Motor to Pulley Ratio Formula Needed

[Ed Redwine]

OK Gang, he is one for all of you engineer types to help me solve. Internet search did not give me what I was looking for. I need a setup to turn an 8" wheel about 500-600 rpm.

What would the speed of this 8 inch wheel be if:
the motor is 1750 rpm:
The motor pulley is 2 in.
The driven pulley is 5 in.
The final sharpening wheel is 8 in.

OK, who can help me with a good formula or rule of thumb: Don't forget, I'm counting on the forum to solve the riddle.
Ed

[Mark N. Akers]

Ed here's a formula that might help, this is close.
Motor RPM X motor pulley diameter/driven pulley.
1750 x 2 = 3500
3500 /5 = 700 rpms
the 8" wheel will be the same speed as the driven pulley (5")as long as they are atttached to the same shaft.
If you drop the motor pulley to a 1.5" that will give you 525 rpms

[Claude]

Mark's formula is correct.

However, the surface feed rate may be more important when sharpening. For example, a 5 inch pulley turning at 700 rpm means the edge of the pulley is traveling at 5 x 3.14 x 700 = 10,990 inches per minute, or 183 inches per second. The 8 inch felt wheel on the same shaft will turn at 700 rpm, but the outer edge will travel at 8 x 3.14 x 700 = 17,584 inches per minute, or 293 inches per second. The point is, if you want to use a large felt wheel (or leather strop, etc., but want it to move slowly, you may have to use a larger secondary pulley.

Just as a point of comparison, the Delta 1 inch belt sander, model 4002, has a feet rate of 36,000 inches per minute, or 600 inches per second.

Claude

[Ed Redwine]

Thanks Mark and Claude. This will get me close. My hang up is the fact that the 8 in wheel and 5 in pulley would not logically (to me anyway) be traveling at the same rate. We've been given a home brew setup that is in serious need to revamping. It runs fast enough, and in the wrong direction...enough to scare even me. So, a rebuild in is order.
Thanks.