On Sep 4, 8:54*am, Cecil Moore wrote:
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

Cecil you misunderstood what I was saying .
When you use a fractional WL then some of the wire cannot radiate,
thus the impedance is reduced from that of a full WL. Power is still
I sqd R Cos phi. where there is a varience. with the R in this case of
a approx 1/2. Gain is not involved, only energy dissipated.
Regards

On Sep 4, 7:23*am, Richard Fry wrote:
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

Gain has nothing to do with energy expended from a radiator. A folded
dipole is of a full WL with an overall dimension of 1/2 WL.

Art Unwin wrote:
Since the radiator is
a full WL that represents a period it is of closed circuit form.

My comments were about a one-wavelength straight
wire dipole in free space. The reflections from
the ends are what result in that large resonant
resistance at the center.

--Vf
Open --If
----------------------fp----------------------
Circuit Vr--
Ir--

Zfp - feedpoint impedance, Vf - forward voltage,
Vr - reflected voltage, If - forward current,
Ir - reflected current

Zfp = (Vf+Vr)/(If-Ir) = thousands of ohms

However, if we fold the 1WL dipole into a circular
1WL loop it is still a standing-wave antenna but
the phase of the reflections is reversed.

Zfp = (Vf-Vr)/(If+Ir) = ~100 ohms.

Where are those reflections coming from in a
circular 1 WL loop? Why is the phase of the
reflections reversed?
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 4, 12:53*pm, Cecil Moore wrote:
Art Unwin wrote:
Since the radiator is
a full WL that represents a period it is of closed circuit form.

My comments were about a one-wavelength straight
wire dipole in free space. The reflections from
the ends are what result in that large resonant
resistance at the center.

* * * * * * * * * --Vf
Open * * * * * * --If
----------------------fp----------------------
Circuit * * * * *Vr--
* * * * * * * * * Ir--

Zfp - feedpoint impedance, Vf - forward voltage,
Vr - reflected voltage, If - forward current,
Ir - reflected current

Zfp = (Vf+Vr)/(If-Ir) = thousands of ohms

However, if we fold the 1WL dipole into a circular
1WL loop it is still a standing-wave antenna but
the phase of the reflections is reversed.

Zfp = (Vf-Vr)/(If+Ir) = ~100 ohms.

Where are those reflections coming from in a
circular 1 WL loop? Why is the phase of the
reflections reversed?
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

Equilibrium means equilibrium thus there are no reflections. Actions
have an equal and opposite reaction. What are you going to draw upon
for an equalizing vector?

On Sep 4, 12:39*pm, Art Unwin wrote:

Gain has nothing to do with energy expended from a radiator. A folded
dipole is of a full WL with an overall dimension of 1/2 WL.

And for equal, matched power applied to their feedpoints, the total of
the energy "expended" (radiated) by both them is identical.

//

Art Unwin wrote:
Equilibrium means equilibrium thus there are no reflections.

No reflections on a standing-wave antenna?
Where do the standing waves come from?
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 4, 3:00*pm, Cecil Moore wrote:
Art Unwin wrote:
Equilibrium means equilibrium thus there are no reflections.

No reflections on a standing-wave antenna?
Where do the standing waves come from?
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

There are no standing waves either

On Sep 4, 3:40*pm, Art Unwin wrote:
On Sep 4, 3:00*pm, Cecil Moore wrote:

Art Unwin wrote:
Equilibrium means equilibrium thus there are no reflections.

No reflections on a standing-wave antenna?
Where do the standing waves come from?
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

There are no standing waves either

Remember my WL is a closed circuit

On Sep 4, 1:25*pm, Richard Fry wrote:
And for equal, matched power applied to their feedpoints, the total of
the energy "expended" (radiated) by both them is identical.
_______________

For clarity I should have included the fact that the total of the
energy "expended" (radiated) when equal and matched powers are applied
to any and all elemental radiators, of whatever form, essentially is
identical.

What is NOT identical across these various radiator forms is the field
intensity in specific directions in the radiation volume.

THAT depends on the design of the radiating structure(s), and the
installation+propagation environment.

This is the reason for the 1.6 dB advantage in the intrinsic, free-
space, peak gain of the full-wave dipole over the 1/2-wave dipole in
the NEC analysis I posted earlier today. Both of these configurations
radiate the same total energy (somewhere).

But along their axes of maximum radiation, the full-wave antenna
produces the greater field intensity of the two, for a given applied
(and matched) power source.

//

Art Unwin wrote:
There are no standing waves either

If there were no standing waves, a current pickup would
read a constant current when moved up and down the conductor,
but it doesn't. A current pickup proves there are standing
waves. You can see it with your own eyes using RF current
meters available from MFJ. If your theory rests on "no standing
waves" being present, it can easily be disproved.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com