Possible Overunity

[singerxyz]

A little encouragement  from an unexpected source-

Ian Lungold a researcher on the Mayan Calendar says that between Nov. 19th, 2007 - Nov. 12th, 2008
will be "The end of manufactured lack" and the arrival of free energy for all.

Interesting!

http://mayanmajix.com/cycles.html
http://www.calleman.com
http://www.mayanmajix.com/

[linda933]

Doors- love it!

My results- hate it :-(

I tested the unit with the resistor as a load i.e. (positive-->resistor<--negative) and the meter connected meter positive to positive, meter neg to neg. I was using 2 nickel-cadmium 9V's because they seemed to produce more current, lasted a shorter time.

Open Circuit
600 VDC 1.6 Amps (2 Amps with fresh battery)

100 Ohm Resistor
7 VDC .8 A

10 Ohm Resistor
3.3 VDC 1.3 A

1 Ohm Resistor
.5 VDC .53 A

My batteries were running down pretty quickly, so the results may actually be slightly better but,
I guess it's back to the drawing board...

Dear Singer,

I think you are still measuring current wrong.  It makes no sense to say you have 1.6A flowing in an open-circuited load!  Makes me think you are still putting the current meter (with its internal low-value shunt resistor) directly across (in parallel with) your output leads.  I was hoping my earlier explanations were sinking in but it looks like not yet. 

Current measurement has to be done by putting the meter in series with the load, not across it.  An open load (infinite resistance) leaves no way to measure current, but that's okay because it will always be zero in an open circuit!

Forget measuring the current.  Measure your resistors each by itself not hooked up to anything with the meter set for ohms measurement.  Write down the results for each resistor.  Then measure the voltage only on each resistor as you load your circuit with it.  Your numbers you give above tell me that you were doing something weird like maybe you got two meters and were trying to measure volts and current at the same time but putting the current meter right across the load.  That would screw up the resistance you have because the meter has a low value internal shunt resistor when in current mode.

I was expecting you would get more voltage across your 100 ohm resistor.  You also didn't test with the 1000 Ohm resistor, which would be interesting to see how much that drops the voltage as compared to open circuit.  I was really hoping that if you didn't get overunity you would at least learn something about measuring basic circuits.  It looks like maybe you're still not understanding yet? 

Linda

[singerxyz]

OK. I just needed this diagram, now I understand. (like the saying goes, that picture could have saved a thousand words!) Basically, I tested everything like the Voltage test in the diagram. I'll re-test tonight.
Thanks again, L

Singer

[singerxyz]

I think I got it right now-

1000 ohm=8.8V .009A
100 ohm=4.2V .04A
10 ohm=2.4V .24A

[pese]

A good way to home in on this magic number to extract maximum power is to notice the following:

As you increase the ohmic value of the load resistor, you will most certainly notice an increase in the output voltage.  It will be zero if you use zero ohms (dead short) and will be maximum if you use an open load (no load...open circuit). 

For most (linear) circuits, and I believe yours will probably fall into this category, the ideal load for maximum power transfer will be very near to whatever load causes the voltage to fall to half its open circuit voltage.  Try to find that value by experimenting.  You will then be able to tell if you have hit the maximum power point by raising and lowering the value slightly from there and observing a power falloff in both directions.

Linda

Somso ,,,

if i put an power load tothe 110volts line , so strom , even tat the voltage go don to 55 volts ..
than i have the maximum "power point" ...
thisway that the whole town will go to darkness ...?

Pese
(smile)

www.pese.cjb.net