--- title: "Transition curve R=300 m, deflection 40°" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:transition-curve-80kmph-radius-300" source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-curve-80kmph-radius-300 site: BCT Engineering --- # Transition curve R=300 m, deflection 40° Surveying II · Chapter 6: Curves - Asked in: 2078 Chaitra, 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/transition-curve-80kmph-radius-300 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Two straights AB and BC intersect at chainage 1+234.5 km, the deflection angle being 40°. It is proposed to insert a circular curve of radius 300 m. A transition curve is to be inserted at each end of circular curve such that the rate of change of radial acceleration is 40 cm/sec³, when the design speed is 80 kmph. Calculate the shift, length of transition curve, chainage of tangent points, junction points and data required to set the transition curves (at left end) and circular curve. ## Short answer Length of transition curve = 91.45 m, shift = 1.162 m, tangent length Ts = 155.34 m. Chainages: T1 = 1079.16 m, S1 = 1170.61 m, S2 = 1288.60 m, T2 = 1380.05 m. Setting-out data for the transition and circular curves are provided in the tables. ## Answer Data: deflection angle $\Delta = 40^\circ$, chainage of IP = 1+234.5 (1234.50 m), $R = 300$ m, $\alpha = 40$ cm/s$^3 = 0.40$ m/s$^3$, $V = 80$ km/h $= 22.222$ m/s. Peg interval taken as 10 m. #### 1. Length of transition curve and shift $$ L = \frac{V^3}{\alpha R} = \frac{22.222^3}{0.40\times 300} = 91.45\ \text{m} $$ $$ s = \frac{L^2}{24R} = \frac{91.45^2}{24\times 300} = 1.162\ \text{m} $$ #### 2. Tangent length, spiral angle and circular arc $$ T_s = (R + s)\tan\frac{\Delta}{2} + \frac{L}{2} = (300 + 1.162)\tan 20^\circ + 45.72 = 155.34\ \text{m} $$ $$ \phi_s = \frac{L}{2R} = 0.15242\ \text{rad} = 8°43'58", \qquad \Delta_c = 40^\circ - 2\phi_s = 22°32'04" $$ $$ L_c = R\Delta_c = 300\times 0.39330 = 117.99\ \text{m} $$ ```text T1 ~~~~spiral~~~~ S1 ____circular____ S2 ~~~~spiral~~~~ T2 |<--- L --->|<------ Lc ------>|<--- L --->| ``` #### 3. Chainages of the junction points | Point | Chainage (m) | |---|---| | T1 = IP − $T_s$ = 1234.50 − 155.34 | 1079.16 | | S1 = T1 + L | 1170.61 | | S2 = S1 + $L_c$ | 1288.60 | | T2 = S2 + L | 1380.05 | #### 4. Setting out the transition curve at the left end (from T1) For a cubic spiral $y = \dfrac{x^3}{6RL}$, the deflection angle from the tangent at T1 to a point at distance $l$ along the curve is $$ \delta = \frac{l^2}{6RL}\ \text{rad} = \frac{573\,l^2}{RL}\ \text{minutes} $$ | Peg chainage (m) | Distance $l$ from T1 (m) | Deflection $\delta=573\,l^2/(RL)$ | Offset $y=l^3/(6RL)$ (m) | |---|---|---|---| | 1080.00 | 0.84 | 0°00'01" | 0.000 | | 1090.00 | 10.84 | 0°02'27" | 0.008 | | 1100.00 | 20.84 | 0°09'04" | 0.055 | | 1110.00 | 30.84 | 0°19'52" | 0.178 | | 1120.00 | 40.84 | 0°34'50" | 0.414 | | 1130.00 | 50.84 | 0°53'59" | 0.798 | | 1140.00 | 60.84 | 1°17'18" | 1.368 | | 1150.00 | 70.84 | 1°44'48" | 2.160 | | 1160.00 | 80.84 | 2°16'29" | 3.209 | | 1170.00 | 90.84 | 2°52'21" | 4.554 | | 1170.61 | 91.45 | 2°54'40" | 4.646 | At the end of the transition ($l = L$): $\delta = 2°54'40" = \phi_s/3$ ✓. #### 5. Setting out the circular curve (Rankine, from S1, 10 m chord) Set up at S1, sight T1 and set the common tangent (at $2\phi_s/3$ from the chord S1T1). Deflection for a chord $c$: $\delta = 1718.9\,c/R$ minutes. | Peg chainage (m) | Chord (m) | Deflection for chord $\delta=1718.9\,c/R$ | Cumulative deflection | |---|---|---|---| | 1180.00 | 9.39 | 0°53'48" | 0°53'48" | | 1190.00 | 10.00 | 0°57'18" | 1°51'06" | | 1200.00 | 10.00 | 0°57'18" | 2°48'23" | | 1210.00 | 10.00 | 0°57'18" | 3°45'41" | | 1220.00 | 10.00 | 0°57'18" | 4°42'59" | | 1230.00 | 10.00 | 0°57'18" | 5°40'17" | | 1240.00 | 10.00 | 0°57'18" | 6°37'35" | | 1250.00 | 10.00 | 0°57'18" | 7°34'52" | | 1260.00 | 10.00 | 0°57'18" | 8°32'10" | | 1270.00 | 10.00 | 0°57'18" | 9°29'28" | | 1280.00 | 10.00 | 0°57'18" | 10°26'46" | | 1288.60 | 8.60 | 0°49'17" | 11°16'03" | Check: the last cumulative deflection 11°16'03" = $\Delta_c/2$ = 11°16'02". **Answer: L = 91.45 m; shift = 1.162 m; $T_s$ = 155.34 m; T1 = 1079.16, S1 = 1170.61, S2 = 1288.60, T2 = 1380.05 m; setting-out data are in the tables.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is L computed for V = 80 km/h and α = 40 cm/s^3? A. L = V^2/(αR) with α = 40 B. L = V^3/(αR) with V = 22.222 m/s and α = 0.40 m/s^3, giving 91.45 m C. L = V/(αR) with V in km/h D. L = αR/V^3 Answer: B. Units must be m and s: V = 22.222 m/s, α = 0.40 m/s^3, so L = 22.222^3/(0.40 × 300) = 91.45 m. (Key point: Length of transition curve) 2. Which formula gives the shift s? A. s = L^2/(24R) = 1.162 m B. s = L^2/(6R) C. s = L/(24R) D. s = R^2/(24L) Answer: A. The shift of the circular curve is s = L^2/(24R), here 91.45^2/(24 × 300) = 1.162 m. (Key point: Shift) 3. What is the angle subtended by the circular arc here? A. Δ = 40° B. Δ + 2φs C. φs D. Δ - 2φs = 22°32'04" Answer: D. The two spirals take φs = L/2R = 8°43'58" each, leaving Δc = 40° - 2φs = 22°32'04" for the arc. (Key point: Central circular arc) 4. What should the deflection at the last peg of the transition (l = L) be? A. φs B. φs/3 C. 2φs/3 D. φs/2 Answer: B. δ = L/6R = φs/3, which is 2°54'40" here and serves as a check on S1. 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