---
title: "Transition curve length from spiral angle 10.32"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-73bh-tc-spiral"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-73bh-tc-spiral
site: BCT Engineering
---

# Transition curve length from spiral angle 10.32

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2073 Bhadra, 8 marks
- Video: [Spiral Curves for Highway Alignment](https://www.youtube.com/watch?v=CCpi0eNks94) — Daniel Findley
- 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-73bh-tc-spiral
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

The angle of intersection between two straights is 145.37°. The spiral angle for each transition curve is 10.32°. Calculate the length of transition curve, combined length of curve and the length of tangent if the radius of the curve is 350 m.

## Short answer

Deflection angle Δ = 34.63°. Transition length Ls = 2Rθs = 126.1 m, total composite length = 337.6 m, and tangent length Ts = 172.7 m.

## Answer

#### Reading of the data
The angle of intersection 145.37° is the interior angle between the two straights, so the deflection angle is $\Delta=180^\circ-145.37^\circ=34.63^\circ$. Radius $R=350$ m, spiral angle $\theta_s=10.32^\circ$.

#### Length of transition curve
For a spiral, $\theta_s=\dfrac{L_s}{2R}$ (radians), so

$$
L_s=2R\theta_s=2(350)(0.18012)=126.08\ \text{m}
$$

#### Combined length of curve
- Shift: $s=\dfrac{L_s^2}{24R}=\dfrac{126.08^2}{24(350)}=1.892$ m
- Central angle of the circular arc: $\Delta_c=\Delta-2\theta_s=34.63-2(10.32)=13.99^\circ$
- Circular arc length: $L_c=R\Delta_c=350(0.24417)=85.46$ m

$$
L=L_c+2L_s=85.46+2(126.08)=337.63\ \text{m}
$$

#### Length of tangent (from PI to the start of the transition)

$$
T_s=(R+s)\tan\frac{\Delta}{2}+\frac{L_s}{2}=(351.892)\tan(17.32^\circ)+63.04=172.74\ \text{m}
$$

**Answer:** $L_s=126.1$ m; combined curve length = 337.6 m; tangent length $T_s=172.7$ m.

(If the given angle were instead the deflection angle, $\Delta=145.37^\circ$, the same $L_s$ results with $\Delta_c=124.73^\circ$, total length 1014.1 m and $T_s=1191.8$ m.)

## Check yourself

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

1. How is deflection angle obtained from 145.37°?
   A. Δ = 180° - 145.37° = 34.63°, since 145.37° is the interior angle
   B. Δ = 145.37°
   C. Δ = 145.37° / 2
   D. Δ = 145.37° - 10.32°
   Answer: A. Angle of intersection means the interior angle between the straights. (Key point: Interior vs deflection)

2. How is Ls found from θs?
   A. Ls = R θs = 63 m
   B. Ls = 2R θs with θs in radians = 126.1 m
   C. Ls = V^3/(CR)
   D. Ls = Rθs^2
   Answer: B. θs = Ls/(2R), rearranged. (Key point: Ls from spiral angle)

3. How is the combined curve length found?
   A. L = R x Δ
   B. L = Lc + Ls
   C. L = Lc + 2Ls, with Lc = R(Δ - 2θs)
   D. L = 2Lc + Ls
   Answer: C. Lc = 350 x 0.24417 = 85.46 m, so L = 85.46 + 252.16 = 337.6 m. (Key point: Combined length)

4. Which formula gives the tangent length?
   A. Ts = R tan(Δ/2) - Ls/2
   B. Ts = (R + s) tanΔ
   C. Ts = R + s + Ls
   D. Ts = (R + s) tan(Δ/2) + Ls/2
   Answer: D. s = Ls^2/(24R) = 1.892 m, giving Ts = 172.7 m. (Key point: Tangent length)

## Close variants set in other exams

- [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×

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