A hydraulics question

So lets say we have a cylinder with a piston in it. The area of the piston face is 1 square unit. The piston is mounted to a shaft with a cross section of .98 square unit. This means that the force that the piston can push using the same pressure fluid will be much greater in one direction than the other. Now lets say we cut a groove into the shaft such that the remaining shaft at the bottom of the groove has a cross section of .1 square unit. Wouldn't that mean that the piston push will equal .9 times the fluid pressure in one direction and 1 times the fluid pressure in the other? Or does the shaft need to be just as small at the end where it passes through the cylinder cap? So that the area of the cylinder cap minus the shaft diameter is the same as the cross section of the grooved area of the shaft. I think it's the latter situation. This is actually more than a thought exercise, I need this to help me decide how to make something. And my poor brain just won't think clearlt this morning. Thanks, Eric

Reply to
Eric R Snow
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It is the latter as it is the area that is reacted on that determines the force and that means the cylinder "cap" area, as you put it.

If you want a ram with equal force or speed in either direction, consider making one that is double ended ie: one that has a shaft either side of the piston.

Tom

Reply to
Tom

Nope.

The diameter of the rod where it attaches to the piston is irrelevant. The only thing that matters is the area of the portion of the rod that's moving in and out of the pressurized chamber above the piston; the area inside the sealing gland in a normal cylinder. In other words, what you're concerned with is the change in volume per unit distance travelled by the piston. Thinking about where the mechanical advantage comes from in a hydraulic jack may help.

Ned Simmons

Reply to
Ned Simmons

The diameter of the shaft (ungrooved) will have no determination on the effective area that the hydraulic fluid acts upon.

Making the groove will increase the force of your return stroke. Indeed, it will also increase the speed at which the cylinder retracts (as opposed to reducing the diameter of the entire shaft) as there is less volume in the cylinder with the larger shaft.

This is kinda how hydraulic multipliers work.

Regards,

Robin

Reply to
Robin S.

Others have answered your question, but don't forget that you have to subtract seal friction from the force calculated from (area) x (pressure), unless you make sure it is dynamic. Pressure gauges are (or at least, used to be) calibrated using dead weights on top of a piston. By spinning the piston slowly (making use of the inertia of the weights on top) you eliminate static friction in the seal.I couldn't find any "design code" for a static ram when I looked a while ago, but if I recall correctly someone was actually getting about 30% less force than they calculated because of static friction.

Reply to
Newshound

I disagree, Ned.

Hydraulic force acts on surfaces which are perpendicular to the axis of movement.

I don't understand why that would matter at all. The force has nothing to do with the cylinder cap. The cylinder cap is not moving.

Volume in hydraulics determines the speed of the cylinder. Assuming the pump can keep up, it has no effect on force. It is the pressure created by the pump, and the area of the surface which is perpendicular to the axis of motion in the cylinder (typically the piston face).

Indeed, any area created by the OP's groove would be added to the force calculation because it is perpendicular to the axis of movement.

Regards,

Robin

Reply to
Robin S.

The pressure in the cylinder acts equally in all directions, and an undercut in the rod will have two opposing areas, so any localized forces cancel one another.

Or look at it as a conservation of energy problem. In a hydraulic (uncompressible working fluid) system, work is equal to pressure x change in volume. The work done by the cylinder rod is distance x force. So,

pressure x change in volume = distance x force

Putting a groove in the rod won't change the the amount of oil that flows in or out of the cylinder for a given change in rod position, so in order for the equation to balance, the relationship between force and pressure can't change as a result of the groove. That relationship is the effective area of the cylinder.

Ned Simmons

Reply to
Ned Simmons

I stand corrected, and I learned something today. Thanks.

Regards,

Robin

Reply to
Robin S.

The groove has two sides, one adds pressure one adds it inversely, cancelling the other out. The only way to increase the pressure is to decrease the size of the rod going through the cap. This increases the sectional area of the piston subject to the hydraulic pressure.

John

Reply to
John

Cutting the groove will not affect the resultant force. Think of it this way: with a groove, the fluid is pushing on a larger piston area, but it's also pushing the opposite way on the "back" side of the groove. The area on the back of the groove is the same as on the front and they cancel.

Bob

Reply to
Bob Engelhardt

LOL

Reply to
Tom

Think "what the piston pushes _against_". The rod-end cap of the cylinder has .02 sq" of surface, so there's .02Xpressure of force available. The butt-end cap has 1sq" of surface.

LLoyd

Reply to
Lloyd E. Sponenburgh

Thanks Everyone who replied to my question. After posting I knew which way was right but I had already pressed the send button. It's funny how sometimes you just can't think straight until you tell the problem to someone else. It is nice to have your thoughts verified though. Cheers, Eric

Reply to
Eric R Snow

A college acquaintance of mine, a math major, refered to this as "bouncing thoughts off a brick wall", and would discuss problems that way. "Look, I know you have no hope of understanding this, but if I try to explain it to you, I'll probably see the answer myself, so do you mind if I just use you as a brick wall for a minute...?" --Glenn Lyford

Reply to
glyford

And it shocks the hell out of them when said "brick wall" not only understands the problem through a different set of 'life filters' that have nothing to do with the field of expertise in question - but still manages to come up with a different way of finding a solution.

"Don't raise the drawbridge, lower the river!" ;-)

Been there, Done that, Read the book, Saw the movie, The book was better...

The Specialist has blinders on and is stuck in a rut of only looking in /one/ place for the solution, and that's the path they've nailed their mind firmly to, and have probably been issued a PhD or other advanced degree in...

But to the Generalist there are often many other equally valid ways to get to the needed solution - you just have to come up with the right one or ones, then follow that path.

One of them is the old NASA Doctrine of "Build 'em, Boost 'em and Bust 'em." Experiment, take your best guesses, build a prototype, and try it. See what didn't work and redesign that part. When they stop breaking, you know you're on the right path.

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Reply to
Bruce L. Bergman

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