Hi to all.. It is well know that a reduction in the diameter of the
wire must be compensated by a higher antenna length to maintain
resonance.

I am looking for an explanation of the reason for this. Why the total
reactance becomes more capacitive? I know math formula showing the
variation of the inductance of the wire vs its diameter, but I a
looking for the real reason, not the mathematical consequence.

I suspect that a higher diam cause higher transormation of AC to
electromagnetic energy on a segment delta(l) so that a shorter
physical length would be needed to include the full electrical 180
degrees of a dipole.. but not really sure of this.

Any comment would be welcome

Thanks, and 73 de Pierre ve2pid

Try increased capacitance from larger wire diameter, larger surface (plate)
area.
Capacitance goes up, inductance (length) has to come down in order to
maintain resonance - LC circuit in standing wave environment.

73 Yuri, K3BU, VE3BMV, VE1BY

"Pierre Desjardins" wrote in message
...
Hi to all.. It is well know that a reduction in the diameter of the
wire must be compensated by a higher antenna length to maintain
resonance.

I am looking for an explanation of the reason for this. Why the total
reactance becomes more capacitive? I know math formula showing the
variation of the inductance of the wire vs its diameter, but I a
looking for the real reason, not the mathematical consequence.

I suspect that a higher diam cause higher transormation of AC to
electromagnetic energy on a segment delta(l) so that a shorter
physical length would be needed to include the full electrical 180
degrees of a dipole.. but not really sure of this.

Any comment would be welcome

Thanks, and 73 de Pierre ve2pid

On Fri, 28 Apr 2006 12:31:42 -0400, Pierre Desjardins
wrote:

Why the total reactance becomes more capacitive?

Hi Pierre,

The total reactance before you shrank the wire diameter was balanced
at zero (presuming a resonant structure). After the wire diameter was
made smaller, for the same length, the inductance was lowered. Less
inductance to balance the existing capacitance leaves an excess
capacitance you observe. Of course, by making the wire thinner also
changes capacitance, the change in inductance moved further.

73's
Richard Clark, KB7QHC

Oh, Richard, Richard...

A smaller wire diameter has MORE inductance, not less, in the same
environment. Think for a moment about coax: reduce the inner
conductor diameter, and the impedance goes up while the propagation
velocity stays the same. That means that C goes down and L goes up.

For Pier something else to ponder is that the change for resonance
(zero reactance) in a half-wave dipole is considerably less than the
change in a full-wave ("anti-resonant") dipole, for the same wire
diameter change.

I don't think that simple concepts of the antenna behaving like a TEM
transmission line are going to cut it here, and I'll wait for a better
explanation than that.

Cheers,
Tom

On 28 Apr 2006 13:28:56 -0700, "K7ITM" wrote:

Oh, Richard, Richard...

Hi Tom,

A smaller wire diameter has MORE inductance, not less, in the same
environment.

Yes, I did invert the relation of thickness to inductance - for a
short wire. However, the feedpoint observation speaks of common
results offering a different perspective. This is the question.

It does not intuitively follow to describe less capacitance for the
same size, but now thinner antenna makes an antenna more capacitive,
does it? [A transform is at work.]

Think for a moment about coax: reduce the inner
conductor diameter, and the impedance goes up while the propagation
velocity stays the same.

This analogy begins to break down for antennas in that as the antenna
grows thinner/thicker, the propagation velocity does change. On the
other hand, and agreeing with your example, Z tracks (lower w/thicker)
with an antenna. This is in conflict.

That means that C goes down and L goes up.
with a proviso:
I don't think that simple concepts of the antenna behaving like a TEM
transmission line are going to cut it here, and I'll wait for a better
explanation than that.

No, it didn't.

For an antenna with an with an element circumference of 0.001
wavelength, the Vf is 0.97 to 0.98. Compared to an antenna with an
element circumference of 0.1 wavelength, the Vf is 0.78 to 0.79.

Velocity factor is a property of the capacitor's insulative medium
(relative permittivity), which has never changed. [I would argue that
the medium has in fact changed by the presence of the radiator, but
that is another thread.]

Large structures near resonance confound small component analytical
results. So, we will both wait for Reggie to explain it in what he
calls english; or for Cecil to explode with a new SWR analysis.

73's
Richard Clark, KB7QHC

I think it's a BIG mistake to be writing about "velocity factor" in
this thread (and perhaps also in some current, related threads). The
reason is that it presupposes behaviour that is just like a TEM
transmission line, and clearly it is not when you get to the fine
details. Until we better understand just what is going on, I propose
that we simply say that resonance occurs for a wire shorter than 1/4
freespace wavelength, when that wire is fed against a ground plane to
which it is perpendicular, and that the thicker the wire, the shorter
it is at resonance when compared with the freespace wavelength. The
effect can be described with an emperical equation, of course. But to
invoke "velocity factor" assumes something about the solution which may
well lead you away from the correct explanation.

I don't really expect many will take this seriously--there seems to be
too much invested in explaining everything in terms of behaviour that
seems familiar. It's a bit like saying a photon is a particle (or a
wave). It is not--it is simply a quantum; and it behaves differently
from particles we know, and behaves differently from waves we know from
our macro-world experience.

The transmission-line analog is a very useful one for practical antenna
engineering, just as considering loading elements as lumped reactances
(perhaps with parasitic lumped reactance and resistance as appropriate)
is useful for practical engineering. But that doesn't mean it fully
explains the behaviour in detail.

Cheers,
Tom

K7ITM wrote:
But to
invoke "velocity factor" assumes something about the solution which may
well lead you away from the correct explanation.

For the feedpoint impedance to be purely resistive, i.e.
resonant, for a standing wave antenna, the reflected wave
must get back into phase with the forward wave. Velocity
factor is a way of explaining how/why that happens. The
diameter of the conductor no doubt appears in the VF
equation.
--
73, Cecil http://www.qsl.net/w5dxp