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07-07-2009, 08:13 PM
 witsend Gold Member Join Date: May 2009 Posts: 1,881
TinselKoala, I will leave my personal comments about you out of this post but will address it later. Here's the thing. You claim that by providing the same amount of power to a control using a continuous power input you gets the same temperature rise as is evident on the circuit using a switching device at 3.7% ON. Well then.

The energy at 3.7% on our circuit is delivered in two phases. The one relates to the ON cycle when the energy is delivered by the battery. The other relates to the OFF cycle when the energy is delivered back to the battery. Take a sample range of the voltages in both cycles, preferably in excess of 1 200 such samples and ideally over a reasonable sample range as your waveform seems to be periodic. If you use a tektronix - I believe you have one - then the sample range could be as great as 10 000 such samples.

You will see that some of the voltage samples will be represented as negative, and some will be represented as positive. Clearly the energy that is positive has come from the battery. And wherever the energy came from that is represented as negative voltage - it did not come from the battery. Then make a sum of all those voltages. Divide that sum by the number of samples. Then do your wattage analysis with that sum. If you're measuring across that shunt - from memory I think it is 0.5 Ohm - then the sum of the voltages divided by the Ohms value of that shunt resistor x battery voltage will give you the wattage that was delivered by the battery. You will find that that sum of the wattage delivered by the battery is LESS than the wattage dissipated at the load.

If you do the analysis like this you will find the gain. If you do not believe it is the correct way to do the analysis then I'm afraid you must argue with the experts.

And that is all that is required to prove the over unity claim. It will not matter what duty cycle you use. It will not matter what frequency you run the test at. The sum over the shunt resistor will always be less than the product over the load resistor. That's strictly in terms of classical analysis of energy delivered by the battery and dissipated at the load. You do not need to be a genius to see that the one will inevitably be greater than the other.

But your instruments need to be accurate enough to take in the full value of the negative voltage. And. If you do a measure of the rate at which your battery delivers its energy you will see that it is consistent with this sum. If you need to see if the battery in fact recharges - then do the two battery test described in this thread. And if you need to finally check the advantage to the battey then run a control along side the test. All these points have been repeated throughout all my postings and you continually choose to ignore them. Why?
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