re: energy producing experiments

[Delburt Phend]

Lets use real numbers from commonly done real experiments.

Four hundred kilograms floating on dry ice can be accelerated by a string that is draped over a pulley. The string is weighted with one kilogram that can drop one meter.

After the one kilogram has dropped one meter the velocity is .2212 m/sec and the momentum is 88.7 kg * m / sec.

Using the despin procedure (as demonstrated by the 36 g and 2000 experiment) you can transfer this momentum to a small one kilograms mass.

A one kilograms mass with 88.7 units of momentum will be moving 88.7 m/sec.

The one kilogram moving 88.7 meters per second can rise 401 meters.

At a height of 401 meters one kilogram has 3933.81 joules of potential energy.

This series of events was started with one kilogram at one meter of height which has 9.81 joules of potential energy.

[Delburt Phend]

How much momentum is produces by a 12000 kg Atwood with 5985 kg on both sides; and the Atwood's uses a 30 meter high stack of 30 one kilogram masses as a drive mass. And how does that momentum compare with the momentum needed to reload the stack after a drop of one meter?

input

You have a 30 meter high stack of 1 kilogram masses spaced one meter apart. To increase the height of this stack by one meter; you only need to remove the bottom kilogram and place it on the top of the stack. To do this you need to give the one kilogram a velocity of 24.26 m/sec. This velocity will require the application of 9.81 newtons for 2.473 sec.

Formulas used are: d =1/2 at²  and d = ½ v²/a and F = ma 

The input force * time is   9.81 N * 2.473 seconds  = 24.26 N * sec; for one kilogram this is 24.26 m/sec

The input momentum is 24.26 m/sec * 1 kg = 24.26 kg*m/sec

Real numbers for output momentum;

A stack of 30 kilograms exerts a force of 30 kg * 9.81 N/kg = 294.3 N; upon a mass of 12000 kg (5985 kg + 5985 kg + 30 kg).

For an acceleration of F = ma:  294.3 N = 12000 kg * a   = .024525 m/sec²

After a drop of one meter the entire 12000 kilograms will be moving .22147 m/sec.

Formula used: d =1/2 v²/a    =   the square root of (1 m * 2 * .024525 m/sec²) = .22147 m/sec.

For a momentum to 2657.67 kg * m/sec;    12,000 kg * .22147 m/sec = 2657.64 kg* m/sec.

This takes 9.03047 sec for the stack to drop one meter; the stack has a force of 294.3 N, for 9.03047 sec * 294.3 N = 2657.66 N * sec.

So we put 24.26 N * sec in; and we get 2657.65 N * sec out, 109.5 times more momentum out than in.

The cylinder and sphere proves that this large slow motion can be given to a small object and in fact the small mass can give the motion back to the large mass. So the circle is complete.

You arrange for the large mass to give a small portion of it motion to the small mass and you have a cycling system, worth 7 trillion dollars.

[Thaelin]

sure glad I didn't want to be a mathematician, this hurts my dendrites just reading it.,

[telecom]

How much momentum is produces by a 12000 kg Atwood with 5985 kg on both sides; and the Atwood's uses a 30 meter high stack of 30 one kilogram masses as a drive mass. And how does that momentum compare with the momentum needed to reload the stack after a drop of one meter?

input

You have a 30 meter high stack of 1 kilogram masses spaced one meter apart. To increase the height of this stack by one meter; you only need to remove the bottom kilogram and place it on the top of the stack. To do this you need to give the one kilogram a velocity of 24.26 m/sec. This velocity will require the application of 9.81 newtons for 2.473 sec.

Formulas used are: d =1/2 at²  and d = ½ v²/a and F = ma 

The input force * time is   9.81 N * 2.473 seconds  = 24.26 N * sec; for one kilogram this is 24.26 m/sec

The input momentum is 24.26 m/sec * 1 kg = 24.26 kg*m/sec

Real numbers for output momentum;

A stack of 30 kilograms exerts a force of 30 kg * 9.81 N/kg = 294.3 N; upon a mass of 12000 kg (5985 kg + 5985 kg + 30 kg).

For an acceleration of F = ma:  294.3 N = 12000 kg * a   = .024525 m/sec²

After a drop of one meter the entire 12000 kilograms will be moving .22147 m/sec.

Formula used: d =1/2 v²/a    =   the square root of (1 m * 2 * .024525 m/sec²) = .22147 m/sec.

For a momentum to 2657.67 kg * m/sec;    12,000 kg * .22147 m/sec = 2657.64 kg* m/sec.

This takes 9.03047 sec for the stack to drop one meter; the stack has a force of 294.3 N, for 9.03047 sec * 294.3 N = 2657.66 N * sec.

So we put 24.26 N * sec in; and we get 2657.65 N * sec out, 109.5 times more momentum out than in.

The cylinder and sphere proves that this large slow motion can be given to a small object and in fact the small mass can give the motion back to the large mass. So the circle is complete.

You arrange for the large mass to give a small portion of it motion to the small mass and you have a cycling system, worth 7 trillion dollars.

Last time I looked at it, they involve law of the energy conservation when transferring linear momentum between 2 bodies ( elastic).
I'm not saying this is correct, but this is what they use in the physics study books.

[Delburt Phend]

Momentum is conserved in all collisions. We don't even need to discuss elastic collisions because momentum is conserved in those collisions as well.

Opponents to energy production from gravity would have you believe that in certain cases energy conservation overrides momentum conservation. This is an illusion and it is totally false. Momentum is conserved in all collisions.

So you have 12,000 kilograms moving .22147 m/sec and it is going to share its momentum of 2657.7 kg * m/sec with one kilogram.  Who in the world would drop the Law of Conservation of Momentum and go to the law of conservation of energy? Are they to pretend that 12 tons moving over 1/5 meters per second can't give 1 kilogram a velocity of 24.26 m/sec? Are they to pretend that 2633.44 units of momentum disappear?

The experiments like https://pisrv1.am14.uni-tuebingen.de/~hehl/despin2kg.mp4 or the 'cylinder and spheres' do not show an inability to transfer the motion of large objects to small objects.