PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
OverUnity Research > Electrical / Electronic Devices > Inductive Kickback Circuits > The Rosemary Ainslie Circuit
2024-11-27, 20:36:11
News: Check out the Benches; a place for people to moderate their own thread and document their builds and data.
If you would like your own Bench, please PM an Admin.
Most Benches are visible only to members.

Pages: 1 ... 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 [24] 25 26 27 28 29
Author Topic: The Rosemary Ainslie Circuit  (Read 477120 times)
Topic Options img
img  PhysicsProf
Re: The Rosemary Ainslie Circuit « Reply #575 on: 2012-02-09, 03:21:43 »
Group: Professor
Hero Member
*****

Posts: 3017
Once again, Mile High makes an unsubstantiated statement, and he fails to quote me fully:

Quote from: MileHigh on 2012-02-08, 22:29:29

To quote PhysicsProf:
Quote
For the output energy (measurement), I like either charging a capacitor and using the above equation, OR using calorimetry.  Here, heating water in a well-insulated container is perhaps the easiest method, unless you have a calorimeter available to you.

PhysicsProf, by the statement above you are revealing (again) that you are not qualified to verify anyone's electrical circuit for over unity.  Sorry but I have to speak the truth.

Therefore Rosemary take note:  Neither Stefan, nor PhysicsProf, nor Aaron for that matter, are qualified to verify your setup for any possible over unity.  That's just the way it is.


MileHigh


Here's the rest of my recommendation, from the same post:
  
Quote
For input energy, I suggest use of a capacitor, as non-leaky as possible, then the input energy can be MEASURED in a straightforward way:

Einput = 1/2 C * (Vfinal**2 - Vinitial**2).

For the output energy, I like either charging a capacitor and using the above equation, OR using calorimetry.  Here, heating water in a well-insulated container is perhaps the easiest method, unless you have a calorimeter available to you.  With water-heating by the load, one can use:

Q = C(H2O) * mass * (Tfinal - Tinitial),
Where C(H2o) = 4.186 J/g-DegC for liquid water.

 Again, I disagree with you Mile High -- I am proposing the use of time-tested methods which I think will work just fine to determine the presence of ou (or not), when the claims are significantly more output energy than input energy as is the case with Rose's claims.

 But perhaps you will explain yourself this time?  I hope you were not just speaking BS out of hubris!

And if you can really explain why you don't like my proposed methods for measuring Einput and Eoutput for Rose's device -- then what method would you propose that would be better?   I look forward to your response, as I always seek the best methods possible.
« Last Edit: 2012-02-09, 03:46:55 by PhysicsProf »
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #576 on: 2012-02-09, 04:31:11 »
Group: Guest
PhysicsProf:

You are making a gross generalization by stating that a capacitor is a "time-tested method" for measuring output power as part of an over unity investigation.  Your statement is simply not true and your statement makes no real sense.  Nor did I need to quote you fully, no subterfuge was intended.

The output of some sort of hypothetical circuit will go into a corresponding hypothetical load.  You are not considering any impedance matching or the fact that the instantaneous impedance and differential impedance of a capacitor is dependent on the current voltage on the capacitor and the size of the capacitor and the impedance will vary as a function of time.  Who knows, you might not even want to have impedance matching.

Who says every alleged over unity circuit wants to drive a capacitor as a load?  Why should a capacitor necessarily be the best load for measuring the output?  What if the best load for the circuit is a 10-ohm resistor?  So what is putting a 100 uF or a 20,000 uF capacitor in place of the 10-ohm resistor going to do for you?  The capacitor will charge up and then look like an open circuit.  When you first connect the capacitor it will look like a short circuit.  And the whole time the best device to measure the output power was a 10-ohm resistor.

You know that using a capacitor to drive a circuit could potentially be a way to measure the power consumption of a circuit and then you erroneously connect the dots and think that you can use a capacitor to measure the power output from a circuit.  My impression is that you are simply not thinking things through because you don't have the knowledge and experience to think things through when it comes to electronics.  Hence, you are not qualified to analyze electronic circuits to determine if they manifest over unity or not.

Sorry, but just keeping it real for the people that want to learn some stuff.  I am uncomfortable in seeing incorrect information put out there for general consumption.  The fact is that only for some limited special cases is a capacitor a useful tool for measuring the output power of a device.  In other words, if I was going to generalize myself, I would say that a capacitor is an almost useless tool for measuring the output power of an allegedly over unity device.

I had no real desire to get into it at this level.  The sentence was highlighted in the hope that you and others would stop and think and perhaps do some research.  It all comes back to one of my central themes that I have just about given up stating:  If you are going to play with and experiment with resistors, capacitors, coils and diodes and you want to get the most out of the experience, you should want to know the fundamentals of how they actually work in the real world.  Open a book or go online and teach yourself to fish.

For example, look at this clip on "Radiant Energy 101."  The person presenting this "teaching" clip doesn't have the slightest idea of what is actually going on in his demo:

[youtube]http://www.youtube.com/watch?v=EVK84Z8BSnc[/youtube]

It's really unfortunate that he doesn't know what's going on.  I used to wish and hope that the forums could be a place were people could have a synergistic learning experience together but it seems like it's not possible.  However, I am still tempted at times to point out major errors.

MileHigh
« Last Edit: 2012-02-09, 04:59:38 by MileHigh »
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #577 on: 2012-02-09, 04:43:03 »
Group: Guest
Referencing that Min2oly clip in the previous posting, an interesting little challenge for the "hard core" experimenters is to really explain what is going on in that clip.

I am not going to engage in this question actively myslef, but if someone can explain it I will post "Bingo!"  The usual suspects like Poynt99 and company can disqualify themselves.

If you are honest with yourself and you can't explain what's going on in that clip and you are an active experimenter them my advice would be to read some books or do some online research.  Pull yourself up by your own bootstraps or work with your friends to figure it out collectively.  Hopefully you will have some fun in the process.

MileHigh
   
img  Magluvin
Re: The Rosemary Ainslie Circuit « Reply #578 on: 2012-02-09, 05:47:39 »
Group: Guest
Its a boost converter.

When the battery is put across the coil, current begins to flow. As it flows, a magnetic field is built in the coil, and as the field builds, more current from the battery is taken.

When he disconnects the battery from the coil, the magnetic field collapses, causing currents to continue to flow through the coil, through the diode and into the cap, as it is the only short path available once the red wire is disconnected.. Mind you the current in the coil never changes direction.  ;) 

Didnt I do pages of this stuff at ou?

Tuff one.  O0

Mags

Mags
   
img  GibbsHelmholtz
Re: The Rosemary Ainslie Circuit « Reply #579 on: 2012-02-09, 06:10:53 »
Group: Guest
Quote from: MileHigh on 2012-02-09, 04:31:11
PhysicsProf:

You are making a gross generalization by stating that a capacitor is a "time-tested method" for measuring output power as part of an over unity investigation.  Your statement is simply not true and your statement makes no real sense.  Nor did I need to quote you fully, no subterfuge was intended.

The output of some sort of hypothetical circuit will go into a corresponding hypothetical load.  You are not considering any impedance matching or the fact that the instantaneous impedance and differential impedance of a capacitor is dependent on the current voltage on the capacitor and the size of the capacitor and the impedance will vary as a function of time.  Who knows, you might not even want to have impedance matching.

Who says every alleged over unity circuit wants to drive a capacitor as a load?  Why should a capacitor necessarily be the best load for measuring the output?  What if the best load for the circuit is a 10-ohm resistor?  So what is putting a 100 uF or a 20,000 uF capacitor in place of the 10-ohm resistor going to do for you?  The capacitor will charge up and then look like an open circuit.  When you first connect the capacitor it will look like a short circuit.  And the whole time the best device to measure the output power was a 10-ohm resistor.

You know that using a capacitor to drive a circuit could potentially be a way to measure the power consumption of a circuit and then you erroneously connect the dots and think that you can use a capacitor to measure the power output from a circuit.  My impression is that you are simply not thinking things through because you don't have the knowledge and experience to think things through when it comes to electronics.  Hence, you are not qualified to analyze electronic circuits to determine if they manifest over unity or not.

MileHigh

There are good and bad aspect of using capacitor as output.  The bad aspect you already pointed out, basically, impedance matching.  But this is why we and engineers are here for, to overcome those problem.  The good thing using cap to measure output is the energy comes straight forward and not arguable.  You dismissed the idea because you see the bad sides when you can use that knowledge to help overcome those problems.  While did not offer a better solution, you just say how unqualify people are.  However, i do recognize the genuine in you that see through the capacitor problem.
   
img  PhysicsProf
Re: The Rosemary Ainslie Circuit « Reply #580 on: 2012-02-09, 06:12:20 »
Group: Professor
Hero Member
*****

Posts: 3017
  MH:   I did not say that the "output-energy" cap would be used until it was charged fully or even very much, nor did I mean that.

 
Let's say, though, that the output cap charges (not necessarily fully charged)  and shows FIVE times the energy RISE compared to the energy DROP in the input cap (no batteries in the system, nor pre-charged caps)... Would that not demonstrate ou?  or have I missed something?

 Further, I ask that you take a moment to address the main method I recommended for measuring the output energy -- calorimetric:

Quote
calorimetry.  Here, heating water in a well-insulated container is perhaps the easiest method, unless you have a calorimeter available to you.  With water-heating by the load, one can use:

Q = C(H2O) * mass * (Tfinal - Tinitial),
Where C(H2o) = 4.186 J/g-DegC for liquid water.


 What do you think about the calorimetric method which I clearly spelled out, for evaluating the output energy, for Rose's device which is claimed to be significantly ou?

« Last Edit: 2012-02-09, 07:32:48 by PhysicsProf »
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #581 on: 2012-02-10, 01:17:53 »
Group: Guest
Magluvin:

You are right but you are just skimming in your details.  Can you add some real substance to what you are saying to spell it out with more clarity for people that don't understand this stuff?

PhysicsProf:

Going back to your full statement:

Quote
For input energy, I suggest use of a capacitor, as non-leaky as possible, then the input energy can be MEASURED in a straightforward way:

Einput = 1/2 C * (Vfinal**2 - Vinitial**2).

For the output energy, I like either charging a capacitor and using the above equation, OR using calorimetry.

You are generalizing about using a capacitor to measure output energy and like I already stated a capacitor is a lousy device to use to measure output energy and will simply not work in the majority of cases.  It will not work for either Ainsley variation.

Your question about the five times energy rise using the output cap method is silly, what do you expect me to say?  It's like if I paint a bench green and then I ask you if the colour of the bench is green, how are you going to reply?  I also never mentioned calorimetry as a way of determining over unity, look at my origial posting and see what I highlighted.

It's ridiculous, you are trying to "push" me to agree with your points that are self-evident and I never raised those issues in the first place.  Stop the nonsensical attempts at misdirection.  What you haven't discussed at all is what I am talking about, a capacitor is in the vast majority of cases is not a useful tool to measure the output power in an electrical circuit that you are trying to analyze for over unity.  Will you concede that point?  The fact that you tried to imply that a capacitor is a good tool to use to measure the output indicates that you are not qualified to do any serious power in vs. power out analysis of electrical circuits.  Just try to be real.

MileHigh
   
img  Magluvin
Re: The Rosemary Ainslie Circuit « Reply #582 on: 2012-02-10, 02:45:12 »
Group: Guest
Quote from: MileHigh on 2012-02-10, 01:17:53
Magluvin:

You are right but you are just skimming in your details.  Can you add some real substance to what you are saying to spell it out with more clarity for people that don't understand this stuff?



Sorry, hit post before I got to say anything.   Reposting

Magsnuts
   
img  Magluvin
Re: The Rosemary Ainslie Circuit « Reply #583 on: 2012-02-10, 03:43:32 »
Group: Guest
Quote from: MileHigh on 2012-02-10, 01:17:53
Magluvin:

You are right but you are just skimming in your details.  Can you add some real substance to what you are saying to spell it out with more clarity for people that don't understand this stuff?



Well how deep does the rabbit hole go?

More surface talk.....

Another possibility is, that I can only assume what orientation the diode is in, which determines the way current can flow through it.

Even if we reverse the diode, we still get our high voltage to the cap, but it is opposite polarity now.

When we charge a coil as shown in the vid, the field builds, and when we disconnect the source from the coil, that field collapses.

If we have a perfect switch used to connect and disconnect the source to the coil, meaning when on its on, but when off it will not allow current to flow, not even HV spark, then when we disconnect(open switch) the "forward" voltage of the collapse has only the path of the diode to charge the cap of my previous description of the vid.

What if we dont have a diode and cap? With our perfect switch, when we open it, the field collapses. And as it collapses down toward zero field, there is no place or path for the voltage potential produced by the collapsing coil to go.

Coils have capacitance. So the combination of coil with built in capacitance we have an LC. Single wire coils have very small capacitance between windings, and bifilar coils have a higher capacitance.

If you charge up an LC and disconnect, it will oscillate. So when we disconnect the source from our coil with no diode and no cap, our coil will oscillate at very high freq due to very small capacitance in our coil. The oscillation dies pretty quickly. No oscillation occurs if we have a path for that potential to the cap. The oscillation will be heavily damped

So if we reconnect our diode and cap, only we reverse the diode, when the source is disconnected from the coil, and the field collapses. But with no where for a continued forward potential current to flow, the coil goes into its oscillation mode, and the field collapses very quickly, and when the field polarity of the coil reverses during oscillation, N S to S N, that is when the potential of the coil has a path for current through the diode and to the cap for storage.

This is what I call back emf.  In my original description, I call it forward emf, meaning that when we disconnect the source from the coil, the collapsing field current is in the same direction as the current produced by the source through the coil.
In this reversed diode description, once the field of the coil reverses within 1/4 wave of LC oscillation, then the output of the coils current is in the opposite direction of the original source current. Back emf.

The output to the cap in either situation will be nearly the same. Possibly a small, tiny, itiy bitty, difference due to the bemf version may have lost some due to the dying oscillation during that 1/4 wave of oscillation mode, being that the oscillation eventually dies. So there is some loss in each cycle.

There are some out there that use the term bemf when actually to be correct in some situations, should use just emf or forward emf, produced by the collapsing field of a coil.

Well, im beat. Dont really have time to edit errors.

Magsleeps soon
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #584 on: 2012-02-10, 04:31:48 »
Group: Guest
I'll make a new thread, that would be more appropriate.
   
img  PhysicsProf
Re: The Rosemary Ainslie Circuit « Reply #585 on: 2012-02-10, 08:13:43 »
Group: Professor
Hero Member
*****

Posts: 3017
  It seems we are making progress -- however slow and tedious!   and full of ad hominems which make the effort hardly worthwhile...  
   It appears there is general agreement that using an input-capacitor provides a reliable means of determining input energy (as I suggested to Rose).  Using a cap to measure output energy is fraught with some difficulty, we agree -- yet may have a place when the output energy in this cap clearly exceeds the energy supplied by the input cap.  That's what I've been saying, and I think there is some agreement on that.  But again I'm hit with personal attacks which seems to be the routine with some.  


 I'm not sure this forum is the place to continue, since ad hominems are evidently allowed -- and abound! -- and distract from the effort to serve mankind (IMO).  I note also the thread in which Sterling Allan is hit with ad hominems (as I have already delineated and objected to.)   There are other examples IIRC.

 Nevertheless,  I will try my question regarding calorimetry one more time.   Then I return to my experiments, noting that debunkers tend to NOT DO EXPERIMENTS, which I find telling.

This was directed to MH:
Quote
I ask that you take a moment to address the main method I recommended for measuring the output energy -- calorimetric:

Quote
Quote
calorimetry.  Here, heating water in a well-insulated container is perhaps the easiest method, unless you have a calorimeter available to you.  With water-heating by the load, one can use:

Q = C(H2O) * mass * (Tfinal - Tinitial),
Where C(H2o) = 4.186 J/g-DegC for liquid water.


 What do you think about the calorimetric method which I clearly spelled out, for evaluating the output energy, for Rose's device which is claimed to be significantly ou?

And could you try to avoid personal attacks in your response this time?  much appreciated.

« Last Edit: 2012-02-10, 08:40:44 by PhysicsProf »
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #586 on: 2012-02-10, 13:32:54 »
Group: Guest
PhysicsProf:

You are just getting the truth from me.  The more honest you are about yourself and the more honest you are in your postings and the less attempts you make at deflection the better a person you will be.

Quote
Using a cap to measure output energy is fraught with some difficulty, we agree

No, we don't "agree."  The truth is that over the past few days you learned that using a capacitor to measure the output energy is fraught with problems.  You have to develop as person to get to the point where you can simply say things like that.  The "professorial swagger" does not work with you because it has to be backed up with substance and you are out of your realm when it comes to electronics.

Quote
I will try my question regarding calorimetry one more time.

Really?  You have followed the Peter Davey water heater thread?  You are even in the process of replicating it.  How may times did I post about using calorimetry and doing it right and taking it seriously in that thread?  Perhaps a dozen times?  So why are you even asking me?  What are your real motivations for posing that question when I never even raised the issue?  I have been talking about electronics the whole time.

Quote
Then I return to my experiments, noting that debunkers tend to NOT DO EXPERIMENTS, which I find telling.

That's a true observation and an ad hominem attack at the same time.  I have thousands of hours working on an electronics bench under my belt.  I was doing similar experiments to what are being done around here 33 years ago.

You know how you said to Rosie Posie that she should try to use a capacitor on the input to measure the power consumption of her circuit?  She is reverently asking you for your guidance because she doesn't understand how a capacitor works.  Over the past two and one-half years I have probably suggested that to Rosemary about 10 times.  The last time I suggested it to her was about a week ago and her response was "utter nonsense."  It's amazing how the same message is responded to depending on the messenger, isn't it?

Again, PhysisProf, just be real and you will be fine.

MileHigh
   
img  PhysicsProf
Re: The Rosemary Ainslie Circuit « Reply #587 on: 2012-02-10, 14:49:59 »
Group: Professor
Hero Member
*****

Posts: 3017
   @MH:  IIRC, you yourself said you would not attempt to do replication experiments on the Davey or other devices currently under testing -- but I would welcome the change. 

Quote
MH:   You have followed the Peter Davey water heater thread?  You are even in the process of replicating it.  How may times did I post about using calorimetry and doing it right and taking it seriously in that thread?  Perhaps a dozen times?  So why are you even asking me?  What are your real motivations for posing that question when I never even raised the issue?

Actually, I think this is an answer of sorts, that you agree with me that the calorimetric method of measuring the output energy is reasonable way to proceed.  My motivation?  to try to reach agreement on the methods, so that if ou is finally measured with agreed-upon methods, you will not simply denigrate the methods used.  We may see if all this effort pans out.  

But you know, this is not really for you MH.  We are doing this research for the benefit of the human family -- at least, that is the motivation for me and many of us in this research field.
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #588 on: 2012-02-10, 22:54:21 »
Group: Guest
PhysicsProf:

Of course I agree with calorimetry testing but you already knew that so there was no point in raising the issue.  But you felt some frustration that I did not reply, didn't you?  I wasn't really thinking about this but I realized afterwards that you got a taste of your own medicine.  What you have to do is respond to issues addressed to you even if it turns out that you are wrong, and you also have to quit pretending that you already knew something after it is taught to you.  You have to have enough character to admit when you are wrong and admit when you learn something.

There is an ironic twist with respect to Rosie's second circuit and the calorimetry testing.  Rosie doesn't have the slightest clue how much power her circuit actually consumes after all these months because she is fixated on her "garbage-in garbage-out" measurements.  So it's impossible to do any calorimetry testing with Rosie's setup because she has no data.

If only one member of her "team" suggested that they swap out one of her batteries for a large capacitor to make an alternative power consumption measurement this whole nonsensical fiasco would never have happened.  Who knows if there really was a "team."  By the same token, not a single member of the "team" realized that the current making the inductive resistor heat up was also passing straight through the function generator.  Rosie Posie refuses to believe that though.

MileHigh
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #589 on: 2012-02-11, 01:32:42 »
Group: Guest
Rosie!

You sent me off on a fun little double-naught spy Google search and I hit pay dirt.

Quoting Rosie:

Quote
Why is it that I'm the ONLY one who is managing any kind of valid argument?

Quote
My dear MileHigh.  That 'signal terminal' as you've taught me to refer to it - from the function generator - includes a ground terminal. The TWO TERMINALS ARE NOT CONNECTED other than through 3 or 4 optional signal signal settings - including, among others, a square wave output signal.  THAT's the signal we apply.  Now.  Follow my lead.  The connection is from that ground terminal - through approximately 1000 Ohms of resistance and wiring - then BACK through about 50 Ohms of resistance in the signal terminal and then to the GATE OF Q1.

I am going to focus on the what's highlighted in the quote above.  The rest of what you say is "mangled."  Heaven knows where you came up with "1000 ohms," did that just parachute into the discussion?

Have you heard of Agilent Technologies Rosemary?  They are probably the largest manufacturer of bench test equipment in the world.

Here is a link to an application note pdf file entitled, "Adding DC Offsets to a Function Generator's Output."

http://cp.literature.agilent.com/litweb/pdf/5990-3466EN.pdf

Here is a quote from the application note:

Quote
Another limitation of using this method is that the current available
to your load is limited to the output current of the function genera-
tor you choose, since the load current must flow through both
the power supply and the function generator. Also, most function
generators have a 50-ohm output impedance, meaning any load
current will flow through this resistance. This resistance will
form a voltage divider with your load impedance, so be sure to
adjust the DC power supply output voltage accordingly.

Did you take note of the red highlighted text in the above paragraph?

MileHigh

P.S.:  I will throw in the Rosie Poseo:  
Quote
Kindest regards MileHigh.  Your writing style is superb.  You're arguments are BASELESS.
   
img  PhysicsProf
Re: The Rosemary Ainslie Circuit « Reply #590 on: 2012-02-11, 03:10:22 »
Group: Professor
Hero Member
*****

Posts: 3017
MH, actually I raised the issue of calorimetry over on Rosie's thread at OU.com several days ago, and I was frustrated yes that neither she nor you would come right out and admit this was a good idea.  But you finally did so:

Quote
MH: PhysicsProf:   Of course I agree with calorimetry testing ... you felt some frustration that I did not reply, didn't you? "

Sure, that was frustrating that you only answered after several tries and I did not know that you would finally agree since you appeared so evasive.

But thanks for replying, I see we are in agreement on that also.  Now if Rosie would just reply!

PS -- I applied calorimetry some time ago (starting about a year ago) to devices I was testing.  Admittedly, I've learned HOW to do this, but I have not personally SEEN ou as of yet by means of this method....    I intend to keep on it, however.
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #591 on: 2012-02-11, 03:17:42 »
Group: Guest
Rosemary:

It's unbelievably exasperating dealing with you so this is probably going to be my last posting for a while.

Let's start with the simplified block diagram:



You said this:

Quote
My dear MileHigh.  OF COURSE the load current must flow through both the power supply and the function generator.  But this little interaction between the two currents is ONLY managed when the circuit is CLOSED that the battery is able to discharge any current at all.  Then INDEED the voltage at the gate of the signal generator increases.  I'll post YET ANOTHER example where the increase in the voltage is clear.  AS EVER. Follow the BLUE TRACE and relate that to the ORANGE trace.  BUT during the period that the circuit is OPEN when the battery cannot supply any current - THEN?  You will notice that the signal at the gate of Q1 defaults to below zero.  THEN it is EVIDENTIAL that the load current is NOT flowing through the function generator from that battery power supply.  ELSE it would be GREATER THAN ZERO.  IT IS NOT.

WHY am I having to show you this?  I'm meant to be the ignoramus here.  Come on MileHigh. You need to do MUCH better than this. And this posted here as your rather optimistic REBUTTAL - which is not a rebuttal at all. Here's that reference for those who don't read OUR.com

The level of disconnect is unbelievable.  There aren't two currents there is only one current.  The signal generator has no "gate."  What's wrong with you?  Can't you even double-check your own prose to see if it at least makes sense?  I am not talking about voltages, we are trying to discuss the current flow.  I am not interested in Q1 at all, it's the "Q2 array" that we are interested when the output waveform from the function generator is negatively offset.  As far as you posting waveforms goes in general, without any simplified schematic showing what the different coloured waveforms represent, your waveforms are meaningless and nobody can understand them.  You get a giant FAIL when you post those DSO screen captures without any simplified reference schematic and some sort of label for each different coloured waveform.

Listen Rosemary, when your function generator outputs a negatively offset voltage, the Q2 array switches on and your circuit goes into oscillation.  When the oscillation waveform is low and the Q2 array switches on then the current supplied by the batteries flows through the Q2 array DRAIN to SOURCE and then THROUGH THE FUNCTION GENERATOR.  It has nothing at all to do with a GATE.

Now watch, I am going to take that quote from the application note and apply it to your circuit:

Quote
Another limitation of using this method is that the current available
to your inductive resistor is limited to the output current of the function generator you
choose, since the inductive resistor current must flow through
the set of batteries and the Q2 array and the function generator. Also, most function
generators have a 50-ohm output impedance, meaning any inductive resistor
current will flow through the Q2 array and this 50-ohm resistance. This 50-ohm resistance will
form a voltage divider with your inductive resistor, so be sure to
adjust the DC power supply output voltage accordingly.

I hope that registers with you Rosie because I am done.

MileHigh
   
img  Magluvin
Re: The Rosemary Ainslie Circuit « Reply #592 on: 2012-02-11, 03:43:45 »
Group: Guest
Mh

Have you guys actually made this experiment on the table or just on a sim?

Using the sim, has there been any power measurements through the sig gen?

I do see a possible path for current to flow in the circuit from the gen through q2 if it is on and the polarity of the gen considered at the time. Or even through q2 diode, considering the polarity of the gen at the time.

Mags
   
img  Magluvin
Re: The Rosemary Ainslie Circuit « Reply #593 on: 2012-02-11, 03:51:15 »
Group: Guest
Another thing is, If I were to use 4 mosfets in parallel, I would use them where q1 is, as q2x4 is set up for high current in the circuit that would have to go through the sig gen.  hmmm

Are we sure this is thee circuit rose uses?  Ill have to see her other circuits to see if there is an error here.

Mags
   
img  MileHigh
Re: The Rosemary Ainslie Circuit « Reply #594 on: 2012-02-11, 04:39:40 »
Group: Guest
Mags:

There is no rhyme or reason behind having the four MOSFETs in parallel.  It's nothing more than a wiring mistake.  The mistake caused an oscillation when the function generator output was negatively offset and then Rose thought that was a "magic oscillation."  Poynt did the simulation and it oscillated just like her circuit.  However, in her myriad of DSO captures, Poynt recently mentioned that some of them show oscillation with a positive offset on the function generator waveform.  So there is some uncertainty as to what data set the paper is showing.

However, I am quite certain that the simulation showed an oscillation with a negative offset on the function generator.

Rosie gets into some kind of discussion about voltages that's impossible for me to follow.

It also appears, although I can't be certain, that she has done a complete about face with respect to the issue of whether current is flowing through the function generator or not.  She was insisting that it was IMPOSSIBLE.  I found the application note and then she said OF COURSE.  Is this more misunderstanding or more mass confusion I don't care at this point.  If it is a complete about face then it's pure creepy Orwellian modus operandum in action.  Something that I find extremely distasteful.

Here is the last point and for all I know it will send poor Rosie into another mass confusion tizzy:  When the function generator is outputting -5 volts, that makes it look like another battery in series with the main battery bank.   In other words, you have a 72-volt battery bank plus 5 volts being supplied by the function generator for a total of 77 volts working to push current through the load resistor.  So when the function generator output goes negative it does two things, 1) it switches on the Q2 MOSFET array, and 2) it acts like another voltage source in series with the battery bank and with a polarity to add to the voltage of the battery bank.

All of this measurement nonsense, the "garbage-in garbage-out" that Rosie clings to for dear life, could have been resolved months ago with one lousy big fat capacitor.

This whole thing is a massive debate centered around garbage measurements on a nonsensical circuit.  That's the sad truth of it all and no matter how you try to get this message through to Rosie Posie you fail.

MileHigh
   
img  giantkiller
Re: The Rosemary Ainslie Circuit « Reply #595 on: 2012-02-11, 04:47:59 »

Group: Elite Experimentalist
Hero Member
*****

Posts: 1593
Frequency equals matter...


Buy me a drink
So why do you all spend the effort to [turn a sow's ear into a silk purse] here?
If she is whacked out there she would be here also... :-\


---------------------------
http://purco.qc.ca/ftp/
   
img  poynt99
Re: The Rosemary Ainslie Circuit « Reply #596 on: 2012-02-11, 13:03:29 »

Group: Administrator
Hero Member
*****

Posts: 3217
It's not as complicated as it may seem...
Mags,

With a positive FG offset, Q1 is active and there are no currents through the FG.

With a negative FG offset, Q2 is active and the current path is through the FG.

The essence of the problem guys is not how or why the thing is oscillating (a primate could get a MOSFET to oscillate), it's regarding the battery voltage measurement. A negative power measurement would indicate a net average power going back to the battery. Now despite this erroneous measurement, we know that is not really what is happening, so what is?

It's the placement of the battery voltage probes! Where they are located in the video demonstration tells us that the wild swing in voltage on the battery probe is "out to lunch". If Rosemary had done precisely what I asked her to do two years ago (or thereby), this would have been over. What she claims to have done was move the probe to one battery, but I doubt she did that properly. In addition, I very much doubt that she disconnected all the other probes attached to the circuit. If this was not done exactly as I asked, then the measurement will still be erroneous.

And that's where we are right now Mags. Rose is as stubborn and rebellious as a young mule, and deep down she knows her circuit doesn't perform as claimed. So we'll never see her perform any requested tests, including yours (which we have also requested a dozen times).

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
img  poynt99
Re: The Rosemary Ainslie Circuit « Reply #597 on: 2012-02-11, 13:18:13 »

Group: Administrator
Hero Member
*****

Posts: 3217
It's not as complicated as it may seem...
Mags,

Here's a simple circuit/test that Rosemary could do that would settle the issue for good. But guess what? She refuses to do it on the actual circuit for which she claims has an infinite COP.

Go figure eh?

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
img  poynt99
Re: The Rosemary Ainslie Circuit « Reply #598 on: 2012-02-11, 14:07:26 »

Group: Administrator
Hero Member
*****

Posts: 3217
It's not as complicated as it may seem...
And guys,

Here's some additional important information that adds more poop to Rosemary's soup;

Around the 3:00 mark of the RATS demonstration video (extremely damning video btw), we see them showing the battery voltage on the Fluke Multimeter while the circuit is running. The reading fluctuates a very tiny bit (a few mV), but indicates a fairly steady 62VDC.

Now, Rosemary CLAIMS that the actual battery voltage is fluctuating wildly from a few volts positive, to over 200V positive! This begs the question of how that meter can display a nice steady 62V (62V IS PERFECTLY IN LINE WITH WHAT ONE WOULD EXPECT WITH 5 BATTERIES!) if the battery voltage is supposedly fluctuating wildly as claimed? C.C

We all know the answer of course (and its relevance), but does Rosemary? Well, we all know the answer to that.

.99



---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
img  GibbsHelmholtz
Re: The Rosemary Ainslie Circuit « Reply #599 on: 2012-02-11, 15:19:20 »
Group: Guest




Hm... I still see the path of least resistance is through Q1 .  When the FG gives a negative signal, it turns on Q2 array.  The current from the battery have two paths: through FG and through Q1 gate to source.  This would also turn on Q1 .  Current would prefer to go through Q1 at this point in my opinion.  Q1 could also use FG path to equalize charges and therefore oscillation occur.  The question is, when in operation, Q1 gets hot or Q2 array gets hot. 
   
Pages: 1 ... 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 [24] 25 26 27 28 29
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-11-27, 20:36:11