If calculating wind load for the purpose of deciding which rotator to use, I
would think the boom length would factor in. Surely a square plate 4'x4'
mounted to the mast would put less stress on a rotator than a 50 foot boom
antenna with the same 16 SF calculation?
At any rate if the difference in the various ways of calculating the wind load
mentioned makes the difference in what you believe will survive, I would think
you are cutting it too close.
Stan, K5GO
Sent from Stan's IPhone
> On Apr 13, 2016, at 7:17 AM, Joe Subich, W4TV <lists@subich.com> wrote:
>
>
>> 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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