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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 634720 times)

Offline Magluvin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #240 on: March 23, 2011, 08:12:42 AM »
Here is an idea based on this knowledge

Imagine the circuit and we connect the switch. We wait till the cap just reaches 5v (batt voltage) we cut the switch off.  We have a diode inserted so that we use the fly wheel effect to charge the cap beyond batt voltage without the batt in the circuit to pull from. We now , with the diode, pull from the other side of the cap instead.

If this works, and the cap receives more voltage than the battery, that extra voltage is free.

We cut the battery from the circuit at 5v on the cap. That 5v on the cap is equal to an amount of energy, the same amount every time.
And that is all the energy that was taken from the batt.
But if that cap ends up with more than 5v as described above, even 5.01v, that cap will be holding more energy than what was taken from the battery, period, no debater can beat this.

Oh but where is the energy coming from?  The merry go round my friends, the merry go round.  ;]  oooo scary.  lol

mags

Free Energy | searching for free energy and discussing free energy


Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #241 on: March 23, 2011, 08:18:21 AM »
Here is an idea based on this knowledge

Imagine the circuit and we connect the switch. We wait till the cap just reaches 5v (batt voltage) we cut the switch off.  We have a diode inserted so that we use the fly wheel effect to charge the cap beyond batt voltage without the batt in the circuit to pull from. We now , with the diode, pull from the other side of the cap instead.

If this works, and the cap receives more voltage than the battery, that extra voltage is free.

We cut the battery from the circuit at 5v on the cap. That 5v on the cap is equal to an amount of energy, the same amount every time.
And that is all the energy that was taken from the batt.
But if that cap ends up with more than 5v as described above, even 5.01v, that cap will be holding more energy than what was taken from the battery, period, no debater can beat this.

Oh but where is the energy coming from?  The merry go round my friends, the merry go round.  ;]  oooo scary.  lol

mags

Here I entirely agree with you.  Maybe Woopy can do this test for us?  It'll be interesting.

 :D
Rosie.

(Go to sleep Mags - or you'll suffer in the morning.  I know that feeling only too well.  LOL)

Offline Magluvin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #242 on: March 23, 2011, 08:34:41 AM »


But again.  I'm not entirely sure that the diode stops a negative flow.  But nor am I sure that its relevant.

Kindest regards,
Rosie
hey Rose
Well in my circuit, the diode never gets hit with bemf because the inductor was never  disconnected from forward flow during the circuit cycle. So the inductor had somewhere to put the charge it was flywheeling, no matter if it was pulling from the battery to charge the cap beyond battery voltage. It had enough energy stored to do so. ;]

But, if we cut the inductor, at the end current is flowing out, clockwise, that flywheel pumps that self capacitance so hard, like a 1000 mile an hour train wreck on a spring, the we get bemf. And why is it so short?  Well we are still disconnected, no where to get charge from at the disconnect, then ""SPARK""  ;]  Got some across the gap. Had to, the pressure was high. ;]  The spike can last only as long as the spark, or for as much as the mass of coil will give up in oscillation if no current can be had across the disconnect..  ;]

Ok   Insomniacs anonymous, over and out.  ;D

Mags

Free Energy | searching for free energy and discussing free energy

Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #242 on: March 23, 2011, 08:34:41 AM »
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Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #243 on: March 23, 2011, 09:06:35 AM »
hey Rose
Well in my circuit, the diode never gets hit with bemf because the inductor was never  disconnected from forward flow during the circuit cycle. So the inductor had somewhere to put the charge it was flywheeling, no matter if it was pulling from the battery to charge the cap beyond battery voltage. It had enough energy stored to do so. ;]

But, if we cut the inductor, at the end current is flowing out, clockwise, that flywheel pumps that self capacitance so hard, like a 1000 mile an hour train wreck on a spring, the we get bemf. And why is it so short?  Well we are still disconnected, no where to get charge from at the disconnect, then ""SPARK""  ;]  Got some across the gap. Had to, the pressure was high. ;]  The spike can last only as long as the spark, or for as much as the mass of coil will give up in oscillation if no current can be had across the disconnect..  ;]

I get it Mags.  I think it's right.  I just need to run it past some of the team.

Ok   Insomniacs anonymous, over and out.  ;D

 ;D

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #244 on: March 23, 2011, 01:29:26 PM »
Guys - It strikes me that I'm getting snarled in defense of protocols that I had thought, by now, were entirely addressed.  This is yet another technique employed by Humbugger et al, to cast aspersions on those test results.  Any and all diversions are being used to take attention from the actual significance of these tests and leave me arguing the correctness of measurements.  And when I do so, then it's too late.  The damage is done.  And there's generalised impression cast over everything  that I know not whereof I speak.

What I may or may not know has no bearing on the report, the demonstration or any claims made.  They are advanced by the 'team' and I'm reasonably satisifed that they are considerably more qualified than Humbugger or Poynty, or MileHigh or any of the others that clamour to deny these claims.   Just know that it was no accident that I left the demonstration to them.  It was intended to remind you all that - while I am not qualified - those that are supporting this evidence most certainly are.  Feel free to discount what I report.  But you'd need strong argument to deny what they, collectively, endorse.   

Meanwhile I'm braced for the inevitable 'attack' on my competence.  I'll address these as they arise.

Rosemary.

Free Energy | searching for free energy and discussing free energy

Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #244 on: March 23, 2011, 01:29:26 PM »
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Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #245 on: March 23, 2011, 01:33:36 PM »
Poynty - I can no longer see your waveforms and i would LOVE to see them.  There are others of us who also cannot open those files.  WOULD YOU PLEASE POST A PICTURE and just size it that it fits this thread.

Kindest regards,
Rosie

ADDED
This is copied over from your forum - written by you Poynty Point.

'Indeed, I am having difficulty figuring out why her shunt trace is at zero. Mine is showing about 1.5V or so, and hey, 0.25/11 x 72 = about 1.6V. It would seem what I'm showing is about right.'

I think what you meant is 0.25/11 x 72 = about 1.6 watts, NOT VOLTS.  In which case can you then explain the temperature over the load which, typically, is at 6 watts or greater at 72 volts applied.

Perhaps someone can unzip the files and resize them. It's a shame that Stefan does not fix the problem so that this is not necessary. I offered him a solution already, as this works fine at OUR.

Regarding the calculation, no I meant Voltage. I am calculating the approximate voltage that should be across the shunt when the FET is ON, and under ideal conditions, those being that the ON resistance of the FET is much much smaller than your CSR of 0.25 Ohms. It is a simple voltage divider between the load resistance (11 Ohms) and the CSR resistance (0.25 Ohms) if we do not include the inductances.

So, 0.25/11 Ohms, times 60 Volts ~ 1.36V. [I used 72 volts last time assuming you had 6 batteries.]

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #246 on: March 23, 2011, 01:41:04 PM »
Perhaps someone can unzip the files and resize them. It's a shame that Stefan does not fix the problem so that this is not necessary. I offered him a solution already, as this works fine at OUR.

Regarding the calculation, no I meant Voltage. I am calculating the approximate voltage that should be across the shunt when the FET is ON, and under ideal conditions, those being that the ON resistance of the FET is much much smaller than your CSR of 0.25 Ohms. It is a simple voltage divider between the load resistance (11 Ohms) and the CSR resistance (0.25 Ohms) if we do not include the inductances.

So, 0.25/11 Ohms, times 60 Volts ~ 1.36V. [I used 72 volts last time assuming you had 6 batteries.]

.99

Poynt - we all know that you can do that resize.  Is there a reason you won't?  Are you keeping this hidden for a reason?  Have you not managed to show that waveform afterall?  Come on Poynty.  How about it?  It's a lame claim if there's no evidence.  And right now it's hidden from view.

Still don't understand your sum.  Where does the 11 Ohm's come from? 

Rosemary

Free Energy | searching for free energy and discussing free energy

Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #246 on: March 23, 2011, 01:41:04 PM »
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Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #247 on: March 23, 2011, 01:42:55 PM »

Still don't understand your sum.  Where does the 11 Ohm's come from? 

Rosemary

What is the resistance value of your load resistor then?

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #248 on: March 23, 2011, 01:45:19 PM »
What is the resistance value of your load resistor then?

.99

11 Ohms is close.  But why are you dividing the shunt value by the ohms value of the resistor?  Are you saying that there's a 1.3 volt across the FET?  Still don't get it Poynty.  The FET voltge is much higher than this during the 'on' period. 

added

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #248 on: March 23, 2011, 01:45:19 PM »
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Offline Magluvin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #249 on: March 23, 2011, 06:11:45 PM »
Hey Rose and Woopy

At lunch so Ill make it short. And sweet.  ;]

Here is a circuit with 2 mods. 

1  a resistor is added to simulate resistance in the inductor.

2 added the diode across the batt/switch for purposes as described last night.  ;]  =]

I ran it on the sim and it works!     ;D

But in the real world the circuit will need more additional circuitry to compare when the cap reaches 5v to cut the switch.

What I did to enable myself to cut the switch manually was slowed down the sim so that I could come very very close to cutting a 5v, and yep that diode allows current in the forward direction to keep flowing into the cap and I was getting over 6v into the cap.
It varied as I could not hit the switch to get a perfect 5v cutoff.

This should be a cop>1.  =]

Woopy  I will look at ways to enable you to automate the cutoff in your setup to do the test tonight.  ;]

Today im not even tired, on 2hours sleep, Im all uppity!  =]  I wonder why?  ;]

Mags
« Last Edit: March 23, 2011, 10:33:27 PM by Magluvin »

Offline neptune

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #250 on: March 23, 2011, 06:26:15 PM »
Hi Rosemary, Poynt 99 does not seem to be about at the moment , so can I offer my opinion on what he is saying ? The Fet is acting as a closed switch , and its resistance is assumed to be negligible . So we just have 2 resistors connected in series across a 60 volt battery , We have the load resistor , 11 ohms . And we have the shunt , at one quarter ohm . Imagine instead that the load resister is 10 ohms .and the shunt is one ohm . Now , if we measure the voltages across each resister in turn , we find the the voltage across the load is ten times the voltage across the shunt .That is how he arrives at his voltage figure , being the voltage across the shunt .So going back to the original values , the load will have 44 times the voltage across it than the voltage across the shunt . There will be approx 58.6 volts across the load and about 1.3 volts across the shunt . 58.6 plus 1.3 = 59.9 volts . Near enough for me , and I failed my maths exams .@Poynt99 feel free to tell me if I am wrong .

Free Energy | searching for free energy and discussing free energy

Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #250 on: March 23, 2011, 06:26:15 PM »
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Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #251 on: March 23, 2011, 07:03:14 PM »
Hi Rosemary, Poynt 99 does not seem to be about at the moment , so can I offer my opinion on what he is saying ? The Fet is acting as a closed switch , and its resistance is assumed to be negligible . So we just have 2 resistors connected in series across a 60 volt battery , We have the load resistor , 11 ohms . And we have the shunt , at one quarter ohm . Imagine instead that the load resister is 10 ohms .and the shunt is one ohm . Now , if we measure the voltages across each resister in turn , we find the the voltage across the load is ten times the voltage across the shunt .That is how he arrives at his voltage figure , being the voltage across the shunt .So going back to the original values , the load will have 44 times the voltage across it than the voltage across the shunt . There will be approx 58.6 volts across the load and about 1.3 volts across the shunt . 58.6 plus 1.3 = 59.9 volts . Near enough for me , and I failed my maths exams .@Poynt99 feel free to tell me if I am wrong .

Ok.  Thanks Neptune.  I'm still not sure why one doesn't take the actual voltage reading.  But it's close - so.  I get it. A kind of check?

Now I'd be very glad if Poynty could print those pictures.  I've finally opened them - but would be glad to get them up here - in case anyone, like me, struggles.  Can you oblige us Neptune?  Someone?

Kindest regards,
Rosie

Offline woopy

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #252 on: March 23, 2011, 07:05:51 PM »
Hi Mag and Rose

i made a quick test with a MO high voltage cap with 0.95 micro F, so it discharge fast enough to see what is going on here.

The battery is at about 4.6 volt and goes to a BAT 43 schotky diode (very few lost in voltage) than to the 220mH inductance(primary of MOT)  ant than to the cap and back to the negative.

I enclose a scope shot .
when i close the circuit the voltage jump to 7.6 volts than the cap descharge down to 4.6 volt (that is the battery voltage, and stay at this value until i open the circuit and the cap goes on descharging.

The best result i got today is a 1.7 time the voltage supplied. In this pix it is 7.7 volts(the freewheeled voltage ) / 4.6 volt (the supplied voltage) = 1.67

Not bad at all.

And in this config (but with a much stronger diode a BYV26D ) , by shorting the cap i can get really high voltage (MORE than 400 volts) impressive.

Now how to use this effect ?

Will test your new circuit ASAP, any idea for the resistor value?

good luck at all

Laurent


Offline penno64

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #253 on: March 23, 2011, 07:21:43 PM »
Last time - I promise

posted feb 19 in shorting coil thread -


********************
Hi All,

How pertinent is this guys video now ? (1 and 2)

http://www.youtube.com/user/NRGFromTheVacuum#p/u/10/2cUS03yNl40

Looks like it's been under our noses all this time and we couldn't see the forest for the trees.

Kindest Regards, Penno
________________________________________________________

Penno, (Garry)

Offline nul-points

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #254 on: March 23, 2011, 07:27:56 PM »
Will test your new circuit ASAP,

Laurent

hi Laurent

nice testing - but you may want to reconsider Mags idea of a diode across the battery/switch - as shown, it's forward biased and if the switch gets closed for too long, then the 'magic smoke' is likely to to get released from either the new diode or the battery!  :)

interesting to see that shorting the capacitor has such a strong effect - especially when the current fashion seems to be 'shorting coils'!!!

cordialement
sandy


http://docsfreelunch.blogspot.com/
 
« Last Edit: March 23, 2011, 07:57:16 PM by nul-points »

 

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