John Martin wrote:
On Aug 30, 10:31 pm, "MG" wrote:

I have a table saw with motor and arbor pulleys 2.5" max dia.

Motor is 3600 rpm ratio 1:1

I need to replace the motor with a 1800 rpm
Ratio 2:1 to keep same blade rpm.

Easy enough, place a 5" pulley on the motor.

But the belt is about 0.5" thick radially, assuming it make contact in the
middle it reduces the effective pulley diameter to 2" so I may need a 4"
inch pulley on the motor to achieve the correct 2:1 ratio.

Can you see my problem?

Where is the actual motion transfer taking place?

Thanks

Mauro


I do see your problem. You allowed 1/2" for the center of the belt on
the 2.5" pulley, and then neglected to do so for the 5" pulley. Why
is that? By your logic, the 5" pulley would actually have an
effective diameter of 4.5", wouldn't it?

And then you should probably allow for the exact spot the belt rides
on the pulleys - it may be a bit over the top of the groove on one and
slightly below on another. And, as Ned mentioned, you should make
your calculations based on about 1/3 the thickness of the belt from
the outside surface.

On the other hand, does it really matter if your saw blade turns at
exactly 3600 RPM? Or at the 3450 RPM that your original motor most
likely runs at? I don't think there will be any trouble if your blade
turns fast or slow by 10 to 15% or so. I wouldn't worry about it, but
I could be wrong.

John Martin


Won't the blade lose power with the pulley reduction? I know it will
run faster than the motor but won't it lose torque? I know a small
pulley driving a big one will gain power to move objects so it stands to
reason that reversing that will give more RPM but less power. Am I wrong
on that?

Jim Chandler