Need help with sprocket ratios?

**Frank S** · Mar 14, 2016, 12:24 AM

Actually your RPMs with the 11/60 sprocket set would be more like 7.5 RPMs
take 41 multiply by 11 the number of teeth in the driving sprocket then divide by the number of teeth in the driven sprocket 60 and it would look like this
1x11=451 /60=7.5167 to put it another way 60/11=5.4545 to 1 or 41/5.4545=7.5167 RPMs
if you have room to install a jack shaft to mount the 60t sprocket on one end then mount a small sprocket on the other end then a large sprocket on the item that you need to turn at 1 to 2 RPM you can accomplish the reduction you need
say you start with your arrangement of 11 and 60 with the 60 mounted on the jack-shaft , put another 11 tooth sprocket on the other end of the jack shaft and a second 60 tooth sprocket on the work piece you final RPMs will be 1.378
one note; with each reduction the torque is increased. your motor is starting out with a 310 in.lb. output torque after the first reduction is approximately 1719 in.lb. after a second reduction to the 1.378 RPM will be 9533 in.lb. or 794 ft.lb.
Sp particular attention must be taken into account for the design horsepower of the chain
Very small motors can become scary strong when reductions are great enough

**Captainleeward** · Mar 14, 2016, 02:12 AM

Well Thank you Frank for that wonderful dissertation on gear ratios I'm impressed.....:O)

**Toolmaker51** · Aug 24, 2016, 10:13 PM

Frank S nailed it with the jack shaft. Way easier than a lot of arrangements to build and service. I'd of liked to help spin the topic; you know, helping Cap'n crunch the numbers.