Testing the TK Tar Baby

[TinselKoala]

Here, maybe this will help to explain to Ainslie just where gmeast's real error is, as well as illustrating that he never acknowledged nor corrected this major error.... he only corrected the minor one of using "3" instead of "4" to represent a 25 percent duty cycle in his (erroneous) second application of the duty cycle to his input average power value.

Of course, since he's thrown out the method altogether, he is now at odds with Ainslie herself, apparently, as well as the rest of reality.

This is a screenshot of his post, now vanished, where he first makes the erroneous calculation arriving at a small average input power to give a large OU result. My comments are in color, his statements and calculations are in black.

[TinselKoala]

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.

[poynt99]

Indeed.

[TinselKoala]

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.

[mrsean2k]

Historical inspiration for the Rosemary Ainslie / GMEast dialogues.


http://www.youtube.com/watch?v=jY-PEeX5xYY