Transition curve, road 8 m, R=300 m, 60°

Data: carriageway width $B = 8$ m, $\Delta = 60^\circ$, $R = 300$ m, chainage of IP = 3605 m, $v = 70$ km/h = 19.444 m/s, rate of change of radial acceleration $C = 0.5$ m/s³.

i) Length of transition curve

$$L_s = \frac{v^3}{C R} = \frac{19.444^3}{0.5\times 300} = 49.01\ \text{m}$$

Adopt $L_s = 50$ m (rounded up to a multiple of 5 m).

ii) Super-elevation

Equilibrium super-elevation for the full speed, taking the road width $B$ as the distance between the outer and inner edges:

$$\frac{h}{B} = \frac{v^2}{gR} = \frac{19.444^2}{9.81\times 300} = 0.1285 \quad\Rightarrow\quad h = 8\times 0.1285 = 1.028\ \text{m}$$

(Slope $1$ in $7.8$. In practice the value is limited to the maximum permitted by the road code, e.g. 7 %, and the balance is taken by side friction.) The super-elevation is attained gradually over $L_s$ so that the outer edge rises uniformly from TS to SC.

iii) Chainages of tangent and junction points

$$\begin{aligned} s &= \frac{L_s^2}{24R} = 0.347\ \text{m},\quad \theta_s = \frac{L_s}{2R} = 4.775^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 198.41\ \text{m}\\ L_c &= \frac{\pi R}{180}(\Delta - 2\theta_s) = 264.16\ \text{m} \end{aligned}$$

iv) Deflection angles for the first two points

Transition curve from TS, $\alpha = 573\,l^2/(R L_s)$ minutes (points at multiples of 10 m):

Circular curve from SC, $\delta = 1718.87\,c/R$ minutes (points at multiples of 20 m):

Answer: $L_s = 50$ m; $h = 1.028$ m; TS = 3406.59, SC = 3456.59, CS = 3720.75, ST = 3770.75 m; deflection angles as tabulated.