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| Bandsaw Motor Update http://w-ww.luthiersforum.com/forum/viewtopic.php?f=10102&t=8177 |
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| Author: | JJ Donohue [ Fri Aug 25, 2006 11:13 am ] |
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I received the original motor today and I had a question as to sizing the pulleys to achieve a target blade speed of 3,000 sfpm. I found 2 different formulae online but I believe the correct one to be the following: SFPM = (s x m) / (d x b) x Pi /12 where: s = motor speed (1725 RPM) m = motor pulley diameter (4") d = drive pulley diameter (6") b = bandsaw wheel diameter (14") Therefore: SFPM = 3,097 Is this the correct formula? If not, then what formula is corect? Thanks |
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| Author: | Alain Lambert [ Fri Aug 25, 2006 11:42 am ] |
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The motor is 1725 rpm the motor pulley is 4" and the drive is 6" therefore, the drive pulley and the band saw wheel turn at 1725 x 4 / 6 = 1150 rpm (turn per minute) The band saw wheel is 14" diameter. the circumference is PixD = 3.1416 x 14 = 43.98 inches or / 12 = 3.6652 ft since it is turning at 1150 rpm and each turn is 3.6652 ft this give 4215 fpm So the formula is probably SFPM = (S x m/d) X (b X Pi /12) Unless I made a mistake (which happens all the time ) |
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| Author: | JJ Donohue [ Fri Aug 25, 2006 12:32 pm ] |
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Thanks Alain...your explanation makes perfect sense. I'll change to the correct formula and figure out the proper pulley combination. |
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| Author: | Phil Marino [ Fri Aug 25, 2006 12:33 pm ] |
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[QUOTE=Alain Lambert] The motor is 1725 rpm the motor pulley is 4" and the drive is 6" therefore, the drive pulley and the band saw wheel turn at 1725 x 4 / 6 = 1150 rpm (turn per minute) The band saw wheel is 14" diameter. the circumference is PixD = 3.1416 x 14 = 43.98 inches or / 12 = 3.6652 ft since it is turning at 1150 rpm and each turn is 3.6652 ft this give 4215 fpm So the formula is probably SFPM = (S x m/d) X (b X Pi /12) Unless I made a mistake (which happens all the time ) [/QUOTE] You got it right, Alain. Phil |
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| Author: | JJ Donohue [ Fri Aug 25, 2006 1:07 pm ] |
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Thanks guys...nice to have some professional "gear-heads" (or pulley-heads) on the forum. Now that I have the right formula, I did an Excel sheet in order to study the various combinations. I'm thinking of setting the pulleys up so that it will have 2 speeds: ...3,161 sfpm (4" motor + 8" drive) ...2,108 sfpm (2" motor + 6" drive) Is this a reasonable combination?...(3,000 for wood and 2,000 for metal, plastic, etc.) |
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| Author: | Phil Marino [ Fri Aug 25, 2006 3:03 pm ] |
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JJ You could even go even slower for your "slow" speed - maybe 2" and 8 " - that would be about 1500 FPM. That's about what I use for aluminum. If you ever want to cut steel, you need to go really slow - like about 150 FPM. You couldn't get that slow by just changing the two pulley sizes. On my bandsaw I put in a "jackshaft" - a third shaft with two pulleys so I get a two-stage reduction. Phil |
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| Author: | JJ Donohue [ Fri Aug 25, 2006 3:39 pm ] |
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Phil: 1500 is a better suggestion. I currently have 6", 8" and 2" pulleys. If I get a 3" I can then go with 3 + 6 = 3100 and 2 + 8 = 1500. Looks like a good plan...thanks. So, I guess I'll need 2 belts...or would you use one belt and have it tighter for the longer distance between pulleys? Also, what about those leather link belts that I've seen advertised. Seems like they might be quickly adjustable for changes. Waddya think? |
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| Author: | Phil Marino [ Sat Aug 26, 2006 12:18 am ] |
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JJ I would stick with standard (black, rubber covered) V-belts. They're cheap, reliable, quiet, and last forever. As to needing another belt length, it depends on how much adustability there is on your bandsaw. It's better not to run the belts either too loose or too tight. Too tight and you can over-stress the motor and bandsaw wheel bearings. Too loose, and you can get slipping and vibration (or worse, throw a belt - that could be dangerous) If you need another belt, Grainger's is a good place (if there's one near you) for belts and pulleys at good prices - and other machine stuff, also. Or, an auto parts store would probably have belts in any length you need. Phil |
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