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Author Topic: re: energy producing experiments  (Read 65970 times)

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #225 on: March 23, 2021, 02:23:41 AM »
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.

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Re: re: energy producing experiments
« Reply #225 on: March 23, 2021, 02:23:41 AM »

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #226 on: April 07, 2021, 12:02:58 AM »
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.

Offline Thaelin

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Re: re: energy producing experiments
« Reply #227 on: April 07, 2021, 01:08:42 AM »
sure glad I didn't want to be a mathematician, this hurts my dendrites just reading it.,

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Re: re: energy producing experiments
« Reply #227 on: April 07, 2021, 01:08:42 AM »
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Offline telecom

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Re: re: energy producing experiments
« Reply #228 on: April 07, 2021, 06:05:49 AM »
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.

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #229 on: April 07, 2021, 11:03:38 PM »
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.

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Re: re: energy producing experiments
« Reply #229 on: April 07, 2021, 11:03:38 PM »
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Offline telecom

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Re: re: energy producing experiments
« Reply #230 on: April 08, 2021, 02:04:37 AM »
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.
I wonder if cylinder and spheres is a special case, bc they look at the collision of 2 balls. I will try thinking how to make some kind of the mechanical device based on this.

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #231 on: April 10, 2021, 11:42:09 PM »
The kinetic energy for the one kilogram is equivalent to 9.81 N applied for 2.47 seconds. 294.3 J

The kinetic energy of the 12000 kg flywheel would be caused by 294.3 N applied for 9.03 seconds. 294.29 J

When motion is transferred from one object to another; the force in the string is equal in both directions; the force is equal to itself.
 
When motion is transferred from one object to another the time over which the force acts is equal for both masses; the time is equal.
 
When the one kilogram mass is unwinding from the flywheel; the force in the string is equal upon the flywheel and upon the one kilogram mass.
 
The time over which this force is applied is equal upon the flywheel and upon the one kilogram mass.

So as the string applies force between the two; the change in the kinetic energy of the flywheel and the change in the kinetic energy of the one kilogram mass will not be the same.  The force of 9.81 N for 2.47 seconds will stop the one kilogram mass but it would hardly faze the flywheel.

Once the flywheel is in motion; that motion that is shared with another mass will not conserve kinetic energy. It will conserve linear Newtonian momentum.

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Re: re: energy producing experiments
« Reply #231 on: April 10, 2021, 11:42:09 PM »
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