Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[Zeitmaschine]

I would not say amplify - it rather builds up over time. LC is not overunity by itself. And yes power comes from radio station.

Then it's time to do some calculations:

Let's say the Crystal Radio uses a ferrite rod 10 cm in length and 1 cm in diameter which receives the energy form the transmitter. The output of the headphone shall be 0.001 watt (this is the energy that makes the air vibrate, so it is real power). Now we take 10,000 of such radios (theoretically) then the output power calculates to 10 watt. Next we calculate the space that is needed for these ferrite rods. 10,000 = 100 × 100 so the area covered with ferrite rods is 100 × 1 centimeter = 1 meter and 100 × 10 centimeter = 10 meter. Hence the area which receives the 10 watt from the transmitter calculates to 10 meter × 1 meter = 10 square meter. To simplify it each single square meter receives 1 watt.

Now 1 square kilometer = 1,000,000 square meter, that means 1 single square kilometer receives 1 million watt (1 megawatt) from the transmitter (radio station). If the radio station can be received in a range of (let's say) as few as 100 square kilometers then the transmitted power calculates to 100 megawatt. I don't know what the power of an ordinary medium wave transmitter could be but for comparison the power delivered to the 180 HAARP antennas is stated to be just 3.6 megawatt.

So please tell me someone what could be wrong with my calculations?

[TinselKoala]

Then it's time to do some calculations:

Let's say the Crystal Radio uses a ferrite rod 10 cm in length and 1 cm in diameter which receives the energy form the transmitter. The output of the headphone shall be 0.001 watt (this is the energy that makes the air vibrate, so it is real power). Now we take 10,000 of such radios (theoretically) then the output power calculates to 10 watt. Next we calculate the space that is needed for these ferrite rods. 10,000 = 100 × 100 so the area covered with ferrite rods is 100 × 1 centimeter = 1 meter and 100 × 10 centimeter = 10 meter. Hence the area which receives the 10 watt from the transmitter calculates to 10 meter × 1 meter = 10 square meter. To simplify it each single square meter receives 1 watt.

Now 1 square kilometer = 1,000,000 square meter, that means 1 single square kilometer receives 1 million watt (1 megawatt) from the transmitter (radio station). If the radio station can be received in a range of (let's say) as few as 100 square kilometers then the transmitted power calculates to 100 megawatt. I don't know what the power of an ordinary medium wave transmitter could be but for comparison the power delivered to the 180 HAARP antennas is stated to be just 3.6 megawatt.

So please tell me someone what could be wrong with my calculations?

You are confusing peak power with average power, and confusing power with energy.

The output of the crystal set is the same as the input in terms of _energy_ (minus a little loss). But the peak power in the output of the crystal set is much larger than the peak power that it receives at the antenna, because it is being converted to electric-field driven oscillations of the piezo earphone element at a much slower oscillation rate, thus its peak power can be much greater. The output of the receiver might have peak powers in the 0.001 Watt range as you say, and this is what you hear. But these peak pulses are being received at over a thousand times faster rate than they are being presented to you at the earphone. At 1000 Hz audio modulated output from a 1,200,000 Hz radio signal, each ONE of the 1000 pulses per second that the headphone displays contains the equivalent power of 1200 incoming pulses from the antenna. So if your earphone is putting out the 0.001 Watt peaks that you are hearing, the actual power at the antenna is 1/1200 of that and when you run through the rest of your calculation, 100MW / 1200 = about 84 kiloWatts, a modest transmitter indeed.
These numbers are just for illustration purposes, I think. The audio power output from a typical crystal set is probably smaller than 0.001 Watt, and the antenna area to be considered might be a lot larger than the ferrite rod in your example.

ETA: I think the crystal radio's piezo earphone element is like a capacitor itself. It effectively integrates the incoming instantaneous power over its time constant and then discharges its received energy as a fast "click" and when these clicks are fast enough you hear them as sound frequencies.

[elementSix]

 8)This GREEN Three Phase Harmonic Filter Reactor Looks very familiar, like TK's in the 2012 video  The Large one is just like that Special Russian Transformer we talked about the other day....

[Zeitmaschine]

But these peak pulses are being received at over a thousand times faster rate than they are being presented to you at the earphone. At 1000 Hz audio modulated output from a 1,200,000 Hz radio signal, each ONE of the 1000 pulses per second that the headphone displays contains the equivalent power of 1200 incoming pulses from the antenna.

Is it actually not the other way around? The initial electric energy which runs the transmitter is provided at a frequency of 50Hz, and this 50Hz frequency then is converted through all steps (transmitting with audio frequency modulation at high frequency and demodulating) back to audio frequency for the headphone? So 50Hz in is converted to a few thousand Hz out?

Just asking. This is very strange stuff ...

[TinselKoala]

Is it actually not the other way around? The initial electric energy which runs the transmitter is provided at a frequency of 50Hz, and this 50Hz frequency then is converted through all steps (transmitting with audio frequency modulation at high frequency and demodulating) back to audio frequency for the headphone? So 50Hz in is converted to a few thousand Hz out?

Just asking. This is very strange stuff ...

Yep, you are right. But the _average_ supplied power from the mains, at 50 Hz, to run an 85 kW transmitter at 1200 kHz will be greater than 85 kW. If you take a look at my wireless power transmission system, I am taking 12 VDC as input, converting it to 800 kHz for the inductive transmission, then receiving that to light one incandescent bulb brilliantly with the 800 kHz direct signal, and also detecting and rectifying it with a diode just like a crystal set, accumulating that into a capacitor and then using the filtered 10 to 40 VDC output from the cap to drive motors and other DC loads. So DC in is converted to DC out, with an intermediate step of making it into 800 kHz AC for the transmission and boost stage. Boost... because if I hook a light bulb directly to the battery of my system, it glows with a certain brightness. If I use the same battery and the same light bulb, but use the battery to power the inductive transmitter and use the inductive receiver to power the light bulb, the bulb glows much more brightly. Of course the transmitter is drawing more power from the battery than the bulb alone would do, but much of this is in fact making it to the bulb.

If you just remember that energy, not power, is the conserved quantity, you won't go wrong. Energy in = energy out, minus losses. Peak power in can be much less than peak power out, if the power is converted to pulses of short duration. But average power, measured correctly, should show the same kind of equality as total energy.