Comment on Intuitive explanation for conductor diameter v. electrical wavelength?

fullsquare@awful.systems ⁨1⁩ ⁨week⁩ ago

openhamprep explanation is wrong

When you consider a segment of wire, it has some inductance and capacitance in that small segment, and the wider it is, the lower inductance and higher capacitance are. In other words, both X_L and X_C go down as diameter increases. Electric field lines within antenna are mostly perpendicular to wire everywhere except on ends, where they spread in all directions, which means that there’s some residual capacitance there that needs to be charged each cycle. This is called end effect and the wider wire is, the more of it you get. In fact, the same aspect ratio (length/diameter) gets you the same end effect ratio each time, and it’s usually in range of 0.98-0.95. If there’s insulation, then insulation has lower velocity of propagation as it has higher permittivity than air, and it is in place with highest electric field intensity, and this can also shorten necessary wire length a couple %. The more insulation there is, the more shortened antenna becomes

But wait, there’s more. We can think of antenna as a lossy resonator, where R is Rrad which for halfwave dipole would be 72 ohms, + loss resistance which is smaller, and neither are changing fast with frequency. What is changing faster is X_L and X_C, and the wider wire is, the smaller both of them are, and this means that you can go further off frequency and still have acceptable SWR, i.e. antenna made with wider wire has more bandwidth. For HF this means using masts or wire cages instead of single wire, but for VHF this gets more practical

Rrad depends on physical antenna length, and if antenna is shortened, then how it was done, but for the range we’re here it won’t change much

you can consider antenna wire as a transmission line, but it will need to include ground as the other side of transmission line, and then impedance is dependent on wire diameter to height over ground ratio. it’s not very useful

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