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Author Topic: GRAVITY: Is it a source of energy?  (Read 10245 times)

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Hello All.

Yes, I did look into the " Aspen effect " my results were inconclusive but the second run up showed a marked decrease in time. I heated everything up above ambient temperature so the rig was actually cooling down.

Cheers Graham.

As a post script, I was wondering if device orientation might play a part? Mine was East West at the time of testing.


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I was told that some very big and well funded labs could find no gain in the Aspen effect ?

I am aware of no specifics regarding the testing ,except that it was claimed to be "exhaustive"


for what its worth ?
   

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Brad, you said:
Well, Brad, you know I am interested in the subject and the purpose of this thread was to arouse interest and arguments.

In post #7 I think I created an experiment that shows you can get more out of what you put into lifting an object, the trick being that you send the weight to the neutral zone then let a different gravitational body of more mass attract the object.

There are probably some good refutations of this argument that I have not seen posted.

I also posted some other arguments. Maybe these arguments are sophomoric, but I am just learning.

Regards

I think some are that stuck in there way's,that they have let schooling get in the way of there education.

Lets have a close look at one of those laws-->energy cannot be created  ???
For this,we're going to the moon  O0. I mean,we have been there before-right?,so no problem there  C.C

Anyway,we're on the moon,and we launch a mass which has a weight of 1000KGs on the moon,toward earth.
That same mass will impact the earth with far greater energy than it took to launch it from the moon.
So the question remain's,if gravity is not a source of energy,where did that extra energy come from?

Now,some here will say that-->well,you just lost some of the moons mass,and that loss is where the energy came from.
This  of course is wrong,as we lost no mass at all--it was just transfer'd
What the moon lost,the earth has now gained.
The moons mass just decreased,and so it's gravitational value must also decrease.
But,at the same time,the earths mass just increased,and so it's gravitational value also increased-->by the very same amount that the moons decreased. The earths to moons gravitational attraction value has not changed

This means that,if we carried out the same experiment again,it would take the same amount of  energy to launch that 1000KG mass from the moon,but the mass would also be slightly larger for that 1000KG weight on the moon,and it would impact the earth with even greater energy,due to the increase in mass,and the increase in earth gravitational value.

But we can stay within earths gravity field,and carry out some mathematical experiments,and still show how gravity can provide energy--or create it,as no gravity will be consumed  :D ,if you can say it like that ?.
What i mean is,energy can only be transformed from one state to another.
How exactly would gravity be transformed or consumed, if the mass remains the same  ???

Anyway,lets get on with the mathematical experiment.


Brad


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Ok,so lets get to our mathematical experiment.

From now on,all masses raised will be in a geostationary position,or orbit,as we will be going pretty high,so as to make the math easier,and show larger results-->much like trying to measure microwatts of power to that of watts of power.

The conservation of energy states that the potential energy (PE)of a mass (M),is equal to the required energy to lift/raise that mass (M) to height (H),and when that mass is lowered to it's starting position,the same amount of energy to that of the PE will be dissipated in one form or another.
!!There will be no energy gain,and no energy loss,but only a transformation from one state to another!!.

Am i right so far?

Example.
If we raise a mass which has a weight of 1000KGs,to a height of 1000 meters,it will take 9800000 Joules of energy to do so,and our PE will be 9800000 Joules.
If we lower that same mass back to it's starting position,we will produce 9800000 Joules of energy in one form or another-->energy is conserved  O0

The same rules apply in any gravitational field of any strength.
Example--we will calculate the same mass in a gravity field that is 1/10th that of earths.
It will take 980000 Joules of energy to raise that same mass to a height of 1000 meters. The PE will be 980000 Joules,and 980000 Joules of energy will be produced in one form or another when that mass is returned to it's starting position.

Even passing through lowering  gravitational values as we gain height above the earths surface,the energy returned as the mass returns to it's starting point,will be the same amount of energy needed to raise it to that height in the first place-->energy is conserved.

Smudge
Please check and confirm the above-->incase i missed something here.

Thanks


Brad



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Ok,so lets get to our mathematical experiment.

From now on,all masses raised will be in a geostationary position,or orbit,as we will be going pretty high,so as to make the math easier,and show larger results-->much like trying to measure microwatts of power to that of watts of power.

The conservation of energy states that the potential energy (PE)of a mass (M),is equal to the required energy to lift/raise that mass (M) to height (H),and when that mass is lowered to it's starting position,the same amount of energy to that of the PE will be dissipated in one form or another.
!!There will be no energy gain,and no energy loss,but only a transformation from one state to another!!.

Am i right so far?

Example.
If we raise a mass which has a weight of 1000KGs,to a height of 1000 meters,it will take 9800000 Joules of energy to do so,and our PE will be 9800000 Joules.
If we lower that same mass back to it's starting position,we will produce 9800000 Joules of energy in one form or another-->energy is conserved  O0

The same rules apply in any gravitational field of any strength.
Example--we will calculate the same mass in a gravity field that is 1/10th that of earths.
It will take 980000 Joules of energy to raise that same mass to a height of 1000 meters. The PE will be 980000 Joules,and 980000 Joules of energy will be produced in one form or another when that mass is returned to it's starting position.

Even passing through lowering  gravitational values as we gain height above the earths surface,the energy returned as the mass returns to it's starting point,will be the same amount of energy needed to raise it to that height in the first place-->energy is conserved.

Smudge
Please check and confirm the above-->incase i missed something here.

Thanks


Brad

Right so far, Brad.
Physics note:
PE = MgH
measured in standard units as Joules = Newton*m
where 1 Newton = N = kg*m/s^2.

Potential Energy = PE  = latent or stored energy, based on an objects POSITION in a field (such as a gravitational or electric field).
   

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You are OK so far Brad.  However you keep expressing a given mass as a weight (like 1000Kg) but weight is a force.  A better statement is to refer to the mass itself as a 1000Kg mass.  Weight is a function of gravity.  It will weigh 1000Kg on the earth but in your 1/10th gravity field it will weigh 100Kg, there will be only a 100Kg force on it.  Of course you know this, so just ignore me.

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You are OK so far Brad.  However you keep expressing a given mass as a weight (like 1000Kg) but weight is a force.  A better statement is to refer to the mass itself as a 1000Kg mass.  Weight is a function of gravity.  It will weigh 1000Kg on the earth but in your 1/10th gravity field it will weigh 100Kg, there will be only a 100Kg force on it.  Of course you know this, so just ignore me.

Smudge

Yes,as the gravitational pull decreases with height,the weight of the mass will decrease,but the mass will remain the same.

This is where the clever math is needed.
If we raise a given mass say only 100 meter's ,where that mass remains in the 9.8m/sec* area,then we use that masses weight to calculate our PE when at rest. If we raise that same mass to a height where the acceleration of G is only .98m/sec,then we have to slowly decrease the value of G as the mass is raised.

So a question would be,do we use an average value of 4.41m/sec to make our calculations ?

Not that it matters much,as the end result is-energy is conserved,and the energy dissipated on the descent back to earth,will be the same as the energy required to raise the mass to that height in the first place-->correct?.

This being the case,there will be no need for calculating needed energy to raise the mass to a given height,no need to calculate the PE,as it will be the same as the energy required to raise the mass to that height,and no need to calculate the dissipated energy as our mass falls back to,and makes contact with earth,as that will be the same value as our PE was.


So,your math skills in this experiment is needed to calculate the energy required to make physical changes within the mass once it reaches it's maximum height,and the stored energy within the mass,once it returns back to the earths surface,and is in a rest state  ;)

So,if what is in blue is correct,then we will move on.


Brad


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Right so far, Brad.
Physics note:
PE = MgH
measured in standard units as Joules = Newton*m
where 1 Newton = N = kg*m/s^2.

Potential Energy = PE  = latent or stored energy, based on an objects POSITION in a field (such as a gravitational or electric field).

Thanks Steve.

Glad you made this point-->Potential Energy = PE  = latent or stored energy, based on an objects POSITION in a field (such as a gravitational or electric field).

Our mass will remain within the earths gravitational field  O0


Brad


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So a question would be,do we use an average value of 4.41m/sec to make our calculations ?
No!  Averaging works OK when the change of force with distance is a linear relationship, but gravity is not linear.  Like the Coulomb force gravity follows an inverse square law.  It is then necessary to resort to integration techniques to get the total energy requirements.

Quote
Not that it matters much,as the end result is-energy is conserved,and the energy dissipated on the descent back to earth,will be the same as the energy required to raise the mass to that height in the first place-->correct?.

This being the case,there will be no need for calculating needed energy to raise the mass to a given height,no need to calculate the PE,as it will be the same as the energy required to raise the mass to that height,and no need to calculate the dissipated energy as our mass falls back to,and makes contact with earth,as that will be the same value as our PE was.
That is correct

Quote
So,your math skills in this experiment is needed to calculate the energy required to make physical changes within the mass once it reaches it's maximum height,and the stored energy within the mass,once it returns back to the earths surface,and is in a rest state  ;)
What physical changes do you have in mind?

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Not that it matters much,as the end result is-energy is conserved,and the energy dissipated on the descent back to earth,will be the same as the energy required to raise the mass to that height in the first place-->correct?.

This being the case,there will be no need for calculating needed energy to raise the mass to a given height,no need to calculate the PE,as it will be the same as the energy required to raise the mass to that height,and no need to calculate the dissipated energy as our mass falls back to,and makes contact with earth,as that will be the same value as our PE was.


Quote
That is correct

Quote
What physical changes do you have in mind?

Well.lets get started  O0


Brad


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The mathematical experiment go's like this-->

We have a sealed cylinder full of water,and with a non compressible hollow ball also inside the cylinder.
The ball is made from aluminum,and has a weight of 10KG,and is filled with air.
The total weight of the mass(our cylinder) is 7500KGs.

1st pic shows the DUT sitting on the earths surface.
We now raise our mass,in a geostationary fashion,until we reach a height where the gravitational value is 1/10th of that on the earths surface-->.98m/sec*

Once this height is reached,the ball is pulled down to the bottom of the tank,and hooked onto the hook-->see pic 2

Once the ball is hooked onto the hook,our mass is returned to earth,where the kinetic energy is dissipated on the way back to earth by way of a parachute to slow it's descent,and heat from atmospheric friction.The remaining kinetic energy is dissipated once the mass impacts earth-->which is only a small impact,so as not to destroy our DUT.
So in other words--a controlled descent  O0.

As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved.

Now,all we have to do is calculate the energy used to pull the ball to the bottom of the tank,when our DUT was in that 1/10th strength gravity field,and calculate the !now! stored energy of the submerged ball back here on earth.

My calculations.

Pulling the ball down to the bottom of the tank ,when the acceleration of G is .98m/sec*

Ball displacement = 449L
Weight of displaced water=44.9KGs
Height=9 meters
Energy required= 396 joules.

Stored energy when DUT is back on earths surface

Ball displacement=449L
Weight of displaced water=449KGs
Height=9 meters
Stored energy=39601 joules.

Energy that came from somewhere=39205 joules.

Is this correct?


Brad

P.S--oops
I forgot to factor in the weight of the ball in both calculations.

Smudge
Could you calculate the right values for both

Cheers
« Last Edit: 2018-06-19, 13:38:55 by TinMan »


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Smudge
Could you calculate the right values for both

You almost got it right.  1 Kg expressed as a force is a force unit that doesn't change with gravity.  It is the force on a 1Kg mass here on earth and has a value of 9.81 Newtons.  In outer space 9.81 Newtons is still a 1Kg force.  So when you expressed your 449L of water as a 449Kg mass you were correct in saying that its weight here on earth is 449Kg force and in your 1/10th gravity is 44.9Kg force.  Those two values have to be changed to Newtons to get the energy in Joules and in both cases you must multiply by 9.81.  So my calculations are:-

Pulling the ball down to the bottom of the tank ,when the acceleration of G is .981m/sec*

Ball displacement = 449L
Weight of displaced water=44.9KGs force =440.47N
Height=9 meters
Energy required= 3964.2 joules.

Stored energy when DUT is back on earths surface

Ball displacement=449L
Weight of displaced water=449KGs force = 4404.7N
Height=9 meters
Stored energy=39642 joules.

Energy that came from somewhere=35678 joules.

Still thinking about where the energy comes from.

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I think it comes down to centre of mass.  Initially the mass centre is below the geometric centre then when it returns to earth the mass centre is above the geometric centre.   If you go into outer space it requires zero energy to move the hollow ball, and the same would apply to moving a solid mass within a hollow cylinder.  We can get similar things with metal magnets where some of the permanent magnetism comes from the spins of conduction electrons.  Those itinerant electrons can be dragged towards one end of the magnet thus altering the magnetization density along the magnet.  This evidences itself when you measure attractive force v. repulsive force, for a given separation distance they are not the same.

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It's that change of centre of mass that means you don't bring the system back to its starting point.  Thus Brad's statement "As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved." is untrue.  Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start.  And that difference accounts for the apparent gain.

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It's that change of centre of mass that means you don't bring the system back to its starting point.  Thus Brad's statement "As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved." is untrue.  Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start.  And that difference accounts for the apparent gain.

Smudge

Smudge

That difference is only the diameter of the ball-->950mm
The energy lost in this case would only be the energy required to raise 449KGs up 950mm,which is 4180.19 joules.

The second thing to remember is this.
When the ball was pulled to the bottom of the tank in the 1/10th gravity region,the center of mass was raised by the very same amount it was lost here on earth-->950mm

Even if we discount the later,and we subtract that 4180.19 joules from our apparent gain of 35678 joules,we still have 31497.81 joules of energy unaccounted for.

Looking at my second point,then i stand by my statement-->As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved
My second point shows that the center of mass is raised and lowered the very same distance

I do hope that this is given the time and calculations it deserves,instead of just falling victim to that !must conform to the laws of physics! syndrome.

This is OverUnityResearch.
We hare here to find and investigate the exceptions to the rules/laws.
Lets not just throw our hands in the air,and say--that must be where the extra energy is coming from.
Lets carry out the math,and see if there is truly an energy gain.

Your math showed a gain,but your accountability for that gain just dose not add up.
449KGs at a height of 950mm dose not net an energy requirement of 35678 joules.


Brad


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Smudge

That difference is only the diameter of the ball-->950mm
Well not exactly but I agree the difference is quite small.
Quote
The energy lost in this case would only be the energy required to raise 449KGs up 950mm,which is 4180.19 joules.
There I must disagree.  We are talking about the centre of mass of the whole thing, so it is the energy required to raise that whole mass that small distance.
Quote
I do hope that this is given the time and calculations it deserves,instead of just falling victim to that !must conform to the laws of physics! syndrome.
I have spent the last 21 years searching for things that do not conform to the accepted laws of physics.  When I see an obvious flaw in someone's reasoning I feel obliged to point this out.  I used the centre of mass argument to demonstrate that the thing returns to earth in a different state to that when it leaves the earth, hence the energy lost in going to that outer space region is not equivalent to the energy gained when it is returned to earth.  You do not have to work out the actual centre of mass to do the calculation.  You just concentrate on the two spherical volumes at the top and bottom of the cylinder (all the rest of the system undergo the same raise and fall hence can be ignored.  On the way up the bottom spherical volume is filled with water while the top one isn't, the energy needed for that amount of water to reach outer space is a given value.  On the way down the the top spherical volume is filled with water but the bottom one isn't. Thus that spherical quantity of water doesn't return to its starting point, it is 9m higher.  The difference in energy between that needed to reach outer space and that gained on return is thus the weight of that volume of water multiplied by 9m.

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I have spent the last 21 years searching for things that do not conform to the accepted laws of physics.  When I see an obvious flaw in someone's reasoning I feel obliged to point this out.  I used the centre of mass argument to demonstrate that the thing returns to earth in a different state to that when it leaves the earth, hence the energy lost in going to that outer space region is not equivalent to the energy gained when it is returned to earth.  You do not have to work out the actual centre of mass to do the calculation.  You just concentrate on the two spherical volumes at the top and bottom of the cylinder (all the rest of the system undergo the same raise and fall hence can be ignored.  On the way up the bottom spherical volume is filled with water while the top one isn't, the energy needed for that amount of water to reach outer space is a given value.  On the way down the the top spherical volume is filled with water but the bottom one isn't. Thus that spherical quantity of water doesn't return to its starting point, it is 9m higher.  The difference in energy between that needed to reach outer space and that gained on return is thus the weight of that volume of water multiplied by 9m.

Smudge

Quote
There I must disagree. We are talking about the centre of mass of the whole thing, so it is the energy required to raise that whole mass that small distance.

Well,i think we both disagree here,and i think your own math disagree's with you as well.
For this we go back to your other posts.

Your calculations in reply 61
Energy that came from somewhere=35678 joules.

Reply 63-->Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start.  And that difference accounts for the apparent gain.

And when you say-->so it is the energy required to raise that whole mass that small distance,im guessing you mean the water only,as the tank it self returns back to the starting point(the ground).

The volume of water in the tank is 7323 liters,and so it's weight is 7323KGs.
To raise a weight of 7323KGs 950mm, takes 68246.7 joules of energy.
The energy you calculated (the apparent gain) was only 35678 joules.

So,something went wrong here  ???

Also,would the cycle not be complete when the ball is released,and the water fell to the bottom of the tank,while at the same time the ball rises back to it's starting position ?.


Brad


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Well,i think we both disagree here,and i think your own math disagree's with you as well.
For this we go back to your other posts.

Your calculations in reply 61
Energy that came from somewhere=35678 joules.
Yes that is the buoyancy energy gained minus the energy needed to pull down the float in your 1/10th gravity.  That 1/10th gravity is a bit of a red herring, you might as well go far away to zero gravity then there is no energy needed to move mass or floats.
Quote
Reply 63-->Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start.  And that difference accounts for the apparent gain.

And when you say-->so it is the energy required to raise that whole mass that small distance,im guessing you mean the water only,as the tank it self returns back to the starting point(the ground).
No I  mean the total device.
Quote
The volume of water in the tank is 7323 liters,and so it's weight is 7323KGs.
To raise a weight of 7323KGs 950mm, takes 68246.7 joules of energy.
The energy you calculated (the apparent gain) was only 35678 joules.
You are using that 950mm there and that is not the distance between geometric centre and mass centre.   That is just your assumption.  As I said, it is not necessary to use that mass centre.  The tank and the water outside of those two spherical volumes can be raised to any height you want,  then back down again, and energy is conserved.  So the mass of that water and of the tank and of the sphere shells is immaterial, we can forget them and just concentrate on those two spherical volumes inside those shells.  Imagine two lightweight spheres joined by a 9m long vertical lightweight rod.  The bottom sphere contains water.  Alongside we have another pair with the top sphere containing water.  I think you will agree that if we raise them both to a distant region where gravity is virtually zero the water in the first one travels 9m more than the water in the second one to get to that zero gravity.  So there is a difference in energy associated with that 9m difference.  If that doesn't convince you I will work out the actual change in mass centre using the details you have provided and do the math for water mass, cylinder mass and spherical shell mass.
Quote
Also,would the cycle not be complete when the ball is released,and the water fell to the bottom of the tank,while at the same time the ball rises back to it's starting position ?.
Yes, that completes the cycle.

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I know Steve had mentioned he was driving cross country through the Midwest USA

and would be crunching numbers here too as internet access and time permited
he was hoping to be close to a internet port by the weekend.
   

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...I have spent the last 21 years searching for things that do not conform to the accepted laws of physics... 
Smudge

Go back in history and read about the how vectors were chosen over quaternions.  This was after Maxwell released his treatise in 1873.   Gibbs and Heavyside tossed half of Maxwell's work.  What we have now is 100+ years of development based on the half that they kept.  What would 100+ years of development of that half give us?

EDIT:
This paper has a nice collections of references to works with quaternions.

https://arxiv.org/pdf/1504.06182.pdf

It's interesting that many of them are related to gravity.
   

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Go back in history and read about the how vectors were chosen over quaternions.  This was after Maxwell released his treatise in 1873.   Gibbs and Heavyside tossed half of Maxwell's work.  What we have now is 100+ years of development based on the half that they kept.  What would 100+ years of development of that half give us?

EDIT:
This paper has a nice collections of references to works with quaternions.

https://arxiv.org/pdf/1504.06182.pdf

It's interesting that many of them are related to gravity.

Through the years,i have found many ways to show the flaws in the conservation of energy,and producing energy at no cost. I even managed to stump the late MarkE once,with the introduction of a venturi in a tank to tank transfer of compressed air.

There are many ways to show work being done without the loss of energy.

Maybe we should open a thread that looks at all this stuff Grumpy ?.


Brad


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Yes that is the buoyancy energy gained minus the energy needed to pull down the float in your 1/10th gravity.  That 1/10th gravity is a bit of a red herring, you might as well go far away to zero gravity then there is no energy needed to move mass or floats. No I  mean the total device.You are using that 950mm there and that is not the distance between geometric centre and mass centre.   That is just your assumption.  As I said, it is not necessary to use that mass centre.  The tank and the water outside of those two spherical volumes can be raised to any height you want,  then back down again, and energy is conserved.  So the mass of that water and of the tank and of the sphere shells is immaterial, we can forget them and just concentrate on those two spherical volumes inside those shells.  Imagine two lightweight spheres joined by a 9m long vertical lightweight rod.  The bottom sphere contains water.  Alongside we have another pair with the top sphere containing water.  I think you will agree that if we raise them both to a distant region where gravity is virtually zero the water in the first one travels 9m more than the water in the second one to get to that zero gravity.  So there is a difference in energy associated with that 9m difference.  If that doesn't convince you I will work out the actual change in mass centre using the details you have provided and do the math for water mass, cylinder mass and spherical shell mass.Yes, that completes the cycle.

Smudge

Smudge

Regardless of whether or not that ball has a resistance placed upon it as it is released and rises,the water will fall back to it's starting position,dissipating the remaining energy to balance out what was needed to raise it in the first place.

It dose not matter if a mass falls from a set height fast over a short period of time,or slow over a long period of time--the end result is the same.


Brad


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Maybe we should open a thread that looks at all this stuff Grumpy ?.
Brad

I did years ago and it didn't go anywhere.  Everyone has their own ideas and little interest in others.
   
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I did years ago and it didn't go anywhere.  Everyone has their own ideas and little interest in others.
Dear G
Could you provide the link to that thread or name of it?
regards


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OK, I have looked more carefully into this system and I believe I have the answer.  If you look at the tube plus ball before it is launched and compare it with how it looks after it returns from outer space (before gaining the buoyancy energy) the two pictures are different.  The hollow ball is in a different position.  Now we assume that these two conditions weigh the same, but that assumption assumes that gravity is uniform in our laboratories.  But gravity isn't uniform, it reduces with distance from the earth following an inverse square law.  The difference in weight between those two pictures is miniscule, but it is there.  Before launch the thing weighs very slightly more than it does after return.  So it takes slightly more energy to get it to outer space than is gained on its return.  But that slightly more applies to the enormous energy consumed over that vast distance, and in fact it is quite significant.  It exactly accounts for the gained buoyancy energy.  Below is my write up proving this by math.  If I have got it wrong please tell me.

Smudge
   
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