A matimatical aproach to overunity

[beno]

Hi,

The reason for this thread was that I posted this in another thread, but it is more appropriate here. So here goes.

Repost below:

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.

Added commment: The circuit is a battery powered circuit, which generate a square-form voltage into some coils - the output is a load resistor. The actual circuit is only partially in my mind. So I can not post it, as I currently does not have it.



Math:
If the input is a squareform, the output would most likely 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

[nul-points]

 

[...]
Added commment: The circuit is a battery powered circuit, which generate a square-form voltage into some coils - the output is a load resistor. The actual circuit is only partially in my mind. So I can not post it, as I currently does not have it.

[...]
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.
[...]
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

hi beno

thanks for doing all this work looking at the possibility that math may show us what we need to focus on to achieve OU

i realise that you don't have an actual circuit or scope to provide some real data, but it may be helpful to folks if they could visualise the waveform(s) you mention in your equations

is there any possibility that you could draw a simple sketch of the relevant waveforms and post a photo of it?

thanks in advance
np


http://docsfreelunch.blogspot.com

[beno]

Hi nul-points,

I can give it a try. Now I'm no artist, so it would not be pretty. But you asked for it 

Best regards

beno

[nul-points]

 

Hi nul-points,

I can give it a try. Now I'm no artist, so it would not be pretty. But you asked for it 

Best regards

beno

that's fine, that will help people to relate to your description, thanks

are the 2 coils wound on the same core like a transformer?

your waveform looks to me more like a Capacitor-Resistor network (series C into parallel R) waveform, ie. two pulses with opposite polarity, each caused by a transition, or edge, of the square wave input signal

is this a waveform you've seen somewhere before? maybe you could give us some more detail about it?

thanks

[onthecuttingedge2005]

a slave will do no work unless it is 'forced' to do work.

the electron is a slave.

this philosophy will be your burden in finding O.U, the electron is a beast of burden that has to be driven to do anything for you.

electron's are inherently lazy(Static) and will remain lazy until 'it' is forced to do some work.

when you try to get an electron to do something you become the slave driver but in the end you will find the slave driver has done more work to get the slave to do something in the electron's world.

the electron refuses to do work unless it is forced to do so.

in the end, you will find this to be true.

Jerry