Topband: FCP

[Tom W8JI]

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

I posted the values. With zero loss wire real part of impedance impeance is 
less than 1 ohm.

> 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.

I believe the way to do this in a model is to use lossless wire. With zero 
loss wire, all losses are caused by fields in the earth, which are either 
E-M radiatioon (very weak force near the antenna, but does not decay other 
than spreading so it goes a long distance), E field (very strong with high 
voltages) and H field (very strong with high current).

We do not want to worry about ratio of loss to radiation resistances in the 
radiator itself, just ground coupling. This is because loss in the radiator 
only confuses the issue.

> 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.

That formula is published all over the world, but is absolutely 
inappropriate for calculation of efficiency except under very special 
conditions. The special conditions where that formula is useful are not met 
in this case, just as they are not met in many other real-world cases.

That's the same formula that consistently gets people in trouble with the 
folded unipole and folded dipole, or when they move mobile antennas around 
on a car and watch base impedance, or add radials and watch base impedance 
to determine efficiency.

If you took the FCP and enclosed it in a very large metallic sphere with no 
loss and measured resistance, all resistances at any point would be due to 
loss in the antenna. If the dome was removed the new value of resistance 
would indicate the contribution of radiation.

The problem is that formula is only reliable when we know where power goes 
after it leaves the antenna, and what resistance at one point in the system 
would be if we stopped all radiation (with a sphere or "Wheeler cap") or 
allowed it to happen.

Since we don't know what portion of applied power heats the earth and 
radiates, BOTH of which change resistance from the conductor loss only case, 
we don't know earth losses.

If anyone doubts the worthlessness of comparing what we call radiation 
resistance to measured resistance at the feedpoint, all they need do is put 
up various radial systems while measuring feed resistance and field 
strength. They will find feed resistance can move all over the place, with a 
fixed radiator and only a ground system change, and field strength changes 
independently of feed resistance, when only the ground system is changed.

Anyone relying on changes in feed resistance to determine efficiency, when 
earth is involved, is kidding himself. This includes evaluating radials or 
ground systems for loss.

So we are left with two things:

1.) We can compare efficiencies in lossless wires at various heights by 
using a model. This does not involve feed resistance. It involves total 
field.

2.) We can measure relative field strength against a *known* reference 
antenna.

>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%.

No, is not. We don't know what it is. Application of a misplaced formula to 
the system will not tell us ground loss, just as changing a ground system 
and watching base impedance will not tell us ground loss.

I can have a system with a vertical and a ground system producing 60 ohms 
feed impedance that has higher efficiency than the very same antenna with a 
new ground producing 37 ohms at the feedpoint.

If we model a zero wire loss dipole and look at "average gain", they will 
find average gain of all lengths are unity in freespace. If we place that 
antenna over earth any change is average gain is 100% from earth losses.

What we find is shorter dipoles are more height-sensitive to earth losses. 
This is because E and H fields are more concentrated in the shorter antenna. 
For example, here are lossless dipoles at ten feet:

1/4 wave long lossless dipole freespace 100% eff 12.7 ohms
45 feet high 39.2% eff  7.6 ohms
10 feet high 3.3% eff  17.8 ohm

1/2 wave dipole 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

Shorter wires are more sensitive to earth proximity and characteristics, and 
the FCP appears to not be exempted from this trend. Also note feedpoint 
resistance does not track earth loses.

My contention is there isn't any magic, other than spreading fields evenly 
over the largest area of lossy media possible. The only way to know field 
strength change is to actually measure field strength. As a matter of fact, 
an FCP dipole ground loss closely compares to a regular dipole the same 
effective length. Farfield radiation cancellation has little to do with 
nearfield loss, and the same is true for radiation resistance.

Changing a ground system can reduce loss, but it is largely because someone 
was doing something wrong with the space available. They somehow 
concentrated E or H fields, or did things to directly conduct current into 
soil at one small point.

73 Tom