I presented something like this before; but some friends asked for a practical application of the machine.
Suppose you have a 19 kilogram rim mounted vertically (horizontal axis) on a quality bearing, and you place a string around the circumference of the rim with a one kilogram mass on the end of the string. F = ma gives you and acceleration rate of 9.81 N / 20 kg = .4905 (or 1/20 * 9.81 m/sec/sec ) = .4905 m/sec/sec.
The distance formula (d = ½ v²/a) says that a velocity of 1.1 m/sec will be achieved after the one kilogram mass has dropped 1.233 meters. You would now have 20 kilograms moving 1.1 m/sec for a momentum of 22 units.
The pictured (gray with tape) cylinder and spheres demonstrates that all the momentum of a spinning cylinder (or rim) can be given to a smaller mass. So you can place all of the 22 units of momentum into a one kilogram mass. For a one kilogram mass to have 22 units of momentum it will have to have a velocity of 22 m/sec. I have done 15 to 1 and 40 to 1 cylinder and spheres mass to spheres mass ratios so I am sure 22 to one will work just as well.
With a velocity of 22 meter per second the one kilogram missile mass will rise 24.66 meters.
Of this 24.66 meters we only need 1.233 meters (of a one kilogram drop) to restart the rim so that it spins up to 1.1 m/sec. The other 23.427 meters can be used to spin a generator.
So we could have a 25 meter tower with a pulley at the top that has a fiber or chain loop with masses attached on only the descending side. You could load up the descending side if you prefer. But you only need one kilogram dropped 1.233 meters to restart the system.
The sky is kind of the limit once you prove that the cylinder and spheres conserves momentum not energy. And that is exactly what the high speed video cam shows.
For those of you that like math; and for a double check. You can us the distance formula in this form: d = 1/2at². The one kilogram mass wrapped over the 19 kg rim pulls upon the rim with 9.81 newton of force for 2.242 seconds. The one kilograms missile flying up at 22 m/sec has 9.81 newtons of force acting upon it for 2.242 sec. F = ma or F = m v/t or Ft = mv. Ft is just as much a conserved quantity as mv.
I can remember throwing with high ratio wheels; and they would throw over the (mature hardwoods) tree tops. It is silly to think that a 20 kg rim moving 1.1 m/sec could not throw 1 kg more than just a little over a meter. The masses I had hanging in the tree tops proves otherwise.
energy producing experiments
Moderator: scott
re: energy producing experiments
I'll take your word for it.. when that mass drops in free fall.The distance formula (d = ½ v²/a) says that a velocity of 1.1 m/sec will be achieved after the one kilogram mass has dropped 1.233 meters. You would now have 20 kilograms moving 1.1 m/sec for a momentum of 22 units.
But your flywheel has Inertia... I estimate it drops at 1/5 of that speed (depends on the size of the wheel, and thus inertia of that wheel)
At least you need to run some cascading mechanism to speed it up, or another kind of energy-input to get your flywheel up to speed.
Any drop of weight (that 1.233 m), will speedup the wheel slightly (+ E.rot).
Any lifting it does (for 1.233 m), will slowdown the wheel slightly (- E.rot).
Plus friction, then it will slow down in the longrun
In the meantime while your wheel has still stored rotational momentum, you can use that as you please of course. But that too will slow down that wheel.
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: energy producing experiments
Inertia and mass are terms that can be used almost interchangeably. Inertia is resistance to motion changes. Inertia is caused by an object having mass. Inertia has no units by which it can be placed in a formula; they use the kilogram as an expression of the quantity of inertia; but the kilogram is a unit of mass.
In the formula F = ma the concept of inertia is covered under the concept of mass. When reference to the concept of mass is made; then the concept of inertia is already taken care of.
The hanging mass of one kilogram places 9.81 newtons of force on 20 units of mass (or 20 units of inertia if you will). The acceleration will be .4905 m/sec/sec. the formula is correct.
I can remember an impact pendulum that could swing for, at least, ten or 15 minutes. The entire cycle of the above machine would be done in about 4 rotations; 2.5 seconds. The machine gives a 2000% increase in energy; bearing friction is not going to be much of an issue.
The machine does not just sit there slowing down it produces an increase of 2000% every 4 seconds.
Look up Atwood’s machines for a experimental proof of wheels complying with F = ma. Rim mass wheels act just like Atwood's.
In the formula F = ma the concept of inertia is covered under the concept of mass. When reference to the concept of mass is made; then the concept of inertia is already taken care of.
The hanging mass of one kilogram places 9.81 newtons of force on 20 units of mass (or 20 units of inertia if you will). The acceleration will be .4905 m/sec/sec. the formula is correct.
I can remember an impact pendulum that could swing for, at least, ten or 15 minutes. The entire cycle of the above machine would be done in about 4 rotations; 2.5 seconds. The machine gives a 2000% increase in energy; bearing friction is not going to be much of an issue.
The machine does not just sit there slowing down it produces an increase of 2000% every 4 seconds.
Look up Atwood’s machines for a experimental proof of wheels complying with F = ma. Rim mass wheels act just like Atwood's.
re: energy producing experiments
yup. correct.The hanging mass of one kilogram places 9.81 newtons of force on 20 units of mass (or 20 units of inertia if you will). The acceleration will be .4905 m/sec/sec. the formula is correct.
And when it goes back up why shouldn't it decelerate with that same amount?
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: energy producing experiments
Because I don’t leave the one kilogram mass attached to the wheel.
If the wheel is required to re-lift the mass because the mass is left attached to the wheel then you are correct. The wheel will re-lift the one kilogram mass for 2.24 seconds and you are right back where you started. But you don't leave the mass attached (in an Atwood's configuration) to the wheel; you throw it.
After a 1.233 meter drop you have 22 units of momentum in the rim and in the dropped mass. You then get out of your Atwood’s configuration and get into a cylinder and spheres configuration and you throw the one kilogram mass instead of leaving it attached.
You can throw one of the 19 kilogram units in the rim and leave the dropped kilogram dangling on the string for a bit if you like. This has been done also (Ohio student); the dangling kilogram is stopped in its tracks. And as soon as the throw is made it restarts the rim.
When one kilogram has all of the 22 units of momentum it is throw up 24.66 meters. So after the first cycle is finished you have one kilogram 1.233 meters below the bearing of the rim and one kilogram 24.66 meters above the bearing of the rim. I can re-lift the lower mass with the upper mass and be ready to go again with 23.43 kg meters of gravitational potential energy (230 joules) in reserve.
You can mechanically arrange the one kilogram to be dropped above the rim and have it join the rim after the drop. Thereupon the dropped mass is immediately thrown by the rim that it has just accelerated.
If the wheel is required to re-lift the mass because the mass is left attached to the wheel then you are correct. The wheel will re-lift the one kilogram mass for 2.24 seconds and you are right back where you started. But you don't leave the mass attached (in an Atwood's configuration) to the wheel; you throw it.
After a 1.233 meter drop you have 22 units of momentum in the rim and in the dropped mass. You then get out of your Atwood’s configuration and get into a cylinder and spheres configuration and you throw the one kilogram mass instead of leaving it attached.
You can throw one of the 19 kilogram units in the rim and leave the dropped kilogram dangling on the string for a bit if you like. This has been done also (Ohio student); the dangling kilogram is stopped in its tracks. And as soon as the throw is made it restarts the rim.
When one kilogram has all of the 22 units of momentum it is throw up 24.66 meters. So after the first cycle is finished you have one kilogram 1.233 meters below the bearing of the rim and one kilogram 24.66 meters above the bearing of the rim. I can re-lift the lower mass with the upper mass and be ready to go again with 23.43 kg meters of gravitational potential energy (230 joules) in reserve.
You can mechanically arrange the one kilogram to be dropped above the rim and have it join the rim after the drop. Thereupon the dropped mass is immediately thrown by the rim that it has just accelerated.
re: energy producing experiments
This is where the system is at highest energy; the cylinder is not spinning.
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re: energy producing experiments
You could place an unknown mass on one end of a light stick and a known mass on the other end. You could then throw the rigid stick up in the air so as to make it spin. The stick will spin about its center of mass. The mass of the unknown mass could then be determined. You do not need gravity to determine mass.
And gravity has noting to do with the conservation of momentum. I assume that this is why Jim says it is not a gravity wheel. Momentum conservation occurs in deep space; just as much as it does in the lab.
I demonstrated the cart wheel to six people today and every one of them caught on, they each could see the principal of momentum conservation. I believe that each agreed that the energy went way up. They could see the pitiful motion of the wheel and the violent motion of the sphere. They talked about how to store the energy and how to cycle. They did not doubt that energy was made. Their advantage was they could see the wheel and touch it and could wrap the weighted string around it and throw the spheres.
We were throwing in a moderate size warehouse, and the top and the sides of the building were being reached.
Some people think we are breaking the Laws or rules; well we break no Newtonian rules or Laws. The second Law is conservation of momentum. A sphere, with the same momentum as the wheel, that has one twentieth the mass has twenty time as much energy.
We know that when small object give their motion of large objects they conserve momentum; both in linear motion and circular motion. Many ballistic pendulum are rotation motion to rotational motion; and momentum is conserved. How can you have one rule for small to large and another rule for large to small.
You can't go back and forth as the gray cylinder does and have two different laws. This gray cylinder achieves two high energy states that are the same; and three low energy states that are the same. It goes back and forth twice; you can't do that with one Law one way and another Law the other way.
And gravity has noting to do with the conservation of momentum. I assume that this is why Jim says it is not a gravity wheel. Momentum conservation occurs in deep space; just as much as it does in the lab.
I demonstrated the cart wheel to six people today and every one of them caught on, they each could see the principal of momentum conservation. I believe that each agreed that the energy went way up. They could see the pitiful motion of the wheel and the violent motion of the sphere. They talked about how to store the energy and how to cycle. They did not doubt that energy was made. Their advantage was they could see the wheel and touch it and could wrap the weighted string around it and throw the spheres.
We were throwing in a moderate size warehouse, and the top and the sides of the building were being reached.
Some people think we are breaking the Laws or rules; well we break no Newtonian rules or Laws. The second Law is conservation of momentum. A sphere, with the same momentum as the wheel, that has one twentieth the mass has twenty time as much energy.
We know that when small object give their motion of large objects they conserve momentum; both in linear motion and circular motion. Many ballistic pendulum are rotation motion to rotational motion; and momentum is conserved. How can you have one rule for small to large and another rule for large to small.
You can't go back and forth as the gray cylinder does and have two different laws. This gray cylinder achieves two high energy states that are the same; and three low energy states that are the same. It goes back and forth twice; you can't do that with one Law one way and another Law the other way.
re: energy producing experiments
The Dawn Mission 3 rpm reverse spin is experimental proof that the law of conservation of energy is false.
At 36 rpm the rocket would have to have a diameter of .53 meters to have a arch velocity of 1 m/sec. This is not an outside diameter this is the average distribution of mass at different radii away from the point of rotation. This might be small but the point being made here would be the same if the diameter were 1.06 meters and the arch velocity 2 m/sec.
Using 1200 kilograms moving 1 m/sec the initial kinetic energy would be 600 joules.
If the Law of Conservation of Energy is true then the final energy would have to be 600 joules.
So ½ * 3kg * X m/sec * X m/sec = 600 joules; which is the square root of 600 * 2 /3 = 20 m/sec.
If the Law of Conservation of Energy were true; the final (or maximum velocity) velocity of the 3 kg would be 20 m/sec.
If momentum is conserved the final velocity is 400 m/sec. 3 kg * 400 m/sec = 1 m/sec * 1200 kg
The 20 m/sec required for Law of Conservation of Energy is a 95% lose of momentum. You have only 5% remaining. You have lost 1140 (3 kg * 380 m/sec) units of momentum. And you have only 60 (3 kg * 20 m/sec) units of momentum remaining.
Per a massive quantity of experiments we know that momentum is the only motion transferred when a small object collides with a larger object. So if the 3 kilograms moving 20 m/sec begins to give its motion back to the rocket body then it can only restore 5% of the original motion. And this full 5% restoration can only be after the 3 kilograms has had its own velocity depleted to the same velocity of the rocket.
There is a point in the Dawn Mission motion transfer when three rpm has been restored to the rocket's rotation and the 3 kilograms is still out on the end of the tether with nearly maximum velocity.
Three rpm is 8.33% of 36 rpm; 3/36 = .0833 or 8.33 %
It is a physical impossibility for 5% of the motion to restore 8.33% of the motion. And this 8.33% restoration occurs when very little of the motion of the 3 kilograms has been removed.
What realty happens is that the rocket is stopped at the point where the 3 kilograms has 400 m/sec velocity. This stop is a little early of 90° to tangent and the 3 kilograms will pull the rocket back into motion before the tether comes to 90°. This will take 8.33% of the total motion; or 99.99 (3 kg * 33.33) units of momentum. The 3 kilogram mass was moving about 367 m/sec at release.
At 36 rpm the rocket would have to have a diameter of .53 meters to have a arch velocity of 1 m/sec. This is not an outside diameter this is the average distribution of mass at different radii away from the point of rotation. This might be small but the point being made here would be the same if the diameter were 1.06 meters and the arch velocity 2 m/sec.
Using 1200 kilograms moving 1 m/sec the initial kinetic energy would be 600 joules.
If the Law of Conservation of Energy is true then the final energy would have to be 600 joules.
So ½ * 3kg * X m/sec * X m/sec = 600 joules; which is the square root of 600 * 2 /3 = 20 m/sec.
If the Law of Conservation of Energy were true; the final (or maximum velocity) velocity of the 3 kg would be 20 m/sec.
If momentum is conserved the final velocity is 400 m/sec. 3 kg * 400 m/sec = 1 m/sec * 1200 kg
The 20 m/sec required for Law of Conservation of Energy is a 95% lose of momentum. You have only 5% remaining. You have lost 1140 (3 kg * 380 m/sec) units of momentum. And you have only 60 (3 kg * 20 m/sec) units of momentum remaining.
Per a massive quantity of experiments we know that momentum is the only motion transferred when a small object collides with a larger object. So if the 3 kilograms moving 20 m/sec begins to give its motion back to the rocket body then it can only restore 5% of the original motion. And this full 5% restoration can only be after the 3 kilograms has had its own velocity depleted to the same velocity of the rocket.
There is a point in the Dawn Mission motion transfer when three rpm has been restored to the rocket's rotation and the 3 kilograms is still out on the end of the tether with nearly maximum velocity.
Three rpm is 8.33% of 36 rpm; 3/36 = .0833 or 8.33 %
It is a physical impossibility for 5% of the motion to restore 8.33% of the motion. And this 8.33% restoration occurs when very little of the motion of the 3 kilograms has been removed.
What realty happens is that the rocket is stopped at the point where the 3 kilograms has 400 m/sec velocity. This stop is a little early of 90° to tangent and the 3 kilograms will pull the rocket back into motion before the tether comes to 90°. This will take 8.33% of the total motion; or 99.99 (3 kg * 33.33) units of momentum. The 3 kilogram mass was moving about 367 m/sec at release.
re: energy producing experiments
Because of the high speed of the 3 kilogram missile of the Dawn Mission; you are left with almost no (only 5%) momentum remaining in the missile if energy is conserved. If you have nothing you can do noting with it; such as restart the rotation of the rocket. Even a restart of 3 rpm is beyond the available quantity of momentum proposed by energy conservation. If energy conservation is true you only have 60 units of momentum and the 3 rpm in reverse cost 100 units of momentum.
The gray cylinder and sphere does a complete restart twice; but the dogma ignores the evidence. After two restarts energy conservation would take 12/240th of a second to cover the same distance that took 3/240th of a second at the start. The speed is still 3/240th at the end; so it is momentum that is traded back and forth not energy.
I don't want to start a conversation about dogma; but it is sad that it is so strong. Energy producing machines can be made for a few dollars and less than an hour of work.
The gray cylinder and sphere does a complete restart twice; but the dogma ignores the evidence. After two restarts energy conservation would take 12/240th of a second to cover the same distance that took 3/240th of a second at the start. The speed is still 3/240th at the end; so it is momentum that is traded back and forth not energy.
I don't want to start a conversation about dogma; but it is sad that it is so strong. Energy producing machines can be made for a few dollars and less than an hour of work.
re: energy producing experiments
If a one kilogram mass is suspended from one side of a 400 kilogram (200 kilograms on each side) Atwood's, the acceleration rate would be .02446 m/sec/sec. If a three kilogram mass is suspended from a 1200 kilogram (600 kilograms on each side) Atwood's the acceleration rate would be .02446 m/sec/sec.
For this one kilogram mass to have a velocity of one meter per second it will have to accelerate the Atwood's for 20.44 meters. That would be a drop of 20.44 meters for only the one kilogram. But at the end of a 20.44 meter drop, for the one kilogram, the whole 401 kilograms will be moving 1 meter per second.
If the moving Atwood's gives all of its motion to the one kilograms mass the mass can not have a velocity greater than 20.44 m/sec or the Law of Conservation of Energy will be violated.
A rim mass wheel accelerates in the same manner as an Atwood's.
So NASA claimed that the 3 kilogram thrown masses of Dawn Mission could not have a velocity greater than 20.44 m/sec. Because NASA claimed that the throw conserved energy. The three kilograms would only have 20.44 m/sec * 3 kg or 61.32 units of momentum. When the three kilograms interacts or collides with the 1200 kilograms only momentum can be conserved. The velocity of the 1200 kilograms at rest can only have a final velocity of 5.1 cm per sec. At 5.1 cm per second the rotational velocity of the 1200 kilogram rocket can only be 1.83 rpm. NASA had a 3 rpm back spin. If the Law of Conservation of Energy is true there is insufficient momentum available to do this back spin.
Angular momentum conservation is even slower yet; at 17.45 m/sec.
And this back spin occurs when everyone knows that very little motion has been removed from the 3 kilograms (“fully unwound�). Only after it rewinds can it begin to loss significant quantities of motion.
Dawn Journal: “After the third stage has finished firing, it remains securely attached to Dawn for another 4 minutes 50 seconds. Although the stage is stabilized by spinning, the spacecraft does not operate that way; yet by this time, they would be spinning together at '36' rpm, too fast for the latter's control system. Therefore, starting 5 seconds before separation, the third stage activates a surprisingly simple system to slow its rotation rate. Wrapped around the Delta are two cables, each 12.15 meters (39 feet 10 inches) long. At the end of each is a 1.44-kilogram (3-pound-3-ounce) weight made of aluminum and tungsten. When the cables are released, the spin causes them to unwind. As they carry the weights farther and farther out, the spin slows down because of the same principle that makes an ice skater spin faster by pulling her arms in or slower by extending them to her sides. After 4 seconds, when they are fully unwound, the cables unhook from the spacecraft. With their weights still attached, they enter independent orbits around the Sun; perhaps one of them will be studied by a future solar system archeologist.
Spinning slowly at 3 rpm in one direction, with xenon inside spinning at 39 rpm in the opposite—the original—direction (because the propellant still lags behind its container), Dawn waits for 8 minutes 20 seconds. That is long enough for the spacecraft and xenon each to slow the other down, and after that, Dawn's systems are ready to go to work.�
Typo: 46 to '36'; several other sources say 36. And the last paragraph states 39 which is 36 + 3
For this one kilogram mass to have a velocity of one meter per second it will have to accelerate the Atwood's for 20.44 meters. That would be a drop of 20.44 meters for only the one kilogram. But at the end of a 20.44 meter drop, for the one kilogram, the whole 401 kilograms will be moving 1 meter per second.
If the moving Atwood's gives all of its motion to the one kilograms mass the mass can not have a velocity greater than 20.44 m/sec or the Law of Conservation of Energy will be violated.
A rim mass wheel accelerates in the same manner as an Atwood's.
So NASA claimed that the 3 kilogram thrown masses of Dawn Mission could not have a velocity greater than 20.44 m/sec. Because NASA claimed that the throw conserved energy. The three kilograms would only have 20.44 m/sec * 3 kg or 61.32 units of momentum. When the three kilograms interacts or collides with the 1200 kilograms only momentum can be conserved. The velocity of the 1200 kilograms at rest can only have a final velocity of 5.1 cm per sec. At 5.1 cm per second the rotational velocity of the 1200 kilogram rocket can only be 1.83 rpm. NASA had a 3 rpm back spin. If the Law of Conservation of Energy is true there is insufficient momentum available to do this back spin.
Angular momentum conservation is even slower yet; at 17.45 m/sec.
And this back spin occurs when everyone knows that very little motion has been removed from the 3 kilograms (“fully unwound�). Only after it rewinds can it begin to loss significant quantities of motion.
Dawn Journal: “After the third stage has finished firing, it remains securely attached to Dawn for another 4 minutes 50 seconds. Although the stage is stabilized by spinning, the spacecraft does not operate that way; yet by this time, they would be spinning together at '36' rpm, too fast for the latter's control system. Therefore, starting 5 seconds before separation, the third stage activates a surprisingly simple system to slow its rotation rate. Wrapped around the Delta are two cables, each 12.15 meters (39 feet 10 inches) long. At the end of each is a 1.44-kilogram (3-pound-3-ounce) weight made of aluminum and tungsten. When the cables are released, the spin causes them to unwind. As they carry the weights farther and farther out, the spin slows down because of the same principle that makes an ice skater spin faster by pulling her arms in or slower by extending them to her sides. After 4 seconds, when they are fully unwound, the cables unhook from the spacecraft. With their weights still attached, they enter independent orbits around the Sun; perhaps one of them will be studied by a future solar system archeologist.
Spinning slowly at 3 rpm in one direction, with xenon inside spinning at 39 rpm in the opposite—the original—direction (because the propellant still lags behind its container), Dawn waits for 8 minutes 20 seconds. That is long enough for the spacecraft and xenon each to slow the other down, and after that, Dawn's systems are ready to go to work.�
Typo: 46 to '36'; several other sources say 36. And the last paragraph states 39 which is 36 + 3
re: energy producing experiments
The gray cylinder and spheres (as shown) has five significant instantaneous orientations in the rotation cycle of the experiment.
The first orientation is the starting position and it is a closed orientation; with the spheres up against the cylinder. The system is released after being spun at a rate where the frame by frame brake down shows that three frames are needed to cover a distance of one tape width. The tape width is about 18 mm so this is a velocity of 6 mm in one 240th of a second. Or 1.44 m/sec for all the 1124 grams. This would be 1.124 kg * 1.44 m/sec = 1.618 units of momentum; and ½ * 1.124 kg * 1.44 m/sec * 1.44 m/sec = 1.164 joules for position or orientation 1.
Shortly after the first is the second orientation where the cylinder is stopped and the spheres have all the motion. To conserve the energy of the first orientation the system will have to have 1.164 joules of energy. Or a velocity of ½ * .2491 kg * 3.06 m/sec * 3.06 m/sec. To conserve momentum the velocity would have to be 6.495 m/sec. Energy conservation requires that over half of the momentum has disappeared. You can not have two velocities there can only be one.
Here lies the problem: we can arrange for the .2491 kg mass to strike the cylinder tangent; by momentarily locking the cylinder and turning the smaller mass. You could arrange the 874.9 grams to be a point mass; and then you would have a 249.1 gram mass striking a 874.9 grams mass at rest. We are absolutely certain (from ballistic pendulums) of the out come of this; and it is the conservation of linear Newtonian Momentum.
For energy to be preserved with this mass difference (between spheres and cylinder) half the momentum disappears every time that all the motion is given to the smaller mass. Momentum conservation will maintain the same quantity of motion. You can only have one velocity so one of these two theories is false.
The third orientation of interest is where the spheres have returned the motion to the cylinder. This occurrences when the tether and spheres are 90° to tangent. With a velocity of only 3.06 m/sec (position 2) the 249.1 g can only give the 1,124 g (249.1 g + 874.9) grams a momentum of .7622 units; for a velocity of .678 m/sec. The velocity measured is again 6 mm per 1/240th of a second. The original momentum has remained with the same velocity (1.44 m/sec). If energy conservation is true the velocity can only be 2.82 mm/240th sec (about 6.37 frames for 18 mm) and this measurement of 6 mm /240th of a second must be in error. And that brings use to orientation 4 and 5.
In orientation four, which is a mirror image of orientation 2, the cylinder is stopped again. If energy is conserved it will be equal to ½ * 1.124 kg * .678 m/sec * .678 m/sec = .2583 joules. The .2491 kg is the only mass in motion; so for the 249.1 g to have .2583 joules it will have to be moving 1.4409 m/sec. For a momentum of .3587 units. Only momentum can be transferred to the larger 874.9 gram mass; and energy conservation tells us we only have .3587 units. Which bring us to position five.
In orientation five the spheres are rotating with the cylinder at about the same velocity; they are flying in position near the cylinder surface as they both rotate together. If energy is conserved the cylinder can only be moving .3587 /1.124 .3119 m/sec or 1.31 mm per 1/240th of a second. It will take 13.7 frames to move 18 mm. The measured velocity at the fifth position is still 18 mm for three frames.
Three frames cannot be mistaken for 13.7 frames. It is only 20 frames from a full rotational speed of the cylinder to a full stop of the cylinder.
Many people want to do science by consensus; but science is done by experiment. In the world of experiments you will have numbers that support or reject theories. The numbers from this experiment prove that the Law of Conservation of Energy is false.
Only momentum can return the motion of the cylinder and that is done twice.
The first orientation is the starting position and it is a closed orientation; with the spheres up against the cylinder. The system is released after being spun at a rate where the frame by frame brake down shows that three frames are needed to cover a distance of one tape width. The tape width is about 18 mm so this is a velocity of 6 mm in one 240th of a second. Or 1.44 m/sec for all the 1124 grams. This would be 1.124 kg * 1.44 m/sec = 1.618 units of momentum; and ½ * 1.124 kg * 1.44 m/sec * 1.44 m/sec = 1.164 joules for position or orientation 1.
Shortly after the first is the second orientation where the cylinder is stopped and the spheres have all the motion. To conserve the energy of the first orientation the system will have to have 1.164 joules of energy. Or a velocity of ½ * .2491 kg * 3.06 m/sec * 3.06 m/sec. To conserve momentum the velocity would have to be 6.495 m/sec. Energy conservation requires that over half of the momentum has disappeared. You can not have two velocities there can only be one.
Here lies the problem: we can arrange for the .2491 kg mass to strike the cylinder tangent; by momentarily locking the cylinder and turning the smaller mass. You could arrange the 874.9 grams to be a point mass; and then you would have a 249.1 gram mass striking a 874.9 grams mass at rest. We are absolutely certain (from ballistic pendulums) of the out come of this; and it is the conservation of linear Newtonian Momentum.
For energy to be preserved with this mass difference (between spheres and cylinder) half the momentum disappears every time that all the motion is given to the smaller mass. Momentum conservation will maintain the same quantity of motion. You can only have one velocity so one of these two theories is false.
The third orientation of interest is where the spheres have returned the motion to the cylinder. This occurrences when the tether and spheres are 90° to tangent. With a velocity of only 3.06 m/sec (position 2) the 249.1 g can only give the 1,124 g (249.1 g + 874.9) grams a momentum of .7622 units; for a velocity of .678 m/sec. The velocity measured is again 6 mm per 1/240th of a second. The original momentum has remained with the same velocity (1.44 m/sec). If energy conservation is true the velocity can only be 2.82 mm/240th sec (about 6.37 frames for 18 mm) and this measurement of 6 mm /240th of a second must be in error. And that brings use to orientation 4 and 5.
In orientation four, which is a mirror image of orientation 2, the cylinder is stopped again. If energy is conserved it will be equal to ½ * 1.124 kg * .678 m/sec * .678 m/sec = .2583 joules. The .2491 kg is the only mass in motion; so for the 249.1 g to have .2583 joules it will have to be moving 1.4409 m/sec. For a momentum of .3587 units. Only momentum can be transferred to the larger 874.9 gram mass; and energy conservation tells us we only have .3587 units. Which bring us to position five.
In orientation five the spheres are rotating with the cylinder at about the same velocity; they are flying in position near the cylinder surface as they both rotate together. If energy is conserved the cylinder can only be moving .3587 /1.124 .3119 m/sec or 1.31 mm per 1/240th of a second. It will take 13.7 frames to move 18 mm. The measured velocity at the fifth position is still 18 mm for three frames.
Three frames cannot be mistaken for 13.7 frames. It is only 20 frames from a full rotational speed of the cylinder to a full stop of the cylinder.
Many people want to do science by consensus; but science is done by experiment. In the world of experiments you will have numbers that support or reject theories. The numbers from this experiment prove that the Law of Conservation of Energy is false.
Only momentum can return the motion of the cylinder and that is done twice.
re: energy producing experiments
I was working with a 16 to 1 cylinder to sphere mass ratio. Under the proposed theory of energy conservation; the momentum of the spheres (when they have all the motion) is one fourth that of the original momentum. This is because four times the spheres velocity times 1/16th of the mass maintains the same quantity of energy as the closed original rotational energy. The closed position with the spheres up against the cylinder; had 16 units of mass and 1 unit of velocity.
This 16 to 1 cylinder and spheres has a 20 mm checker board marking for a frame by frame evaluation of the speed. In one release it takes 4 frames to cross the 20 mm. After a stop and restoration of motion it takes about 5.5 frames to cross the 20 mm from side to side. This is now a noticeable difference between start and finish; whereas the gray cylinder had no noticeable difference between start and finish. I suspect that this difference is due to air resistance; because the velocities are much higher than the gray cylinder.
It is also important to note that the cylinder hits the ground before the spheres have completed the transfer of motion to the cylinder, because the spheres are still extended containing a good deal of motion.
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The important thing is that there is just to much motion remaining in the cylinder to have a 75% loss of momentum.
I have another one that is an 18 to 1 mass ratio between the total (cylinder and spheres) and the spheres. It starts with a 4 frame crossing of the 20 mm square; and ends with a 5 frame crossing. This occurs after a complete transfer of motion to the spheres and then back again to the cylinder and spheres.
If energy was conserved the crossing of the black 20 mm square would require at least 17 frames.
This 16 to 1 cylinder and spheres has a 20 mm checker board marking for a frame by frame evaluation of the speed. In one release it takes 4 frames to cross the 20 mm. After a stop and restoration of motion it takes about 5.5 frames to cross the 20 mm from side to side. This is now a noticeable difference between start and finish; whereas the gray cylinder had no noticeable difference between start and finish. I suspect that this difference is due to air resistance; because the velocities are much higher than the gray cylinder.
It is also important to note that the cylinder hits the ground before the spheres have completed the transfer of motion to the cylinder, because the spheres are still extended containing a good deal of motion.
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The important thing is that there is just to much motion remaining in the cylinder to have a 75% loss of momentum.
I have another one that is an 18 to 1 mass ratio between the total (cylinder and spheres) and the spheres. It starts with a 4 frame crossing of the 20 mm square; and ends with a 5 frame crossing. This occurs after a complete transfer of motion to the spheres and then back again to the cylinder and spheres.
If energy was conserved the crossing of the black 20 mm square would require at least 17 frames.
Re: re: energy producing experiments
I wonder what would be the other means of catching this kinetic energy and converting into the potential energy rather than throwing the missile upward?pequaide wrote: Suppose you have a 19 kilogram rim mounted vertically (horizontal axis) on a quality bearing, and you place a string around the circumference of the rim with a one kilogram mass on the end of the string. F = ma gives you and acceleration rate of 9.81 N / 20 kg = .4905 (or 1/20 * 9.81 m/sec/sec ) = .4905 m/sec/sec.
The distance formula (d = ½ v²/a) says that a velocity of 1.1 m/sec will be achieved after the one kilogram mass has dropped 1.233 meters. You would now have 20 kilograms moving 1.1 m/sec for a momentum of 22 units.
The pictured (gray with tape) cylinder and spheres demonstrates that all the momentum of a spinning cylinder (or rim) can be given to a smaller mass. So you can place all of the 22 units of momentum into a one kilogram mass. For a one kilogram mass to have 22 units of momentum it will have to have a velocity of 22 m/sec.
With a velocity of 22 meter per second the one kilogram missile mass will rise 24.66 meters.
Will the ballistic pendulum do the job?
Or may be a spring?
re: energy producing experiments
Throwing up allows for the use of an Atwood's, which is used to increase the time over which the force acts. (step two)
I now have a 4.5 to one mass ratio in white; standard PVC pipe. It shows no slow down after the spheres have all the motion and then return the motion to the cylinder. Momentum simply is not lost; as proposed by energy conservation.
I now have a 4.5 to one mass ratio in white; standard PVC pipe. It shows no slow down after the spheres have all the motion and then return the motion to the cylinder. Momentum simply is not lost; as proposed by energy conservation.
re: energy producing experiments
The white pipe 4.5 total (cylinder and spheres 597 g) mass to a 1 spheres mass (132.6 g) also does a (starting) 4 frame crossing of the 20 mm square just like the thicker more massive gray pipe. If you use a shorter tether it will stop twice, before it hits the ground. After two stops of the cylinder and two returns of motion to the cylinder the crossing still takes only 4 frame. If energy were conserved it would take 17 frames; or the motion would nearly disappear.
This cylinder has one inch steel spheres which are drilled and more readily available. These are cheep and can be sent to individuals that would like to investigate them.
This cylinder has one inch steel spheres which are drilled and more readily available. These are cheep and can be sent to individuals that would like to investigate them.