COP 17 Heater | Rosemary Ainslie

[jas_bir77]

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).

[Aaron]

@Jas

Hi Jas,

Sounds like it worked out. Glad to hear it!

[Aaron]

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 »

• Quote

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.

[Joit]

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.

.

[witsend]

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.

[gotoluc]

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

[rave154]

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

[witsend]

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.

[Michael John Nunnerley]

Brilliant as always

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

[witsend]

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.

[Hoppy]

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

[poynt99]

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

[witsend]

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.

[witsend]

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.

[witsend]

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.