---
title: "Rankine's method table, 150° intersection, 3°"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:rankine-simple-curve-150-3deg"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-simple-curve-150-3deg
site: BCT Engineering
---

# Rankine's method table, 150° intersection, 3°

Surveying II · Chapter 6: Curves

- Asked in: 2073 Magh, 6 marks
- Video: [Lecture 03 Rankine's Method Numerical Solving](https://www.youtube.com/watch?v=vuXu115TQzk) — Sachin Salve
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-simple-curve-150-3deg
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Compute and tabulate the data required for setting out a simple circular curve by Rankine's Method from the following information:
Angle of intersection = 150°00'
Chainage of point of intersection = 1585.00 m
Degree of curve = 3°
Peg interval = 20 m, Normal chord = 20 m

## Short answer

Deflection angle Δ = 30°, degree of curve 3° for 20 m chord gives radius R = 180×20/(π×3) = 381.97 m. Tangent T = 102.35 m, curve length L = 200.00 m. T₁ chainage 1482.65 m, T₂ 1682.65 m. Deflection angles are computed per chord using δ = 1718.87c/R minutes, with final cumulative angle 15°00′00″.

## Answer

Angle of intersection (interior) = 150°, so deflection angle $\Delta = 180^\circ - 150^\circ = 30^\circ$. The degree of curve is taken for the normal chord (arc) of 20 m.

#### Elements
$$
\begin{aligned}
R &= \frac{180\,C}{\pi D} = \frac{180\times 20}{\pi\times 3} = 381.97\ \text{m}\\
T &= R\tan\frac{\Delta}{2} = 381.97\tan 15^\circ = 102.35\ \text{m}\\
L &= \frac{\pi R\Delta}{180} = 200.00\ \text{m}
\end{aligned}
$$

$$
\begin{aligned}
\text{Chainage of T}_1 &= 1585.00 - 102.35 = 1482.65\ \text{m}\\
\text{Chainage of T}_2 &= 1482.65 + 200.00 = 1682.65\ \text{m}
\end{aligned}
$$

#### Rankine's deflection angles
$\delta = 1718.87\,c/R$ minutes for a chord $c$. First sub-chord $c_1 = 17.35$ m (to the next 20 m peg), last sub-chord $c_n = 2.65$ m.

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 1482.65 | - | - | 0°00′00″ |
| 1500.00 | 17.35 | 1°18′04″ | 1°18′04″ |
| 1520.00 | 20.00 | 1°30′00″ | 2°48′04″ |
| 1540.00 | 20.00 | 1°30′00″ | 4°18′04″ |
| 1560.00 | 20.00 | 1°30′00″ | 5°48′04″ |
| 1580.00 | 20.00 | 1°30′00″ | 7°18′04″ |
| 1600.00 | 20.00 | 1°30′00″ | 8°48′04″ |
| 1620.00 | 20.00 | 1°30′00″ | 10°18′04″ |
| 1640.00 | 20.00 | 1°30′00″ | 11°48′04″ |
| 1660.00 | 20.00 | 1°30′00″ | 13°18′04″ |
| 1680.00 | 20.00 | 1°30′00″ | 14°48′04″ |
| 1682.65 | 2.65 | 0°11′56″ | 15°00′00″ |

Check: final deflection $= \Delta/2 = 15^\circ00'00''$ (computed 15°00′00″).

**Answer:** $R = 381.97$ m, $T = 102.35$ m, $L = 200.00$ m, T₁ at 1482.65 m, T₂ at 1682.65 m; deflection angles as tabulated.

## Check yourself

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

1. Interior angle 150 deg means deflection angle
   A. 30 deg
   B. 150 deg
   C. 15 deg
   D. 60 deg
   Answer: A. Delta = 180 - 150 = 30 deg. (Key point: Deflection angle)

2. With a 20 m arc, R is
   A. 1718.87/3 = 572.96 m
   B. 180 x 20/(pi x 3) = 381.97 m
   C. 20 x 3 = 60 m
   D. 3 x 180/pi
   Answer: B. For a 20 m arc, R = 180C/(pi D). (Key point: Radius)

3. What are the first and last sub-chords?
   A. 20 m and 20 m
   B. 2.65 m and 17.35 m
   C. 17.35 m (to 1500) and 2.65 m (from 1680)
   D. 10 m and 10 m
   Answer: C. T1 = 1482.65 so the first peg is 1500, and T2 = 1682.65. (Key point: Sub-chords)

4. Last cumulative deflection must be
   A. 30 deg
   B. 7.5 deg
   C. 90 deg
   D. Delta/2 = 15 deg
   Answer: D. The final chord deflection equals half the deflection angle. (Key point: Check)

## Related in Surveying II

- [Setting out simple curve by Rankine's method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asimple-curve-rankine-method-sketch) — 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
- [Rankine's data, 8° curve, 187.62 m long chord](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asimple-curve-8deg-18762-long-chord) — Surveying II, chapter 6
- [Simple curve data, roads intersecting at 150°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asimple-curve-two-roads-150-4deg) — Surveying II, chapter 6
- [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
- [Simple circular curve by deflection angle method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asimple-curve-deflection-angle-method-sketch) — Surveying II, chapter 6

## Related in other subjects

- [Simple circular curve elements and chainages](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-64s-simple-curve) — Transportation Engineering I, chapter 3
- [Simple Rankine plant 2 MPa / 20 kPa: efficiency, power](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Arankine-2mpa-20kpa-50kgs) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Vertical curve chainage and elevations, R 8000 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-69bh-vc-chainage) — Transportation Engineering I, chapter 3