Experimental Proof That Energy Is Not A Conserved Quantity.

[DelburtPhend]

Of the three theories of motion only Linear (arc speed) Newtonian Momentum predicts that the spheres will have enough motion to restore all of the rotational speed back to the cylinder.

 

What you see in the video is the full restoration of motion to the cylinder: twice.

 

The Dawn Mission analogy starts with 400 units of linear momentum; and yes I am naive enough to think that 400 kilogram moving one meter per second is 400 units of momentum.

 

Your math gives you only 39.9 units of momentum for a perfectly elastic collision; which is not true it will not even give you that. But even your improper math gives you only 39.9 units of momentum when the spheres are extended and 39.9 is not 400.

 

Your math says angular momentum gives you only 14.18 units of momentum when the spheres are extended: and 14.18 is not 400.

 

Your math says kinetic energy is satisfied with 20 units of linear momentum when the spheres are extended and 20 is not 400.

 

Collision experiments are not an apparition; in ballistic pendulums; 14.18 produces 14.18; 39.9 produces 39.9: 20 produces 20; and 400 produces 400.

 

Linear Newtonian momentum predicts that the spheres will have a velocity of 400 m/sec when the spheres are fully extended. And when the spheres re-wrap around the cylinder the full 400 units of momentum are returned to the cylinder as shown in the experiment.

 

Only linear Newtonian momentum predicts that the rotation of the cylinder will be returned to the original rate. The videoed experiment ends the possibility that Energy is a conserved quantity.

[DrKrettin]

 

 

All good stuff. Well....until next time - assuming he doesn't come back with something reasonable.

 

He didn't.

[DelburtPhend]

 

The only thing that makes motion is Force: F.

 

Newton's explanation for the quantity of motion was F = ma

 

Acceleration is velocity over time: so, F = mv/t

 

Multiplying both side by t we get Ft = mv;

 

Newton's explanation of motion was Ft or mv;

 

Now the professors of the world of physics say that mv can't be used to evaluate motion. That is identical to saying motion can't be used to evaluate motion.

 

Instead the inappropriate use of two other formulas have taken momentum's place. In an absolute sense Newtonian Physics has been thrown out.

 

If the motion of a balanced flywheel is caused by 20 newtons applied for 20 seconds; then that is the quantity of motion that the flywheel has. It is totally none functional to say that it is zero.

 

High school students on spinning chairs and ice skaters are not experiments. To qualify as an experiment you would have to know the radius and mass and the rate of spin. These events are innuendo and I think they are the only things ever cited for angular momentum conservation. There are no angular momentum conservation experiments done in the lab.

 

Linear Newtonian momentum conservation will also increase rotation rates for students on chairs and ice skaters; so nothing is proven by these events. So you are pitting zero experiments against a plethora of experiments for linear Newtonian momentum conservation.

 

Energy conservation acknowledges losing heat; so that leaves that out. 

[DelburtPhend]

http://www.bing.com/videos/search?q=momentum+conservation+videos&qpvt=momentum+conservation+videos&view=detail&mid=228F4110630B8B981825228F4110630B8B981825&FORM=VRDGAR

 

 

 

I am not picking on anybody but this is how it is taught; but this gentleman makes two (in my opinion) very obvious errors.

 

I will let you try and pick them out; and then I will give my evaluation later.

[DelburtPhend]

First he states that angular momentum works when there is no external torque; Torque alludes to the fact that the point mass is being accelerated; And he is saying that the point mass is not being accelerated by an external force. In space gravity gives massive amounts of force that accelerate the satellite. But in the lab there is none. So does it work with and/or without the application of outside force?

 

The second error is even more glaring. He clearly states that the linear momentum increases from 3 m/sec to 6 m/sec. And that there is no application of outside force. Clearly impossible.

[DelburtPhend]

There it is: you can't go from 3 m/sec to 6 m/sec without the application of outside force. There is a good reason that there are no experimental proofs for angular momentum conservation, because it has never worked in the lab.

 

That leaves us with only Linear Newtonian Momentum to explain the slow motion video of the, 1 / 4.5 cylinder and spheres, de-spin device. And Newtonian Physics predicts a 450% increase in energy when the spheres have all the motion.

[DelburtPhend]

Franklin Hu has a good experiment here; https://www.youtube.com/watch?v=9MGQJar8dNg

 

I was prevented from an evaluation on another site.

 

Franklin uses the correct formula for angular momentum; and he realizes that the Linear or Newtonian velocity must double if angular momentum is to be conserved. Because he states that the rate of rotation must quad. If the linear velocity of the batteries had remained constant they would go around the half sized circle only twice as fast.

 

Because: The radius in the formula (L = rmv) has halved then the linear velocity must double; 1/2 r * m * 2 v for angular momentum to be conserved. At twice the arc velocity they go around the half size circle four times as fast.

 

But isn't it possible that the pulling of the batteries (in) is actually a force that has accelerated the batteries to double their original speed.

 

If the pulling is not a force then you have a free increase in momentum: which is a violation of Newton's Laws of Motion. I shall assume that a pull is a force and Newton is still correct.

 

If the batteries had wrapped around an immovable post there would have been no linear (arc) acceleration: because there would have been no outside force. And I don't think there is any Earth motion involved; because the batteries would not lose motion as they wrap around the post.

[exchemist]

Is this your website https://delburtphend.home.blog/2018/09/17/energy-from-gravity-2/ and was it the https://www.physicsforums.com you were posting on and what prevented the evaluation?

Best of luck. 

[exchemist]

Oh! not again :)  I had guessed there might be something a bit odd, that is why I asked the question "what prevented evaluation". I did not look too hard   and became a very suspicious :) . 

Haha. If you skim through the thread you will see a number of us manfully tried to explain angular momentum and energy conservation to this person, at some length, almost exactly a year ago.

 

I suppose, as it's Christmastime again, our friend may again be at a loose end, or may have been released once more from the psychiatric institution he normally resides in. 

 

I quite enjoyed the process, at the time, but I'm not sure I would want to go through it all over again. 

 

This poster would not last more than a few days in any moderated serious science forum.

[Dubbelosix]

Conservation breakdown could happen if a virtual particle avoids collapsing back to radiation by avoiding contact with an antiparticle. This is how Hawking radiation works and gives rise to the information paradox. How you detect this in spacetime is uncertain or how you measure a difference in energy is also uncertain other than suggesting the metric tensor has to vary (as Sean Carrol has argued). The beginning of spacetime may have given rise to particles, in irreversible ways through a short non-conserved particle production phase.

[DelburtPhend]

It should also be noted that if there is a doubling of the linear velocity of the batteries then the kinetic energy has also increased by a factor of four. This Hu experiment would then be a violation of Newtonian Physics and The Law of Conservation of Energy: if there really was no application of outside force. The batteries are moving around a half size circle at four times as fast; that means that the linear velocity has doubled.

This Hu battery experiment shows that angular momentum needs an outside force (such as gravity from the Sun or an experimenters hand) to be correct.

If you gentleman are smarter than I: then why did you miss the fact that the Hu experiment has a quad increase in energy?

A one meter diameter steel cylinder one meter high would have a mass of about 6 tons.

A 100 meter high stack of these 6 ton cylinder would have a mass of 600 metric tons.

This 600 tons could accelerate a 577 metric ton Atwood’s machine for an acceleration of 5 m/sec/sec.

After the 600 ton stack has dropped .1 meter the entire 1177 ton mass would be moving 1.0 meters per second.

This is 1177 ton moving 1 m/sec for 1.177 million units of momentum.

A mass moving 44.295 m/sec will rise 100 meters. Six thousand kilograms moving 44.295 m/sec has 265,770 units of momentum.

265,770 is less than a fourth of the total momentum (1,177,000) for a .1 m drop of the 600 ton tower.

The tower can drop one meter before the 6 ton cylinder, rising from the bottom, can be installed at the top.

So only 1/4 of the momentum from 1/10 of the drop is needed to restore the original configuration of the power plant. The other 97.5% is free energy.

This is about 5.74 million joule in less than 2 second.

Edited December 18, 2018 by DelburtPhend

[OceanBreeze]

It should also be noted that if there is a doubling of the linear velocity of the batteries then the kinetic energy has also increased by a factor of four. This Hu experiment would then be a violation of Newtonian Physics and The Law of Conservation of Energy: if there really was no application of outside force. The batteries are moving around a half size circle at four times as fast; that means that the linear velocity has doubled.

This Hu battery experiment shows that angular momentum needs an outside force (such as gravity from the Sun or an experimenters hand) to be correct.

If you gentleman are smarter than I: then why did you miss the fact that the Hu experiment has a quad increase in energy?

A one meter diameter steel cylinder one meter high would have a mass of about 6 tons.

A 100 meter high stack of these 6 ton cylinder would have a mass of 600 metric tons.

This 600 tons could accelerate a 577 metric ton Atwood’s machine for an acceleration of 5 m/sec/sec.

After the 600 ton stack has dropped .1 meter the entire 1177 ton mass would be moving 1.0 meters per second.

This is 1177 ton moving 1 m/sec for 1.177 million units of momentum.

A mass moving 44.295 m/sec will rise 100 meters. Six thousand kilograms moving 44.295 m/sec has 265,770 units of momentum.

265,770 is less than a fourth of the total momentum (1,177,000) for a .1 m drop of the 600 ton tower.

The tower can drop one meter before the 6 ton cylinder, rising from the bottom, can be installed at the top.

So only 1/4 of the momentum from 1/10 of the drop is needed to restore the original configuration of the power plant. The other 97.5% is free energy.

This is about 5.74 million joule in less than 2 second.

 

 

Are you some kind of bad elf that only shows up at Christmas? :xmas_grinch:

 

I am not even going to bother correcting all of your wrongness this time around.

I will just point out how you start off wrong

You wrote:

 

This 600 tons could accelerate a 577 metric ton Atwood’s machine for an acceleration of 5 m/sec/sec.

 

In fact, the acceleration of the masses on an Atwood Machine is calculated as follows:

 

a = [(M1-M2)*g] / (M1+M2)

 

In your example M1 is 600,000 kg, M2 is 577,000 kg

 

a = [(600,000 – 577,000) kg * 9.8 m/s^2] / ( 600,000 + 577,000) kg = 0.1915 m/s^2

 

Your value of 5 m/s^2 is bollacks and so is everything that follows on from that.

[DelburtPhend]

It is 600/1177 * 9.81 m/sec/sec = 5      288.5 ton on each side of the Atwood's

[OceanBreeze]

It is 600/1177 * 9.81 m/sec/sec = 5      288.5 ton on each side of the Atwood's

 

Learn how an Atwood machine works then get back to us. On second thought, forget about getting back to us.

[exchemist]

Are you some kind of bad elf that only shows up at Christmas? :xmas_grinch:

 

I am not even going to bother correcting all of your wrongness this time around.

I will just point out how you start off wrong

You wrote:

 

This 600 tons could accelerate a 577 metric ton Atwood’s machine for an acceleration of 5 m/sec/sec.

 

In fact, the acceleration of the masses on an Atwood Machine is calculated as follows:

 

a = [(M1-M2)*g] / (M1+M2)

 

In your example M1 is 600,000 kg, M2 is 577,000 kg

 

a = [(600,000 – 577,000) kg * 9.8 m/s^2] / ( 600,000 + 577,000) kg = 0.1915 m/s^2

 

Your value of 5 m/s^2 is bollacks and so is everything that follows on from that.

It's a sort of Christmas quiz, evidently: "Bozo the Elf has a silly proposal for free energy. See how many major errors in mechanics you can spot." 

[DelburtPhend]

The formula to determine acceleration has three variable: Force; mass and acceleration. F = ma

 

But I am not using the formula to determine acceleration I have already determined that the acceleration is going to be 5 m/sec/sec. As they say in math; (a) is not a variable it is given.

 

The force is also given; because the accelerating mass is given to be 600 metric tons. This is 600,000 kilograms; and gravity gives use 9.81 newtons of force for each kilogram: so we have 5,886,000 newtons.

 

Now we come to the variable; mass (m). The mass being accelerated has two parts: the drive mass (600 ton) and the balanced mass of the Atwood's (n; the unknown).

 

So we have 5,886,000 newtons = (600 tons + n tons) * 5 m/sec/sec.        5,886,000 / 5 = 1,177,200 kilograms.

 

Of the 1,177,200 kilograms 600,000 are the drive mass; that leaves 577,200 to be the balanced mass of the Atwood's.

 

Now we are familiar with the construction of the Atwood's; half of the balanced mass hangs from one side of the pulley and half the mass hangs from the other side. 577,200 / 2 = 288.6 metric tons hangs from both sides of the balanced Atwood's

 

This would mean that M2 is 288.6 metric tons. And M1 is (600 + 288.6) 888.6 metric tons. (M1 and M2 are the hanging masses).

 

Of course I do not go through all this math each time. Because this short formula also work: accelerating mass / total mass * g = a . And solving for total mass we have: 9.81 m/sec/sec *600 ton / 5 m/sec/sec = 1177.2 tons; 600 of this total is the drive mass; so the Atwood's has a balanced mass of 577.2 ton. A 577.2 metric ton Atwood's.

[DelburtPhend]

The stack of one hundred 6 ton masses produces a minimum of 14 time more momentum than that which is needed to reload the stack. A 6 ton mass at the bottom can be returned to the top with 265,767 units of momentum; but when the stack is dropped one meter in produces a minimum of 3,722,000 units of momentum.

 

Ballistic pendulums show that small masses can not give there energy to a large object. If a one kilogram mass combines with a 3.5 kilogram mass at rest it will lose 77% of the energy to heat; and it can't come back.

 

Linear Newtonian momentum is conserved in ballistic pendulums.

 

The Hu experiment (batteries on a close-hanger) proves that an accelerating force must be added for angular momentum to be conserved.

 

Linear Newtonian Momentum does not need outside force to be conserved.

 

So if you have an experiment that transfers the motion back and forth from 1 unit of mass to 4.5 units of mass (without the application of outside force) what is the only type of motion that is being conserved.