Generating more meaningful debate
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re: Generating more meaningful debate
The weights could pull out using centrifugal force in a T shape and then push in a weight. Then the further to the rim weights are stored energy from the centrifugal force. The velocity of the wheel will not change if the weights evenly shift in and out in the same amount however with more weight closer to the rim, is that stored energy from the centrifugal force by making the wheel heavier? Maybe Bessler's giant pendulum outside of his wheel was meant to add momentum to this wheel that stores energy by being heavier when in motion.
"It's not the size of the dog in the fight, it's the size of the fight in the dog." - Mark Twain
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KE = [1/2 mv^2] = 3/2 KT is calculus.I can understand that. Conservation of Angular Momentum might be a little more complicated. There are 2 different aspects to angular momentum. With an over balanced wheel there are 2 opposing weights.
If one is at 6 o'clock (180°) and the other is at 12:00 o'clock (0°), then as the wheel rotates clockwise the bottom weight moving from 6:00 o'clock to 9:00 o'clock moves towards the axle while the descending weight that is swinging down moves to the right of the center line through the axle.
To simplify as in algebra, x + .5x /2 < x. If x is mass times distance = torque then the numbers given are for a rotation of 90°.
Then the next 90° of rotation is 0.5x < x. This allows for 180° of rotation as being .75x + .5x/2 < x or .625x < x.
that would allow for net force and acceleration to be calculated. I guess algebra works well enough. Anyway that is what conservation of angular momentum seems to allow for and is what I will be attempting to demonstrate.
If one is at 6 o'clock (180°) and the other is at 12:00 o'clock (0°), then as the wheel rotates clockwise the bottom weight moving from 6:00 o'clock to 9:00 o'clock moves towards the axle while the descending weight that is swinging down moves to the right of the center line through the axle.
To simplify as in algebra, x + .5x /2 < x. If x is mass times distance = torque then the numbers given are for a rotation of 90°.
Then the next 90° of rotation is 0.5x < x. This allows for 180° of rotation as being .75x + .5x/2 < x or .625x < x.
that would allow for net force and acceleration to be calculated. I guess algebra works well enough. Anyway that is what conservation of angular momentum seems to allow for and is what I will be attempting to demonstrate.
Re: re: Generating more meaningful debate
preoccupiedpreoccupied wrote:john.smith is James Lingaard since he is insulting ab hammer already. Or everybody who loves MT 20 hates AB Hammer. I'm not a forensic scientist but I think everybody can agree that within a reasonable degree of scientific certainty john.smith is james lingaard now. He just went on a rant insulting AB Hammers ability to earn a living quoting the bible and deleted it.
Don't worry about it, I have known for a long time. I posted to show I myself have been caught by wanting something to much. Before I started receiving disability I was desperate due to my health problems and fear of the future. That is not a problem now. I have had both of my knees operated on and it improved my standing ability but the damage was to bad and to long to ever reach a normal status. My last truck repair cost me almost $1,500 and I still have to work despite my pain to pay bills. But I need to have 30% more to get my 100% Disability and if received I'll have plenty of time for wheel work and test. I'm a bit ill today or I would probably not of done any post.
My advice for those trying to figure out Bessler is. If how you look at Bessler doesn't work? Change the way you look at Bessler for new direction.
"Our education can be the limitation to our imagination, and our dreams"
So With out a dream, there is no vision.
Old and future wheel videos
https://www.youtube.com/user/ABthehammer/videos
Alan
So With out a dream, there is no vision.
Old and future wheel videos
https://www.youtube.com/user/ABthehammer/videos
Alan
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re: Generating more meaningful debate
At E1, what you're missing is that the retraction disc allows the wheel t be a bound system with the earth. This si what allows for the earth's momentum to be transferred to the wheel. Conservation of Momentum allows for this.
If the weight moving from 6:00 o'clock to 9:00 o'clock used the wheels energy to retract then the momentum of the wheel would remain unchanged.
Simple question for you, can a 375 gram weight be moved 15 cm's without using any energy ? The obvious answer is no. work is mass x distance. The retraction disc in my opinion at the moment will allow for a 375 gram weight to be moved 15 cm's and possibly create energy doing this. That is where the laws of physics would seem to be violated but conservation of angular momentum does seem to allow for it.
And Bessler's quote once again; Around the firmly placed horizontal axis is a rotating disc (low or narrow cylinder) which resembles a grindstone. This disc can be called the principle piece of my machine
http://besslerwheel.com/writings/das_triumphans.html
Later if my demonstration is successful I will go into why Bessler said it rotated. It is what he built was complex. I will be buying the wood in the morning to build the frame.
Ab Hammer, why I have Bessler's wheel and you don't is as you said, you want a 100% disability rating. I have had it much worse than you yet will say your family does not want you being successful so your family can have a better life. Still, for what I've been through not sure if I'll be very thankful.
A bit much just to have a life. And I guess when I see how much you want and how little you do I guess I am supposed to be glad I had it so bad so I would stay motivated. Just not sure if Bessler's wheel is worth what I've been put through.
p.s., have always thought it would take one person working alone because as Mr. Vibrating shows (nothing personal) people do have different opinions. at the same time a working build will get me surgery that I've been needing. And with everyone else, the same things that were said a year ago are being repeated. No progress has been made.
If the weight moving from 6:00 o'clock to 9:00 o'clock used the wheels energy to retract then the momentum of the wheel would remain unchanged.
Simple question for you, can a 375 gram weight be moved 15 cm's without using any energy ? The obvious answer is no. work is mass x distance. The retraction disc in my opinion at the moment will allow for a 375 gram weight to be moved 15 cm's and possibly create energy doing this. That is where the laws of physics would seem to be violated but conservation of angular momentum does seem to allow for it.
And Bessler's quote once again; Around the firmly placed horizontal axis is a rotating disc (low or narrow cylinder) which resembles a grindstone. This disc can be called the principle piece of my machine
http://besslerwheel.com/writings/das_triumphans.html
Later if my demonstration is successful I will go into why Bessler said it rotated. It is what he built was complex. I will be buying the wood in the morning to build the frame.
Ab Hammer, why I have Bessler's wheel and you don't is as you said, you want a 100% disability rating. I have had it much worse than you yet will say your family does not want you being successful so your family can have a better life. Still, for what I've been through not sure if I'll be very thankful.
A bit much just to have a life. And I guess when I see how much you want and how little you do I guess I am supposed to be glad I had it so bad so I would stay motivated. Just not sure if Bessler's wheel is worth what I've been put through.
p.s., have always thought it would take one person working alone because as Mr. Vibrating shows (nothing personal) people do have different opinions. at the same time a working build will get me surgery that I've been needing. And with everyone else, the same things that were said a year ago are being repeated. No progress has been made.
Last edited by john.smith on Wed Oct 04, 2017 11:00 pm, edited 1 time in total.
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To me, the most meaningful area for debate concerns the energy source, ie. the symmetry break.
In all of classical mechanics, there is only one, single, exploit capable of generating both excess energy and momentum. It is the only form of interaction possible from which a thermodynamic gain (or loss) can be calculated; the only tangible mathematical solution that can even be contemplated realistically.
Gravity and mass are constant, and any closed-loop trajectory through static fields yields zero net energy, therefore there is no form of leverage, whether angular or linear or any combination, that can trade force for excess displacement or vice versa; in a vertical rotating system all cycling weights must travel equal distance up as down, and so gravitational input and output energies can only ever be equal.
There is simply nothing you can do with these simple, fundamental maths, to create or destroy energy - gravitational potential energy is equal to mass times gravity times height, regardless of time, because neither gravity, mass or distance (ie. space itself) change over time - they're all static fields..
And Bessler himself quite cleary refutes any such possibility as a futile endeavour.
However it is equally clear that his wheels - especially the one-way wheels - were perpetually over-balancing. The only way a statorless vertical wheel can remain under static torque is if that force was due to an OB weight; releasing the wheel to freely rotate thus allowed this weight to get lower, outputting its GPE; the fact that the wheels had no stator against which to apply torque, leaves an output of GPE as the only real option.
Yet there is no way this output GPE can itself be directly converted back into equal input GPE, much less an excess - indeed, as noted above, since his wheels immediately spun up to speed, we can be certain that most of this output GPE is in fact converted straight into rotational kinetic energy (RKE), so there is definitely no direct conversion of output GPE back into input GPE, as by any conventional form of leverage..
Yet, as also noted above, the GPE must nonetheless be restored; if weights keep falling then they must also be rising at some point..
Therefore, the trick must involve conversion of RKE, or else some component or derivative of it, back into GPE; in short, the symmetry break has to involve processes integral to the generation and management of RKE interactions. Specifically, we know we have this initial conversion of GPE to RKE, and then a conversion of RKE back into GPE; so the opportunity for gain must arise at some stage during this process of conversion to RKE, or re-conversion back into GPE..
We can put a slightly finer point on this and state that we require an interaction that either converts a given GPE into excess RKE, or else converts a given RKE into excess GPE.
However, as noted above, the very notion of "excess GPE" is troublesome, if not self-contradictory - if the weights are travelling equal distance up as down, and both gravity and rest mass are constants, then there can never be any more or less GPE than unity - its value remaining constant regardless of all other factors. GPE-in can only ever equal GPE-out.
Which leaves us with just one possibility; excess RKE.
And this is the only viable exploit, which i alluded to initially above - excess RKE is mathematically tenable. Unavoidable in fact... provided one can muster an effective exception to Newton's 3rd law.
Quite simply, if we can perform a reactionless acceleration from within an already-moving system, then we create a symmetry break between the input energy we've spent to accelerate that mass, compared to its actual rise in kinetic energy as measured by a static observer external to the moving system.
To paint a very simple picture, suppose you're floating through space at some modest speed, and you throw a tennis ball forwards, towards your direction of travel; ordinarily, this would also propel yourself backwards, imparting equal opposite negative momentum to that positive momentum you imparted to the ball: a static observer would see your ball accelerate, but would also see you decelerate... and your net system momentum would remain constant.
Now however suppose what happens if you can somehow throw the ball without incurring that equal opposing deceleration? So, you impart exactly the same amount of input energy as before, throwing the ball just as hard, but this time, you're not propelled backwards, and all of your input energy instead is imparted to the ball...
..the static observer would see your net system momentum rise, by the amount you've imparted to the ball, without the negative component you'd otherwise have forfeited yourself. But furthermore, they'd also see that the kinetic energy of the ball had risen by an amount greater than that you yourself have expended!
For instance, you could accurately apportion your input energy by pre-loading it into a spring. Upon release, the static observer would expect to see your net system KE rise by precisely that amount of PE, while the net system momentum remained constant. But with no counter-force to the acceleration, what they would instead see is that the KE of your projectile rises by more than the energy loaded into your spring.
The reason for this divergence is that kinetic energy scales as the half square of rising velocity, and so while a 1 kg-m/s acceleration from zero to 1 meter / sec only costs 1/2 a Joule, that same kilogram-meter-second of momentum costs a whopping 10.5 Joules if you're already at 10 meters / sec and accelerating up to 11 meters / sec; same amount of momentum increase, but twenty times the energy cost, and value..
Yes, you read that right - a half Joule of energy spent in a reactionless acceleration at an ambient speed of 10 m/s is worth 10.5 Joules, in other words, 20x over-unity.
And so by elimination of all other possibilities, we can logically deduce with near-certainty that the solution we are looking for - the only workable solution, even in principle, that can generate both excess momentum and energy - is an effective reactionless acceleration.
This is what Bessler was doing. By default; there is no other possibility.
An input GPE was used to generate angular momentum. Aboard this angular momentum, an acceleration then occurred, but without applying the usual equal opposing deceleration back to the system. Thus, the kinetic energy value of that accelerated mass was greater than the GPE that caused the initial rotation, and so was able to restore it, with excess energy and momentum left over.
Clearly, as evinced by the fact that he never demonstrated a horizontal wheel, gravity's role is not simply to provide a stable currency of input and output loads, which otherwise could've been met by springs; it must therefore be a key component of the effective N3 violation.
In short, there is something you can do in a gravity field that results in an effective thwarting of Newton's 3rd law, as it would normally apply.
This might only be a temporary side-stepping of N3, just a foot in the door, but enough to get one OU step ahead, spending say 1 Joule each input but garnering a 1.5 Joule output. Thus the system only ever dips a toe into becoming a runaway momentum, in a 'three steps forward, two steps back' kind of way..
..or else, it might be a full-blown runaway momentum; by which i mean that net system momentum builds up over successive reactionless accelerations. In such a scenario, an input of 1 J at 1 RPM might yeild 1.5 J, but as the system accelerates, that same 1 J input at higher speed might be worth 150 J or more..
In other words, the exploit might only allow us to get ahead by half of an interaction, so we only temporarily gain excess momentum and energy, by the same amount each time, each successive cycle, just enough to recoup a reliable gain... or else, it might be cumulative, and so the gain margins snowballing with rising speed, until some other limiting factor kicks in.. you could intepret some figurative resemblence to "Jacob's ladder" in either case, as a kind of cascade of sequential or even nested reactionless accelerations...
As an example of the kinds of angles to be investigating, consider the following two propositions:
1) In an interaction (momentum exchange) between the inertia of a rest mass vs that of an MoI; the change in momentum of a rest mass is relative to its velocity FoR, whereas angular momentum is absolute (as manifested in centrifugal / centripetal forces). In other words, an interaction (mutual acceleration or deceleration) between angular vs linear inertias might present a basis for thermodynamic decoupling of respective KE's as a function of relative vs intrinsic inertial forces..
2) The excess kinetic energy generated by a reactionless acceleration evolves with the square of excess velocity, whereas the energy of a gravitational interaction scales linearly with displacement. Hence, if the opportunity for the former condition is somehow met by the latter (for example gravitationally countering a recoil, say), an input / output disunity may arise.
With respect to the first proposal, consider that most of Bessler's imagery, from the Kassel engravings to most of Machinen Tractate, deals pretty much exclusively with angular vs linear interactions. It's all cranks, rocker arms and conrods, the latter portions of MT almost obsessing about playing angular against linear motions, and back into angular again. This is perhaps the overriding theme of all his images, which most share in common. It's thus not really all that controversial to suppose that angular vs linear inertias were very much something of a preoccupation for him..
With regards to the second, this is one example of a regime that might offer both an asymmetry as well as the self-limiting factor Bessler's wheels exhibited; the constant force of gravity would represent an ever-smaller force/time window within which to counter accelerating impulse reactions.
Whatever, these are just two example investigations; the goal is the same - what we're looking for, what Bessler definitely had, was some form of reactionless acceleration.
If you can accelerate a mass without suffering an equal opposite counter-acceleration, then that there mass is OU, right here, right now. You can harvest it immediately and directly, in any form you please. Convert it to GPE, or load a spring, whatever.
The bottom line is that it'll likely cost some energy to implement a reactionless acceleration, since motion is required, after all.. However, because KE=1/2mV^2, and conservation of momentum and energy depend upon Newton's 3rd law holding, the fruits of that endeavour are a guaranteed gain in energy and momentum that can be empirically calculated as a direct function of the momentum asymmetry.
This is not simply an opinion piece, but a statement of facts; the lay of the land, and our only play to get an objective upper hand. Classical energy and momentum can be created and destroyed, by implementing an effective reactionless acceleration, even if only under temporary conditions; a cost is implicit, but the benefits truly, mathematically outweight those costs, and by a margin only limited by practical constraints... remember, a reactionless 10 kg-m/s acceleration could cost as little as 5 Joules to perform, and yet be worth 100 Joules or 1,000 J when harvested immediately afterwards, depending only on the current speed of the net system.. the faster it goes, the greater the gain margin, per the rising discrepancy between relative vs absolute velocities of the interacting internal angular inertias.
Whatever other clues anyone else may have found with regards to the principle, this is, by logical deduction, the path along which the purpose of any prospective mechanism must align. There simply is no other summit to aim for, much less a route up it. Reactionless accelerations are the only possible explanation for the properties of Bessler's wheels.
In all of classical mechanics, there is only one, single, exploit capable of generating both excess energy and momentum. It is the only form of interaction possible from which a thermodynamic gain (or loss) can be calculated; the only tangible mathematical solution that can even be contemplated realistically.
Gravity and mass are constant, and any closed-loop trajectory through static fields yields zero net energy, therefore there is no form of leverage, whether angular or linear or any combination, that can trade force for excess displacement or vice versa; in a vertical rotating system all cycling weights must travel equal distance up as down, and so gravitational input and output energies can only ever be equal.
There is simply nothing you can do with these simple, fundamental maths, to create or destroy energy - gravitational potential energy is equal to mass times gravity times height, regardless of time, because neither gravity, mass or distance (ie. space itself) change over time - they're all static fields..
And Bessler himself quite cleary refutes any such possibility as a futile endeavour.
However it is equally clear that his wheels - especially the one-way wheels - were perpetually over-balancing. The only way a statorless vertical wheel can remain under static torque is if that force was due to an OB weight; releasing the wheel to freely rotate thus allowed this weight to get lower, outputting its GPE; the fact that the wheels had no stator against which to apply torque, leaves an output of GPE as the only real option.
Yet there is no way this output GPE can itself be directly converted back into equal input GPE, much less an excess - indeed, as noted above, since his wheels immediately spun up to speed, we can be certain that most of this output GPE is in fact converted straight into rotational kinetic energy (RKE), so there is definitely no direct conversion of output GPE back into input GPE, as by any conventional form of leverage..
Yet, as also noted above, the GPE must nonetheless be restored; if weights keep falling then they must also be rising at some point..
Therefore, the trick must involve conversion of RKE, or else some component or derivative of it, back into GPE; in short, the symmetry break has to involve processes integral to the generation and management of RKE interactions. Specifically, we know we have this initial conversion of GPE to RKE, and then a conversion of RKE back into GPE; so the opportunity for gain must arise at some stage during this process of conversion to RKE, or re-conversion back into GPE..
We can put a slightly finer point on this and state that we require an interaction that either converts a given GPE into excess RKE, or else converts a given RKE into excess GPE.
However, as noted above, the very notion of "excess GPE" is troublesome, if not self-contradictory - if the weights are travelling equal distance up as down, and both gravity and rest mass are constants, then there can never be any more or less GPE than unity - its value remaining constant regardless of all other factors. GPE-in can only ever equal GPE-out.
Which leaves us with just one possibility; excess RKE.
And this is the only viable exploit, which i alluded to initially above - excess RKE is mathematically tenable. Unavoidable in fact... provided one can muster an effective exception to Newton's 3rd law.
Quite simply, if we can perform a reactionless acceleration from within an already-moving system, then we create a symmetry break between the input energy we've spent to accelerate that mass, compared to its actual rise in kinetic energy as measured by a static observer external to the moving system.
To paint a very simple picture, suppose you're floating through space at some modest speed, and you throw a tennis ball forwards, towards your direction of travel; ordinarily, this would also propel yourself backwards, imparting equal opposite negative momentum to that positive momentum you imparted to the ball: a static observer would see your ball accelerate, but would also see you decelerate... and your net system momentum would remain constant.
Now however suppose what happens if you can somehow throw the ball without incurring that equal opposing deceleration? So, you impart exactly the same amount of input energy as before, throwing the ball just as hard, but this time, you're not propelled backwards, and all of your input energy instead is imparted to the ball...
..the static observer would see your net system momentum rise, by the amount you've imparted to the ball, without the negative component you'd otherwise have forfeited yourself. But furthermore, they'd also see that the kinetic energy of the ball had risen by an amount greater than that you yourself have expended!
For instance, you could accurately apportion your input energy by pre-loading it into a spring. Upon release, the static observer would expect to see your net system KE rise by precisely that amount of PE, while the net system momentum remained constant. But with no counter-force to the acceleration, what they would instead see is that the KE of your projectile rises by more than the energy loaded into your spring.
The reason for this divergence is that kinetic energy scales as the half square of rising velocity, and so while a 1 kg-m/s acceleration from zero to 1 meter / sec only costs 1/2 a Joule, that same kilogram-meter-second of momentum costs a whopping 10.5 Joules if you're already at 10 meters / sec and accelerating up to 11 meters / sec; same amount of momentum increase, but twenty times the energy cost, and value..
Yes, you read that right - a half Joule of energy spent in a reactionless acceleration at an ambient speed of 10 m/s is worth 10.5 Joules, in other words, 20x over-unity.
And so by elimination of all other possibilities, we can logically deduce with near-certainty that the solution we are looking for - the only workable solution, even in principle, that can generate both excess momentum and energy - is an effective reactionless acceleration.
This is what Bessler was doing. By default; there is no other possibility.
An input GPE was used to generate angular momentum. Aboard this angular momentum, an acceleration then occurred, but without applying the usual equal opposing deceleration back to the system. Thus, the kinetic energy value of that accelerated mass was greater than the GPE that caused the initial rotation, and so was able to restore it, with excess energy and momentum left over.
Clearly, as evinced by the fact that he never demonstrated a horizontal wheel, gravity's role is not simply to provide a stable currency of input and output loads, which otherwise could've been met by springs; it must therefore be a key component of the effective N3 violation.
In short, there is something you can do in a gravity field that results in an effective thwarting of Newton's 3rd law, as it would normally apply.
This might only be a temporary side-stepping of N3, just a foot in the door, but enough to get one OU step ahead, spending say 1 Joule each input but garnering a 1.5 Joule output. Thus the system only ever dips a toe into becoming a runaway momentum, in a 'three steps forward, two steps back' kind of way..
..or else, it might be a full-blown runaway momentum; by which i mean that net system momentum builds up over successive reactionless accelerations. In such a scenario, an input of 1 J at 1 RPM might yeild 1.5 J, but as the system accelerates, that same 1 J input at higher speed might be worth 150 J or more..
In other words, the exploit might only allow us to get ahead by half of an interaction, so we only temporarily gain excess momentum and energy, by the same amount each time, each successive cycle, just enough to recoup a reliable gain... or else, it might be cumulative, and so the gain margins snowballing with rising speed, until some other limiting factor kicks in.. you could intepret some figurative resemblence to "Jacob's ladder" in either case, as a kind of cascade of sequential or even nested reactionless accelerations...
As an example of the kinds of angles to be investigating, consider the following two propositions:
1) In an interaction (momentum exchange) between the inertia of a rest mass vs that of an MoI; the change in momentum of a rest mass is relative to its velocity FoR, whereas angular momentum is absolute (as manifested in centrifugal / centripetal forces). In other words, an interaction (mutual acceleration or deceleration) between angular vs linear inertias might present a basis for thermodynamic decoupling of respective KE's as a function of relative vs intrinsic inertial forces..
2) The excess kinetic energy generated by a reactionless acceleration evolves with the square of excess velocity, whereas the energy of a gravitational interaction scales linearly with displacement. Hence, if the opportunity for the former condition is somehow met by the latter (for example gravitationally countering a recoil, say), an input / output disunity may arise.
With respect to the first proposal, consider that most of Bessler's imagery, from the Kassel engravings to most of Machinen Tractate, deals pretty much exclusively with angular vs linear interactions. It's all cranks, rocker arms and conrods, the latter portions of MT almost obsessing about playing angular against linear motions, and back into angular again. This is perhaps the overriding theme of all his images, which most share in common. It's thus not really all that controversial to suppose that angular vs linear inertias were very much something of a preoccupation for him..
With regards to the second, this is one example of a regime that might offer both an asymmetry as well as the self-limiting factor Bessler's wheels exhibited; the constant force of gravity would represent an ever-smaller force/time window within which to counter accelerating impulse reactions.
Whatever, these are just two example investigations; the goal is the same - what we're looking for, what Bessler definitely had, was some form of reactionless acceleration.
If you can accelerate a mass without suffering an equal opposite counter-acceleration, then that there mass is OU, right here, right now. You can harvest it immediately and directly, in any form you please. Convert it to GPE, or load a spring, whatever.
The bottom line is that it'll likely cost some energy to implement a reactionless acceleration, since motion is required, after all.. However, because KE=1/2mV^2, and conservation of momentum and energy depend upon Newton's 3rd law holding, the fruits of that endeavour are a guaranteed gain in energy and momentum that can be empirically calculated as a direct function of the momentum asymmetry.
This is not simply an opinion piece, but a statement of facts; the lay of the land, and our only play to get an objective upper hand. Classical energy and momentum can be created and destroyed, by implementing an effective reactionless acceleration, even if only under temporary conditions; a cost is implicit, but the benefits truly, mathematically outweight those costs, and by a margin only limited by practical constraints... remember, a reactionless 10 kg-m/s acceleration could cost as little as 5 Joules to perform, and yet be worth 100 Joules or 1,000 J when harvested immediately afterwards, depending only on the current speed of the net system.. the faster it goes, the greater the gain margin, per the rising discrepancy between relative vs absolute velocities of the interacting internal angular inertias.
Whatever other clues anyone else may have found with regards to the principle, this is, by logical deduction, the path along which the purpose of any prospective mechanism must align. There simply is no other summit to aim for, much less a route up it. Reactionless accelerations are the only possible explanation for the properties of Bessler's wheels.
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Re: re: Generating more meaningful debate
preoccupied wrote:You are quoting formulas. Well when you learn pre algebra you will piece things together using blocks to understand what an equation is. If you have a lever and you put velocity closer to the axle, then you have the exact same amount of speed traveling a shorter distance. What you want is speed to equal power, to turn a shorter distance traveled on a wheel that is faster to also be more leverage because it is faster.
If a weight moves towards the rim from position 5 to position 6 while another weight moves from position 4 to position 3, is it the same as moving a weight from position 6 to position 7 while moving a weight from position 4 to position 3? How would you go about it? Would you move a single weight incrementally towards the rim while moving several weights a short distance towards the axle. Or would you move a single weight closer and closer to the axle and several weights a short distance towards the rim?
The weights store centrifugal force in a T shape by using the T shape to pull the weights further from the axle towards the rim by gear teeth moving along a track. The movement along the track pulls an equal amount of weight near the axle inward by a pulley. This makes the wheel heavier. If you move a weight form position 4 to 3 and a weight to position 5 to 6 and then move a weight to position 4 to 3 and a weight to position 6 to 7, the new position 6 to 7 is a heavier position change than the previous 4 to 3 and 5 to 6, however both changes do not change the velocity of the wheel because a weight is moving in and out the same amount. It should be true that a heavier wheel takes longer to stop turning as a flywheel and can withstand a load longer. A spring will pull all mechanics back to their original position after the pendulum stops adding more force to the wheel. The pendulum has a spring that recharges it. It's twisted. When the pendulum is going in one direction the spring will lift it higher and when it's going in the other direction it will do nothing to the pendulum. It's a ratchet. The heavy flywheel will recharge the spring on the pendulum until it's at maximum and the pendulum will increase the velocity of the flywheel and store more centrifugal force in the weights. When the weight of the flywheel is greater than the velocity necessary to move one set of weights in and out the perpetual motion machine will work on its own forever. This is what Bessler must have meant by a preponderance of force in his German writing. I know that there was a heated debate at least on Jim_Mich's end of conversations disputing if Bessler's writings meant a preponderance of Weight or a preponderance of Force. This Bessler Wheel that I'm sharing is a preponderance of force, in the form of stored energy in the form of flywheel increased weight from centrifugal force pulling weights inwards and outwards at the same time.preoccupied wrote:The weights could pull out using centrifugal force in a T shape and then push in a weight. Then the further to the rim weights are stored energy from the centrifugal force. The velocity of the wheel will not change if the weights evenly shift in and out in the same amount however with more weight closer to the rim, is that stored energy from the centrifugal force by making the wheel heavier? Maybe Bessler's giant pendulum outside of his wheel was meant to add momentum to this wheel that stores energy by being heavier when in motion.
"It's not the size of the dog in the fight, it's the size of the fight in the dog." - Mark Twain
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re: Generating more meaningful debate
Re.the topic of this thread,i can understand what A.B. Hammer and others are saying some monetary award for our labours would be nice and i know there are many carpet beggars looking at this site,but what we should do I.M.O..is show or explain our"failures",its possible with this bigger audience some one will come forward privately or otherwise with a solution.....A cut Rose displayed in a vase,admired and seen by many is better than one that dies in the bud and seen by no-one.......i say the sooner the better for everyone J.C.,for a more interesting Forum..
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Mr. Vibrating made a mistake. When a wheel works perpetually it is considered that the earth looses that amount of momentum. In other words scientists will accept that the earth itself has slowed. This is conservation of momentum. There is no conservation of thermodynamics. Thermodynamics does not cover an over balanced wheel that uses weights.rasselasss wrote:An excellent explanation Mr.Vibrating,the best i have seen yet on this Bessler Forum..i am totally in agreement..
There is a trick to using what Bessler knew but it seems most in here are trained to react to a post. I haven't seen anyone take the time to consider why a weight moves inward when nothing is pulling it in. And this would be why I'm building it. Building takes time and requires a person to slow down and think about what they're doing.
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re: Generating more meaningful debate
Reminds me of "Funeral Blues"..Stop all the clocks by W.H.Auden.
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Is this because you agreed with Mr. Vibrating and I said he is wrong ?
What Mr. Vibrating said;
Gravity and mass are constant, and any closed-loop trajectory through static fields yields zero net energy, therefore there is no form of leverage, whether angular or linear or any combination, that can trade force for excess displacement or vice versa; in a vertical rotating system all cycling weights must travel equal distance up as down, and so gravitational input and output energies can only ever be equal.
That statement should remind you of the poem that you referenced. https://allpoetry.com/Funeral-Blues
People can read it for themselves. Mr. Vibrating stated that perpetual motion is impossible unless it is this; https://www.youtube.com/watch?v=oEkK87m-2B8 or this; https://www.youtube.com/watch?v=HYnJEJN94Rs
With the atmos clock you might need to be aware of Boltzmann's constant; https://www.khanacademy.org/science/in- ... s-constant
what E1 doesn't understand about conservation of angular momentum is what everybody else has missed as well, if anything he brought up something that needs more or better explaining. I'll give it a try.
Inertia is mv^2/r. This simply means that if it's raius is reduced by 1/2 then it's inertia increases by a factor of 2. An example of this is a 1kg weight moving at 2 m/s with a radius of 1m. It has 2kgf/m/s. If it's radius is reduced by 1/2 to 0.5m
then it's inertia is 1kg * (2m/s^2 )/ 0.5 = 4kgf/m/s.
Why does the inertial force double when the radius is divided by 0.5 ? This is because the velocity of the weight doubles as well. This allows for a weight to increases it's force f = mv from f = mv to f = m(2v). Since the weight is moving in opposition to gravity it's relativistic mass would decrease. At the same time it's distance from the axis of rotation is decreasing as well allowing for an overbalance to be realized.
I do have another project that I am pursuing along with Bessler's wheel. it's an experiment in atmospheric chemistry that if successful would answer some questions that climate scientists have. This is one example; The projection of CCl4 remains more uncertain than projections for other ODSs due to our incomplete understanding of the current CCl4 budget (likely a missing source; see Chapter 1).
They don't know where it comes from. I think I do. At the same time I also think I know how Bessler's wheel worked/works.
p.s., ODS is ozone depleting substances
What Mr. Vibrating said;
Gravity and mass are constant, and any closed-loop trajectory through static fields yields zero net energy, therefore there is no form of leverage, whether angular or linear or any combination, that can trade force for excess displacement or vice versa; in a vertical rotating system all cycling weights must travel equal distance up as down, and so gravitational input and output energies can only ever be equal.
That statement should remind you of the poem that you referenced. https://allpoetry.com/Funeral-Blues
People can read it for themselves. Mr. Vibrating stated that perpetual motion is impossible unless it is this; https://www.youtube.com/watch?v=oEkK87m-2B8 or this; https://www.youtube.com/watch?v=HYnJEJN94Rs
With the atmos clock you might need to be aware of Boltzmann's constant; https://www.khanacademy.org/science/in- ... s-constant
what E1 doesn't understand about conservation of angular momentum is what everybody else has missed as well, if anything he brought up something that needs more or better explaining. I'll give it a try.
Inertia is mv^2/r. This simply means that if it's raius is reduced by 1/2 then it's inertia increases by a factor of 2. An example of this is a 1kg weight moving at 2 m/s with a radius of 1m. It has 2kgf/m/s. If it's radius is reduced by 1/2 to 0.5m
then it's inertia is 1kg * (2m/s^2 )/ 0.5 = 4kgf/m/s.
Why does the inertial force double when the radius is divided by 0.5 ? This is because the velocity of the weight doubles as well. This allows for a weight to increases it's force f = mv from f = mv to f = m(2v). Since the weight is moving in opposition to gravity it's relativistic mass would decrease. At the same time it's distance from the axis of rotation is decreasing as well allowing for an overbalance to be realized.
I do have another project that I am pursuing along with Bessler's wheel. it's an experiment in atmospheric chemistry that if successful would answer some questions that climate scientists have. This is one example; The projection of CCl4 remains more uncertain than projections for other ODSs due to our incomplete understanding of the current CCl4 budget (likely a missing source; see Chapter 1).
They don't know where it comes from. I think I do. At the same time I also think I know how Bessler's wheel worked/works.
p.s., ODS is ozone depleting substances
re: Generating more meaningful debate
Besides that shortsightedness being your own problem... one first need [m·v²/r] to deflect momentum towards the center and keep it going in circles. That's not inertia but a force.john.smith wrote:I haven't seen anyone take the time to consider why a weight moves inward when nothing is pulling it in.
...
Inertia is mv^2/r
This force is a structural one and external to that weight. Inertia is the resistance of changing velocity. This centripetal is needed as a cause to effect in such change.
For a meaningful dialog, what are those units?inertia is 1kg * (2m/s^2 )/ 0.5 = 4kgf/m/s
Inertia [ kg·m²]
Force [N] = [kg·m/s²]
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
Re: re: Generating more meaningful debate
Well, well Mr. Smith. Please allow me to be the first to congratulate you! This calls for a very special tootsie roll.john.smith wrote:. . . why I have Bessler's wheel and you don't . . .
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¯\_(ツ)_/¯ the future is here ¯\_(ツ)_/¯
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© 2023 Walter W. Clarkson, LLC
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I believe Mr V is spot on here but what he is actually saying is that we need to find a way to eliminate back emf in a mechanical system. Not trying to take anything away from the thought process here as I believe it is excellent but the concept has been around for quite awhile.MrVibrating wrote:To me, the most meaningful area for debate concerns the energy source, ie. the symmetry break.
In all of classical mechanics, there is only one, single, exploit capable of generating both excess energy and momentum. It is the only form of interaction possible from which a thermodynamic gain (or loss) can be calculated; the only tangible mathematical solution that can even be contemplated realistically.
Gravity and mass are constant, and any closed-loop trajectory through static fields yields zero net energy, therefore there is no form of leverage, whether angular or linear or any combination, that can trade force for excess displacement or vice versa; in a vertical rotating system all cycling weights must travel equal distance up as down, and so gravitational input and output energies can only ever be equal.
There is simply nothing you can do with these simple, fundamental maths, to create or destroy energy - gravitational potential energy is equal to mass times gravity times height, regardless of time, because neither gravity, mass or distance (ie. space itself) change over time - they're all static fields..
And Bessler himself quite cleary refutes any such possibility as a futile endeavour.
However it is equally clear that his wheels - especially the one-way wheels - were perpetually over-balancing. The only way a statorless vertical wheel can remain under static torque is if that force was due to an OB weight; releasing the wheel to freely rotate thus allowed this weight to get lower, outputting its GPE; the fact that the wheels had no stator against which to apply torque, leaves an output of GPE as the only real option.
Yet there is no way this output GPE can itself be directly converted back into equal input GPE, much less an excess - indeed, as noted above, since his wheels immediately spun up to speed, we can be certain that most of this output GPE is in fact converted straight into rotational kinetic energy (RKE), so there is definitely no direct conversion of output GPE back into input GPE, as by any conventional form of leverage..
Yet, as also noted above, the GPE must nonetheless be restored; if weights keep falling then they must also be rising at some point..
Therefore, the trick must involve conversion of RKE, or else some component or derivative of it, back into GPE; in short, the symmetry break has to involve processes integral to the generation and management of RKE interactions. Specifically, we know we have this initial conversion of GPE to RKE, and then a conversion of RKE back into GPE; so the opportunity for gain must arise at some stage during this process of conversion to RKE, or re-conversion back into GPE..
We can put a slightly finer point on this and state that we require an interaction that either converts a given GPE into excess RKE, or else converts a given RKE into excess GPE.
However, as noted above, the very notion of "excess GPE" is troublesome, if not self-contradictory - if the weights are travelling equal distance up as down, and both gravity and rest mass are constants, then there can never be any more or less GPE than unity - its value remaining constant regardless of all other factors. GPE-in can only ever equal GPE-out.
Which leaves us with just one possibility; excess RKE.
And this is the only viable exploit, which i alluded to initially above - excess RKE is mathematically tenable. Unavoidable in fact... provided one can muster an effective exception to Newton's 3rd law.
Quite simply, if we can perform a reactionless acceleration from within an already-moving system, then we create a symmetry break between the input energy we've spent to accelerate that mass, compared to its actual rise in kinetic energy as measured by a static observer external to the moving system.
To paint a very simple picture, suppose you're floating through space at some modest speed, and you throw a tennis ball forwards, towards your direction of travel; ordinarily, this would also propel yourself backwards, imparting equal opposite negative momentum to that positive momentum you imparted to the ball: a static observer would see your ball accelerate, but would also see you decelerate... and your net system momentum would remain constant.
Now however suppose what happens if you can somehow throw the ball without incurring that equal opposing deceleration? So, you impart exactly the same amount of input energy as before, throwing the ball just as hard, but this time, you're not propelled backwards, and all of your input energy instead is imparted to the ball...
..the static observer would see your net system momentum rise, by the amount you've imparted to the ball, without the negative component you'd otherwise have forfeited yourself. But furthermore, they'd also see that the kinetic energy of the ball had risen by an amount greater than that you yourself have expended!
For instance, you could accurately apportion your input energy by pre-loading it into a spring. Upon release, the static observer would expect to see your net system KE rise by precisely that amount of PE, while the net system momentum remained constant. But with no counter-force to the acceleration, what they would instead see is that the KE of your projectile rises by more than the energy loaded into your spring.
The reason for this divergence is that kinetic energy scales as the half square of rising velocity, and so while a 1 kg-m/s acceleration from zero to 1 meter / sec only costs 1/2 a Joule, that same kilogram-meter-second of momentum costs a whopping 10.5 Joules if you're already at 10 meters / sec and accelerating up to 11 meters / sec; same amount of momentum increase, but twenty times the energy cost, and value..
Yes, you read that right - a half Joule of energy spent in a reactionless acceleration at an ambient speed of 10 m/s is worth 10.5 Joules, in other words, 20x over-unity.
And so by elimination of all other possibilities, we can logically deduce with near-certainty that the solution we are looking for - the only workable solution, even in principle, that can generate both excess momentum and energy - is an effective reactionless acceleration.
This is what Bessler was doing. By default; there is no other possibility.
An input GPE was used to generate angular momentum. Aboard this angular momentum, an acceleration then occurred, but without applying the usual equal opposing deceleration back to the system. Thus, the kinetic energy value of that accelerated mass was greater than the GPE that caused the initial rotation, and so was able to restore it, with excess energy and momentum left over.
Clearly, as evinced by the fact that he never demonstrated a horizontal wheel, gravity's role is not simply to provide a stable currency of input and output loads, which otherwise could've been met by springs; it must therefore be a key component of the effective N3 violation.
In short, there is something you can do in a gravity field that results in an effective thwarting of Newton's 3rd law, as it would normally apply.
This might only be a temporary side-stepping of N3, just a foot in the door, but enough to get one OU step ahead, spending say 1 Joule each input but garnering a 1.5 Joule output. Thus the system only ever dips a toe into becoming a runaway momentum, in a 'three steps forward, two steps back' kind of way..
..or else, it might be a full-blown runaway momentum; by which i mean that net system momentum builds up over successive reactionless accelerations. In such a scenario, an input of 1 J at 1 RPM might yeild 1.5 J, but as the system accelerates, that same 1 J input at higher speed might be worth 150 J or more..
In other words, the exploit might only allow us to get ahead by half of an interaction, so we only temporarily gain excess momentum and energy, by the same amount each time, each successive cycle, just enough to recoup a reliable gain... or else, it might be cumulative, and so the gain margins snowballing with rising speed, until some other limiting factor kicks in.. you could intepret some figurative resemblence to "Jacob's ladder" in either case, as a kind of cascade of sequential or even nested reactionless accelerations...
As an example of the kinds of angles to be investigating, consider the following two propositions:
1) In an interaction (momentum exchange) between the inertia of a rest mass vs that of an MoI; the change in momentum of a rest mass is relative to its velocity FoR, whereas angular momentum is absolute (as manifested in centrifugal / centripetal forces). In other words, an interaction (mutual acceleration or deceleration) between angular vs linear inertias might present a basis for thermodynamic decoupling of respective KE's as a function of relative vs intrinsic inertial forces..
2) The excess kinetic energy generated by a reactionless acceleration evolves with the square of excess velocity, whereas the energy of a gravitational interaction scales linearly with displacement. Hence, if the opportunity for the former condition is somehow met by the latter (for example gravitationally countering a recoil, say), an input / output disunity may arise.
With respect to the first proposal, consider that most of Bessler's imagery, from the Kassel engravings to most of Machinen Tractate, deals pretty much exclusively with angular vs linear interactions. It's all cranks, rocker arms and conrods, the latter portions of MT almost obsessing about playing angular against linear motions, and back into angular again. This is perhaps the overriding theme of all his images, which most share in common. It's thus not really all that controversial to suppose that angular vs linear inertias were very much something of a preoccupation for him..
With regards to the second, this is one example of a regime that might offer both an asymmetry as well as the self-limiting factor Bessler's wheels exhibited; the constant force of gravity would represent an ever-smaller force/time window within which to counter accelerating impulse reactions.
Whatever, these are just two example investigations; the goal is the same - what we're looking for, what Bessler definitely had, was some form of reactionless acceleration.
If you can accelerate a mass without suffering an equal opposite counter-acceleration, then that there mass is OU, right here, right now. You can harvest it immediately and directly, in any form you please. Convert it to GPE, or load a spring, whatever.
The bottom line is that it'll likely cost some energy to implement a reactionless acceleration, since motion is required, after all.. However, because KE=1/2mV^2, and conservation of momentum and energy depend upon Newton's 3rd law holding, the fruits of that endeavour are a guaranteed gain in energy and momentum that can be empirically calculated as a direct function of the momentum asymmetry.
This is not simply an opinion piece, but a statement of facts; the lay of the land, and our only play to get an objective upper hand. Classical energy and momentum can be created and destroyed, by implementing an effective reactionless acceleration, even if only under temporary conditions; a cost is implicit, but the benefits truly, mathematically outweight those costs, and by a margin only limited by practical constraints... remember, a reactionless 10 kg-m/s acceleration could cost as little as 5 Joules to perform, and yet be worth 100 Joules or 1,000 J when harvested immediately afterwards, depending only on the current speed of the net system.. the faster it goes, the greater the gain margin, per the rising discrepancy between relative vs absolute velocities of the interacting internal angular inertias.
Whatever other clues anyone else may have found with regards to the principle, this is, by logical deduction, the path along which the purpose of any prospective mechanism must align. There simply is no other summit to aim for, much less a route up it. Reactionless accelerations are the only possible explanation for the properties of Bessler's wheels.
If we can eliminate back emf in any electro-mechanical system we are automatically in overunity territory.
I think that Bessler's clue that everything turned with the wheel is one of the most significant clues in that he is calling for a statorless design. The instant we include a stator, we have introduced back emf in our design in that the equal and opposite back reaction is provided by a stator attached to earth. Even a gravity anchored stator such as in MT13 provides the mechanism by which back emf can be introduced into the system so by calling for a statorless design, Bessler has eliminated 99.999% of all other PM designs - including mine unfortunately.
So the minute we add a earth grounded stator to our design we can be sure we're on the road to failure.
This clue from Bessler I believe is the STRONGEST CLUE TO ITS LEGITIMACY as this amounts to a very sophisticated understanding of the idea that a force is always countered by an opposing force and must be engineered such that the back emf does not oppose wheel rotation.
The Paul Babcock Motor is one that is claimed to work on the principle of elimination of back emf. Back emf is still created but the idea is to OPPOSE THE BACK EMF IN SPACE rather than being GROUNDED TO EARTH. In this way back emf reactions cannot oppose rotation of the wheel. No, I am not Paul Babcock and have no business relationship.
https://www.youtube.com/watch?v=8WOu9uBmPN8
In my own magnetic motor project I basically use Pauls concept but have replaced all the electronics with simple permanent magnets but the overall concept is the same.
Now if I could just come up with a mechanical design that adheres to this principle!
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Re: re: Generating more meaningful debate
Would you understand it better if I said 9.8 n-m's * 2 m/s^2 ?ME wrote:Besides that shortsightedness being your own problem... one first need [m·v²/r] to deflect momentum towards the center and keep it going in circles. That's not inertia but a force.john.smith wrote:I haven't seen anyone take the time to consider why a weight moves inward when nothing is pulling it in.
...
Inertia is mv^2/r
This force is a structural one and external to that weight. Inertia is the resistance of changing velocity. This centripetal is needed as a cause to effect in such change.For a meaningful dialog, what are those units?inertia is 1kg * (2m/s^2 )/ 0.5 = 4kgf/m/s
Inertia [ kg·m²]
Force [N] = [kg·m/s²]
Inertia is mv^2/r
9.8 newtons = 1 kg 9.8 n-m's = 1 kg @ 1 meter
If the weight is moving at 2 meters per second @ 1 meter from the axle then it's velocity increases to 4 meters per second if it's radius is reduced by 1/2 to 50 cm's from the axle.
So when you say > Besides that shortsightedness being your own problem <
Am not sure why then insult when I have put a lot of work int this. Shortsightedness would be on those who cannot accept that I am right because I am in agreement with physics.