Testing the TK Tar Baby

[TinselKoala]

Waveform comparisons.

Figure 1: The Ainslie paper 1, Figure 4 waveform shot. Note the "shunt" and the "battery" waveforms and their vertical gain settings.

Figure 2: The Tar Baby "shunt" (CVR) and battery waveforms, taken with DC bias drive as shown in the latest video posted above.

[TinselKoala]

This is the only example required to show the oscillation
waveform detail as this frequency and phase relationship is
seen to persist in all variations to the offset, the duty cycle and
the and the (sic) applied voltage at the source. The evidence is that
the oscillation will persist with the provision of a constantly
applied negative charge at Q1. There is a precise 180-degree
anti-phase relationship between RSHUNT and Battery voltages
that is self-reinforcing, extending as it does, for the full
duration of the cycle while the signal at the gate of Q1 is
negative. The significance of this is more fully described
under Discussion and the evidence is that current is not, in
fact, being discharged by the battery supply during this
oscillation phase.

-- from the Ainslie Paper 1, page 4.


May I please have the OverUnity Prize now?



Oh....wait..... I have not yet proven that my batteries do not discharge, have I.

[MileHigh]

Hi TK:

Very nice waveform clip.  You can see how it resembles the NERD waveforms and it does actually appear to look like that there is more power being returned to the battery.

I don't think you can make estimates of energy out and energy in just by multiplying out approximate total areas like you stated.  Just like you use RMS when you look at an AC voltage going into a load resistor you have to factor in the fact that you are multiplying two variables together.  So one variable is doing a "weighting" on the other variable.

For what it's worth the energy per "energy cycle" is:  (out or in to battery)

E(out of battery) = (for t=0 to t=<end of power out>) Integral [ i(t) * v(t) ] dt

The NERD team DSO is doing that calculation.

A few thoughts:

When you look at just a few cycles the amount of current flow out of the battery is going to be approximately equal to the amount of current flow in.  Just doing the crude "area integration by eyeballing it" it sure looks like that.  So if you did your old trick where you photograph the waveform and then print it out on your printer, and then carefully cut out the "current out" and "current in" sections" and then weigh them - they will weigh approximately the same.

The reason for this is deceptively simple.  The delta between current out and current in when you are looking at a microscopic part of the AC waveform is very small.  It's so small that as a result it is hidden in the "background noise."  Of course your analog meter or even the LEDs of DOOM should confirm that the batteries are actually discharging.

More to come, standard disclaimers apply.

MileHigh

[TinselKoala]

It sure is quiet in here. Sort of like the stunned silence in a theater after some horribly moving and emotional movie has just ended.

At the time the photo of Tar Baby's traces was taken the load was at slightly over 51 degrees C. That's the temperature of the 250 mL of mineral oil that the resistors are submerged in. The specific heat of mineral oil is 1.67 Joules per gram per degree C, and the density is about 0.83 gm/mL. I don't know exactly what time I started, but let's say it took 100 minutes to get the oil from ambient at  22 C to the 51 degree point. So that's a 29 degree rise in 100 minutes.

Doing the math..... there are 250 ml x 0.83 gm/ml == 208 gm oil in there. So 208 x 1.67 x 29 == a little over 10,000 Joules to heat the oil, plus any losses through the insulation during that period. So it took at least 10,000 Joules and probably a little more. What's the average power? Since a Watt is One Joule PER second, we take the number of Joules and DIVIDE by the number of seconds to determine the Joules per second or Watts. 100 minutes is 6000 seconds. And 10,000 Joules / 6000 seconds is..... about 1.67 Watts average power, if I did the math right. No big deal, the oscillations easily pass that much power, obviously.


Or.... to "Do The Math"  the Ainslie way..... there are 5 A-H at 48 volts in my battery. That's 5 x 60 x 60 x 48 == 864,000 Joules. The test used... by her calculation method... since .... 10000 Joules per second x 100 minutes of the test period = 1,000,000 Joules...... in that test alone the battery capacity was far exceeded... and yet all the batteries still measure over 12 volts each.

So.  Do the math.  4.18 x 900 grams x (82 - 16) 66 degrees C = 248 292 joules per second x 90 minutes of the test period = 22 342 280 joules.  Then ADD the last 10 minutes where the water was taken to boil and now you have 4.18 x 900 grams x (104 - 16) 88 degrees C = 331 156 joules per second x 10 minutes = 3 310 560 Joules.  Then add those two values 22 342 280 + 3 310 560 = 25.6 Million Joules.

Of course this is wrong, and she has admitted that this is wrong, but she still hasn't corrected it or retracted the claim that the test used more than the battery's capacity.

Therefore.... as far as I can tell... ALL my evidence that can be compared to Ainslie's evidence directly indicates that Tar Baby is just as much overunity as Ainslie's device, just as I have claimed all along.

Of course, since Ainslie has not shown a Dim Bulb test... or any of a lot of the other tests I've performed... those are not fair comparisons and shouldn't be considered. Right?

So just going by what she HAS shown in real data...... I'm showing the same things, and so the same conclusions should be drawn about both devices, based on what's been shown and fairly compared.

[TinselKoala]

Hi TK:

(snip)

For what it's worth the energy per "energy cycle" is:  (out or in to battery)

E(out of battery) = (for t=0 to t=<end of power out>) Integral [ i(t) * v(t) ] dt

The NERD team DSO is doing that calculation.

(snip)

Oh yeah... I seem to recall some dude on YT trying to explain all that stuff, some years ago. Clearly over my head, so I just do Dim Bulb tests now.

http://www.youtube.com/watch?v=EeIVSiEZDnc

But is that _really_ what the NERDs are doing with that DSO? I think they are dumping the data...some of it.... to the spreadsheet and using a home-grown numerical integration approximation on it. Certainly, what they are _showing_ on their scope traces is garbage multiplied by noise and integrated over nonsense. I've certainly never seen a properly-obtained power trace integral that looked like theirs. Plus they forgot to put the value of their shunt resistor into the calculations on the DSO.