---
title: "Transition curve road 8 m, R=330 m, 60°"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-road-8m-330m-deflection-60"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-road-8m-330m-deflection-60
site: BCT Engineering
---

# Transition curve road 8 m, R=330 m, 60°

Surveying II · Chapter 6: Curves

- Asked in: 2078 Baisakh
- 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-330m-deflection-60
- 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 330 m, the chainage of the intersection point being 3605.0 m. A transition curve of such a length that the rate of gain of radial acceleration is 0.5 m/s³, when the speed is 50 km/hr. Find out:
a) Length of the transition curve.
b) Superelevation
c) Chainage of all junction points
d) Layout the transition curve by deflection angle method taking peg interval as 5 m.

## Short answer

Length of transition curve = 16.24 m. Superelevation = 0.0596 (1 in 16.8), outer edge rise = 0.477 m. Chainages: T1 = 3406.34 m, S1 = 3422.57 m, S2 = 3751.91 m, T2 = 3768.15 m. Layout by deflection angles from T1 with 5 m pegs, first deflection 0°01′26″.

## Answer

Data: width of road $B = 8$ m, $\Delta = 60^\circ$, $R = 330$ m, chainage of IP = 3605.0 m, $\alpha = 0.5$ m/s$^3$, $V = 50$ km/h $= 13.889$ m/s, peg interval 5 m.

#### (a) Length of the transition curve

$$
L = \frac{V^3}{\alpha R} = \frac{13.889^3}{0.5\times 330} = 16.24\ \text{m}
$$

#### (b) Superelevation

For full balancing of the centrifugal force, $\tan\theta = e = \dfrac{V^2}{gR}$:

$$
e = \frac{13.889^2}{9.81\times 330} = 0.0596 \approx 1\text{ in }16.8
$$

Rise of the outer edge over the width of the road:

$$
h = e\times B = 0.0596\times 8 = 0.477\ \text{m}
$$

#### (c) Chainages of the junction points

$$
s = \frac{L^2}{24R} = 0.0333\ \text{m},\qquad T_s = (R+s)\tan 30^\circ + \frac{L}{2} = (330 + 0.0333)(0.57735) + 8.119 = 198.664\ \text{m}
$$

$$
\phi_s = \frac{L}{2R} = 1°24'35",\qquad \Delta_c = 60^\circ - 2\phi_s = 57°10'51",\qquad L_c = R\Delta_c = 329.338\ \text{m}
$$

| Point | Chainage (m) |
|---|---|
| T1 = IP − $T_s$ | 3406.336 |
| S1 = T1 + L | 3422.574 |
| S2 = S1 + $L_c$ | 3751.912 |
| T2 = S2 + L | 3768.149 |

#### (d) Layout of the transition curve by deflection angles (5 m pegs, from T1)

$$
\delta = \frac{573\,l^2}{RL}\ \text{minutes}
$$

| Peg chainage (m) | Distance $l$ from T1 (m) | Deflection $\delta=573\,l^2/(RL)$ | Offset $y=l^3/(6RL)$ (m) |
|---|---|---|---|
| 3410.00 | 3.66 | 0°01'26" | 0.002 |
| 3415.00 | 8.66 | 0°08'02" | 0.020 |
| 3420.00 | 13.66 | 0°19'58" | 0.079 |
| 3422.57 | 16.24 | 0°28'12" | 0.133 |

At the end of the curve $\delta = \phi_s/3 = 0°28'12"$. The second transition is set out in the same way from T2 towards S2 (measured backwards along the tangent), and the circular curve from S1 by Rankine's deflection angles $\delta = 1718.9\,c/R$ minutes.

**Answer: L = 16.24 m; superelevation = 0.477 m (1 in 16.8); T1 = 3406.34 m, S1 = 3422.57 m, S2 = 3751.91 m, T2 = 3768.15 m.**

## Check yourself

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

1. On what basis is the transition length L = V^3/(alpha R) found?
   A. From the permissible rate of gain of radial acceleration (0.5 m/s^3)
   B. From the road width and superelevation only
   C. From the deflection angle only
   D. From the chainage of the IP
   Answer: A. L = V^3/(alpha R) comes from limiting the rate of change of radial acceleration; here L = 16.24 m. (Key point: Length of transition)

2. In L = V^3/(alpha R), what must V be in?
   A. km/h as given
   B. m/s (50 km/h = 13.889 m/s)
   C. m/min
   D. km/s
   Answer: B. With alpha in m/s^3 the speed must be in m/s, so 50 km/h is converted to 13.889 m/s. (Key point: Units in L formula)

3. How is superelevation found for full balancing of centrifugal force?
   A. e = V/(gR), rise = 0.05 m
   B. e = gR/V^2, rise = 16.8 m
   C. e = V^2/(gR) = 13.889^2/(9.81 x 330) = 0.0596; rise = e x B = 0.477 m
   D. e = V^2/R, rise = e x R
   Answer: C. tan(theta) = V^2/(gR) = 0.0596, and the outer edge rise over the 8 m width is 0.0596 x 8 = 0.477 m. (Key point: Superelevation)

4. What is the total tangent length Ts used to locate T1?
   A. Ts = R tan(Delta/2) only
   B. Ts = (R + s) tan(Delta) + L
   C. Ts = R tan(Delta/2) - L/2
   D. Ts = (R + s) tan(Delta/2) + L/2
   Answer: D. Shift s moves the circle inward, so Ts = (R + s)tan(Delta/2) + L/2 = 198.66 m. (Key point: Tangent length)

## 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=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 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 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

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