Konstantin Meyl -- Tesla Scalar Wave Theory

[Groundloop]

Consider this....
Will adding or removing charge to this 'resonant motion' capacitor change the resonant frequency of motion or just make it more complex?

WW,

I know that the Coulomb force is quite strong and will dampen the oscillation.
If we add charge at the bottom of the capacitor plate stroke then the Coulomb force will increase.
But the opposite will happen at the at the top stroke where we remove 1/2 of the energy. If the Coulomb force
doubles at the bottom it will be 1/2 as much at the top so the result will be zero.  Am I right?

GL.

[humbugger]

Hum,

Just thinking.......................

If we use a pizo electric speaker and mount a capacitor plate on top of it then we get a movable capacitor plate.
We charge the plate when the distance is smallest and take out 1/2 the energy when the distance is doubled.
The work we pay to do this is the same as we get out. (With some small losses.) But, we know that if we use
the resonant frequency of the weight of the capacitor plate, then we only need to put in approx. 2% of the initial
work to move the plate. Will this work?

GL.

I expect that the doubling of energy stored when the distance is doubled is directly related to the physical work done to move the plates apart, which will be a function of the charge and area.  The opposite charges attract, so moving the plates apart requires energy.  Resonance will not help, because the mechanical work extracted from the vibrating piezo will lower the resonant Q and damp the resonance, as in all systems where work is extracted.

This is another example of trying to extract real power from a resonant circuit or object without damping the resonance...it cannot be done.  Your figure of 2% will rise dramatically because the plates will resist being separated.

Humbugger

[Groundloop]

I expect that the doubling of energy stored when the distance is doubled is directly related to the physical work done to move the plates apart, which will be a function of the charge and area.  The opposite charges attract, so moving the plates apart requires energy.  Resonance will not help, because the mechanical work extracted from the vibrating piezo will lower the resonant Q and damp the resonance, as in all systems where work is extracted.

This is another example of trying to extract real power from a resonant circuit or object without damping the resonance...it cannot be done.  Your figure of 2% will rise dramatically because the plates will resist being separated.

Humbugger

I agree that we need to do work to lift the plate. But we get that work back when the plate goes back again.
It is like a spring, we only need to use work to pull the spring apart and we get all the work back when
the spring is released. You must see the whole cycle not 1/2 of the cycle. Is it not so?

GL.

[humbugger]

I agree that we need to do work to lift the plate. But we get that work back when the plate goes back again.
It is like a spring, we only need to use work to pull the spring apart and we get all the work back when
the spring is released. You must see the whole cycle not 1/2 of the cycle. Is it not so?

GL.

I think you are forgetting that the "spring" is much weaker on the pullback stroke because you have removed charge at the top of the stroke.  Anyway, the answer is no, I don't think it will "work" if by "work" you mean the ability to take out more energy than you put in all told.  But that's just my old-fashioned opinion.

Humbugger

[Groundloop]

I think you are forgetting that the "spring" is much weaker on the pullback stroke because you have removed charge at the top of the stroke.  Anyway, the answer is no, I don't think it will "work" if by "work" you mean the ability to take out more energy than you put in all told.  But that's just my old-fashioned opinion.

Humbugger

Yes, I now see that the Coulomb force will be 1/2 on the return stroke so the result force
will be 1/2 of the Coulomb force in total. There will be small losses in the capacitor charge
so we will also need to recharge the capacitor for losses. But all in all, all we need to
input is enough energy to keep the capacitor plate in weight resonance. Will this give
us more energy out than we did put into the system? You say no, I say maybe because
we only need to pay for the first initial charge of the capacitor and can repeat the cycle
a lot of times before we need to top up the capacitor charge again. So we end up with the
question, do we need to use more energy to fight the 1/2 Coulomb force up stroke than
we get back from the double capacitor energy at the double distance? Anybody up for
doing math on a real life capacitor? Plate=round radius 50mm, plate distance 0,1mm at
closest and 0,2mm distance at top stroke. Voltage on plates 12 volt. Dielectric= Air.

GL.