---
title: "Transition curve road 8 m, R=300 m, IP 3605"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-road-8m-300m-3605-50kmh"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-road-8m-300m-3605-50kmh
site: BCT Engineering
---

# Transition curve road 8 m, R=300 m, IP 3605

Surveying II · Chapter 6: Curves

- Asked in: 2071 Bhadra, 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-road-8m-300m-3605-50kmh
- 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 3605 m. A transition curve is to be used at each end of circular curve of such a length that the rate of gain of radial acceleration is 0.5 m/s³, when the speed is 50 km/h. Find out:
i) length of transition curve
ii) superelevation
iii) chainage of all tangent points and junction points
iv) Calculate the first two deflection angles for transition curve, and first two deflection angles for circular curve. Take peg interval = 10 m for transition curve and 20 m for circular curve.

## Short answer

Transition length 20 m, superelevation 0.524 m. Chainages: TS 3421.76, SC 3441.76, CS 3735.92, ST 3755.92 m. Transition deflections: 0°06′29″ at 8.24 m, 0°31′46″ at 18.24 m from TS; circular deflections from SC: 1°44′29″ for first chord, cumulative 3°39′05″ after 20 m chord.

## Answer

Data: carriageway width $B = 8$ m, $\Delta = 60^\circ$, $R = 300$ m, chainage of IP = 3605 m, $v = 50$ km/h = 13.889 m/s, rate of change of radial acceleration $C = 0.5$ m/s³.

#### i) Length of transition curve
$$
L_s = \frac{v^3}{C R} = \frac{13.889^3}{0.5\times 300} = 17.86\ \text{m}
$$

Adopt $L_s = 20$ 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:

$$
\frac{h}{B} = \frac{v^2}{gR} = \frac{13.889^2}{9.81\times 300} = 0.0655
\quad\Rightarrow\quad h = 8\times 0.0655 = 0.524\ \text{m}
$$

(Slope $1$ in $15.3$. 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
$$
\begin{aligned}
s &= \frac{L_s^2}{24R} = 0.056\ \text{m},\quad \theta_s = \frac{L_s}{2R} = 1.910^\circ\\
T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 183.24\ \text{m}\\
L_c &= \frac{\pi R}{180}(\Delta - 2\theta_s) = 294.16\ \text{m}
\end{aligned}
$$

| Point | Chainage (m) |
|---|---|
| TS = IP − T_s | 3421.76 |
| SC = TS + L_s | 3441.76 |
| CS = SC + L_c | 3735.92 |
| ST = CS + L_s | 3755.92 |

#### 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 |
|---|---|---|---|
| 3430.00 | 8.24 | 6.48′ | 0°06′29″ |
| 3440.00 | 18.24 | 31.76′ | 0°31′46″ |

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 | 18.24 | 1°44′29″ | 1°44′29″ |
| 3480.00 | 20.00 | 1°54′35″ | 3°39′05″ |

**Answer:** $L_s = 20$ m; $h = 0.524$ m; TS = 3421.76, SC = 3441.76, CS = 3735.92, ST = 3755.92 m; deflection angles as tabulated.

## Check yourself

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

1. For v = 13.889 m/s, C = 0.5 m/s^3 and R = 300 m, which expression gives Ls?
   A. v C R = 2083 m
   B. v^3 C/R = 0.0045 m
   C. v^2/(C R) = 1.29 m
   D. v^3/(C R) = 17.86 m
   Answer: D. Ls = v^3/(C R) = 2679.4/150 = 17.86 m, then adopted as 20 m. (Key point: Length of transition)

2. What is the super-elevation for v = 50 km/h on R = 300 m and width 8 m?
   A. 1.028 m
   B. 5.24 m
   C. 0.524 m
   D. 0.0655 m
   Answer: C. h = B v^2/(gR) = 8 x 0.0655 = 0.524 m. (Key point: Super-elevation)

3. How is the total tangent length Ts of a composite curve written?
   A. Ts = (R + s) tan(Delta/2) + Ls/2
   B. Ts = R tan(Delta/2) + Ls
   C. Ts = (R - s) tan(Delta/2) - Ls/2
   D. Ts = R tan(Delta/2)
   Answer: A. The circular arc is shifted inward by s, so the tangent is (R + s)tan(Delta/2) plus half the spiral length. (Key point: Shift and tangent length)

4. For the first peg at 3430 m, why is l = 8.24 m and not 10 m?
   A. The peg interval is 8.24 m
   B. TS is at 3421.76 m, so l is measured from TS to the peg
   C. l is measured from SC
   D. l is taken as half of the chord
   Answer: B. The deflection alpha = 573 l^2/(R Ls) uses l measured from TS: 3430 - 3421.76 = 8.24 m. (Key point: Deflection measurement)

## Close variants set in other exams

- [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=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×

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- [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=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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- [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
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