Jon,
 
> FWIW, another reality check for the input power consumed by the MAHG is to
> first calculate the energy in Joules per pulse and multiply by the rep rate.
> IE, the input energy is 166 * 12 * (1/50) * .05 = 1.992J per pulse.
> Therefore, the input power = 1.992 * 50 = 99.6 watts. With the equipment JLN
> has at his disposal, I would assume the peak current measurement, duty cycle
> and line frequency, are all within 5%.
 
As Terry mentioned, on the original page
 http://jlnlabs.imars.com/mahg/tests/index.htm

He says that he measures rms volts times rms amps and gets real power.
....and he says he does it with THREE different meters, analog ammeter included!!
 
 ERGO....
 
He must surely measure at the battery itself, no? Why wouldn't he - cheap and effective. That is supposed to be the  real beauty of using a battery PS in the first place.... so why not take full advantage? You seem to be saying that he would have missed this simple expedient... but that seems unlikely to me.
 
However we will not know for sure, unless we get some response. Hopefully it will be soon. I suspect that there are several people making plans now for a replication.

....how hard can it be to just put an analog ammeter on the  battery - and then nobody can complain about a claim of 4-5 watts P-in, right ?  If and when, or course, you are showing 400 milliamps or thereabouts on the ammeter... or maybe I should ask Steve Lawrence IF this is where the internal resistance of the battery might be a major factor?
 
At any rate, I just cannot believe he would have overlooked this measurement with his ammeter.OTOH it _has_ been a bad day...
 
Jones 

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