[TowerTalk] W6NL 40m Moxon

[Ken K6MR]

Ok, this even makes mathematical sense (as it should):

Assume the wind angle is with respect to the antenna direction. Total area = Element area * COS (angle) + Boom area * SIN(angle)

First derivative is: EA*SIN(angle) - BA*COS(angle) = 0

A little manipulation gives: angle (max area) = Arctan(BA/EA)

Sanity check: if boom is really big w/r to elements, angle approaches  90 degrees. If elements are big, angle approaches zero. As you noted, if boom = elements, angle = 45 degrees.

Thanks for the thought provoking discussion. This should have been obvious but never took the time to work it out.  The software does all the work these days and pencil and paper are pretty much dead.

Ken K6MR



From: Joe Subich, W4TV
Sent: Sunday, April 10, 2016 10:37
To: towertalk at contesting.com
Subject: Re: [TowerTalk] W6NL 40m Moxon


On 4/10/2016 12:05 PM, Ken K6MR wrote:
> Thanks Joe. I was thinking about wind vectors but didn’t work it through.

I used 45 degrees here because the element and boom components were
nearly the same.  If one had long boom 6 or 10 meter antennas it would
probably be necessary to solve for the angle of maximum force somewhere
between broadside to the boom and 45 degrees.  With short, thin
elements a long boom is a much larger contributor than the elements.

73,

   ... Joe, W4TV