# [TowerTalk] Elevated vertical dipoles was Re: R7000

Tue Sep 16 09:36:14 EDT 2008

```jimlux wrote:
> Joe Subich, W4TV wrote:
>>> My test was with the capacitive hat (the short radial rods)
>>> at the lower end of the antenna about 12" off ground.
>> That's a rather rough test for the R7000 or any "no radial
>> vertical."  The radials are "hot" and couple rather heavily
>> to the soil.  Performance should improve significantly by
>> moving the base up to 15 feet or so (1/8 wave at the lowest
>> operating frequency).
>>
>>
>>
> I don't know that the fractional wavelength is the right way to think
> about the spacing.  I'd say it's more a function of the physical size of
> the radiator.  (without actually doing any analysis)
>
> My thinking runs along these lines..
>
> If you think about a physically small dipole or loop, what you're
> concerned about is how much of the near field interacts with the (lossy)
> soil. The near field is where the energy is stored in the antenna system
> (compared to that energy radiated away). A physically smaller antenna
> will have a smaller near field (i.e. higher absolute field in a smaller
> volume)..
>
>
<capacitor analogy snipped>
> Same for an antenna. (maybe... I'll have to think about it)
>

OK, now that the first cup of coffee is circulating... some more
thoughts, and it gets more complex.

Here's a vastly simplified model...

Consider a half wave dipole, say, for 30MHz, that's 5m long.  Further,
assume that the energy stored in the near field is entirely contained
within a bubble/sphere around the antenna, say, 10 meters in radius (2x
antenna length).. with this model, if the antenna is more than 10 meters
off the ground, there are no ground losses, and if it's closer, there's
more and more loss, the closer you get, as the bubble intersects the
earth. If the antenna were, say, 5 meters from the ground, the bottom
1/4 (by latitude, not volume) of the sphere is in the "lossy
ground"(obviously, there's not a hard boundary in real life)

Now, consider an antenna that's half as long (2.5m) with a lossless
matching network (so the energy going in is the same). If we use the
sphere with 2xlength radius model, the bubble is now 5m in radius.  So,
you could get it within 5m of the ground with no loss. And, if you got
to 2.5m, you'd have the same bottom of the 1/4 sphere intersecting the
ground and causing loss. That's my original thinking...

However, physically small antennas have relatively more energy stored in
their near field than larger antennas (at the same frequency).  In the
bubble analogy, I originally thought that the stored energy in the 5m
bubble is the same as the stored energy in the 10m bubble.  But that's
not really the case.  There will actually be MORE energy in the 5m
bubble than in the 10m bubble.

So, if you're in the 1/4 of the sphere in the ground regime, you're
actually going to lose more with the small antenna than the large one.
I suppose this is no different than the ohmic losses in the antenna
itself going up, because the IR losses go up in a small antenna. (for
exactly the same reason.. more energy in near field=more current in antenna)

It might be that it exactly balances out.. the higher energy density in
the small antenna means you have to be farther away before you get to
the same boundary flux.

Time to do some modeling..

Jim
```