Cool thats what I was hoping!!
The top gear(driven clutch) is a 21 tooth gear and the bottom gear(drive shaft) gear is a 39 tooth. I will be using 27" quad tires with 12" rims. As far as power goes I want the bottom end torque and power instead of high top end speed. If I can get 60mph top speed with bottom end power that would be fine with me.
As said before there will be a chain from the drive shaft to the ideler shaft and them another chain from the idler shaft to the rear axle sprocket. The rear axle sprocket has to be a max of 12 " in diameter. I hope this is the info that you need. And thanks for the help guys I realy appreciate it!
Ok,
So first I went here to look up your 12" max sprocket size for the rear:
http://martinsprocket.com/2001/SecEa.pdf#E43A 50 size chain sprocket with 58 teeth has an outside diameter of 11.9 inches. Depending on your clearance you may want a tooth or two less but we will go with 58 for the calcs.
I don't know anything about sleds.... Does the output shaft of the chain box directly drive the sprocket that drives the track? The reason I asked that and the top speed of the sled is because that would be the easiest way to figure the output speed of the chain box.
In order to figure it without that info then we would need to know the top speed of the engine as well as the final ratio of the CVT when it is shifted into high.
There are some things we can figure and I will guess on the rest and you can fill in the correct numbers later.
If your tires are 27" you take the diameter times pi (3.14) to find the rollout = 84.78"
If you want a top speed of 60 mph then we need to convert that to rpm to figure out how fast the drive axle needs to be turning. So 60MPH divided by 60 minutes in an hour = 1mpm times 5280 ft/mile = 5280 ftpm times 12 inches per foot = 63360 inches per minute.
So 60 MPH = 63360 inches per minute. Now we know the rollout is 84.78 inches (the distance we travel for every rotation of the tire). So we divide 63360 by 84.78 and we come up with 747 RPM as the desired top speed of the rear axle or tire.
Your chainbox gearing of 21/39 gives a reduction of .538.
As I said I know nothing of sleds so... I will assume that the motor runs 8000 RPM and that the CVT is 1:1 when it is shifted to high.... That would mean that the input to the chain box is turning 8000 RPM. With the reduction of .538 (8000 X .538) The output of the chain box will be turning 4304 RPM.
So 4304 RPM at the chainbox and we want 747 RPM at the rear axle. 4304/747 = 5.76. We need a 5.76:1 ratio between the chainbox and the rear. With the 58 tooth sprocket that we calculated before 58/5.76 = 10 It shows that we would need a 10 tooth sprocket at the output of the chain box to make it work if we don't use a double reduction. 10 tooth is to small so it is good that you have the double reduction.
With the double reduction you would just need both ratios to = 5.76 when multiplied together. So if we used the 58 tooth on the rear axle and a 20 tooth driving it on the jackshaft that would be 58/20 = 2.9:1 So 5.76/2.9 = 1.98 We would need another 1.98 in reduction from the chaincase to the jackshaft. I don't know how big of a sprocket you can fit under the motor on the jackshaft but if you went with a 40 tooth then you would need 40/1.98 = 20.2 or a 20 tooth on the output of the chaincase. You wouldn't want to use a 40/20 thought for the reasons that Fabr cited. (wear patterns) 38/19 would be better...
So that is all probably clear as mud.
That is all based on guesses for the motor RPM and clutch ratio. If you can give better information on those numbers or tell me the size of the track drive sprocket and the top speed of the sled then I could get you more accurate numbers. Also It would be good to determine the max size of each sprocket location. Rear is 12", what is the max size on the jackshaft and what would be the max size on the chaincase output. You might as well get them as big as possible to reduce pressure and increase wear life.
