Rectifier Design with Regulators Surge/In-Rush Currents
Using Q = C * V, differentiate with respect to T (time) and you get

qQ/dT = C * dV/dT + V * dC/dT  [ uses d(U*V) = U * dV/dT + V * dU/dT ]

We assume dC/dT = 0 (that C is constant) and set dQ/dT = I current. Find

I = C * dV/dT

Now you have full-wave-rectifiers probucing double the power line frequency as ripple. 

Let this ripple be 120 Hz. Let the load current (out of the cap) be 1 amp,
assume your regulator will significantly attenuate the 120 Hz ripple.

The math is I = C * dV/dT which we re-arrange to be C = I * dT/dV, and we substitute

C = 1amp * (1/120 second) / 1 volt = 1/120 farad = 8,333 microfarad=ad

However, the rectifier PEAK current is also one of your concerns.

If you have huge ripple (the entire half cycle), the recharge time is 1/4 cycle.
And your rectifier current will be about 2x the load current (out of regulator).

If you have small ripple, the recharge time is reduced. Unfortunately the peak of the
sinewave is not a triangle wave, or we could predict the recharge time based on the fraction of ripple/peak, where 1v ripple/20v transformer output == 5% and you'd have some confidence the peak was 1/0.05 = 20X the average.

If you have very small rectifier on-time (less than 1 millisecond), the regulator will be less successful in reducing the ripple because of the very fast edge of the re-charge time. And your very fast current transient thru the diodes will cause Magnetic Interference all around. 

Thus small inductors in series with the diodes are possibly useful.

Or try 1 ohm resistors. Remember the peak currents are LARGE, even if the duty cycle is low (5%).