Cadman’s Hydrostatic Displacement Engine

[Cadman]

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

[Grumage]

DreamThinkBuild...... Possibly?

His 3D renditions are always pleasing to look at.   

Cheers Graham.

[Floor]

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

[Cadman]

Dear Cadman
... Prove me wrong...

See CK, what did I tell you? 

Dear Floor,
You're overlooking three most important things.
1, the fluid level in the tank remains constant.
2, the work to raise the pistons is a continuous input from gravity, as is the work to lower them.
3, the fluid is never pumped up past the bottom of the displacer piston during the up stroke.

All events during the up stroke happen at the same time.
All events during the down stroke happen at the same time.

My valve doesn't function well? Just because I'm not happy with the design yet doesn't mean it doesn't function well, and there is plenty of energy to operate it.

Regards,
Cadman

[citfta]

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