---
title: "Transition curve for 230 m radius"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-81c-tc-230"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-81c-tc-230
site: BCT Engineering
---

# Transition curve for 230 m radius

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2081 Chaitra, 5 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-81c-tc-230
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A section of major highway passing has a horizontal curve of radius 230 m. Design the length of transition curve. Use coefficient of longitudinal and lateral friction value as 0.35 and 0.15 respectively.

## Short answer

For a horizontal curve of radius 230 m with f_long=0.35 and f_lat=0.15, design speed is determined as 80 km/h with e=7% (max). Transition curve length is calculated by three criteria: rate of change of centrifugal acceleration (92.4 m), rate of superelevation introduction (49.9 m), and IRC empirical formula (75.1 m). The maximum value 92.4 m is rounded up to 95 m as the adopted transition length.

## Answer

**Given:** $R = 230$ m, $f_{long} = 0.35$, $f_{lat} = 0.15$. The design speed is not given, so it is found from the maximum superelevation.

#### Step 1: Design speed and superelevation
Taking the maximum superelevation $e = 7\%$ (IRC/NRS for plain and rolling terrain):
$$
V = \sqrt{127R\,(e+f)} = \sqrt{127(230)(0.07+0.15)} = 80.2\ \text{km/h}
$$
So the design speed $V = 80$ km/h; required $e = V^2/225R = 12.4\%$ is more than 7%, so adopt $e = 7\%$ and check $f = \dfrac{80^2}{127\times230} - 0.07 = 0.149 < 0.15$. OK.

#### Step 2: Length of transition curve ($v = 80/3.6 = 22.22$ m/s)
**(a) Rate of change of centrifugal acceleration**
$$
C = \frac{80}{75+V} = \frac{80}{155} = 0.516\ \text{m/s}^3\ (0.5 \le C \le 0.8)
$$
$$
L_s = \frac{v^3}{CR} = \frac{22.22^3}{0.516\times230} = 92.4\ \text{m}
$$

**(b) Rate of introduction of superelevation** (1 in 150, plain/rolling; $W = 7.0$ m, rotation about the centre line with camber $n = 2.5\%$):
$$
L_s = N\,\frac{(e+n)W}{2} = 150\times\frac{(0.07+0.025)(7.0)}{2} = 49.9\ \text{m}
$$

**(c) IRC empirical formula**
$$
L_s = \frac{2.7V^2}{R} = \frac{2.7(80)^2}{230} = 75.1\ \text{m}
$$

#### Step 3: Adopt the greatest
$L_s = 92.4$ m, rounded up: **$L_s = 95$ m**.

**Answer: length of transition curve = 95 m** (design speed 80 km/h, $e=7\%$).

## Check yourself

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

1. When speed is not given, how is the design speed found for R = 230 m with e = 7% and f = 0.15?
   A. V = sqrt(127 R (e + f)) = about 80 km/h
   B. V = sqrt(225 R e) = about 60 km/h
   C. V = R/(e + f)
   D. It cannot be found
   Answer: A. V = sqrt(127 x 230 x 0.22) = 80.2 km/h, so 80 km/h is adopted. (Key point: Finding the design speed)

2. Which formula gives Ls from the rate of change of centrifugal acceleration?
   A. Ls = v^3/(C R)
   B. Ls = v^2/(C R)
   C. Ls = C R/v^3
   D. Ls = N e W/2
   Answer: A. Ls = v^3/(CR) = 22.22^3/(0.516 x 230) = 92.4 m with C = 80/(75+V). (Key point: Centrifugal acceleration criterion)

3. Three criteria gave 92.4 m, 49.9 m and 75.1 m. Which is adopted?
   A. The smallest, 49.9 m
   B. The average
   C. The largest, 92.4 m, rounded to 95 m
   D. 75.1 m, the empirical value only
   Answer: C. The largest value satisfies all criteria, rounded up to 95 m. (Key point: Governing criterion)

4. With e limited to 7% at 80 km/h, R = 230 m, the friction required is
   A. 0.07
   B. 0.149, which is less than 0.15, so OK
   C. 0.35
   D. 0.22
   Answer: B. f = V^2/(127R) - e = 6400/29210 - 0.07 = 0.149 < 0.15. (Key point: Check of friction)

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