[TowerTalk] wind load vs Rohn specs
Bill Coleman AA4LR
Wed, 13 Sep 2000 10:33:07 -0400
On 9/12/00 4:21 PM, Kurt Andress at K7NV@contesting.com wrote:
>> >Antenna area measurements should be made with a calculator. Simply summing
>> >the Length x diameter of the members is more accurate.
>In my mind, the statement means that we take element #1 and sum up the
>areas of all its sections, then we do the same for every other element.
>add all the individual element areas together to get a total area for the
>elements. The boom is done just like one of the elements.
Ah ha! This is EXACTLY the Vertical projection.
The Vertical projection makes more sense to me. It accounts for the drag
of all portions of the antenna, but it leaves out vertical members such
as the boom-to-mast plate, or maybe truss support members or wires above
the boom or elements.
>I've never ever meant to imply that the leading elements shield the others
>the wind. I don't believe that is possible for the normal yagi element
Well, actually they do, to some degree. But it's probably too hard to
calculate accurately, so why not just assume that all members are in
clear air, even those that obviously cannot be (such as the boom versus
the elements). This gives a kind of "worst case" area, except for the
vertical members it leaves out.
Or is the intent to add the area of the vertical members as well? Such a
sum would be neither the Vertical or the Horizontal optical projection.
Rather, it is a sum of the projections of the unassembled components.
>> Could you dig up some references?
>Communications Quarterly, Spring, 1993, Determination of Yagi Wind Loads
>Using the "Cross-Flow Principle", by Dick Weber, K5IU
I'm sure this will answer all my questions. I am currently searching for
a copy of it. Thank you.
>The essence of the article is that the forces acting on the antenna
>normal to their major axis. This means all element loads act parallel to
>the boom loads act parallel to the elements.
Intersting. In aerodynamics, all drag is considered parallel to the
airstream. Forces normal to the airstream are lift.
>Since, the loads on an antenna are developed this way, there just is no
>them to generate a peak load somewhere in between 0 & 90 degress azimuth.
OK, I can accept this. But I don't see how this has an effect on the
computation of this projected area number, unless we have the elements or
the boom "shadowing" other pieces.
>> So far we've talked about yagi's. What about other configurations, like
>> quads? How do we compute their "projected area"?
>I'd be doing it the same way. Sum of all wires and spreaders in the 0
>boom the same as above.
So, in the zero direction, do we take the area of the entire boom, or
merely it's cross-section. (assuming the elements are much longer than
>Well, I guess you have missed the 20 or 30 posts I've made in the last 2-3
>on this matter.
No, I haven't missed them at all, Kurt. I've gone back and read them and
tried to understand them.
>Please explain your new method of dealing with the problem, exactly how
>take my 1" diameter x 72" long piece of tubing and arrive at its "Flat Plate
>Then I can think about what I'd do with that number.
As I understand it, flat plate equivalent is an old method. You don't
need a wind tunnel, just a wind balance, perhaps on a moving vehicle.
Bascially, you characterise the drag of an object by equivocating it to a
flat plate of a given size. You mount the object on the balance, then
construct a flat plate that properly balances the object in the wind. The
result is the area of the flat plate (which is assumed to have an aspect
ratio close to 1)
The Wright Brothers used one at the turn of the century.
Such a number is a very accurate representation of the drag presented by
Bill Coleman, AA4LR, PP-ASEL Mail: firstname.lastname@example.org
Quote: "Boot, you transistorized tormentor! Boot!"
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