A matimatical aproach to overunity

[beno]

Hi,

@nul-points: Yes it is a wave form I have seen before... That is currently all I can say to you right now. Regarding the coils, yes in my head they are on the same core. But do please keep in mind - this is currently 'only' an image in my head. Howevere the spike is the key, the driving of the current.  That spike is most easyly achived by using either square-forms or switches. But this is 'soly' based upon some image in my head of how some of Teslas circuits could have been made.


@onthecuttingedge2005: I know that this might be the case with the slave driver doing more work than the slave (the electron). Convetional knowledge says so. You know... Power normally do not comes from nothing, unless you are on the quantum level, and that level will normaly be closed to conventional circuits.

But if so, then why are we here in this forum then, if we all just accept the conventional knowledge?

Now, I do know that currently, I have only a theory based upon the theoretically wave form of a coil. The rest of the circuit is 'missing' - and perhaps you are right that the power-consumption here is greather than the gain in the coils (I do not think so). But if it is so, then I will be tempted to say that all the talk about overunity, is talking about a dream that will never come to reality. Other than in circuits than in some way is drawing its power from the sun (like wind,solar and so on) or the magnetic core in the earth itself.

The spike is the overlord hammering fear into the minds of the electrons, so they are driven by themself (or perhaps persuaded to do the work). Without the need of a 'slave driver'


Best regards

beno

[quantumtangles]

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

This is creative and interesting jazz maths. It is nice to see it here.


As I understand it, this boils down to dv/dt = I


By changing voltage rapidly per unit time, you increase current. Examples of discrete devices capable of changing voltage rapidly over time might include certain 555 timers and certain transistors (depending upon their specifications). I like your idea not least because I thought of this myself in or about February 2012 (long after you thought of it). However I realised that continuous power is required, which is to say that one cannot (I respectfully suggest) fragment the power output into discrete packages because in that event, the Sine wave would not be a Sine wave. It would not be a wave at all


So I like your line of thinking, but how does one generate continuous output by pimping dv/dt with discrete components such as transistors or 555 timers? In passing, half a sine wave (DC) would still work if we used a bridge rectifier to convert AC by doubling the hertz above the flat line (for positive voltage values. Accordingly, manipulation of the wave is always possible, provided we can get continuous output in the first place (provided there is a wave to manipulate).


By way of practical example, suppose we have a simple circuit with a 9v DC battery power source, and we connect this to a voltage regulator and a 555 timer chip. In theory, we may think we can change the voltage so rapidly that current will increase enormously. But I have tried this and it did not work. One always needs approximately 2v more (2v more going into the voltage regulator) than the regulated output voltage. I think it does not work both for this reason and also because one is essentially trying to create discrete pulses incapable of constituting a wave. I iterate I have not yet obtained an oscilloscope so I have been running blind in terms of wave generation thus far and this may be worthy of further examination.






If ignorance was a criminal offence, I would be serving a life sentence

[beno]

Hi quantumtangles,

Now I have not had the time to build the circuit in my mind, which perhaps could do the trick. But yes, your are right.. one could use a 555. Actually the circuit in my mind will use a 555.

However the waveform should not be Sinus-formed. As I see it must be a squareform (with a biased DC voltage), perhaps 12 volt +/- 5 V (now this might not be the correct values, perhaps the more correct values is 7 +/- 5 V).

I also think there must be at least one coils/core's involved. Perhaps two coils on the same core, or two cores with one or more coils.


As I think:

If one would take the energy from the earth's/sun energy-field, par example from the Schumann cavity, one should use sine-waves. And I would think it would be of critical importance, that one use n-times the schumann cavity frequency (or perhaps with some minor error - to prevent perfect aligned standing waves in the circuit)

If however you'll want to draw from other sources - then you should other waveforms. From my current point of view, a squareform should most likely be preferred.

But let me point out, that in my current point of view, we are not likely to make a overunity device from a universal perspective, just a device which draws power from other sources than the battery   And therefore from a local point of view, seems to be an overunity-device [Actually a solar-cell is also from a local point of view .. an overunity device].

Which other sources you might ask? Well it could be the Barkhausen effect or other sources.

Best regards

beno

[quantumtangles]

Interesting Beno.


In February 2012, I contacted a friend with a PhD in Electrical Engineering and asked various questions about amplifying current per I=dv/dt. I sent him the email when I first thought of your idea (in February 2012), a considerable time after you first came up with it.


I do not have permission to reproduce the email in full, but in expurgated terms the responses were as follows:

When I pointed out that the 555 timer could switch on and off perhaps ten times a second, (leading me to think current could be amplified significantly in consequence of rapid voltage changes, the academic responded as follows:


"I am not sure what you mean by this. Current and voltage are continuous variables and switching on/off would result in discrete variables which are not useful. If you consider a Sine wave then the instantaneous value (dv/dt) is the latest value of Sine with respect to time in a very short time span or you can divide the time axis into very small pieces and the value within a particular piece is dv/dt or di/dt."

In passing, it seems to me it is not necessary to switch the voltage completely off. In other words, the 555 timer could be used to vary the voltage (between 0.001 volts and 9 volts for example) rather than turn it off completely, thus ensuring a continuous (wave) function.


Second Question:

Using the example of a 9 volt battery connected to a 555 timer capable of changing the voltage in 0.1 seconds (ten times a second), I asked whether dt would really be 0.1 (seconds) and whether this would necessarily involve a total voltage change of 18 volts or 9 volts (I initially thought voltage would swing from -9v to +9v in 0.1 seconds and accordingly that there would be a dv of 18v. This was my question to him:

Dt would be 0.1 seconds (voltage would swing from -9v to +9v in 0.1 seconds). Is dv = 18v (-9v to +9v) or is dv = 9v?

If dv = 18

I = 18/0.1
I = 180 amps

But even if dv = 9

I = 9/0.1
I = 90 amps.

He responded as follows:

"18V is your range not the instantaneous value; everywhere you see 'dv' or 'di' It means that the value of voltage or current at a very short span is desired, and depending on the shape of your current or voltage it can take any value between -9 and 9. Therefore, dv cannot be 18. Also, you can not use the above expression where you only have resistors."


In summary, the academic did not object to the general idea that rapid changes in voltage would lead to significantly increased current. His practical objection was that doing so would lead to non-continuous output incapable of constituting a wave function. I do not understand why this should necessarily be the case in all circumstances, so I am going to test it. For example, as I said before, the 555 timer could be used to vary voltage between pre-set parameters of say 0.001 volts and 9 volts, thus ensuring continuity of the wave function.


However we do need to pay particular attention to his comment about resistors for reasons I am coming to. Equivalent series resistance is actually quite important but I will get back to it momentarily.

Going back to what I was saying before, I plan to try the 555 timer again but this time use an oscilloscope to observe the shape of the resulting wave. The objective is to obtain a continuous sine or square-wave using a 555 timer chip to change the voltage as rapidly as possible and then measure output current.


A potential practical hurdle which I think the academic was referring to (in terms of resistance), and that may be of interest to you, is as follows:


A power source of some sort will be required to carry out the experiments. Let us say we decide to use AA duracell batteries connected in series (one on top of the other as one would find in a battery powered torch). See Afrotechmods on Youtube.


Conventional AA duracell batteries contain vast amounts of energy. Two 1.5v AA batteries in series = 3v x 2.5Ah = 7.5Wh = 27 kJ of energy


Note that 1Wh = 3.6kJ


This is a huge capacity, but there is a non-ideal property called equivalent series resistance which means that AA batteries have some sort of internal resistance, otherwise known as equivalent series resistance or series impedance.


Batteries, capacitors and other electronic components almost all have this small internal resistance, which limits the amount of current that can flow. For a typical AA battery, the Equivalent Series Resistance is 120 milli-ohms. Not very much, and it seems fairly irrelevant. But when we do the maths, it starts to get relevant when we put a load on the battery.


For Conventional/Normal use of an AA battery (in a Hi-fi remote control for example)


V = IR


Vdrop = 0.02A x 0.12 ohms = 2.4mV (This is the voltage drop).


Vunderload = 1.5v - 2.4mv = 1.4976 volts


Power = I(squared)R
P= 0.02A x 0.12 ohms = 48uW of heat

So what. Who cares. This is a tiny amount of heat generated under normal operating conditions for a AA battery.


But when we pimp the load, a completely different picture emerges. Suppose we manage to get our 555 timer to churn out 10amps from a AA battery without melting, watch what happens to the battery.


V = IR


Vdrop = 10A x 0.12 ohms = 1.2 volts


Vunderload = 1.5v - 1.2v = 0.3 volts


Power = I(squared)R


P= 10A(squared) x 0.12 ohms = 12 Watts of heat


So the typical AA battery is going to light up like a bulb filament if we try to draw 10 amps from it (which is to say, even if we manage to get our 555 timer to increase current using rapid changes in voltage per unit time (I= dv/dt) without turning into a crisp, we are still going to need a bigger boat. We are going to need an ultra-capacitor. This is because instead of there being a 2.4mv drop across the resistive element of the AA battery, there would be a 1.2v drop across the resistive element, and the AA battery will essentially melt.


For the above reasons, we may start leaning in the direction of a 555 timer combined with an ultra-capacitor supplied by a 12v car battery (or a bench power supply, though it would probably take half an hour or so to charge an ultra-capacitor even with a bench power supply).


Then we have to ask whether the 555 timer will melt due to high current or whether we can somehow keep it away from the current amplifying element of the circuit (which it is itself a part of). So what we are looking for here, if we use conventional thinking, are some fairly exotic components to test your hypothesis. There may be a way of performing the experiments with less exotic components but I have not figured out how to do that yet.


What we are really trying to achieve here is to use sophisticated components (such as 555 chips and transistors), designed for completely different purposes, to cause rapid changes in voltage and accordingly to generate large current amperages. It is a counter-intuitive idea (to use such components to enhance current).


Ultimately it may be possible, using this technique, to power heavy duty devices for short periods with rechargeable battery arrays, perhaps even with there being sufficient time between battery connection and discharge to recharge other batteries in the array (using solar panels etc). But for sure it would not be OU (which personally I consider to be impossible). Some external energy source is always needed imo.

Please let me know if you suggest alternative components for the experiments.


Kind regards,

[beno]

Hi quantumtangles,

In my comming experiments, I would start with using square-forms (I think it would be needed to have a DC-bias). You know: Vhigh = 12 volt , Vlow=7 volt (the actual values must be determined by experiments, but I'll start with a low voltage, perhaps even down to a couple of volts). The reason for using square-forms, is I do not belive we'll get high enough dI/dt using sine-waveforms. But I might be wrong on this !  The key-question is how does the current/voltage look in the coils? But this can be determined with our oscilloscopes   After all, the ultimate proof is in the circuit, not in the mind, or in some calculations.

I also consider either using one or more batteries/capacitors as the supply, I have also thought about using a car-battery. But I will not start with a car-battery, just to play it safe so I don't fry to many components .  But we are having the same thoughts

And yes the 555 is perhaps not the ideal switching component, and perhaps should only be used as a timer-component. I'll start with using it as a switching component, but perhaps I'll continue with FET's. Like in a switch-mode power-supply.

And we also agree that energy will always come from somewhere, the only question is from where.  However I do belive that this could obtain a COP around 1, and even above 1 [as seen from a local perspective - not a universal perspective], but I don't expect that this is going to be a high-power circuit, perhaps 5-30 watt [however if we succeed, someone might build a high-power circuit].
So yes, this is only for fun

But to answer your question: No, I do not think that I have any alternative components to add to your experiments (because it seems that we are already thinking on simliar circuits). The only difference is the input waveforms, but the 555 can be used for both waveforms.

Kind regards

beno