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Comments posted on various websites relating to amateur radio posit that the groundwave is unimportant for skywave paths. Is this concept correct?

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    $\begingroup$ I'm not sure why you'd ask this question. A "skywave path" is one which by definition utilizes primarily the skywave mode of propagation. So all other modes of propagation (groundwave, line of sight, ...) would have to be unimportant. $\endgroup$ – Phil Frost - W8II Jul 15 '18 at 12:47
  • $\begingroup$ I suspect that some are taking this question as "does ground wave propagation affect sky wave propagation, should we try to make sure it is also good for the best results?". $\endgroup$ – Mike Waters Jul 23 '18 at 19:20
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Typically groundwave and skywave paths cover completely different distances and will not interfere, so coupling there is not likely.

Additionally, an antenna with good skywave propagation may have a higher gain at elevation angles appropriate for skywave than its gain at angles appropriate for groundwave propagation, or both angles could have similar gains, so the relationship between skywave and groundwave strength is unclear and may or may not be related. With the right antenna, the groundwave elevation angle might even be a null with no radiation at all.

Modeling of the specific antenna involved would be necessary to determine if there is any relationship between groundwave and skywave propagation for that antenna.

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    $\begingroup$ What's "the groundwave angle"? $\endgroup$ – Phil Frost - W8II Jul 20 '18 at 12:47
  • $\begingroup$ A good question...that deserves its own question. $\endgroup$ – user10489 Jul 22 '18 at 12:58
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While the groundwave may not be present at a given receive location, this does not mean its existence and value are unimportant. They ARE important especially for vertical monopoles, as, for horizontal plane distances near the radiator they are an indication of the overall performance of the transmit antenna system — including its production of the radiation useful for skywave service.

Below is a graphic illustrating the reality that (other things equal), as the groundwave is improved by reducing the ohmic loss in the path to r-f ground for that antenna system, so also is the skywave improved. The greater the groundwave radiated/launched by a vertical monopole, the greater the skywave.

Discussion of the graphic presentation:

Z = Elevation in meters above Earth, and E = Field Intensity in millivolts/meter existing at elevation Z at a horizontal distance of 1,000 meters from the base of the radiator (level Earth). The notation at the upper right corner of the graph shows that it lies on the 45° departure angle from the base of the monopole (1000 feet vertical height AGL @ 1000 feet horizontal distance from the monopole base).

Most of the radiation at/above/below that departure angle reaches the ionosphere, decaying at a 1/r rate — which rate applies to sky waves, not ground waves.

It is necessary to include the surface (ground) wave in a "Method of Moments" antenna system analysis such as possible using NEC4x in order for all of the fields launched by a vertical monopole in the elevation sector of interest to appear correctly, leading to an accurate understanding of these principles.

R. Fry, CPBE

enter image description here

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    $\begingroup$ With neither axis labeled I'm not sure what, if anything, this graphic is illustrating. $\endgroup$ – Phil Frost - W8II Jul 14 '18 at 23:37
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    $\begingroup$ Please edit your answer to include the information. $\endgroup$ – Kevin Reid AG6YO Jul 15 '18 at 4:07
  • $\begingroup$ Although the axes of the graphic were labeled, I edited my answer as requested in the comment next above. $\endgroup$ – Richard Fry Jul 15 '18 at 9:24
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    $\begingroup$ It's certainly a very nonconventional labeling. Best I can tell, your description is backwards. "E" is the elevation above ground, which is why the vertical axis has units "m." and there's the note "Elev Angle 45°" at the top. And Z is the field intensity in mV/m. However, how does this graph illustrate the point? Where does it indicate "skywave" versus "groundwave"? $\endgroup$ – Phil Frost - W8II Jul 15 '18 at 12:44
  • $\begingroup$ The correct explanation was provided. How others comprehend it is beyond my control, sorry. $\endgroup$ – Richard Fry Jul 15 '18 at 17:13
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An interesting conclusion is possible with reference to the NEC4.2 study posted in this thread.

  1. Maximum field intensity from both of the monopoles compared shows that their greatest relative fields (E/Emax) exist in the horizontal plane. Certainly these fields must be included in the ground- (or surface-) waves radiated by these systems.
  2. The ratio of the field intensities radiated in the horizontal plane at a horizontal distance of 1 km from these two systems is 50.2/14.6 = 3.44 (approx).
  3. Radiated field intensity varies as the square root of radiated power.
  4. Therefore the antenna system using 16 buried radials each 50' in length is radiating 3.44², or about 11.8X the power radiated by the system using only a single 8' buried ground rod (with other things equal).
  5. Following the process of items 1-4 above for the fields in the NEC4.2 graphic at a height of 300 meters AGL and a horizontal distance of 1000 meters (a "takeoff angle" of ~16.7°), the field ratio is about 45/12 = 3.75X.
  6. The E-field ratio for these conditions is somewhat greater at 16.7° than at 0° because the surface wave field is attenuated at a rate greater than 1/r, while the fields radiated toward elevation angles producing skywaves always decay at a 1/r rate.

A conclusion from this: The greater the radiated groundwave fields for these conditions, the greater the radiated skywave fields.

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    $\begingroup$ Similar results could be obtained comparing any antenna with and without a 10 dB attenuator. Not sure any relationship between groundwave and skywave propagation can be inferred from that. I only see that improving antenna efficiency increases the field strength. $\endgroup$ – Phil Frost - W8II Jul 24 '18 at 14:33
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    $\begingroup$ Also, perhaps you could edit your existing answer rather than adding a new one? $\endgroup$ – Phil Frost - W8II Jul 24 '18 at 15:37

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