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  #1  
Old 01-05-2008, 11:44 PM
TJ Smith TJ Smith is offline
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Belt speed

Does anyone have the belt speed formula for getting the different speeds with a three steep pully? rpm x what
Thanks in advance
TJ


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Old 01-06-2008, 08:32 AM
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skipknives skipknives is offline
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I don't have a formula but if i can ramble a little maybe i can help
First, seperat from your motor you need to mark an old belt,,turn the input shaft on the sander once,,then mesure how many belt-feet you have gained,,
Second,,you need to figure the ratio between each pully,,(for me) i marked the pully stack on the motor,,hand turned the motor once and mesured the inches the sanding belt moved on that adjustment,,
When i had all that on paper i multaplyed "X" motor speed.

PS some sanders are front wheel drive and change with different size contact wheels.
I hope i helped a little
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Old 01-06-2008, 08:47 AM
Bob Hartman Bob Hartman is offline
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Maybe something here will help.

http://www.baumhydraulics.com/pages.php?pageid=4

Bob
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Old 01-06-2008, 09:57 AM
logem logem is offline
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Quote:
Originally Posted by TJ Smith
Does anyone have the belt speed formula for getting the different speeds with a three steep pully? rpm x what
Thanks in advance
TJ
Are you using a grinder of design similar to a KMG? If so, then this should give you belt speed in inches per minute:

Motor RPM x (motor pully dia/pully2 diameter) x (drive wheel diameter x 3.14149) = belt speed

Divide this by 12 to get feet per minute and then divide by 60 to get feet per second.

---------------------------------------------

For example:

If motor rpm is 1725

v-belt is on large side of 4,3,2 pully at the motor shaft and small side of 4,3,2 pulley on drive wheel shaft.

drive wheel is a 6 inch wheel

belt speed = [1725 x (4/2) x (3.1415 x 6)] / (12 x 60) = 90 ft/second

I hope this makes sense to you.

Mike L.

Last edited by logem; 01-06-2008 at 09:59 AM.
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Old 01-06-2008, 10:03 AM
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Ray Rogers Ray Rogers is offline
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Basically, you find the rotational speed of the driven wheel (the wheel the belt rides on) which will be the circumference of the wheel (3.14 x diameter) times the wheel's rpm. This will be inches per minute so divide by 12 to get fpm.

The problem gets more wrinkles from the number of pulleys between the motor and the final driven wheel. If your 3-pulley set up is attached to the motor shaft directly and those 3 pulleys drive one other pulley directly attached to the final driven wheel then the forula above will work by figuring the ratio of the diameters of the driving pulley (one of the 3) to the driven pulley (attached to the driven wheel) then multiplying that ration times the motor rpm to get the rpm of the driven wheel.

If the 3 pulley set up is at one end of a shaft and the other end is another pulley being driven by a pulley attached to the motor then you need to figure in the ratio of those two pulleys too. In other words, figure the speed of each shaft individually at each connection between the motor and the final driven wheel to finally arrive at the rpm of the driven wheel .

Bottom line is, it depends on how you make the connection between the motor and the driven wheel. The ration of a driven pulley diameter to the driving pulley diameter can be used to figure the rpm of the driven pulley. Do this through as many steps as necessary to find the rpm of the wheel the belt actually rides on. The diameter (in inches) of that wheel x the rpm of that wheel x 3.14 is the number of inches the belt will travel in one minute.....


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Old 01-06-2008, 11:44 AM
TJ Smith TJ Smith is offline
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Ok
I have a KMG My drive wheel is turning at 1750 and is the standard 4 inch wheel.
I have a 3450 rpm motor with a 2 inch pully with a 4 inch pully on the drive shaft.
I think this gives me 1750 at the shaft with the equivilent of 1.5 hp.
My drive wheel is standard or 4 inches .
This should give me 4 x 3.14 = 12.56 divide by 12 =1.046 x 1750 =1830.5.
How do I end up with surface feet per minute???
Thanks for the help
TJ


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Last edited by TJ Smith; 01-06-2008 at 11:49 AM.
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Old 01-06-2008, 01:17 PM
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Ray Rogers Ray Rogers is offline
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I think you already did it: 1830.5 sfpm Taking you at your word that the driven wheel (the one the belt rides on) is turning at 1750 rpm and is 4" in diameter then you already have the answer ...


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Old 01-06-2008, 02:10 PM
cdent cdent is offline
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TJ, I think the calculating is just a tad off. I think your drive shaft is turning at 1725 or half of your motor rpm's (with that pulley set up). Not a big deal, though I might be wrong.

Take care, Craig
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Old 01-06-2008, 03:02 PM
Rob Frink Rob Frink is offline
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Here's a quikie method.

The KMG's 4" dia drive wheel is about 1 foot in circumference. ( 4" x pi= about 12"= 1 foot)

This means the the belt travels 1 foot for every revolution. If you spin the drive shaft at 1700 revs per min...then your belt moving about 1700 feet per min.

-Rob


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Old 01-06-2008, 08:28 PM
TJ Smith TJ Smith is offline
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Thanks guys
My drive shaft rpm is 1725 so that works out
Thanks again for the help.
TJ


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