If a two kilogram mass moving 2 m/sec were to combine with a one kilogram mass at rest nearly all experimenters would use the Law of Conservation of Momentum to calculate the new velocity. I would. And the new velocity would be 1.33 m/sec.
Now how can the 2 kilograms give all of its energy to the 1 kilograms when some of the energy is already lost. After transferring only a third of the motion you have already gone from 4 joules of energy to 2.65 joules.
Now suppose that the three kilograms combines with a fourth kilogram at rest, and then a fifth. Now a smaller object (2 kg) has shared its motion with a larger object (3 kg) and we know the math for that. It is the Law of Conservation of Momentum. How can part of the transfer use one formula and the next part another formula.
energy producing experiments
Moderator: scott
re: energy producing experiments
I was searching the web for an Atwood’s machine and I ran across Wubbly’s spring stopped Atwood machine; posted on December 2009.
Wubbly’s final conclusion was that momentum was an accounting tool; which I assume he means that momentum is a meaningless mathematical expression. He concluded that the same amount of energy (the extension of the spring) can stop different quantities of momentum. The different momentums were stopped by the same amount of energy; but it was not the same amount of Force.
If the spring is being stretched out over a period three times as long; then the force that is delivered by the spring is three times as great. The Force delivered is proportional to the momentum stopped. And if we use the spring to restart the motion then the force delivered by the spring would be proportional to the momentum produces. And this is an important point.
The spring can be loaded with a 2.2 kilogram mass moving 1.71 m/sec; but it can then unload on a 13.77 kilogram mass and it will give that mass a final velocity of .68 m/sec. Or 3.77 units of momentum can produce 9.37 units of momentum.
The initial drive mass takes more time to get the mass in motion, and the spring takes more time to get the motion back out. The motion conserved for each run is momentum. The momentum produced in each run is different, but the formula going along for the ride is 1/2mv².
So let’s take the 13.77 kilogram mass moving .68 m/sec and transfer all the motion to a 2.2 kilogram mass. If we are able to transfer 3.8 units of the 9.37 units of momentum to the 2.2 kilograms we then would have an energy producing device. If the Law of Conservation of Momentum is true then this transfer of only 40% of the momentum should be easy. Or you could look on the other side; that 60% of the motion would have to be lost if you were to be unable to make an energy producing device.
I have made many of these experiments and the momentum loss is in the range of 5% not 60%. And this 5% would be consistent with bearing resistance and/or air resistance.
This experiment proves that different quantities of momentum can produce the same quantity of energy. And it proves that the same quantity of energy can produce different quantities of momentum. We already know that the same quantity of momentum can produce different quantities of energy. 13.77 kilograms moving .68 m/sec has 9.37 units of momentum; if the 2.2 kilograms had 9.37 units of momentum it would have 6.25 times as much energy as the 13.77 kilograms (19.95 joules to 3.18 joules).
I have never seen another law exercise authority over the Law of Conservation of Momentum.
This has all been discussed before in Atwood Analysis, but there may be newcomers.
Negative response is not needed; we have heard it all before.
Wubbly’s final conclusion was that momentum was an accounting tool; which I assume he means that momentum is a meaningless mathematical expression. He concluded that the same amount of energy (the extension of the spring) can stop different quantities of momentum. The different momentums were stopped by the same amount of energy; but it was not the same amount of Force.
If the spring is being stretched out over a period three times as long; then the force that is delivered by the spring is three times as great. The Force delivered is proportional to the momentum stopped. And if we use the spring to restart the motion then the force delivered by the spring would be proportional to the momentum produces. And this is an important point.
The spring can be loaded with a 2.2 kilogram mass moving 1.71 m/sec; but it can then unload on a 13.77 kilogram mass and it will give that mass a final velocity of .68 m/sec. Or 3.77 units of momentum can produce 9.37 units of momentum.
The initial drive mass takes more time to get the mass in motion, and the spring takes more time to get the motion back out. The motion conserved for each run is momentum. The momentum produced in each run is different, but the formula going along for the ride is 1/2mv².
So let’s take the 13.77 kilogram mass moving .68 m/sec and transfer all the motion to a 2.2 kilogram mass. If we are able to transfer 3.8 units of the 9.37 units of momentum to the 2.2 kilograms we then would have an energy producing device. If the Law of Conservation of Momentum is true then this transfer of only 40% of the momentum should be easy. Or you could look on the other side; that 60% of the motion would have to be lost if you were to be unable to make an energy producing device.
I have made many of these experiments and the momentum loss is in the range of 5% not 60%. And this 5% would be consistent with bearing resistance and/or air resistance.
This experiment proves that different quantities of momentum can produce the same quantity of energy. And it proves that the same quantity of energy can produce different quantities of momentum. We already know that the same quantity of momentum can produce different quantities of energy. 13.77 kilograms moving .68 m/sec has 9.37 units of momentum; if the 2.2 kilograms had 9.37 units of momentum it would have 6.25 times as much energy as the 13.77 kilograms (19.95 joules to 3.18 joules).
I have never seen another law exercise authority over the Law of Conservation of Momentum.
This has all been discussed before in Atwood Analysis, but there may be newcomers.
Negative response is not needed; we have heard it all before.
re: energy producing experiments
There are three motion formulas: 1/2mv², mv, r * mv. One common experiment of the ballistic pendulum tells us a lot: angular momentum is not conserved; kinetic has 95%, or 85%, 75% etc missing; and linear momentum holds. Linear momentum holds all the time every time. Then why don't you use the formula that works.
Build your machines with the understanding that mv works and your machines will work.
Build your machines with the understanding that mv works and your machines will work.
re: energy producing experiments
A video I have seen recently kind of reminded me of this thread:
https://www.youtube.com/watch?v=vWVZ6APXM4w
How is it that the block reaches the same height whilst at the same time having more, rotational, kinetic energy. The argument in the youtube comments is amusing, if material deformation is the cause of this then nature sure has a dirty mind in providing just the right amount of deformation in order for the block to reach the same height.
https://www.youtube.com/watch?v=vWVZ6APXM4w
How is it that the block reaches the same height whilst at the same time having more, rotational, kinetic energy. The argument in the youtube comments is amusing, if material deformation is the cause of this then nature sure has a dirty mind in providing just the right amount of deformation in order for the block to reach the same height.
re: energy producing experiments
That was interesting broli.
My guess is that it has to do with Pressure.
Force = Pressure x Area & P = F / A.
The rotating block has far more total energy than the control, yet the elasticity of the bullets & wood blocks is the same.
So what could be different to allow a greater transfer of energy from the bullet ?
It must be that deformation was less in the rotating block impact IMO - that means the bullet followed a slight curved path on entry [internal reference frame & probably undetectable to the naked eye] - this means the bullet presents a slightly greater surface contact area as it penetrates.
Since the Force is the same for both speeding bullets but one presents a greater contact area as it penetrates then its pressure must fall IINM.
i.e. P = F / A therefore greater contact Area means less Pressure.
Less Pressure means less deformation & penetration into the wood - Less deformation [or less internal energy wastage] means a greater transfer of energy.
That's my guess anyway & it seems the Laws of Nature were built by an accountant with a sometimes sense of humour.
Of course I could be wrong.
My guess is that it has to do with Pressure.
Force = Pressure x Area & P = F / A.
The rotating block has far more total energy than the control, yet the elasticity of the bullets & wood blocks is the same.
So what could be different to allow a greater transfer of energy from the bullet ?
It must be that deformation was less in the rotating block impact IMO - that means the bullet followed a slight curved path on entry [internal reference frame & probably undetectable to the naked eye] - this means the bullet presents a slightly greater surface contact area as it penetrates.
Since the Force is the same for both speeding bullets but one presents a greater contact area as it penetrates then its pressure must fall IINM.
i.e. P = F / A therefore greater contact Area means less Pressure.
Less Pressure means less deformation & penetration into the wood - Less deformation [or less internal energy wastage] means a greater transfer of energy.
That's my guess anyway & it seems the Laws of Nature were built by an accountant with a sometimes sense of humour.
Of course I could be wrong.
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getterdone
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re: energy producing experiments
I was thinking that because of the thickness of the block, once it starts spinning , perhaps aero-dynamics play a part in the uplift motion.
I'm always wrong about this physic stuff tho lol
I'm always wrong about this physic stuff tho lol
Beer is the cause and the solution of all my problems.
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Furcurequs
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Cool.
This kind of problem brings out the inner geek in me. Okay, or maybe I'm just a total geek. Wait, or is that "nerd"? I get those terms confused.
Anyway...
At least before detailed computer modeling, a problem like this probably would have just been solved with an understanding of the conservation of momentum laws. Using them we can know what conditions we must have just after the collision without having to concern ourselves with the gory details of the actual collision itself.
In both scenarios presented in the video there is obviously going to be an inelastic collision in which the bullet imbeds into the block, so we can know that the linear momentum in the vertical of the "center of mass" of the total "block, bullet and nails system" immediately after the collision has to be equal to the the linear momentum in the vertical of just the bullet right before the collision (since the block and nails begin at rest, of course).
We are to assume, apparently, that the bullets in both scenarios have essentially the same mass and velocity and energy when fired and the blocks and nails are supposed to be of the same mass, too. So, if that's the case then we can see we would have the same total mass, linear momentum, velocity of the center of mass and thus translational kinetic energy in the "block, bullet and nails systems" after both of the collisions and thus in both scenarios they would be carried to the same height and so we can likely answer the question posed as they would accept as correct.
When the bullet is fired offset to the center of mass of the block and nails system, however, some of the (macroscopic) kinetic energy that would have been lost to heat and deformation before in the in-line collision instead goes into the rotational kinetic energy of the block, bullet and nails system itself. If we had actual measurements and numbers, these things could be quantitatively calculated, of course, rather than just seen qualitatively.
Actually, when considering the motion of just the center of mass of the complete system after the second collision, the rotational kinetic energy isn't really unlike the system's other internal kinetic energies - such as the thermal vibrations of the molecules inside and whatnot. We just can see the rotational motion macroscopically (and could still access that rotational kinetic energy macroscopically, too, if we wanted of course).
Anyway, when the bullet is aimed in line with the center of mass of the system of block and nails, at the moment of collision the bullet is met with the linear inertia of the entire mass giving a nice head on collision. When the collision is offset, the collision is "softer" (I don't know if that's a proper physics term, but I'm going to use it) and so some of the (macroscopic) kinetic energy that would otherwise have been lost as heat/deformation remains macroscopic and goes into the rotation. All the mass of the block isn't accelerated so suddenly because it is easier to rotate the block around its center of mass than it is to linearly move the entire mass (through the center of mass). To do the actual calculations, though, we would have to also concern ourselves with the conservation of ANGULAR momentum and moments of inertia and stuff.
Of course, it probably doesn't hurt to try to understand the gory details at the moment of collision, too, but I would personally choose to avoid the gory calculations and try to just understand it on an intuitive level and/or defer to a computer simulation for modeling in such detail.
Ruggerodk, sorry I didn't get back to you in the other thread. My chronic pain problems got the better of me for a little while.
Fletcher, thanks for your input in that thread, too. I believe you pretty much covered what I might have said. ...though I may add a little something later if I can get to it.
Dwayne
This kind of problem brings out the inner geek in me. Okay, or maybe I'm just a total geek. Wait, or is that "nerd"? I get those terms confused.
Anyway...
At least before detailed computer modeling, a problem like this probably would have just been solved with an understanding of the conservation of momentum laws. Using them we can know what conditions we must have just after the collision without having to concern ourselves with the gory details of the actual collision itself.
In both scenarios presented in the video there is obviously going to be an inelastic collision in which the bullet imbeds into the block, so we can know that the linear momentum in the vertical of the "center of mass" of the total "block, bullet and nails system" immediately after the collision has to be equal to the the linear momentum in the vertical of just the bullet right before the collision (since the block and nails begin at rest, of course).
We are to assume, apparently, that the bullets in both scenarios have essentially the same mass and velocity and energy when fired and the blocks and nails are supposed to be of the same mass, too. So, if that's the case then we can see we would have the same total mass, linear momentum, velocity of the center of mass and thus translational kinetic energy in the "block, bullet and nails systems" after both of the collisions and thus in both scenarios they would be carried to the same height and so we can likely answer the question posed as they would accept as correct.
When the bullet is fired offset to the center of mass of the block and nails system, however, some of the (macroscopic) kinetic energy that would have been lost to heat and deformation before in the in-line collision instead goes into the rotational kinetic energy of the block, bullet and nails system itself. If we had actual measurements and numbers, these things could be quantitatively calculated, of course, rather than just seen qualitatively.
Actually, when considering the motion of just the center of mass of the complete system after the second collision, the rotational kinetic energy isn't really unlike the system's other internal kinetic energies - such as the thermal vibrations of the molecules inside and whatnot. We just can see the rotational motion macroscopically (and could still access that rotational kinetic energy macroscopically, too, if we wanted of course).
Anyway, when the bullet is aimed in line with the center of mass of the system of block and nails, at the moment of collision the bullet is met with the linear inertia of the entire mass giving a nice head on collision. When the collision is offset, the collision is "softer" (I don't know if that's a proper physics term, but I'm going to use it) and so some of the (macroscopic) kinetic energy that would otherwise have been lost as heat/deformation remains macroscopic and goes into the rotation. All the mass of the block isn't accelerated so suddenly because it is easier to rotate the block around its center of mass than it is to linearly move the entire mass (through the center of mass). To do the actual calculations, though, we would have to also concern ourselves with the conservation of ANGULAR momentum and moments of inertia and stuff.
Of course, it probably doesn't hurt to try to understand the gory details at the moment of collision, too, but I would personally choose to avoid the gory calculations and try to just understand it on an intuitive level and/or defer to a computer simulation for modeling in such detail.
Ruggerodk, sorry I didn't get back to you in the other thread. My chronic pain problems got the better of me for a little while.
Fletcher, thanks for your input in that thread, too. I believe you pretty much covered what I might have said. ...though I may add a little something later if I can get to it.
Dwayne
re: energy producing experiments
I'd reckon it has something to do with the DePalma discovery of rotating ball trajectory:
http://www.evert.de/eft907e.htm
http://www.evert.de/eft907e.htm
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Furcurequs
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Here's the followup video with the official explanation for the bullet and block experiment:
https://www.youtube.com/watch?v=BLYoyLcdGPc
Dwayne
https://www.youtube.com/watch?v=BLYoyLcdGPc
Dwayne
re: energy producing experiments
.
Zoelra Posted: Sat Aug 31, 2013 5:58 am Post subject: The freefall of mass due to gravity is the gateway to PM : -
" Talking about the 3rd derivative brings back memories of the Dean Drive, the Fourth Law of Motion, and Surge as it was called. Great reading.
http://www.rexresearch.com/dean/davis4.htm "
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Yes , it is a great read ! You really get the feeling you are in the presence of an exceptionally clear thinker .
After reading it and then watching the “Bullet in the Block “ experiment I realised Davis’s ‘Fourth Law of Motion’ theory could apply to the spinning block extra rotational motion energy .
Very interesting stuff . More “grist� for your mill Pequaide ! : )
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Zoelra Posted: Sat Aug 31, 2013 5:58 am Post subject: The freefall of mass due to gravity is the gateway to PM : -
" Talking about the 3rd derivative brings back memories of the Dean Drive, the Fourth Law of Motion, and Surge as it was called. Great reading.
http://www.rexresearch.com/dean/davis4.htm "
-----------------------------------------------------------------
Yes , it is a great read ! You really get the feeling you are in the presence of an exceptionally clear thinker .
After reading it and then watching the “Bullet in the Block “ experiment I realised Davis’s ‘Fourth Law of Motion’ theory could apply to the spinning block extra rotational motion energy .
Very interesting stuff . More “grist� for your mill Pequaide ! : )
.
Have had the solution to Bessler's Wheel approximately monthly for over 30 years ! But next month is "The One" !
re: energy producing experiments
What is amusing is that the bullet and block experiment is a discussion of energy conservation. The experimenter states that 97% of the energy is lost, so the spinning block losses a tiny bit less energy; big deal. So what?
What is interesting however is that momentum is conserved. As always.
If the block were suspended from a string (as many are) all the motion would be rotational. It is rotating around the other end of the string. And of course again; momentum is conserved; and it is not angular.
It is sad to see how people protect a pet theory: and there is not one scintilla of experimentation to support it.
What is interesting however is that momentum is conserved. As always.
If the block were suspended from a string (as many are) all the motion would be rotational. It is rotating around the other end of the string. And of course again; momentum is conserved; and it is not angular.
It is sad to see how people protect a pet theory: and there is not one scintilla of experimentation to support it.
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Furcurequs
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- Joined: Sat Mar 17, 2012 4:50 am
re: energy producing experiments
Pequaide,
So, just how much kinetic energy do you suppose would be lost firing the bullet into your suspended block?
It would essentially be the same loss, huh?
Angular and linear momentums would both be conserved in the collision and the angular momentum of the bullet moving along a linear path can be calculated around the pivot point of the string just before the collision since it is a cross product.
Dwayne
So, just how much kinetic energy do you suppose would be lost firing the bullet into your suspended block?
It would essentially be the same loss, huh?
Angular and linear momentums would both be conserved in the collision and the angular momentum of the bullet moving along a linear path can be calculated around the pivot point of the string just before the collision since it is a cross product.
Dwayne
I don't believe in conspiracies!
I prefer working alone.
I prefer working alone.
re: energy producing experiments
If you looked at both blocks I think you would find one with the bullet embedded farther.
What goes around, comes around.
re: energy producing experiments
Furcurequs quote "the angular momentum of the bullet moving along a linear path"
Wow. So it is: pick a point; any point--- unbelievable.
I will let you all argue about it, I know what is conserved.
Wow. So it is: pick a point; any point--- unbelievable.
I will let you all argue about it, I know what is conserved.