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?
You're on the right track. The pitch diameter of the pulleys is
measured at the neutral axis of the belt. If you imagine the belt
being made up of many thin layered loops, the loop that doesn't change
length as the belt is flexed is the neutral axis. The reinforcing
cords in a v-belt define the neutral axis, which is perhaps 1/3 of the
belt thickness from the top surface of a typical v-belt.
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.
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
If the motor is the same HP, it will produce twice the torque at half
the speed. The "gearing" will bring it back to the same speed and
As for the pulleys, pretty close just measuring either the root or the
crest of the pulley on both - and if they are the same width the speed
ratio will be the same as the measurement ratio. (within a VERY small
percentage of error)
Posted via a free Usenet account from http://www.teranews.com
"These pitch diameters equal approximately the outside diameter of the
sheave minus an amount equal approximately to the belt thickness plus
from 1/16 to 1/8 inch, depending upon the belt size. For example, if
the belt is size E or 1 1/2 by 1 inch, the pitch diameter would equal
( approximately at least) the outside sheave diameter minus 1 1/8
Approximately approximation of an approximate answer!
This was the first thing I learned after I bought this book for five
bucks ten years ago, something that I had pondered for many years.
To measure the blade's pulley "pitch diameter": make a chalk mark on
the pulley and belt just where the belt "exits" the pulley, rotate the
pulley one revolution so the belt's chalk mark moves away from the
pulley, and measure the distance on the belt from its chalk mark to the
pulley's chalk mark. This is the pitch circumference, divide by pi for
the pitch diameter.
Buy a new pulley whose pitch diameter is twice this. Pulleys *are* sold
by pitch diameter. Maybe not the ones on the rack at Ace Hardware, but
the ones at MSC, McM-C, etc.
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