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Re: [TowerTalk] Why isn't end to end the best method for least interfere

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Why isn't end to end the best method for least interference between two 20 m dipoles?
From: jimlux <jimlux@earthlink.net>
Date: Mon, 26 Mar 2018 20:31:29 -0700
List-post: <mailto:towertalk@contesting.com>
On 3/26/18 6:21 PM, Kipton Moravec wrote:
Below are the results we did with 2 tuned 20m inverted V dipoles at 35 feet, in a test for interference during Field Day. The results for Orthogonal was the same, independent of which side was transmitting.

-------------------------------

With the antennas at ≈ 350 ft apart. An experiment was undertaken to
determine the greatest isolation.
• The source was an HF radio operating at about 5 watts.
• The receiver was a spectrum analyzer.
Antenna 1   Element North/South
Antenna 2   Element North/South
Orientation   Broadside
Received Signal   -5 dBm

Antenna 1    Element East/West
Antenna 2    Element East/West
Orientation    End to End
Received Signal    -10 dBm

Antenna 1     Element East/West
Antenna 2    Element North/South
Orientation    Orthogonal / Perpendicular
Received Signal    -22 dBm

• This as not the expected results. But did show that the optimum
orientation was orthogonal and it did provide about 17 dB of isolation.
• Further testing will be done to determine if these results are
repeatable.

------------------------------

We expected the end to end orientation to have the most attenuation.

Can someone explain why we got what we got?  Is there reference material out there that explains this?



testing at 14 MHz (roughly)?

Inverted V isn't a horizontal dipole.
The sloped element is partly vertical, so "end to end" is sort of like a set of verticals (considering the V pol radiation). Think about it as a pair of verticals some short distance apart (maybe 0.1 lambda) fed 180 degrees out of phase. That's a lot like a W8JK.

I think if you model the V, and look at the Vpol radiation off the end of the dipole, it's not zero.

Free space loss (between isotropes) for 100 meters separation and 14 MHz would be:
32.4 + 20*log10(0.1) + 20*log10(14) = 32.4 -20 + 23 = 35 dB loss.
With 5 W radiated, that's 37dBm, so you'd expect about -2dBm.

Your parallel V's is in that ballpark (if you get within 3 dB, you're doing great)


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