gkrouse
You left out the best answer.
(As many as you can fit into it and have room for the movement)
i have solved it! Please help contribute to getting this out to the world!
Moderator: scott
re: i have solved it! Please help contribute to getting thi
"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
re: i have solved it! Please help contribute to getting thi
I think the best answer is: it doesn't matter. How many does yours use?
Re: re: i have solved it! Please help contribute to getting
Another reason is it's Glenn R Rouse, and Glenn wouldn't be doing computer simulations or any programming, as this guy has in his profile.justsomeone wrote:https://futurenet.club/p/gkrouse#!
This must be him. He's can't be trying to hide his identity if this info matches his profile here and one quick Google of gkrouse.....
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re: i have solved it! Please help contribute to getting thi
If I understand correctly, a working prototype must be presented within a year or a patent application will be denied. I believe this is what happened to RAR. Because of this I am building/testing/tweaking before applying for the patent. If I am wrong, please correct me.
Oh, as far as number of weights goes, it depends upon the strength of the frame, strength of the floor, available space, and cost. Those are my limits.
Oh, as far as number of weights goes, it depends upon the strength of the frame, strength of the floor, available space, and cost. Those are my limits.
Congratulations gkrouse!! It had to be obvious that what most people were trying did not work, and that it would take a different thinking to achieve it. Look forward to your reveal.
My opinion - Any even number of 'weights' will work as they work in pairs. 'Weights' is a very misleading word, cylindrical weights is better.
My opinion - Any even number of 'weights' will work as they work in pairs. 'Weights' is a very misleading word, cylindrical weights is better.
re: i have solved it! Please help contribute to getting thi
The least you can expect from someone who claims to have rediscovered the Bessler wheel is that he can show a spinning wheel.
If even this minimum requirement is not met, one can dismiss such an assertion as nonsense.
Again and again such assertions are made here without even the slightest proof. If at least a fake wheel would be shown, but not even that. It is boring.
If even this minimum requirement is not met, one can dismiss such an assertion as nonsense.
Again and again such assertions are made here without even the slightest proof. If at least a fake wheel would be shown, but not even that. It is boring.
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The Toys Page tells us four are needed.
Besides the two on each hammer toy, note that the 'staff & chain', items A and B, indicate a series of five reactionless accelerations, culminating in "something extraordinary".
As noted previously, the number of cycles required to reach unity using reactionless accelerations coupled with inelastic collisions can be crudely surmised as the sum of the ratio of the inertia between the interacting bodies being applied for this purpose..
..for example, if we're using two equal inertias (so, two equal weights or two equal angular inertias, or else a combination of each), then we need to complete two full cycles to reach unity, because their ratio is 1:1, and 1 + 1 = 2.
If, OTOH, our scheme requires two inertias to be in a 2:1 ratio - or equivalently, 3 equal inertias, but with a collision between one and the other pair - then the system will reach unity after 3 cycles, because 2 + 1 = 3.
So, if Bessler's showing us "something extraordinary" that arises after5 cycles... then the system's 'unity threshold' must be four cycles... which means the ratio of the two inertias involved in each 'bang' must be 3:1... IOW, 4 equal inertias, but with a 3:1 collision; that is, either three equal masses are accelerated reactionlessly then collided with a fourth, or else one is accelerated then collided with the other three..
Alternatively, there is another general regime that also reaches unity at four cycles, and OU at the fifth: this involves systems in which equal momentum and counter-momentum are initially produced - so, at cost of input energy - and then the counter-momentum is somehow got rid of / sunk to earth / destroyed by magic, leaving just the 'primary' momentum. Because getting rid of half our momentum means likewise sacrificing its corresponding KE, then assuming a 1:1 inertia ratio, from an initial 75% loss, efficiency would rise by 25% per cycle to reach unity at 4 cycs and OU at 5; in this case, the 'number of weights used' is kind of incidental - 2 to 4 would be the minimum for the momentum-gain part of the mechanism. The number of GPE's used to feed it input energy could be kind of arbitrary, that's even assuming it needs gravity in the first place..
Pretty sure gkrouse wouldn't have a clue what i was talking about tho..
Besides the two on each hammer toy, note that the 'staff & chain', items A and B, indicate a series of five reactionless accelerations, culminating in "something extraordinary".
As noted previously, the number of cycles required to reach unity using reactionless accelerations coupled with inelastic collisions can be crudely surmised as the sum of the ratio of the inertia between the interacting bodies being applied for this purpose..
..for example, if we're using two equal inertias (so, two equal weights or two equal angular inertias, or else a combination of each), then we need to complete two full cycles to reach unity, because their ratio is 1:1, and 1 + 1 = 2.
If, OTOH, our scheme requires two inertias to be in a 2:1 ratio - or equivalently, 3 equal inertias, but with a collision between one and the other pair - then the system will reach unity after 3 cycles, because 2 + 1 = 3.
So, if Bessler's showing us "something extraordinary" that arises after5 cycles... then the system's 'unity threshold' must be four cycles... which means the ratio of the two inertias involved in each 'bang' must be 3:1... IOW, 4 equal inertias, but with a 3:1 collision; that is, either three equal masses are accelerated reactionlessly then collided with a fourth, or else one is accelerated then collided with the other three..
Alternatively, there is another general regime that also reaches unity at four cycles, and OU at the fifth: this involves systems in which equal momentum and counter-momentum are initially produced - so, at cost of input energy - and then the counter-momentum is somehow got rid of / sunk to earth / destroyed by magic, leaving just the 'primary' momentum. Because getting rid of half our momentum means likewise sacrificing its corresponding KE, then assuming a 1:1 inertia ratio, from an initial 75% loss, efficiency would rise by 25% per cycle to reach unity at 4 cycs and OU at 5; in this case, the 'number of weights used' is kind of incidental - 2 to 4 would be the minimum for the momentum-gain part of the mechanism. The number of GPE's used to feed it input energy could be kind of arbitrary, that's even assuming it needs gravity in the first place..
Pretty sure gkrouse wouldn't have a clue what i was talking about tho..
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re: i have solved it! Please help contribute to getting thi
As far as I know it can run with only 2 weights, rolling cylinders.
But i prefer 8 so that the wheel will accelerate very fast to its natural frequency.
Bessler said something like that,
Combined with the 8 special impacts which were heard, we get the turn.
To make it clear, the impacts does not drive the wheel, the impacts are there to arrange the position of the cylindrical weights.
As Bessler said one is going to the rim the other to the axle. They act in pairs.
But i prefer 8 so that the wheel will accelerate very fast to its natural frequency.
Bessler said something like that,
and many pieces of lead.eine Herde die herumgeht
Combined with the 8 special impacts which were heard, we get the turn.
To make it clear, the impacts does not drive the wheel, the impacts are there to arrange the position of the cylindrical weights.
As Bessler said one is going to the rim the other to the axle. They act in pairs.
Best regards
Georg
Georg
re: i have solved it! Please help contribute to getting thi
This guy and his claim still around?
"Orffyreus commented that when the secret is revealed, he is afraid that people will complain that the idea is so simple it is not worth the asking price."
re: i have solved it! Please help contribute to getting thi
Claimants come and claimants go.TGM wrote:This guy and his claim still around?
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Walter Clarkson
© 2023 Walter W. Clarkson, LLC
All rights reserved. Do not even quote me w/o my expressed written consent.
¯\_(ツ)_/¯ the future is here ¯\_(ツ)_/¯
Advocate of God Almighty, maker of heaven and earth and redeemer of my soul.
Walter Clarkson
© 2023 Walter W. Clarkson, LLC
All rights reserved. Do not even quote me w/o my expressed written consent.