Hi
I (we?) have no idea of the application, but if the system is always going
to operate at low speeds, what about a fat flywheel on the drive motor to
smooth things out?
Regards
Roland
On 23/04/2008, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
> Hi
> I want to describe my understanding for low feed system.
> I will use direct drive mounting and that mean 5 revolutions for 1 inch of
> travel per minute.
> 5 revolution times 360 degree is 1800 degree per 1 inch per minute.
> 0.0001 travel per minute means system will have 1.8 degree per minute.
> My system generate 8192 pulses per quadrate so 4 times 8192 =32768pulses
> per 360 degree of revolution.
> So, if I divide 32768 by 360 I will have pulses per 1 degree of
> revolution, = 91 pulse per degree.
> For 1.8 degree per minute I will have 1.8x91= 163.84 pulses per minute.
> Finally, in the case of feed 0.0001 inch per minute I will have 163.84
> pulses per minute.
> Per second it will be 163.84/60 = 2.731 pulses per second.
> So, is this enough pulses?
> If I put 1250 K encoder, will it solve the problem on the way in obtaining
> stable 0.0001 per minute feed?
>
> Is 1250 K pulse is it per rev or it is per 90degree or revolution?
> Thanks
> Aram
>
>
>
> > Hi Aram,
> > I did find a high res encoder at 1250K counts/rev. I'm not really
> > certain that is ppr or in quadrature.
> >
> > Price is in the $1000 range.
> >
> > http://www.opticalencoder.com/pdf/CP-850-950-HHC.pdf
> >
> > Dave
> >
> > On Apr 22, 2008, at 11:09 AM, [EMAIL PROTECTED] wrote:
> >
> >> Hi
> >> I have 3 questions.
> >> First if to make motor work on very low feed – 0.001 per minute
> >> need only
> >> high resolution encoder than may be needed belt reduction to the
> >> encoder
> >> shaft only, to get more pulses and not to whole motor shaft. Is this
> >> correct?
> >> Second, what is reasonable low feed with direct drive to motors
> >> that have
> >> 8192x4 puls per revolution?
> >> Third, how much those encoder, with 2 250 000 lines, may cost ?
> >> Thanks
> >> Aram
> >>
> >>
> >>> [EMAIL PROTECTED] wrote:
> >>>> Hi
> >>>> I want to build system to tool grinder and that means that my system
> >>>> should behave stable on very low RPM.
> >>>> Minimum PRM 0.0001 per minute. I want to use direct drive with ball
> >>>> screw
> >>>> 5 rev per inch or 1 rev for 5 mm in case of metric system.
> >>>> My motors have 8192 pulse per ….. it takes 4x8192 per revolution.
> >>>> It is
> >>>> AG
> >>>> industrial from servo dynamics.
> >>>> Any fundamentals ideas?
> >>>> Should I use higher resolution encoder?
> >>> If you really need .0001 Rev/Minute, that is 32768 counts *
> >>> .0001 = 3.3 encoder counts/MINUTE, or 18 seconds between each
> >>> encoder count. You can't get smooth motion like that. Of
> >>> course, .0001 RPM x 5 TPI on the screw is a movement rate of
> >>> .00002 IPM. Do you truly need it this slow?
> >>>
> >>> To get smooth motion, you really want an encoder count rate of
> >>> maybe 15 counts/second, or even better, 60. So, for 15 cts/sec
> >>> at .0001 RPM, you need 15 * 60 * 10000 = 9 000 000 counts/rev,
> >>> or 2 250 000 pulses/rev. This will be a pretty expensive
> >>> encoder. (Multiply by 4 for 60 counts/sec.)
> >>>
> >>> Jon
> >>>
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