--- title: "Derive tangential angle 1718.87 C/R" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:tangential-angle-171887-derivation" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tangential-angle-171887-derivation site: BCT Engineering --- # Derive tangential angle 1718.87 C/R Surveying II · Chapter 6: Curves - Asked in: 2065 Chaitra (old course), 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/tangential-angle-171887-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Derive the expression that the tangential angle for points on the circular curve is equal to $1718.87 \times \frac{C}{R}$ and also express about the deflection angles for laying out of circular curve. ## Short answer Central angle for chord C is C/R rad. Tangential (deflection) angle is half: C/(2R) rad. In minutes, δ = (C/(2R))×(180/π)×60 = 1718.87 C/R minutes. To lay out, set theodolite at T1, zero on IP, turn cumulative deflection angles Δi = sum of δ, fix points by chords; final deflection = Δ/2. ## Answer #### Derivation of tangential angle Consider a circular curve of radius $R$ with centre $O$ and a chord $T_1P_1$ of length $C$ starting at the tangent point T₁ (T₁ is the PC). ```text T1 ------------> tangent \ δ \ \__ P1 \ / O ``` For a short chord the arc and chord are nearly equal, so the angle at the centre is $$ \beta = \frac{C}{R}\ \text{radian} $$ The angle between a tangent and a chord is half the central angle subtended by the chord (tangent–chord theorem). So the tangential (deflection) angle is $$ \delta = \frac{\beta}{2} = \frac{C}{2R}\ \text{radian} $$ Converting radians to minutes (1 radian $= \dfrac{180}{\pi}\times 60 = 3437.747'$): $$ \delta = \frac{3437.747}{2}\cdot\frac{C}{R} = 1718.87\times\frac{C}{R}\ \text{minutes} $$ #### Deflection angles for laying out the curve - The **tangential angle** $\delta_i$ for a chord $c_i$ is $1718.87\,c_i/R$ minutes. For the first sub-chord $c_1$, a full chord $c$ and the last sub-chord $c_n$ the angles are $\delta_1 = 1718.87c_1/R$, $\delta = 1718.87c/R$, $\delta_n = 1718.87c_n/R$. - The **deflection angle** of a point is the angle at T₁ between the tangent and the chord to that point; it is the sum of the tangential angles up to that point: $\Delta_1=\delta_1$, $\Delta_2 = \delta_1 + \delta$, and so on. - The theodolite is set at T₁ with zero on the IP. At each point, turn the telescope by the cumulative deflection angle and measure the chord from the previous point. - Check: the last deflection angle is $\Delta/2$, where $\Delta$ is the deflection angle of the curve. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a short chord C on a circle of radius R, the central angle (radians) is taken as A. C/R B. 2C/R C. C^2/R D. C/(2R) Answer: A. Arc ≈ chord, and arc = R x angle, so the central angle is C/R radian. (Key point: Central angle of short chord) 2. Which theorem relates the tangential angle to the central angle? A. The angle between tangent and chord equals twice the central angle B. The angle between tangent and chord equals half the central angle C. The angle in a semicircle is 90° D. The angle between tangent and chord equals the central angle Answer: B. δ = β/2 = C/(2R) radian. (Key point: Tangent-chord theorem) 3. Where does 1718.87 come from? A. 180/π = 57.3 rounded B. 1718.87 = 30 x 57.3 C. (180/π) x 60 / 2 = 3437.747/2 D. π x 60 x 30 Answer: C. One radian = 3437.747 minutes; halving gives 1718.87 minutes per unit C/R. (Key point: Origin of 1718.87) 4. What is the deflection angle of the second point from T1? A. 2δ2 B. δ2 - δ1 C. δ2 alone D. δ1 + δ2 (sum of tangential angles) Answer: D. Deflection angles are cumulative sums of tangential angles measured from the tangent at T1. (Key point: Cumulative deflection) 5. The last deflection angle on the curve should equal A. Δ/2 B. 2Δ C. Δ D. Δ/3 Answer: A. The chord from T1 to T2 makes Δ/2 with the tangent. 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