That *is* a rabbit hole..
Get yourself a copy of PHARLAP from Australia and you can go wild ray tracing
the ionosphere.
https://www.dst.defence.gov.au/our-technologies/pharlap-provision-high-frequency-raytracing-laboratory-propagation-studies
Yes, the mode splitting is super important when it comes to things like the
"bright line" at the distant edge of the skip zone. And of course, the
polarization sense flips with every sea/ground surface reflection so multi hops
gets more confused.
There are a fair number of people who have done polarization diversity
receiving. A big X is popular (because it's easier to build than a big cross,
and the two sloped elements have similar ground effects)
On Thu, 16 Apr 2026 11:06:23 -0700, David Gilbert via TowerTalk
<towertalk@contesting.com> wrote:
Exactly so. That's a great comparison. That diffuse refractive
ionospheric volume is also variable in density and position ... maybe
not as extreme as the my local terrain or the big rocks below the
surface of my lot, but the analogy holds.
Your comment on polarization brings up an interesting question. I know
that both vertically and horizontally polarized waves end up being
elliptical as they get refracted by the ionosphere, but is one or the
other treated more efficiently as it passes through? I asked that
question to ChatGPT and got the answer below. I know nothing about
O-mode versus X-mode, so I guess I have some studying to do.
But maybe when we think of using both horizontal and vertical
polarization to better receive what really is an elliptical signal, the
difference in the path absorption is also having an effect for
transmit. I separately asked ChatGPT what the difference in ionospheric
coupling efficiency was IGNORING GROUND LOSSES, and it replied that the
difference was typically about 3 dB but could sometimes be a bit more
than that.
73,
Dave AB7E
From ChatGPT:
When an HF wave enters the ionosphere, it doesn’t just gradually become
elliptical. It *splits into two characteristic propagation modes*:
* *Ordinary wave (O-mode)*
* *Extraordinary wave (X-mode)*
These modes are defined by the plasma + Earth’s magnetic field
interaction (Appleton–Hartree physics). Each mode:
* Has a *different refractive index*
* Follows a slightly *different path*
* Experiences *different absorption*
Both modes are generally *elliptically polarized*, regardless of what
you launched from the ground.
------------------------------------------------------------------------
Why launch polarization still matters
Even though everything becomes elliptical up there, your *initial
polarization determines how efficiently you excite those two modes*.
Horizontal polarization (typical dipole)
* Couples *strongly into both O and X modes*
* Tends to produce *more balanced mode excitation*
* Results in *better overall returned power*, especially at mid-latitudes
Vertical polarization (typical vertical antenna)
* Often couples *less efficiently into one of the modes*
* More energy can be lost due to:
o *D-layer absorption*
o Ground losses on launch
* Can result in *lower skywave efficiency*, especially at lower
takeoff angles over average الأرض
------------------------------------------------------------------------
The big practical effect: absorption differences
The *D-layer* (especially during daytime) is where much of the loss happens.
* The *O-mode is usually more strongly absorbed*
* The *X-mode penetrates better*
If your antenna launches energy in a way that favors the more absorptive
mode, you lose signal.
👉 Horizontal antennas tend to distribute energy in a way that:
* Ensures *some energy survives in the lower-loss X-mode*
* Reduces total absorption loss
On 4/16/2026 9:08 AM, Jim Lux wrote:
>
> In some sense, it's like modeling ionospheric "reflection" as a single
> height, when in reality it's a diffuse refractive effect spanning many
> 10s-100s of km.
> That said, I think one can come up with a frequency specific "single useful
> number" given an assumed soil profile. You'd get an "effective depth" and a
> rolled up reflection coefficient (which would be angle dependent).
>
> I suspect, also, that for H-pol, it's WAY simpler than for V-pol, just
> because the "first surface" is so reflective at low angles.
>
>
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