Experiments

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What law would they cite though? If you live under a HV line, I can't see anyone taking an antenna on your property seriously. I know you are right, but I am wondering on my part of the exact verbiage.
Theft by taking. The exact verbage would vary state to state but I'm sure there is a generic version to go along with fundamental behavioral guidelines. I'm not a lawyer but do have family attorneys so will ask them if I remember the next time we are together.

I could ask our company attorney but it might cost me a couple hundred bucks.:D
 
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We also believe that fluorescent tube doesn't light from electrical current inside, but from the phosphors being directly excited.
I respectfully doubt that, since to the best of my knowledge phosphors respond to impact from an electron or from absorbing an incoming photon, not to quasi-static electric or magnetic fields (quasi-static being low frequency compared to RF).
It is easy enough to conduct the experiment: just use a broken fluorescent tube and compare the results to an intact tube.
 
I spoke with my EE friend about this quite a bit and, without going into detail, we came to the conclusion that some of the energy absorbed by new load is intercepted from existing loss pathways, and some is taken from the power line. The proportion depends on relative position in space and the relative "impedances" of the various pathways.

Picture a steel bar held a given distance from a bar magnet, and you insert another piece of steel between them. What happens to the lines of flux through the original steel bar? Do they increase, or is some of the existing flux redirected to the new steel? We concluded that the answer is some of each, and feel the electric field behaves similarly.

We also believe that fluorescent tube doesn't light from electrical current inside, but from the phosphors being directly excited.

Close, but not 100%. I went back to one of my physics books. My "caused by mag field" comment earlier was not correct.

There is a current that flows through the bulb. The only analogy I can give is like you taking a plastic straw that has tiny funnels on each end and placing it parallel to a running stream of water. The air gap resistance between lines and ground are some #, and now the bulb becomes less ohms over its length vs air of same length, etc. The electrons can more more freely through the bulb in that electric field, etc.

Inserting the bulb into the alternating high magnitude electric field causes electrons to flow through the bulb which hit the mercury gas atoms to excite them, and in that process (power line leakage current passes through the bulb) the ions (atoms) change electrical state and give off UV, that UV then strikes the phosphorus that is inside the bulb, the phosphorus is a catalyst which absorbs the UV energy and then gives off some photons in visible spectrum. That's how the light happens.

As for the power discussion. Since the whole process of mercury excitation and emitting photons in visible band does also include exothermic heat (aka energy loss), power is being used! The bulbs will likely shine brighter as the rel humidity increases.

The bulb in the E field is a small capacitor, thus it does change the impedance the source sees. The bulb has basically increased the leakage current by some small amount, etc. It's very analogous to your metal in mag field example, but in this case adding a metal rod (instead of bulb) in the E field will have same affect as bulb.


Not all energy dies a heat death. If I use it to elevate a mass it does not. If I use it to store energy in a molecular bond it does not (like manufacturing a product or similar). Some energy gets converted into other than thermal energy.
Almost all energy conversion involves heat energy which is typically lost from the system. The entropy of the universe is always (always) increasing.
 
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... use a broken fluorescent tube and compare the results to an intact tube.

I would have assumed the phosphors needed the gas inside the intact tube.

I might also assume the test would be the orientation of the tube in the field. If just phosphors, orientation shouldn't matter, but if it's current flow then I would assume it would.
 
I would have assumed the phosphors needed the gas inside the intact tube.

I might also assume the test would be the orientation of the tube in the field. If just phosphors, orientation shouldn't matter, but if it's current flow then I would assume it would.
In open air it wont work, unless "lightning" strikes it. A broken bulb will not act like an old evacuated CRT. The advancement of CRT to phosphorus bulb came with the aid of mercury gas (and the like).
 
In open air it wont work, unless "lightning" strikes it. A broken bulb will not act like an old evacuated CRT. The advancement of CRT to phosphorus bulb came with the aid of mercury gas (and the like).
Correct. The initial photons are generated by collisions within the mercury vapor moving under the influence if the electric field. All the phosphor does is let you trade in photons of a wavelength you do not want (like UV) for visible light of whatever color you want.
 
Theft by taking. The exact verbage would vary state to state but I'm sure there is a generic version to go along with fundamental behavioral guidelines. I'm not a lawyer but do have family attorneys so will ask them if I remember the next time we are together.

I could ask our company attorney but it might cost me a couple hundred bucks.:D


It would be difficult to prove in court though. My money is that if the POCO tried to make a case, an attorney would argue it with ease.
 
Correct. The initial photons are generated by collisions within the mercury vapor moving under the influence if the electric field. All the phosphor does is let you trade in photons of a wavelength you do not want (like UV) for visible light of whatever color you want.
I am not sure what you mean by initial photons. When the internal ends heat to allow electrons to flow they smash into mercury gas atoms, this process causes excitation, and when the gas atoms then change energy level (loses energy) the photons of UV are given up, which then speed off and hit the phosphorous inside.

As for stealing the power. I suspect that adding 1,000 flourescent bulbs under one section of hi voltage lines will have a very small, possibility not measurable, impact on impedance. A bulb is perhaps adding just single digit pico-Farad worth of capacitance. Put 1,000 bulbs between every tower along the whole line route, then perhaps the source wants to throw up ;)
 
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By initial photons, I mean the ones that end up exciting the phosphors. The ones that in a clear glass (or quartz to pass the UV) fluorescent tube would be the only light given off.
 
By initial photons, I mean the ones that end up exciting the phosphors. The ones that in a clear glass (or quartz to pass the UV) fluorescent tube would be the only light given off.
Ah, i see.
 
I hope it settles around 70F as that is a good all-around temperature.

Sorry to be the bearer of bad tidings, but it's going to be colder than that. A whole LOT colder. The current average temperature of the known universe is about 3K° (-454°F). By the time the last proton decays in about 10110 years, given the continued expansion, it might be even lower. Of course, by that time the concept of temperature will be meaningless since everything (a soup of photons and leptons) will be at the same temperature.
 
Sorry to be the bearer of bad tidings, but it's going to be colder than that. A whole LOT colder. The current average temperature of the known universe is about 3K° (-454°F). By the time the last proton decays in about 10110 years, given the continued expansion, it might be even lower. Of course, by that time the concept of temperature will be meaningless since everything (a soup of photons and leptons) will be at the same temperature.
The grass is always greener on the other side of the universe. I heard that from a bovine friend of mine.
 
190110-0932 EST

Has anyone here actually taken a fluorescent bulb out to a 345 kV 3 phase distribution line and made any observations?

A very interesting discussion is at https://en.wikipedia.org/wiki/Electric_power_transmission .

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I have used a VDG machine many years ago to do this. At best you get a faint glow unless you get close enough to the collector for an arc, then the tube flashes brighter. One end of the tube is in your hand with electrodes touching. If I were going to stand under a power line with a fluorescent tube, I would wait for a new moon. Compact fluorescents glow with a VDG machine also, both tubes work best with the room light off at night. One might take an umbrella covered with aluminum foil & attach that to the top end of the tube for better results.


If you have any of Ben Franklins DNA you might want to stroll around during a thunderstorm with your new Fluorescent tube handled umbrella & see how that lights up your path.
 
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