[TowerTalk] Wind loading

[Grant Saviers]

A simplified calculation to add the wind forces is by computing the 
moment from each diameter and then adding them together.  Multiply the 
height in feet above ground to the center of each diameter times the 
total wind force on that diameter.  Add them together. That yields the 
overturning moment at the whip base.  Then you need the same moment from 
a free standing base support to keep the whip upright - so much weight x 
feet out, easiest if using 4 spokes with weights at their ends.  That 
will have a safety factor since the pivot point for overturning is one 
spoke length out on the other side.  Wind splitting two legs has the 
weight and pivot distance 0.707 times the spoke length, but now there 
are two weights resisting the wind moment, so all is ok.

Another tool is YagiMech from DX Eng which will calculate the stress in 
each element and check if it exceeds the strength of the 6061. It is for 
elevated yagis and thus uses a uniform wind velocity for all element 
diameters.  A vertical experiences a wind velocity that varies with 
height above ground as the K7NV formulas show.

Grant KZ1W

On 8/30/2016 11:33 AM, Dick Blumenstein wrote:
> Hi Gary-
>
> I found this link from K7NV 
> <http://k7nv.com/notebook/topics/windload.html>. Hopefully, this will 
> help. I would assume you would do a force calculation for each 
> diameter piece and add the forces up of all the different pieces.
>
> Dick, K0CAT
>
> =============================
>
>
> Gary Smith wrote on 8/30/2016 2:20 PM:
>> I'm having to play games with setting up eight, short, active
>> vertical elements on an incredibly rocky area. I am not able to drive
>> in ground rods, much less the base of the antennas. I'm coming up
>> with a plan to make wooden bases for them and hold them down with
>> rocks.
>>
>> Since I live on the ocean's edge and have to contend with hurricane
>> force winds every so often, I'm trying to figure out the wind loading
>> of the verticals I'm considering. I'll buy the aluminum sections from
>> DXE and prefer to use larger sections just because of branches
>> falling at the marsh edge, as they always do. But thicker pieces will
>> have more wind loading and that is a problem.
>>
>> If I use the thinnest configuration I'll have about 22 feet tall made
>> of .375, .5, .625 & .750 sections.
>>
>> I asked what the wind loading would be and nobody I talked to was
>> able to give me a clue. Does anyone know a formula to figure the wind
>> loading of tapered elements?
>>
>> Thanks & 73,
>>
>> Gary
>> KA1J
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