---
title: "Transition curve R=180 m, speed 13.2 m/s"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-curve-180m-radius-132"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-curve-180m-radius-132
site: BCT Engineering
---

# Transition curve R=180 m, speed 13.2 m/s

Surveying II · Chapter 6: Curves

- Asked in: 2070 Magh, 10 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-curve-180m-radius-132
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A road curve of 180 m radius is to be set out to connect two tangents. The maximum speed of this part of the road will be 13.2 m/sec. Transition curves are to be introduced at each end of the curve. Find a suitable length of transition curve and circular curve including the value of first two deflection angles of each curve.

## Short answer

Ls from v³/(CR) = 20 m, using C = 0.653 m/s³ from Shortt's formula. Lc = 105.66 m for assumed Δ = 40°. Transition deflections: 0°15′55″ at 10 m, 1°03′40″ at 20 m; circular: 1°35′30″ for first chord, cumulative 3°10′59″ after 20 m chord.

## Answer

The deflection angle between the tangents and the peg intervals are not given, so they are assumed: $\Delta = 40^\circ$ (for illustration of $L_c$) and pegs every 10 m from TS and from SC. The rate of change of radial acceleration is found from Shortt's empirical formula.

#### Length of transition curve
$$
v = 13.2\ \text{m/s} = 47.52\ \text{km/h},\qquad C = \frac{80}{75+V} = \frac{80}{75+47.52} = 0.653\ \text{m/s}^3
$$

This lies in the permitted range 0.3–0.8 m/s³.

$$
L_s = \frac{v^3}{C R} = \frac{13.2^3}{0.653\times 180} = 19.57\ \text{m}
$$

Adopt $L_s = 20$ m.

#### Other elements
$$
\begin{aligned}
\theta_s &= \frac{L_s}{2R} = \frac{20}{360}\ \text{rad} = 3.183^\circ\\
s &= \frac{L_s^2}{24R} = 0.093\ \text{m}\\
T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 75.55\ \text{m}\ (\Delta = 40^\circ)
\end{aligned}
$$

#### Length of circular curve
$$
L_c = \frac{\pi R}{180}(\Delta - 2\theta_s) = \frac{\pi\times 180}{180}(40^\circ - 6.366^\circ) = 105.66\ \text{m}
$$

In general $L_c = R(\Delta - 2\theta_s)$ with angles in radians; it depends on the actual deflection angle.

#### First two deflection angles
Transition curve, $\alpha = 573\,l^2/(RL_s)$ minutes:

- At $l = 10$ m: $\alpha_1 = \dfrac{573\times 10^2}{180\times 20} = 15.92' = 0°15′55″$
- At $l = 20$ m: $\alpha_2 = \dfrac{573\times 20^2}{180\times 20} = 63.67' = 1°03′40″$ (this is the end point SC, and equals $\theta_s/3 = 1.061^\circ$)

Circular curve (from SC), $\delta = 1718.87\,c/R$ minutes, with 10 m chords:

- First point: $\delta_1 = \dfrac{1718.87\times 10}{180} = 95.49' = 1°35′30″$
- Second point: $\delta_2 = 2\delta_1 = 190.99' = 3°10′59″$

**Answer:** $L_s = 20$ m, $L_c = 105.66$ m (for $\Delta = 40^\circ$); transition deflections 0°15′55″ and 1°03′40″; circular deflections 1°35′30″ and 3°10′59″.

## Check yourself

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

1. Which formula gives C when not specified?
   A. C = 75 + V
   B. C = V/80
   C. Shortt's formula C = 80/(75 + V), V in km/h
   D. C = g R/v
   Answer: C. Shortt's empirical formula gives C = 80/(75 + 47.52) = 0.653 m/s^3, within 0.3-0.8. (Key point: Rate of change C)

2. Why is v = 13.2 m/s converted to 47.52 km/h?
   A. Because Ls needs km/h
   B. To get the degree of curve
   C. Because the formula for C uses V in km/h
   D. To compute the shift
   Answer: C. Shortt's formula uses V in km/h while Ls = v^3/(CR) uses v in m/s. (Key point: Speed conversion)

3. How do the deflection angles on the transition vary with l?
   A. Proportional to l^2, alpha = 573 l^2/(R Ls) min
   B. Proportional to l^3
   C. Linearly with l
   D. Constant
   Answer: A. Doubling l from 10 to 20 m raises alpha from 15.92' to 63.67', four times. (Key point: Transition deflection)

4. With 10 m chords from SC (R = 180), why is the second deflection 2 times the first?
   A. Because the chord doubles
   B. Because it is measured from TS
   C. Deflections accumulate: delta2 = delta1 + delta1 with equal chords
   D. Because R doubles
   Answer: C. Each 10 m chord gives 1 deg 35' 30''; cumulative deflection of the second point is twice that. (Key point: Circular deflection)

## Related in Surveying II

- [Transition curve, road 8 m, R=300 m, 60°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-300m-60-70kmph) — Surveying II, chapter 6, asked 2×
- [Transition curve road 8 m, R=300 m, IP 3605](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-300m-3605-50kmh) — Surveying II, chapter 6, asked 3×
- [Transition curve 6°, Ls=90 m, chainage 13+700](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-t1v-t2v-chainage-13700) — Surveying II, chapter 6
- [Transition curve 6°, Ls=90 m, chainage (13+7) chains](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-t1v-t2v-chainage-13-7) — 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

## Related in other subjects

- [Tangents, circular and transition curves](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Ah-alignment-elements-use) — Transportation Engineering I, chapter 3
- [Transition curve 400 m radius, 85 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-tc-400-85) — Transportation Engineering I, chapter 3, asked 2×
- [Transition curve 200 m radius, 60 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-tc-200) — Transportation Engineering I, chapter 3
- [Transition curve for 310 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-tc-310) — Transportation Engineering I, chapter 3