AKA:low signal loss(frictionless) communicator
High voltage effect:Use the Equation:
WATT/AMP=VOLT If you measure the AMPS in a cross section of the thick part of wire and compare it with the AMPS in the cross section of the thin part of wire. Then you will find that in the thin part of wire there is a
HIGHER AMPS PER /WIDTH OF CONDUCTOR which will induce lower VOLTAGE to maintain the same WATTAGE. And in turn the
lOWER AMPERAGE PER/ WIDTH OF CONDUCTOR of the thick part of wire you will have a higher VOLTAGE induced to maintain the same WATTAGE along the width of the wire.
So along the length of the wire there is no voltage change but along the width if the wire there is. And the electromagnetic flux produced reflects that fact.
LATER ADDED: Instead of alternating current this is an alternating voltage circuit. Which may or may not be able to transmit a carrier wave just as well as an AC carrier wave.
Super high frequency effect: As electrons pass by each bulge in the wire, in a row, the electrons across the bulge produce a high voltage electromagnetic wave and then the electron moves forward to the next bulge in the wire, and produces a second high voltage wave in the series, and you are effectively making an ultra high frequency radio wave which is probably even higher than the frequency of light. But the high voltage is what makes the low attenuation possible(voltage being the force behind the signal).
LATER ADDED: To amend this paragraph, different frequencies go through different substances with different levels of attenuation. And I am not sure if high voltage even has an effect on the attenuation of the signal.Frequency cancellation effect: When the peaks and valleys of a separate frequency than the frequency of the wire bulges, enters the wire, the peak of that frequency will induce a current on the first bulge, but because it isn't the same frequency as the wire bulges, an opposite polarity will be induced on the next bulge before the current from the first bulge reaches the next bulge. And the valleys of the signal frequency will cancel itself out by 2 opposing currents equaling zero current. Instead of adding to the signal strength on the next bulge, which only the correct frequency can do.Also: You may need a diode in the wire for one direction of current flow, and as a high frequency filter(or RC constant) . But it may or may not be needed, because your ear can't hear that high of a frequency anyway.
---------------------------------
I was recently informed by Roy that the voltage along a wire is determined by the resistance or the current , so the voltage is constant along the wire. And you cannot increase the voltage along the wire. And I agree. But I say that a thickening of a wire will cause the electromagnetic flux's voltage to be boosted, so that along the wire the voltage is the same but cutting across the thickness of the wire the voltage would be boosted. Meaning higher voltage radio signals even though the circuit's voltage, itself, remains unchanged along the length of the wire.
This is the same principle that transformers use, but they use large coils wound around an iron core. But the new thick wire transformer uses the same amount of copper as a spool of wire does, except you don't have to wind it and it's easier to make a secondary circuit,only meaning that now you have a high voltage low amperage electromagnetic flux radio signal, instead of the high amperage version of using a conventional copper spool around a core.
The only problem that needs to be addressed is that the signal out of the thick wire would be extremely weak even if it was a zero loss signal. So some how you need to either send it at higher wattage or boost it to higher wattage when you receive it to make an audible sound in your communicator. But if you found a way to have a powerful signal without the use of a battery that would be the best case scenario.
Later added: To boost the signal strength in order for the signal to travel farther and also for the sound to be louder on the receiving end, just have a longer spool of "
bumpy wire"(wire that is variably thick and thin in diameter, along it's length, at set intervals, for a set frequency). This in effect would be the same as having a bigger antenna. Or how they make radios without batteries by just having a long spool of wire as the antenna and also power supply
I heard once that at high frequencies the body acts as a resistor to the flow of electricity and so cannot get shocked no matter how high the amperage and voltage. This may or may not be a fact, but if it is a proven fact, and is true for electricity, would it be also true for electromagnetic waves at high frequency? Because if the opposite is true of there being no resistance at high frequencies of radio waves, zero attenuation signal transmission.
(
LATER ADDED): Sort of in the same way as how 2 air particles split in 2 directions as they flow around a wing but they both enter at the same time and must also exit at the same time. Which means that if you have spread out particles on the top half of wing then you have lower pressure on that side.
(SIDE NOTE)The Equation of flight: Time/ Distance=Air pressure
Where Time: is the particles must end at the same exact place that they began after passing the wing
Where Distance: That the particles travel is determined by the shape of the wing
Where air pressure: Is the lifting force of the wing.
