--- title: "Degree of curve, elements of composite curve" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:degree-of-curve-composite-elements" source: https://www.bctengineer.com/q/semester_4/surveying_ii/degree-of-curve-composite-elements site: BCT Engineering --- # Degree of curve, elements of composite curve Surveying II · Chapter 6: Curves - Asked in: 2065 Chaitra (old course), 7 marks - Video: [How to design a Combined Curve?|Design parameters for combined curve.](https://www.youtube.com/watch?v=Uv4QGr7brWw) — Civillogy (Study of Civil Engineering) - 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-composite-elements - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question What is degree of curve? Find the elements of a composite curve including sketch. ## Short answer Degree of curve D is the central angle subtended by a 30 m arc; R = 1718.87/D. A composite curve has a circular arc between two transition spirals. Key elements: spiral angle θs = Ls/(2R), shift s = Ls²/(24R), tangent length Ts = (R+s)tan(Δ/2)+Ls/2, circular length Lc = πR(Δ-2θs)/180. ## Answer #### Degree of curve The degree of curve $D$ is the central angle subtended by a standard arc of length $C$ (30 m). $R = \dfrac{180\,C}{\pi D} = \dfrac{1718.87}{D}$. A larger $D$ gives a sharper curve. #### Elements of a composite curve A composite curve is a circular arc of radius $R$ with a transition (spiral) curve of length $L_s$ at each end. ```text V (IP) / \ /Δ \ / \ TS / \ ST o___.___.___o SC CS \ (R) / \ | / \ O / ``` | Element | Formula | |---|---| | Length of transition | $L_s = v^3/(CR)$ | | Spiral angle | $\theta_s = L_s/(2R)$ rad (= $90L_s/\pi R$ degrees) | | Shift of the circular curve | $s = L_s^2/(24R)$ | | Tangent length (IP to TS or ST) | $T_s = (R+s)\tan\dfrac{\Delta}{2} + \dfrac{L_s}{2}$ | | Length of circular arc | $L_c = \dfrac{\pi R}{180}(\Delta - 2\theta_s)$ | | Total length | $2L_s + L_c$ | | Deflection angle at SC | $\alpha_s = \theta_s/3$ | | Deflection angle to a point at distance $l$ | $\alpha = 573\,l^2/(RL_s)$ minutes | | Offsets of SC | $x \approx L_s$, $y = L_s^2/(6R)$ | | External distance | $E_s = (R+s)\sec\dfrac{\Delta}{2} - R$ | Chainages: TS = IP − $T_s$, SC = TS + $L_s$, CS = SC + $L_c$, ST = CS + $L_s$. The circular arc is shifted inwards by $s$ from its original position, and each transition is placed so that about half of its length lies on each side of the original tangent point (hence the $L_s/2$ in $T_s$). ## 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. Angle between the two tangents B. Angle at the tangent point between tangent and long chord C. Central angle subtended by a standard arc length (30 m) D. Deflection angle of the whole curve Answer: C. D is the central angle subtended by a standard arc C (30 m metric), so R = 1718.87/D. (Key point: Degree of curve) 2. If the degree of curve increases, the curve becomes A. flatter with a larger radius B. longer in tangent length only C. unchanged in radius D. sharper with a smaller radius Answer: D. R = 1718.87/D, so larger D means smaller R and a sharper curve. (Key point: Sharpness) 3. What is the shift s of the circular curve in a composite curve? A. Ls^2/(24R) B. Ls^3/(CR) C. Ls^2/(6R) D. Ls/(2R) Answer: A. The circle must be shifted inward from the tangent by s = Ls^2/(24R) to make room for the transition. (Key point: Shift) 4. How are the circular arc length and total length of a composite curve found? A. Lc = πR(Δ + 2θs)/180; total = 2Ls + Lc B. Lc = πR(Δ - 2θs)/180; total = 2Ls + Lc C. Lc = πR(Δ - θs)/180; total = 2Ls - Lc D. Lc = πRΔ/180; total = Ls + Lc Answer: B. Each spiral absorbs θs of the total deflection Δ, leaving Δ - 2θs for the arc; the total adds both spirals. (Key point: Length of arc and total length) 5. Why does the tangent length Ts contain the term Ls/2? A. Because the shift equals Ls/2 B. Because the deflection angle at SC is Ls/2 C. Because half of the spiral lies on each side of the original tangent point D. Because the spiral is half the length of the arc Answer: C. The transition is placed so half its length lies on either side of the original T point, which adds Ls/2 to the tangent. 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