RE: [CAD_CAM_EDM_DRO] Positioning

There's a formula to calculate holding torque, see down the bottom of this
page:
http://www.nookindustries.com/LinearLibraryItem/Ball_Screw_Design_Considerat
ions#figure1





The diameter of the screw doesn't matter, only the pitch. You're turning
linear motion to rotary, basically pushing something up a ramp. Higher
pitch = steeper ramp.





Tony








From: CAD_CAM_EDM_DRO@yahoogroups.com
[mailto:CAD_CAM_EDM_DRO@yahoogroups.com]








Hi Jon and all:
The total force to move the table is very low if it is floating on a bed
of oil.
The force to crank the table is very low when you have ball screws of the
larger
diameter like is on a Bridgeport mill. The cutter force will easily back
drive the
ball screw if you do not have a locking force on the table.
Jon, could explain how you came about the .0318 drum diameter? I know
the force is low, but how did you arrive at this diameter. The method to
calculate
this skip past me at this time. Maybe I have been reading too much about
turning!!!

GOD'S BLESSINGS

Bill Thomas
P.S. Jon how is your sons doing?





On Tue, 3 Jun 2014 14:19:25 -0400, 'CS Mo' cs@... [CAD_CAM_EDM_DRO]
wrote

>
>
> Jon, I think you're forgetting that there is already a lot of friction in

the system. You'd have to calculate the force required to move the table and
subtract that from your value. With the screws disconnected I think you'll
find that it is pretty hard to move the table by pushing it (x or y) by
hand.

>
> --CS
>
> >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
> >
> >
> >------------------------------------
> >Posted by: Jon Elson <elson@...>
> >------------------------------------
> >
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