Thermodynamics

[replaced]

This is a false argument that "skeptics" have tried to make because they
don't even have the frames of reference through which these systems
operate.

You are arguing against perpetual motion (as in your other thread) that
has absolutely nothing to do with non-equilibrium systems that have a very
extended time until dissipation because of free potential coming from the
environment. In this case, that is gravitational potential.

For there to be a net energy gain does not mean it has to bounce higher
each time - that is ridiculous to think there is a perpetual motion argument
in an environment full of resistances such as a bouncing ball. Perpetual
motion claims have no place in a grown up discussion about non-equilibrium
thermodynamics / open dissipative systems and to argue against perpetual
motion is such an example is just as ridiculous as arguing for one.

For there to be a net energy gain apparent, you add up all the joules of
work done from the time you initially lift the ball, let go and by the time it
comes to a rest, you add up all work demonstrated and divide it by how
much you input by the initial lift. No amount of your smoke and mirrors
are going to change this basic fact - and it has nothing to do with
having to bounce higher each time and has nothing to do with perpetual
motion.

A rubber ball bounces because of the potential elastic energy in the ball. Take a metal ball and drop it and it won't bounce as much. All that is happening when the ball goes back up is convert the elastic energy from when the ball hit the ground back to kinetic, and during this process friction is constantly lowering the kinetic and potential energy of the ball.

Non-equilibrium system STILL contribute to entropy of the universe as a
matter of indisputable fact but the dissipation can be greatly delayed as
constant environmental free input comes into the system to contribute
to more work above and beyond what was initially put into it to get it
going.

Environmental free input? Why not be specific on exactly what you are referring to.

Here is a bare basic definition of an open dissipative system that is
so simple a child should understand it - the branch of thermodynamics
that explain these open dissipative systems is Non-Equilibrium
Thermodynamics or N.E.T. No magic and nothing to do with perpetual
motion.

For anyone that can't comprehend this, it means the environmental input
helps prevent it from coming into balance!

Environmental input is so small that you couldn't possibly do anything useful with it. Eventually, in a finite amount of time, the entire universe will be in 100% complete equilibrium.

While the universe won't be in complete balance, you can't perform work with something that is generally balanced, such as a drained battery.

The universe will eventually be in complete balance. It'll happen after the protons decay and black holes evaporate via hawking radiation.

For these simple systems, they will come into balance - after demonstrating
MORE work than was input. Entropy is greatly extended.

You lift the ball and let go - work is performed on each cycle. The loss
of dissipation when it hits the ground is not the only work done, that is
only what is wasted. But there is a small amount of waste in the lift
against air friction, etc... but almost insignificant compared to the impact
dissipation.

That is not overunity because what the ball is doing is transferring the kinetic energy to potential energy caused by the elasticity of the rubber when it hits the ground, then back to kinetic as the "rubber band" in the ball unwinds. The amount of heat being produced by the rubber ball doing this will never power something to lift the ball to the same height, because the amount of kinetic energy as heat (individual atoms shaking) is EQUAL to the ball when it was first dropped. All you've done was transfer the entire kinetic energy of the ball to kinetic energy spread out in countless atoms. Heat is the kinetic energy of individual atoms/molecules.

On each lift - the environmental input is fresh gravitational potential that
helps to do what???

"avoid thermodynamic equilibrium"

Because it is an open dissipative system that can be explained simply by
non-equilibrium thermodynamics that completely ALLOWS free input
to perform more work extending the period before it comes into equilibrium.

The bouncing ball performs 8 times the work compared to what was input,
violates nothing, is overunity and still has NOTHING to do with perpetual
motion.

Your belief is rooted in conservation of energy and a belief that everything
conforms to classical thermodynamics. Both are wrong and neither apply
to a bouncing ball. Nice try but you aren't even aware of the distinctions
between two separate branches of thermodynamics yet you think you
can explain open dissipative systems.

Ok, let me some it up. As the ball bounces on the earth, momentum from the ball is being transferred to individual atoms which we call heat. To say that more momentum is being transferred than the momentum of the ball when it first hit the ground is ridiculous.

[Aaron]

cold effects

Orion,

I believe you. The basics schematics are there but I can't guarantee
anyone will get the same as me. This project was repeated with more
"standard" setups than any project I've seen to make sure everyone
was trying to build it exact. Results still varied.

As soon as I saw a 2C drop, for a long time, I stopped trying to make
heat and focused on these cold effects.

But the Ed Gray motor also had claims of getting cold and icing up
in parts - Mark McKay can give insights into those claims if they
were even claimed at all.

Floyd Sweet's vacuum triode amplifier same claims of icing up when
wires were shorted, etc...

What I experienced was weaker than all of the other claims I have seen.

[LetsReplicate]

How much work does it take to lift a 5 gram ball to 1 meter on the surface of the Earth?

The same amount of work you get out of it plus the losses to entropy. My point was that when the ball is actually dropped, it has 0 power. It also has 0 power at all of the peak points, as well as when it bounces. So your calculations are false because they are missing a dimension. EVERYTHING is physics and engineering is based on vectors and failure to understand that leads people down the garden path of fantasy.

[Aaron]

bouncing ball

A rubber ball bounces because of the potential elastic energy in the ball. Take a metal ball and drop it and it won't bounce as much. All that is happening when the ball goes back up is convert the elastic energy from when the ball hit the ground back to kinetic, and during this process friction is constantly lowering the kinetic and potential energy of the ball.

Environmental free input? Why not be specific on exactly what you are referring to.



Environmental input is so small that you couldn't possibly do anything useful with it. Eventually, in a finite amount of time, the entire universe will be in 100% complete equilibrium.



The universe will eventually be in complete balance. It'll happen after the protons decay and black holes evaporate via hawking radiation.



That is not overunity because what the ball is doing is transferring the kinetic energy to potential energy caused by the elasticity of the rubber when it hits the ground, then back to kinetic as the "rubber band" in the ball unwinds. The amount of heat being produced by the rubber ball doing this will never power something to lift the ball to the same height, because the amount of kinetic energy as heat (individual atoms shaking) is EQUAL to the ball when it was first dropped. All you've done was transfer the entire kinetic energy of the ball to kinetic energy spread out in countless atoms. Heat is the kinetic energy of individual atoms/molecules.



Ok, let me some it up. As the ball bounces on the earth, momentum from the ball is being transferred to individual atoms which we call heat. To say that more momentum is being transferred than the momentum of the ball when it first hit the ground is ridiculous.

All you're doing is trying to justify what you already believe instead of
paying attention to what is actually being said or what is actually happening.

Potential elasticity of the ball? The comparison to the metal ball and rubber ball show me exactly where your method of analysis is and why
you believe what you believe.

The rubber ball is a device with a higher recovery efficiency than a metal
ball synonymous with higher recovery methods in an electromagnetic
energizer compared to a regular electric motor.

Bringing up a smoke and mirrors comparison between a metal ball and a
rubber ball as evidence as anything is pixie dust. It proves you do not
comprehend my statement that the rubber ball is 87% efficient, which
means that on each bounce 13% is dissipated on impact. Furthermore,
you ignore the fact that I say it doesn't bounce higher each time and
arguing that it doesn't means you are making one false argument after
the other. I explicitly stated the ball does NOT bounce higher on each
bounce. You are either intentionally trying to misdirect people's attention
away from the facts I stated or you simply are incapable of really
comprehending the facts.

The efficiency of a system has to be high enough to to work properly
in order to get over 1.0 cop, which is more work than you put into it.
Efficiency is DIFFERENT than coefficient of performance and you have
demonstrated clearly that you don't know the difference.

The metal ball may be 5% efficient and a rubber ball that I tested is
83% efficient. That means one bounces 5% to the height of the distance
it was dropped from and the other bounces 83% to the height that it
was dropped from.

If a refrigerator compressor had too low of an efficiency like 5-10%
efficiency (analogous with the metal ball), it wouldn't be able to have
more joules of work done in heat movement compared to what is drawn
from the wall - and the refrigerator would be useless. It wouldn't cool
anything.

The reason refrigerators do work and are over 1.0 cop (showing more
work done than is input) is because the system that the compressor
can run at a high enough efficiency (rubber ball) to show over 1.0 cop).

If we ran each distinct point brought up between you and I though
a legal type process, virtually everything you are saying would be
dismissed as making no sense because you make false arguments based
on illogical interpretations of what I said in addition to blatant examples
of arguing something that I already stated. You would be thrown out
of the courtroom and your license yanked and would be held in contempt
of court.

You say: "friction is constantly lowering the kinetic and potential energy of the ball. "

This proves 100% beyond a shadow of a doubt you are not able to
comprehend anything I said. I agree it lowers kinetic and potential energy
of the ball - on each cycle, which is an intrinsic fact embedded in the fact
that I stated the ball I tested is 83% efficient meaning that on each
successive bounce it is reduced by 17%!!! 83% efficient means that
on each bounce the kinetic and potential energy is reduced (constantly
lowering) and the ball bounces lower each time!!!

That means it is able to recover 83% of the height to establish a new
potential difference between the ball and ground and that separation
in potential differences will determine not how many joules are stored
as potential (a false concept) but will determine how many joules of
energy gravitational potential will contribute to push the ball to the
ground - dissipate 17% and recover to 83% of the previous height
in order to establish a new separation of potential difference (a dipole)
that again breaks the symmetry of polarizes the aether to push the ball
down again with another 17% of that being dissipated in impact, repeat.

Bringing up these points in your false argument when I was the one that
brought them up to begin with shows that you believe the earth is flat
and you want to convince me (us) that it is flat instead of actually having
any willingness to see that it is not. If you're not here just to blindly
ramble on about what you think you already know and are here to possibly
see that you could be wrong (I believed what you believed in the past
until I knew better) - then prove it by spelling out how many joules of
energy are demonstrated (not even including the dissipation on impact)
in the following examples:

5 gram ball
lifted to 1 meter =
lifted to 83 cm =
lifted to 69 cm =
lifted to 57 cm =
lifted to 47 cm =
lifted to 32 cm =
lifted to 32.6 cm =
lifted to 27 cm =
lifted to 22.45 cm =
lifted to 18.64 cm =

If you have any intellectual honesty whatsoever, please post
how many joules of work are necessary to lift a 5 gram ball to
the above listed heights.

Environmental free input = gravitational potential, which is NOT static
but is active and dynamic and is coming in the downward direction pushing
things to the ground - there is no gravitational pull - it is a downward push.
That downward push of the aether on an object is what pushes it to the
ground.

You claim environmental input is so small?? Try telling that to the solar and
wind industry, universities, major corporations, the govt, etc... light and
wind and even heat for heat pumps such as geothermal are all environmental
input that is so HUGE that the COP of these systems blow away all
conventional gasoline generators, natural gas heat, nuclear, etc...

It is erroneous to claim the universe will be in 100% equilibrium. There are
many things that have shattered a big bang theory as well as the claim it is coming
to a stop as the observations show at the farthest edges of the universe, the
stars/galaxies are accelerating away faster and faster instead of slowing
down, which is the opposite of entropy and decay and opposite of
coming into equilibrium - it is doing the OPPOSITE. Even if the big bang
actually happened, that acceleration rips to shreds the claim that it is
coming into equilibrium - completely shredding the belief in conventional
thermodynamics and physics as having any place in reality or this universe.

Hawkings is another cheerleader for the establishment like Einstein. Neither
describe reality.

You claim: "The amount of heat being produced by the rubber ball doing this will never power something to lift the ball to the same height, because the amount of kinetic energy as heat (individual atoms shaking) is EQUAL to the ball when it was first dropped"

Again, you don't show what you understand, you demonstrate what you
don't. Will never power anything to lift the ball to the same height? That
is laughable. How many times do I have to say it doesn't? If you can't
comprehend what 83% efficiency means - you aren't even qualified
to argue.

Again, the ball in and of itself is demonstrating work being
lifted to a height and is demonstrating work by dissipating potential as
work on the impact. You are 100% resistant to the fact that if you add
up all the joules of measurable work in the lift and impacts on every single
bounce, it is MORE than the input that lifted it to the initial meter.

[Aaron]

@Lets Replicate

The same amount of work you get out of it plus the losses to entropy. My point was that when the ball is actually dropped, it has 0 power. It also has 0 power at all of the peak points, as well as when it bounces. So your calculations are false because they are missing a dimension. EVERYTHING is physics and engineering is based on vectors and failure to understand that leads people down the garden path of fantasy.

You are unwilling to spell it out in an equation.

I'll ask one final time if you are capable of answering the question, which
you are only evading with rhetoric.

How many joules of energy (work - the active process) are dissipated
when lifting a 5 gram ball to 1 meter?

Please don't confuse people with ramblings of jargon and nonsense.

Just answer the simple question:

How many joules of energy (work - the active process) are dissipated
when lifting a 5 gram ball to 1 meter?

[replaced]

Again, the ball in and of itself is demonstrating work being
lifted to a height and is demonstrating work by dissipating potential as
work on the impact. You are 100% resistant to the fact that if you add
up all the joules of measurable work in the lift and impacts on every single
bounce, it is MORE than the input that lifted it to the initial meter.

I will now prove you completely wrong by making one sentence.

If you were to put a turbine or something to extract the kinetic energy of the ball, you WILL slow the ball down.

So go ahead and add up the joules of each bounce, because if you were to go ahead and actually try to extract the kinetic energy from the ball, I guarantee you that the ball would either not bounce as high than if you did not touch it. If you were to put a turbine in front of the ball, you would slow down the ball and move the turbine because some of the momentum from the ball would be transferred to the turbine so it can move.

Also, when you push an object in the opposite direction, there is no net energy increase. Virtual particles such as electrons and positrons can spontaneously come in and out of existence because when they collide they completely cancel each other out with the momentum that they have. Net energy increase doesn't occur if 2 objects come out of existence with opposite velocity and equal mass.

A 25 cent bouncing ball is no different than Newton's Cradle. No free energy here. It's just that there is a mechanism in place to transfer the momentum between 2 objects.

Each time the rubber ball bounces, the kinetic energy not converted to friction is still in the ball. Eventually, all of the kinetic energy is transferred to heat when the ball stops moving. The kinetic momentum in the form of heat would be generally equal to the amount of kinetic momentum of the ball just as it touches the ground.

Another interesting fact is that the kinetic energy ball bouncing on the earth creates NO momentum on the earth. This is because the amount of "force" that the earth is pulling on the ball is equal to the ball pulling on the earth. The ball curves spacetime as well and it too attracts all the atoms of earth. So every time the ball hits the earth, the earth moves in the opposite velocity to the ball, there is no energy created. The movement of the earth because of the ball hitting the earth displaces the kinetic energy of the ball going the opposite direction.

[Aaron]

@replaced

I will now prove you completely wrong by making one sentence.

If you were to put a turbine or something to extract the kinetic energy of the ball, you WILL slow the ball down.

The impact on the ground already does that - but you still don't understand
the distinctions. You're still trying to prove the Earth is flat.

I'll ask again, but you too will probably just evade this question a 2nd time -
you already proved to be incapable of addressing my other points that
clearly show you are incapable of intellectual honesty:

5 gram ball
lifted to 1 meter =
lifted to 83 cm =
lifted to 69 cm =
lifted to 57 cm =
lifted to 47 cm =
lifted to 32 cm =
lifted to 32.6 cm =
lifted to 27 cm =
lifted to 22.45 cm =
lifted to 18.64 cm =

If you have any intellectual honesty whatsoever, please post
how many joules of work are necessary to lift a 5 gram ball to
the above listed heights.

I dare you - list the amount of work in joules DISSIPATED to lift
a ball of 5 grams to the above heights. If you cannot do it, you don't
even have a 9th grade education in elementary physics and are trying
to make a grown up argument with a child's mentality.

[replaced]

5 gram ball

Minimum amount of Joules required to lift a ball 1 meter from sea level:

lifted to 1 meter > 0.05 Joules
lifted to 83 cm > 0.0415 Joules
lifted to 69 cm > 0.0345 Joules
lifted to 57 cm > 0.0285 Joules
lifted to 47 cm > 0.0235 Joules
lifted to 32 cm > 0.016 Joules
lifted to 32.6 cm > 0.0163 Joules
lifted to 27 cm > 0.0135 Joules
lifted to 22.45 cm > 0.011225 Joules
lifted to 18.64 cm > 0.00932 Joules

[Aaron]

@replace

5 gram ball

Minimum amount of Joules required to lift a ball 1 meter from sea level:

lifted to 1 meter > 0.05 Joules
lifted to 83 cm > 0.0415 Joules
lifted to 69 cm > 0.0345 Joules
lifted to 57 cm > 0.0285 Joules
lifted to 47 cm > 0.0235 Joules
lifted to 32 cm > 0.016 Joules
lifted to 32.6 cm > 0.0163 Joules
lifted to 27 cm > 0.0135 Joules
lifted to 22.45 cm > 0.011225 Joules
lifted to 18.64 cm > 0.00932 Joules

Thank you!

You inserted the word minimum but that is fine. Those numbers
wouldn't account for air resistance, etc... that the ball could encounter
and we're not operating in a vacuum. But for the most part, yes, I
come up with the same numbers.

Let's just take 1 example:

lifted to 47 cm > 0.0235 Joules

The equation that states 0.005 kg X 9.8 X 0.47 meters = 0.0235 joules.
Close enough.

My next question to continue an honest discussion is this:

To lift the 5 gram ball to 47 cm, are 0.0235 joules of energy (work)
actually being dissipated in order to lift the 5 gram ball
to 47 cm?

Your answer would be yes since I stated:

"list the amount of work in joules DISSIPATED to lift
a ball of 5 grams to the above heights."

But I just want clarification.

[replaced]

Thank you!

You inserted the word minimum but that is fine. Those numbers
wouldn't account for air resistance, etc... that the ball could encounter
and we're not operating in a vacuum. But for the most part, yes, I
come up with the same numbers.

Let's just take 1 example:

lifted to 47 cm > 0.0235 Joules

The equation that states 0.005 kg X 9.8 X 0.47 meters = 0.0235 joules.
Close enough.

My next question to continue an honest discussion is this:

To lift the 5 gram ball to 47 cm, are 0.0235 joules of energy (work)
actually being dissipated in order to lift the 5 gram ball
to 47 cm?

Your answer would be yes since I stated:

"list the amount of work in joules DISSIPATED to lift
a ball of 5 grams to the above heights."

But I just want clarification.

No.

0.235 joules of energy are not being dissipated. What is happening is that 0.235 joules of energy is pushed on the earth in the opposite velocity when 0.235 joules of energy is applied to the ball.

When you lift the ball up, you push the earth down.

What is happening as the ball bounces up and down is that 0.235 joules of energy we can manipulate/use turns into 0.235 joules of energy we cannot use because it gets dispersed amongst trillions and trillions of atoms. The energy never goes away or disappears.

[Farmhand]

Hi guy's, Here's a layman's point of view. Just my "opinion"

In the bouncing ball example. The ball is only "lifted" once, after it is
dropped and hit's the ground it rebounds. The way I see it the ball is not
much different from an LC circuit, it's the same energy being transformed from
potential to kinetic until dissipated by the losses. When the ball is first lifted
extra potential energy is given to it, when it is dropped the potential is
converted to kinetic until it hits the ground then as it compresses the kinetic
is converted to potential until the point of maximum compression where it has
regained most of the potential it was given, the "same potential energy" then
it converts back to kinetic as it decompresses to bounce back up, when at
the top of it ascent it has then regained most of the "same" potential energy
it had fully compressed as "Height" rather than compression.



I don't see what the big deal is. It's not over C.O.P. 1 or Over Unity because
no useful work is taken from the ball and no extra energy is added to the ball.

It's just a ball bouncing.

There is a situation or two where if the ball hit a rock and bounced sideways
off a big cliff falling much further so to attain terminal velocity then if the
increased energy of the ball was harvested, but that would only work once
and only if the ball was already in existence and the energy to create the ball
and get it to the location was not a factor.

Cheers

[Aaron]

lifting a ball doesn't dissipate energy?

No.

0.235 joules of energy are not being dissipated. What is happening is that 0.235 joules of energy is pushed on the earth in the opposite velocity when 0.235 joules of energy is applied to the ball.

When you lift the ball up, you push the earth down.

What is happening as the ball bounces up and down is that 0.235 joules of energy we can manipulate/use turns into 0.235 joules of energy we cannot use because it gets dispersed amongst trillions and trillions of atoms. The energy never goes away or disappears.

Well it is 0.0235 instead of 0.235 but your point still stands.

To eliminate unnecessary variables for the point, lets say we are in a
vacuum and there is no air resistance to add any resistance to the ball
either dropping or bouncing up.

If you the ball goes from the ground level and goes to 47cm above the
ground level, is any work happening? If the ball goes from ground level
to 47cm does ANY dissipation happen?

IF there is any work happening by lifting the ball, what is it? You already
quoted 0.0235 joules of work are done to lift the ball to 47cm.

Now you are saying that much work is not being done to lift the ball.

So are you saying the ball went to 47cm without performing any work?

Please clarify.

A 5 gram ball lifted to 47cm... 0.0235 joules of work is performed or
not?

1. Is 0.0235 joules of work performed to lift the ball to 47cm?
2. Is 0 joules of work performed to lift the ball to 47cm?
3. Is there less than 0.0235 but greater than 0 joules of work performed to lift the ball to 47cm?

You say 0.0235 joules of work is done to lift the ball to 47cm yet you say
that much work isn't dissipated. If it isn't dissipated it isn't work is it?

And if it is not work, then that many joules of energy is not able to be
calculated to lift 5 grams to 47cm is there?

Any clarification is appreciated.

[replaced]

Well it is 0.0235 instead of 0.235 but your point still stands.

To eliminate unnecessary variables for the point, lets say we are in a
vacuum and there is no air resistance to add any resistance to the ball
either dropping or bouncing up.

If you the ball goes from the ground level and goes to 47cm above the
ground level, is any work happening? If the ball goes from ground level
to 47cm does ANY dissipation happen?

IF there is any work happening by lifting the ball, what is it? You already
quoted 0.0235 joules of work are done to lift the ball to 47cm.

Now you are saying that much work is not being done to lift the ball.

So are you saying the ball went to 47cm without performing any work?

Please clarify.

A 5 gram ball lifted to 47cm... 0.0235 joules of work is performed or
not?

1. Is 0.0235 joules of work performed to lift the ball to 47cm?
2. Is 0 joules of work performed to lift the ball to 47cm?
3. Is there less than 0.0235 but greater than 0 joules of work performed to lift the ball to 47cm?

You say 0.0235 joules of work is done to lift the ball to 47cm yet you say
that much work isn't dissipated. If it isn't dissipated it isn't work is it?

And if it is not work, then that many joules of energy is not able to be
calculated to lift 5 grams to 47cm is there?

Any clarification is appreciated.

Yeah it is 0.0235 instead of 0.235, i didn't catch that.

The confusion comes from the thinking that the Earth itself doesn't cancel out the momentum of the ball. No energy is dissipated, it is only converted.

The word "work" is like a secondary quality because it is used to describe a process.

1. Is 0.0235 joules of work performed to lift the ball to 47cm?

0.0235 joules is transferred/converted to move the ball. So yes, 0.0235 joules of work performed to lift the ball 47cm. But work is just a word used to describe energy we can use. Energy isn't create or destroyed.

2. Is 0 joules of work performed to lift the ball to 47cm?

No. But 0 joules are lost or created.

To eliminate unnecessary variables for the point, lets say we are in a
vacuum and there is no air resistance to add any resistance to the ball
either dropping or bouncing up.

The variable that i think that needs to be eliminated is the Earth. The earth is not an immovable object, it is moving as the ball bounces on it and as the earth is attracted and moves toward the ball as the ball is in the air. when you lift a rubber ball, the entire earth moves toward it at the same force as the rubber ball is moving toward the earth, which this effect is displaced by your feet pushing down on the ground on the earth.

Now get 2 balls with both having the same mass and earth like gravitation between each other and have one ball bounce off the other. You'll notice that BOTH balls will move away from each other at the first bounce and the end result is both balls moving at half the velocity of the initial hit from the first ball.

The end result of a rubber ball hitting the earth is the earth moving in the same velocity as it did before.

[Aaron]

@Farmhand

In the bouncing ball example. The ball is only "lifted" once, after it is
dropped and hit's the ground it rebounds. The way I see it the ball is not
much different from an LC circuit

I don't see what the big deal is. It's not over C.O.P. 1 or Over Unity because
no useful work is taken from the ball and no extra energy is added to the ball.

My point of view is also "laymans" because technical jargon and complex
formulas are irrelevant to see things as they are. I'm not implying you
agree with me by stating this.

If after the initial lift, the ball is bounced to 83% or 83 cm, that means
that you only have to input 17% to make up the loss to get it back to
the meter. From then on, only putting in 17% will get you an entire
1 meter of lift until the end of time. In many circuits, all you have to input
is enough to make up the loss while it outputs the full amount (just for
inputting the loss) - same difference.

I never said the ball is giving you useful work to power lights, etc... but
the point is - how much work is demonstrated in measurable joules of
energy and if you add it all up, is it more than we put in? In a bouncing
ball, the intended work is to have a ball that bounces. That is very useful
work for a ball that is intended to bounce.

The term "useful" is subjective and only applies to what we personally want
work for and has nothing to do with the fact that other work could be
happening.

An incandescent bulb is 10% efficient - we only say that because it gives
us 10% light and 90% heat so the efficiency is 10%. The useful work
we WANT is light so it is 10%. But the fact is that it is 100% efficient
because we get 10% light and 90% heat. Whether or not we want the
heat is irrelevant - 100% of the input we put into the bulb is converted
to work and work is work measurable in joules of energy.

If it is a resistive heating element that gives no light, it is 100% efficient
because the useful work we WANT is heat - therefore all of the power
we put in is converted to heat, then it is 100% efficient.

But of course in a resistive heating element (non inductive), there is no
rebounding, resonance, bouncing back, etc... it is a straight short circuit
that has no chance for recovery. I know an incandescent bulb filament
is an inductive resistors but for all practical purposes, we're not getting
any recovery and it might as well be a straight wire resistor.

Again, "useful" work is subjective and only describes work we want and
doesn't account for the fact that there is other work being done regardless
of if we want it done or not.

The "energy" added to the ball is free gravitational input that comes in
to the system at the peak of each bounce. It is useful because it makes
the ball bounce, which is what a rubber ball is designed to do.

If we have an inductive resistor like this:



The skeptics have said over and over in the Ainslie threads that because
this is a nichrome wire resistive heating element that EVERYTHING that
you put into it is dissipated into heat as it is a resistor.

However, this is false because when you turn off the power, you get an
inductive spike back that can go to a battery or fill a cap completely
destroying their argument that everything was turned to heat. That is
fresh regenerated potential that came back by the magnetic field collapsing
that can do more work. If you pulse this resistor at 1 mhz, you are getting
1 million regenerated inductive spikes coming back for all the 1 mhz
pulses into it to make heat.

[Aaron]

work dissipates energy

No energy is dissipated, it is only converted.

The word "work" is like a secondary quality because it is used to describe a process.

1. Is 0.0235 joules of work performed to lift the ball to 47cm?

0.0235 joules is transferred/converted to move the ball. So yes, 0.0235 joules of work performed to lift the ball 47cm. But work is just a word used to describe energy we can use. Energy isn't create or destroyed.

2. Is 0 joules of work performed to lift the ball to 47cm?

No. But 0 joules are lost or created.

You say no energy is dissipated - it is only converted. But as it is "converted"
in your belief system, is there not a decreasing amount of potential
to do more work after each conversion? If so, that is the same as
dissipation of energy with a diminishing amount of potential available
at each "conversion."

I get your viewpoint that you think energy is neither created nor destroyed
in this process. Whether or not I agree is irrelevant because the fact of
the matter is that work is performed and you agree to that.

But work is dissipation. And when energy is apparently happening, the
potential is dissipating. I say back into the aether - you say it is
transforming. In both cases, it doesn't change the fact that work is
happening as the ball is lifting.

So regardless of if you or I think energy is dissipating into the aether or
is just transforming into another state or converting, still, 0.0235 joules
of work are demonstrated and nothing will change this.

And from multiple bounces, you even admit there are quit a few joules
of energy or work demonstrated in all the ball lifts over a certain quantity
of bounces:

------------------------------------------

Originally Posted by replaced
5 gram ball

Minimum amount of Joules required to lift a ball 1 meter from sea level:

lifted to 1 meter > 0.05 Joules
lifted to 83 cm > 0.0415 Joules
lifted to 69 cm > 0.0345 Joules
lifted to 57 cm > 0.0285 Joules
lifted to 47 cm > 0.0235 Joules
lifted to 32 cm > 0.016 Joules
lifted to 32.6 cm > 0.0163 Joules
lifted to 27 cm > 0.0135 Joules
lifted to 22.45 cm > 0.011225 Joules
lifted to 18.64 cm > 0.00932 Joules

------------------------------------------

The first bounce to 83cm and second bounce to 69 cm alone are
doing more joules of work than the joules of work necessary to lift
the ball to the initial meter and we haven't even added work done
on the impact. The impact on the bouncing ball example is only
the loss or "undesired" work - but work nevertheless.

The lift of the ball is the light in an incandescent bulb and the impact
heat dissipation is the heat production in an incandescent bulb.

Both the light and lift are work and both the heat and impact are work.
NOT potential work but actual work in and of itself. The height that the
ball is before it drops is just an indication of how much free gravitational
potential will come into the system with an ability to perform work as any
resistance is encountered.