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I have always struggled to get a good physical understanding of Electromagnetic waves. I've done the whole Maxwell's equation approach, wrote boundary element solver tool but cannot rationalize that with a phenomenological understanding of E/M. The waves propagate radially from a broadcast antenna, so you have a time/distance varying electric field in a radial direction. A full-wave horizontal rod with feed taps at 1/4 and 3/4 of its length seems like it should be the best receiver. Help me get a better understanding, if I am wrong.

diagram

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  • \$\begingroup\$ What is the effective aperture of your antenna in the orientation it is placed. \$\endgroup\$
    – Andy aka
    Commented Feb 21, 2020 at 8:15

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It is true that the variations in field strength propagate radially from the transmitter.

However, as your diagram shows, the fields themselves are vector fields, and the orientation of the vectors (both electric and magnetic) is at right angles to the direction of propagation. They have no effect on a conductor oriented radially.


In a comment, you mention:

Now, I'm thinking about the receiving antenna as a wave guide.

Yes, that's a good analogy. In fact, that's the principle behind the Beverage antenna alluded to by hotpaw2. It works only because it is physically close to the ground, which makes it part of a transmission line — a kind of wave guide. The propagating E field induces a voltage between the wire and the ground. This wave of voltage moves along the wire at roughly the same speed as the radio wave, reinforcing signals coming from the correct direction. Signals arriving from other angles interfere with themselves destructively.

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  • \$\begingroup\$ Thanks, @Dave. Good thoughts to enlighten my mind. Now, I'm thinking about the receiving antenna as a wave guide. \$\endgroup\$ Commented Feb 21, 2020 at 14:52
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The waves are not only varying as they propagate radially, but, for a vertical transmit antenna, the EM field will vary vertically as well.

That's because a vertical antenna has finite height, thus the top and bottom are not radiating at the same phase (due to the speed of light of propagation from the bottom to the top of the antenna). Thus the radiating EM field will be vertically polarized, and the highest gain will be with an antenna of that same polarization (thus accelerating receive electrons in the same orientation as the accelerating electrons generating the EM waves).

There will be some reception with a radially oriented antenna, that's the basic theory behind a long wire Beverage antenna. But the gain will be lower because the pair of antennas will be trying to accelerate transmit and receive electrons orthogonally to one another; and the field propagating from the top and bottom of the transmit antenna will be seen out of phase and thus partially cancel at the receive location.

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  • \$\begingroup\$ Thanks, @hotpaw2. This is helpful! \$\endgroup\$ Commented Feb 21, 2020 at 14:48
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First, because the broadcast tower sends out vertically polarized waves and your receiver antenna is horizontally polarized.

Second, because the directivity of a lambda dipole is an —8— with the dipole as its mirror plane.

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  • \$\begingroup\$ Thanks, @Janka. I appreciate the thoughts. I understood directivity to express whether the antenna receives better in one (compass) direction than any other. In my flawed thinking I thought this antenna would be very directional - that is, that its orientation radial to the orientation of the broadcast tower would be a large maxima in reception. \$\endgroup\$ Commented Feb 21, 2020 at 14:57
  • \$\begingroup\$ Dave Tweed explained it very simple: In the radial direction, the antenna cannot pickup energy from the wave. \$\endgroup\$
    – Janka
    Commented Feb 21, 2020 at 15:49

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