Transition deflection angle derivation and laying out

Derivation of $\alpha = \dfrac{573\,l^2}{RL}$ minutes

For a transition curve the radius at distance $l$ from the start (TS) varies inversely with $l$: $r\,l = RL$, where $R$ is the radius of the circular curve and $L$ the length of the transition. The angle between the tangent at TS and the tangent at a point at distance $l$ is

$$\phi = \int_0^l \frac{dl}{r} = \int_0^l \frac{l\,dl}{RL} = \frac{l^2}{2RL}\ \text{radians}$$

Taking $x \approx l$ along the first tangent, the ordinate is $y = \int \phi\,dx = \dfrac{l^3}{6RL}$. The deflection angle of the point from TS:

$$\alpha \approx \frac{y}{x} = \frac{l^2}{6RL}\ \text{radians} = \frac{3437.75}{6}\cdot\frac{l^2}{RL}\ \text{minutes} = \frac{573\,l^2}{RL}\ \text{minutes}$$

At the end of the transition, $\alpha_s = \phi_s/3$.

Laying out the transition curve

1. Calculate $L_s$, $s$, $T_s$ and the chainages of TS, SC, CS and ST.
2. Locate TS by measuring $T_s$ back from the IP.
3. Set up the theodolite at TS, sight the IP with the plates at zero.
4. For each peg at distance $l$ from TS, compute $\alpha$ and set it on the circle; measure the chord length from the previous point, and fix the point where tape and line of sight meet.
5. The last angle to SC should equal $\theta_s/3$ (check). The second transition is set out in the same way from ST.