On Sat, 15 Aug 2009 17:21:51 -0400, Phisherman
wrote:

I'd be concerned about going to a 2 HP from a 1/2 HP. Power is a
small part of a better-running bandsaw

Motor power is also a small part of the load. The blade tensioning
load is by far the largest force applied to the bandsaw structure.

For rough, back of the envelope figuring, you can assume approximately
800 to 900 pounds of vertical load on the wheel bearings per inch of
blade width from the tensioned blade. That value is based on the
12,500 psi tensile stress recommended by Suffolk Machinery,
manufacturers of the Timberwolf line of blades, and a blade thickness
of .035". Those values, 12,500 psi x .035, yields about 438 pounds of
tension load per inch of blade width. The bearing load is twice that
because the blade wraps 180 degrees around the wheel. The load would
be less for a thinner blade and greater for a thicker blade or a
higher tensioning stress. If you're making the calculation for your
particular saw, you should base the tensioning load on the largest
blade the saw can use both in thickness and width..

For a 14" bandsaw you can add about 20 pounds of vertical load per
motor horsepower. That's based on about 3 foot pounds of rated torque
per horsepower for a 1725 RPM motor ( Torque in foot pounds = 5252 x
HP/RPM). NEMA standards call for a breakdown torque between 200% to
300% of rated torque but most motors achieve better than that. So
assuming 400%, that's a maximum torque of about 12 foot pounds reacted
through a 7" (for a 14" bandsaw) lever arm. That's about 20.6 pounds
per HP. That load is reacted by the table structure and the lower
wheel bearing. For saws smaller than 14", the added load per
horsepower would be greater due to the shorter moment arm. And for
larger saws, the added load per horsepower would be smaller because of
the longer moment arm.

For an upgrade that increases the motor power by 1.5 HP (0.5 to 2.0
HP), the OP could expect the maximum added load to be 1.5 x 20 = appx.
30 pounds. If the largest blade the saw can properly tension is 3/4" x
0.035, that's an increase from about 680 to about 710 or about 4.4%.

I'll leave it as an exercise for the reader to evaluate the
significance of that increased load. Note that any increase in load
occurs only during a cut that would otherwise stall the smaller motor.

I invite the reader to review my figures and logic and please point
out any errors in either.

Tom Veatch
Wichita, KS
USA