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