Jim Weir or other qualified persons: a tangent on the 2 radio 1 antennathread

In article ,
Brian Whatcott wrote:
On Mon, 14 Jun 2004 15:17:48 GMT, Dave S
wrote:
//
My question is, given the limited "resistance" of some of the radio
components (and the ability to tolerate less than a watt input if I
paraphrased it correctly) I am wondering just 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

You are not the only one who has experienced breakthrough
near a big transmitter tower. 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

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
0.995 milliwatts per SQUARE YARD of surface area, at 1 nautical mile.

0.995 milliwatts/square yard is the same energy density that a _FIVE_WATT_
transmitter creates at a distance of 20 yards.

Does anybody worry about 5 watts @ 20 yards? Assuming you don't have a
pacemaker, that is. grin


The above is -not- 'fair' to the big transmitter sites, however. 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. One of the stations in downtown Chicago announces
itself at at least 8 megawwatts (ERP) -- might be 9 megawatts, memory
isn't giving a firm answer on -that- point. grin

8 megawatt ERP is 160 times the effective energy of a 50kw output.
Or about 53Mwatt/sq.yd at 1 statute mile (40mw/sq.yd at 1 naut. mi.)
Roughly equivalent to a FIFTY WATT transmitter at 40 yards. (many taxicab
companies use 30-watt VHF radios in the vehicles, and it usually doesn't
affect the FM receiver in the cab itself -- with maybe _two_ yards between
the tx and rx antennas.)

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%. If it was, there
would be a 'dead zone' behind _every_ receiver. 'gain' figures for a
3-element beam antenna suggest that capture efficiency for a single
element is on the order of _one_ percent.

Which would equate to 5 microwatts of power, and an induced voltage of about
15 millivolts. _Not_ threatening to the 'health' of the equipment, but
definitely strong enough to produce enough 'distortion' in a 1st RF amp
stage to create enough 'in-band' signal to pass through the rest of the
receiver.

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

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.