It is actually kind of easy (if you have a calcualator) I will draw you a diagram to illustrate it if you will give me about 15 minuits.
Thank you for your patience.
Its not the sise of the motor that matters so much as the size of the pully you attach to its output shaft. So if you want the motor being turned to spin at 1750 rpms. You will need to put one size on the drive that when thedrive is spinning at full rpms will equal 1750 rpms on the driven motor. If you could provide the rpms of the drive motor and the pullyt circumfrence I will see if I can work this out for you.
Ok I hope i got this right. Its been a long time since geometry class.
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ok just to clairify the motor has a 9" output pully and it is turning 1750 rpm's the sheave at the other end of the belt is on a shaft and is 14 or 16 feet in circumfrence ot 14/16 inches in circumfrence. It is the rpm's at the sheave that you need to determin. Correct?
ok I presume it is inches instead of feet on the sheave. Give me a few minuits and Ill be back
For the 14" sheave the shaft RPS's will be 0.64285714285714285714285714285714:1
So we take that number and and make it a fraction which equals 0.64285714285714285714285714285714/100, or 0.64285714285714285714285714285714% of 1750.
Now devide 1750 by 0.64285714285714285714285714285714% and we get 11.249999999999999999999999999998.
Move the decimal 2 places to the right, and you have 1124.9999999999999999999999999998 RPMs at the end sheave.
Or 625.000000000000000000000000001RPMS less than the 9" input pully
For the 16" sheave the shaft RPM's will be 0.5625:1 Which ends up being 0.5625%
Devide 1750 by 0.5625% = 9.84375 Move decimal = 984.375 RPM's at end sheave.
So the out puts for a 14" shaft are 1125 RPM and for the 16" shaft they are 983 RPM's.
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