---
title: "Rankine's deflection angle data, 145°, 5°"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:rankine-deflection-145-5deg-lc10"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-deflection-145-5deg-lc10
site: BCT Engineering
---

# Rankine's deflection angle data, 145°, 5°

Surveying II · Chapter 6: Curves

- Asked in: 2072 Magh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-deflection-145-5deg-lc10
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Compute the data for setting out a simple circular curve by Rankine's deflection angle method from the following informations:
Angle of intersection = 145°0'
Chainage of point of intersection = 1580 m
Degree of curve = 5°
Least count of the theodolite = 10"
Peg interval = 30 m

## Short answer

Deflection angle Δ = 35°, degree 5°/30 m arc gives R = 1718.87/5 = 343.77 m. T = 108.39 m, L = 210.00 m. T1 = 1471.61 m, T2 = 1681.61 m. Deflection angles by δ = 1718.87c/R min, rounded to 10″; final cumulative = Δ/2 = 17°30′00″.

## Answer

Angle of intersection (interior) = 145°, so deflection angle $\Delta = 180^\circ-145^\circ = 35^\circ$. Degree of curve 5° for a 30 m arc.

#### Elements
$$
\begin{aligned}
R &= \frac{1718.87}{D} = \frac{1718.87}{5} = 343.77\ \text{m}\\
T &= R\tan\frac{\Delta}{2} = 343.77\tan 17.5^\circ = 108.39\ \text{m}\\
L &= \frac{\pi R\Delta}{180} = 210.00\ \text{m}\\
\text{Ch. of T}_1 &= 1580 - 108.39 = 1471.61\ \text{m}\\
\text{Ch. of T}_2 &= 1471.61 + 210.00 = 1681.61\ \text{m}
\end{aligned}
$$

#### Deflection angles
$\delta = 1718.87\,c/R$ minutes, rounded to the least count of 10″. Peg interval 30 m: first sub-chord $= 28.39$ m, last sub-chord $= 1.61$ m.

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 1471.61 | - | - | 0°00′00″ |
| 1500.00 | 28.39 | 2°22′00″ | 2°22′00″ |
| 1530.00 | 30.00 | 2°30′00″ | 4°52′00″ |
| 1560.00 | 30.00 | 2°30′00″ | 7°22′00″ |
| 1590.00 | 30.00 | 2°30′00″ | 9°52′00″ |
| 1620.00 | 30.00 | 2°30′00″ | 12°22′00″ |
| 1650.00 | 30.00 | 2°30′00″ | 14°52′00″ |
| 1680.00 | 30.00 | 2°30′00″ | 17°22′00″ |
| 1681.61 | 1.61 | 0°08′00″ | 17°30′00″ |

Check: final deflection $= \Delta/2 = 17^\circ30'00''$; the table gives 17°30′00″.

**Answer:** $R = 343.77$ m, $T = 108.39$ m, $L = 210.00$ m; T₁ = 1471.61 m, T₂ = 1681.61 m; deflection angles as tabulated.

## Check yourself

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

1. The angle of intersection is 145 degrees. What deflection angle is used for the curve elements?
   A. 35 degrees
   B. 72.5 degrees
   C. 145 degrees
   D. 17.5 degrees
   Answer: A. Deflection angle = 180 - 145 = 35 degrees; 17.5 degrees is Delta/2 used in the tangent length. (Key point: Deflection angle)

2. How is R found from the degree of curve D = 5 degrees for a 30 m arc?
   A. R = 30/D = 6 m
   B. R = 180/D = 36 m
   C. R = 1718.87 x D = 8594 m
   D. R = 1718.87/D = 343.77 m
   Answer: D. R = 180 x 30/(pi D) = 1718.87/D. (Key point: Radius from degree)

3. With T1 at 1471.61 m and pegs at multiples of 30 m, what is the first sub-chord?
   A. 71.61 m
   B. 30 m
   C. 1.61 m
   D. 28.39 m
   Answer: D. The first peg is at 1500, so the first sub-chord is 1500 - 1471.61 = 28.39 m; the last one is 1.61 m. (Key point: Sub-chords)

4. What should the final cumulative deflection angle equal as a check?
   A. 2 x Delta = 70 degrees
   B. Delta/2 = 17 degrees 30 minutes
   C. D = 5 degrees
   D. Delta = 35 degrees
   Answer: B. The deflection to T2 from the tangent at T1 is half the central angle, Delta/2. (Key point: Field 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
- [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
- [Rankine's method table, 150° intersection, 3°](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Arankine-simple-curve-150-3deg) — 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
- [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
- [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

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