Originally Posted by tonyel  What sort of forces are generated by routing aluminium? I'm not really worried about getting to high speeds, as this is only a hobby type machine. I'm thinking of doing a little FE analysis to see what sort of deflections I'll get in the structure, but I don't really know what kind of forces are involved. My guess is that it wouldn't be more than my weight (about 70kg, since you ask...). |
I did some research into this but there is little hard information. One approach is to look at the chip size and work that back to the cutter size, rotational and linear speeds needed and then equate that to the required energy which then translates to torque which based on cutter size translates to the cutting force. Its but a crude measure but allow a frig factor of 2 you can get some estimate.
So lets say a 10mm, 2 flute cutter milling aluminium taking a .5mm cut. The spindle speed required is 1000 * cutting speed)/(pi * cutter dia) = 320 * 100/10 = 3200rpm
Lets say you gear the screws down 2:1 and spin them at 2rps = 120rpm = 3000mm/m (motor = 4rps = 800 full steps/sec, full step = 25/400 = 0.0625mm, so microstep at 1/8 give better than 0.01mm resolution)
At 3000mm/min feed on a 2flute cutter the feed per tooth is 0.5mm/min and the removal rate Q = 30cc/min
Power = Q * k where k=12 for aluminium = 360W (k rises to over 17 for feeds below .2mm/min which is why you want high feed rates to reduce power input - non-intuitive or what!)
Torque = Power * 60/ (2 * pi *revs * eff) where eff is cutter efficiency of typically 80%
= 360 * 60 / (2 *3.14 * 3200 *.8) = 2.25Nm
which acting at 5mm radius = 2.25/.005 = 450N so you need to be allowing for up to 900N force on the workpiece from the cutter... so 90Kg, you weren't far off...
The electrical input will need to be Power/(cutter eff * motor eff) = 360/(.8 *.6) = 720W (so forget a Dremel!)
Don't take this verbatim, I may have made a boo boo... here is some useful
guidance to milling aluminum