Ok. I haven't had much time to work on the FL583, but I started playing with my two clutch options today. I found a Pre 71 Polaris Driven WILL fit an FL350 shaft. Its larger in diameter, however, it actually has LESS room for the belt to travel than the stock driven. Sooooo no real point in using that one. Other than maybe on a stock FL to really increase the torque.
This brings me too my math delemia. Let me preface that by saying, I can fab just about anything, but I tend to do it on the bench, by site. Not really good with calculations. Sooooooo I have figured out a way to run the large Skidoo driven. I cut the jack shaft (factory driven shaft) off the donor sled. Turns out, the inside of the shaft was already a 1/2 (once you got past the threaded end). So I ran that up to the machine shop, and their going to use a lathe and drill out the inside to match the shaft on the fl350. IF I use that clutch, I'll then drill two holes, run a dowel through the sleave and the shaft, weld it, grind it smooth, basiclly turning the shaft to a 1 inch.
Now, for my question. I am trying to figure out just how much greater the torque on the trans would be. I thought I had a good way of doing that, but I was wrong. I marked each clutch, and marked a hole on the axle. I figured I'd count how many turns of the clutch it took to turn the axle once. Of course, both came up with 10 to 1. I now realize, thats not going to work lol. Sooooooo anyone got a formula I can use? Do I divide the diameter of the stock driven into the sled driven?
And also, it looks like at idle, I might need some sort of runner to keep the belt off the rear axle. Any ideas?? Heres pictures of the trans with the Stock Driven, and one with the SkiDoo driven. The skidoo is 10.6 inches in diameter (The sled HAD a larger driven, but I traded strait up for a stock skidoo driven) And yes, the skidoo driven is on backwards. Just because I don't have the spacer made yet, and I was just testing the diameter.