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
I screwed up in my previous answer. I figured only part of the problem. This will give you a better answer.
14,844 from the calculator vs. 14,836 for your # crunching Yeah, be more careful next time - you were off by .05% ;) R
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
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 ;-)
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
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.
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.
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.
>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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That is due to the leverage. You have a "negative" ratio.
And the cyl might have a 18" stroke, and the loader might lift the bucket 180 inches. This means, forgetting the weight of the arms and buckets for a minute, you would only have a 1480 lb lifting capacity.
10:1 lift ratio means 1:10 load ratio.
Not true. Not at all. You need to study your basic physics - simple levers, classes and ratios.
I'm not sure what point you are trying to make, if any. In a typical loader design, the cylinders act through a lever an fulcrum with a negative mechanical force advantage. Even if the cylinder develops a force of say 10 tons, the force available at the end of the loader arms will be significantly less. Usually the geometry of a loader or backhoe design has the maximum mechanical advantage available in curling the bucket.
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
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
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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