Quote from: TinselKoala on May 09, 2012, 12:16:22 PM

WHY DO YOU THINK YOU HAVE TO DO ALL THAT SPREADSHEET STUFF? It's because you are taking INSTANTANEOUS power values and computing an average over time, then you are attempting to integrate that average power data over that time to get a TOTAL ENERGY FLOW in Joules. And you don't even realize it.

IF by that 'spread sheet stuff' you're referring to integrated power analysis - then INDEED.  We use multiple representative samples to calculate the amount of energy delivered over a period of time.  When we've established a fair sample and when we've computed the power from EACH of those samples then we DIVIDE THAT SUM by the NUMBER OF SAMPLES.  Then - and ONLY then  - and INDEED THEN - we have the ACTUAL WATTAGE.

What you're trying to argue is that 1/6th of the ON time is a REPRESENTATIVE SAMPLE.  It is NOT.  It is only representative of 1/6th of a duty cycle.  IF you DO NOT factor in that 1/6th PERIOD then you are also NOT giving an accurate representation of WATTAGE.  You are giving us something that is not comparable to anything at all.

Again and this is getting tedious in the extreme
Rosie Pose

Quote from: Rosemary Ainslie on May 09, 2012, 12:29:31 PM
IF by that 'spread sheet stuff' you're referring to integrated power analysis - then INDEED.  We use multiple representative samples to calculate the amount of energy delivered over a period of time.  When we've established a fair sample and when we've computed the power from EACH of those samples then we DIVIDE THAT SUM by the NUMBER OF SAMPLES.  Then - and ONLY then  - and INDEED THEN - we have the ACTUAL WATTAGE.

What you're trying to argue is that 1/6th of the ON time is a REPRESENTATIVE SAMPLE.  It is NOT.  It is only representative of 1/6th of a duty cycle.  IF you DO NOT factor in that 1/6th PERIOD then you are also NOT giving an accurate representation of WATTAGE.  You are giving us something that is not comparable to anything at all.

Again and this is getting tedious in the extreme
Rosie Pose

Ainslie, do you accept that during the ON time of the duty cycle only, the current is 320 mA and the voltage of the batteries is 62 volts?

YES, or NO.

Quote from: TinselKoala on May 09, 2012, 12:33:38 PM

Ainslie, do you accept that during the ON time of the duty cycle only, the current is 320 mA and the voltage of the batteries is 62 volts?

YES, or NO.

YES.  UNEQUIVOCALLY

added

Quote from: Rosemary Ainslie on May 09, 2012, 12:29:31 PM
IF by that 'spread sheet stuff' you're referring to integrated power analysis - then INDEED.  We use multiple representative samples to calculate the amount of energy delivered over a period of time.  When we've established a fair sample and when we've computed the power from EACH of those samples then we DIVIDE THAT SUM by the NUMBER OF SAMPLES.  Then - and ONLY then  - and INDEED THEN - we have the ACTUAL WATTAGE.

What you're trying to argue is that 1/6th of the ON time is a REPRESENTATIVE SAMPLE.  It is NOT.  It is only representative of 1/6th of a duty cycle.  IF you DO NOT factor in that 1/6th PERIOD then you are also NOT giving an accurate representation of WATTAGE.  You are giving us something that is not comparable to anything at all.

Again and this is getting tedious in the extreme
Rosie Pose

No, Ainslie, that is NOT AT ALL what I am arguing. NOBODY said anything about "representative sample." WE ARE TRYING TO DISCUSS THE ON TIME ONLY because you made a specific claim related to that ON time only.

YOU CLAIMED that the current during the on time was minuscule, and we have shown that it is NOT, and in fact accounts for considerable power. YOU have complicated and obfuscated the issue by your hallucinations and misrepresentations like the above, because you desperately need to cover up the fact that you are WRONG in your contentions. Note the claim about the ON time current in your blog posts below. YOU ARE WRONG, by your own data.

Quote from: Rosemary Ainslie on May 09, 2012, 12:12:46 PM
Guys, picowatt and TK included,

Here's the thing.  IF one determines the wattage delivered or dissipated - one takes the sum of a fair sample range - and then one divides that sum by the number of samples in that range.  That gives you a 'fair' average - which represents the WATTS delivered - with that number being 'representative'.  In effect it IS an average of a sample range of voltage measurements that DO, in fact, represent an INSTANTANEOUS power measurement.  But the qualification is that it includes a full and representative sample range before one can arrive at that the required number.  Then.  That AVERAGE value is the ACTUAL rate of wattage.  And that's then a product represented as a second - over the TIME that the energy has been delivered or dissipated - to give the Joules or POWER.  Which ALWAYS incorporates TIME.  One small representative sample in one small section of one part of the duty cycle is NEVER, EVER, a WATT.  Nor can it be called a WATT.  Then.  The product of this number is applied to the value over time to represent the WORK that has been performed related to Joules.  The instantaneous power calculated in 1/6 of a duty cycle is NOT representative of WATTS.  It is merely representative of instantaneous power within a small fraction of the applied duty cycle. 

Those them are the facts guys.  And no amount of invective or protest is likely to alter those them facts.

Regards,
Rosie

And again, what is this the formula for?   (V)x(A) =??

Quoting you,

"it is merely representative of instantaneous power within a small fraction of the applied duty cycle"

Correct, but what is the unit of measurement used for that instantaneous power??

I agree with your statement regarding average power dissipation, however, during the portion of the cycle wherein the FG is positive in that 'scope shot, 20watts is being dissipated.

PW