---
title: "Transition deflection angle derivation and laying out"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-deflection-angle-derive-laying-out"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-deflection-angle-derive-laying-out
site: BCT Engineering
---

# Transition deflection angle derivation and laying out

Surveying II · Chapter 6: Curves

- Asked in: 2076 Bhadra, 4 marks
- 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-derive-laying-out
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Derive the formula for deflection angle (α) in a transition curve, $\alpha = \frac{573\,l^2}{RL}$ minute and explain the laying out procedure.

## Short answer

From r l = RL, φ = l²/(2RL); using α ≈ y/x = l²/(6RL) rad = (3437.75/6)·l²/(RL) min = 573 l²/(RL) min. Laying out: compute Ls, s, Ts; set theodolite at TS, sight IP, set computed α for each peg, measure chord; check αs = θs/3 at SC.

## Answer

#### Derivation of $\alpha = \dfrac{573\,l^2}{RL}$ minutes
For a transition curve the radius at distance $l$ from the start (TS) varies inversely with $l$: $r\,l = RL$, where $R$ is the radius of the circular curve and $L$ the length of the transition. The angle between the tangent at TS and the tangent at a point at distance $l$ is

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

Taking $x \approx l$ along the first tangent, the ordinate is $y = \int \phi\,dx = \dfrac{l^3}{6RL}$. The deflection angle of the point from TS:

$$
\alpha \approx \frac{y}{x} = \frac{l^2}{6RL}\ \text{radians} = \frac{3437.75}{6}\cdot\frac{l^2}{RL}\ \text{minutes} = \frac{573\,l^2}{RL}\ \text{minutes}
$$

At the end of the transition, $\alpha_s = \phi_s/3$.

#### Laying out the transition curve
1. Calculate $L_s$, $s$, $T_s$ and the chainages of TS, SC, CS and ST.
2. Locate TS by measuring $T_s$ back from the IP.
3. Set up the theodolite at TS, sight the IP with the plates at zero.
4. For each peg at distance $l$ from TS, compute $\alpha$ and set it on the circle; measure the chord length from the previous point, and fix the point where tape and line of sight meet.
5. The last angle to SC should equal $\theta_s/3$ (check). The second transition is set out in the same way from ST.

## Check yourself

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

1. What condition defines the transition curve in the derivation?
   A. Radius is constant
   B. Radius is proportional to l
   C. Radius varies as l^2
   D. Radius r varies inversely with distance l: r l = RL
   Answer: D. Curvature increases linearly from zero, so r l = RL. (Key point: Radius law)

2. Integrating dl/r gives the angle φ between tangents as
   A. l^2/(2RL) radians
   B. l^2/(RL) radians
   C. l^3/(6RL) radians
   D. l/(RL) radians
   Answer: A. φ = ∫ l dl/(RL) = l^2/(2RL). (Key point: Tangent angle)

3. The deflection angle α from TS is
   A. l^3/(6RL) radian
   B. about y/x = l^2/(6RL) radian, equal to 573 l^2/(RL) minutes
   C. l^2/(2RL) radian
   D. 3 times the tangent angle
   Answer: B. With y = l^3/(6RL) and x ≈ l, α = l^2/(6RL) rad = 3437.75/6 = 573 per unit l^2/(RL) in minutes. (Key point: Deflection angle)

4. The last deflection angle to SC is
   A. θs
   B. 2θs/3
   C. θs/3
   D. θs/2
   Answer: C. αs = φs/3 is the check at SC. (Key point: Check at SC)

## Related in Surveying II

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- [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

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