--- title: "Transition curve 400 m radius, 85 kmph" subject: "Transportation Engineering I" chapter: 3 key: "semester_6/transportation_engineering_i:n-tc-400-85" source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-tc-400-85 site: BCT Engineering --- # Transition curve 400 m radius, 85 kmph Transportation Engineering I · Chapter 3: Geometric Design of Highway - Asked in: 2076 Baisakh, 8 marks; 2075 Bhadra, 8 marks (asked 2 times) - Video: [Unit 2. L. No.7: Numerical problems on Superelevation (contd...)](https://www.youtube.com/watch?v=-oWcMGV5MnA) — venkatesha murthy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-tc-400-85 - Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3 ## Question A national highway curve of 400 m radius is to be set out to connect two straights. The maximum speed of moving vehicles on this curve is restricted to 85 Kmph. Transition curve are to be introduced at each end of curve. Calculate: (i) A suitable length of transition curve (ii) The necessary shift of circular curve (iii) The chainage at the beginning and end of curve Given that angle of intersection = 125°25' Rate of change of centrifugal acceleration = 0.52 m/s³ Chainage at the point of intersection = 1075.5 m ## Short answer Transition curve length Ls = v³/(cR) = 63.3 m, adopt 65 m. Shift s = Ls²/(24R) = 0.44 m. Curve begins at chainage 836.4 m (TS) and ends at 1282.5 m (ST). ## Answer **Given:** $R = 400$ m, $V = 85$ km/h, $c = 0.52$ m/s³, intersection angle (interior) $= 125^\circ 25'$, PI chainage $= 1075.5$ m. **Reading:** the "angle of intersection" is the interior angle between the straights, so the deflection angle $\Delta = 180^\circ - 125^\circ25' = 54^\circ35' = 54.583^\circ$. #### (i) Length of transition curve $v = 85/3.6 = 23.61$ m/s. $$ L_s = \frac{v^3}{cR} = \frac{(23.61)^3}{0.52\times400} = \frac{13162.6}{208} = 63.3\ \text{m} $$ Check with the IRC empirical formula (plain/rolling): $2.7V^2/R = 2.7(85)^2/400 = 48.8$ m $<63.3$ m. The rate of superelevation criterion is not requested (no width data). Adopt **$L_s = 65$ m**. #### (ii) Shift $$ s = \frac{L_s^2}{24R} = \frac{65^2}{24\times400} = \frac{4225}{9600} = 0.44\ \text{m} $$ #### (iii) Chainages Spiral angle $\phi_s = \dfrac{L_s}{2R} = 0.0813$ rad $= 4.66^\circ$. $$ \begin{aligned} T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} \\ &= (400.44)\tan(27.292^\circ) + 32.5 = 206.6 + 32.5 = 239.1\ \text{m} \end{aligned} $$ Length of circular curve: $$ L_c = R\left(\Delta - 2\phi_s\right) = 400\,(0.95269 - 0.16250) = 316.1\ \text{m} $$ | Point | Chainage (m) | |---|---| | Beginning of curve (TS) $= 1075.5 - 239.1$ | **836.4** | | Start of circular curve (SC) $= 836.4+65$ | 901.4 | | End of circular curve (CS) $= 901.4+316.1$ | 1217.5 | | End of curve (ST) $= 1217.5 + 65$ | **1282.5** | **Answer:** $L_s = 65$ m; shift $= 0.44$ m; curve begins at chainage 836.4 m and ends at 1282.5 m (total length $316.1 + 130 = 446.1$ m). *If the 125°25' were taken as the deflection angle instead, $T_s = 808.6$ m, the curve would begin at 266.9 m and end at 1207.5 m.* ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The deflection angle from the interior intersection angle 125 deg 25' is A. 125 deg 25' B. 54 deg 35' C. 35 deg 25' D. 90 deg Answer: B. Deflection = 180 - 125 deg 25' = 54 deg 35'. (Key point: Deflection angle) 2. Which formula gives the length of transition from rate of change of centrifugal acceleration? A. Ls = v^3/(c R) B. Ls = v^2/(c R) C. Ls = c R/v^3 D. Ls = L^2/(24R) Answer: A. Ls = v^3/(cR) = 23.61^3/(0.52 x 400) = 63.3 m, adopted 65 m. (Key point: Length of transition) 3. The shift of the circular curve is s = Ls^2/(24R) = A. 1.76 m B. 4.40 m C. 0.044 m D. 0.44 m Answer: D. 65^2/(24 x 400) = 0.44 m. (Key point: Shift) 4. The total tangent distance is Ts = A. (R+s) tan(Delta/2) + Ls/2 B. R tan(Delta/2) C. (R+s) tan(Delta) + Ls D. (R-s) tan(Delta/2) - Ls/2 Answer: A. Ts = 400.44 tan(27.29 deg) + 32.5 = 239.1 m. 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