Help! Inertia / Forces
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Posted by: newpunkrkr ® 02/21/2006, 09:44:46 Author Profile eMail author Edit

I posted this before but didn't any help. I'm not a student doing homework problems. I'm production manager for a steel brokerage and am building some machinery. Any help or advice would be greatly apprecaited. Thanks in advance! I've got an aluminum rectangular mass with two steel pins that I need to rotate at a high speed. It is 26" x 4.5". The aluminum weighs 46lbs, and the pins are 8.5lbs each. The pins are centered 11.25" left and right off center. I need to spin it 90 degrees in .450 sec (start and stop), via a 1 7/16" shaft mounted below in center. I'm having trouble calculating the moment of inertia and the forces/torque required to preform that operation. I looked at the calculators on this page but they are all for cylinders, so I'm unsure on the conversion to my application. Any help or suggestions to get me in the right direction would be appreciated. Build it, Break it, and then Build it Better

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Re: Help! Inertia / Forces
Re: Help! Inertia / Forces -- newpunkrkr	Post Reply	Top of thread	Forum

Posted by: zekeman ® 02/21/2006, 11:43:41 Author Profile eMail author Edit

The moment of inertia of the rectangular mass is: W/12(d^2+b^2) and the two pins are approximately 2*w1*r^2 where W= 46 w1= 8.5 b=26 d= 4.5 r=11.25 I get I=2669+2151=4820 lb in^2 If you want ro rotate this mass in .45 sec, the most efficient way is to accelerate it at a constant rate for 1/2 the time and decelerate it for the remainder bringing the mass to rest in .45 sec. The average angular vbelocity would be (p1/2)/t rad/sec and the peak is double or pi/t=6.98 rad/sec Since you accomplish this in .45/2 seconds the acceleration is 6.98/.225=31 rad/sec^2 From basic mechanics the minimum torque required excluding the driver would be: I*30.2=4820*30.2/g=4820*31/32.2=4645 inch pounds. To be more accurate you would have to include the moment of inertia of the driver and the pins about the1r own centerline. How do you plan to accomplish this?

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- zekeman	Post Reply	Top of thread	Forum

Posted by: newpunkrkr ® 02/21/2006, 12:41:41 Author Profile eMail author Edit

My calculations matched yours. And I used: 2*(Icom + M h^2), to adjust the Inertia for the pins adjusted on center. Does that make sense. also, for "I*30.2=4820*30.2/g=4820*31/32.2=4645 inch pounds" - where did the 32.2 come from? Thank you for your help Build it, Break it, and then Build it Better

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- newpunkrkr	Post Reply	Top of thread	Forum

Posted by: newpunkrkr ® 02/21/2006, 12:46:54 Author Profile eMail author Edit

I retract that question... I realised it's gravity. I didn't see the g in there. Thanks. Build it, Break it, and then Build it Better

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- zekeman	Post Reply	Top of thread	Forum

Posted by: newpunkrkr ® 02/21/2006, 11:54:30 Author Profile eMail author Edit

The idea is to spin it with a Servo via Belt Drive. We require +/- .001 tolerance in the stop locations around the circumfrence, so chain and direct drive gears seem out of the question due to play and wear issues. Any thoughts? Build it, Break it, and then Build it Better

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- newpunkrkr	Post Reply	Top of thread	Forum

Posted by: zekeman ® 02/21/2006, 12:57:38 Author Profile eMail author Edit

I think you had better rethink that .001" tolerance at the pin locations. A belt drive will never do it, since it will be encumbered with slack and wear problems, etc. If you really need that tolerance then other very expensive solutions are available. If you have the bucks you can get a solution.

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- zekeman	Post Reply	Top of thread	Forum

Posted by: zekeman ® 02/21/2006, 11:48:08 Author Profile eMail author Edit

Correction The result is ok but I erroneously used 30.2 instead of 31 for the acceleration in some of the steps.

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Re: Help! Inertia / Forces
Re: Help! Inertia / Forces -- newpunkrkr	Post Reply	Top of thread	Forum

Posted by: swearingen ® 02/21/2006, 10:05:50 Author Profile eMail author Edit

Can you provide a dimensioned picture of the item? I'll be able to give you an answer, but I'm having trouble visualizing what this thing looks like.

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- swearingen	Post Reply	Top of thread	Forum

Posted by: newpunkrkr ® 02/21/2006, 10:40:09 Author Profile eMail author Edit

Thanks - attached is a picture.   Build it, Break it, and then Build it Better

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- newpunkrkr	Post Reply	Top of thread	Forum

Posted by: ipajewsk ® 02/21/2006, 14:30:59 Author Profile eMail author Edit

As was ly mentioned, I think that your location tolerance is going to be a bigger problem than the motion timing. How do you plan on stopping the motion? Are you planning on simply switching the voltage polarity to your motor? Using a brake or a clutch? The analysis that was done a couple of posts above this assumes a simple 1/2 accel, 1/2 decel curve. That may, or may not, apply depending on how you plan on stopping this motion. For instance, you may have a small delay caused by your control system actuating the stopping mechanism (whatever it may be).

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Re: Help! Inertia / Forces
Re: Re: Help! Inertia / Forces -- ipajewsk	Post Reply	Top of thread	Forum

Posted by: newpunkrkr ® 02/21/2006, 14:50:48 Author Profile eMail author Edit

Servo Motors are able to contol their own accel and decel / start and stop via their drive, so no clutch or break is needed. I don't know if you are familiar with them, but here is a link that explains them briefly: https://zone.ni(dot)com/devzone/conceptd.nsf/webmain/A18266D91803B4D18625685D006EC4E8#4 Thanks again to all that helped with this. Build it, Break it, and then Build it Better Modified by Administrator at Tue, Feb 21, 2006, 15:14:46

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