---
title: "Rankine's table, IP 2+450 km, 30°, 3° curve"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:rankine-ip-2450-deflection-30-3deg"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-ip-2450-deflection-30-3deg
site: BCT Engineering
---

# Rankine's table, IP 2+450 km, 30°, 3° curve

Surveying II · Chapter 6: Curves

- Asked in: 2081 Chaitra, 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-ip-2450-deflection-30-3deg
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

The chainage of IP is 2+450 km. Deflection angle is 30°. The curve is 3° with the fixed arc length of 25 m. The peg interval is 30 m. Calculate the setting out table by Rankine's deflection angle method.

## Short answer

Using D = 3° for a 25 m arc, R = 477.46 m, T = 127.94 m, L = 250.00 m. Chainages: T1 = 2322.06 m, T2 = 2572.06 m. Deflection angles δ = 1718.87c/R are tabulated; final check is Δ/2 = 15°00′00″.

## Answer

Chainage of IP $= 2+450$ km $= 2450$ m, $\Delta = 30^\circ$, degree of curve $D = 3^\circ$ for a fixed arc of 25 m, peg interval 30 m.

#### Elements
$$
\begin{aligned}
R &= \frac{180\,C}{\pi D} = \frac{180\times 25}{\pi\times 3} = 477.46\ \text{m}\\
T &= R\tan 15^\circ = 127.94\ \text{m}\\
L &= \frac{\pi R\Delta}{180} = 250.00\ \text{m}\\
\text{Ch. T}_1 &= 2450 - 127.94 = 2322.06\ \text{m}\\
\text{Ch. T}_2 &= 2322.06 + 250.00 = 2572.06\ \text{m}
\end{aligned}
$$

#### Setting-out table (theodolite at T₁)
$\delta = 1718.87\,c/R$ minutes:

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 2322.06 | - | - | 0°00′00″ |
| 2340.00 | 17.94 | 1°04′34″ | 1°04′34″ |
| 2370.00 | 30.00 | 1°48′00″ | 2°52′34″ |
| 2400.00 | 30.00 | 1°48′00″ | 4°40′34″ |
| 2430.00 | 30.00 | 1°48′00″ | 6°28′34″ |
| 2460.00 | 30.00 | 1°48′00″ | 8°16′34″ |
| 2490.00 | 30.00 | 1°48′00″ | 10°04′34″ |
| 2520.00 | 30.00 | 1°48′00″ | 11°52′34″ |
| 2550.00 | 30.00 | 1°48′00″ | 13°40′34″ |
| 2572.06 | 22.06 | 1°19′26″ | 15°00′00″ |

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

**Answer:** $R = 477.46$ m, $T = 127.94$ m, $L = 250.00$ m, T₁ = 2322.06 m, T₂ = 2572.06 m; deflection angles as tabulated.

## Check yourself

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

1. For a 3° curve with 25 m arc, R is
   A. 286.48 m
   B. 1718.87/3 = 572.96 m
   C. 477.46 m
   D. 572.96 m
   Answer: C. R = 180 x 25/(π x 3) = 477.46 m; the 25 m arc is used, not 30 m. (Key point: Radius from degree)

2. With R = 477.46 m and Δ = 30°, T and L are
   A. 95.5 m and 125 m
   B. 477.46 m and 250 m
   C. 250 m and 127.94 m
   D. 127.94 m and 250.00 m
   Answer: D. T = R tan 15° = 127.94 m; L = πRΔ/180 = 250.00 m. (Key point: Tangent and curve length)

3. The chainage of T1 is
   A. 2322.06 m
   B. 2572.06 m
   C. 2450 m
   D. 2577.94 m
   Answer: A. T1 = IP - T = 2450 - 127.94 = 2322.06 m. (Key point: Chainage of T1)

4. The final cumulative deflection angle should be
   A. 30°
   B. 15°
   C. 7.5°
   D. 3°
   Answer: B. Δ/2 = 15°. (Key point: Final deflection check)

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