[TowerTalk] wind load vs Rohn specs

[Tower2sell@aol.com]

Bill,

You bring up some interesting points that require engineering judgement to answer. First let me explain that the EPA method that is in ANSI/TIA/EIA-222-F is a simplified method and a more exact methods can be used. A more exact method is a wind tunnel test or a method that takes shielding, orientaion and shape of members into account. The code method ignores these in order to present a simplified approach. 

Now for the EJ on elements that are shielded; depending on the diameter one could consider an element shielded if it is within a certain distance to the shielding member. Now this distance will vary from engineer to engineer, and this aspect is not covered in the code and is left to a more exact method.

One question that has always bothered me was why are flat elements still are considered flat if they are say skewed 30 degrees to the wind. The current code does not consider lift and drag similar to airplane wings.

One of the big changes in the code from rev C to rev D E & F, was that rev C used the FLAT PLATE AREA (FPA)  or the assumption that the antenna was flat and had the conversion terms in the formulas. Then the code changed for D and assumed the antenna area was round (EPA) and a different conversion was used. The problem was that few of the antenna manufacturers changed to the new code. Some publish a thrust value at a rated wind speed. This can be converted to the new 

EPA = Thrust/.00256xV^2),

where V is the rated or published wind speed (not your design wind speed).

You can then get PA = EPA/1.2 and use the ROHN values for "round" projected area. 

I hope this helps,

Tower2sell@aol.com



In a message dated Tue, 12 Sep 2000 10:48:00 AM Eastern Daylight Time, Bill Coleman AA4LR <aa4lr@radio.org> writes:

<< On 9/12/00 9:38 AM, Tower2sell@aol.com at Tower2sell@aol.com wrote:

>The answer is -- Use The Projected Area in the direction of the wind. If 
>the wind is blowing horizontal - use the horizontal projection. You may 
>have to rotate the antenna to find the maximum horizontal projected area.

You see, this confuses me.

Consider. We have a two element antenna on a 10 foot boom. Each element 
is 50 feet of 1 inch tubing.

I'll conceed that the drag broadside to the array is going to be less 
than the drag head on, so we'll deal with the drag head on.

>From your note, you imply that the drag of the antenna is equivalent to 
the drag area of a single element -- since the second element is 
completely hidden in our horizontal projection.

I don't see how this can be right. Naturally, the downwind element will 
not have as great a drag as the upwind one, since it is in disturbed air. 
But it won't have zero drag, either! And its drag will vary depending on 
a number of factors, including it's spacing from the first element.

Further, if we contemplate antennas having more elements, and longer boom 
lengths (less than 50 or so feet), their horizontal projections will 
remain the same, but the drag will continue to increase, right up to near 
the point where the antenna becomes a solid raft of tubes from one end of 
the boom to another, at which point we'll see a dramatic DECREASE in drag.

I just can't understand how a horizontal projection predicts the drag of 
the antenna to any degree. It can't pass the thought-problem test at all.



Bill Coleman, AA4LR, PP-ASEL        Mail: aa4lr@radio.org
Quote: "Boot, you transistorized tormentor! Boot!"
            -- Archibald Asparagus, VeggieTales

 >>



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