---
title: "Transition curve 200 m radius, 60 kmph"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-70m-tc-200"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-70m-tc-200
site: BCT Engineering
---

# Transition curve 200 m radius, 60 kmph

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2070 Magh, 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-70m-tc-200
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A horizontal curve of 200 m radius is to be set out connecting two straights. The maximum speed of the vehicle on the curve is restricted to 60 kmph. Transition curves are to be introduced at each end of the circular curve. If the angle of intersection and chainage at the point of intersection is 125°30' and 1063 m calculate: (a) Length of transition curve for comfort condition (b) Length of combined curve (c) Chainage of beginning and end of curve.

## Short answer

Horizontal curve radius 200 m, speed 60 km/h, deflection 54.5°, PI chainage 1063 m. Transition length 40 m by comfort criterion. Combined curve length 230.2 m. Curve starts at chainage 939.8 m and ends at 1170.1 m.

## Answer

#### Reading of the data
The angle of intersection 125°30′ is taken as the interior angle between the straights, so the deflection angle is $\Delta=180^\circ-125^\circ30'=54.5^\circ$. $R=200$ m, $V=60$ km/h, PI at chainage 1063 m.

#### (a) Length of transition curve (comfort condition)
$C=\dfrac{80}{75+V}=\dfrac{80}{135}=0.593$ m/s³, $V=16.67$ m/s:

$$
L_s=\frac{V^3}{CR}=\frac{16.67^3}{0.593(200)}=39.06\ \text{m}
$$

Adopt $L_s=40$ m.

#### (b) Length of combined curve
- Shift: $s=\dfrac{L_s^2}{24R}=\dfrac{40^2}{24(200)}=0.333$ m
- Spiral angle: $\theta_s=\dfrac{L_s}{2R}=0.1000$ rad $=5.730^\circ$
- Arc angle: $\Delta_c=54.5^\circ-2(5.730^\circ)=43.041^\circ$
- Arc length: $L_c=R\Delta_c=150.24$ m

$$
L=L_c+2L_s=150.24+2(40)=230.24\ \text{m}
$$

#### (c) Chainages

$$
T_s=(R+s)\tan\frac{\Delta}{2}+\frac{L_s}{2}=(200.333)\tan(27.25^\circ)+20=123.18\ \text{m}
$$

- Beginning of curve = 1063 − 123.18 = **939.82 m**
- End of curve = 939.82 + 230.24 = **1170.06 m**

**Answer:** $L_s=40$ m; combined curve length = 230.2 m; curve starts at 939.8 m and ends at 1170.1 m.

## Check yourself

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

1. How is Δ obtained from the 125°30' angle?
   A. Δ = 180° - 125°30' = 54.5°
   B. Δ = 125.5°
   C. Δ = 125.5/2
   D. Δ = 90 - 125.5
   Answer: A. It is the interior angle between the straights. (Key point: Interior angle)

2. How is Ls found?
   A. Ls = V^2/CR
   B. Ls = V^3/(CR) with C = 80/(75+V) = 0.593 giving 39 m, adopt 40 m
   C. Ls = C V^3/R
   D. Ls = R/V
   Answer: B. V = 16.67 m/s. (Key point: Comfort length)

3. How is the combined curve length found?
   A. RΔ = 190 m
   B. Lc + Ls
   C. Lc + 2Ls with Lc = R(Δ - 2θs) = 150.24, total 230.24 m
   D. 2Lc + 2Ls
   Answer: C. θs = 5.73°. (Key point: Combined length)

4. How is the beginning chainage found?
   A. PI plus Ts
   B. PI minus R tan(Δ/2)
   C. PI minus L
   D. PI chainage minus Ts = (R+s)tan(Δ/2) + Ls/2, giving 939.82 m
   Answer: D. End = 939.82 + 230.24 = 1170.06 m. (Key point: Chainages)

## Related in Transportation Engineering I

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## Related in other subjects

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