Here is an example based on the Biogon 38mm:

Assuming it has a fall off as good as Cos^3

The size of a 6x6 negative is actually 55mmx55mm, its diagonal is 77.77mm
and there is arcTan(77.77/(2x38)) degrees from the center to the corners of
the film, that is 45.66 degrees or a total angle of view of 91.32 degrees
from corner to corner.

The fall off at the corners would be:

Theta = 45.66
Cos(45.66) = 0.6989

Fall-off in Stops = 3.322 x Log(1 / Cos^3 of the angle Theta)

Fall-off in Stops = 3.322 x Log(1 / (0.6989)^3)
Fall-off in Stops = 3.322 x Log(1 / 0.3414)
Fall-off in Stops = 3.322 x Log(2.929)
Fall-off in Stops = 3.322 x 0.46672

Fall-off in Stops = 1.55

There are ONE and a HALF stops of fall off at the corners of the film with
respect to the center of the film!!

Guillermo