In article ,
Brian Whatcott wrote:
On Sun, 20 Jun 2004 10:40:40 +0000,
(Robert Bonomi) wrote:

In article ,
Brian Whatcott wrote:
On Mon, 14 Jun 2004 15:17:48 GMT, Dave S
wrote:
//
/// how much energy the radio
system is being exposed to flying by the transmitting elements a mile
away laterally, and how prudent that is for the longevity of the
components. Lets use 50,000 watts if that is appropriate for the example.

Dave

Here's a rough, rough estimate of
intercepted power.
If 50 kw were distributed through a spherical surface of 1 mile in
radius, what would the power intercepted by one square yard?
(arbitrary cross-section value for a 1/4 wave whip...)
power times Antenna cross-section / Extended surface area
[4/3 pi r squared] = 4 milliwatts
[Brian]

Correction: surface area of a sphere is 4 pi r squared
(volume is 4/3 pi r cubed)

1.294 milliwatts per SQUARE YARD of surface area, at 1 statute mile

/// It's true, they they are limited to 50kw 'out the back of the
transmitter' ,
*BUT* 'gain' antennas are almost universally deployed by VHF (and above)
stations. An 'effective radiated power' in the several _megawatt_ range
is not uncommon.
///
A typical VHF aircraft antenna is, electrically, about 4/3 of a yard long.
if it is 1/4" in diameter, it presents a maximum cross-section of just
about 1/100 of 1 square yard. Which, at 100% capture/conversion efficiency
would pick up just under 0.5milliwatts of energy. v^2 would be 0.025 -- the
peak voltage would be about 0.158 V.

Capture/conversion efficiency is nowhere *near* 100%. ///

[Robert]


I am glad SOMEONE knew the formula for the surface of a sphere.
That's a correction factor of X3 Then things go a little askew.
"Cross-section" is not a term denoting actual area, but equivalent
radio cross-section. As in "The Stealth bomber had a radar cross
section of 1.2 square feet"

Of course, the surface area that _reflects_ a signal back to the source
has absolutely *nothing* to do with how much signal is -absorbed- by a
receiving antenna.

In fact, the _more_ signal that is absorbed, the *less* that is reflected
back to the source. The Stealth technology utilizes _that_, plus 'carefully
shaped' surfaces to reflect remaining signal energy in a direction _other_
than back to the signal source.

Just as cross-sections can be reduced, cross-sections can be increased
(at a given frequency) , for example, . by a broadside dipole array.

Oh my, a 'Microsoft tech support' response -- "technically accurate,
but useless in application".

_How_many_ 'broadside dipole arrays' are in use, *airborne* in civilian
light aircraft?

The 'effective' relative cross-sectional area of such an antenna is
given _directly_ by the 'gain' of the antenna, basis a standard 1/2
wave dipole. 3db gain == double the effective cross-section, 6db == 4x
the effective cross-section, etc.



The antenna under discussion is that used for VHF reception in a typical
civilian light aircraft. A quarter-wavelength whip -- 'normal' to a
ground-plane. Giving an 'effective' size-equivalent of a 1/2 wavelength
dipole. If the whip is a 'coil-loaded' unit, to give a reduced physical
size, then actual signal capture is reduced below that of a full-length
1/4-wave unit.

*IF* one is using a 'gain' antenna -- say a "5/8-wave" unit -- it is
trivial to factor in the additional 'effective' cross-section; by
simply using the 'gain' of the antenna. The '5/8 wave' whip is the
-only- commonly-used _omni-directional_ 'gain' antenna in common use
on VHF frequencies. With a theoretical 3db gain, it has an 'output
voltage level' that is, at best, only about 40% higher than the 1/4-wave whip.

That's the major problem with your input, in fact.

The _big_ problem with my analysis is the execrable 'capture efficiency'
of a standard 1/4-wave whip (or 1/2-wave dipole). I simply don't have
a good handle on _how_poor_ that efficiency actually is. The lack of
any measurable radio 'shadow' behind a broadside dipole array, and the
'gain' of a 3-element beam, vs a single 1/2-wave dipole element suggests
that the 'efficiency' value is _very_ low.