Reverse recovery current in slow mains rectifier diodes?

Hello,
Do you know what is the reverse recovery time (trr) of the diodes in the DBLS209G mains rectifier bridge?

If it were connected in a circuit as in the attached (LTspice sim also attached), then would you expect to see much of a reverse recovery “snap-off” of the reverse current?
…..By this, I mean as described here by Power.com...............................
(i presume the slow reduction to zero current and commutation at the mains zero cross means that reverse recovery "snap off" will be very minimal?...thus the recomendations of power.com will not be relevant to us?)

EMC problem due to slow diodes in mains rectifier…..
**broken link removed**

DBLS209G Rectifier bridge datasheet
https://www.mouser.com/ds/2/395/DBLS201G SERIES_H13-523528.pdf

What I will say next is subject to discussion because is a very very wild guess.

I have calculated the time it takes for the diode to reach the reverse current given in figure 2.
Rated peak reverse voltage=1400 V
Reverse voltage applied to 1 diode = 339 V => percent of the Rated peak reverse voltage = 24,21 % => from fig 2, I get Reverse current (at 125ºC because it is the worst case) ~15 uA.

Now, I calculate the current it takes for a straight line with the slope of a sine wave in "pi" to reach that current.
current(t)=339/450 * sin(ω*t) => slope in Π/ω is : dCurrent(t)/dt (Π/ω)= -339/450 *ω

Tangent line to the sine wave in Π/ω equation is: y=-339/450 *ω * (t-Π/ω) (taking the origin at t=0s ). with origin at t=0,01 s it is: y=-339/450 *ω*t
So, y=-15 uA => time to reach that reverse current: -15 uA = -339/450 *ω *t => t=63.38 ns

Obviously the peak reverse current is higher than the supposed -15 uA so the real time is expected to be higher.. Anyway, this is just a very big wild guess.

Can you measure your reverse recovery time in real life ?