Failure in Calculating DC Pulse Power

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  • Last Post 22 December 2022
Chris posted this 16 December 2022

My Friends,

Some Researchers are using an Equation to calculate Input Power, and Ignoring the Oscilloscope Math.

The Calculator is provided here: Vishay or Here 

The Math is:

PIn = I * V * sqrt( Duty_Cycle ) / sqrt( 3 )

 

Now this is in the Public arena, with Respect to Jagau, I disagree with this approach of measurement for many reasons which I will discuss.

First, Jagau is My Friend, and I have great Respect for him, but I feel this issue needs to be discussed and corrected, with Respect to Jagau!

Jagau has provided a Completed Equation:


PIn= 0.292a x 24 x sqrt 0.0.382 / sqrt 3 = 2.533 watts

 

Jagau has provided several Images of his Scope:

 

Jagau has given this image as his Output Power:

 

Jagau has stated:

Pin= I*V*sqrt(Duty_Cycle)/sqrt(3)

Pin= 0.292a x 24 x sqrt 0.0.382 / sqrt 3 = 2.533 watts

So for  Pout of 4.032 watts  i need only 2.533 watts at input, Still a pretty good COP

 

Of course Mr Has Nothing All Day Long, has stated:

I checked this from the beginning and the results were the same for the manual calculation and for the online calculator.

If you want to understand more just check Dr. Nastase's papers online.

So for me this is already clear: Jagau's calculation method is valid.

It was also verified and confirmed by other members of our core team before Jagau came public with it. Including Yoel. Jagau is a very careful researcher. He asked for double-checks here and also the experts in real life before deciding to move the info in our public section.

So just let's move on, sorry for being blunt but this discussion already took too much time, if there are some who are still not convinced that's not our responsability to convince them all.

Regards,

Fighter

 

In a VERY Desperate attempt to convince everyone he ( Fighter ) is a Genius and its his magnificent invention that has saved the world from Certain Doom and Destruction!

Fighter is only trying to deceive everyone including himself! Dr. Nastase's papers do not hold water and should NOT be used for Accurate Measurements!

IMPORTANT:

All these Equations are based around the PBE, or Power Balance Equation: PIn = POut, which does not hold when Power is Generated! Or when Power is Lost in the Machine!

 

Why is the Scope and PSU no where near the Calculations?

You Can NOT use these equations with these machines, and what's more, its already been stated and shown why in other threads here on this Forum:

 

If Output Power is not Equal to Input Power, then these Equations are Wrong and should not be used! Also RMS Calculations can NOT be used for DC Pulses! Again we have covered this already!

@Jagau, this is why your Scope and your PSU do not agree with your Calculations, I am sorry My Friend! We need to be Accurate and Correct!

@21.4% Duty Cycle you can very easily calculate the Current Mean: ≅0.500A * 2 Div's / 2 = ≅0.500A @21.4% Duty Cycle means Current is On for 21.4% of a Cycle and off for: 78.6% of the time, which gives us: ≅0.200A or ≅200ma, average over 1 Cycle and that's approximately what the PSU shows:

 

NOTE: The Scope shows 282mv which is also wrong!

I hope you don't take this the wrong way! I feel that this has gone way too far and we need to point out their Mistakes!

 

Why fighter is pushing Fakery?

Fighter and his "Team" are absolutely dead wrong and should not be misleading people like they are!!!

 

Conclusion

Jagau's Frequency: 1.044kHz and Duty Cycle: 21.4% is a simple and straight forward Power to measure via any Accurate method, and the PSU does show a reasonably accurate measurement in this case! Especially compared to other methods!  

NOTE: I did also point out these In-Accuracies in the Thread: Chris's Waveform Experiment, RMS vs Average which spawned from the thread: Melnichenko's Effect

Dr. Nastase's papers do not hold water and should NOT be used for Accurate Measurements! We have shown this proof here on this thread and in others!

Best Wishes,

   Chris

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scalarpotential posted this 16 December 2022

Heya Chris,
did you know that for repetitive signals with the same period both in phase "Average power = Irms x Vrms" gives the same numbers as the math + average method of a DSO?
You did agree that the latter method is right, but you believe RMS V and I are useless, they are only useless if the value isn't multiplied with another RMS value, such as VRMS^2/R, IRMS^2R and VRMSIRMS. Both claims can be proven.

Just a FYI, no arguing intended.

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Chris posted this 16 December 2022

Hey Scalarpotential,

If you can present some proof?

did you know that for repetitive signals with the same period both in phase "Average power = Irms x Vrms" gives the same numbers as the math + average method of a DSO?

 

I have presented proof:

These Data Points Draw the following DC Pulsed Waveform:

 

The Peak Voltage here is 1.0 Volts, so the peak voltage value is controlled and can not fluctuate. This is true for Positive Values Only, as we have no Negative Values.

The Calculated Mean or Average and RMS is:

  • Square Mean: 0.5
  • Square RMS: 0.707106781186548

 

It is clear that the Mean or Average is NOT: 0.707106781186548 Mathematically, the Average is { 1, 0 } = 1 + 0 = 1 / 2 = 0.5! 

I gave all the data, including the math, in the thread, if anyone wants to verify it they are welcome!

Jagau is my Friend, I have great respect for Jagau so it saddens me that I read this:

Hi all
I didn't think I was creating so much resentment towards me. There are at least 3 forums that talk about me but not in a good way, it's not at all what I expected from a forum for exchanging ideas, we are all entitled to our opinions but it must be in the respect.
I therefore withdraw from all the forums in which I participate and will perhaps resume after the holidays if it is worth it and if the heart still tells me to continue sharing. It's a bit too much for my sanity.
jagau

 

When one uses a Method of Calculation Power, in the Public Domain, then its the duty of all good researchers to point out mistakes in that Method if there are any!

Jagau is not the inventor of these Methods, he merely uses them, under the pretense they are Correct, and they are NOT! It is very clear when one does a deep dive into them!

We all make mistakes, I consider Jagau a First Class Researcher, and we have all learnt something important.

Best Wishes,

   Chris

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scalarpotential posted this 16 December 2022

If you take the signal you provided and a 1 ohm load, calculate the average power absorbed by the resistor, just use any method that is intuitive and logical to you.

rms is a shortcut, it's extracted from an average power product and is compatible with every formula of the ohm's law chart. you can have an average voltage of 0,5v, average current of 0,5a, and an average power of 0,5w which is the product of VxI (or vv/r or iir), that would make v=0,707 and i=0,707, or sqrt(0,5) x sqrt(0,5), this is rms, square because of vxv/r or ixixr, over the complete duty cycle, or integral over T dt, then mean and squareroot.. I can't explain it well enough, it probably doesn't make sense, but it's interesting and it works, just go through the math. Try your own method first at the start of this post.

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Chris posted this 16 December 2022

Hello scalarpotential,

I am sorry, from what I understand about your post, I disagree with your statements.

Ohms Law for Instantaneous Power in a Resistive Load: P = E x I

 

So I do not understand why, and for what reason, you're attempting to Complicate what is not Complicated!

Best Wishes,

   Chris

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scalarpotential posted this 16 December 2022

If you have 1v at 50% dc, and a 1 ohm resistance, what is the power in the resistor? Can you calculate it? And 25%?

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Chris posted this 16 December 2022

Hey scalarpotential,

If you have 1v at 50% dc, and a 1 ohm resistance, what is the power in the resistor? Can you calculate it?

 

Yes, this is very easy, and its very accurate: Average Voltage x Average Current = 0.5 Watts

Assuming the accuracy of the Resistance is accurate and the Voltage is accurate. There is no: RMS here and no need to apply RMS for DC Pulsed Circuits! Its Wrong to apply RMS as I have already shown!

Also:

sqrt(0,5) x sqrt(0,5), this is rms

 

No, this is not RMS!

RMS is:

 

Youre confusing Square Root with RMS!

Best Wishes,

   Chris

scalarpotential posted this 16 December 2022

What's the value of average voltage & current?

If you plug in 1v 50% dc into the equation shown in the image, let's take for n=100, then sum of 50 times 1x1, 50 times 0x0, divided by 100, if you wourk it out you get 1v*sqrt(0,5).

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Chris posted this 16 December 2022

Scalarpotential,

What's the value of average voltage & current? If you plug in 1v 50% dc into the equation shown in the image, let's take for n=100, then sum of 50 times 1x1, 50 times 0x0, divided by 100, if you wourk it out you get 1v*sqrt(0,5).

 

Again:

Also:

sqrt(0,5) x sqrt(0,5), this is rms

 

No, this is not RMS!

RMS is:

 

Youre confusing Square Root with RMS!

 

Squaring a Number, or taking the Square Root of a Number is NOT RMS!

You're getting confused by Math!

Best Wishes,

   Chris

 

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scalarpotential posted this 16 December 2022

Hey scalarpotential,

If you have 1v at 50% dc, and a 1 ohm resistance, what is the power in the resistor? Can you calculate it?

 

Yes, this is very easy, and its very accurate: Average Voltage x Average Current = 0.5 Watts

Assuming the accuracy of the Resistance is accurate and the Voltage is accurate.

Can you show the value of Vavg and Iavg?

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Chris posted this 16 December 2022

Scalarpotential,

I have presented proof:

These Data Points Draw the following DC Pulsed Waveform:

 

The Peak Voltage here is 1.0 Volts, so the peak voltage value is controlled and can not fluctuate. This is true for Positive Values Only, as we have no Negative Values.

The Calculated Mean or Average and RMS is:

  • Square Mean: 0.5
  • Square RMS: 0.707106781186548

 

It is clear that the Mean or Average is NOT: 0.707106781186548 Mathematically, the Average is { 1, 0 } = 1 + 0 = 1 / 2 = 0.5! 

 

Can you not read?

What is the RMS value of a periodic signal?  When a periodic signal is generated by a source connected to a load, a resistor for example, the RMS value is the continuous signal, the DC value which would deliver the same power to the load as the periodic signal.

Ref: Adrian S. Nastase

 

It is mostly ignored, a pure DC Signal, On and then Off, is still a DC Signal and not an AC Signal! For the On Time, there is Pure DC, which delivers the same power to the load, as DC does! Believe it or not!

If you persist with this silly non-sense that has been proven wrong many times here in this thread, we will have to evaluate your membership here, as this is not the first time we have seen argument over this topic!

I have proven you wrong, can you not see this?

Again:

Also:

sqrt(0,5) x sqrt(0,5), this is rms

 

No, this is not RMS!

RMS is:

 

Youre confusing Square Root with RMS!

 

Squaring a Number, or taking the Square Root of a Number is NOT RMS!

You're getting confused by Math!

 

 

Jagau's PSU tells the whole story:

 

Which I calculated!

Or are you here just to Cause Trouble? Create Confusion over Simple things? Because your contributions have been Zero since you became a Member Here!

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Chris posted this 17 December 2022

My Friends,

This latest round of Non-sense from Fighter and his team discredits the Energy Community more than ever:

Really ? You don't understand ? We had another one whining for weeks that he doesn't agree with the calculation. Do you guys really think we gonna let our forum get full with spam for each one coming here and complaining he doesn't agree with the calculation method ? We don't have time for that, not gonna waste our time with each one and his spam here.

It's not our obligation and we not gonna try to convince anyone. We offer documentation and even an official calculator for that calculation method. If they get it, it's fine, if they don't get it, it's also fine. If they don't agree, that's their problem.

We have better things to do like research. Which anyone coming here to spam our forum should actually do.

Talking is easy, coming with new things it's the thing requiring work and time.

So stop wasting our members time. Else we gonna address these things in a more radical way.

Fighter

 

So Fighter and his team, the same team that is no longer backing him up, by the way, are insisting on using a Calculation, in place of Measurement, or he will erase you and your opinion from his website, a website that has lost all credibility, by Forcibly Faking Numbers, to support his toy that mostly is good for nothing! The Fake ZPM!

No wonder no one takes the Energy Community seriously!

Fighter and his little crowd of buddies are faking results to get traffic and publicity! No one is backing up their Calculations with ANY Evidence! 

In point of Fact, Evidence all points to the Opposite! Calculations being Wrong!

It is very easily shown, both on the scope and also on the PSU, the Numbers they are insisting on, are Wrong and Fake! 

I had hoped Jagau would have come forward and either shared his effort to either Verify his Calculation with Emperical Evidence, or Confirm the Calculation: PIn = I * V * sqrt( Duty_Cycle ) / sqrt( 3 ), is Wrong and does not hold Water. I totally understand Jagau's position, one of Frustration and Anger, but Truth always wins out in the End!

A deceitful act against the public, and this is what this is turning out to be! This thread gives an honest and factual approach to what has been done! Truth is Fighter and "his" zpm are being Forcibly Faked for Publicity! It is Beyond Stupidity to carry on with this sort of Non-Sense!

I 100% disagree with this sort of behaviour!

We really need to be honest and truthful to ensure we succeed in the end!

I have removed everything to do with Fighter from my Website, as I want nothing to do with his Hoaxing Non-Sense!

Here at Aboveunity.com, we have been 100% honest and upfront with everything, some providing accurate AU with Measurements! Showing REAL Scope shots with Real Data! I have encouraged all Successes to be deliberately Vague and Share Little, so people like Fighter and his buddies get no where! Fighter has gained one thing, becoming the latest Community Hoaxer!

 

Jagau's PSU tells you the Truth! 4.3218 Watts NOT 2.533 Watts! 

The Scope and the PSU will show any Negative Power! Three methods I can prove there is ZERO Negative Power here!

PIn = 4.3218 Watts POut = 4.032 Watts = COP = 0.9329!

Fighters Fakery non-sense is absolutely Disgraceful! ZERO Credibility for him and his buddies!

Disgraceful! Bloody Disgraceful!

 

Some see the Truth Already and some people will learn this the hard way no doubt!!!

I have learnt something, a few I once trusted, I no longer do!!!

Bloody Disgraceful!

 

Best Wishes,

   Chris

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Chris posted this 22 December 2022

My Friends,

Perhaps we should discuss the following snippet from the above quote:

Really ? You don't understand ?

 

Fighter and his band of Disgraced and Banned Ex-Members are pushing the Negative Power theory!

This is 100% Laughable and falls into the stupidity basket because:

1 x 1 = 1

Positive Voltage x Positive Current = Positive Power!

-1 x -1 = 1

Negative Voltage x Negative Current = Positive Power!

 

There is NO Negative Power Here! This is CLEARLY Seen on Jagau's Scope! So again they are Faking Data to try to Fool You!!!

So WHO is it that does not understand! Hmm what a Low IQ Buffoon!!!

I have shared with you in great detail how Negative Power Works here: Negative Power

Best Wishes,

   Chris

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Ere many generations pass, our machinery will be driven by a power obtainable at any point of the universe. This idea is not novel. Men have been led to it long ago by instinct or reason. It has been expressed in many ways, and in many places, in the history of old and new. We find it in the delightful myth of Antheus, who drives power from the earth; we find it among the subtle speculations of one of your splendid mathematicians, and in many hints and statements of thinkers of the present time. Throughout space there is energy. Is this energy static or kinetic? If static, our hopes are in vain; if kinetic - and this we know it is for certain - then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature.

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