---
title: "Transition curve 220 m, pavement 7 m"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-77c-tc-220"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-77c-tc-220
site: BCT Engineering
---

# Transition curve 220 m, pavement 7 m

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2077 Chaitra, 8 marks
- Video: [#highways, #curve, #transition, Design of Transition curve on a highway, Transition on curves.](https://www.youtube.com/watch?v=0Mp2UzTLOss) — Transportation Engg Talks | Satish IITR
- 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-77c-tc-220
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

Calculate the length of transition curve and shift, if the design speed is 60 kmph. The radius of circular curve is 220 m. An allowable rate of change of centrifugal acceleration is 60 cm/sec³. Allowable rate of change of superelevation is 1 in 120. The width of the pavement is 7.00 m. Assume rotation of Pavement about the inner edge.

## Short answer

Length of transition curve is 60 m, shift is 0.68 m. Superelevation limited to 0.07 governs.

## Answer

#### Superelevation
$e=\dfrac{V^2}{225R}=\dfrac{60^2}{225(220)}=0.0727$, which exceeds the maximum 0.07 (IRC/NRS), so $e=0.07$. Friction needed: $\dfrac{V^2}{127R}-e=0.059<0.15$, so it is safe.

#### Length of transition curve
1. **Rate of change of centrifugal acceleration** ($C=0.6$ m/s³, $V=60/3.6=16.67$ 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), pavement rotated about the **inner edge**, so the outer edge must be raised by $eW=0.07\times7.0=0.49$ m relative to the inner edge:

$$
L_s=N\,e\,W=120(0.07)(7.0)=58.8\ \text{m}
$$

The greater value governs: adopt **L_s = 60 m**.

#### Shift

$$
s=\frac{L_s^2}{24R}=\frac{60^2}{24(220)}=0.682\ \text{m}
$$

**Answer:** length of transition curve = 60 m; shift = 0.68 m. (IRC empirical value $2.7V^2/R=44.2$ m is smaller, so it does not govern.)

## Check yourself

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

1. Why is e = 0.07 used instead of V^2/(225R) = 0.0727?
   A. Because the calculated value exceeds the maximum permitted superelevation of 0.07
   B. Because the friction factor must be zero
   C. Because 0.0727 is below the minimum allowed
   D. Because transition curves require e = 0.07 exactly
   Answer: A. e is capped at e_max = 0.07; friction then supplies the rest (0.059 < 0.15), so it is safe. (Key point: Superelevation limit)

2. Which formula gives the transition length from rate of change of centrifugal acceleration?
   A. Ls = V^2 / (127 R)
   B. Ls = V^3 / (C R) with V in m/s
   C. Ls = N e W
   D. Ls = L^2 / (24 R)
   Answer: B. Ls = 16.67^3 /(0.6 x 220) = 35.07 m. (Key point: Comfort criterion)

3. With 1 in 120 rate and rotation about the inner edge, how is Ls found?
   A. Ls = 120 x 0.07 x 3.5 = 29.4 m
   B. Ls = 0.07 x 7 / 120
   C. Ls = N e W = 120 x 0.07 x 7 = 58.8 m, because the outer edge is raised by eW
   D. Ls = 120 x 7 = 840 m
   Answer: C. Raise of outer edge relative to inner is eW = 0.49 m, which at 1 in 120 needs 58.8 m. (Key point: Superelevation introduction)

4. How is the final Ls chosen and the shift computed?
   A. The smallest value (35 m) is adopted; s = Ls/(24R)
   B. The average of criteria; s = Ls^2/(8R)
   C. 58.8 m exactly; s = Ls^2/(2R)
   D. The greatest of the criteria is adopted (60 m); shift s = Ls^2/(24R) = 0.68 m
   Answer: D. The longest length satisfies every criterion; shift = 3600/5280 = 0.68 m. (Key point: Governing length and shift)

## Related in Transportation Engineering I

- [Transition curve length and shift, 7.2 m pavement](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-62j-tc-shift) — 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 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
- [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

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