Thanks for the link. This is definitely not an intuitive result. The good thing
is that the loads are smaller using this procedure.
Yagi Stress calculates these loads, but I’ve never done pencil and paper to
understand how it does it. Kurt notes that YS is using this method so the
numbers I have for normal yagis are good. Unfortunately Kurt didn’t allow for
unusual designs like the W6NL Moxon.
Ken K6MR
From: gboutin@infinichron.com<mailto:gboutin@infinichron.com>
Sent: Wednesday, April 13, 2016 05:17
To: towertalk@contesting.com<mailto:towertalk@contesting.com>
Cc: k6mr@outlook.com<mailto:k6mr@outlook.com>
Subject: Re: [TowerTalk] W6NL 40m Moxon
Ken,
This isn't exactly new. I think you'll find what you're after by reviewing
K7NV's summary post from back in 1998.
http://lists.contesting.com/archives//html/Towertalk/1998-08/msg00499.html
<http://lists.contesting.com/archives/html/Towertalk/1998-08/msg00499.html>
>> Gerald, VE1DT
OK Jim, I need that explained. If the wind is at 45 degrees to the boom, how
is
there not a contribution of force by both the boom and the elements?
Ken K6MR
From: Jim Thomson<mailto:jim.thom@telus.net>
Sent: Tuesday, April 12, 2016 21:11
To: towertalk@contesting.com<mailto:towertalk@contesting.com>
Subject: [TowerTalk] W6NL 40m Moxon
Date: Sun, 10 Apr 2016 07:55:37 -0400
From: "Joe Subich, W4TV" <lists@subich.com>
To: towertalk@contesting.com
Subject: Re: [TowerTalk] W6NL 40m Moxon
> 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?
You don't need to be concerned with equal areas. The force (load) on
the element will be broadside to the element and will vary with the
angle of the wind to the element. The force (load) on the boom will
be broadside to the boom and vary with the angle of the wind to the
boom. Those forces add as a vector sum which is maximized when the
wind is at 45 degrees to the boom/elements.
The area of largest force is:
sin(45) * (2.88 + 2.76) + cos(45) * (3.67 + 1.24 + 1.20)
= 0.707 * (5.64) + 0.707 * (6.11)
= 8.31 sq. ft.
The boom value will need to be increased somewhat to account for a boom
to mast plate and the element values increased to account for element
mounting hardware, particularly if the original U channel is used. I'd
allow between 8.5 and 9 sq. ft in any tower/mast loading calculation.
73,
... Joe, W4TV
## whoa. I was sure that all went out the window years ago. These days,
its either the boom area OR the total ele area...which ever is the
greatest.
That's how f12, optibeam, m2 and everybody else now does it.
## called the cross flow principle..which is how they designed the Eifel
tower.
The force coming off say an ele, is always at right angles to the ele,
regardless
of the angle the wind. Its all well documented.
Jim VE7RF
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