Roy Lewallen wrote:

Have you tried doing this comre it before.

I ran this with a 15' 9" boom 6m beam at 20 feet over real high accuracy
ground, which is at a typical height for a height for a home TV antenna.

The vertically polarized version had a peak of 11.5 dBi at 9 degrees
while the horizontal had 16.0 at 13 degrees. At 9 degrees the
horizontal still had 15.0 dBi, clearly the winner.

Given that this was very close to the ground in terms of wavelength, it
may not be a good example.

tom
K0TAR

tom wrote:
Roy Lewallen wrote:

Have you tried doing this comre it before.

For the record, I did not write that. I always comre it after.

Roy Lewallen, W7EL

I ran this with a 15' 9" boom 6m beam at 20 feet over real high accuracy
ground, which is at a typical height for a height for a home TV antenna.

The vertically polarized version had a peak of 11.5 dBi at 9 degrees
while the horizontal had 16.0 at 13 degrees. At 9 degrees the
horizontal still had 15.0 dBi, clearly the winner.

Given that this was very close to the ground in terms of wavelength, it
may not be a good example.

tom
K0TAR

On Sat, 20 Jun 2009 16:13:07 -0400, wrote:

I switched to 527
because I can actually see a channel 23 transmitting antenna from my
window.

Hi Pat,

If you think this experience contradicts EZNEC (or conventional
teachings which it concurs with), then the confusion comes from the
sense of "very far away." Truly, if you can see the Channel 23
transmitting antenna, then it is not that far away in the scheme of
things. It merely points out that you have not correctly modeled your
experience.

For those who may not missed my original post, I find it hard
to believe a horizontal dipole tuned to the right frequency (near 1:1
SWR with 75 ohm source) would not be able to hear a signal coming from
zero degrees elevation. In the real world, there are all sorts of
reflections off of all sorts of things that will make it work, but is
it true that there should be no signal if everything was ideal?

What is going to be a reflector to a source that is bore-sight with
the horizon?

Richard explained the attenuation of the E-field. That makes sense to
me, but doesn't really explain the other nulls at 6 degrees elevation
and every 6 degrees above that.

You didn't ask about that.

There are strong positive lobes at 3
degrees and every 6 above that. The plot looks like a nice flower :-)
I would think that attentuation of the E-Field would explain zero
degrees, but as elevation increased, the attenuation would decrease.

You have, again, lost sight of the meaning of "very far away."

The EZNEC plot looks more like it is showing additive and subtractive
combining of the signal.

It is. What you see is called the Fresnel Zone if you were line of
sight.

Here, even if you can "see" the Channel 23 transmitting antenna, then
its various reflections could add up to ZERO. This, again, confounds
expectation, but it is the experience of every mobile operator who
encounters "picket fencing."

Another reply mentioned a different program
that calculated ground wave in addition to skywave. Maybe that is
what I am missing. I normally think of ground wave as why VLF, LF,
and MF signals travel further than line of sight, though. Does ground
wave have a significant effect at VHF/UHF?

Yes, it is dead within a mile for Channel 23.

I'm still confused,

and so are a number of your respondents.

73's
Richard Clark, KB7QHC

Roy Lewallen wrote:
tom wrote:
Roy Lewallen wrote:

Have you tried doing this comre it before.

For the record, I did not write that. I always comre it after.

Roy Lewallen, W7EL


I noticed that after I sent it. I have no idea what I may or may not
(this being Windows after all) have done to make that weird edit occur.

Sorry Roy.

tom
K0TAR

wrote:
Frank wrote:

NEC will calculate "Space wave plus surface wave" if required.

Frank
Have you tried doing this calculation with a horizontally polarized VHF
antenna? What did you find?

Roy Lewallen, W7EL

I'm not sure what you mean.

My comment was directed to Frank. I'm hoping he'll try running the
calculation, or has already. He'll discover that:

1. A horizontally polarized wave does not propagate via a surface wave, and
2. The attenuation of surface waves increases with frequency. At VHF,
it's essentially zero at any useful distance.

Either of these is adequate to explain the result he'll see, or has
seen, that the result with surface wave is the same as the result
without surface wave, for this situation.

Richard and Dave have answered your question -- the reason for the zero
field at zero elevation angle EZNEC result is that the model ground is
perfectly flat and infinite in extent, and the observation point is
essentially at an infinite distance. This is useful for evaluating sky
wave propagation at long distances, but not for line-of-sight
propagation where Earth curvature can be a factor. Free space analysis
is better for this. Programs which calculate reflection coefficients for
various ground shapes aren't very useful at VHF and above, because local
features, especially in an urban environment, cause reflections that are
often much more significant than the ones calculated by the program. All
you can do with these programs in a situation like yours is to get a
general idea of what's happening.

EZNEC seems to say that a horizontally
polarized dipole seems to have zero gain (-99.99DBi) at zero degrees
elevation regardless of the frequency. So far, I have only tried 14
(the 20 meter example that came with EZNEC) Mhz, 491 Mhz (TV channel
17 center), and 527 MHz (TV channel 23 center). I switched to 527
because I can actually see a channel 23 transmitting antenna from my
window. For those who may not missed my original post, I find it hard
to believe a horizontal dipole tuned to the right frequency (near 1:1
SWR with 75 ohm source) would not be able to hear a signal coming from
zero degrees elevation. In the real world, there are all sorts of
reflections off of all sorts of things that will make it work, but is
it true that there should be no signal if everything was ideal?

If by ideal you mean that the ground is perfectly flat and infinite in
extent, yes.

Richard explained the attenuation of the E-field. That makes sense to
me, but doesn't really explain the other nulls at 6 degrees elevation
and every 6 degrees above that. There are strong positive lobes at 3
degrees and every 6 above that. The plot looks like a nice flower :-)
I would think that attentuation of the E-Field would explain zero
degrees, but as elevation increased, the attenuation would decrease.
The EZNEC plot looks more like it is showing additive and subtractive
combining of the signal.

That's exactly what EZNEC is doing. It calculates the sum of two "rays"
-- idealized straight line radiation from the antenna. One is the
"direct ray", which goes directly from the antenna to the observation
point. (Actually, the fields from all segments are individually
calculated and summed.) The second is the "reflected ray", which
reflects from the ground plane to arrive at the same distant point. Some
geometric investigation will show that the reflected ray at an elevation
angle of -x degrees will add to the direct ray at an elevation angle of
+x degrees at the distant point. The reflection coefficient of the
ground, which is determined by the frequency, wave polarization, and
ground conductivity and permittivity, is used to determine the amplitude
and phase of the reflected ray. The reflection coefficient of low angle
horizontally polarized waves is nearly -1 for any reasonable ground
characteristics, so the low angle pattern is nearly the same for a
horizontally polarized antenna over the Earth as for one over a
perfectly conducting ground. If you look the the distance from the
antenna to a distant point via the direct route and compare it to the
distance to the same point via the reflected route, you'll see that the
difference between the two routes (that is, the distances traversed by
the two "rays") monotonically increases with increasing elevation angle.
At zero elevation angle, the two distances are the same. Since the
reflected ray undergoes a phase reversal (as Richard explained),
expressed as a -1 reflection coefficient, the direct and reflected rays
cancel at the distant point, resulting in zero field strength. As the
elevation angle increases, the difference between the two two ray paths
increases, resulting in imperfect cancellation. At some elevation angle,
assuming the antenna is high enough off the ground, the difference in
path lengths will be exactly a half wavelength. At this angle, the two
rays will reinforce at the distant point. At a higher angle, they'll be
exactly one wavelength different, and the rays will cancel again. And so
forth. The path distance difference goes from zero at zero elevation
angle to a maximum straight up of twice the antenna height above ground,
so the number of "flower petals" you see depends on the height, in
wavelengths, of the antenna above ground.

Another reply mentioned a different program
that calculated ground wave in addition to skywave. Maybe that is
what I am missing. I normally think of ground wave as why VLF, LF,
and MF signals travel further than line of sight, though. Does ground
wave have a significant effect at VHF/UHF?

No, it doesn't. Which is why I posted the query to Frank.

I'm still confused,
Pat, N8CQV

Hope this helps. There's additional information in the _ARRL Antenna
Book_ and other sources which explains how the patterns of antennas are
affected by ground.

Roy Lewallen, W7EL

Hi, Richard. Thanks for sticking with me throught this. I have added
comment and more questions below. 73, Pat

On Sat, 20 Jun 2009 17:08:25 -0700, Richard Clark
wrote:

On Sat, 20 Jun 2009 16:13:07 -0400, wrote:

I switched to 527
because I can actually see a channel 23 transmitting antenna from my
window.

Hi Pat,

If you think this experience contradicts EZNEC (or conventional
teachings which it concurs with), then the confusion comes from the
sense of "very far away." Truly, if you can see the Channel 23
transmitting antenna, then it is not that far away in the scheme of
things. It merely points out that you have not correctly modeled your
experience.
So,EZNEC models "very far away" and, in my example, the channel 23
tower is 5.1 miles away (about 10,000 wavelengths) so I need a
different model? A different program? I think you are correct in
saying that is my main confusion.


For those who may not missed my original post, I find it hard
to believe a horizontal dipole tuned to the right frequency (near 1:1
SWR with 75 ohm source) would not be able to hear a signal coming from
zero degrees elevation. In the real world, there are all sorts of
reflections off of all sorts of things that will make it work, but is
it true that there should be no signal if everything was ideal?

What is going to be a reflector to a source that is bore-sight with
the horizon?

Richard explained the attenuation of the E-field. That makes sense to
me, but doesn't really explain the other nulls at 6 degrees elevation
and every 6 degrees above that.

You didn't ask about that.
Sorry about that. The null at zero degrees is what surprised me so I
figured that if I understood it, the rest would make sense.

There are strong positive lobes at 3
degrees and every 6 above that. The plot looks like a nice flower :-)
I would think that attentuation of the E-Field would explain zero
degrees, but as elevation increased, the attenuation would decrease.

You have, again, lost sight of the meaning of "very far away."
I don't doubt that!

The EZNEC plot looks more like it is showing additive and subtractive
combining of the signal.

It is. What you see is called the Fresnel Zone if you were line of
sight.

Here, even if you can "see" the Channel 23 transmitting antenna, then
its various reflections could add up to ZERO. This, again, confounds
expectation, but it is the experience of every mobile operator who
encounters "picket fencing."
Good example. I started in this hobby 41 years ago on 6 meter AM.
However, I always thought that picket fencing was caused by
reflections from various objects (power lines, airplanes, metal
fences, water towers, etc,etc) rather than the radiation patterns of
the antennas.


Another reply mentioned a different program
that calculated ground wave in addition to skywave. Maybe that is
what I am missing. I normally think of ground wave as why VLF, LF,
and MF signals travel further than line of sight, though. Does ground
wave have a significant effect at VHF/UHF?

Yes, it is dead within a mile for Channel 23.

I'm still confused,

and so are a number of your respondents.

That may be true, but I appeciate them trying to help.

Since my last post, I changed the polarization of my EZNEC dipole to
virtical. I expected a nice donut shaped pattern, but instead saw
another flower shaped pattern with deep nulls at various elevations
including zero degrees. When I select free space, I get the donut.

I truely believe EZNEC gives valid results when provided with a proper
model. So, either I am not providing a good model to EZNEC (likely)
or a simple virtical dipole radiates very little at zero degrees
(-90dBi) and a lot at one degree (7.33 dBi), etc, etc (which seems
less likely to me).

73's
Richard Clark, KB7QHC

NEC will calculate "Space wave plus surface wave" if required.

Frank

Have you tried doing this calculation with a horizontally polarized VHF
antenna? What did you find?

Roy Lewallen, W7EL

As an example I tried the following:

CM Horizontal Dipole
CE
GW 1 11 0 0 3 0 1.5 3 0.01
GE 1 -1 0
GN 2 0 0 0 13.0000 0.0050
FR 0 1 0 0 100 1
EX 0 1 6 0 1 0
RP 1 1 360 0000 3 0 1.00000 1.00000 10000
EN

At 10,000 meters, 3 meters off the ground:
E(theta) is maximum off the ends at 3.5E-8 V/m
E(phi) is maximum off the sides at 2.6E-7 V/m

This agrees with the expected horizontal polarization off
the sides, and vertical polarization off the ends.

Frank, VE6CB

Frank wrote:
NEC will calculate "Space wave plus surface wave" if required.

Frank
Have you tried doing this calculation with a horizontally polarized VHF
antenna? What did you find?

Roy Lewallen, W7EL

As an example I tried the following:

CM Horizontal Dipole
CE
GW 1 11 0 0 3 0 1.5 3 0.01
GE 1 -1 0
GN 2 0 0 0 13.0000 0.0050
FR 0 1 0 0 100 1
EX 0 1 6 0 1 0
RP 1 1 360 0000 3 0 1.00000 1.00000 10000
EN

At 10,000 meters, 3 meters off the ground:
E(theta) is maximum off the ends at 3.5E-8 V/m
E(phi) is maximum off the sides at 2.6E-7 V/m

This agrees with the expected horizontal polarization off
the sides, and vertical polarization off the ends.

Frank, VE6CB

Now what's the result without the surface wave?

Roy Lewallen, W7EL

On Sat, 20 Jun 2009 17:33:19 -0700, Roy Lewallen
wrote:
snip
Richard and Dave have answered your question -- the reason for the zero
field at zero elevation angle EZNEC result is that the model ground is
perfectly flat and infinite in extent, and the observation point is
essentially at an infinite distance. This is useful for evaluating sky
wave propagation at long distances, but not for line-of-sight
propagation where Earth curvature can be a factor. Free space analysis
is better for this. Programs which calculate reflection coefficients for
various ground shapes aren't very useful at VHF and above, because local
features, especially in an urban environment, cause reflections that are
often much more significant than the ones calculated by the program. All
you can do with these programs in a situation like yours is to get a
general idea of what's happening.

I think I understand now.

snip
but is it true that there should be no signal if everything was ideal?

If by ideal you mean that the ground is perfectly flat and infinite in
extent, yes.
Got it

snip
The EZNEC plot looks more like it is showing additive and subtractive
combining of the signal.

That's exactly what EZNEC is doing. It calculates the sum of two "rays"
-- idealized straight line radiation from the antenna. One is the
"direct ray", which goes directly from the antenna to the observation
point. (Actually, the fields from all segments are individually
calculated and summed.) The second is the "reflected ray", which
reflects from the ground plane to arrive at the same distant point. Some
geometric investigation will show that the reflected ray at an elevation
angle of -x degrees will add to the direct ray at an elevation angle of
+x degrees at the distant point. The reflection coefficient of the
ground, which is determined by the frequency, wave polarization, and
ground conductivity and permittivity, is used to determine the amplitude
and phase of the reflected ray. The reflection coefficient of low angle
horizontally polarized waves is nearly -1 for any reasonable ground
characteristics, so the low angle pattern is nearly the same for a
horizontally polarized antenna over the Earth as for one over a
perfectly conducting ground. If you look the the distance from the
antenna to a distant point via the direct route and compare it to the
distance to the same point via the reflected route, you'll see that the
difference between the two routes (that is, the distances traversed by
the two "rays") monotonically increases with increasing elevation angle.
At zero elevation angle, the two distances are the same. Since the
reflected ray undergoes a phase reversal (as Richard explained),
expressed as a -1 reflection coefficient, the direct and reflected rays
cancel at the distant point, resulting in zero field strength. As the
elevation angle increases, the difference between the two two ray paths
increases, resulting in imperfect cancellation. At some elevation angle,
assuming the antenna is high enough off the ground, the difference in
path lengths will be exactly a half wavelength. At this angle, the two
rays will reinforce at the distant point. At a higher angle, they'll be
exactly one wavelength different, and the rays will cancel again. And so
forth. The path distance difference goes from zero at zero elevation
angle to a maximum straight up of twice the antenna height above ground,
so the number of "flower petals" you see depends on the height, in
wavelengths, of the antenna above ground.
I noticed that. It makes sense now.
snip


Hope this helps.
Roy Lewallen, W7EL
It did. Richards and the others helped a lot, but your explanation
made the light bulb turn on ;-) It also explains why I saw a similar
result with a virtical dipole.

Thanks again to all who helped.

73,
Pat

At 10,000 meters, 3 meters off the ground:
E(theta) is maximum off the ends at 3.5E-8 V/m
E(phi) is maximum off the sides at 2.6E-7 V/m

This agrees with the expected horizontal polarization off
the sides, and vertical polarization off the ends.

Frank, VE6CB

Now what's the result without the surface wave?

Roy Lewallen, W7EL

Good point: E(theta) off the ends 1.25E-7 V/m
E(phi) off the sides 1.04E-7 V/m

Frank