The Rosemary Ainslie Circuit

[MileHigh]

I just read the last couple of pages of the OU thread and I want to curl up into a ball.  You've got Magluvn who was more or less on the right track in understanding how a discharging coil works losing it and getting the over unity giddies.  Rosemary looks at his comments about Woopy's little demo clip and demonstrates that she still hasn't the slightest clue herself.

This is all very Salvador Dali and I am going insane!  lol   

[humbugger]

All along, I thought the extra energy was supposed to be from the slow but sure breakdown of the load resistor material.  

Yes, but keep this key bit of Rosemary's advice in mind whenever you think about the load resistor and it's physical arrangement:

Whatever else you use - try and keep a threaded rod number of sorts.

I swear to God that is a direct quote from a post over on that other forum with her giving Neptune advice on the load resistor for his replication.  There was no explanatory context.  Poor Neptune...

Humbug

[poynt99]

Has anyone yet been able to figure out how this shunt voltage measurement can be showing 0V while the FET is ON?  

.99

[humbugger]

Yes, I think so.  If you look closely, I think you'll see that the gate voltage is very close to zero or at least below 3~4 Volts during the so-called MOSFET "on" time.  In other words, the MOSFET is not on at all.  Apparently, the oscillation tendency is killed at the first hint of MOSFET enhancement.

This also clearly explains Rosemary's two "modes", the low power one (shown here) where there is no ON time really and she gets 5W in the load and the 44W high power mode where half the time the MOSFETss are on and half the time they oscilate.  Both of our sims show this second higher power mode, so we see the current draw during the actual on time.

The shoe fits quite well.  If you go back and look at Rose's various scope shots, you'll see that many of them show positive shunt current during the non-oscillating periods and some don't.  She mixes them in randomly, giving no clues.  Probably because she has no idea that's why she sees (but lies about and twists) positive battery drain during the higher power modes and nothing but noise (which she interprets as battery charging) during the "on" times of the low power mode where she has adjusted the sig-gen offset such that the high gate drive just bearly touches Vgs ON.

Hum

[poynt99]

It does look as if the Gate voltage is slightly negative, so I would agree. Very odd gate wave form there. I think our models are incomplete, but I don't really want to spend too much more time on trying to perfect the wave forms...it could take quite a while, and I have already tried a few different things with no luck.

I suppose the Ainslie circuit topic will be quiet for a while until someone else replicates it. The scopes are returned, and my impression is there will be no further testing by Rose.

.99

[humbugger]

Yeah, the thread over there seems to have been taken over by Magluvin with his "new" revelations about "FEMF" as MH was saying.

The MOSFET gate doesn't have to be negative, you know of course, just below the Vgs threshold, usually 3-4V positive.  This stuff is what I was trying to show you in post #314 of this thread with the 7.2V offset you kept saying was meaningless.  So I gave up.  See also posts 316 (yours) and 317 (mine).

Hum

[poynt99]

In that post, #314, I see that the shunt indicates current, and that the FET is definitely ON. (I still see no significance of the 7.2V offset, as I have the little tick to the left side to tell me where the zero reference is, and that is all that is important. I can count graticules from that zero ref point, and I don't need to know that it is offset 7.2V from the scope zero reference line)

Did you happen to notice the Math vertical setting? 10kW, with a 30.8kW offset! There is a mighty big but thin spike there if you look closely. The mean of the math shows 41.5W. Hmm,,,, what does Rose say about that input power?

.99

[poynt99]

I guess she says, "yeah, but it's negative 41.5W!"

.99

[poynt99]

Here is an observation, again referencing the scope shot in #314:

Battery mean = +73.3V
Shunt mean = +179mV

That is a positive input power of 13.12W. That means it is source power, not re-charging power.

.99

EDIT:

Correction, since the shunt is a 0.25 Ohm (not 1 Ohm), the input power actually calculates to about 52.5W.

[humbugger]

Maybe you knew where zero was before i drew my long line through the little mark but then why did you keep saying you couldn't understand why the current was zero (and Stefan and Rose kept saying that too, regarding that picture).  That's what frustrated me.

Also, please note that what looks like  a big W is not a W at all and does not mean Watts.  It is VV or V*V, meaning Volts times Volts...the math multiply trace.  How would the scope know that one of the traces was from a current shunt and represented current?  How would it know what the shunt value was to get a number in Watts?

I suspect that the LeCroy is not multiplying the averaged values of the two traces, but is multiplying the point-by-point samples and then taking the maen of that.  That's why there is disagreement between the product of the mean values shown for each trace and the math mean.  As far as measuring the input power, with a virtually DC voltage in this case, the old-fashioned way of just getting the mean for the shunt voltage, factoring the shunt resistance in and multiplying by battery voltage is the right way to do it, I think.  When she's in the high power mode, doing this with her shunt violtage mean gives the same results I got: 50W input power for 44W output heat.  That's based on 75% efficiency half the time (oscillating) and 95+% efficiency the other half, while the MOSFETs are actually turned on.

I posted all this over there at OU and I think I posted it here too...not sure.  Stefan has been holding my posts for days but finally putting them up, much to Rosie's dismay.  He even finally posted that long one I copied here that you said he wasn't going to put up.  He just did that today, in fact.  Weird guy, that Stefan.

Hum

[humbugger]

Here is an observation, again referencing the scope shot in #314:

Battery mean = +73.3V
Shunt mean = +179mV

That is a positive input power of 13.12W. That means it is source power, not re-charging power.

.99

Yeah..the shit is all over the map when Rosie reports it.  I'm sure she would have said the load power was 40W or 400W or something if she was forced to accept a positive battery draw,

The big demo part 2 where she shows the load getting 44W clearly shows 50W being pulled from the battery if you multiply the 243mV shunt average (positive) and the 50.3V positive battery voltage but she prefers to quote the V*V number which showed -5.43.

She and her team also obviously think that VV is a W and means Watts.  Common sense and reading the manual and knowing that the scope has no idea one trace represents current nor any idea of the shunt value would preclude the scope from calculating Watts.

The scope math V*V trace and reporte dnumber are just totally off base obviously.  It gives a negative number when both the arguments means are positive numbers.  Rosie's entire "proof" is based on this one glaring misunderstanding and gross scope math error in every case.  She and her team and 99% of the readers have no basis in scope use knowledge nor common electronics sense, as you know.  This has been an extreme example of what can happen when monkeys with theories get hold of instruments far beyond their ability to understand or use properly.

Thee way she displays the scope traces is the reason.  If she were to have made properly-sampled measurements on a single zoomed-in and well settled typical oscillation cycle trace, the answers would be entirely different.  As it is, her sampling is way undersampled and therefore, along with the wild Vbatt singing of 100V or more and the shunt inductance and all the associated phase shifts...the V*V numbers are totally meaningless.

HUm

[poynt99]

Maybe you knew where zero was before i drew my long line through the little mark but then why did you keep saying you couldn't understand why the current was zero (and Stefan and Rose kept saying that too, regarding that picture).  That's what frustrated me.

Of course I knew where zero was, the tick shows you this. Use the reported offset number if you wish, but it is not necessary, and furthermore, it will not always be there. They could have accomplished the very same thing just by moving the trace vertical with the vertical position dial.

Also, please note that what looks like  a big W is not a W at all and does not mean Watts.  It is VV or V*V, meaning Volts times Volts...the math multiply trace.  How would the scope know that one of the traces was from a current shunt and represented current?  How would it know what the shunt value was to get a number in Watts?

It does mean watts, if you take into account the value of the shunt resistor, which in this case is 0.25 Ohms. So take the reported VV and multiply by 4x.

I suspect that the LeCroy is not multiplying the averaged values of the two traces, but is multiplying the point-by-point samples and then taking the maen of that.

Yes, that is correct.

When she's in the high power mode, doing this with her shunt violtage mean gives the same results I got: 50W input power for 44W output heat.  That's based on 75% efficiency half the time (oscillating) and 95+% efficiency the other half, while the MOSFETs are actually turned on.

I recalculated my input power above, as I forgot to take into account the 4x multiplier, and it comes out to 52.5W, which is close to what you said.

.99

[humbugger]

Yes sir ee

By the way, after Stefan Hartman chalked up the -5.43VV error to the offset messing up the math (which I knew was wrong but said nothing lest I be banned again) I went and downloaded a copy of the LeCroy manual for that scope.

The "Position" control is labelled "Offset" and has no bearing on the math at all.  There is no seperate means of adding offset.

Hum

[poynt99]

Yes sir ee

By the way, after Stefan Hartman chalked up the -5.43VV error to the offset messing up the math (which I knew was wrong but said nothing lest I be banned again) I went and downloaded a copy of the LeCroy manual for that scope.

The "Position" control is labelled "Offset" and has no bearing on the math at all.  There is no seperate means of adding offset.

Hum

The Tek scopes have an entry for offset (not the same as vertical position dial) and so I assumed the LeCroy did also. Guess I assumed wrong.

.99

[GibbsHelmholtz]

The big demo part 2 where she shows the load getting 44W clearly shows 50W being pulled from the battery if you multiply the 243mV shunt average (positive) and the 50.3V positive battery voltage but she prefers to quote the V*V number which showed -5.43.

MEAN[bat] * MEAN[shunt] does not necessary means MEAN[bat * shunt]

average batter can be positive, average shunt can be positive, average power can still be negative.

[humbugger]

MEAN[bat] * MEAN[shunt] does not necessary means MEAN[bat * shunt]

average batter can be positive, average shunt can be positive, average power can still be negative.

Oh, really?  Please describe or show one example of a DC battery with a positive mean current being drawn where the power is negative (net charge to the battery).  I'll be waiting with bated breath.  And it's a good thing too because I just ate a garlic pizza.


Humbugger

[GibbsHelmholtz]

Oh, really?  Please describe or show one example of a DC battery with a positive mean current being drawn where the power is negative (net charge to the battery).  I'll be waiting with bated breath.  And it's a good thing too because I just ate a garlic pizza.


Humbugger

If I have the mojo to say it, then I must have a back up plan.  What's in it for me?  How about this.  I show you, you do a sim for the prof. 

[Allcanadian]

@Humbugger

When she's in the high power mode, doing this with her shunt violtage mean gives the same results I got: 50W input power for 44W output heat.  That's based on 75% efficiency half the time (oscillating) and 95+% efficiency the other half, while the MOSFETs are actually turned on.

If standard theory holds then the circuit must be 100% efficient as all the input electrical energy would be converted to heat somewhere in the circuit.
I think the single biggest issue in the forums in regards to interpreting oscilloscope traces is that people are trying to interpret an inductive discharge just as they would an AC waveform which is a mistake. When we see an AC waveform the positive voltage above the zero boundary shows the forward direction of current and the voltage below the zero boundary shows a reversal of current due to a reversal of the source voltage. The voltage indicated below the zero boundary is not negative electricity as others have supposed but simply a reversal of the source voltage.

With an inductive discharge things are a little different, when we charge an inductance the indicated voltage polarity across the coil is the same as the source voltage and the direction of current is forward. Then when we disconnect the source voltage the inductance discharges with an opposite voltage polarity but the current flow remains in the forward direction. This is where some people have made a mistake in assuming that the current has reversed just because the voltage polarity has but this is not the case, the current remains in the forward direction the same as when it started. One easy way to prove this is to place a diode in series with the inductance like we often see in boost converters , if the current has reversed then why would it flow through the output diode but it does dispite the fact that the voltage polarity across the inductance has reversed.

I know this phenomena all to well and it has led to premature balding and a compulsive need to howl at the moon because when you want to separate a forward source current from an inductive discharge current both following a singular path then diodes do not work. I have had to resort to biased mosfets to detect and switch the current however then a new problem pops up because the inductive voltage rise is way faster than the mosfet can switch and this leads to high voltage transients saturating the circuitry. I have even had cases where simply adding one single ultra-fast diode in series with an inductive discharge has led to a voltage rise of over 200v simply because the damn thing conducts about as fast as paint dries.

I guess my point here is that most of what I have seen from the "real" scope traces in the forums has led me to believe that the data is being misinterpreted in many cases by both the user and the oscilloscope, there is no negative electricity, no negative current, nor a reversal in current charging the source. This is why I switched over to fundamental physics where voltage is replaced with differential charge density and current with the motion of charges due to a differential charge density. This way a person can depict or plot the circuit parameters before testing and have at least half an idea what we are actually measuring at any given time. As usual I will end with the standard wishy washy disclaimer that I am not implying that this is happening in Rosemary's circuit but it may be a possiblity.
I just hope she doesn't get her Zipons crossed and vaporize all of us, you know what happened when the ghostbusters crossed there Proton beams -- liquifaction of the earths crust, solar flares, cats sleeping with dogs, it's all bad .
Regards
AC

[exnihiloest]

...
Then when we disconnect the source voltage the inductance discharges with an opposite voltage polarity but the current flow remains in the forward direction. This is where some people have made a mistake in assuming that the current has reversed just because the voltage polarity has but this is not the case, the current remains in the forward direction the same as when it started.
...

This is perfectly correct. I would just add a comment for a better understanding.

The direction of the induced current is always in agreement with Lenz's law: it opposes the field variation, so we could think that the current is of opposite sign when the magnetic field is built than when it collapses. This is true, the current changes sign, but only when the field is an external varying magnetic field, not generated by the inductance itself.
To build the field from the inductance itself, we must use a generator whose the current overcomes the opposite current that would be generated by induction from the field variation only. So the inductance switched on leaves us with a current of same direction as when it is switched off and there is no more generator, the field collapse becoming the only current source by induction.
A field variation due to a current in an inductance opposes the current variation that creates it, but not the absolute value of the current which can be of any sign.

More simply to not forget the rule, we just have to remember that for the same current, the voltage is of opposite sign in a generator than in a load (general rule). When the inductance is switched on, it is the load, the current from the generator build the magnetic field. When the inductance is switched off, it becomes the generator due to the collapse of the magnetic field inducing the current, so the voltage reverses.

[humbugger]

If I have the mojo to say it, then I must have a back up plan.  What's in it for me?  How about this.  I show you, you do a sim for the prof. 

No deal.  Why should I agree to do work in return for you showing the proof of your own assertion?  It was not I who made the assertion; it will not be I who works to obtain its proof.   

Humbugger

[FuzzyTomCat]

Hi all,

Just wanted to bring this to light ... before it gets buried as everything else from RA .....

____________________________________________________________________________________________________________

http://www.overunity.com/index.php?topic=10407.msg279127#msg279127
Channel (1) 5.00V
Channel (2) 100V
Channel (3) 50V
Channel (4) 100mV

200K Data Dump   02/22/2011

____________________________________________________________________________________________________________

http://www.overunity.com/index.php?topic=10407.msg279128#msg279128

2nd example.
Settings the same as previous.

Channel (1) 5.00V
Channel (2) 100V
Channel (3) 20V
Channel (4) 100mV

500K Data Dump  02/22/2011

____________________________________________________________________________________________________________

http://www.overunity.com/index.php?topic=10407.msg279129#msg279129

3rd example
Settings same as previous

Channel (1) 2.00V
Channel (2) 100V
Channel (3) 10V
Channel (4) 100V

10K Data Dump  03/02/2011

____________________________________________________________________________________________________________

http://www.overunity.com/index.php?topic=10407.msg279130#msg279130

4th example
Settings same as previous

Channel (1) 1.00V
Channel (2) 50.0V
Channel (3) 10.0V
Channel (4) 100mV

500K Data Dump  02/09/2011

____________________________________________________________________________________________________________

All oscilloscope settings are the same ??

All the oscilloscope screen shot dates are from the 03/12/2011 demo ??

All the oscilloscope Data Dumps are the same size ??  ( 10K to 500K ) .... what custom spreadsheet software was used on the oscilloscope Data Dumps "above" 100K in size ??

THIS IS THE DOCUMENTED PROOF OF RESULTS EVERYONE EXCEPTS

This posted information at OU.com is a minimum pathetic !!

Regards,
Fuzzy

[GibbsHelmholtz]

No deal.  Why should I agree to do work in return for you showing the proof of your own assertion?  It was not I who made the assertion; it will not be I who works to obtain its proof.   

Humbugger

Hahaha,
I like you Humbugger. Very nice logic.  You're a perpetual motion machine.

[humbugger]

Hahaha,
I like you Humbugger. Very nice logic.  You're a perpetual motion machine.

I like you too, Gibbs, and appreciate your impeccable logic.  However, my declining your offer does not alleviate your responsibility to answer my challenge to your assertion.   

Humbugger

[GibbsHelmholtz]

I like you too, Gibbs, and appreciate your impeccable logic.  However, my declining your offer does not alleviate your responsibility to answer my challenge to your assertion.   

Humbugger

Humbugger walks by a hot dog seller.
- Selling good hotdogs
- Really? Prove it.
The seller happens to believe in perpetual motion.
-Alright, enjoy your hot dog.

[humbugger]

Ahhh, yes, the old switcheroo!

First question:  I know the bottom trace is supposed to represent mean power.  The two upper traces are mean voltage and mean current?  In the same circuit?  Quite the circuit there that shows negative current flow while the voltage is positive and positive current flow when the voltage is zero.  Suggest to me a real circuit that acts like that, if you would.

The other problem I have is that in the context of Rosemary's device, the voltage is a steady omnipresent DC battery voltage, as I stated in my challenge.

Humbugger

P.S.  You're buying on that hot dog.