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Horizontal Dipole - zero degrees elevation
Hello all you antenna experts,
I have been a ham and an engineer for a long time, but I have never delved into antenna theory. So, consider me a newbie for this question. I recently was helping a friend set up a crude antenna to connect to his DTV converter box following the June 12th conversion to DTV. I explained that a dipole was a simple antenna that could be used. He was very interested in the subject so the conversation soon turned to theory. I explained the 468/freq formula and eventually mentioned EZNEC. I have never used EZNEC myself so I downloaded the demo version and now have all of 24 hours experience with it. I started with the included backyard 20 meter dipole. I was surprised that there was no radiation toward the horizon. I figured that I was too close to the ground so I changed the frequency to 491 MHz (RF TV channel 17) and shortened the dipole accordingly. Still no zero degree radiation. I raised the dipole to 100' - then 1000' - then 10000' - still no radiation at zero degree elevation. I then found this in the help section: ----------------------- Because the far field sky wave from a horizontally polarized source is zero at a zero elevation angle for any ground type, and a vertically polarized source produces zero sky wave for any finite-conductivity ground, attempts to calculate a 2D pattern without the ground wave component under these conditions will result in an error message. ----------------------- Sure enough, changing to free space instead of a real ground changed the pattern to what I would have expected. I would have thought being many wavelengths above ground would be just as good as free space, but EZNEC doesn't think so. Am I missing something? Does a horizontal dipole really have a problem seeing a broadcast TV transmitter out on the horizon? Thanks. ...Pat |
Horizontal Dipole - zero degrees elevation
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Horizontal Dipole - zero degrees elevation
The horizontally polarized transmission has its E-Field parallel to
earth. Earth is a conductor (albeit a poor one, but in comparison to free space, it is quite a short circuit). That E-Field's two potentials are being laid across that conductor during the wave propagation to that far point where EZNEC then sums up all field contributions to present you with the lobe characteristic. It stands to reason that at that great distance, the wave will have attenuated considerably - hence the low value. Removing the short circuit (going to free space) removes this attenuation. 73's Richard Clark, KB7QHC NEC will calculate "Space wave plus surface wave" if required. Frank |
Horizontal Dipole - zero degrees elevation
Frank wrote:
The horizontally polarized transmission has its E-Field parallel to earth. Earth is a conductor (albeit a poor one, but in comparison to free space, it is quite a short circuit). That E-Field's two potentials are being laid across that conductor during the wave propagation to that far point where EZNEC then sums up all field contributions to present you with the lobe characteristic. It stands to reason that at that great distance, the wave will have attenuated considerably - hence the low value. Removing the short circuit (going to free space) removes this attenuation. 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 |
Horizontal Dipole - zero degrees elevation
On Sat, 20 Jun 2009 12:28:30 -0700, Roy Lewallen
wrote: Frank wrote: The horizontally polarized transmission has its E-Field parallel to earth. Earth is a conductor (albeit a poor one, but in comparison to free space, it is quite a short circuit). That E-Field's two potentials are being laid across that conductor during the wave propagation to that far point where EZNEC then sums up all field contributions to present you with the lobe characteristic. It stands to reason that at that great distance, the wave will have attenuated considerably - hence the low value. Removing the short circuit (going to free space) removes this attenuation. 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 I'm not sure what you mean. 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? 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. 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? I'm still confused, Pat, N8CQV |
Horizontal Dipole - zero degrees elevation
wrote in message ... On Sat, 20 Jun 2009 12:28:30 -0700, Roy Lewallen wrote: Frank wrote: The horizontally polarized transmission has its E-Field parallel to earth. Earth is a conductor (albeit a poor one, but in comparison to free space, it is quite a short circuit). That E-Field's two potentials are being laid across that conductor during the wave propagation to that far point where EZNEC then sums up all field contributions to present you with the lobe characteristic. It stands to reason that at that great distance, the wave will have attenuated considerably - hence the low value. Removing the short circuit (going to free space) removes this attenuation. 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 I'm not sure what you mean. 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? 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. 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? I'm still confused, Pat, N8CQV what you are missing is the 'real world'. eznec is probably modeling over a perfectly flat infinite surface. In the far field in a perfect world the signal along the surface is a combination of ground wave and sky wave, the ground wave decays rapidly with distance leaving the sky wave which will always be very small along the surface. now remember, the frame of reference is at ground level, not the antenna height, so zero degrees is along the infinite flat surface. And there is nothing in there that models where the other antenna is... it just creates a picture of how the strength of the fields are at a given elevation/azimuth angle from the reference point. Now, in the real world... the ground is never level, even on the ocean where it may look flat it curves down in every direction... so the horizon is not at zero degrees for any antenna above ground level, over flat ground its always below horizontal, on a hill or from a tower its even more negative, and in a valley it can be way above horizontal. both of those cause those predicted patterns to be changed a bit. Also, if the other antenna is not at ground level then it is at some positive angle above horizontal... or if its far enough away maybe a negative angle. You should also note that many broadcast antennas for TV and FM are designed with a tilt to send the signal down toward the ground, especially if they are on high hills or big towers, they would have weak signals near the tower if they didn't tilt it down... of course they don't care about beyond the horizon stuff anyway. to confuse things even more on vhf/uhf frequencies signals are easily reflected from hills, buildings, and other objects... they are also bent by changes in air temperature and humidity. So in many cases a vhf/uhf signal sent out toward the horizon may get bent down toward the ground and go beyond the horizon by quite a distance, see 'tropospheric ducting' for more info. hf signals of course get refracted back down toward the ground by the ionosphere, so for very distant stations the arrival angle can still be quite high, and almost never straight from the horizon. |
Horizontal Dipole - zero degrees elevation
wrote in message ... On Sat, 20 Jun 2009 12:28:30 -0700, Roy Lewallen wrote: Frank wrote: The horizontally polarized transmission has its E-Field parallel to earth. Earth is a conductor (albeit a poor one, but in comparison to free space, it is quite a short circuit). That E-Field's two potentials are being laid across that conductor during the wave propagation to that far point where EZNEC then sums up all field contributions to present you with the lobe characteristic. It stands to reason that at that great distance, the wave will have attenuated considerably - hence the low value. Removing the short circuit (going to free space) removes this attenuation. 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 I'm not sure what you mean. 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? 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. 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? I'm still confused, Pat, N8CQV p.s. you want to see more real world get one of the terrain analysis programs that lets you place real antennas over real non-flat ground and see where the signals really go. |
Horizontal Dipole - zero degrees elevation
On Sat, 20 Jun 2009 21:13:51 GMT, "Dave" wrote:
what you are missing is the 'real world'. eznec is probably modeling over a perfectly flat infinite surface. In the far field in a perfect world the signal along the surface is a combination of ground wave and sky wave, the ground wave decays rapidly with distance leaving the sky wave which will always be very small along the surface. now remember, the frame of reference is at ground level, not the antenna height, so zero degrees is along the infinite flat surface. And there is nothing in there that models where the other antenna is... it just creates a picture of how the strength of the fields are at a given elevation/azimuth angle from the reference point. Dave, Your paragraph above helped. For VHF and above, in the real world, am I better off using EZNEC's "free space" setting instead of real ground? I know at HF frequencies, where antennas are often close to the ground, it makes a big difference, but could free space be a better approximation of VHF antenna many wavelengths off the ground? Pat |
Horizontal Dipole - zero degrees elevation
wrote in message ... On Sat, 20 Jun 2009 21:13:51 GMT, "Dave" wrote: what you are missing is the 'real world'. eznec is probably modeling over a perfectly flat infinite surface. In the far field in a perfect world the signal along the surface is a combination of ground wave and sky wave, the ground wave decays rapidly with distance leaving the sky wave which will always be very small along the surface. now remember, the frame of reference is at ground level, not the antenna height, so zero degrees is along the infinite flat surface. And there is nothing in there that models where the other antenna is... it just creates a picture of how the strength of the fields are at a given elevation/azimuth angle from the reference point. Dave, Your paragraph above helped. For VHF and above, in the real world, am I better off using EZNEC's "free space" setting instead of real ground? I know at HF frequencies, where antennas are often close to the ground, it makes a big difference, but could free space be a better approximation of VHF antenna many wavelengths off the ground? Pat yes, it will probably be a better approximation, especially if the two antennas are in sight of each other. the ground effect is mostly appropriate for hf and at very long distance. vhf has many other effects that cause reflections and ducting that kind of over ride the ground image model that most modeling programs use. |
Horizontal Dipole - zero degrees elevation
wrote in message ... On Sat, 20 Jun 2009 21:13:51 GMT, "Dave" wrote: what you are missing is the 'real world'. eznec is probably modeling over a perfectly flat infinite surface. In the far field in a perfect world the signal along the surface is a combination of ground wave and sky wave, the ground wave decays rapidly with distance leaving the sky wave which will always be very small along the surface. now remember, the frame of reference is at ground level, not the antenna height, so zero degrees is along the infinite flat surface. And there is nothing in there that models where the other antenna is... it just creates a picture of how the strength of the fields are at a given elevation/azimuth angle from the reference point. Dave, Your paragraph above helped. For VHF and above, in the real world, am I better off using EZNEC's "free space" setting instead of real ground? I know at HF frequencies, where antennas are often close to the ground, it makes a big difference, but could free space be a better approximation of VHF antenna many wavelengths off the ground? Pat Hi Pat I recently re-entered the Ham community after being away from electronics for 40 years. I was extreemely resistant to accepting the accuracy of computer modeling. As I have become more familiar with computer modeling data, I now respect its value very much. Before acceptance of the accuracy of computer modeling, I recently made my own complex impedance measurement equipment for 2 meters. I also made actual radiation pattern measurements using polar orbiting satellites at the "range illuminator". Your question about how well the data from EZNEC matches "actual" radiation pattern is probably related to our not being able to feed the appropriate information into the computer modeling program. You are probably interested in 'just learning' how accurate EZNEC is.for predicting antenna sensitivity toward the horizon when the signal is horizontally polarized. But, if you want to get actual radiation pattern data from any given antenna at VHF where polar orbiting satellites are sending a beacon signal, you can simply record the receiver RSSI while the satellite passes overhead. I have some EZNEC and actual patterns measured using the 137 MHz signals from NOAA satellites. My data wont convince you that EZNEC is quite valuable for predicting the performance of YOUR dipole. But, if you have interest in knowing more about what I have done for recording antenna pattern data using polar orbiting satellites, contact me. I suspect you already know more about what you need than anything I can add. Jerry KD6JDJ |
Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
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 |
Horizontal Dipole - zero degrees elevation
Frank wrote:
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 Ok, how do I get those, and select those, results? I looked for it, but couldn't find it in EZNEC. tom K0TAR |
Horizontal Dipole - zero degrees elevation
"tom" wrote in message . net... Frank wrote: 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 Ok, how do I get those, and select those, results? I looked for it, but couldn't find it in EZNEC. tom K0TAR It comes from the "RP" card; line 8 below. I do not know how to implement this line of code in EZNEC. I am sure there are plenty of people In this group that know how to do it. Frank VE6CB 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 |
Horizontal Dipole - zero degrees elevation
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Horizontal Dipole - zero degrees elevation
On Sat, 20 Jun 2009 23:19:34 -0700, Richard Clark
wrote: On Sat, 20 Jun 2009 20:50:59 -0400, wrote: 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. Hi Pat, You can use the "near field" table to then build a second antenna (representing your own at the proper distance) with a load, and then look at the power developed to that load to see how things perform. This goes beyond the normal use, but for those who will slog through it, there's a way. snip 73's Richard Clark, KB7QHC Thanks again, Richard. I just purchased an EZNEC license so I will try that to further my learning. (I had been using the demo version). 73, Pat |
Horizontal Dipole - zero degrees elevation
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