Re: [CAD_CAM_EDM_DRO] Positioning

On 06/03/2014 12:38 PM, 'CS Mo' cs@...
[CAD_CAM_EDM_DRO] wrote:

>> I have read on another forum that retrofitting ball screws to a
>> convention mill, say a Bridgeport, reduces the power needed to move the
>> table - both by the operator and by the cutter. One comment was to the
>> effect that with ball screws, hand machining became quite difficult.
>> What I am interested in is how does one determine the necessary holding
>> torque of stepper motors (or other drivers) to avoid the table being
>> moved by the cutter?
>>
>>

it is a bit difficult. But, I tried to figure this out,
once. Assuming the worst
case is a slow-turning cutter (so it has a lot of belt
reduction in the spindle
drive) you could take the motor torque (you can get an HP to
torque
conversion online, all you need to know is RPM). Then
figure out the
belt ratio by the speed reduction. So, if it was a 1725 RPM
motor and
you had 300 RPM on the spindle, that would be a 5.75
reduction in speed,
so you get a 5.75 multiplication in torque. Assume the
motor was 2 Hp, 1725
RPM, that gives 6.09 ft-Lbs. Converting to inch-Lbs is 73.08
now, multiply by 5.75 for the belt reduction, and you have
420.2 In-Lbs.
So, if you had a 2" diameter cutter, it would have a radius
of one
inch, and so be exerting 420 Lbs of cutting force against a
side-cutting
surface. Now, to convert that to the torque exerted on a
leadscrew to
counteract this force. These screw conversions are
complicated, but
for a 5 TPI ballscrew, it is a linear advance equivalent to
a drum of .0318"
radius pulling on a string. So, 420 Lbs * .0318 = 13.4
in-Lbs of torque.
So a motor with a holding torque of 13.4 X 16 = 214 In-Oz could
resist this force. BUT, you need additional torque so the
motor can
move while resisting this force, so I'd at LEAST double that
for steppers.

Jon