That chart is an example one, to give you an ides as to how the CVT would work.
For a specific application, we need to go through the steps. I'm going to assume that your wheel size is 26 inches. If this is not the case, you will need to adjust the calculations with your wheel size.
First, you state that you want the bike to have a top speed of about 32 MPH, and you are using a Honda GHX-50. Per the engine curves, below, the GXH-50 has maximum torque output at about 4700 RPM, and maximum HP output at about 7000 RPM, and an upper RPM limit of about 7500 RPM. The first step is to calculate the gear ratio to achieve the desired upper limit for speed, at max RPM. With
GearRat, I can calculate this required total gear ratio to be 18.13, to achieve 32 MPH at 7500 RPM. Let's round this off to 18.1 for the rest of the calcs.
A ratio of 18.1, when you use the second type of X2 CVT (which has a gearbox with a 3.18 ratio) means that you still need a further reduction of 5.7. If you have a 12 T sprocket on the output of the CVT, you would need a 68 Tooth sprocket on the rear axle to achieve the required reduction. Note that either of these sizes of sprocket are hard to locate, and further, it is very desirable to have a freewheel sprocket in the final drive chain. You can't find 68T left-hand thread freewheel sprockets, but, you CAN find them down to about 16 teeth, in the standard english 1-3/8 inch, 24 threads-per-inch mounting.
A 16 tooth drive sprocket would need a 91 tooth driven sprocket. Again, this would have to be a custom sprocket. And, it would be huge. The alternative is to add a jackshaft between the CVT and the rear sprocket.
Lets assume a jackshaft, with a 16 tooth freewheel output sprocket, and a 14 Tooth sprocket on the output of the CVT. Playing around with the numbers on GearRat, and a 29T sprocket on the input of the jackshaft, and a 44T sprocket at the rear axle gets you the right ratios. (CVT1.JPG)
With the gearing settled, we can calculate the upper 'bend' or breakpoint in the chart. This point is the speed, using a total gear ratio of 18.1 (per GearRat) and the max torque RPM of 4700. The speed is 20.1 MPH. (CVT2.JPG)
Next, the lower breakpoint can be calculated, which is the max torque speed (where the CVT belt drive has it's maximum ratio (about 2.2:1) and the engine is at maximum torque (4700 RPM.) Plug in the 2.2 belt reduction in the CVT, to obtain this speed, which is 9.1 MPH. (CVT3.JPG)
Finally, the bottom end of the system RPM-speed curve can be found, by plugging in the RPM when the clutch engages. If we assume this to be about 3000 RPM, the corresponding speed speed would be 5.8 MPH (CVT4.JPG)
That's it then. It looks as if this approach should meet your goals...
Refer to the last image (Chart.JPG) for the RPM-Speed curve for
your proposed configuration.
