Caveat Lector wrote:
Here is a site for examples of capture areas of antennas

http://www.sommerantennas.com/gain.html


Have you got a link to a similar site that covers ferrite rod antennas?

Caveat Lector wrote:
Here is a site for examples of capture areas of antennas

http://www.sommerantennas.com/gain.html

John Popelish wrote:
Have you got a link to a similar site that covers ferrite rod antennas?

I hope this question isn't taken the wrong way, but why would you want
a similar site? That one is terrible for accuracy. The poor fellow who
wrote that page doesn't even know what capture area is.

Wouldn't it be better to find a totally different type of site, one
that at least has some technical accuracy?

73 Tom

wrote:
Caveat Lector wrote:

Here is a site for examples of capture areas of antennas

http://www.sommerantennas.com/gain.html


John Popelish wrote:
Have you got a link to a similar site that covers ferrite rod antennas?


I hope this question isn't taken the wrong way, but why would you want
a similar site? That one is terrible for accuracy. The poor fellow who
wrote that page doesn't even know what capture area is.

I meant a site that addressed (correctly, one would hope) that concept
of capture area for rod antennas.

Wouldn't it be better to find a totally different type of site, one
that at least has some technical accuracy?

Okay, I'll take one of those. :-)

What information are you looking for, capture area or effective height?
Capture area determines how many watts you'll get into a conjugately
matched load connected to the antenna. Effective height determines how
many volts you'll get from an open circuited antenna. The two aren't
directly related. For more information about the two topics, do a
groups.google.com search for postings I've made on those topics in this
newsgroup.

As I've posted here quite a number of times before, the capture area of
a lossless infinitesimally short dipole is very nearly as great as that
of a half wave dipole, in their most favored directions. (The difference
is about 10%, and it's due to the slight pattern shape difference caused
by different current distributions). So except for loss the capture area
of a ferrite rod antenna is within 10% of that of a dipole. But loss in
a ferrite rod antenna will reduce the capture area considerably. 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

Roy Lewallen, W7EL

John Popelish wrote:
wrote:
Caveat Lector wrote:

Here is a site for examples of capture areas of antennas

http://www.sommerantennas.com/gain.html


John Popelish wrote:
Have you got a link to a similar site that covers ferrite rod antennas?


I hope this question isn't taken the wrong way, but why would you want
a similar site? That one is terrible for accuracy. The poor fellow who
wrote that page doesn't even know what capture area is.

I meant a site that addressed (correctly, one would hope) that concept
of capture area for rod antennas.

Wouldn't it be better to find a totally different type of site, one
that at least has some technical accuracy?

Okay, I'll take one of those. :-)

Roy Lewallen wrote:
What information are you looking for, capture area or effective height?

Since I wasn't even aware that effective height applied to rod
antennas (or exactly what effective height means) I guess I was
thinking of capture area.

Capture area determines how many watts you'll get into a conjugately
matched load connected to the antenna.

That's it.

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?

The two aren't
directly related. For more information about the two topics, do a
groups.google.com search for postings I've made on those topics in this
newsgroup.

Thanks.

As I've posted here quite a number of times before, the capture area of
a lossless infinitesimally short dipole is very nearly as great as that
of a half wave dipole, in their most favored directions. (The difference
is about 10%, and it's due to the slight pattern shape difference caused
by different current distributions). So except for loss the capture area
of a ferrite rod antenna is within 10% of that of a dipole. But loss in
a ferrite rod antenna will reduce the capture area considerably.

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.

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?

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.

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

Roy Lewallen wrote:
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.

But I can design the coil to be self resonant or not, just by
adjusting the surface area of the wire, or the spacing. It is non
intuitive that if I peak the coil this way, and obtain more voltage,
it is a different case than if I peak the coil with cable capacitance,
or an additional capacitor. I guess I really don't comprehend the
point of this value.

(Snip excellent review of basic lossless radiator. Thank you.)

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.

I understand what you are saying.

(snip)
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.

Well, now I can calculate the effective height of my antennas, even
though I am not sure what it has to do with height.


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.

I am also experimenting with designs that do not necessarily have a
small, coil, close to the rod. (My interest in the discussion of
extended coils is showing.) One of the possibilities that shows a
significant increase in tuned Q is an hour glass shaped coil (small
diameter in the center, but sweeping to a larger diameter at the
ends). I have been asked to try putting a rod through the center of a
flat spiral coil. It seems to me that, at some extreme, the above
formula will fail, because it assumes that essentially all the signal
energy exiting the coil was collected by the rod, and that the signal
the coil would collect by itself would be insignificant. But if my
coils get large enough, they become loop antennas in their own right,
and the rod, though it may have a significant length and area, is only
a part of what is happening. In other words, the mueff can get pretty
small, even though the rod has significant dimensions. I guess, what
I am asking are what assumptions about coil dimensions (that are not
explicitly referenced in the formula) are being made in the above formula?

"John Popelish" wrote in message
...
Caveat Lector wrote:
Here is a site for examples of capture areas of antennas

http://www.sommerantennas.com/gain.html


Have you got a link to a similar site that covers ferrite rod antennas?

Try URL:

http://www.st-andrews.ac.uk/~jcgl/Sc...rt7/page5.html

Some others using Google
--
CL -- I doubt, therefore I might be !