On Sun, 16 Jan 2011 18:04:22 -0600, wrote:

On Sun, 16 Jan 2011 23:40:21 +0000 (UTC),
(Larry W) wrote:

In article , 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



If your bore diameter and shaft diameters are correct, the force on the
piston side of the cylinder will be about 14800 pounds, or 7.4 tons. If
it is a double acting cylinder, the force on the rod side will be about
8240 pounds, or 4.1 tons. Hope this is not a homework problem.

I have a loader on my tractor with two fairly large cylinders. I can
lift a maximum of 800lbs, or lift a 1000lb hay bale a couple inches
off the ground, but that's all.
That is due to the leverage. You have a "negative" ratio.