On 22 Sep, 20:36, Walter Maxwell wrote:
On Sat, 22 Sep 2007 19:19:18 -0700, wrote:
On Sep 22, 5:40 pm, art wrote:

O.K. I may have muddied things. I hold to the fact that a one
wavelength dipole will always radiate at a higher efficiency than a
1/2 wave dipole.

If it does, I doubt it's enough to measure on the air..

The example I gave as for an instance was a quad
versus a 1/2 wave dipole.
This is readily seen by any operator empirically.

I've never seen it here. For that reason, I hardly use loops.
Neither vertically oriented, or horizontal as for NVIS use.
I don't see them as worth the extra trouble. Being I tested
them on 75m using NVIS paths, a noticable difference in
efficiency should have been readily apparent. It wasn't.
In fact, I usually has slightly better performance using the
dipoles, which I think was due to the bulk of the max current
portions of the antenna being higher above ground in general.
The loop sagged a bit in areas, and wasn't all that high above
ground. The more wire near the ground, the more ground loss
in general.

Mathematically it is
proven that way also even tho both are in accordance to Maxwell's
laws.

Where is the math? You should find a very slight difference
at best..
It's common knowledge that even a short piece of wire 1/10
of a wave long will radiate nearly all the power that is applied
to it.
You can go lots shorter than that if you want.
If even a short piece of wire will radiate nearly all the power
applied to it, what is the point on harping about some magical
properties of a full wave length of wire?
Art, you are starting to bark at the moon I'm afraid...

I was going to comment on some of your other posts, but I
think I'll spare you the increase in blood pressure.
All I can say is that you are starting to wander off in
mumbo jumbo land again..
Replacing known science with conjured mumbo jumbo is no
way to live.
MK

Art, it distresses me to read the misleading statements you profess to be true in your posts.

There is no difference in the 'efficiencies' between a full-wave and a half-wave dipole. Let's assume the wire
size and conductivity of each dipole is such that we can say they both radiate 98 percent of the power
delivered to them. Let's also say that the same amount of power is delivered to both dipoles. What now is the
difference in the radiation between the two dipoles?

snip

Walt. I have no problem with Maxwells laws but I do have a problem
with a mathematical stunt to measure radiation
of a half wave dipole based on one having to accept that at all points
on the radiator the current is sinosoidal.
There is no distinct analysis with the specifics of radiation. period.
Using a parallel circuity ala tank circuit
insights are produced that radiation is created by the shorted energy
containers of capacitance and inductance which is an intrinsic part of
any radiator in distributed form. The tank circuit is well documented.
So the question now becomes, at least for full understanding of
radiation is how does a half wave radiator follow the format shown by
a
spark plug with a flyback transformer or a full wave radiator or even
the blast from a nuclear bomb?
Yes, a half wave antenna also has distributed inductance and
capacitance energy containers but how is the mechanism
shown by the tank circuit implimented when the arrangement is not in
equilibrium?
Now I think I know what happens but I am interested in contrary
thoughts from those skilled in the arts without the
retoric. As an aside, why does computor programs drift away from
planar form radiators as well to full wave radiators
when the computor is asked to compute for maximum gain? Is this like
the quadratic equation with four answers where you get to determine
the imaginary answers? I would suggest for starters those that are
skilled in the art quantify the energy used for end effect radiation
either in the normal atmosphere or in the rarified atmosphere of Quito
Equador when Tesla type emmissions are visable? Without assumptions
ofcourse!!!!

Regards
Art