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Why is this Experiment violating the conservation of Energy?


Maximillian

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Please can someone use available equations to prove the paper wrong especially the energy needed to give the ball for it to rise to the water surface?


 

 

Because it won't rise to 10m.

What will stop it from rising to 10m with a density less than that of the air assuming the atmosphere has a height of 10m or what will stop the ball rising to the top of the water when its density is less than that of the water?

Edited by Maximillian
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Simple. You can do 2 things to lower the ball's density.

 

1) Expel mass

 

2) Increase the radius

 

Both work against the hydrostatic pressure of the water column above it. So yes, the ball will rise again, but the potential energy gained is equal to the energy expanded to reduce the density. Your idea boils down to blowing up a balloon at the bottom of a swimming pool after waiting for it to drop to the bottom first. Try to blow one up under water if you ever have the opportunity.

Edited by Fuzzwood
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Please can someone use available equations to prove the paper wrong especially the energy needed to give the ball for it to rise to the water surface?

 

 

You have been shown, repeatedly, where the error is and how to do the calculation correctly. Perhaps you need to take one of those examples and say exactly which bit you don't understand.

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Why won't you either do a proper energy audit or a work energy analysis and post it here/ So far you have made assumptions - that everyone here knows and has pointed out - that have rendered your efforts incomplete and your answer wrong.

 

Also bear in mind you are challenging one of the bedrocks of modern science - do you really think free energy would be that easy; in the words of Oliver Cromwell "I beseech you, in the bowels of Christ, think it possible that you may be mistaken"

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How much Energy do We need to increase the volume of a ball by 0.0002m3 at constant mass at 10m below the water surface?

 

 

That's not the issue. The issue is that you're ignoring half the problem: the ball does not start off at the bottom of the column of water. It starts at the top. That's where the comparison takes place: is there any energy that can be extracted through a complete cycle. And the answer is no. The ball has the same energy at the start as at the finish.

 

By your analysis a battery is a perpetual motion device because you would be ignoring the energy cost of charging it in the first place.

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That's not the issue. The issue is that you're ignoring half the problem: the ball does not start off at the bottom of the column of water. It starts at the top. That's where the comparison takes place: is there any energy that can be extracted through a complete cycle. And the answer is no. The ball has the same energy at the start as at the finish.

 

By your analysis a battery is a perpetual motion device because you would be ignoring the energy cost of charging it in the first place.

Please can you help me with the answer am trying to compare it to mine.

Zero displacement obviously means no energy can be extracted.

Imagine am a fish all I want is to use little energy to enable me reach the top of the water.

I use 20J which should put me 2m above the water bottom but it takes me 10m at the surface.

Edited by Maximillian
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Zero displacement obviously means no energy.

Imagine am a fish all I want is to use little energy to enable me reach the top of the water.

I use 20J which should put me 2m above the water bottom but it takes me 10m at the surface.

 

How did the fish get from the top to the bottom in the first place?

Zero displacement obviously means no energy.

 

So you agree that there is no net energy change when your ball starts at the top is taken to the bottom and then returns to the top. Good.

 

But you have expended 20J to achieve zero energy change. A net loss of energy.

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How did the fish get from the top to the bottom in the first place?

 

Its like asking how did humans move from living in the sky to living on land, they have always been there.

 

How did the fish get from the top to the bottom in the first place?

 

So you agree that there is no net energy change when your ball starts at the top is taken to the bottom and then returns to the top. Good.

 

But you have expended 20J to achieve zero energy change. A net loss of energy.

Yes, in a closed loop I have expended 20J to gain Zero energy.

Imagine the fish gets to the top using 20J instead of 100J and never returns, what happens to the conservation of energy?

Edited by Maximillian
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Yes, in a closed loop I have expended 20J to gain Zero energy.

 

And that is what you described in your opening post.

 

Now you are just changing the argument for no obvious reason. If you don't take into account all the energy (what did the fish eat, where did the water come from) and start with the equivalent of a charged battery (as in swansont's earlier example) then you won't have zero net energy expenditure.

 

This has gone beyond pointless. Your initial result was obviously wrong. You seem to now understand why it was wrong. What is the point of carrying on?

 

p.s. Where did you file that patent?

Edited by Strange
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Imagine the fish gets to the top using 20J instead of 100J and never returns, what happens to the conservation of energy?

 

 

Most of the energy went into the height of the water, which drops down as the fish goes up, as has been explained to you several times, and some is retained as KE of the fish (or object)

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I used 20J of pressure energy to make a magnet in water gain 100J of gravitational potential energy on reaching the water surface.

It starts falling down with a kinetic energy of 100J through a wire coil, by the time it has landed down it has produced for me electrical energy of 100J that I store in batteries.

I threw the ball up with 20J of energy but I now have 100J of electricity stored in my batteries.
Have I not gone through a complete cycle and still have more energy?

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I used 20J of pressure energy to make a magnet in water gain 100J of gravitational potential energy on reaching the water surface.

It starts falling down with a kinetic energy of 100J through a wire coil, by the time it has landed down it has produced for me electrical energy of 100J that I store in batteries.

I threw the ball up with 20J of energy but I now have 100J of electricity stored in my batteries.

Have I not gone through a complete cycle and still have more energy?

You do not gain 100 J of gravitational potential energy.

How many times do I have to tell you this?

Is it that you don't understand, or are you trolling?

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I used 20J of pressure energy to make a magnet in water gain 100J of gravitational potential energy on reaching the water surface.

It starts falling down with a kinetic energy of 100J through a wire coil, by the time it has landed down it has produced for me electrical energy of 100J that I store in batteries.

I threw the ball up with 20J of energy but I now have 100J of electricity stored in my batteries.

Have I not gone through a complete cycle and still have more energy?

 

 

You won't get 100 J of energy in the battery. You will get less than 10 on each trip. That's how much energy is available for extraction in this scenario.

 

Back when I was actively blogging I tried to drop a magnet through a coil to light up an LED. Couldn't do it, even with a spinning magnet, which causes the flux to change faster. A magnet going through water would drop really slowly, and cause even less of an EMF in the coil. Any energy successfully extracted will cause it to drop even slower, owing to Lenz's law.

 

But you are welcome to try the experiment.

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Have I not gone through a complete cycle and still have more energy?

 

 

There is a fundamental principle called conservation of energy. If you think you are getting more energy out than you put in then you have made an error.

 

It is as simple as that. You just need to understand where your error is.

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Finally my equations have been approved to be correct https://answers.yahoo.com/question/index?qid=20160914073523AAhRFp3.

In this case g was 9.8ms-2 therefore the energy was 19.6J input which makes the ball less dense and by all means gets to the water surface gaining a gravitational potential energy of 100J.

You say am starting from a charged battery true but if that charged battery has 20J of electrical energy it will never output more than 20J of whatever kind of energy. You don't need a closed cycle to prove the conservation of energy, so what happens if that cycle is never closed does it mean energy conservation will be violated. What you need is to measure the total energy of an isolated system now then measure it in the future, it should be the same. What is for sure is we input 20J of pressure energy and we have outputted 100J of gravitational potential energy. To conserve energy the water has to provide the extra 80J but its definitely not from us.

Edited by Maximillian
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Finally my equations have been approved to be correct https://answers.yahoo.com/question/index?qid=20160914073523AAhRFp3.

In this case g was 9.8ms-2 therefore the energy was 19.6J input which makes the ball less dense and by all means gets to the water surface gaining a gravitational potential energy of 100J.

You say am starting from a charged battery true but if that charged battery has 20J of electrical energy it will never output more than 20J of whatever kind of energy. You don't need a closed cycle to prove the conservation of energy, so what happens if that cycle is never closed does it mean energy conservation will be violated. What you need is to measure the total energy of an isolated system now then measure it in the future, it should be the same. What is for sure is we input 20J of pressure energy and we have outputted 100J of gravitational potential energy. To conserve energy the water has to provide the extra 80J but its definitely not from us.

I didn't look at the link- this is a slow computer + it would take too long.

The equation PE= mgh is still wrong (A I explained earlier)

You won't get the right answer by using the wrong equation.

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I used 20J of pressure energy to make a magnet in water gain 100J of gravitational potential energy on reaching the water surface.

It starts falling down with a kinetic energy of 100J through a wire coil, by the time it has landed down it has produced for me electrical energy of 100J that I store in batteries.

I threw the ball up with 20J of energy but I now have 100J of electricity stored in my batteries.

Have I not gone through a complete cycle and still have more energy?

How is it going to fall down again if you spent 20J to let it rise?

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The system works and I have scientifically proven that.

Whether you accept it or not it speaks for itself.

I have to admit that your responses help to prove that it actually works.

Where is the working device? Experiment is the only thing that "proves" anything in science.

 

All you've shown here is that you don't understand basic physics.

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