|
How 'bout you model it with your concept of an "artificial ground", and
let us know the result? You can measure the voltage with EZNEC by connecting the two points to be measured with a wire and inserting a zero amplitude current source in the wire. The source will act like an open circuit, and the voltage will be reported in the Source Data output. After you've determined the voltage relative to your "artificial ground", modify the "artificial ground" and note the effect on the voltage. Then see if you can figure out what the voltage is between the "artificial ground" and the Earth. Or, give us your justification for assuming that it's zero. If it is zero, via what path? As Tom has been saying, the voltage between two points depends on the path you take between them. Roy Lewallen, W7EL Cecil Moore wrote: Ian White, G3SEK wrote: In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. How about using an artificial ground at the measurement point? -- 73, Cecil, W5DXP |
Roy Lewallen wrote:
After you've determined the voltage relative to your "artificial ground", modify the "artificial ground" and note the effect on the voltage. Then see if you can figure out what the voltage is between the "artificial ground" and the Earth. Or, give us your justification for assuming that it's zero. If it is zero, via what path? As Tom has been saying, the voltage between two points depends on the path you take between them. Wow, you sure ASSume a lot from a simple question. Let's turn it around and you guys prove that the voltage at the ends of a dipole is less than or equal to the feedpoint voltage even though a florescent light bulb is brighter at the ends. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
Wow, you sure ASSume a lot from a simple question. Let's turn it around and you guys prove that the voltage at the ends of a dipole is less than or equal to the feedpoint voltage even though a florescent light bulb is brighter at the ends. -- 73, Cecil http://www.qsl.net/w5dxp What causes the light to light up, Cecil, an E field, or a V field? You can't seem to get over the fact that the voltage between two points in a time varying E field may not be unique and thus not measurable. Go read the book I told you to read, Cecil, and you'll understand why. 73, Tom Donaly, KA6RUH |
Cecil Moore wrote in message ...
Tom Bruhns wrote: If you tell me there is a large voltage along a good conductor, then I know there is a very large heat dissipation in that wire. There are large voltages along my open-wire feedline when the SWR is high, but very low heat dissipation in that wire. Hint: think standing waves on the antenna wire. It's TEM line, right? The voltages are practically all ACROSS the line, between the conductors. There is very little voltage ALONG the conductors, just I*R (and note the directions for _that_). Go look up Faraday's Law of Magnetic Induction, and refresh your understanding of Kirchoff's Voltage Law and Ohm's Law as well. Those three pretty much let you figure it all out. Cheers, Tom |
Tdonaly wrote:
Cecil wrote, Bend the ends of a resonant dipole around close to each other and measure the voltage with a shielded differential RF voltmeter. For 100 watts input, you will get almost 1000 volts RMS between the ends, a far cry from the ~70 volts RMS at the center feedpoint. You missed the point, again, Cecil. Carry on. Nope, you missed the point. This ain't rocket science. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tdonaly wrote:
What causes the light to light up, Cecil, an E field, or a V field? You can't seem to get over the fact that the voltage between two points in a time varying E field may not be unique and thus not measurable. Go read the book I told you to read, Cecil, and you'll understand why. Why not just tell me to read the Bible where God is the cause of everything - same difference. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tom Bruhns wrote:
There is very little voltage ALONG the conductors, ... Depends upon how long the conductors are. The difference along a 1/4WL conductor is known to be minimum VS maximum assuming a minimum at one end. Are you saying that EZNEC doesn't display the current distribution on an antenna when I press the 'i' key? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tdonaly wrote:
What causes the light to light up, Cecil, an E field, or a V field? According to quantum electrodynamics, fields don't exist. So are you talking about photons or virtual photons above? In either case, voltage can still be measured by a voltmeter. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
Why not just tell me to read the Bible where God is the cause of everything - same difference. -- 73, Cecil http://www.qsl.net/w5dxp That's about the kind of reply I expected, which is too bad. It sort of reminds me of the Catholic clerics who refused to look through Galileo's telescope. 73, Tom Donaly, KA6RUH |
Cecil wrote,
Tdonaly wrote: What causes the light to light up, Cecil, an E field, or a V field? According to quantum electrodynamics, fields don't exist. So are you talking about photons or virtual photons above? In either case, voltage can still be measured by a voltmeter. -- 73, Cecil http://www.qsl.net/w5dxp I see you've gone into objection-stopper mode: write something whether it makes sense or not. Evidently, you don't know what makes the little light light up, or you'd answer the question. 73, Tom Donaly, KA6RUH |
He's truly a master, isn't he?
Roy Lewallen, W7EL Cecil Moore wrote: Tdonaly wrote: What causes the light to light up, Cecil, an E field, or a V field? According to quantum electrodynamics, fields don't exist. So are you talking about photons or virtual photons above? In either case, voltage can still be measured by a voltmeter. |
Cecil, I'm sorry you don't understand that in the presence of
time-varying fields, the potential between two points depends on the path you take. Grasping that concept can be very empowering in understanding what's going on in antennas, and in transmission lines, and in inductors and transformers. I can only hope that some lurkers have benefitted from the discussion. Cheers, Tom Cecil Moore wrote in message ... Tom Bruhns wrote: There is very little voltage ALONG the conductors, ... Depends upon how long the conductors are. The difference along a 1/4WL conductor is known to be minimum VS maximum assuming a minimum at one end. Are you saying that EZNEC doesn't display the current distribution on an antenna when I press the 'i' key? |
Tdonaly wrote:
I see you've gone into objection-stopper mode: write something whether it makes sense or not. Sorry, my neighbor forced some Amaretto upon me yesterday. The fact that the voltage is hard to measure doesn't prove that it doesn't exist. Simply loop the ends of the dipole close together and measure the voltage with an RF voltmeter. It's a no brainer. These are copper wires. No need to discuss fields. During 1/2 cycle, there is an excess of electrons on one end. During the next 1/2 cycle, there is an excess of electrons on the other end. A voltmeter measures excesses of electrons between two wires very well. Or take a look at the end feedpoint impedance of a 1/2WL monopole. It's a no brainer to discover that the voltage is high and the current is low. So what is the feedpoint voltage for a 1/2WL end-fed monopole? EZNEC says it is around 3500 ohms giving a feedpoint voltage of about 550v for 100w compared to a 1/4WL monopole's feedpoint voltage of about 60v. Do you not understand that the voltage to current ratios vary from high to low at 1/4WL intervals along a wire antenna? So please tell us how photons manage to light up a florescent light bulb. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Roy Lewallen wrote:
He's truly a master, isn't he? Ask an irrelevant question - obtain an irrelevant answer. :-) Tdonaly wrote: What causes the light to light up, Cecil, an E field, or a V field? According to quantum electrodynamics, fields don't exist. So are you talking about photons or virtual photons above? In either case, voltage can still be measured by a voltmeter. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tom Bruhns wrote:
Cecil, I'm sorry you don't understand that in the presence of time-varying fields, the potential between two points depends on the path you take. I know that, Tom, but we are talking about measuring the RF voltage between two copper wires one inch apart. The path is well defined. It is a no brainer. There is no need for obfuscation. The measurement proves the voltages at the ends of a dipole to be at least a magnitude higher than the voltage at the feedpoint. Are you not aware of how the ratio of voltage to current varies over 1/4WL of a wire antenna? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
Tdonaly wrote: I see you've gone into objection-stopper mode: write something whether it makes sense or not. Sorry, my neighbor forced some Amaretto upon me yesterday. The fact that the voltage is hard to measure doesn't prove that it doesn't exist. Simply loop the ends of the dipole close together and measure the voltage with an RF voltmeter. It's a no brainer. These are copper wires. No need to discuss fields. During 1/2 cycle, there is an excess of electrons on one end. During the next 1/2 cycle, there is an excess of electrons on the other end. A voltmeter measures excesses of electrons between two wires very well. Or take a look at the end feedpoint impedance of a 1/2WL monopole. It's a no brainer to discover that the voltage is high and the current is low. So what is the feedpoint voltage for a 1/2WL end-fed monopole? EZNEC says it is around 3500 ohms giving a feedpoint voltage of about 550v for 100w compared to a 1/4WL monopole's feedpoint voltage of about 60v. Do you not understand that the voltage to current ratios vary from high to low at 1/4WL intervals along a wire antenna? So please tell us how photons manage to light up a florescent light bulb. -- 73, Cecil http://www.qsl.net/w5dxp After all this time and all these posts, you're still trying to use circuit theory to explain electromagnetic phenomena in situations where length and time, as parameters, just can't be ignored. Secondly, you haven't even demonstrated that you took time to digest and understand Tom's original post, which you responded to with a knee-jerk "Brain fart." That's o.k. for you. If you really want to believe in a simplified view of electronics that's fine. But, for the rest of us, all the books, the research, the mathematics, and the thinking of people from Faraday to Feynman, actually mean something. 73, Tom Donaly, KA6RUH |
Tom Bruhns wrote: Cecil, I'm sorry you don't understand that in the presence of time-varying fields, the potential between two points depends on the path you take. I know that, Tom, but we are talking about measuring the RF voltage between two copper wires one inch apart. The path is well defined. That's one path. It is a no brainer. There is no need for obfuscation. Universally recognized principles of electromagnetics are obfuscation? The measurement proves the voltages at the ends of a dipole to be at least a magnitude higher than the voltage at the feedpoint. You changed the geometry. But even if you hadn't, you might be able to say the changing electrical fields are greater at the ends of a dipole, but not the voltages, because the voltages aren't uniquely defined. Are you not aware of how the ratio of voltage to current varies over 1/4WL of a wire antenna -- 73, Cecil http://www.qsl.net/w5dxp There's not much point in arguing with you Cecil, since you don't want to countenance the more sophisticated ideas of some of the other posters to explain what's going on at the ends of a dipole. That's too bad. You'll give some people the impression that things are as simple as you say they are when things are not simple at all. If they were, even an old hick like me could become an engineer, and the job wouldn't pay much at all. 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
But, for the rest of us, all the books, the research, the mathematics, and the thinking of people from Faraday to Feynman, actually mean something. I don't recall Faraday or Feynman ever saying that the impedance of an antenna wire is constant up and down the wire. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tdonaly wrote:
Universally recognized principles of electromagnetics are obfuscation? Complicating a simple measurement task beyond belief is obfuscation. You changed the geometry. But even if you hadn't, you might be able to say the changing electrical fields are greater at the ends of a dipole, but not the voltages, because the voltages aren't uniquely defined. When a dipole is bent into an open loop, the relative voltage between the ends is uniquely defined just like the voltage across a transmission line is uniquely defined. If I poke two wires through two holes in a faraday cage and ask you to measure the 10 MHz voltage between them with 10% accuracy, would you say it can't be done? You'll give some people the impression that things are as simple as you say they are when things are not simple at all. The measurement may be extremely challenging, but the *concepts* are simple. All you need to do is note the similarity of the transmission line impedances on an SWR circle to a wire antenna. If the spacing on a transmission line is an appreciable percentage of a wavelength, the transmission line will radiate. That's all a center-fed wire antenna is - a transmission line with large spacing between the conductors and it radiates. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil wrote,
Tdonaly wrote: But, for the rest of us, all the books, the research, the mathematics, and the thinking of people from Faraday to Feynman, actually mean something. I don't recall Faraday or Feynman ever saying that the impedance of an antenna wire is constant up and down the wire. -- 73, Cecil http://www.qsl.net/w5dxp Interesting. Still trying to divert the argument to your own frame of reference. Nice try. Hasta la vista, Cecil - at least until you write something else I disagree with. 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
Interesting. Still trying to divert the argument to your own frame of reference. Still trying to engineer solutions instead of murmuring mantras about how impossibly difficult everything is. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Here's an interesting quote from _Transmission Lines, Antennas, and Wave
Guides_, by King, Mimno, and Wing: "The amplitude of the current is not the same at different points along a conductor, because electric charge is deposited all along the surface of the conductor. Superficially it may appear that an antenna consisting of a straight conductor that is an appreciable fraction of a wavelength long and with a generator at its center may be looked upon simply as an open-end transmission line with the parallel conductors bent to lie along the same axis instead of being parallel to each other. Although there is considerable similarity between the two cases from the point of view of the approximate distribution of current, *they are nevertheless fundamentally different*. The transmission line may be analyzed to a good approximation in terms of ordinary electric-circuit theory, because equal and opposite currents are very close together. This is not true of the antenna and ordinary electric-circuit theory cannot be applied. *It is fundamentally incorrect to treat a center-driven antenna as though it were the bent-open ends of a two-wire line.* Circuits that satisfy the condition for the near zone, either because they are sufficiently small or because they have equal and opposite currents everywhere so close together that the currents in widely separated parts of the circuit exert a negligible effect on one another, are analyzed correctly by the methods of ordinary electric-circuit theory. All other circuits must be investigated in terms of electromagnetism. This nearly always involves a study of the electromagnetic field as a useful intermediate step in determining distributions of current and charge." The above text surrounded by asterisks is printed in italics. The quote begins on p. 85 of the paperback 1965 Dover reprint. I highly recommend reading Chapter II, of which the quote is a part. Roy Lewallen, W7EL Cecil Moore wrote: Tdonaly wrote: Universally recognized principles of electromagnetics are obfuscation? Complicating a simple measurement task beyond belief is obfuscation. You changed the geometry. But even if you hadn't, you might be able to say the changing electrical fields are greater at the ends of a dipole, but not the voltages, because the voltages aren't uniquely defined. When a dipole is bent into an open loop, the relative voltage between the ends is uniquely defined just like the voltage across a transmission line is uniquely defined. If I poke two wires through two holes in a faraday cage and ask you to measure the 10 MHz voltage between them with 10% accuracy, would you say it can't be done? You'll give some people the impression that things are as simple as you say they are when things are not simple at all. The measurement may be extremely challenging, but the *concepts* are simple. All you need to do is note the similarity of the transmission line impedances on an SWR circle to a wire antenna. If the spacing on a transmission line is an appreciable percentage of a wavelength, the transmission line will radiate. That's all a center-fed wire antenna is - a transmission line with large spacing between the conductors and it radiates. |
"Cecil Moore" wrote in message ... Tdonaly wrote: But, for the rest of us, all the books, the research, the mathematics, and the thinking of people from Faraday to Feynman, actually mean something. I don't recall Faraday or Feynman ever saying that the impedance of an antenna wire is constant up and down the wire. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- ARGH! another thread that will never die... time to 'ignore' this one also. don't you guys realize you will never agree on anything on here? you talk in circles around and around the same dogmatic circles never seeing that you are all saying the same thing in different terms and arguing around nits that don't matter in the real world. |
Roy Lewallen wrote:
Here's an interesting quote from _Transmission Lines, Antennas, and Wave Guides_, by King, Mimno, and Wing: *It is fundamentally incorrect to treat a center-driven antenna as though it were the bent-open ends of a two-wire line.* Funny, I thought Maxwell's equations worked for either case. Did you also know that *It is fundamentally incorrect to treat a wolf like a dog.* If you slowly increase the spacing and angle between the two conductors of a transmission line, at exactly what spacing and angle does it magically cease being a transmission line and become an antenna requiring a completely different treatment? Please be specific as to the exact spacing and angle at which it becomes "fundamentally incorrect" to treat the configuration as a transmission line. Incidentally, I don't usually use circuit theory for transmission lines. Your current distribution curve displayed by EZNEC for a 1/2WL dipole looks exactly like Fig 1, page 2-2, ARRL Antenna Book, 15th edition. I thought you or Walt probably wrote that section. If you had displayed both the current and the voltage distribution in EZNEC, what would it look like for a 1/2WL dipole? We can easily measure the impedance at any point along a 1/2WL section of transmission line with reflections or the feedpoint impedance at any point along a 1/2WL dipole. As you know, those impedances are the ratio of net voltage to net current. At the impedance minimum on a transmission line, the voltage is minimum and the current is maximum. Same for a feedpoint impedance minimum on an antenna. At the impedance maximum on a transmission line, the voltage is maximum and the current is minimum. Same for a feedpoint impedance maximum on an antenna. We can deduce the ratio of the voltage to the current from the feedpoint impedance - or is that one of the rules of physics that a transmission line obeys but an antenna disobeys? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Roy Lewallen wrote in message ...
Here's an interesting quote from _Transmission Lines, Antennas, and Wave Guides_, by King, Mimno, and Wing: "The amplitude of the current is not the same at different points along a conductor, because electric charge is deposited all along the surface of the conductor. Superficially it may appear ... Here's another interesting quote from exactly the same source, page 71, the very first two paragraphs of King's "Antennas" chapter. CHAPTER II ANTENNAS Electric Circuit Theory and Electromagnetic Theory. -- In order to understand the behavior of antennas and of electric circuits at ultra-high frequencies, it is essential to recognize that phenomena of a vastly more general nature are involved than are encountered in conventional electric networks. Attention is seldom called to the fact that electric-circuit theory which proceeds from Kirchhoff's laws is a highly specialized form of a more general theory. In some respects, the situation is like that in mechanics, in which the simple law of gravitation due to Newton may be looked upon as a special case of a more general law formulated in the theory of relativity. Much as Newtonian mechanics is adequate for the mechanical engineer, ordinary electric-circuit theory is accurate for the requirements of electrical power engineering and for many requirements in communications. But even as Newton's laws of motion are inadequate in dealing with atomic phenomena and some astronomical problems, so ordinary electric-circuit theory fails when applied to antennas and to most circuits that are to be used at ultra-high frequencies. The reason is that the conditions that limit the generality of Newton's laws on the one hand, or the theorems of electric-circuit theory on the other, are not satisfied. For those who have assumed that Kirchhoff's laws are perfectly general, a series of surprises is in store. They may, in fact, feel like Alice when the Red Queen was annoyed by her reluctance to believe "six impossible things before breakfast." But presently they may return through the Looking Glass and discover that they have been living in the one-dimensional Wonderland of electric-circuit theory and that Nature is as simple as this suggests only in sufficiently small spaces. It is difficult to understand the structure of general electromagnetism without first learning the appropriate symbolism, that of mathematics. But if one is willing to accept some things on faith and to meet others with an open, perhaps even an adventurous mind, a degree of familiarity with many electromagnetic phenomena can be acquired from a qualitative discussion. |
David Robbins wrote:
ARGH! another thread that will never die... time to 'ignore' this one also. don't you guys realize you will never agree on anything on here? you talk in circles around and around the same dogmatic circles never seeing that you are all saying the same thing in different terms and arguing around nits that don't matter in the real world. I think that's the point. Cecil and his buds are doing this for entertainment. - Mike KB3EIA - |
Mike Coslo wrote in message .net...
David Robbins wrote: ARGH! another thread that will never die... time to 'ignore' this one also. don't you guys realize you will never agree on anything on here? you talk in circles around and around the same dogmatic circles never seeing that you are all saying the same thing in different terms and arguing around nits that don't matter in the real world. I think that's the point. Cecil and his buds are doing this for entertainment. :-) Gee, I think Cecil is trying to carry this one by himself. I don't see any of "his buds" on the list of contributors. Certainly Tom, Ian, Roy and I are in very close agreement about this topic. It is worth a step back from all this, though. If the lurkers can get the single idea that it's the _current_ distribution on an antenna structure that ultimately matters, not any voltages (whether we can agree about them or not), then they'll have gained something of value. If they also pick up some simple things like "electric fields are always very nearly perpendicular to the surface of good conductors" and "the potential between two points in general depends on the path you take," that would be a nice plus. If they get curious and go learn more about those things, that would be wonderful. Cheers, Tom |
Mike Coslo wrote:
I think that's the point. Cecil and his buds are doing this for entertainment. Sure beats mowing the yard. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Tom Bruhns wrote:
:-) Gee, I think Cecil is trying to carry this one by himself. I don't see any of "his buds" on the list of contributors. Certainly Tom, Ian, Roy and I are in very close agreement about this topic. I agree that *measuring* the voltage is a challenge. It is worth a step back from all this, though. If the lurkers can get the single idea that it's the _current_ distribution on an antenna structure that ultimately matters, ... It's hard to develop that H-field without an E-field. The feedpoint impedance of a standing wave antenna gives a deductive clue to the magnitude of the E-field and it is not the same as a traveling wave antenna. It also varies up and down the standing wave antenna. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil Moore wrote:
Roy Lewallen wrote: Here's an interesting quote from _Transmission Lines, Antennas, and Wave Guides_, by King, Mimno, and Wing: *It is fundamentally incorrect to treat a center-driven antenna as though it were the bent-open ends of a two-wire line.* Funny, I thought Maxwell's equations worked for either case. Did you also know that *It is fundamentally incorrect to treat a wolf like a dog.* If you slowly increase the spacing and angle between the two conductors of a transmission line, at exactly what spacing and angle does it magically cease being a transmission line and become an antenna requiring a completely different treatment? Please be specific as to the exact spacing and angle at which it becomes "fundamentally incorrect" to treat the configuration as a transmission line. Incidentally, I don't usually use circuit theory for transmission lines. That question is a lot like asking for the exact speed an object has to be moving before non-relativistic becomes invalid, or how small before quantum theory has to be used. I maintain that the authors of that book know more than you do about the topic by at least an order of magnitude -- more likely about three. If you really want to know the answer to your silly question, you should study what they've written and try to understand it, rather than posting it as a question to me on this newsgroup. . . . Roy Lewallen, W7EL |
Roy Lewallen wrote:
I maintain that the authors of that book know more than you do about the topic by at least an order of magnitude -- more likely about three. That doesn't prove you understand what they said. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil, W5DXP wrote:
"We can deduce the ratio of the voltage to current from the feedpoint impedance." What`s the difference between a 1/4-wave of transmission line and a 1/2-wave dipole? The transmission line is designed for low radiation and the antenna is designed for high radiation. Current will be turned around by the open-circuit at the end of the transmission line. At the end of the antenna, displacement current will flow between oppositely polarized parts of the antenna, the earth, and other surrounding objects, and some conduction current will be redirected in the opposite direction on the antenna. Impedance of the transmission linee is uniform in its presentation to either one of the two traveling waves traversing the line, forward or reflected.. SWR changes along an antenna due to radiation, inefficiencies, irregularities, etc. An average characteristic (surge) impedance is used for antenna calculations. SWR for the usual open-circuit 1/2-wave dipole: SWR = Avg. Zo of the antenna / radiation resistance. Avg. Zo = 276 log 1/periphery of antenna conductor log is the base 10 unit periphery is measured in the same units as length, usually in wavelengths. Loaded antennas tend to be high Q resonant circuits and this limits their bandwidths. Low resistance allows a large current as reactance is cancelled at resonance. The large current causes large voltage drops across the antenna inductance and capacitance. I`ve seen mobile antennas which were ready to fire up coronas at their tips at the slightest excuse. This demonstrates high reactance (energy storage) and low resistance (radiation and loss). There are several variables which determine the voltage and current at the end of the dipole. The current does not drop to zero simply because the conduction in the forward direction stops. Displacement current continues the flow to some extent. The dipole has a Q, and the wave traveling on the dipole generates a voltage to current ratio related to the size and configuration of the antenna. You can predict that forward voltage on a transmission line will double at its open-circuit end, and you can predict the forward voltage. I think prediction of voltage at the open-circuit end of an antenna is harder. Best regards, Richard Harrison, KB5WZI |
Roy Lewallen wrote:
I maintain that the authors of that book know more than you do about the topic by at least an order of magnitude -- more likely about three. If you really want to know the answer to your silly question, you should study what they've written and try to understand it, rather than posting it as a question to me on this newsgroup. Hmmmmmm, perhaps you and your authors would understand the subject better by reading and understanding a good book on quantum electrodymanics. You will not fully understand fields until you accept the fact that photons can have four, not two, polarizations. I maintain that the developers of quantum electro-dynamics know more than you and your authors combined. You have earlier rejected the latest physics theories. I doubt that you will understand fields until you fully understand virtual photons. (See, two can play your silly argumentum ad verecundiam game.) :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Harrison wrote:
The current does not drop to zero simply because the conduction in the forward direction stops. Displacement current continues the flow to some extent. Many electrons do not flow through air unless the air is ionized. "Displacement current" is not understood very well and is a sort of virtual current. The actual electron flow is taking place in the rest of the circuit. When a lot of electrons flow between the plates of a capacitor, for instance, the voltage capability of the capacitor has been exceeded and it is in failure mode. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil, W5DXP wrote:
"When a lot of electrons flow between the plates of a capacitor, for instance, the voltage capability of the capacitor has been exceeded and it is in a failure mode." Yes. Conduction current flows when electrons penetrate the dielectric. Capacitors consist of conductors separated by insulation. This dielectric bars conduction up to some voltage limit where the insulation breaks down. This insulation can be a high vacuum, but at some voltage, a spark can jump the gap even in a vacuum. We have vacuum capacitors, both fixed and variable. We have vacuum tubes which have measured interelectrode capacitances. We have radio waves which propagate as the result of alternating magnetic and electric fields. These propagate in the high vacuum of deep space where corpuscular particles may be speeding through like infinitesimal bullets, but evidence is that no matter is required to allow influence by electric and magnetic fields. An alternating electric field, even in a vacuum where no electrons are in motion, produces an alternating magnetic field which generates an alternating electric field, and so on and so forth ad infinitum, ad nauseam. We know this works as we use radio signals in deep space and view light from the sun and other cosmic sources. It was J.C. Maxwell who speculated that displacement current produces a magnetic field the same as conduction current does, to explain how radiation occurs. Best regards, Richard Harrison, KB5WZI |
Cecil, W5DXP wrote:
"We can deduce the ratio of voltage to current from the feedpoint impedance." Kraus says: "The current I1 at a distance from the nearest current maximum, as shown in Fig. 6-11 is given by I1 = Io cos beta x where I1 = terminal current, Io = maximum current When x=0, R1 = Ro; but when x=lambda/4, R1 = infinity if Ro is not zero. However, the radiation resistance measured at a current minimum (x=lambda/4) is not infinite as would be the calculated from: R1 = Ro/ cos squared (beta x), since an actual antenna is not infinitesimally thin and the current at a minimum point is not zero. Nevertheless, the radiation resistance at a current minimum may in practice be very large, i.e., thousands of ohms. I think the voltage may be measured at the end of a dipole using a diode probe spaced from the dipole by an insulator. The ground return can be provided by the exterior of a 1/2-wavelength coax which is well grounded. For calibration, the easily determined voltage at the dipole feedpoint can be indicated by the diode probe. The coupling can be kept small if there is sufficient antenna power. The antenna does not need to be reconfigured into a halo to measure its voltage at its open-circuit end. Best regards, Richard Harrison, KB5WZI |
You do realize, of course, that the reading you get will depend on how
you've oriented the coax. Which orientation would provide the "correct" answer? And why? Roy Lewallen, W7EL Richard Harrison wrote: . . . I think the voltage may be measured at the end of a dipole using a diode probe spaced from the dipole by an insulator. The ground return can be provided by the exterior of a 1/2-wavelength coax which is well grounded. . . . |
Roy, W7EL wrote:
"Which orientation would provide the "correct" answer?" Maybe none. We know what the correct answer is at the feedpoint. By trial and error we can find the most stable and least critical placement for our test lead. We can calibrate our voltage indicator with a sample at the feedpoint under the conditions that prevail. A test lead perpendicular to the balanced antenna suffers the least interference. It`s subject to imperfections and it`s subject to improvements to overcome the imperfections. In AM broadcasting we get away with hanging a sampling loop on each tower of a directional array to monitor both tower current and phase. It works well enough. Do you have a better idea than a diode probe to sample voltage at the end of a dipole? Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Do you have a better idea than a diode probe to sample voltage at the end of a dipole? I use a 4th dimension ground. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Do you have a better idea than a diode probe to sample voltage at the
end of a dipole? =============================== The only people person who needs to know the volts at the ends of an antenna conductor such as a 1/2-wave dipole is the antenna designer. He has to concern himself with insulators, breakdown voltages, the effects of corona discharge and other such mundane (to other people) things. Antenna designers, of which extremely few are needed in this small world, being practical, sensible people who are obliged by professional self-discipline to keep economy in time and materials foremost in their minds, do not waste valuable resources researching the 'end-voltage' question, purchasing latest space-age technology equipment just for a one-off job, and employing teams of incompetent but highly-paid assistants and workmen to make one solitary measurement. Of course they don't ! In a few seconds they just do a little school arithmetic : - Volts at end of dipole = Q * Vin / 2 Where Vin is centre-fed dipole feedline volts and Q is the dipole's resonant Q factor. Q is dipole inductive reactance divided by radiation resistance. According to the ARRL handbook, Heaviside, Terman, me and countless others, Q = Omega*L/R Richard, you are familiar with these elementary notions. Do you not feel a little saddened, like me, that amateurs, even future professional engineers, are handicapped by having unnecessarily complicated, incorrect even, ideas knocked into their heads by do-gooders on this newsgroup, trying to appear knowledgeable, who should know better ? I hasten to distinguish between accumulated PRACTICAL EXPERIENCE and TECHNICAL BAFFLEGAB. The latter is easily recognisable. === Reg, G4FGQ |
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