I won't pretend to understand Ken's law. Does it prove that an overbalanced wheel is possible ? Or does it dress up the self evident fact (that if it were possible to keep a wheel overbalanced that it would turn ) in formulae and fancy words ?
Prehaps a better law would be : If you think you have run out of ideas consider that you may just have run out of ways to prove your one idea does not work.
Pair of Pairs
Moderator: scott
re: Pair of Pairs
Ken
I hope you are not putting this law thing in your book!
Seems a bit strange for someone who claims to understand physics.
I think you are keeping real physicists away from the forum.
Can you prove any of your law?
I hate to recommend this be put into the fraud topic, but wow this is a little off base.
Have you found something?
I hope you are not putting this law thing in your book!
Seems a bit strange for someone who claims to understand physics.
I think you are keeping real physicists away from the forum.
Can you prove any of your law?
I hate to recommend this be put into the fraud topic, but wow this is a little off base.
What achievements? You can not show us anything Bessler Achieved.if it was not for Johann Bessler's achievements, then I would never have sought a mathematical expression for the new law to start with.
Have you found something?
JB Wheeler
it exists I think I found it.
it exists I think I found it.
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LustInBlack
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re: Pair of Pairs
This calls for a Moderator ! 8]
re: Pair of Pairs
I do not think a moderator is the way to go.
IMO you should drop the whole conversation and do research. Using the trial and error method of getting a wheel that works. Then you can debate if it makes or breaks any known laws and theory's.
Why debate this now when there is no substantial proof either way. I say quit discussing and get back to work on designing. All you are doing is clouding an issue that is of no importance.
Gravity causes things to fall, Gravity with leverage will lift things. Gravity can be used in the vertical mode to over come CF and augment CP. Gravity will assist CF combined with inertia and kinetic forces on the descending side. UP is in, down is out. Out leverage is greater than in. Out inertia and kinetics are greater than in. Gravity can be utilised with other forces to move weights down and out, lifting a mass just as heavy as that in motion.
And what if the CG is always between the axis (not axle) and descending rim. How can you form a new law of physics without having physical proof of its existence. If I should perchance be the one to bring forth a working wheel, You will not know the exact mechanism. But I will supply specifications as to size,weight,RPM, and usable torque output.
Does this form a fourth law? If so it is not a new law, it has been around as as a known fact for years. Why predict anything, if it works, install a Prony brake to it and read the foot pounds of force produced. Or attach an auto alternator to it and apply a measurable load on it.
And now you are getting me carried away!. Right now I would settle for a wheel that runs longer than the initial force taken to start it.
Ralph
IMO you should drop the whole conversation and do research. Using the trial and error method of getting a wheel that works. Then you can debate if it makes or breaks any known laws and theory's.
Why debate this now when there is no substantial proof either way. I say quit discussing and get back to work on designing. All you are doing is clouding an issue that is of no importance.
Gravity causes things to fall, Gravity with leverage will lift things. Gravity can be used in the vertical mode to over come CF and augment CP. Gravity will assist CF combined with inertia and kinetic forces on the descending side. UP is in, down is out. Out leverage is greater than in. Out inertia and kinetics are greater than in. Gravity can be utilised with other forces to move weights down and out, lifting a mass just as heavy as that in motion.
If my derivation is accurate (and I believe it is), then we can now predict the power output for a one-directional wheel simply by knowing its rotational rate, the total mass of the weights driving it, and the distance that the CG of those weights is maintained to one side of the wheel's axle. We do not need to know the exact mechanism needed to achieve the effect.
And what if the CG is always between the axis (not axle) and descending rim. How can you form a new law of physics without having physical proof of its existence. If I should perchance be the one to bring forth a working wheel, You will not know the exact mechanism. But I will supply specifications as to size,weight,RPM, and usable torque output.
Does this form a fourth law? If so it is not a new law, it has been around as as a known fact for years. Why predict anything, if it works, install a Prony brake to it and read the foot pounds of force produced. Or attach an auto alternator to it and apply a measurable load on it.
And now you are getting me carried away!. Right now I would settle for a wheel that runs longer than the initial force taken to start it.
Ralph
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LustInBlack
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re: Pair of Pairs
Ralph, just to make clear that I acknowledge the first syntax of your post as a reply to me.. But the rest as a reply to Ken ..
re: Pair of Pairs
Lustinblack,
Yes you got that right! But take note, I did recognize the " ! 8]" at the end of your statement. :0)
Ralph
Yes you got that right! But take note, I did recognize the " ! 8]" at the end of your statement. :0)
Ralph
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ken_behrendt
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re: Pair of Pairs
Gentlemen...
What I am calling "Bessler's 4th Law of Motion" is extremely simple to understand and I am surprised that it did not occur to me many years ago.
Imagine the following "thought experiment". Suppose we had a working overbalanced gravity wheel. Also suppose that we had two intelligent butterflies that we could place inside of the drum of the wheel.
The first butterfly is placed at the point in space inside the rotating drum at which the offset CG of the wheel's weights remains fixed as the wheel rotates. To remain at that location, the butterfly must beat its wings continuously so that he can remain airborne. The second butterfly, however, has it much easier, he is placed on the wheel's axle.
Now, assume that the first butterfly looks horizontally to his left and his right and can measure the magnitudes of the vertical velocities of the moving weights within the wheel. He will tell us that the weights appear to be ascending on one side of the wheel at the exact same rate that they descend on the other side of the wheel. In other words, he notes that the vertical components of the velocities are equal in magnitude but opposite in sign.
However, we will obtain a quite different report from the second butterfly located on the axle as he looks out horizontally to his left and right. He will report that the magnitude of the vertical velocity of the descending weights appears to be larger than that of the ascending weights. When he takes the average of these two vertical velocities, he notes that the average magnitude of the vertical velocity of the weights with respect to the wheel's axle is negative in sign. In other words, on average, with respect to the wheel's axle, the weights appear to be always descending!
From these two sets of observations, the two butterflies will reach different conclusions about what is happening. The first butterfly that hovers at the fixed CG location within the rotating gravity wheel drum will conclude that, on average, the weights revolving around his position have no net vertical velocity. Because the vertical velocities he observes are equal in magnitude, but opposite in sign, he calculates that the average vertical velocity is zero. So, he concludes that, for calculational purposes, on average, the weights neither rise nor fall and there can therefore be no change in the gravitational potential energy of the weights with respect to the location of their CG around which they rotate and at which he is hovering.
The second butterfly, however, reaches a quite different conclusion. He has determined that, with respect to the wheel's axle, the weights, on average, appear to be dropping at all times and that the rate at which they drop is constant.
Since the second butterfly knows how long it takes any weight to orbit the axle once, he concludes that, on average, it is as though all of the weights around the axle, in essence, drop a distance which is equal to the product of their average falling velocity multiplied by the time it takes the wheel to complete a single rotation.
From this he sees that it is as though all of the weights inside the wheel's drum drop through this distance per wheel rotation. Being a student of gravity physics, the second butterfly goes on to calculate from this information the amount of gravitational potential energy being lost by the weights per wheel rotation. He also now understands where the energy that the gravity wheel appears to mysteriously output comes from.
Since the weights are attached to the wheel which forces them to follow the horizontally eccentric orbit that they have around the axle, the gravitational potential energy these weights continuously lose must be transferred to the wheel as a whole and that is why the wheel accelerates from a standstill and can continuously perform external work.
This second butterfly is also familiar with the basics of Einsteinian Relativity Theory. He concludes that all of the energy that the gravity wheel outputs must be derived directly from the rest masses of the subatomic particles from which the weights in the wheel are composed. He predicts that, over long stretches of time, as the wheel continues to output energy to its environment, the total mass and weight of the weights within the wheel will slowly diminish.
By way of analogy, skeptics who claim that weights moving about a closed path in a gravity field can not undergo a net change in gravitational potential energy are somewhat like the first butterfly in the above thought experiment...they are not viewing the problem from the correct perspective. The actual situation, however, is the one that is perceived by the second butterfly and which is formalized in the mathematical expression of what I call Bessler's 4th Law of Motion.
Well, this is a super simple explanation of the new law of motion. Basically, it allows one to rationalize how weights, moving around a closed path in a planet's gravity field, could continuously output energy. This bizarre effect arises from the situation within an overbalanced wheel in which one has two separate centers of rotation...one at the CG of the rotating weights and the other at the axle of the wheel about which all parts rotate. This law is distinct from Newton's first three laws of motion and does not conflict with any of them. However, it only applies to overbalanced gravity wheels and, quite unfortunately, does not give us the details of the mechanism required to achieve the effect.
This post has gotten much longer than I originally intended, but let me very briefly respond to some of the comments:
LIB wrote:
Wheeler asked:
Ralph wrote:
ken
What I am calling "Bessler's 4th Law of Motion" is extremely simple to understand and I am surprised that it did not occur to me many years ago.
Imagine the following "thought experiment". Suppose we had a working overbalanced gravity wheel. Also suppose that we had two intelligent butterflies that we could place inside of the drum of the wheel.
The first butterfly is placed at the point in space inside the rotating drum at which the offset CG of the wheel's weights remains fixed as the wheel rotates. To remain at that location, the butterfly must beat its wings continuously so that he can remain airborne. The second butterfly, however, has it much easier, he is placed on the wheel's axle.
Now, assume that the first butterfly looks horizontally to his left and his right and can measure the magnitudes of the vertical velocities of the moving weights within the wheel. He will tell us that the weights appear to be ascending on one side of the wheel at the exact same rate that they descend on the other side of the wheel. In other words, he notes that the vertical components of the velocities are equal in magnitude but opposite in sign.
However, we will obtain a quite different report from the second butterfly located on the axle as he looks out horizontally to his left and right. He will report that the magnitude of the vertical velocity of the descending weights appears to be larger than that of the ascending weights. When he takes the average of these two vertical velocities, he notes that the average magnitude of the vertical velocity of the weights with respect to the wheel's axle is negative in sign. In other words, on average, with respect to the wheel's axle, the weights appear to be always descending!
From these two sets of observations, the two butterflies will reach different conclusions about what is happening. The first butterfly that hovers at the fixed CG location within the rotating gravity wheel drum will conclude that, on average, the weights revolving around his position have no net vertical velocity. Because the vertical velocities he observes are equal in magnitude, but opposite in sign, he calculates that the average vertical velocity is zero. So, he concludes that, for calculational purposes, on average, the weights neither rise nor fall and there can therefore be no change in the gravitational potential energy of the weights with respect to the location of their CG around which they rotate and at which he is hovering.
The second butterfly, however, reaches a quite different conclusion. He has determined that, with respect to the wheel's axle, the weights, on average, appear to be dropping at all times and that the rate at which they drop is constant.
Since the second butterfly knows how long it takes any weight to orbit the axle once, he concludes that, on average, it is as though all of the weights around the axle, in essence, drop a distance which is equal to the product of their average falling velocity multiplied by the time it takes the wheel to complete a single rotation.
From this he sees that it is as though all of the weights inside the wheel's drum drop through this distance per wheel rotation. Being a student of gravity physics, the second butterfly goes on to calculate from this information the amount of gravitational potential energy being lost by the weights per wheel rotation. He also now understands where the energy that the gravity wheel appears to mysteriously output comes from.
Since the weights are attached to the wheel which forces them to follow the horizontally eccentric orbit that they have around the axle, the gravitational potential energy these weights continuously lose must be transferred to the wheel as a whole and that is why the wheel accelerates from a standstill and can continuously perform external work.
This second butterfly is also familiar with the basics of Einsteinian Relativity Theory. He concludes that all of the energy that the gravity wheel outputs must be derived directly from the rest masses of the subatomic particles from which the weights in the wheel are composed. He predicts that, over long stretches of time, as the wheel continues to output energy to its environment, the total mass and weight of the weights within the wheel will slowly diminish.
By way of analogy, skeptics who claim that weights moving about a closed path in a gravity field can not undergo a net change in gravitational potential energy are somewhat like the first butterfly in the above thought experiment...they are not viewing the problem from the correct perspective. The actual situation, however, is the one that is perceived by the second butterfly and which is formalized in the mathematical expression of what I call Bessler's 4th Law of Motion.
Well, this is a super simple explanation of the new law of motion. Basically, it allows one to rationalize how weights, moving around a closed path in a planet's gravity field, could continuously output energy. This bizarre effect arises from the situation within an overbalanced wheel in which one has two separate centers of rotation...one at the CG of the rotating weights and the other at the axle of the wheel about which all parts rotate. This law is distinct from Newton's first three laws of motion and does not conflict with any of them. However, it only applies to overbalanced gravity wheels and, quite unfortunately, does not give us the details of the mechanism required to achieve the effect.
This post has gotten much longer than I originally intended, but let me very briefly respond to some of the comments:
LIB wrote:
Yes...Bessler's 4th Law of Motion says that a chronically overbalanced wheel MUST output energy. Thus, we are assured that if we do, finally, manage to build such an overbalanced wheel, then it will work. Our "quest" is not doomed to failure because it is a physically impossible task right from the start. On this Discussion Board we have, over the last year or so, seen several designs for overbalancing gravity wheels. But, not one of them (mine included!) was truly chronically overbalanced. They all either could not maintain the imbalance or, if they did, also contained not too obvious sources of counter torque that rendered them useless.What you are saying, is basicly that you provide more than faith in the phenomenon, you provide proof..
Wheeler asked:
I decided to name this new law of motion after Bessler because he is the only one that I am reasonable sure did manage to build devices that demonstrate this law in action. That is, he seems to most definitely have constructed chronically overbalanced gravity wheels capable of outputting energy continuously to their environments.What achievements? You can not show us anything Bessler Achieved.
Ralph wrote:
I agree that the most important thing is to find the mechanism that Bessler had that allowed his wheels to turn. Once that is done, then I am very confident that it will be seen that they do not violate any of the laws of motion...including this 4th one!IMO you should drop the whole conversation and do research. Using the trial and error method of getting a wheel that works. Then you can debate if it makes or breaks any known laws and theory's.
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
Vaver = -2(√2)πdωcosφ
re: Pair of Pairs
Ken,
If you want anyone to understand your hogwash "4th Law of Motion" then you must explain it much more clearly. Either throw out the Greek symbols or else define what they mean. I speak, read and write English, NOT Greek.
Your first equation I interpret as...
a^2 = r^2 + d^2 - 2*r*d*Cos(90+AngleTheta)
This is a version the classic formula for finding a third side of a triangle when two sides and the angle between them are known. From page 176 of my 1935 ninth edition Machinery's Handbook the basic formula is...
c = Sqr(a^2 + b^2 + 2*a*b*Cos(C) where a,b,c are triangle sides and C is the angle between sides a and b. This I fully understand.
Next you make some assumptions which I feel you can not make. You assume that the maximum and minimum vertical velocities of the weights will be perfectly horizontal to the wheel's axle. You also assume that the CG will be perfectly horizontal to the axle. And you use the Greek lower case omega letter in you formula. You don't state what this letter stands for. It appears to be the wheel's rotational velocity? The first part of your two formulasÂ’s read...
2*Pi*(r-d) and 2*Pi*(r+d) which calculates the circumference of circles that have radii of (r-d) and (r+d). When circumference is multiplied by rotational velocity (such as RPM) then you will get plane velocity.
Next you have V with a line on top. The line on top signifies a sum. So you have summed the up velocity with the down velocity. If I were writing a Visual Basic program to do the calculations then Visual Basic wouldn't allow me to use such a symbol. So I would most likely name the variable SumV or such so as to remind me that it is the sum of two velocities Vasc and Vdes. So your formula might be written as...
SumV = -2*Pi*d*RPM
The problem is that the following formulas are not equivalent...
(2*Pi*(r-d)*RPM) + (-2*Pi*(r+d)*RPM) doesn't equal -2*Pi*d*RPM
To prove this I'll assign real values of r = 7, d = 3 and RPM = 10. The two formulas will then be...
Vasc = 2*3.1415+(7-3)*10 = 251.327
Vdes = -2*3.1415+(7+3)*10) = -628.319
And the sum of Vasc and Vdes will be...
SumV = Vasc + Vdes = -376.991
Compare this with...
SumV = -2*3.1415*3*10 = -188.496
The second results are only half the first, so something is screwed up with Ken's formulas! Ken's 4th Law of Motion is pure hogwash (in my opinion).
Now letÂ’s get into Ken's logic behind these equations. He is showing that the weights pivot around the wheel's axle while moving in and out from that axle. He then shows the weight rising slowly and descending quickly. This causes a problem because soon all the weights will be at the bottom since they're not moving back to the top as quickly as they fall.
I still say that my opinion is that that Ken's 4th Law of Motion is hogwash.
Ken's last three formulas then try to build on his earlier formulas. One needs to build a solid foundation before proceeding any further.
These are just my opinions.
If you want anyone to understand your hogwash "4th Law of Motion" then you must explain it much more clearly. Either throw out the Greek symbols or else define what they mean. I speak, read and write English, NOT Greek.
Your first equation I interpret as...
a^2 = r^2 + d^2 - 2*r*d*Cos(90+AngleTheta)
This is a version the classic formula for finding a third side of a triangle when two sides and the angle between them are known. From page 176 of my 1935 ninth edition Machinery's Handbook the basic formula is...
c = Sqr(a^2 + b^2 + 2*a*b*Cos(C) where a,b,c are triangle sides and C is the angle between sides a and b. This I fully understand.
Next you make some assumptions which I feel you can not make. You assume that the maximum and minimum vertical velocities of the weights will be perfectly horizontal to the wheel's axle. You also assume that the CG will be perfectly horizontal to the axle. And you use the Greek lower case omega letter in you formula. You don't state what this letter stands for. It appears to be the wheel's rotational velocity? The first part of your two formulasÂ’s read...
2*Pi*(r-d) and 2*Pi*(r+d) which calculates the circumference of circles that have radii of (r-d) and (r+d). When circumference is multiplied by rotational velocity (such as RPM) then you will get plane velocity.
Next you have V with a line on top. The line on top signifies a sum. So you have summed the up velocity with the down velocity. If I were writing a Visual Basic program to do the calculations then Visual Basic wouldn't allow me to use such a symbol. So I would most likely name the variable SumV or such so as to remind me that it is the sum of two velocities Vasc and Vdes. So your formula might be written as...
SumV = -2*Pi*d*RPM
The problem is that the following formulas are not equivalent...
(2*Pi*(r-d)*RPM) + (-2*Pi*(r+d)*RPM) doesn't equal -2*Pi*d*RPM
To prove this I'll assign real values of r = 7, d = 3 and RPM = 10. The two formulas will then be...
Vasc = 2*3.1415+(7-3)*10 = 251.327
Vdes = -2*3.1415+(7+3)*10) = -628.319
And the sum of Vasc and Vdes will be...
SumV = Vasc + Vdes = -376.991
Compare this with...
SumV = -2*3.1415*3*10 = -188.496
The second results are only half the first, so something is screwed up with Ken's formulas! Ken's 4th Law of Motion is pure hogwash (in my opinion).
Now letÂ’s get into Ken's logic behind these equations. He is showing that the weights pivot around the wheel's axle while moving in and out from that axle. He then shows the weight rising slowly and descending quickly. This causes a problem because soon all the weights will be at the bottom since they're not moving back to the top as quickly as they fall.
I still say that my opinion is that that Ken's 4th Law of Motion is hogwash.
Ken's last three formulas then try to build on his earlier formulas. One needs to build a solid foundation before proceeding any further.
These are just my opinions.
re: Pair of Pairs
Ken wrote;
I think he is now actually saying it is not important that he says he has a law, but that the real important thing is to find the solution to Besslers wheel.
It seems like Ken has softened his Law a bit by this statement.I agree that the most important thing is to find the mechanism that Bessler had that allowed his wheels to turn.
I think he is now actually saying it is not important that he says he has a law, but that the real important thing is to find the solution to Besslers wheel.
JB Wheeler
it exists I think I found it.
it exists I think I found it.
re: Pair of Pairs
Finding the mechanism that allowed his wheels to turn is easy. It was tapered pins resting in a bushings. These pins was centered in the axle.
It is not what allowed his wheels to turn, but what made them continue turning. That interest's me.
Ralph
It is not what allowed his wheels to turn, but what made them continue turning. That interest's me.
Ralph
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ken_behrendt
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re: Pair of Pairs
Jim wrote:
Wheeler wrote:
Ralph...
Bessler's wheels could continue to turn because his mechanism managed to chronically maintain the imbalance of the wheel. And this new law of motion assures us that whenever that situation is achieved (and by whatever mechanism does it!), then the system MUST continuously output usable energy.
Well. I am positively dumbfounded that you fellows are having such difficulty in following my simple derivation of what I am calling "Bessler's 4th Law of Motion". I tried in that derivation to used standard mathematical notation and the Greek letters normally used in mechanical engineering to denote various angles and the wheel's rotational rate or frequency in the diagram that accompanied my derivation in my last post on my "...Updates" thread down in the Community Buzz forum.
Let me try again and continue using the previous simple example of the two butterflies located within the rotating drum of an overbalancing gravity wheel in which its internal weight shifting mechanisms manage to keep the CG of the weights positioned at a horizontal distance, d, from the wheel's axle. As everyone may recall, the first butterfly was placed at the location of the CG of the weights and he had to beat his wings continuously to remain hovering there as the wheel's drum rotated past him. The second butterfly was located on the axle of the wheel and, in this example, should be considered as stationary.
When the first butterfly at the CG then looks horizontally toward the ascending side of the wheel, he will note that the maximum vertical ascending velocity of the weights is given by:
Vmaxasc = 2πrω
where π or pi is the usual transcendental number used extensively in geometry problems with a value of about 3.1415927... , r is the distance from the location of the CG of the weights to any particular weight, and the Greek letter ω is a standard symbol for the rotational rate or frequency of a wheel and represents rotations per unit of time.
and, when he looks horizontally toward the descending side of the wheel, he will note that the maximum vertical descending velocity is given by:
Vmaxdes = - 2πrω
Once he has this information, he then wants to determine the average value of these two maximum vertical velocities. This average value was represented in my Community Buzz thread "...Updates" thread by that capital V with the bar over it. If I had thought the use of that standard symbol for the arimetical average of two numbers would have caused confusion, then I would have called it something else. So, let's do that now. Instead of V "bar", let's just call it Vaver.
So, the first intelligent butterfly will write:
Vaver = { (Vmaxasc) + (Vmaxdes) } / 2 = { (2πrω) + (- 2πrω) } / 2
Vaver = { (2πrω) - (2πrω) } / 2 = 0 /2 = 0
Vaver = 0
From this, the first butterfly hovering at the CG of the weights that rotate around him concludes that the vertical velocities of the weights horizontal to him as they both ascend and descend on opposite side of him are equal and opposite in magnitude and, thus, their average value is zero.
This first butterfly then says that because there is no net vertical motion of the gravity wheel's weights in the Earth's gravity field, then there can therefore be no net change in the gravitational potential energy of the weights and, as the skeptics had been telling him all along, such a system of weights can not output any energy to the gravity wheel within whose hollow spinning drum he hovers. He is, however, puzzled as to why the wheel was able to accelerate from a standstill and is now operating various energy consuming devices in its environment.
Now for the second butterfly located on the axle of the rotating gravity wheel. What he sees happening is quite different from what the first butterfly observes.
The second butterfly also looks out horizontally toward the ascending and descending sides of the wheel, but he does it from the location of the wheel's axle and not from that of the weights' CG. What does the second butterfly note?
First, he notes that the maximum vertical ascent velocity of the weights, Vmaxasc, is given by:
Vmaxasc = 2π(r-d)ω
where d is the distance, measured horizontally, from the wheel's axle to the CG where the first butterfly is still hovering.
Then, he notes that the maximum vertical descent velocity of the weights, Vmaxdes, is given by:
Vmaxdes = - 2π(r+d)ω
When he then proceeds to calculate the average value of these two maximum vertical velocities, as seen from the wheel's axle, he obtains:
Vaver={ (Vmaxasc) + (Vmaxdes) } / 2 ={ (2π(r-d)ω) + (- 2π(r+d)ω) } / 2
Vaver = { (2π(r-d)ω) - (2π(r+d)ω) } / 2
Vaver = {2πω(r-d-r-d)} / 2 = πω(-2d)
and, finally, what I consider to be "Bessler's 4th Law of Motion":
Vaver = - 2πdω
Thus, the second butterfly concludes, correctly, that the average vertical velocity of the weights, with respect to the wheel's axle, is a negative or downward one! He further notes that this situation will exist in ANY overbalanced wheel that has the CG of its driving weights horizontally offset a distance d from the wheel's axle. And, of course, the wheel must have a non-zero value for the rotation rate ω
Now, the second butterfly on the wheel's axle wonders what the consequences of this situation will be. Being a student of gravity physics, he knows that when a weight of mass, M, drops from an height of h above the ground down onto the ground, it will lose an amount of gravitational energy, ΔPEgrav, which is given by the expression:
ΔPEgrav = Mg Δh
in which g is the acceleration due to gravity near the Earth's surface (about 32 feet per second per second).
The second axle sitting butterfly now realizes something very interesting. He knows that the driving weights, with a total mass of M, rotating about the axle of the overbalanced gravity wheel that he sits in have a average velocity that is always downward...just as though they were dropping in a gravity field although their average downward velocity remains constant for a given wheel rotation rate and is not undergoing acceleration as would be the case for objects in freefall. He also knows the amount of time it takes the wheel to complete a single rotation. That is simply the reciprocal of its rotation rate, ω, or 1/ω. He then uses this to derive a simple expression for the continuous loss of gravitational potential energy by the weights in the wheel he resides in per single wheel rotation. The expression he derives is:
ΔPEgrav = Mg(- 2πdω)(1/ω)
ΔPEgrav = - 2πMgd
Now he has an expression which tells him that the wheel will, on average, continuously lose gravitational potential energy during each and every one of its rotations even though the weights it contains are moving around a closed path within the wheel! He realizes that this situation will take place in any type of system in which there are two separate centers of rotation. In the case of an overbalanced gravity wheel, the two centers of rotation are at the location of the CG of its weights and at the wheel's axle.
Now that he knows how much gravitational potential energy is lost per wheel rotation, finding the constant power output, P, of the wheel at any given rotation rate, ω, is a simple matter. He need only divide the expression for the loss of gravitational potential energy per wheel rotation by the time required for the wheel to complete a single rotation which, as mentioned above, is simply the reciprocal of its rotation rate or 1/ω. Doing this yields:
P = ΔPEgrav / (1/ω) = (- 2πMgd)/(1/ω)
P = - 2πMgdω
Well, fellow Wheel Wizards that is basically all you need to know about Bessler's 4th Law of Motion. It's really very simple once you get the hang of it.
If, however, the fixed CG of the weights is not located directly horizontal to the axle of the wheel, then there will be a reduction in the average downward velocity of the weights with respect to the axle.
In this case, it will be necessary to use a modified version of Bessler's 4th Law of Motion to account for the situation. In this case, one must know the distance from the wheel's axle to the CG of the weights, d, AND the "dip" angle, φ of the CG below a line passing horizontally through the axle. The modified version of Bessler's 4th Law of Motion then becomes:
Vaver = - 2πdωcosφ
I think if the reader will follow this derivation and also view what I posted on my "...Updates" thread over in the Community Buzz forum, then he will gain a greater appreciation of what was going on inside of Bessler's wheels. No, understanding this new law of motion will not give you his secret mechanism on a silver platter...you will still have to work VERY hard to find that. But, it will, at a minimum, show you that what you quest for is not physically impossible as the the many skeptics of this subject constantly tell us.
For the reader who still can not follow the simple step by step derivation I have given above, I have attached a nice chart below which summarizes the important formulas associated with Bessler's 4th Law of Motion
ken
No...that is not true. In the simplest type of overbalancing wheel I used in my derivation, only the maximum vertical descending and ascending velocities of the weights were horizontal to the wheel's axle. The minimum (actually zero) vertical velocities of the weights were vertical to the offset CG of the weights. I selected the case where the CG is only horizontally displaced from the wheel's axle, because this is the simplest case to use in deriving Bessler's 4th Law of Motion.Next you make some assumptions which I feel you can not make. You assume that the maximum and minimum vertical velocities of the weights will be perfectly horizontal to the wheel's axle.
That is a standard mathematical symbol to indicate the simple arithmetical average of several values of a variable. In the derivation I gave in my "...Updates" thread over in the Community Buzz forum, it denotes the arithmetic average of the apparent maximum vertical descent and ascent velocities of the weights as they pass through positions that are horizontal to either the CG of the weights or the axle of the wheel.Next you have V with a line on top. The line on top signifies a sum. So you have summed the up velocity with the down velocity.
No, Jim, the magnitudes of the vertical velocities of the weights relative to the wheel's axle will continuously vary as they orbit about the axle. Weights can not pile up at the bottom of the wheel because they are attached, via their individual shifting mechanisms, to the wheel and are carried along with it. The magnitude and sign of their vertical velocity components with respect to the wheel's axle will be determined by their location in space relative to the axle.He then shows the weight rising slowly and descending quickly. This causes a problem because soon all the weights will be at the bottom since they're not moving back to the top as quickly as they fall.
Wheeler wrote:
I still consider this to be a genuine "law" of motion and something which should be of interest to all gravity wheel researchers...whether or not they are trying to replicate Bessler's devices. But, it is only a mathematical model of what is happening inside of an overbalanced gravity wheel that allows it to continuously output energy. We still need the mechanism that does this and obtaining that mechanism is still my #1 priority. But, it's nice to know that our search is possible and that when, finally, we are successful. We will have a way to rationalize its performance.It seems like Ken has softened his Law a bit by this statement.
I think he is now actually saying it is not important that he says he has a law, but that the real important thing is to find the solution to Besslers wheel.
Ralph...
Bessler's wheels could continue to turn because his mechanism managed to chronically maintain the imbalance of the wheel. And this new law of motion assures us that whenever that situation is achieved (and by whatever mechanism does it!), then the system MUST continuously output usable energy.
Well. I am positively dumbfounded that you fellows are having such difficulty in following my simple derivation of what I am calling "Bessler's 4th Law of Motion". I tried in that derivation to used standard mathematical notation and the Greek letters normally used in mechanical engineering to denote various angles and the wheel's rotational rate or frequency in the diagram that accompanied my derivation in my last post on my "...Updates" thread down in the Community Buzz forum.
Let me try again and continue using the previous simple example of the two butterflies located within the rotating drum of an overbalancing gravity wheel in which its internal weight shifting mechanisms manage to keep the CG of the weights positioned at a horizontal distance, d, from the wheel's axle. As everyone may recall, the first butterfly was placed at the location of the CG of the weights and he had to beat his wings continuously to remain hovering there as the wheel's drum rotated past him. The second butterfly was located on the axle of the wheel and, in this example, should be considered as stationary.
When the first butterfly at the CG then looks horizontally toward the ascending side of the wheel, he will note that the maximum vertical ascending velocity of the weights is given by:
Vmaxasc = 2πrω
where π or pi is the usual transcendental number used extensively in geometry problems with a value of about 3.1415927... , r is the distance from the location of the CG of the weights to any particular weight, and the Greek letter ω is a standard symbol for the rotational rate or frequency of a wheel and represents rotations per unit of time.
and, when he looks horizontally toward the descending side of the wheel, he will note that the maximum vertical descending velocity is given by:
Vmaxdes = - 2πrω
Once he has this information, he then wants to determine the average value of these two maximum vertical velocities. This average value was represented in my Community Buzz thread "...Updates" thread by that capital V with the bar over it. If I had thought the use of that standard symbol for the arimetical average of two numbers would have caused confusion, then I would have called it something else. So, let's do that now. Instead of V "bar", let's just call it Vaver.
So, the first intelligent butterfly will write:
Vaver = { (Vmaxasc) + (Vmaxdes) } / 2 = { (2πrω) + (- 2πrω) } / 2
Vaver = { (2πrω) - (2πrω) } / 2 = 0 /2 = 0
Vaver = 0
From this, the first butterfly hovering at the CG of the weights that rotate around him concludes that the vertical velocities of the weights horizontal to him as they both ascend and descend on opposite side of him are equal and opposite in magnitude and, thus, their average value is zero.
This first butterfly then says that because there is no net vertical motion of the gravity wheel's weights in the Earth's gravity field, then there can therefore be no net change in the gravitational potential energy of the weights and, as the skeptics had been telling him all along, such a system of weights can not output any energy to the gravity wheel within whose hollow spinning drum he hovers. He is, however, puzzled as to why the wheel was able to accelerate from a standstill and is now operating various energy consuming devices in its environment.
Now for the second butterfly located on the axle of the rotating gravity wheel. What he sees happening is quite different from what the first butterfly observes.
The second butterfly also looks out horizontally toward the ascending and descending sides of the wheel, but he does it from the location of the wheel's axle and not from that of the weights' CG. What does the second butterfly note?
First, he notes that the maximum vertical ascent velocity of the weights, Vmaxasc, is given by:
Vmaxasc = 2π(r-d)ω
where d is the distance, measured horizontally, from the wheel's axle to the CG where the first butterfly is still hovering.
Then, he notes that the maximum vertical descent velocity of the weights, Vmaxdes, is given by:
Vmaxdes = - 2π(r+d)ω
When he then proceeds to calculate the average value of these two maximum vertical velocities, as seen from the wheel's axle, he obtains:
Vaver={ (Vmaxasc) + (Vmaxdes) } / 2 ={ (2π(r-d)ω) + (- 2π(r+d)ω) } / 2
Vaver = { (2π(r-d)ω) - (2π(r+d)ω) } / 2
Vaver = {2πω(r-d-r-d)} / 2 = πω(-2d)
and, finally, what I consider to be "Bessler's 4th Law of Motion":
Vaver = - 2πdω
Thus, the second butterfly concludes, correctly, that the average vertical velocity of the weights, with respect to the wheel's axle, is a negative or downward one! He further notes that this situation will exist in ANY overbalanced wheel that has the CG of its driving weights horizontally offset a distance d from the wheel's axle. And, of course, the wheel must have a non-zero value for the rotation rate ω
Now, the second butterfly on the wheel's axle wonders what the consequences of this situation will be. Being a student of gravity physics, he knows that when a weight of mass, M, drops from an height of h above the ground down onto the ground, it will lose an amount of gravitational energy, ΔPEgrav, which is given by the expression:
ΔPEgrav = Mg Δh
in which g is the acceleration due to gravity near the Earth's surface (about 32 feet per second per second).
The second axle sitting butterfly now realizes something very interesting. He knows that the driving weights, with a total mass of M, rotating about the axle of the overbalanced gravity wheel that he sits in have a average velocity that is always downward...just as though they were dropping in a gravity field although their average downward velocity remains constant for a given wheel rotation rate and is not undergoing acceleration as would be the case for objects in freefall. He also knows the amount of time it takes the wheel to complete a single rotation. That is simply the reciprocal of its rotation rate, ω, or 1/ω. He then uses this to derive a simple expression for the continuous loss of gravitational potential energy by the weights in the wheel he resides in per single wheel rotation. The expression he derives is:
ΔPEgrav = Mg(- 2πdω)(1/ω)
ΔPEgrav = - 2πMgd
Now he has an expression which tells him that the wheel will, on average, continuously lose gravitational potential energy during each and every one of its rotations even though the weights it contains are moving around a closed path within the wheel! He realizes that this situation will take place in any type of system in which there are two separate centers of rotation. In the case of an overbalanced gravity wheel, the two centers of rotation are at the location of the CG of its weights and at the wheel's axle.
Now that he knows how much gravitational potential energy is lost per wheel rotation, finding the constant power output, P, of the wheel at any given rotation rate, ω, is a simple matter. He need only divide the expression for the loss of gravitational potential energy per wheel rotation by the time required for the wheel to complete a single rotation which, as mentioned above, is simply the reciprocal of its rotation rate or 1/ω. Doing this yields:
P = ΔPEgrav / (1/ω) = (- 2πMgd)/(1/ω)
P = - 2πMgdω
Well, fellow Wheel Wizards that is basically all you need to know about Bessler's 4th Law of Motion. It's really very simple once you get the hang of it.
If, however, the fixed CG of the weights is not located directly horizontal to the axle of the wheel, then there will be a reduction in the average downward velocity of the weights with respect to the axle.
In this case, it will be necessary to use a modified version of Bessler's 4th Law of Motion to account for the situation. In this case, one must know the distance from the wheel's axle to the CG of the weights, d, AND the "dip" angle, φ of the CG below a line passing horizontally through the axle. The modified version of Bessler's 4th Law of Motion then becomes:
Vaver = - 2πdωcosφ
I think if the reader will follow this derivation and also view what I posted on my "...Updates" thread over in the Community Buzz forum, then he will gain a greater appreciation of what was going on inside of Bessler's wheels. No, understanding this new law of motion will not give you his secret mechanism on a silver platter...you will still have to work VERY hard to find that. But, it will, at a minimum, show you that what you quest for is not physically impossible as the the many skeptics of this subject constantly tell us.
For the reader who still can not follow the simple step by step derivation I have given above, I have attached a nice chart below which summarizes the important formulas associated with Bessler's 4th Law of Motion
ken
- Attachments
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- This summarizes the details of Bessler's 4th Law of Motion
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
Vaver = -2(√2)πdωcosφ
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John Collins
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re: Pair of Pairs
Sorry Ken , but I think you're spouting complete garbage, but that's just my opinion.
John Collins
John Collins
re: Pair of Pairs
Ken. I have a sandwich and my friend and I are both staring at it. I'm taking bites from the sandwich. He doesn't know this because I'm doing this at an angle where he can't see where I'm taking the bite from, so he doesn't get jealous. Even though I've just taken a bite, it's a magic sandwich and it gains energy in some unknown and magical way and then regrows the part that was bitten off almost instantly. My friend never sees this happening so to him the sandwich always stays the same. I have another friend beside me and from our perspective the sandwich is always losing energy/mass, yet he doesn't really know why I can never finish the sandwich and make it disappear. He's starting to get paranoid and is suspecting the other friend has something to do with it.
What's really the difference between this and your hypothesis?
What's really the difference between this and your hypothesis?
re: Pair of Pairs
Ken,
You're making a multitude of errors in both logic and in how you use your formulas. Also you make many assumption that I'm not sure if even you know that you're making them.
You're trying to make each weight move at two different velocities at the same time. Such cannot be done. First you calculate the velocity as if the weights rotate around the CG at a constant rotational speed. Then you calculate their velocity as if they rotate around the axle at a constant rotational speed. It's like taking two gears with differing number of teeth and trying to make them mesh with both rotating at a same speed. It cannot be done.
I nominate Ken as the supreme hogwash slinger. How can Bessler wheel followers be taken seriously with the likes of this? First it was rest mass being converted into energy and now this nonsense. And Ken has the nerve to put Bessler's name on this hogwash.
You're making a multitude of errors in both logic and in how you use your formulas. Also you make many assumption that I'm not sure if even you know that you're making them.
You're trying to make each weight move at two different velocities at the same time. Such cannot be done. First you calculate the velocity as if the weights rotate around the CG at a constant rotational speed. Then you calculate their velocity as if they rotate around the axle at a constant rotational speed. It's like taking two gears with differing number of teeth and trying to make them mesh with both rotating at a same speed. It cannot be done.
I nominate Ken as the supreme hogwash slinger. How can Bessler wheel followers be taken seriously with the likes of this? First it was rest mass being converted into energy and now this nonsense. And Ken has the nerve to put Bessler's name on this hogwash.