Pair of Pairs

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re: Pair of Pairs

Post by 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.



Mass being total energy density.
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.
What are you saying here in the underlined part and after?
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.
Okay well lets look at that then. Tiny but large enough that Bessler's wheel ran on it?
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.
Ken if a drop in the gravitational field equals a loss in mass then a rise in the field equals a gain in mass.

Now Ken this is your quote.
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.
Think about that when you say an object accelerating by gravity is losing mass.

One final thing Ken. Einstein's equivalence principle states that it is impossible to distinguish between a uniform acceleration and a uniform gravitational field. Since your using Einstein you have to validate this, but using this and your point of view, any object experiencing acceleration on a horizontal plane that is the same rate as gravity will also lose mass. Contradicts the fact that any object being made to accelerate has an increase in mass doesn't it.
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re: Pair of Pairs

Post by Fletcher »

edit - time for a change.
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re: Pair of Pairs

Post by ken_behrendt »

Michael wrote (with reference to the very tiny mass losses occurring within a running overbalanced wheel):
Okay well lets look at that then. Tiny but large enough that Bessler's wheel ran on it?
Exactly. As my solution to the hypothetical problem of my last post showed, Bessler's wheels were only losing mass in the picogram range per wheel rotation. That's one trillionth of a gram of mass! But, as we learn from Einstein's Mass-Energy Equivalence equation, that tiny amount of mass can, indeed, provide enough energy to power a gravity wheel.
Now Ken this is your quote.

Quote:
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.


Think about that when you say an object accelerating by gravity is losing mass.


At first glance this seems paradoxical. If an object drops in a gravity field and loses mass that is converted into the kinetic energy that the object displays and an increase in the kinetic energy of an object also increases its mass, then shouldn't the mass of the object remain the same?

The answer that resolves this apparent paradox is that are low velocities compared to that of light such as arise when objects fall through the Earth's atmosphere, practically all of the kinetic energy that derives from their loss of mass goes into increasing their velocity rather than their mass again. As the velocity of the object approaches that of light, this situation changes drastically so that almost all of the kinetic energy supplied in an effort to accelerate the object will go into increasing its mass.

For the purposes of performing calculations with a chronically overbalanced gravity wheel, we can safely assume that all of the kinetic energy derived from the loss of the rest masses of its weights goes into increasing its velocity and none of this energy goes back into increasing the masses of its weights again. It must also be remembered that, once the gravity wheel reaches its terminal rotation rate and has its maximum amount of kinetic energy, any additional energy derived from the mass of its weights will not be transfered to the wheel as a whole, but, rather, to the surrounding atmosphere to overcome air resistance. Must energy will also be lost by infrared energy emission from the wheel's various bearings as they heat up from friction.
One final thing Ken. Einstein's equivalence principle states that it is impossible to distinguish between a uniform acceleration and a uniform gravitational field. Since your using Einstein you have to validate this, but using this and your point of view, any object experiencing acceleration on a horizontal plane that is the same rate as gravity will also lose mass. Contradicts the fact that any object being made to accelerate has an increase in mass doesn't it.
Einstein's Equivalence Principle also assures us that there is no difference between gravitational and inertial mass. When I made reference to this famous principle, I was trying to say that if we could reduce the mass of a weight by using it inside of a working overbalanced gravity wheel, then we would notice that it lost its weight (due to gravitational mass) as well as its inertial properties (which are due to inertial mass).

When an object accelerates horizontal to the direction of action of a gravity field it will not lose any rest mass. Such an acceleration can not be provided by the gravity field, but would have to be produced by an outside energy source. That outside source of thrust (perhaps a rocket engine) would be what was losing mass...not the object accelerated (unless, of course, the velocity of that object and an attached source of thrust began to approach that of light).





Well, in my last post here I presented a problem concerning Bessler's giant 12 foot diameter Kassel wheel and wanted to show how Bessler's 4th Law of Motion and some expressions derived from it could be used to solve that problem. Basically, I gave some hypothetical parameters for the wheel which seemed "reasonable" to me and then proceeded to calculate the amount of rest mass lost by the great wheel's weights per wheel rotation, the number of rotations needed to completely extract all of the mass energy from the weights in one of the two component one-directional wheels that may have made up the larger composite two-directional Kassel wheel, and the amount of time, in years, for this process to be completed.

The results of the calculations were rather surprising because they showed that the weights driving the great wheel only lost about 2 picograms per wheel rotation or about 2 trillionths of a gram of mass per rotation. Then, in order to completely extract the energy associated with the mass of one of the great wheel's two component one-directional wheels, one would have to allow it to run at 26 rpm for over 2 billion years! This would not be possible unless, as Einstein's famous equation indicates, mass was not a form of very highly condensed energy.

However, in reviewing the calculations I performed in the last post, I noticed that I had neglected to do any calculations of the constant power output that the Kassel wheel would have had according to Bessler's 4th Law of Motion based on the hypotheical parameters I attributed to it. So, for the remainder of this, hopefully, shorter post, I will address this problem and give a calculation of the power output of the Kassel wheel.


The constant power output, P, of an overbalanced wheel whose CG is directly horizontally to the left or right of its axle is given by:

P = - 2πMgdω

where M is the total mass of the weights driving the wheel, d is the horizontal distance of the CG of the weights from a vertical line passing through the wheel's axle, and ω is the rotational rate of the wheel in rotations per unit of time.

When the CG of the weights is below the axle of the wheel and forms a "dip" angle of φ with a horizontal line passing through the axle, the power output equation above is modified to:

P = - 2πMgdωcosφ

Since, from the previous problem's given parameter's, we know that, although below the wheel's axle, the CG of the 128 lbs of weight within one of the two composite one-directional wheels that drove the great wheel was only displaced horizontally 1 inch or 2.54 cm from a vertical line passing through the axle, we can use the first equation for constant power output and write:

P = -2(3.1415927)(58 059.823 gm)(980.655 cm-sec¯²)(2.54 cm)(0.433333 sec¯¹)

or, since the units of erg are gm-cm²-sec¯², we can solve the above and give the power in units of erg-sec¯¹:

P = -3.93756 x 10^8 erg-sec¯¹

Knowing that there are 10,000,000 or 10^7 ergs per joule lets us rewrite this as:

P = -39.3756 joule-sec¯¹

And, since a watt equals one joule per second, this becomes:

P = - 39.3756 watts

Also, since one horsepower is equal to 746 watts, we can also express the Kassel's constant power output while rotating at its maximum terminal rotation rate of 26 rpm as:

P = - 0.0527823 horsepower


There have been several other power estimations of the Kassel wheel which have been approximately in the range of tens of watts (I think I once calculated 25 watts by using a different mathematical approach than that used here). It should be remembered that in the previous problem I presented, I only guessed at what "reasonable" values to assign to the great wheel's internal parameters.

The P output I calculated above would be the minimum power level of Bessler's largest wheel. As a load was applied to the wheel, it would immediately slow down from the maximum rotation rate of 26 rpm and this would then reduce the CF that was disabling its weight shifting mechanisms. As a result of the then improved performance of the weight shifting mechanisms, the CG of the driving weights would slowly rotate up to a higher location relative to the axle (put not above or even level with the axle) and that would cause the horizontal displacement distance, d, of the CG to increase.

If d was doubled or tripled by the great wheel slowing down and reducing the CF acting on its weight shifting mechanisms, then the constant power output expression derived from Bessler's 4th Law of Motion indicates that the power output of the wheel would consequently increase by a factor of two or three times. Thus, the actual maximum power output that the Kassel wheel could deliver to perform external work might have been anywhere from 80 to 120 watts.

Although this is only enough power to light a single light bulb, it is amazing to think that the great wheel could have provided it for over two billion years for each direction that the wheel was run in!



ken
Last edited by ken_behrendt on Tue Jul 04, 2006 6:58 pm, edited 1 time in total.
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 »

Exactly. As my solution to the hypothetical problem of my last post showed, Bessler's wheels were only losing mass in the picogram range per wheel rotation. That's one trillionth of a gram of mass! But, as we learn from Einstein's Mass-Energy Equivalence equation, that tiny amount of mass can, indeed, provide enough energy to power a gravity wheel.
So Ken are you now saying Bessler's wheel was nuclear powered?
The answer that resolves this apparent paradox is that are low velocities compared to that of light such as arise when objects fall through the Earth's atmosphere, practically all of the kinetic energy that derives from their loss of mass goes into increasing their velocity rather than their mass again.


A thousand and one physicists would disagree.
As the velocity of the object approaches that of light, this situation changes drastically so that almost all of the kinetic energy supplied in an effort to accelerate the object will go into increasing its mass.
Pray tell exactly where the the horizon this drastic shift takes place?
For the purposes of performing calculations with a chronically overbalanced gravity wheel, we can safely assume that all of the kinetic energy derived from the loss of the rest masses of its weights goes into increasing its velocity and none of this energy goes back into increasing the masses of its weights again.
Ken I repeat again; Energy is the Product of the Velocity of a Mass.


Ken you keep stating that gravitation is a transfere or transmutation ( this is what Fletcher was getting at with the alchemy post ) of mass into energy which then drives the mass to a faster velocity. In reality this says nothing Ken. What are the mechanisms that take the mass, unleash some of the energy that binds the matter together, and then uses that energy, in what way-again what are the mechanisms, to accelerate it to another object? Furthermore from what parts of the mass is this energy taken from?




Since your using Einstein you have to validate this, but using this and your point of view, any object experiencing acceleration on a horizontal plane that is the same rate as gravity will also lose mass. Contradicts the fact that any object being made to accelerate has an increase in mass doesn't it.

Einstein's Equivalence Principle also assures us that there is no difference between gravitational and inertial mass. When I made reference to this famous principle,
I am assuming you don't mean the equivalence principle
I was trying to say that if we could reduce the mass of a weight by using it inside of a working overbalanced gravity wheel, then we would notice that it lost its weight (due to gravitational mass) as well as its inertial properties (which are due to inertial mass).
When an object accelerates horizontal to the direction of action of a gravity field it will not lose any rest mass. Such an acceleration can not be provided by the gravity field, but would have to be produced by an outside energy source. That outside source of thrust (perhaps a rocket engine) would be what was losing mass...not the object accelerated (unless, of course, the velocity of that object and an attached source of thrust began to approach that of light).


Then Ken, by your theory we actually can tell the difference between a constant acceleration and a gravity field.

What about this Ken?

Ken if a drop in the gravitational field equals a loss in mass then a rise in the field equals a gain in mass.
Last edited by Michael on Tue Jul 04, 2006 7:18 pm, edited 2 times in total.
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re: Pair of Pairs

Post by ken_behrendt »

Michael asked:
So Ken are you now saying Bessler's wheel was nuclear powered?
No more so than any falling object in a planet's gravity field is nuclear powered.

It must be remembered that in a nuclear reactor, the lose in mass that takes place principally affects only the nuclei of the fissionable isotopes involved.

When weights drop in a gravity field or in a running gravity wheel, the loss of mass is equally distributed amongst all of the subatomic particles in the matter that makes up the weight.


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 more so than any falling object in a planet's gravity field is nuclear powered.
Then what powers the wheel Ken?
When weights drop in a gravity field or in a running gravity wheel, the loss of mass is equally distributed amongst all of the subatomic particles in the matter that makes up the weight.


How Ken, do the electrons go to a different orbit?
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re: Pair of Pairs

Post by james kelly »

Me thinks , perhaps, some of us belong in booby hatch!
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re: Pair of Pairs

Post by ken_behrendt »

Michael...

The most important concept to emerge from relativity theory is that mass and energy are the same thing. The physics of the 18th and 19th centuries considered these two things to be quite different and that view is now realized to be incorrect.

A gravity wheel is powered by the kinetic energy delivered to it from the mass that is continuously lost from its weights (and any parts attached to them) as it rotates. That energy then accelerates the wheel up to its maximum terminal velocity and, if sufficient, can power outside devices.

These picogram quantity losses in mass per rotation that occur inside of a running gravity wheel do not affect the arrangement of electrons within the individual atoms that make up the driving weights of the wheel.

I have calculated that the Kassel wheel could have operated in one direction for slightly over 2 billion years before all of the mass energy of its weights was depleted. However, I assumed that there is not other process going on which could somehow replenish this mass energy as the great wheel rotated. Perhaps there is, but we will only know about it once we can actually build these devices and see exactly what happens to their weights after they have been in operation for long periods of time.


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 primemignonite »

Of course, all of this argumentation and exposition is massively impressive, to say the least. If what Behrendt proposes somehow serves to lead us onto discovering Bessler's interior arrangement with glorious self-motivation resulting, then the work of having abided it will have been truly 'worth it'. Worth it to the expert argumentors, as well as to we mere simpleton on-lookers, no?

However, that having been said, boldly announcing forth as the "Fourth Law . . . etc.", does this not connote an incompleteness or even defectiveness on the part of Newton and his tripartite masterpiece of long-standing? Yes, possibly. Dare I add that drawing moustaches onto Newton seems to me to be an awfully risk-fraught to say nothing of 'cheeky' endeavor, or am I somehow missing something here?

Honestly speaking, I myself would not have the requisite temerity nor gonads to do like-wise, even IF possessing of Behrendt's un-challengeable brilliance and pregnantly beautiful wordsmanship, so obvious and apparent to all.

I ask: would any others? Alas, I suggest "NO!", would be the resounding answer.

In this raw display of expounded courage, full of valorous discretion or not, Behrendt doubtless' stands both high and alone, this pinnacle we find him on being almost palpable to us, even though not quite yet fully seen. We SHALL, each and every one of us, be watching!

I know, many are staggered and lowered by it all as I certainly am, but most peculiarly, at the same time, one is somehow miraculously raised up and almost re-made by it's display. It DOES tend to rub off - such unvarnished creative excellence.

More!

As-ever, wondering and admiring in a fog of violet ecstasy . . .

James

P.S. WHERE is Jonathan when he is needed most? He has that delicate 'just so" touch, this so as to add required deflation, when and where it may be necessary.
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re: Pair of Pairs

Post by ken_behrendt »

James wrote:
However, that having been said, boldly announcing forth as the "Fourth Law . . . etc.", does this not connote an incompleteness or even defectiveness on the part of Newton and his tripartite masterpiece of long-standing?
Newton's laws of motion were the bedrock upon which the physics of the next two centuries was constructed. Without it, there would have been no Einstein and our "modern" day physics would still be at the level it was during the Renaissance.

I like to think that if Newton had met Bessler and been allowed to examine and test his fabulous wheels, then he (Newton) would have formulated the 4th law very quickly thereafter. Newton would have realized that Bessler's wheels contained weights that, although they were rising and falling through the same vertical distance during each wheel rotation, were still managing to output gravitational energy. That effect is a clear violation of the conservation of gravitational potential energy for bodies moving along a closed path whose plane was perpendicular to the Earth's surface.

The purpose of the 4th Law is to allow us to rationalize how this could be possible. The next issue that needs to be resolved is whether or not it is mechanically possible to actually construct a chronically overbalanced gravity wheel. Unfortunately, the new law says nothing about whether or not this is possible or how to go about doing it. But, it does assure us that, if we do achieve it, then it MUST work.

Apparently, there is an average vertical velocity of the weights inside of an overbalanced wheel which is always negative or downward and this unusual condition allows the weights and the Earth's gravity field to interact exactly the same as though the weights were dropping straight down toward the center of the Earth outside of the wheel instead of just rotating about a closed path within it.

So, with an overbalanced wheel, one gets the best of two worlds. One obtains the energy provided by weights continuously falling in a gravity field, yet the weights never impact the ground because they are forced to continuously move around the rim of the wheel.


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: re: Pair of Pairs

Post by primemignonite »

james kelly wrote:Me thinks , perhaps, some of us belong in booby hatch!
This statement of Kelly's may be the most accurate thing yet put forward on this thread. I'll seriously consider it as a workable hypothesis.

To that I will add this cherry: Behrendt surely gets an A+ 'for effort'; but so far, as to the 'accomplishment' part, we have yet to see anything by which to judge. If such should ever appear, that too will deserve no other than the identical, high grade, as either it IS the thing itself, or it ISN'T!

You have the undoubted lead; so, we are all patiently awaiting results, KB . . .

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re: Pair of Pairs

Post by Michael »

Ah Ken, you've really opened up a prickly can of worms here haven't you.
I am quite aware of the energy mass equivilence principle. Are you aware that by treating gravitation as nothing but this principle in relation to two distict masses, you are in fact treating those two distinct masses as a single mass? And if two distinct masses behave and are a single mass under this principle, then all masses in the universe are in fact one mass. This opens up whole area of discussion that is to big for here.

Examples;
Well if just one mass, why are there measurable distinct masses, and why aren't they all together in one clump.

Answer; the big bang.

Question; What caused the big bang?

Etc etc, not to mention most of this is just theory.

Instead of getting into all of this Ken can you answer the question you haven't answered yet. A weights descent into the gravitational field equals a loss of mass but a gain in velocity is balanced by a weights rise in the gravitational field as gaining mass, wouldn't that be right?
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re: Pair of Pairs

Post by primemignonite »

As James Kelley placed his finger on that very real possibility from before, now Michael does also with this pertinent and timely query:

"Instead of getting into all this Ken can you answer the question you haven't answered yet. A weights descent field equals a loss of mass but a gain in velocity is balanced by a weights rise in the gravitational field as gaining mass, wouldn't that be right?"

Exactly!

Might this be the beginning of the demise of that fourth Newtonian appendage, so latterly and rudely attached?

I am sure that none object to new hypotheses being propounded, quite the opposite being the case no doubt, but, adding to the great Englishman's tried-and-true Law so soon before demonstrations have been made and seen, does seem just a bit premature. Maybe I am just too much the old fogey and cynic to be able to get past this.

More reasons as to why it might be right to do after all, are herby humbly solicited.

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re: Pair of Pairs

Post by ken_behrendt »

Michael wrote:
Instead of getting into all of this Ken can you answer the question you haven't answered yet. A weights descent into the gravitational field equals a loss of mass but a gain in velocity is balanced by a weights rise in the gravitational field as gaining mass, wouldn't that be right?
If I appear not to have answered it, that is because I am not sure exactly what you are trying to say.

Let me restate the basics of what relativity theory says about falling and rising weights.

When a weight falls downward in a gravity field, then the kinetic
energy that the weight acquires comes from the loss in mass experienced by all of the subatomic particles inside of its atoms. With the physics of the 19th century one would say that the object's kinetic energy increased because its gravitational potential energy decreased.

And, conversely, when a weight rises upward in a gravity field, any kinetic energy that the weight initially possessed is, eventually, completely converted into an increase in all of the subatomic particles from which it is composed. And, again with the physics of the 19th century one would say that the object's kinetic energy decreased because its gravitational potential energy increased.

I'm assuming here that there is no atmosphere present that would "steal" some portion of the object's kinetic energy during its descent or ascent. And, I am also assuming that the velocity of the object is not close to that of light as might be the case if it was falling toward the event horizon of a black hole.



James wrote:
You have the undoubted lead; so, we are all patiently awaiting results, KB . . .
I, however, continue to impatiently await results. The mystery of Bessler's wheels has dragged on now for almost 3 centuries and, despite the new law of motion, we only have the vaguest idea of how he achieved working overbalanced gravity wheels.

Now that I have the presentation of the 4th law of motion out of the way, I intend to resume my efforts to find Bessler's mystery mechanism. Perhaps with my latest design that I will shortly be testing, I will either have it or, hopefully, be a step closer.

Knowing how weights moving about a closed path inside of an overbalanced gravity wheel can output kinetic energy is something that has perplexed me for decades and which, I am sure, has prevented the vast majority of the scientific orthodoxy from considering these devices worthy of any research effort being expended upon them. With the 4th Law of Motion, we at least know that such a system of weights moving in a closed path CAN output energy, although it is done at the expense of the mass of the weights involved. But, in the final analysis, we still need a working mechanism that produces the effect and not just an equation to describe it...


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 »

When a weight falls downward in a gravity field, then the kinetic
energy that the weight acquires comes from the loss in mass experienced by all of the subatomic particles inside of its atoms. With the physics of the 19th century one would say that the object's kinetic energy increased because its gravitational potential energy decreased.

And, conversely, when a weight rises upward in a gravity field, any kinetic energy that the weight initially possessed is, eventually, completely converted into an increase in all of the subatomic particles from which it is composed. And, again with the physics of the 19th century one would say that the object's kinetic energy decreased because its gravitational potential energy increased.
Okay then Ken since any downward movement that creates a loss is balanced by any upward movement that creates a gain and you are cycling weights how do you propose to make an engine?
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