Hydraulic Math

LeakyBoot

Member
3 inch cylinder with 1 and 1/4 shaft Dia. If a gauge is installed in port of shaft end of cylinder and 1500 PSI applied to piston end, what will the gauge read? Will the reduced piston area of the shaft side of the piston give an increase in pressure when pushed along by the full area of the 3 inch piston?
 
Oil pressure is what ever the pump and relief are capable of. As far as push pull pressure, return force is calculated by square inches of piston times oil pressure, so return will be less because you deduct the square inches of the rod from the piston.
 
Yes. The displacement of the rod reduces the containment by its cross-section. Similar to the way a larger one way cylinder of 10 sq. inches area would develop pressure in a 5 sq. inch area cylinder if pushing on it. Jim
 
The pressure on the rod end will be roughly 2.4 times the base end assuming the cylinder is not at the end of stroke.
It is proportional to the difference in surface area on each side of the piston.
 
Another way of looking at it. A cylinder (with one rod end) will produce more pressure on push than pull, under the same applied pressure. Jim
 
Farmer Boy is right. This is one reason to never run hydraulic valves in series. If one cylinder has a load trying to push a cylinder out and you further extend that cylinder while using a downstream valve that load on that second valve can have a really high pressure applied to it.
 
I don't think so, the smaller volume will translate into less speed, but the pressure will be the same.
 
If pressure is applied to the piston end, I must assume that the end with the shaft is connected to the valve, the valve should be open and the pressure should be Zero if the valve is operating properly. Maybe I did not read the question as some others did.
 
Leakyboot,

The "pounds per square inch" on the piston side will be the same on the rod side. The rod side has less square inches, thus a greater pressure.

D.
 
Reason for question is I was thinking of inserting a double acting cylinder inline between hyd power supply going to a single acting cylinder, bleeding all air from the system and hoping to boost the 1500 psi start pressure up by a few hundred pounds. Using a cylinder with a larger rod dia should help even more. ??
 
dennis min,

I am absolutely NOT an authority on hydraulics, but I like to try to figure out these puzzles. I tend to agree with your thinking here.

Now a question : IF the piston were positioned in the stroke such that there were equal amounts of "oil space" cubic inches on each side..... it seems a gauge placed on each side should read the same. What do you think ?
 
Farmer boy has it but in your situation you need to consider the volume requirement of the single acting cylinder. The double-acting will multiply the pressure but will the rod side have enough fluid volume to push the single acting to it's full stroke? You need a long enough cylinder to do it.
As far as running cylinders in series, it's done a lot. Automotive floor lifts, for one. It's the only way to keep two cylinder acting together. They will get out of sync in time because of pressure differences causing fluid bypass, so there has to be a way to equalize the two sides.
 
No. He's putting supplying 1500 psi to the piston end of cylinder, and the rod end is smaller. It more or less works like an intensifier piston in a HEUI injector. With the way he's asking the question, there is no speed. The cylinder is locked in place because there's no where for the oil to flow.
 
3 squared times Pi equals 28.2744 times 1500 equals 42,411.6
1.25 squared times Pi equals 4.90875 subtract that from 28.2744 equals 23.36565
then take 42411.6 and divide by 23.36565 to get farmboy's correct answer.
Pi simplified is 3.1416
Now, if you turn that shaft into a piston, it would develop 8,640 PSI of pressure. We used air over hydraulic pumps where I worked at to get 20,000 PSI water from 100 PSI air.
 
Thanks for all the replies. I think I will try it and see if I get a positive result. I have a 10 ft disc mower that the tractor I like to use it on just does not have the psi to lift the bar into travel position. The New Idea uses a small dia cylinder, single acting, and on retract to lift the bar. So it takes a lot of psi. All I really need to do in remove the hose from the rod end of the extra larger cylinder and install a female quick coupler to plug in the hose from the mower.
 
Just have to use a long enough cyl that won't run out of fluid before the job is done. I think one term used is "intensifier"
 
Pete Md,

Yep, I re-read my response and it sure does contradict itself. Maybe I'd be better saying the force is equal on both sides. The difference is the area to which the pressure is applied. I maybe had better off saying the force is what we're looking at.

500 psi on one inch of area is equal to 50 psi on 10 inch area. Hydraulically speaking...

Sorry for the errant response.

D.
 
I have an old ford 5000 that wouldn?t lift my 14 foot disk I put a bigger cylinder on that disk it was slower lifting but I was able to disk with that tractor
 
You have the right answer. Pressure x area = force. The pressure is dependent on the load and goes up to relief at full load. The down side has less area, so the force is less for any given pressure right up to relief pressure. The resisting force up to relief is what will be determining the pressure the pump needs to put out.
 
Your theory may work. But I don't see it as a practical solution. There will be problems with oil loss and trapped air.

Everything hydraulic can be calculated with pressure/force charts.

What you can do is take the results from the test, convert the results to the diameter cylinder you need for a permanent fix.
 
I would be surprised if the operator's manual for your mower didn't specify how much hydraulic pressure is required to operate it. That would give you a figure to work with in your calculations.

Your idea is interesting. Be sure and let us know how your experiment works out.
 
(quoted from post at 16:13:19 04/15/18) Your theory may work. But I don't see it as a practical solution. There will be problems with oil loss and trapped air.

Everything hydraulic can be calculated with pressure/force charts.

What you can do is take the results from the test, convert the results to the diameter cylinder you need for a permanent fix.

It is complicated by the fact that he said lift was by retracting the working cylinder. So he would be tying the cylinders rod end to rod end.

For those who are having trouble with a cylinder making more pressure, you can think about the cylinder having 1,500 pounds/sq in on the piston, 14.7 pounds/sq in air pressure on the rod, and the oil around the rod has to be high enough to make up for the low pressure on the rod.

I've seen systems which apply pressure to both sides on a cylinder to extend it fast under low load. Then all you have is pressure on the rod and as the piston goes out the excess oil goes back into the head end and increases the speed dramatically.
 

This thread taught me to only trust jon f mn. Several others confused pressure with force. The pressure is the same regardless of the area it is acting on. The force acting on the piston is a function of the area the pressure is acting on times the pressure. Yes, the force is different, but the pressure, expressed lb per sq inch, will be the same. The force, expressed in pounds, will be different depending on the area that the pressure is acting upon.
 
(quoted from post at 23:19:24 04/15/18)
This thread taught me to only trust jon f mn. Several others confused pressure with force. The pressure is the same regardless of the area it is acting on. The force acting on the piston is a function of the area the pressure is acting on times the pressure. Yes, the force is different, but the pressure, expressed lb per sq inch, will be the same. The force, expressed in pounds, will be different depending on the area that the pressure is acting upon.

Actually, regarding the original post, and what he is intending to do, farmer boy and leaky boot had the best answers.

The O.P. already had it figured out, the end with the smaller surface area due to the rod WILL produce higher output pressure than what the tractor by itself is capable of and act as a "pressure intensifier".

This may just work!
 
The volume of oil doesn't factor into the calculations - it is only the area upon which the pressure is acting that matters. So, as long as it isn't bottomed out, the position of the piston within the stroke doesn't matter. For the piston to be in equilibrium there needs to be equal forces pressing on it from both sides. Since the rod side of the piston has a smaller area than the base side the pressure must be proportionally higher.
 
Leakyboot, Yes, what you are proposing works. We have done it with a Class rake that we could not lift with our older tractor.
 

You haven't mentioned what tractor you want to use with the mower or it's hydraulic capabilities.
I find it hard to believe NI would have built the mower so that it could only be lifted by the most modern high pressure hydraulic systems.
There forth I'm questioning the condition of the tractors hydraulic pump and relief valve.
If the tractors hydraulics are in good condition wouldn't it be simpler if room allows to swap out the mowers lift cylinder for one with a larger diameter barrel.
Don't have my book in front of me that tells what the increase is on lifting force if one changed from a 3" dia cylinder to say a 3 1/2 or even 4" keeping the rod dia the same.

My first move would be to install a pressure gauge inline and see how much actual pressure the tractor is putting against the cylinder. If I had another tractor that would raise the mower bar I'd connect to it's hydraulics and using the inline gauge I'd see how much pressure it actually takes to lift the bar.
This would give one a number needed in order to formulate what is needed to make up for the tractors lack in hydraulic capability.
 
(quoted from post at 16:25:21 04/15/18) I have an old ford 5000 that wouldn?t lift my 14 foot disk I put a bigger cylinder on that disk it was slower lifting but I was able to disk with that tractor

It's possible the disc had the wrong cylinder installed but do a pressure test on the tractor.
My 5000 has not trouble lifting my 10 ft disc or 15 ft batwing mower.
In recent years I've noticed my 4000 was struggling to lift things it used to do with ease, after a few test I found the original 69 hydraulic pump had finally worn out.
 
Most of the people replying here are confused about the definition of pressure and the difference between force and pressure. Pressure is "force per unit area" e.g. pounds (force) per square inch (area). For a hydraulic system that is at equilibrium (not moving) the pressure is the same throughout. The force exerted is greater at the end of a cylinder with smaller diameter (because the cross-sectional area is smaller), but the pressure is the same at each end.
 

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