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Radiating Efficiency
"Frank's" wrote W7EL tells us that EZNEC doesn't display the surface wave which obviously contains power. Would that affect the efficiency using the integration technique? ====================================== The surface or ground wave is the most important fraction of total radiated power. The correct radiation pattern of a vertical antenna shows maximum radiation along the ground. Angle of maximum radiation = 0 degrees. When deducing efficiency, to omit power radiated along the ground from the hemispherical integration will result in serious error. ( Efficiency by NEC4 ) / ( Efficiency by formula ) = 0.38 I can imagine other losses in addition to loss in the radials but to have the other losses several times greater is a bit much. Where are these large losses? Are they in the soil surface - the only other candidate? ---- Reg. |
Radiating Efficiency
This thread is getting interesting...
Reg, your point that the resonant radial effects are more pronounced at higher frequencies makes me think that a small experiment might be in order. A 21MHz or so ground mounted monopole with a small radial field would be a minimal investment in materials to do a BLE style empirical investigation, at least compared to that at 3MHz. It wouldn't be a minimal investment in time but might be an interesting experiment. Given that it would be an epic challenge to measure field strength at a grid of points on a hemisphere surrounding the monopole, I imagine a single point field strength measurement would be an acceptable metric? Thoughts on doing the measurement? I guess BLE would be a good guide for the empiricist. Dan |
Radiating Efficiency
"Reg Edwards" wrote in message ... "Frank's" wrote W7EL tells us that EZNEC doesn't display the surface wave which obviously contains power. Would that affect the efficiency using the integration technique? ====================================== The surface or ground wave is the most important fraction of total radiated power. The correct radiation pattern of a vertical antenna shows maximum radiation along the ground. Angle of maximum radiation = 0 degrees. When deducing efficiency, to omit power radiated along the ground from the hemispherical integration will result in serious error. ( Efficiency by NEC4 ) / ( Efficiency by formula ) = 0.38 I can imagine other losses in addition to loss in the radials but to have the other losses several times greater is a bit much. Where are these large losses? Are they in the soil surface - the only other candidate? ---- Reg. Correct Reg, I had thought the modified RP card considered ground wave, but it does not. I am working on a combined integration including the surface wave, which should provide a more accurate indication of the overall efficiency. I will also attempt the analysis of various lengths of radial wires. Frank |
Radiating Efficiency
Dan wrote -
A 21MHz or so ground mounted monopole with a small radial field would be a minimal investment in materials to do a BLE style empirical investigation, at least compared to that at 3MHz. It wouldn't be a minimal investment in time but might be an interesting experiment. Given that it would be an epic challenge to measure field strength at a grid of points on a hemisphere surrounding the monopole, I imagine a single point field strength measurement would be an acceptable metric? Thoughts on doing the measurement? I guess BLE would be a good guide for the empiricist. ======================================== The results of such experiments would probably generate far less interest than BLE's original work and would not be of great use. But somebody might possibly create a reputation out of it. To repeat BLE's experiments at HF, and to be of use, would require measurements to be made over a wide range of frequencies, over various lengths of radials, over various numbers of radials, over a range of soil resistivities and over a range of soil permittivities. Complexity and cost would be enormous. Financial returns would be relatively small. Who would invest in such a project? How many people wishing to erect a 28 MHz vertical over a set of radials need more knowledge than what already exists. Even CB-ers could do it! It would be far more economic, with a guessed understanding of how radials work, to write and dedicate a computer program to do the job. With a little more tidying-up, program Radial_3 would do. And it's FREE to USA citizens. No licence required! ;o) ---- Reg. |
Radiating Efficiency
The surface or ground wave is the most important fraction of total
radiated power. The correct radiation pattern of a vertical antenna shows maximum radiation along the ground. Angle of maximum radiation = 0 degrees. When deducing efficiency, to omit power radiated along the ground from the hemispherical integration will result in serious error. ( Efficiency by NEC4 ) / ( Efficiency by formula ) = 0.38 I can imagine other losses in addition to loss in the radials but to have the other losses several times greater is a bit much. Where are these large losses? Are they in the soil surface - the only other candidate? ---- Reg. Correct Reg, I had thought the modified RP card considered ground wave, but it does not. I am working on a combined integration including the surface wave, which should provide a more accurate indication of the overall efficiency. I will also attempt the analysis of various lengths of radial wires. Frank Reg, with the antenna we have established as our test model: All wires #14, monopole 9 m, 36 X 10m radials 25 mm below ground. Ground parameters Er = 16, resistivity 150 ohm-m. I have calculated the radiation efficiency including the surface wave. What is interesting is that the surface wave contributes very little at elevation angles over 10 degrees. For 100 W input the total radiated power, not considering the surface wave, is 31 W. When the surface wave is included the total radiated power is 36 W. Frank |
Radiating Efficiency
"Frank's" wrote Reg, with the antenna we have established as our test model: All wires #14, monopole 9 m, 36 X 10m radials 25 mm below ground. Ground parameters Er = 16, resistivity 150 ohm-m. I have calculated the radiation efficiency including the surface wave. What is interesting is that the surface wave contributes very little at elevation angles over 10 degrees. For 100 W input the total radiated power, not considering the surface wave, is 31 W. When the surface wave is included the total radiated power is 36 W. ========================================== Frank, So the missing 64 watts, if not dissipated in the radials must be dissipated in the soil under the antenna. To prove something, what is the efficiency when the ground is sea water. Resistivity = 0.22 ohm-metres and permittivity = 80. PLEASE CONFIRM THE NUMBER OF RADIALS. I thought you were unable to model 36 radials. Were you using only 1 radial? With only 1 radial my program makes efficiency = 28 percent. Which would be satisfactory agreement with your 36 % ) Or did you substitute the radial system with a lumped resistance of 5 ohms which I suggested could be used when estimating efficiency? ---- Reg. |
Radiating Efficiency
Frank's wrote:
The surface or ground wave is the most important fraction of total radiated power. The correct radiation pattern of a vertical antenna shows maximum radiation along the ground. Angle of maximum radiation = 0 degrees. When deducing efficiency, to omit power radiated along the ground from the hemispherical integration will result in serious error. ( Efficiency by NEC4 ) / ( Efficiency by formula ) = 0.38 I can imagine other losses in addition to loss in the radials but to have the other losses several times greater is a bit much. Where are these large losses? Are they in the soil surface - the only other candidate? ---- Reg. Correct Reg, I had thought the modified RP card considered ground wave, but it does not. I am working on a combined integration including the surface wave, which should provide a more accurate indication of the overall efficiency. I will also attempt the analysis of various lengths of radial wires. Frank Reg, with the antenna we have established as our test model: All wires #14, monopole 9 m, 36 X 10m radials 25 mm below ground. Ground parameters Er = 16, resistivity 150 ohm-m. I have calculated the radiation efficiency including the surface wave. What is interesting is that the surface wave contributes very little at elevation angles over 10 degrees. For 100 W input the total radiated power, not considering the surface wave, is 31 W. When the surface wave is included the total radiated power is 36 W. Frank Good. Now all you have to do is verify your calculations experimentally. 73, Tom Donaly, KA6RUH |
Radiating Efficiency
Reg, with the antenna we have established as our test model:
All wires #14, monopole 9 m, 36 X 10m radials 25 mm below ground. Ground parameters Er = 16, resistivity 150 ohm-m. I have calculated the radiation efficiency including the surface wave. What is interesting is that the surface wave contributes very little at elevation angles over 10 degrees. For 100 W input the total radiated power, not considering the surface wave, is 31 W. When the surface wave is included the total radiated power is 36 W. Frank Good. Now all you have to do is verify your calculations experimentally. 73, Tom Donaly, KA6RUH You got that right. I need a serious VNA so bad, I can taste it! 73, Frank |
Radiating Efficiency
Tom Donaly wrote: Frank's wrote: The surface or ground wave is the most important fraction of total radiated power. The correct radiation pattern of a vertical antenna shows maximum radiation along the ground. Angle of maximum radiation = 0 degrees. When deducing efficiency, to omit power radiated along the ground from the hemispherical integration will result in serious error. ( Efficiency by NEC4 ) / ( Efficiency by formula ) = 0.38 I can imagine other losses in addition to loss in the radials but to have the other losses several times greater is a bit much. Where are these large losses? Are they in the soil surface - the only other candidate? ---- Reg. Correct Reg, I had thought the modified RP card considered ground wave, but it does not. I am working on a combined integration including the surface wave, which should provide a more accurate indication of the overall efficiency. I will also attempt the analysis of various lengths of radial wires. Frank Reg, with the antenna we have established as our test model: All wires #14, monopole 9 m, 36 X 10m radials 25 mm below ground. Ground parameters Er = 16, resistivity 150 ohm-m. I have calculated the radiation efficiency including the surface wave. What is interesting is that the surface wave contributes very little at elevation angles over 10 degrees. For 100 W input the total radiated power, not considering the surface wave, is 31 W. When the surface wave is included the total radiated power is 36 W. Frank Good. Now all you have to do is verify your calculations experimentally. 73, Tom Donaly, KA6RUH That's right. Without verification by direct measurements any program like this is guesswork. 73 Tom |
Radiating Efficiency
"Reg Edwards" wrote in message ... "Frank's" wrote Reg, with the antenna we have established as our test model: All wires #14, monopole 9 m, 36 X 10m radials 25 mm below ground. Ground parameters Er = 16, resistivity 150 ohm-m. I have calculated the radiation efficiency including the surface wave. What is interesting is that the surface wave contributes very little at elevation angles over 10 degrees. For 100 W input the total radiated power, not considering the surface wave, is 31 W. When the surface wave is included the total radiated power is 36 W. ========================================== Frank, So the missing 64 watts, if not dissipated in the radials must be dissipated in the soil under the antenna. To prove something, what is the efficiency when the ground is sea water. Resistivity = 0.22 ohm-metres and permittivity = 80. PLEASE CONFIRM THE NUMBER OF RADIALS. I thought you were unable to model 36 radials. Were you using only 1 radial? With only 1 radial my program makes efficiency = 28 percent. Which would be satisfactory agreement with your 36 % ) Or did you substitute the radial system with a lumped resistance of 5 ohms which I suggested could be used when estimating efficiency? Reg, The antenna has thirty-six 10 m radials, 25 mm below ground. The input impedance at 8.07 MHz is 36.3 - j 2.0. I used 0.2 ohm-m for sea water resistivity as it seems to the accepted standard. The radiation efficiency, not including the surface wave, is 93.6 %. I can add the surface wave if you are interested, but it is a little tedious to convert the far field surface wave, in cylindrical coordinates, to normalized spherical coordinates; which is the form NEC produces for the sky-wave components. Frank |
Radiating Efficiency
Frank, the reason I asked what the efficiency is when using sea water
was to prove your own efficiency calculations. With sea water the efficiency should be very near to 100 percent. You get 93% WITHOUT taking the surface wave into account. To make youself happy you could include the surface wave. Program Radial_3, with sea water, makes efficiency 98 percent which I'm confident is near enough correct. Most of the loss is in the HF resistance of the 14-gauge antenna wire. ================================================== ====== It's beginning to look as though the oft-quoted formula - Efficiency = Rrad / ( Rrad + Rloss ) is very much in error. This formula is quoted in all the ARRL books and other learned magazines. I will correct my program to agree better with NEC4 even though the absolute value of efficiency is not important and is used only as an indication to maximise effectiveness of the radial system. I await your experiments to determine the impedance Zo of ONE radial wire and the approximate distance at which it occurs. You can do N = 36 radials at a later date. Thank you very much. ---- Reg. |
Radiating Efficiency
On Thu, 27 Jul 2006 19:01:14 +0100, "Reg Edwards"
wrote: It's beginning to look as though the oft-quoted formula - Efficiency = Rrad / ( Rrad + Rloss ) is very much in error. Hi Reggie, In fact the entire enquiry has justified it. How do you come to the opposite conclusion? 73's Richard Clark, KB7QHC |
Radiating Efficiency
"Reg Edwards" wrote in message
... Frank, the reason I asked what the efficiency is when using sea water was to prove your own efficiency calculations. With sea water the efficiency should be very near to 100 percent. You get 93% WITHOUT taking the surface wave into account. To make youself happy you could include the surface wave. Program Radial_3, with sea water, makes efficiency 98 percent which I'm confident is near enough correct. Most of the loss is in the HF resistance of the 14-gauge antenna wire. ================================================== ====== It's beginning to look as though the oft-quoted formula - Efficiency = Rrad / ( Rrad + Rloss ) is very much in error. This formula is quoted in all the ARRL books and other learned magazines. I will correct my program to agree better with NEC4 even though the absolute value of efficiency is not important and is used only as an indication to maximise effectiveness of the radial system. I await your experiments to determine the impedance Zo of ONE radial wire and the approximate distance at which it occurs. You can do N = 36 radials at a later date. Thank you very much. Reg, The efficiency, including the surface wave, is 96%. There is also a 2% copper loss. With perfect conductors the efficiency would then be 98%. All figures from 100 W input. I have not forgotten the single radial computation. Frank |
Radiating Efficiency
Frank,
The efficiency formula is incomplete rather than being in error. In the case of radial systems it could be something like - Efficiency = Rr / ( Rr + Rradials + Rsoilsurface ) . When soil resistivity becomes very small, efficiency approaches 100 percent and the error when compared with NEC4 reduces to zero. The error is therefore a function of soil resistivity. ---- Reg. |
Radiating Efficiency
Good. Now all you have to do is verify your calculations
experimentally. 73, Tom Donaly, KA6RUH ======================================== When 2 + 3 + 1 = 6 inches, do you always have to verify it with a wooden ruler? ;o) ---- Reg. |
Radiating Efficiency
"Reg Edwards" wrote in message
... Frank, The efficiency formula is incomplete rather than being in error. In the case of radial systems it could be something like - Efficiency = Rr / ( Rr + Rradials + Rsoilsurface ) . When soil resistivity becomes very small, efficiency approaches 100 percent and the error when compared with NEC4 reduces to zero. The error is therefore a function of soil resistivity. ---- Reg. Sounds reasonable Reg. Frank |
Radiating Efficiency
Reg Edwards wrote:
Frank, The efficiency formula is incomplete rather than being in error. In the case of radial systems it could be something like - Efficiency = Rr / ( Rr + Rradials + Rsoilsurface ) . When soil resistivity becomes very small, efficiency approaches 100 percent and the error when compared with NEC4 reduces to zero. The error is therefore a function of soil resistivity. ---- Reg. So, Rloss = Rradials + Rsoilsurface. The components of Rloss change depending on the antenna being studied. But, Efficiency is still % = (Rr/(Rr + Rloss))*100 /s/ DD |
Radiating Efficiency
Reg Edwards wrote:
Good. Now all you have to do is verify your calculations experimentally. 73, Tom Donaly, KA6RUH ======================================== When 2 + 3 + 1 = 6 inches, do you always have to verify it with a wooden ruler? ;o) ---- Reg. 2 + 3 + 1 doesn't ever equal 6 inches. 2 inches + 3 inches + 1 inch does, however, and if you build something that is 2 inches + 3 inches + 1 inch, you'll most likely verify it with a ruler just to see how close you got. 73, Tom Donaly, KA6RUH |
Radiating Efficiency
I await your experiments to determine the impedance Zo of ONE radial
wire and the approximate distance at which it occurs. You can do N = 36 radials at a later date. Thank you very much. Reg, I started to do an analysis of a single radial monopole. Beginning with a 1 meter length radial I compared the results with "radial_3". There is such a huge discrepancy between the programs I wondered if I had made an error someplace. Just to confirm the antenna dimensions: 9 m monopole, with a 1 m radial, 25 mm below ground. All wires # 14 AWG, and the test frequency 8.07 MHz. Ground Er = 16, resistivity 150 ohm-meters. "radial_3": Antenna input Z = 34.21 - j 26.2, and; radial input Z = 82.7 - j 62.3. "NEC 4.1": antenna input Z = 137.8 - j 81.8, and; radial input Z = 101.4 - j 79.1. Interesting to note that the input impedance determined by radial_3 is very close to an ideal monopole above a perfectly conducting ground. Frank |
Radiating Efficiency
Frank, I am not interested is what happens at very short lengths or
what Radial_3 makes of it. To determine Zo, start around 10 metres. If very little happens to input impedance between 10 and and 15 metres then you already have Zo = Zin = Ro + jXo. Neither am I interested in efficiency or antenna input impedance.. The problem of Efficiency has already been sorted out. All I wish to know is Zin = Zo of a single radial, at various lengths greater than about 10 metres, of diameter = 1.64mm, depth = 25mm, ground resistivity = 150 ohm-metres, permittivity = 16, at a frequency of about 8.07 MHz. That is the input impedance of one radial when the attenuation is about 20dB or greater. To summarise, I wish to know Zo = Ro + jXo for one radial. ---- Reg. |
Radiating Efficiency
Reg, this whole thread started because bizarre things happen with
Radial_3 with short radials. You can go ahead and put a caveat in it if you like, but it's wrong at short lengths and that fact needs to be made clear to users of the program, before they install a radial system of 120 2m radials on 80m and find that it's a terrible ground system. The question that I want answered is how to optimize my radial system, and I think at this point, I shall be consulting a document entitled "Maximum Gain Radial Ground Systems for Vertical Antennas" by K3LC. Anyone have other suggestions for sources of material for optimum radial selection given a certain length of wire? 73, Dan Reg Edwards wrote: Frank, I am not interested is what happens at very short lengths or what Radial_3 makes of it. To determine Zo, start around 10 metres. If very little happens to input impedance between 10 and and 15 metres then you already have Zo = Zin = Ro + jXo. Neither am I interested in efficiency or antenna input impedance.. The problem of Efficiency has already been sorted out. etc. |
Radiating Efficiency
The question that I want answered is how to optimize my radial system,
and I think at this point, I shall be consulting a document entitled "Maximum Gain Radial Ground Systems for Vertical Antennas" by K3LC. Anyone have other suggestions for sources of material for optimum radial selection given a certain length of wire? 73, Dan Rudy Severns' article; "Verticals, Ground Systems and Some History", in the July 2000 QST, is worth a read. Available on the ARRL web site, or, I also have the pdf. Frank |
Radiating Efficiency
To determine Zo, start around 10 metres.
If very little happens to input impedance between 10 and and 15 metres then you already have Zo = Zin = Ro + jXo. Neither am I interested in efficiency or antenna input impedance.. The problem of Efficiency has already been sorted out. All I wish to know is Zin = Zo of a single radial, at various lengths greater than about 10 metres, of diameter = 1.64mm, depth = 25mm, ground resistivity = 150 ohm-metres, permittivity = 16, at a frequency of about 8.07 MHz. That is the input impedance of one radial when the attenuation is about 20dB or greater. To summarise, I wish to know Zo = Ro + jXo for one radial. Reg, the radial impedance rapidly converges to 101.6 + j 21.1. 10 m -- radial Z = 102 + j 20.99 12 m -- radial Z = 101.3 + j 21.1 14 m -- radial Z = 101.65 + j 21.32 16 m -- radial Z = 101.7 + j 21.1 18 m -- radial Z = 101.615 + j 21.1 20 m -- radial Z = 101.61 + j 21.11 Frank |
Radiating Efficiency
Reg, the radial impedance rapidly converges to 101.6 + j 21.1. 10 m -- radial Z = 102 + j 20.99 12 m -- radial Z = 101.3 + j 21.1 14 m -- radial Z = 101.65 + j 21.32 16 m -- radial Z = 101.7 + j 21.1 18 m -- radial Z = 101.615 + j 21.1 20 m -- radial Z = 101.61 + j 21.11 ============================================== Frank, Excellent results! The radial has already converged on Zo = 102 + j21 at a distance of 10 metres. Just where Radial_3 predicts it should. The magnitude of Zo is within 20 percent of NEC4 and the impedance angle is in the right ball-park with the correct sign. Now work downwards from 10 metres, to about 7.5 metres, the 3/4-wave resonant point, to find the point where Zin has truly diverged from Zo. You will have to work in gradually smaller increments. Could you go down to the 1/2-wave resonant point at about 4 metres? You will now be able to see what I'm heading for. By the way, how much hard labour is all this causing you? Don't try to tell me what you are actually doing because I havn't the foggiest idea. ---- Reg, |
Radiating Efficiency
I've explained this to Reg many times before, but somehow it doesn't
seem to sink in. Here's the explanation again. Reg Edwards wrote: "Frank's" wrote W7EL tells us that EZNEC doesn't display the surface wave which obviously contains power. Would that affect the efficiency using the integration technique? ====================================== The surface or ground wave is the most important fraction of total radiated power. It's important only to AM broadcasters and others communicating at low frequencies, and to HF operators interested in working distances of only a few miles. At HF, it decays to a negligibly low value within a few miles, and so is no importance at all beyond that distance. The correct radiation pattern of a vertical antenna shows maximum radiation along the ground. Angle of maximum radiation = 0 degrees. Let's consider that for a moment. The surface wave decays rapidly. So the "correct" radiation pattern as described by Reg changes dramatically with distance. At HF, the pattern at one mile will be strikingly different from the pattern at 20 miles; the first will be maximum at zero elevation angle, the latter won't. So if you want this kind of "correct" pattern, you'll have to specify the distance from the antenna. Once you're a few miles away, at HF, the pattern becomes constant with distance, because the surface wave has decayed to essentially zero. Then you have the pattern which is useful in determining communication beyond a few miles. (This is the pattern reported by EZNEC and NEC as the "far field" pattern.) Reg's "correct" pattern isn't useful for anything but short distance communication, and isn't valid except at the distance specified. When deducing efficiency, to omit power radiated along the ground from the hemispherical integration will result in serious error. Well, it kind of depends what you lump in with losses when calculating efficiency. The classical formula for antenna efficiency, Rrad/(Rrad + Rloss), generally applies only to the antenna itself and near field losses such as ground system losses. So power in the surface wave is considered to be "radiation", and if you want to calculate this efficiency by dividing the power in the radiated field by the power from the sources, you would have to include the surface wave in the calculation. However, for people communicating more than a few miles, the surface wave power, which is dissipated in the ground within a few miles of the antenna, is just as lost as power dissipated in the wires or the ground system. So if you consider "radiated power" to be power radiated beyond a few miles and "loss" to be the rest, then you can lump the surface wave power into the "loss" portion. If you do that, the efficiency is correctly reported by EZNEC or NEC's average gain function. ( Efficiency by NEC4 ) / ( Efficiency by formula ) = 0.38 I won't comment on that because I haven't a clue where it came from. I can imagine other losses in addition to loss in the radials but to have the other losses several times greater is a bit much. Where are these large losses? Are they in the soil surface - the only other candidate? The surface wave power is indeed dissipated in the soil within a few miles of the antenna. Roy Lewallen, W7EL |
Radiating Efficiency
Roy, there you go again - confusing the issue by re-stating the
bleeding obvious. What happens to the ground wave AFTER it has been radiated is not relevant to the efficiency problem. ( Which has now been sorted out anyway.) The losses we are concerned with all occur in the near field. They are - (A) Loss in the radials and loss in the soil in the vicinity of the radials, which is represented by Rradials. It is the input resistance of the radials. It can be determined by measurements. Rradials is the value used in the usual formula - Efficiency = Rr / ( Rr + Radials ), which, in the present context, is incorrect. (B) Loss in the soil surface and soil NOT in the vicinity of radials, but still in the near field, represented by Rsoil. Frank, using NEC4, has managed to seperate losses A and B although not the resistance of Rsoil. So, after finding a value for Rsoil, a more accurate formula is - Efficiency = Rr / ( Rr + Radials + Rsoil ) Where Rr is the radiation resistance referred to the base of the antenna. Note that Rsoil cannot be measured but can be deduced from the actual efficiency. Both Rradials and Rsoil are functions of the same soil characteristics. ---- Reg, |
Radiating Efficiency
On Sat, 29 Jul 2006 08:10:13 +0100, "Reg Edwards"
wrote: Efficiency = Rr / ( Rr + Radials ), which, in the present context, is incorrect. Hi Reggie, Of course it is incorrect, it is a definition of your own invention. Efficiency = Rr / ( Rr + Radials + Rsoil ) This is merely an elaboration of the commonplace Efficiency = Rrad / ( Rrad + Rloss ) which you declared is very much in error. The error is yours, and your bafflegab that has flowed from this pronouncement of an "error" has been in an effort to cover your tracks. The wonderment you are met with is that you failed to acknowledge that Rloss is commonly accepted to mean more than Ohmic loss. Reggie, the archive is rich with this discussion, and no one but yourself makes the mistake of presuming Rloss has ever meant to be confined to the Ohmic loss of the radiator's metal parts. 73's Richard Clark, KB7QHC |
Radiating Efficiency
On Sat, 29 Jul 2006 11:03:36 -0700, Richard Clark wrote:
On Sat, 29 Jul 2006 08:10:13 +0100, "Reg Edwards" wrote: Efficiency = Rr / ( Rr + Radials ), which, in the present context, is incorrect. Hi Reggie, Of course it is incorrect, it is a definition of your own invention. snip Hi Reg, By any remote chance have you read any of the posts that followed the one you initiated in 'radial attenuation', just below the one we're in now? If not, please do so, and then savor the crow on your plate concerning your somewhat 'stiff' position on BLE's failure to determine the ground characteristics that lie beneath the radials. Thank you. Walt, W2DU |
Radiating Efficiency
On Sat, 29 Jul 2006 16:37:05 -0400, Walter Maxwell
wrote: [snip] ....... then savor the crow on your plate ...... Thank you. Walt, W2DU Ah yes Walter, crow on the plate I have had the misfortune of tasting that old bird on many occasions. I found it very tough and not at all palatable. G Danny, K6MHE |
Radiating Efficiency
"Reg Edwards" wrote in message ... Reg, the radial impedance rapidly converges to 101.6 + j 21.1. 10 m -- radial Z = 102 + j 20.99 12 m -- radial Z = 101.3 + j 21.1 14 m -- radial Z = 101.65 + j 21.32 16 m -- radial Z = 101.7 + j 21.1 18 m -- radial Z = 101.615 + j 21.1 20 m -- radial Z = 101.61 + j 21.11 ============================================== Frank, Excellent results! The radial has already converged on Zo = 102 + j21 at a distance of 10 metres. Just where Radial_3 predicts it should. The magnitude of Zo is within 20 percent of NEC4 and the impedance angle is in the right ball-park with the correct sign. Now work downwards from 10 metres, to about 7.5 metres, the 3/4-wave resonant point, to find the point where Zin has truly diverged from Zo. You will have to work in gradually smaller increments. Could you go down to the 1/2-wave resonant point at about 4 metres? You will now be able to see what I'm heading for. By the way, how much hard labour is all this causing you? Don't try to tell me what you are actually doing because I havn't the foggiest idea. ---- Reg, This is fairly trivial Reg. It takes me about 90 seconds to run the program, analyze the data, and record the results for each length. I consider this a learning experience. Some of your requests have forced me to read the NEC manual and other books I have on modeling. As a preliminary run I have gone overboard, just to see the overall trend. The fact is I see nothing dramatic happening until the radial gets very short. Possibly you can see regions where I need to concentrate. Obviously most of the steps are very large, and I may have missed something. I would have expected to see a phase reversal though. 9m Zin = 101.8 + j 21.7 8m Zin = 100.5 + j 21.5 7m Zin = 100.5 + j 19.0 6m Zin = 105.1 + j 17.8 5m Zin = 110.5 + j 26.1 4m Zin = 97.0 + j 40.2 3m Zin = 70.5 + j 25.9 2m Zin = 67.2 + j 19.6 I did try steps of 0.1 m from 8 m to 6.7 m, and saw nothing but a progressive trend. Frank |
Radiating Efficiency
This is fairly trivial Reg. It takes me about 90 seconds to run
the program, analyze the data, and record the results for each length. I consider this a learning experience. Some of your requests have forced me to read the NEC manual and other books I have on modeling. As a preliminary run I have gone overboard, just to see the overall trend. The fact is I see nothing dramatic happening until the radial gets very short. Possibly you can see regions where I need to concentrate. Obviously most of the steps are very large, and I may have missed something. I would have expected to see a phase reversal though. 9m Zin = 101.8 + j 21.7 8m Zin = 100.5 + j 21.5 7m Zin = 100.5 + j 19.0 6m Zin = 105.1 + j 17.8 5m Zin = 110.5 + j 26.1 4m Zin = 97.0 + j 40.2 3m Zin = 70.5 + j 25.9 2m Zin = 67.2 + j 19.6 I did try steps of 0.1 m from 8 m to 6.7 m, and saw nothing but a progressive trend. Frank ========================================= Frank, I'm pleased to hear this does not involve you in a lot of labour. Thanks for the additional useful information. Go back to program Radial_3 for a few minutes and insert our standard inputs. Set the number of radials equal to One. Slowly vary length between 1 and 10 metres while observing Rin + jXin of the radial system. Vary length to find maxima and minima in the value of Rin. Max and min are more pronounced at the shorter lengths due to lower attenuation. Remember, the radial ( transmission line ) is open-circuit at the other end. There is a minimum of Rin when the radial is 1/4-wave resonant at 2.4 metres. There is a maximum of Rin when the radial is 1/2-wave resonant at 4.8 metres. There is another minimum of Rin, but less prominent, when the radial is 3/4-wave resonant at 7.3 metres. As length and attenuation along the line increase, the variations of Rin about its mean become smaller. Eventually, of course, it converges on Ro, the characteristic impedance. ( Ro is also computed but remains constant as length is varied.) There would be a full-wave resonance at approximately 10 metres but it is damped-down into the noise by the attenuation of about 20 dB at that length. Now, what I would like you to do is search for the maxima and and minima in Rin, with their lengths. using NEC4. At some places you may have to use increments of 0.1 metres. If you find any max and minima the values of Rin + jXin will be different from my program and the lengths at which they occur may also differ. I would like to use the information to improve the accuracy of my program on the assumption that NEC4 is more correct when calculating buried radials. ( In this investigation, you may think it peculiar that lengths as small as 0.1 metres should be significant at 8 MHz. This is due to the very low velocity of propagation along buried radials. Program Radial_3 estimates VF.) ---- Reg. |
Radiating Efficiency
You have the right reference, K3LC. I think that's the one used in
Devolder's book, Low Band DX'ing. (I don't have it in front of me at the moment) 73, ....hasan, N0AN wrote in message oups.com... Reg, this whole thread started because bizarre things happen with Radial_3 with short radials. You can go ahead and put a caveat in it if you like, but it's wrong at short lengths and that fact needs to be made clear to users of the program, before they install a radial system of 120 2m radials on 80m and find that it's a terrible ground system. The question that I want answered is how to optimize my radial system, and I think at this point, I shall be consulting a document entitled "Maximum Gain Radial Ground Systems for Vertical Antennas" by K3LC. Anyone have other suggestions for sources of material for optimum radial selection given a certain length of wire? 73, Dan Reg Edwards wrote: Frank, I am not interested is what happens at very short lengths or what Radial_3 makes of it. To determine Zo, start around 10 metres. If very little happens to input impedance between 10 and and 15 metres then you already have Zo = Zin = Ro + jXo. Neither am I interested in efficiency or antenna input impedance.. The problem of Efficiency has already been sorted out. etc. |
Radial attenuation
Dear Rich,
Try pulling the other leg - it has bells on it! Punchinello |
Radiating Efficiency
Go back to program Radial_3 for a few minutes and insert our standard
inputs. Set the number of radials equal to One. Slowly vary length between 1 and 10 metres while observing Rin + jXin of the radial system. Vary length to find maxima and minima in the value of Rin. Max and min are more pronounced at the shorter lengths due to lower attenuation. Remember, the radial ( transmission line ) is open-circuit at the other end. There is a minimum of Rin when the radial is 1/4-wave resonant at 2.4 metres. There is a maximum of Rin when the radial is 1/2-wave resonant at 4.8 metres. There is another minimum of Rin, but less prominent, when the radial is 3/4-wave resonant at 7.3 metres. As length and attenuation along the line increase, the variations of Rin about its mean become smaller. Eventually, of course, it converges on Ro, the characteristic impedance. ( Ro is also computed but remains constant as length is varied.) There would be a full-wave resonance at approximately 10 metres but it is damped-down into the noise by the attenuation of about 20 dB at that length. Now, what I would like you to do is search for the maxima and and minima in Rin, with their lengths. using NEC4. At some places you may have to use increments of 0.1 metres. If you find any max and minima the values of Rin + jXin will be different from my program and the lengths at which they occur may also differ. I would like to use the information to improve the accuracy of my program on the assumption that NEC4 is more correct when calculating buried radials. ( In this investigation, you may think it peculiar that lengths as small as 0.1 metres should be significant at 8 MHz. This is due to the very low velocity of propagation along buried radials. Program Radial_3 estimates VF.) Reg, I have done some computations around a 2 m radial length. I noticed that I had made an error in my previous calculation at exactly 2 m. 1.8 m -- Radial Z = 70.17 - j 24.1 1.9 m -- Radial Z = 68.5 - j 19.0 2.0 m -- Radial Z = 67.2 - j 14.2 2.1 m -- Radial Z = 66.2 - j 9.5 2.2 m -- Radial Z = 65.5 - j 5.0 2.3 m -- Radial Z = 65.1 - j 0.6 2.4 m -- Radial Z = 65.0 + j 3.6 2.5 m -- Radial Z = 65.2 + j 7.8 2.6 m -- Radial Z = 65.6 + j 11.7 I can keep going if you think that these are the results you expected. I am tempted to continue, in steps of 0.1 m, and plotting the results on the Smith Chart. I expect the data to rapidly spiral to the center of a chart normalized at 101. 6 ohms. Frank |
Radiating Efficiency
1.8 m -- Radial Z = 70.17 - j 24.1
1.9 m -- Radial Z = 68.5 - j 19.0 2.0 m -- Radial Z = 67.2 - j 14.2 2.1 m -- Radial Z = 66.2 - j 9.5 2.2 m -- Radial Z = 65.5 - j 5.0 2.3 m -- Radial Z = 65.1 - j 0.6 2.4 m -- Radial Z = 65.0 + j 3.6 2.5 m -- Radial Z = 65.2 + j 7.8 2.6 m -- Radial Z = 65.6 + j 11.7 I can keep going if you think that these are the results you expected. I am tempted to continue, in steps of 0.1 m, and plotting the results on the Smith Chart. I expect the data to rapidly spiral to the center of a chart normalized at 101. 6 ohms. Frank ====================================== Frank, Very interesting! I am plotting graphs of R and jX versus length of radial. Joining up the dots I expect to see shallow sinewaves superimposed on fairly level mean values. At present there is a definite shape of curve appearing in the reactance values while the resistance values are still fairly level. As length increases I expect to see something similar happening to the resistance curve which seems to be in between peaks and troughs. From now on it seems safe for you to increase length in increments of 0.2 metres. There is no danger of missing peaks and troughs in the curves. Please keep up the good work. ---- Reg |
Radiating Efficiency
"Reg Edwards" wrote in message ... 1.8 m -- Radial Z = 70.17 - j 24.1 1.9 m -- Radial Z = 68.5 - j 19.0 2.0 m -- Radial Z = 67.2 - j 14.2 2.1 m -- Radial Z = 66.2 - j 9.5 2.2 m -- Radial Z = 65.5 - j 5.0 2.3 m -- Radial Z = 65.1 - j 0.6 2.4 m -- Radial Z = 65.0 + j 3.6 2.5 m -- Radial Z = 65.2 + j 7.8 2.6 m -- Radial Z = 65.6 + j 11.7 I can keep going if you think that these are the results you expected. I am tempted to continue, in steps of 0.1 m, and plotting the results on the Smith Chart. I expect the data to rapidly spiral to the center of a chart normalized at 101. 6 ohms. Frank ====================================== Frank, Very interesting! I am plotting graphs of R and jX versus length of radial. Joining up the dots I expect to see shallow sinewaves superimposed on fairly level mean values. At present there is a definite shape of curve appearing in the reactance values while the resistance values are still fairly level. As length increases I expect to see something similar happening to the resistance curve which seems to be in between peaks and troughs. From now on it seems safe for you to increase length in increments of 0.2 metres. There is no danger of missing peaks and troughs in the curves. Please keep up the good work. ---- Reg 1.8 m -- Radial Z = , I have reached 7.4m, but there does not seem any point in continuing. Let me know what you think. 1.8 m -- Radial Z = 1.8 m -- Radial Z = 1.8 m -- Radial Z = |
Radiating Efficiency
"Reg Edwards" wrote in message ... 1.8 m -- Radial Z = 70.17 - j 24.1 1.9 m -- Radial Z = 68.5 - j 19.0 2.0 m -- Radial Z = 67.2 - j 14.2 2.1 m -- Radial Z = 66.2 - j 9.5 2.2 m -- Radial Z = 65.5 - j 5.0 2.3 m -- Radial Z = 65.1 - j 0.6 2.4 m -- Radial Z = 65.0 + j 3.6 2.5 m -- Radial Z = 65.2 + j 7.8 2.6 m -- Radial Z = 65.6 + j 11.7 I can keep going if you think that these are the results you expected. I am tempted to continue, in steps of 0.1 m, and plotting the results on the Smith Chart. I expect the data to rapidly spiral to the center of a chart normalized at 101. 6 ohms. Frank ====================================== Frank, Very interesting! I am plotting graphs of R and jX versus length of radial. Joining up the dots I expect to see shallow sinewaves superimposed on fairly level mean values. At present there is a definite shape of curve appearing in the reactance values while the resistance values are still fairly level. As length increases I expect to see something similar happening to the resistance curve which seems to be in between peaks and troughs. From now on it seems safe for you to increase length in increments of 0.2 metres. There is no danger of missing peaks and troughs in the curves. Please keep up the good work. ---- Reg 1.8 m -- Radial Z = , I have reached 7.4m, but there does not seem any point in continuing. Let me know what you think. It does not appear to be behaving as I would expect of a transmission line; but then I have no experience of transmission lines immersed in a lossy material. Also note the jump in impedance at 5.6 m. I double checked the result, and it appears to be correct. 2.8 m -- Radial Z = 67.5 + j 19.2 3.0 m -- Radial Z = 70.6 + j 25.9 3.2 m -- Radial Z = 74.7 + j 31.6 3.4 m -- Radial Z = 79.8 + j 36.0 3.6 m -- Radial Z = 85.5 + j 39.0 3.8 m -- Radial Z = 91.4 + j 40.4 4.0 m -- Radial Z = 97.1 + j 40.2 4.2 m -- Radial Z = 102.0 + j 38.6 4.4 m -- Radial Z = 105.9 + j 35.9 4.6 m -- Radial Z = 108.6 + j 32.7 4.8 m -- Radial Z = 110.1 + j 29.3 5.0 m -- Radial Z = 110.5 + j 26.1 5.2 m -- Radial Z = 110.1 + j 23.3 5.4 m -- Radial Z = 109.2 + j 17.2 5.6 m -- Radial Z = 107.9 + j 19.5 5.8 m -- Radial Z = 106.5 + j 18.4 6.0 m -- Radial Z = 105.1 + j 17.7 6.2 m -- Radial Z = 103.8 + j 17.5 6.4 m -- Radial Z = 102.7 + j 17.6 6.6 m -- Radial Z = 101.7 + j 17.9 6.8 m -- Radial Z = 101.0 + j 18.4 7.0 m -- Radial Z = 100.5 + j 19.0 7.2 m -- Radial Z = 100.2 + j 19.6 7.4 m -- Radial Z = 100.1 + j 20.2 .. .. 8.0 m -- Radial Z = 100.5 + j 21.4 .. .. 9.0 m -- Radial Z = 101.8 + j 21.7 These data are so close to the center of a Smith Chart normalized to 101.6 + j 21. Not sure how you normalize with a complex number, but assume it is with the magnitude of Z. Frank |
Radiating Efficiency
"Frank's" wrote in message news:H6Mzg.180983$771.142858@edtnps89... "Reg Edwards" wrote in message ... 1.8 m -- Radial Z = 70.17 - j 24.1 1.9 m -- Radial Z = 68.5 - j 19.0 2.0 m -- Radial Z = 67.2 - j 14.2 2.1 m -- Radial Z = 66.2 - j 9.5 2.2 m -- Radial Z = 65.5 - j 5.0 2.3 m -- Radial Z = 65.1 - j 0.6 2.4 m -- Radial Z = 65.0 + j 3.6 2.5 m -- Radial Z = 65.2 + j 7.8 2.6 m -- Radial Z = 65.6 + j 11.7 I can keep going if you think that these are the results you expected. I am tempted to continue, in steps of 0.1 m, and plotting the results on the Smith Chart. I expect the data to rapidly spiral to the center of a chart normalized at 101. 6 ohms. Frank ====================================== Frank, Very interesting! I am plotting graphs of R and jX versus length of radial. Joining up the dots I expect to see shallow sinewaves superimposed on fairly level mean values. At present there is a definite shape of curve appearing in the reactance values while the resistance values are still fairly level. As length increases I expect to see something similar happening to the resistance curve which seems to be in between peaks and troughs. From now on it seems safe for you to increase length in increments of 0.2 metres. There is no danger of missing peaks and troughs in the curves. Please keep up the good work. ---- Reg 1.8 m -- Radial Z = , I have reached 7.4m, but there does not seem any point in continuing. Let me know what you think. It does not appear to be behaving as I would expect of a transmission line; but then I have no experience of transmission lines immersed in a lossy material. Also note the jump in impedance at 5.6 m. I double checked the result, and it appears to be correct. 2.8 m -- Radial Z = 67.5 + j 19.2 3.0 m -- Radial Z = 70.6 + j 25.9 3.2 m -- Radial Z = 74.7 + j 31.6 3.4 m -- Radial Z = 79.8 + j 36.0 3.6 m -- Radial Z = 85.5 + j 39.0 3.8 m -- Radial Z = 91.4 + j 40.4 4.0 m -- Radial Z = 97.1 + j 40.2 4.2 m -- Radial Z = 102.0 + j 38.6 4.4 m -- Radial Z = 105.9 + j 35.9 4.6 m -- Radial Z = 108.6 + j 32.7 4.8 m -- Radial Z = 110.1 + j 29.3 5.0 m -- Radial Z = 110.5 + j 26.1 5.2 m -- Radial Z = 110.1 + j 23.3 5.4 m -- Radial Z = 109.2 + j 17.2 5.6 m -- Radial Z = 107.9 + j 19.5 5.8 m -- Radial Z = 106.5 + j 18.4 6.0 m -- Radial Z = 105.1 + j 17.7 6.2 m -- Radial Z = 103.8 + j 17.5 6.4 m -- Radial Z = 102.7 + j 17.6 6.6 m -- Radial Z = 101.7 + j 17.9 6.8 m -- Radial Z = 101.0 + j 18.4 7.0 m -- Radial Z = 100.5 + j 19.0 7.2 m -- Radial Z = 100.2 + j 19.6 7.4 m -- Radial Z = 100.1 + j 20.2 . . 8.0 m -- Radial Z = 100.5 + j 21.4 . . 9.0 m -- Radial Z = 101.8 + j 21.7 These data are so close to the center of a Smith Chart normalized to 101.6 + j 21. Not sure how you normalize with a complex number, but assume it is with the magnitude of Z. Reg: It seems that the Smith Chart must be normalized with a complex number. As expected, with a very high loss transmission line, the impedance rapidly spirals towards the complex Zo. From the data it is not clear what is really happening, but on the Smith Chart it becomes very clear. The curve crosses the "Real" axis at: 2.8 m, 5.8 m, and about 9 m. At this point the data are so close to the Smith Chart center that any more results are irrelevant. Even without a Smith Chart, normalization of the data will clearly reveal where the quarter wave multiples are located. Frank |
Radiating Efficiency
Frank,
I don't understand Smith Charts. But you can tell me what conclusions the charts lead you to believe about the behaviour of a lossy transmission line of various lengths. I am concerned with reducing the uncertainty of my program from the information contained in the graphs of input impedance versus length which I am now able to complete. joke I don't suppose you would like to repeat the excercise using a ground resistivity of 1000 ohm-metres. Or repeating at 20 MHz. joke ---- Reg. "Frank's" wrote in message news:1qOzg.181008$771.118115@edtnps89... "Frank's" wrote in message news:H6Mzg.180983$771.142858@edtnps89... "Reg Edwards" wrote in message ... 1.8 m -- Radial Z = 70.17 - j 24.1 1.9 m -- Radial Z = 68.5 - j 19.0 2.0 m -- Radial Z = 67.2 - j 14.2 2.1 m -- Radial Z = 66.2 - j 9.5 2.2 m -- Radial Z = 65.5 - j 5.0 2.3 m -- Radial Z = 65.1 - j 0.6 2.4 m -- Radial Z = 65.0 + j 3.6 2.5 m -- Radial Z = 65.2 + j 7.8 2.6 m -- Radial Z = 65.6 + j 11.7 I can keep going if you think that these are the results you expected. I am tempted to continue, in steps of 0.1 m, and plotting the results on the Smith Chart. I expect the data to rapidly spiral to the center of a chart normalized at 101. 6 ohms. Frank ====================================== Frank, Very interesting! I am plotting graphs of R and jX versus length of radial. Joining up the dots I expect to see shallow sinewaves superimposed on fairly level mean values. At present there is a definite shape of curve appearing in the reactance values while the resistance values are still fairly level. As length increases I expect to see something similar happening to the resistance curve which seems to be in between peaks and troughs. From now on it seems safe for you to increase length in increments of 0.2 metres. There is no danger of missing peaks and troughs in the curves. Please keep up the good work. ---- Reg 1.8 m -- Radial Z = , I have reached 7.4m, but there does not seem any point in continuing. Let me know what you think. It does not appear to be behaving as I would expect of a transmission line; but then I have no experience of transmission lines immersed in a lossy material. Also note the jump in impedance at 5.6 m. I double checked the result, and it appears to be correct. 2.8 m -- Radial Z = 67.5 + j 19.2 3.0 m -- Radial Z = 70.6 + j 25.9 3.2 m -- Radial Z = 74.7 + j 31.6 3.4 m -- Radial Z = 79.8 + j 36.0 3.6 m -- Radial Z = 85.5 + j 39.0 3.8 m -- Radial Z = 91.4 + j 40.4 4.0 m -- Radial Z = 97.1 + j 40.2 4.2 m -- Radial Z = 102.0 + j 38.6 4.4 m -- Radial Z = 105.9 + j 35.9 4.6 m -- Radial Z = 108.6 + j 32.7 4.8 m -- Radial Z = 110.1 + j 29.3 5.0 m -- Radial Z = 110.5 + j 26.1 5.2 m -- Radial Z = 110.1 + j 23.3 5.4 m -- Radial Z = 109.2 + j 17.2 5.6 m -- Radial Z = 107.9 + j 19.5 5.8 m -- Radial Z = 106.5 + j 18.4 6.0 m -- Radial Z = 105.1 + j 17.7 6.2 m -- Radial Z = 103.8 + j 17.5 6.4 m -- Radial Z = 102.7 + j 17.6 6.6 m -- Radial Z = 101.7 + j 17.9 6.8 m -- Radial Z = 101.0 + j 18.4 7.0 m -- Radial Z = 100.5 + j 19.0 7.2 m -- Radial Z = 100.2 + j 19.6 7.4 m -- Radial Z = 100.1 + j 20.2 . . 8.0 m -- Radial Z = 100.5 + j 21.4 . . 9.0 m -- Radial Z = 101.8 + j 21.7 These data are so close to the center of a Smith Chart normalized to 101.6 + j 21. Not sure how you normalize with a complex number, but assume it is with the magnitude of Z. Reg: It seems that the Smith Chart must be normalized with a complex number. As expected, with a very high loss transmission line, the impedance rapidly spirals towards the complex Zo. From the data it is not clear what is really happening, but on the Smith Chart it becomes very clear. The curve crosses the "Real" axis at: 2.8 m, 5.8 m, and about 9 m. At this point the data are so close to the Smith Chart center that any more results are irrelevant. Even without a Smith Chart, normalization of the data will clearly reveal where the quarter wave multiples are located. Frank |
Radiating Efficiency
"Reg Edwards" wrote But you can tell me what conclusions the charts lead you to believe about the behaviour of a lossy transmission line of various lengths. Sooo, are we investigating behaviour of lossy transmission lines heating the worms, aka wire in a dirt, or behaviour of radial within the vertical antenna, aka the other half of a "dipole" and it's participation in forming the radiation pattern and effciency? Helooooo?!? What's next? Shortening the radiator? As a test case, see how "inefficient" is the 3/8 wave vertical, with 360 3/8 wave radials on 160m. If your "software" tels me that it will be more efficient with 1 m radials, you are in for serious reality check. Yuri, K3BU |
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