---
title: "Transition curve 625 m radius, 90 kmph"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-75b-tc-625"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-75b-tc-625
site: BCT Engineering
---

# Transition curve 625 m radius, 90 kmph

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2075 Baisakh, 8 marks
- Video: [Length of Transition curve l Numerical l Transportation Engineering l CE](https://www.youtube.com/watch?v=eE5IFBSdOIQ) — ONLINETYARI CLASSES
- 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-75b-tc-625
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A horizontal curve of 625 m radius is to be set out to connect two straight of a national highway. The speed of the vehicle is restricted to 90 Kmph. Calculate
a) length of transition curve
b) the chainage of beginning and end of the curve given that, angle of intersection = 130°24', rate of change of centrifugal acceleration = 0.25 m/s³
c) chainage of point of intersection = 1092.500 m

## Short answer

L_s = 100 m. Chainage TS = 753.40 m, ST = 1394.45 m (deflection angle 49°36′, shift 0.667 m).

## Answer

#### Reading of the data
The angle of intersection 130°24′ is taken as the interior angle between the straights, so the deflection angle is $\Delta=180^\circ-130^\circ24'=49^\circ36'=49.6^\circ$. (With $\Delta=130.4^\circ$ the tangent length would exceed the chainage of the PI and give a negative starting chainage, so this reading is the consistent one.)

#### (a) Length of transition curve
$V=90/3.6=25$ m/s, $C=0.25$ m/s³:

$$
L_s=\frac{V^3}{CR}=\frac{25^3}{0.25(625)}=100.0\ \text{m}
$$

Adopt $L_s=100$ m.

#### (b) Chainages of beginning and end of the curve
- Shift: $s=\dfrac{L_s^2}{24R}=\dfrac{100^2}{24(625)}=0.667$ m
- Spiral angle: $\theta_s=\dfrac{L_s}{2R}=0.0800$ rad $=4.584^\circ$
- Central angle of circular arc: $\Delta_c=\Delta-2\theta_s=49.6^\circ-9.167^\circ=40.433^\circ$
- Circular arc length: $L_c=R\Delta_c=441.05$ m
- Total length of combined curve: $L=L_c+2L_s=641.05$ m
- Tangent length from PI:

$$
T_s=(R+s)\tan\frac{\Delta}{2}+\frac{L_s}{2}=(625.667)\tan(24.80^\circ)+50=339.10\ \text{m}
$$


#### (c) Chainages (PI at 1092.500 m)
- Beginning of curve (TS) = 1092.500 − 339.10 = **753.40 m**
- End of curve (ST) = 753.40 + 641.05 = **1394.45 m**

**Answer:** $L_s=100$ m; curve begins at 753.40 m and ends at 1394.45 m.

## Check yourself

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

1. Why is the deflection angle 49°36' used instead of 130°24'?
   A. 130°24' is the interior angle between straights, so Δ = 180° - 130°24'
   B. Because 130°24' is the spiral angle
   C. Because deflection is always half the intersection angle
   D. Because 130°24' is already the deflection angle
   Answer: A. Using 130.4° would make the tangent too long and the starting chainage negative. (Key point: Deflection angle)

2. What formula gives Ls here?
   A. Ls = V^2/(127R) = 1.0 m
   B. Ls = V^3 / (C R) = 25^3 /(0.25 x 625) = 100 m
   C. Ls = R/θs
   D. Ls = 2.7V^2/R
   Answer: B. V = 90/3.6 = 25 m/s, giving 100 m. (Key point: Transition length)

3. How is the central angle of the circular arc found?
   A. Δc = Δ + 2θs
   B. Δc = Δ/2
   C. Δc = Δ - 2θs, since each spiral uses up angle θs = Ls/2R
   D. Δc = Δ - θs
   Answer: C. θs = 4.584°, so Δc = 49.6 - 9.167 = 40.43°. (Key point: Circular arc angle)

4. What is the tangent length formula for the combined curve?
   A. Ts = R tan(Δ/2)
   B. Ts = (R + s) tan Δ
   C. Ts = Ls + R
   D. Ts = (R + s) tan(Δ/2) + Ls/2
   Answer: D. With s = 0.667 m, Ts = 339.1 m, so TS = 1092.5 - 339.1 = 753.4 m. (Key point: Tangent length)

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