--- title: "Degree of curve, tangential angle formula" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:degree-of-curve-tangential-angle-derivation" source: https://www.bctengineer.com/q/semester_4/surveying_ii/degree-of-curve-tangential-angle-derivation site: BCT Engineering --- # Degree of curve, tangential angle formula Surveying II · Chapter 6: Curves - Asked in: 2071 Magh, 8 marks - Video: [Advanced Geomatics: Degree Of Curvature](https://www.youtube.com/watch?v=vki6Jf1HPG4) — Simmy Sigma - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/degree-of-curve-tangential-angle-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Explain degree of curve with neat sketch. Derive the formula of tangential angle, $\alpha = \frac{90^\circ C}{\pi R}$ and deflection angle $\Delta_n = \delta_1 + \delta_2 + \delta_3 + \dots + \delta_n$. ## Short answer Degree of curve D is the central angle subtended by a standard chord C (e.g. 30 m), giving R = 1718.87/D m. The tangential angle α = 90°C/(πR) is half the central angle of chord C, from the tangent-chord theorem. The deflection angle Δn is the sum of tangential angles of successive chords: Δn = δ1+δ2+...+δn. ## Answer #### Degree of curve The **degree of curve** $D$ is the central angle subtended by a standard chord (arc) of fixed length $C$ (30 m in the metric system, 100 ft in the foot system). A sharper curve has a larger degree and a smaller radius. ```text O /|\ / D \ / | \ A---+---B chord/arc C = 30 m ``` Relation to the radius: central angle in radians $= C/R$, so $$ D = \frac{180^\circ}{\pi}\cdot\frac{C}{R}\quad\Rightarrow\quad R = \frac{180\,C}{\pi D} = \frac{1718.87}{D}\ \ (C = 30\ \text{m}) $$ #### Derivation of tangential angle $\alpha = \dfrac{90^\circ C}{\pi R}$ ```text T1 ---------------- IP \ a \___ P1 \ \ C \ \ O (centre) ``` Let T₁P₁ be a chord of length $C$ at the start of the curve. The tangent at T₁ is perpendicular to the radius $OT_1$. The chord is short, so it is nearly equal to its arc, and the central angle subtended by it is $$ \angle T_1OP_1 = \frac{C}{R}\ \text{radians} = \frac{180}{\pi}\cdot\frac{C}{R}\ \text{degrees} $$ The angle between a tangent and a chord equals half the angle subtended by the chord at the centre (tangent–chord theorem). Hence the tangential angle $$ \alpha = \frac{1}{2}\cdot\frac{180^\circ C}{\pi R} = \frac{90^\circ C}{\pi R} $$ In minutes, $\alpha = 1718.87\,C/R$. #### Derivation of the deflection angle $\Delta_n$ Deflection angle of the $n$th point is the angle at T₁ between the initial tangent and the chord T₁P$_n$. For successive chords with central angles $\beta_1, \beta_2, \dots, \beta_n$, the arc T₁P$_n$ subtends $\beta_1+\beta_2+\dots+\beta_n$ at the centre. By the same theorem $$ \Delta_n = \frac{1}{2}(\beta_1+\beta_2+\dots+\beta_n) = \delta_1+\delta_2+\delta_3+\dots+\delta_n $$ where $\delta_i = \beta_i/2 = 1718.87\,c_i/R$ minutes. Thus the deflection angle of any point is the sum of the tangential angles of all chords up to it. For the whole curve $\Delta_{last} = \Delta/2$, which is used as the check in the field. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the degree of curve D? A. The deflection angle of the whole curve B. The angle between the two tangents C. The central angle subtended by a standard chord/arc of fixed length (30 m) D. The angle at the IP Answer: C. D is the angle at the centre subtended by a standard length (30 m metric), measuring sharpness. (Key point: Definition) 2. How are D and R related for a 30 m standard length? A. R = 1718.87 x D, so a larger D means a larger radius B. R = 1718.87/D, so a larger D means a smaller radius C. R is independent of D D. R = 30 D Answer: B. D = (180/pi)(C/R) so R = 1718.87/D; sharper curves have larger D and smaller R. (Key point: Radius relation) 3. Which principle gives the tangential angle alpha = 90 C/(pi R)? A. The radius bisects the tangent B. The angle between tangent and chord equals the central angle C. The tangent equals the chord D. The angle between tangent and chord equals half the central angle subtended by the chord Answer: D. Central angle = (180/pi)(C/R); half of it is the tangent-chord angle 90C/(pi R). (Key point: Tangent-chord theorem) 4. Why is the deflection angle of the nth point the sum of the tangential angles delta1 + ... + deltan? A. Tangential angles are always equal B. Each chord is equal in length C. The theodolite adds angles automatically D. The arc to point n subtends the sum of central angles at O, and the deflection is half of it Answer: D. Delta_n = half(beta1 + ... + betan) = delta1 + ... + deltan. 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