COP 17 Heater | Rosemary Ainslie

[Hoppy]

Luc

Could you please post a scope shot of the waveform across both the input and load resistors?

Hoppy

[Hoppy]

I agree with Aaron that there is no excuse for rudeness just because a difference of opinion is expressed. Skeptics should not be trying to convert those they consider misguided by resorting to rudeness. An opinion should be expressed politely and not forced down the throats of others. I have different ideas about the nature of radiant energy and am happy that I know the cause of the effects in Luc's demo. This does not mean that I'm necessarily right, just that my conclusions are based on a different set of principles to some others. I realise that to express one's beliefs to another with radically different beliefs can and often does lead to confrontation, so we all need to respect the views of others even if we think they are misguided.

This forum is simply a medium to exchange ideas. It cannot and does not need to reach firm conclusions by consensus. Its simply a talking shop, so lets get things into perspective.


Hoppy

[SkyWatcher]

Hi folks, glad thats over. I ran a few more tests, this time @ 24V input. Also the diode im using is an NTE5817. So I used my high tech. temperature measurement device again, my thumbs and I put a thumb on the input side resistor and the output side resistor. @ 30 seconds the output resistor was too hot to touch and the input resistor was imperceptibly warm. I then measured with a volt meter (dmm) the voltage drop across the input resistor which showed .47V and the output resistor measured 1.45V and the coil measured .1V. This explains why the input resistor was not very warm because only .221W was across it and also explains why the output resistor was so hot because it had 2.1W across it. Now this seems very efficient to me, I guess the reasons we will all be discussing im sure.
peace love light

[Hoppy]

Hi folks, glad thats over. I ran a few more tests, this time @ 24V input. Also the diode im using is an NTE5817. So I used my high tech. temperature measurement device again, my thumbs and I put a thumb on the input side resistor and the output side resistor. @ 30 seconds the output resistor was too hot to touch and the input resistor was imperceptibly warm. I then measured with a volt meter (dmm) the voltage drop across the input resistor which showed .47V and the output resistor measured 1.45V and the coil measured .1V. This explains why the input resistor was not very warm because only .221W was across it and also explains why the output resistor was so hot because it had 2.1W across it. Now this seems very efficient to me, I guess the reasons we will all be discussing im sure.
peace love light

Where is the remaining 23.43V dropped on the input side?

Hoppy

[SkyWatcher]

Hi folks, I forgot to mention the freq. and duty cycle. The freq. was 1.3khz @ 9% duty cycle. Hi hoppy, I would assume over the coil although since it is only 9% duty cycle would the circuit components even contain 24 volts since i cant see whats going on in the coil without a scope, I know the books say so. Hope I understand what your asking correctly. Are these results interesting is the question, I am trying to figure that out myself, was hoping others could shed light on it as gotoluc is wondering as well. This is what i see, We have 2 resistors and a coil in series when switched is closed, when the switch opens the coil and whatever field exists in the inductive resistor discharge within themselves through the diode and possibly some oscillations occurring and this heats the resistor in the output to a high level and is separate from the input. The question is why are temperatures between the resistors so far apart. It seems maybe something extra is occurring in the coil resistor diode loop. Well those are my thoughts so far, still thinking about this myself.
peace love light

[gotoluc]

Luc

Could you please post a scope shot of the waveform across both the input and load resistors?

Hoppy

Hi Hoppy,

did you see the scope shot across the input resistor at this post: http://www.energeticforum.com/renewa...html#post59949

I don't feel comfortable having my probe across the output resistor unless it is a X100 probe since the peaks can fry my USB scope. Also I've tried it many times in the past to scope inductive flyback and it makes the USB interface disconnect just about every time.

If someone know of a safe way please post the technique.

Luc

[gotoluc]

Hi everyone,

as I mentioned in an earlier post above, I submitted some questions to user: poynt99 an EE of the Overunity Forum and he has now answered.

You can click this link to read: Effects of Recirculating BEMF to Coil


But this was my real bottom line question and he answered it as I understand it.


Hi .99,

Thank you for taking the time to answer the questions.

This question was the most important to me and you answered it as I understand it.
My Question:

so if we have 10 volts in a circuit under load and the filament starts to glow and we reduce the duty cycle since the circuit is a pulse circuit and we then raise the voltage to 100 volts and the bulb starts to glow to the same level as when 10 volts was running through the circuit, do you really believe that we now have more Energy at the bulb because the voltage was raised 10 time higher then before???

poynt99 Answer:

No. Assuming that you were able to set the bulb intensity roughly the same as before, then the average current and voltage in the circuit would also be measured as roughly the same as before.


My reply to poynt99:

Your answer is worth Gold to me and I will make a new video to show you why.

Thanks again for sharing.

Luc

That to me is GREAT news and I will make a new video to demonstrate to all of you why.

Luc

[Allcanadian]

@All
Concerning Tinselkoala and his merry gang of critics, I think it's important to consider the fact that one person's failure does not necessarily effect anything you may do. We are individuals as such we determine our own success not anyone else. As well, they have admitted their failure to produce results, would you listen to or hire a lawyer who has never won a case? Would you go to a doctor who has never healed anyone and who's patients have all died, LOL? Consider the history of inventors who have defied a world full of critics by succeeding in what everyone considered impossible. It would seem obvious that if we want to succeed we should listen to person's who have found success and not person's who have failed to produce the desired results.
Regards
AC

[DrStiffler]

Hi folks, glad thats over. I ran a few more tests, this time @ 24V input. Also the diode im using is an NTE5817. So I used my high tech. temperature measurement device again, my thumbs and I put a thumb on the input side resistor and the output side resistor. @ 30 seconds the output resistor was too hot to touch and the input resistor was imperceptibly warm. I then measured with a volt meter (dmm) the voltage drop across the input resistor which showed .47V and the output resistor measured 1.45V and the coil measured .1V. This explains why the input resistor was not very warm because only .221W was across it and also explains why the output resistor was so hot because it had 2.1W across it. Now this seems very efficient to me, I guess the reasons we will all be discussing im sure.
peace love light

@SkyWatcher

Hello!

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?

If so and we look at steady state (DC) and not complex wave I get the following results.

If we have a (DC) (which it is not) but lets use it to make it simple.

1.45(Volts) / 10(Ohms) = 0.145 Amps or 145mA

The energy across the resistor would be (I^2*R) 0.145^2 * 10 = 0.2103J
which is 210mW.

*Note I am sorry if I missed the values you are using if different from what I am showing, but plug in your values and see what comes out.

Your input

If input resistor is 0.25 Ohm the current would be 0.47/0.25 = 1.958 Amps

So if the input 0.47V drop a constant DC (make it simple)

If you have 1.958Amp through the resistor in steady state the dissipation of the resistor would be 1.958^2 * 0.25 = 0.9588J or 959mW

This should be pretty warm to the feel after a minute or two.

I would like to do the numbers if you could supply me a bit more info? Oh yea we can figure it on DC, but we know we have a complex wave so we will need to integrate to get close to the truth.

[witsend]

Hi gotoluc, I can't quite get my head around your question (edit) to poynt99? Maybe your video will clarify?

And Aaron, many thanks for taking action. Was beginning to feel pummelled. It's much appreciated. I got it that henieck was earnest. But woke up - eventually.

[DrStiffler]

Common measurements

@All
In order for the thread to not go on into infinity and nothing really answered, we all need to decide on a common circuit. Now if all can not use the Inventors circuit for what ever reason, then it should be possible to at least setup a table of procedures so that every on can present his/her data in the same way.

If we do this it will be possible for all to see at once the energy balance of a circuit. Maybe some of the replicators are not able to do integration either manually or from equipment, but if scope shots using the same scan parms then there are a number of people that are here or at least watching that can do it from the properly presented data. I'm sure Dr. Lindemann and Arron can if asked and presented the correct info.

So many times these things go on for ever and no one setups up the correct procedure for reporting. This is where everyone does it different and everyone is confused.

May some one will offer to present a common diagram that all can follow or at least a block diagram which can call for numbers from measurement and all can at least supply numbers in a consistent way.

Anyway lets get this done, people waste a ton of time and money on parts and maybe even equipment and the answer never seems conclusive.

[witsend]

Dr Stiffler, your requirements are noted. Has to be a good thing. I must say I'm often puzzled as to what circuit is being referenced, and where. Bearing in mind that parts may vary - circuits may differ - can you perhaps propose a couple of circuits. Mine - obviously - is the subject of this thread, but Peter's is probably a great improvement and then gotoluc's seems to offer such extraordianry proof. Perhaps 3 basic - and then some clear reference as to which of the three are being used? I'd be very glad to include gotoluc's as his video reference seems to indicate a remarkable gain on a frequency that is also, apparently, not in resonance.

Can we impose on you to suggest a schedule of some sort? Or could you ask someone in the thread may have the time to attend to it? I'm sure all would be happy to conform and then, as you say, less confusion required in evaluating those results.

kind regards,
Rosemary

[witsend]

Gotoluc - I took the trouble to read your link reference. I had no idea that you were not qualified. How do you manage to do so much without it? You continually amaze me.

The question regarding voltage and amperage - if I may try and answer it - and, as I believe it's understood classically - the electromagnetic interaction is just that. Never the one without the other. .99 pointed to the fact that you can get a measurable voltage at a supply source - but without a current flow - it's just that. Potential energy waiting to discharge.

It discharges as current. That's the electric component in the 'electromagnetic' interaction. The magnetic component is the extruded field that is measureable outside the body of the component parts as a result of current flow. So. If you measure voltge in a circuit you are measuring the polarity of the extruded magnetic fields. Which means that current is flowing. That measurement is enabled by your voltmeter. And because it can show the polarity of the magnetic field it can also show you the directional change in current - or not - whichever is appropriate.

The problem with ammeters, as I understand it, is that they do not usually distinguish between the polarity of the extruded magnetic field and the very best ammeters are not able to measure well at the high frequencies generated by our switching circuits.


If I'm wrong here, hopefully one of our boffins will point it out. But the point is that current and voltage are really just two sides of the same coin. The one points to the other. And neither can happen independently - unless, as .99 said, it is simply a measure of a battery or a supply source terminal not yet connected to a circuit to enable that current to flow. So if you are reading voltage on a circuit then there is a measurable current flow - and if you are reading current flow - then there will also be a measurable voltage. Never one without the other.

That, in any event, is as I understand it.

As it relates to your circuit - the voltage across the resistor at the input is an actual measure of the current flow from the supply source. Whatever that value of that voltage, then the wattage dissipated MUST equal the voltage from the supply source, divided by the Ohms value of the resistor itself * time. That is the actual energy delivered by the supply source. It cannot, magically, be more or less than is measured.

[Hoppy]

Hi Hoppy,

did you see the scope shot across the input resistor at this post: http://www.energeticforum.com/renewa...html#post59949

I don't feel comfortable having my probe across the output resistor unless it is a X100 probe since the peaks can fry my USB scope. Also I've tried it many times in the past to scope inductive flyback and it makes the USB interface disconnect just about every time.

If someone know of a safe way please post the technique.

Luc

OK, I understand. Try scoping across a 1R or less resistor in series with both 22R resistors, or take the 22R resistors out completely and replace with the low value resistors. Use non-inductive resistors. The type you are using are inductive as they are wire wound and are not really suitable. By comparing input and output waveforms, you will see the relative power levels by the area enclosed by the waveforms. Integration is needed to work this out accurately with complex waveforms.

Hoppy

[RAMSET]

Expedite??

Rosemary,
Perhaps you can ask the moderator to help
Poynt99 is anxious to inter the discussion here and is having difficulty posting

Hope you can help

Chet