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

On Tue, 2003-08-12 at 09:13, Les Newell wrote:
> Hi Kim,
>
> > 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.
>
> With no cutting load there is a load of 500lbs on each nut. The important
> thing to remember is that the load on each nut is in opposite directions
> therefore the torque generated by the preload on one nut is cancelled out by
> the other nut.

It depends which direction the cut is going. In one cutting direction
there is going to be a load of 500 pounds from the cut AND 500lbs from
the preload, for a total of 1000 pounds of load on the ballscrew.


> If we now take the case of full cutting load on the unsprung nut: there is a
> load of 1000lbs on the unsprung nut and 500 lbs on the sprung one. The force
> that the motor has to generate is the difference between these two values.

In one direction it is. In the other direction it is the sum of the
two.

> In the other direction there is a force of 500lbs on the sprung nut and 0lbs
> on the unsprung one. The difference is still 500lbs.
>
> Adding preload does slightly increase the friction in the nuts but this is a
> prettly small loss and probably not worth worrying about.

HUH ? Am I missing something here ? 500 pounds of axial preload on a
ballnut is a very considerable load, especially for people using smaller
motors.


> Les
>
>
>
> >
> > Comments below.
> >
> > 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 ?
> >
> >
>
>
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Kim Lux <lux@...>