Does anyone know a formula for working out how to lower the RPM between two pulleys. I want to end up with about 40 RPM. Most reasonably priced motors (240v) seem to be about 1425 RPM. Thanks.
If you want to be posh, gates do design-flex drive design software, and you pump in the shaft sizes, fixing method, spacings the lot and it spits out actual part numbers for all the components! http://www.gates.com/designflex/index.cfm?location_id=809
How to do it depends on the application, and how much money you want to spend, as said a reduction drive gearbox or motor/ gearbox combination is one way, an inverter is another, pulleys, chain drive and sprockets, or a planetary speed reducer are all options, 35.6 : 1 reduction is quite a drop but can be done, you'll have loads of torque available
Driver pulley divided by the driven pulley x input rpm. (you can use circumference or diameter of the pulley, gear for the calculation as its all related to pi)
This is the base formula used which can be transposed any way you wish and being able to stay away from ratios if you want to. Its generally an easier way to get an output rpm which is what you are interested in.
Thanks TIG Paul. At that price it's cheap enough to buy and have a play. I don't understand all the terms yet. Free input???. No idea. Thanks again for the help.
I dont know if you have a design but this may give you an idea or two, probably a bit bigger than your thinking of, but you can see the shaft mounted motor gearbox at the right hand end of the driveshaft, Ive also made benchtop rumblers using the same principle, powered by low voltage drill motors.
I see the principal now. I was thinking of a small blue gas bottle for the drum with a shaft welded to each end, but this design has it on rollers which is the way to go to make emptying it simpler. Thanks for the explanation. Did you make that one?.