---
title: "Transition curve and shift for 250 m radius"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-79c-tc-250"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-79c-tc-250
site: BCT Engineering
---

# Transition curve and shift for 250 m radius

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2079 Chaitra, 8 marks
- Video: [Design SECRET to Perfect Transition Curve (Spiral Curve) - Mastering Road Curves](https://www.youtube.com/watch?v=JVqkgftaHNk) — Civil Guru Solutions
- 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-79c-tc-250
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A horizontal curve of radius 250 m in a two-lane highway passing through rolling terrain in heavy rainfall area is designed with an equilibrium superelevation of 0.065. Design the length of a transition curve and the shift required to connect this curve to the straight section of the highway. Assume rate of introduction of superelevation of 1 in 150.

## Short answer

From equilibrium e=0.065, design speed is 60 km/h. Transition length Ls=40 m (govered by empirical formula 2.7V²/R). Shift s=0.27 m.

## Answer

#### Assumptions
- In the IRC method the design superelevation $e=\dfrac{V^2}{225R}$ fully balances the centrifugal force at 75% of the design speed, so it is the equilibrium superelevation for mixed traffic. With $e=0.065$: $V=\sqrt{225(250)(0.065)}=60.5\approx 60$ km/h, so the design speed = **60 km/h**.
- Two-lane carriageway $W=7.0$ m, pavement rotated about the **centre line**, rate of introduction 1 in $N=150$.
- Extra widening for 2 lanes ($l=6.1$ m): mechanical $W_m=\dfrac{nl^2}{2R}=0.149$ m, psychological $W_{ps}=\dfrac{V}{9.5\sqrt R}=0.399$ m, so $W_e\approx 0.55$ m.

#### Length of transition curve (largest of three criteria)
1. **Rate of change of centrifugal acceleration** (IRC): $C=\dfrac{80}{75+V}=0.593$ m/s³

$$
L_s=\frac{(V/3.6)^3}{CR}=\frac{16.67^3}{0.593(250)}=31.3\ \text{m}
$$

2. **Rate of introduction of superelevation** (rotation about the centre line):

$$
L_s=\frac{N\,e\,(W+W_e)}{2}=\frac{150(0.065)(7.55)}{2}=36.8\ \text{m}
$$

3. **IRC empirical** (plain/rolling): $L_s=\dfrac{2.7V^2}{R}=\dfrac{2.7(60)^2}{250}=38.9$ m.

Adopt **$L_s$ = 40 m**.

#### Shift

$$
s=\frac{L_s^2}{24R}=\frac{40^2}{24(250)}=0.267\ \text{m}
$$

**Answer:** transition length $L_s=40$ m; shift $s=0.27$ m.

*Note:* if 0.065 is taken as the equilibrium superelevation at the full design speed ($e=V^2/127R$), then $V=45.4$ km/h. The three criteria then give 12.1 m, 36.3 m and 22.3 m, so the superelevation criterion governs and $L_s=40$ m and $s=0.27$ m are unchanged.

## Check yourself

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

1. How is design speed found from e = 0.065 and R = 250 m?
   A. V = sqrt(225 R e), about 60 km/h
   B. V = sqrt(127 R e), about 45 km/h
   C. V = e R
   D. V = 225 e / R
   Answer: A. Equilibrium e = V^2/225R gives V = sqrt(225 x 250 x 0.065) = 60 km/h. (Key point: Design speed from equilibrium superelevation)

2. With rotation about the centre line, Ls for rate 1 in N is
   A. N e (W + We)
   B. N e (W + We)/2
   C. e (W + We)/N
   D. N/(e W)
   Answer: B. Only half the width is raised relative to the centre line: 150 x 0.065 x 7.55/2 = 36.8 m. (Key point: Superelevation rate criterion)

3. Criteria gave 31.3 m, 36.8 m and 38.9 m. The adopted Ls is
   A. 31.3 m
   B. 36.8 m
   C. 40 m, greatest rounded up
   D. 75 m
   Answer: C. The largest (38.9 m) governs and is rounded up to 40 m. (Key point: Adopted length)

4. The shift s = Ls^2/(24R) for Ls = 40 m, R = 250 m is
   A. 0.027 m
   B. 2.67 m
   C. 0.267 m
   D. 6.4 m
   Answer: C. s = 1600/(24 x 250) = 0.267 m. (Key point: Shift)

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