---
title: "Composite curve design, IP 1150 m, Rankine"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:composite-curve-rankine-ip-1150"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/composite-curve-rankine-ip-1150
site: BCT Engineering
---

# Composite curve design, IP 1150 m, Rankine

Surveying II · Chapter 6: Curves

- Asked in: 2068 Bhadra, 9 marks
- 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-ip-1150
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

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 = 1150 m. Also calculate the setting out data of circular curve by Rankine's method. Take peg interval = 20 m.

## Short answer

R=100.68 m, Ls=50 m, Ts=83.725 m, Lc=55.432 m. Chainages: TS 1066.275 m, SC 1116.275 m, CS 1171.707 m, ST 1221.707 m. Rankine deflection angles from SC: first sub-chord 3.725 m, δ=1°03′36″; then two 20 m chords, δ=5°41′27″ each; final sub-chord 11.707 m, δ=3°19′52″; cumulative deflection at CS=15°46′22″.

## Answer

#### Given data
$\Delta = 60^\circ$, $v_{max} = 40$ km/h $= 11.111$ m/s, centrifugal ratio $v^2/(gR) = 1/8$, $C = 0.30$ m/s³, chainage of IP $= 1150$ m, peg interval 20 m.

#### Radius of the circular curve
$$
\frac{v^2}{gR} = \frac{1}{8} \Rightarrow R = \frac{8v^2}{g} = \frac{8\times 11.111^2}{9.81} = 100.68\ \text{m}
$$

Adopt $R = 100.68$ m. (Rounding down to 100 m would make the centrifugal ratio slightly greater than the permitted 1/8.)

#### Length of transition
$$
L_s = \frac{v^3}{CR} = \frac{11.111^3}{0.30\times 100.68} = 45.42\ \text{m}
$$

Adopt $L_s = 50$ m.

#### Elements of the composite curve
$$
\begin{aligned}
s &= \frac{L_s^2}{24R} = \frac{50^2}{24\times 100.68} = 1.035\ \text{m}\\
\theta_s &= \frac{L_s}{2R} = \frac{50}{2\times 100.68} = 0.24831\ \text{rad} = 14.227^\circ\\
T_s &= (R+s)\tan 30^\circ + \frac{L_s}{2} = 101.715\times 0.57735 + 25 = 83.725\ \text{m}\\
L_c &= R(\Delta - 2\theta_s) = 100.68\left(1.04720 - 0.49662\right) = 55.432\ \text{m}
\end{aligned}
$$

| Point | Chainage (m) |
|---|---|
| TS = 1150 − 83.725 | 1066.275 |
| SC = TS + 50 | 1116.275 |
| CS = SC + 55.432 | 1171.707 |
| ST = CS + 50 | 1221.707 |

#### Setting out the circular curve by Rankine's method (theodolite at SC)
$\delta = 1718.87\,c/R$ minutes for chord $c$; the first sub-chord runs to the next 20 m peg and the last sub-chord ends at CS.

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 1116.275 | - | - | 0°00′00″ |
| 1120.000 | 3.725 | 1°03′36″ | 1°03′36″ |
| 1140.000 | 20.000 | 5°41′27″ | 6°45′03″ |
| 1160.000 | 20.000 | 5°41′27″ | 12°26′30″ |
| 1171.707 | 11.707 | 3°19′52″ | 15°46′22″ |

Check: last cumulative deflection $= (\Delta - 2\theta_s)/2 = (60^\circ - 28.454^\circ)/2 = 15.773^\circ$ = 15°46′22″.

For reference, the transition (set out from TS), $\alpha = 573l^2/(RL_s)$ minutes:

| Point (chainage, m) | l from TS (m) | α (min) | α (° ′ ″) |
|---|---|---|---|
| 1066.275 | 0.000 | 0.00′ | 0°00′00″ |
| 1080.000 | 13.725 | 21.44′ | 0°21′27″ |
| 1100.000 | 33.725 | 129.46′ | 2°09′28″ |
| 1116.275 | 50.000 | 284.56′ | 4°44′34″ |

(At SC, $\theta_s/3 = 4.742^\circ$ = 4°44′33″; the 1″ difference comes from rounding 572.96 to 573.)

**Answer:** $R = 100.68$ m, $L_s = 50$ m, $T_s = 83.725$ m, $L_c = 55.432$ m; TS = 1066.275 m, SC = 1116.275 m, CS = 1171.707 m, ST = 1221.707 m; Rankine deflection angles as tabulated.

## Check yourself

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

1. How is R found from the centrifugal ratio 1/8?
   A. R = 8 v^2/g
   B. R = v^2/(8g)
   C. R = g/(8 v^2)
   D. R = 8 g/v^2
   Answer: A. v^2/(gR) = 1/8 gives R = 8v^2/g = 100.68 m, adopted as 100 m. (Key point: Radius)

2. Which combination gives Ls?
   A. Ls = v^3/(C R) = 45.72 m, adopted 50 m
   B. Ls = C R/v^3
   C. Ls = v^2/(C R)
   D. Ls = R theta
   Answer: A. v = 11.111 m/s, C = 0.3 and R = 100 give 45.72 m, rounded up to 50 m. (Key point: Transition length)

3. How is Lc found?
   A. Lc = Ls/2
   B. Lc = R(Delta + 2 thetas)
   C. Lc = R Delta
   D. Lc = R(Delta - 2 thetas) = 54.72 m
   Answer: D. thetas = 0.25 rad, so Lc = 100(1.0472 - 0.5) = 54.72 m. (Key point: Circular arc length)

4. What should the last cumulative deflection of the circular curve (theodolite at SC) equal?
   A. (Delta - 2 thetas)/2 = 15 deg 40' 34''
   B. thetas
   C. Delta - 2 thetas
   D. Delta/2 = 30 degrees
   Answer: A. The circular arc has central angle Delta - 2 thetas, so the final deflection is half that. (Key point: Rankine check)

## Close variants set in other exams

- [Composite curve design by Rankine's method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acomposite-curve-rankine-design-2079j) — 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
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- [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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