Well - to greatly simplify the answer - it is a function of the shear strength of the material being cut, the shear angles of the cutter and just a few other variables....
It would likely be easier to look at the power available of a commercial machine and attempt to equate it to your application - not only the spindle power but power supplied to the axii as well and calculate in the advantage of the leadscrew then subtract the friction of the guiding mechanisms.
There are power requirement numbers on the net for cutting various materials. Google is your friend for this.
Might not be the answers you were looking for, and definately far from complete, but you asked for any help!
Scott
Consistency is a good thing....unless you're consistently an idiot.