I'll say ...
Right now it's mostly trying to figure how to properly size a hydraulic motor ... so I can convert a rear-mount PTO snowblower to a front-end loader mounted hydraulically-driven snowblower.
I have an Eaton "L2" series pump - actually a model 25503 - which has a 2.05 cu. in. displacement, is rated for 3600 psi continuous and a speed of 2750 rpm, with an output of 22 GPM @ 3000 psi @ 2750 rpm.
That's good - because the mid-PTO of my tractor is rated at 2550 rpm at PTO rpm.
The question is: What should the displacement of (driven) hydraulic motor be, given the flow rate of the pump and the rpm I can run it at, to achieve a driven speed of 540 rpm on the snowblower ...
Taking into account frictional/efficiency losses of course.
If you have any insight into this, please drop me a PM.
Yow would have better luck finding a solution on the truck talk forum down here in the soapbox you will just have the EO linch mob AKA LBD and LOS wanting to blame Obama or wanting you to shoot it