Transition curve R=300 m, deflection 40°

Data: deflection angle $\Delta = 40^\circ$, chainage of IP = 1+234.5 (1234.50 m), $R = 300$ m, $\alpha = 40$ cm/s$^3 = 0.40$ m/s$^3$, $V = 80$ km/h $= 22.222$ m/s. Peg interval taken as 10 m.

1. Length of transition curve and shift

$$L = \frac{V^3}{\alpha R} = \frac{22.222^3}{0.40\times 300} = 91.45\ \text{m}$$ $$s = \frac{L^2}{24R} = \frac{91.45^2}{24\times 300} = 1.162\ \text{m}$$

2. Tangent length, spiral angle and circular arc

$$T_s = (R + s)\tan\frac{\Delta}{2} + \frac{L}{2} = (300 + 1.162)\tan 20^\circ + 45.72 = 155.34\ \text{m}$$ $$\phi_s = \frac{L}{2R} = 0.15242\ \text{rad} = 8°43'58", \qquad \Delta_c = 40^\circ - 2\phi_s = 22°32'04"$$ $$L_c = R\Delta_c = 300\times 0.39330 = 117.99\ \text{m}$$

   T1 ~~~~spiral~~~~ S1 ____circular____ S2 ~~~~spiral~~~~ T2
   |<--- L --->|<------ Lc ------>|<--- L --->|

3. Chainages of the junction points

4. Setting out the transition curve at the left end (from T1)

For a cubic spiral $y = \dfrac{x^3}{6RL}$, the deflection angle from the tangent at T1 to a point at distance $l$ along the curve is

$$\delta = \frac{l^2}{6RL}\ \text{rad} = \frac{573\,l^2}{RL}\ \text{minutes}$$

At the end of the transition ($l = L$): $\delta = 2°54'40" = \phi_s/3$ ✓.

5. Setting out the circular curve (Rankine, from S1, 10 m chord)

Set up at S1, sight T1 and set the common tangent (at $2\phi_s/3$ from the chord S1T1). Deflection for a chord $c$: $\delta = 1718.9\,c/R$ minutes.

Check: the last cumulative deflection 11°16'03" = $\Delta_c/2$ = 11°16'02".

Answer: L = 91.45 m; shift = 1.162 m; $T_s$ = 155.34 m; T1 = 1079.16, S1 = 1170.61, S2 = 1288.60, T2 = 1380.05 m; setting-out data are in the tables.