Motion Control Suggestions
cdragon
Posts: 5
I'm more of a electronics/software kind of guy, but I'm working on a project that involves some moving parts this time.· What I need to do is control the rotation of a disc (4-5 pounds) so that it stops 10 times a rotation in the same·precise locations every rotation.· The RPM will be around 300.
I'm looking for suggestions, should I use servos or steppers?
Any suggestions would be appreciated.
Thanks.
-Marc
I'm looking for suggestions, should I use servos or steppers?
Any suggestions would be appreciated.
Thanks.
-Marc
Comments
I'm sure there is a solution. I'm not sure it's clear about the speed. You said that the 4-5 pound disk will rotate at 300 rpm but also that it will stop 10 times on each rotation. That amounts to 3000 stops and starts every minute. That means you would have to move a 4-5 pound inertial load 36 degrees and stop it accurately in 20 milliseconds. That doesn't even take into account how long it is stopped.
Is there more info you could supply that would help someone understand your goal?
Lee
I have to agree with Lee, this is just a bit confusing as to what's actually happening and when. Can you break it down, step-by-step as to what you need to have happen, in sequence, how much time you have to do each step, and what speeds must be attained/maintained during the process. The process itself seems like some wire or rod extrusion processes I've seen.
My gut feeling at the moment, just based on what you'd said so far, is that you will probably need more than one element in the system that causes rotation. One certainly will be a motor, but I suspect the other may be something like a high-speed, rotary solenoid with mechanical linkage, for the exact positioning and rapid "hole" advancement.
I also suspect that you may need to use a clutch and a brake on this system, so the motor itself doesn't need to stop and start all the time, but rather can run continuously, with the clutch/brake bringing the disc to a halt as needed.
Regards,
Bruce Bates
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200 steps per rev / 10 positions = 20 steps per postion
2000 steps per second = 2 steps per msec
20 steps / 2 spmsec = 10 msec move per position
If you drive the disk from a central shaft, the interia is really working against you.
You may have to have a substantially large motor to get the starts and stops.
You may have to use PID, an encoder, and/or an electric brake·to get even close.
And you still may have a lot of misses [noparse][[/noparse]overshoot and undershoot].
On the other hand, if you have a stepper driving a ring gear on the outer rim; it all may be quite simple.
With the stepper on the outer ring, your math shifts into a much finer resolution due to the gearing.
And your motor is working with leverage against the inertia.
I can't do the math for you without actual gear sizes, but I think you would be more satisfied with the results.
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"When all think alike, no one is thinking very much.' - Walter Lippmann (1889-1974)
······································································ Warm regards,····· G. Herzog [noparse][[/noparse]·黃鶴 ]·in Taiwan