---
title: "Function of transition curve, deflection angle"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-curve-function-deflection-angle"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-curve-function-deflection-angle
site: BCT Engineering
---

# Function of transition curve, deflection angle

Surveying II · Chapter 6: Curves

- Asked in: 2074 Bhadra, 5 marks
- Video: [Deflection Angle in Transition Curve | Hindi | Part-5 | Curves in Surveying |](https://www.youtube.com/watch?v=tAVMhkYDGB8) — Civil Inside
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-curve-function-deflection-angle
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

State the function of transition curves. Derive the expression for deflection angle in transition curve that $\alpha = \frac{573\,l^2}{RL}$ min where, $l$ = chord length, $R$ = Radius, $L$ = Length of transition curve.

## Short answer

Transition curves gradually change radius from infinity to R, allowing smooth introduction of centrifugal force and super-elevation without sudden jerk, improving comfort and safety. The deflection angle from the tangent to a point at length l is α = 573 l²/(RL) minutes, derived from the cubic parabola geometry where α = φ/3 and φ = l²/(2RL) radians.

## Answer

#### Functions of a transition curve
A transition (easement) curve is introduced between a straight and a circular curve. Its functions are:

1. To change the radius gradually from infinity (straight) to $R$ (circular curve), so the centrifugal force builds up gradually.
2. To allow gradual introduction of super-elevation (and widening), so there is no sudden jerk.
3. To keep the rate of change of radial acceleration within a comfortable limit.
4. To improve the appearance and safety of the alignment, and to reduce the risk of overturning or side-slip.

#### Derivation of $\alpha = \dfrac{573\,l^2}{RL}$ minutes
A transition curve is designed so that the radius at any point is inversely proportional to the distance $l$ from the start (TS): $r \propto 1/l$. At the end of the transition ($l = L$) the radius is $R$, so

$$
r\,l = R\,L \quad\Rightarrow\quad r = \frac{RL}{l}
$$

Let $\phi$ be the angle between the tangent at TS and the tangent at a point P at distance $l$. For a small element $dl$, $d\phi = dl/r = l\,dl/(RL)$. Integrating from 0 to $l$:

$$
\phi = \frac{l^2}{2RL}\ \text{radians}
$$

Taking x along the initial tangent, $dy/dx = \tan\phi \approx \phi = x^2/(2RL)$ for small angles and $x \approx l$. Integrating,

$$
y = \frac{x^3}{6RL}\approx \frac{l^3}{6RL}\qquad(\text{cubic parabola})
$$

The deflection angle $\alpha$ of the chord from TS to P, measured from the initial tangent, satisfies

$$
\tan\alpha \approx \alpha = \frac{y}{x} = \frac{l^2}{6RL}\ \text{radians} = \frac{\phi}{3}
$$

Converting radians to minutes (1 rad = 3437.75′):

$$
\alpha = \frac{3437.75}{6}\,\frac{l^2}{RL} = \frac{573\,l^2}{RL}\ \text{minutes}
$$

At the end of the transition, $\alpha_s = 573L/R$ minutes $= \theta_s/3$ (one-third of the spiral angle).

## Check yourself

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

1. What change of radius does a transition curve provide?
   A. Gradually from infinity at the straight to R at the circular curve
   B. Suddenly from R to infinity
   C. Constant radius R throughout
   D. From zero to R
   Answer: A. This lets centrifugal force build up gradually. (Key point: Radius change)

2. What is another function of a transition curve?
   A. Increases the speed limit by itself
   B. Allows gradual introduction of superelevation and widening
   C. Removes the need for superelevation
   D. Makes the total length shorter
   Answer: B. Superelevation is run out over the transition length. (Key point: Superelevation)

3. What condition is assumed in the derivation?
   A. Radius is proportional to l
   B. Radius is constant
   C. Radius at any point is inversely proportional to the distance l from TS, so r l = R L
   D. r l^2 = R L
   Answer: C. With r l = RL, d(phi) = l dl/(RL). (Key point: Basic assumption)

4. Which relation between the deflection angle alpha and spiral angle phi results?
   A. alpha = phi/2
   B. alpha = 2 phi
   C. alpha = phi
   D. alpha = phi/3, with phi = l^2/(2RL)
   Answer: D. y/x = l^2/(6RL) = phi/3, and converting to minutes gives 573 l^2/(RL). (Key point: Result)

## Related in Surveying II

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