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

[Omnibus]

If your scope allowed to do some more fancy things you could cut on tweaking the parameters a lot. For instance by multiplying the current and voltage channel, taking the mean of that, and dividing that result by the mean of the current squared resistance. This gives you and instant cop value on screen. Since you're dealing with such low capacitance you could make your own simple variable capacitor and adjust it while running or adjust resistance while running. This will probably give some specific RC formula for each frequency that will give maximum cop.

Unfortunately, can't do with this scope. Would be interesting to be able to do it, though.

As far as capacitance dependence, I did that but haven't posted it yet. With 100Ohms, starting from around 300pF and lowering the cap value -- the OU starts positive (several times) and, as you lower the pF, there's a cutoff point (around 60pF) where the slope of the input energy starts to be negative. Most interesting.

[Omnibus]

I say the same for months. The answer is "no, it's not ok", because any useful work is also viewed as R*I^2 term from a generator. Then another mean is required to distinguish losses from useful work.

Among these means, measuring the heat is not enough with an idling engine, because the mechanical energy goes back to heat that add to joule or core losses. Then measuring heat measures joule losses+useful work returning to heat, i.e it does not distinguish between two.

Steorn's scam is based on this principle.

The only way is to measure a significant useful work done outside the calorimeter while measuring also the heat inside. If the energy sum is superior to R*I^2 therefore there is OU.

You'd do better to stay out of this discussion. You're not only not a scientist and therefore have no basis to judge whether or not others are but you're a confused individual who very laboriously strives to disrupt useful discussions and that onlyclogs them.

[Omnibus]

@Omega_0

The real question you should ask is whether its ok to deduct the I^2R term from the Pin, without first measuring the heat from first principle. It is a big assumption and a bit of gamble. Of course, I'm doing the same, following Steorn's method. So I'm also faced with this question. Steorn did the calorimetry but its a bit shady like all their stuff.

That would only be a problem where there's a possibility for parasitic capacitance to be present, that is, when coils are involved. @broli was emphasizing on that repeatedly and I agree with him (recall, however, that I did studies which proved that even such a conjecture isn't viable). However, if one makes sure that all the current passes through the active resistance calorimetry becomes unnecessary. If all the current passes through the dissipative resistance the dissipated power will inevitably be I^2R -- the first Joule's law is still valid. That's the beauty of the present contraption -- parasitic capacitance in this case is out of the question and therefore all of the current goes through R. OU calculated the way it's done here is guaranteed, at that by purely electrical measurements (let alone that with those low currents calorimetry would be way less accurate).

[Omnibus]

@teslaalset,

@ Omnibus, Broli,
I followed this thread partly. A few thoughts that I have with this RC example:
- What exactly is the output terminal in this circuit?
- It might be good to do a simulation by performing an 'old school' calculation and compare them with the real measurement results, so we could have a discussion on why there are differences between them.
I could do the calculation, but need to know the output terminal in this circuit case.

It would be great if you could do a sim with MathCAD or Mathematica. Unfortunately, I have them in New York and, as you know now I'm in Massachusetts, and don't have them here. Plot the waves with the parameters from the screen shots and and do the algebra and the integration. It's very curious that the effect could have been predicted purely theoretically, based on the properties of capacitors and resistors (RC circuit so far is only discussed in terms of being of filter but the power balance in such circuit has never been studied). That's interesting beyond words.

[broli]

Here's a simple LC tank I talked about

$ 1 1.0800000000000002E-9 22.512744558455275 50 4.37 43
r 304 96 512 96 0 0.1
s 208 96 304 96 0 1 false
w 512 368 304 368 0
c 512 96 512 368 0 1.0E-9 -7.34294588127192
l 304 96 304 368 0 4.8E-5 -0.08282465215064858
v 208 368 208 96 0 0 40.0 5.0 0.0 0.0 0.5
r 304 368 208 368 0 45.0
w 576 96 512 96 0
w 512 368 576 368 0
r 576 96 576 240 0 100.0
c 576 240 576 368 0 6.0E-11 -7.806576575717537
o 6 64 0 33 7.136238463529799E-5 9.134385233318143E-5 0 -1
o 3 64 0 34 40.0 0.05 1 -1
o 3 64 0 33 20.0 0.2 2 -1
o 9 64 0 33 0.3125 0.05 3 -1
h 1 4 3

copy that code and go to http://www.falstad.com/circuit/... then go to "file->import" and paste it.

If you hover your mouse on the traces it should show over which component it's measuring. I fine tuned the components so it oscillates at around 700kHz. The main LC tank current starts at 104.5mA pk-pk, which depends on the initial charge of the inductor through the battery and resistor. The RC circuit passes 6.7mA pk-pk.

This circuit should be a good starting point for pro simulations.

@omnibus:

What happens to the negative slope if the resistance of R was reduced? I could have sworn I just read what you said about it but couldn't find it anymore.