was hydraulic questions more info

The two diameters I mentioned are the piston dia. How do I tell the bore size , as it is different than piston dia. correct?

The bore size is only a few thousandths different than the piston dia. In fact, if a cylinder is called out as a 4" cylinder then the bore will be 4 inches and the piston slightly smaller. ERS

Seems I screwed up again, it was the "rod" that I measured, and incorrectly called the piston. I looked in a catalog at some cylinders of the same orientation and looked at the rod size compared to the bore, and it is very different. I think Im dealing with a 4" bore and the other is probably 5 or 6" bore. All Im trying to do is find out what GPM the pump should be. I tried finding the valves on Parkers site and got errors. I tried searching for the mfg. company and nothing turned up. Can you guys tell me how to figure GPM with the info I have? I couldnt find anything telling me the force required to bend 1" rebar. I have emailed some companys with no answers yet. Thanks guys, sorry to be such a pain Craig

Ok, a few basics:

Force and Pressure: Maximum force is simply the operating pressure times the area of the PISTON (not the rod) Measure the outside diameter of the cylinder, figure that the wall is about 1/4" or so. So 4-1/2" outside diameter can be assumed to be 4" Your pump can be set up with a pressure relief at all sorts of different points, 1000psi is fairly low,

3000 psi is right up there, maybe use 2000 or 2500 psi as a reasonable point. So a 4" cylinder has 12.78 square inches and 2500 psi and would produce 31,950 pounds of force.

GPM and speed: 1 gallon is 234 cubic inches so 1 gpm and a 4" cylinder with 12.78 square inches would move 18" per minute 234/12.78=18). (Pretty slow!)

Hp required: the hp forumla is gpm*psi/1740 so the example above would take 1*2500/1740= 1.43 hp For a gas engine you want the hydrualic rating to be something like 3/4 the gas engine rating. That would calculate to 4.72 GPM MAX at 2500 psi for your 9hp engine.

Last thing is using 2 stage pumps. These have two sections, you use both sections of a dual pump unit at low loads to give you high speeds, then when the load increases one section drops out and you get high pressure at much lower GPM rating. Sounds complicated, works well, especially for smaller engines.

Take a look at item # 1012 at

This one will do

2.9 gpm at 2500psi so it would drive a 4" cylinder at 54" per minute under full load, would do a 12" stroke for the bending in about 15 seconds. Bigger cylinder on the shear section would go slower but not as far so it would cycle down in perhaps the same 15 seconds. Return stroke would be at 11gpm so it would retract in 4 or 5 seconds. Item # 1056 is slightly bigger would take more hp to drive, would run the cylinders faster. They both have a 2" square bolt pattern that mates up with the standard adapters.

I suspect that your pumps are in the 3 to 4 gpm range. YMMV

m> Seems I screwed up again, it was the "rod" that I measured, and

Roy, Thank you for helping out a hydraulicly challenged individual! I really appreciate you taking the time to explain this. I racked my brain with a couple books and a Surplus Center catalog as a refrence and tried all different formulas, but I had no way of knowing if I was even in the ballpark. (I kept coming up with one pump they had that was

4GPM, but calculated that is would be closer to 5.4 GPM with the engine speed.) Thanks again for your help, Craig

On 6 Apr 2005 18:32:05 -0700, "monkers" wrote something ......and in reply I say!:

Operate the cylinder by hand, or with the help of some hand-operated tools etc. See how much oil comes out/goes in if you take it right out, fill it up, the push it right back. There is your capacity. If you measure the stroke, you can also then work out the bore.

GPM will depend then entirely on how fast you want the machine to operate.

Bending 1" rebar. The force required will depend a lot on the configuration of the machine, and how sharply it bends the bar etc. The further the piston is from the place where the rebar is actually bent, the less force is required.

There are ways to get an idea of the forces, but the above would need to be known. My ghuess is that a 4" cylinder running at the standard

2500 PSI should be way enough to bend 1" bar. But again, if you placed the push right at the fulcrum that the bar reasted on, you would have buckley's....

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On 6 Apr 2005 18:32:05 -0700, "monkers" wrote something ......and in reply I say!:

More ideas. They make hydro log splitters. I looked around and found one that has a 9HP motor, which produces 16GPM. This will cycle a 24" x 4" cylinder with a 2" rod. This is close to what you describe, and a bit of math can work out the differences.

It claims 20 tons of force. At 2500 PSI and a 4" bore that seems a bit optimistic, #(2 ^2 * pi(3.14) = 12.6 in2 * 2500 PSI = 31500 lbs force / 2240 lbs/ton = 14 tons)

But some systems run at 3000 PSI. Even then I still get only 17 tons.

20 tons would need 3500 lbs.

Now it probably uses a double-speed pump. But without that, the cycle time for _push_ would be:

4" * 24" capacity = 301 cu in = .175 cuft = 1.15 gals (imperial)

At 16 GPM, therefore I reckon about 5 seconds.

Then there is another 4-5 secs to come back again (the rod makes the capacity smaller on the pull stroke).

If you use the 3.8 litre US "wet" gallon (I have never quite sat well with all that wet and dry stuff ), the pump is actually only making

12GPM imperial, and the above times would be about 6.5 seconds push, 5 seconds pull.

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