--- title: "Transition curve R=2000 m, chainages, deflection table" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:transition-curve-2000m-160kmh-chainage-table" source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-curve-2000m-160kmh-chainage-table site: BCT Engineering --- # Transition curve R=2000 m, chainages, deflection table Surveying II · Chapter 6: Curves - Asked in: 2066 Magh (old course), 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/transition-curve-2000m-160kmh-chainage-table - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Two tangents which deflect at an angle of 37°46' are to be connected by a circular curve of 2000 m radius with a transition curve at either end. The chainage of the point of intersection is (3436+26) chains. Find the chainages of the beginning and end of the three curves and draw a table of the deflection angles for chords of 15 m for each transition curve. Assume velocity = 160 km/hr, rate of change of radial acceleration = 0.3 m/sec³, chain used was 30 m length. ## Short answer Ls=150 m, Ts=759.26 m, Lc=1168.31 m. Chainages: TS 102346.74 m, SC 102496.74 m, CS 103665.04 m, ST 103815.04 m. Transition deflection angles from TS every 15 m: 0°00′26″ at 15 m, increasing to 0°42′58″ at 150 m (equal to θs/3). ## Answer #### Given data $\Delta = 37^\circ46' = 37.7667^\circ$, $R = 2000$ m, $v = 160$ km/h $= 44.444$ m/s, $C = 0.3$ m/s³, chain = 30 m. Chainage of IP $= (3436+26)$ chains, read as $3436\times 30 + 26 = 103106$ m. #### Length of transition $$ L_s = \frac{v^3}{CR} = \frac{44.444^3}{0.3\times 2000} = 146.32\ \text{m} $$ Adopt $L_s = 150$ m (ten chords of 15 m). #### Elements $$ \begin{aligned} s &= \frac{L_s^2}{24R} = \frac{150^2}{24\times 2000} = 0.469\ \text{m}\\ \theta_s &= \frac{L_s}{2R} = 0.0375\ \text{rad} = 2.1486^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 759.26\ \text{m}\\ L_c &= R\,(\Delta - 2\theta_s)\frac{\pi}{180} = 1168.31\ \text{m} \end{aligned} $$ #### Chainages | Point | Chainage (m) | |---|---| | Beginning of first transition (TS) = IP − T_s | 102346.74 | | End of first transition (SC) = TS + 150 | 102496.74 | | End of circular curve (CS) = SC + 1168.31 | 103665.04 | | End of second transition (ST) = CS + 150 | 103815.04 | #### Deflection angles for the transition (chords of 15 m) $\alpha = \dfrac{573\,l^2}{R L_s}$ minutes, measured at TS from the tangent: | Point (chainage, m) | l from TS (m) | α (min) | Deflection α | |---|---|---|---| | 102361.74 | 15 | 0.43′ | 0°00′26″ | | 102376.74 | 30 | 1.72′ | 0°01′43″ | | 102391.74 | 45 | 3.87′ | 0°03′52″ | | 102406.74 | 60 | 6.88′ | 0°06′53″ | | 102421.74 | 75 | 10.74′ | 0°10′45″ | | 102436.74 | 90 | 15.47′ | 0°15′28″ | | 102451.74 | 105 | 21.06′ | 0°21′03″ | | 102466.74 | 120 | 27.50′ | 0°27′30″ | | 102481.74 | 135 | 34.81′ | 0°34′49″ | | 102496.74 | 150 | 42.98′ | 0°42′58″ | Check: at SC, $\alpha = \theta_s/3 = 0.7162^\circ$ = 0°42′58″. The second transition is set out from ST with the same angles, taking $l$ from ST backwards. **Answer:** $L_s = 150$ m, $T_s = 759.26$ m, $L_c = 1168.31$ m; TS = 102346.74, SC = 102496.74, CS = 103665.04, ST = 103815.04 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is chainage (3436+26) chains converted? A. 3436 x 26 B. 3436 x 30 + 26 = 103106 m C. (3436 + 26)/30 D. 3436 + 26 = 3462 m Answer: B. The chain is 30 m, so 3436 chains is 3436 x 30 m, plus 26 m. (Key point: Chainage conversion) 2. Why is Ls adopted as 150 m while the calculation gave 146.32 m? A. Because R is reduced B. To make theta zero C. Because C is increased D. Rounded up to ten 15 m chords, a convenient length Answer: D. Rounded up so the transition divides into ten 15 m chords and C is not exceeded. (Key point: Transition length) 3. What does alpha equal at SC as a check? A. thetas B. 2 thetas/3 C. Delta/2 D. thetas/3 = 0 deg 42' 58'' Answer: D. alpha at l = Ls is 573 Ls/R = thetas/3 = 0.7162 degrees. (Key point: Deflection check) 4. How is the second transition set out? A. Using the circular curve method B. It cannot be set out C. From TS using larger angles D. From ST with the same angles, l measured backwards from ST Answer: D. The two spirals are symmetric, so the same angles apply from ST. 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