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high earth resistance
All,
This question is: How can the ground system for a vertical be improved? The RF ground calculates to about 52 Ohms. It consists of 16 radials varying in length from 24 to 45 feet. Four are buried and 12 are on ground. The location is the Pacific Northwest part way up a hillside. The surface is grass, the subsurface is clay. The value varies a little with rain, it has measured as low as 46 to 48 Ohms. Just as additional information, the antenna is a loaded vertical, overall about 4 meters. The coil is about 35% from the base. The antenna is raised about 1 meter. I am aware the configuration is not optimum and a top hat would help, those changes will be made independently. Several articles indicate this radial system should yield well under 20 Ohms, some estimates are under 8 Ohms. At this point my first target is under 20 Ohms. Little that I do seems to affect it. My question is: What am I missing? The problem is the value does not appear to change very much. I started with 4 radials and have been adding them periodically. Those changes do not seem to have a measurable effect. (However I must admit that earlier numbers not as good as the current measurement. I believe the current measurements are better then 10%.) Thanks - Dan |
high earth resistance
dansawyeror wrote in
: Dan, You refer to calculation of ground resistance and later to measurements. It would help if you explained what you measured, the frequency of measurement, and how you measured it, then how you calculated your results. Perhaps the regimen of that might help you to an answer. Owen All, This question is: How can the ground system for a vertical be improved? The RF ground calculates to about 52 Ohms. It consists of 16 radials varying in length from 24 to 45 feet. Four are buried and 12 are on ground. The location is the Pacific Northwest part way up a hillside. The surface is grass, the subsurface is clay. The value varies a little with rain, it has measured as low as 46 to 48 Ohms. Just as additional information, the antenna is a loaded vertical, overall about 4 meters. The coil is about 35% from the base. The antenna is raised about 1 meter. I am aware the configuration is not optimum and a top hat would help, those changes will be made independently. Several articles indicate this radial system should yield well under 20 Ohms, some estimates are under 8 Ohms. At this point my first target is under 20 Ohms. Little that I do seems to affect it. My question is: What am I missing? The problem is the value does not appear to change very much. I started with 4 radials and have been adding them periodically. Those changes do not seem to have a measurable effect. (However I must admit that earlier numbers not as good as the current measurement. I believe the current measurements are better then 10%.) Thanks - Dan |
high earth resistance
Owen,
I did not want to concentrate on that; however it is a fair question. The frequency is 3970 kc and the methodology is based on use of an hp network analyzer. The analyzer has both polar and Cartesian displays. The outputs were cross checked.f The feed line was calibrated using a 25 Ohm termination to 180 degrees. The return measured about -9.4db (close to what is predicted with line loss). This was checked against an open circuit confirmed the calibration by a reading 1 degree and 1db. (very close to what is predicted) For the actual measurement the return loss was measured at just over -26db at 0 degrees. (no feed-point network) Three models were used to calculate the input resistance of the antenna. They all predicted an antenna resistance of about 3 to 4 Ohms. (the suggested input network is 120 pf) To focus on the question is: Why is the ground resistance so high? It is not important at this stage to determine its precise value. The point is it is high enough to cause a return of 0 degrees. This puts the 'system' at over 50 degrees. Even if the antenna were 6 to 8 Ohms the ground loss would be at least 42 to 44 Ohms. - Dan Owen Duffy wrote: dansawyeror wrote in : Dan, You refer to calculation of ground resistance and later to measurements. It would help if you explained what you measured, the frequency of measurement, and how you measured it, then how you calculated your results. Perhaps the regimen of that might help you to an answer. Owen All, This question is: How can the ground system for a vertical be improved? The RF ground calculates to about 52 Ohms. It consists of 16 radials varying in length from 24 to 45 feet. Four are buried and 12 are on ground. The location is the Pacific Northwest part way up a hillside. The surface is grass, the subsurface is clay. The value varies a little with rain, it has measured as low as 46 to 48 Ohms. Just as additional information, the antenna is a loaded vertical, overall about 4 meters. The coil is about 35% from the base. The antenna is raised about 1 meter. I am aware the configuration is not optimum and a top hat would help, those changes will be made independently. Several articles indicate this radial system should yield well under 20 Ohms, some estimates are under 8 Ohms. At this point my first target is under 20 Ohms. Little that I do seems to affect it. My question is: What am I missing? The problem is the value does not appear to change very much. I started with 4 radials and have been adding them periodically. Those changes do not seem to have a measurable effect. (However I must admit that earlier numbers not as good as the current measurement. I believe the current measurements are better then 10%.) Thanks - Dan |
high earth resistance
dansawyeror wrote in
: .... I think a summary is that at 3970KHz, the feedpoint Z looks like about 45 +j0 and you reckon the radiation resistance should be around 4+j0, suggesting the earth system contributes around 40 ohms of resistance. Observations at a single frequency provide a limited view of what might be happening. Elevated radials should exhibit a clear resonance, and will offer the lowest impedance at resonance. Buried radials will not exhibit such a clear resonance in lossy soil, and using your wire to form more short radials might give better performance than few longer radials. Radials lying on the ground are likely to be somewhere in between. You may observe resonance, and in that case the ground system impedance will be optimised by "tuning" those radials (which will probably be a good bit shorter than formula length for free space radials). Owen |
high earth resistance
On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror
wrote: To focus on the question is: Why is the ground resistance so high? Hi Dan, Focus on Why? Why? This is something that you have absolutely no control over without a huge investment in new dirt several 10s of meter deep out to at least as far as the antenna is tall. You want less resistance? then dig a hole and fill it with quartz sand. Build a dune to drive the loss down further. A suitable substitute is to lay out a ground field (radial wires). The more the better. Here is the Why? you should be asking (Why more wires?). The path through the earth is shortened between adjacent ground wires. Less path, less loss. The earth current travels not IN toward the center as the radials do. The earth current travels ACROSS or circumferentially towards the radial wires. This is due to the phase lag between the induced earth current and the radial current. The greater the distance between radials, the more path loss from the average distance between the radials, to the radials. As this is very difficult to treat in words alone, it is undoubtedly confusing in the description above. 73's Richard Clark, KB7QHC |
high earth resistance
Richard Clark wrote:
The path through the earth is shortened between adjacent ground wires. Less path, less loss. The earth current travels not IN toward the center as the radials do. The earth current travels ACROSS or circumferentially towards the radial wires. This is due to the phase lag between the induced earth current and the radial current. The greater the distance between radials, the more path loss from the average distance between the radials, to the radials. As this is very difficult to treat in words alone, it is undoubtedly confusing in the description above. Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Roy Lewallen, W7EL |
high earth resistance
To focus on the question is: Why is the ground resistance so high? It
is not important at this stage to determine its precise value. The point is it is high enough to cause a return of 0 degrees. This puts the 'system' at over 50 degrees. Even if the antenna were 6 to 8 Ohms the ground loss would be at least 42 to 44 Ohms. Simple answer- add more radials. Especially if the impedance is varying noticeably depending on ground moisture, you need more radials. Another possibility is that if there is some other conductor in the near field (tower, house, other wires, etc), the impedance might not be what you expect. Tor N4OGW |
high earth resistance
Well, I can't explain why lagging or leading current and voltage would change the physical direction of propagation of the EM wave front through the ground... But I do know that Dan does not have enough radials.. He needs at least another 16 and better yet would be in the range of 50 total... denny / k8do |
high earth resistance
On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror wrote:
Owen, I did not want to concentrate on that; however it is a fair question. The frequency is 3970 kc and the methodology is based on use of an hp network analyzer. The analyzer has both polar and Cartesian displays. The outputs were cross checked.f [snip] To focus on the question is: Why is the ground resistance so high? It is not important at this stage to determine its precise value. The point is it is high enough to cause a return of 0 degrees. This puts the 'system' at over 50 degrees. Even if the antenna were 6 to 8 Ohms the ground loss would be at least 42 to 44 Ohms. How is the antenna loaded and what is the Q of the loading coil? Coil losses are almost certainly 5 ohms and could easily be as high as 35 ohms. bart |
high earth resistance
Richard,
What coupling does this imply? Is it direct as is contact or is it a field coupling as capacitive or inductive? The question is: what would insulation or corrosion due to buried radials? - Dan Richard Clark wrote: On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror wrote: To focus on the question is: Why is the ground resistance so high? Hi Dan, Focus on Why? Why? This is something that you have absolutely no control over without a huge investment in new dirt several 10s of meter deep out to at least as far as the antenna is tall. You want less resistance? then dig a hole and fill it with quartz sand. Build a dune to drive the loss down further. A suitable substitute is to lay out a ground field (radial wires). The more the better. Here is the Why? you should be asking (Why more wires?). The path through the earth is shortened between adjacent ground wires. Less path, less loss. The earth current travels not IN toward the center as the radials do. The earth current travels ACROSS or circumferentially towards the radial wires. This is due to the phase lag between the induced earth current and the radial current. The greater the distance between radials, the more path loss from the average distance between the radials, to the radials. As this is very difficult to treat in words alone, it is undoubtedly confusing in the description above. 73's Richard Clark, KB7QHC |
high earth resistance
On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen
wrote: Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Hi Roy, Brown, Lewis and Epstein. 73's Richard Clark, KB7QHC |
high earth resistance
On Mon, 09 Apr 2007 08:21:36 -0700, dansawyeror
wrote: What coupling does this imply? Is it direct as is contact or is it a field coupling as capacitive or inductive? Hi Dan, Yes to all three. The question is: what would insulation or corrosion due to buried radials? Not much, practically; unless the wire is extremely thin, and the currents are large. Loss is in the earth. You can, of course, build a very crummy radial system if you try hard. For instance, using expensive piano wire in place of cheap house wiring. 73's Richard Clark, KB7QHC |
high earth resistance
Richard Clark wrote:
On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen wrote: Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Hi Roy, Brown, Lewis and Epstein. Which page? Roy Lewallen, W7EL |
high earth resistance
On Mon, 09 Apr 2007 09:08:09 -0700, Roy Lewallen
wrote: Richard Clark wrote: On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen wrote: Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Hi Roy, Brown, Lewis and Epstein. Which page? Hi Roy, It is distributed through the discussion. Pg. 757 (at the top of the page introduces): "These losses are due to conduction of earth currents through a high resistance earth..." "Where there are radial ground wires present, the earth current consists of two components, part of which flows in the earth itself and the remainder of which flows in the buried wires." "...all the various components differ in phase." This establishes the relationship and distinction in the various currents. It is the current in the earth that is the topic of discussion here. That current is out of phase with respect to the currents (at the same radial distance) found in the buried wires. No wires, no phase issue. No phase issue, and earth currents would be radial. Now, to distinguish this from circumferential is not to say this is absolute (it does not follow an arc of constant radius). This is extended to coverage at the bottom of page 758: "The actual earth current and the current flowing in the radial wires are given...." [formula shown in the original] "From (8) [that formula] we see that the earth current proper leads the current in the wires by 90 electrical degrees." At a radius, the earth phase and the wire phase exhibit a potential difference which results in conduction that is not strictly radial (the term circumferential through the combination of vectors might be replaced with spiral, or diagonal). The earth's resistance comes into play at page 760: "When the earth is of good conductivity [a paradox ensues], the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." and hence the advice for replacing dirt with sand OR providing more closely spaced radials, closer in. "Thus the regions of high current density are subjected to still more current with higher losses in these regions." 73's Richard Clark, KB7QHC |
high earth resistance
Richard,
I disagree with your conclusion that currents flow circumferentially. It does not say so in the paper, and I don't believe it can be inferred from what is said in the paper. If you were to draw a diagram showing the currents produced by the phase shift between earth and radial currents, you'd find that the net current resulting from this phase shift is purely radial, not circumferential. If the currents flow circumferentially, do they flow clockwise or counterclockwise, and at what magnitude relative to the radial currents? Surely there's some reference which shows this calculation which you could direct me to or, if not, you could show how the calculation is done and what the result is. Roy Lewallen, W7EL Richard Clark wrote: On Mon, 09 Apr 2007 09:08:09 -0700, Roy Lewallen wrote: Richard Clark wrote: On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen wrote: Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Hi Roy, Brown, Lewis and Epstein. Which page? Hi Roy, It is distributed through the discussion. Pg. 757 (at the top of the page introduces): "These losses are due to conduction of earth currents through a high resistance earth..." "Where there are radial ground wires present, the earth current consists of two components, part of which flows in the earth itself and the remainder of which flows in the buried wires." "...all the various components differ in phase." This establishes the relationship and distinction in the various currents. It is the current in the earth that is the topic of discussion here. That current is out of phase with respect to the currents (at the same radial distance) found in the buried wires. No wires, no phase issue. No phase issue, and earth currents would be radial. Now, to distinguish this from circumferential is not to say this is absolute (it does not follow an arc of constant radius). This is extended to coverage at the bottom of page 758: "The actual earth current and the current flowing in the radial wires are given...." [formula shown in the original] "From (8) [that formula] we see that the earth current proper leads the current in the wires by 90 electrical degrees." At a radius, the earth phase and the wire phase exhibit a potential difference which results in conduction that is not strictly radial (the term circumferential through the combination of vectors might be replaced with spiral, or diagonal). The earth's resistance comes into play at page 760: "When the earth is of good conductivity [a paradox ensues], the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." and hence the advice for replacing dirt with sand OR providing more closely spaced radials, closer in. "Thus the regions of high current density are subjected to still more current with higher losses in these regions." 73's Richard Clark, KB7QHC |
high earth resistance
On Mon, 09 Apr 2007 12:12:32 -0700, Roy Lewallen
wrote: I disagree with your conclusion that currents flow circumferentially. It does not say so in the paper, and I don't believe it can be inferred from what is said in the paper. Hi Roy, To insist that the paper be complete where the reader has the competence to understand what is implied; well, that goes beyond standard practice. Further, the implication is hardly momentous when the force of the writing is in demonstrating (not finding) a solution to loss. Their style is clearly descriptive, not pedantic. One very simple observation drawn directly from the text at page 760: "When the earth is of good conductivity, the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." How is it THIS current is traveling radially, the same direction as the wires, both leaving the wire (an orthogonal aspect) and yet moving in the same direction. This is a contradiction to the geometry of the description if we are to abide by your rejection of my "interpretation." Their (not my) statement, supported by their other text, hardly makes sense otherwise. Current only flows along a potential gradient and the phase shift between (by their own distinctions) wire and ground constitutes such a gradient. It is a vastly more speculative "interpretation" to suggest the current leaves the wire to travel in the same direction and the authors definitely don't say that, do they? Common sense would dictate a fairer interpretation that conforms to phases and the distinctions (separation of currents) they drew from them. 73's Richard Clark, KB7QHC |
high earth resistance
So, does the current go clockwise or counterclockwise? How much goes
that way compared to the radial component? Where can I find a quantitative or explicit statement of your interpretation? Roy Lewallen, W7EL Richard Clark wrote: On Mon, 09 Apr 2007 12:12:32 -0700, Roy Lewallen wrote: I disagree with your conclusion that currents flow circumferentially. It does not say so in the paper, and I don't believe it can be inferred from what is said in the paper. Hi Roy, To insist that the paper be complete where the reader has the competence to understand what is implied; well, that goes beyond standard practice. Further, the implication is hardly momentous when the force of the writing is in demonstrating (not finding) a solution to loss. Their style is clearly descriptive, not pedantic. One very simple observation drawn directly from the text at page 760: "When the earth is of good conductivity, the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." How is it THIS current is traveling radially, the same direction as the wires, both leaving the wire (an orthogonal aspect) and yet moving in the same direction. This is a contradiction to the geometry of the description if we are to abide by your rejection of my "interpretation." Their (not my) statement, supported by their other text, hardly makes sense otherwise. Current only flows along a potential gradient and the phase shift between (by their own distinctions) wire and ground constitutes such a gradient. It is a vastly more speculative "interpretation" to suggest the current leaves the wire to travel in the same direction and the authors definitely don't say that, do they? Common sense would dictate a fairer interpretation that conforms to phases and the distinctions (separation of currents) they drew from them. 73's Richard Clark, KB7QHC |
high earth resistance
Severns' article "Verticals, Ground Systems and Some History",
July 2000, p. 39, quotes the following: "As indicated in Figure 1, the tangential component of the H field (H(phi)) induces horizontal currents (Ih) flowing radially and the normal component of the E field (Ez) induces vertically flowing currents (Iv). The paper is available for download from www.arrl.org. Frank "Roy Lewallen" wrote in message ... Richard, I disagree with your conclusion that currents flow circumferentially. It does not say so in the paper, and I don't believe it can be inferred from what is said in the paper. If you were to draw a diagram showing the currents produced by the phase shift between earth and radial currents, you'd find that the net current resulting from this phase shift is purely radial, not circumferential. If the currents flow circumferentially, do they flow clockwise or counterclockwise, and at what magnitude relative to the radial currents? Surely there's some reference which shows this calculation which you could direct me to or, if not, you could show how the calculation is done and what the result is. Roy Lewallen, W7EL Richard Clark wrote: On Mon, 09 Apr 2007 09:08:09 -0700, Roy Lewallen wrote: Richard Clark wrote: On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen wrote: Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Hi Roy, Brown, Lewis and Epstein. Which page? Hi Roy, It is distributed through the discussion. Pg. 757 (at the top of the page introduces): "These losses are due to conduction of earth currents through a high resistance earth..." "Where there are radial ground wires present, the earth current consists of two components, part of which flows in the earth itself and the remainder of which flows in the buried wires." "...all the various components differ in phase." This establishes the relationship and distinction in the various currents. It is the current in the earth that is the topic of discussion here. That current is out of phase with respect to the currents (at the same radial distance) found in the buried wires. No wires, no phase issue. No phase issue, and earth currents would be radial. Now, to distinguish this from circumferential is not to say this is absolute (it does not follow an arc of constant radius). This is extended to coverage at the bottom of page 758: "The actual earth current and the current flowing in the radial wires are given...." [formula shown in the original] "From (8) [that formula] we see that the earth current proper leads the current in the wires by 90 electrical degrees." At a radius, the earth phase and the wire phase exhibit a potential difference which results in conduction that is not strictly radial (the term circumferential through the combination of vectors might be replaced with spiral, or diagonal). The earth's resistance comes into play at page 760: "When the earth is of good conductivity [a paradox ensues], the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." and hence the advice for replacing dirt with sand OR providing more closely spaced radials, closer in. "Thus the regions of high current density are subjected to still more current with higher losses in these regions." 73's Richard Clark, KB7QHC |
high earth resistance
The article is in QST.
"Frank's" wrote in message news:6qxSh.56971$__3.40608@edtnps90... Severns' article "Verticals, Ground Systems and Some History", July 2000, p. 39, quotes the following: "As indicated in Figure 1, the tangential component of the H field (H(phi)) induces horizontal currents (Ih) flowing radially and the normal component of the E field (Ez) induces vertically flowing currents (Iv). The paper is available for download from www.arrl.org. Frank "Roy Lewallen" wrote in message ... Richard, I disagree with your conclusion that currents flow circumferentially. It does not say so in the paper, and I don't believe it can be inferred from what is said in the paper. If you were to draw a diagram showing the currents produced by the phase shift between earth and radial currents, you'd find that the net current resulting from this phase shift is purely radial, not circumferential. If the currents flow circumferentially, do they flow clockwise or counterclockwise, and at what magnitude relative to the radial currents? Surely there's some reference which shows this calculation which you could direct me to or, if not, you could show how the calculation is done and what the result is. Roy Lewallen, W7EL Richard Clark wrote: On Mon, 09 Apr 2007 09:08:09 -0700, Roy Lewallen wrote: Richard Clark wrote: On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen wrote: Indeed it is. Can you point me to a reference where I can get a more detailed explanation of this circumferential current and its cause? Hi Roy, Brown, Lewis and Epstein. Which page? Hi Roy, It is distributed through the discussion. Pg. 757 (at the top of the page introduces): "These losses are due to conduction of earth currents through a high resistance earth..." "Where there are radial ground wires present, the earth current consists of two components, part of which flows in the earth itself and the remainder of which flows in the buried wires." "...all the various components differ in phase." This establishes the relationship and distinction in the various currents. It is the current in the earth that is the topic of discussion here. That current is out of phase with respect to the currents (at the same radial distance) found in the buried wires. No wires, no phase issue. No phase issue, and earth currents would be radial. Now, to distinguish this from circumferential is not to say this is absolute (it does not follow an arc of constant radius). This is extended to coverage at the bottom of page 758: "The actual earth current and the current flowing in the radial wires are given...." [formula shown in the original] "From (8) [that formula] we see that the earth current proper leads the current in the wires by 90 electrical degrees." At a radius, the earth phase and the wire phase exhibit a potential difference which results in conduction that is not strictly radial (the term circumferential through the combination of vectors might be replaced with spiral, or diagonal). The earth's resistance comes into play at page 760: "When the earth is of good conductivity [a paradox ensues], the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." and hence the advice for replacing dirt with sand OR providing more closely spaced radials, closer in. "Thus the regions of high current density are subjected to still more current with higher losses in these regions." 73's Richard Clark, KB7QHC |
high earth resistance
Owen Duffy wrote in
: dansawyeror wrote in : ... I think a summary is that at 3970KHz, the feedpoint Z looks like about 45 +j0 and you reckon the radiation resistance should be around 4+j0, suggesting the earth system contributes around 40 ohms of resistance. Observations at a single frequency provide a limited view of what might be happening. Dan, I asked your for the details of your antenna and measurements, and how you did your calculations, but I am still left wondering how you have what appears to be a purely resistive feedpoint impedance and a radiation resistance of 4 ohms. The second implies a short vertical, and if that is the case, the first implies some form of loading... but you didn't mention loading of any kind. Loading, if you have used it, may introduce an equivalent series resistance at the feedpoint. Once again, a dansawyer problems leaves us guessing to fill in the missing dots before attempting to joint them up to make a picture. Often, solving a problem is about being able to draw the picture, once the picture is draw, the answer becomes trivial. Owen |
high earth resistance
I asked your for the details of your antenna and measurements, and how
you did your calculations, but I am still left wondering how you have what appears to be a purely resistive feedpoint impedance and a radiation resistance of 4 ohms. The second implies a short vertical, and if that is the case, the first implies some form of loading... but you didn't mention loading of any kind. Loading, if you have used it, may introduce an equivalent series resistance at the feedpoint. Once again, a dansawyer problems leaves us guessing to fill in the missing dots before attempting to joint them up to make a picture. Often, solving a problem is about being able to draw the picture, once the picture is draw, the answer becomes trivial. Owen I was also confused by the base of the antenna being 1 meter above ground, and the radials lying on the ground, or buried. If the base of the antenna is 1 meter high, then any connection to the radials is part of the radiation system. Why would you feed the antenna 1 meter up, and not at the base? The antenna is therefore a ground mounted 5 meter vertical. NEC predicts an input impedance of 4 - j 1300 with 36 ten meter radials 1" below an average ground. Loading coils will of course add to the input impedance. The measured data are suspect. It would be interesting to know the length, and type, of coax connecting to the network analyzer. The return loss of 25 ohms at the end of a piece of coax will be 9.5 dB. Unless the coax is cut to a precise known length it is unlikely that the phase angle of the return loss will be zero. Frank. |
high earth resistance
On Mon, 09 Apr 2007 13:27:21 -0700, Roy Lewallen
wrote: So, does the current go clockwise or counterclockwise? How much goes that way compared to the radial component? Where can I find a quantitative or explicit statement of your interpretation? Hi Roy, Probably in a library. Field work seems to resolve issues too. It may even prove your speculation in contradiction to mine. Outside of these authors, we both seem to be shy of "authoritative references" to parse that Byzantine statement of theirs. I can only further speculate that BL&H were remiss in specifically quantifying loss (you aren't asking me for numbers you are already aware of, are you?), while offering numerous formulaic relationships of loss against many factors. If we look at their data and observe that adding radials lowers loss, but not by any precise relationship, are we left without quantifiable proof, or the obvious implication of strong correlation? Was there deceit in their arriving at some conclusions through inference? As Reggie would note, they didn't actually measure earth at all! Such a retort was met with indignity in the past, is it now their impeachment? However, as to counter/anti/clockwise, What impels current to follow any such presumption? There are two sides to every wire laying in a plane and phase mappings for earth currents that are symmetrical about them. To anticipate your challenging me on that statement (clearly BL&H never, explicitly say this), I can only offer a modest sense of observing the bleeding obvious. Myself, I don't find BL&H so obscure to impose this remarkable characteristic that current leaves the wire on only one side. Brown, Lewis and Epstein were REPORTING, not inventing, nor offering pedant readings of scripture. Scribes, such as we are, are free to interpret within the bounds of their own data, assumptions, and conclusions. I've offered mine that conforms to many of their points. If you have your own, you must survive by the same strictures. Given the specific contention, I am especially intrigued in how you would answer why the current departed the wire, and where it goes in light of a potential map created by the phase shifts. 73's Richard Clark, KB7QHC |
high earth resistance
Richard Clark wrote:
On Mon, 09 Apr 2007 13:27:21 -0700, Roy Lewallen wrote: So, does the current go clockwise or counterclockwise? How much goes that way compared to the radial component? Where can I find a quantitative or explicit statement of your interpretation? Hi Roy, Probably in a library. Field work seems to resolve issues too. It may even prove your speculation in contradiction to mine. Outside of these authors, we both seem to be shy of "authoritative references" to parse that Byzantine statement of theirs. I can only further speculate that BL&H were remiss in specifically quantifying loss (you aren't asking me for numbers you are already aware of, are you?), while offering numerous formulaic relationships of loss against many factors. If we look at their data and observe that adding radials lowers loss, but not by any precise relationship, are we left without quantifiable proof, or the obvious implication of strong correlation? Was there deceit in their arriving at some conclusions through inference? As Reggie would note, they didn't actually measure earth at all! Such a retort was met with indignity in the past, is it now their impeachment? However, as to counter/anti/clockwise, What impels current to follow any such presumption? There are two sides to every wire laying in a plane and phase mappings for earth currents that are symmetrical about them. To anticipate your challenging me on that statement (clearly BL&H never, explicitly say this), I can only offer a modest sense of observing the bleeding obvious. Myself, I don't find BL&H so obscure to impose this remarkable characteristic that current leaves the wire on only one side. Brown, Lewis and Epstein were REPORTING, not inventing, nor offering pedant readings of scripture. Scribes, such as we are, are free to interpret within the bounds of their own data, assumptions, and conclusions. I've offered mine that conforms to many of their points. If you have your own, you must survive by the same strictures. Given the specific contention, I am especially intrigued in how you would answer why the current departed the wire, and where it goes in light of a potential map created by the phase shifts. 73's Richard Clark, KB7QHC Ok, I understand that. Your answers to the two questions I asked are that you don't know and you don't know. In the absence of any evidence, I'll continue to disbelieve there's a circumferential component of the current. Roy Lewallen, W7EL |
high earth resistance
Frank's wrote:
Severns' article "Verticals, Ground Systems and Some History", July 2000, p. 39, quotes the following: "As indicated in Figure 1, the tangential component of the H field (H(phi)) induces horizontal currents (Ih) flowing radially and the normal component of the E field (Ez) induces vertically flowing currents (Iv). The paper is available for download from www.arrl.org. Thanks for the reference, but it makes no mention of circumferential currents. At first glance, figure 1 seems to show a circumferential Ih, but the text (as you quoted) clearly calls this a radial current -- the circle in figure 1 turns out to be Hphi, circling the vertically directed current. Roy Lewallen, W7EL |
high earth resistance
Roy Lewallen wrote:
So, does the current go clockwise or counterclockwise? How much goes that way compared to the radial component? Where can I find a quantitative or explicit statement of your interpretation? I think, based on the excerpt Richard has provided, it does both. Imagine a leaky hose with water diffusing into the surroundings. So, if you were to integrate over the entire width,or over any region which is symmetric over the wire, the *net* is entirely radial, but if you look at a small region, directly adjacent to the wire, there will be current diverging from the wire as you move outward (assuming current flow is outward... obviously, on the opposite half cycle, it converges toward the wire, as it moves generally inward)... I suspect one could also analyze it as a wave propagating away from teh wire in the lossy surrounding medium, where the medium has a lower propagation velocity than in the wire. (e.g. imagine a waveguide made with the walls being soil) Another sort of "hydraulic" model would be if you represented the radials as below grade drainage ditches which have a lot more pitch than the surrounding soil, so the water tends to flow diagonally down the ditch walls. The interesting question would be whether this is important at all.. One might go through lots and lots of analysis, worrying about the small incremental effects of non-radial current, and find that the inherent variations in soil properties are orders of magnitude larger. Sounds like a good exercise for a graduate level E&M or calculus class.. you could cast it as a similar exercise in heat flow.. both temperature and electrostatic fields satisfy Laplace's equation. |
high earth resistance
"Roy Lewallen" wrote in message ... Frank's wrote: Severns' article "Verticals, Ground Systems and Some History", July 2000, p. 39, quotes the following: "As indicated in Figure 1, the tangential component of the H field (H(phi)) induces horizontal currents (Ih) flowing radially and the normal component of the E field (Ez) induces vertically flowing currents (Iv). The paper is available for download from www.arrl.org. Thanks for the reference, but it makes no mention of circumferential currents. At first glance, figure 1 seems to show a circumferential Ih, but the text (as you quoted) clearly calls this a radial current -- the circle in figure 1 turns out to be Hphi, circling the vertically directed current. Roy Lewallen, W7EL Yes, that is exactly the way I interpreted the article which is based on the B, L, and E paper. Frank |
high earth resistance
Richard Clark, KB7QHC wrote:
"Brown, Lewis and Epstein." Ed Laport worked with B,L,&E at RCA. I don`t have my copy of his "Radio Antenna Engineering" at hand, but do recall Ed`s exhortations not to interconnect radials beyond an antenna`s drivepoint because it encourages hysteresis (circulating) currents which do nothing to improve antenna action but do increase loss. All the useful ground currents at a tower are radial with respect to the tower`s base. Best regards, Richard Harrison, KB5WZI |
high earth resistance
Bart,
Thanks for the thought. Modeling predicts it is much lower then that; however I don't want to overlook anything. The Q looks very good on the network analyzer, however I have never run the math. The coil is about 3.3 inches in diameter. I will let you know. Dan Bart Rowlett wrote: On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror wrote: Owen, I did not want to concentrate on that; however it is a fair question. The frequency is 3970 kc and the methodology is based on use of an hp network analyzer. The analyzer has both polar and Cartesian displays. The outputs were cross checked.f [snip] To focus on the question is: Why is the ground resistance so high? It is not important at this stage to determine its precise value. The point is it is high enough to cause a return of 0 degrees. This puts the 'system' at over 50 degrees. Even if the antenna were 6 to 8 Ohms the ground loss would be at least 42 to 44 Ohms. How is the antenna loaded and what is the Q of the loading coil? Coil losses are almost certainly 5 ohms and could easily be as high as 35 ohms. bart |
high earth resistance
Owen,
The antenna is a short loaded vertical. The base is about 1.1 meters long and 90mm in diameter, the coil is about 160mm long and about 80mm in diameter, the top is 3 meters. The coil wire is 12 gage and the spacing is about .5. As a model cross check, the impedance of the coil measures about 60 uH. - Dan Owen Duffy wrote: Owen Duffy wrote in : dansawyeror wrote in : ... I think a summary is that at 3970KHz, the feedpoint Z looks like about 45 +j0 and you reckon the radiation resistance should be around 4+j0, suggesting the earth system contributes around 40 ohms of resistance. Observations at a single frequency provide a limited view of what might be happening. Dan, I asked your for the details of your antenna and measurements, and how you did your calculations, but I am still left wondering how you have what appears to be a purely resistive feedpoint impedance and a radiation resistance of 4 ohms. The second implies a short vertical, and if that is the case, the first implies some form of loading... but you didn't mention loading of any kind. Loading, if you have used it, may introduce an equivalent series resistance at the feedpoint. Once again, a dansawyer problems leaves us guessing to fill in the missing dots before attempting to joint them up to make a picture. Often, solving a problem is about being able to draw the picture, once the picture is draw, the answer becomes trivial. Owen |
high earth resistance
dansawyeror wrote in
: Owen, The antenna is a short loaded vertical. The base is about 1.1 meters long and 90mm in diameter, the coil is about 160mm long and about 80mm in diameter, the top is 3 meters. The coil wire is 12 gage and the spacing is about .5. As a model cross check, the impedance of the coil measures about 60 uH. So, what do you think its impedance would be? 1500+j?? This is probably accounting for somewhere between 5 and 15 ohms of additional resistance, depending on Q. Owen |
high earth resistance
Owen Duffy wrote in news:Xns990EA1652F44Enonenowhere@
61.9.191.5: So, what do you think its impedance would be? 1500+j?? Duh, ??+j1500 is more like it. Same outcome, it is probably accounting for 5 to 15 ohms of resistance that you haven't included in your calcs. Owen |
high earth resistance
On Mon, 09 Apr 2007 17:04:35 -0700, Jim Lux
wrote: The interesting question would be whether this is important at all.. Hi Jim, Additional current through earth brings no net positive result and the question asked where the source of loss resides. Being unable to quantify temperature is no reason to keep picking up the wrong end of a soldering iron. "I don't believe it's hot" has rarely offered salve for burns. ;-) 73's Richard Clark, KB7QHC |
high earth resistance
Owen,
Are you referring to the coil alone? You are correct the coil accounts for a significant part, probably between 4 and 8 Ohms. I am measuring real values between 52 and 56 Ohms with 0j. This varies mainly with moisture. So let's say the total antenna is 12 Ohms. If the measured value is 52 that says the ground is 40. This puts us back to the original question: why is the ground so high? - Dan The circuit Owen Duffy wrote: Owen Duffy wrote in news:Xns990EA1652F44Enonenowhere@ 61.9.191.5: So, what do you think its impedance would be? 1500+j?? Duh, ??+j1500 is more like it. Same outcome, it is probably accounting for 5 to 15 ohms of resistance that you haven't included in your calcs. Owen |
high earth resistance
dansawyeror wrote in
: Owen, Are you referring to the coil alone? You are correct the coil accounts for a significant part, probably between 4 and 8 Ohms. I don't recall that you have told us where the coil is located, and I am not going to build a model to discover something that is relevant and that you could have told us. You have told us the coil had an "impedance" of 60uh. If it had an inductance of 60uh, it would have an impedance of 1500/Q+j1500 where Q at 3.9MHz for a practical coil is likely to be above 100 and less than 300. The resistive component of the coil has to be referred to the feedpoint so that you can deduct it from the total feedpoint Z. The best way to do that is your NEC model. You should be able to form a better range for the equivalent coil loss at the feedpoint than you stated above (since you seem to have the means of measuring the coil and inserting the values in your model). I am measuring real values between 52 and 56 Ohms with 0j. This varies mainly with moisture. So let's say the total antenna is 12 Ohms. If the measured value is 52 that says the ground is 40. This puts us back to the original question: why is the ground so high? You also previously said "... Even if the antenna were 6 to 8 Ohms the ground loss would be at least 42 to 44 Ohms." I read this to mean total system R is 50 ohms. In this post it is reported between 52 and 56. The other issue that Frank raised is the elevated feedpoint and whether you modelled that correctly. The radiation resistance you quoted seems (without checking) reasonable for a short monopole over ideal ground, but one expects it would be higher for an elevated feed point. Have you modelled the antenna you built, or have you build an antenna you cannot model accurately and are applying model results incorrectly to the thing you have built? Though you see only one question, "why is ground 40 ohms", you haven't disclosed enough information in your posts to convince me that it is 40 ohms. If you ask the wrong question, you might not get a useful answer. Is ground 40 ohms? Owen |
high earth resistance
I believe everything is stated:
The antenna is a short loaded vertical. The base is about 1.1 meters long and 90mm in diameter, the coil is about 160mm long and about 80mm in diameter, the top is 3 meters. The coil wire is 12 gage and the spacing is about .5. As a model cross check, t he impedance of the coil measures about 60 uH. The order is base, coil, and top. - Dan Owen Duffy wrote: dansawyeror wrote in : Owen, Are you referring to the coil alone? You are correct the coil accounts for a significant part, probably between 4 and 8 Ohms. I don't recall that you have told us where the coil is located, and I am not going to build a model to discover something that is relevant and that you could have told us. You have told us the coil had an "impedance" of 60uh. If it had an inductance of 60uh, it would have an impedance of 1500/Q+j1500 where Q at 3.9MHz for a practical coil is likely to be above 100 and less than 300. The resistive component of the coil has to be referred to the feedpoint so that you can deduct it from the total feedpoint Z. The best way to do that is your NEC model. You should be able to form a better range for the equivalent coil loss at the feedpoint than you stated above (since you seem to have the means of measuring the coil and inserting the values in your model). I am measuring real values between 52 and 56 Ohms with 0j. This varies mainly with moisture. So let's say the total antenna is 12 Ohms. If the measured value is 52 that says the ground is 40. This puts us back to the original question: why is the ground so high? You also previously said "... Even if the antenna were 6 to 8 Ohms the ground loss would be at least 42 to 44 Ohms." I read this to mean total system R is 50 ohms. In this post it is reported between 52 and 56. The other issue that Frank raised is the elevated feedpoint and whether you modelled that correctly. The radiation resistance you quoted seems (without checking) reasonable for a short monopole over ideal ground, but one expects it would be higher for an elevated feed point. Have you modelled the antenna you built, or have you build an antenna you cannot model accurately and are applying model results incorrectly to the thing you have built? Though you see only one question, "why is ground 40 ohms", you haven't disclosed enough information in your posts to convince me that it is 40 ohms. If you ask the wrong question, you might not get a useful answer. Is ground 40 ohms? Owen |
high earth resistance
dansawyeror wrote in
: The order is base, coil, and top. That wasn't clear to me, it looked to me like you were feeding it between the base and the top. I think Frank may have formed the same view. Owen |
high earth resistance
It's feed at the bottom.
The coil description should be inductance not impedance. - Dan Owen Duffy wrote: dansawyeror wrote in : The order is base, coil, and top. That wasn't clear to me, it looked to me like you were feeding it between the base and the top. I think Frank may have formed the same view. Owen |
high earth resistance
On Mon, 09 Apr 2007 16:42:10 -0700, Roy Lewallen wrote:
Brown, Lewis and Epstein were REPORTING, not inventing, nor offering pedant readings of scripture. Scribes, such as we are, are free to interpret within the bounds of their own data, assumptions, and conclusions. I've offered mine that conforms to many of their points. If you have your own, you must survive by the same strictures. Given the specific contention, I am especially intrigued in how you would answer why the current departed the wire, and where it goes in light of a potential map created by the phase shifts. 73's Richard Clark, KB7QHC Ok, I understand that. Your answers to the two questions I asked are that you don't know and you don't know. In the absence of any evidence, I'll continue to disbelieve there's a circumferential component of the current. Roy Lewallen, W7EL Roy, it seems to me everyone has missed an important point concerning a circumferential component of the current. We know that the current flowing on the radial wires is radial in direction. What seems to be missed is the current that returns to earth between the wire radials. That current is going to flow in the direction of the lowest resistance. As such it's not going to flow radially alongside the currents flowing on the wire, because the radial resistance of earth between the radial wires is much greater than the resistance of the wires. Consequently, currents reaching earth between the wires will find a lower resistance by traveling toward the nearest radial wire instead of continuing in a perfectly radial direction. This new direction of current flow will not necessarily perfectly circumferential, but will certainly be somewhere between radial and circumferential. Walt, W2DU |
high earth resistance
On Thu, 12 Apr 2007 18:55:30 GMT, Walter Maxwell
wrote: This new direction of current flow will not necessarily perfectly circumferential, but will certainly be somewhere between radial and circumferential. Hi Walt, So as BL&H report without too much pain: at page 760: "When the earth is of good conductivity, the current leaves the wires and enters the earth closer to the antenna than it does when the earth is a poor conductor." Now, as to your comment That current is going to flow in the direction of the lowest resistance. It is awfully damned hard to beat the least resistance path of copper over earth. And yet BL&H offer us this observation I requote above. What will trump a higher resistance path is greater potential difference and proximity. Note that BL&H are quite specific about proximity to the antenna, and hence it follows that the separation between radials is closer there, than further out from the antenna. Certainly I can find no where to quote this observation of growing closeness from BL&H for Roy's consideration, but I trust my common sense of geometry here too, and I will proceed. BL&H report (without going into the how, or how much): "From (8) [that formula] we see that the earth current proper leads the current in the wires by 90 electrical degrees." such that at "that" radial distance, there must exist the greatest circumferential potential difference between the wire and the earth currents which is clearly mandated by phase. If a potential gradient along the circumference is greater than that along the radial, and the distance along the circumference is smaller than the distance along the radial; then it stands to reason why BL&H even offer to comment "current leaves the wire." Current through earth is largely lost to heat although I do have a fractal antenna that uses earth current to optimize its low angle launch characteristics. Ultimately this reduces to the rather pedestrian observation that more radials closer in serve efficiency - observed and reported by BL&H. 73's Richard Clark, KB7QHC |
high earth resistance
Walter Maxwell wrote:
Roy, it seems to me everyone has missed an important point concerning a circumferential component of the current. We know that the current flowing on the radial wires is radial in direction. What seems to be missed is the current that returns to earth between the wire radials. That current is going to flow in the direction of the lowest resistance. As such it's not going to flow radially alongside the currents flowing on the wire, because the radial resistance of earth between the radial wires is much greater than the resistance of the wires. Consequently, currents reaching earth between the wires will find a lower resistance by traveling toward the nearest radial wire instead of continuing in a perfectly radial direction. This new direction of current flow will not necessarily perfectly circumferential, but will certainly be somewhere between radial and circumferential. Walt, I hadn't missed that phenomenon, but didn't mention it because it doesn't produce a circumferential current. If you look at the current flowing from the earth to each radial wire, you'll see that the sum of these currents will be purely radial, assuming that the system is symmetrical, i.e., radials are equally spaced and equal length, the ground is homogeneous, and the radiator is vertical. Consider a bit of current returning between two radials, which is a little closer to the radial on the right. It'll detour to the right, giving it a rightward component as well as an inward radial component. But for every such bit of current, there's another one the same distance from the radial to the left which will have leftward and inward radial components. The radial components are in the same direction (inward) so will add but the circumferential ones (leftward and rightward) cancel, leaving a net radial current flow. You can say that the returning currents bend to the right or left as they propagate toward the antenna base, but not that there's a systematic circumferential current flow -- no current crosses from radial to radial in a clockwise or counterclockwise circular pattern like Richard implied. I recall reading a paper which showed that connecting radials with circumferential wires actually degrades a ground system's effectiveness, but I wasn't able to lay my hand on it when I looked. Roy Lewallen, W7EL |
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