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Old January 29, 2013, 12:36 PM   #12
Join Date: March 4, 2005
Location: Ohio
Posts: 13,714

So you are saying you've got a can about 6.6 times the diameter of the drive shaft being turned by the pulley, giving you a second reduction.

5.25/2×6.6=17.325 so 1725/17.325 = 99.6 rpm (close enough to 100)


I get just 76.6 RPM for 1 g in a 12 inch diameter drum. You can prove this to yourself by swinging any object on a 6 inch string in circles in the vertical plane. It takes just over 1 circle per second to keep the object from starting to fall at its apogee.

Where length is in inches:
ac = 1g = 32.17 ft/s² = 386 in/s² = v²/r = (2πr/s)²/r = r(2π/s)² = r4π²/s² = 39.48r/s²

s = √(39.48r/g) = √(39.48r/386 in/s²) = √(0.1023r) = √0.6137s² = 0.783 s/rev

rpm = 60 s/min / 0.783 s/rev = 76.6 rev/min
I've made up a table here:

I don't really know where greatest cleaning speed could be predicted to land. I assume the commercial tumbler makers have figured this out, perhaps by trial and error. My Thumbler B liners are 7" across in the inside corners, and 6⅜" across the flats. That's 100 rpm for 1g across the corners and 105 rpm across the flats to hit 1g. Thumbler makes them with a standard motor for 30 rpm and a high speed motor that turns them at 40 rpm. So I think that suggests Maybe 1/4 to 1/2 the rpm on my table will be the range you normally want to fall into.

If you have a smooth sided drum with no ribs or flats, you are more likely to want to go closer to the numbers on the table to get the content raised before it just slips down the sides.
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File Type: gif Drum RPM for 1 g.gif (7.7 KB, 214 views)
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Last edited by Unclenick; January 29, 2013 at 12:59 PM.
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