Cecil Moore wrote:
There are certain special-case antennas where folding occurs
without introducing destructive interference, e.g. a 1/2WL
folded dipole or a full-wave quad where the wires are a

Cecil, you should know by now that a half wave dipole of any type
couldn't be all that efficient or effective. Art says so.

tom
K0TAR

tom wrote:
Cecil, you should know by now that a half wave dipole of any type
couldn't be all that efficient or effective. Art says so.

Art might be quick to point out that there is one
wavelength of wire in a 1/2WL folded dipole. :-)
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

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

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

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

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.

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.

//