Author Topic: Testing the TK Tar Baby (Read 2008080 times)

Groundloop · « **Reply #135 on:** March 30, 2012, 05:32:22 PM »

Quote from: MileHigh on March 30, 2012, 03:36:30 PM

Awesome TK, you got it!

The moral of the story is that when you short one charged capacitor to a discharged capacitor and lose one-half of your energy it's identical to an inelastic collision between a moving mass and a stationary mass.

In both cases you produce heat and that accounts for the 'missing' energy.

MileHigh

MileHigh,

How much energy do we loose when we charge a capacitor from a coil?
(The capacitor is a part of the coil, speaking about the LC nature of a coil here.)

First we apply a voltage (and current) over the coil. We remove the voltage
and the voltage over the coil goes to zero while the ampere in the coil goes to infinite.
This happen because the C part of the coil is fully discharged. Then the voltage
polarity flips and the current goes to zero while the voltage goes to infinite.
(Speaking about an no loss ideal coil here.)
In real life we have some resitive losses in a coil so we get a dampened oscillation.

So the question is, how much energy do we loose at each capacitor charge and discharge
for each cycle of dampened oscillation? And what causes the dampening? The coil resistance
or the capacitor charge or discharge losses? Do we loose half of the energy at each capacitor charge
at each cycle?

GL.

MileHigh · « **Reply #136 on:** March 30, 2012, 05:49:58 PM »

TK:

Quote

But how do the systems know to lose exactly half their energies to heat, and keep half in KE or capacitance?

You mentioned conservation of momentum. And when we look at a the cap version we know that there is a conservation of charge. Lo and behold, charge is equivalent to momentum.

Mass akin to Capacitance, M same as C
Momentum (Mass x Velocity) akin to Charge, MV same as Q
Velocity akin to Voltage, V same as V

C = Q/V, so Q = CV

Same as M = MV/V, so MV = MV

Conservation of charge or conservation of momentum dictates that 1/2 of the energy must be lost as heat for energy to be conserved.

MileHigh

TinselKoala · « **Reply #137 on:** March 30, 2012, 07:19:04 PM »

Right.. !! And thanks for forcing the clarification. It's been a long time since I've thought about the very basics, and it's always good to review and check one's work and understanding of it.

NOW....

Early tests with the 5 x IRFPG50 in place indicate no major differences from the 830s so far.... except as predicted the frequency of the oscillations has decreased significantly.... significantly in TWO ways -- and in the ability to partially turn on the Q1 transistor.

The first way of course is the magnitude of the decrease, which is as predicted knowing the difference in the mosfet's various capacitances, especially the gate. The second is what it reveals about the NERD team's device.

The delayed timebase of the HP180 is telling me that there are just under 17 periods PER 7 microseconds. Doing the math we get just a tad over 2.4 MHz. Which is exactly twice the 1.2 MHz mentioned on the NERD RATs video.
ETA: Philips says 2156 kHz.

Coincidence? Bizarre unaccounted for exact doubling of the frequency due to more random wire lengths? OR..... another MISTAKE in instrument interpretation by the NERDS?
I don't recall ever seeing an expansion of their oscillation trace so that one could actually determine anything about it.

One more difference I've noticed: It is easier to get the Q1 mosfet to turn on partially by mismanaged offset or frank bipolar pulsing.

So.... Again, I would really like to see _evidence_ in the form of analyzable data, that show the NERDs getting substantial load heating or drain current when the oscillations are occurring and a strict negative-going gate drive pulse is used.

Right now it appears that I can either get realistic load heating by allowing mosfets to turn fully or partially on during the antiphase from the oscillations, OR I can use a strict, non-offset unipolar negative going pulse to insure that only Q2 mosfets are involved and oscillating.

So.... for me to go further I need to know in which mode to operate. The heating mode draws over 1.5 amps on my inline ammeter, the purely oscs mode draws less than 200 mA depending on gate drive amplitude but usually less than 100 mA.

I'm going back over the actual data from the NERDs (scope traces, what dumps I can find) to see if there is real evidence that they are heating their load _without_ drain current flow indicated by a drain trace voltage drop. Any help here would be very much appreciated.

ETA: Heh... when I first stuck all the 5 mosfets in there and turned it on, at first it looked "normal", just like what I'd seen before, until I tried the bipolar pulsing and IT DIDN'T WORK to turn Q1 on !! Frantic scrambles, checking everything... sure enough, I had put Q1 in it socket adapter "backwards" -- so that it in fact WAS in strict parallel with the other four. Insert facepalm here, with lulz.

TinselKoala · « **Reply #138 on:** March 30, 2012, 08:18:57 PM »

Scoping across the battery.

Upper A trace is the common mosfet drains, at 100 V/div, baseline indicated.
Lower B trace is across the battery terminals, at 20 V/div, baseline indicated.

Main timebase is 0.2 milliseconds / division , delayed expanded timebase 1 microsecond / division.

Inline DMMs indicate 37.8 V on the battery, 100 mA draw. (Batteries were freshly charged overnight; present no-load voltage 37.9, load temperature (from previous tuning !!) at 110 F.)

MileHigh · « **Reply #139 on:** March 30, 2012, 08:19:53 PM »

Groundloop:

The amount of energy you lose per cycle of oscillation is dependent on the resistance in the wires. Beyond that, you have a lot of misconceptions about inductors.

I suggest that you read the thread linked below about how coils work when they discharge their stored energy. I urge you to try to understand it completely. People on the free energy forums experiment with coils for years without actually understanding how they work. That should change and the more people that understand the more peer pressure there will be on others to understand.

http://www.overunityresearch.com/index.php?topic=1312.0

MileHigh

TinselKoala · « **Reply #140 on:** March 30, 2012, 08:52:40 PM »

Quote from: MileHigh on March 30, 2012, 08:19:53 PM

Groundloop:

The amount of energy you lose per cycle of oscillation is dependent on the resistance in the wires. Beyond that, you have a lot of misconceptions about inductors.

I suggest that you read the thread linked below about how coils work when they discharge their stored energy. I urge you to try to understand it completely. People on the free energy forums experiment with coils for years without actually understanding how they work. That should change and the more people that understand the more peer pressure there will be on others to understand.

http://www.overunityresearch.com/index.php?topic=1312.0

MileHigh

Don't forget radiation. The LC circuit will also lose power to radiation, like a radio transmitter.

The most common mistake I see on "the free energy forums" is mistakenly thinking that each cycle of an inductive - capacitive ringdown is somehow new or different energy. If you look at the waveform made of the instantaneous multiplication of the current and voltage during an inductive ringdown you'll see a similar waveshape, but the area under the positive loops wrt the baseline represents the "positive" energy flowing during that time period, and the area above the negative loops wrt the baseline the "negative" energy. But it's the same energy ! And it's all contained in the _FIRST_ cycle's time integral. The first negative going excursion will be a little smaller in area as the energy sloshes back, because some of it is lost to resistance and radiation. The second positive cycle's time integral will be a bit smaller still, ditto. Lather rinse and repeat until the oscillation is damped out totally back to baseline and all that energy _in the first cycle_ is finally dissipated as losses.
Some people close to our hearts have made this mistake (adding up all the energies in successive cycles of a ringdown), and for all I know are still making it.

TinselKoala · « **Reply #141 on:** March 30, 2012, 09:29:49 PM »

At 0:09 of the NERD RATs video we encounter the following tossed salad using all English words:

"Heat dissipated at the resistor element relates to +- five watts."

Does this mean "Power dissipated at the resistor element equalled more or less 5 Watts"? Or "...... positive and negative five Watts"?

See the video here:
http://www.youtube.com/watch?v=fyOmoGluMCc

If "more or less" what is the same symbol meaning in the part saying " On this test the battery supply voltage was at +- 60 volts. "
Here if the supply was at "less than" 60 volts... well, that means one or more of the batteries must have been below 12 volts. Or if the supply was "positive and negative" 60 volts... that's acceptable, barely. But what then of the +- five watts? Negative five watts?

OK... so if the "heat related to five watts, more or less".... just how was this determined? Normally one would simply calculate I = V/R to determine the current, and P = I^2 x R to determine the power dissipated in the resistor, regardless of how hot it got (because THAT variable is influenced by a lot more factors than just the Joule heating caused by the current.)

Accepting the five watts figure, and working backwards, we find that with a load resistance of 10 ohms, typical of water heater elements, we get I^2 = P/R, or I^2 = 5/10 or 0.5. Taking the square root of both sides, we find I = just over 700 milliAmps is required to dissipate 5 Watts in a 10 Ohm load. Since V=IR, we then find that a DC potential of only 7 volts is actually required to do this, so there must be other things happening in the circuit to prevent the load from "seeing" the 60 volt supply for 100 percent of the time at full strength. Mosfet manufacturers and circuit designers use what is called "PWM" or pulse width modulation circuitry to limit the power throughput of the mosfets in just this manner. Using pulses that only allow the mosfet to turn partially on (as here) or fully on (in a commercial PWM) for short periods of time, the _average_ power, that does the work in a motor for example, is limited and controlled.

Anyway, an average DC current of 700 mA will match their power dissipation figure given. And a fully charged battery pack of 6, 12 volt 60 or 50 or 40 Amp-Hour batteries will be able to provide that for a _long_ time before going below 12 volts each. 5 Watts is 5 Joules PER second. There are (12 x 40 x 60 x 60) or 1 728 000 Joules per battery, times 5 batteries, for a total of 8 640 000 Joules in the 60 volt supply. Dividing by 5 Joules PER second, we find 1 728 000 seconds, or 28800 minutes, or 480 hours, or, at 5 hours per working day, 96 working days, or, at three typical school working days per week, 32 weeks, or, at 18 weeks per semester, almost full time for two semesters, before the batteries need go below 12 volts each.

Groundloop · « **Reply #142 on:** March 30, 2012, 09:38:13 PM »

Quote from: MileHigh on March 30, 2012, 08:19:53 PM

Groundloop:

The amount of energy you lose per cycle of oscillation is dependent on the resistance in the wires. Beyond that, you have a lot of misconceptions about inductors.

I suggest that you read the thread linked below about how coils work when they discharge their stored energy. I urge you to try to understand it completely. People on the free energy forums experiment with coils for years without actually understanding how they work. That should change and the more people that understand the more peer pressure there will be on others to understand.

http://www.overunityresearch.com/index.php?topic=1312.0

MileHigh

MH,

Actually, I do understand how coil works. But you avoided to answer my questions because that would have
shown the readers here that the 50% loss of charging a capacitor does not hold water when you charge
the capacitor from a coil. And I'm NOT talking about a capacitor soldered to a coil. I'm talking about the
capacitance you get by making a coil, many wires in close proximity to each other. And where are my misconceptions?
In a coil of wire you get a LC circuit. The only way that LC circuit can oscillate is by transfer of energy between the
inductive part of the coil and the capacitive part of the coil and back again.

Here is a snip that explains what I tried say:
"If a charged capacitor is connected across an inductor, charge will start to flow through the inductor, building up a magnetic field around it, and reducing the voltage on the capacitor. Eventually all the charge on the capacitor will be gone and the voltage across it will reach zero. However, the current will continue, because inductors resist changes in current, and energy to keep it flowing is extracted from the magnetic field, which will begin to decline. The current will begin to charge the capacitor with a voltage of opposite polarity to its original charge. When the magnetic field is completely dissipated the current will stop and the charge will again be stored in the capacitor, with the opposite polarity as before. Then the cycle will begin again, with the current flowing in the opposite direction through the inductor."

Maybe my English was not good enough but that was what I tried to say. :-)

So a simple LC circuit shows that the 50% loss in charging a capacitor is not true in all cases.

GL.

Groundloop · « **Reply #143 on:** March 30, 2012, 09:39:47 PM »

Quote from: TinselKoala on March 30, 2012, 08:52:40 PM

Don't forget radiation. The LC circuit will also lose power to radiation, like a radio transmitter.

The most common mistake I see on "the free energy forums" is mistakenly thinking that each cycle of an inductive - capacitive ringdown is somehow new or different energy. If you look at the waveform made of the instantaneous multiplication of the current and voltage during an inductive ringdown you'll see a similar waveshape, but the area under the positive loops wrt the baseline represents the "positive" energy flowing during that time period, and the area above the negative loops wrt the baseline the "negative" energy. But it's the same energy ! And it's all contained in the _FIRST_ cycle's time integral. The first negative going excursion will be a little smaller in area as the energy sloshes back, because some of it is lost to resistance and radiation. The second positive cycle's time integral will be a bit smaller still, ditto. Lather rinse and repeat until the oscillation is damped out totally back to baseline and all that energy _in the first cycle_ is finally dissipated as losses.
Some people close to our hearts have made this mistake (adding up all the energies in successive cycles of a ringdown), and for all I know are still making it.

TK,

I'm aware of the fact that all the energy comes from the first initial charge of the coil.
But that was not my question. See post above.

GL.

TinselKoala · « **Reply #144 on:** March 30, 2012, 09:47:01 PM »

Quote from: Groundloop on March 30, 2012, 09:39:47 PM

TK,

I'm aware of the fact that all the energy comes from the first initial charge of the coil.
But that was not my question. See post above.

GL.

You asked your question of MH, so I didn't answer it.
My post doesn't say that "all the energy comes from the first initial charge of the coil" although that is true and I'm glad you realize it.
It says that all this energy sloshes back and forth between the coil and the capacitor, that the amount of energy can be calculated by integrating the FIRST positive-going waveform of the instantaneous power trace, and that subsequent integrals of waveform areas will reflect losses (or gains if they increase) that accumulate until all the energy, if not replaced somehow, is lost. I also said that some people will try to add up all the positive going areas and claim that this represents some kind of "energy" figure that makes sense. Actually what must be added up is the _incremental loss_ in area for each period. The sum of these incremental losses is.... well, isn't it obvious?

TinselKoala · « **Reply #145 on:** March 30, 2012, 10:03:48 PM »

All I can find in the first video about heating the Ainslie load is the statement of "6 or 5 Watts" and the apparent citation of a load equilibrium temperature of 51 degrees or so. Without knowing the intimate details of the load's environment this is of course meaningless and useless.

When I tell you that Tar Baby's load is immersed in 250 ml of mineral oil, and I give a time-temperature curve while it's handling current, and a decreasing time-temperature curve when it's not, then you may calculate away and tell precisely how much power Tar Baby's load is dissipating from the equilibrium temperature reached. And you have some chance of being able to repeat the conditions and the measurements for yourselves.

If somebody shows you a coffee pot with a water heater element dangling into it, and a meter that reads 51.1 degrees.... all you know OR CAN KNOW is that the meter reads 51.1 degrees. Other than that, they might as well be making oxtail soup.

Presumably, allegedly, the heat load was calibrated by fiddling around with a DC power supply somehow and heating up the water under the same conditions with DC power, and then the figure of 6 or 5 watts was calculated. I am afraid I have no confidence in the ability of the NERDs to carry out such a complex and "fraught" calibration procedure. But nevertheless I am willing to accept their 6 watts figure... as it supports my case entirely.

Notice the common OU effect: when the effect is first announced it is so powerful that it will boil water, melt lead, vaporise solder, make oxtail soup for an army of RATs.... but when measurements become more precise and harder to ignore, the power levels inevitably decrease, just like a battery running out of ju

TinselKoala · « **Reply #146 on:** March 30, 2012, 10:17:39 PM »

OK, now let's look at this.

I've been running since early this morning on the IRFPG50 mosfets, making the oscillations, no LEDs anywhere in the circuit. So it would take a Chief Inspector Clouseau to determine just where the NERD RATs device is different than mine... but I am assured by the NERDs that Tar Baby is NOT an Ainslie replication.
(Probably for the same reason my Mylow replication wasn't a Mylow replication: it didn't supply free energy. But that's neither here nor there.)

My load is still sitting at 110 F, my inline ammeter indicating 100 mA, my batteries indicating 37.6 volts under the load of driving the circuit. Even with 5 A-H batteries and no recharging this could go on for a while.

If their load was dissipating 5 watts, and that means a current of 700 mA, then if I set TarBaby to produce a drain of 700 mA on the inline meter, and monitor load temperature, what will happen?

And what will it mean? Tar Baby's load is pretty well insulated, unlike hers, and we are also using mineral oil, which has a specific heat of 1.67 as contrasted with water's 4.18. Will Tar Baby be able to reach the lofty equilibrium temperature of 51.1 degrees C? Will Tar Baby melt into a puddle of asphalt, tar, plastic and blackened mosfet bits? Stay tuned to this channel, we'll be back after Your Local News and Weather Together, with traffic reports from outer space.

TinselKoala · « **Reply #147 on:** March 30, 2012, 10:38:04 PM »

My load's now at over 140 F and still climbing, mosfet temps have finally stabilized so I don't have to keep tweaking the drive to keep drain current from going over 800 mA... right now it's at 670 mA....

I mean, if they can get all excited about mere load temperatures, they should really be worried, right about now. My batteries are still at over 12 volts eeeee -ach!

(Q1 runs hot under this condition, so, because I don't have any spare PG50s, I decided to put a small fan directed at its heatsink to try to keep it cool. Probably not necessary but I'm a belt and suspenders kind of Fuhrer, er, fellow.)

TinselKoala · « **Reply #148 on:** March 30, 2012, 10:57:51 PM »

Here's another thing I don't quite understand (although I have my "theories".)

The NERD RATs device has massive heatsinks on all the Q2 mosfets and the Q1 mosfet isn't heat sunk at all, really. And the circuit and the analyses and the sims and my experience live on camera have all indicated that the Q2 mosfets are the only ones oscillating, usually, not really switching, and the Q1 mosfet isn't working at all, and as I have shown it may even be removed entirely without affecting the "known" parameters of the measurements. And when I set Tar Baby to a current of 700 mA, this is done by allowing Q1 to turn on, and it gets hot.

So why the evidence for hot Q2 mosfets in the NERD device? My Q2s hardly get warm at all , unless I force them to turn on fully and carry a lot of current. And at the same time make Q1 stay off or maybe just oscillating.... hmmmmmmmmm........ exceedingly strange.

(ETA: Load temp at 160 F and slowly rising, drain current 700 mA, batt voltage 36.7, Q1 hot and fan-cooled, Q2s barely warm.)

MileHigh · « **Reply #149 on:** March 30, 2012, 11:36:25 PM »

Groundloop:

Your first statement in your last posing is correct. Your quote is correct. But most of what you stated in your previous posting was incorrect. Feel free to start a thread about that if you want, TK is back in replication mode.

My final comment would be connecting two caps together compared to an LC circuit is like comparing apples and oranges. So naturally there will be differences. The intention for me was to explain how the energy "disappears" when you connect two capacitors together. It's an "inelastic collision" between two capacitors "moving" at different voltages.

Back to that old RAT magic....

MileHigh

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