Okay, I found a paper copy of the explanaion I recieved. (I need to work on my organization skills)

In this example we will move 3" at 10IPM so it will take 18seconds or .3 minutes. (You can use a stopwatch to verify)

the formula for inverse time is

1/(length/feedrate)

this means...
1/ (3/10)
1/.3
F=3.33

1. G0X0Y0
2. G1X3.0G93F3.3

when you add an A axis move to this the machine will break down each axis separately and make sure each axis finishes in the time allowed.

In typical simultaneous 4 and 5 axis motion the cam will break down the moves to about .003" long. In this case at 10IPM, the move would take .0003 minutes. Since most controls don't like the tiny numbers, they came up with using the "Inverse" of the time. So 1/.0003=F3333, It is common to see Inverse time feeds in the range of F100-F10000. Some controls have a setting for max inverse feed rate you will have to check yours.

P.S. remember to program a G94 to return your feedrates to IPM!!!

hope this helps,
As far as the cad points you have calculated for the radius "interpolation" , how small are your moves? How critical is the radius form?