Fwd: Re: [ATM] Determining Stepper Motor Size

[mlbrown at everstrive.com (Matthew L. Brown)]

Unfortunately, steppers aren't quite so simple.  The torque is highly 
dependant on the velocity and the drive electronics, and full vs half 
stepping mode.  Usually the torque quoted is the holding ("detent") torque.

As the speed goes up, the torque goes down.  See, for example:
http://www.usdigital.com/products/ms23/index.shtml

Now, as you can see, the low speed torque, which'll be breaking the static 
friction, will be one thing to watch.  The max running speed will be that 
which equals the turning friction.  You can estimate both with a set of 
weights on your tube (so much weight at such a radius).

=Matt


>Delivered-To: mlbrown@0
>From: "Frank" <frank@katestone.com.au>
>To: "Stanley A. Gorodenski" <stan_gorodenski@asualumni.org>
>Subject: Re: [ATM] Determining Stepper Motor Size
>Date: Wed, 18 Feb 2004 11:29:53 +1000
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>Hi Stan
>
>I don't think you got many (any??) replies to your email so here's
>my 2c worth.....
>
>It's not clear to me what you're trying to do so I'll just state some
>facts and maybe we can go from there.
>
>* A perfectly balanced telescope has 2 properties which affect its
>ability to be rotated about an axis - its moment of inertia and the
>static and dynamic friction in the bearings. Usually the dynamic
>friction is smaller than the static friction - once you get it moving
>it's easier to keep it moving.
>
>* If the bearings were perfect (no static or dynamic friction) a stepper
>would make the scope rotate (actually ACCELERATE)  about the
>axis at a rate (A) determined by the moment of inertia (I) of the scope
>and the torque (T) of the stepper according to T = IA (the rotational
>equivalent of F = ma). So even a very small motor can rotate a large
>scope about its axis if the bearings are good. (From memory
>(questionable) , the 200" uses a 0.5 hp motor to rotate it).
>
>* Perfect bearings don't exist so the stepper must deliver enough
>torque to overcome the static friction friction (a torque) to get the
>scope rotating. Once it's rotating, the stepper needs to overcome/cancel
>the dynamic friction and the scope will rotate at a constant rate (angular
>velocity). If the stepper can provide more torque than the torque due to
>the dynamic friction in the bearings, then the angular velocity will
>increase
>(ie we have angular acceleration).
>
>The torque that your stepper delivers (75 oz-in) can be viewed as follows:
>Imagine a horizontal axis of 1" radius. Wind some string around it and
>hang a 75oz weight off the string. Gravity will pull down on the weight
>and the weight (through the string) will try to rotate the axis. The torque
>applied to the axis by the weight/string mechanism is 75 oz-in.
>
>Would this be enough to turn your scope about the axis?
>
>Hope this helps
>Cheers
>Frank Q
>
> > I am attempting to determine if a particular stepper motor, 75 oz-in
> > torque, will be sufficient to be able to slew in declination, given a
> > particular reduction ratio. Because a telescope tube assembly is,
> > ideally, perfectly balanced, aside from friction it seems to me the
> > major determining factor in sizing a stepper motor should be the
> > tangential acceleration (in beginning physics text books this is
> > considered to be a constant velocity vector rather than an acceleration)
> > to get to a particular slew rate (the rate may be 4 degrees per second).
> > Once the tangential acceleration is determined, then one can compute the
> > force this represents which then would directly be used to compute
> > torque. This value would then be compared to the torque rating of the
> > stepper motor. Is this the correct procedure to determine the stepper
> > motor torque needed? If so, what is the generally accepted acceleration
> > value used?
> >
> > Stan
> >
>
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