Transition curve for 230 m radius

Given: $R = 230$ m, $f_{long} = 0.35$, $f_{lat} = 0.15$. The design speed is not given, so it is found from the maximum superelevation.

Step 1: Design speed and superelevation

Taking the maximum superelevation $e = 7\%$ (IRC/NRS for plain and rolling terrain):

$$V = \sqrt{127R\,(e+f)} = \sqrt{127(230)(0.07+0.15)} = 80.2\ \text{km/h}$$

So the design speed $V = 80$ km/h; required $e = V^2/225R = 12.4\%$ is more than 7%, so adopt $e = 7\%$ and check $f = \dfrac{80^2}{127\times230} - 0.07 = 0.149 < 0.15$. OK.

Step 2: Length of transition curve ($v = 80/3.6 = 22.22$ m/s)

(a) Rate of change of centrifugal acceleration

$$C = \frac{80}{75+V} = \frac{80}{155} = 0.516\ \text{m/s}^3\ (0.5 \le C \le 0.8)$$ $$L_s = \frac{v^3}{CR} = \frac{22.22^3}{0.516\times230} = 92.4\ \text{m}$$

(b) Rate of introduction of superelevation (1 in 150, plain/rolling; $W = 7.0$ m, rotation about the centre line with camber $n = 2.5\%$):

$$L_s = N\,\frac{(e+n)W}{2} = 150\times\frac{(0.07+0.025)(7.0)}{2} = 49.9\ \text{m}$$

(c) IRC empirical formula

$$L_s = \frac{2.7V^2}{R} = \frac{2.7(80)^2}{230} = 75.1\ \text{m}$$

Step 3: Adopt the greatest

$L_s = 92.4$ m, rounded up: $L_s = 95$ m.

Answer: length of transition curve = 95 m (design speed 80 km/h, $e=7\%$).