Transition curve in hilly region, 60 m radius

Given

$V=40$ km/h, $R=60$ m, $e=0.07$, $W=7.0$ m, extra widening $W_e=1.2$ m, rate $1$ in $N=60$, rotation about the centre line, deflection angle $\Delta=60^\circ$.

(a) Length of transition curve (greatest of three)

1. Comfort: $C=\dfrac{80}{75+40}=0.696$ m/s³, $V=11.11$ m/s

$$L_s=\frac{V^3}{CR}=\frac{11.11^3}{0.696(60)}=32.9\ \text{m}$$

2. Superelevation (rotation about centre line):

$$L_s=\frac{N\,e\,(W+W_e)}{2}=\frac{60(0.07)(8.2)}{2}=17.2\ \text{m}$$

3. IRC empirical (mountainous/steep terrain): $L_s=\dfrac{V^2}{R}=\dfrac{40^2}{60}=26.7$ m.

Adopt L_s = 35 m.

(b) Total curve length

• Shift: $s=\dfrac{L_s^2}{24R}=\dfrac{35^2}{24(60)}=0.851$ m
• Spiral angle: $\theta_s=\dfrac{L_s}{2R}=0.2917$ rad $=16.71^\circ$
• Arc angle: $\Delta_c=60^\circ-2(16.71^\circ)=26.58^\circ$
• Arc length: $L_c=R\Delta_c=27.83$ m

$$L=L_c+2L_s=27.83+2(35)=97.83\ \text{m}$$

(c) Total tangent length

$$T_s=(R+s)\tan\frac{\Delta}{2}+\frac{L_s}{2}=(60.851)\tan30^\circ+17.5=52.63\ \text{m}$$

This is the tangent length from the PI to each end of the curve ($2T_s=105.26$ m for both sides).

Answer: $L_s=35$ m; total curve length = 97.8 m; tangent length $T_s=52.6$ m on each side.