I think he's saying that the calculation for the feed per tooth does not include diameter. I'm sure he is NOT trying to say that cutter diameter does not affect the OPTIMUM chip load, as clearly a 1/8" end mill can not tolerate chip loads that might be desirable on a 1" cutter.
Jon
3000 RPM and .5" gives about 400 SFPM, that's reasonable. 144 IPM, 3000 RPM and 4 flutes gives about .012" chip load, that's a good deal higher than I typically run. At full width (plowing) and .25" depth of cut, this would remove about 18 cu inches/minute of material. The "power factor" for aluminum is 3.0, so this should require 54 Hp! Won't work on a Bridgeport!
Jon
Jon, I was looking at this:
The depth of cut is also a function of the diameter. I normally would not cut deeper than about 1/2 the diameter if cutting through the stock. (You're removing the same thickness of metal per tooth with each size, but less total material with the shorter depth of cut with the smaller end mill.
If you are just shaving an edge, you can get away with a bit more, but remember that the smaller the end mill diameter, the greater the deflection -- and thus the poorer the cut and the greater the chance you will break the tool.
So -- if you keep the depth of cut reasonable for the size of the end mill, with the exception that you could wind up with too little chip clearance with a tiny end mill. Watch for that.
Good Luck, DoN.
Before spending lots of time, take a look at GWizard at
Oh, it entirely depends on how that number is used, and whether it is a multiplier or divisor. On this particular slide rule, it is used as a divisor (general HP value/P) so it is actually pretty close if your .25 figure is used as a multiplier.
Jon
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