Stepper Motor Suggestions Requested
Hello,
I need to turn an 18" wooden disc with a stepper motor (connected to the center of the disc, similar to a turntable) controlled by a BS2. The stepper will not be supporting the weight of the disc. I need fairly high arc resolution.
I've not used stepper motors before, so I thought the robots thread would be a good one for suggestions, since you all work with steppers.
I have no idea of what torgue would be needed or the size of the stepper. Could someone educate me on this? I don't want to buy more than I need.
Thanks
I need to turn an 18" wooden disc with a stepper motor (connected to the center of the disc, similar to a turntable) controlled by a BS2. The stepper will not be supporting the weight of the disc. I need fairly high arc resolution.
I've not used stepper motors before, so I thought the robots thread would be a good one for suggestions, since you all work with steppers.
I have no idea of what torgue would be needed or the size of the stepper. Could someone educate me on this? I don't want to buy more than I need.
Thanks
Comments
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
The turning speed is not an issue, so long as it is reasonable. For example, if it would turn a 16" disc one revolution in 10-15 seconds that would be OK.
The disc would weigh 1-2 pounds and be in a fairly humidity constant invoronment, so changing humidity should not be an issue. If this weight is not reasonable, I could always fabricate a lighter weight metal disc.
I = 1/2 * m * (r/12)^2
Where:
I is the Inertia in pounds per square feet or torque
m is the mass in pounds
r is the radius in inches
I = 1/2 * 2lb * (9in/12)^2
I = 1lb * (9/12)^2
I = .5625 lb-ft^2
As a rule of thumb, I would probably double this value. So to start you need to find a geared stepper capable of providing at least 1.125 pounds per square feet of torque. The reason I say geared stepper, is the statement you made wanting a high angular resolution. ...Best way to do this is and increase your output torque at the same time is to use a geared stepper motor.
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Reference:
http://www.upei.ca/~physics/p261/projects/flywheel1/flywheel1.htm
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 6/5/2007 8:47:38 PM GMT
excuse me if this sounds like a nit, but I don't think that's quite correct (although your end result does seem entirely reasonable).
The moment of inertia of the disk is in pounds-square-feet, not pounds per square foot (personally I'd prefer SI units, but whatever...
More important, it's not the same as torque. Torque is in pound-feet (or Nm...) and is related to the moment of inertia by the angular acceleration in an analogous way to the relationship between linear force, mass and acceleration:
T=Ia
where T is the torque, I is the moment of inertia and a is the angular acceleration in radians per second squared. Your rule of thumb is saying that the minimum acceptable torque for driving a flywheel is that which is sufficient to accelerate it from rest at 2 radians per second squared.
I think that's right. Apologies for any errors...
Regards
Duncan
Ok, so it looks like the formula isn't quite finished... From the previous post, I have provided the inertia (I)
Using the following formula with a revolution once every 10 seconds...
T = (I
Correct me if I'm wrong, but don't you still need the torque from the Stepper to overcome the initial or static inertia? It's analogous to rolling a basket ball verses rolling a bowling ball. Even if your intent is to·move each object at the same speed, one's going to be more reluctant than the other to move in the intended direction. This 'static inertia' I think is what the motor, Stepper or whatever,·should be rated for.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 6/6/2007 3:12:52 AM GMT
yes, the moment of inertia is the inertia that you need to overcome at startup. I don't think there is such a thing as "static inertia"; it's the same irrespective of the angular velocity, but doesn't matter once the system has reached its terminal angular velocity after switch on.
The formula I gave earlier is for angular acceleration, not angular velocity. It isn't saying anything about the terminal angular velocity.
The angular velocity will be dependent on the characteristics of the motor and whatever torque is created in the direction opposite to rotation as a result of friction and inertia.
I'm not sure how that would work for a stepper motor, but for a DC motor it's going to work something like this:
There is a relationship between angular velocity and torque. At the instant the system is switched on, the angular velocity is zero, and the torque exerted by the motor is going to be the stall torque. That will determine the initial angular acceleration, in combination with the moment of inertia of the load.
The equilibrium condition will be reached once the angular velocity reaches a value such that the motor torque and frictional torque are equal for that particular angular velocity. For a spinning disk, that should be fairly close to the "no load" angular velocity of the motor. At that time, the system is in equilibrium, there is no angular acceleration and the moment of inertia of the disk is no longer important.
At all points in between, the load torque is a combination of the frictional torque and the reaction to acceleration of the disk due to its moment of inertia.
You know, this would be a lot easier with a whiteboard... not sure if any of that is terribly clear
Regards
Duncan
[noparse][[/noparse]edit] ...but yes, the inertia is what you're going to care about in terms of the rating of the motor (and there is only one inertia, you just don't care about it once the system is up to speed, so that is what you are calling "static inertia"). The issue is that, while the motor is stalled or rotating at low angular velocity, there is no back EMF (I hate that term, but whatever) which means that a large amount of current is going to be dumped into the motor, putting it at risk of burning out before it gets up to speed. You need enough stall torque capability for the motor to accelerate the load out of the "danger zone" quickly enough that the motor doesn't burn out before that happens. If there is no other loading on the disk, the motor torque doesn't really matter much apart from that. Once the motor gets up to some reasonable angular velocity, the motor back EMF increases, reducing the current draw of the motor and preventing it from overheating.
Post Edited (Uncanday) : 6/6/2007 4:18:55 AM GMT
I am not a mathematician, but here is a link or two that could help you with your project.
http://www.homeshopcnc.com/page5.html· for the stepper motor, I would suggest a nema 17.
http://www.grizzly.com/products/h5685·for the rotatory table, just install the wooden disk on top.· this will turn it.
http://forums.parallax.com/showthread.php?p=641371·I made this·BS2 Electronics wiring diagram and quick program to control it.
if this is out of your league, contact me; i will give you other suggestions.
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Fernando Gomez
revinc.us
gomez-rivera.com
Never compare yourself with anyone else, there will always be someone bigger·or·smaller·than you.
Post Edited (willy1067) : 6/10/2007 2:00:18 AM GMT