Art might be quick to point out that there is one
wavelength of wire in a 1/2WL folded dipole. :-)
_______________-

But does Art realize that a 1/2-wave dipole is a fractional wavelength
radiator that couldn't possibly have his definition of "equilibrium,"
yet it has the same measured pattern and gain as a 1/2-wave folded
dipole?

RF

On Sep 3, 9:13*am, Richard Fry wrote:
Art might be quick to point out that there is one
wavelength of wire in a 1/2WL folded dipole. :-)

_______________-

But does Art realize that a 1/2-wave dipole is a fractional wavelength
radiator that couldn't possibly have his definition of "equilibrium,"
yet it has the same measured pattern and gain as a 1/2-wave folded
dipole?

RF

Yes, that is correct, but the power used on a 1/2 wave dipole is half
that of a full wave.
A closely folded dipole radiates the same as a Quad . View Cebik's
comments on this

On Sep 3, 10:26*am, Art Unwin wrote:

Yes, that is correct, but the power used on a 1/2 wave dipole is
half that of a full wave.

A full wave what? Are you calling a folded 1/2-wave dipole a full-
wave antenna?

For equal, matched power applied either to a 1/2-wave dipole or to a
folded 1/2-wave dipole, and although their feedpoint currents will be
different, both configurations will generate the same values of field
intensity.

Also how do you explain this, given that the 1/2-wave dipole by your
definition does not have "equilibrium?"

RF

Dale Parfitt wrote:
Thank you Cecil,
That's all I was looking for.

You're welcome and I agree with 95% of what W8JI says.
(For instance, he is mistaken about the delay through
a 100T, 10TPI, 2" diameter 75m loading coil.)

Some may or may not understand why random folding
of antenna radiators tends to change the radiating
conductors into non-radiating conductors. (The same
effect is at work in loading coils.)

When two conductors are carrying differential coherent
currents with no common-mode current, there is negligible
radiation when the two conductors are parallel to
each other and the spacing is a very small fraction of
a wavelength. It's called a transmission line and most
of the losses at HF are I^2*R. Usually, one of the goals
of a transmission line is not to radiate. Transmission line
fields tend to cancel in the near field due to destructive
interference.

A single straight wire in free space is a very efficient
radiator because interference occurs mostly in the far
field. Fold it back upon itself and unless the
second conductor is positioned perfectly, there will exist
differential currents between the two conductors which
will tend to cancel the radiation - leaving mostly I^2*R
losses at HF.

Small folded/loaded antennas tend to cancel the radiating
fields. The only other avenue for a lot of the energy is
conversion to heat.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 3, 10:26*am, Art Unwin wrote:

Yes, that is correct, but the power used on a 1/2 wave dipole
is half that of a full wave.

On the chance that you meant a full-wave dipole in your quote above, I
did a quick comparison of one with a 1/2-wave dipole (linked below).

The peak, intrinsic gain of the full-wave is about 1.6 dB greater than
the 1/2-wave -- which is due to the narrower lobe it produces. This
has nothing to do with "equilibrium."

The 2,082 -j583 ohm input Z of the full-wave version is not user
friendly. But if zero-loss matching networks are used at the
feedpoint of both antennas, then for EQUAL applied power to each, the
peak field intensity produced by the full-wave dipole would be about
1.6 dB (20%) greater than from the 1/2-wave version.

If the power applied to the full wave dipole was 1.6 dB less than
applied to the 1/2-wave dipole, then their measured peak fields would
be identical. But that is not a power reduction of one half (3 dB),
as in your statement, Art.

http://i62.photobucket.com/albums/h8...CompareArt.jpg

RF

Cecil Moore wrote:
Dale Parfitt wrote:
Thank you Cecil,
That's all I was looking for.

You're welcome and I agree with 95% of what W8JI says.
(For instance, he is mistaken about the delay through
a 100T, 10TPI, 2" diameter 75m loading coil.)

Some may or may not understand why random folding
of antenna radiators tends to change the radiating
conductors into non-radiating conductors. (The same
effect is at work in loading coils.)

Any relation to the loosely wrapped "coils" of shortened verticals like
bug catchers?

- 73 de Mike N3LI -

Richard Fry wrote:
If the power applied to the full wave dipole was 1.6 dB less than
applied to the 1/2-wave dipole, then their measured peak fields would
be identical. But that is not a power reduction of one half (3 dB),

Increase the length of the one wavelength dipole to a
1.25WL EDZ and the maximum gain indeed does increase by
~3dB over a 1/2WL dipole. Consider that the highest gain
for a single-wire antenna with a figure-8 radiation pattern
occurs with a feedpoint impedance of ~175-j1000 ohms, i.e.
the antenna wire, by itself, is *non-resonant*. A parasitic
element 1.25WL long would have a negligible effect on an
antenna system. :-0

Consider that if one disconnects the feedline from a 1/2WL
center-fed dipole, the two remaining 1/4WL wires separated
by an insulator are *non-resonant*. Breaking guy wires into
1/4WL separated by insulators is one way of avoiding
resonance. :-)
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

Michael Coslo wrote:
Any relation to the loosely wrapped "coils" of shortened verticals like
bug catchers?

Take a look at the geometry. Assuming that the current on one
side of a turn on the coil is equal to the current on the
opposite side of the coil but traveling in the opposite direction,
one can see why those two currents per turn resemble transmission
line currents (differential) rather than antenna currents (common-
mode). (A 10" coil wound with 1/2WL of wire radiates roughly the
same amount of energy as a 10" straight wire.)

Physically large air-core loading coils can be modeled as a
transmission line with a Z0 and VF (delay).

http://hamwaves.com/antennas/inductance.html

The delay through the coil can be calculated by knowing the
Beta = _____ rad/m "Axial propagation factor of n=0 sheath
helix waveguide mode at the design frequency"

The VF of W8JI's 100T, 10TPI, 2" dia test coil calculates
out to be ~0.03 resulting in a ~25 nS (~37 deg) delay
through the coil at 4 MHz.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

Cecil Moore wrote:
Michael Coslo wrote:
Any relation to the loosely wrapped "coils" of shortened verticals
like bug catchers?

Ack! I meant to write Hamsticks, not bug catchers! Sum daze I am in a daze!

- 73 de Mike N3LI -

On Sep 4, 7:04*am, Cecil Moore wrote:
Dale Parfitt wrote:
Thank you Cecil,
That's all I was looking for.

You're welcome and I agree with 95% of what W8JI says.
(For instance, he is mistaken about the delay through
a 100T, 10TPI, 2" diameter 75m loading coil.)

Some may or may not understand why random folding
of antenna radiators tends to change the radiating
conductors into non-radiating conductors. (The same
effect is at work in loading coils.)

When two conductors are carrying differential coherent
currents with no common-mode current, there is negligible
radiation when the two conductors are parallel to
each other and the spacing is a very small fraction of
a wavelength. It's called a transmission line and most
of the losses at HF are I^2*R. Usually, one of the goals
of a transmission line is not to radiate. Transmission line
fields tend to cancel in the near field due to destructive
interference.

A single straight wire in free space is a very efficient
radiator because interference occurs mostly in the far
field. Fold it back upon itself and unless the
second conductor is positioned perfectly, there will exist
differential currents between the two conductors which
will tend to cancel the radiation - leaving mostly I^2*R
losses at HF.

Small folded/loaded antennas tend to cancel the radiating
fields. The only other avenue for a lot of the energy is
conversion to heat.
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

Cecil I must respectively disagree. Your arguement is based on the
presence of common mode current. When there is a state of equilibrium
there is no vector that represents common mode. Since the radiator is
a full WL that represents a period it is of closed circuit form. In
such a case any radiator bend is accompanied by a bend that is equal
and opposite per Newtons laws. The moment you introduce common mode
currents you have strayed from the concepts of equilibrium, where all
forces are accounted for. Maxwells laws are based on the position that
all forces involved are accounted for where the summation of such
equals zero.
Regards