Spindle torque required

[m_c]

Here's one to get the brain cells working.

If you have an aluminium part, approximately 15mm diameter mounted on a spindle (we'll call it spindle one, or S1), and you want to machine said part with a 6mm cutter mounted on a seperately powered spindle (spindle two /S2), how many Nm will S1 need to ensure that the cutter in S2 won't spin it in ways you don't want it.
And assume that S2 would have enough power to snap the cutter long before it stalled.

Answers on a postcard please.

[irving2008]

Crudely, as long as S1 has more torque than 15/6 of S2 it'll stay put... ish.. it all depends on DOC and feed rate as well as S2 spindle speed, power input and cutting efficiency of tool...

For some theory look here: aluMATTER | Aluminium | Milling | Milling: Overview (http://aluminium.matter.org.uk/content/html/eng/default.asp?catid=126&pageid=2144416174) then take a wild guess :)

[m_c]

Interesting link that.

I've just had a quick google, and one of the first hits was Cutting Force During Milling - CNCzone.com-The Largest Machinist Community on the net! (http://www.cnczone.com/forums/mechanical_calculations_engineering_design/100340-cutting_force_during_milling.html)
Which for a 1/4" cutter gives a maximum measured force of 700N.

Now taking Torque = Force x distance = 700N x 0.015m = 10.5Nm

Which isn't quite high enough to totally kill my simple idea, but would involve a stupidly powerful motor, so back to the complex idea.