--- title: "Short note: Rankine's method of curve setting" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:rankine-method-circular-curve" source: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-method-circular-curve site: BCT Engineering --- # Short note: Rankine's method of curve setting Surveying II · Chapter 6: Curves - Asked in: 2075 Baisakh, 4 marks - Video: [Deflection Angle Method/Rankine's Method of Setting out of Simple Circular Highway Curve](https://www.youtube.com/watch?v=am3MYbQApz8) — CIVIL Solved - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/rankine-method-circular-curve - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Write a short note on setting out of circular curve by Rankine's method. ## Short answer Rankine's method sets out a circular curve using a theodolite at the first tangent point, sighting the intersection point with vernier zero. Each point is fixed by turning the cumulative deflection angle (sum of tangential angles δ = 1718.87c/R minutes for chord c) and measuring the chord from the previous point; the final cumulative angle must equal Δ/2 as a check. ## Answer **Rankine's method** (also called the deflection angle method or tangential angle method) sets out a circular curve using a theodolite and a tape. The curve is located by a series of short chords, and each point is fixed by the angle at the tangent point between the tangent and the chord to that point, together with the length of the chord. #### Principle The angle between the tangent and a chord equals half the central angle subtended by that chord. For a chord $c$ on a curve of radius $R$: $$ \delta = \frac{90^\circ\,c}{\pi R}\ \text{(degrees)} = \frac{1718.87\,c}{R}\ \text{(minutes)} $$ The deflection angle of any point from the first tangent is the sum of the tangential angles up to that point: $\Delta_n = \delta_1+\delta_2+\dots+\delta_n$. #### Data to compute 1. $R$, $T = R\tan(\Delta/2)$ and $L = \pi R\Delta/180$. 2. Chainages of the first tangent point T₁ = IP − T and last tangent point T₂ = T₁ + L. 3. First sub-chord $c_1$ (from T₁ to the next full peg), full chords $c$, and last sub-chord $c_n$ (to T₂). 4. Tangential angle of each chord and the cumulative deflection angles. #### Field procedure 1. Locate T₁ and T₂ by measuring $T$ back and forward from the IP along the tangents. 2. Set up the theodolite at T₁, level it, and with the vernier at zero sight the IP. 3. Set the first deflection angle $\delta_1$ and measure the sub-chord $c_1$ from T₁ with the tape; the arrow end gives point 1. 4. Set the cumulative angle $\Delta_2$; swing the tape from point 1 with length $c$ until it cuts the line of sight, fixing point 2. Repeat for all points. 5. The last cumulative angle must equal $\Delta/2$, and the last point must coincide with T₂ (check). If the error is small, adjust the last points. ```text IP /\ / \ T1 / \ T2 o' 'o \ . . / 1 2 3 4 ``` **Merits:** accurate, needs only one instrument setup and is suitable for long and moderate curves. **Limitation:** errors accumulate along the curve; obstacles on the curve require shifting the instrument. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the principle of Rankine's method? A. The tangent-chord angle equals half the central angle subtended by the chord B. It equals the full central angle C. It equals twice the central angle D. It equals the angle subtended at the IP Answer: A. This tangent-chord property gives delta = 1718.87 c/R minutes. (Key point: Principle) 2. How is the deflection angle of a peg from the first tangent found? A. The tangential angle of its own chord only B. Sum of the tangential angles of all chords up to that point C. Delta minus the previous angle D. Half of its chord length in degrees Answer: B. Delta_n = delta1 + delta2 + ... + delta_n. (Key point: Deflection angle of a point) 3. How is the work checked? A. Last angle equals Delta B. Sum of angles equals 90 degrees C. Last cumulative angle equals Delta/2 and the last point coincides with T2 D. Last chord equals normal chord Answer: C. Both conditions confirm the computed and set out curve agree. (Key point: Check) 4. What is a limitation of this method? A. It needs two theodolites B. It cannot be used for long curves C. It does not need any angle measurement D. Errors accumulate along the curve, and obstacles need instrument shifting Answer: D. Each point is fixed from the previous one, so errors accumulate. (Key point: Limitation) ## 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 - [Short note: Setting out vertical curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-setting-out-note) — Surveying II, chapter 6 - [Setting out right hand side composite curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aright-hand-composite-curve-setting-out) — Surveying II, chapter 6 - [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× - [Composite curve design, IP 1150 m, Rankine](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Acomposite-curve-rankine-ip-1150) — Surveying II, chapter 6, asked 2× - [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 ## Related in other subjects - [Setting out of a building](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Asetting-out-building) — Building Technology, chapter 2