Re: HF antenna by feeding existing old TV coax shield?
Jef - N5JEF
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It's not going to be a very good antenna, but as Greg indicated, all wires can radiate, and all antennas "work" as we see from all the anecdotal "evidence" posted by hams.
Fundamentally, there is nothing wrong with using the shield of an existing coaxial cable as the radiating element of an antenna. Essentially, it's just a conductor. It certainly has no "shielding nature" carried over from its previous role.Being attached to a wooden structure will have only a small effect at HF (except for the likelihood of instability due to charge leakage at the relatively high voltage end of the wire being close to (wet?) wood.).
***The big question here is what band(s) you might want to operate with***. You say nothing about what frequency or frequencies, and that has _everything_ to do with what performance you might expect.
As with any antenna, the concerns are radiation pattern, radiation efficiency, and impedance matching. In this case, it really all comes down to radiation pattern. Radiation efficiency is not going tobe an issue because the wire is long relative to an HF wavelength so the Q is low and and there will be no losses due to circulating currents as you would have with a very short whip or loop antenna. Impedance matching will depend entirely on what frequency and what connection point on the antenna, but can be handled with any good antenna coupler (but not with a 64:1 transformer, which only makes sense with a high impedance (end-fed half wave) load.)
So in regard to radiation pattern, there are a few key factors to consider here, and they all depend on frequency/wavelength.
Why do we care about the locations of the current maxima? Because that's where the antenna radiates from. Radiation is always only from locations of accelerating charge.
Where are the current nodes? Well, it's easy to estimate. You know the end of the antenna is low current (thus high voltage) because there is nowhere further for the current to go (except for some "displacement" current as charge build up there, so the impedance ends up being about 4500 ohms rather than infinite. So if that's the voltage node, the current node is going to be about a quarter wave closer to the source, repeating every half wave. There will be some velocity factor effect, but it can be ignored since this is all just estimating.
So, for example, if you were considering operating on 40 meters, there would be a current maximum about 33 feet from the end, and another current maximum about 66 feet back from that, and so on...
Then you ask yourself, "What's the polarization going to be around those higher current regions?" Is the wire mostly horizontal, mostly vertical, or a mix?
Then you ask yourself, "What's the ground reflection going to do around those higher current regions?" If the current maximum is about a quarter wave above the ground reflection plane, then the radiation is going to tend to be near vertical. If the current maximum is about a half wave above the ground reflection plane, the radiation is going to tend to be lower angle. If the current maximum is more than about one wavelength above the ground the radiation will tend to split into alternating lobes of minimal and maximal radiation. If the antenna is free space you get a nice clean theoretical radiation pattern—which is almost never the case at HF.
With these basic understandings, and knowing what wavelength you want to operate at, you can get a very good idea of what to expect from your wire antenna at HF.
Again, the _current nodes_ radiate. The voltage nodes are important because they complete the wave, so they affect tuning, and impedance matching, and can be sources of arcing and instability.
- Jef N5JEF
On Fri, Oct 9, 2020 at 10:40 PM Greg D <ko6th.greg@...> wrote: