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

[Omega_0]

@Omni,

First of all, we don't have full info. I don't know whether the battery remains disconnected for rest of the 75% cycle or gets fed in via diodes. The whole description of setup and circuit, components and their values was omitted from the demo (may be intentionally?). So I'm only guessing that the fall in input is due to energy being fed back into battery.

Regarding the net input loss being V*I - I^2R or something else, honestly I have no idea. Theoretically, it should include other losses like hysteresis loss etc, but the situation here is complicated because of things such as presence of a strong magnet, changing inductance and core saturation. Steorn says nothing about where the energy is going and whether the scope shows true energy loss....

That's why I said, this demo is inconclusive and real proof will come from replicators.

[Omnibus]

@Omega_0,

That was my initial inclination too -- to consider the falling part of the curve as returning energy to the battery. That appears to be one of the most interesting features of this device. Of course, that's a detail and in time we'll know what the real story is. At present, and in general, I don't think it has any importance as to whether or not the device is OU. That energy has to be subtracted from the I*V during the duty cycle in any event.

I don't think that it matters what other terms are subtracted from I*V other than I^2R once we have the traces and the values of in and out which you've so correctly shown in your post. Replicators need to show only that run of the traces and OU will be independently proved beyond any doubt. Details such as the above are for later.

Yours was one of the best posts on the topic. Thanks.

[Omega_0]

Moreover, he is claiming overunity (108%) at the input side also, which complicates matter further. You won't see the classical energy loss at the input if he is right.

The output is straightforward (surprisingly they didn't make it convoluted), but the input trace is beyond me. Need more info or Sherlock Holmes to understand it.

[Omnibus]

I think your interpretation of the input trace is pretty decent for the purposes of determining whether or not there's OU which is the important part. As to why its shape is exactly the one we see is only a detail to be studied later on.

[teslaalset]

I did a further analysis of the blue curve that corresponds to the driving coil power (picture posted on page 151, by Omega_O).
The blue curve can be re-calculed to time by differentiating the curve.
If I am correct on this it will be like below graph.

The area A corresponds to the pulsing period of the coil.
This shape is formed by the rectangular voltage shape and the current curve that is exponential with time constant LR.

The area B is something to discuss about.
In my view it looks like this part is used to de-magnitize the coil and the inductance is almost zero (saturated coil with very low permeability)
The negative spike is caused by the high voltage peak that is caused by switching off the voltage abrupt.

At first I thought that area B is caused by pulsing the pulse with a reverse (negative) voltage. But that would cause negative voltage times negative current and thus a positive power area instead of a negative one.
So what is causing area B?