Transition curve R=2000 m, chainages, deflection table

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

Deflection angles for the transition (chords of 15 m)

$\alpha = \dfrac{573\,l^2}{R L_s}$ minutes, measured at TS from the tangent:

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.