Hi MrAl,
The formula is several formulae joined together.
First we need to know how long the capacitor needs to supply.
The arcsin term gives the angle (in radians) from the zero crossing point when the voltage on the capacitor will be less than the voltage on the AC side, i.e. when Vin - Vr = Vc. Divide the result by 2pi×f and you have the time from the zero crossing to when the capacitor stops suppling current.
The 1/(4pi×f) part assumes that the capacitor starts discharging at 90 degrees. It gives the time from 90 degrees to the zero crossing point, one quarter of the AC cycle. This of course isn't completely accurate it'll be a bit less than that.
The I/Vr term calculates the capacitor size required assuming a constant current is drawn by the regulator given Vr is the voltage drop with the time given by the remainting terms.
[latex]C = \frac{I \times t}{V}[/latex]
If a resistive load were used the I/Vr term would be replaced by the Vc = Vs×e^(-t×RC) with Vr being Vs - Vc and rearranged to make C the subject.
The above formula also doesn't take into account of the diode's forward voltage.
For interest's sake I've added this to the formula below.
[latex]C = \frac{I}{V_R} \times \left(\frac{1}{4f} + \frac{ arcsin{\left(\frac{V_{F}}{V_{IN}} \right)}+arcsin{\left(\frac{V_{IN}-V_R}{V_{IN}} \right)}}{2\pi f} \right)\\
C=\text{Minimum capacitance required}\\
V_R= \text{Maximum ripple}\\
V_{IN}= \text{Input voltage, excluding diode losses}\\
I=\text{Current drawn by regulator}\\
f= \text{Frequency}\\
[/latex]
My latest spreadsheet includes this more accurate formula. I'll post it when I feel ready to.