[TowerTalk] W6NL 40m Moxon

[Ken K6MR]

I’ve wondered this myself but never bothered to model it. Here are some rough figures from Yagi Stress:

Reflector element without the tee: 2.88 sq ft.
Driven element without the tee: 2.76 sq. ft.

Reflector tee (each): 0.62 sq. ft.
Driven tee (each) 0.60 sq. ft.

Boom: 3.67 sq. ft.

Now the hard part: how do they add? I’m an EE so my logic may be totally flawed. YS doesn’t allow for the tees in a single model so I can’t have it do everything at once.
Since the tees are parallel with the boom, I would think you would just add the area of the tees to the area of the boom which should be the projected area with the wind blowing 90 degrees to the antenna.
That would give 6.11 sq. ft. for that case. Wind straight on would be just the elements: 5.64 sq. ft.

I suppose with some trig you could figure the total projected area at various wind angles. One going down and one going up. Would some angle that gives equal projected area from each figure be the max?

My first one is built from scratch and mounted to the side of the tower so I wasn’t concerned with wind area. The XM240 is at the top of a mast so when I convert that one I’ll have to be more careful.

Ok you mechanical types: does this look reasonable??

Ken K6MR



From: Kevin Stover<mailto:kevin.stover at mediacombb.net>
Sent: Friday, April 8, 2016 19:18
To: towertalk at contesting.com<mailto:towertalk at contesting.com>
Subject: [TowerTalk] W6NL 40m Moxon

I need to figure out the wind area of the W6NL 40 Moxon antenna.
The antenna starts like as a Cushcraft XM-240 which they say is 5.5 sqft.
The Moxon is going to be more but how much? I'm guesstimating 8 sqft.

--
R. Kevin Stover
AC0H
ARRL
FISTS #11993
SKCC #215
NAQCC #3441


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