As a competent and experienced engineer, it should then be simple for
you to answer the following:

What is the gain difference, in dB, between a dipole resonant at 97.5
MHz (the geometric center of the FM band) which is 1 mm diameter and one
which is 1 cm diameter? Feel free to assume that the conductor is
perfect, or use copper if you prefer.

Also feel free to calculate the antenna Q and "antenna potential",
although the question here is about gain.

Roy Lewallen, W7EL

Richard Harrison wrote:
Roy, W7EL wrote:
"That`s interesting.(I don`t know why you want fat. It will give you
lower gain.) How much lower? Why?"

It`s a fact. Fat antennas have more bandwidth, and that is inversely
proportional to Q. Teducing antenna Q, by fattening the antenna, reduces
the antenna potential by about the same factor.

Best regards, Richard Harrison, KB5WZI

On Mon, 31 Jan 2005 15:14:49 -0800, Roy Lewallen
wrote:

As a competent and experienced engineer, it should then be simple for
you to answer the following:

What is the gain difference, in dB, between a dipole resonant at 97.5
MHz (the geometric center of the FM band) which is 1 mm diameter and one
which is 1 cm diameter? Feel free to assume that the conductor is
perfect, or use copper if you prefer.

Also feel free to calculate the antenna Q and "antenna potential",
although the question here is about gain.

Roy Lewallen, W7EL

Richard Harrison wrote:
Roy, W7EL wrote:
"That`s interesting.(I don`t know why you want fat. It will give you
lower gain.) How much lower? Why?"

It`s a fact. Fat antennas have more bandwidth, and that is inversely
proportional to Q. Teducing antenna Q, by fattening the antenna, reduces
the antenna potential by about the same factor.

Best regards, Richard Harrison, KB5WZI


Hi Roy,

What an unusual demand to throw in the face of someone who agrees with
you: no difference in gain. Richard's quote is merely your ironic
question to Buck's quote (already discounted by Buck).

However, for Brad's interest (and conforming to his original design,
not of 1cM but more like 170mm diamter) the Q for the fatter dipole is
indeed much less (in fact it covers the entire FM band into a 50 Ohm
load between 2:1 VSWR points) where the thin dipole (1mm) is something
less than 6MHz. Bandwidth (and inferentially Q) differential 4:1
which would translate the input V to the tips to something less (at
the same proportion) than that experienced with the thin dipole (which
for a recieve antenna is a strange characteristic to focus upon).

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Mon, 31 Jan 2005 15:14:49 -0800, Roy Lewallen
wrote:


As a competent and experienced engineer, it should then be simple for
you to answer the following:

What is the gain difference, in dB, between a dipole resonant at 97.5
MHz (the geometric center of the FM band) which is 1 mm diameter and one
which is 1 cm diameter? Feel free to assume that the conductor is
perfect, or use copper if you prefer.

Also feel free to calculate the antenna Q and "antenna potential",
although the question here is about gain.

Roy Lewallen, W7EL

Richard Harrison wrote:

Roy, W7EL wrote:
"That`s interesting.(I don`t know why you want fat. It will give you
lower gain.) How much lower? Why?"

It`s a fact. Fat antennas have more bandwidth, and that is inversely
proportional to Q. Teducing antenna Q, by fattening the antenna, reduces
the antenna potential by about the same factor.

Best regards, Richard Harrison, KB5WZI



Hi Roy,

What an unusual demand to throw in the face of someone who agrees with
you: no difference in gain. Richard's quote is merely your ironic
question to Buck's quote (already discounted by Buck).

However, for Brad's interest (and conforming to his original design,
not of 1cM but more like 170mm diamter) the Q for the fatter dipole is
indeed much less (in fact it covers the entire FM band into a 50 Ohm
load between 2:1 VSWR points) where the thin dipole (1mm) is something
less than 6MHz. Bandwidth (and inferentially Q) differential 4:1
which would translate the input V to the tips to something less (at
the same proportion) than that experienced with the thin dipole (which
for a recieve antenna is a strange characteristic to focus upon).

73's
Richard Clark, KB7QHC

If you were agreeing with me, Richard (Harrison), I apologize. It wasn't
apparent to me with my poor language skills. Thanks to Richard (Clark)
for effectively applying his superior parsing skills to the problem.

There are only two ways to change the free space gain of an antenna --
change the efficiency, or change the pattern. Those are all the choices
you've got. A fat antenna is certainly no less efficient than a skinny
one -- in fact, it'll be more efficient. But the difference in this case
would be so small as to be unmeasurable. There would be some very slight
change in pattern between a fat antenna and a slim one, but again the
change would be negligibly small.

Considering only free space performance to remove the additional
variable of ground reflection, and assuming that an antenna is
essentially 100% efficient, it's impossible to design an antenna that
has gain in its best direction which is any less than 2.15 dB below that
of a half wave dipole. The lowest possible gain of any efficent antenna
is the isotropic, at 0 dBi.

Roy Lewallen, W7EL

Roy, W7EL wrote:
"There are only two ways to change the free space gain of an
antenna---change the efficiency , or change the pattern."

I agree. Terman defines "directive gain" and "power gain" which involve
the pattern and efficiency of an antenna.

The isotropic antenna is by definition omnidirectional. All others are
more directional and thus have gain in their best direction.

The power ratio of a 1/2-wave resonant conductor radiates in its best
direction 1.64 times the power per unit area from an isotropic antenna.
This is a simple power ratio, not dB. This is from Terman`s Table 23-1
in his 1955 edition.

My original posting in this thread was based on the fact that antenna
voltage distribution depends on constructon and frequency. Voltage
amplitudes at all points on the sntenna increase when the rms voltage at
any point rises.. Radiation and reception from an antenna are a function
of antenna voltage. This is unrelated to directive gain.

A higher antenna Q results in higher voltage. The dipole we discussed
was resonant. We`ve seen the textbook curves for resonant circuits which
pften show impedance versus frequency, and we have tuned lumped and
distributed versions. A high Q series resonant circuit has little
resistance to limit current at resonance. A high Q parallel resonant
circuit has little conductance to limit voltage.
My posting said: "Reducing antenna Q by fattening the antenna, reduces
the antenna potential by about the same factor."

A higher Q antenna results in more voltage, more radiation, and more
reception. It also has less bandwidth.

I usually read Roy`s postings because they are interesting and I often
learn something from them. They are greatly appreciated by me and many
others have said they appreciate them too. Some aren`t even EZNEC users,
so there is still room for growth.

On the issue of antenna Q, I recall a Yagi design article which advised
against large diameter parasitic elements as they would have
insufficient Q and not perform properly. That seemed strange to me at
the time but maybe there was something to it.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote
My original posting in this thread was based on the fact that antenna
voltage distribution depends on constructon and frequency. Voltage
amplitudes at all points on the sntenna increase when the rms voltage at
any point rises.. Radiation and reception from an antenna are a function
of antenna voltage. This is unrelated to directive gain.
__________________

TV and FM broadcast transmit antenna designs include many using 1/2-wave,
resonant dipoles with "fat" radiators -- however they all have published
gains based on the standard gain of a dipole (1.64X that of an isotropic
radiator). There is no difference in gains between slender and fat radiator
designs in the broadcast industry.

Examples on request.

RF

Richard Fry wrote:
. . .There is no difference in gains between slender
and fat radiator designs in the broadcast industry.

Rest assured, there's no difference in any other industry either.

Roy Lewallen, W7EL

Why does everybody insist on OVER-COMPLICATING this simple problem?

The Q of a resonant 1/2-wave dipole is given by -

Q = Omega * L / 2 / R

Where L is the end-to-end wire inductance and R is the radiation resistance
of about 71 ohms.

Just the same 'formula', in fact, as any other tuned circuit or transmission
line. Resonant rise in voltage and current, and bandwidth, etc., all
follow.

It's so simple it doesn't occur to Terman and other 'beings' to mention it.
----
Reg

Richard Harrison wrote:
. . .
A higher Q antenna results in more voltage, more radiation, and more
reception. It also has less bandwidth.
. . .

There's a problem here. Let's say we begin with a smallish loop antenna,
one small enough so it has essentially equal current around the
perimeter. We'll make it using a perfect conductor. Put 100 watts into
it; since it has no loss, 100 watts will be radiated, distributed in a
dipole-like pattern.

Now reduce the size of the loop. The Q will increase.

You've said that because of the increase in Q, it will have more
radiation. My question is, does that greater radiation result from a
more directional pattern, or from more power being radiated? If the
former, why would the smaller loop have a sharper pattern than the
larger one (considering the assumption made about the initial loop
size)? If the latter, we've really stumbled onto something here -- more
than 100 watts out with 100 watts in. Perpetual motion, here we come!

Roy Lewallen, W7EL