On Jul 15, 3:01*pm, Owen Duffy wrote:
"J.B. Wood" wrote in news:i1monr$2od$1
@ra.nrl.navy.mil:

On 07/15/2010 04:14 AM, Owen Duffy wrote:
I note some variation in the use of the term 'Radiation Resistance' (Rr)
that suggests that it has different meanings to different folk.

snip
Hello, and I don't find any ambiguities in any of my various EM and
antenna theory textbooks. *FWIW, from the IEEE Standard Dictionary of
Electrical and Electronics Terms:

"Radiation resistance (antenna). The radio of the power radiated by an
antenna to the square of the rms antenna current referred to a specified
point. *Note: *This term is of limited utility in lossy media."

Hmmm. The last statement suggests that, as defined, it is not clear and
unambiguous in the real world because the real world involves "lossy
media".

The "reference to a specified point" suggests that if one gives a value for
Rr, it is necessary to also state the reference point. Is that what it
means?

This is exactly the lack of clarity that is troubling me.

So if we're looking at free (in vacuo) space the radiation resistance is
simply a "load" resistance component that accounts for where the
radiated power goes. *The radiation resistance doesn't include any other
resistive losses in the antenna structure/proximity operating
environment that may also be dissipating source power introduced at the
feedpoint of the antenna.

This does not address the issue of ground reflection that I mentioned.

*An aerodynamic analogy would be the
distinction between "induced" drag (the price paid for "lift") and
"parasite" drag, which are both components of the total drag.
Sincerely, and 73s from N4GGO,

I am not an aerodynamics type, so drawing that analolgy only helps to
confuse. You might as well use optics!

I know you are trying to be helpful John, but the IREE definition doesn't
seem to clarify the issue.

To put some numbers on my first example, if I have an NEC model of a centre
fed half wave dipole with zero conductor losses, mounted over real (ie
lossy) ground, and feedpoint R at resonance is say, 60 ohms, and total
power in the *far field* divided by I^2 is say, 50 ohms, is Rr 50 ohms? Is
the power "radiated" from such a dipole ONLY the power that makes it to
'distant space', or is radiated power input power less dipole conductor
losses?

The IREE definition suggests that I need also to state that Rr is 50 ohms
at the centre, and the term is is of "limited utility" (not unambiguously
clear?) because of the lossy ground reflections.

If indeed the term Radiation Resistance is only applicable in lossless
scenarios as suggested by the IREE dictionary, what it a clear and
unambiguous language for the real world?

Cheers
Owen

In real world terms radiation resistance is measured by the vector
that overcomes radiation resistance or the conveyance of
communication. This compels the measurement of that which is
accelerated as it is an action and reaction type vector. If one
doesn't have a measurement of the mass that is being accelerated then
radiation resistance itself cannot be supplied. What happens
to the accellerated mass has no connection what so ever to the
accelleration vector.To find the accelerating vector one must first
determine the efficiency of the apparatus used and this will vary
dependent on the method used to produce the accelerating vector so
that one can determine the losses. So if we cannot identify that
vector which creates acceleration of charge where the charge is the
measurement of radiation one must first determine what creates
radiation so that the radiation unit can be measured. The bottom line
is
that one must use a superconductor where only the accelerating vector
comprises of the impedance
seen by the time varying current and where the
resistance of the radiating member is divorced from the equation as is
coupling losses in the absence of a magnetic field.
Art