Thanks Joe. I was thinking about wind vectors but didn’t work it through.
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@mediacombb.net>
Sent: Sunday, April 10, 2016 07:34
To: towertalk@contesting.com<mailto:towertalk@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@mediacombb.net>
>> Sent: Friday, April 8, 2016 19:18
>> To: towertalk@contesting.com<mailto:towertalk@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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> _______________________________________________
>
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>
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--
R. Kevin Stover
AC0H
ARRL
FISTS #11993
SKCC #215
NAQCC #3441
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