Calculating this is a mess. But a dirty way of doing it is mark the engine sprocket at the vertical. Mark the driveshaft at the top. Rotate the engine sprocket 1 full revolution and how far the driveshaft rotates.
Now, I'm not 100% certain how to run through all the sprocket sizes but......
The engine sprocket is 17t (t = teeth) and the one it drives is 20t so that has a final ratio of 1.176:1 So the engine sprocket will turn 1.176 time to 1 rev of the 20t
So the 20t is back to back to the 30t so that is a final ratio of 1.5:1 So 1 rev of 30t should be 1.5 of the 20t (again you can mark it and check that)
So the 30t is connected to the 19t on the PTO....Final ratio of 1.579:1 so the 30t will rotate 1 time to 1.579 times of the 19t sprocket. So I "THINK" the drive shaft is rotating roughly half a rev (or maybe a full rev?) faster than the engine sprocket. So at a lower RPM the impeller will be spinning faster. If the logic in my brain is working properly. But again a simple line on the sprocket and shaft will get you that.
Now, what I'm not 100% sure of is...Is one engine revolution = to what on the engine sprocket? And is that relative. So, if that is a 6 speed box what is the final ratio and how is that related to RPM.
This is all just a mess that typed from my brain. To calculate the ratio just divide the big sprocket by the smaller sprocket. 20/17=1.1764
I could also be way off in left field. Anyone is more than welcome to tell me I'm way wrong and I will not doubt you.:banghead: