On 4/13/2016 7:54 AM, Jim Thomson wrote:
## if say the yagi pointing due east..and the wind is coming from the
SW....heading to the NE, all the forces off all the els will be due
> east. All the forces off the boom will be to the due north.
With quartering winds (winds not perpendicular to either the boom or
mans), you have forces on the mast/tower to the north due to the boom
and forces on the mast/tower to the east due to the elements. The net
force on the tower/mast will be the *combination* of those forces to
the northeast (vector addition). That's no different that the vector
addition of forces on the tower due to tension in the guy wires.
I'm no mechanical or structural engineer but you simply can not treat
each load as if the others do not exist. At the very least there is a
superposition of forces - assuming that they are not dynamic and add in
some exponential manner.
73,
... Joe, W4TV
On 4/13/2016 7:54 AM, Jim Thomson wrote:
## if say the yagi pointing due east..and the wind is coming from the
SW....heading to the NE, all the forces off all the els will be due east. All
the forces off the boom will be to the due north.
If you have an equal amount of boom on either side of the mast, and each ele
has an equal amount of metal on either side of the boom, and the boom is
mounted directly on top of the mast,
there will be zero wind milling, it will be perfectly tq balanced. The number and
placement of the else on the boom doesn’t even enter into the tq equation.
You can have a 40 ft boom,
with hust the REF mounted..and it wont windmill.
## as far as calculating ant windload, its either the total windload of the
boom..or the el, which ever is greater.
http://lists.contesting.com/_towertalk/2003-02/msg00151.html
Gone are the days of squaring the ele area, then also squaring the boom area,
summing both, then taking the sq rt of the entire mess.
Dick Weber K5IU, PE, wrote about this topic in great detail many years ago.
Wind angle has nothing to do with it.
Jim VE7RF
From: Ken K6MR
Sent: Tuesday, April 12, 2016 9:53 PM
To: Jim Thomson ; towertalk@contesting.com
Subject: RE: [TowerTalk] W6NL 40m Moxon
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
Sent: Tuesday, April 12, 2016 21:11
To: 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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