Hello, look at this video
http://www.youtube.com/watch?v=0MuojD4YFnE&feature=mh_lolz&list=FLO_rs91s3TuY
I've found it interesting. Do you think is it a fake? I've not found this "lazarev' circular" in the web.
Greetings, andrea
Looks cool.. No idea how it works.
If there is no temperature difference between top (i.e. condensation part) and bottom (i.e. evaporation part), then it is a fake.
Quote from: blueplanet on July 04, 2011, 07:31:38 AM
If there is no temperature difference between top (i.e. condensation part) and bottom (i.e. evaporation part), then it is a fake.
The cappilary effect does not require an imbalance of temperature.
the evaporation and condensation cycle plays no part in this.
see here for a detailed explaination and mathematics
http://en.wikipedia.org/wiki/Capillary_action (http://en.wikipedia.org/wiki/Capillary_action)
this can be used to gain energy from the rising water, but to do this of great magnitude,.. you would need a 3 acre field covered with hundreds of thousands of tiny pipes, each rising an inch or two, then another layer repeated above, and stack them up to a substantial height, and a volume of water that could actually perform work.
there doesnt seem to be a maximum height, because altitude doesnt prevent the effect.
it works at sea level and even inside a submarine, it works on top of a mountain, or on a plane, or even out in space.
theres no power involved, its like water spreading across the surface of a table.
Its just a matter of the diameter of the pipe, compared to the fluidity of the liquid.
a thick oil can be made to capilate a wider pipe
as long as the pipe is shorter than the maximum liquid column that pipe can support with that liquid.. then it can drip out into a higher reservoire, and gain potential energy in the system.
Quote from: sm0ky2 on July 04, 2011, 03:50:49 PM
The cappilary effect does not require an imbalance of temperature.
the evaporation and condensation cycle plays no part in this.
see here for a detailed explaination and mathematics
http://en.wikipedia.org/wiki/Capillary_action (http://en.wikipedia.org/wiki/Capillary_action)
this can be used to gain energy from the rising water, but to do this of great magnitude,.. you would need a 3 acre field covered with hundreds of thousands of tiny pipes, each rising an inch or two, then another layer repeated above, and stack them up to a substantial height, and a volume of water that could actually perform work.
there doesnt seem to be a maximum height, because altitude doesnt prevent the effect.
it works at sea level and even inside a submarine, it works on top of a mountain, or on a plane, or even out in space.
theres no power involved, its like water spreading across the surface of a table.
Its just a matter of the diameter of the pipe, compared to the fluidity of the liquid.
a thick oil can be made to capilate a wider pipe
as long as the pipe is shorter than the maximum liquid column that pipe can support with that liquid.. then it can drip out into a higher reservoire, and gain potential energy in the system.
Yes, this could solve the question. But the tube doesn't seem so thin, isn't it? A youtube user's citation, in that video, address to the property of salted water mixed with pure water explained (for example) in the book Etidorhpa. But it seems to me that this is an example of capillary action.
Please post experimental evidence of your own that proves this idea is working.
We would be very grateful if you can demonstrate that this system is running continuously for at least one day.
The whole world will be better off if you can harness slightly more than zero watts from this system.
We don't want just empty talk.
Quote from: sm0ky2 on July 04, 2011, 03:50:49 PM
The cappilary effect does not require an imbalance of temperature.
the evaporation and condensation cycle plays no part in this.
see here for a detailed explaination and mathematics
http://en.wikipedia.org/wiki/Capillary_action (http://en.wikipedia.org/wiki/Capillary_action)
this can be used to gain energy from the rising water, but to do this of great magnitude,.. you would need a 3 acre field covered with hundreds of thousands of tiny pipes, each rising an inch or two, then another layer repeated above, and stack them up to a substantial height, and a volume of water that could actually perform work.
there doesnt seem to be a maximum height, because altitude doesnt prevent the effect.
it works at sea level and even inside a submarine, it works on top of a mountain, or on a plane, or even out in space.
theres no power involved, its like water spreading across the surface of a table.
Its just a matter of the diameter of the pipe, compared to the fluidity of the liquid.
a thick oil can be made to capilate a wider pipe
as long as the pipe is shorter than the maximum liquid column that pipe can support with that liquid.. then it can drip out into a higher reservoire, and gain potential energy in the system.
Quote from: sm0ky2 on July 04, 2011, 03:50:49 PM
...
as long as the pipe is shorter than the maximum liquid column that pipe can support with that liquid.. then it can drip out into a higher reservoire, and gain potential energy in the system.
I have a problem with that. The liquid does not drip out unless energy is expended for that.
@ all,
Regarding the movie, please help me understand it correctly:
I see a bottom layer of liquid then an empty space filled of course with some gas (probably a mixture of air and vapors) then at an arbitrary height a capillary layer is mechanically fixed then finally on top of which there is another layer of liquid. One end of the pipe lies in the bottom layer of liquid.
Is this correct?
I hope not, because if it is, the device should obviously work as advertised. The top layer of liquid exerts some pressure (mass x g / surface) on the gas in-between the liquid layers. The gas passes the pressure in the bottom layer of liquid and because no other freedoms exist in the device, the liquid goes up the pipe. Yup, it doesn’t seem necessary the pipe be capillary; any pipe will do it.
Why the capillary layer? Two folds: to hold the liquid in place (to prevent it from falling down but slowly dripping) while passing the pressure to the gas.
IMHO the device is ingenious but still the full credits shall go to Heron of Alexandria (http://en.wikipedia.org/wiki/Heron%27s_fountain) â€" it’s the same device and it is his invention, really. Nevertheless, a large quantity of liquid falls from top layer to the bottom layer while a smaller quantity is pumped a bit higher or, if used as waterworks, the device works for a wile until all liquid slowly drains to the bottom. Then, after it’s initial energy (potential gravitational energy of the top layer of liquid) has been consumed, it needs “refuelingâ€. ;)
Did I see something wrong?
Best regards,
Tinu
while the device does resemble Heron's Fountain,
what I found, is the source of the information, describes a quite different concept.
here's the read for those interested.
http://www.bibliotecapleyades.net/tierra_hueca/etidorhpa/EtidorhpaHome.htm (http://www.bibliotecapleyades.net/tierra_hueca/etidorhpa/EtidorhpaHome.htm)
Observation 1: The tube looks to be too large in diameter for capillarity to occur
Observation 2: There is capillarity in the wood. This combined with the viscosity
of the fluid will prevent air from penetrating from below the wood to above it.
Observation 3: Gravity and capillarity are pulling the upper water through the wood, increasing the
volume of water in the lower level. This in turn is increasing the air pressure in the lower level.
Guess: This increase in air pressure in the lower level is pushing the fluid up the tube.
Test: If the wooden disc were perforated by a second tube, the top end of which
would be well above the upper water level (a snorkel), and the lower end terminating
well above the lower water level. This would allow equalization of the air pressure between
above the wood and below the wood. If the device continues to function, under these condition,
that would be interesting.
cheers
floor
ok, after reading the book in it's entirety (great read by the way)
it seems the device is described in detail, and even replicated in the text by those discussing it.
According to the literature, It is the effects of the difference in specific gravity between two liquids.
those referenced in the book were fresh water, with salt water poured on top.
The layer dividing the two is not necessary, as it was used in the book to simulate "real earth" conditions,
and the experiment was performed again, without using the layer inbetween.
According to the book, the heavier salt water pushes the fresh water up the tube, and the effect continues until salt water comes out of the tube at the top, then it stops, as the system balances itself out.
I have not yet experimented along these lines myself, it seems logical, but the book is a rather eccentric science fiction novel, that combines science, technology, religion, theology, and a lot of theories in which modern humanity does not subscribe.
Quote from: sm0ky2 on July 04, 2011, 03:50:49 PM
The cappilary effect does not require an imbalance of temperature.
the evaporation and condensation cycle plays no part in this.
see here for a detailed explaination and mathematics
http://en.wikipedia.org/wiki/Capillary_action (http://en.wikipedia.org/wiki/Capillary_action)
this can be used to gain energy from the rising water, but to do this of great magnitude,.. you would need a 3 acre field covered with hundreds of thousands of tiny pipes, each rising an inch or two, then another layer repeated above, and stack them up to a substantial height, and a volume of water that could actually perform work.
there doesnt seem to be a maximum height, because altitude doesnt prevent the effect.
it works at sea level and even inside a submarine, it works on top of a mountain, or on a plane, or even out in space.
theres no power involved, its like water spreading across the surface of a table.
Its just a matter of the diameter of the pipe, compared to the fluidity of the liquid.
a thick oil can be made to capilate a wider pipe
as long as the pipe is shorter than the maximum liquid column that pipe can support with that liquid.. then it can drip out into a higher reservoire, and gain potential energy in the system.
Wouldn't the capillary effect also apply to the dripping tip of the pipe too - preventing the water to escape?
Vidar
Quote from: Low-Q on March 16, 2015, 04:55:18 PM
Wouldn't the capillary effect also apply to the dripping tip of the pipe too - preventing the water to escape?
Vidar
Low-Q - in most situations you are right. For that reason, I do not think what is shown in the video will work with water. That may be why he used gasoline......
What is interesting, however, is that his description of the effect, does not concern capillary action at all.
The effect he references ( in the book Etidorpha) explains an effect of the specific gravity of liquids.
And this effect can be verified by experiment. simply place 2 liquids at different heights, and of different specific gravity, the "heavier" one above the lower one, and a differential pressure occurs. If a pathway exists, the heavier liquid has enough potential energy from gravity, to lift the "lighter" liquid to a height, even above that of the heavier liquid.
This effect occurs, both through a capillary, or through a much larger tube.
The guy in the video, describes the capillary, as providing a return path for the liquid, but what I did NOT see, was two liquids of different specific gravity, as would be required for this description to fit. It seemed, to me, that the entire mass of both upper liquid and lower liquid, were the same gasoline.
Therefore (unless there is hoaxing involved), Either the size of capillary for gasoline, must be much larger than water would act through...
OR it is an effect of Air Pressure, as one person above suggested.
It is my understanding that capillary action is a function of viscosity/surface tension/cohesion/adhesion, and NOT specific gravity, as explained by the videos author.
After years of contemplation and study of several of my own devices and those of others:
I have come to two conclusions:
1) The operating principal is governed by the pressure (yes caused by the specific gravity of each fluid (gas or liquid or combination, as defined by Archimedean displacement).
1 fluid displacing the other, while subsequently increasing the molar count variable,
Resulting in a higher pressure below. Causing the water to be ejected above.
Even at a greater starting height.
2) There is a short and long version of the Etidorpha text.
And the long version goes into much greater detail concerning this device.
One of the purposes of the specific gravity layer and/or the substrate layer is to suffice as a filtration system, to prevent the exchange of the compressing fluid.
additionally there is a second substrate layer above the working chamber
(Being the chamber with the stored gravitational potential energy)
A complete scientific analysis was performed, the result of which accounts for 100% of the potential energy, the difference on energy owing to the drop in fluid level of the upper volume. This drop (across a relatively larger surface area) is equal to the e=mgh of the volume of water which transitions upwards.
Interestingly enough, this mathematical analysis does not account for an anomalous energy quotient.
That being contained in the momentum of the falling fluid.
Allowing for the maximum resolution of e = 2e (?)
To be clear, the upper chamber which is being dripped into:
Shares an open air volume with the chamber at the bottom.
The pressure equivalent simulating a proportionally smaller "atmospheric pressure" measurement.
in this way the compressing fluid acts as a spring.
As pressure builds up from displacement, the fluid is displaced.
The substrate keeps the compressible fluid (being less dense) from passing into the working chamber.
and the same for the compressible fluid in the working chamber itself.
As pressure builds up inside the lower chamber, it pushes on the entire surface area of the more dense fluid. Within the tube going up, the pressure only exerts on the small surface area of the tube.
This is where pressure and gravity differ.
Whereas gravity pulls equally on all of the mass of the fluid anywhere in the system.
In a single fluid system, all pressure is equal and there is no flow.
With 2 fluids of different specific gravity, one will always displace the other.
Add a dynamic change in pressure,
Now you have a flow. Much like what occurs inside mountains or elevated lands where we see a 'natural spring' pumping water out above ground.
In the natural case, and in most replicated devices, the compressible fluid is air.
And water is used as the heavier fluid. But this does not really matter.
The minute losses in these systems do not account for the gravitational momentum.
The "drop" in height, which subsequently also = mgh.
So you see, if we place a pelton wheel in the lower chamber to take the momentum from the falling water,
Archimedean displacement still occurs.
And we have =2e
There are many replications of pressure based 'heron's fountain' type devices
Physics classes do an excellent job of describing the energy in the system.
Now, combine this effect with a mechanism to extract the kinetic energy
from the water's momentum.
Example:
https://youtu.be/-KtFZMN7_Bw (https://youtu.be/-KtFZMN7_Bw)
Conclusion: maximum ideal case output is = 2e
This appears to be a mathematically sound model of overunity,
based on the two very different aspects of gravity.
Acceleration and Displacement
Quote from: sm0ky2 on November 21, 2022, 06:23:47 PM
Interestingly enough, this mathematical analysis does not account for an anomalous energy quotient.
That being contained in the momentum of the falling fluid.
Allowing for the maximum resolution of e = 2e (?)
when will be pictures ?
My analysis is theoretical at this point.
I have not constructed a physical device that utilizes both gravitational principles simultaneously
Ive made a video with sketches and a brief explanation:
https://youtu.be/qPFJgIWZe-E (https://youtu.be/qPFJgIWZe-E)
Link to TK's version of a complex design::
https://youtu.be/L3UGTyc36f4 (https://youtu.be/L3UGTyc36f4)
I'll give another example, so this becomes more understandable.
This is an entirely different system.
Which is governed by the same equation.
https://youtu.be/s5eIRjmor1w (https://youtu.be/s5eIRjmor1w)
What you are seeing here is a change in external pressure
causing a change in internal pressure.
The energy in this system is consumed by compressing the air in the eye dropper.
This causes a change in density making it less buoyant.
This energy is returned by releasing the external pressure,
and the eye dropped becomes buoyant once again.
Pay attention::::
The momentum of the falling eye dropper is NOT considered.
I propose that it is possible to separate the acceleration force of gravity
from the buoyancy/displacement effect.
It is this, that may allow for gravitational energy to be extracted
Cool,
Thanks for sharing your thoughts!
Really does seem Novel yet such a simple application
Of a huge potential ( gravity component which was being wasted)
A possible "perpetuity " ??
Respectfully
Chet K
Ps
TinselKoala's link again ( one part of your concept
https://m.youtube.com/watch?v=L3UGTyc36f4&feature=youtu.be (https://m.youtube.com/watch?v=L3UGTyc36f4&feature=youtu.be)
im looking beyond a perpetual fountain,
This is an aspect of gravity itself, from which
we could obtain infinite energy
Take the cartesian diver, for example:
A bottle of any height follows the same equation.
It's a change in pressure equivalent to that which causes the air to compress.
Like a buoyancy switch
When its 'on', no force is placed on the bottle.
Apply force x for a distance y and buoyancy turns "off"
The switch function is conservative
(turns itself back on)
However, the PE<->KE field conservation of gravity is violated
We get the PE side for free.
At some height h :
The force at which the diver reaches the bottom is greater than the applied force
that activated the switch.
E = 2e
With the cartesian diver, harnessing this seems straightforward
Add a small magnet to the diver and allow it to rise and fall through a coil
Now let's improve the efficiency of switch function
— take a spring, or a static force (clamp) and press on the bottle
just LESS than enough to activate the buoyancy switch.
Now our energy in/out is minimized.
This will emphasize the gravitational output.
Now that we've isolated the mathematical work function of gravity
We can engineer any number of devices to take advantage of this principle.
Lets use Heat as our switch in this next one:::
See
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5942801/ (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5942801/)
So we have a helium balloon, lifting a small magnet suspended over a coil
The internal pressure is so close to the switching condition,
that a small drop in outside temperature compresses the gas making it non buoyant
raise the temperature back up and it floats again.
(hot air baloon, nothing special right?)
What happens at the coil?
How high does the balloon need to float before this energy exceeds the change in temperature?
I have devised an experiment that definitely proves my theory:
Feel free to try this at home.
If you find an experimental result that defies my theory,
Please share a description and your results.
As follows:
Take a closed vessel containing two fluids of different densities
One suspended vertically within the gravitational field.
At any pressure you choose
The momentum of the denser fluid will be exactly the same
when it drops from a given height.
Meaning, no matter the pressure system, a change in pressure, resulting in a change in height
Manifests this quotient of energy we refer to as gravitational potential.
E=mgh
In any conservative pressure system (that is one which returns the applied force back to the environment which applied it)
This energy quotient is extra internal work performed by the system.
We would expect this to manifest itself as internal heat and thereby raise the pressure accordingly.
However this does not reflect in experimental results.
The entirety of this momentum is observed outside of the system, minus standard losses.
Therefore: a change in internal energy of a 2-fluid pressurized system
Results in a second energy, derived from gravity.
The two hold a proportionality akin to voltage and current
The pressure state representing a static value
And the gravitational momentum representing a time derived motion.
An acceleration force, applied over a distance, of one fluid transitioning through another.
In nature, we would view the pressure system as being the energy that causes a rainstorm
And the gravitational component being the combined momentum of every drop of rain when it hits the ground
As you can see these are two greatly different numbers.
There is a 3rd component that that maintains the proportionality
The "resistance" component
This is the fluidic resistivity factor
At a given pressure, 2 fluids have a factor, similar to friction,
Wind resistance, etc.
Which controls the transition rate,
This is the "g" constant.
At standard pressure (atm), using air as the first fluid,
This is 9.8 m/s/s
Much like electricity, fluids take the path of least resistance.
For this reason, the density of the lighter fluid affects the constant.
In the absence of gravity:
The pressure system continues to operate exactly as prescribed.
All energies are still accounted for.
We simply no longer have the gravitational component.
The Cartesian Diver fails to dive, even though the air compresses.
This proves that gravity can be used as a power source.
A more practical abuse
Would be to place a pelton wheel on a siphon
The siphon doesnt care about the rate of flow
(beyond that which is required for its' operation)
This momentum is a factor of the rate of flow.
The faster your siphon, the greater the accumulation of energy.
The energy required to initiate the siphon process is the difference in height
of the water volumes.
Which is already accounted for in the change in fluid heights of the two containers.
Ergo: the momentum is free
we set the voltage, we get our voltage back
We control the current = overunity
If this applies in the physical world,
It must also apply to electricity in some form
a capacitor, discharged through a variable resistor
heat monitor of the resistor may lead to more insight
https://www.hwcv.org/empty-pagef738a1ef (https://www.hwcv.org/empty-pagef738a1ef)
Why do you forget about that ? I think it looks like something.
Further verification
Thank you kolbacict
page missing??
Quote from: Thaelin on November 28, 2022, 05:48:54 AM
page missing??
Page is there, however there is a 404-error having to do with the frontpage image
So the main picture on the webpage doesnt show up, just returns a 404.
The rest of the site works.
Clicking the button, then the link that says machine
Takes you to a video (in danish i think?)
Which shows the principle behind using capillary to control buoyancy
Energy in and out of a Cartesian Diver is exactly like a spring
We get back almost what we put in.
Gravitational Energy is extra
New video
https://youtu.be/AMgUzpwRe6Y (https://youtu.be/AMgUzpwRe6Y)
Examples:
https://youtu.be/sNOXFiJ4IDU (https://youtu.be/sNOXFiJ4IDU)
https://youtu.be/QukS1pxoezY (https://youtu.be/QukS1pxoezY)
https://youtu.be/ljvp-iR18Ko (https://youtu.be/ljvp-iR18Ko)
https://youtu.be/s5eIRjmor1w (https://youtu.be/s5eIRjmor1w)
https://youtu.be/TMju6WzDnHI (https://youtu.be/TMju6WzDnHI)
As we can see, by analyzing the gravitational momentum component
Of any 2-fluid system
A change in pressure results in a change in internal energy
greater than the energy input into the system by the pressure change alone.
And that this energy is directly derived from gravity
Using the electroactive properties of a gas,
Such as NO2
we could potentially control buoyancy using electricity
A 'reverse Cartesian Diver':
Eye dropped is filled with slightly less than a buoyant volume of Air
and HV electricity is passed into the bulb
Air ionizes and the plasma expands,
causing the eye dropper to become buoyant and float to the top.
Let's look at another aspect of this situation:
Suppose we have a sealed vessel containing:
a volume of water
a volume of air
and a flexible container of air (baloon)
Above the water, we have a set volume and pressure
submerge the baloon, and the water pressure (displacement)
collapses the baloon, thereby decreasing the overall volume.
The pressure of the air in the upper portion of the volume:
Decreases by the displacement of the expanded baloon.
Increases by the displacement of the collapsed baloon.
Does this result in a lower total internal pressure?
and if so, does this change in pressure equate to the change in
potential (buoyant) energy of the baloon?
Or: does the drop in pressure cause the displacement of the water
to increase by this same difference?
I used the extreme case of a baloon, so as to draw comparison to
the same situation using a non flexible container (hard ball of same mass and volume)
We see the rise in water volume (displacement) differs greatly.
So if gravity is conservative: where does this excess energy come from?
The energy required to submerge either vessel (of the same initial volume)
is the same. The potential energy, however is less in the collapsed baloon.
The change in pressure of the upper volume is different in both cases as well.
As also is the change in pressure of the submerged air vessels.
The flexible baloon compresses greatly, the non-flexible baloon does not compress.
In the non-flexible case: total internal pressure increases by a factor of
the effect of gravity on the displaced mass of water and the height of displacement.
In the flexible-case, we subtract the compression ratio of the baloon at the submerged depth.
Input energy is the same in both cases, and the difference in potential energy is exactly equal to
the energy required to recompress the upper air volume.
Water, being an incompressible fluid, maintains its' volume, even though the pressure has increased.
Air pressure in the flexible case, should therefore also increase, however a drop in pressure indicates
that the temperature must have dropped within the upper air volume.
For this to be possible, the temperature of the submerged volume must increase.
When we analyze the temperature difference, this is precisely equal to the difference in
potential energy between the two submerged vessels.
Conclusion: the extractable portion of the gravitational force does not affect our
interpretation of the PVT equivalency.
Lets push further:
We have a vessel and a volume of water.
We perform electrolysis and convert this into HHO:
Using a known energy value
Causing the buoyant state
We allow the vessel to rise through the atmosphere.
At a significant height, we perform reverse electrolysis
thereby reclaiming our electrical energy and accounting for
all losses in the system.
Now we have excess energy = Mass of Water * Height * Gravity
Here we see that the energy required to achieve the buoyant state
is completely detached from the gravitational energy in the system.
It is important to understand here that Displacement
due to Buoyancy: is an effect of the gravitational force
NOT the cause.
Why is there a difference in force?
What IS the difference in force?
Why is buoyant force different than gravity?
And can we use this to our advantage?
1) there is a difference in force related to Density.
Objects that are less dense than the medium experience buoyancy
Objects that are more dense than the medium experience gravity.
2) for objects more dense than the fluid, the effects of gravity are proportional to
the difference in density. For solids in air (at 1 atm), there is not a large difference
between the least dense (non buoyant) solid and the most dense solid.
Therefore, gravitational force through the atmosphere is (almost) the same for every object.
However, when the medium is more dense (such as a liquid), changes in density of the object
greatly vary gravitational acceleration.
Buoyant Force, because air isnt very dense to begin with, the changes are far more drastic.
The less dense an object is (buoyant), the faster it rises through the gravitational field.
3) This is a more complex topic: buoyant force is caused by a difference in gravitational force.
When we calculate gravity, we use both interacting masses to determine field strength.
Similar to interactions between 2 charged particles.
Say, for instance, we have an electric field, and a series of charges of different values:
While all of our charges are attracted to the electric field source, they are each attracted
by a force proportional to their individual charge.
If charge density were high enough, we would have an 'electric buoyancy', similar to
what happens in a gravitational field. Particles with leas charge would experience a force
AWAY from the field source.
While particles with a greater charge would experience a (non buoyant) force TOWARDS
the field source.
Where this line is drawn depends (above critical density of the system) upon the proportionality
of the constituents.
Here on Earth, the atmosphere holds the highest proportion.
Therefore in Earth's gravity, we draw the line at the density of air.
4) yes, yes we absolutely CAN!
I'm open to any opposing theories, constructive criticism,
Or idea that could advance this area of science.
Here's another gravity engine i designed
https://youtu.be/jv9lQVZ4tUM (https://youtu.be/jv9lQVZ4tUM)
Hi Sm0ky2,
Very clever design in your video! Would you be willing to share more specifically how you envision moving the solenoid back and forth with minimal energy expenditure? That solenoid is going to have to compress a fair amount of the air to increase the pressure enough to get the cartesian diver to overcome the resistance in the sprocket/chain mechanism.
PS I've enjoyed reading all your posts on this thread. BTW, do you have any thoughts on the Plankz Buoyancy Engine?
Quote from: Beginners Mind on December 20, 2022, 01:47:24 AM
Hi Sm0ky2,
Very clever design in your video! Would you be willing to share more specifically how you envision moving the solenoid back and forth with minimal energy expenditure? That solenoid is going to have to compress a fair amount of the air to increase the pressure enough to get the cartesian diver to overcome the resistance in the sprocket/chain mechanism.
PS I've enjoyed reading all your posts on this thread. BTW, do you have any thoughts on the Plankz Buoyancy Engine?
Yes so:
Of course there are efficiency improvements that could be made:
For example we could reduce friction by using a belt/pulley system,
with a nylon or similar water-resistant belt, and the upper pulley fixed to
the drive shaft.
The compression mechanism itself, wether an electric solenoid is used, or
something like a linear motor, etc.: is synonymous to a spring.
In that the energy used to compress the gas is returned upon decompression.
Reducing the energy input, essentially to frictional losses. This is Key.
The precise amount of compression required is a factor of the volume of buoyant gas
in the upper (flexible) portion of the bobber, the coefficient of the flexible material,
and the mass-ratio for the desired output energy.
This is a volumetric compression ratio based on the mass of displaced water between
the compressed and expanded states. The greater the compression ratio: the "heavier"
the bobber becomes during the non-buoyant stage, and the more gravitational force is
imparted onto the bobber as it falls. The same applies to the buoyant force, when the gas expands.
Height of the container provides a longer cycle-time, meaning the force acts on the drive shaft for
a longer period of time, each half-cycle, for (roughly) the same compression force. (+in / -out)
The important thing to take away from this is the fact that energy in and energy out (at the solenoid),
balance each other out. The output energy obtained at the drive shaft is purely gravitational.
(minus system losses)
P.S.:
My examination of Plank's ideas, the best attempts i have seen use a compressed gas,
introduced beneath the water chamber. This removes the loss of water
(which, is my understanding of the technical problems in Plank's original design)
I have NOT given this situation a full analysis, in terms of the energy required to compress
the air, beyond the pressure exerted on the bottom of the water tank, in comparison to the
Buoyant force obtained over the height of the water column. However, in practical applications
this seems to require more energy in, than out.
If i were to make an attempt at this, i would use flexible (expandable) vessels to harness the
compressed air, thereby regaining some of the pressure at the top side. There are of course technical
difficulties in terms of valves, etc. to make such a thing possible.
Quote from: sm0ky2 on December 18, 2022, 07:42:49 PM
I'm open to any opposing theories, constructive criticism,
Or idea that could advance this area of science.
Here's another gravity engine i designed
https://youtu.be/jv9lQVZ4tUM (https://youtu.be/jv9lQVZ4tUM)
Hi sm0ky2
like your idea, have even believed in it for a while but I see a problem with retrieving back energy inputted in solenoid. As diver sinks deeper, air cap is further compressed by water pressure. At the bottom acts on air cap not only pressure exerted by solenoid but also whole water column above diver. This means air cap has smaller volume than it had at the top. Smaller air cap volume means water level is lower than it was when diver was at the top. And lower water level means less pressure which means less energy pushing solenoid back to initial position.
To better illustrate issue with this design:
1. bottle has initial pressure P1 of the air above the water.
2. Solenoid increases that pressure to P2
3. Diver loses buoyancy and descent to bottom
4. Hope you see that air cap on diver is exposed to higher pressure at the bottom than it was on top
5. Less air cap volume causes water level to drop causing air pressure in the top of bottle to drop to P3
I think as P3 is less then P2 we cannot fully retrieve energy inputted to solenoid.
Quote from: MT on December 20, 2022, 05:04:29 PM
As diver sinks deeper, air cap is further compressed by water pressure. At the bottom acts on air cap not only pressure exerted by solenoid but also whole water column above diver. This means air cap has smaller volume than it had at the top. Smaller air cap volume means water level is lower than it was when diver was at the top.
If we take two divers in one aquarium. And connect them mechanically with the necessary interaction.
When one sinks and contracts, the other will rise and expand. The water level at the top will not change. 8)
Sm0ky2 - Thanks for your reply. Not withstanding the further comments about possible problems recapturing energy from the solenoid, I'm still puzzled how nearly all the energy powering a solenoid in this situation can be theoretically recaptured. The plunger would need to be held either in or out, depending on the solenoid's orientation, for a period of time during which there will be I^2R losses in the coil that can't be recaptured, won't there? Please correct my thinking if I misunderstand. Thx.
Quote from: kolbacict on December 21, 2022, 12:49:06 AM
If we take two divers in one aquarium. And connect them mechanically with the necessary interaction.
When one sinks and contracts, the other will rise and expand. The water level at the top will not change. 8)
Nice try :) but not sure you can easily workaround it. If solenoid is engaged in one aquarium then both of them will be sinking.
And how do German "rosсh" installations known work?
Or they don't work?
The capacity for compression, of the Cartesian diver, can be limited to a maximum by its design.
Even so, the amount of energy required to re- expand the diver increases with any increase
in the depth of its dive. This is because at greater depths, greater pressure exists outside the
diver.
Quote from: MT on December 21, 2022, 01:58:07 AM
Nice try :) but not sure you can easily workaround it. If solenoid is engaged in one aquarium
then both of them will be sinking.
Two tanks would then be required and instead, a pressure sealed axle passes through and between
the two tanks. Two mechanically linked, insert able pistons and cylinders (one for each aquarium),
would then be required.
Quote from: Willy on December 21, 2022, 12:22:58 PM
Two tanks would then be required and instead, a pressure sealed axle passes through and between
the two tanks. Two mechanically linked, insert able pistons and cylinders (one for each aquarium),
would then be required.
Yes, there are maybe many options.
But It seems to me that if such serious people as Rosch did not succeed, if this is a fake, then there is nothing to look for in this direction.
Quote from: MT on December 20, 2022, 05:04:29 PM
Hi sm0ky2
like your idea, have even believed in it for a while but I see a problem with retrieving back energy inputted in solenoid. As diver sinks deeper, air cap is further compressed by water pressure. At the bottom acts on air cap not only pressure exerted by solenoid but also whole water column above diver. This means air cap has smaller volume than it had at the top. Smaller air cap volume means water level is lower than it was when diver was at the top. And lower water level means less pressure which means less energy pushing solenoid back to initial position.
Yes you are correct. The ratio at which this changes is (approx) 880 milligrams per cubic centimeter
of air volume differential between the upper and lower maximum.
Therefore, there is a maximum height at which the weight of the water column begins to affect
the ability to decompress the air volume. This is one of the maximum system parameters.
And is defined by the elasticity of the flexible part of the bobber. (spring constant)
The compression method must match the decompression capacity.
Quote from: kolbacict on December 21, 2022, 12:49:06 AM
If we take two divers in one aquarium. And connect them mechanically with the necessary interaction.
When one sinks and contracts, the other will rise and expand. The water level at the top will not change. 8)
Pressure in a sealed vessel is equal throughout the vessel
One cannot sink while the other rises
Quote from: Willy on December 21, 2022, 12:22:58 PM
Two tanks would then be required and instead, a pressure sealed axle passes through and between
the two tanks. Two mechanically linked, insert able pistons and cylinders (one for each aquarium),
would then be required.
I like this idea:
Working pressure from one chamber acts on the pressure in the opposing chamber
In this method, it may be possible to have a tandem system.
In such: the decompression energy from one chamber could add to the compression energy of the other, thereby lowering the energy required to compress the air.
The equilibrium state would be approx half-compression in both.
The piston would only have to half-compress either chamber while energy is extracted from both.
Hmm ;)
Quote from: kolbacict on December 21, 2022, 02:11:52 PM
Yes, there are maybe many options.
But It seems to me that if such serious people as Rosch did not succeed, if this is a fake, then there is nothing to look for in this direction.
The mathematics implies that there is significant energy that could be extracted.
Remember with solar and wind: our early designs were far less efficient than the ones in use today.
Also: the failures of one group bears little on the successes of another.
Regardless of how serious or well funded the one may be.
Quote from: kolbacict on December 21, 2022, 08:18:36 AM
And how do German "rosсh" installations known work?
Or they don't work?
Not sure if i have a reference to their work.
Is this the "rosch" type device?
https://youtu.be/8Zenn0VD2Bo (https://youtu.be/8Zenn0VD2Bo)
If so: this is one of the compressed air powered Plank's Engines.
Quote from: Beginners Mind on December 21, 2022, 01:19:34 AM
Sm0ky2 - Thanks for your reply. Not withstanding the further comments about possible problems recapturing energy from the solenoid, I'm still puzzled how nearly all the energy powering a solenoid in this situation can be theoretically recaptured. The plunger would need to be held either in or out, depending on the solenoid's orientation, for a period of time during which there will be I^2R losses in the coil that can't be recaptured, won't there? Please correct my thinking if I misunderstand. Thx.
In practice, a true electric solenoid may be impractical.
This was used mostly as an analog. A mechanical gearing system or spring based system, or even pneumatic systems may be more feasible for the compression/decompression mechanism.
To actually recover the energy using an electric solenoid would result in great electrical losses do to the decompression rate being rather slow.
https://youtu.be/tvSo7CXQQ7c
Quote from: sm0ky2 on December 22, 2022, 06:57:55 AM
Pressure in a sealed vessel is equal throughout the vessel
One cannot sink while the other rises
Can you make a diver so that when the water pressure rises, he floats instead of sinking?
To invert the scenario, implies an inversion of the compression.
Such a bobber would use vacuum-based displacement,
where-in external pressure activates a secondary internal actuator
Increasing the physical volume of the air chamber,
In this bobber the flexible material would be replaced by an expanding rigid chamber
that could support lower internal pressure.
(vacuum-spring?)
The force required to do this would be substantially large.
It may be mechanically possible, but wouldn't make sense to build such a thing for this purpose
In air buoyancy would be more suitable for vacuum-based buoyant devices
Quote from: sm0ky2 on December 22, 2022, 09:10:52 AM
The force required to do this would be substantially large.
It may be mechanically possible, but wouldn't make sense to build such a thing for this purpose
In that case we could to use single aquarium without partitions.
And the water level would not change from above. :)
Think of it like this:
When you expand a baloon under water:
Such as is done in emergency life jackets, with a CO2 cartridge
You are providing an inner pressure, forcing against the mass of that volume's
worth of water.
If you sucked all the air out of a military submarine,
while underwater
It would crush like you stepped on an aluminum can.
Your're fighting the entire atmosphere
The Pistol Shrimp uses this to its' advantage,
as does a particular high-efficiency water heater
But its not of a lot of use for direct gravitational harvesting
outside of some nano-tech we haven't invented yet
Its' called a cavitation-gravitation oscillator
Which uses the difference between 0 ATM and
The 0 state of a denser than air medium
The compress an incalculable mass of incomprehensible atoms
Which in turn radiate all sort of grabbables
Quote from: sm0ky2 on December 22, 2022, 07:05:39 AM
In such: the decompression energy from one chamber could add to the compression energy of the other, thereby lowering the energy required to compress the air.
The equilibrium state would be approx half-compression in both.
The piston would only have to half-compress either chamber while energy is extracted from both.
Hmm ;)
Very cool
In most cases, the height of the water tank is not large enough to make a
substantial change in the pressure of the bobber
When pressure inside the tank is returned to its' rest state,
the air inside the bobber is more than capable of re-expanding itself
Real-world example:
Sink a raft in the ocean
You will need to go down several hundred feet before it
irretrievably collapses under the pressure of the ocean.
Fresh water is less dense so it would take an even greater depth
YES there is a change in pressure by depth
And we can calculate exactly how much this changes.
Doesnt affect the action of the Cartesian Diver until very great
heights are reached within the water column.
The ranges we would use for any sort of application,
fall well within normal operating conditions
Quote from: sm0ky2 on December 23, 2022, 11:25:52 AM
YES there is a change in pressure by depth
And we can calculate exactly how much this changes.
Doesnt affect the action of the Cartesian Diver until very great
heights are reached within the water column.
The ranges we would use for any sort of application,
fall well within normal operating conditions
Was searching a bit diving websites and yes it should not have much impact for devices with practical dimensions like 1-2 meters of height. According to Boyle's law, doubling pressure halves gas volume at constant temp. And pressure at water increases slowly, 1/10th of bar per meter.
So a diver starting at surface with ~1bar pressure and 4liters of ideal gas volume would need to go 10meters deep to see its volume halved. At 30meters it would be just 1 liter.
That shrinking works in our advantage when diver is sinking as volume is lower and is thus even less buoyant and similarly during rising it works in our disadvantage.
Question, have you or maybe somebody else already did some or even complete energy calculations of this design on a concrete example?
I mean first initial energy of whole system with diver at top. Then:
1. energy needed to compress
2. work bober can produce on its way down
3. energy that can be retrieved during decompression of whole system with diver down
4. work bober can do on its way up.
For example 1m3 aquarium, 10kg diver with 12 liter of ideal gas.
I have not made any sort of prototype
Here is the basic physics.
Displacement of the motive mechanics will need to be accounted for
https://physics.anu.edu.au/engage/outreach/_files/Cartesian%20Divers.pdf (https://physics.anu.edu.au/engage/outreach/_files/Cartesian%20Divers.pdf)
Here is a more in-depth analysis
Which shows irreversible sinking as a factor of bobber diameter.
https://iypt.ru/wp-content/uploads/2020/10/A-lot-of-good-physics-in-the-Cartesian-diver.pdf (https://iypt.ru/wp-content/uploads/2020/10/A-lot-of-good-physics-in-the-Cartesian-diver.pdf)
Thank you for links. Just find the same idea was already suggested in 2016...
https://overunity.com/16357/cartesian-diver-generator/
Thank you for your find!
I had not come across that thread when I created this one.
It seems they never got past the depth thing
It seems there are 2 solutions:
We can ignore it, and keep our test devices within the height constraints
Or:
We can enclose the buoyant vessel.
At some point we might just buy a submarine and tie it to a generator cable
Because we are reinventing so many wheels
However, in milling several of those wheels in my head i have a few ideas:
Heres one:
If PV truly = nRT
And we maintain a relatively constant temperature....
An air hose could protrude up from the bottom of the tank and inject air into the "eye dropper"
then allow that air to then escape.
The pressure of the working volume would catch all the energy in/out
and we would have a direct buoyancy switch without needing to compress the entire outer chamber
They are surely more ways how to utilize potential gain but unfortunately there is no gain. Was doing a bit of calculations.
This idea is based on assumption that we can retrieve all energy that we spent on compression. Yes, we can retrieve all energy during decompression when the diver is at the top but thats not possible when he is at the bottom.
Problem is with air bubble in the diver. Suppose that diver is at the top and we compress its air bubble so its volume got one liter smaller spending energy E. With diver at the top the same amount of E is needed to regain that one liter of volume. But when diver is at the bottom with higher water pressure using only E will not regain one liter of volume, it will be less. To get it inflated to original volume extra work is needed and this work is exactly (in ideal system with no losses) what could be possibly gained from diving motion to bottom.