward) for drainage; no superelevation inwards. 9. **Sight distance** at entries and in the circulating carriageway. 10. **Channelising islands, pedestrian refuges, markings, signs and lighting.** **Capacity check:** the practical capacity of the weaving section (IRC:65, Wardrop/TRRL formula) is $$ Q_p = \frac{280\,w\left(1+\dfrac{e}{w}\right)\left(1-\dfrac{p}{3}\right)}{1+\dfrac{w}{l}}\ \text{PCU/h} $$ where $e = (e_1+e_2)/2$, $w$ = weaving width (m), $l$ = weaving length (m), and $p$ = proportion of weaving traffic to total traffic in the weaving section. The formula is valid when $0.4 \le e/w \le 1.0$, $0.12 \le w/l \le 0.4$, $0.4 \le p \le 1.0$ and $w$ lies between 6 and 18 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. According to the answer (IRC:65), what design speeds are adopted for rotaries? A. 60 km/h urban and 80 km/h rural B. 30 km/h urban and 40 km/h rural C. 40 km/h urban and 30 km/h rural D. 20 km/h urban and 20 km/h rural Answer: B. IRC:65 takes about 30 km/h for urban and 40 km/h for rural rotaries. (Key point: Design speed) 2. What is the minimum weaving length relative to the weaving width? A. Equal to the weaving width B. 2 times the weaving width C. At least 4 times the weaving width D. 10 times the weaving width Answer: C. IRC:65 requires weaving length of at least 4 times the weaving width, with w/l between 0.12 and 0.4. (Key point: Weaving length) 3. How is the width of the weaving section obtained? A. w = e1 + e2 B. w = (e1 + e2)/2 + 3.5 m C. w = (e1 + e2)/2 - 3.5 m D. w = 3.5 m fixed Answer: B. The weaving width is the mean entry width plus 3.5 m, with a minimum of 7 m. (Key point: Weaving width) 4. In the practical capacity formula, which quantity do the limits 0.4 to 1.0 apply to besides e/w? A. Proportion p of weaving traffic B. Ratio w/l C. Central island radius ratio D. Cross fall Answer: A. The formula is valid for 0.4 <= e/w <= 1.0, 0.12 <= w/l <= 0.4 and 0.4 <= p <= 1.0. 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