John Popelish wrote:
Roy Lewallen wrote:
. . .
Effective height determines how many volts you'll get from an open
circuited antenna.

Does that include an antenna that has been brought to resonance with an
appropriate capacitive load?

No. "Open circuited" means that there's nothing connected across the
feedpoint.

. . .
So if a very small rod antenna had a lossless core that could handle any
flux level, and was wound with superconductor, it could couple into the
same volume of space as a 1/2 wave dipole? Amazing.

Yes, exactly the same volume of space, although capture area isn't a
measure of this. Capture area is an area, not a volume, and it's
different in all directions, just like gain. In fact, there's a 1:1
correspondence between capture area and gain, they're just different
ways of expressing the same thing. In the absence of loss, all the power
applied to *any* antenna will radiate, so the integral of the power
density over all directions is the same for all lossless antennas -- the
integral of the power density will equal the applied power, since
there's no dissipation (and in the far field the power density is simply
E^2 / Z0 = H^2 * Z0 where Z0 is the impedance of free space and E and H
are the electric and magnetic field strengths respectively). The
reciprocal of this principle is that the integral of capture areas in
all directions (what you're calling the "volume" the antenna is
"coupling" to) is the same for all lossless antennas.

But more directly to the point, your tiny theoretical rod antenna would
have a gain of about 0.45 dB less than a half wave dipole, and its
capture area would be correspondingly smaller -- about 10%. This is
assuming you're looking in the best direction for each antenna. Because
the total radiated power or integral of the capture areas must be the
same for the two antennas, this means that the tiny antenna has to have
more gain or capture area than the dipole in some other directions. And
indeed it does -- the tiny antenna has slightly fatter lobes than the
half wave dipole. This is a good experiment to run with EZNEC or NEC-2.
The EZNEC demo is adequate. Use free space, set wire loss to zero, and
compare the gains and patterns of a half wave dipole to a very short
one. The Average Gain tells you the ratio of total radiated power to
input power, and it should equal one if the program is doing its
calculations correctly.

If you're interested in knowing how much power you can get from a
ferrite rod, then, what you need to know is its efficiency, which is a
function of wire length, number of turns, and the antenna feedpoint
impedance. I don't have the time right now to work it out for you.

The effective height of a ferrite rod antenna is approximately:

(2 * pi * mueff * N * A) / lambda

where

mueff = effective relative permeability of the rod (mainly a
function of rod length)
N = number of turns
A = rod cross sectional area
lambda = wavelength

I can apply this formula directory to what I am experimenting with,
except that I have to approximate mueff. I am making the rod by
stacking ferrite beads, with various gaps between them. Can I
approximate mueff by taking the ratio of coil inductance with and
without the rod?

Yes. That's exactly what it is.

And, what if the rod area is not constant all along the rod? Since my
rods are assembled from pieces, I have a lot of freedom in this direction.

That one I don't know the answer to.

Roy Lewallen, W7EL