Composite curve design by Rankine's method

Data: deflection angle $\Delta = 60^\circ$, design speed $V = 40$ km/h $= 11.111$ m/s, centrifugal ratio $P/W = 1/8$, rate of change of radial acceleration $\alpha = 0.30$ m/s$^3$, chainage of IP = 1+030 m (1030.00 m), peg interval 20 m.

1. Radius, length of transition and shift

Centrifugal ratio $= \dfrac{V^2}{gR}$:

$$R = \frac{V^2}{g\times(1/8)} = \frac{11.111^2}{9.81/8} = 100.68\ \text{m}$$ $$L = \frac{V^3}{\alpha R} = \frac{11.111^3}{0.30\times 100.68} = 45.42\ \text{m}$$ $$s = \frac{L^2}{24R} = \frac{45.42^2}{24\times 100.68} = 0.854\ \text{m}$$

2. Tangent length and spiral angle

$$T_s = (R + s)\tan\frac{\Delta}{2} + \frac{L}{2} = (100.68 + 0.854)\tan 30^\circ + 22.71 = 81.33\ \text{m}$$ $$\phi_s = \frac{L}{2R}\ \text{rad} = 12°55'24", \qquad \Delta_c = \Delta - 2\phi_s = 34°09'13"$$ $$L_c = R\Delta_c = 100.68\times 0.59609 = 60.01\ \text{m}$$

3. Chainages

4. Setting out the circular curve by Rankine's method (peg interval 20 m)

Set the theodolite at S1, backsight T1, and set the tangent to the circular curve (turn the telescope to the common tangent, which makes $2\phi_s/3$ with the chord S1T1 for a cubic spiral). Then set out the circular curve from S1 by deflection angles from this tangent:

$$\delta = \frac{1718.9\,C}{R}\ \text{minutes}$$

The first chord is a sub-chord up to the next 20 m peg.

Check: the final cumulative deflection $= \Delta_c/2 = 17°04'36"$, and it equals 17°04'37".

Answer: R = 100.68 m, L = 45.42 m, shift = 0.854 m, $T_s$ = 81.33 m, $L_c$ = 60.01 m; the Rankine deflection angles are tabulated above.