Transition curve road 8 m, R=330 m, 60°

Data: width of road $B = 8$ m, $\Delta = 60^\circ$, $R = 330$ m, chainage of IP = 3605.0 m, $\alpha = 0.5$ m/s$^3$, $V = 50$ km/h $= 13.889$ m/s, peg interval 5 m.

(a) Length of the transition curve

$$L = \frac{V^3}{\alpha R} = \frac{13.889^3}{0.5\times 330} = 16.24\ \text{m}$$

(b) Superelevation

For full balancing of the centrifugal force, $\tan\theta = e = \dfrac{V^2}{gR}$:

$$e = \frac{13.889^2}{9.81\times 330} = 0.0596 \approx 1\text{ in }16.8$$

Rise of the outer edge over the width of the road:

$$h = e\times B = 0.0596\times 8 = 0.477\ \text{m}$$

(c) Chainages of the junction points

$$s = \frac{L^2}{24R} = 0.0333\ \text{m},\qquad T_s = (R+s)\tan 30^\circ + \frac{L}{2} = (330 + 0.0333)(0.57735) + 8.119 = 198.664\ \text{m}$$ $$\phi_s = \frac{L}{2R} = 1°24'35",\qquad \Delta_c = 60^\circ - 2\phi_s = 57°10'51",\qquad L_c = R\Delta_c = 329.338\ \text{m}$$

(d) Layout of the transition curve by deflection angles (5 m pegs, from T1)

$$\delta = \frac{573\,l^2}{RL}\ \text{minutes}$$

At the end of the curve $\delta = \phi_s/3 = 0°28'12"$. The second transition is set out in the same way from T2 towards S2 (measured backwards along the tangent), and the circular curve from S1 by Rankine's deflection angles $\delta = 1718.9\,c/R$ minutes.

Answer: L = 16.24 m; superelevation = 0.477 m (1 in 16.8); T1 = 3406.34 m, S1 = 3422.57 m, S2 = 3751.91 m, T2 = 3768.15 m.