Topband: Term "cancel fields", apology

[Guy Olinger K2AV]

Hi Tom,

You are very close, just a couple additional details.

First, NEC4 can add up the field vectors from the various wires in the
FCP and determine that in the far field, those fields add up to a big
ball of nulls.  I have to assume that, or why are we using it?

In the near field, given that the fields from one side of center
produce a net out of phase with the field summary from the other side,
there is a parallax in the near field distances to source, and the
fields no longer summarize down to a value appropriate to the NEC4 -30
dBi radiation max. There is enough remnant field in the dirt after
summary to allow the FCP to see a load.  But relative to some other
counterpoises in use, there is a lot less remnant field after summary
using the FCP.

So the current has to go way up to put the transmitter's full power
into the area of uncancelled field. Figure two in the PDF article
shows the results of field summary in the ground medium, point by
point. The solid line in the drawing is the uncancelled field after
summary down in the dirt underneath your "FCP dipole". The other two
traces are what some are calling crossed radials at 1/4 and 1/8 wave.
The N and S radials are tied together at the center, and the E and W
radials are tied together. The feedpoint is between the N/S center and
the E/W center.

An FCP is a warped concoction. Two FCP's in series is even more
warped. It's an "antenna" that can't antenna. Like trying to p**
through a hypodermic needle.  Yes, technically it's an orifice, yes,
something wet comes out, but clearly it's not intended to behave like
a hose.

An FCP dipole is a bunch of wire that is essentially prevented by the
folding from radiating in a far field way.  -30 dBi qualifies for
"essentially". The only thing left for it to do is heat up the wire or
the ground, if it's close enough. It SHOULD calculate inefficient!
But what we do in reality is use one FCP in series with a vertical
radiator of some sort.

The reason I have done in the past what you are describing with the
FCP dipole was to determine the level of ground coupling and monkey
around with the shape, to make it use more and more current to couple
the ground to dissipate the power.

In your model, what is the source impedance of your FCP dipole at 8
feet above ground?  I presume it's very low.

In the NEC4 model of the pair, over average ground, zero resistance
wire and 8 feet high I get -30 dBi, 1.51 ohms -j 263, and the current
is 31.5 amps if you drive it with 1500 watts.  Only one amp of that is
going to radiation in the far field.

Since the efficiency is radiated power/(radiated power + loss power),
the efficiency of the pair is EXTREMELY low, close to a tenth of a
percent, as you have already posted. But it has to run at 31.5 amps to
couple all that power into the ground, using TWO FCP's.

The two FCP are in series, and they are generally used singly. So a
single FCP will be 0.75 ohms at 44.7 amps for the same 1500 watts,
heating the ground, with the miniscule RF radiation at -30 dBi.

Getting to the real world and using #12 wire, we get 2.25 ohms - j130
for the single FCP.  Since nothing has changed except the wire
resistance, we now know that 0.75 ohms is due to coupling ground, and
1.5 ohms is due to the wire, made worse than the natural resistance of
0.29 ohms by the folding effect aka linear loading. ***In an FCP of
number 12 wire, there is more loss in the wire than in the ground.***
Why we go ahead and say use #12 is a separate discussion.

Now, to put it in perspective, connect the feed end of the FCP to a
resonant vertical wire with 35 ohms radiation resistance.

Just looking at the resistive components, the combo now has vertical
35 ohm + FCP 2.25 ohms = 37.25 ohms. The efficiency is 35/37.25 or
94%. We're not driving the FCP at 45 amps, we're driving it at six or
seven amps, and the current limiting factor is the vertical radiator.

With my up 90 out 105 inverted L over an FCP, with a drive R of 125
ohms, we have an efficiency of 125/127.25 or 98%.

73, Guy

On Wed, Aug 1, 2012 at 8:48 PM, Tom W8JI <w8ji at w8ji.com> wrote:
> Hi Guy,
>
>> This is a non-issue, Tom.
>
> I think it is a critical issue, because it demonstrates the difference
> between EM radiation and induction fields that only store and return energy
> to the system.
>
>> It stems from what I meant when I said "cancel fields".
>>
>> Apparently some considerable number, even the majority, think this
>> phrase means cancel fields entirely.  To others this means
>> cancellation is taking place at some level.
>
> Then the question arises, how much does it reduce fields?
>
> As far as I know, all we can do is redistribute those fields as evenly as
> possible over the widest area possible in the lossy media to reduce losses.
>
>> The FCP radiates at -30 dBi according to NEC4. That's really low for
>> an antenna element. But, it DOES radiate.
>
> But that has little to do with local losses in the soil.
>
>> According to NEC4, using area integration of field squared data from
>> the counterpoises only, excluding the vertical radiator, the FCP only
>> has 8.2% of the power loss in the dirt as a pair of 1/4 wave raised
>> radials.
>
> What do you think of this example using a lossless half wave dipole antenna
> over the same soil:
>
> Dipole with lossless wire:
>
> freespace 100% eff  and 72.3 ohm
> 45 feet 46% eff and 37.6 ohms
> 10 feet 6% efficiency and 54 ohms
>
> Comparing an FCP dipole:
>
> Freespace 99.2% eff 0.03 ohms
> 45 feet 20% eff and 0.04 ohms
> 10 feet 0.4% 1 ohm
>
> Here is my confusion:
>
> The theory  (correct me if I misunderstood it)  is folding the element
> reduces fields in the earth compared to spread-out elements, and
> concentrating the field in a small area also reduces earth losses. I
> understand the article to say reducing the area exposed to fields reduces
> loss.
>
> If that is the case, why is a full-size lossless-conductor dipole's
> efficiency higher (earth losses lower) than a dipole made from two right
> angle FCP's? Both have equal soils and both are lossless internally, so all
> losses are by fields in the earth.
>
> What I understood, and always found true, is spreading fields out over a
> larger cross section of lossy media reduced losses. This is because current
> and electric field density is less.
>
> Why does a 260 ft wide lossless dipole have less earth loss than a lossless
> compact folded antenna at the same height? What am I missing?
>
> 73 Tom
>
>
>
>
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