On Jan 16, 12:46*pm, wrote:
On Sun, 16 Jan 2011 11:30:21 -0500, "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

*Pressure pushing or pulling? If pushing the shaft diameter does not
have any bearing on the force delivered. The gallons per minute
likewize have no effect on the force delivered 3 inch bore is 7 square
inches of area, so the push force delivered is 7X2100= 14,700 lbs
force.

If it is a double acting cyl, the pull-back area is 7-3.14=3.86 sq
inches, and the pull-back force is 3.86X2100=8906 lbs force.

Correct. To get a bit pedantic: The 'power' is usually limited by
the valve. When it comes to logsplitters the valve is usually set at
2,500 - 2,750 psi. The next class action suit should be for the
manufacturers of log splitters. Their force claims (usually 22 ton
for a 3" cylinder) are way over inflated.

Harry K