PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[ltseung888]

BTW -- the Tseung device (attachment) uses a lot of Pinput -- measured by two 10,000 uF caps, I find about 90mW for Pinput.  BUT -- what counts in the next "contest" is the RATIO of Pinput (measured used the cap/time method) and Poutput (measured in the cal'r), that is, n = Pout/Pin.

Dear Prof. Jones,

Thank you for showing the FLEET prototype.  It is almost one year since we demonstrated a FLEET prototype with peak-to-peak Output Power > Input Power on July 13, 2010 in Hong Kong.  I placed a one year review article at my bench in reply 26:
http://www.overunityresearch.com/index.php?topic=755.msg15473#new

I am training a 13 year old student, Michael Du, to do the prototypes for me.  He is also learning the Physics and Mathematics â€" starting from Newton’s Laws of Motion, kinetic theory of gases, resonance etc.  It will take him sometime.  With God’s Blessing, he has produced a FLEET prototype based on Multiple LCR resonance with an air core toroid. 

I shall order the oscilloscopes, signal generators and other test equipment and electronics.  Their spare guest room and/or garage is likely to be the laboratory.  When the equipments arrive, I shall be able to re-demonstrate my best FLEET prototype with peak-to-peak COP > 200. 

I do appreciate the excellent work you have done.   Some groups in Hong Kong and in US (e.g. Mr. Harvey Gramm, the moderator in my debate with Poynt99) indicated that they achieved resonance or results similar to the Steven Mark TPU.  Theoretically, I know that it is possible.  Hopefully, with the help of Michael, we can show that result in Irvine, California in the near future.

May God guide us in our efforts to benefit the World.  Amen.

[xee2]

@ JouleSeeker

Are there going to be any on line reports from the free energy conference you went to?

[beno]

First of all let me start with an excuse if it is not correct that I'm posting it here. Perhaps I should have started another thread. If so do please accept my appology.


Now the reason for this thread was that I asked this simple question: Is it at all possible to to reach overunity? A gut-reaktion would most likely be no - energy does not come out of nothing. But what about from at matimatical point of view?  I'll here show that the answer is Yes - but there might be something wrong in the calculations below.


Do please feel free to show where my calculation goes wrong - but here goes...


Background:
This theoretical approach involves an ideal-battery [fixed voltage, draws only current], two coils (wound on some core), and minimum a load transistor on the "secondary"
I was thinking about a pulsating input on a 2 magnetic coupled coils in some circuit, and there should be a load transistor on the secondary.

Now I do know that, a real circuit would need alot more components but just to get us starting.



Math:
If the input is a squareform, the output would most likley be given by:

  V(t)=Vdc +- Vp*exp(-t/tau)   That is exponential decreasing over time, but down to the DC value which is the value which we are shitching around..
   Where: t is time, Vdc is the base voltage, Vp is the peak voltage (above/belov Vdc) - one could call it delta-voltage - caused by switching a coil on and off, tau = L / R, Where R is the load-transistor. Please note that this wave-form is purly theoretical, I have not yet put a coil and a scope to the test, as I currently has no oscilloscope.

   That is a peak value which decrease over time, and in the the next part of the duty-cycle the voltage is reversed.


Therefore the current over time would be something like:

  I(t) = V(t)/R = (Vdc +- Vp*exp(-t/tau))/R

Power input durring a cycle f=1/dT and therefore: dT=1/f

P=Vdc*I(t)      for a battery with fixed voltage - this is of cause an ideal battery which does not have much todo with a real battery

Pin = [0 to t/2] Vdc*{(Vdc - Vp*exp(-t/tau))/R} + [t/2 to t] Vdc*{(Vdc + Vp*exp(-t/tau))/R}   where [] means integration over time - sorry for the missing integration sign.
    = [0 to t/2] Vdc*{(Vdc - Vp*exp(-t/tau))/R} + [t/2 to t] Vdc*{(Vdc + Vp*exp(-t/tau))/R}   where [] means integration over time - sorry for the missing integration sign.

If we say that the both the negative and positive amount equals each other (that it is just inverted) - they will cancel out each other. Therefore in an ideal world it would be:

    Pin = (Vdc*Vdc/R)*dT ; dT is delta time

The output on the load resistor is:

Pout    = [0 to t] (V(t)*V(t))/R

Or put in another way:

Pout   = [0 to t/2] (Vdc - Vp*exp(-t/tau))*(Vdc - Vp*exp(-t/tau))/R + [t/2 to t] (Vdc + Vp*exp(-t/tau))*(Vdc + Vp*exp(-t/tau))/R

Now we know that: (a-b)*(a-b) + (a+b)*(a+b) = a*a + b*b - 2*b*a + a*a + b*b + 2*b*a = 2*a*a + 2*b*b  This gives:

Pout   = [0 to t] [2*Vdc*Vdc + 2*(Vp*exp(-t/tau))^2]/R
Pout   = 2*Vdc*Vdc*dT/R + [0 to t] [(2*(Vp*exp(-t/tau)))^2/R]               where [] means integration over time - sorry for the missing integration sign.

                                  2*Vdc*Vdc*dT/R + [0 to t] [(2*(Vp*exp(-t/tau)))^2/R]
Efficiency = Pout/Pin = -------------------------------------------------------         where [] means integration over time - sorry for the missing integration sign.
                                               (Vdc*Vdc/R)*dT
           
                         [0 to t] [(2*Vp*exp(-t/tau))^2/R]          [0 to t] [(2*Vp*exp(-t/tau))^2/R]
            =  2 + ----------------------------------- = 2 + -----------------------------------
                              (Vdc*Vdc/R)*dT                                    (Vdc*Vdc/R)*dT


So this means that it should be possible to reach overunity (if we can keep the second part under -1) - but under which circumstances?
To investigate this we need to solve the integral. But something interesting would most likely happen around dT=tau.


Because: a*[0 to t] exp(-T*t) = a/T*(1-exp(-T(t/2))) This means:


[0 to t] [(2*Vp*exp(-t/tau))^2/R] = ((2*Vp)/(1/tau))*(1-exp(-(1/tau)*(t/2))^2/R
              = {2*Vp*tau*(1-exp(-(dT/(2*tau))))}^2 / R

Because tau = L/R this gives     = (4*Vp*Vp*L*L*R*(1-exp(-(dT/(2*tau))))^2) which again gives:


                                      (4*Vp*Vp*L*L*R*(1-exp(-(dT/(2*tau))))^2)
Efficiency = Pout/Pin =  2 + -------------------------------------------
                                                    (Vdc*Vdc/R)*dT


If the above caculations are correct, then we have an efficiency of around: 2231 for the following values:

  Frequency: 50 kHz , means dT=0,00000009 sek (1/f)
  Vp : 0,5 V      Now I do not know how big this value actually is.
  Vdc : 2,5 V
  L : 90 uH
  R : 1 kOhm

Now this is theory and the rest of the circuit do needs power to maintain opperation, but it should be enough to drive the circuit. Otherwise one might try to increase the Vdc


Best Regards

beno

[JouleSeeker]

@Lawrence,  so sorry to hear about your eye problems...  I do hope you will find improvement. You wrote, "peak-to-peak Output Power > Input Power"  and I have studied what you mean by this.  IIRC, to calculate output power, you took the peak-to-peak output voltage  times the peak-to-peak output current measured across a resistor. Please correct me if wrong.  Unfortunately, this does not give a reliable measure of the output power, as we discussed at length on the OUR forum.

After much discussion, I instead used a Tek 3032 scope to provide
Pout (t) = V(t) * I(t), showing the waveform, then let the Tek calculate the MEAN output power.  Even this, I now believe, has potential for measurement error, hence the move to put the device in a calorimeter where the total output energy can reliably be determined.

The input energy is easier, obtainable from driving the circuit with a known capacitor, from a known start voltage to a known stop voltage, and using E = 1/2 C*V*V.  We will then determine the efficiency using Eout/Ein.


@Xee2:  the conference I was invited to in Idaho was primarily related to home preparedness (given conditions in the US and world-wide).  The people there were very interested in home-energy production, and the latest work in alternative energy production.
Unfortunately, it turned out that time was limited as I was the "keynote" speaker on Friday evening, following a healthy list of speakers, and about one hour was shaved off my allotted time as other speakers went over, so I could not say much that I wanted to say. There will be other opportunities...  I think there will be a summary on youtube; will watch for that.

@beno -- you wrote, "Perhaps I should have started another thread."  I think so in this case... but -- if you will summarize for us how your "theory" translates into experiments that could be performed (based on your equations) -- then THAT would be welcomed.  What experiment-modifications are you proposing? (If more general and theoretical than that, I would indeed recommend another thread).
   I would suggest that the valuable theories are those which lead to or guide experiments.

Thanks @all for comments.

[beno]

@Jouleseeker: I'll start another tread - but I'll propose to change first of all the "input waveform" into the coils. Sorry for my comments into your thread.