Milling aluminum and cooling

[irving2008]

Power(kW) = KQ/1000, where K = cutting power per cc (17 for ali), Q = volume removal rate....assume 70% efficiency in spindle, 0.7 * 0.6 * 1000 = 17 x Q => Q= 24cc/min, add safety factor of 50% then in reality you'll get somewhere around 12cc/min at a typical feed rate of 0.25mm/tooth.

0.25mm/tooth on a 2 flute cutter at 10000rpm (spindle speed drops under load) = 0.25 * 2 *10000 = 5000mm/min, thats your target feed speed to avoid overheating tool by rubbing, but in reality you'll struggle to acheive that so a comprise will be needed - lets say you can do 500mm/min....

12cc = 1200mm3/min, if the cutter is endmilling 3mm wide at 500mm/min then the depth of cut = 1200/(3 x 500) = 0.8mm

for 12mm thick ali you'll need 15 passes for the depth, times however many for width...

a larger dia cutter you'll need to reduce the depth of cut proportionally, or risk burning out motor/stalling it/breaking cutter or usually all 3... larger cutters will need a slower spindle speed... if you can vary it...

however start with a lower feed rate... increase DOC from .2mm until machine complains (and it will) then back off 25% - you'll soon learn limits of machine and cutter (and you'll break a few so get some in)