Composite curve design, IP 1150 m, Rankine

Given data

$\Delta = 60^\circ$, $v_{max} = 40$ km/h $= 11.111$ m/s, centrifugal ratio $v^2/(gR) = 1/8$, $C = 0.30$ m/s³, chainage of IP $= 1150$ m, peg interval 20 m.

Radius of the circular curve

$$\frac{v^2}{gR} = \frac{1}{8} \Rightarrow R = \frac{8v^2}{g} = \frac{8\times 11.111^2}{9.81} = 100.68\ \text{m}$$

Adopt $R = 100.68$ m. (Rounding down to 100 m would make the centrifugal ratio slightly greater than the permitted 1/8.)

Length of transition

$$L_s = \frac{v^3}{CR} = \frac{11.111^3}{0.30\times 100.68} = 45.42\ \text{m}$$

Adopt $L_s = 50$ m.

Elements of the composite curve

$$\begin{aligned} s &= \frac{L_s^2}{24R} = \frac{50^2}{24\times 100.68} = 1.035\ \text{m}\\ \theta_s &= \frac{L_s}{2R} = \frac{50}{2\times 100.68} = 0.24831\ \text{rad} = 14.227^\circ\\ T_s &= (R+s)\tan 30^\circ + \frac{L_s}{2} = 101.715\times 0.57735 + 25 = 83.725\ \text{m}\\ L_c &= R(\Delta - 2\theta_s) = 100.68\left(1.04720 - 0.49662\right) = 55.432\ \text{m} \end{aligned}$$

Setting out the circular curve by Rankine's method (theodolite at SC)

$\delta = 1718.87\,c/R$ minutes for chord $c$; the first sub-chord runs to the next 20 m peg and the last sub-chord ends at CS.

Check: last cumulative deflection $= (\Delta - 2\theta_s)/2 = (60^\circ - 28.454^\circ)/2 = 15.773^\circ$ = 15°46′22″.

For reference, the transition (set out from TS), $\alpha = 573l^2/(RL_s)$ minutes:

(At SC, $\theta_s/3 = 4.742^\circ$ = 4°44′33″; the 1″ difference comes from rounding 572.96 to 573.)

Answer: $R = 100.68$ m, $L_s = 50$ m, $T_s = 83.725$ m, $L_c = 55.432$ m; TS = 1066.275 m, SC = 1116.275 m, CS = 1171.707 m, ST = 1221.707 m; Rankine deflection angles as tabulated.