Transition curve R=180 m, speed 13.2 m/s

The deflection angle between the tangents and the peg intervals are not given, so they are assumed: $\Delta = 40^\circ$ (for illustration of $L_c$) and pegs every 10 m from TS and from SC. The rate of change of radial acceleration is found from Shortt's empirical formula.

Length of transition curve

$$v = 13.2\ \text{m/s} = 47.52\ \text{km/h},\qquad C = \frac{80}{75+V} = \frac{80}{75+47.52} = 0.653\ \text{m/s}^3$$

This lies in the permitted range 0.3–0.8 m/s³.

$$L_s = \frac{v^3}{C R} = \frac{13.2^3}{0.653\times 180} = 19.57\ \text{m}$$

Adopt $L_s = 20$ m.

Other elements

$$\begin{aligned} \theta_s &= \frac{L_s}{2R} = \frac{20}{360}\ \text{rad} = 3.183^\circ\\ s &= \frac{L_s^2}{24R} = 0.093\ \text{m}\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 75.55\ \text{m}\ (\Delta = 40^\circ) \end{aligned}$$

Length of circular curve

$$L_c = \frac{\pi R}{180}(\Delta - 2\theta_s) = \frac{\pi\times 180}{180}(40^\circ - 6.366^\circ) = 105.66\ \text{m}$$

In general $L_c = R(\Delta - 2\theta_s)$ with angles in radians; it depends on the actual deflection angle.

First two deflection angles

Transition curve, $\alpha = 573\,l^2/(RL_s)$ minutes:

• At $l = 10$ m: $\alpha_1 = \dfrac{573\times 10^2}{180\times 20} = 15.92' = 0°15′55″$
• At $l = 20$ m: $\alpha_2 = \dfrac{573\times 20^2}{180\times 20} = 63.67' = 1°03′40″$ (this is the end point SC, and equals $\theta_s/3 = 1.061^\circ$)

Circular curve (from SC), $\delta = 1718.87\,c/R$ minutes, with 10 m chords:

• First point: $\delta_1 = \dfrac{1718.87\times 10}{180} = 95.49' = 1°35′30″$
• Second point: $\delta_2 = 2\delta_1 = 190.99' = 3°10′59″$

Answer: $L_s = 20$ m, $L_c = 105.66$ m (for $\Delta = 40^\circ$); transition deflections 0°15′55″ and 1°03′40″; circular deflections 1°35′30″ and 3°10′59″.