Buoyancy-Based Power Generation - Full Disclosure

[sm0ky2]

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.

[sm0ky2]

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.

[sm0ky2]

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:

[sm0ky2]

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.

[mscoffman]

@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