It appears to me that we can convert from frd_sigma if we know the f number of the input beam light. In the code, frd_sigma is just the fraction of an image length corresponding to the convolution (with a factor of sqrt(2) due to.how scipy.special.erf works). If effective_illum_radius as passed to the code represents the fraction of the image length that the pupil image takes up, then the ratio of frd_sigma/effective_illum_radius is the conversion factor between the angular size of the pupil image and the angular size of the FRD (i.e., the 'actual' value of FRD). I believe that generally f/2.8 is used in the fibers (though you have the correct values for each individual measurement), corresponding to a numerical aperture of 0.17857 (=1/[2*2.8]).

It's simplistic, but I think in general "actual" frd in radians = (Numerical Aperture)*sqrt(2)*frd_sigma/(effective_illum_radius).

As a sanity check, if effective_illum_radius ~ 1 and frd_sigma~0.05-0.1, then for f/2.8 the frd ranges from 12.6 milliradians to 25 milliradians, which sounds a little high but reasonable depending on the experimental setup.

I'll follow up with lab data to verify.

FRD is incorporated into all analysis up to this point and in the foreseeable future with this equation.