high earth resistance
 

Owen Duffy wrote in news:Xns990EA1652F44Enonenowhere@
61.9.191.5:

So, what do you think its impedance would be? 1500+j??

Duh, ??+j1500 is more like it.

Same outcome, it is probably accounting for 5 to 15 ohms of resistance that
you haven't included in your calcs.

Owen

high earth resistance
 

On Mon, 09 Apr 2007 17:04:35 -0700, Jim Lux
wrote:

The interesting question would be whether this is important at all..

Hi Jim,

Additional current through earth brings no net positive result and the
question asked where the source of loss resides.

Being unable to quantify temperature is no reason to keep picking up
the wrong end of a soldering iron. "I don't believe it's hot" has
rarely offered salve for burns. ;-)

73's
Richard Clark, KB7QHC

high earth resistance
 

Owen,

Are you referring to the coil alone? You are correct the coil accounts
for a significant part, probably between 4 and 8 Ohms.

I am measuring real values between 52 and 56 Ohms with 0j. This varies
mainly with moisture.

So let's say the total antenna is 12 Ohms. If the measured value is 52
that says the ground is 40. This puts us back to the original question:
why is the ground so high?

- Dan

The circuit

Owen Duffy wrote:
Owen Duffy wrote in news:Xns990EA1652F44Enonenowhere@
61.9.191.5:

So, what do you think its impedance would be? 1500+j??

Duh, ??+j1500 is more like it.

Same outcome, it is probably accounting for 5 to 15 ohms of resistance that
you haven't included in your calcs.

Owen

high earth resistance
 

dansawyeror wrote in
:

Owen,

Are you referring to the coil alone? You are correct the coil accounts
for a significant part, probably between 4 and 8 Ohms.

I don't recall that you have told us where the coil is located, and I am
not going to build a model to discover something that is relevant and
that you could have told us.

You have told us the coil had an "impedance" of 60uh. If it had an
inductance of 60uh, it would have an impedance of 1500/Q+j1500 where Q at
3.9MHz for a practical coil is likely to be above 100 and less than 300.
The resistive component of the coil has to be referred to the feedpoint
so that you can deduct it from the total feedpoint Z. The best way to do
that is your NEC model.

You should be able to form a better range for the equivalent coil loss at
the feedpoint than you stated above (since you seem to have the means of
measuring the coil and inserting the values in your model).

I am measuring real values between 52 and 56 Ohms with 0j. This varies
mainly with moisture.

So let's say the total antenna is 12 Ohms. If the measured value is 52
that says the ground is 40. This puts us back to the original
question: why is the ground so high?

You also previously said "... Even if the antenna were 6 to 8 Ohms the
ground loss would be at least 42 to 44 Ohms."

I read this to mean total system R is 50 ohms.

In this post it is reported between 52 and 56.

The other issue that Frank raised is the elevated feedpoint and whether
you modelled that correctly. The radiation resistance you quoted seems
(without checking) reasonable for a short monopole over ideal ground, but
one expects it would be higher for an elevated feed point. Have you
modelled the antenna you built, or have you build an antenna you cannot
model accurately and are applying model results incorrectly to the thing
you have built?

Though you see only one question, "why is ground 40 ohms", you haven't
disclosed enough information in your posts to convince me that it is 40
ohms. If you ask the wrong question, you might not get a useful answer.

Is ground 40 ohms?

Owen

high earth resistance
 

I believe everything is stated:

The antenna is a short loaded vertical. The base is about 1.1

meters long and 90mm in diameter, the coil is about 160mm long

and about 80mm in diameter, the top is 3 meters. The coil wire

is 12 gage and the spacing is about .5. As a model cross check, t

he impedance of the coil measures about 60 uH.

The order is base, coil, and top.

- Dan


Owen Duffy wrote:
dansawyeror wrote in
:

Owen,

Are you referring to the coil alone? You are correct the coil accounts
for a significant part, probably between 4 and 8 Ohms.

I don't recall that you have told us where the coil is located, and I am
not going to build a model to discover something that is relevant and
that you could have told us.

You have told us the coil had an "impedance" of 60uh. If it had an
inductance of 60uh, it would have an impedance of 1500/Q+j1500 where Q at
3.9MHz for a practical coil is likely to be above 100 and less than 300.
The resistive component of the coil has to be referred to the feedpoint
so that you can deduct it from the total feedpoint Z. The best way to do
that is your NEC model.

You should be able to form a better range for the equivalent coil loss at
the feedpoint than you stated above (since you seem to have the means of
measuring the coil and inserting the values in your model).

I am measuring real values between 52 and 56 Ohms with 0j. This varies
mainly with moisture.

So let's say the total antenna is 12 Ohms. If the measured value is 52
that says the ground is 40. This puts us back to the original
question: why is the ground so high?

You also previously said "... Even if the antenna were 6 to 8 Ohms the
ground loss would be at least 42 to 44 Ohms."

I read this to mean total system R is 50 ohms.

In this post it is reported between 52 and 56.

The other issue that Frank raised is the elevated feedpoint and whether
you modelled that correctly. The radiation resistance you quoted seems
(without checking) reasonable for a short monopole over ideal ground, but
one expects it would be higher for an elevated feed point. Have you
modelled the antenna you built, or have you build an antenna you cannot
model accurately and are applying model results incorrectly to the thing
you have built?

Though you see only one question, "why is ground 40 ohms", you haven't
disclosed enough information in your posts to convince me that it is 40
ohms. If you ask the wrong question, you might not get a useful answer.

Is ground 40 ohms?

Owen

high earth resistance
 

dansawyeror wrote in
:

The order is base, coil, and top.

That wasn't clear to me, it looked to me like you were feeding it between
the base and the top. I think Frank may have formed the same view.

Owen

high earth resistance
 

It's feed at the bottom.

The coil description should be inductance not impedance.

- Dan

Owen Duffy wrote:
dansawyeror wrote in
:

The order is base, coil, and top.

That wasn't clear to me, it looked to me like you were feeding it between
the base and the top. I think Frank may have formed the same view.

Owen

high earth resistance
 

On Mon, 09 Apr 2007 16:42:10 -0700, Roy Lewallen wrote:


Brown, Lewis and Epstein were REPORTING, not inventing, nor offering
pedant readings of scripture. Scribes, such as we are, are free to
interpret within the bounds of their own data, assumptions, and
conclusions. I've offered mine that conforms to many of their points.
If you have your own, you must survive by the same strictures. Given
the specific contention, I am especially intrigued in how you would
answer why the current departed the wire, and where it goes in light
of a potential map created by the phase shifts.

73's
Richard Clark, KB7QHC

Ok, I understand that. Your answers to the two questions I asked are
that you don't know and you don't know. In the absence of any evidence,
I'll continue to disbelieve there's a circumferential component of the
current.

Roy Lewallen, W7EL

Roy, it seems to me everyone has missed an important point concerning a circumferential component of the
current.

We know that the current flowing on the radial wires is radial in direction. What seems to be missed is the
current that returns to earth between the wire radials. That current is going to flow in the direction of the
lowest resistance. As such it's not going to flow radially alongside the currents flowing on the wire, because
the radial resistance of earth between the radial wires is much greater than the resistance of the wires.
Consequently, currents reaching earth between the wires will find a lower resistance by traveling toward the
nearest radial wire instead of continuing in a perfectly radial direction. This new direction of current flow
will not necessarily perfectly circumferential, but will certainly be somewhere between radial and
circumferential.

Walt, W2DU

high earth resistance
 

On Thu, 12 Apr 2007 18:55:30 GMT, Walter Maxwell
wrote:

This new direction of current flow
will not necessarily perfectly circumferential, but will certainly be somewhere between radial and
circumferential.

Hi Walt,

So as BL&H report without too much pain:
at page 760:
"When the earth is of good conductivity,
the current leaves the wires and enters the earth closer to
the antenna than it does when the earth is a poor conductor."

Now, as to your comment
That current is going to flow in the direction of the
lowest resistance.

It is awfully damned hard to beat the least resistance path of copper
over earth. And yet BL&H offer us this observation I requote above.

What will trump a higher resistance path is greater potential
difference and proximity. Note that BL&H are quite specific about
proximity to the antenna, and hence it follows that the separation
between radials is closer there, than further out from the antenna.
Certainly I can find no where to quote this observation of growing
closeness from BL&H for Roy's consideration, but I trust my common
sense of geometry here too, and I will proceed.

BL&H report (without going into the how, or how much):
"From (8) [that formula] we see that the earth
current proper leads the current in the wires
by 90 electrical degrees."
such that at "that" radial distance, there must exist the greatest
circumferential potential difference between the wire and the earth
currents which is clearly mandated by phase. If a potential gradient
along the circumference is greater than that along the radial, and the
distance along the circumference is smaller than the distance along
the radial; then it stands to reason why BL&H even offer to comment
"current leaves the wire."

Current through earth is largely lost to heat although I do have a
fractal antenna that uses earth current to optimize its low angle
launch characteristics.

Ultimately this reduces to the rather pedestrian observation that more
radials closer in serve efficiency - observed and reported by BL&H.

73's
Richard Clark, KB7QHC

high earth resistance
 

Walter Maxwell wrote:

Roy, it seems to me everyone has missed an important point concerning a circumferential component of the
current.

We know that the current flowing on the radial wires is radial in direction. What seems to be missed is the
current that returns to earth between the wire radials. That current is going to flow in the direction of the
lowest resistance. As such it's not going to flow radially alongside the currents flowing on the wire, because
the radial resistance of earth between the radial wires is much greater than the resistance of the wires.
Consequently, currents reaching earth between the wires will find a lower resistance by traveling toward the
nearest radial wire instead of continuing in a perfectly radial direction. This new direction of current flow
will not necessarily perfectly circumferential, but will certainly be somewhere between radial and
circumferential.

Walt,

I hadn't missed that phenomenon, but didn't mention it because it
doesn't produce a circumferential current. If you look at the current
flowing from the earth to each radial wire, you'll see that the sum of
these currents will be purely radial, assuming that the system is
symmetrical, i.e., radials are equally spaced and equal length, the
ground is homogeneous, and the radiator is vertical. Consider a bit of
current returning between two radials, which is a little closer to the
radial on the right. It'll detour to the right, giving it a rightward
component as well as an inward radial component. But for every such bit
of current, there's another one the same distance from the radial to the
left which will have leftward and inward radial components. The radial
components are in the same direction (inward) so will add but the
circumferential ones (leftward and rightward) cancel, leaving a net
radial current flow. You can say that the returning currents bend to the
right or left as they propagate toward the antenna base, but not that
there's a systematic circumferential current flow -- no current crosses
from radial to radial in a clockwise or counterclockwise circular
pattern like Richard implied.

I recall reading a paper which showed that connecting radials with
circumferential wires actually degrades a ground system's effectiveness,
but I wasn't able to lay my hand on it when I looked.

Roy Lewallen, W7EL