changing chain drive to belt drive

using a belt drive would undoubtedly be easier to power my bike than a chain drive. I don't have the know-how to mount a new sprocket on the chainsaw. How would I change the sprockets on my rear tire to something that can be spun using a belt?

The easiest bolt on system would be to purchase a Whizzer Rear Wheel Sheave Approx $30.00 and then the mounting hardware to Mount it the sheave to the spokes of your rear wheel Approx $50.00 There is alot more to all of this than just buying the parts and mounting them. You need to be sure you have the clearence for what ever parts you buy? and also you need to be sure that you can achieve the proper amount of reduction to make the bike go?

ocscully

THanks for the info. I've been winging it mostly. But had to restart because I didn't think things through. How would I go about reducing?

Most of the motor kits sold for 26 in. wheel bicycles seem to come with an over all reduction of approx. 17:1 - 20:1 . The chinese 2-stroke kits/motors have a 4.2:1 primary reduction built into the motor then an 11t output sprocket to a 44t driven sprocket at the rear wheel. The 4-stroke kits usually come with a primary reduction gearbox with a reduction in the neighborhood of 3.7:1 then an 11t output sprocket to a 56t rear driven sprocket for an overall reduction of 18.8:1. Reduction is just a math problem. Divide the larger dia. driven sprocket/pulley by the smaller dia. driver sprocket/pulley. With sprockets you count teeth with pulleys you measure dia. So say you had a 2 in. dia pulley on your motor you would need a 40 in. dia pulley at the rear wheel to get an overall reduction of 20:1 The driven pulley is larger than the rear wheel. Hence jackshafts or reduction gearboxs come into play. Due a search for the Ratio Calculator here on this site, its a very useful tool to work alot of this out. The Dia. of the whizzer Sheave is 17 in. - 18 in. I believe?

ocscully