--- title: "Circular curve with cubic spiral transitions R=450" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:cubic-spiral-transition-450m-18-36" source: https://www.bctengineer.com/q/semester_4/surveying_ii/cubic-spiral-transition-450m-18-36 site: BCT Engineering --- # Circular curve with cubic spiral transitions R=450 Surveying II · Chapter 6: Curves - Asked in: 2075 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/cubic-spiral-transition-450m-18-36 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question It is required to join two straights having a total deflection angle 18°36' by a central circular curve of radius 450 m with two ends cubic spiral transition curves. The design velocity is 70 kmph and rate of change of radial acceleration is 30 cm/sec³. Chainage of IP = 2524.20 m. Take peg interval for circular and transition curve = 20 m for both. ## Short answer Transition length Ls = 55 m. Shift = 0.280 m, spiral angle = 3.501°. Tangent length Ts = 101.24 m, circular arc Lc = 91.08 m. Chainages: TS = 2422.96, SC = 2477.96, CS = 2569.05, ST = 2624.05 m. ## Answer A combined curve (transition – circular arc – transition) is designed with a cubic spiral. Data: $\Delta = 18^\circ36' = 18.6^\circ$, $R = 450$ m, $v = 70$ km/h, $C = 0.30$ m/s³, chainage of IP = 2524.20 m, peg interval 20 m. #### Step 1: Length of transition $$ v = \frac{70}{3.6} = 19.444\ \text{m/s},\qquad L_s = \frac{v^3}{C R} = \frac{19.444^3}{0.30\times 450} = 54.46\ \text{m} $$ Adopt $L_s = 55$ m (rounded up to a convenient value). #### Step 2: Shift, spiral angle and tangent length $$ \begin{aligned} s &= \frac{L_s^2}{24R} = \frac{55^2}{24\times 450} = 0.280\ \text{m}\\ \theta_s &= \frac{L_s}{2R} = \frac{55}{900}\ \text{rad} = 3.501^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = (450+0.280)\tan 9.3^\circ + 27.5 = 101.24\ \text{m} \end{aligned} $$ #### Step 3: Length of circular arc and chainages $$ L_c = R(\Delta - 2\theta_s)\frac{\pi}{180} = 450\times\frac{\pi}{180}(18.600^\circ - 7.003^\circ) = 91.08\ \text{m} $$ | Point | Chainage (m) | |---|---| | TS = IP − T_s = 2524.20 − 101.24 | 2422.96 | | SC = TS + L_s | 2477.96 | | CS = SC + L_c | 2569.05 | | ST = CS + L_s | 2624.05 | #### Step 4: Setting-out data Transition (from TS, angles measured from the tangent at TS), $\alpha = 573\,l^2/(R L_s)$ minutes: | Point (chainage, m) | l from TS (m) | α (min) | α (° ′ ″) | |---|---|---|---| | 2422.96 | 0.00 | 0.00′ | 0°00′00″ | | 2440.00 | 17.04 | 6.72′ | 0°06′43″ | | 2460.00 | 37.04 | 31.76′ | 0°31′45″ | | 2477.96 | 55.00 | 70.03′ | 1°10′02″ | Circular curve: set the theodolite at SC and backsight TS, then turn through $2\theta_s/3$ (and plunge) to set the common tangent at SC. The deflections $\delta = 1718.87\,c/R$ minutes are measured from this tangent, and the last cumulative deflection should be $(\Delta - 2\theta_s)/2 = 5.7986^\circ$: | Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection | |---|---|---|---| | 2477.96 | - | - | 0°00′00″ | | 2480.00 | 2.04 | 0°07′47″ | 0°07′47″ | | 2500.00 | 20.00 | 1°16′24″ | 1°24′10″ | | 2520.00 | 20.00 | 1°16′24″ | 2°40′34″ | | 2540.00 | 20.00 | 1°16′24″ | 3°56′58″ | | 2560.00 | 20.00 | 1°16′24″ | 5°13′21″ | | 2569.05 | 9.05 | 0°34′34″ | 5°47′55″ | The second transition is set out from ST towards CS in the same way, with $l$ measured back from ST. **Answer:** $L_s = 55$ m, $T_s = 101.24$ m, $L_c = 91.08$ m, TS = 2422.96, SC = 2477.96, CS = 2569.05, ST = 2624.05 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the transition length found? A. Ls = v^3/(C R) with v in m/s and C the rate of change of radial acceleration, 54.46 m (adopt 55 m) B. Ls = v^2/(C R) C. Ls = v R/C D. Ls = C R/v^3 Answer: A. Limiting the rate of change of radial acceleration gives Ls = v^3/(CR); here v = 19.444 m/s. (Key point: Transition length) 2. What are the spiral angle and shift? A. theta_s = Ls/R and s = Ls^2/(8R) B. theta_s = Ls/(2R) and s = Ls^2/(24R) C. theta_s = Ls/(6R) and s = Ls^2/(6R) D. theta_s = 2Ls/R and s = Ls/24R Answer: B. For the cubic spiral theta_s = Ls/2R rad and shift s = Ls^2/24R = 0.28 m. (Key point: Spiral angle and shift) 3. How is the length of the central circular arc found? A. Lc = R Delta B. Lc = R(Delta + 2 theta_s) C. Lc = R(Delta - 2 theta_s) in radians D. Lc = R(Delta - theta_s) Answer: C. Each transition uses up theta_s of the deflection, so the arc deflection is Delta - 2 theta_s, giving 91.08 m. (Key point: Circular arc length) 4. Which sequence gives the chainage of ST? A. ST = IP + R tan(Delta/2) B. ST = IP + Ls C. ST = SC + Ls D. TS = IP - Ts, SC = TS + Ls, CS = SC + Lc, ST = CS + Ls Answer: D. Adding the three lengths to TS gives ST = 2624.05 m. 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