Alternative 3rd function?

[orangetree]

Perfect vacuum? Not possible since you have finite space + pre-existing particles

Since there is obviously no such thing as "negative pressure", the cavitating (not sure if this is specifically a correct usage btw, but seems close enough) space will just ~approach zero ~ and probably not very closely.

How far will it travel? How it moves is simply sum of forces. assume hard zero on the cavitating end. add the force on the rod, and the force from pressure on the other end. Assume a non accelerating reference system for good measure! piston moves opposite from the greater force.

 

[npalen]

Perfect vacuum? Not possible since you have finite space + pre-existing particles

Since there is obviously no such thing as "negative pressure", the cavitating (not sure if this is specifically a correct usage btw, but seems close enough) space will just ~approach zero ~ and probably not very closely.

How far will it travel? How it moves is simply sum of forces. assume hard zero on the cavitating end. add the force on the rod, and the force from pressure on the other end. Assume a non accelerating reference system for good measure! piston moves opposite from the greater force.

So lets assume a 2" diameter piston for simplicity so the area is 3.14 sqin.
How much force would be required to pull the piston 1" creating "negative pressure"?

Edit: Would the force required eventually be theoretically infinite or would the maximum force be 46.182 pounds and never exceeded regardless of how far the piston was "pulled"?

 

[ovrszd]

In the latest "theme" of this thread about over-pressurization from grapple mis-use......one instance comes to mind for me. With the grapple lid open but not open all the way....and bucket down-turned with the tines of the lid trying to pull on something (or even push).

Yep. And this added load can be transmitted to the curl cylinders of the FEL. If those cylinders are in an extended position it's very possible to bend the rams. I did this on a previous 45hp tractor/FEL.

A Grapple, Tree Puller or any 3rd function attachment that has the ability to "grip" something changes the dynamics of the load on the hydraulics and super structure of the implement and tractor involved.

 

[orangetree]

So lets assume a 2" diameter piston for simplicity so the area is 3.14 sqin.
How much force would be required to pull the piston 1" creating "negative pressure"?

Edit: Would the force required eventually be theoretically infinite or would the maximum force be 46.182 pounds and never exceeded regardless of how far the piston was "pulled"?

Yes, the lowest the pressure on the 'cavitation' side could have is zero. std 1atm is ~14.7psi. Ignoring friction, ambient pressure on the rod area itself, asymmetries, etc, it's only the fluid pressure on the opposing side exerting force on the plunger. At a minimum that fluid is vented to air (eg, at 14.7psi absolute), so that's the only other force on the piston. acceleration = force / mass, so the piston accelerates (eg moves) in the direction (net) force is. and that force is, at a minimum, 14psi exerted on 3.14sqin ~= 45lbs.

in real life, I'd expect also quite a bit of friction (these seals can withstand thousands of psi, after all), so the actual force required to "try this" by plugging one 2" cylinder port in your hand is likely much higher But if you have kids, grab a pneumatic lego cylinder and try it. Not much force required to "pull a vacuum" in that little cylinder.

note that in everyday life we typically talk about pressures relative to atmosphere - ie "gauge pressure" - because it's the most relevant. This discussion is talking about vacuum, so it makes more sense to talk absolute pressure. Just keep in mind that's a different reference than the way you presumably are used to thinking about it.

 

[npalen]

Yes, the lowest the pressure on the 'cavitation' side could have is zero. std 1atm is ~14.7psi. Ignoring friction, ambient pressure on the rod area itself, asymmetries, etc, it's only the fluid pressure on the opposing side exerting force on the plunger. At a minimum that fluid is vented to air (eg, at 14.7psi absolute), so that's the only other force on the piston. acceleration = force / mass, so the piston accelerates (eg moves) in the direction (net) force is. and that force is, at a minimum, 14psi exerted on 3.14sqin ~= 45lbs.

in real life, I'd expect also quite a bit of friction (these seals can withstand thousands of psi, after all), so the actual force required to "try this" by plugging one 2" cylinder port in your hand is likely much higher But if you have kids, grab a pneumatic lego cylinder and try it. Not much force required to "pull a vacuum" in that little cylinder.

note that in everyday life we typically talk about pressures relative to atmosphere - ie "gauge pressure" - because it's the most relevant. This discussion is talking about vacuum, so it makes more sense to talk absolute pressure. Just keep in mind that's a different reference than the way you presumably are used to thinking about it.

So are we saying that the 45lbs stays constant regardless of how far the piston is "pulled"? Also, how far does the piston have to move before it generates the 45lbs force requirement and zero PSI absolute on the blind (sealed) side of the piston? Would be interesting to see this on a graph.

 

[LD1]

So are we saying that the 45lbs stays constant regardless of how far the piston is "pulled"? Also, how far does the piston have to move before it generates the 45lbs force requirement and zero PSI absolute on the blind (sealed) side of the piston? Would be interesting to see this on a graph.

Disregarding seal friction.....the fact that it takes FAR FAR more than 45lbs to extend a cylinder is because of the piston seals.

Under ideal circumstances.....the ram CANNOT move either direction....because the fluid is trapped on either side of the piston and blocked off by the spool valve.

 

[npalen]

Hydraulic accumulators have been mentioned and are also widely used in industrial applications where continuous clamping of components is needed when machining parts, for example.

It would be interesting to experiment with a long coil of hose plumbed to the base side of the grapple clamping cylinder(s) to act as an accumulator. The hose would expand when pressure is applied to the cylinder and releasing the valve spool would trap the oil in the expanded hose holding clamping force on the grapple.

Leakage by the valve spool would start reducing the force but that is true with any accumulator device.

We used to have issues with hose expansion and contraction on large tillage equipment with master/slave type cylinder lift systems. The problem came when multiple cylinders that were synchronised perfectly would soon be out of phase due to expanded hoses pushing oil back past a leaky valve spool.

Edit: A single braid larger diameter hose would work best as an accumulator in the above grapple clamping application. A 3/4" diameter single braid hydraulic hose, for example, would have much more "accumulation" than, say, a 3/8" diameter double braid hose.
Would need to check the hose pressure rating, however.

 

[orangetree]

Disregarding seal friction.....the fact that it takes FAR FAR more than 45lbs to extend a cylinder is because of the piston seals.

Under ideal circumstances.....the ram CANNOT move either direction....because the fluid is trapped on either side of the piston and blocked off by the spool valve.

AFAICT this is just a thought exercise, trying to add some intuitive understanding if one hasn't taken a statics or fluids course.

Agreed, in the real world there is fluid on the other side of that cylinder, and that fluid is constrained by a closed valve ... you are not going to move that cylinder by hand

 

[orangetree]

It would be interesting to experiment with a long coil of hose plumbed to the base side of the grapple clamping cylinder(s) to act as an accumulator. The hose would expand when pressure is applied to the cylinder and releasing the valve spool would trap the oil in the expanded hose holding clamping force on the grapple.

I couldn't find estimates of the elasticity of the container(s) here, so i just took an estimate of fluid compressibility. Do you have any educated-WAGs at expansion of the hoses? estimating strain in the cylinder itself would be straightforward enough just looking at hoop stress, if I knew the wall thickness ... but that's all beyond the scope of this I think A working estimate from a shop that can test it would be fantastic, though. We're really just looking at Fermi estimation here as there are so many variables, and (at least speaking for myself) not an engineer in this field.

Leakage by the valve spool would start reducing the force but that is true with any accumulator device.

I'd rely on this as a feature, not a bug. IMHO I don't think I want my CUT/hobbyist tractor storing residual clamping force. For my tractor, I feel "what you see is what you get" is the safest - I don't want it "rebounding"; that sounds like a recipe for disaster. EG if i overload the grapple and it opens: great, stay open. I have a nifty lever that can re-close it.

So that's a problem with accumulators; but they seem be the only solution here that doesn't involve a complex, valved reservoir: to drain the excess when force-compressing the blind end, or to makeup the missing fluid when force extending to the rod end.

 

[orangetree]

So are we saying that the 45lbs stays constant regardless of how far the piston is "pulled"? Also, how far does the piston have to move before it generates the 45lbs force requirement and zero PSI absolute on the blind (sealed) side of the piston? Would be interesting to see this on a graph.

~45lbs is simply atmospheric pressure, applied to a to a ~3inch diameter circle. Yes, that stays ~constant. We *are* starting to get to spherical cows, and TBC that's not my area

How far does it have to move? That depends entirely on the assumptions we have to apply to simplify the problem enough to post these ~qualitative answers
> Pressure = APysicsConstant * Temperature_Kelvin / Volume
So if you double the volume, you half the pressure. This is "ideal gas law" and it's very accurate estimate for "everyday" temps & pressures.

So the 45lbs exists no matter what on the end we're assuming is fluid "free" to flow in and out (eg NOT up against a valve like LD1 points out). And lets say the cylinder is bottomed-out blind, with "negligible" fluid on the blind end. That fluid on the blind end is also starting atmosphere (if the rod is motionless and our above assumption hold).

So you start to pull. The blind end cavitates. You assumed it has "negligible" contents, so we moved from "zero" volume to "more than zero" - an infinite increase. The force would basically jump up immediately to that ~45psi.

Ignoring static & dynamic friction, and lots of other stuff. But yes, with respect to the pressures acting on the 3sqin of piston, it's that simple.