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Overunity Machines Forum



Testing the TK Tar Baby

Started by TinselKoala, March 25, 2012, 05:11:53 PM

Previous topic - Next topic

0 Members and 148 Guests are viewing this topic.

TinselKoala

Quote from: Rosemary Ainslie on May 09, 2012, 12:35:55 PM
YES.  UNEQUIVOCALLY

added
What, then, is the power during that time, unequivocally?

What are you arguing about? PW and I are talking, and have been talking, about the ON time exclusively. And that is because we are examining the veracity of a claim YOU MADE ABOUT THE ON TIME CURRENT in those blog posts above.

A claim that you have just now admitted is wrong.

Quote...indicating that the current flow from the battery DURING THE ON TIME was next to nothing.

320 mA at 62 volts is NOT "next to nothing".

Rosemary Ainslie

Quote from: picowatt on May 09, 2012, 12:38:26 PM
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

Actually TK you are STILL wrong.  You are talking about the POWER that is being dissipated at the element resistor.  You CANNOT separate that POWER from its context.  You CLAIM that there is an instantaneous power dissipation based on 20 watts determined by the current flow and the voltage - V x I.

NOW.  Any determination of the ACTUAL wattage delivered or dissipated is a determinant of the POWER.  And POWER IS DEFINED AS DETAILED IN THE DOWNLOAD FROM WIKI. THAT sum requires that average over time - and THAT'S the ONLY WAY TO DETERMINE WATTAGE. 

Rosie Pose

powercat

Quote from: PhiChaser on May 09, 2012, 01:10:26 AM
They don't need my help, you can tar yourself just fine Rosemary.
In fact, every time you refuse to engage these guys on a technical level you tar yourself.
Every time you slander TK about his videos you tar yourself.

All anyone has to do is ask you to do a Dim Bulb Test and you slip slide away down Zipon Avenue and mutter about sauce.
And you again tar yourself...
If that doesn't work they can ask you about power calculations but you seem to have found the loophole there by 'putting the witness on trial' as PW put it.
And still you continue to tar yourself... 

You really do believe that you are right though, don't you?
After seeing all the evidence and data and discussion amassed here regarding negative power mean circuits that YOU STILL believe you're right and they are wrong?!?
That is some serious denial there.
Like EPIC denial.
Bigger than that river in Africa even...  :o
Don't go away though Rosie, we all like your company!
Always good to have a 'free thinker' in the group. 
;D

Regards,
PC

Yes this is ridiculous she is completely nuts, their are no "Guys" their haven't been any since the beginning of the thread,  NO supporters
So despite the overwhelming evidence why is she the only one in denial ?

Delusional disorder
http://www.webmd.com/schizophrenia/delusional-disorder
QuoteDelusional disorder most often occurs in middle to late life and is slightly more common in women than in men.
A person with this type of delusional disorder has an over-inflated sense of worth, power, knowledge, or identity. The person might believe he or she has a great talent or has made an important discovery.

When logic and proportion Have fallen
Go ask Alice When she's ten feet tall

TinselKoala

Quote from: Rosemary Ainslie on May 09, 2012, 12:45:04 PM
Actually TK you are STILL wrong.  You are talking about the POWER that is being dissipated at the element resistor.  You CANNOT separate that POWER from its context.  You CLAIM that there is an instantaneous power dissipation based on 20 watts determined by the current flow and the voltage - V x I.

NOW.  Any determination of the ACTUAL wattage delivered or dissipated is a determinant of the POWER.  And POWER IS DEFINED AS

(http://upload.wikimedia.org/wikipedia/en/math/c/6/f/c6f4fcdcefb75e1417669dfbc912d208.png)THAT sum requires that average over time - and THAT'S the ONLY WAY TO DETERMINE WATTAGE. 

Rosie Pose

YOU LIE AGAIN AINSLIE. I have said that that power is dissipated IN THE CIRCUIT ELEMENTS, not the load resistor. When will you learn to stop putting words in people's mouths that they did not say?

And you really should stop digging yourself in deeper. You still have no concept and you don't even understand the WIKI that you are parroting.



Rosemary Ainslie

Quote from: TinselKoala on May 09, 2012, 12:40:41 PM
What, then, is the power during that time, unequivocally?

What are you arguing about? PW and I are talking, and have been talking, about the ON time exclusively. And that is because we are examining the veracity of a claim YOU MADE ABOUT THE ON TIME CURRENT in those blog posts above.

A claim that you have just now admitted is wrong.


320 mA at 62 volts is NOT "next to nothing".

AGAIN.  You are trying to IMPLY that watts can be separated from POWER.  IF you are trying to determine the amount of energy dissipated at the load resistor then you are REFERENCING A POWER MEASUREMENT.  This is ALWAYS BASED ON WATTS and those WATTS NEED TO BE DETERMINED OVER TIME.  AGAIN.  Here's that equation.

Rosie Pose