STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Omega_0]

That energy stored is your initial condition in the equation I have posted earlier. It just determines the position of the graph not the slope. Since we have enough math now I'm eager to make the comparison between experiments.

broli,
IMO, the initial condition will remain constant throughout the wave cycle, but the energy stored varies with voltage across the cap, which will obviously not stay constant during a full cycle. Take a look at the file attached in my previous post. The Ec is not a simple bias.

[Omnibus]

Now, here are data for 800kHz, 1Ohm and 50pF. What we see here is one full cycle and for that one cycle we see, in agreement with @broli's results, practical coincidence of input and output energies at the end of each cycle (see the third graph with the expanded y-axis). The intimidating curves of the input energy are nothing more than the typical behavior of a reactive element consuming and returning energy to the source within each half cycle. What is remarkable, however, is the negative slope of the input energy best-fitting line which is the input power. Because the magnitude of this slope is very small, we see practical coincidence of the input and the output energy for one full cycle. However, had the full cycles been significantly more than one the input energy best fitting line and that of the output energy will cross each other and input and output energies will no longer coincide. You can play with the R and C values and you'll see that with some the slope of the input energy-time best-fitting line is lower than that of the output energy. In this instance, however, it is dramatic, not only is it lower but that input slope is negative. This means that from the very beginning, even for one full cycle where the integrated input and output powers coincide at each half-cycle, input energy is only returned to the source.

You can play now with the phase angle and you can see that if you start adding even very small values, say 0.01rad, to the phase angle you'll begin noticing a discrepancy in energies even for this one full cycle in favor of the output energy.

I think it is along these lines that we can look for understanding as to what's going on in the experiments. Obviously, experimentally, conditions are created to have the I-V phase shift, which according to theory would yield OU, even more expressed. That's for same R in both the input and output equations.

Another way of looking at the experimentally observed OU is to see it as being due to the input energy experiencing an R of a much lower value than the R which  generates the output energy.

Of course, we still need to analyze carefully the experimental data but the very fact that theory predicts negative slope of the input power for some combination of R, C and f makes one lean more towards 'yes' with regard to this 'yes' or 'no' question concerning the reality of OU. Of course, as I emphasized earlier, the real theoretical proof would be if the above negative slope can be demonstrated analytically because one may always say that numerical methods are prone to subtle errors due to the discrete character of the calculations. On the other hand, numerical methods are the basis of contemporary engineering, so it's hard to draw the line.

[broli]

broli,
IMO, the initial condition will remain constant throughout the wave cycle, but the energy stored varies with voltage across the cap, which will obviously not stay constant during a full cycle. Take a look at the file attached in my previous post. The Ec is not a simple bias.

I have already posted the analytical equation for energy derived from the current and voltage equation for an RC circuit. This inherently includes the behavior of energy stored in the capacitor. It's up to you to choose an initial condition. You could start with a fully charged capacitor and use that as as your initial energy at t=0 or start with a completely uncharged cap and use 0J as your initial energy for t=0. Either way all the initial energy does is add/subtract some constant value to your graph and move it up/down. It doesn't change the slope. In a circuit with R=0 the slope would be 0 and this is what the equation shows too.

Below you can see the energy wave at R=0 and R=100.

[Omega_0]

I'd like to add something to my observation which I was thinking but don't have a strong argument for it.
There is a lot of difference between energy returning back to a voltage source and simple change of direction of voltage at the source.

In first case we have genuine energy gain which recharges the source (if its a battery or cap) and this energy or power can be safely taken as negative. The negative sign here has physical meaning with regard to energy. In second case there is a simple reversal of direction of current, the energy always flows out of the source. There is no recharging or gain. The negative sign here has a meaning with regard to the direction only.

In case of a sine wave, the sign changes periodically, it does not mean that energy is flowing in and out of the source. Energy always flows out in such cases regardless of the sign of the wave. So a cap connected to a sine wave source will be charged in one polarity during half period and reverse polarity during the other. Whatever energy goes into the cap is returned back in exact equal amount. And it always passes through R, heating it up. So energy wise the consumption is always positive.

If, lets suppose, we change the dielectric matter of the cap with one of a higher permitivity, just after the first half cycle,  it will store more energy as its capacity will increase and it will return this excess energy to the source (which is now at lower voltage), leading to a gain. Now its not really possible in a closed system. But its possible in open system. Fortunately no one has proved that the universe is a closed system, there is always a possibility that the dielectric will derive the excess energy from quantum vacuum or any such weird unknown thing.

I guess this is happening in case of steorn Eorbo, where a change in permeability results in excess of energy. I have gotten some interesting results with my build, please check the thread.

[Omnibus]

@Omega_0,

Just looked at the thread you opened. You've done a fantastic job. Great results. You are the first one to independently confirm the claims made by Steorn. Also, I should add in agreement with you, that Steorn deserve a lot of credit for inspiring these studies.

On the technical side, the spent energy is only the energy that has been dissipated. It can never be negative. On the other hand, with reactive loads you have energy flowing in and out of the source with no remaining trace in the surroundings. Think of this process as being similar to an electrochemical cell which can also be an electrolyzer if the process is reversed.  You can see the humongous input peaks in the figs I just posted. These are energies to no avail. The only part that matters in terms of output is the energy at the end of each half-cycle. @broli has illustrated it very well in his Mathematica calculation. As he noted, you can only see a non-zero slope in this energy-time plot when there is solely a resistor. If there's cap only the overall slope of the curve is zero -- no energy dissipated. Of course, if the overall slope (despite fine structure of the curve) is negative, as I have shown to be the case, purely theoretically, with some values of R, C and f, then, in addition to the energy dissipated because of current flowing through R and obtaining it as an output, we have only energy returned to the source. OU in its fullest. Mind you, here we don't have to invoke cosmic forces, energy from vacuum, zpe and what not. It's a pure and simple constructive characteristic of a given device.


P.S. It seems to me it's better to continue the discussion in this thread to have it all together.