I am certain it is the "Admin" login (Rose) that has been deleting posts other than the ones Greg deleted.

The "users online" viewing has been disabled for all regular members Glen. Same with PM's.

Well, there you have it then. Not content with being banned from all other forums, she decides to kill her own forum as well, by deleting the only posts and thread of significance, and blocking non-members and members alike from seeing the relevant stats. Meanwhile the two active posters seem to be both of them rather paranoid and excitable types.... I'm sure they are feeling right at home with Ainslie.

All those months of work, on my part with Tar Baby, on .99's part with sims and the careful patient explanations, poynt by gentle poynt.... picowatt's analyses, his slow boat from Singapore with his mosfets, Glen's reviews of past work, the efforts and comments of our "lurkers" and other contributors.... all of it down the tubes. Ainslie will just wait for the dust to settle, and then she'll pop up again in some other venue, lying about her "proof" of overunity and suckering another bunch of sincere and hopeful idiots into doing her work for her.

You forgot to mention the "three readers of this thread". Oh, that was the other thread. ;]

Not all down the tubes. What you guys have done is pulled many of us out of the sinkhole we were sucked into. She is now isolated. No readers, and 1 member that is mostly posting about other technology and the other, he just might be Rose's long lost twin. They both lie, cheat, and, lie. lol  like they are connected in some way.

Hey, I cant see her ever being able to present anything here again. Nor some other places.

The word is out. ;] And the word is strong.

MaGs

Hearing things over and over and over and over...
Edited to substitute a link to the animated .gif
surprised I can make this change, after this much time.
http://imageshack.us/a/img21/9339/brokenrecordm.gif

TK,

Yeah I am aware of those 'admissions', as ambiguous as they are. But ultimately he never admitted that the correct Pin is 10.5W for his circuit, and that is the bottom line. Rose agreed with and duplicated his erroneous computation, therefore she too is incorrect.

Ainslie said,

Poynty - I KNOW what you're getting to.  Who would NOT?  Here's the thing...

          5 Amps - 5 minutes
          0 Amps - 5 minutes
          ________________
          = 0.25 Amps per every 10 minutes.

<blockquote>

.99 said:

Correct Rose. You agree then that the average current is 0.25 Amperes. And you would also agree that the average current would be 0.125 Amperes if the switch was ON for 2.5 minutes, and OFF for 7.5 minutes, correct?

And I don't get this. How do you arrive at 0.25 amps as an average current here, if your input current is 5 amps at a duty cycle of 50 percent?

Ah... I see it now. It's another "mindreading" typo error from Ainslie.... she should have "0.5" amps where she has "5" amps. This is why LEADING ZEROS before a decimal point are indeed important. You read her mind successfully and realized she meant one-tenth of what she actually wrote. Silly me.... I just see what's actually written, and I see that 5 amps at a duty cycle of 50 percent gives me 2.5 amps average, not 0.25 amps.





</blockquote>

Ah, how we complicate things and go astray.

You can have input _voltage_ and still have zero input _power_, because the power is determined by the current that the voltage is able to push through the circuit. This is where the improper mental models conflict with reality and cause the naive and inexperienced to wander blindly down deadend paths in darkness and despair, trying to get a zipon-powered overunity flashlight to turn on.

You are an humble grad student tasked with monitoring the input power to a black box. You have a voltmeter and an ammeter hooked up between the always-on, regulated power supply set to 10 volts, and the black box sitting over there on the lab bench. And you have a clipboard, a pencil, and a stopwatch. Since you are a very fast writer, you record V and I values once per second. And your record for the first ten seconds looks like this:

1   10 V   10 A
2   10 V   10 A
3   10 V   10 A
4   10 V   10 A
5   10 V   10 A
6   10 V     0 A
7   10 V     0 A
8   10 V     0 A
9   10 V     0 A
10 10 V     0 A

Got the picture, gmeast?  The average power is the average voltage times the average current. The average voltage is 10 volts -- the peak voltage times the duty cycle OF THE VOLTAGE, which is 100 percent,  and the average current is the peak current times the duty cycle OF THE CURRENT. When you compute the average power here, you do NOT then come back again and divide by 2 for the "duty cycle" of 50 percent.


Doing the problem EXACTLY as gmeast has done above, except using the current as given:

The Frequency  = 0.1 Hz (one pulse or full cycle in ten seconds is 0.1 cycle per second or 0.1 Hz)
V = 10 Volts
A = 10 Amps
Duty cycle  = 50 %

1/(2 x 0.1 Hz) = duration of a single pulse = 5 seconds
10 V x 10 A = 100 Watts per pulse
100 Watts per pulse x 5 seconds = 500 Watt-seconds per pulse

There are 0.1 pulses per second, so
(500 Watt-seconds/pulse) x (0.1 pulse/second) = 50 Watts.
In the latter calculation I have essentially added together all the identical packages of instantaneous power but they are spaced apart over time.

(Now.... do I stop here, or do I now divide AGAIN by 2, since I have a duty cycle of 50 percent? Gmeast proceeds to apply the duty cycle again, as I have shown in the image above.)

Since I have a duty cycle of 50 percent, my instincts tell me to divide again by 2, giving me an average input power of 25 Watts.


Which is of course wrong. But that's the way gmeast thinks it should be done, and even Ainslie knows better than that, I think.

http://www.youtube.com/watch?v=SCxypoN8-xc

Note: at about 3:00 I say "fifteen hundred kiloHertz" when I actually should have said "fifteen hundred Hertz" or "one point five kiloHertz". The FG is set to 1.5 kHz for this stage of the demo.