Calculate BS Blade speed?

[scoly]

My BS motor operates at 1750 rpm's and has three pulleys (4", 3.25 , and 2.25). The belt attaches to a 7" pulley for the wheels. What is the calculation for the blade speed on each pulley and which would be the recommended speed for woodworking?
Thanks
Steve

[LCHIEN]

I answered this a couple of months ago in this thread (page 4). The example was for the HF 4-speed BS. Just use the equations.

http://www.bt3central.com/showthread...nd+blade+speed

The key text:
But here is how pulleys work:

Shaft one is the motor. Shaft two is the center, Shaft 3 is the axle of the bandsaw wheel. RPM1= RPM of the shaft 1 RPM2=rpm of shaft 2 etc.
D1=dia of motor pulley
D2 = dia of driven pulley on shaft 2 (e.g the pulley attached to the belt to the motor)
D3=dia or driving pulley on shaft 2 (e.g. the pulley attached to the belt to the wheel)
D4 = dia of driven pulley on shaft 3
D5=dia of bandsaw wheel (e.g. 14")

The Motor turns at 3400 RPM
The shaft1 and pulley 1 turns at 3400 RPM
Shaft 2: RPM2= 3400 x D1/D2
Pulley 3 turns at RPM2 since its fixed to the shaft.
Pulley 4 and shaft 3 turn at RPM3 = RPM2 x D3/D4 = 3400 x D1 x D3 / (D2 x D4)

Band saw wheel turns at RPM3.
Linear velocity of the blade = RPM3 x 3.14 x D5/12" (in feet per minute. fpm)

= (3400 x D1 x D3/(D2 x D4)) x 3.14 x 14 / 12 feet per minute.
= 12,461 x D1 x D3 / (D2 x D4) fpm

If you eliminate the Shaft 2 an connect the motor pulley of D1 to the pulley on shaft 3 with D4 then the speed of the blade will be
= 12,461 x D1/D4 feet per minute


IIRC for a bandsaw cutting wood the desired speed is approx 3000 fpm.

so D1 should be about 1/4th the diameter of D4 in the one-belt system.

If you really want to be accurate, the pulley dia. whould be measured at the middle of the belt thickness, not at the outer edge. But that error is probably not more than a couple of percent, and your 1750 RPM speed is not exact but dependent upon load. Probably only 1750 not cutting. Slower when cutting.

[scoly]

Thanks for the link. Guess I worded the question wrong. The pulley attached to the motor has three sizes for the belt. Didn't know there were bs's with three seperate pulleys.
Couldn't figure out how to use the formula so did a search and found this:
"
Motor RPM multiplied by the Motor pulley diameter divided by the Driven pulley diameter times the Bandsaw wheel diameter times pi (3.1416) divided by 12 = S.F.P.M.
Motor RPM X Motor pulley diameter / Driven pulley diameter X Bandsaw wheel diameter x 3.1416 / 12 = S.F.P.M."

Came up with using the 3.25" pulley will yield 2978 sfpm.
Thanks again,
Steve