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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 670075 times)

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #600 on: April 13, 2011, 06:43:04 PM »
And this is the smallest number I can show with two complete waveforms on the one and an incomplete on the other.  Sorry about that.

Hope that puts this to bed FOREVER.

 ;D

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #601 on: April 14, 2011, 04:37:58 AM »
These worthy posts from Hum have most likely been missed (due to a delay in the admin posting them), so I shall reference them here.

http://www.overunity.com/index.php?topic=10407.msg281320#msg281320
http://www.overunity.com/index.php?topic=10407.msg281327#msg281327

.99

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #602 on: April 14, 2011, 08:56:30 AM »
And this is for Mookie, Pickle, MileHigh and those many others who share Poynty's mandate.  When and If I see a public retraction of those allegations against my sanity and my competence - when the three of you desist from confusing science with its proponents - then, indeed, I will be very happy to enter into a dialogue and may even find the time to answer those irrelevant questions of yours.

Rosemary

Offline eisnad karm

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #603 on: April 14, 2011, 09:07:44 AM »
can my alter ego and the others who without good reason have there status returned as well as being moderated sucks as it takes days sometimes for a post to appear.
Kind Regards
(It is ok Bill no need to look up my I.P, I admit who I am)

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #604 on: April 15, 2011, 02:55:17 AM »
Guys - I see that there's some confusion with the terms 'a mouth full of teeth'.  What's meant by this is that - if the mouth only has teeth then one has - by implication - lost the 'tongue'.  God forbid - as has been implied - that I am 'combative'.  That better describes the 'reach' of Pickle et al - who persistently 'hit' below the belt. 

But to get back on topic - I really need to address the endless claim that I must replace the battery with capacitors.  I would remind you all that I have never used capacitors and know nothing about them - other than it's essentially a two terminal device with two plates separated by a dialectric.  That's Wiki's definition - or something along those lines.  In any event - it is able to store charge - presumably by varying that dialectric.  In this, it is also, essentially the same principle as a battery - I would have thought?  If anyone can improve on this definition then I'd be glad to hear of it.

But here's the thing.  What got me started on this circuit - was the need to prove the thesis - which means that I've got to get back there even if I simply do this is broad brush stroke.  The idea is that we have an energy supply source - in this case, a battery - that has an innate voltage imbalance.  In other words the electrolytic mix of the battery ensures that it does not lose that voltage imbalance until it has discharged sufficient current to vary the innate charge condition of that mix.  It then discharges that potential difference in the form of current flow that is well able to move through the inductive/conductive circuit material.  Which also means that without that circuit as a path for current flow then the battery retains its charge (losses through the air excepted as they're neglible).

But, in doing so, it also TRANSFERS that innate imbalance to inductive/conductive circuit components that - prior to that discharge WERE balanced.  That's known.  Nothing esoteric here.  Now.  The intention of the circuit is this.  It applies a switch - so that the amount energy that is induced in the INDUCTIVE/CONDUCTIVE material - has the 'time' afforded it by that switch - to discharge that IMPOSED imbalance.  And it can ONLY discharge this if there is a path made available for it to do so.  That path is provided by the innate 'body diode' of the MOSFET transistor. 

BUT - in discharging it's energy it is then again PERFECTLY BALANCED - in a way that battery never is.  The battery can only become balanced when ALL it's molecules have been reordered.  That takes considerably more time than is offered by the switch.  Therefore we have two things happening.  The imbalance is transferred to the circuit material through current flow.  Then there is the required break in the delivery of that current.  The circuit material then DISCHARGES that imbalance.  In doing so it RECHARGES the battery thereby restoring the battery's imbalance.  In the same way, a plug or grid supply would offer a continual voltage imbalance.  But the induced voltage over the circuit and its sundry components is able to entirely discharge that induced voltage imbalance - given sufficient time at the switch - to do so.

Effectively one needs that relationship.  One must be able to transfer that potential difference to an entirely 'balanced' inductive/coductive component - and then allow time for that inductive/conductive component to discharge that induced potential difference back to the supply.  The supply NEVER entirely loses potential difference.  In fact, when energy is routed back through the supply it restores that potential difference.  But the circuit material DOES lose its potential difference to establish its prior 'balanced' condition. 

My suspicion is that a cap would discharge to an entirely 'balanced' condition.  In which case we would not be able to perpetuate the required voltage imbalance at the source.  But I'm working in the dark as I've never tried it.  If this is wrong - and if the cap NEVER discharges to zero - then it is feasible.  What's needed is a supply that DOES NOT discharge to zero. 

Anyway.  I'm open to correction.  But that's the reason I'm not sure that capacitors would work.

Kindest regards,
Rosemary     

added.

I've just re-read this.  Not sure if it's clear.  Effectively one needs a continual imbalanced source and circuitry that is first balanced and then imbalanced.  And balance here used in the sense that it's charge imbalance measured as potential difference.
« Last Edit: April 15, 2011, 03:33:23 AM by Rosemary Ainslie »

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #605 on: April 15, 2011, 04:17:58 AM »
And guys, this is that circuit that I referred to some time back.  I have drawn it with a switch.  It needs a transistor - ideally a MOSFET - but I have no idea how this is to be done - what supply to use - or where to establish the common ground for those switches. 

But if anyone here is clever enough to put this together - then here's the thing.  It would be a far better way to prove the principle that I'm trying to bring into focus.  Effectively, in this arrangement - the one battery will always discharge to recharge either itself or the other battery. 

If, as is claimed by the thesis, that charge is entirely retained, in other words, if current flow ONLY ever returns to the source and IF there is no material loss at that supply - then this would unarguably be a closed system. 

And as a reminder.  The idea is that in the transfer of energy through a circuit NO material is lost or gained by the suppy.  In other words, current flow belongs to its source and only ever returns there.  In the same way - current flow that is induced from the circuit material ALSO ony returns back to that material.  The heat that is dissipated on circuit components is a different discussion.  I'll get there.

Be very glad if someone can resolve the principles in the circuit.  It's been something that I've been toying with for about 2 years. 

Kindest regards,
Rosemary

Hope it's clear.  I can't seem to resize my pictures any more.

I'm trying this again.  I think I've resized it.   
« Last Edit: April 15, 2011, 02:32:11 PM by Rosemary Ainslie »

Offline nul-points

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #606 on: April 15, 2011, 05:59:43 AM »
this is that circuit that I referred to some time back.  I have drawn it with a switch.  It needs a transistor - ideally a MOSFET - but I have no idea how this is to be done
...
 It would be a far better way to prove the principle that I'm trying to bring into focus.  Effectively, in this arrangement - the one battery will always discharge to recharge either itself or the other battery. 
...
Be very glad if someone can resolve the principles in the circuit.  It's been something that I've been toying with for about 2 years. 
...
Rosemary

hi Rosemary

i don't think that your new circuit will operate as you expect

because the circuit is circular, the Diode/Switch pairs and the remaining individual components can be moved around the ring to any position and the circuit will still be exactly equivalent

therefore your circuit (on the left, below) can be arranged, for example, as the other two circuits below

it might be easier to see from the arrangement on the right:-

a) if the switches always operate in anti-phase, then the only current which will ever flow is due to the combined voltage of the two cells (or batteries) flowing through the combined series resistance (R1 + R2 + reverse impedance of either D1 or D2) <<depending which switch is closed

ie. B1 & B2 will take a long time to discharge because basically the circuit is not doing a great deal (eg. the cells are merely discharging via diode leakage current, regardless of duty cycle or switch rate)


b) if the switches should ever be allowed to operate in-phase, then when both switches are ON the current will be (V1+V2)/(R1+R2);
  and when both switches are OFF the current will be similar to (a) (except the diode leakage will be due to two diodes in series)

ie. the discharge behaviour of B1 & B2 will depend on the duty cycle and the switch rate


** ignoring milliohm resistance in the switch on-states in both (a) & (b)


hope this helps
np

http://docsfreelunch.blogspot.com
 
« Last Edit: April 15, 2011, 06:23:43 AM by nul-points »

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #607 on: April 15, 2011, 07:09:35 AM »
Hi Nul-Points.  Many thanks for that - but I was hoping you or someone would design it with a transistor - something appropriate to allow that dual path.  I agree it won't work with a reed switch or somesuch unless one can get a difference in the two battery supply voltages. 

Can I impose on you to try a design on this principle?  Is it even possible?  I just don't know enough about switches to answer this myself.

But either way - many thanks for this work.  I'll get back here and try and explain the point of it all - but it will be later today.

Kindest regards,
Rosemary

Offline nul-points

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #608 on: April 15, 2011, 12:17:30 PM »
Hi Nul-Points.  Many thanks for that - but I was hoping you or someone would design it with a transistor - something appropriate to allow that dual path. 
...
I agree it won't work with a reed switch or somesuch unless one can get a difference in the two battery supply voltages. 
...
Can I impose on you to try a design on this principle?  Is it even possible?  I just don't know enough about switches to answer this myself.
...
Rosemary

hi Rosemary

here's one possibility below (RA-2c)

i'm not sure why you think there will be any effective difference in behaviour between RA-2b and RA-2c ?


if i correctly understand the weight of all your previous explanations, so far,  then the diodes provide the '2nd path' in your 'dual path' expectations of switched current

if that is so, then for your purposes, circuits RA-2b and RA-2c should effectively provide the same action

the net effect is the same - both B1 & B2 would discharge over time, regardless of what was happening at S1 & S2 - or Sig 1 & Sig 2


a "difference in the two battery supply voltages" is not relevant to the operation of your circuit (RA-2a), or the equivalents which i've shown (RA-2b & RA-2c), since the two cells ("batteries"), as you can see, effectively form one true battery of voltage (V1 + V2)


hope this helps
np

http://docsfreelunch.blogspot.com
 
 

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #609 on: April 15, 2011, 02:17:24 PM »
Hi again nul-points.  You've sort of put two batteries in series.   And then given two loads with two switches.  Is that all I've managed in that design?  LOL.  If so, let me try this again.  I'll get back here when I've redrawn it.

Many thanks for your efforts there by the way.  But I'm still trying to point to something.  I think I must make a small adjustment there.  I need to explain it more fully by getting it workable with those reed switches.  Then - possibly you'll be able to put in the required transistors.

Hang ten.  It takes me ages to draw - photo - then down load. 

Kindest as ever,
Rosie

Offline poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #610 on: April 15, 2011, 02:43:22 PM »
I had posted this in February. Maybe it's not what you had in mind.

.99

Offline nul-points

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #611 on: April 15, 2011, 03:19:02 PM »
Hi again nul-points.  You've sort of put two batteries in series.   And then given two loads with two switches.  Is that all I've managed in that design?  LOL.  If so, let me try this again.  I'll get back here when I've redrawn it.
...
Rosie

hi Rosemary

yes, that's effectively what you drew above:
(hence my comment that your circuit may not operate as you expect)

    two cells in series, two resistors in series, two switches in series

it doesn't matter what order these occur in the ring circuit
(the diodes across the switches always need to have the same polarity sense, of course)

you also indicated that you were looking for a solid-state solution - hence no relays in the equivalent circuit i posted

no worries

all the best
np


http://docsfreelunch.blogspot.com
 

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #612 on: April 15, 2011, 03:36:02 PM »
Poynty - I'm not sure if that cuts it.  I'll need to print it out.  Can't quite read it off the computer.  Unless you can do it a tad smaller.  But I should be able to get it printed today.  Meanwhile here's another shot at showing what I'm trying to get to. 

Sorry about the smudges.  I think this may just work with reed switchings and switching in antiphase.  Anyway - it's what I'm trying to point to.  Just think of the current switching around the middle like two inverted 's's or alternatively think of a swastika going first one way and the then the other. 

Sorry about the smudges.  Not the best drawing in the world.

Kindest regards,
Rosemary

Offline Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #613 on: April 15, 2011, 03:48:23 PM »
Sorry about that drawing.  Golly.  It's now got HUGE.  And it's so smudged.  :o

Anyway.  Here's what I'm trying to do.  The positive current from either battery always first feeds into the positive terminal of the other then finds it way back to the negative terminal of the source.  Then the negatively induced current during the 'off time' first gets routed to the negative terminal of the other battery before it returns to the positive terminal. 

That way - as I see it - every 'on' will result in a recharge as will every 'off'.   I'm open to correction here.  And I have no idea if it's even feasible.  It's just a principle that haunts me.  It seems to answer that closed system that you all look for.  I'm entirely satisfied that it will do no more nor less than our own circuit.  But it answers the logic better.  And then, only if it's doable at all.

Anyway.  I'll await comments and see if I can wrap my head around Poynty's design.  Thanks for that.

Kindest regards,
Rosie

Poynty - please get those fatuous and entirely erroneous comments of FuzzyTomCat off that thread.  It's already been polluted out of mind.

Thanks
R

Offline nul-points

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #614 on: April 15, 2011, 06:18:31 PM »
Here's what I'm trying to do.  The positive current from either battery always first feeds into the positive terminal of the other then finds it way back to the negative terminal of the source.  Then the negatively induced current during the 'off time' first gets routed to the negative terminal of the other battery before it returns to the positive terminal. 

That way - as I see it - every 'on' will result in a recharge as will every 'off'.   I'm open to correction here.  And I have no idea if it's even feasible.  It's just a principle that haunts me.  It seems to answer that closed system that you all look for.  I'm entirely satisfied that it will do no more nor less than our own circuit.  But it answers the logic better.
...
Rosie

ah Rosemary - your previous circuit was better!  ;o)


i believe that you have a clear goal in mind - but unfortunately not a clear path ahead as to how it might be achieved

current will flow through a circuit path only when there is a net potential difference applied across it


as an exercise in becoming more familiar with circuit design, i can recommend a next step for you, after an initial sketch attempt at expressing a new idea:

consider each source of energy in turn, and, with a few helpful example component values to make the maths go more smoothly, get a feel for the main current paths, values, and hence voltage drops around the various series & parallel paths between the +ve & -ve of that supply

this approach will help you develop a better 'feel' for how the various circuit elements are likely to work together - and hopefully save you wasting too much effort on the non-viable circuit ideas

all the best
np


http://docsfreelunch.blogspot.com