Re: [CAD_CAM_EDM_DRO] Bent Nook ballscrews ?????

On Mon, 2003-08-11 at 22:24, Jon Elson wrote:
> Kim Lux wrote:
>
> >A substantial amount of force would have to be greater than the pushing
> >capacity of the motor so that there was never a force that would
> >displace the nuts against each other. If the spring is weaker than the
> >pushing force of the motor, then it will displace the backlash under a
> >heavy cut.
> >
> >As soon as the nuts are adjusted for the maximum force the motor is
> >capable of moving, there isn't any force left to move the cutter.
> >
> >
> No, you're confusing drag with linear thrust.

No, I'm pretty sure I'm not. Lets say we have an axis that needs, on
occasion, 500 lbs of cutting force. In order to ensure that we never
had backlash, we'd have to put a preload and the backlash nut of 500
lbs, to keep the backlash zero in the presence of our load. That then
puts the load on the ballscrew at 1000 lbs, ie 500lbs for the cutting
load and 500 lbs for the backlash.

> If the system is designed to
> apply 1000 Lbs of linear thrust, you want somewhat over 1000 Lbs
> of preload force between the two ballnuts.

Yes. Thus the total load on that ballscrew is 2000 lbs. It takes twice
the motor to turn the preloaded system.

> On a good ground screw,
> 1000 Lbs of preload will cause a very mild drag, maybe 10 Oz-In
> when new, and decreasing as the screw and nuts wear in a little.

Where do you get your information from ? A 3/4" 5 TPI ballscrew has a
force to torque ratio of about 31 lbs force per in pound of torque.
1000lbs/31 inlbs/lb = 32ish inlbs of torque or 516 oz in of torque, just
to turn the backlash preload on the nuts.

> You don't have to adjust the nuts until they bind up. That could be
> 5000 - 10,000 Lbs of preload, and will quickly destroy the balls.

HUH ?


>
> Jon
>
>
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Kim Lux <lux@...>