Is this a miracle or a scam?

[MarkE]

i guess its not their loss to not show proof,
if never actually having a working device.


because they not really your friends.

they do not care.

'they' seem more dishonest and 'you' know it.

were is the storefront so people can purchase one.

moving on.

you might try learning how to install solar panels in order to make a living.

Ironically for Peter, every efficient solar PV inverter in the world performs maximum power point tracking which transforms the effective load impedance in order to get the most power possible under any instant condition from the solar panel string / array.

[peterqv98]

To tinman:
This is very interesting! Would you be so polite to explain in detail what have you done several years ago? Because the boost converter is not able to generate the black box effect if the four bulbs are identical.

[tinman]

To tinman:
This is very interesting! Would you be so polite to explain in detail what have you done several years ago? Because the boost converter is not able to generate the black box effect if the four bulbs are identical.

A boost converter will do exactly that if the bulbs are placed as they have them in the black box video.
The two globes on the input side are in series with another resistor,that being the inductor.
Where as the two globes on the output side will be parallel across that inductor.

[MarkE]

A boost converter will do exactly that if the bulbs are placed as they have them in the black box video.
The two globes on the input side are in series with another resistor,that being the inductor.
Where as the two globes on the output side will be parallel across that inductor.

It is all just Ohm's Law.  Any power converter that has a well filtered input transforms the impedance of the output as seen by the input.  Taking a first approximation where the converter is considered 100% efficient, at any given output voltage to input voltage ratio, the reflected impedance Zout_reflected is:  Zout_reflected = (Vin/Vout)² * Zout.  So, if for purposes of argument one had two one Ohm resistors: Rin and Rout and a 12V supply then without a DC-DC converter:

V_RIN = V_ROUT = 6V
P_RIN = V_RIN²/R_IN = 6²/1 = 36W
P_ROUT = V_ROUT²/R_OUT = 6²/1 = 36W

But stick a 2:1 voltage converter in the way, and the reflected impedance of R_OUT at the input becomes 4 Ohms.  Loop current drops from 6A to 2.4A.  The voltage across R_IN drops to 2.4V, while the voltage across the power converter input becomes 9.6V.  The power dissipated by R_IN becomes 5.76W.  The voltage at the output of the voltage converter is half the input: 4.8V, and the power dissipated by R_OUT becomes 23.04W, identically 4X the power dissipated by R_IN.

[peterqv98]

Hi everyone ,
I am glad that the discussion tends to become already more constructive.
To tinman: Actually I did not understand again what exactly have you done several years ago. Would you be so polite to send some drawings?
To MarkE: But your perfect mathematical calculations have nothing to do with the black box performance. Yet any mathematical model has to be related, more or less, to physical reality. Please reconsider carefully again you calculations and you will easily see that some of your basic assumptions are entirely wrong.