On 8/31/06, Red <RedHaines@centurytel.net> wrote:
> Hi, Dan;
>
> At what value does the Reynolds number, product of diameter and wind
> velocity, increase and require a different coefficient of drag?
Red,
By my calculations the drag on pretty much any antenna element is
going to be turbulent, and certainly so for a tens of miles per hour
wind.
The kinematic viscosity of air is around 1.3*10^-5 m^2/s... even with
a tiny breeze of 2mph, and a 2 inch antenna boom the Reynolds number
is, (courtesy of google calculator that does the unit conversions for
you):
((2 miles per hour) * 2 inches) / ((1.3e-5 (m^2)) / s) = 3 493.78954
which is likely to be well developed turbulence...with a 10 inch tube
in a 2mph breeze, the Reynolds number is about 17000.
I think that it's very likely that all of the formulas you find here
and there for antenna windload calculation assume fully developed
turbulence. For example, a 1 inch rod in a 55mph wind gives:
(55 miles per hour) * 1 inches) / ((1.3e-5 (m^2)) / s) = 48 039.6062
Another concern might be vortex shedding resonances... the 10 inch
tube will shed vortices on alternate sides in a breeze... if the
frequency at which it does so is near the natural modes of the
structure, it could really get going. The spiral winding on car
antennas is designed to break the vortex shedding up to avoid
"singing" of the antenna... you can see the same thing on some tall
smoke stacks... there's a spiral winding around them... but they're
usually masonry and bricks and mortar aren't known for their
flexibility or graceful failure mode.
73,
Dan
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