Transition curve for NH, 200 m radius

Design speed is not given. A radius of 200 m cannot be used at the 80 km/h ruling speed ($R_{min}\approx229$ m), so the minimum design speed for a National Highway in rolling terrain, 65 km/h, is adopted.

Assumptions

• Design speed $V=65$ km/h; carriageway width $W=7.0$ m (two lanes, IRC/NRS 2070); National Highway in rolling terrain; pavement rotated about the centre line; rate of introduction of superelevation 1 in $N=150$ (IRC: 1 in 150 for plain and rolling terrain).

Superelevation and extra widening

• $e=\dfrac{V^2}{225R}=\dfrac{65^2}{225(200)}=0.0939>0.07$, so e = 0.07. Friction needed $=\dfrac{V^2}{127R}-e=0.096<0.15$ (safe).
• Mechanical widening $W_m=\dfrac{nl^2}{2R}=\dfrac{2(6.1)^2}{2(200)}=0.186$ m; psychological widening $W_{ps}=\dfrac{V}{9.5\sqrt R}=0.484$ m; total $W_e=0.67\approx0.7$ m.

Length of transition curve (greatest of three)

1. Rate of change of centrifugal acceleration: $C=\dfrac{80}{75+65}=0.571$ m/s³

$$L_s=\frac{(65/3.6)^3}{CR}=\frac{18.06^3}{0.571(200)}=51.5\ \text{m}$$

2. Rate of introduction of superelevation:

$$L_s=\frac{N\,e\,(W+W_e)}{2}=\frac{150(0.07)(7.0+0.7)}{2}=40.4\ \text{m}$$

3. IRC empirical formula (plain/rolling terrain):

$$L_s=\frac{2.7V^2}{R}=\frac{2.7(65)^2}{200}=57.0\ \text{m}$$

The greatest value governs. Answer: adopt L_s = 60 m (calculated 57.0 m). The shift is $s=\dfrac{L_s^2}{24R}=0.75$ m.