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Reg wrote,
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 All this post serves to do is confirm the suspicion in many minds, Reg, that you haven't ever taken the trouble to study electromagnetics. Might I suggest that, the next time you read Heaviside's work, you put forth the effort necessary to understand it. The simple approximations that you rely on so heavily for your own programs may be good enough for the rough work you demand of them, but, they are not a statement of reality, and they only work given the narrow set of conditions which, for some reason, you always leave unstated. Tom Donaly, KA6RUH |
Dear Tom,
The proof of ANYTHING lies in the EATING. Your appetite is lacking. As with the one million hoodwinked housewives who can't be wrong. Reg. |
The point is that there's no "correct" answer. The voltage between two
points separated in space with the presence of a varying magnetic field can be pretty much anything you'd like it to be -- the value depends, as others have pointed out, on the path taken between the points. Only Reg is able to suspend the principles of electromagnetics and confidently compute a single answer. Roy Lewallen, W7EL Richard Harrison wrote: 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 |
what?
there are multiple answers? how many? "Roy Lewallen" wrote in message ... The point is that there's no "correct" answer. The voltage between two points separated in space with the presence of a varying magnetic field can be pretty much anything you'd like it to be -- the value depends, as others have pointed out, on the path taken between the points. Only Reg is able to suspend the principles of electromagnetics and confidently compute a single answer. Roy Lewallen, W7EL Richard Harrison wrote: 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 |
For Reg, only one. For me, an infinite number.
Roy Lewallen, W7EL H. Adam Stevens wrote: what? there are multiple answers? how many? "Roy Lewallen" wrote in message ... The point is that there's no "correct" answer. The voltage between two points separated in space with the presence of a varying magnetic field can be pretty much anything you'd like it to be -- the value depends, as others have pointed out, on the path taken between the points. Only Reg is able to suspend the principles of electromagnetics and confidently compute a single answer. Roy Lewallen, W7EL |
In article , Roy Lewallen
wrote: For Reg, only one. For me, an infinite number. Roy Lewallen, W7EL H. Adam Stevens wrote: what? there are multiple answers? how many? "Roy Lewallen" wrote in message ... The point is that there's no "correct" answer. The voltage between two points separated in space with the presence of a varying magnetic field can be pretty much anything you'd like it to be -- the value depends, as others have pointed out, on the path taken between the points. Only Reg is able to suspend the principles of electromagnetics and confidently compute a single answer. Roy Lewallen, W7EL Roy, the good point you are making is often hard to get across. I tried for about forty years to teach multivariable calculus to EE students and others. Sometimes I succeeded. But some students couldn't get over the fact that in one dimension, integrals are independent of the path, and so they expected that to be true in two dimensions, or in three. It makes things a lot simpler if you believe that all line integrals are independent of the path, since from that it follows that integrals around loops are always zero. That makes things like Green's Theorem and Stokes' Theorem simpler too, since (if you grant that surface integrals involved are also zero) they simply say that 0 = 0. Simple isn't always right. Maybe in some other universe, all fields are conservative, but not in this one. David, ex-W8EZE and retired math prof -- David or Jo Anne Ryeburn To send e-mail, remove the letter "z" from this address. |
David or Jo Anne Ryeburn wrote:
Simple isn't always right. Maybe in some other universe, all fields are conservative, but not in this one. It might help if the language was made more understandable. Isn't the problem in getting an absolute ground reference point at the end of the dipole? It is possible to model a 3D antenna with a 4th dimension ground "plane". -- 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! =----- |
In article , Cecil Moore
wrote: David or Jo Anne Ryeburn wrote: Simple isn't always right. Maybe in some other universe, all fields are conservative, but not in this one. It might help if the language was made more understandable. Isn't the problem in getting an absolute ground reference point at the end of the dipole? It is possible to model a 3D antenna with a 4th dimension ground "plane". Nope, that's not the problem. The language I am using is standard. To understand it, get out your multivariable calculus text, if you've kept it, or go to the library and borrow a good one. (Shameless plug: Howard Anton's Calculus book is a good one -- I helped debug it before publication.) Look up "independence of path" in the section on line integrals, and look up the related topic of "conservative vector fields". *That's* the problem, and not anything about ground references. What Roy is talking about is an important special case of lack of independence of path; there are many more examples. David -- David or Jo Anne Ryeburn To send e-mail, remove the letter "z" from this address. |
No, that's not the problem. The problem is that you're looking for a
single voltage between two points separated in space. There is no single value for that voltage. If you made some kind of "artificial ground" close to the antenna, then there are an infinite number of possible voltages between it and the Earth. But I'm sure you can work it out as easily with virtual photons and 4 dimensional geometry as Reg does with grade school arithmetic. It's quite simple, really, if you don't need to have the right answer. Roy Lewallen, W7EL Cecil Moore wrote: David or Jo Anne Ryeburn wrote: Simple isn't always right. Maybe in some other universe, all fields are conservative, but not in this one. It might help if the language was made more understandable. Isn't the problem in getting an absolute ground reference point at the end of the dipole? It is possible to model a 3D antenna with a 4th dimension ground "plane". |
Roy Lewallen wrote:
If you made some kind of "artificial ground" close to the antenna, then there are an infinite number of possible voltages between it and the Earth. In four dimensions, the ground is not "artificial" and is at the same potential as the Earth. -- 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! =----- |
David or Jo Anne Ryeburn wrote:
wrote: It might help if the language was made more understandable. Isn't the problem in getting an absolute ground reference point at the end of the dipole? Nope, that's not the problem. The language I am using is standard. For you, maybe, but not for the average amateur radio operator. Instead of trying to demonstrate how smart you are, how about trying to explain the concept in words that the average amateur radio operator can understand? Feynman took that approach in his book, _QED_. Again, the problem goes away if you use a 4 dimensional model and define the 4th dimensional path distance to ground as zero. -- 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: In four dimensions, the ground is not "artificial" and is at the same potential as the Earth. What is that supposed to mean? 73, Jim AC6XG |
Cecil Moore wrote in message ...
David or Jo Anne Ryeburn wrote: wrote: It might help if the language was made more understandable. Isn't the problem in getting an absolute ground reference point at the end of the dipole? Nope, that's not the problem. The language I am using is standard. For you, maybe, but not for the average amateur radio operator. Instead of trying to demonstrate how smart you are, how about trying to explain the concept in words that the average amateur radio operator can understand? How about: 1. Kirchoff's Voltage Law: The sum of the EMFs around a closed loop equals the sum of the voltage drops around a loop. (It's commonly MISstated as the sum of voltage drops around a loop is zero; it's much better to distinguish between EMFs and voltage drops, as there are EMFs which CANNOT be localized to one point in the loop.) 2. Faraday's Law of Magnetic Induction: For any closed loop, there is an EMF proportional to the rate of change of magnetic flux enclosed by the loop. (This is one EMF which cannot be localized--it just drives the whole loop.) 3. Ohm's Law: Voltage drop is proportional to current times resistance (so in a good conductor with modest current, there is very little voltage drop). From those, it should be obvious that the voltage you measure between any two points depends on the path you take, in the presence of a time-varying magnetic field (and perhaps at other times as well). Didn't you previously post that you understood that? As David has suggested, that set of three laws isn't the only way to think about it, but I think they should serve the typical ham fairly well; no integrals needed. Cheers, Tom |
Jim Kelley wrote:
Cecil Moore wrote: In four dimensions, the ground is not "artificial" and is at the same potential as the Earth. What is that supposed to mean? Solve the problem using a 4D math model. The 3D variable path problem goes away. -- 73, Cecil, W5DXP |
Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: In four dimensions, the ground is not "artificial" and is at the same potential as the Earth. What is that supposed to mean? Solve the problem using a 4D math model. The 3D variable path problem goes away. If you've solved it, you should post the solution. (Otherwise, it's just sounds like you're trying to impress somebody.) 73, Jim AC6XG |
On Thu, 23 Oct 2003 10:45:21 -0700, Jim Kelley
wrote: If you've solved it, you should post the solution. The promise of that (following submission to a spectrum of vanity publishers) could last two years alone! 73's Richard Clark, KB7QHC |
Roy Lewallen, W7EL, posted:
No, that's not the problem. The problem is that you're looking for a single voltage between two points separated in space. There is no single value for that voltage. If you made some kind of "artificial ground" close to the antenna, then there are an infinite number of possible voltages between it and the Earth. Well, I chose one particular configuration and one particular integration path because I was curious about the original question - something about how much voltage would the end insulator have to handle for 100 watts of radiated power. I chose a vertical, half-wave monopole fed against perfect ground and looked at the driving source data with EZNEC. The feed- point impedance was 2188 +j66 ohms and a driving current of .213 amps produced a radiated power of 100 watts and a feedpoint voltage of 466 volts. 1500 watts scales that up to 1805 volts. Symmetry about the ground would increase that to 3600 volts for the free-space case. That is what I would adopt as my design-to target for end insulators. I know it's crude, but I was just looking for a ballpark figure. Jim, K7JEB |
Roy Lewallen, W7EL, posted:
No, that's not the problem. The problem is that you're looking for a single voltage between two points separated in space. There is no single value for that voltage. If you made some kind of "artificial ground" close to the antenna, then there are an infinite number of possible voltages between it and the Earth. As another case study, I analyzed a half-wave, inverted-V antenna over perfect ground with the ends of the antenna close to the ground. I placed high-resistance (1e12-ohm) loads between these open ends and ground as voltmeters. The angle of the inverted-V was 90 degrees, placing the apex of the antenna 0.35 wavelength above ground. The feedpoint impedance was trimmed up to be 19 + j0.4 ohms. Going right for the maximum, 1500 watts into this impedance requires an 8.9 amp current source. The voltage to ground across the load on the antenna endpoint is 5300 volts for 1500 watts of radiated power. This scales back to 1370 volts for 100 watts. Again, this is just a special case of the general problem. But it has a configuration that is easy to implement in the EZNEC program and is quite relevant to typical ham-radio, low-band dipole installations. If anyone wants to go over this data, or that for the vertical, half-wave monopole, drop me an e-mail at k7jeb (at) qsl.net and I will send over the .EZ files. Jim, K7JEB |
Roy Lewallen wrote:
"The problem is that you`re looking for a single voltage between two points in space. There is no single value for that voltage." Keith Henney in his 1950 "Radio Engineering Handbook on page 638 wrote: "A thin-wire dipole gives an end potential of about 3,900 volts for 1000 watts antenna input for a height of 1/4-wave. It will be higher for smaller heights, and falls to a minimum of about 1700 volts as height increases to 3/4-wave; beyond this point it settles down to the free-space value of about 3000 volts. Potentials vary as the square root of the power ratio and as the inverse square root of the capacitance per unit length. For a potential of 3900 volts on a wire 0.101 in. in diameter (No.10 B&S), the radial gradient is of the order of 31 kv per cm." At frequencies where the antenna has high reactance and low resistance, a few watts of power produce very high potentials. At high altitudes, high potentials can easily produce corona and flashover. Apply more watts and produce corona and flashover at sea level. One way to measure the voltage at the end of a dipole might be to increase input to the dipole until orona cccurs and then scale that to the power of interest. Best regards, Richard Harrison, KB5WZI |
K7JEB wrote:
I know it's crude, but I was just looking for a ballpark figure. Hi Jim, all of the estimates have been in the same ballpark except for the ones that indicate the voltage doesn't appreciably change from the center feedpoint of a 1/2WL dipole to the open ends. Looks like antennas and transmission lines obey Maxwell's equations after all. :-) -- 73, Cecil, W5DXP |
Richard Harrison wrote:
At high altitudes, high potentials can easily produce corona and flashover. That's why the cubical quad was invented. The voltage at the voltage maximum points is in the ballpark of half that of a dipole and those points can be insulated. -- 73, Cecil, W5DXP |
Looks perfectly reasonable to me, Jim. Another perfectly reasonable thing to
ask is "what is the electric field strength near the antenna elements?" because if the field strength is too great, you can get corona. And EZNEC (NEC2, etc) will give you at least an estimate of that value. As you say, it can be interesting to know if an insulator will be adequate, and that's a question well worth asking. Cheers, Tom Jim, K7JEB, wrote: Well, I chose one particular configuration and one particular integration path because I was curious about the original question - something about how much voltage would the end insulator have to handle for 100 watts of radiated power. I chose a vertical, half-wave monopole fed against perfect ground and looked at the driving source data with EZNEC. The feed- point impedance was 2188 +j66 ohms and a driving current of .213 amps produced a radiated power of 100 watts and a feedpoint voltage of 466 volts. 1500 watts scales that up to 1805 volts. Symmetry about the ground would increase that to 3600 volts for the free-space case. That is what I would adopt as my design-to target for end insulators. I know it's crude, but I was just looking for a ballpark figure. |
Jim, K7JEB, wrote:
.... Again, this is just a special case of the general problem. But it has a configuration that is easy to implement in the EZNEC program and is quite relevant to typical ham-radio, low-band dipole installations. It's also easy to get the electric field strength near the antenna from EZNEC. I expect the field to be highest near wires, because of the shape the field must take near the wires, so that's where I'd look first to get an idea about possible breakdown of the air or insulators. Cheers, Tom |
Really???
K9CUN |
What we know as and call "voltage" is the potential difference between two
points. Give me access to the two points and an ideal voltmeter (AC or DC) and I'll measure it for you. I am assuming sinusoidal steady state AC, isn't everyone? 73 de Jack, K9CUN |
Just one caution about this. I don't think EZNEC or other NEC-based
programs will give accurate field strength any closer to a wire than a few wire diameters. So it probably wouldn't be good for predicting the likelihood of corona discharge and the like. It might be possible, though, to model a wire as a number of very much smaller parallel wires arranged in a circle, to get good field strength values closer to the real wire. I wouldn't completely trust the results, though, until at least a few test cases were run which could be compared to theoretical values. Roy Lewallen, W7EL K7ITM wrote: Jim, K7JEB, wrote: ... Again, this is just a special case of the general problem. But it has a configuration that is easy to implement in the EZNEC program and is quite relevant to typical ham-radio, low-band dipole installations. It's also easy to get the electric field strength near the antenna from EZNEC. I expect the field to be highest near wires, because of the shape the field must take near the wires, so that's where I'd look first to get an idea about possible breakdown of the air or insulators. Cheers, Tom |
Roy, W7EL wrote:
"I don`t think EZNEC or other NEC-based programs will give accurate field strength any closer to a wire than a few wire diameters." The Antenna Section in Keith Henney`s "Radio Engineering Handbook" was written by Edmund A. Laport, Chief Engineer of RCA`s International Division at the time (around 1950). On page 637 Ed writes: "Where high power is to be transmitted, or at high altitudes, antenna insulation and conductor designs require care to details. For h-f use, only radial potential gradients need to be considered. At high altitudes, pluming may occur with consequent damage to the system. Fortunately in practice, high power is generally used with directive antennas, and the power is divided among several dipole sections thus tending to minimize the problem. A thin-wire dipole gives an end potential of about 3,900 volts rms for 1000 watts input for a height of 0.25 wavelength. It will be higher for smaller heights, and falls to a minimum of about 1,700 volts as height increases to 0.75 wavelength; beyond this point it settles down to the free-space value of about 3,000 volts. Potentials vary as the square root of the power ratio and as the inverse square root of the capacitance per unit length. For a potential of 3,900 volts on a wire 0.101 in. in diameter (No, 10 B&S), the radial gradient is of the order of 31 kv per cm. As a rough approximation for a cage, the gradient for one wire is divided by the number of wires in the cage." The multiwire observation is important because, if the potential gradient at any point in air becomes greater than 30,000 volts/cm., the air becomes ionized and sparking or corona discharge will occur. On page 645, Ed writes: "With 100-kw carrier input (to an 8-dipole array), the end potential on each dipole is 7,500 volts rms. A 3000-ft. 580-ohm two-wire balanced feeder used with this antenna had an efficiency of 67 per cent. I am comfortable with Ed`s experience which can be scaled for the power and configuration of the antenna. Best regards, Richard Harrison, KB5WZI |
"So what's the voltage across the 1k resistor?"
------------------------------------------------------ It's whatever my ideal voltmeter reads when it's connected across the ends of the 1-kOhm resistor. Jack |
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