"Dean Hoffman" wrote in message
...
RBM wrote:
I'm trying to determine the force in tons of a hydraulic cylinder. Being
mathematically challenged, even having a formula would be like reading
Chinese to me.
The cylinder has an outer diameter of 3.5", the shaft is 2", as best I
can figure from information derived from the net, the bore would be 3".
It has a stroke of 18". It will be fed from a pump delivering 8 GPM @
2100 PSI.

Any help would be appreciated


The things that matter are the PSI and the diameter of the bore.
Figure the surface area of the internal diameter of the cylinder.
3.14 X half the bore X half the bore. Multiply that by the PSI.
I come up with 14,836#. That is, of course, under ideal circumstances.
I think the old hydraulic control valves used to have a pressure relief
set at 2000 psi.

OK, that actually sounds reasonable at something around 7 tons. This was on
a log splitter with a 5 HP engine, but I have no idea what it's pump put
out, but it's a really small pump. I'm rearranging it to go on a small skid
steer with auxiliary hydraulics.

Thanks