[TowerTalk] W6NL 40m Moxon

[Jim Thomson]

##  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 at 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 at contesting.com
Subject: [TowerTalk] W6NL 40m Moxon

 

Date: Sun, 10 Apr 2016 07:55:37 -0400
From: "Joe Subich, W4TV" <lists at subich.com>
To: towertalk at 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   



_______________________________________________



_______________________________________________
TowerTalk mailing list
TowerTalk at contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk