Testing the TK Tar Baby

[TinselKoala]

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]

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]

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]

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]

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?