Rotational torque required ?

[CarbonJock]

Hello : I would like to rotate an Aluminum Disc horizontally , 12.0" Dia. 6.25lb. an sets on a 3.375" needle thrust bearing dry lube .This bears all weight and centers the disc , thus stabilizing the disc .

The Disc will rotate 30 degrees stopping for a variable * time interval .So Rpm will be Approximately 0.75 an lower . The Disc makes 12 stops per completed cycle an time interval or duration on station will vary between 5-15 Seconds , although once set will not be change until Disc cycles are completed .

My question is : What's the Minimal Torque and Holding Torque required to preform this task . Thanks in advance .

[PinkertonD]

First off, a drawing or accurate sketch would help.

You have not defined a load for the Driving Torque to handle, or for anything that may be going on at the stations.
Is there any loading at the station? Jars being filled etc?
Are all stations being loaded or just one?
What is the total load when all stations are completed?
You have not defined a load that you are "Holding Torque" against.

For me at least there is nowhere near enough information supplied to help you. This is does not sound like a a simple question and answer thing as I see it so far. Sounds more like a system that needs an overall designer.

Just guessing, mind you.

[CarbonJock]

Don't complicate your mind with details , the Disc is 12.0" Dia. weighs 6.25 lb makes 30 deg. rotation an stops , starts repeats for a 12 stop cycle completion.
Disc Load bears on bearing surface . Only torque required is the Disc momentum an holding on station .

[PinkertonD]

A tiny thing called Inertia will play quite a significant role in determining your desired results. There was no material thickness stated to asses the distribution of the stated weight. I am far too lazy to search out and then calculate a thickness for that stated size and weight of material.

I am surprised you ask a question on an Engineering forum and then decide what the Engineers need in order to resolve your problem. Interesting approach.

Without any details to further complicate my mind, I will have to pass.

[Kelly_Bramble]

Don't complicate your mind with details , the Disc is 12.0" Dia. weighs 6.25 lb makes 30 deg. rotation an stops , starts repeats for a 12 stop cycle completion.
Disc Load bears on bearing surface . Only torque required is the Disc momentum an holding on station .

You're over simplifying -

First Google "Rotational Inertia" here's a basic equation, however this in inadequate.

T = [ N x WR2 ] / [ Ta x 308 ]

Where:

T = Time to Start (seconds)
N = Velocity at load (rpm)
Ta = Average Torque During start (ft-lbs)
W = Weight (lbs)
R = Radius of Gyration (ft^2)
[WR2 = Rotating Inertia (lbs-ft^3)]
308 = Constant derived converting minutes to seconds, mass from weight, and radius to circumference

Other derivations:

simple-rotational-inertia.gif

[Kelly_Bramble]

Other quick tools - you will need algebra to derive the torque...

http://www.engineersedge.com/motors/...tric_motor.htm

[zeke]

The quick uncomplicated answer is zero, since you haven't told us the that cycle time should be minimal.
If cycle time is minimal and you have the resources, the torque must be very large and the quickest way to go station to station is by bang-bang method.
So you see, the answer you get depends on what your design parameters are.
If you want answers you must supply information first.
We are not in the business of guessing.

[CarbonJock]

Fellow engineers you're all trying to make something of nothing really !. The Diameter of the disc has been established ,an weight of the disc has been established .
For torque an inertia purposes the thickness has NO Bearing Period . I simply asked for " Minimal Torque " , an Bang Bang isn't an option .

A smooth steady rotation , stopping for 5 seconds on station and then continuing 12 more times per cycle . The ONLY LOAD IS THE DISC . A clue : The holding torque will need to be more than the inertia torque . This is Physics 55.5 folks ...

[Kelly_Bramble]

Fellow engineers you're all trying to make something of nothing really !. The Diameter of the disc has been established ,an weight of the disc has been established .
For torque an inertia purposes the thickness has NO Bearing Period . I simply asked for " Minimal Torque " , an Bang Bang isn't an option .

A smooth steady rotation , stopping for 5 seconds on station and then continuing 12 more times per cycle . The ONLY LOAD IS THE DISC . A clue : The holding torque will need to be more than the inertia torque . This is Physics 55.5 folks ...

If you know the answer... what are you doing here?

No, the answer is not simple and you're welcome...

how-to-calculate-the-torq.pdf

[RWOLFEJR]

Hello : I would like to rotate an Aluminum Disc horizontally , 12.0" Dia. 6.25lb. an sets on a 3.375" needle thrust bearing dry lube .This bears all weight and centers the disc , thus stabilizing the disc .

The Disc will rotate 30 degrees stopping for a variable * time interval .So Rpm will be Approximately 0.75 an lower . The Disc makes 12 stops per completed cycle an time interval or duration on station will vary between 5-15 Seconds , although once set will not be change until Disc cycles are completed .

My question is : What's the Minimal Torque and Holding Torque required to preform this task . Thanks in advance .

- You're going to need some sort of radial bearing in addition to your needle thrust bearing. Use of a needle thrust for radial isn't going to get it.

- You say "So (implying a for certain) rpm will be .75 an (or?) lower" so apparently you are saying that your max rpm is .75?

- How many stops and how long has no bearing on torque requirement. The rpm... weight... rim speed... velocity change... that's what matters.

So you have 6-1/4 lbs. rotating on a 6" radius at .75 rpm max and need to stop, hold, and start it right? With no other forces acting on this... and no suggestion as to how quickly you want to start and stop? Yet you say no "bang bang?" Without knowing the velocity change you desire short of "bang bang" it's a big mystery.

But you later imply you have done the math and know what you need Sooooo looks like as soon as you add a radial bearing to your arrangement you'll be set...!

Good luck,
Bob

[PinkertonD]

This is Physics 55.5 folks ...

You have an interesting approach to seeking help with a problem.

[CarbonJock]

Unprofessional comment(s) removed.....

[Kelly_Bramble]

OK, I'm closing this thread and Banning Carbonjack for 90 days....