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Sizing a Servo

04/25/2009 8:07 AM

I have a ring I want to drive with a servo, but I'm not sure how to calculate the torque needed to drive the ring.

The ring weighs 6.7kg, it is 656.74mm O.D., it has external gear teeth at 638mm O.D. with pitch circle diameter of 636mm and 636 teeth. There will be approximately 100grams of hardware mounted on the ring, divided into 4 equal parts spaced evenly on a hole pattern with a diameter of 612mm to the centers.

The pinion on the drive motor is expected to be .5kg with an O.D. of 42mm and have 40 teeth.

The motor ratings are given in Nm for maximum torque, and Joules for rotor moment of inertia.

I need help understanding the thought process to solve this.

Any help is greatly appreciated.

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#1

Re: Sizing a Servo

04/25/2009 8:59 AM

Once you've got to steady state (i.e. constant rotation rate), the masses, moments of inertia etc. don't come into it - the torque will just be down to friction (which you haven't given).

Inertia only comes into the equation when you're accelerating/decelerating, so you need to give some details about how fast you want to rotate and how long you want it to take to get to speed. (You also need to at least approximate the friction, to see whether it is significant relative to the accelerating torque).

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#2

Re: Sizing a Servo

04/25/2009 10:40 AM

Sorry, I should have known better.

Please assume friction is zero. Very good bearings are being used. But, the ring is used to cut through material which will put approximately 5kg of opposite force on the ring. The ring needs to turn 1/4 turn in 1 second. It will be driven by a servo motor, stop and go, one quarter turn at a time.

The ring and motor will also be mounted on a linear actuator in order to cut on the fly at linespeed of 0.65 in/sec. This would product some extra forces I guess but it shouldn't be a show stopper in my opinion.

Thanks.

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#3
In reply to #2

Re: Sizing a Servo

04/25/2009 12:32 PM

Also need to know a bit about the construction of the ring - is it a disc of uniform thickness? If not, what?

Is the 5kgf tangential to the periphery ring? If not, at what radius from the ring center of rotation?

Is the ring balanced? If so, I don't think the superimposed linear motion will affect the torque calcs.

It would be easier to sort out the geometry & hence calculate the forces & torques if you could post a sketch, or give a link to one on e.g. flickr.

I'd like to help, but need to know more!

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#4
In reply to #3

Re: Sizing a Servo

04/25/2009 5:00 PM

It is a ring slide from Bishop Wisecarver. Supported by 4 "V" bearings. Wires form a cross in the center and cut through the product sitting in one quadrant. Like a cheese slicer. The center of the product is approximately 5 inches from the center of the ring. The ring itself is kind of "T" shaped. Bearings could support it from the inside or the outside or both. I chose the outside. The ring cross section from edge to edge (ID and OD) is 44.74mm. From front to back is 15.5mm at widest points.

Thanks for your help!!

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#5
In reply to #4

Re: Sizing a Servo

04/26/2009 7:37 AM

OK, I've set up an Excel spreadsheet to do the calcs. I don't suppose it'll be very legible, but this is what it looks like:

I've allowed 10° for the ring to accelerate, in the hope that it'll be up to speed before it starts cutting. I've also assumed that the effective cutting radius is 1/2 the disk radius. Some other assumptions in there, too, but I think they're reasonable.

The bottom line results (all referred to motor or gearbox shaft driving pinion) are:

Max. rpm: 1166

Torque to accelerate ring: 1580 Nm***

Torque to accelerate pinion: 0.26 Nm

Torque to cut product: 124 Nm

I think the calcs are OK, but I'd like someone to check them before anyone spent good money on motors/gearboxes/drives etc. I'm happy to send the spreadsheet over to anyone who wants a look (pm me your e-mail address).

I've got to go out in a while, but if you're on facebook, you could join the CR4 group & maybe we could fix a mutually suitable time for a live chat.

Hope this helps.

*** That figure looks pretty huge, & definitely needs checking! I've have another try later. J.

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#6
In reply to #5

Re: Sizing a Servo

04/26/2009 8:27 AM

*** DON'T PANIC!

Driving daughter to drama rehearsal - realized error of my ways (multiplied torque by gear ratio, instead of dividing ).

Revised spreadsheet:

The bottom line results (all referred to motor or gearbox shaft driving pinion) are (now):

Max. rpm: 1166

Torque to accelerate ring: 6.25 Nm

Torque to accelerate pinion: 0.26 Nm

Torque to cut product: 0.49 Nm

This lot sounds a bit more reasonable!

(BTW I've neglected friction & gearing efficiency throughout).

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#9
In reply to #6

Re: Sizing a Servo

04/26/2009 8:36 AM

I'm very interested in the thought process and formulas used in order to solve the problem.

Can you share?

Thanks.

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#10
In reply to #9

Re: Sizing a Servo

04/26/2009 9:19 AM

It's a bit long-winded, but here's a summary of some of the important bits:

Assuming the ring (with mass Mr) is a uniform annulus with outer radius Rro and inner radius Rri, the moment of inertia is given by:

Jr = Mr x (Rro2 + Rri2)/2

To find the torque to accelerate, we need the angular acceleration. We have:

Arc = ωc x tc ........(1)

where Arc is the angle the ring moves at constant angular velocity,

ωc is the angular velocity during the constant phase,

and tc is the time for the constant phase.

Ara = (ωc / 2) x ta ..........(2)

where Ara is the angle the ring moves while accelerating (or decelerating)

and ta is the time to accelerate.

The total time, tt, is given by:

tt = tc + 2 x ta ..........(3)

Rearranging (1) and (2) for tc and ta , substituting into (3), and then rearranging for ωc , we get:

ωc = (Arc + 2 x 2 x Ara)/tt ........(4)

I've assumed 10° for acceleration (and another 10° for deceleration).

Taking tt = 1 second, converting the angles to radians and substituting into (4), we get ωc.

The angular acceleration, α, is given by:

α = ωc2 / (2 x Ara)

The torque to accelerate, Ta, is given by

Ta = Jr x α

The torque at the pinion drive is just this figure divided by the gear ratio.

For the cutting torque, I just took the cutting force, multiplied by g to get Newtons, and multiplied by the radial cutting distance. This is again divided by the gear ratio to get the final result.

Hope this makes sense. As I say, I'm happy to e-mail you the spreadsheet if you want. I also have a document describing the moment of inertia, angular equations of motion and torque formulae used, which I could send as an MS word doc or a pdf.

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#11
In reply to #10

Re: Sizing a Servo

04/27/2009 10:27 AM

Would somebody else please give this man an additional GA! Excellent work John.

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#12
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Re: Sizing a Servo

04/27/2009 11:22 AM

Aw, shucks !

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#7
In reply to #5

Re: Sizing a Servo

04/26/2009 8:28 AM

I can barely read your spreadsheet, but I think you have .25 sec for total move time.

Actually, the ring will move 1/4 turn in a full second.

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#8
In reply to #7

Re: Sizing a Servo

04/26/2009 8:36 AM

Good eyesight!

New revision:

Pinion 291 rpm

Torque to accelerate ring: 0.39 Nm

Torque to accelerate pinion: 0.017 Nm

Torque to cut product: 0.49 Nm

Sorry.

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