---
title: "Transition curve length and shift, 7.2 m pavement"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-62j-tc-shift"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-62j-tc-shift
site: BCT Engineering
---

# Transition curve length and shift, 7.2 m pavement

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2062 Jestha (old course), 8 marks
- Video: [Finding Length and Shift of Transition Curve as Per IRC Method| Solved Problem](https://www.youtube.com/watch?v=y8o1lecv13o) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-62j-tc-shift
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Calculate the length of the transition curve and the required shift, if the design speed is 60 kmph, the radius of the circular curve is 220 m. And allowable rate of change of centrifugal acceleration is 60 cm/sec³. Allowable rate of change of super elevation is 1 in 120. The pavement width including extra widening is 7.2 m.

## Short answer

Transition length governed by centrifugal acceleration (35.1 m) and superelevation (30.2 m); adopt 40 m. Shift = Ls^2/(24R) = 0.303 m.

## Answer

#### Assumptions
Pavement width (with extra widening) $W=7.2$ m, rotated about the **centre line** (rotation is not stated). $e=\dfrac{V^2}{225R}=0.073$, limited to the maximum $e=0.07$.

#### Length of transition curve
1. **Rate of change of centrifugal acceleration** ($C=0.6$ m/s³, $V=16.67$ m/s):

$$
L_s=\frac{V^3}{CR}=\frac{16.67^3}{0.6(220)}=35.07\ \text{m}
$$

2. **Rate of introduction of superelevation** (1 in 120, centre-line rotation):

$$
L_s=\frac{N\,e\,W}{2}=\frac{120(0.07)(7.2)}{2}=30.24\ \text{m}
$$

The greater value governs, so adopt **L_s = 40 m**.

#### Shift

$$
s=\frac{L_s^2}{24R}=\frac{40^2}{24(220)}=0.303\ \text{m}
$$

**Answer:** L_s = 40 m (calculated 35.1 m); shift = 0.30 m. (If the pavement were rotated about the inner edge, $L_s=NeW=120(0.07)(7.2)=60.5$ m.)

## Check yourself

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

1. Which formula gives the transition length from the rate of change of centrifugal acceleration C?
   A. Ls = V R/C
   B. Ls = V^3/(C R)
   C. Ls = C R/V^3
   D. Ls = V^2/(C R)
   Answer: B. Ls = V^3/(CR), with V in m/s and C in m/s^3, here 16.67^3/(0.6 x 220) = 35.07 m. (Key point: Centrifugal acceleration criterion)

2. The allowable rate is 60 cm/sec^3. What value of C is used in the formula?
   A. 60 m/s^3
   B. 6 m/s^3
   C. 0.6 m/s^3
   D. 0.06 m/s^3
   Answer: C. 60 cm/s^3 = 0.6 m/s^3, so units must be in metres before using the formula. (Key point: Unit conversion of C)

3. With the pavement rotated about the centre line, which expression gives Ls from the rate of introduction of superelevation (1 in N)?
   A. Ls = N e W
   B. Ls = N e W / 2
   C. Ls = e W / N
   D. Ls = N W / (2e)
   Answer: B. Only half the width is raised relative to the centre line, so Ls = N e W/2 = 120 x 0.07 x 7.2 / 2 = 30.24 m. (Key point: Superelevation rate criterion)

4. Which Ls is adopted and what is the shift?
   A. 30.24 m; shift 0.17 m
   B. 35.07 m; shift 0.30 m
   C. 40 m; shift 0.30 m
   D. 60.5 m; shift 0.69 m
   Answer: C. The greater value (35.07 m) governs and is rounded to 40 m; s = Ls^2/(24R) = 1600/5280 = 0.303 m. (Key point: Adopted length and shift)

## Related in Transportation Engineering I

- [Transition curve 220 m, pavement 7 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-77c-tc-220) — Transportation Engineering I, chapter 3
- [Transition curve 220 m radius, 65 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-68bh-tc-65) — Transportation Engineering I, chapter 3
- [Transition curve 400 m radius, 85 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-tc-400-85) — Transportation Engineering I, chapter 3, asked 2×
- [Transition curve 200 m radius, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-tc-200) — Transportation Engineering I, chapter 3
- [Transition curve for 300 m radius, 7 m pavement](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-64p-tc-300) — Transportation Engineering I, chapter 3
- [Transition curve 625 m radius, 90 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-75b-tc-625) — Transportation Engineering I, chapter 3

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