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Inductive Resistor Open source development of highly efficient inductive resistor circuits.

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  #1381 (permalink)  
Old 07-29-2009, 07:22 AM
jas_bir77 jas_bir77 is offline
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Thumbs up suscibtion activated

hi, arron i think i was able to activate my subscription of this thread after trying it for sometime
thanks anyway.

(will mail you clarifications again if i do not receive the notifications, since i am not sure if i have my subscription activated or not).
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  #1382 (permalink)  
Old 07-29-2009, 07:24 AM
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@Jas

Hi Jas,

Sounds like it worked out. Glad to hear it!
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Old 07-29-2009, 07:57 AM
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TK's results

Here is TK's results from OU.com

Re: Claimed OU circuit of Rosemary Ainslie

« Reply #794 on: Today at 07:48:52 AM »
The paper will contain lots of stuff like this, and my raw data and spreadsheet analyses will be made available to anyone that wants to see them. And I take pictures of the pertinent waveforms, too.

The below comments apply to the Ainslie circuit, driven by the Datapulse 101 at 4.5 percent ON, clean 5 ns risetime pulses. The gate atten is turned slightly down to make a current "B" waveform similar to Aaron's, with a nice smooth voltage ringdown on the load point "A". Battery voltage went from 25.1 at the start to 24.9 at the end, and after 15 minutes sitting came back up to 25.1.

Now, as anyone can see from the graph below, the Ainslie circuit does not put as much of the input power to the load, as does a DC supply at about the same average power.

So the only hope for OU at these parameters is if the difference is put entirely back into the battery, and somehow is multiplied by at least a factor of ten.
And the only way that's going to happen is if someone does the multiplication themselves.
In error, of course. Or in willful misunderstanding.

http://www.overunity.com/index.php?a...=3657 3;image
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  #1384 (permalink)  
Old 07-29-2009, 09:55 AM
Joit Joit is offline
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The Picture shows actually, with a Batterie the consumption is lower as with a DC Supply. just that.
But he can not confirm, that he have a high COP.
For the Voltage at the Batteries, someone still can guess, its only Amperage, what going out from it, and Voltage settle back.

But for me, i have usual a decrease of Voltage too, when i take out Current from the Batterie.
And by TK i never wanna feel secure with him, you never know what comes at next,
like show one Time a better Result, to refute it with the next Post.
For the Rest i save my Comments about it.



MH About the Caps, i tried them anywhere, like in overall,
with different Sizes from 47µf up to 2200µF at the Circuit.
Putting unpoled in Serie is stopping the Spike too, and just take the Voltage out from the Spike, perfect, when you want to smooth something out.
That actually shows, that you did not play a lot with Spikes around,
and only have your Theories about.

.
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  #1385 (permalink)  
Old 07-29-2009, 10:13 AM
witsend witsend is offline
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Joit, I don't think any of us were expecting to have TK validate our results. But they've been documented and I'm sure he'll make the data available on request.

maybe now we can get on with our own work here. I think Aaron's getting some decent scopemeters to get better measurements. And we'll see what our little thread will produce by way of results. But it's very important to check that we do things properly from here on. Because we need to keep it very open to srutiny. We'll need advice on test protocols and such like.
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  #1386 (permalink)  
Old 07-29-2009, 03:59 PM
gotoluc gotoluc is offline
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Hi everyone,

I made a quick change to the circuit this morning. I removed the 3 batteries in parallel on the flyback side and replaced it with a 30,000uf capacitor to give a large storage so we can tune the duty cycle and get our 12.95vdc with load (bulb) attached to have a comparison to the previous test 11.

It was so fast to see the results that even after 5 minutes it was obvious to me that there is a real humongous efficiency bonus using the batteries.

Video link: YouTube - Effect of Recirculating BEMF to Coil test 12

I had to stop the test after 1 hour and 40 minutes since the voltage on the source batteries started making large drops to 35vdc and climb back to 38vdc and back down again. I don't know what is up with that but anyways it's clear that there is an astronomical efficiency boost using the batteries in parallel on the inductive flyback side.

I hope others can replicate this so we can advance to an even more efficient way to use this effect. I'm now realizing that if I would of tested this method in my resonance experiments I may of had some better results. So many things to test now.

Talk to you all soon.

Luc
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  #1387 (permalink)  
Old 07-29-2009, 04:24 PM
rave154 rave154 is offline
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GOTO,

another great video.

i think a next possible step would be combining the tesla switch circuit with your circuit to try and get the best of both worlds in one circtuit maybe?

Rosemary,

PWM = Pulse Width modulation, basically the 555 circuit driving the pulses to the mosfet.

As for it being normal for there to be a 'pause' for a neon to light up, i dont know if its normal, but every other time ive used a neon with other circuits there doesnt appear to be a noticable 'pause' and the pause im getting with this circuit is VERY noticable as its almost a full second.

The 'flash" i refered to happens when i disconnect the negative lead of the battery that is powering the 555 part of the circuit ( not the coil ), it happens once only for each disconnection.The flash DOES NOT happen if say, while the circuits are running i discconect the SOURCE lead of the mosfet from ground, disconnect the PIN-3 output of the 555 from the mosfect, disconnect the positive of the coil from the positive rail....it only happens upon disconnection of the negative ( and positive ) lead/s of the 555 circuit from its battery.

dont worry about not getting my "lingo" because its just the way i tend to explain things, if i dont know the correct technical term for something i tend to describe things in an odd way sometimes.

David. D
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  #1388 (permalink)  
Old 07-29-2009, 04:24 PM
witsend witsend is offline
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Hi gotoluc. Not sure if you're around still to see this post. HOW INTERESTING IS THAT? It seems that you've hit on something. I wonder if the trick is to use batteries as a better way to prove values. It seems to be the case.

Well done Luc. I think we all have a better idea how to do the base testing now. That video was such clear evidence of the gain that you've been seeing without quite being able to measure.

Yet again - many thanks, have a wonderful break and hurry back. We're still in great need of your input.
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  #1389 (permalink)  
Old 07-29-2009, 04:26 PM
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Michael John Nunnerley Michael John Nunnerley is offline
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Brilliant as always

Quote:
Originally Posted by gotoluc View Post
Hi everyone,

I made a quick change to the circuit this morning. I removed the 3 batteries in parallel on the flyback side and replaced it with a 30,000uf capacitor to give a large storage so we can tune the duty cycle and get our 12.95vdc with load (bulb) attached to have a comparison to the previous test 11.

It was so fast to see the results that even after 5 minutes it was obvious to me that there is a real humongous efficiency bonus using the batteries.

Video link: YouTube - Effect of Recirculating BEMF to Coil test 12

I had to stop the test after 1 hour and 40 minutes since the voltage on the source batteries started making large drops to 35vdc and climb back to 38vdc and back down again. I don't know what is up with that but anyways it's clear that there is an astronomical efficiency boost using the batteries in parallel on the inductive flyback side.

I hope others can replicate this so we can advance to an even more efficient way to use this effect. I'm now realizing that if I would of tested this method in my resonance experiments I may of had some better results. So many things to test now.

Talk to you all soon.

Luc
Nice one Luc,

This is showing that there has to be an energy convertion, be it a chemical like with the batteries or as I am doing at the moment, but not yet finished, a type of motor invertor or convertor as the case may be. I will know when I have finished constructing and start some tests.

Speak to all when I can

Mike
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  #1390 (permalink)  
Old 07-29-2009, 04:32 PM
witsend witsend is offline
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Quote:
Originally Posted by rave154 View Post
GOTO,

dont worry about not getting my "lingo" because its just the way i tend to explain things, if i dont know the correct technical term for something i tend to describe things in an odd way sometimes.

David. D
Hi Rave154. Thanks for explaining PWM. I tell people often. You cannot overestimate how little I know. I actually don't know the answer to your questions. Wish I could help. I've got a shrewd idea that Aaron could. Or someone?

And it certainly isn't your lingo. I'm sure everyone bar me knows what you're saying. The day may yet come when I can understand you guys. But that day is not exactly around the corner. Sorry. At least I gave it my best shot.
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  #1391 (permalink)  
Old 07-29-2009, 04:56 PM
Hoppy Hoppy is offline
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Quote:
Originally Posted by gotoluc View Post
Hi everyone,

I made a quick change to the circuit this morning. I removed the 3 batteries in parallel on the flyback side and replaced it with a 30,000uf capacitor to give a large storage so we can tune the duty cycle and get our 12.95vdc with load (bulb) attached to have a comparison to the previous test 11.

It was so fast to see the results that even after 5 minutes it was obvious to me that there is a real humongous efficiency bonus using the batteries.

Video link: YouTube - Effect of Recirculating BEMF to Coil test 12

I had to stop the test after 1 hour and 40 minutes since the voltage on the source batteries started making large drops to 35vdc and climb back to 38vdc and back down again. I don't know what is up with that but anyways it's clear that there is an astronomical efficiency boost using the batteries in parallel on the inductive flyback side.

I hope others can replicate this so we can advance to an even more efficient way to use this effect. I'm now realizing that if I would of tested this method in my resonance experiments I may of had some better results. So many things to test now.

Talk to you all soon.

Luc
Luc

Your parallel set of three batteries in the previous tests were supplying power to the lamp load (7W). This latest test shows the battery bank replaced with a capacitor which has stored energy at only a tiny fraction of the level stored in the battery bank. Surely then, it is not difficult to understand why the primary set of three series connected batteries will discharge faster when it has to supply the full lamp load and hold the capacitor at a constant voltage, as the load is now not being shared between the two sets of batteries.

Hoppy
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  #1392 (permalink)  
Old 07-29-2009, 06:12 PM
poynt99 poynt99 is offline
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Quote:
Originally Posted by Hoppy View Post
Luc

Your parallel set of three batteries in the previous tests were supplying power to the lamp load (7W). This latest test shows the battery bank replaced with a capacitor which has stored energy at only a tiny fraction of the level stored in the battery bank. Surely then, it is not difficult to understand why the primary set of three series connected batteries will discharge faster when it has to supply the full lamp load and hold the capacitor at a constant voltage, as the load is now not being shared between the two sets of batteries.

Hoppy
That's 100% correct.

Think about this a bit guys.

.99
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  #1393 (permalink)  
Old 07-29-2009, 07:31 PM
witsend witsend is offline
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following quotes from Hoppy
(I rescued this point from some posts back?)

Its not a case of breaching battery resistance. The battery resistance is very low and offers an ideal path for the discharging inductor.
Ok. You're saying that the supply battery establishes a voltage across the inductor that exceeds the voltage of the second battery.

The voltage will fall almost instantaneously to a value slightly above the battery terminal voltage;
I assume you mean that the voltage drop is measured across the inductor when the switch is open and for some reason this voltage is then determined at a value dependant on the second battery. And one can then see the recharge value as a voltage across the terminals of the second battery.

You need to see this initially high voltage as a conversion of low voltage / tension to higher voltage / tension.
Why tension? Voltage is voltage - unless you need to give it a property beyod voltage?

It actually posseses less energy than that used to initially charge the inductor.
What voltage are you talking about here Hoppy? I can't understand you. Voltage is voltage. It has as much energy as is measured. It doesn't hide some and show some - like some weird kind of 'peek a boo' show - now you see me now you don't.

The loss is in heat due to the resistance of the wire forming the inductor
What loss? I'm referring to the second battery recharging. What loss?

It may appear to be more because the average discharge voltage across the battery is high enough to charge it, whereas the original voltage level before conversion is too low.
If this means anything at all it entirely eludes me. A discharge voltage recharges.

The available current will be less at discharge because the initial peak voltage was a lot higher. The available power on discharge will therefore also be less.
Since when is current lower in the event that the voltage is higher? Are you trying to proportion the actual amount of energy transferred to some ratio between voltage and current. Voltage and current are precisely proportionate if they are measured on a circuit with a path for the current to flow.

The 1000 ohm resistor is not at all critical. It can be less or more in value. Its just to show that the discharge voltage can be a lot higher than your primary battery voltage. Experimenting in this way will reveal a lot about how your circuit works and it will eventually become apparent to you that there is no free lunch in a circuit of this type.
I'm willing to discuss this but not based on this argument. Mostly because I simply cannot follow it.

I think we need to get back to the point here. We have a circuit where the second battery is disconnected from the supply battery but we're testing the amount of energy available from the inductor which may be stored, per classicists or regenerated per new age. We want to establish which is correct.

So If a second battery is recharged - and there is evident dissipation of energy from a resistor - then where did the extra energy come from to recharge the second battery? The first counter argument to recharging the battery was based on the fact that there was no extra energy. The battery delivered (as an example) 5 watts to the load resistor - spent 3 watts in heat and simply took back 2 watts to recharge. Zero gain. Alternatively, the argument was - the circuit delivered 5 watts to the load, spent 3 watts in heat and recharged a second battery with 2 watts. Still no gain.

Our argument is this. The battery delivered 5 watts of energy - spent 3 watts on the load - sent 2 watts back to recharge the battery and then took those same 2 watts of energy back to the resistor to add another 2 watts in heat. Definitely a gain. The only difference is that I've grossly understated the amount of energy that went back to the battery to recharge it. The rate of charge on the second battery proves this.

I've done my best to get my head around your classical concepts. How about trying to wrap you head around mine? I at least can show that the numbers support this claim. It's not a free lunch. It's just much cheaper than we've had to pay to date.
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  #1394 (permalink)  
Old 07-29-2009, 07:35 PM
witsend witsend is offline
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Quote:
Originally Posted by poynt99 View Post
That's 100% correct.

Think about this a bit guys.

.99
Are you back in the battle ground here Poynt? Or just keeping a fatherly eye on procedures? Nice to see you if the former.
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  #1395 (permalink)  
Old 07-29-2009, 07:39 PM
witsend witsend is offline
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Quote:
Originally Posted by Hoppy View Post
Luc

Your parallel set of three batteries in the previous tests were supplying power to the lamp load (7W). This latest test shows the battery bank replaced with a capacitor which has stored energy at only a tiny fraction of the level stored in the battery bank. Surely then, it is not difficult to understand why the primary set of three series connected batteries will discharge faster when it has to supply the full lamp load and hold the capacitor at a constant voltage, as the load is now not being shared between the two sets of batteries.

Hoppy
You're entirely missing the point here Hoppy. The batteries were dissipating energy at the light and yet recharging. In the second example, no recharge cycle and the actual rate of discharge could be seen over the supply batteries.

EDIT - In point of fact he showed the rate of discharge of the supply batteries when the recycled current was not used to recharge. In effect the benefit in the recharge cycle was simply 'thrown away'. He could just have easily put the light across the supply source batteries and systematically depleted each one.

ANOTHER EDIT - and the evident rate at which those batteries would deplete was clearly far greater than was evident over the 24 odd hour test that he ran the test with virtually no visible loss of energy anywhere.

Last edited by witsend : 07-29-2009 at 08:12 PM.
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  #1396 (permalink)  
Old 07-29-2009, 08:03 PM
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misc comments

Batteries need to be used plain and simple. The best results are with batteries. Anyone that recommends otherwise is simply not doing the experiments themselves.

As a note, I experimented with a few setups with the inductive resistors.

2 identical resistors tightly next to each other wired Tesla bifilar pancake style gives a bit bigger spike for the same input, but the heat is spread over more wire for a lower peak temperature. So it doesn't seem to be an advantage so far. Maybe at longer times the temp will get up but there will probably be a bit bigger charge to the batt. Maybe will do a longer test later.

2 identical resistors simply in parallel would bring the 10 ohm resistors to 5 ohm of resistance. More heat but more wattage too so no advantage here.

Single coil seem to be best so far per energy entered for highest temperature plus charge back to battery.

----------------------

I'm using a bigger inductive resistor than I had in my vids/pics.

The one now is 10 ohm 100 watts Ohmite resistor with 24 uH of inductance. That is 3 times the inductance of the one Rosemary used.

With a resistor this much bigger, it can easily keep up with the frequency and the off on is as good as the smaller one. I have a 96 uH one that is 10 ohm 250w and will test later but I don't think there is any problem keeping up with switching speeds.

The 24uH one I'm testing now seems to be getting to a lower temp compared to the much smaller resistor. The ratio of the spike to the input pulse appears to be about the same as the smaller one.

With the diode removed, the ringing is distorted and with the diode, the ringing is nice, solid and sharp.

I have 171 ohms at the gate with a Bourne precision 1k pot that is a 10 or 15 turn one. Extremely precise.

Anyway, just a few comments.
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  #1397 (permalink)  
Old 07-29-2009, 08:09 PM
Joit Joit is offline
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Nothing 100% correct here.

An easy Proof would be, to replace the Cap with a small Batterie, and i bet, you dont have that big or any drop from Voltage at the Source.
Or place 3 from these Caps at Parallel over a Diode and charge them.
These Diode should anyway isolate the Cap from the Rest from the Circuit.
After the classical Theorie, a Minus Pole should anyway dont have charge.
So its contradictory that it should affect the Circuit, another Hole at classical Theorie, what goes up. What is here 100% correct?

It only would need one from you ..... Guys, to replicate it, instead trowing out some Theses all Time.
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  #1398 (permalink)  
Old 07-29-2009, 08:11 PM
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field collapse

Looking to explore some ideas on this field collapse.

I labeled this pic with areas A, B, and C.

B and C are just the 2 strongest "time constants" on the collapse after the A point.

B is where it starts to dive sharp and C is where it really lets go.

Any comments on what A is doing?

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  #1399 (permalink)  
Old 07-29-2009, 08:23 PM
rave154 rave154 is offline
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a thought

Hi all,

again im not sure wether this is the right thread to be posting this, but, since this circuit is using pulsed DC and coils and flyback voltage......here goes...

think about the following setup...

a 1-1 transformer, same grade of wire and same number of turns on both primary & secondary with a good quality core to boot.

Q 1)... if i pulse the primary with a known voltage and known current draw for a known time, therebye i can calculate how much i have "spent" so to speak........ how much do i "get" out of the secondary??...... i have been told, read, that it is around 90% of whatever is put into the primary

ie the standard transformer is about 90% efficient, we can quibble about wether it is 92% or 89% another day..does 90% sound fair?

Q 2) this one is the BIGGY i guess....forgetting about the secondary for a moment and concentrating just on the primary, with its pulse of known voltage, known current draw for a known time, when this pulse is turned off.......how much is the flyback "worth" as a % of the pulse that was put IN??????

answers to these 2 Q's will be VERY greatly appreciated

David. D
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  #1400 (permalink)  
Old 07-29-2009, 08:54 PM
Hoppy Hoppy is offline
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Quote:
Originally Posted by witsend View Post
You're entirely missing the point here Hoppy. The batteries were dissipating energy at the light and yet recharging. In the second example, no recharge cycle and the actual rate of discharge could be seen over the supply batteries.

EDIT - In point of fact he showed the rate of discharge of the supply batteries when the recycled current was not used to recharge. In effect the benefit in the recharge cycle was simply 'thrown away'. He could just have easily put the light across the supply source batteries and systematically depleted each one.

ANOTHER EDIT - and the evident rate at which those batteries would deplete was clearly far greater than was evident over the 24 odd hour test that he ran the test with virtually no visible loss of energy anywhere.
Rosemary,

I'm going to throw in the towel because I now realise that I'm not going to get these 'basics' across to you and Luc.

Hoppy
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  #1401 (permalink)  
Old 07-29-2009, 09:04 PM
Hoppy Hoppy is offline
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Quote:
Originally Posted by witsend View Post
following quotes from Hoppy
(I rescued this point from some posts back?)

Its not a case of breaching battery resistance. The battery resistance is very low and offers an ideal path for the discharging inductor.
Ok. You're saying that the supply battery establishes a voltage across the inductor that exceeds the voltage of the second battery.

The voltage will fall almost instantaneously to a value slightly above the battery terminal voltage;
I assume you mean that the voltage drop is measured across the inductor when the switch is open and for some reason this voltage is then determined at a value dependant on the second battery. And one can then see the recharge value as a voltage across the terminals of the second battery.

You need to see this initially high voltage as a conversion of low voltage / tension to higher voltage / tension.
Why tension? Voltage is voltage - unless you need to give it a property beyod voltage?

It actually posseses less energy than that used to initially charge the inductor.
What voltage are you talking about here Hoppy? I can't understand you. Voltage is voltage. It has as much energy as is measured. It doesn't hide some and show some - like some weird kind of 'peek a boo' show - now you see me now you don't.

The loss is in heat due to the resistance of the wire forming the inductor
What loss? I'm referring to the second battery recharging. What loss?

It may appear to be more because the average discharge voltage across the battery is high enough to charge it, whereas the original voltage level before conversion is too low.
If this means anything at all it entirely eludes me. A discharge voltage recharges.

The available current will be less at discharge because the initial peak voltage was a lot higher. The available power on discharge will therefore also be less.
Since when is current lower in the event that the voltage is higher? Are you trying to proportion the actual amount of energy transferred to some ratio between voltage and current. Voltage and current are precisely proportionate if they are measured on a circuit with a path for the current to flow.

The 1000 ohm resistor is not at all critical. It can be less or more in value. Its just to show that the discharge voltage can be a lot higher than your primary battery voltage. Experimenting in this way will reveal a lot about how your circuit works and it will eventually become apparent to you that there is no free lunch in a circuit of this type.
I'm willing to discuss this but not based on this argument. Mostly because I simply cannot follow it.

I think we need to get back to the point here. We have a circuit where the second battery is disconnected from the supply battery but we're testing the amount of energy available from the inductor which may be stored, per classicists or regenerated per new age. We want to establish which is correct.

So If a second battery is recharged - and there is evident dissipation of energy from a resistor - then where did the extra energy come from to recharge the second battery? The first counter argument to recharging the battery was based on the fact that there was no extra energy. The battery delivered (as an example) 5 watts to the load resistor - spent 3 watts in heat and simply took back 2 watts to recharge. Zero gain. Alternatively, the argument was - the circuit delivered 5 watts to the load, spent 3 watts in heat and recharged a second battery with 2 watts. Still no gain.

Our argument is this. The battery delivered 5 watts of energy - spent 3 watts on the load - sent 2 watts back to recharge the battery and then took those same 2 watts of energy back to the resistor to add another 2 watts in heat. Definitely a gain. The only difference is that I've grossly understated the amount of energy that went back to the battery to recharge it. The rate of charge on the second battery proves this.

I've done my best to get my head around your classical concepts. How about trying to wrap you head around mine? I at least can show that the numbers support this claim. It's not a free lunch. It's just much cheaper than we've had to pay to date.
Rosemary,

With respect I'm having much trouble understanding your thinking, so as I said in my previous post, I need to throw in the towel to avoid annoying those that wish to believe that your circuit is special. I hope that one day you will be able to convince the academic world and prove me wrong and I mean this sincerely.

Best wishes Hoppy
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  #1402 (permalink)  
Old 07-29-2009, 09:06 PM
Joit Joit is offline
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Aaron i could figure, this small one comes from the Timer, that it cuts faster off then the Mosfet acts, i actually did not see this before but from the Value it looks like low Voltage.
You maybe could try to give the Timer some more Volt, if this small one change. The timer should spit out more voltage at the Output too, and then should change the Spike, if its 'his fault'.


Rave, i try to answer it a bit.
But Q1 and Q2 are pretty similar. What i ve seen, at a Bedini act the whole Magnetfield from both Coils, so if you put something at the other end on it, you will see it at the main Rotor.

How much you can induct into another Coil depends at your Coils.
Earlier made Transformer had no loss or less loss at Induction.
Today, they are cheap made and have a weldseam through the Ironsheets.
That cause eddy current, what lowers the Induction.
Ans still the Question, to use Aircoils, or one with a Core, what increase the Inductionfield from a Coil.
The Power what you get from the Feedback should be about 90-100% too, but it still depends, what Coil you use, and how much eddy current is in there.
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  #1403 (permalink)  
Old 07-29-2009, 09:33 PM
witsend witsend is offline
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Hoppy - here's the thing. We have a circuit that appears to achieve results that fly in the face of. The circuit is being developed, admitedly only by Aaron - to replicate ours. The advantage here is that the circuit will, this time, have better representation per video and such like so that it will obviate those points that I did not, unfortunately take care of.

However, while we wait for shipment of meters and fine tuning and so forth, I should imagine it would take another two to three weeks, my object is to introduce, if possible, the concept that this 'new age' energy source, is not some wild and unexpected event. It is simply the result of inductive laws doing what inductive laws require. There's no magic. There's no improbable, extraordinary variation to known physics. Only a change to predicted outcomes.

My own model - that saw this moment in the collapse of magnetic fields - is entirely different to classical. My current flow depended on the flow of those little dipoles. But when I presented the question I was surprised to see that this regenerated cycle of current had not, in fact, ever been considered. I could never understand why. It seemed required. I then built the circuit - obviously with help - and presented my findings to a lot of engineers. This was exhaustively studied. it appeared that the numbers on the circuit did indeed conflict with prediction. Now I felt I was on safe territory.

The history of this is tedious. The fact is that - to this day - I have never heard a 'dye hard' academic argue the 'regenerated cycle of current'. But I have seen them use the most extraordinary explanations required - even to see it as stored energy. So. If an appeal to you to understand the 'induced - regenerated current flow - simply encourages you to throw in the towel - and you're not even an academic - can you imagine the uphill ahead of us? Why can it not be considered? Just as a hypothetical postulate?

It immediately seems to evoke a 'let me keep out of this argument' or 'I will not be persuaded by argument'. Yet I must buy into yours? Hear me at least. I've heard you.

Last edited by witsend : 07-30-2009 at 02:32 PM. Reason: spelling
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  #1404 (permalink)  
Old 07-29-2009, 09:51 PM
witsend witsend is offline
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No answer? You must surely admit that it would be nice to see voltage generating current flow with the required polarity consistent with the known physics. No need to wrap the mind around anything other than the energy equalling volts times amps. And the relief in following the path of the current - strictly in terms of what is best understood. Then the fact is that this hypothesis suggests that current is again given a polarity - in conjunction with the applied voltage. No complicated reasons required to explain holes, and stray capacitance and goodness knows what. Now it simply goes through a diode in relation to the polarity of the current. And the fact that - if it does go back to the positive terminal through the battery it will inevitably recharge it exactly as it would discharge going to the negative terminal. Suddenly physics makes sense again.

The only difference is that we have more energy than previously expected. But not even that's a train smash. We're still able to confirm that charge is wholly conserved. And that - in turn - has echoes of 3rd Laws. So. Perhaps the entire electromagnetic interaction only needs to be accepted into another category of our conservation of Energy Laws.

Last edited by witsend : 07-29-2009 at 10:18 PM. Reason: developed argument
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  #1405 (permalink)  
Old 07-29-2009, 09:53 PM
gyula gyula is offline
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Quote:
Originally Posted by rave154 View Post
Hi all,

again im not sure wether this is the right thread to be posting this, but, since this circuit is using pulsed DC and coils and flyback voltage......here goes...

think about the following setup...

a 1-1 transformer, same grade of wire and same number of turns on both primary & secondary with a good quality core to boot.

Q 1)... if i pulse the primary with a known voltage and known current draw for a known time, therebye i can calculate how much i have "spent" so to speak........ how much do i "get" out of the secondary??...... i have been told, read, that it is around 90% of whatever is put into the primary

ie the standard transformer is about 90% efficient, we can quibble about wether it is 92% or 89% another day..does 90% sound fair?

Q 2) this one is the BIGGY i guess....forgetting about the secondary for a moment and concentrating just on the primary, with its pulse of known voltage, known current draw for a known time, when this pulse is turned off.......how much is the flyback "worth" as a % of the pulse that was put IN??????

answers to these 2 Q's will be VERY greatly appreciated

David. D
Hi David,

The answer for your first Q is what you seem also to establish: it can be around 90-95%, in case of well designed pulse transformer.

On answering your second question, let's say you use 25V input battery voltage, you have about 10uH inductance embedded in the wire resistor and the current you switch with the MOSFET be 2.5A peak value, ok? (assuming the resistor part is 10 Ohm, ok? 25V/10=2.5A)
Now the flyback pulse you may get depends mainly on the switching time, how fast or slow you are able to switch off the MOSFET? Most power MOSFETs are able to switch off at 30-40 nsec or so but it needs professional circuit layout and dedicated driving circuit to get that fast speed.
Suppose you are able to switch it off in 100 nsec, ok?
The formula is Vf=L*(dI/dt) where Vf the received flyback pulse amplitude, L is the inductance (10uH), dI is the current change from 2.5 to zero (2.5A), dt is the switch off time (100ns). These data give 250V flyback pulse amplitude, ten times higher than the 25V battery voltage.

This result is a pulse amplitude, ok? Very much depends on mainly the switching speed, (see the formula) the drive pulse into a MOSFET gate should be 'razor' shape, with some ten nanosec rise and fall times, right? This cannot be provided by an NE555 type IC (the bipolar monolithic 555 IC has 100 nsec rise and 100 nsec fall time when the loading capacitor is 15pF at pin 3 output ( see data sheet page 6 for a Texas Instruments NE555:
http://focus.ti.com/lit/ds/symlink/ne555.pdf )
And a power MOSFET's input capacitance normally 1-2 nanoFarad, this loads down the 555's output, slowing down the switching...

rgds, Gyula
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  #1406 (permalink)  
Old 07-29-2009, 10:03 PM
witsend witsend is offline
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And at the risk of prolonging a rather contentious argument related to this, I would point out that the only effort on the table to debunk this experiment has been offered without any substantial argument based on data extrapolated from the equipment available. It was concluded with reference to a graph that, in turn, had no reference whatsoever to the battery draw down rate nor any attempt at measuring the actual energy delivered as a comparative value. Apples and pears.

So. We've got the paper with the data shown. We've got the apparatus to measure it - en route. All that's left between now and then is the hope to advance the argument in support of the experimental evidence.

I've tried before to advance the fact without the argument. It's then argued on the basis of evident errors in measurement. That's not going to go away. it'll be used again and again, unless the argument itself, the bare bones of the supporting thesis be considered and that the measurements support that thesis. Then, maybe - we can make inroads into this classical mindset.

Last edited by witsend : 07-29-2009 at 10:20 PM.
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  #1407 (permalink)  
Old 07-29-2009, 10:11 PM
rave154 rave154 is offline
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GYULA, thanks for the specs, i appreciate the input :-)

i guess my main question was........ how much "oomph" is the flyback worth....in comparison to the "oomph" that created it in the first place?.

ie - not just comaring voltage to voltage....... but total ooomph IN(initial pulse )....to total oomph OUT (the flyback)

an example....... if i send a pulse in.....12V, 100Ma, for say...... 1/10th of s second....... then..i store the flyback in a cap....

now....i could have taken that initial pulse......and used it instead to light a 12v bulb rated at 100mA......and i could have lit that bulb for 1/10th of a second...... if i now take the cap.....and put the bulb across it, how long will it light for?..... 90% of 1/10th of a second?......80% of 1/10th of a second?
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  #1408 (permalink)  
Old 07-29-2009, 10:35 PM
Joit Joit is offline
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@Witsend
I think, there are two Camps, the Current Generator and Enduser for Current.
Using the Spikes is done from a lot of People with her different Concepts.
It starts with H. Ford with his Magento, what he did use at his Lizzy. and goes over few other Inventors like Muller probatly Tesla J Bedini still Adams,
and countless others, you find alot of this Inventions at Pure Energy Systems News (PESN).

The other Camp, for enduser build Circuits, to use the Current, what is been generated, and in there, i dont think, you can do usefull Things with Spikes,
like Computers or any other Electrical Devices.
They are educated, to avoide this Spikes, because they are often to powerfull, and mostly damage Parts of the Circuits.

They had learned, you can generate Current with a Dynamo, and use it, its free even too, so far you do pay for it.
And that it seems that the Curcuit dont give better Results on a Dynamo, say Grid, seems be shown.

I could figure, there would be a lot use of recirculating the Spikes through anything, what need to make Heat, well, it need to be reasearched, and well, Current is seems right now anyway 'free' for some Peoples.

Even the fact, that you can make Heat with this Circuit and got a lesser Drop from Energy what you need for this should be a highlight,
but when some dont wanna see what is there, noone can help them.

I would say, leave it for now, lets see, if we can get better Results, and dont get distracted.
Some may dont wanna see it, others make a Business out from it.
You see it with this pulsed charger, what comes on the Market suddenly,
or that you can 'Refresh' AAA Batteries with special Charger.
No Doubt, they come from Investigations, what others Peoples did, and not herself.
J.Bedini said once, if he would had listen to a Guy, what told him,
'Its maybe not a good Idea to charge Batteries with AC' then he would still be there, where a lot Peoples still stuck/are.
Slowly i really understand him, when he says, he got Problems with this Board here, and i understand a lot other Peoples,
when they say, Humans are not Ready for such Devices.
The best Example, why, is shown here, and seems, how much higher the Claims, how much higher the Attemps to debunk it, and lesser the sobber Mind, to think about it.
R.I.P V. Schauberger, which was brought from his Engineer Fellows or should i say Engineer Circle into the Nuthouse because of his Ideas.

And btw, if this is a Application or a Patent, in both Case everyone should be glad,
that it is him allowed to rebuild it, and not restricted, to build any similar.
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  #1409 (permalink)  
Old 07-29-2009, 10:45 PM
poynt99 poynt99 is offline
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Quote:
Originally Posted by witsend View Post
Are you back in the battle ground here Poynt? Or just keeping a fatherly eye on procedures? Nice to see you if the former.
I thought I was posting in Luc's " Effect of Recirculating BEMF to Coil" thread. Sorry for posting. This was intended to pertain to Luc's thread only.

.99
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  #1410 (permalink)  
Old 07-29-2009, 10:49 PM
witsend witsend is offline
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thanks Joit. You're right of course. People will think what they need to and words aren't going to change it. I think your motto is good. Theorising is just hot air.

I'll take comfort when our results are out. I think I need to depend on people like jibbguy et al to spread the word. Maybe this time I don't need to talk theory. Maybe people will see it for themselves. Anyway. It's better to hope than not.

Thanks again Joit. You're always there! That's also a comfort. I must say I'm feeling depressed today. Maybe because the TK saga is over and I'm still feeling bruised.


Last edited by witsend : 07-30-2009 at 04:50 PM.
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