Hydro Differential pressure exchange over unity system.

[TinselKoala]

MrWayne: thank you again for answering.  No, you are not really in my "OUIdiots" file... you are still one level above that in the directory tree.


Crossheads are used whenever a straight push needs to have some out-of-line motion at the end of travel. Perhaps the most familiar place we see crossheads being used are in railway steam locomotives, where the straight push of the piston needs to be sent off-axis to the driving wheel. It is sort of like an external wrist pin; it acts to relieve the side loads on the piston and cylinder and conrod packings by moving them to easily lubricated sliding guides.
Another improvement in geometry I was thinking might be useful for approximately the same purpose would be to adopt the shape of an oil pumpjack's head, which serves to change the straightline travel of the pumpshaft going into the well, into the arc motion of the jack's rocker arm without putting side loads on the pump shaft. I know you have plenty of pumpjacks around so you know what I mean here.


OK, so the hydraulic reservoir that is at or near ambient pressure is the small white one at the top, and its rams that pump fluid from it back into the pressure side are the two rams at the very top, also seen in that same image?
And that big tank at the bottom is a water tank, and there are no other reservoirs (except the pressurized hydraulic accumulator)?

And how does the hydraulic fluid get from the motor's outlet back up to that reservoir? Is there enough pressure in what is being supplied to the motor to drive it back up there or is it assisted in some way to get back to the reservoir which is much higher than the motor?

On another issue... I have been looking for the patent. I see some drawings from it, but I've been back through this entire thread and over both your mrwaynesbrain website and the PESN page and I can't find a listing for the patent. I probably just missed it... I am literally half blind these days, my left eye is failing... so could you please link to it again? And it is an actual, fully granted patent, right? Not "just" an application, I hope.

[TinselKoala]

I called it a Riser, because it moves and the Tank does not move.  Since my initial state is sunk, it is also called a Riser, because the "power" stroke is rising up.  I guess I could have called it a sinker, since it sinks with no inputs.  I don't even remember if I picked up the term from someone else.  I would not call it a floater though... LOL.  What difference does it really make?  A rose by any other name.

There are two ways of looking at it.  Is it floating up, or a pneumatic cylinder?  This is a hybrid device.  It has aspects of both, and that is what makes it harder to understand.  Looking at the transfer curves, depending on the state, it could look like either.  That is part of my problem -- trying to make sure I have the theory correct much less getting rid of typos in my formulas.

It doesn't have closefitting seals that don't allow fluids like air or water to pass, does it? It sure looks to me like it's floating and the pneumatic part is only raising the effective water level, and when the riser--your term-- is locked, the water rises up around it, effectively creating the same kind of "virtual water weight" that I showed in my first little video. Right? Therefore it is not a pneumatic cylinder, which requires sealing and does not allow water to come up past it. Nor is it some kind of hybrid. It is a floater; it rises because it displaces liquid, the amount of liquid displaced can be varied by locking it in place and pushing up on the bottom piston, and when it's unlocked it becomes a simple floater again, rising to the correct height , and with the correct buoyant force, for the amount of water its wetted volume displaces.
Right?

And I am afraid I don't understand how it is supposed to sink through water, since it has extended volume but only weighs, or is counterbalanced to weigh, 0.1 pounds in your sim.
Or was is only 0.01, I don't have it right in front of me.

There are roses that have thorns, and there are thornless roses. They might smell the same and be called the same... but you will find out just how different they are when you try to grasp one.

[see3d]

It doesn't have closefitting seals that don't allow fluids like air or water to pass, does it? It sure looks to me like it's floating and the pneumatic part is only raising the effective water level, and when the riser--your term-- is locked, the water rises up around it, effectively creating the same kind of "virtual water weight" that I showed in my first little video. Right? Therefore it is not a pneumatic cylinder, which requires sealing and does not allow water to come up past it. Nor is it some kind of hybrid. It is a floater; it rises because it displaces liquid, the amount of liquid displaced can be varied by locking it in place and pushing up on the bottom piston, and when it's unlocked it becomes a simple floater again, rising to the correct height , and with the correct buoyant force, for the amount of water its wetted volume displaces.
Right?

And I am afraid I don't understand how it is supposed to sink through water, since it has extended volume but only weighs, or is counterbalanced to weigh, 0.1 pounds in your sim.
Or was is only 0.01, I don't have it right in front of me.

There are roses that have thorns, and there are thornless roses. They might smell the same and be called the same... but you will find out just how different they are when you try to grasp one.

Well it does not matter to me which way it is approached, as long as it gives the right answer.  I approached it from the point of view of forces rather than displacement.  One can be translated it to the other in practice.  I would like to see it calculated both ways, and both matching.  Tried that once, but had trouble with the displacement version.  If you show me the exact formula for your way, I will plug it into the sim program and see how it compares.

Sinking:  In the down state, there is no H0, so no buoyancy force on the pod.  There is some buoyancy force for the Riser wall that is sunk to the bottom of the H1-H2 water pocket. The Riser would in a practical sense have more than enough weight to sink, so it is counterbalanced to 0.01 pounds to just barely sinking (1% of the load weight).  It takes an input force (even without an output load weight), to raise it up.  Therefore, remove the input force, and it returns to the sunk state.  Here is the new picture I am adding to the PDF:

[johnny874]

 tk,
the reason air has high velocity is it has low pressure once it leaves tne nozzle.
that is a principle they use in converting high pressure steam into tne rpm's of the steam turbine.

[see3d]

Here is the problem I am having with my reworking of my formula:

I start with a riser locked down  and no input force.  I apply an input force and the piston will move up until it encounters an opposite force equal to the input force.  When the piston moves up, it forms head H0 which compresses the air.  I first determine the air PSI that will exert the same force over the area of H0.  I can see no way around this fundamental linear relationship, so this is where I start. 

AirPSI = InputForce / H0Area.
0.125 = 1.00 / 7.987

I reason that the H2-H1 head differential must also support the same PSI calculating from the exit back. 

H2 - H1 = InputForce * H1Area / (H0Area * H2Area * WaterWeight)
5.210 - 2.251 = 1.00 * 9.684 / (7.987 * 11.380 * 0.036)
2.959 = H2 - H1

The two formulas are linked by:
  The constant water volume of the H1, H2 Tank pocket
  The linear reduction in air volume with increased PSI

However, when I do an internal consistency check for the force at the bottom of the Riser wall, I get two different numbers depending if I calculate it from H2 or H1 + air PSI.  I have not found where my reasoning is incorrect yet. 
Calculated from the H1 side, I get 0.206 PSI at the bottom of the Tank. 
From the H2 side, I get 0.188 PSI at the bottom of the tank. 


I either have something messed up in my conceptual thinking, or I just have a typo in my formula that I have not found.

Here is the picture: