Appearantly the PDF did not serve its purpose very well.  Simpler is not what I requested, but rather what I requested was explain it in steps and in detail, including the purpose for and the result of each event / motion.

Yes I viewed the links.  Assuming that I know how a hydraulic cylinder works, the actions of your device and why it is supposed to work are still not clear to me.

                                 floor

Sorry Floor,

The pdf is sufficient for understanding by anyone ‘skilled in the art’, and not very skilled at that. This is basic hydraulics and physics.
My free time is very limited and I cannot spend it in time consuming unproductive effort.

If anyone else wishes to accommodate you that’s fine with me.

Respectfully
Cadman

Dear Cadman

I will be thrilled if the design you present actually works as you have stated, but please don't under estimate the value of a good critic.  Also please don't mistake my skepticism for cynicism.  However still it remains, that the burden of proof is upon you, and such proof does of necessity include concise explanation.

In the Cadmans hydro static displacement engine.PDF, you state that ...

"Either size of pipe and either weight of liquid, it’s the same PSI. That isn’t much pressure but if you apply it to a piston with an area of 28 square inches you have a piston force of 60.676 lbs. That is significant."

But as anyone skilled in the art can tell you, there is the same amount of work done in lifting an automobile 20 inches with the large diameter output cylinder of a hydraulic jack as is input by the jack's user in pumping the jack handle a large distance.
  and

1. Yes, a fluid seeks it own level (basic physics).

2. A small force over a large displacement can be swapped for a large force over a small displacement in hydraulic devices.  This is similar to mechanical leverage (basic physics).
...

"Pressurized fluid from the feed pipe enters the Pressure chamber A, forcing the piston up. Fluid above the power piston transfers to the isolation tank  .....   at the same rate as an equal volume
of fluid is siphoned from the vented isolation tank into the Transfer chamber B."

Simplified.... fluid has fallen in the feed pipe side of a U  (shaped) tube while rising in the piston
side of that tube.

Its rise is arrested when the top of the power piston contacts the gland (seal around the piston rod).  Other wise, the the fluid's rise on the piston side, would ONLY stop, once it has risen to a height equal to the fluid level of the feed tube supply tank. 

Except that (assuming the combined weight of the power piston, piston rod and displacement piston is greater than an equal volume of the fluid) the fluid will rise on the piston side of the U tube, to a height which is less than the height of the fluid level of the feed tube supply tank. This
is because the combined weight of two pistons and their connecting rod limits that rise. They
alter the other wise balance in pressure / weight between the two sides of the U tube.  The piston side exerts greater pressure than does the feed tube side. The volume of this lesser amount of
fluid rise, will be exactly equal to the fluid volume displaced upwardly, when the
displacement piston falls to its starting position.

This is why your detent valve doesn't function well.  There is no excess energy to operate it.

Just my opinion. Prove me wrong. I might be wrong.

Thanks for your time and effort.
  best wishes
         floor

Hi Cadman,

Don't be discouraged by the naysayers.  If you have done the experiments and proven to yourself that the idea has merit that is really all that matters.  I am very involved in a couple of large projects right at the moment but intend to pursue your idea when I have time.  I am sure there are others that also believe you might have a workable idea.

Respectfully,
Carroll

Hi Carroll,

I think Floor is sincere. Not a naysayer, just healthy skepticism.
I repeated my experiment yesterday. It's confirmed.

Regards,
Cadman

I'm sorry.  My post was misunderstood.  It wasn't directed toward Floor.  I am aware of some others that have been very negative about your idea.  It was those I was referring to.  My apologies to Floor.  I also greatly respect his work and in fact am trying to replicate some of it for my own testing.

Glad to hear you have successfully verified your results.   

Sincerely,
Carroll

Dear Floor,
This is for you. I unexpectedly find myself at a stop on my build so here is the analysis you asked for. I hope it brings understanding. If I have made any mistake in the math please point it out.

The water weighs ~8.345 lbs per gallon. 231 cubic inches per gallon.
The area of a circle is Pi times the radius squared.
The area of the 6” piston is 28.2743 sq. inches. Call it 28.27.
The area of the 3/4” diameter rod is 0.4417 sq. inches. Call it 0.45.
The area available is the piston area minus the rod area. 28.27−.45 = 27.82 sq. in. This area is for both pistons. Actually the area of the bottom piston increases back to 28.27 as soon as the rod loses contact with the valve, but I’m going to ignore that and use the lesser area.
The stroke is 6”.
The operational sequence begins with the piston assembly at the bottom of the stroke, the feed line open and the bypass line closed.

The height of the water column in the feed pipe is 63” to the bottom of the 6” piston. That gives a working pressure of 2.2759 psi.
The force available from the piston is the area of the piston times the psi applied to it. 27.82 × 2.2759 = 63.315538.
Call it 63.3 lbs of force.

The weight of water in a vertical tube is the area of the tube diameter times the tube height, divided by 231, times the water weight per gallon.
The displacer piston is 40” tall. There is 1.5 in dia passage bored through the piston from top to bottom with a check valve on the top. The fluid in this passage weighs 2.5535 lbs.
Call it 2.6 lbs.

The displacer piston itself is of light weight semi-hollow construction and weight is 5 lbs.

There are 3 other water columns above the pressure piston, 6” 0.5” and 7”. 13.5” total height. Their combined weight is 13.7892 lbs.
Call it 13.78 lbs.

The 0.75” chromed rod weighs 1.5 lbs/ft and is 72 " or 6 ft long. Weight is 9 lbs.

The pressure piston is 1” thick plastic and steel assembly. Piston weight is 2 lbs.
Lets toss in another 1 lb for seals, collars and misc hardware.
Weight is 3 lbs.

Total weight rod, pistons, hardware and water is  2.6 + 5 + 13.78 + 9 + 3 = 33.38 lbs.

At the bottom of the stroke we have 63.3 lbs of force lifting 33.38 lbs.
The surplus is 63.3 – 33.38 = 29.92 lbs. This is at the beginning of the up stroke.

As the piston rises through it’s 6” stroke the height of the water column in the feed pipe relative to the pressure piston decreases from 63” to 57”. That means there is a gradual decrease in the force applied to the pressure piston, it drops to 2.0591 psi. However the water volume above the displacer piston also drops to 0 as it spills into the tank. That lowers the combined water column height to 7.5”. Total water weight decreases to 7.6606 lbs, a difference of 13.7892 – 7.6606 = 6.1286 lbs

So we now have 2.0591 x 27.82 = 57.2841 lbs of force.
A total lift weight of 33.38 – 6.1286 = 27.2514 lbs
Surplus is 57.2841 – 27.2514 = 30.0327 lbs.

So with this engine configuration there is an almost constant surplus force throughout the up stroke of ~30 lbs. This force shifts the valve which blocks the feed pipe and opens the pressure area at the bottom of the piston to the top of the piston which removes the pressure holding up the piston assembly, and the down stroke begins.

The down stroke action is almost a no-brainer. We have a piston assembly dead weight of 5 + 9 + 3 = 17 lbs. A constant force supplied by gravity. With the valve shifted the pressure on the top and bottom of the pressure piston equalizes and the water flows from below the piston to the top of the piston, it sinks. Gravity is pulling it down.
The water below the displacer can not reverse flow back into the lower section because of the check valve in the line. The weight of the piston assembly causes the displacer to sink in the upper cylinder section also. The fluid trapped below the displacer piston finds a path upward through the displacer and it’s check valve, which is now free to open, and ends up residing above the displacer when the piston assembly has sunk to the bottom of the stroke. The water has been displaced to a higher elevation by gravity. The speed which this occurs can be slowed by flow restriction but the force of the piston assembly supplied by gravity is always available up to a maximum of 17 lbs.

The cylinder is now at the bottom of it’s stroke and it’s 17 lbs of weight will shift the valve and the whole process can repeat.

Respectfully,
Cadman

PS. To any naysayer. I have been designing and building industrial hydraulic machines for a living for over 30 years.
You have an idea, you work it out on the cad, study the motion, apply the forces, and if it looks good you build a prototype. The prototype is to test the machine and discover any unforeseen problems. If the problems can be corrected you continue testing and improving until it’s ready for production.

This one looks good.

This is how it’s done in the real world.

Well, the math behind my idea doesn’t lie.

To anyone who is negative about my idea please post your rebuttal and supply the math to back it up. Prove your contention with the math.

There, the gauntlet is thrown.

Cadman

Hi Cadman.

I have been following your work on another forum with much interest.

I looked at your PDF diagram's,and read your calculations.

I hate to be the bearer of bad new's,but i think you missed a vital bit of math.
You need to take a closer look at what is happening in the chamber under your displacement piston.

I have run some quick calculations,and posted them on the attached diagram of your engine.

Anyway,good luck,and keep up the good work.


Brad

Hello Brad,

Thank you for taking the time to offer your analysis. Please note that my posted calculations were for a slightly different but significant configuration than the one in the pdf.

If I understand your analysis correctly you are referring only to the displacer section of the engine.

I have previously taken account of the force below the displacer and discarded it, for these reasons. That is a static condition when the piston assembly is at rest. As soon as the assembly begins it’s upward movement that upward force on the displacer disappears. I noted this in the pdf, that internal water column can not flow and becomes part of the piston weight. Just like sucking water up a straw and putting your finger over the end then lifting the straw up. Additionally, the in-flowing liquid beneath the displacer becomes a weight to be lifted.

On the down stroke when the piston’s check valve opens there is nothing to sustain pressure against the bottom of the displacer piston. That check valve is functionally a bypass valve during the down stroke. I think Archimedes' Principle applies here. The object (displacer) will be buoyed up by a force equal to the weight of the water displaced. The piston assembly is not going to float on the liquid beneath it, it’s too heavy.

Aside from all of that, I did not allow for piston buoyancy in the force calculations so that is an error. Thank you for making me realize that.

Resectfully,
Cadman

Hi Brad,

I think you need to read his PDF.  The drawing in your post is missing a couple of things that are explained in the PDF.  And the drawing in the PDF is a little more clear for understanding the operation.  The main thing I think you missed is that when the piston gets to the top of it's stroke the valve supplying head pressure to the bottom of the piston is turned off.  And another valve is turned on to allow the water from below the piston to move to the top of the piston.  Cadman can and should of course correct anything if I got it wrong.

Take care,
Carroll

Hi Carroll and Cadman.

If i have read the PDF correctly,the water in the bottom of the upper piston displacement chamber has to rise 40" up through the displacement tube in the upper displacement piston during the down stroke-dose it not ?
The head pressure of a 40" column of water is roughly 1.5psi.
The area of the piston is 26.05 square inches.
26.05 x 1.5 = 39.075 pounds.
This is the downward force require to be placed on that piston in order to get the water to rise 40".

If i have read the PDF wrong,i do apologize,but i see no other way for the water rising that 40" to get to the top of the upper displacement piston,and then into the water reservoir.

See added picture.


Brad