Circular curve with cubic spiral transitions R=450

A combined curve (transition – circular arc – transition) is designed with a cubic spiral. Data: $\Delta = 18^\circ36' = 18.6^\circ$, $R = 450$ m, $v = 70$ km/h, $C = 0.30$ m/s³, chainage of IP = 2524.20 m, peg interval 20 m.

Step 1: Length of transition

$$v = \frac{70}{3.6} = 19.444\ \text{m/s},\qquad L_s = \frac{v^3}{C R} = \frac{19.444^3}{0.30\times 450} = 54.46\ \text{m}$$

Adopt $L_s = 55$ m (rounded up to a convenient value).

Step 2: Shift, spiral angle and tangent length

$$\begin{aligned} s &= \frac{L_s^2}{24R} = \frac{55^2}{24\times 450} = 0.280\ \text{m}\\ \theta_s &= \frac{L_s}{2R} = \frac{55}{900}\ \text{rad} = 3.501^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = (450+0.280)\tan 9.3^\circ + 27.5 = 101.24\ \text{m} \end{aligned}$$

Step 3: Length of circular arc and chainages

$$L_c = R(\Delta - 2\theta_s)\frac{\pi}{180} = 450\times\frac{\pi}{180}(18.600^\circ - 7.003^\circ) = 91.08\ \text{m}$$

Step 4: Setting-out data

Transition (from TS, angles measured from the tangent at TS), $\alpha = 573\,l^2/(R L_s)$ minutes:

Circular curve: set the theodolite at SC and backsight TS, then turn through $2\theta_s/3$ (and plunge) to set the common tangent at SC. The deflections $\delta = 1718.87\,c/R$ minutes are measured from this tangent, and the last cumulative deflection should be $(\Delta - 2\theta_s)/2 = 5.7986^\circ$:

The second transition is set out from ST towards CS in the same way, with $l$ measured back from ST.

Answer: $L_s = 55$ m, $T_s = 101.24$ m, $L_c = 91.08$ m, TS = 2422.96, SC = 2477.96, CS = 2569.05, ST = 2624.05 m.