Everyone please download the attached pdf before reading any further. Document © 2019 Cadman Weyland.
After you are satisfied that it is genuine please re-post it to all of your forums, if that hasn't already been done, and email it to all of your acquaintances who are interested in free energy. This is a design for a self running engine that does not require fossil fuel, electricity, or any fuel that must be purchased or created. Nor does it use exotic physics, technology, or materials.
I originally intended to release this all at once with the design drawings, specs, parts sources and part numbers as well as a set of .stl files for those with 3D printers (not required). However, I feel as if my hand is being forced so I am releasing what I have now so anyone with basic fabricating skills can build their own.
The operating principles for this engine were elucidated by Archemedies and Blaise Pascal centuries ago and used to be a standard part of the course for first year students of Natural Philosophy.
Please see the following for a little background information.
My uereka moment came here
http://physics.kenyon.edu/EarlyApparatus/Fluids/Hydrostatic_Paradox/Hydrostatic_Paradox.html
Those first two devices pretty much demonstrate it all.
Also see
https://www.youtube.com/watch?v=uIbX4TSguTI&list=PLxA8_oP9TgCbUrQCISTT8IGviVjpO8vyp&index=13
https://www.youtube.com/watch?v=8ma4kW3xVT0
Study the drawings and do the math. Whether you count the liquid being raised as weight or counter hydraulic pressure, the final result is a self running engine with significant power.
Cadman
Dear Cadman.
I'd like to take the opportunity to thank you on behalf of this community for your presentation.
Here in the UK we can add an approximate extra 1.7 Lbs to the weight for an Imperial Gallon. ( 4.54 L/Kg )
Cheers Graham.
I have one question regarding your idea. i understand the sequence of the upstroke(power stroke) but am having a little difficulty with the return downstroke, which relies on gravity. To function correctly both pistons and the connecting rod must have good seals with the cylinder walls and the rod gland. this means some type of o ring or u cup or even metal sealing rings, which in turn will impose considerabe drag or friction on the sliding parts. My question then, does gravity have enough pull to overcome the fiction of the sealing parts and weight of piston assembly and force needed to shift the bottom valve position? From my experience hydraulics and pneumatics require considerable spring force to return a piston to its starting point, on one way acting systems.
thanks!
Quote from: Grumage on June 16, 2019, 08:49:30 AM
Dear Cadman.
I'd like to take the opportunity to thank you on behalf of this community for your presentation.
Here in the UK we can add an approximate extra 1.7 Lbs to the weight for an Imperial Gallon. ( 4.54 L/Kg )
Cheers Graham.
Thank you for your kind words Graham. Of course it doesn't matter which measurement system is used, it's all the same force in the end.
Quote from: vince on June 16, 2019, 04:51:32 PM
I have one question regarding your idea. i understand the sequence of the upstroke(power stroke) but am having a little difficulty with the return downstroke, which relies on gravity. To function correctly both pistons and the connecting rod must have good seals with the cylinder walls and the rod gland. this means some type of o ring or u cup or even metal sealing rings, which in turn will impose considerabe drag or friction on the sliding parts. My question then, does gravity have enough pull to overcome the fiction of the sealing parts and weight of piston assembly and force needed to shift the bottom valve position? From my experience hydraulics and pneumatics require considerable spring force to return a piston to its starting point, on one way acting systems.
thanks!
Hello Vince,
Thanks for your interest. O-rings or metal piston rings are not needed and should not be used for the very reason you say. Besides, metal rings and PVC pipe do not work well together. A 20 lb assembly has plenty force if you are careful with clearances and keep things a little loose. This engine does not have high rod travel speed or high fluid pressures.
I use cup piston seals 3D printed with petg and have also used a flat rubber gasket material cut just slightly larger than the bore. Both work well. The cup should just touch the wall all around at the top edge with little compression. I lube the piston, rod, and seals with a moly grease when assembling. You will be surprised how easy it slides. Speaking of lube, the fluid is 50/50 water anti-freeze and I'm thinking about adding some powdered teflon to the fluid. I think the very best seals would be a leather cup soaked in the fluid prior to assembly. The rod seals I use are plain old buna-N u-cups but I have also used cut rubber sheet.
The snap action slide valve I designed takes about 6 lbs of force to shift and I'm not satisfied with that design yet.
The hydraulics and pneumatics you are used to have small diameter plumbing relative to the piston bore don't they. You need large diameter plumbing with this engine to minimize flow restriction.
Regards
Cadman
Edit: piston seals are cup seals, not u-cup.
Hi Cadman,
Thanks for sharing your idea. Would have an 'unusual' question: what piston speed could be roughly expected if one maintains the sizes mentioned in the pdf file example (say 6" PVC cylinder)?
I know you wrote this machine does not have high rod travel speed and I also think the speed depends on the lubrication and the fluid too. So just a rough speed estimation would do.
Thanks
Gyula
Hi Gyula,
That's impossible to say at this point, I haven't finished the 6". I will be very happy if the piston speed of this 6" bore & stroke engine could average 6" per second. More likely it will be quite a bit less since I am currently limited to 1.5" ID plumbing. I can say this much, it all depends on the flow velocity you can achieve. Large diameter plumbing, fluid rotation while filling and discharging the cylinder, large radius bends, and high flow valves will be just as important as in a high performance ICE. Maybe more important.
If it matters, what I have built up until now is a mock up using 4" PVC pipe, garden hoses, hand operated valves (vise grips and wood blocks) and weights. Just enough to verify the concept is valid. A 4" is too small for any kind of power. The 6" under construction is for experiments, testing and development.
What I want eventually is a 14" bore or 12" square, 3 to 6" stroke, 4" ID plumbing, 60 cycles per minute, 2 cylinder engine with a crankshaft and flywheel. 8)
Regards,
Cadman
I know this is possible. Haven't studied your PDF yet.
Congratulations, I hope.
floor
A quick and dirty evaluation.
All the sixes, in Imperial measurements.
Six Feet head of water. 2.6 PSI
Six inch diameter piston. 28.27 square inches multiplied by 2.6 equals 73.5 Pounds pressure.
Six inch stroke. Volume of displaced fluid ( water ) 169.17 Cubic Inches. Mass 2.78 Kg or 6.116 Pounds.
Taking Cadman's value of 20 Pounds for the power piston and displacer piston assembly we then add the 6.116 Pounds ( working fluid to be returned ) giving 26.116 Pounds total.
By subtracting the 26.116 from the input pressure of 73.5 pounds we have a net gain of 47.4 Pounds
( 21.5 Kg ) losses excluded.
That's " my " take on Cadman's presentation. Mathematics isn't my best area of expertise so please feel free to contradict.
It's the clever use of the " displacer " piston to transfer the working fluid back to the header tank, my hat is off to you Cadman.
I can think of several means of reducing the losses like simple sliding gates for the control valves and having several vertical tubes through the displacer that can be covered by a thin Rubber flap at the top.
Thanks again Cadman.
Cheers Graham.
https://web.archive.org/web/20070225160446/http://www.theverylastpageoftheinternet.com/ (https://web.archive.org/web/20070225160446/http://www.theverylastpageoftheinternet.com/)
"MAIN MENUE": E. L. S. A. gravity mill
Quote from: Grumage on June 20, 2019, 07:13:18 AM
A quick and dirty evaluation.
All the sixes, in Imperial measurements.
Six Feet head of water. 2.6 PSI
Six inch diameter piston. 28.27 square inches multiplied by 2.6 equals 73.5 Pounds pressure.
Six inch stroke. Volume of displaced fluid ( water ) 169.17 Cubic Inches. Mass 2.78 Kg or 6.116 Pounds.
Taking Cadman's value of 20 Pounds for the power piston and displacer piston assembly we then add the 6.116 Pounds ( working fluid to be returned ) giving 26.116 Pounds total.
By subtracting the 26.116 from the input pressure of 73.5 pounds we have a net gain of 47.4 Pounds
( 21.5 Kg ) losses excluded.
That's " my " take on Cadman's presentation. Mathematics isn't my best area of expertise so please feel free to contradict.
It's the clever use of the " displacer " piston to transfer the working fluid back to the header tank, my hat is off to you Cadman.
I can think of several means of reducing the losses like simple sliding gates for the control valves and having several vertical tubes through the displacer that can be covered by a thin Rubber flap at the top.
Thanks again Cadman.
Cheers Graham.
Thanks Graham, and hats off to Archemedies and Pascal too.
In my calculations, I doubled the displaced weight at BDC, half for the displaced liquid above the pressure piston and half for the liquid above the displacer piston. You're spot on about several vertical tubes. I am using four 0.75" tubes that almost exactly equals the area of my 1.5" plumbing, with a flat disk valve at the top.
Quote from: lancaIV on June 20, 2019, 07:37:01 AM
https://web.archive.org/web/20070225160446/http://www.theverylastpageoftheinternet.com/ (https://web.archive.org/web/20070225160446/http://www.theverylastpageoftheinternet.com/)
"MAIN MENUE": E. L. S. A. gravity mill
That isn't the same thing but I wonder how often this has been thought of before now. I suspect this engine fits into the 'forgotten technology' category.
Cadman
@ Cadman
The reason I have said that I think such a device may work
can be found here. @
https://overunity.com/18234/new-version-of-flotation-device/msg535333/#msg535333
I KNOW that device works (not just think it might or believe it will).
I have now spent some time looking at / trying to
understand your device. I have a request. Will you
/ some else, (volunteer) do some diagrams / a flow chart of your device.
Please explain each step in the actions, with drawings
and explanations, (on the same page as the drawings).
1.
a. what happens / where.
b. what consequences
c. for what purpose
2.
a. what happens / where.
b. what consequences
c. for what purpose
and so on.
floor
Floor,
That was the purpose of the pdf. I really don't know how much simpler I could explain it. Did you see the page I linked to at Kenyon?
Assuming you know how a hydraulic cylinder works? Look at it this way, the gland is a fixed divider that separates the upper and lower parts. The lower part is a hydraulic cylinder. The upper part works like an old fashioned well pump.
Cadman
Hi Cadman,
I've been studying your design but I can't get a picture in my head how it's supposed to work.
An animation would help if that's possible.
Looking forward to see this thing in action.
artv
Quote from: Cadman on June 21, 2019, 10:32:05 AM
Floor,
That was the purpose of the pdf. I really don't know how much simpler I could explain it. Did you see the page I linked to at Kenyon?
Assuming you know how a hydraulic cylinder works? Look at it this way, the gland is a fixed divider that separates the upper and lower parts. The lower part is a hydraulic cylinder. The upper part works like an old fashioned well pump.
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
Please explain each step in the actions, with drawings
and explanations, (on the same page as the drawings).
1.
a. what happens / where.
b. what consequences
c. for what purpose
2.
a. what happens / where.
b. what consequences
c. for what purpose
and so on.
fllor
Quote from: Floor on June 21, 2019, 05:01:36 PM
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
DreamThinkBuild...... Possibly?
His 3D renditions are always pleasing to look at. :)
Cheers Graham.
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
Quote from: Floor on June 22, 2019, 12:22:45 PM
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
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
Quote from: citfta on June 23, 2019, 07:34:09 AM
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
Quote from: Cadman on June 23, 2019, 08:26:24 AM
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
Sorry, my mistake. I too respect his work, although I have trouble understanding it sometimes. :)
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
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.
Nice. One could consider magnetic coupling to avoid the internal shaft gland seal between the pistons. Magnets embedded in the piston edges, magnets external to the shell, with guide rods and linear bearings. Probably many ways to achieve this.
Quote from: Cadman on June 23, 2019, 09:26:27 AM
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
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
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
Thank you Cadman, etal.
respectfully
floor
Quote from: citfta on June 24, 2019, 10:26:04 AM
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
Hi Brad,
You may have a point. This is going to require an experiment to sort out.
If you took a 6" ID tube 5 ft tall closed at the bottom, filled it with 7" of water, and dropped a 5.95" x 40" 17 lb slug into it, I just can't believe that the slug would float and not sink to the bottom of the bucket even if the slug had a 1.5" hole in it.
Ahh, I think I may have it. We are not taking atmospheric pressure into account.?
Well, we'll see what's what. The proof is in the pudding, eh?
Regards,
Cadman
Quote from: Cadman on June 24, 2019, 11:34:04 PM
Hi Brad,
You may have a point. This is going to require an experiment to sort out.
If you took a 6" ID tube 5 ft tall closed at the bottom, filled it with 7" of water, and dropped a 5.95" x 40" 17 lb slug into it, I just can't believe that the slug would float and not sink to the bottom of the bucket even if the slug had a 1.5" hole in it.
Ahh, I think I may have it. We are not taking atmospheric pressure into account.?
Well, we'll see what's what. The proof is in the pudding, eh?
Regards,
Cadman
Your pipe inside a pipe is a good test.
I think you will be surprised at the result.
Like you say,absolute answers are found by testing the actual device.
Brad
For your entertainment. :)
https://www.youtube.com/watch?v=aTOJpDBS1Bw
Thank you Graham!!
If the volume of the displacement tube weights more than the equal volume in water (it sinks in a tub full of water) then it works ...... Only it don't work with that much weight because the displacement tube could just be water instead!
Quote from: citfta on June 25, 2019, 04:06:22 PM
Thank you Graham!!
Graham made an error in his calculations-but easy to make.
Brad
Quote from: tinman on June 25, 2019, 09:30:59 PM
Graham made an error in his calculations-but easy to make.
Brad
Thanks Brad, indeed there was.
The bore diameter is 3/4" in the video equating to 8:1 ratio. So the piston length should have been 6".
Cheers Graham.
Quote from: Grumage on June 25, 2019, 03:17:01 PM
For your entertainment. :)
https://www.youtube.com/watch?v=aTOJpDBS1Bw
Thank you very much Graham!
As happy as I am to see that, I believe there is an error, but one that could easily be correted.
This is how I figure it: and please someone correct me if this is wrong
area of your 3/4" piston with a 1/8" hole is 0.4295 sq in
area of my 6" piston with a 3/4" rod is 26.4126 sq in
that gives a ratio of 61.4962 to 1
the scaled piston assembly weight would be 0.2764 lbs or 125.39 grams. (17 / 61.4962)
I get 5" piston length at 1:8 scale. (40 /8)
And BTW, you don't need to grease the rod, that just adds resistance. It will either displace the water or not.
If my figures are correct, would it be too much to ask for you to drill partial holes around your center hole in your rod until it weighs 125 grams, then rerun your test?
It would be very appreciated and save me a lot of work if you can.
Regards,
Cadman
PS steel weighs 0.283 lbs per cu in. or 128.37 g
The answer is very simple. The displacer is exactly what it must do (in this case ) displace water.
In the vertical position it will sink (displace water) only until the water it displaces equals the weight of the displacer.
Not hard to understand.
To displace water for the entire length of the displacer would require it to weight as much as it's exact volume in water.
If it does not, it will sink only to a point where the water it displaces is equal in weight to the displacer.
I think this will be a problem unless it can be done in an additional step just to raise the displacer.
Quote from: Cadman on June 26, 2019, 10:21:37 AM
Thank you very much Graham!
As happy as I am to see that, I believe there is an error, but one that could easily be correted.
This is how I figure it: and please someone correct me if this is wrong
area of your 3/4" piston with a 1/8" hole is 0.4295 sq in
area of my 6" piston with a 3/4" rod is 26.4126 sq in
that gives a ratio of 61.4962 to 1
the scaled piston assembly weight would be 0.2764 lbs or 125.39 grams. (17 / 61.4962)
I get 5" piston length at 1:8 scale. (40 /8)
And BTW, you don't need to grease the rod, that just adds resistance. It will either displace the water or not.
If my figures are correct, would it be too much to ask for you to drill partial holes around your center hole in your rod until it weighs 125 grams, then rerun your test?
It would be very appreciated and save me a lot of work if you can.
Regards,
Cadman
PS steel weighs 0.283 lbs per cu in. or 128.37 g
You're most welcome. :)
Brad and I have been in conversation about the issue already.
So the piston needs to be 5" long and approximately 125 g in weight?
Cheers Graham.
5" for 1:8 scale. The 125 g is according to my figures trying to scale the piston weight so the psi under the piston would be scaled.
I think my figures are right but would appreciate a confirmation. My head is quite foggy today.
Respectfully
Cadman
Post script.
Aren't we forgetting about the fact that the displacer piston is attached to the power piston assembly?
Wasn't the figure of 20 Lbs mentioned?
Cheers Graham.
In the analysis I posted, 17 lbs included the whole assembly.
If you want to use 20 lbs, that's OK at this point.
Maybe that's what it needs to be. It's a work in progress.
Cheers
Cadman
PS. We want to use the lightest piston assembly that will do the job. The lighter it is, the more power we'll have on the up stroke.
Just an Idea, as an aside, not to distract from the good progress.
In the drawing the fluid is flowing down from the reservoir to the power piston
it would then lift the displacement piston represented as the weight. Don't know how
it would then reverse.
Quote from: lumen on June 26, 2019, 12:21:33 PM
The answer is very simple. The displacer is exactly what it must do (in this case ) displace water.
In the vertical position it will sink (displace water) only until the water it displaces equals the weight of the displacer.
Not hard to understand.
To displace water for the entire length of the displacer would require it to weight as much as it's exact volume in water.
If it does not, it will sink only to a point where the water it displaces is equal in weight to the displacer.
I think this will be a problem unless it can be done in an additional step just to raise the displacer.
That is not correct.
Here we are dealing with hydraulic pressures and forces,not buoyancy forces.
Brad
Hi Brad,
No, I think all the hydraulic math is good and works but the displacer weight (buoyancy problem) was never correctly accounted for.
Like a 3000 ton ship will displace 3000 tons of water to stay afloat. The problem is worked in reverse to find the required weight for the displacer to sink.
What is the weight of the water your displacing......so the displacer must weight that much or more to sink, or it will float.
You are both right. It finally came to me, how to say what makes the difference for the displacer, why the water will move upwards.
Pressure differential.
Both the piston and water column exert force and create the pressure. Less pressure above the piston means the lighter body, the water column through the displacer, will be forced up by the pressure below the displacer piston. This not the same situation as the water feeding the pressure piston.
Regards
Cadman
I'd like all readers to think about this word, I don't think its been mentioned as yet.
" DENSITY "
;)
Cheers Graham.
More entertainment.
:)
https://www.youtube.com/watch?v=gx0U0Wu9quQ
Quote from: Grumage on June 27, 2019, 03:47:38 PM
More entertainment.
:)
https://www.youtube.com/watch?v=gx0U0Wu9quQ
That is great! I guess that puts that issue to rest. :)
Thank you sir!
Good morning All.
Wait, there's more....
My PC keyboard has developed a form of Dyslexia, adding extra letters to typed text so I was unable add to yesterday's post.
Just before my son and I filmed the demonstration we needed to add an extra bit of length to the plastic ball point pen ink tube. My son decided to try the assembly with the whole length of additional tubing. A whole " Foot ( 12" ) " was allowed to sink down the cylinder. For a brief moment we saw the piston sink and stop at a point where the pressure had equalised. The Buoyancy point!
Cheers Graham.
Quote from: lumen on June 27, 2019, 12:07:17 AM
Hi Brad,
No, I think all the hydraulic math is good and works but the displacer weight (buoyancy problem) was never correctly accounted for.
Like a 3000 ton ship will displace 3000 tons of water to stay afloat. The problem is worked in reverse to find the required weight for the displacer to sink.
What is the weight of the water your displacing......so the displacer must weight that much or more to sink, or it will float.
I agree mostly.
Hydrostatic Force is the force due to the pressure of fluid at rest. Whereas
buoyant force is the upward force exerted by any fluid upon a body placed in it.
Hydraulics also incur frictional forces,where as the buoyant force dose not-other that water surface tension.
But as you say,if we use the buoyant force,we can make accurate calculations.
Cadmans piston weighs 5lb.
It is 6 inches in diameter,and 40 inches high--so lets do the math ;)
Cadmans piston has a volume of 1130.97 cubic inches minus the displacement tube,which has a volume of 70.68 cubic inches. The total volume is 1060.29 cubic inches.
There is 60 cubic inches in a liter,so Cadmans displacer piston will displace 17.67 liters of water.
This is very close to 17.67 kg's or 38.95 pounds.
From that we subtract the 5 pounds for the pistons weight,and we are left with 33.95 pounds.
He had 17 pounds in total,and from that we subtract our 5 pounds we have already accounted for. This leaves us with 12 pounds of gravitational force.
Once again,we seem to be short 21.95 pounds.
Now,if you have a look at my calculations i posted before in this thread ,in the pic below,using hydraulic force calculation's,you will note that i came to a short fall in weight of 22 pounds.
Hows that for close enough ;)
Brad
Here is a reply I made to Brad at OUR.
Sorry Brad, but you are overlooking a very simple thing that Cadman pointed out at OU.com. The pressure on the bottom of the piston is going to push the water up the tube just like Graham's video shows. And that is going to allow the piston to sink because the density of the piston is greater than the density of the water as Graham has hinted at in the thread at OU.com. I guess only a full scale test will prove this but common sense says that a solid object with a density greater than the density of water will sink if the water can somehow get past the object. And if I recall correctly Cadman has said he intends to put more than one tube up through the displacement piston. So that would also change the calculations in favor of the piston sinking.
Respectfully,
Carroll
Brad
Only have a minute for a quick post.
Actually the engine is only displacing a column of water 6" high, which weighs 6.26 lbs. Even if you add in the 1.5" x 40" column in the piston it would only be 9.2 lbs.
Cheers
Cadman
Quote from: Grumage on June 28, 2019, 06:21:33 AM
Wait, there's more....
Decided to try the assembly with the whole length of additional tubing. A whole " Foot ( 12" ) " was allowed to sink down the cylinder. For a brief moment we saw the piston sink and stop at a point where the pressure had equalised. The Buoyancy point!
Cheers Graham.
Thanks Graham,
Now you can see the problem showing up.
It's not just a sinking weight, it's displacing water and causing pressure to build and stop it from displacing additional water.
tinman,
Thanks for doing the math. I think we both knew that the result was the same calculating pressure or buoyancy.
Quote from: lumen on June 28, 2019, 10:39:20 PM
Thanks Graham,
Now you can see the problem showing up.
It's not just a sinking weight, it's displacing water and causing pressure to build and stop it from displacing additional water.
tinman,
Thanks for doing the math. I think we both knew that the result was the same calculating pressure or buoyancy.
Hi Lumen.
Yes indeed, we can see the problem. However that problem occurred at twice the value of the working parameters shown by Cadman.
As I see it the major problem will be the size chosen for a prototype. At small scale frictional losses will become its " Achilles heel " particularly with piston seals.
In this scenario
" BIG " will definitely be
" BETTER. "Cheers Graham.
Been studying the design.
Here are some observations / opinions.
One more thing cadman,
congratulations
floor
One more, one more thing.
The check valve at the bottom of the displacement piston
needs to be simply closed / not functioning as a check valve during the
lifting of the displacement piston.
The device works because :
1. With that displacement piston valve closed, during the piston lifting
gravity is assisting that lifting.
2. Gravity does all of the work of lowering the displacement piston.
3. The combination of 1. and 2. is what enables the higher lifting of the fluid,
when the piston falls.
regards
Another observation.
It is best that the valve at the bottom of the displacement piston, is closed upon
its arrival the bottom of its decent. Other wise fluid will fall in the displacement
piston tube, while rising (due to hydro static pressure) in the supply tank .
An un needed waste.
@Cadman
Here are some proofs of your principle.
Please find the attached PDF file "Cadman from floor.PDF" below.
Floor,
Thank you for sharing your proofs. That took quite a bit of effort to produce.
Sincerely
Cadman
Cadman
Your very welcome. Thank you for your design !
I buggered up a couple of the drawings in that "Cadman from floor 3" PDF
sorry ! too much rushing around lately.
Here is the fixed version ""Cadman from floor 3b" PDF
regards
floor
Nicely done Floor.
This illustrates the difference between the two conditions existing in the up-stroke vs. the down-stroke.
Something that I have failed to convey properly in all of my posts.
Now all that remains is to demonstrate it with the working prototype. Making progress there too.
Thank you!
Respectfully,
Cadman
F.Y.I.
Cadman,
These might be similar to your concept development and may possibly contain some valuable design and construction insight.
(Hydraulic) RAM PUMPS [a.k.a. Hydram]
Design and Construction of a Hydraulic Ram Pump
(including a bit of history and analysis)
Shuaibu Ndache MOHAMMED
Department of Mechanical Engineering,
Federal University of Technology, Minna, Nigeria
http://lejpt.academicdirect.org/A11/get_htm.php?htm=059_070 (http://lejpt.academicdirect.org/A11/get_htm.php?htm=059_070)
AbstractThe Design and Fabrication of a Hydraulic Ram Pump (Hydram) is undertaken. It is meant to lift water from
a depth of 2m below the surface with no other external energy source required. Based on the design the
volume flow rate in the derived pipe was 4.5238 × 10-5 m3/s (2.7 l/min), Power was 1.273 kW which results
in an efficiency of 57.3%. The overall cost of fabrication of this hydram shows that the pump is relatively
cheaper than the existing pumps.
Keywords Hydram; Pump; Volume Flow Rate; Power; Efficiency; Impulse Valve; Delivery Valve.
Home Built Hydraulic Ram Pumps
http://www.inthefieldministries.org/jscalhou/Home%20Built%20Hydraulic%20Ram%20Pumps.pdf (http://www.inthefieldministries.org/jscalhou/Home%20Built%20Hydraulic%20Ram%20Pumps.pdf)
"Complete detailed and illustrated instructions from locally available pumping parts - 1" Hydraulic RAM"
How to build a RAM PUMPFrench River Springs
Published on Jul 7, 2018
https://www.youtube.com/watch?v=enBEMgDR3-A (https://www.youtube.com/watch?v=enBEMgDR3-A)
How to Make a Large Size Free Energy Water Pump - Ram Pump
Tradisional Channel
Published on Dec 28, 2018
https://www.youtube.com/watch?v=Esln877z4IE (https://www.youtube.com/watch?v=Esln877z4IE)
Hydraulic RAM Pumps - Amazon (examples, descriptions, prices)
Small - $125US - - - Large - $185US
https://www.amazon.com/Hydraulic-Ram-Pump-1-Medium/dp/B072J2Q7L6 (https://www.amazon.com/Hydraulic-Ram-Pump-1-Medium/dp/B072J2Q7L6)
Great work and very interesting, thanks!
FIN
Thanks
The pleasure is mine. I've gotten pretty fast at kicking out drawings and generally
enjoy doing them. Thanks again for your inspired design.
best wishes
Dam Wrong / bad drawing (above).
Here it is (below).
Hello All.
Since Cadman's original presentation I've been thinking a lot about it. May I offer " Concept 2 " for consideration?
Cycle description and drawing are shown from the " at rest " position.
Let's assume a working head of 2 meters. To the top of the cistern. The concertina is made to have an internal surface area of
6 square inches ( Cadman's piston ) this should provide a force of approximately 75 Lbs. The displacement cylinder is made to be directly connected to the top of the concertina but of a diameter that can just accommodate the volume of water from it. This should be around 6.1 Lbs in weight but because of the small diameter only create a couple of pounds pressure on the transfer valve. The displacement cylinder is allowed to move up and down through a central seal within the cistern. My single drawing shows the device at the start of the cycle. As water is admitted to the underside of the concertina the whole assembly will rise 150 mm
through the cistern. The displacer piston is rigidly fixed to a framework that carries the whole assembly. The top of the displacer piston has a simple " Leather cup washer " that will allow the water to pass it on the upstroke but then be retained on its return. On the return stroke the fluid now above the seal will naturally spill over the edge of the displacement cylinder back into the cistern.
My idea has just 2 areas of friction, one the sliding seal between displacement cylinder wall and cistern floor and the other is the flap valve ( cup washer ) mounted to the displacement piston top.
Cheers Graham.
Edit.
Drawing to follow pending resize. It seems I cannot resize the picture using my iPad so I'll post it anyway and re edit the page later today. My apologies for the inconvenience.
To clarify.
I have not shown the admission valve which will open and close at each end of the stroke.
The " bias weight " will obviously affect the " net " energy available but is needed to reset the system for the next upstroke. Its weight will have to be determined by experiment.
Once the water has passed by the upper cup washer on the displacement piston its mass is now supported by the framework and no longer seen by events below.
The displacement piston is made from a " high density " material, preferably Stainless Steel for longevity.
Cheers Graham.
I did an experiment similar to this years ago. I later had the idea that oil rises to the top in water, and it worked. I did not complete the experiment - however if you incorporate oil rising you may get a buoyancy advantage. ie oil inside a ping pong ball should rise and can push a piston against gravity. Then you have the problem of the return of course.
OR...
Maybe instead of the circular bias weight, you could just use a light weight displacer and a leaver from the large cylinder to push it down and displace water, then as more water is let in, you can use the displacer as a float to push the cylinder back down as it floats up.
@SolarLab
Quote from: SolarLab on July 02, 2019, 01:48:39 PM
F.Y.I.
Cadman,
These might be similar to your concept development and may possibly contain some valuable design and construction insight.
(Hydraulic) RAM PUMPS [a.k.a. Hydram]
and so on.....
Its not a ram pump. Ram pumps are very cool, but also ram pumps have been around
for a long time.
Cadmann's design is very much dissimilar, because...
ram pumps require an uphill sourceof water (usually a stream). They allow one to divert
and lift A SMALL PERCENTAGE of that down flowing water to a point which is higher than
the original water source, but this is only BECAUSE MOST OF THE WATER FROM SOURCE
POINT CONTINUES TO FLOW DOWN HILL (BELOW THE RAM PUMP).
Cadman's design functions based upon an entirely different approach / principle.
floor
Graham,
Are the concertina and displacement piston both 6" diameter and the displaced volume equal to the concertina volume?
Cheers
Cadman
Hi Cadman.
The displacement cylinder is directly connected to the top of the concertina with the addition of a non return valve at its base. Its diameter is much smaller perhaps 3/4" ( 20mm ) at most. As your device is based around Pascal's law where the pressure increases by virtue of surface area obviously a small diameter tube will see around 2 PSI at most. This means that the force available from the large surface area of the concertina should have little problem moving the volume of water ( 6.1 Lbs ) from the concertina ( power cylinder ) into the displacement cylinder.
I chose the concertina over a piston and cylinder for several reasons. The major one being to reduce friction between piston seal and cylinder wall.
An effective concertina could be crafted from a Rubber inner tube, Quad bike/Garden tractor size. If sliced horizontally on its inner diameter the opening could be clamped down to the base, bottom/floor of the device. The upper edge might prove a little more difficult to achieve requiring two discs that clamp the Rubber together and also carry the 2 metre long vertical displacer cylinder.
Concept 2 is what it is, purely conceptual. I've thrown it out here for evaluation, nothing more.
Cheers Graham.
Addendum.
I forgot to mention that the fixed displacement piston is of high density ( Steel ) that has a diameter of 1/2" ( 12mm ) and is solid. The water flows around the 1/8" 3mm gap via displacement. A simple Leather cup washer ( bicycle pump style ) is fitted at the top so that on the upstroke the water can pass by it easily but is then held in suspension and just flows over the edge of the cylinder wall on the return stroke. Refilling the cistern.
Cheers Graham.
Basics of hydraulics.
Quote from: Floor on July 06, 2019, 10:07:59 AM
Basics of hydraulics.
Indeed.
But our ideas employ displacement, what are you trying to explain please?
Cheers Graham.
Graham,
(Floor posted while I was writing this. I'll let him explain but he's pointing out basic pressure multiplication or division with one piston having unequal surface area on each of it's faces. Also difference in motions between unequal piston diameters.)
----------------
My analysis.
Double check my figures, but I think they are correct.
Assume a working head of 2 meters
pressure at bottom = 2.844 psi dropping to 2.627 psi after up-stroke
The concertina is made to have an internal surface area of 6 square inches
2.764" diameter = 6 sq. in.
The fixed displacement piston ... has a diameter of 1/2"
1/2" dia = 0.196 sq. in.
The displacement cylinder ... diameter is much smaller perhaps 3/4"
3/4" dia = 0.441 sq. in
Weight of liquid above concertina in the displacement cylinder minus the displacement piston volume
1.923 lb.
Up-force from concertina with transfer valve closed
15.765 lbs at the top of the stroke
Up-force from concertina with transfer valve open
0 lbs
My conclusion:
The inlet valve in the base is not needed.
The transfer valve is the one to be opened or closed at the end of the strokes.
If the transfer valve is the one shifted, then this will work, provided the total weight of the concertina piston plus the displacement cylinder plus the weight of liquid above the concertina is less than the up-force from the concertina (15.765 lbs).
With the transfer valve opened the liquid will seek the level of the head through the displacement cylinder with no upward force exerted on the piston. The concertina / displacement cylinder assembly is free to sink to the bottom of the stroke.
With the transfer valve closed the liquid is separated into two volumes and the head pressure will exert influence on the concertina piston bottom surface and provide 15+ lbs of lift. The liquid above the concertina piston becomes a static weight. If the static weight (liquid plus mechanical weight) is less than the lift force the concertina piston + displacement cylinder will rise.
The issue to contend with is to have a head reservoir of sufficient volume to in order to minimize the loss of head height while filling the concertina since it's volume is much greater than the displaced volume.
Cheers
Cadman
Quote from: Grumage on July 06, 2019, 10:44:53 AM
Indeed.
But our ideas employ displacement, what are you trying to explain please?
Cheers Graham.
Its great to see the diversity of ideas and designs flowing.
There is more than one way to skin a cat.
I myself, have no intentions of presenting a design or design variations of Cadman's device.
My intention has been to present explanations of the principles involved. This, in order
to clarify the underlying physics / principles (useful to we, the less knowledgeable in this area).
The various devices / designs / ideas being presented involve both displacement and hydraulics.
Example: your bellows are a hydraulic pressure (PSI) device. Gravity acting upon the supply
side tube, acts similarly to the smaller diameter syringe in my drawings (above). While the bellows
are acting similarly to the larger diameter syringe in my drawings (above).
best wishes
floor
Quote from: Cadman on July 06, 2019, 11:09:11 AM
Graham,
The issue to contend with is to have a head reservoir of sufficient volume to in order to minimize the loss of head height while filling the concertina since it's volume is much greater than the displaced volume.
Cheers
Cadman
Just cheat it for now (use a float valve to maintain head height) instead of...
https://overunity.com/18243/cadmans-hydrostatic-displacement-engine/msg535881/#msg535881
floor
Now that I've had time to really look at it, I think Graham's design is more elegant than my own.
Way to go Graham!
Regards
cadman
Quote from: Cadman on July 06, 2019, 11:09:11 AM
Graham,
(Floor posted while I was writing this. I'll let him explain but he's pointing out basic pressure multiplication or division with one piston having unequal surface area on each of it's faces. Also difference in motions between unequal piston diameters.)
----------------
My analysis.
Double check my figures, but I think they are correct.
Assume a working head of 2 meters
pressure at bottom = 2.844 psi dropping to 2.627 psi after up-stroke
The concertina is made to have an internal surface area of 6 square inches
2.764" diameter = 6 sq. in.
The fixed displacement piston ... has a diameter of 1/2"
1/2" dia = 0.196 sq. in.
The displacement cylinder ... diameter is much smaller perhaps 3/4"
3/4" dia = 0.441 sq. in
Weight of liquid above concertina in the displacement cylinder minus the displacement piston volume
1.923 lb.
Up-force from concertina with transfer valve closed
15.765 lbs at the top of the stroke
Up-force from concertina with transfer valve open
0 lbs
My conclusion:
The inlet valve in the base is not needed.
The transfer valve is the one to be opened or closed at the end of the strokes.
If the transfer valve is the one shifted, then this will work, provided the total weight of the concertina piston plus the displacement cylinder plus the weight of liquid above the concertina is less than the up-force from the concertina (15.765 lbs).
With the transfer valve opened the liquid will seek the level of the head through the displacement cylinder with no upward force exerted on the piston. The concertina / displacement cylinder assembly is free to sink to the bottom of the stroke.
With the transfer valve closed the liquid is separated into two volumes and the head pressure will exert influence on the concertina piston bottom surface and provide 15+ lbs of lift. The liquid above the concertina piston becomes a static weight. If the static weight (liquid plus mechanical weight) is less than the lift force the concertina piston + displacement cylinder will rise.
The issue to contend with is to have a head reservoir of sufficient volume to in order to minimize the loss of head height while filling the concertina since it's volume is much greater than the displaced volume.
Cheers
Cadman
Hi Guys.
It's a pity we can't converse by other means, I hate typing!
I don't think we can dispense with the bottom admission valve as we want the concertina and displacement cylinder to collapse back to rest for the next cycle.
However I have realised there might be a flaw in my idea. Upon opening the transfer valve ( top of stroke ) the bias weight along with the weight of the displacement cylinder should transfer the water into the smaller diameter cylinder above ( the collapse ) We close the transfer valve and open the admission valve. The concertina rises along with the displacement cylinder the water is now having to pass the sides of the displacer piston and through the cup washer. We open the transfer valve again, is this where things get stuck? Do we need to allow air in to get the concertina assembly to collapse for the reset? Hmmm....
Cheers Graham.
Quote from: Grumage on July 06, 2019, 12:52:38 PM
Hi Guys.
It's a pity we can't converse by other means, I hate typing!
I don't think we can dispense with the bottom admission valve as we want the concertina and displacement cylinder to collapse back to rest for the next cycle. However I have realised there might be a flaw in my idea. Upon opening the transfer valve ( top of stroke ) the bias weight along with the weight of the displacement cylinder should transfer the water into the smaller diameter cylinder above ( the collapse ) We close the transfer valve and open the admission valve. The concertina rises along with the displacement cylinder the water is now having to pass the sides of the displacer piston and through the cup washer. We open the transfer valve again, is this where things get stuck? Do we need to allow air in to get the concertina assembly to collapse for the reset? Hmmm....
Cheers Graham.
Yeah, I wish we could converse too. It would make things a lot easier.
Think about it without the admission valve.
Picture it like Floor did with the U tube. With the transfer valve open there is an unblocked path on the left from the top of the reservoir to the bottom of the concertina, through the concertina, and back up to the reservoir. The concertina is just like a dense restrictor orifice at this time sitting at the bottom of the right half of the U tube.
Now close the transfer valve and the fluid above the concertina is separated from the fluid in the left half of the U tube. Being lighter than the left fluid column the concertina, which now has a solid piston top, along with the displacement cylinder plus the smaller amount of fluid above it will rise as the two halves in the U tube seek a balance.
Open the transfer valve now that the assembly is elevated and the concertina reverts to a restrictor and being more dense and heavier than the liquid it will sink to the bottom. Your own video proved that.
Cheers
Cadman
I like Grumage's bellows usage. Maybe bellows type boots (used to cover
front axles on trucks / autos) could be adapted to this purpose.
@Floor since you grasp Grumage latest concept could you draw a cyclic diagram of that concept? I think many would struggle with all the textual descriptions here. I can even do an 3d animation if needed but need to understand it better first.
Quote from: broli on July 08, 2019, 03:32:01 AM
@Floor since you grasp Grumage latest concept could you draw a cyclic diagram of that concept? I think many would struggle with all the textual descriptions here. I can even do an 3d animation if needed but need to understand it better first.
Dear broli.
I'll try to explain concept 2 verbally again, my drawing abilities are very limited in this modern age. When I was taught it was with paper and pens. :) Your offer of 3D sounds wonderful, bravo.
My single drawing shows the device at rest with water from the upper cistern piped down to the underside of the bellows/concertina, admission valve closed.
Upon opening of the admission valve the water fills the bellows causing it to rise ( transfer valve is shut ) we have allowed it to rise by 150mm or 6" lifting the displacement cylinder vertically through a sliding seal in the centre of the cistern 2 meters above.
We close the admission valve and open the transfer valve. The combined weight of the " bias " and cylinder now causes the water to flow into the displacement cylinder and travel upward round the sides of the solid Steel displacer piston. The bellows have now returned to rest as in my picture.
The next upward stroke see's the bellows fill and rise again, as we have 75Lbs of force available acting against just the volume of water plus that of the bias and cylinder. The displacer which is rigidly fixed from above the device, displaces the water through a cup washer at the top.
We now open the transfer valve and the bellows and cylinder collapse once again. As the cylinder collapses any water above the displacer piston falls over the sides to refill the cistern.
Back at rest again and ready for the next cycle.
For an automatic reciprocating engine we shall have to design some mechanism to operate the two valves.
But....
This all seems far too easy, what has been missed, am I dreaming?
Cheers Graham.
Quote from: Grumage on July 06, 2019, 07:28:15 AM
Addendum.
I forgot to mention that the fixed displacement piston is of high density ( Steel ) that has a diameter of 1/2" ( 12mm ) and is solid. The water flows around the 1/8" 3mm gap via displacement. A simple Leather cup washer ( bicycle pump style ) is fitted at the top so that on the upstroke the water can pass by it easily but is then held in suspension and just flows over the edge of the cylinder wall on the return stroke. Refilling the cistern.
Cheers Graham.
Thanks Grumage, wasn't totally sure.
@all readers
One more proof / drawing, below. The last of that part of my endeavor
see the "floor Cadman 3F-10n.PDF"
@Broil
Yes I will do a flow diagram. But don't jump on the 3d until
I have made certain with Grumage that my interpretations are correct.
thanks
floor
Hello All.
I felt that perhaps my description of operation was lacking one detail, I'd forgotten to mention that the system had been previously purged of any air.
Sorry for any inconvenience.
Cheers Graham.
Although the proofs I presented come at Cadman's design in a sort of backwards and upside down
manner, they definitely verify his design as being able to use gravity as an energy source.
Here is my summation.
floor
You want O.U.
Cadman's design is a go.
Study and understand it, and you will see this for your self.
floor
Two of Cadman's devices could resupply each others supply / origin reservoirs, if their operation are 180 out as synchronization.
floor
Correct me if I am misstaken
floor
a few more
Hi all, very interesting concepts in this thread. Did someone try to build a real model?
Regards
Hi andrea
Cadman and Grummage were doing builds or at least partial builds, at that time.
I don't know what Cadman and Grummage came to as conclusions.
I concluded that the method (by drawings only) came to exactly unity only
(before considerations of losses).
I assume the others came to the same conclusion ?
Notes..
Its hard to say which is more offensive and which is more difficult.
Playing the critique, or the optimist.
Either way, the O.U. search / game leads to lots of fails.
but
It also leads to lots of learning, if one can become good at
assimilating disappointment.
best
wishes
Don't throw in the towel just yet. I am in the middle of my final build, stalled due to cold weather.
Your analyses is correct as far as it goes, but you don't have all the information. Not everything has been made public.
@ Cadman
Very cool and thanks.
@ andrea
welcome to the O.U. forum
floor
The self rotating hydrostatic engine was developed by John W. Keely of mid 1800's. It was shown to have rotating valve creating water hammer internally with about 50lbs force generated. it was shown with flywheel, adjustable timing, and would saw wood as power take off on end shaft. (Name- Hydro Pneumatic Pulsating Vacuu Engine.) A Large and also small brass model were built. Patent application readable notes all match.
It's time to throw in the towel on this.
The variable position check valve needed for this to work has too many problems with it and the seals I have are inadequate.
It's too much trouble to continue so it's getting shelved so I can get on with other things.
Quote from: Cadman on February 01, 2021, 11:33:26 AM
...but you don't have all the information. Not everything has been made public.
Any chance of getting the secret sauce info so that your work may live on?
There is no secret sauce, just an unusual application of Pascal's Principle.
There is a build topic at Overunity Research.
This post is the mechanism
https://www.overunityresearch.com/index.php?topic=3974.msg93205#msg93205
BTW, it was a mistake to try to build this out of PVC and plastics.
Attached is the spreadsheet for the calculations with some additional information.
Maybe it will give someone an idea to use.
Cadman
Great. Thanks for sharing your work on this.