use the high Ke from the speeding bullet fired from a rifle suspended on a pendulum rod to hit a second ballistic pendulum of the same mass as the rifle - complete the thought process - will the ballistic pendulum swing up & gain Pe equivalent to the Ke of the bullet plus highest Pe of swinging rifle ? - if Pe is greater you have created an OU model.
Heavens why would you transform it back down? When you increase its mass by uniting with the pendulum you are you know. You want the mass to rise t its fullest height and use it in a controlled descent. Then you have your over unity.
Manipulating Momentum
Moderator: scott
re: Manipulating Momentum
Not knowing is not the problem. It is the knowing of what just isn't so.
It is our responsibilities, not ourselves,that we should take seriously.
It is our responsibilities, not ourselves,that we should take seriously.
Re: re: Manipulating Momentum
Trevor - take your meds, there's a good boy ...Trevor Lyn Whatford wrote:...the fact is that there is more work done against gravity than my lifting of the weight therefore gravity is doing more work than I did, as gravity is doing work to stop it from swinging at the same time there is friction losses so a percentage of the work done by gravity to stop the swing has been converted into nonconservative forces, so gravity cannot conserve all of its input, as soon as I let go of the pendulum gravity took over so any friction loss from then on was due to the work from gravity, not forgetting that as I lifted the pendulum gravity was working against me as much as I was working against gravity so that work was equal ...
Here's what happens in your pendulum:
You lift the Mass up - your muscles are doing Work against the Force of gravity. Another way to express this is that You input some Potential Energy - with reference to the bottom of the pendulum swing.
You let the Mass fall down - the Mass accelerates due to the Force of gravity. The Potential Energy is converted into Kinetic Energy. So at the bottom of the pendulum swing you now have zero Potential Energy. However - you do have a lot of Kinetic Energy. But this is slightly less than the Potential Energy you provided with your muscle power.
But what happens as the pendulum swings back up again? The Mass experiences the force of gravity acting in the opposite direction, which decelerates the mass. The Kinetic energy is converted back into Potential Energy. So by the time the Mass reaches it's highest swing - Velocity reaches zero, and Potential Energy is at maximum. However, due to losses, the potential Energy is not quite as high as the previous height.
This repeats until all the energy has been dissipated. That is what a damped oscillation is.
In a closed system such as this - energy is conserved. This is very basic physics, easily proven by experiment, and experienced daily in normal industry and practical applications.
If you can't understand very basic physics, you are screwed and nobody will give you any respect for your ideas.
I am a Bessler believer - and I would like to believe that gravity can become a source of energy. So I am open to the idea that energy maths may be flawed - and there may be some clever mechanism that can exploit the available force. But there is absolutely no way you can say that a simple pendulum is doing more Work than the Energy which you input into it. To state this simply means that you have not understood the definition of Work or Energy. Do your homework - you are not greater than Newton, so don't deceive yourself.
Anything not related to elephants is irrelephant.
re: Manipulating Momentum
Because ... Work Done [Joules] = Force [Newton's] x Distance [meters]
Force [Newton's] = Mass [kg's] x Acceleration [m/s^2]
The point is there is NO Excess of Usable Joules !
Imagine it this way using the rifle example.
You know the Input Energy in Joules [convert grains of charge]
You know the velocity & mass of the bullet.
You know how high the momentum kicks the rifle suspended by a pendulum rod up, therefore you know the Pe gain at top of arc [this is the same as the Ke of the rifle].
You know the mass of the ballistic pendulum.
Calculate hoiw high it swings upwards & convert to Pe Joules [same as the Ke it was able to use].
Is there a gain in height anywhere to indicate OU ?
Where did the rest of the Ke of the bullet go ? - it was dissipated in the deformation of the bullet & the ballistic mass as an increase in atomic internal energy.
-------------------------------------------------------------
Put even more simply - have tow identical rifles on pendulums facing each other 10 feet apart - fire rifle one at the other.
The bullet enters the barrel of rifle two & stays in their.
How high did it swing & was there a gain in Pe etc to indicate OU ?
Answer: NO.
BUT .. we had all this energy of the fast moving bullet !
Force [Newton's] = Mass [kg's] x Acceleration [m/s^2]
The point is there is NO Excess of Usable Joules !
Imagine it this way using the rifle example.
You know the Input Energy in Joules [convert grains of charge]
You know the velocity & mass of the bullet.
You know how high the momentum kicks the rifle suspended by a pendulum rod up, therefore you know the Pe gain at top of arc [this is the same as the Ke of the rifle].
You know the mass of the ballistic pendulum.
Calculate hoiw high it swings upwards & convert to Pe Joules [same as the Ke it was able to use].
Is there a gain in height anywhere to indicate OU ?
Where did the rest of the Ke of the bullet go ? - it was dissipated in the deformation of the bullet & the ballistic mass as an increase in atomic internal energy.
-------------------------------------------------------------
Put even more simply - have tow identical rifles on pendulums facing each other 10 feet apart - fire rifle one at the other.
The bullet enters the barrel of rifle two & stays in their.
How high did it swing & was there a gain in Pe etc to indicate OU ?
Answer: NO.
BUT .. we had all this energy of the fast moving bullet !
Pequaide - you should know that I am one of your biggest fans. But Fletcher is correct in challenging you. It's for your own good - because you have been experimenting and theorizing for years, with still no satisfactory outcome.
I appreciate your theory that suggests that energy maths is flawed. IF all the momentum of a heavy mass can be transfered to a small mass then it is very clear that the velocity of the small mass will increase proportionally. And it is equally clear that velocity squared will go up exponentially.
But IF energy maths is flawed as you say, then your arguments that use energy maths are also potentially flawed.
If you are trying to prove your theory - you can't use the same equations that you are trying to discredit. That is where it is 100% essential that you demonstrate an experiment that shows a palpable increase in energy. Either lifting mass higher than it's starting position, or sustaining rotation with some load (even if it's just a small friction loss).
Simply connecting your Atwoods experiment to your rotary trebuchet should prove (or disprove) your theory. I am at a loss why you haven't done this. I can only assume that it simply doesn't work, because somewhere there is a flaw in your theory.
Feel free to prove me wrong.
(If I seem to have become a little negative these days, some of it is because my city has been trashed with an earthquake, and i'm expecting it to be trashed even further in the next couple of months. But I have also had some other disappointments as far as gravity wheel inventions go - and I am very aware how easy it is to deceive yourself - even with the best of intentions).
I appreciate your theory that suggests that energy maths is flawed. IF all the momentum of a heavy mass can be transfered to a small mass then it is very clear that the velocity of the small mass will increase proportionally. And it is equally clear that velocity squared will go up exponentially.
But IF energy maths is flawed as you say, then your arguments that use energy maths are also potentially flawed.
If you are trying to prove your theory - you can't use the same equations that you are trying to discredit. That is where it is 100% essential that you demonstrate an experiment that shows a palpable increase in energy. Either lifting mass higher than it's starting position, or sustaining rotation with some load (even if it's just a small friction loss).
Simply connecting your Atwoods experiment to your rotary trebuchet should prove (or disprove) your theory. I am at a loss why you haven't done this. I can only assume that it simply doesn't work, because somewhere there is a flaw in your theory.
Feel free to prove me wrong.
(If I seem to have become a little negative these days, some of it is because my city has been trashed with an earthquake, and i'm expecting it to be trashed even further in the next couple of months. But I have also had some other disappointments as far as gravity wheel inventions go - and I am very aware how easy it is to deceive yourself - even with the best of intentions).
re: Manipulating Momentum
Greendoor, I think what Peq and Kirk is saying, is that either one formula or the other is correct. If you calculate KE with respect to momentum, something appears to not add up. So, it appears there is a problem with one or the other formulas for momentum or energy, or if the formulas are correct, then one of them is not conservative. To prove what is wrong, you have to, as fletcher says, produce a (number of) experiment(s), and use all the current physics equations to find and prove which one is incorrect.
Fletcher, Kirks idea isn't to convert momentum to KE back to momentum in the method you propose. It is more like firing the bullet into the sky, and using a giant lever to lift the gun up with the height (and weight) of the bullet, compared to the height that could be achieved by firing the gun at the ground and letting the recoil lift the gun.
Fletcher, Kirks idea isn't to convert momentum to KE back to momentum in the method you propose. It is more like firing the bullet into the sky, and using a giant lever to lift the gun up with the height (and weight) of the bullet, compared to the height that could be achieved by firing the gun at the ground and letting the recoil lift the gun.
You can't say that either MV or 0.5MVV have to be wrong. There are clearly cases where both are right. Most of modern industry depends on this being the case.
Both Pequaide and Kirk seem to be depending on essentially the total transfer of momentum from a large to a small mass to violate the energy law. Pequaide considers that bringing a flywheel to a halt is proof of a complete transfer of momentum - but is it really? How do we know that the force isn't wasted somewhere - perhaps to earth?
Kirk points to the Pelton wheel as proof of total conversion of momentum to energy. But at best a Pelton wheel is 100% efficient as far as Energy transfer - I don't believe their is any OU possible in a Pelton wheel.
I think Kirks logic is the easiest to refute. He states that a mass bouncing off a heavier mass is proof of a total transfer of momentum - but if the mass actually bounces off, then obviously it is impossible for all the momentum to have been transfered ...
It's all empty rhetoric without a simple experiment that proves negative entropy ...
Both Pequaide and Kirk seem to be depending on essentially the total transfer of momentum from a large to a small mass to violate the energy law. Pequaide considers that bringing a flywheel to a halt is proof of a complete transfer of momentum - but is it really? How do we know that the force isn't wasted somewhere - perhaps to earth?
Kirk points to the Pelton wheel as proof of total conversion of momentum to energy. But at best a Pelton wheel is 100% efficient as far as Energy transfer - I don't believe their is any OU possible in a Pelton wheel.
I think Kirks logic is the easiest to refute. He states that a mass bouncing off a heavier mass is proof of a total transfer of momentum - but if the mass actually bounces off, then obviously it is impossible for all the momentum to have been transfered ...
It's all empty rhetoric without a simple experiment that proves negative entropy ...
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nicbordeaux
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re: Manipulating Momentum
Wow, a little of the fog of madness which descended on the entire forum has dissipated. I thought I was logging into a bogus forum of something, you know, like the MIB or the aliens were messing with my computer lol.
FWIW, a silly old video http://www.youtube.com/watch?v=P6vYzx_Vc4s (yup, I know there is no rigourous protocol and hand launch, so what, the experiment told me the basics of what I wanted to know. You set up a rigourous protocol and spend 1 K on every idea, you go broke real fast, and you don't move along very quick)
I think Bill has pointed out that certain impact points on the radius will be worse than others.
FWIW, a silly old video http://www.youtube.com/watch?v=P6vYzx_Vc4s (yup, I know there is no rigourous protocol and hand launch, so what, the experiment told me the basics of what I wanted to know. You set up a rigourous protocol and spend 1 K on every idea, you go broke real fast, and you don't move along very quick)
I think Bill has pointed out that certain impact points on the radius will be worse than others.
If you think you have an overunity device, think again, there is no such thing. You might just possibly have an unexpectedly efficient device. In which case you will be abducted by MIB and threatened by aliens.
Re: re: Manipulating Momentum
Of course- that is the ke energy. That is the harvest. What you are refusing to acknowege is ke is all over the map wth varying EQUIVALENT momentums and we have a method to interchange them with good efficiency. By exchanging the momentum of a 1 pound mass falling 1 foot with a quarter pound mass which then rises 16 feet we gain the ability to do work. So if my quarter pound encounters a ballistic pendulum and has no more momentum than the 1 pound falling 1 foot you now believe there is no over unity? Fletch, you are an engineer. I predict it will be harder for you to see this as it flies in the face of a lifetimes conditioning. I became physically dizzy when I first had the realization of the process. Yes, we have to fix the maths.Fletcher wrote:Because ... Work Done [Joules] = Force [Newton's] x Distance [meters]
Force [Newton's] = Mass [kg's] x Acceleration [m/s^2]
The point is there is NO Excess of Usable Joules !
Imagine it this way using the rifle example.
You know the Input Energy in Joules [convert grains of charge]
You know the velocity & mass of the bullet.
You know how high the momentum kicks the rifle suspended by a pendulum rod up, therefore you know the Pe gain at top of arc [this is the same as the Ke of the rifle].
You know the mass of the ballistic pendulum.
Calculate hoiw high it swings upwards & convert to Pe Joules [same as the Ke it was able to use].
Is there a gain in height anywhere to indicate OU ?
Where did the rest of the Ke of the bullet go ? - it was dissipated in the deformation of the bullet & the ballistic mass as an increase in atomic internal energy.
-------------------------------------------------------------
Put even more simply - have tow identical rifles on pendulums facing each other 10 feet apart - fire rifle one at the other.
The bullet enters the barrel of rifle two & stays in their.
How high did it swing & was there a gain in Pe etc to indicate OU ?
Answer: NO.
BUT .. we had all this energy of the fast moving bullet !
Not knowing is not the problem. It is the knowing of what just isn't so.
It is our responsibilities, not ourselves,that we should take seriously.
It is our responsibilities, not ourselves,that we should take seriously.
re: Manipulating Momentum
The early cylinder and spheres experiments were done in freefall; they were not touching the earth. And these experiments did not even have a bearing to produce friction. These were the ones that were strobe light photographed and video taped. I also posted a picture of a throwing Atwood’s not to long ago. But let’s look at the rifle.
The rifle’s motion is cause by powder burning, so we do not know the input energy. So let’s just drop the rifle one meter and use that as the input energy.
Let’s put one rifle (5 kilogram) on the end of a one meter lever arm; the rifle is at rest at 9 o’clock. We us the same pivot point and put another rifle at 12 o’clock on the end of a one meter lever arm. We then let the 5 kilogram rifle at 12 o’clock swing down to 3 o’clock and we lock it into the other lever that was at rest at 9 o’clock.
The dropped rifle will be moving 4.429 m/sec when it reaches 3 o’clock (49.05 joules: 1/2mv²). When it locks in with the other rifle we will have 10 kilograms moving 2.21 m/sec (24.52 joules). This would leave us with a 10 kilogram balanced wheel with its rim mass moving 2.2147 m/sec.
Now we transfer all the momentum to a one kilogram bullet. The bullet will have to be moving 22.147 m/sec (245.24 joules) and it will rise 25 meters (25 m * 9.81 N = 245.25 J). The one kilogram bullet at 25 meter could lift the rifle 5 times higher than what it was dropped.
A 10 kilogram rim mass wheel moving 2.2147 m/sec could have an input energy as low as 24.52 joules. We used 49 joules and still came out with 5 times the energy. The energy increases are huge. I don’t know why people are troubled when the experiment is started by hand. Ten kilograms moving 2.2147 m/sec is going to be 24.52 joules no matter how you start it.
Now as far as the two formulas (mv and 1/2mv²); I accept both formulas, both have there uses. But one of them is not conserved. As you see in this experiment kinetic energy is all over the place (24.52 J; 49.05 J; 245.2 J) as momentum (from Newton’s Three Laws of Motion) remains constant.
The rifle’s motion is cause by powder burning, so we do not know the input energy. So let’s just drop the rifle one meter and use that as the input energy.
Let’s put one rifle (5 kilogram) on the end of a one meter lever arm; the rifle is at rest at 9 o’clock. We us the same pivot point and put another rifle at 12 o’clock on the end of a one meter lever arm. We then let the 5 kilogram rifle at 12 o’clock swing down to 3 o’clock and we lock it into the other lever that was at rest at 9 o’clock.
The dropped rifle will be moving 4.429 m/sec when it reaches 3 o’clock (49.05 joules: 1/2mv²). When it locks in with the other rifle we will have 10 kilograms moving 2.21 m/sec (24.52 joules). This would leave us with a 10 kilogram balanced wheel with its rim mass moving 2.2147 m/sec.
Now we transfer all the momentum to a one kilogram bullet. The bullet will have to be moving 22.147 m/sec (245.24 joules) and it will rise 25 meters (25 m * 9.81 N = 245.25 J). The one kilogram bullet at 25 meter could lift the rifle 5 times higher than what it was dropped.
A 10 kilogram rim mass wheel moving 2.2147 m/sec could have an input energy as low as 24.52 joules. We used 49 joules and still came out with 5 times the energy. The energy increases are huge. I don’t know why people are troubled when the experiment is started by hand. Ten kilograms moving 2.2147 m/sec is going to be 24.52 joules no matter how you start it.
Now as far as the two formulas (mv and 1/2mv²); I accept both formulas, both have there uses. But one of them is not conserved. As you see in this experiment kinetic energy is all over the place (24.52 J; 49.05 J; 245.2 J) as momentum (from Newton’s Three Laws of Motion) remains constant.
there is no ou in a Pelton wheelgreendoor wrote:You can't say that either MV or 0.5MVV have to be wrong. There are clearly cases where both are right. Most of modern industry depends on this being the case.
Both Pequaide and Kirk seem to be depending on essentially the total transfer of momentum from a large to a small mass to violate the energy law. Pequaide considers that bringing a flywheel to a halt is proof of a complete transfer of momentum - but is it really? How do we know that the force isn't wasted somewhere - perhaps to earth?
Kirk points to the Pelton wheel as proof of total conversion of momentum to energy. But at best a Pelton wheel is 100% efficient as far as Energy transfer - I don't believe their is any OU possible in a Pelton wheel.
I think Kirks logic is the easiest to refute. He states that a mass bouncing off a heavier mass is proof of a total transfer of momentum - but if the mass actually bounces off, then obviously it is impossible for all the momentum to have been transfered ...
It's all empty rhetoric without a simple experiment that proves negative entropy ...
you need better reading comprehension. When there is total transfer at v/2 the ball becomes stationary. It's all relative to the frame of reference. The ball stopped because all its momentum is now in the wheel and the wheel is MOVING. Please study the operation of the Pelton wheel. It is a mature product and well understood. First question - why is the velocity of the impinging stream twice the wheel velocity? If you dont understand that you dont understand how you achieve max transfer to a moving mass. So far I have the distinct impression you dont understand that.
Not knowing is not the problem. It is the knowing of what just isn't so.
It is our responsibilities, not ourselves,that we should take seriously.
It is our responsibilities, not ourselves,that we should take seriously.
re: Manipulating Momentum
Hello, boys!
Pls be patient, but I have something to say and I'm not sure if this will have to see with this thread - hope you'll get my mind and my English
Remember those 'special guys' that come to TV as someone with muscles and zen mind power, enough to break 12 bricks with only one hand or head hit... an astonishing and impressive stuff!
He prepares to the brick pile that are supported by the two smaller sides.
The special guy cares to assemble also TWO wooden 1cm sticks in-between each pair of bricks, both at same narrow sides.
Then he comes and breaks all the bricks and show how his hand, arm or head are still fine.
My vision is this: when he puts all those wooden sticks he is avoiding that all bricks will come to sum their resistances one to the other!
As part of his trick, when he shuts the first brick he brings or forces his arm or hand to downwards in a continuous trajectory moving! 8)
Now comes the momentum: for sure, this means that he doesn't break to 12 bricks BUT he breaks to 12 bricks ONE by ONE at different sequential moments, or times... almost as inertia.
The break of just the first brick in the top - with a 'dry' shut - will ask for the same force but at 1/12 of the time, what means less work.
(maybe this is some more gasoline to the fire - or not? 8)
It would be priceless to see that special hero to kick his head to the pile but without those sticks! 8)))
Best!
Muliro
Pls be patient, but I have something to say and I'm not sure if this will have to see with this thread - hope you'll get my mind and my English
Remember those 'special guys' that come to TV as someone with muscles and zen mind power, enough to break 12 bricks with only one hand or head hit... an astonishing and impressive stuff!
He prepares to the brick pile that are supported by the two smaller sides.
The special guy cares to assemble also TWO wooden 1cm sticks in-between each pair of bricks, both at same narrow sides.
Then he comes and breaks all the bricks and show how his hand, arm or head are still fine.
My vision is this: when he puts all those wooden sticks he is avoiding that all bricks will come to sum their resistances one to the other!
As part of his trick, when he shuts the first brick he brings or forces his arm or hand to downwards in a continuous trajectory moving! 8)
Now comes the momentum: for sure, this means that he doesn't break to 12 bricks BUT he breaks to 12 bricks ONE by ONE at different sequential moments, or times... almost as inertia.
The break of just the first brick in the top - with a 'dry' shut - will ask for the same force but at 1/12 of the time, what means less work.
(maybe this is some more gasoline to the fire - or not? 8)
It would be priceless to see that special hero to kick his head to the pile but without those sticks! 8)))
Best!
Muliro
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nicbordeaux
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- Location: France
Re: re: Manipulating Momentum
pequaide wrote:
... Now as far as the two formulas (mv and 1/2mv²); I accept both formulas, both have there uses.
But one of them is not conserved.
As you see in this experiment kinetic energy is all over the place (24.52 J; 49.05 J; 245.2 J) as momentum (from Newton’s Three Laws of Motion) remains constant.
Conservation of Energy & Conservation of Momentum are axioms.
There is no Conservation of Kinetic Energy Law that I'm aware of & as I've said repeatably - BUT there is a CoE Law [total system based on Laws of Thermodynamics] - this allows Ke to vary in physical contact & exchange process.
To both Kirk & Pequaide.
What I'm attempting to promote here [perhaps badly] is the Mechanical Energy Budgeting Approach - mechanical OU is postulated on an abundance of excess mechanical energy able to do real Work from a mechanical system & restore itself or equivalence [it is part of the greater thermodynamics].
Ke's vary thru processes & exchanges - I say again that the only convincing & authoritative test is to take the Ke [Capacity to do Work] & harvest it then use it to do mechanical Work - if you can then do Work above your Input Energy & show an equivalence to starting conditions then you have shown OU - IOW's it is useful !
This requires a Mechanical Energy Budget for the Complete System & not just for Stages where the loop has not been closed or shown how it could be closed.
DEMONSTRATE that & you will have the worlds attention !
Half measures are not convincing enough, IMO.
Last edited by Fletcher on Sat Mar 05, 2011 10:01 pm, edited 1 time in total.
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nicbordeaux
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OK, who is going t build Kirk's wheel? It's all very well saying he's right, but if you sincerely believe that, and why not, you must be prepared to help with the build. After all, the claim is PM, or at the very least, free energy. Kirk asked Bill to set up a factory turning the devices out. Who is going to put his bucks where his finger is ? (for those who type with one finger).
Re: re: Manipulating Momentum
Kirk - I had this revelation some years ago, and proceeded to stink up this forum with endless babbling that came to zero fruition.Kirk wrote:What you are refusing to acknowege is ke is all over the map wth varying EQUIVALENT momentums and we have a method to interchange them with good efficiency. ... I became physically dizzy when I first had the realization of the process. Yes, we have to fix the maths.
Until you can demonstrate an experiment that proves this theory that the other theory is wrong, your theory is still just a theory.
Kirk - unless i'm mistaken, you are getting excited because you believe you have the key to OU - and that the Pelton wheel is an example of the 100% momentum transfer that is required to achieve this. Or have I misunderstood your intention?Kirk wrote:there is no ou in a Pelton wheel
you need better reading comprehension. When there is total transfer at v/2 the ball becomes stationary. It's all relative to the frame of reference. The ball stopped because all its momentum is now in the wheel and the wheel is MOVING. Please study the operation of the Pelton wheel. It is a mature product and well understood. First question - why is the velocity of the impinging stream twice the wheel velocity? If you dont understand that you dont understand how you achieve max transfer to a moving mass. So far I have the distinct impression you dont understand that.
If all you are suggesting is that a Pelton wheel is an example of 100% (unity) transfer of energy, then I apologise - because that is exactly what it is. Wikipedia is my source of information on the Pelton Wheel. It's very efficient, but not OU.
I can't see the relationship between the Pelton wheel and bouncing balls of a flywheel that is rotating at half speed. I wish you would explain why you think this is the same ...
You say the ball drops down at 8 fps. The wheel is rotating away (down) at 4 fps. So the ball impacts the wheel at a relative speed of 4 fps. Durng the time of contact, the impulse (force x time ) is shared equally between the ball and the wheel. The wheel is 100 times more massive, therefore it accelerates much less than the ball - so obviously it continues in the same direction only slightly faster. The ball however bounces backwards, after coming to a complete stop.
I do NOT see why you think the ball will come to a halt - it's got to bounce back. And in bouncing back - it has momentum. Where is the complete transfer of momentum that you are talking about?
As I understand it, a Pelton wheel does not work like this. The water does not bounce back they way you are suggesting. The water is constrained into making a u-turn.
Before we rush into suggesting that any maths if faulty, you need to demonstrate an observation of an experiment that demonstrates a need to re-evalute the maths.
I guess the honeymoon period is over for me.
Anything not related to elephants is irrelephant.