Derive deflection angle of transition curve

Transition curve (cubic spiral)

A transition curve connects a straight with a circular curve so that the radius changes gradually from $\infty$ to $R$. The radius at a distance $l$ from the start is inversely proportional to $l$:

$$\rho \propto \frac{1}{l} \quad\Rightarrow\quad \rho\, l = RL$$

where $R$ is the radius of the circular curve and $L$ is the length of the transition curve.

        y
        |        . P(x, y)
        |     .  
        |  .    deflection angle = delta
   T1 --+-----------------------> x   (tangent)

Derivation

Take T1 (start of the curve) as the origin and the tangent as the x-axis. For a point P at a distance $l$ along the curve, the tangent angle (spiral angle) $\phi$ is given by $d\phi = \dfrac{dl}{\rho} = \dfrac{l\,dl}{RL}$:

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

The coordinates of P are, for small $\phi$ ($\cos\phi \approx 1$, $\sin\phi \approx \phi$):

$$dx = dl,\quad dy = \phi\,dl = \frac{l^2}{2RL}\,dl$$ $$x \approx l, \qquad y = \int_0^l \frac{l^2}{2RL}\,dl = \frac{l^3}{6RL}$$

This is the equation of a cubic parabola.

The deflection angle $\delta$ at T1 between the tangent and the line T1P:

$$\tan\delta = \frac{y}{x} = \frac{l^3}{6RL\cdot l} = \frac{l^2}{6RL}$$

As $\delta$ is small, $\tan\delta \approx \delta$ (in radians):

$$\delta = \frac{l^2}{6RL}\ \text{radians}$$

Converting radians to minutes: $1\ \text{rad} = \dfrac{180}{\pi}\times 60 = 3437.75'$:

$$\delta = \frac{3437.75}{6}\cdot\frac{l^2}{RL} = 572.96\,\frac{l^2}{RL} \approx \boxed{\frac{573\,l^2}{RL}\ \text{minutes}}$$

Hence proved.

At the end of the transition ($l = L$): $\delta_L = \dfrac{L}{6R}$ rad $= \dfrac{\phi_s}{3}$, where $\phi_s = \dfrac{L}{2R}$ is the spiral angle. This is the usual check on the last peg.