---
title: "Composite curve design by Rankine's method"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:composite-curve-rankine-design-2079j"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/composite-curve-rankine-design-2079j
site: BCT Engineering
---

# Composite curve design by Rankine's method

Surveying II · Chapter 6: Curves

- Asked in: 2079 Jestha, 8 marks
- Video: [Transition Curve | Elements of Transition Curve | Advantage | Prashant YT | Surveying | BE/Diploma](https://www.youtube.com/watch?v=k4IiiB17yfA) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/composite-curve-rankine-design-2079j
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

For design a composite curve with the following data: Deflection angle = 60°, maximum speed of vehicle = 40 km/hr, centrifugal ratio = 1/8, rate of change of radial acceleration = 0.30 m/sec³, chainage of IP = 1 + 030 m. Calculate the setting out data of circular curve by Rankine's method. Take peg interval = 20 m.

## Short answer

R = 100.68 m, L = 45.42 m, s = 0.854 m, T_s = 81.33 m, L_c = 60.01 m. Chainages: T1=948.67 m, S1=994.09 m, S2=1054.10 m, T2=1099.52 m. Rankine deflections from S1: 1°40′55″ (5.91 m), 5°41′28″ (20 m) twice, 4°00′46″ (14.10 m), total 17°04′37″.

## Answer

Data: deflection angle $\Delta = 60^\circ$, design speed $V = 40$ km/h $= 11.111$ m/s, centrifugal ratio $P/W = 1/8$, rate of change of radial acceleration $\alpha = 0.30$ m/s$^3$, chainage of IP = 1+030 m (1030.00 m), peg interval 20 m.

#### 1. Radius, length of transition and shift

Centrifugal ratio $= \dfrac{V^2}{gR}$:

$$
R = \frac{V^2}{g\times(1/8)} = \frac{11.111^2}{9.81/8} = 100.68\ \text{m}
$$

$$
L = \frac{V^3}{\alpha R} = \frac{11.111^3}{0.30\times 100.68} = 45.42\ \text{m}
$$

$$
s = \frac{L^2}{24R} = \frac{45.42^2}{24\times 100.68} = 0.854\ \text{m}
$$

#### 2. Tangent length and spiral angle

$$
T_s = (R + s)\tan\frac{\Delta}{2} + \frac{L}{2} = (100.68 + 0.854)\tan 30^\circ + 22.71 = 81.33\ \text{m}
$$

$$
\phi_s = \frac{L}{2R}\ \text{rad} = 12°55'24", \qquad \Delta_c = \Delta - 2\phi_s = 34°09'13"
$$

$$
L_c = R\Delta_c = 100.68\times 0.59609 = 60.01\ \text{m}
$$

#### 3. Chainages

| Point | Chainage (m) |
|---|---|
| T1 (start of first transition) = IP − $T_s$ | 948.67 |
| S1 (end of transition, start of circular curve) = T1 + L | 994.09 |
| S2 (end of circular curve) = S1 + $L_c$ | 1054.10 |
| T2 (end of second transition) = S2 + L | 1099.52 |

#### 4. Setting out the circular curve by Rankine's method (peg interval 20 m)

Set the theodolite at S1, backsight T1, and set the tangent to the circular curve (turn the telescope to the common tangent, which makes $2\phi_s/3$ with the chord S1T1 for a cubic spiral). Then set out the circular curve from S1 by deflection angles from this tangent:

$$
\delta = \frac{1718.9\,C}{R}\ \text{minutes}
$$

The first chord is a sub-chord up to the next 20 m peg.

| Peg chainage (m) | Chord (m) | Deflection for chord $\delta=1718.9\,c/R$ | Cumulative deflection |
|---|---|---|---|
| 1000.00 | 5.91 | 1°40'55" | 1°40'55" |
| 1020.00 | 20.00 | 5°41'28" | 7°22'23" |
| 1040.00 | 20.00 | 5°41'28" | 13°03'51" |
| 1054.10 | 14.10 | 4°00'46" | 17°04'37" |

Check: the final cumulative deflection $= \Delta_c/2 = 17°04'36"$, and it equals 17°04'37".

**Answer: R = 100.68 m, L = 45.42 m, shift = 0.854 m, $T_s$ = 81.33 m, $L_c$ = 60.01 m; the Rankine deflection angles are tabulated above.**

## Check yourself

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

1. How is R found from centrifugal ratio 1/8 and V = 40 km/h (11.111 m/s)?
   A. R = V^2 / (g × 1/8) = 100.68 m
   B. R = V / (g × 1/8)
   C. R = g × (1/8) / V^2
   D. R = V^3 / (α × 1/8)
   Answer: A. Centrifugal ratio = V^2/(gR), so R = 11.111^2/(9.81/8) = 100.68 m. (Key point: Radius from centrifugal ratio)

2. Which expression gives the total tangent length Ts?
   A. Ts = R tan(Δ/2)
   B. Ts = (R + s) tan(Δ/2) + L/2
   C. Ts = (R - s) tan(Δ/2) + L
   D. Ts = R tan(Δ/2) + s
   Answer: B. The shifted circle has radius R plus the shift s, and the transition adds L/2, giving Ts = (R+s)tan(Δ/2) + L/2 = 81.33 m. (Key point: Tangent length of composite curve)

3. S1 is at chainage 994.09 m and pegs are at 20 m intervals. What is the first chord?
   A. 20.00 m
   B. 5.91 m, a sub-chord to the peg at 1000
   C. 14.10 m
   D. 45.42 m
   Answer: B. The first chord is a sub-chord to the next full peg: 1000.00 - 994.09 = 5.91 m. (Key point: First chord from S1)

4. What should the final cumulative deflection at S2 equal?
   A. Δ/2 = 30°
   B. φs = 12°55'24"
   C. Δc/2 = 17°04'36"
   D. 2φs/3
   Answer: C. The circular arc subtends Δc = Δ - 2φs at the centre, so the cumulative deflection at its end is Δc/2 (17°04'37" obtained). (Key point: Check on Rankine deflections)

## Close variants set in other exams

- [Composite curve design, IP 1150 m, Rankine](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acomposite-curve-rankine-ip-1150) — Surveying II, chapter 6, asked 2×

## Related in Surveying II

- [Rankine's table, IP 2+450 km, 30°, 3° curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Arankine-ip-2450-deflection-30-3deg) — 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
- [Rankine's deflection angle data, 145°, 5°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Arankine-deflection-145-5deg-lc10) — Surveying II, chapter 6
- [Curve by tangential angle, deflection 48°30'](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atangential-angle-curve-radius-r-48) — Surveying II, chapter 6
- [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×
- [Composite curve by deflection angle method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acomposite-curve-deflection-angle-method) — Surveying II, chapter 6, asked 5×

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