---
title: "Transition curve, road 8 m, R=300 m, 60°"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-road-8m-300m-60-70kmph"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-road-8m-300m-60-70kmph
site: BCT Engineering
---

# Transition curve, road 8 m, R=300 m, 60°

Surveying II · Chapter 6: Curves

- Asked in: 2072 Asoj, 10 marks
- Video: [How to setout Combined road curve #civilengineering](https://www.youtube.com/watch?v=kQhbW1i1V_o) — Engr Suliman Khan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-road-8m-300m-60-70kmph
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A road 8 m wide is to deflect through an angle of 60° with the center line radius of 300 m, the chainage of intersection point being (3+605) Km. A transition curve is to be used at each end of the circular curve of such a length that the rate of change of radial acceleration is 50 cm/sec³, when the speed of design vehicle is 70 Kmph, find out:
i) Length of transition curve
ii) Super elevation
iii) Chainage of tangent points and junction points
iv) Deflection angles for first two points of transition curves and circular curve.
Take peg interval for transition curve = 10 m and circular curve = 20 m.

## Short answer

Ls = 50 m (from v³/(CR)), superelevation h = 1.028 m. Chainages: TS 3406.59, SC 3456.59, CS 3720.75, ST 3770.75 m. Transition deflections: 3410 m 0°00′27″, 3420 m 0°06′52″; circular curve from SC: 3460 m cum. 0°19′31″, 3480 m cum. 2°14′06″.

## Answer

Data: carriageway width $B = 8$ m, $\Delta = 60^\circ$, $R = 300$ m, chainage of IP = 3605 m, $v = 70$ km/h = 19.444 m/s, rate of change of radial acceleration $C = 0.5$ m/s³.

#### i) Length of transition curve
$$
L_s = \frac{v^3}{C R} = \frac{19.444^3}{0.5\times 300} = 49.01\ \text{m}
$$

Adopt $L_s = 50$ m (rounded up to a multiple of 5 m).

#### ii) Super-elevation
Equilibrium super-elevation for the full speed, taking the road width $B$ as the distance between the outer and inner edges:

$$
\frac{h}{B} = \frac{v^2}{gR} = \frac{19.444^2}{9.81\times 300} = 0.1285
\quad\Rightarrow\quad h = 8\times 0.1285 = 1.028\ \text{m}
$$

(Slope $1$ in $7.8$. In practice the value is limited to the maximum permitted by the road code, e.g. 7 %, and the balance is taken by side friction.) The super-elevation is attained gradually over $L_s$ so that the outer edge rises uniformly from TS to SC.

#### iii) Chainages of tangent and junction points
$$
\begin{aligned}
s &= \frac{L_s^2}{24R} = 0.347\ \text{m},\quad \theta_s = \frac{L_s}{2R} = 4.775^\circ\\
T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 198.41\ \text{m}\\
L_c &= \frac{\pi R}{180}(\Delta - 2\theta_s) = 264.16\ \text{m}
\end{aligned}
$$

| Point | Chainage (m) |
|---|---|
| TS = IP − T_s | 3406.59 |
| SC = TS + L_s | 3456.59 |
| CS = SC + L_c | 3720.75 |
| ST = CS + L_s | 3770.75 |

#### iv) Deflection angles for the first two points
Transition curve from TS, $\alpha = 573\,l^2/(R L_s)$ minutes (points at multiples of 10 m):

| Point (chainage, m) | l (m) | α (min) | α |
|---|---|---|---|
| 3410.00 | 3.41 | 0.44′ | 0°00′27″ |
| 3420.00 | 13.41 | 6.86′ | 0°06′52″ |

Circular curve from SC, $\delta = 1718.87\,c/R$ minutes (points at multiples of 20 m):

| Point (chainage, m) | Chord c (m) | δ | Cumulative |
|---|---|---|---|
| 3460.00 | 3.41 | 0°19′31″ | 0°19′31″ |
| 3480.00 | 20.00 | 1°54′35″ | 2°14′06″ |

**Answer:** $L_s = 50$ m; $h = 1.028$ m; TS = 3406.59, SC = 3456.59, CS = 3720.75, ST = 3770.75 m; deflection angles as tabulated.

## Check yourself

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

1. Which formula gives the length of transition curve from the rate of change of radial acceleration C?
   A. Ls = v/(C R)
   B. Ls = v^3/(C R)
   C. Ls = v^2/(C R)
   D. Ls = C R/v^3
   Answer: B. Radial acceleration builds from 0 to v^2/R over time Ls/v at rate C, giving Ls = v^3/(C R). (Key point: Length of transition)

2. The computed Ls is 49.01 m. What is done with it?
   A. Used as 49.01 m exactly
   B. Adopted as 50 m, rounded up to a multiple of 5 m
   C. Doubled to 98 m
   D. Rounded down to 45 m
   Answer: B. Ls is rounded up to a convenient multiple of 5 m so the rate of change of radial acceleration does not exceed the limit. (Key point: Rounding Ls)

3. How is the equilibrium super-elevation h found for the full design speed?
   A. h = B g R/v^2
   B. h = B v^2/(g R)
   C. h = B v/(g R)
   D. h = v^2/(B g R)
   Answer: B. The ratio h/B equals v^2/(gR), the centrifugal ratio, so h = B v^2/(gR). (Key point: Super-elevation)

4. Which sequence correctly gives the chainages of the junction points?
   A. TS = IP - Ls, SC = TS + Ts, CS = SC + Lc, ST = CS + Ts
   B. TS = IP - Ts, SC = TS + Lc, CS = SC + Ls, ST = CS + Lc
   C. TS = IP + Ts, SC = TS - Ls
   D. TS = IP - Ts, SC = TS + Ls, CS = SC + Lc, ST = CS + Ls
   Answer: D. The composite curve starts at TS = IP - Ts and then adds the spiral, circular arc and second spiral lengths in order. (Key point: Chainage of points)

## Close variants set in other exams

- [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×

## Related in Surveying II

- [Transition curve road 8 m, R=330 m, 60°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atransition-road-8m-330m-deflection-60) — Surveying II, chapter 6, asked 2×
- [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
- [Combined curve table, R=300 m, Ls=60 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acombined-curve-transition-60m-r300) — 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
- [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

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