Topband: K2AV Counterpoise

[k8bhz at alphacomm.net]

Hi Dan,

I can see where you've reached your conclusion, but your comment that 
"Rudy seems to say..." is a good qualifier.

Let's get back to basics, say with a classic 1/4 WL vertical. The 
theoretical impedance against an in infinite ground is 36 ohms. From 
free space modeling, one quickly realizes that this is nothing more half 
of a free space dipole, which it is. Next, power being fed into the base 
is simply: P = I squared X R radiation, at resonance. Very simply, more 
current means more power. Now with any radial system, screen, or ground 
rod, or counterpoise, that current into the base HAS to come from the 
ground system. There can be NO discontinuity here...it's not 
electrically possible.

Now let's look at the ELECTRICAL length of a radial. The one given is 
that the far end is open-circuited, I think we all agree so far...That 
means that for standing wave purposes, current here must be zero (the 
wire stops!), so impedance is highest. At the antenna end of the radial, 
impedance reaches a minimum at 1/4 ELECTRICAL WL, and increases back to 
the reflected maximum at 1/2 ELECTRICAL W/L. Notice please that 
ELECTRICAL WL is extended by Vp, same as shortening a PHYSICAL piece of 
coax to make a phasing line or tuned stub. To recap this, the impedance 
presented at the antenna end of a radial varies from a minimum at 1/4 
ELECTRICAL WL to a maximum at 1/2 ELECTRICAL WL. (And multiples of these 
lengths, of course).

The error of your conclusion comes from the fact that Rudy is comparing 
equal currents INTO the radials, Which means that the POWER into the 
antenna is NOT constant. POWER is at a minimum to achieve the fixed 
current value at 1/4 WL, and goes to a maximum at 1/2 WL. A verification 
is when he increases radial length and the current maximum becomes 
greater than at the base of the antenna....more power into the antenna, 
more power in the radial standing wave current peak, and more losses.

In our real world, our transmitters/amplifiers are fixed in power 
output, not infinitely variable, so as the combined vertical/ground 
system impedance goes up, current decreases. Recall, radiated power is P 
= I squared X R radiation. This obviously reaches a minimum as the 
impedance hits it's maximum, at 1/2 ELECTRICAL WL. Rudy's 
summations/conclusions reflect this.

Sorry for being so long-winded.

Brian  K8BHZ


On 10/26/2017 11:47 PM, Dan Maguire via Topband wrote:
> K8BHZ wrote:
>>>> The length to avoid is nothing more than a half wavelength, which translates the same impedance from end to end
>>>> i.e., the high Z open end translates to a high Z antenna base end. This results in minimum radial current.
> I'm not so sure I buy that and I don't think N6LF does either.  If you look at the section "An Explanation for the Dips in Ga" (Part 1, QEX pg 40) in Rudy's document
>
> http://rudys.typepad.com/files/qex-mar-apr-2012.pdf
>
> you'll find this: ["L" is the variable for radial length]
>
> <quote N6LF>
> ... For the same current at the feed point, with longer radials the currents are much higher as we go out from the base. We would expect these higher currents to increase both E and H-field intensities at ground level under the radials. ... Since the power dissipation in the soil will vary with the square of the field intensity, it’s pretty clear why the efficiency takes such a large dip when the radials are too long.
> </quote>
>
> So Rudy seems to be saying that the increased loss is due to higher radial currents, not minimum radial currents.
     Below is a different animation, this time showing E-field intensity 
as the radial length changes. For this animation the radial height was 
10 ft so the dip occurs at ~0.45 WL (frame 9). That corresponds to the 
highest radial current and the max E-field.
>
>
> Dan, AC6LA