I believe the maths for diameter change is derived from Pythagoras, and should be:
2 x ( sqr ( radius^2 - sag^2 ) )
so for a target of 1 inch that's being worked on a lathe with 7 thou of sag, the actual diameter will be 1 inch where there's no sag, and where there IS sag,
2 x ( sqr ( 0.5^2 - 0.007^2 ) )
which my Babbage engine gives as 0.9999019952 inches.... Not a whole lot of error there. Even 20 thou droop gives a cut OD of 0.999199 - less than a thou for a 1 inch cut. For (say) a 4 inch diameter, the 7 thou sag will give 0.025 thou error, and the 20 thou sag will give 0.2 thou error.
You're right, Doc - as long as the ways are reasonably good, it really doesn't matter for all but the fine work.