RadioBanter - View Single Post - Aerial grounding and QRM pick-up: theory & practice

Walter Maxwell · September 23rd 07, 04:36 AM posted to rec.radio.amateur.antenna

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?

The only difference is in the SHAPE of the radiation patterns--the full-wave dipole will have a somewhat
narrower lobe in the direction broadside to the dipole than that of the half-wave dipole, therefore deriving
slightly more gain IN THAT DIRECTION than that of the half-wave, but with less gain than the half-wave in all
other directions. Consequently, the total integrated power in either radiation pattern will be exactly the
same!!!

If you want to express the mathematics of the conditions I described here according to J.C.Maxwell's
equations, you will find that Maxwell's equations fit the conditions EXACTLY.

Walt, W2DU