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[Red_Sunset]

Exactly! And that is why - as I mentioned earlier -  the pivot should only move when the conversion process of gravity into rotation has finished (in the vertical position).
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That peak correlates with the horizontal (weightless) position of a pendulum (at which the lever »fires«).
Thus, a lot more (odd coincidences) to investigate, I think.

Zeit,
I am not sure that I am understanding you correctly:   

What do you mean with " the pivot should only move ....................in the vertical position"
*  Move into which direction?, logically I would assume that with the pendulum in the vertical position, the pivot could move better 'down' than 'up'.

What do you mean with " at which the lever fires" 
*  Do you mean this " as opposition to 'down' on the vertical",  this being 'up' ?

Red

[Zeitmaschine]

What do you mean with " the pivot should only move ....................in the vertical position"

When the pendulum is in its vertical position (swings through its vertical position) then the conversion of energy from potential to kinetic has ended while the conversion from kinetic to potential is just about to start. All energy is stored in the horizontal velocity of the mass of the pendulum (mass × velocity = stored kinetic energy). So I can't see how the down movement of the pivot at that vertical position could ever draw energy out of the pendulum.

What do you mean with " at which the lever fires"

When the pendulum rests at its vertical position (not swinging) then the lever has an overbalance on the pendulum's side because of the weight of the resting pendulum (normal state). When the pendulum swings then it is weightless at both of its horizontal positions. At that moment the lever »fires«, it goes down on its opposite side because without the counter weight of the pendulum that side is now overbalanced.

The same applies to  the spark: The spark fires because of an imbalance (of electrons) between the transformer coil and the piece of metal, which occurs most intensely on each peak of the sine wave. The magnetic field of the transformer moves the electrons back and forth in the secondary coil (like a pendulum) without doing any real work (also like the pendulum) but creating thereby a spark, like the swinging pendulum creates a moving lever, without doing any real work (friction disregarded).


This is stuff for brilliant minds.

[Zeitmaschine]

Another add-on (for even more brilliant minds):

As a matter of fact, when the pendulum is in its weightless position (horizontal) then we can move its pivot around as we wish and we can even stop it completely in that position (although losing its weightlessness), because at that point all energy is stored as gravity potential indefinitely (like a capacitor stores an electric field indefinitely). After releasing it, it will keep on swinging, like nothing has happened.

So the only thing in question can be: Can we move down the pivot, when the pendulum swings through its vertical position, without taking energy out of it? When we move down the pivot at vertical (6 o'clock), then we move the pivot in direction of the acting centrifugal force. Therefore, does a pendulum convert part of its energy into centrifugal force? Or does it just convert between potential and kinetic energy? Even if so, then force is not energy. What would happen, if we would move the pivot of a pendulum - heavy weight, slowly swinging, suspended on a one meter string - down two meters, exactly when it swings through its 6 o'clock position? Would it keep on swinging without energy loss?

What would happen if we drop the pendulum at 6 o'clock? At that point all energy is stored in its horizontal velocity. Then will the pendulum's bob keep on moving in horizontal direction (no air drag) while it accelerates downwards? If we stop the pendulum's freefall after awhile, will it continue to swing?

[Red_Sunset]

Another add-on (for even more brilliant minds):

As a matter of fact, when the pendulum is in its weightless position (horizontal) then we can move its pivot around as we wish and we can even stop it completely in that position (although losing its weightlessness), because at that point all energy is stored as gravity potential indefinitely (like a capacitor stores an electric field indefinitely). After releasing it, it will keep on swinging, like nothing has happened.

So the only thing in question can be: Can we move down the pivot, when the pendulum swings through its vertical position, without taking energy out of it? When we move down the pivot at vertical (6 o'clock), then we move the pivot in direction of the acting centrifugal force. Therefore, does a pendulum convert part of its energy into centrifugal force? Or does it just convert between potential and kinetic energy? Even if so, then force is not energy. What would happen, if we would move the pivot of a pendulum - heavy weight, slowly swinging, suspended on a one meter string - down two meters, exactly when it swings through its 6 o'clock position? Would it keep on swinging without energy loss?

What would happen if we drop the pendulum at 6 o'clock? At that point all energy is stored in its horizontal velocity. Then will the pendulum's bob keep on moving in horizontal direction (no air drag) while it accelerates downwards? If we stop the pendulum's freefall after awhile, will it continue to swing?

For a brilliant mind, there is nothing like a practical test to confirm a proposition.
The pendulum definitely recovers the energy it dropped at vertical, when the swing is at or near horizontal and the pivot moves up.
Moving the pivot up will directly impact/reduce  the angular swing at this point. Effective reducing the potential energy invested in he swing.  At this point, the pendulum angular momentum has the most torque ( & most potential energy / dgr), so it has the most too loose per degree at this position.

At vertical , extracting the centrifugal force comes directly out of the increased available energy as provided by the drop & cord length.  The energy zero reference position is the starting position (horizontal, pivot up).   
In practice, it is impossible to instantaneously drop the weight at vertical, therefore it is impossible to have that movement not impacting the swing. The same applies to the horizontal for pivot up.  (although it could be curtailed within agreeable boundaries).
A practical execution does tend to put limitations on theoretical proposals.

Red_Sunset

[Zeitmaschine]

For a brilliant mind, there is nothing like a practical test to confirm a proposition.

This is the crucial point! I'm afraid I can't practically construct a pendulum-lever setup that works with mathematical precision. But if it don't, then the erratic moving lever will slow down the pendulum and this will prove nothing.

Because of that, I would rather like to construct the electric version of that mechanic pendulum-lever setup. Instead of using gravity as energy source, the electric version should use the electric field that surrounds us like gravity does. So when gravity pulls down the lever on one side and hence moves up the pivot on the other side, because of the temporary weightlessness of the pendulum's bob, then what would be the equivalent of this in an electric LC circuit? And indeed there should be an electric equivalent, regardless for now whether there is overunity involved or not.

Ideas?