Jim Smith wrote:
>I ran across something which appears to give a quantitative answer to
>how far apart to stack beams, whether for the same or different bands.
>It involves working out the "radius of effective aperture" of each beam
>and adjusting the distance between any 2 beams to be equal to or greater
>than the sum of the radius of effective aperture for the 2 beams.
>
This is one of those old approximate concepts which gives a useful
general understanding, but it is no substitute for modern modeling
techniques.
For a general discussion, see:
http://www.ifwtech.co.uk/g3sek/stacking/stacking2.htm
You can make some use of the concept when stacking long yagis for
VHF/UHF, but even when done correctly it is a very poor guide compared
with modeling.
For HF antennas the aperture concept is a very poor guide to practical
stacking distances... and this particular example is useless because
it's full of errors.
>The following are for units of feet, not metres.
>
>First one calculates the "Effective Aperture" (A) for a beam from
>
>A = Wavelength squared X the antenna gain relative to a dipole (ratio
>not dB) and divided by 4 Pi
>
That should be gain relative to isotropic, so it starts out with a
fundamental mistake... but it isn't very meaningful anyway, for two
reasons:
1. Aperture does not have sharp physical boundaries because it's only a
fuzzy idea. The only time you see a recognizable physical boundary to an
antenna's aperture is in something like a microwave dish (and even there
the edge isn't completely sharp).
2. This is especially true of electrically small antennas such as
dipoles and the few-element beams we use at HF.
>Then one calculates the radius of the effective aperture (R) from
>
>R = A divided by [(H + 2) X Pi]
>
>Where H is the height of the antenna in feet. Lord only knows what the
>2 is doing in there.
>
That is complete nonsense. Height above ground doesn't enter into the
aperture concept of stacking at all.
So don't trust what comes next.
>So, for a 40m beam operating at 7150 with a 4.1 dBd gain at 70 ft
>
>A = 137.7^2 X 2.57 / (4 X 3.14) = 3878 sq ft
>
>R = 3875 / (72 X 3.14) = 17.15 ft
>
>Similarly a 6m beam operating at 51 MHz with a 7.8 dBd gain at 86 ft
>
>A = 179 sq ft
>
>R = 0.65 ft
>
> From which the minimum desirable vertical spacing between a 40m and a
>6m beam is:
>
>17.15 + 0.65 = 17.8 ft
>
>So one would, in principle, move the 6m beam up to the 88 ft level and
>recalculate.
>
>Similarly, two 6m beams should be spaced about 1.3 ft. Sounds a little
>close to me but I don't know much about this stuff.
>
>Is this method actually useful or is it all hogwash?
The 6m example should tell you. There is a useful grain of truth in the
basic concept, but the numbers are - as you rightly suspected - hogwash.
--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek
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