Transition curve and shift for 250 m radius

Assumptions

• In the IRC method the design superelevation $e=\dfrac{V^2}{225R}$ fully balances the centrifugal force at 75% of the design speed, so it is the equilibrium superelevation for mixed traffic. With $e=0.065$: $V=\sqrt{225(250)(0.065)}=60.5\approx 60$ km/h, so the design speed = 60 km/h.
• Two-lane carriageway $W=7.0$ m, pavement rotated about the centre line, rate of introduction 1 in $N=150$.
• Extra widening for 2 lanes ($l=6.1$ m): mechanical $W_m=\dfrac{nl^2}{2R}=0.149$ m, psychological $W_{ps}=\dfrac{V}{9.5\sqrt R}=0.399$ m, so $W_e\approx 0.55$ m.

Length of transition curve (largest of three criteria)

1. Rate of change of centrifugal acceleration (IRC): $C=\dfrac{80}{75+V}=0.593$ m/s³

$$L_s=\frac{(V/3.6)^3}{CR}=\frac{16.67^3}{0.593(250)}=31.3\ \text{m}$$

2. Rate of introduction of superelevation (rotation about the centre line):

$$L_s=\frac{N\,e\,(W+W_e)}{2}=\frac{150(0.065)(7.55)}{2}=36.8\ \text{m}$$

3. IRC empirical (plain/rolling): $L_s=\dfrac{2.7V^2}{R}=\dfrac{2.7(60)^2}{250}=38.9$ m.

Adopt $L_s$ = 40 m.

Shift

$$s=\frac{L_s^2}{24R}=\frac{40^2}{24(250)}=0.267\ \text{m}$$

Answer: transition length $L_s=40$ m; shift $s=0.27$ m.

Note: if 0.065 is taken as the equilibrium superelevation at the full design speed ($e=V^2/127R$), then $V=45.4$ km/h. The three criteria then give 12.1 m, 36.3 m and 22.3 m, so the superelevation criterion governs and $L_s=40$ m and $s=0.27$ m are unchanged.