[TowerTalk] W6NL 40m Moxon

Sun Apr 10 13:37:18 EDT 2016

On 4/10/2016 12:05 PM, Ken K6MR wrote:
> Thanks Joe. I was thinking about wind vectors but didn’t work it through.

I used 45 degrees here because the element and boom components were
nearly the same.  If one had long boom 6 or 10 meter antennas it would
probably be necessary to solve for the angle of maximum force somewhere
between broadside to the boom and 45 degrees.  With short, thin
elements a long boom is a much larger contributor than the elements.

73,

   ... Joe, W4TV

>
> With these numbers it should be simple to add a torque compensator to make up for the higher torque generated by the reflector’s larger wind load. With only a 22 ft. boom the rotator needs should be minimal. I also had to add a small counterweight to the reflector end of the boom to compensate for coax and balun weight.
>
> Ken K6MR
>
>
>
> From: Kevin Stover<mailto:kevin.stover at mediacombb.net>
> Sent: Sunday, April 10, 2016 07:34
> To: towertalk at contesting.com<mailto:towertalk at contesting.com>
> Subject: Re: [TowerTalk] W6NL 40m Moxon
>
> Thanks for the input Ken and Joe.
> I used 9ft in the ARRL mast calculator and end up with 131 mph survival
> wind speed on a 3" x .25 wall 87,000 ksi mast.
> The antennas are an Optibeam 11-5 at 70' and the Moxon 15ft above, and
> still under max load of the 70' Rohn 55 tower. Now I've got to figure
> out what to turn it with.
>
> On 4/10/2016 6:55 AM, Joe Subich, W4TV wrote:
>>
>>
>>> 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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>
> --
> R. Kevin Stover
> AC0H
> ARRL
> FISTS #11993
> SKCC #215
> NAQCC #3441
>
>
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