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

# Derive deflection angle of transition curve

Surveying II · Chapter 6: Curves

- Asked in: 2077 Chaitra, 4 marks; 2073 Magh, 4 marks; 2072 Magh, 4 marks (asked 3 times)
- 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-deflection-angle-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Prove the deflection angle of transition curve is $\alpha = \frac{573\,l^2}{RL}$ mins where symbols have their usual meaning.

## Short answer

For a cubic spiral with ρl = RL, the spiral angle φ = l²/(2RL), coordinates x ≈ l, y ≈ l³/(6RL). The deflection angle δ = y/x ≈ l²/(6RL) radians, which converts to 573 l²/(RL) minutes.

## Answer

#### Transition curve (cubic spiral)

A transition curve connects a straight with a circular curve so that the radius changes gradually from $\infty$ to $R$. The radius at a distance $l$ from the start is inversely proportional to $l$:

$$
\rho \propto \frac{1}{l} \quad\Rightarrow\quad \rho\, l = RL
$$

where $R$ is the radius of the circular curve and $L$ is the length of the transition curve.

```text
        y
        |        . P(x, y)
        |     .  
        |  .    deflection angle = delta
   T1 --+-----------------------> x   (tangent)
```

#### Derivation

Take T1 (start of the curve) as the origin and the tangent as the x-axis. For a point P at a distance $l$ along the curve, the tangent angle (spiral angle) $\phi$ is given by $d\phi = \dfrac{dl}{\rho} = \dfrac{l\,dl}{RL}$:

$$
\phi = \int_0^l \frac{l\,dl}{RL} = \frac{l^2}{2RL}
$$

The coordinates of P are, for small $\phi$ ($\cos\phi \approx 1$, $\sin\phi \approx \phi$):

$$
dx = dl,\quad dy = \phi\,dl = \frac{l^2}{2RL}\,dl
$$

$$
x \approx l, \qquad y = \int_0^l \frac{l^2}{2RL}\,dl = \frac{l^3}{6RL}
$$

This is the equation of a cubic parabola.

The deflection angle $\delta$ at T1 between the tangent and the line T1P:

$$
\tan\delta = \frac{y}{x} = \frac{l^3}{6RL\cdot l} = \frac{l^2}{6RL}
$$

As $\delta$ is small, $\tan\delta \approx \delta$ (in radians):

$$
\delta = \frac{l^2}{6RL}\ \text{radians}
$$

Converting radians to minutes: $1\ \text{rad} = \dfrac{180}{\pi}\times 60 = 3437.75'$:

$$
\delta = \frac{3437.75}{6}\cdot\frac{l^2}{RL} = 572.96\,\frac{l^2}{RL} \approx \boxed{\frac{573\,l^2}{RL}\ \text{minutes}}
$$

**Hence proved.**

At the end of the transition ($l = L$): $\delta_L = \dfrac{L}{6R}$ rad $= \dfrac{\phi_s}{3}$, where $\phi_s = \dfrac{L}{2R}$ is the spiral angle. This is the usual check on the last peg.

## Check yourself

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

1. Which relation defines the radius ρ at distance l from the start in the proof?
   A. ρ is proportional to l, so ρ/l = RL
   B. ρ is constant and equal to R
   C. ρ is proportional to 1/l, so ρ·l = RL
   D. ρ is proportional to l^2
   Answer: C. The radius changes from infinity to R, so it is inversely proportional to l: ρ l = R L. (Key point: Basic condition of transition curve)

2. Integrating dφ = dl/ρ = l dl/(RL) gives what tangent angle at distance l?
   A. φ = l^2/(2RL)
   B. φ = l/(RL)
   C. φ = l^3/(6RL)
   D. φ = l^2/(6RL)
   Answer: A. φ = ∫ l dl/(RL) from 0 to l = l^2/(2RL). (Key point: Spiral angle)

3. Using dy = φ dl and x ≈ l, what is y?
   A. y = l^2/(2RL)
   B. y = l^3/(3RL)
   C. y = l^3/(6RL), a cubic parabola
   D. y = l^3/(2RL)
   Answer: C. y = ∫ l^2/(2RL) dl = l^3/(6RL), the cubic parabola. (Key point: Offset y)

4. How does the 573 arise in δ = 573 l^2/(RL) minutes?
   A. δ = l^2/(6RL) rad, and 1 rad = 3437.75' so 3437.75/6 = 573
   B. δ = l^2/(2RL) rad times 1718.9
   C. δ = l^2/(6RL) rad times 57.3
   D. δ = l^2/(3RL) rad times 1718.9
   Answer: A. tan δ = y/x = l^2/(6RL) ≈ δ in radians; multiplying by 3437.75 minutes per radian gives 572.96 ≈ 573. (Key point: Converting to minutes)

## Related in Surveying II

- [Function of transition curve, deflection angle](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-curve-function-deflection-angle) — Surveying II, chapter 6
- [Transition deflection angle derivation and laying out](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-deflection-angle-derive-laying-out) — Surveying II, chapter 6
- [Transition curve R=300 m, deflection 40°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-curve-80kmph-radius-300) — Surveying II, chapter 6
- [Transition curve R=2000 m, chainages, deflection table](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-curve-2000m-160kmh-chainage-table) — Surveying II, chapter 6
- [Composite curve by deflection angle method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acomposite-curve-deflection-angle-method) — Surveying II, chapter 6, asked 5×
- [Simple circular curve by deflection angle method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asimple-curve-deflection-angle-method-sketch) — Surveying II, chapter 6

## Related in other subjects

- [Transition curve length from spiral angle 10.32](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-73bh-tc-spiral) — Transportation Engineering I, chapter 3, asked 2×
- [Transition curve from spiral angle 8.35](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-69bh-tc-spiral) — Transportation Engineering I, chapter 3, asked 2×
- [Transition curve in hilly region, 60 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65c-tc-hilly) — Transportation Engineering I, chapter 3
- [Composite curve with 48 degree deflection](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-72m-composite-60) — Transportation Engineering I, chapter 3