Power/Force of hydraulic cylinder???

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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
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RBM wrote:

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.
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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
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RBM wrote:

I screwed up in my previous answer. I figured only part of the problem. This will give you a better answer. http://tinyurl.com/4slybla
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14,844 from the calculator vs. 14,836 for your # crunching Yeah, be more careful next time - you were off by .05% ;)
R
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wrote:

14,844 from the calculator vs. 14,836 for your # crunching Yeah, be more careful next time - you were off by .05% ;)
R
Close enough for Govment work
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You have the force answer, if your pump really puts out 8 GPM at 2100 PSI it will fill that cylinder in around 2 seconds.
Is that a Campbell Hausfeld pump? We know how they lie ;-)
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It's the auxiliary hydraulic pump built into a Thomas 85 skid steer. Thomas was built in Canada, don't know who's pump they use, but I really doubt it's a crappy C/H
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snipped-for-privacy@aol.com wrote:

What this means is that the splitter should function well with the skid steer at or near idle speed.
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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 3.86X210006 lbs force.
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On Jan 16, 12:46pm, snipped-for-privacy@snyder.on.ca wrote:

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
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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.
--
Often wrong, never in doubt.

Larry Wasserman - Baltimore Maryland - lwasserm(a)sdf. lonestar. org
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On Sun, 16 Jan 2011 23:40:21 +0000 (UTC), snipped-for-privacy@sdf.lNoOnSePsAtMar.org (Larry W) wrote:

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.
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I had a small compact Ford tractor. The cylinders on the loader were considerably smaller that this, which has a 3.5" diameter. Mine could only pick up about 600 pounds. You also have to keep in mind that your cylinders aren't only picking up the hay bale, they're also picking up the weight of the bucket and arms of the loader.

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RBM wrote: ...

They're also not pointing vertically directly below the load, either; only the force in the direction of lift is actually vertical.
--
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levers, classes and ratios.
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On Jan 16, 6:37pm, snipped-for-privacy@snyder.on.ca wrote:

He is correct. The force that counts is that applied at the load. That depends on all the various levers, angles, etc. but it still comes out to "what is the actual vertical force on the load". It will be considerably less than the force measured 'at the cylinder pivots'.
Harry K
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I also read this statement to suggest that he was thinking in terms of vector mechanics such that only the vertical component of the cylinder's force vector was applied to the load. If so. this has very little to do with this type of system. The cylinder could be horizontal, sideways of pointing down and still supply (almost) exactly the same force at the load.
On second reading, it's really hard to tell exactly what he meant by that second clause...
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Larry Fishel wrote:

Well, all of the above... :)
The poster who started this subthread to which I was responding noted that altho had sizable cylinder on a tractor loader that didn't have anything close to the lift force that one might expect simply from the bore/pressure of the cylinder and the responder directly ahead mentioned only the weight of the loader itself as a confounding factor...
A typical small/medium tractor loader is generally a very simple mechanism a la the one shown at the link below--
<http://www.americanlisted.com/idaho_12/garden_house_22/ford_8_n_loader_tractor_3375_14793104.html
While there are complex mechanisms found on either very large and/or expensive loaders and/or the compact loaders such as the Bobcat and ilk, a basic loader of the above type doesn't have mechanical advantage--the upward component at the bottom is fairly small and the relatively long moment arm beyond the lift point (necessary to get reasonable lift height, etc.) requires quite a bit to counteract.
I was simply trying to be concise at, perhaps, the cost of some clarity in pointing out that mechanics are a part as well as weight and cylinder force...as per usual on usenet, the opportunity to seek advantage and display perceived intellectual prowess at expense of others is, apparently, irresistible force for some. :(
--
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dpb wrote: ...

OK, I went and dug up the operator's manual off the JD148 that's on the 4440 as the loader tractor here...
<http://www.tractordata.com/farm-tractors/000/0/9/96-john-deere-4440-photos.html
It's hydraulics are 2250 psi, lift cylinders are 2.5" bore/1.75" rod -- from which one can estimate max lift force at the lift pin of sotoo 11000 lbf. The published breakout lift force is "only" 4700 lbf in comparison. It'll handle 2000-lb round bales two-high loading semi's easily, however as far as the weight; wish the reach were about 6" more, though, as one has to be sure to have the fork as near the bottom as possible and if they're just a little loosely wrapped they can rotate some so maneuvering can sometimes slow one down loading. When there are nearly a thousand to go; any time wasted, even a few seconds/bale, is sorely trying to the soul... :(
I'll try to post somewhere a picture from last summer's haying altho don't think I stopped to take pictures while actually loading out; had my hands full of controls at that point... :)
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