Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Groundloop]

Hi Groundloop.  The short answer is that I don't know.  The whole thesis requires an intimate relationship between the mass of the circuit and the mass of the supply.  What I do know is that we've had similar results with a single battery - so.  It's more than possible.  But our own circuit does NOT oscillate with a single battery supply.  Which also means that this mass ratio may have to be factored in.  We can certainly get some significant results with just two batteries - but there again - we don't get that high dissipation which, in my opinion - is what needs to be tested to the duration.  But frankly - I"m not interested in this side of things.  We included our battery draw down rate in a previous submission - the one published in Quantum.  There we ran two tests concurrently with a control dissipating the same level as our test.  The control was dead - at 10 volts per battery - when the experimental batteries had not even dropped by a quarter volt.  We included those results and Professor Jandrell - the reviewer - categorically refused to allow any reference to battery draw down rates.  Apparently it establishes NOTHING conclusively.  And my interests are ONLY to get this to the academic forum - somehow.  And they ONLY want classical measurement protocols.  Quite apart from which any evaluation of the battery performance actually DOES need chemistry experts.  We're none of us qualified to comment - not on our team.

But it should not be required.  Think back.  Everyone looked at the evidence of battery performance on Bedini's tests and then they wanted MEASUREMENTS.  Now that they've got MEASUREMENTS they want battery performance.  It should be MORE than enough to point to the fact that our batteries have now lasted 9 months without a recharge.  And they're ALL at the kick off voltages.  What more evidence is there required? 

In any event - I'll do those tests gladly.  But provided ONLY that it EITHER is considered conclusive by all EXPERTS - or that someone pays me to do them.  I'm not picking up another tab unless there's some value.  These last three years have cost an awful lot of money.  And I've dedicated my time here with none of the benefits  that I normally get in trading.  And it's been a thankless task.  A baptism by fire.  And the only plus is that it's attracted the attention of some academics that we could, at least, get it to a technical college for progress.  The rest of the experience has been decidedly futile.  I realise now that there's way more ego here than anything else.  I just wish I had not learned so much about so many.  I preferred it when I thought these forums were crammed by genuine enthusiasts.  I now feel that they're just a bunch of old lady bashers - and that they selectively choose their evidence on the basis of the masculinity or otherwise of the members.  Which makes me somewhat underqualified.  Not a happy time Goundloop.

Kindest regards,
Rosemary

Rosemary,

Thank you for taking time to answer my question. So you need a rock solid power supply to make it work.
Will it work if we rectify the mains and make a rock solid power supply? I'm thinking of a "black box" solution
where we can plug the system between a space heater and the mains. Then we can use a "Kill'oWatt"
meter to measure the power usage into the circuit. Now if the space heater is marked 2000 Watt/h and the
"Kill'oWatt" meter is shown less that that, then you have proved that your circuit has a practical application
benefit. What do you think of such a solution?

Regards,
Groundloop.

[Rosemary Ainslie]

Lol you already have the batteries and the circuit/ function generator. Why would it cost you a dime

Why indeed?  You make a good point HAPPY.  We'll need to monitor those tests.  But that's the easy part.  I'll do the first solid 8 hour night shift - as that's likely to be the least popular.  Then I'll get someone to do the next 8 hour shift - and a third person to do the third 8 hour shift in every 24 hours.  We'll all have to 'hole in' because we won't be able to take a break for lunch or tea - or anything at all.  But that's no problem.  Provided only that we get to the solution.

Now.  Conservatively speaking on the water to boil test - we're running that battery current at an outside maximum of 0.1 volts over 0.9 Ohms x 60 volts x 18% duty cycle which represents the ONLY measurable discharge from the battery - and that during the ON period of each switching cycle.  That comes to a wattage discharged at 1.19 watts.  The capacity of each of those batteries is let's say 10 amp hour MAX.  Therefore each battery's maximum wattage potential is 120 watts - being 10 amps x 1 hour.  We use 5 in series to get it to that 'water to boil' point.  The batteries are in series.  So.  1.19/6 = 0.199 watts per battery.  Again.  Each battery has a capacity of 10 amps x 1 hour = 120 watts.  This means that each battery will last 120/0.19 = about 630 hours or so.  Theoretically.  5 batteries will therefore last a mere 3150 hours or so. 3150 hours / 24 means the test should be proved after a test run of only 131 days.  So now the three of us will be able to give you a conclusive result over a little under a 4 months.  That's assuming that any energy at all is being discharged at the battery.  Because we can't find that in the results.

Alternatively - let us assume that we're dissipating not less than 100 watts as measured in the heat discharged.  That would be the amount of energy needed to get a little under 1 liter of water to boiling point.  Now we get to the following sum.  The battery's capacity is 120 watts.  We've got 6 in series giving us a maximum capacity of 720 watts.  We're dissipating 100 watts as evident in the temperature of the water.  Therefore in a mere 7 hours and 12 minutes we would have ENTIRELY discharged all those batteries. 

Now - this is the question.  Have we run that test period?  We've been running those tests for about 5 hours a day for a period of 9 months.  NOTA BENE.  We have NEVER recharged those batteries except for the 2 that caught fire.  We've tested varying extremes of temperature over that entire time.  We have now also tested them on different loads and different driving signals.  THERE IS ABSOLUTELY NO EVIDENCE OF ANY LOSS OF VOLTAGE OVER THOSE BATTERIES.

And again.  A 7.5 hour test period is DOABLE.  I will happily oblige.  But we'll still be left with a ZERO DISCHARGE.  I'll report this.  No-one will believe it.  They'll ask for double.  Then they'll ask for triple.  And so it goes.  Finally they'll ask for 130 days.  If no-one is satisfied, by now that there is a zero discharge to the battery, then THERE IS ABSOLUTELY NOTHING THAT WILL SATISFY ANYONE EVER THAT THE BATTERIES ARE NOT DISCHARGING.

I am on record.  If you want me to run that test for 10 days continuously - and if you will pay for the monitoring THEN I WILL DO IT.  Because I cannot expect 2 people to sacrifice 8 hours a day for a 10 day period to come and sit tight and see that nothing catches fire.  I will need to pay them.  And frankly - even at the end of that monitoring period I am ABSOLUTELY SATISFIED that NO-ONE WILL ACCEPT THAT EVIDENCE. 

SO.  DO NOT ASK ME TO RUN THAT BATTERY CONTINUOUSLY.  We've done this.  Conservatively speaking - over a 9 month period we've dissipated not less than 4500 watts calculated at a conservative 100 watts a day x 5 days a week x 9 months.  And ALL THOSE BATTERIES ARE ALL STILL FULLY CHARGED.

Rosemary 

[Magluvin]

And it has to be constantly monitored. There could be a fire, etc.  Its not a flashlight were talking about here. ;]

Most probably think you can just check in every week for 6 months, no trouble at all.  sheesh.  ;]

Hows it going Rose?  Just keep on keepin on.  ;]

Mags

[Rosemary Ainslie]

And it has to be constantly monitored. There could be a fire, etc.  Its not a flashlight were talking about here. ;]

Most probably think you can just check in every week for 6 months, no trouble at all.  sheesh.  ;]

Hows it going Rose?  Just keep on keepin on.  ;]

Mags

Hi Mags.  Always nice to see you around.  Take care of yourself, and thanks for the support.  Always in need of it

Rosie

[Rosemary Ainslie]

Rosemary,

Thank you for taking time to answer my question. So you need a rock solid power supply to make it work. Will it work if we rectify the mains and make a rock solid power supply? I'm thinking of a "black box" solution where we can plug the system between a space heater and the mains. Then we can use a "Kill'oWatt" meter to measure the power usage into the circuit. Now if the space heater is marked 2000 Watt/h and the "Kill'oWatt" meter is shown less that that, then you have proved that your circuit has a practical application benefit. What do you think of such a solution?

Regards,
Groundloop.

There's nothing wrong with your suggestion - as ever GL.  Except that we don't have the tolerances in standard MOSFETS.  There is no amount of paralleling that will accommodate those high voltages that are induced in the oscillating cycles.  Get past this problem and we're there.  We don't even have to speculate on the outcome.  It is most CERTAINLY result in a benefit.  We've tested this through a variac with VERY good results.  But then the objections were based on the amount of energy drawn by the variac. 

Groundloop - you're the only one who's really applying himself to a solution and I'm awfully grateful for the efforts.  I need to get to the nitty gritties of the technology and I'm not sure that you want to bend your mind around this.  If you've got the time and the patience please read the following carefully.

We RELY on a current that has a FIXED direction.  Now.  When you apply a rectified AC signal here's what happens.  The sine wave moves above and below zero.  The assumption is that the positive voltage generates a current that moves clockwise through the circuit.  And, correspondingly, the negative voltage moves the current flow counter clockwise through the circuit.  So.  In your mind's eye - see this as an applied current in one direction - changing back to zero and then moving in the opposite direction.  Effectively you have the application of two current flows from every one cycle.

Then you rectify this.  Now what happens?  Here the negative component of that sine wave is only REDIRECTED.  It's STILL a negative flow.  But instead of presenting at the positive or drain of the circuit - it's routed through the 'back door' so to speak - that it presents at the drain as a positive.  It has NOT followed the same path as the applied positive.  It's still flowing in a different path.  In effect the source of the positive and the source of the negative current flows are still different. 

What the thesis relies on is this.  Every current flow has a distinct CHARGE.  A positive current flow leads with the positive - something like this >-+ -+ -+.  And correspondingly a negative current flow leads with a negative charge - like this <-+ -+ -+.  Now.  Put those two together as they present in real time as in a sine wave.  It would be >-+ -+ -+>then<-+ -+ -+<.  You see for yourself that there's no conflict of like charges.  NOW.  Back to our circuit and let's factor in the charge at the gate.  There's an applied current coming from the battery.  It's justification is >-+ -+ -+>  In order for this to find an appropriate charge at the bridge the functions generator would need to APPLY a '-+' because IF that charge was presented '+-' then the two positives '-+ +-' would repel.  But to present that '-+' the functions generator is actually presenting a NEGATIVE charge that simply 'reads' as a positive charge relative to the system.  So.  Now we correctly should see that as a charge coming from the generator <-+ .  It's leading with a NEGATIVE signal.  The battery current sees this applied charge as a PERFECTLY POSITIVE charge alignment to link up and flow to its own source or negative rail. But the DIRECTION of that charge from the FG has reversed.

THEN.  The charge presented at the gate reverses.  It moves from a <-+ to a >+-.  NOTA BENE AGAIN. The direction of charge from the FG has reversed relative to supply.  The justification of the current from the battery is still >-+.  It meets with the applied <+- charge.  And then it is repelled.  It SHUTS OFF the supply.  The battery can no longer discharge.  AGAIN.  We now have >-+ -+ -+ and then <+-. 

Now there's nothing to stop the  potential difference on the circuit components from discharging their own voltage - that potential difference established during the time that the current flowed from the battery.  It discharges to zero.  The discharge involves changing magnetic fields.  Changing magnetic fields induce electric fields.  This induces a reverse polarity which - in turn - produces a reverse current flow.  And this current flow is coming from the inductive components of the circuit material.  This now presents as a NEGATIVE VOLTAGE with a NEGATIVE AND REVERSE CURRENT FLOW.  So what we now have is this.  <-+ -+ -+. Bear in mind the presentation of charge at the gate of that MOSFET.  It is still  <+-.  So there is NO WAY the induced current can flow through that Q1 MOSFET.  Here again is the configuration of charge.  From the battery <-+ -+ -+ and <+- at the gate.  Which is where the body diode of Q2 kicks in.

The system is now discharging current with the following justification.  <-+ -+ -+ and the body diode of Q2 is perfectly polarised to allow this flow of current. So.  In discharging that current it is also moving from a high negative voltage back to zero.  In the process of discharging it represents changing magnetic fields.  And changing magnetic fields induce changing electric fields.  So.  This current then first DISCHARGES to zero in a counter clockwise direction.  Then it moves through zero to a positive value relative to zero.  Then it again becomes >-+ -+ -+.  But it's passage through Q1 is still blocked because - relative to this directional flow the charge at that gate is +-. However it's passage through that body diode of Q2 is enabled provided that the charge presentation stays the same but that the directional flow is reversed.  That much explains the paths for the flow of this current.

But here's the thing.  Why does that second discharge - that cycle that is triggered by the negative voltage from the system - then result in an INCREASE in the voltage across the battery?  Under normal circumstances a positive current flow will represent a REDUCTION to the battery voltage.  Instead of reducing the battery voltage now climbs from it's previous value of, say 60 volts - to anything up to and greater than twice its rated voltage potential to about 150 volts or more?  Here's why.

IF indeed we have two energy supply systems working in anti phase to each other - then a discharge of energy from the one will represent a recharge of energy to the other.  Therefore, correctly, when the second supply source is discharging it's potential difference - during that negative flow of current - then this will present an increased negative potential to the entire system.  The probe across the battery is measuring the REAL potential difference on the ENTIRE circuit.  It first measured a total of +60 volts.  Now it's factoring in the applied potential difference at its ground.  Ground now measures something less than -60 volts.  And the resulting voltage then moves from plus to +60 to something considerably less than it's earlier voltage.  Notwithstanding which, the current that is measured to flow through that battery is, nonetheless, flowing in a counter clockwise direction and thereby indicating a 'recharge' cycle to that battery supply source. 

And correspondingly when that new cycle kicks in.  When the negative voltage AGAIN moves into it's positive cycle, as required by Inductive Laws, with a current flow that is now 'permitted' or allowed for, at the gate of Q2 - then the voltage over the battery climbs from it's previous value of say +60 to something more than twice it's value as it also includes the positive value of voltage or potential difference that is now available from the circuit material. 

So.  The apparent anomalies are perfectly resolved PROVIDED ONLY that one FIRST assumes that current flow can be positive or negative, that the charge of the material comprising current flow can also be positive or negative and finally that there is a second or supplementary supply source of potentials and energy from the circuit material itself.  And as this impacts on conventional or mainstream thinking - it's only departure is as it relates to the nature and properties of current flow.  And since that has NOT, historically, been resolved, then this all should assist in resolving those question.

Golly.  This is probably a very long post.  I don't think I'll risk a preview to prove this because I don't feel up to shortening it.

Regards,
Rosemary