Pair of Pairs

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re: Pair of Pairs

Post by Michael »

Either the 4th Law of Motion is real and my interpretation of it is basically correct. Or, Johann Bessler spent about two decades perpetrating one of the biggest frauds in the history mankind!
I think John Worrel Keely probably has the reigning title for money made and time spent on that one.
Of course there's also Dennis Lee...Joeseph Newman...Tom Beardon...

By your law Ken your saying that eventually all the weight will loose mass and become nonexistent right?
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re: Pair of Pairs

Post by ken_behrendt »

Michael asked:
By your law Ken your saying that eventually all the weight will loose mass and become nonexistent right?
Not exactly. Whenever an object drops in a gravitational field, it loses a tiny bit of its mass that is converted into the kinetic energy that the object acquires. And, on the other hand, whenever an object is hurled upward in a gravitational field, the kinetic energy imparted to it is drained off and then converted back into mass inside of the subatomic particles of the object.

I'm assuming in this example, that the gravity field is provided by some sort of airless body like an asteroid or moon. On Earth, much of the kinetic energy that shows up in a dropping object near the planet's surface is lost in moving air molecules out of the way so that only a portion of the energy gotten from the loss of its mass goes into increasing the velocity of the object. The same applies when an object is hurled up into the air surrounding the Earth. Much of its initial kinetic energy is wasted moving air molecules out of the way so that only a portion goes into increasing the mass of the object.

Now in the case of a continuously running, chronically overbalanced gravity wheel, since the average vertical velocity of its rotating weights is always downward, these weights will continuously lose a very, very tiny percentage of their mass with each wheel rotation.

Over time, the wheel's weights will have lost enough mass and weight so that the torque they provide the wheel with is insufficient to overcome the air resistance and bearing friction acting on the wheel's parts. At this point, the wheel will stop rotating...it would have long ago ceased being able to do any additional external useful work in its environment.

What would happen if, at this point, one was to open the gravity wheel's drum up and extract some of the "depleted" weights? Well, their are two possible observations that might occur.

a.) The weights, even if made of lead, would feel as light as feathers while retaining all of their other normal physical, chemical, electrical, and magnetic properties. And, the weights would not disappear as they lost mass. They would still contain the same number of atoms and subatomic particles as they did when they where first put into the wheel (minus, of course, any radioacitve atoms whose nuclei decayed while the wheel was running). The weights would also have the exact same shape, and volume that they had when first installed in the wheel. They, thus, do not become "non-existent". It would be more like they were becoming "gravitationally invisible" as far as the gravity fields of the universe were concerned.

b.) Or, one would note that the extracted weights appeared to be completely normal with their full original weights and masses.

How could this second outcome happen?! Well, perhaps as the weights continuously lose mass to provide the energy that powers the wheel and anything attached to it, the weights can then simultaneously somehow extract energy from the motion of the Earth or our solar system or even the Milky Way galaxy itself which is then used to replenish the masses of the weights from nanosecond to nanosecond! It this proves to be the case, then such a wheel could, if its parts held up, literally run on forever.

Right now, I can not predict exactly what will happen after a Bessler type overbalanced gravity wheel is allowed to run continuously for millenia. However, after we finally manage to achieve these devices we will eventually find the answer to what will finally happen to their weights as the energy associated with their masses is continuously drained from them.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: Pair of Pairs

Post by Michael »

Respectfully Ken,
Not exactly. Whenever an object drops in a gravitational field, it loses a tiny bit of its mass that is converted into the kinetic energy that the object acquires.
Jonathan had a problem with this and I have a problem with this. Why do you think this? I'm trying to understand are you saying the amount of kinetic energy the object gains is traded off for an equal amount of energy that is normally a part of the objects latent mass?
Over time, the wheel's weights will have lost enough mass and weight so that the torque they provide the wheel with is insufficient to overcome the air resistance and bearing friction acting on the wheel's parts. At this point, the wheel will stop rotating...it would have long ago ceased being able to do any additional external useful work in its environment.
Hold on, I posted what I did before I read this part of your letter. So in fact you are saying the weights will loose mass over time, regardless if they completely disappear ( I wrote that to define the process not neccesarily whether they would reach that point because, as you say friction will stop any further motion advancement ).
a.) The weights, even if made of lead, would feel as light as feathers while retaining all of their other normal physical, chemical, electrical, and magnetic properties. And, the weights would not disappear as they lost mass. They would still contain the same number of atoms and subatomic particles as they did when they where first put into the wheel (minus, of course, any radioacitve atoms whose nuclei decayed while the wheel was running). The weights would also have the exact same shape, and volume that they had when first installed in the wheel. They, thus, do not become "non-existent". It would be more like they were becoming "gravitationally invisible" as far as the gravity fields of the universe were concerned.
Okay Ken, I don't even want to begin to touch on this other than to say, no sir-no way.
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re: Pair of Pairs

Post by ken_behrendt »

Michael asked:
Jonathan had a problem with this and I have a problem with this. Why do you think this? I'm trying to understand are you saying the amount of kinetic energy the object gains is traded off for an equal amount of energy that is normally a part of the objects latent mass?
Regardless of whether Jonathan, you, or anybody else has a "problem" with this, it is an accepted consequence of the revelations of Einsteinian relativity theory. This conversion of the mass of falling objects into the kinetic energy that subsequently builds up in them has been taught to physics students in universities for many decades now as I showed in a previous thread.

To put it as simply as possible, where there is mass, there is energy and where there is energy, there is mass. Mass and energy are, according to relatvity theory, the SAME thing! When energy suddenly appears somewhere, then mass must disappear from somewhere. And, when mass appears somewhere, energy must disappear from somewhere. Many of the mobilists on Earth right now are still using the physics of the 18th and 19th centuries...maybe it's time they "upgraded" their concepts of the physical universe to include the physics of the 20th century!


With regard to a massless object still retaining its other physical, chemical, electrical, and magnetic properties, you wrote:
Okay Ken, I don't even want to begin to touch on this other than to say, no sir-no way.
Surprisingly, aside from gravitational and inertial properties, most of the other properties of matter do not really depend that much upon the masses of the atoms and molecules from which they are constructed.

The vast majority of the properties of matter depend upon the electrostatic forces of attraction and repulsion existing between electrically charged subatomic particles in their atoms and the differences in energy between the various quantized energy levels that the particles can exist in. These, however, are not dependent upon the masses of the particles involved to any great degree and will still persist even if the subatomic particles involved become completely massless. I think that a hunk of massless lead will look much like it did before it became massless. Of course, at this point, I can only conjecture about this because I have never actually seen a piece of massless lead. But, its existence is theoretically possible.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: Pair of Pairs

Post by Michael »

Ken explore it a little with you. You know how kinetic energy works of course, right?
Whenever you double an objects velocity the amount of energy it is carrying quadruples. If you agree to this say yes.
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re: Pair of Pairs

Post by Fletcher »

Ya know Ken, when you first put this theory up months ago I thought .. following your logic Bessler could've made a fortune afterall (but not in his life time). He found the perfect way to transmute metal by altering the atomic number of lead. That's if the relationship exists between a metals atomic number & its mass which I believe I was taught at school IINM & I don't think that theory has been overturned, yet, but you're obviously working on it. If he could get his wheel to run backwards (which he could) he could change lead into gold, easy, & visa versa, though who would want to do that except a plumber & perhaps Ken.
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re: Pair of Pairs

Post by bluesgtr44 »

Man, I really wanted to leave this one pretty much alone....just can't.

Ken, we are all aware of how PPM is a closed loop. My apologies, but that is how I see your logic on this...a closed loop. You leave out so many other possibilities that it just stymies me. I cannot for the life of me see how you can remain so myopic on your approach to this...I would like to say approach(es) but, it always leads back to the same old road.

I admire your perserverance as to how Bessler was able to achieve this...I just cannot understand how you can carry it to the level of a "law". You have a theory, Ken....a theory. And you present it rather well for a forum of Besslerites.

Your "law" automatically assumes that the actual wheel or drum is doing the lifting on the ascending side...why?
so long as they keep away from the center of gravity. To this end they are enclosed in a structure of framework, and coordinated in such a way that not only are they prevented from attaining their desired equilibrium or ‘point of rest’, but they must ever seek it, thereby developing an impressive velocity which is proportional to their mass and to the dimensions of their housing.
What I get from this...the covered drum is not the enclosure...the structure or framework is inside the wheel. Coordinated...a designated path the weights follow, probably because they are encased. Prevented...why? They are in that dad gummed enclosure and they can't get out! ....they must forever seek it....the forces know where they want to go...something is stopping them, redirecting them, forcing them to traverse a different path than the one they know leads to rest.

I am not knocking your equations as it relates to the "law"....only ytour insistance that this is the only way Besslers wheel could have worked. I know there are other possible theories....theories!


Steve
Finding the right solution...is usually a function of asking the right questions. -A. Einstein
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re: Pair of Pairs

Post by Fletcher »

And so forged in the mighty mines or Mordor, the dwarfs did his bidding & created the 4th Law, more powerful than any other Law & never before the like of it seen by the mere eyes of men. One mighty Law, a Law to bind them all :)
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re: Pair of Pairs

Post by Michael »

:) Lol
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re: Pair of Pairs

Post by ovyyus »

Ken wrote:Whenever an object drops in a gravitational field, it loses a tiny bit of its mass that is converted into the kinetic energy that the object acquires. And, on the other hand, whenever an object is hurled upward in a gravitational field, the kinetic energy imparted to it is drained off and then converted back into mass inside of the subatomic particles of the object.
If the falling object is a single atom, then how can that single atom convert any of its mass into kinetic energy without undergoing fundamental change?
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re: Pair of Pairs

Post by ken_behrendt »

Hi, guys, and thanks for the "spirited" feedback. I'm somewhat pressed for time at the moment, so let me try to briefly respond to all as best I can.

Michael wrote:
Whenever you double an objects velocity the amount of energy it is carrying quadruples. If you agree to this say yes.
No! That is only true when the velocity of the object is small compared to the high velocity of light (which IIRC is about 29 979 245 800 cm-sec¯¹ or about 186,000 mile-hr¯¹). To accelerate an object from, say, 90,000 mile-hr¯¹ to double the speed at 180,000 mile-hr¯¹ will, according to relativity theory, require its kinetic energy to much more than quadruple.

But, for objects moving at low velocities with respect that of light, their kinetic energy will approximately quadruple for every doubling of their velocity.


Fletcher wrote:
Ya know Ken, when you first put this theory up months ago I thought .. following your logic Bessler could've made a fortune afterall (but not in his life time). He found the perfect way to transmute metal by altering the atomic number of lead.
Neither my recently announced "Bessler's 4th Law of Motion" or the changes in mass associated with changes in elevation in a planet's gravity field will cause transmutation of elements! Although a piece of matter continuously loses rest mass as it drops in a gravity field, that does not change the number of subatomic particles within the individual atoms of which it is composed. Each of the subatomic particles within an atom of lead will be very slightly less massive, the their number does not change and the lead atom is still a lead atom.

I got a chuckle out of your reference to the Lord of the Rings story..."...the dwarfs did his bidding & created the 4th Law, more powerful than any other Law & never before the like of it seen by the mere eyes of men." LOL

Although I sincerely feel this new law of motion is important, it is still far more limited in its application than Newton's famous three laws of motion (some of which he "borrowed" from those earlier "giants" like Galileo upon whose "shoulders" he stood). Indeed, Bessler's 4th Law of Motion only pertains to chronically overbalanced gravity wheels.


Steve wrote:
I admire your perserverance as to how Bessler was able to achieve this...I just cannot understand how you can carry it to the level of a "law". You have a theory, Ken....a theory. And you present it rather well for a forum of Besslerites.

Your "law" automatically assumes that the actual wheel or drum is doing the lifting on the ascending side...why?
I call it a "law of motion" because it describes the process that releases energy from the moving weights in any overbalanced gravity wheel. This law makes no statement about whether this is mechanically possible to achieve or about the exact mechanical means by which it can be achieved. All it does is indicate that IF one can construct a truly overbalanced gravity wheel, then that wheel MUST output energy which will accelerate the wheel and, if there is enough outputted, perform "useful" work in the environment of the wheel.

Because of Bessler's work I am, of course, firmly convinced that there is no "if" about the mechanical possibility of the matter...he did, in fact, achieve this. But, unfortunately, we still do not know all of the details of the mechanism he used to achieve a truly chronically overbalanced wheel.


Bill wrote:
If the falling object is a single atom, then how can that single atom convert any of its mass into kinetic energy without undergoing fundamental change?
Changing the mass of an atom does, indeed, make minute changes in some of its quantum properties, but these are only of interest to atomic physicists. In most cases, with the exception of the weights in a running gravity wheel, the changes that take place in the masses of the atoms that compose objects go completely unnoticed. For example, everytime a person rises to a standing position from a seated position, all of the atoms in that person's body experience a very tiny increase in the rest masses of the subatomic particles from which they are composed. And, when a standing person sits down, the reverse happens and all of the subatomic particles in his body experience a very tiny decrease in their rest masses. However, this has no measurable affect, fortunately, on the various chemical processes taking place within each cell of his body.

However, I have often wondered what would happen if we could remove all of the rest mass energy from an object. Would it still retain its normal physical and chemical properties in this state? Obviously, the object would not retain its normal gravitational and inertial properties. Such a massless object would not be able to lose any further rest mass energy and, thus, it could not drop in a gravity field! If held in the air and released, the buoyant forces acting on it would actually make it rise like a helium filled balloon!

One of the important consequences of relativity theory is the so-called "Equivalence Principle". Because of it, one can conclude that any object that loses its gravitational mass must also lose its inertial mass! Thus, our hypothetical massless object must not only be weightless, but also "inertialess".

If the object was located in the "vacuum" of space and a small rocket engine (also massless) was attached to it to accelerate the object, then it would almost instantly accelerate to a velocity far in excess of the velocity of light! Fortunately, the velocity would not become infinite because, in reality, space is not a perfect vacuum, but, rather, contains a very tenuous amount of monoatomic hydrogen and helium gas that would impose a drag on the accelerating massless object and, thereby, force it to only achieve a finite terminal velocity. But, that terminal velocity could be hundred or even thousands of times that of light!



Well, I've got to rush out now and join a friend for some pizza at a local restaurant before they close up for the night.

I've been doing some further research with Bessler's 4th Law of Motion and managed to derive some addition equations that may be of interest to those following the saga of the announcement of this new law of motion. When these equations are applied to Bessler's Kassel wheel, they make some predictions about its performance that are truly astounding.

If I get a chance, I'll post the equations tomorrow along with the predictions they make for the 12 foot diameter Kassel wheel. The results are really remarkable!



ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: Pair of Pairs

Post by Michael »

No! That is only true when the velocity of the object is small compared to the high velocity of light (which IIRC is about 29 979 245 800 cm-sec¯¹ or about 186,000 mile-hr¯¹). To accelerate an object from, say, 90,000 mile-hr¯¹ to double the speed at 180,000 mile-hr¯¹ will, according to relativity theory, require its kinetic energy to much more than quadruple.

But, for objects moving at low velocities with respect that of light, their kinetic energy will approximately quadruple for every doubling of their velocity.
Ken I'd like to see your source for that But regardless,

you admit the generally the Law of kinetic energy is true, that calculation of kinetic energy requires the velocity of a mass. If you change the velocity-or the mass, you change the final statement.

From the position of kinetic energy your view would have the mass at a loss. Consider these examples;


A mass is accelerating horizontally at a rate that is proportionate to the acceleration of gravity. Since the object is not falling against the gravity gradient there is no trade of mass for energy, with gravity. As the object doubles in it's velocity it's energy is going to quadruple.

Under Ken's View.
Now we are going to drop a like mass and have it pulled by gravity. As that mass is pulled into the gravitaional gradient it's going to loose mass but gain energy.
Problem. Energy is a result of a masses velocity. Change mass or velocity, change energy. Still going with Ken's thoughts, the weight will loose mass. Kens view is it will gain energy from gravity. The truth though is it must loose energy as well because it has lost mass.

IMPORTANT PART.

Under the Law of kinetic energy ( not Ken's view ) both objects will have the same energy for the same velocities since the weights are identical. With Kens view the gravitational one will be different, it has to be. Ken, when the object looses mass it will loose energy. If you think the energy gain it gets is exactly balanced with the mass it has lost, this has to show up in a change of the objects velocity.The object will have to have a larger velocity, and it doesn't.
Last edited by Michael on Mon Jul 03, 2006 4:15 am, edited 5 times in total.
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re: Pair of Pairs

Post by Patrick »

Whenever an object drops in a gravitational field, it loses a tiny bit of its mass that is converted into the kinetic energy that the object acquires.
This statement is simply false.
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Re: re: Pair of Pairs

Post by Fletcher »

ken_behrendt wrote:Bill wrote:
If the falling object is a single atom, then how can that single atom convert any of its mass into kinetic energy without undergoing fundamental change?
Changing the mass of an atom does, indeed, make minute changes in some of its quantum properties, but these are only of interest to atomic physicists. In most cases, with the exception of the weights in a running gravity wheel, the changes that take place in the masses of the atoms that compose objects go completely unnoticed. For example, everytime a person rises to a standing position from a seated position, all of the atoms in that person's body experience a very tiny increase in the rest masses of the subatomic particles from which they are composed. And, when a standing person sits down, the reverse happens and all of the subatomic particles in his body experience a very tiny decrease in their rest masses. However, this has no measurable affect, fortunately, on the various chemical processes taking place within each cell of his body.
Ken .. see the bolded portions above. In the context of what you have just written about the theory of a gain or loss in mass when an object gains or looses gravitational potential then there would appear to be an inconsistency to your argument.

It appears that the theory does not require a comparison of relative vertical velocity so much as vertical displacement of the weights. You also mention nothing about horizontal displacement so one assumes that it is inconsequential to the argument & does not materially effect the conclusions you draw.

Now relating this back to your theory of loss of rest mass being the motive force behind an OOB gravity wheel. Since weights travel in a closed system around a closed path always reaching the same gravitational potential each cycle, would they not loose & then gain the same mass each cycle & therefore there would be no net loss of mass ? The "what goes up, must come down" theory.

Additionally your theory of a 4th Law, although not completely underpinned by the depleted rest mass hypothesis, does predict the path the weights must follow & their relative vertical velocities IINM. This would seem at odds with the motive force theory where a weights vertical velocity & horizontal displacement do not factor (just the gravitational potential argument). How do you reconcile the apparent inconsistencies between the 2 theories that you have so neatly cobbled together ?

P.S. I realize you, like everyone else, must attempt to rationalize how an OOB gravity wheel could possibly work.
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re: Pair of Pairs

Post by ken_behrendt »

Michael and Fletcher...

I had a LOT of trouble trying to understand what you were either trying to ask me or tell me. But here's my best answers to the parts of your posts that I think I understood:

Michael...

It is a well-known fact that, according to the Special Theory of Relativity, an object with some initial mass can not ever reach the velocity of light as it is accelerated toward that velocity. The reason is that, according to the equations of Einstein's revolutionary theory, the object would have to be supplied with an infinite amount of energy to achieve light velocity because, as energy is supplied to the object to accelerate it, more and more of that energy will go into increasing the mass of the object as its velocity increases toward that of light. As the mass of the object continues to increase, even more energy must be supplied per extra mile per second of increased velocity obtained.

Studying the equations that govern this effect shows that the increased mass of the object is given by the expression:

M (increased) = M (original) / [ (1 - {(v^2)/(c^2)}^)1/2 ] which shows that, as v begins to equal c in value, the denominator of the right hand quotient goes toward a value of zero. Well, when one divides M (original) by a number that is approaching zero, then M (increased) will go toward an infinite value!

The point is that the usual expression for the kinetic energy of an object must be, according to relativity theory, modified to take this effect of mass increase into account. When that is done, it is seen that the doubling of velocities so that the final velocity is close to that of light will result in having to supply the object with much more than quadruple the kinetic energy it had when it had its original velocity.

You then state:
Problem. Energy is a result of a masses velocity. Change mass or velocity, change energy. Still going with Ken's thoughts, the weight will loose mass. Kens view is it will gain energy from gravity. The truth though is it must loose energy as well because it has lost mass.
The falling object will loose mass by converting it into kinetic energy that is then used to accelerate the object. This effect will take place even if the object, say, due to air resistance, falls at a constant velocity. In this case, the energy being derived at the expense of the mass of the object is not increasing the velocity of the descending object, but, rather, is being used to push the air along its descent path out of the way.

I think that what is confusing you about this matter is that you do not realize that the amount of mass lost by a falling object is VERY tiny. I present a problem at the end of this post in which I calculate the mass lost by weights which will show you how very small the mass loss really is.


Fletcher wrote:
Since weights travel in a closed system around a closed path always reaching the same gravitational potential each cycle, would they not loose & then gain the same mass each cycle & therefore there would be no net loss of mass ? The "what goes up, must come down" theory.
That is the whole point of the 4th Law of Motion. It shows how it is possible for weights moving around a closed path to continuously lose gravitational potential energy or, as I like to say, mass energy. This law states that in a running overbalanced gravity wheel the average vertical velocity component of the weights will always be negative in value which is equivalent to saying that, on average, the weights are dropping.

The Earth's gravity field will then produce the same reduction in rest mass in the weights as they move around their closed path inside of the overbalanced gravity wheel as it would if they were outside and actually dropped straight down toward the Earth along a path whose length equalled that of the perimeter of their closed orbit within the gravity wheel! This is really an amazing relevation and I wonder why it had not occurred to me years ago. Anyway, if this is still confusing to you, then I suggest you study my earlier post where I attached my "Bessler's 4th Law of Motion Demonstrator". You will see from the sinusoidal graph produced by a single moving weight within an overbalanced gravity wheel that, indeed, it's average vertical velocity component during a single rotation is downward.

You then wrote:
This would seem at odds with the motive force theory where a weights vertical velocity & horizontal displacement do not factor (just the gravitational potential argument).
I am not sure what you mean by the "motive force theory". My new law of motion only uses the properties of the vertical motions of the weights in an overbalanced wheel to rationalize the ability of the wheel to output energy and then use this energy to accelerate itself and, if there is enough of an output to more than overcome air resistance and bearing friction, perform useful work in the wheel's environment.

I think like Michael, you should study the material that I am now going to add to this post.




Well, I've been playing around with the new law of motion like a kid playing with a new toy and want to show off some interesting new expressions that I derived from it.

But, first, let me briefly recap Bessler's 4th Law of Motion and some of the relationships immediately associated with it.

The primary equation of the 4th Law of Motion is:

Vaver = - 2πdω

where Vaver is the average vertical velocity component of the weights as they move around their orbit inside the wheel, π is, of course, pi which is about 3.1415927, d is the horizontal distance from an overbalanced gravity wheel's axle to the CG of its rotating weights, and ω is the rate of rotation of the wheel in rotations per unit of time. This equation assumes that the CG of the weights is directly horizontally to the left or right of the wheel's axle. If, however, the CG is below the height of the axle so that it makes an angle, φ, with the horizontal, then a modified form of the previous expression must be used to compensate for this:

Vaver = - 2πdωcosφ

From the fundamental equation that is applied when the CG is directly horizontally to the left or right of the overbalanced wheel's axle, I was able to derive an expression for the change in gravitational potential energy, ΔPEgrav, that the weights inside of the wheel's drum experience per wheel rotation. That expression was:

ΔPEgrav = - 2πMgd

where M is the total mass of the weights that drive the wheel, and g is the acceleration due to Earth's average surface gravity which is about 980.665 cm-sec¯².

And, from this, I showed that the continuous power output of the wheel, P, when it is rotating, is given by:

P = - 2πMgdω

And, for the case where the CG is below the height of the axle, this is modified to give:

P = - 2πMgdωcosφ



I then wondered if I could derive any other useful expressions from these and came up with the following.

As everybody knows by now, I promote the idea that the energy that an overbalanced gravity wheel outputs actually comes from the loss of the rest masses of its weights as they rotate around the axle during the wheel's operation. Equating Einstein's famous mass-energy equivalence equation with my expression for the change in the gravitational potential energy of a weight moving vertically in a planet's gravity field yielded an expression for the amount of rest mass lost by the driving weights per rotation of the overbalanced gravity wheel that they attached to. That expression is:

ΔM = -4πMgd/c²

where c is, of course, the velocity of light in a vacuum which is equal to 29 979 245 800 cm-sec¯¹

From this it's a simple matter to obtain an expression for the total number of rotations, Σ, that the gravity wheel can complete before it has extracted all of the mass energy of its weights. That expression is:

Σ = c²/4πgd

And, finally, an expression for the total time, T, required for the wheel to complete the Σ rotations so that it's driving weights will be completely massless:

T = c²/4πgdω


Once I had these extra expressions, I decided to apply them to a "typical" overbalanced gravity wheel type problem. I decided to make up a fictional scenario involving Bessler's 12 foot diameter two-directional Kassel wheel to see if these equations could give me some insights into it's potential. What I found was most interesting!



PROBLEM: A document is found that appears to have been written by a close friend of Count Karl. The writer claims that he was once confidentially told by the count that Bessler's two-directional wheel was actually made up of two one-directional wheels contained within a single drum and that the great wheel was driven in either direction by only one of the two wheels it contained while the other wheel was disabled by having a mechanism lock its weights into positions that placed their CG at the axle of their component wheel.

Furthermore, each one-directional wheel contained 32 4 lb. weights and the CG of the driving wheel, when the drum was turning at its maximum terminal rotation rate of 26 rpm, dropped below the level of the axle so that it was only displaced 1 inch horizonally from a vertical line passing through the axle (note: since we know how far the CG is displaced horizontally from the axle, we need not use the modified equations that contain the dip angle, φ).

From this information, one must determine the amount of mass lost, per wheel rotation, by the weights of the one-directional wheel driving both component one-direction wheels and the drum that contains them, the total number of rotations that will be required to completely drain the mass energy from this enabled one-directional wheel's weights, and the total time, in years, required to do this.

Assume that the wheel is indestructible and that, despite constant air resistance and bearing friction, the terminal rotation rate of the wheel remains constant until the weights of the single component one-directional wheel driving it in any one direction are completely drained of the energy associated with their rest masses.



SOLUTION: We start by converting the weight of a component one-directional wheel's weights from lbs. to gms (remember that the total weight of the two component one-directional wheels' weights inside of the Kassel wheel's hollow drum will be double this amount).

32 x 4 lbs. = 128 lbs.

128 lbs. = 58 059.823 gm

Then, using the expression for the loss of mass of the weights per wheel rotation, we write, after using a value of 2.54 cm for a horizontal displacement distance, d, of 1 inch:

ΔM = -4πMgd / c²

ΔM = -4(3.1415927)(58 059.823 gm)
(980.665 cm-sec¯²)(2.54 cm) /
(29 979 245 800 cm-sec¯¹)²

ΔM = - 2.02208 x 10¯¹² gm

Thus, after a single rotation, the weights that drove the Kassel wheel in one direction would have lost only about 2 picograms of mass!

The total number of rotations of the wheel in either direction that are required to completely exhaust the mass energy of the weights of the one component wheel driving the hollow drum is given by:

Σ = c² / 4πgd

Σ = (29 979 245 800 cm-sec¯¹)² / 4(3.1415927)(980.665 cm-sec¯²)(2.54 cm)

Σ = 2.87129 x 10^16 rotations

Thus, the wheel will have to complete about 28 712.9 million million rotations to completely use up the mass energy of the weights in either one of its two one-directional component wheels.

Finally, we determine the time required to complete this enormous number of rotations. After determining that a wheel rotation rate, ω, of 26 rpm is equal to 0.4333333 rotations per second, we write:

T = c² / 4πgdω

T = (29 979 245 800 cm-sec¯¹)² / 4(3.1415927)(980.665 cm-sec¯²)(2.54 cm)(0.4333333 sec¯¹)

T = 6.62605 x 10^16 sec.

and, since one year equals 31 556 926 seconds, we find that the total time required is:

T = 2.0997 x 10^9 years.

That's slightly over 2 BILLION years before either of the two one-directional wheels within the Kassel wheel's drum would be unable to maintain the constant 26 rpm terminal rotation rate of the great wheel!

Such a timespan is actually on the order of the time required for sentient life to evolve on the surfaces of terrestrial planets! And, after that time period had expired, one could just give the giant wheel a push in the other direction and it would then continue to rotate in that new direction for another 2 billion years!

So, we see that Bessler's 4th Law of Motion leads us to the conclusion that an overbalanced gravity wheel is not really perpetual in an absolute sense, but, with a running time in the billions of years, it certainly is "perpetual" for all practical purposes.

Would it be possible to construct a single machine that could, in fact, run non-stop for 2 billion years without the need for maintenance and without suffering a critical malfunction?

As an optimist, I think it might be possible, but it would have to be a very special piece of machinery that was carefully located where it would not be disturbed for the billions of years that it continuously operated. I'm envisioning a machine made almost completely of precious metals that would use the best non-volatile silicone lubricants on boron nitride bearings and would then be enclosed in a virtually indestructible acrylic plastic chamber from which all air had been flushed out and replaced with nitrogen gas at ambient pressure.

This chamber and the great wheel it protected would then be carefully located in some sort of mined out cavern below the surface of land that would experience little or no seismic activity or flooding during the billions of years of theoretical running time. Over that vast expanse of time, even the slow drift of the Earth's tectonic plates would have to be taken into account as continents collided, mountain ranges formed, and oceans floors expanded.

Perhaps some automated mechanism could be attached to the giant wheel that would sense when it had stopped rotating in one direction and would then give it a short duration torque (provided by a dropping weight) in the other direction to again cause the great wheel to continue its continuous motion for another 2+ billion years. That would then mean that it would take a total of 4+ billion years to completely drain all of its weights in both of its component one-directional wheels of their last erg of mass energy.

Would the wheel last that long?

Probably not. Approximately 4 billion or so years from now our Sun will swell up into a Red Giant as it completes its life cycle. In the process, the inner terrestrial planets of our solar system will be vaporized. Of course, by then all of what humanity evolves into will have been long gone in search of fresh worlds to colonize in other star system of our galaxy or even in other neighboring galaxies.

Perhaps the great wheel would still be quietly spinning away as the walls of the cavern surrounding it turned to vapor and were blown away into space by the expanding surface of our dying Sun...



Anyway, I hope that everybody found this interesting and gained some new insights into the subject of overbalanced gravity wheels from it. I know there are a lot of mobilists out there that are, unfortunately, "allergic" to things mathematical, so I tried to keep my presentation as simple, yet accurate, as possible. Although, the mathematics are rather elementary, they may, at first, be a little intimidating to a person who is not used to solving math problems.

The important thing to remember is what Bessler's 4th Law of Motion says about the average vertical velocity of the weights inside a running chronically overbalanced wheel. Basically, this new law of motion tells us that, because the average vertical velocity component of the rotating weights in an overbalanced wheel is negative, the weights interact with the Earth's gravity as though they were constantly dropping.

This action causes them, by the physics of the 19th century, to continuously lose gravitational potential energy and, by the physics of the 20th century, to continuously lose rest mass. This is where the energy that the wheel outputs comes from and, as I've shown above, although huge, it is not an inexhaustible supply of energy.



ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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