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"
Just as cross-sections can be reduced, cross-sections can be increased
(at a given frequency) , for example, . by a broadside dipole array.
That's the major problem with your input, in fact.

Brian W