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## Solid States Devices => Tesla Technologgy => Topic started by: d3x0r on June 07, 2014, 09:14:40 AM

Title: Coils as momentum
Post by: d3x0r on June 07, 2014, 09:14:40 AM

I wish I hadn't gotten myself banned from physicsforums.com ; they seemed fairly knowledgable; only took 2 days too....

So the theory is, if I have two infinite capacitances on the ends of a coil (or 1 even and just make a LC loop standalone), if I could get a current to flow in the coil, then that current will flow forever, until the voltage of the capacitors is high enough to oppose the flow.. . but if there is no differential to start with, and the capacitance never fills up, the coil would keep conducting.  True or false?

if I take a coil and use an external magnet to induce a current in a single direction and stop the magnet in place, this should allow the induced current to flow itself... ?

So I did; I took a 2000 turn 3.3mH air core coil and a 6" magnet and a few capacitors.

The first cap is 9.9nF (12kV?)
the second cap is 100nF (400V?)
the third cap is 6800uF (40V?)

the voltage rating doesn't really matter, I'm not going to exceed a few volts with a hand generator like this.

If it's the voltage difference only that stops the current flow, then a similar motion will generate a comparable voltage; unless voltage is somehow a exponential value of resistance such that 2V is resistance 4x, and 3V is resistance 40x.

(compiling accompanying video; will edit with update)

just musing... Tesla said 'there's no limit to the amount of power you can store in a capacitor'; that can be true if you know the momentum.
wish a I had some of those farad caps to test more.

Current * Inductance : Velocity * Mass

So, I was wondering, how fast does a coil's current increase for a certain voltage X applied to it?  I guess it becomes a more complex question because of the resistance of the coil also?

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

So in the experiment I was able to charge 9.9nF to voltage 2.2V
I was able to charge 109.9nF to voltage 2V
I was able to charg 1000109nF to voltage 2V  (1000uF + )
I was able to charge 6800109nF to voltage 1V   (6800uF )  (I think here the resistance of the wire is a total cap on the current that can flow)

In 200ms; my motion from top to bottom is apparently about 200ms.... give or take a little but the same peeks in the middle.

While I was measuring the 200ms time, I inserted a resistor, my first was 10ohm with no effect, my secnod was 100ohm with no effect, but I left it in.  (no effect on 9.9nF charge rate or level) but there was a notable effect at 6.8mF the voltage only charged to .25V and was very subdued... which leads me to go back to - the resistance of the coil is going to limit how much it can charge/conduct....

A higher inductance coil allows greater charge... that mobious coil is way better than the solenoid I ran... but maybe if there was a pancake instead?  (I don't have a high inductance pancake), the large mobius is about 3.3mH also...

(I should do some math on total joules stored in the caps)

2V 9.2nF
Energy E = 18.4n Joules

2V 108nF
Energy E = 216n Joules

1V 6.8mF
Energy E = 3.4m Joules

3.4m Joules/sec = 3.4mW ; that's pretty straight forward... so these are jouls in 200ms or *5 of the above = watts....

9.2nF  0.092mW
108nF 1.080mW
6.8mF 17mW

Title: Re: Coils as momentum
Post by: d3x0r on June 07, 2014, 10:24:12 AM
In a simulator... a huge capacitance, and a switch to get momentum in the coil... which can be used to charge into the cap...

I note that when I start it, I use a lot of power very quickly, then when switching to the capacitor there's a power disappation of the same but over a much longer time than initial;  Until the cap is at like 25% of the voltage source...

At some point if the capacitor has a greater potential difference than the initial circuit, the coil will instantly lose all momentum and reverse.

Title: Re: Coils as momentum
Post by: MarkE on June 07, 2014, 11:10:43 AM

I wish I hadn't gotten myself banned from physicsforums.com ; they seemed fairly knowledgable; only took 2 days too....

So the theory is, if I have two infinite capacitances on the ends of a coil (or 1 even and just make a LC loop standalone), if I could get a current to flow in the coil, then that current will flow forever, until the voltage of the capacitors is high enough to oppose the flow.. . but if there is no differential to start with, and the capacitance never fills up, the coil would keep conducting.  True or false?
False.  If you put together a series branch:  C - L - C, the loop will behave the same as C/2 - L.  Assuming ideal capacitors and inductor: the impedance of the network is very high at low frequencies:  2/jwC, and very high at high frequencies:  jwL.  At jW = (LC)-0.5 the impedance drops theoretically to zero.  In real life, there will be parasitic series resistance, parallel capacitance across the inductor, leakage through the capacitors etc that will take away from the ideal.  You postulate what happens if the capacitors are really, really large.  The answer is that the impedance will be low down to very low frequencies.
Quote

if I take a coil and use an external magnet to induce a current in a single direction and stop the magnet in place, this should allow the induced current to flow itself... ?
That's how generators work.  Once you stop moving the magnet, the induced voltage goes away and the current will decay according the the circuit L/R time constant.
Quote

So I did; I took a 2000 turn 3.3mH air core coil and a 6" magnet and a few capacitors.

The first cap is 9.9nF (12kV?)
the second cap is 100nF (400V?)
the third cap is 6800uF (40V?)

the voltage rating doesn't really matter, I'm not going to exceed a few volts with a hand generator like this.

If it's the voltage difference only that stops the current flow, then a similar motion will generate a comparable voltage; unless voltage is somehow a exponential value of resistance such that 2V is resistance 4x, and 3V is resistance 40x.
I don't know what you are after mixing up the ideas of voltage and resistance.
Quote

(compiling accompanying video; will edit with update)

just musing... Tesla said 'there's no limit to the amount of power you can store in a capacitor'; that can be true if you know the momentum.
wish a I had some of those farad caps to test more.
The assumption behind that statement is that the capacitor does not break down no matter what the applied voltage.  We know that assumption is false.
Quote

Current * Inductance : Velocity * Mass
Yes, it takes real energy to change the current in an inductor.  If you have a linear inductor, solving the integral for the energy is easy:  0.5*L*I[sup2[/sup].
Quote

So, I was wondering, how fast does a coil's current increase for a certain voltage X applied to it?  I guess it becomes a more complex question because of the resistance of the coil also?
For time spans much shorter than L/R, Idelta ~= V*T/L.  For time spans much longer than R/L, I ~= V/R.
Quote

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

So in the experiment I was able to charge 9.9nF to voltage 2.2V
I was able to charge 109.9nF to voltage 2V
I was able to charg 1000109nF to voltage 2V  (1000uF + )
I was able to charge 6800109nF to voltage 1V   (6800uF )  (I think here the resistance of the wire is a total cap on the current that can flow)

In 200ms; my motion from top to bottom is apparently about 200ms.... give or take a little but the same peeks in the middle.

While I was measuring the 200ms time, I inserted a resistor, my first was 10ohm with no effect, my secnod was 100ohm with no effect, but I left it in.  (no effect on 9.9nF charge rate or level) but there was a notable effect at 6.8mF the voltage only charged to .25V and was very subdued... which leads me to go back to - the resistance of the coil is going to limit how much it can charge/conduct....
Yes at frequencies well below jw = R/L, the resistance dominates, and for frequencies well above jw = R/L, the inductive reactance dominates.
Quote

A higher inductance coil allows greater charge... that mobious coil is way better than the solenoid I ran... but maybe if there was a pancake instead?  (I don't have a high inductance pancake), the large mobius is about 3.3mH also...
More inductance means more impedance at any given frequency.  It also means more energy storage at any given current.  Be wary of saturation.
Quote

(I should do some math on total joules stored in the caps)

2V 9.2nF
Energy E = 18.4n Joules

2V 108nF
Energy E = 216n Joules

1V 6.8mF
Energy E = 3.4m Joules

3.4m Joules/sec = 3.4mW ; that's pretty straight forward... so these are jouls in 200ms or *5 of the above = watts....
Average, yes that is correct.
Quote

9.2nF  0.092mW
108nF 1.080mW
6.8mF 17mW
Title: Re: Coils as momentum
Post by: d3x0r on June 07, 2014, 12:00:44 PM
Once you stop moving the magnet, the induced voltage goes away and the current will decay according the the circuit L/R time constant.

Well... I guess my aparatus isn't inclined to demonstrate that it's not 'once you stop moving the magnet' but additionally after the current built in the coil decays; with 6.8mF and 1mH (approx) it's only millseconds to reach max charge and decay (ideally)....and since I'm using a gross movement covernig 200ms it's hard to see the fractional time;
Modified sim to use a current source of 1A instead of a voltage source, so I could get the current moving, and then throw the switch, so I could confirm more acurately that ideally it's a 4ms delay after removing the EMF source.

Edit: added a 10ohm resistor inline to measure the voltage across it as a CSR... So yes, when I stop, the current incrase stops, and starts to decrease... but the voltage still increases after...

Additionally, usually in the generator, the magnet doesn't 'stop' it instantly starts to receed... which causes a current in the opposing direction already, which would cancel any such effect.

I don't know what you are after mixing up the ideas of voltage and resistance.

that's actually impedance on the cap; and it's not the case, it's a linear relationship...

If you have a linear inductor, solving the integral for the energy is easy:  0.5*L*I[sup2.For time spans much shorter than L/R, Idelta ~= V*T/L.  For time spans much longer than R/L, I ~= V/R.Yes at frequencies well below jw = R/L, the resistance dominates, and for frequencies well above jw = R/L, the inductive reactance dominates.More inductance means more impedance at any given frequency.  It also means more energy storage at any given current.  Be wary of saturation.
THe experiment was done without cores; does that mean saturation is basically instantaneous because of tiny permeability?

----
I further tried some simulations, and chained another inductor and switch to the original cap, and then a cap for that... and basically what gets put into the first is transferred to the second... something like a slinky in action...

It's only the initial input that can change it.

Most circuits are driven by potential differences in the capacitances, which is what impedes the flow in the coil....

if I use a 1H coil to charge a 1mF cap, it goes to like 30V, but then the next 1H coil will only get accelerated back to the current that the cap captured from the first coil... (acceleration is relative to current)

---
Just even with this perspective there's no additional features available because of the methods available to form current in a coil....

Title: Re: Coils as momentum
Post by: TinselKoala on June 07, 2014, 02:14:27 PM
Air (vacuum) cores do not saturate, that's the whole point of using them.