The secret to Overunity

[TinselKoala]

h)  The meter ACTUALLY says 1.287 V, NOT 12.87 V.       

[sm0ky2]

h)  The meter ACTUALLY says 1.287 V, NOT 12.87 V.       

Thanks TK


If that was fully charged at the time, a 0.5v drop
across the inductor and led, it probably won't last very long.


Edit: there is another side to the "actual voltage" when measuring that way.


Which is discussed from every angle in the many joule thief threads.


Ultimately the duty-cycle is what you should be looking at.
Dual scope, one across a resistor between battery and load
and one across a resistor between load and neg of battery.
Then you get a glimpse in real-time of what goes in and out.
And from the return side, you can know the "in", by plugging the data sheet
values into the equations to get an idea of the internal battery current.
You can know if it is charging or depleting


It is better for your battery to use a controlled charging circuit separate from the load.
You can replace the AA with a supercapacitor.
This eliminates the problem of damaging the battery.


You can also instantly know with a super cap, wether your energy is being provided or consumed.
Their internal resistance is unidirectional.
Meaning the load has a consistent Ohms. (they try to anyways)
While the charge resistance follows more the curve of a capacitor.
In this way it acts as a battery for a load, but can charge very quickly,
Even while providing energy to the load.


You can even place a switch somewhere in the circuit
So you can turn it on/off
And check the voltage in the rest state.


Or hell, this is the 21st century, find your block diagram on the internet
Get its ieee #, and pull up the data sheet


It will tell you more than you want to know about your circuit.

[Void]

I'm gonna go with G here

did you measure the charging current at the battery?
You probably want to back the V down a bit
closer to the charging voltage of your AA
led should still light up 7.5-8v,maybe lower depending on the diode.

12v (with enough current) will charge a AA, but you risk some events.

The real test would be how much is actually being returned to the battery.
For that you need an oscilloscope to see the frequency, pulse width, off-time, etc.

Or run it for some time, then turn it off and test the battery.
Or you could just wait......  that may take a while, some JT's can run for months like that.

Well!!  For a guy that has no problem hitting others when they say that is possible to get self runners, I SHOULD give you the same treatment... But I will not..  Give us more date on how you have conduct your experiment and we can conclude on results..

h)  The meter ACTUALLY says 1.287 V, NOT 12.87 V.       

Ha ha. Yeah, I was just having a little bit of fun.
There are of course a few things that are quite wrong are not so good with my test demo.
The most obvious being that the circuit is only pulling a very small current compared to the
Ah capacity of a typical AA batery. The particular battery I used in the demo is a rechargeable
type that is rated for 2000 mAh! A short 5 minute demo of a circuit that appears to be drawing
only around 1mA is of course way too short to be able to try to draw any sort of reasonable conclusions
about whether the battery can remain charged up or not.

Also, as TK pointed out, the battery voltage can't be 12.87V as was indicated in the video and in
my comment, so although that could maybe just be a typo/mistake, it needs some clarification. Also
when you are applying switching spikes to a capacitor or battery, you can't trust the digital meter to
be reading exactly right, as the high frequency switching spikes can cause a digital voltmeter to
read off. It is better to disconnect the battery for a moment and take a quick voltage reading with
the circuit switched off to make sure the digital voltmeter reading is not being thrown off by the
high frequency switching spikes. In this case it was reading close enough, but it was somehwat
susceptable to the high frequency switching spikes, as when I moved my hand close around the meter
its voltage reading would sometimes change by a few hundredths of a volt or so.

Also, the viewer of the video doesn't know how accurate the anaolg ammeter is, so that is another problem.
In this particular case the ammeter spring mechanism has been bent from being accidentally hit
with much too high of an over current pulse a few times, so the meter doesn't zero properly any more.
The meter is still useable to get an idea of the current, but it reads about 0.5mA too high. You have to
subtract 0.5mA from the meter reading to get the approximate actual current reading. The actual current
draw from the battery in that test was therefore actually closer to around 600uA, so the actual current
draw was even less than you might think if you just watch the video.

There are other things that are not so good, with a fairly messy circuit and cluttered video viewing area,
and with test lead wires running off the screen so you can't see exactly where they are going. In this
case however there really wasn't any tricks. The circuit was really only powered from the single AA battery
as I said, but because my test run was way too short for the circuit current draw compared to the
battery's current capacity of 2000mAh, you just can't tell at all whether the battery can really hold
its voltage or not. In this particular case with the circuit only drawing about 600uA, I should
at the least let the test setup run continuously for a few days to get any idea if the battery voltage
really might be holding steady or not. A continuous test run for a full week or even more would be an even
better test run period for this type of very low current draw setup.

Anyway, this should be pretty obvious to people who have been in the OU game for any length of time,
but less experienced people might possibly miss some of the problems with that demo. It was at least good
that I was making an effort to measure the circuit supply battery voltage and battery current draw in my
demo video, but some effort should really be made to show that the meters are reading accurately before
starting the test run, and care should be taken to make sure that a possible mistake such as I mentioned
above is not being made in reading the meters.

Good luck with your experimenting guys!

All the best...

[sm0ky2]

http://www.youtube.com/watch?v=AX-jrlGC-aA

Do you think we can transfer this type of electrostatic linear action
Into a rotary force, such as in a sterling engine?
Basically, operate a crank with it


Or an inductive solenoid, could possibly generate electricity from the
physical motion caused by the charge force.


It is important to note that the force presented by two opposing charges
Is not directly coupled to the power transfer that causes this force.
Meaning, the force exists when no charge is moving.
No current.
No current, no power consumption. Yet there is force.


If we ignore coronal losses due to insulator inefficiencies
The power consumption of this type of device is reduced to
The capacitance of the moving conductor, and the magnitude
of the charge.


The power production (or work) is force over distance.
This is not a 1:1 power conversion
But rather a relativistic transformation
of charge to mechanical energy.


Consider the Newtonian example of real world scenario
What is the "quantity" of energy required to physically
cause this oscillation?


Now what is the solenoid equivalent?


And how does that situation compare to the capacitance
discharge cycle, shown in this video?


We see here, that the energy difference must come from the static field.
The charge


But we do not calculate a depletion of the charge
In fact, in most conditions, this type of action reinforces the charge.


The only depletion (aside from the insulative loses) is through the capacitance
cycle. - a set quantity of energy transferring at a set rate.
The frequency of this discharge define the time over which the force is applied
across a given distance.


This can be compared to the solenoid example, if we remove the inductive constant.


The electrostatic circuit is not a single conductive circuit.
And therefore, Ampere's Law does not apply to the charge force.

[sm0ky2]

We can now USE Ampere's Law, to extract this energy.
By simply magnetizing the capacitive sphere, or attaching
The moving sphere to a mechanical rod that moves a magnet.


A coil placed appropriately between the two stationary charges
can be used as an inductive pick-up coil to siphon energy
directly from the charge force.


While this will affect the velocity of the moving conductor
(and thereby reducing the frequency proportional to the load)
It does not affect the charge or the force.
Or the quantity of energy being transferred by the capacitance.