Transition curve road 8 m, R=300 m, IP 3605

Data: carriageway width $B = 8$ m, $\Delta = 60^\circ$, $R = 300$ m, chainage of IP = 3605 m, $v = 50$ km/h = 13.889 m/s, rate of change of radial acceleration $C = 0.5$ m/s³.

i) Length of transition curve

$$L_s = \frac{v^3}{C R} = \frac{13.889^3}{0.5\times 300} = 17.86\ \text{m}$$

Adopt $L_s = 20$ 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:

$$\frac{h}{B} = \frac{v^2}{gR} = \frac{13.889^2}{9.81\times 300} = 0.0655 \quad\Rightarrow\quad h = 8\times 0.0655 = 0.524\ \text{m}$$

(Slope $1$ in $15.3$. 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

$$\begin{aligned} s &= \frac{L_s^2}{24R} = 0.056\ \text{m},\quad \theta_s = \frac{L_s}{2R} = 1.910^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 183.24\ \text{m}\\ L_c &= \frac{\pi R}{180}(\Delta - 2\theta_s) = 294.16\ \text{m} \end{aligned}$$

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: $L_s = 20$ m; $h = 0.524$ m; TS = 3421.76, SC = 3441.76, CS = 3735.92, ST = 3755.92 m; deflection angles as tabulated.