COP 17 Heater | Rosemary Ainslie

[Aaron]

Poynt99 can post

Chet,

Poynt99 can post now.

[gotoluc]

Hi everyone,

I totally agree with Dr. Stiffler that a testing base line needs to be established.

I had the same idea in mind and that is why I first needed to confirm what I understood of using a fine filament bulb in series at the input as visual indicator of energy entering the circuit.

I have found that using meters no mater how expensive they are have been giving me false results. Once you makes a change to the circuit, like frequency and or voltage you want to re-tune the duty cycle to send back exactly the same amount of energy to compare the previous test. I know many will say your eye cannot be that accurate. That is right! ... if the intensity of the bulb is high... but if you keep the bulb intensity to the point where it just starts glowing and not go above that you will see that it has an amazing accuracy to show any change in the circuit. Just stay with this technique I just explained and it should work well enough. Also, know that you are not measuring anything with the bulb, it is just a visual indicator of change or a way to re-tune duty cycle after you make a change in order to send back exactly the same amount of energy in the circuit to be able to observe if the change you made is giving you more or less work for the same amount of energy. That is all you want to do at this time. Forget Over Unity proof.

You will understand more after seeing the video's below how I have been using this technique.

Link to test 5: YouTube - Effect of Recirculating BEMF to Coil test 5

Link to test 6: YouTube - Effect of Recirculating BEMF to Coil test 6

Thanks

Luc

[DrStiffler]

Hi everyone,

I totally agree with Dr. Stiffler that a testing base line needs to be established.

I had the same idea in mind and that is why I first needed to confirm what I understood of using a fine filament bulb in series at the input as visual indicator of energy entering the circuit.

I have found that using meters no mater how expensive they are have been giving me false results. Once you makes a change to the circuit, like frequency and or voltage you want to re-tune the duty cycle to send back exactly the same amount of energy to compare the previous test. I know many will say your eye cannot be that accurate. That is right! ... if the intensity of the bulb is high... but if you keep the bulb intensity to the point where it just starts glowing and not go above that you will see that it has an amazing accuracy to show any change in the circuit. Just stay with this technique I just explained and it should work well enough. Also, know that you are not measuring anything with the bulb, it is just a visual indicator of change or a way to re-tune duty cycle after you make a change in order to send back exactly the same amount of energy in the circuit to be able to observe if the change you made is giving you more or less work for the same amount of energy. That is all you want to do at this time. Forget Over Unity proof.

You will understand more after seeing the video's below how I have been using this technique.

Link to test 5: YouTube - Effect of Recirculating BEMF to Coil test 5

Link to test 6: YouTube - Effect of Recirculating BEMF to Coil test 6

Thanks

Luc

@gotoluc

May I ask a question? The first video you say 800ft of wire is ~63uH?

This is very strange,I get 65uH from 21ft.? Did you mean something else?

[gotoluc]

@gotoluc

May I ask a question? The first video you say 800ft of wire is ~63uH?

This is very strange,I get 65uH from 21ft.? Did you mean something else?

Oops! did I say micro Henries. It is 63.5 milli Henries.

Thanks for catching that.

Luc

[SkyWatcher]

Hi folks, Hi DrStiffler, you said "I have tried to work through your numbers and seem to be missing something.
When you talk about the (Load) resistor are you using a 10 Ohm resistor?".
No I used 1 ohm, 5W resistors, one on the input and one on the output, essentially the same as Rosemary's circuit with the ammeter removed and replaced with the 1 ohm, 5W. same as Gotoluc's circuit and when my dmm is set on dc volts those are the numbers it was showing.
peace love light

[gotoluc]

Hi everyone,

a YouTube user posted a comment which explains very well and my reply may also help some.

User: Can101276
I like your unconventional testing, it is crude, however it will alow you to see work being done and see how your changes effect the work performed.

Correct me if im wrong, but what I got out of your experiment was that you can get more work performed with high voltage, high frequency pulses, than with lower frequency, higher duty cycle pulses at the same voltage.
So in dummy speak, more work for same watts!!!
You are on the right track.

My reply:
Yes Can101276, you are understanding this correctly. However there maybe limits as my higher voltage tests over 250vdc the efficiency starts going down but that maybe an electronic component issue causing this. I also see efficiency starting to go down when frequencies over 5KHz are used. This could have something to do with the coil characteristics. More testing will be required to understand this better.

Luc

[Hoppy]

Hi everyone,

a YouTube user posted a comment which explains very well and my reply may also help some.

User: Can101276
I like your unconventional testing, it is crude, however it will alow you to see work being done and see how your changes effect the work performed.

Correct me if im wrong, but what I got out of your experiment was that you can get more work performed with high voltage, high frequency pulses, than with lower frequency, higher duty cycle pulses at the same voltage.
So in dummy speak, more work for same watts!!!
You are on the right track.

My reply:
Yes Can101276, you are understanding this correctly. However there maybe limits as my higher voltage tests over 250vdc the efficiency starts going down but that maybe an electronic component issue causing this. I also see efficiency starting to go down when frequencies over 5KHz are used. This could have something to do with the coil characteristics. More testing will be required to understand this better.

Luc

Luc, may I ask how are you producing 250V? I thought you only had a 120V grid supply, bridge rectifier and smoothing cap. The maximum DC voltage you should get from a 120V AC full wave rectified and smoothed supply is about 169V.

Hoppy

[SkyWatcher]

Hi folks, I ran a few more tests last night with the same circuitry except for the light bulb and tried some different circuit variables as Gotoluc is showing. I used a tail light bulb as the input resistor so I can visually see the energy flowing through the bulb and for every test the bulb filament illumination was the same, again as gotoluc is experimenting with. These were the tests based on holding the 1 ohm, 5W output resistor with thumb and when it became too hot to touch any longer.
1) 1.3 Khz @ 9% duty cycle @ 24V required 38 seconds
2) 4 khz @ 7.6% duty cycle @ 24V required 33 seconds
3) 1.3 khz @ 4.8% duty cycle @36V required 26 seconds
4) 2.7 khz @ 4.8% duty cycle @ 36V required 22 seconds
5) 1.4 khz @ 3% duty cycle @ 48V required 22 seconds
6) 2.8 khz @ 3.2% duty cycle @ 48V required 16 seconds
7) 4.2 khz @ 2.9% duty cycle @ 48V required 14 seconds
Well For whatever its worth there ya go.
peace love light

[witsend]

Hi everybody. Delighted to hear that Aaron's going to 'tidy up' the thread. Many thanks. I love reading back over the various points, and will be able to do so without the need to skip through volumes of extraneous nonsense.

Regarding gotoluc's circuit - I'm still struggling to understand it from the video. Sorry. I just do not find my way around these things with the ease that you guys have. I'm referring to the circuit with 'no bulb' and same resistive loads, only because Hoppy recommended 'less variables' and because I understand it better.

So - it's basically my circuit but the 'inductive resistive load' is replaced with a single wound inductor in series with a 22Ohm resistor, call it R1. The base of the inductor goes to the switch. At the junction of the inductor and the switch is the 'feed back' diode which has a second 22 Ohm resistor in series. Call that resistor R2. This, 'fly back' or 'feed back' diode, in turn goes back to - either 1) the junction of the wire between R1 and the inductor or 2) straight to the source - parallel to the R1?

Is this right? Not sure if you're there gotoluc - but am posting this and will continue with the points in the next post. I'd be glad if you can say which of the two options or even if all of it is wrong. Don't worry about the source input - I get it that it's rectified DC and stored in a cap. The cap, presumably is then the energy supply source. So that would be the point that the 'fly back diode' would intersect on the 2nd option?

[gotoluc]

Luc, may I ask how are you producing 250V? I thought you only had a 120V grid supply, bridge rectifier and smoothing cap. The maximum DC voltage you should get from a 120V AC full wave rectified and smoothed supply is about 169V.

Hoppy

Most variac can give higher voltage than 120vac. Mine outputs up to 178vac so I can actually get about 230vdc at full output.

Luc

[gotoluc]

Hi Rosemary,

in video test 5 I carefully explained the routing of the flyback diode (it starts on negative side of coil) and the resistor is next on the positive side of the diode which is pre the return to the positive of the coil.

I hope this clears it enough. I'm sure others can help you also, since I will be out for the day till maybe very late. Eastern Standard Time

Luc

[witsend]

Thanks gotoluc. Enjoy the day. Maybe you can read the analysis when you get back.

Many thanks
Rosemry

[witsend]

Therefore - in my view - it will be an impossibly difficult task to argue the benefit in stored energy on this particular circuit. I first thought you were using that flyback diode directly to the cap. But I now need to point out the actual point of conflict between 'us' and the classicist. In this example we are arguing that we only need a drawdown of 8% of the energy supplied to give back four times greater efficiency. That argument is quashed on many levels but possibly the strongest being that the utility supplier will also argue that while we may only need 8% of the applied voltage he gives us - he cannot thereby reduce that supply to 8%. He's got to keep it at a constant level. In other words he is expected to apply the full 120 volts over that entire duty cycle period and he'd prefer to be paid for whatever it does cost to generate that potential difference at 120 volts as opposed to 120*8% = 9.6volts. So we lose out there as well.

I would be really interested to see results on the flyback being returned to the cap, as I first presumed. However, I would like to refer back to my own circuit. Before I do this, I must also point out that Peter has given you guys a variation on my circuit using a cap in parallel to the load resistor. It was proposed as a more efficient system. I do not know as I've never tested it. But one of my colleagues has. He reported an extraordinary 10 minutes where the cap ran the system with a disconnected battery. But he was never able to repeat that result. And he is now, unfortunately in Durban - some distance away. And I understand his work demands are such that he is not able to spend as much time on the circuit. The point being that if we are going to see that magical closed system then Peter is spot on. It will be with the use of that cap and some really fine tuning to get it right. Such takes time and dedication.

In the meanwhile - let me try and explain the only benefit of my circuit to the general cause is that it delivers overunity results that can be measured strictly in terms of classical analysis. In other words we do not need to get into that endless circuitous?? argument that loops?? back on itself without a definitive conclusion.

We only use a resistor. Therefore if the battery has 'stored' extra energy on that load - then it can be measured. And better yet, if there is any energy generated during the Off period by the resistor - then that energy can be returned to the battery. Unlike our utility supplier - we can now definitively and accurately measure the gains versus the losses to determine a result that even our classicists cannot argue with. In fact they've told us how to measure it. BUT the measure of that returned energy - while conforming to classical requirement - also needs some reasonably sophisticated storage oscilloscopes to PROVE the returning energy. The sum of the energy measured over both the On and Off period of the duty cycle ALWAYS (within certain frequency parameters) is less than the product of both cycles. The product relates to the energy dissipated at the load. The sum or difference between the two cycles relates to the energy delivered by the battery supply source.

Then I have much more to say about this system as it relates to an application to a utility supply source but that can be the subject of another post. I think this is enough to go on with.

And gotoluc - I need to thank you because your hard work has not gone wasted. You see how important it is to address the different circuits to better explain the question.

[witsend]

Hi Hoppy - was hoping you'd answer. I'm going to study your revised numbers. Thanks for that. I realise I use some pretty sketchy values. Bear with me though. I'm trying to argue your point.

EDIT - many thanks Hoppy. I read it through and see numerous errors on my part. But you've explained it all. My point however is that the circuit does not prove the gain. Have you read our paper? Can you see where the fault is with our experiment? I'd love to have your input on this - if you've got time.

Kind regards,
Rosemary

[witsend]

OK witsend, lets use your method. Firstly the 120V supply will drop 88V across R1 and 32V across the coil (based on the coil resistance being rounded to 8R). Taking R1, 88V * 8% = 7V. P = Vsq / R = 2.2W. Taking R2, the inductor is the generator and therefore the source of the output voltage, akin to the supply battery. This is very important to understand because it explains the increased heat in R2. The full voltage stored in the inductor will be developed across R2 and the diode. The diode forward voltage drop is less than 1 volt, so can be ignored for this exercise. Taking 88V across 22R, P= 88V sq / 22 = 352W * 8% = 28.1W. In practice, much less power will be available from the inductor because of switching losses, so the real power dissipated in the resistor will probably be around 15W - 20W.

On the input side we have neglected the power available to charge the inductor. This is 32V sq / 8R = 128W * 8% = 10.2W. If we add this to the
2.2W dissipated in R1, this gives 12.4W. The difference between the power dissipated in R2 and R1 represents the switching losses.

The input resistor is cooler than the output resistor because the input power has to be shared between R1 and the charging inductor. This is not the case at the output, because the inductor has already recieved its energy from the input and can therefore deliver virtually all its stored energy to R2.

Hoppy

Actually - does this not, in fact represent a gain? If the input is 10.2 for the inductor and 2.2 for R1 then total input is only 12.4 and there is clearly more than this dissipated even allowing for the switching losses? What have I missed?