Sm0ky2's Technological Comporium:
Chapter 2: Energy Generation
Section A: Generators
Generator 44: Buoyancy Controlled Generator (BCG)
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Notice of Disclosure: Aug. 17, 2010
This device and all of its uses are hereby disclosed freely, and open-source to the general public, all areas of commerce and industry, as well as for governmental, private, and institutional use. This open-source technology can be used free and without restriction, with no patents, or proprietorship of any kind, explicit, implicit or otherwise.
Disclosure of this device, its construction, its function, and its components is hereby fully disclosed, no information is to be withheld by its inventor.
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The Buoyancy Controlled Generator (BCG):
[(device overview) What is the BCG?]
The BCG is a buoyant-based power generation system. A controlled-displacement-bouyancy is implemented to cause motive force in the direction of generation. In this system, power is generated during two of a three-stage cycle. Two types of generation take place;
1) buoyant-generation, 2) gravitational generation
Buoyant Generation: This is power generated by the buoyant force, or upward force caused by buoyancy. ( UP side of the cycle)
Gravitational Generation: This is power generated by the gravitational force. (DOWN side of the cycle)
The 3rd stage of the cycle is the Reload-Stage. where mobile units are reloaded into the buoyancy chamber.
There is also a 'transition' at the top end of the cycle, where the mobile units transition from buoyant to gravitational motivation.
In a cyclic nature, mobile units are transitioned up and down to facilitate power generation.
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How does the BCG work?
The Buoyancy Controlled Generator operates using Buoyancy Units (B-Units), which are modified to ride on a pre-destined track.
The B-unit, its working principal, and its construction are outlined in the next post.
The B-Unit is a paid of concave "saucers", with the ability to flatten out, or bow-upwards in the shape of a 'bowl'.
this motion is controlled by a hydraulic cylinder in the center of the unit. Displacement, caused by the transformation of the saucers, causes the B-Unit to become buoyant in a fluid or Air.
When the hydraulic cylinder actuates, the B-Unit expands, increasing its displacement within the fluid or gas/Air.
When the unit contracts, it decreases its displacement.
This increase & decrease of mass-displacement is designed in such a way that the unit and all associated mass, becomes buoyant and non-buoyant, respectively.
Drawing 1a:
The BCG consists of two tanks of water: Primary and Secondary.
The Primary tank is where Buoyant-Power-Generation takes place.
The Secondary tank is for the recovery, and reloading of the modified B-units.
The B-Units are modified and/or attached to a housing, allowing them to ride along a track. This track forms a circular path through the verticle-length of the primary and secondary tanks, and through the air-space between the tops of the two tanks.
Drawing 1b:
Situated along the track, are four main components.
A) Buoyant Generator: This generator is actuated by the buoyant force (upwards) imparted upon the B-Units. This side of the track is refered to as the 'Long Stroke' - in which the B-Units move slowly, but with more torque, than the 'Short Stroke'.
B) Gravitational Generator: This generator is actuated by the gravitational (downwards) acceleration imparted upon the B-Units from gravity. This side of the track is refered to as the 'Short Stroke' - in which the B-Units drop quickly through the gravitational field.
C) Robotic Arms: These are used to transport the B-Units during the Loading, Re-Loading, and Transition stages of the cycle.
D) Unit Reloader: When the system is in the Re-Load Stage, the Reloader stores the supply of B-Units to be used for power generation. Units are transfered from the Recovery Tank, into the Unit Reloader. The Loader Arm takes the units, one at a time and moves them into position to begin the cycle.
Drawing 1c
The cycle begins with the Loader Arm - positioning Buoyancy Unit at the start point of the Long Stroke
The control unit sends a wireless signal to the transciever inside the B-Unit. this does two things, one it checks the B-status.
This is the buoyancy of the unit, relative to a "zero-B", or 0 force state. (weightlessness)
The control unit then instructs the B-Unit how to adjust its buoyancy to begin the Long Stroke.
B-unit rises, generating power along its path to the top of the Primary Tank. Where the Transition Arm moves the B-Unit to the Short-Loader at the starting point of the Short Stroke.
The Short-Loader Arm positions the B-Unit at the drop-point, and the control unit sends the command for the B-Unit to set the buoyancy chamber to minimum displacement.
B-Unit falls, generating power along its path to the Recovery Tank.
Drawing 2a
Recovery and Reloading:
Recovery takes place when the B-Unit reaches the bottom of the Short Stroke. The B-Unit touches down into the water in the Secondary Tank. When all B-Units have completed their cycle, and have been recovered, the system goes into Reload Mode.
Reloading:
When the control unit recieves the recovery response from all actibe B-Units, it begins the reloading process.
The robotic arm at the top (Transition Arm) releases the top portion of the track, and raises up about 3 inches.
At the same time, another arm at the recovery point, disengages the track and lifts the Short Side of the track 3 inches, in synch with the Transition Arm.
The Tank-Doors then close. Tank doors are positioned at the top of the water-line on both tanks, and seal completely air-tight.
The door at the bottom, connecting the two tanks is then slowly opened. This creates a pressure equilibrium between the two tanks, and B-units can then be safely transfered into the Reloader.
The lower track extends slightly, connecting the track into the Reloader.
Once reloading is complete, the track disconnects, the lower door closes, the Tank Doors slowly open, and the track is lowered and reconnected. Then the system is ready to begin another cycle.
Drawings 2b, and 2c
That pretty much sums up the description of the generator.
Although specific components, or materials used to build this generator may have current or future patents, with respect to their particular industry, the technology behind this device is completely unpatented, and has been in use for hundreds of years.
So, give a 6-gun salute to the Hindenburg and hand your child a helium baloon. This is Sm0ky2 with a "Hail Mary" pass to the Free Energy community. Touchdown, with 3 months left in the 2nd quarter.
This thread has officially begun::::::::::::::::::::::::::::::::::::::
You sure have put some work behind it.
But wouldn't it take energy to expand the floaters under water.
...and how would you keep the water from leaking down into the small water container? the water would drain from the big water container.
Quote
But wouldn't it take energy to expand the floaters under water?
etc...
Yep, I'm wondering what SmOky2 smokes these days, too...
This concept is over-complicated, and absolutely non-working, at least as a "FE" device...
But it's still nice to see someone dedicating such a marvelous idea to humanity. :-)
Useful info for the inventor:
http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm
Quote from: spinn_MP on August 17, 2010, 03:50:34 PM
etc...
Yep, I'm wondering what SmOky2 smokes these days, too...
This concept is over-complicated, and absolutely non-working, at least as a "FE" device...
But it's still nice to see someone dedicating such a marvelous idea to humanity. :-)
Useful info for the inventor:
http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm
If you have something critical to say then say it. Don't relegate the criticism to the clown Simanek. @sm0ky2's device is problematic but not because of your sorry input.
Quote from: Omnibus on August 17, 2010, 04:15:13 PM
If you have something critical to say then say it. Don't relegate the criticism to the clown Simanek. @sm0ky2's device is problematic but not because of your sorry input.
Piss off, OmniBot! Go play with that "OU" "SMOT" of yours...
Quote from: spinn_MP on August 17, 2010, 04:20:57 PM
Piss off, OmniBot! Go play with that "OU" "SMOT" of yours...
Go away, troll.
Quote from: Omnibus on August 17, 2010, 04:30:26 PM
Go away, troll.
From all the lectures you've have been given for free from all the people, that's the only phrase you can remember? "GO AWAY, TROLL?"
Btw, "Omnibus" is a synonim for "Troll", check the google...
Quote from: spinn_MP on August 17, 2010, 04:47:23 PM
From all the lectures you've have been given for free from all the people, that's the only phrase you can remember? "GO AWAY, TROLL?"
Btw, "Omnibus" is a synonim for "Troll", check the google...
Hear it again -- go away mean little twerp.
Fighting already? we're on the first page !!!! come on...
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The B-Unit is a well tested and well proven device.
It can be scaled to lift any mass, the equation is simply determined by the volume of the fluid displaced by the buoyancy chamber.
Mass of B-Unit + Load = Mass of the displaced fluid.
This is a 0 Force , or Zero-B state. Weightlessness, and the default starting point of the B-Unit. This is controlled by a hydraulic cylinder actuated generally by 2 methods, though others have been applied in differing circumstances.
Method 1: electronic solenoid. This is not very cost efficient, but allows for quicker system response.
Method 2: worm-gear. This is much more cost efficient, in terms of resistive pressure vs energy input, however the buoyancy system takes more time to transition between states, which controls altitude, acceleration, and verticle direction.
The electrical cost of actuating the hydraulic cylinder is directly related to the buoyant force. The proportion does not reduce to an energy equilibrium, as depending on the fluid and depth/pressure, these two values can be equal to, greater than or less than one another.
but by analyzing data, a relationship can be drawn, such that an increase in resistive pressure relates directly to an increase in buoyant force. And inversley, a decrease in resistive pressure, relates directly to a decrease in buoyant force.
The denser the fluid, the greater the resistive presure. This is the force exerted on all sides of the buoyancy chamber (B-Unit).
This force resists the expansion of the chamber, E= f * d
Distance is determined by the desired buoyant force
This is the work done by the hydraulic cylinder.
In the Air, the difference in resistive pressure between an altitude of 0 to 200 ft (the lower limit of FAA aircraft control), is negligible.
The difference in resistive pressure between 0 to 200ft under water, is thousands of times greater. But so is the buoyant force.
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The energy available from buoyant force imparted onto the B-Unit is determined by the distance of the 'lift'. Other than the direct proportionality, there is no E = E correlation betwee these two energy values. simply because the distance of the "lift" is determined by the container of the fluid, and or operational parameters set in place by the user. While the energy required to create buoyant force, is determined only by the pressure of the fluid at the start of the "lift".
This system is a complexity of several systems interlinked. Utilizing multiple force-vectors, from completely different systems.
a thermodynamic analysis of this as a whole, is not possible, because it is not a "closed system". Those rules simply don't apply here.
If you look at the B-Unit itself, as a point, lets say 200ft under water. Now, the energy required to expand the unit is exactly equal to the energy obtainable by allowing the water pressure to collapse the unit. Those two values are thermodynamically equivalent.
This example also applies to the Emergency LifeJackets, used by Divers. Under 200ft of water, the energy required to inflate the life jacket is equal to ( or less than) the energy contained in the tiny CO2 cartridge. However:: This has nothing to do with the energy gained by the diver as his inflated life jacket jettisons him to the surface.
have the diver hold onto a tension-cable anchored to the ocean floor, and you can measure this energy directly. E = f * d
Distance, in this case being 200 feet.
you can see, that while the energy value in different types of life jackets may vary,
this is only a proportional change to the buoyant force.
the total work ( force over distance) caused by buoyancy is many times greater than the work done by charging while CO2 cartridge or the work done by the compress gas to inflate the life jacket.
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The determining factor for wether or not the B-Unit is being unsed in an "overunity" fashion, is simply the distance of the lift.
once you surpass the work-force if the transition, at height-x,
anything higher than that is "free energy".
It doesnt defy any laws of physics, infact the physics supports this entire system. It must be treated as several independent systems. The robotics and hydraulic systems are small, and consume a set ammount of energy per cycle.
The power generation is a completely different system, determined by the height of the "lift" and "fall".
you adjust the height, you adjust the power output.
power input changes proportionately on the "lift" side.
but this hydraulic actuation, is fractionally smaller.
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Like the braking system on older cars, before they started using the engine to assist you.
your one foot isnt putting nearly enough energy into that heavy car to STOP it.
but by pushing with a great ammount of hydraulic-leverage you are able to tightly clamp the brake pads onto the rotor causing enough friction to stop the car. These are two completely seperate systems. one system increases and decreases the frictional drag on the rotors. how much energy is consumed by the rotating axle with this increase in friction has nothing to do with the energy required to cause it. the other system, is the car's engine using gasoline to make it move forward. this system is offset by the greatly increased friction of the brakes.
but the energy values (input to the brakes, and output from the engine) have only a factional / proportional relationship, because they are interfaced components of two entirely seperate systems.
If the two were equivalent, you would need a whole other gear-set for yoru transmission, for instance, if you were to link a clutchable "reverse" gear, whos purpose was to fight against the inertia of the car, you could stop just as fast as you accelerate.
but starting and stopping would consume just as much gas.
What else slows the car down? friction. hmm, we know how to add friction. this costs much less energy than "reversing" the driving process.
lets just give the car a friction-switch. we'll call this the "braking system".
Everyone accepted this system with open arms. Hey thats great! we can stop the car with a simple push of our foot, or a pull of a lever.
What i didnt hear them saying was that the laws of physics were broken, or einsteins would start falling from the sky, or that we were in a state of thermodynamic crisis. Why not? because those laws only apply to a closed system. Not two interfaced, variable, systems.
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Now about the Water Leakage comment.
The door between the two tanks is sealed during the generation cycle. no water passes through the bottom.
At the top, there would be minimal 'drippage' from wet units transfering over to the short Side. this water-runn-off is channelled into the recovery tank, and the water-level of the recovery tank drains off to a drainage pipe, above the water-line.
so yes there will be manageble 'water-loss', which should be taken into consideration, but is not prohibitive by any means.
During the Reloading Process, both tanks are completely sealed off from the outside by the Tank Doors on top, and the door between the two is unsealed.
There is minimal water transfer during this phase, because of the pressure-equilibrium formed between the tanks.
The actual ammount of water that is transfered varies on the air-content, and respective compressibility factor of the water in the secondary tank, and the pressure exterted on it by the water in the primary tank. In any case, it is a small, and managable ammount of water transfer.
Resistive Force on the B-Unit requires us to impart an increased ammount of energy in order to expand the unit, and have a positive B-Value. (negative weight)
In theory, we could store this energy, and extract it out at the end of the "lift", before reaching the top of the buoyant fluid.
This would be Energy in = Energy out, minus losses.
Now,. look down, because you have just increased your potential energy within the gravitational field by a factor of X
and it is about to be converted to kenetic energy as you fall.
(moment of silence)................
Resistive Force is exerted on all sides of the B-Unit, as a function of ambient fluid pressure.
We negate the force on the sides, and consider only the 'effective' radial force vectors that affect input energy.
mostly Top, and Bottom
and from the surface area of the B-Unit, we can determine the work force for a theoretical input energy.
In practice this input energy is greater, due to inefficiencies in the system.
Potential & Kenetic Energy in Force Vector Power Generation:::
The potential energy at any point in a vector force field, such as a magnetic, gravitational, or buoyant field, is determined by the distance between the start and ending points of its motion through the field. In a gravitational field we call this "height".
E = mgh can also be represented as E = b'ad
where:
b' is the objects "buoyancy factor" within the fluid, which is always taken as an |absolute value|
a is the accelleration factor of the field, this is a dynamic force, this becomes more apparent when working with dense fluids. in air, it is marginal at low altitudes.
d is the distance between two points within the field.
This Kenetic energy cannot be converted 100% without inhibiting motion through the field, except in the unique instance of 'free-fall' impact - which is not considered here.
We therefore, consider the minimumal contraints of required force, necessary to move the B-Unit distance x, over time t.
subtract this from the available field force acceleration (a)
And are left with the available production factor.
This production factor is the interfacing component between the buoyant force and the actual power generation.
which is again completely different system, we'll touch more on that later.
heres a visual of the potential energy scale through a vector force field.
Drawing AppC3:
If we consider an ideal 'perfect' electromagnet.
Where input energy, relates exactly to the energy contained within the field.
An object being drawn into the field has a thermodynamic relationship (conservatism) within the force vector. What i mean by this, is it requires just as much energy to pull the object back out of the field as you gained by allowing it to be attracted to the magnet.
Now, when you compare the Force is the field, as it pertains to acceleration imparted upon the object, this is directly proportional to the energy if the field (input).
However, these two values do not relate on thermdynamic grounds. They are seperate systems.
The energy obtained by 'dropping' an object through a small portion of the field is tiny compared to the vast ammounts of electrical energy required to create and sustain an EMF strong enough to move that object.
While in theory, you could drop an infinite number of small objects, covering every point throughout the EM-field, and conserve all of the energy of the field.... This poses mechanical difficulties, which make it impracticle.
@sm0ky2;
Nope, it won't work. An object is buoyed up by the weight of the fluid
displaced...but to expand an object at depth requires one to lift the
entire weight of the fluid column above it. Fluidics works just like levers.
By the way, I like using a free piston in a cylinder rather than lenticular
metal disks. Vacuum is better than a gas because it has no expansion
coefficients in either pressure or temperature. It makes for easier calcs.
And liquid metal mercury is funner to think about than H2O.
:S:MarkSCoffman
The solution to this is to utilize the energy of "static force fields".
These are fields that exist in our environment, which do not require constant energy input to sustain them.
Such fields are:
permanent magnetic fields, gravitational fields, and buoyant fields.
there are others, but lets just stick to the basics here.
magnetism is not used here, with the exception of motors, which are a well established technology.
The two static fields that are used in the BCG are buoyancy in water (or other fluid), and the earth's gravitational field.
I just wanna comment the initial ideas in this thread. It will take energy to push the "spheres" of air inside the bottom of the big tank.
Leak of water is just an engeneering problem and cannot be used to deny that this idea will work. If we only focus on the forces step by step in one complete loop it would be seen clearer. The path inside the big tank could also be a sealed plastic membrane, or a very flexible hose (It will be friction, but look away from friction at this point), which alows to freely transform as the spheres rises inside the tank - then the problem of any leak will be solved.
I think however it will not work, but maybe. It will at least take energy to push the spheres the first half way into the bottom of the big tank. You might get free energy provided by the water pressure to push the remaining half sphere into the tank, so hopefully the sum of energy required to completely put the sphere of air inside the tank will be zero. Then the buoyancy effect will take it from there and make an inifinte loop of rotation...
The "spheres" or whatever items used to rise inside the water tank will always be the weight of displaced water lighter than the items that is outside. So these rising items could be made of anything from air to lead - doesnt matter.
I'll make drawings of some ideas that had been interesting to test out.
Try it, it is an interesting idea to explore.
Vidar
I had a similar half baked thought to the original idea of this thread. Putting a pipe with no water straight up and down in it into a tank of water so the buoyancy weight would drop to the bottom using the speed of gravity to allowing it to shoot buoyancy weight out of the air filled pipe. Then somehow have a separate wheel that would capture the buoyant weight to turn the wheel. Then it would need a surface wheel would sweep the buoyancy weight back into the pipe. My original thought was dropping a buoyancy weight down a pipe in water allowing it to shoot out the bottom and it would then float to the surface.
lets get something straight, even if you managed to get it right, which you wont, you would have barely enough energy to make it work, you would have little change upon the current power sources. nuclear energy tech alone would rip you a new ass-hole. sometimes you just don't see the full picture.
I am sorry, anyone who plays with more energy than you becomes closer at becoming god like. it is based on how much energy you can play with. I think you need to higher your standards greatly.
you wouldn't want somebody like me to beat you to it would you?
Jerry 8)
I think if it works even a little it would make a nice novelty item.
Interesting work. Thanks for sharing it Smokey.
Systems such as the system you propose (using buoyancy and gravity) are fascinating.
I made a modest stab at such a system myself, but abandoned it because the mathematics of the system indicated it could not constitute a useful machine.
Specifically, I once considered using castor oil as a working fluid in two linked tanks (the primary tank containing seawater so essentially buoyancy and gravity as with your invention).
I abandoned the idea because oil was to fall onto a turbine in a mostly empty cylinder, and then float up to the surface of a connected neighbouring cylinder filled with seawater (due to its relatively lower density...961 kg/m3 for castor oil and 1020 kg/m3 for seawater).
However the energy 'gained' through oil floating to the surface of the seawater was 'lost' later on because castor oil (compared to seawater) is a less dense working fluid and therefore the same flow rate provided less Force in Newtons to the turbine.
It all seemed to balance out even before taking into account system inefficiencies and flow rate/viscosity issues, so I abandoned the idea completely.
In passing, I had planned to use compressed air to move castor oil from the turbine tank back into the base of the primary tank where it would float to the surface and be 'siphoned' into the turbine tank.
Turning again to your invention, Low-Q makes an interesting objection that may have traction, namely the idea that work must be done to swing the buoyant objects into the primary tank from beneath. That makes some sense as a candidate objection and I would be most interested in your views on this.
Certainly we must take into account all forces and torques, even forces we may not want to take into account ;D
Thanks so much for sharing your work.
Interesting and thoughtful.
ok, we shall now adress the concerns noted during the Reload Phase of the cycle. I have seen mentioned a few times now, the energy requirements to pass the B-unit into the lower portion of the large tank ; this leads me to believe that i was not clear about the state of the device during this phase.
Reload Phase:::
After all B-units have risen to the top of the bouyancy chamber, generated power on their "rise" and "fall" cycles, and returned to the bottom of the small tank, the B-units are returned to their "0-state".
Meaning, that within the working fluid (water) they are at a state of weightlessness.
This is achieved by depressing the inner solenoid and compressing the halves of the B-unit, such that the verticle bouyant force = 0
Both tank lids are sealed during the reloading phase, therefore the tank pressure is equalized between both large and small tanks, prior to the inner door opening.
With pressure sabalized, the inner door open, and all B-units at their "0-state", the energy required to pass the B-units from the reloading chamber, into their starting position is negligible. There is no resisting force required to move them.
Once the B-units are in place, the inner door is sealed again, and the tank lids are opened - During this stage, as the tank lids are opened the large tank will be at a slightly less-than normal pressure, and "repressurize" to atmospheric pressure. The lower tank will be at a slightly higher than normal pressure, and de-pressurize back to ambient pressure. This is where a small ammount of water is transfered between the two tanks.
This is caused by the compressability of air at the top of the lower tank while the lid is closed, and the inner door opened. (Reload phase)
This takes the system back to the initial "starting" phase.
B-units are at the "0-State", virtual weightlessness.
The first units' solenoid causes the B-unit to expand, becomming bouyant and the generation process begins once again.
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So, to clarifiy, there is no significant energy requirements to force the bouyant chambers under the water,
because they are at a zero-bouyancy state during the reloading phase.
Thanks for explaining the reload phase Smoky.
I would find it easier to understand the reload phase and indeed the entire system if we had a mathematical model.
A hypothetical machine of specified dimensions. This would enable us to calculate all forces acting on the system and also, we would know the power output in watts of the system.
I know comprehensive mathematical models of thermodynamic systems can be difficult to construct, but if you were to provide us with the dimensions (the size of each component), we could have a go at preparing a mathematical model for discussion and analysis.
It is an interesting system and it would be a thousand pities if we failed to examine it carefully using well established equations of motion and fluid dynamics.
So please let us have some system specifications (how big do you want this machine to be) so we can prepare a mathematical model taking into account gravity, buoyancy, viscosity, B unit relative smoothness, temperature, density, air resistance at operating temperature, power output in watts and power input in watts for any parts that require assistance during the power generating cycle.
This is not to say that our mathematical model will be absolutely correct right away (often there are forces one has failed to take into account) but it should be a good starting point for a more detailed analysis of the machine.
Over to you Smoky. Please provide dimensions for your system (simply the size of each component and the materials from which the components are constructed).
For example, water may have various densities in kg/m3. Would you like us to assume it uses fresh water of density 1000 kg/m3 or seawater of density 1020 or 1030 kg/m3?
It can be as big or small as you like, but mathematical focus requires us to have dimensions and materials information.
Thanks for explaining the reload phase of the cycle.
Kind regards,
By way of example of required specifications, please let me know:
1. The diameter and height in metres of the primary tank and lower vessels.
2. The density in kg/m3 of the water in the primary/lower vessels. Also the flow rate in m3/s of any water lost from the unit (during phase transition) which will have to be replaced.
3. The material from which the B units are made (particularly the density and elasticity of the material) as well as their dimensions when 'inflated' and 'deflated', so we can calculate area in square metres and volume in cubic metres during each phase and the pressure in Pascals (= Newtons per square metre) and therefore the force in Newtons required to 'inflate' them.
4. The precise specifications of the motor powering the robot arm (power consumption in watts, angular velocity in radians per second or RPM, high torque or low torque, stepper motor, conventional motor etc).
5. The material from which the robot arm is manufactured so as to calculate forces applied to the robot arm motor during operation when moving B units and when repositioning itself to move other B units when not under full load. By what mechanism is the inner door opened and closed?
6. The speed in m/s of the B units during their up and down cycles.
7. The materials from which the connecting track (the track on which the B units travel) is made and the dimensions of the connecting track (to calculate weight when in and out of the water).
8. The total length of the track on which the B units will travel.
9. The mechanism by which the B units will be attached to the track on which they travel (eg are there any metal parts whose weight must be taken into account).
10. If the B units travel at different speeds (during the up and down cycles) how is this achieved by way of a single connecting track to which the B units are attached? Is there more than one track, in which event what are the specifications of the second track.
11. If the track is a loop, what is the estimated angular velocity of the loop (or in RPM how many times does the circuit of B units complete a full rotation in one minute?).
11. How is the motive force of the B units transferred to a shaft, pulley or other power transfer mechanism?
12. For the dimensions of the machine you will select, what type of alternator motor would you like to connect to the system (in terms of operating RPM, required torque in N.m and output in kW?
There will be other questions, but this is a preliminary list to give an idea of the information we will need to calculate the power output in watts of the machine.
The power output can then be compared with power consumed by the robot arm motor, air resistance, water resistance, other friction etc to check whether we can make a net gain in energy output (which is the whole point of the machine).
With full disclosure this may be calculated and discussed. If you can only provide some of this information, please provide as much as you can and I will do the calculations based on the information provided and best estimates for the missing parameters.
Kind regards,
Nicely said QT,
i will do my best to define as many of those things as possible.
The "test unit" B-Units have all been roughly the same
the interior low-pressure chamber has a volume of
~0.07 Cubic Meters - this is based solely on test units, which have been constructed with a set parameter of low-pressure chamber-seals and inner hydraulic cylinders
The exterior size of the B-unit affects its mass, but not its relative bouyancy. The reason for this, is that the working fluid acts directly on the low-pressure chamber by design. This allows for an unrestricted freedom when choosing an exterior housing.
The most promising of these housings, will be composed of 3 layers of hexagon-plates, that interlock in a way that the entire casing becomes smaller as the B-Unit contracts, and becomes larger as it expands. i'll post the info on that as it progresses.
The size and dimensions of the Two tanks are determined by several system requirements, but are completely variable to suit the needs of the system.
There are certain perameters that must be met. The tank lids must seal at or below the specified water level, to minimize water transfer.
and the tank lids and inner door must meet the safety standards of the pressure imparted onto the lower tank when the inner door is opened.
The individual B-units ride along guide-rails that form a track through the system. the top of these rails are disengaged to accomodate the tank sealing. - this action can be a part of the tank-lid closing.
the interface for power generation might take one of several different forms.
a Coil / Magnet interface would allow for direct power generation, or possibly a geared mechanism to drive a rotary generator, connectimg arms that could attach to the B-Units during their entire cycle or at selected times for power generation. There are advantages and disadvantages to each or other approaches, depending on the needs of the system, and each will have its own associated energy costs and electrical production rates.
The phsycal force available, from gravity should be a simple calculation.
the Bouyant force, however, is completely variable, and user defined.
Displacement of the low-pressure chamber is roughly 85% of the ideal state under tap-water, and 82% in air.
giving us an underwater displacement of 0.0595 cubic meters at full expansion.
this may vary slightly with the density of the working fluid.
and potentially increase with technological advancement, such as the aforementioned exterior casing changes.
the inner pressure of the chamber can be calculated by the force to mass ratio that was used to determine the "effective displacement" .
essentially, even though the B-unit takes up more volume,. it is only displacing 0.0595 m^3 of water-mass. if that makes sense....
this is due to the design, that allows the exterior of the low-pressure chanber to "breathe", regardless of the exterior casing design.
even in the flying-hovercraft models, the B-unit still must breathe in this manner for the displaced-mass-bouyancy mechanics to function as intended.
The low-pressure chamber is not a true "vacuum". and the inner pressure is a function of the change in inner volume, and the elasticity of the low-pressure chamber seal. (rubber is used currently)
As far as the function of the opening/closing of the tanks and inner door, these can be the physical operation of the last B-unit during its' ascent, or decent, or a portion of the energy generated can be used for these functions.
the total power generated will be a function of the bouyancy force used in the b=units, and the height of the large tank.
the height of the small tank is a fuction of the total number of B-units being used, and the space requirements needed to accomodate them.
The B-units will be at a Full-Negative Bouyant State during their descent, and thus gravity has its maximum acceleration force available, dissapated as a function of the downward power generation aperatus and linking mechanism.
To increase the speed and/or power of the upwards generation, you simple raise the B-unit to a higher bouyant state.
The aperatus to move B-units from the smaller chamber to the larger one, while the tanks are sealed (reload), has not been defined yet. But as a comparative example, we could use the stepper motor and arm mechanism found in a commercial Dishwashing Machine (resturant)
although our actual power requirements to move the B-units may be somewhat less than those standards.
The parameters of the actual system, are likely to be defined after selecting the power generation system and linking mechanisms.
from there you can determine your power requirements on the up and down strokes, and well as any additional moving track components that may add more drain on the system.
from there you can determine the bouyant force required with the given working-fluid, and subsequent Positive-Bouyant-State that you must program the B-unit to change to during the start of the "rise" cycle.
Also, the physical size of the B-unit can be scaled to meet virtualy any requirements.
hovercraft will carry a much larger B-unit than the test models.
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the energy available vs the energy required to initiate a bouyant state are two entirely different systems.
Examine the mass of a diver, fully strapped with gear, soaking wet, submerged under 80 feet of ocean water. and how much energy it would require to launch this diver to the surface at ~12 feet per second. This can be easily converted into a value of "displaced mass" needed to accomplish this task.
This value is then used to determine the size of the emergency life-vest, and its associated standard-pressure CO2 cartridge.
now,. compare the energy-value of the CO2 cartridge, to the energy of lifting the diver you just calculated a minute ago....
I understand the importance of hammering out all of the known energy values at each stage of this type of system,
but you must know what values you are attempting to compare.
because just like the diver and his pull-chord vest, that he hopes he never has to use....
The Bouyancy-Based Power Generation technology uses this vast difference in energy to produce a positive gain.
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Heres another example, place a rubber-bulbed eyedropped full of air into a plastic bottle full of water and screw on the cap.
now set-up a lever system to push on te side of the bottle with equal force each time.
and you push and the eye dropped sinks, not just sinks, but sinks with force, kerklunk on the bottom.
release the pressure and it shoots up to the top, again with force.
you are pressing lightly and only for a short time, after that you just hold the lever still and it maintains the pressure change in the bottle.
but the eye-dropper is able to produce usable energy both down and up, hitting the bottom with force, and bobbing at the top with force.
plus the pressure in the bottle returns your input energy back through the lever when you release it.
This shows that the two energy systems are not related, although they are interlinked.
one changes the pressure, and thereby the displacement
and the other is the bouyant result of that displacement.
Two entirely different systems. energy in and out does not correlate.
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Now with this in mind,. we can select arbitrary dimensions, fluid densities, Generator resistances, ect.
the mass of the B-unit can be defined as needed, the only restraint is that its mass must correspond to the range of B-states.
meaning the from the Zero-B-state to the Maximum displacement, within the given fluid pressures. If the total-mass of the B-unit exceeds the displaced mass, it will not reach a Zero-B-State.
Height of the lower tank, number of B-units, ect. will determine maximum system capabilities, but the for calculations, one B-unit will suffice.
physics implies that the linear translation of the B-unit costs no energy, but in practice we must account for water-resistance, gravitational effects of its bouyant-state as friction on the track-rails.
so through the reloading process, as well as the linear-translation necessary at the top of the track, we must account for this as resistance (load) on our robotic arms, this will define the parameters of the motors needed, ect.
to make it simple, you could theorize a 5-volt generation system, that directly drives all the components in the system, the small motors, the controller circuits, any meters, sensors, timers, ect.
energy costs to move, seal, and lock the upper tank lids is a big unknown... there are not many systems like this in use to use as an example. i have seem some that use this principle, but the only one that comes to mind, is an aquatic reserve, that has an upper-pool linked to a lower pool. and to transfer animals from one to another the tanks must be sealed.
theres the weight of the lids, thickness, tensil-strength, ect. and rubber seal? pressurized locking mechanisms??
friction, resistance from the water.. so many unknowns there..
we can start with just defining the outlying parametres and requirements, then build this thing from there, adding / altering each component in the system and its energy drain on the total generating power available to us.
I have chosen parameters for your invention to get some feel for the power output during the downward cycle of the B units.
Primary tank height = 30m
Diameter = 1m
Contains fresh water of density 1000 kg/m3
Volume = 23.56 m3
B units x 200 (= 100 B units on each side of the system connected by a loop of buoyant lightweight material).
Let us also assume the diameter of the circle represented by all the lightweight buoyant material holding the B units together in a chain has a circumference of 314.159m.
If the circumference of the chain holding the B units has a circumference of 314.159m, then:
C= 314.159m
C = 2 x pi x r
314.159 = 2 x pi x r
314.159 = 2 x pi x r
314.159 = 6.28319 x r
r = 314.159 / 6.28319
radius = 50m
diameter = 100m
Downward cycle of the B units
We must assign a weight to each B unit if we are to calculate the force that will be applied by the B units in the all important downward part of the cycle.
Too much weight and we will slow the machine down on the upward buoyant cycle. Too little weight and the machine will not generate much energy on the downward part of the cycle.
If each B unit weighs 1 kg, then having 100 B units pulling downwards due to gravity on the right hand side of the primary tank at any one time will amount to a weight of 100kg moving downwards due to gravity.
Let us assume for the moment the system can move at enormous speed, and that all 100 of the B units can descend downwards to the base of the system within one second.
This would mean that one revolution of the entire chain of B units (including those moving upwards due to buoyancy) would take place every two seconds. I know this may be difficult to achieve in practice, but lets go for gold here.
If one complete revolution of the B units takes 2 seconds = 30 revolutions per minute (30 RPM)
On the downward part of the cycle, the following energy would be generated due to gravity:
Force = mass (kg/s) x acceleration m/s/s
F = m*a
F = 100kg/s x 9.81 m/s/s
F = 981 Newtons
Applying this force in Newtons and converting it to mechanical power output in watts for the downward part of the cycle, using an efficiency co-efficient of 0.85 to take B unit friction (and only B unit chain friction) into account:
P mech (watts) = Force (981N) x pi x 0.85 (eff) x 30rpm x 100m (diameter B chain) / 60
Pmech watts = 130.98 kW
So this would be the power output at 30RPM if the system could rotate without having to fight through the primary tank (if it could maintain this mass flow circulation without needing buoyancy to lift the B units).
It should be noted that the assumptions made include an assumption of high speed in the B unit chain (30RPM is a high speed) and that in practice the overall angular velocity of the B units (as they ascend through the water in the primary tank) must be lower.
How much lower remains to be calculated, but that figure will affect our RPM and therefore our cyclical power output.
Now we need to do the maths relating to the upward cycle of the B units.
What shape are the B units? They should be shaped to reduce water resistance on the upward cycle (air resistance will be negligible in the downward cycle but we can calculate all of these things).
How can the B units enter the base of the primary tank without causing water loss? The base of the primary tank will also be under pressure.
Pressure at the base of the primary cylinder is as follows:
P = h x g x d
P = height(m) x gravity(9.81 m/s/s) x density (kg/m3)
P = 30m x 9.81 m/s/s x 1000 kg/m3
P = 294,300 Pascals Gauge
Pressure due to the height of the column of fresh water in the primary tank (294,300 Pa Gauge) + atmospheric pressure (101,325 Pa) = 395625 Pa Absolute, so it will rush outwards forcefully unless prevented from doing so. This is just under 4 atmospheres of pressure.
A candidate solution (to the problem of how the B units can enter the base of the primary tank without water loss) is this.
An airlock at the base of the primary tank could be pressurised using a float activated air compressor to maintain a pressure of 500,000 Pascals inside the air lock. The compressor would only activate if water entered the air lock.
There would be two flaps in the airlock. An outer and inner gate.
The B units would enter the airlock from outside the primary tank. The outer flap (like a cat flap) would close when the B unit entered to keep air pressure as high as possible.
Then the inner flap of the airlock would open allowing the B unit to enter the high pressure water at the base of the primary cylinder.
The water would be unable to flow out into the airlock chamber because the pressure in the airlock would be maintained to 500,000 Pa by the air compressor.
The Abac Genesis air compressor consumes 11kW, but produces 800,000 Pa at a rate of about 0.02 m3/s.
0.02 m3/s is equivalent to 20 cubic litres per second, so the right configuration of B units just might work (though I have not done any buoyancy maths yet).
So the B units should be able to enter the primary tank through an airlock without water escaping.
The water at pressure of just under 400,000 Pa would not be able to escape from the primary tank into an airlock pressurised to 500,000 Pa. Simply not possible.
The output nozzle of the air compressor might also be directed to inject air into the B units if needed to give them extra buoyancy. Again I have not looked at buoyancy maths yet, but this is a candidate solution to the water loss problem.
theres no real need to complicate the reload procedure....
instead of an airlock, we can just pressurize the smaller tank, prior to opening the inner door.
we can compare the energy it takes to pressurize the smaller tank,
with the energy it takes to pump the small ammount of water back to the top of the larger tank. Whichever is smaller, would be the most efficient method of dealing with water loss.
another tactic would be to allow for an air-gap at the top of the larger tank. Since both tanks are at a pressure-equibrillium, the smaller tank would be more resistant to further compression of the water, and the water displaced by the B-units would compress the air at the top of the large tank. this will further decrease the water lost in the process.
I agree. It makes no difference whether you use an air lock or pressurise the smaller tank. There should be no water loss if this is done correctly. Both ideas amount to the same thing with different labels.
However, you probably have to leave the primary tank open to atmospheric pressure. If you seal it, the first problem is that the chain keeping the B units together will have to leave the tank through some sort of seal (I think this would be difficult if not impossible).
But my main objection to sealing the primary tank is that there is no reason to seal it, and pumping pressure into sealed tanks is a bad idea if there is no reason for it.
Leaving the primary tank open to atmospheric pressure would be fine. Lower pressure in this tank could not be a bad thing.
You mention pumping water up to the top of the primary tank from the base. This will consume franchise sized amounts of energy. The general rule with pumping water upwards is that you will spend 5 times as much energy pumping it uphill than you will ever get out of it at the bottom of the hill. So anything and I mean anything is better than pumping water uphill in an energy generator.
But this is all detail. My main concern is the speed in m/s at which the buoyant B units will ascend the primary tank.
The speed, when we do the maths, could be as low as 0.5 m/s, in which event the RPM will be low, and therefore the power output will be low.
How would you propose to try and increase the upward velocity of the B units to prevent low RPM?
The circuit of B units is a circuit, and thus what might have been high downward velocity due to gravity will be retarded by low upward velocity due to buoyancy and resistance. It will be upward velocity due to buoyancy and low velocity due to resistance.
This is the key to whether or not the system is viable. Upward speed of the B units against the resistance of the water.
It really is all about buoyancy velocity in terms of power output.
You can remove the problem of slow upward velocity.
Timing how long it takes for the B units to rise in the primary tank, (let us say it takes the B units 5 times as long to rise as it does for them to fall when unconnected to the rising units) you can use a greater number of chains of B units, and connect only the downward moving belts of units to the driveshaft.
In other words, all the myriad chains of B units are constantly moving up and down, but only downward moving units are connected to the drive-shaft.
Upward moving units still float slowly up, but are disconnected from the drive-shaft when they are doing so. This prevents the floaters slowing down the fast movers (the gravity units).
By calculating the time difference between rising and falling B units (before they are connected together) simply add extra chains of B units and configure the system so that at least one set of B units is continuously falling (when connected to the drive shaft) at the same time as x number of slowly ascending B units are in various stages of 'getting to the top' of the cylinder and ready to fall.
If the time lag is x 5, then five times as many are needed. If the relative time lag is 7 units of time, then 7 times as many unit chains are needed etc.
There would be a continuous downward avalanche of B units.
Then you only get downward gravity action on the drive-shaft, and high RPM (provided a mechanism to disengage all ascending units 'from' and engage all descending units 'to' the drive-shaft can be engineered).
It would be like a gear box that engages only downward moving belts of B units.
Sometimes I amaze even myself ;D
PS I multiplied by gravity twice in my earlier Pmech equation having already used gravity to calculate force using Newton's equation. I don't think that's right. I must stop improvising equations. To solve for the error, divide 130kw by 9.81 m/s/s = 13.25kW Not sure any more. Too tired.
While, knowing the mass-displacement of the bouyancy chamber, while in the working fluid, at a known pressure CAN be calculated.
In practice, the bouyant-force is measured, then the calculations are performed in reverse, to obtain the "effective displacement" volume.
This is because of the unknown pressure expansion curve of the low-pressure chamber seal. At different inner and outer pressures, rate and amplitude of expansion, the inner seal expands differently.
[note, this is a convex expansion function which can vary greatly with the use of different materials]
The B-unit must be first placed in the fluid, and the automated Zero-B state is iniated. This creates a "weightless" condition, and from there the B-unit scales the Bouyant state: positive (up force) and negative (down force gravity). We can measure bouyant force in either verticle direction at any desired state. With the measured verticle force, we can establish the displacement with respect to the mass of the B-unit, and its initial volume displacement. It was mentioned before, the energy requirements to run the systems of the B-unit itself. To answer that question, the circuitry and logic chips used in the B-unit control systems consume less energy than a digital wristwatch.
The two power consuming functions are the wireless transmitter (optional) and the hydraulics used to alter the bouyant-state.
The latter accounting for 99% of the energy input, and also presents an engineering hurdle that must be overcome.
the test units are not large enough to carry their own hydraulics in air.
The mass per unit volume of a denser fluid such as water make this problem dissapear, even at test unit size, however, it would be more beneficial to displace additional fluid-mass to lift the additional mass and still provide usable bouyant force. that will get rid of the hose that tethers the B-unit's caliper to the ground-based main cylinder.
There is much room for design improvements, such as the use of plastic cylinders, or a lightweight aluminum tubing that has been investigated. low-density hydraulic fluids, ect.
The worm-gear actuator has been recently improved, it now uses a 5v motor, and is much smaller / lighter
the original test unit used a 12v car window motor and actuator, to depress and retract, the hydraulic lever using a 6v util bat.
Now this is done using a wormgear on a much smaller compound-lever, lowering the size of the motor/actuator
and its energy use/ battery size.
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The actual power generation should take place with each invididual unit. This eliminates a majority of the timing concerns, and cooridinating all parts of the system. Link the interface to a large flywheel, and a 1-way "freewheel" greaing to a much smaller driveshaft. simlar to a bicycle.... the moving B-unit gets the flyhweel moving, and this power is translated to the generator shaft at much faster RPM. additional flywheel mass can be added to the shaft itself, which keep the generator in operation constantly, although the B-units are adding energy periodically.
Think of those toys, where you pull a zip-string, and the driveshaft spins, then a freewheel gear and spring combination wind the zip-cord back up.
something kind of like that, but more cyclical, probably two-stage.
up and down.
the up-gear would be larger, to make use of the slower speeds and greater upward (bouyancy dependent) torque,
the down-gear would be smaller than the up, to make use of gravity's constant speed, and mass-dependent torque.
Using that type of generation interface, then you can control the number of units rising and falling independently, as well as the unload and reload phases. That may be one benefit of an "air-lock", that you can separate x-number of B-Units, to reload, without disrupting the main track with upper tank seals. Just pump the water back to the main tank, as necessary.
If the generation interface were to be a continious loop it could be geared like a bicycle-chain, with a clutch that releases during the unload-reload phases, and changes to a larger gear during the "up" phase.
so at any given time, you can have whatever number of up-units, and down-units adding energy to the driveshaft. in theory, you could just store whatever ammount of energy into the mass of the fly-wheel, and generate power at a constant rate via the drive shaft.
Here is an example of how something like this could be done.
a mechanism to engage/disengage the interface with any individual B-Unit can link the units to a belt or chain, possible multiple B-units driving a single chain, but not necessarily.
as the chain goes around the loop, it drives a geared wheel, which in turn rotates a much smaller driveshaft. multiple drive chains exert power onto the driveshaft at any given time, and this is translated to a flywheel.
a cone-belt transmission on the driveshaft, allows power to be taken from the Flywheel at a controllable rate, regardess of the fly-wheels velocity.
so a clutch at the top and bottom of the loop will disengage the slower "up" gear, allow for the horizontal transition, then engage the faster "down gear", B-unit falls, the lower clutch disengages the down-gear, and the B-unit is reloaded, then clutch engages the up-gear again.
this can be done with a notch-type design cut into the upper and lower track, that guide the chains onto and off of a set of 3 gears, up, down and freespin. i drew a crude picture of some of these possible components...
in the example with the scuba diver, and his emergency lifevest
That little CO2 cartridge is packing approx ~ 881.3 Joules of energy
take your boat, out into the ocean, 80-90 feet or so.
strap a bunch of heavy gear to your body, and swim down the the bottom,
Now give your friend your favorite wenching system, loaded with 881.3 Joules of energy.
and you will understand why scuba gear now includes a BCD (bouyancy control device).
dont forget to pull the chord on your CO2 cartridge lifevest, before you drown.
Are you building your machine smoky?
I am waiting for a mathematician's feedback in relation to my own machine before a build attempt. Makes sense to do the maths first don't you think?
Quote from: quantumtangles on June 11, 2011, 04:57:27 PM
Are you building your machine smoky?
I am waiting for a mathematician's feedback in relation to my own machine before a build attempt. Makes sense to do the maths first don't you think?
the B-units i built several years ago flew in air, carrying ther own hydraulic cylinder (not the pump), they work.
Underwater, the bouyancy mathematics are the same, and the logic-chip that controls the B-Unit works in water or air (or any fluid/gas).
there was problems with tensil-strength and outside pressure that collapsed the water unit in on itself, so the unit's casing has to be redesigned to make it stronger. Creating a low-pressure chamber in air is a lot easier than doing it underwater, but it can be done. I have seen floating concrete blocks what were "evacuated" hollow centers, plugged with rubber. the weight of the B-unit is not the problem, and the pressure needed for the hydraulics to expand it is still in range of the pumps we used.
its the casing that needs to withstand the waterpressure, and inner low-pressure.
the math with the energy involved is straight forward.
the math with pressure inside/out and tensil strength of the casing goes far beyond my bachelor's engineering education... thats going to require some professional assistance to do the math on that one. Probably the same guys that design Submarines.
i found a work-around solution that can withstand far greater pressures than are required, but as for the math on that i dont know enough to even try..
i can do water pressure at depth-X, but when you add in a low pressure chamber it gets far more complicated. because there is not outward pressure on the inner surface of the chamber, like there is with a pressurized-gas chamber.
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In air, we work in terms of Bars, 1 bar for a collapsed B-unit, 2 Bars when it is expanded (maximum bouyant state).
A 0-B state is approx. 1.3 Bars
or in terms of the hydraulic system 130,000 Pa. which is roughly 1/10th of the pressure applied to the brakes on your car when you stop. This is for a low-pressure chamber volume of 0.07 m^3 .
This creates a weightless state, and is (roughly) the starting point of the logic-chip controller when it boots up, or the command is given to return to the 0-B state.
Larger B-units will require more hydraulic pressure to achieve bouyancy. The technology was originally designed for use in a flying craft, but there wasnt enough interest in the device to continue development.
Relating the pressure to actual input energy, in terms of electricity from the battery, is better measured than calculated. because of all the components involved in the electric motor, lever, hydraulic actuator, master-cylinder compression ratio, ect., It can be calculated using standard conversion for pressure, but in practice, the system experiences great losses, and thus the battery consumption is far more than the energy to create said pressure. This, as i noted previously, does not relate directly to bouyant force, created by the device.
the logic system is 5v at very low current, and its energy usage has been completely ignored in terms of "input energy", because the value is somewhat negligible.
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My interest in using this as an energy generation system was derrived from a simple equation: mgh
once this balances out with the energy of the hydraulic system:
leveraged force x distance of lever depression
= (mass of the B-unit) x (9.8 m/s) x (height)
everything else is essentially "free energy".
Bouyancy continues indefinitely, within the limitations of our atmospheric pressure, which is theoretically over a mile high.
From looking at your drawings, it looks to me as this has already been built. Here is the video I am referring to:
http://youtu.be/UGPEOlQ2_uA
will try to embed...
<iframe width="425" height="349" src="http://www.youtube.com/embed/UGPEOlQ2_uA" frameborder="0" allowfullscreen></iframe>
hey smoke, thought this may be of interest, was watching myth busters and saw and interesting fact that actually changes the perception of many vintage bouancy machines, you know the one where hollow balls are fed into the bottom of a tank, weel they were all debunked because it was said you could not get the balls into the bottom of the tank for the same reason they float up in the tank, mythbusters proved this false, not intentionally regarding the machines, but the pumped ping pong balls down through the water into a sunken boat to see if it would float like the 70 year old donald duck cartoon and it worked refloating the boat, but they pumped the balls down with water and used the viscosity and cohesion of the water holding onto the balls, you see because they were in a tube they were only subject to the pressure within the tube not the whole ocean, and as it was mosttly filled with air (the balls) the pump was quite small and managed easilly. interesting side information from their film that may be of some use in designing such machines, hope you can use it
Archer
ps the viscosity,cohesion and pressue variance regarding the tube is just my analisys of what was happening it iis not in the film, so it may not be exactly the reason, but my physics and logic says that appears most likely why it works
There are so-called magnetic fluids. This is used in separators of non-magnetic materials.when a magnetic field is applied to a magnetic fluid, its density increases.the body at the bottom floats up.
R. Rozencveig speaks about it
How much energy needs to be spent to create a field and how much will we get from increasing the potential energy of the body?