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Horsepower calculation. Bearing 1

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whlock

Mechanical
Feb 18, 2013
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When one using a bearing to drive a certain weight of material, do u just use the normal force multiply by the coefficent of friction and then use this force to figure out the torque?

I am using a bearing to drive some weight thru 6.67' distance, the bearing has pitch diameter of 3".
I obtained the following using the above info

FPM= 40
RPM= 51

Then here is the point I start getiing confused,

Force = coefficent of friction x weight
MY weight is 1500 lbs & coefficent of friction is 0.03

F = 1500 x 0.03 = 4.5 lbs

Then I use this force to figure out my torque
T = F x radius = 4.5 x .125 ft = .563 lb ft

and with that I figure horse power

HP = (.563 x 51)/ 5250 = .005 HP which seems really low to me...

Did I not do it correctly?
 
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I can't figure out "using a bearing to drive a certain weight of material". Maybe it's friday night but can you explain what this is?

I use bearings to allow shafts to rotate with as low an amount of friction as possible....

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way
 
The bottom line answer is no. The driving load is dictated by the payload; bearing friction are parasitic losses and possible stiction that may affect startup. Of course, your "When one using a bearing to drive a certain weight of material" is unclear, since bearings are passive devices and cannot drive loads.

TTFN
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Possibly the OP means / meant - using a bearing to support a certain weight.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
 
You can get your PHD if you figure out how to get the right torque.
In some cases the torque of the seals is greater than the frictional
torque. The amount of clearance or whether it is preloaded causes
great variation as well. Is this a thrust or radial bearing, etc,
is it a roller, ball or pad bearing?
 
This is definitely a case where the OP is asking for a simple answer to a very complex situation. If whlock would be satisfied with a rough estimate of the mechanical losses involved in his bearing system, then he has all of the information he needs. But as dinjin points out, if a more precise answer is required then the problem gets quite complicated. Inertias must be considered, the fluid mechanics in the bearing contact over the entire event must be considered, the kinematics of the system must be considered, etc.
 
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