Re: stepper motor inertia

Phil,
The mistake was that for a cylinder I= m r^2/2 ; I forgot the /2 when
I copied out of my notebook.
So the constants combined together are pi 7.8/2 = 12.2 .
Thanks for spotting the inconsistancy.

It looks possible that a motor operated at maximum torque, calculated
from the torque/speed curve, could move about 3X faster over long
distances than a motor with constant acceleration (speed) up to some
speed limit.
Won't someone who writes stepper motor software (Art?) code this and
find out if I'm right or wrong?

regards-
Elliot

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Philip
Burman" <philip.burman@...> wrote:

>
> Hi Elliot,
>
> I think there is an error in your calc.
>
> Your constant 12.2 should be Pi*density which is 24.50.
>
> Regards
> Phil
>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "optics22000" <elliot@>
> wrote:
>
> >
> > the moment of inertia of a cylinder rotated around its axis is
> > I= m r^2
> > m= pi r^2 l density
> > density steel is about 7.8 g cm^-2
> > so I=12.2 r^4 l
> >
> > Conside a 1/2" diameter leadscrew 1' long,
> > I=60 gm cm^2 ; this is much less than the rotor inertia
> >
> > regards-
> > Elliot
> >
>