More on servo motor selection

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From drew@... Mon Sep 10 19:24:00 2001 -0700
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Hi all,

I did some more thinking about what I expect from the little CNC mill
I'd like to play with and I think that a .100" deep full width cut with
a .500" dia. endmill at 30 ipm seems to be a resonable max. That cut
removes .1 x .5 x 30 = 1.5 cu. in. of material per minute. At Jon's
suggested rate of 3 cu. in. per hp for aluminum I need .5 hp to sustain
the above rate. 1 hp = 746 watts so I need 373 watts. So given Mariss'
calculation that, what seems to be the better motor I mentioned earlier
(see message #30095), can generate 60 oz. in @ 2600 rpm which equals
about 115 watts, I need at least a 3 to 1 reduction between the motor
and the table. The table I'm using has a 10 pitch leadscrew so is this
more than enough to cover the required reduction. In other words, does
the leadscrew turn the 115 watts of the motor into 1150 watts? If I
do add say a 2 to 1 reduction between the motor and the leadscrew do
I now have over 3000 watts? With a 2 to 1 reduction @ 2600 rpm I can
still run the table at 130 ipm which seems plenty to me right now.

So how does all this sound? Am I way off in left field confusing electrical
hp with mechanical hp?

Jon E., could you explain the process of computing the forces at the
table for computing the
Drew Rogge
drew@...