---
title: "Rankine's data, 8° curve, 187.62 m long chord"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:simple-curve-8deg-18762-long-chord"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/simple-curve-8deg-18762-long-chord
site: BCT Engineering
---

# Rankine's data, 8° curve, 187.62 m long chord

Surveying II · Chapter 6: Curves

- Asked in: 2079 Chaitra, 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/simple-curve-8deg-18762-long-chord
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A simple circular 8° curve with fixed 30 m arc length is to be set with the 187.62 m long chord. The chainage of intersection point is 2552.00 m. The regular peg interval is 30 m. Compute the necessary data to set out the curve in field by Rankine's method.

## Short answer

R = 214.86 m, Δ = 51°46′33″ from long chord, T = 104.27 m, L = 194.16 m, chainages T1 = 2447.73 m, T2 = 2641.89 m. Deflection angles δ = 1718.87c/R tabulated; final cumulative 25°53′16″ ≈ Δ/2.

## Answer

Degree of curve $D = 8^\circ$ for a 30 m arc, long chord $= 187.62$ m, chainage of IP $= 2552.00$ m, peg interval 30 m.

#### Step 1: Radius and deflection angle
$$
R = \frac{1718.87}{8} = 214.86\ \text{m}
$$

$$
\sin\frac{\Delta}{2} = \frac{\text{LC}}{2R} = \frac{187.62}{2\times 214.86} = 0.43661
\ \Rightarrow\ \frac{\Delta}{2} = 25.8879^\circ,\ \Delta = 51.7759^\circ = 51°46′33″
$$

#### Step 2: Tangent length, curve length, chainages
$$
\begin{aligned}
T &= R\tan\frac{\Delta}{2} = 214.86\tan 25.8879^\circ = 104.27\ \text{m}\\
L &= \frac{\pi R\Delta}{180} = 194.16\ \text{m}\\
\text{Ch. T}_1 &= 2552.00 - 104.27 = 2447.73\ \text{m}\\
\text{Ch. T}_2 &= 2447.73 + 194.16 = 2641.89\ \text{m}
\end{aligned}
$$

#### Step 3: Rankine's deflection angles
$\delta = 1718.87\,c/R$ minutes. First sub-chord $= 12.27$ m, last sub-chord $= 1.89$ m.

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 2447.73 | - | - | 0°00′00″ |
| 2460.00 | 12.27 | 1°38′11″ | 1°38′11″ |
| 2490.00 | 30.00 | 4°00′00″ | 5°38′11″ |
| 2520.00 | 30.00 | 4°00′00″ | 9°38′11″ |
| 2550.00 | 30.00 | 4°00′00″ | 13°38′11″ |
| 2580.00 | 30.00 | 4°00′00″ | 17°38′11″ |
| 2610.00 | 30.00 | 4°00′00″ | 21°38′11″ |
| 2640.00 | 30.00 | 4°00′00″ | 25°38′11″ |
| 2641.89 | 1.89 | 0°15′05″ | 25°53′16″ |

Check: final deflection $= \Delta/2 = 25.8879^\circ$ = 25°53′17″; computed 25°53′16″. Field check: the long chord T₁T₂ should measure 187.62 m.

**Answer:** $R = 214.86$ m, $\Delta = 51°46′33″$, $T = 104.27$ m, $L = 194.16$ m, T₁ = 2447.73 m, T₂ = 2641.89 m.

## Check yourself

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

1. For an 8° curve with 30 m arc, R is
   A. 1718.87 m
   B. 143.24 m
   C. 429.72 m
   D. 214.86 m
   Answer: D. R = 1718.87/8 = 214.86 m. (Key point: Radius)

2. How is Δ found from the long chord?
   A. sin(Δ/2) = LC/(2R)
   B. Δ = LC/R radian
   C. tan(Δ/2) = LC/R
   D. cos(Δ/2) = LC/R
   Answer: A. Long chord = 2R sin(Δ/2), giving Δ/2 = 25.888° and Δ = 51°46′33″. (Key point: Deflection angle from long chord)

3. T1 is at 2447.73 m and pegs are 30 m apart, so the first sub-chord is
   A. 2.27 m
   B. 12.27 m
   C. 17.73 m
   D. 30 m
   Answer: B. Next full peg is 2460; 2460 - 2447.73 = 12.27 m. (Key point: First sub-chord)

4. The final deflection angle and field check are
   A. Δ = 51°46′ and tangent T = 187.62 m
   B. Δ/2 and sum of chords = LC
   C. Δ/2 = 25°53′17″ and long chord T1T2 = 187.62 m
   D. 90° and L = LC
   Answer: C. Last angle should be Δ/2, and the measured T1T2 should equal the given long chord. (Key point: Final check)

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