Greg_g: The math isn't too complicated.
A 4" cylinder has an area of about 12.57 sq.in. a 24" cylinder thus has slightly over 300 cu.in. volume.
An 8 gpm pump moves 1848 cu.in. /min or 30.8 cu.in./sec. The extension stroke would thus take about 9.7 sec.
My post had assumed 7 gpm, so your times are a bit faster than my original numbers, but not enough to make much of a difference. The point, however, remains the same, if you want the 10sec. + or - cycle times that one of the good Timberwolf splitters, for example, can produce, you are going to need to use a pump that can output about 15-16 gpm.
My B7800 (30 pto hp) only outputs about 9.5 gpm for both power steering and hydraulic system (about 7 or less for the hydraulics alone), and my memory is that you have to get up to the Kubota M Series before you get that kind of hydraulic pump output.
However, even my B7800 has plenty of pto hp to run a pump that will produce output much more than 16 gpm. On the homemade splitter that I am building, I am using a Barnes 2-stage that outputs 28 gpm at low pressures (up to about 750 psi) and about 7 gpm between that and 3000psi and only requires about 16 pto hp.
With a 5" cylinder I would only be using high pressure when dealing with a knot or twist or when using a 4-way or 6-way wedge on pretty large logs. I figure that my average cycle time (except perhaps when dealing with very large or difficult logs) will be under 10 sec.
All that said, if you are getting satisfactory cycle times using your tractor hydraulics that's fine and there is no reason to go to a pto-driven pump.
But, H**l, I always overdesign everything anyway; I built the doghouse for our newfie and black lab with 2x6 wall and ceiling framing, a partition that shelters the inner room from wind, snow and rain, and full insulation in walls, ceiling and floor slab. ...wife claims that it is better built than our house, which will be a comfort if she banishes me to it. /forums/images/graemlins/grin.gif