Quote from: Rosemary Ainslie on May 09, 2012, 01:48:53 PM
OK - here's the thing.

IF the battery was delivering 20 watts and that was to be taken as a representative sample of the energy delivered by the battery then it would need to be delivering that amount of energy continuously - for the duration of each duty cycle.  Then - without interruption - there would be 20 watts delivered by the battery during the ENTIRE duty cycle.

HOWEVER.  It does not.  I delivers 20 watts during 1 sixth of the cycle.  Then it delivers NOTHING during another 5/6ths of each cycle.  Assume then that in that brief 1/6th of that specific period of time it was heating the element to the full extent of all of those 20 watts.  Then during 5/6ths of that same cycle when NO ENERGY was being delivered then it would, self-evidently be COOLING DOWN - because there was and is no further application of heat.  Then it again delivers 20 watts and then cools down - and so on.  In order to determine the ACTUAL energy INCLUDING the period when that energy transfer was interrupted - it would need to be determined over the time period when it was both ON and OFF.  Which is closer to 3.33 watts.   

That's it folks.  That's the whole of TK's argument which is PATENTLY and BLANTANTLY wrong.  And if you've bought into this EvolvingApe - then you too have been duped.  And this is YET another example of TK's misdirections.  Sadly - I think he's convinced himself that he's even got an argument.  More's the pity.  But it's certainly in keeping with this thread standard.

Rosie Posie

OK, HERE'S THE THING.

NOBODY BUT YOU is talking about AVERAGE POWER AT ALL.

The point is that YOU CLAIMED that there was negligible tiny current flowing during the ON PHASE. THE ON PHASE. THE ON PHASE.

And I have demonstrated that YOUR CLAIM IS WRONG and the CONCLUSION BASED ON IT IS FALSE.

DURING THE ON PHASE. DURING THE ON PHASE. Can you not understand English?

There is 320 mA, AT LEAST, flowing during the ON phase. AND THERE IS ALSO CURRENT FLOWING DURING THE OFF PHASE. Your load is heating because it is dissipating some power during BOTH THE PHASES in that scope shot. BOTH ON AND OFF PHASES are giving you current.

However, I am content for the moment to have PROVEN YOU WRONG, by your own admission, about the current flowing DURING THE ON PHASE. DURING THE ON PHASE.

Which is what we were discussing prior to your MISINFORMATION, MISDIRECTION, MISUNDERSTANDING, and MENDACITY. Not to mention your profound disrespect for your betters and their education and experience, which you do not even come close to fathoming, much less sharing.

And here's where her 3.33 figure comes from. Look at the period. Take the "on" portion and look at it relative to the OFF portion. The On portion is about 1/6 the Off portion. So Ainslie thinks the duty cycle is... what?

What is the duty cycle, Ainslie? Is it 1/6 ON?

20 Watts times 1/6 is indeed an average of 3.33 Watts. Isn't it.

QuoteI (sic) delivers 20 watts during 1 sixth of the cycle.  Then it delivers NOTHING during another 5/6ths of each cycle.

See what happens when you just punch numbers into a calculator without understanding their meanings or units?

Quote from: evolvingape on May 09, 2012, 01:30:26 PM
http://en.wikipedia.org/wiki/Scientific_misconduct

Data sharing

"Kirby Lee and Lisa Bero suggest, "Although reviewing raw data can be difficult, time-consuming and expensive, having such a policy would hold authors more accountable for the accuracy of their data and potentially reduce scientific fraud or misconduct."


In the spirit of open source I am convinced Rosemary would be willing to release the hundreds and hundreds of scope shots that she has stored in her private collection, never made available for public review.

RM

Rosemary,

I have been following this discussion since this thread was started. The quality has been astounding, we are clearly dealing with professional analysis here from time served experts who have learned their craft over many years of study and practical application. (Not including you in that overview). The only person who seems to not grasp the basic's of the content under discussion is you. TK's argument is clear and concise, and also correct. I do not need you to advise me of potential duping, and rest assured, should I require advice on such a matter you are most definately not on the list of knowledgeable people to approach for a qualified opinion.

Please advise me how I can access your full and complete raw data, for analysis, I wish to study it. As you are 100% committed to open source this should not be a problem, and I expect you can fulfill this simple request promptly.

Thankyou,

RM

Quote from: evolvingape on May 09, 2012, 01:30:26 PM
http://en.wikipedia.org/wiki/Scientific_misconduct

Data sharing

"Kirby Lee and Lisa Bero suggest, "Although reviewing raw data can be difficult, time-consuming and expensive, having such a policy would hold authors more accountable for the accuracy of their data and potentially reduce scientific fraud or misconduct."


In the spirit of open source I am convinced Rosemary would be willing to release the hundreds and hundreds of scope shots that she has stored in her private collection, never made available for public review.

RM

Now you can understand why Ainslie refuses to do math on demand. EVERY POST of hers that I can find that includes any calculation with more than two variables and one operation is almost CERTAIN to contain one or more egregious errors, and SHE KNOWS IT.

So she dares not engage in any mathematical discussions. Just as in the present example, she always reveals her incompetence and ignorance.

The duty cycle is closer to 1/8 than to 1/6, Ainslie, because it is the RATIO of the ON time to the TOTAL PERIOD TIME. But you would need to understand ratios and proportions to be able to grasp that.

Quote from: TinselKoala on May 09, 2012, 02:10:29 PM
And here's where her 3.33 figure comes from. Look at the period. Take the "on" portion and look at it relative to the OFF portion. The On portion is about 1/6 the Off portion. So Ainslie thinks the duty cycle is... what?

What is the duty cycle, Ainslie? Is it 1/6 ON?

20 Watts times 1/6 is indeed an average of 3.33 Watts. Isn't it.

See what happens when you just punch numbers into a calculator without understanding their meanings or units?

I agree with you that the ON period of the duty cycle is - in fact - closer to 12.5% ON.  In which case that 20 watts dissipated that you calculated needs to be revised downwards to 2.5 watts - is the first point.  NOW.  IF that 20 watts is then NOT applied for the remaining 87.5% of an ensuing period of each duty cycle - HOW THEN DOES THAT ELEMENT RESISTOR CONTINUE TO DISSIPATE 20 watts?  IF our STANDARD PROTOCOLS are even half way correct the assumption is that it does NOT dissipate energy if there is no energy applied.  THEREFORE?  THAT 20 Watts becomes considerably less than 20 watts.  In fact it's reduced by 87.5% less than 20 watts - IN FACT.  Therefore, 20 watts IS NOT dissipated at the element resistor EVER.  ONLY 2.5 watts is dissipated.  That's based YOUR argument.  And I AGREE.  WHOLEHEARTEDLY.   

IF, however, you are arguing that 20 watts is CONTINUALLY BEING DISSIPATED despite the removal of energy during 87.5% of each period of each duty cycle - then I DO NOT SUPPORT YOUR ARGUMENT.  OR ANY PART OF IT.  And nor would any self-respecting power engineer.  Therefore - IN FACT - at no stage is there the dissipation of 20 watts of energy at that element resistor.  OR - if there is - THEN it is NOT coming from the ON period of the duty cycle.

Quite apart from which Leon,  we've measured 20 watts of energy applied to the that load - in detailed schedules.  And the temperature at 20 watts is NOWHERE NEAR to the 200 degrees centigrade that we measure as heat on that resistor in our water to boil test.

Rosie Posie