"Richard Harrison" wrote in message
...
Mike, N3LI wrote:
"Why would the velocity be less at increased (antenna element) width?"

Let B = the phase velocity on the antenna element, in radians per unit
length. 2pi/B = wavelength on the element.
Therefore, 2pi/B=velocity of phase propagation.
Due to the behavior of of open-circuited transmission lines and
open-circuited antennas:
B=2pif times sq.rt. of LC radians / unit length.

2 pi f / B = velocity of propagation.

It is intuitive that a fat antenna element has more L & C than a thin
element and thus a lower velocity of propagation.

Best regards, Richard Harrisob, KB5WZI


If B is the 'phase velocity' then surely it's also the 'velocity of phase
propagation' so that's not 2pi/B? The velocity of propagation of an
electromagnetic wave, be it the phase or group velocity, is most-strongly
dependent on the permittivity and permeability of the medium in which the
wave is propagating.

Another clue to resolution of this issue is that the terminal impedance of
an open circuit stub has a slope with respect to frequency that depends on
the characteristic impedance of the stub. For the case of a cylindrical
dipole, the characteristic impedance is not constant but increases
progressively along the lengths of the limbs, but the slope of the terminal
impedance can be related to an 'average characteristic impedance' and the
value of this parameter depends on the average distributed inductance and
capacitance, as has been said here before. Z = sqrt(L/C) in general terms.

The self-inductance of a conducting cylinder has been shown fairly
rigorously by some (e.g. Rosa, and used by Terman) to be inversely
proportional to its radial 'thickness' but this appears to be a contentious
issue and can lead the unwary user into a paradox!

Chris