shaft speed reduction formula

Submitted by Larry on 04/02/2004 at 18:56. ( larryjar@northerntel.net ) 65.19.233.198

Hi
What is the formula to reduce speed of a cylinder?
The motor is 1/4hp turning 1725 rpm with a 1/2" shaft and want to turn another shaft approx 55 to 60 rpm. what size pullys do I use?
Thank You
as ever
larryjar

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use ratios

This response submitted by skinner on 04/02/2004 at 19:22. ( skinner@srcaccess.net ) 204.117.52.153

you probably wont like the answer I am going to give you but here it goes. It is going to be hard to get that 1725 down to 50 but what you have to do is figure if you have a one inch pulley on the motor and you were driveing a 12 inch pulley that would be a 1:12 ratio so dived 12 into 1725 and that will be your answer (143.75) what you probably will need to do is to have the motor drive a shaft that is set on two bearings (evaporative a/c bearing are cheap and will work fine) with a large pulley say 12 and a 1 inch then have the small pulley drive another larger pulley on the tumbler this way you can gear it down twice. Just remember to devide the small pulley into the large pulley the take the answer of that and devide into the motor speed do that for each time you change pulley size between the motor and the tank. Sorry if I confused you but I am not very good at expalining things just email me if you have any questions.


The formula.

This response submitted by Gerry on 04/02/2004 at 23:15. ( info@fleshing.ca ) 216.106.104.34

Hi Larry,

Skinner is on the right track, but as he admits, his explanation is a little difficult to understand. Her is the formula:

RPM of motor X by diameter of pulley in inches, devide by other diameter of pulley in inches equales the second shaft speed. If you need to reduce more, add an other shaft and connect by V belt to a small pulley.
Example: (1725 X 2 ÷ 10 = 345) if you wish to slow it down more,
(345 X 2 ÷ 12 = 57.5 rpm)

Hope this will help all of those trying to make there own drum. For most efficient results, your drum must turn 14 to 18 rpm. A third shaft is the easiest way to get there. (57.5 X 2 ÷ 7 = 16.4 rpm.)

Good luck,

Gerry


Larry, I have a question

This response submitted by George on 04/02/2004 at 23:20. ( georoof@aol.com ) 64.12.116.134

Why don't you just buy the proper motor and gear reduction box? Though I agree with Skinner (my math skills are terrible anyway), I THINK pulleys are sold by their DIAMETER and the ratio has to be determined by their CIRCUMFERENCE. In other words, a one inch pulley would use the formula C=pi x D or a 1 inch pulley would have a circumference of 1 x 3.1417 or 3.1417inches and a 12 inch would be 37.7 inches. Math whizzes, help me out here. It's only been 40 years since I graduated from high school.


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