On 07/20/2020 11:49 PM, Tom Smart wrote:
So my 3500 rpm rated motor at 180vdc would be 2916 rpm at 150 vdc and 2333 rpm
at 120vdc?
Yes, exactly.
If my ballscrew is 5tpi then at rated voltage i would get 700 ipm at 150vdc 583
ipm and at 120vdc 466ipm feeds?
Yes.
If i keep the amps at the rating of 9.1 I should keep my 31.3 IN-LBS or 2.6
FT-LBS?
So 2.6 / .0813 would be 81.76 lbs of linear force?
That denominator s .0318 but your result is correct.
So if my table weighs 200lbs my acceleration would be .41 G?
I'm wondering if I've done a calculation wrong or is this a good setup?
Well, 81 Lbs sounds pretty weak for a milling machine. Is
the motor directly driving the leadscrew or is there a belt
reduction? My Bridgeport setup is kind of anemic, but it
has a 2.5:1 belt reduction on the motor. That gets me to 23
In-Lb at the leadscrew, for 700 Lbs of linear force.
Now, the 9.1 A rating of the motor, is that a continuous
(stall) rating or a peak rating? If that is stall, then the
motor can handle more current during acceleration peaks, at
least twice, possibly 4 X.
That will help.
Jon
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