It's coming to the gallery shortly.
Just to make sure my numbers make sense, I took the stall torque of the small CIM (343.4 oz-in) doubled it, and converted it to lbs (42.9250 lb-in). Then I divided it by 4mm converted to inches to get the force at 4mm (272.57375 lbs). I put a pin constraint on each of the 4, 4-40 holes (at .455 radius... taken from the whitepaper) and applied the calculated load to the keyway.
Does that sound right? Since I'm right out of high school I'd like it if someone verified this.... The sketch is in the NBD whitepaper, and took me 10 minutes from sketch to analysis.
EDIT:
It's uploaded.