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### Author Topic: re: energy producing experiments  (Read 128654 times)

#### Delburt Phend

• Full Member
• Posts: 221
##### 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.

#### Delburt Phend

• Full Member
• Posts: 221
##### 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.

#### Thaelin

• TPU-Elite
• Hero Member
• Posts: 1079
##### 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.,

#### telecom

• Hero Member
• Posts: 560
##### 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.

#### Delburt Phend

• Full Member
• Posts: 221
##### 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.

#### telecom

• Hero Member
• Posts: 560
##### 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.

#### Delburt Phend

• Full Member
• Posts: 221
##### 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.

#### Delburt Phend

• Full Member
• Posts: 221
##### Re: re: energy producing experiments
« Reply #232 on: April 17, 2021, 10:06:16 PM »
This is long but some may have interest; it produces energy.

What would be the acceleration; if an Atwood’s had a drive mass of 100 g at a 1 cm radius: and a balanced mass of 100 grams on each side, at a 10 cm radius?

100 g at 10 cm would have the same torque as 1000 grams at 1 cm. So you could have 1 kg at 1 cm on one side of the Atwood's with the 100 grams of drive mass. And on the other side you could have 100 g at 10 cm; and the Atwood’s would accelerate at the same rate.

In fact you could have all the mass at the 1 cm radius by trading the remaining 100 grams on the 10 cm radius for 1000 grams at the 1 cm radius. Now you have 2000 grams of balanced mass at the 1 cm radius with the 100 extra grams that would be the source of the accelerating force.

So you would have an accelerating force of: F = ma, .981 N = 2.1 kg * a: .981 N / 2.1 kg = .467 m/sec²

Okay: keep in mind that 1000 g at 1 cm is equivalent to 100 g at 10 cm. Their angular acceleration will be the same.

Also keep in mind that I am purposefully manipulating machines to my advantage; to make energy from the gravitational field. I will shift motion from one machine to another to achieve this purpose.

The acceleration of the Atwood's drive mass of 100 grams at a radius of 1 cm is .467 m/sec².

After a drop of one meter for only this 100 gram drive mass: the drive mass will be moving: .9664 meters per second.  d = ½ v²/a

If the 2000 grams at 1 cm were present they would also be moving .9664 m/sec; but the 2000 grams is not present; what is present (as stated in the question) is (2 * 100 g at 10 cm)  200 g at 10 cm.

At 10 cm the two 100 gram masses are moving 9.664 m/sec.

At 9.664 m/sec the two masses have an energy of: ½ *.2 kg * 9.664 m/sec * 9.664 m/sec =  9.339 joules of energy. The drive mass of .1 kg is moving .9664 m/sec for .04669 joules. For a total final energy of 9.3857 joules

The drive mass of .1 kg dropped 1 meter for an input energy of; 9.81 N/ kg * .1 kg * 1 meter = .981 joules.  From; Nm;  or ½ * .1 kg * 4.429 m/sec * 4.429 m/sec = .981 joules.

So you see that the output energy is 9.3857 joules and the input energy was only .981 joules.

When released both these 100 grams masses, on the end of the tube, will rise 4.76 meters. And you started with 100 grams at one meter.

Mechanical questions: Release the masses at highest velocity while they are moving up. Catch them on a ledge and feed them back down through.

A comment from someone else:

So if your calculations appear to show a net gain in energy you must be making a mistake or there is some mass flow down which you have not accounted for.

It’s not my job to show what the error is; it’s up to you to demonstrate that the machine or mechanism you propose actually works like you claim. That means you have to build it; proving physics is wrong by using its equations to show it is, doesn’t make sense unless - maybe - you can mathematically prove you can create energy.
So until you build one there is no reason to believe you have created what is, in essence, a perpetual motion machine of the first kind (i.e. an effective efficiency larger than 100%).

Phend
This one is 957%. And this one has only one meter of working height; without a stack, or a large flywheel.

“Not your job” Oh you would be more than willing to point out the error; but you can’t find one. You can obviously do the math because you came up with the correct acceleration. You just can’t find an error.

I have tested all the steps in these machines with real experiments; and they all work.

Those that will accept a paradigm shift will become billionaires.

#### Delburt Phend

• Full Member
• Posts: 221
##### Re: re: energy producing experiments
« Reply #233 on: April 23, 2021, 12:11:30 AM »
This is from someone of the oppositional view; but it is an excellent description of the cylinder and spheres experiments on youtube.

"Fine, I’ll humor you by doing some theoretical qualitative analysis and comparing it to some of your video evidence with cylinders and spheres.

The cylinder and spheres is a yo-yo de-spin.

Let me tell you what I think will happen in a few conceptually simple cases.
If the wire-length is just right, it will be unwound and perpendicular to the cylinder exactly when the cylinder is no longer spinning.

This is the “ideal” case if you want to de-spin a satellite completely but hardly ever is achieved because things never happen exactly like you planned.

If the wire is too long, the cylinder will come to a stop before the wire is completely unwound and (completely) perpendicular to the cylinder. This will cause the cylinder to reverse its spinning - because the weights/string cause a torque - to some degree.

It entirely depends on the exact value of some parameters and variables -e.g. mass of spheres, moment of inertia of cylinder, position of the wire when the cylinder stops - what will happen next.

The balls could reverse the cylinder again and re-wrap, or start re-wrapping and then reverse the spin.

We also can’t tell what happens when a wire is too short. There are just too many variables and parameters for anyone to take a guess what will exactly happen.

At least one of your own experiments actually has the cylinder reverse its spin twice before or while the wires appear to re-wrap in the opposite direction (“132 grams / 1320 grams”).

Another configuration completes such a re-wrap and then appears to start de-wrapping again before hitting the floor (“spheres 305 g cylinder 972 gra”/”newtonian momentum energy increases”).

The “restart of despin” actually just shows a reverse in spin, and the “Double yo-yo despin” does a re-wrap and starts de-wrapping again.

I won’t do any calculations here, first of all because you would not care, and second because the exact relationship between the variables and parameters make this much more complicated than the above descriptions might suggest.
It is possible mind you, just not something I am interested in doing if you’re not.

And NASA already did it (e.g. THEORY AND DESIGN CURVES FOR A YO-YO DE-SPIN MECHANISM FOR SATELLITES) so why bother.

So my theoretical qualitative “prediction” matches with most of your video evidence.

It would be interesting if you would try out the same cylinder and spheres but with different wire lengths to check if my intuitions are right."

#### Delburt Phend

• Full Member
• Posts: 221
##### Re: re: energy producing experiments
« Reply #234 on: July 05, 2021, 04:31:19 PM »
Were the forces mentioned in the Third Law of Motion the same force that was defined in the Second Law of Motion?

Danny Kodicek
, Educational Software Developer and Teacher
Yes, all three laws of motion are essentially the definition of what we mean by force. Collectively, they also comprise the law of conservation of momentum: the first law tells us that momentum can only be changed by the action of a force; the second tells us the effect of force on momentum; and the third says that whenever a force changes the momentum of one object, another force must change the momentum of some other object by an equal amount in the opposite direction, thus leaving the total momentum unchanged.

Larry Rothstein
, Engineering at SAIC (1981-present)
In a way yes, and in a way no. The Second Law of Motion describes the motion of a body subjected to an External force. In this case it it the net Force on the body, or the sum of all External Forces. Note the emphasis on External. For example, you push a ball and the ball accelerates and flies through the air.
The Third Law of Motion describes the reaction to the External Force. For example, when your foot is in contact with the ball, the ball is pushing back on your foot. The ball accelerates and the your foot decelerates a small amount

Delburt Phend
32m ago
Note the answers to the 2nd and 3rd Laws question. What does this mean?
It means that the force in the tether (in the despins) causes an equal change of linear Newtonian momentum for the released mass and the satellite. F = ma is a linear (curved lines also) equation.

Dawn Mission yo-yo despin is an example of a despin.

#### Delburt Phend

• Full Member
• Posts: 221
##### Re: re: energy producing experiments
« Reply #235 on: October 25, 2021, 06:37:27 PM »
What is the acceleration of a one kg rim mass wheel; that has a one meter diameter and is accelerated by 100 kg suspended by a string from a 1 cm shaft?

#### Delburt Phend

• Full Member
• Posts: 221
##### Re: re: energy producing experiments
« Reply #236 on: November 11, 2021, 09:57:51 PM »
A massless one meter beam balance has 4 kg on one end and 1 kg on the other. It is rotating about its center of mass at 6.19 radians/sec. What is the arc speed of the one kg?

The mass relationship is 4 kg to 1 kg so when the beam is rotating about it center of mass: the 4 kg will have a radius of 20 cm and the 1 kg will have a radius of 80 cm.

At 6.19 radians per second the 4 kg is moving .2 m * 6.19 rad/sec = 1.238 m/sec around the arc of the circle.

At 6.19 radians per second the 1 kg is moving .8 m * 6.19 rad/sec = 4.952 m/sec around the arc of the circle.

At 4.952 m/sec the 1 kg can rise 1.25 m.

One kilogram in a modified Atwood’s will accelerate a 4 kg block to 1.98 m/sec after the 1 kg has dropped 1 meter.

Okay lets go through this again: You drop a one kilograms mass to accelerate a 4 kilogram block in a modified Atwood’s.

This gives you a final velocity of: the square root of (1 m * 2 * 9.81 m/sec² * 1/5) = 1.9809 m/sec

And this is 9.9045 units of momentum. Because this is 5 kg moving 1.9809 m/sec

A balanced beam that has 4 kg on one end and 1 kg on the other end; and is moving 6.19 rad/sec also has 9.9045 units of momentum.

The one kilogram on the beam has half the momentum (4.95 units) and it can rise 1.25 m.

It was dropped 1 m.

If the one kilogram had all the momentum it would rise 5 m.

Physics with George 5：Rotation about the center of mass of a system like the Moon-Earth - YouTube

#### sm0ky2

• Hero Member
• Posts: 3947
##### Re: re: energy producing experiments
« Reply #237 on: November 15, 2021, 03:50:18 AM »
2 things dilbert

1 - to get your 9 units you had to spend 10 getting the 5kg up to 60rpm

2 - to transfer all of the momentum into the 1kg mass:
you have to stop the 4kg mass…..

#### kolbacict

• Hero Member
• Posts: 1252
##### Re: re: energy producing experiments
« Reply #238 on: November 15, 2021, 01:01:27 PM »
will anytime any picture here ?

#### Delburt Phend

• Full Member
• Posts: 221
##### Re: re: energy producing experiments
« Reply #239 on: November 15, 2021, 05:48:29 PM »
smOky2 quote: 1 - to get your 9 units you had to spend 10 getting the 5kg up to 60rpm

No: it only cost 4.429 units of momentum. If you drop a one kilogram mass one meter it will have a velocity of 4.429 m/sec. To throw a one kilogram mass up one meter it will cost you 4.429 units of momentum. The one kilogram was dropped one meter.

smOky2 quote: 2 - to transfer all of the momentum into the 1kg mass:
you have to stop the 4kg mass…..

This is true. We can place this rotating beam in a horizontal plane. The 4 kilograms will have lost 7.9236 units of momentum: but it is in the same plane as that plane where it started; and it is at rest which is also how it started.

Now we have one kilogram that adds those 7.92 units of momentum to 1.98 and we have one kilogram moving (7.92 = 1.98) 9.9 m/sec and it can rise   d = ½ v²/a;  5 meters.      It was dropped one meter.

It seems possible that we can place a 4 kg block on one end of a horizontal beam and a 1 kg mass on the other end of a horizontal beam. Even though both masses are moving; I think this is doable.

As for pictures;  “Delburt Phend youtube ” shows rotating objects giving all their motion to smaller mass spheres.

And the ‘MIT Atwood’s machine’ has an input momentum of .0443 units and an output of .4624 units of momentum.    .01 kg * 4.429 m/sec    1.11 kg * .41666 m/sec

If this MIT Atwood’s was a modified Atwood’s you would have 1.100 kg moving .4166 m/sec (on a plane) and one meter away there would also be .01 kg moving .4166 m/sec. If these two masses were placed on the ends of a balanced beam; and the beam were rotating about its center of mass, then the 1.100 kg would be moving .2102 m/sec and the .01 kg would be moving 23.135 m/sec. At 23.135 m/sec the .01 kg will rise 27.255 m.

The .01 kg will rise 27.255 m and it was dropped 1 meter.

So one way to make energy is to combine a modified Atwood’s with a  rotating balanced beam.