[TowerTalk] W6NL 40m Moxon

Sun Apr 10 20:11:12 EDT 2016

All the more reason to be sure that the antenna is torque balanced. Those shifting winds can beat a rotator to death (not a prop motor, but all regular ones). I can loosen the bolts on my reasonably big yagis on a windy day and they stay put. It also reduces the torque required to turn to merely that necessary to overcome inertia.

I use Yagi Stress to calculate the torque compensator without really knowing what it was doing. This was a nice review of where the magic numbers come from.

Ken K6MR

From: Hardy Landskov<mailto:n7rt at cox.net>
Sent: Sunday, April 10, 2016 15:09
To: 'Joe Subich, W4TV'<mailto:lists at subich.com>; towertalk at contesting.com<mailto:towertalk at contesting.com>
Subject: Re: [TowerTalk] W6NL 40m Moxon

Hi All,
My 2 cents. I have seen the wind suddenly die--then change direction--then change again. It's like a box of chocolates....
N7RT/4

-----Original Message-----
From: TowerTalk [mailto:towertalk-bounces at contesting.com] On Behalf Of Joe Subich, W4TV
Sent: Sunday, April 10, 2016 7:56 AM
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

On 4/10/2016 12:51 AM, Ken K6MR wrote:
> 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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