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# New Book

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### Author Topic: re: energy producing experiments  (Read 20632 times)

#### telecom

• Sr. Member
• Posts: 412
##### Re: re: energy producing experiments
« Reply #75 on: March 29, 2017, 03:57:58 AM »
I think you have the concept; but I prefer to look at it as a function of time; telecom.

A one kilogram missile moving 19.81m/sec will rise 20 meters.  From d = ½ v²/a. This will take 2.019 second.  From d = ½ at².

A kilogram mass applies 9.81 newtons of force. This is 9.81 newton applied for  2.019275 seconds.  = 19.81 N seconds; N*s

Form a chain of twenty one kilogram masses; they are one meter apart, 20 meters high. This vertical chain could be dropped one meter. The original configuration can be restored if only one kilogram is raised 20 meters.

Connect this 20 kilogram chain to a 80 kilogram flywheel. It will take 1.009637 seconds for the chain to drop one meter while it spins the wheel. But this is 20 kilograms dropping one meter for (20 kg * 9.81 N/kg)  196.2 N applied for 1.009637 seconds for 198.09 N seconds.

The output momentum is ten times that of the input momentum.

The output momentum is 100 kilograms moving 1.9809 m/sec for 198.09 kg*m/sec.

The input momentum was 1 kg moving 19.81 m/sec for 19.81 kg*m/sec.

The input energy is 196.2 joules.

The output energy is 19,620 joules.

In the 20 kilogram; twenty meter; vertical chain arrangement one kilogram applies its 9.81 newtons for 1.0096 seconds 20 times before it needs to be returned to its original vertical position at the top of the chain. The other 19 kilograms in the chain always assist in the application of force but each one kilogram applies its 9.81 N for 1.0096 seconds 20 times.

The descending mass is 9.81 N applied for 20.19274 seconds. This is 198.09 Ns.

The ascending mass is only 9.81 N applied for 2.0193 second for 19.809 Ns.

This time difference multiplies the momentum by 10 times and the energy by 100 times.

But doesn't it require each of the masses to be lifted 1 m for the original position?
Or I'm missing something?
Regards

#### Free Energy | searching for free energy and discussing free energy

##### Re: re: energy producing experiments
« Reply #75 on: March 29, 2017, 03:57:58 AM »

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #76 on: March 29, 2017, 11:27:50 PM »
Stack 20 dominoes on top of each other on their sides. Hold the second domino from the bottom and slide out the bottom domino. Place the removed domino on the top of the stack.

We now have the starting position; of 20 dominoes that are at a distance of one domino width from the table. We can lower the stack of twenty dominoes by one domino width so that the stack again rests on the table. We can remove the bottom domino again; and place it on the top of the stack. So by raising one domino twenty domino widths we can continually restore the original configuration. (This is quite a trick with twenty; but works easily with six)

All the one kilogram masses in the chain drop one meter; but only one needs to be raised 20 meters to restore the driving chain (stack) to its original configuration. It takes a whole lot less momentum to raise one kg twenty meters than to raise 20 kg one meter. It takes 19.81 (19.81 m/sec * 1 kg) units of momentum to raise one kilogram 20 meters: and 88.59 (4.429 m/sec * 20 kg) units of momentum to raise 20 kilograms one meter. 19.81 / 88.59 = 22.36%

You are actually using 19.81 units of momentum to make 88.59. The momentum increase is 88.59 / 19.81 447%. When you transfer all the motion of the 88.59 units of momentum to one kilogram you get (½ * 1 kg * 88.59 m/sec * 88.59 m/sec) 3924 joules of energy: for an energy increase from 196.2 J to 3924 joules; 2,000%.

If you attach  the chain to a 80 kilogram flywheel you are then using 19.81 units of momentum to make 198.1.

#### telecom

• Sr. Member
• Posts: 412
##### Re: re: energy producing experiments
« Reply #77 on: March 30, 2017, 08:48:38 PM »
Stack 20 dominoes on top of each other on their sides. Hold the second domino from the bottom and slide out the bottom domino. Place the removed domino on the top of the stack.

We now have the starting position; of 20 dominoes that are at a distance of one domino width from the table. We can lower the stack of twenty dominoes by one domino width so that the stack again rests on the table. We can remove the bottom domino again; and place it on the top of the stack. So by raising one domino twenty domino widths we can continually restore the original configuration. (This is quite a trick with twenty; but works easily with six)

All the one kilogram masses in the chain drop one meter; but only one needs to be raised 20 meters to restore the driving chain (stack) to its original configuration. It takes a whole lot less momentum to raise one kg twenty meters than to raise 20 kg one meter. It takes 19.81 (19.81 m/sec * 1 kg) units of momentum to raise one kilogram 20 meters: and 88.59 (4.429 m/sec * 20 kg) units of momentum to raise 20 kilograms one meter. 19.81 / 88.59 = 22.36%

You are actually using 19.81 units of momentum to make 88.59. The momentum increase is 88.59 / 19.81 447%. When you transfer all the motion of the 88.59 units of momentum to one kilogram you get (½ * 1 kg * 88.59 m/sec * 88.59 m/sec) 3924 joules of energy: for an energy increase from 196.2 J to 3924 joules; 2,000%.

If you attach  the chain to a 80 kilogram flywheel you are then using 19.81 units of momentum to make 198.1.
I think I finally understood this, but this looks like a perpetual energy making machine to me,
unless I'm hallucinating!
Where is the catch then???

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #78 on: March 30, 2017, 10:09:43 PM »
Yes: that is what it is; 'a perpetual energy making machine'

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #79 on: March 30, 2017, 10:26:55 PM »
The picture is of another picture off of a monitor; sorry, rather fuzzy. I am just trying to figure this picture stuff out. It is a picture of the cylinder; rotationally stopped.

I did learn how to make my own slow motion videos.

#### Free Energy | searching for free energy and discussing free energy

##### Re: re: energy producing experiments
« Reply #79 on: March 30, 2017, 10:26:55 PM »

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #80 on: January 23, 2018, 01:34:46 AM »

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #81 on: March 31, 2018, 03:07:38 AM »

It is actually impossible that Newton can be wrong in this energy source:

I have one inch spheres that have a mass of 152 grams, they are tungsten spheres and should weight a little more. But I will do some math with what is real. (152 grams)

If I accelerate one of these spheres to 20 m/sec it will rise 20.387 meters. This 20.387 m is 802.6 inches.

This means I can stack 802 of the 152 grams spheres on top of each other. This is 121.9 kilograms. And I can drop the entire stack one inch (2.54 cm).

When the stack is dropped 2.54 cm it will have a final velocity of: the square root of (.0254 m * 2 * 9.81 m/sec/sec) = .7059 m/sec.  This is a momentum of 121.9 kg * .7059 m/sec = 86.054 units

The sphere needs only .152 g * 20 m/sec = 3.04 of these 86 units to travel back up to the top and reconfigure the pre-drop arrangement.

The 121.9 kilograms is equal to a rim of the same mass; and moving at the same speed of .7059 m/sec around the arch of the circle.
121.9 kg * .7059 m/sec = 86

Or it could be a 60.95 kilogram rim moving 1.4118 m/sec around the arch of the circle.

Or 30.475 kg moving 2.836 m/sec.

Or 15.2375 kg moving 5.647 m/sec.

Or 7.615 kg moving 11.29 m/sec.

Or 3.809 kg moving 22.588 m/sec. = 86.046 units;   this 22.588 m/sec will send 3.8 kilograms up 26.0 meters.

When this 22.588 m/sec for 3.8 kilograms is given to .152 kilograms it will send it to the next county.

You would hear a whir and you would never see the sphere again; if you were lucky enough not to get hit.

#### Free Energy | searching for free energy and discussing free energy

##### Re: re: energy producing experiments
« Reply #81 on: March 31, 2018, 03:07:38 AM »

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #82 on: April 07, 2018, 03:51:36 PM »

It takes (19 frames) to go from 1.2 m/sec of rotation of the cylinder and spheres to the first stop of the cylinder. It takes the same amount of time (19 frames) to go from the last stop, of the cylinder's rotation, to the last full return of the cylinder's rotational motion. This motion appears to be the same 1.2 m/sec (from counting the frames needed to cross the black square). The 19 frames confirms that the rotational velocity is indeed the same. The accelerations are of the same magnitude (19 frames) therefore linear Newtonian momentum is conserved.

It would be moving about a fourth that fast (83 frames) if energy was conserved and the alleged heat was lost. It would take 83 frame to go from the last stop to the last restart instead of 19 frames.  The entire experiment is only 75 frames.

A mass moving on the end of a string can wrap around a stationary post. The string will become shorter and the radius will be reduced; but the linear Newtonian momentum will remain the same.

In this (two 86 gram bolts) cylinder and sphere experiment the total mass of a spinning object is reduced from 1448g (cylinder and spheres) to 304g (spheres); but the linear Newtonian momentum remains the same.

In this (two 86 gram bolts taped to the cylinder) cylinder and sphere experiment the energy increases from (.5 * 1.448 kg * 1.2 m/sec 1.2 m/sec) = 1.0425 joules to (.5 * .304 kg * 5.716 m/sec * 5.716 m/sec) = 4.966 joules: but the linear Newtonian momentum remains the same.

This is an unlimited source of free energy. Because you can transform a 400 kilogram rim moving 1 m/sec into a 1 kilogram rim moving 400 m/sec; for an increase from 200 J to 80,000 J.

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #83 on: April 10, 2018, 12:39:29 AM »
I made a 10 (1296g) to 1 (132 g) cylinder and spheres and it follows the same pattern as the 4.5 to 1 and others cylinder and spheres. It takes 25 frames to stop the cylinder's spin and it takes 25 frames to fully restart the spin of the cylinder. There are three frames needed to cross the black square at the release; and it takes three frames to cross from one side of the black square to the other side after 50 frames (after a full stop and a full restart).

If energy were conserved when the cylinder was stopped it would only have one third of the linear Newtonian momentum needed to return the cylinder to full rotation. That means it would take 75 frames to return the less than one third of the motion. It takes 25.

If energy were conserved when the cylinder was stopped it would only have one third of the linear Newtonian momentum needed to return the cylinder to full rotation. That means it would take 9.5 frames to cross the black square from side to side after the restart. It takes 3.

The Linear Newtonian momentum formula (mv) would be satisfied with a velocity increase of 10. This is an energy increase to 1000%.

The kinetic energy formula (½ * m * v * v) would be satisfied with a velocity increase of the square root of ten: 3.16.  Ballistic pendulum experiments prove that only Linear Newtonian Momentum is conserved as the small mass spheres collide with the cylinder; kinetic energy is never conserved.

The cylinder and spheres event keeps its Linear Newtonian Momentum.

This means that a 400 kilogram rim moving 1 m/sec would throw off weighted strings of 40 kilograms moving 10 m/sec. This 40 kilograms could throw off weighted strings of 4 kilograms moving 100 m/sec. This 4 kilograms could throw off 1 kilogram moving 400 m/sec.   Now we have ½ * 1 kg * 400 m/sec * 400 m/sec = 80,000 joule and you started with 200 joules.

#### telecom

• Sr. Member
• Posts: 412
##### Re: re: energy producing experiments
« Reply #84 on: April 13, 2018, 04:23:06 AM »
Is the only way to capture this excess energy by sending the projectile upward?
And then using it as a potential energy?

#### Free Energy | searching for free energy and discussing free energy

##### Re: re: energy producing experiments
« Reply #84 on: April 13, 2018, 04:23:06 AM »

#### Low-Q

• Hero Member
• Posts: 2583
##### Re: re: energy producing experiments
« Reply #85 on: April 13, 2018, 02:01:47 PM »
Is the only way to capture this excess energy by sending the projectile upward?
And then using it as a potential energy?
Just have in mind that the potential energy is not more than the energy you put in to "create" this potential energy.
If you throw a ball upwards, its mass acceleration while the ball is still in the hand, require a given amount of energy to achieve a given velocity.
This velocity is the reason why the ball reach a given hight. The mass is the same all the way, and the velocity of the ball when it comes back and hits your hand, is the same, or less due to air resistance.
So no gain in potential energy.

Vidar

#### telecom

• Sr. Member
• Posts: 412
##### Re: re: energy producing experiments
« Reply #86 on: April 13, 2018, 06:41:59 PM »
Yes, but there is a gain in kinetic energy which is transferred into potential.
How are you going to harvest kinetic energy?

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #87 on: April 14, 2018, 02:21:31 AM »
Telecom's question: How are you going to harvest kinetic energy?

I am going to let the technology used in hydroelectric plants harvest the kinetic energy.

Lets throw a 147 gram ball up into the air 20 meters. That will require a velocity of (d = ½ v²/a) 19.81 m/sec. This is (1/2mv²) 28.83 joules.

We will wrap a 20 meter string around a (147 gram *39) 5733 gram rim mass wheel. We will attach the ball to the string and drop the ball 20 meters.

The ball and rim will accelerate at (9.81m/sec/sec / 40)  .24525 m/sec/sec.

The ball and rim will have a final velocity (after the ball has dropped 20 meters) of (d = ½ v²/a) 3.13 m/sec.

This is 5880 grams moving 3.13 m/sec. This is 18.4 units of momentum.

We will now transfer this 18.4 units of momentum to the 147 gram ball by using the cylinder and spheres.

The ball will now be moving 125 m/sec.

This is (1/2mv²) 1151 joules of energy; and you started with 28.8 joules.

And it is better to stack the balls and drop the stack of 20 balls.

The mass of 147 grams is about that of a baseball; so this can be, and is, applied to a real experiment.

At Sault St. Marie the Saint Marys River drops 2000 metric ton of water 7 meters every second. This is 30 MW

So we will increase the drop to double the drop (50 meters) at Niagara: 100 meters.

You could drop 2000 metric tons of mass with a head of 200,000 metric tons every second.

Less than 3% of the energy produced would be needed to reload the system and bring one 2000 metric ton unit back to the 100 meter top. This would generate at least 500 MW.

#### Delburt Phend

• Newbie
• Posts: 46
##### Re: re: energy producing experiments
« Reply #88 on: April 16, 2018, 11:51:25 PM »
There a three formulas that describe motion. Modern physics claims that all three formulas are conserved quantities. It is claimed that all three are simultaneously conserved. A mathematical evaluation of the three formulas would quickly reveal that they cannot all be conserved: the evaluation would show that only one of the three formulas could be conserved.

The formulas are mv: 1/2mv²: and mvr. These are linear Newtonian momentum: Kinetic Energy: and Angular Momentum.

So in a closed system where we have motion interactions all three formulas would allegedly remain the same.

First let’s take the ballistic pendulum experiments. A 1 kilogram mass moving 50 m/sec collides with a 19 kilogram mass at rest. The resultant motion is a 20 kilogram mass moving 2.5 m/sec. The mv is conserved. The 1/2mv² loses 95% of the physical motion to unrecoverable heat. Now heat is considered motion but a return to the original condition of 1 kilogram moving 50 m/sec would expose this heat content to be a myth. There is no mechanism for recovery of the heat. The cylinder and spheres proves that the yo-yo despin can be returned to its original state; of the slowly spinning satellite.  This totally eliminates 1/2mv² as a conserved quantity. A number is rarely equal to its square.

The mvr can be exposed by interrupting the string of a rotating mass on the end of a string. Rotate a soft ball on the end of a string, have someone interrupt the string somewhere down its length. As the ball begins rotating in the smaller circle angular momentum is decreased; linear Newtonian momentum remains the same.  Mutilating a conserved number by different r's will not give you the same conserved quantity.

Of the three laws only one is conserved; mv.