Claimed OU circuit of Rosemary Ainslie

[ramset]

I hope your done [for the moment]
Look up please

[henieck]

Rosemary from energeticF.:

”Then comes the tricky part. The duty cycle changes. The battery is, in effect, no longer able to deliver any current. The fields on the resistor collapse to zero. And the strength of that collapse relates to the applied energy from the On cycle. Energy is energy. What comes in must go out.”

- can anybody ask her please how she explains the fact â€" that when I measure the coil â€" it conforms to the Ohms Law and there is never any extra energy entering the coil over what Ohm’s Law “permits” â€" therefore magnetization of the coil gets for free. In other words: both an ordinary resistor and her resistive coil conforms to the Ohm’s Law â€" so the magnetization was suppose to be for free! From zipoons!

This is getting really hilarious now. Hey, what happened to the zipons and the whole theory?

Rosemary is clearly making some progress â€" she doesn’t say anymore that the coil gets magically energized by itself or by zipons There was a big argument about that with me. She maintained just few days ago, that while measuring the resistor- it conforms to Ohms law â€" and that since there is no additional energy above that Law entering the coil = therefore the magnetization was far free. So there is a huge change in her brain taking place - but somehow it wasn’t sufficient to destroy the whole concept yet… That was the main base for the entire concept of free energy. It magical magnetization was the place where the free energy was suppose to magically enter the system. Now she is changing her model somewhat. Fascinating…Talking about being aware of one’s own processes â€" I wonder whether she is aware of her last change. Normally when somebody realizes that reality differs form the beliefs gets anxious. Maybe this is why she had to walk the dog today…;)

Rosemary:

„It takes full advantage of that 'moment' and developes a spike to carry the full force of the energy applied in the On time and kicks it back in nano seconds as a 'spike' that is always evident 'beween' the on and off period. That's the counter electromotive force.”

- it is always good to know something new. It is about the other CMF â€" collapsing magnetic field, it is not the same as BEMF. Counter electromotive force fights back the applied current- energy from collapsed magnetic field occurs when current stop to flow. http://en.wikipedia.org/wiki/Counter-electromotive_force


Rosemary:

“We don't have to go into too much detail about the 'inductive component' as the inductive componenet here is in the resistor itself. Whatever it measures as 'heat' WILL also reflect the energy that was delivered.

- this is fundamentally false. It even contradicts the previous statement from the same letter â€"“And the strength of that collapse relates to the applied energy from the On cycle”. The way she thinks does not cease to amaze me. She clearly uses the unique human mind’s feature that it can believe in two contradictory statements at the same time : all energy delivered goes into heat â€" and:  “the strength of that collapse relates to the applied energy from the On cycle. Energy is energy. What comes in must go out.”



Rosemasy:

“But that energy is evidently also measurable on the shunt. And where does it go? It goes first through the battery, thereby recharging it - and then to the load thereby heating it up.”

- you have 10$ in the battery, take it and put into the resistive coil. Let us say 2$ turn into heat and 8$ turn into magnetic field (what is the exact ratio depends upon the resistive coil characteristics). When all 10$ are spent then the switch opens. What is heat â€"is not mutually convertible back to electricity in this setup-  but due to the magnetic field collapse you can have all the 8$ back in electric form again. This goes into the battery. End of the “off“ cycle. Therefore there is less in the battery, but taking additional 2$ from the battery you can make the new cycle again. What is tricky is that normally without the feedback the circuit would only take 5$ during on cycle (4 recoverable and 1$ for the heat). The circuit tries to automatically compensate for any load â€" so when you connect the battery â€" it takes all the collapsing energy â€" and as much you have to give in addition on the front end during the next cycle.

Now if you think about heat as a taxation (for poor conductivity) â€" that it is obvious that there is more tax form 2$ than from 1$. (one resistor hotter than the other one).This is how I see it…

[ramset]

henieck
http://www.metacafe.com/watch/2820531/don_smith_free_energy/

You asked whats new 

Also   

RAMSET 
An engineers opinion  [TK}
Aaron with reference to your question and diagram: imagine a voltmeter or oscilloscope connected at the mosfet drain. Or at the top of the resistor where it connects to the positive rail. When the gate signal is ON and the mosfet is conducting, what is the voltage at these points? When the gate signal is OFF and the mosfet is NOT conducting, what is the voltage at these points

When the mosfet is conducting the voltage at those points is LOW, not high. When the voltage at those points is HIGH the mosfet is not conducting, it is off. Joit's trace shows the voltage going HIGH at the mosfet drain for short periods. The transistor is OFF at these times. just hook up a bulb like I have done. Try it!


.
Slow the freqs down, and think about what you are seeing. Carefully.

Aaron the drawing is fine, and yes, as shown the mosfet STATE (on or off) clearly exactly follows the gate drive state: ON is ON, for sure. But that's not the issue: the issue is HOW LONG it's on, and how that on time is measured. If you are looking at the Point A in Ainslie's circuit or the drain of the mosfet in your diagram, the VOLTAGE that the scope is measuring--what it uses to give the duty cycle figure...that voltage is HIGH when the MOSFET ( and the light) IS OFF. Use that exact circuit and put a meter in the exact place Rosemary does. Push that button and tell us what the voltage is on your meter.


Spiritual Entrepreneur
    @@Aaron said
simple question
Ramset,

That is right but that does not answer the specific questions.

A closed switch is like taking a voltmeter and putting both leads next together on the same wire, there will be no potential difference even though current is flowing and there is voltage moving. That much is common sense. Actually, there will be small milivolt reading because there is a small potential difference but for all practical purposes, there is no voltage.

If the DRAIN has a 3.7% duty cycle, IS IT or IS IT NOT conducting current from a power source for 3.7% of the time per pulse?

It really is a simple yes or no question.



The ball is in your court TK

[henieck]

ramset, I can see the both threads now, thanks. I just can’t send posts there.

[TinselKoala]

Thanks, chet.  Aaron's question has an easy answer: NO. When the drain signal is at battery voltage (high) the mosfet is OFF and the load is non-conducting. Clearly. You can wire up that circuit, using a real bulb instead of an LED, and if your voltmeter is sensitive enough to pick up the small voltage drop, you will see it. What looks like a big signal on the scope traces is really a very small fluctuation sitting on top of a large DC offset---the battery voltage. This is why 'AC coupling' must be used to resolve it. If you use DC coupling, if the screen shows a line at battery voltage the fluctuation will be too small to resolve--the voltage doesn't drop very much when the mosfet is ON. But it drops, for sure. If an oscilloscope is relatively dumb, like the Fluke 199, it needs to be TOLD whether you are calling the "high" signal "OFF" or "ON" to give a duty cycle output. There are 2 separate controls in the FLuke that must be set properly: the trace invert function AND the duty cycle definition. Only ONE of the Four possible combinations of these controls is correct in this experiment.
Fortunately the LeCroy is smarter than that. And fortunately the Philips analog scope does not even have a trace invert function, and it is a true 2 beam scope, so no fancy fiddling is done with the signal, as in the 2-trace (not 2 beam) Tek 2213 scope, which does have a Ch2 trace invert function.

@Chet: thanks for posting those things over there. It's pretty ridiculous I know. Re Zoltan: I've known about him for some time now. It appears to me that he too does not understand that energy is the area under an instantaneous power curve. But I have not tested his circuit, nor do I have plans to.
Re Don Smith...him too I have been aware of for a while.
In the past I have tried to stay out of the electric and electronic OU discussions unless I see some obvious error, or an incredible claim, or both at once like here. It's not really my field (although I will pit my knowledge against the likes of Ainslie, Joit, and Aaron any day any where) and I don't like to get entangled doing remedial EE101.
I will take a look at those things you asked me to, later on. I just have too much going on right now to give it proper attention. For example, I mistakenly put the coffee pot in the refrigerator this morning.
Keep up the good work. I hope you don't wind up getting tarred with the same brush they are chasing me with.
--TK