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Tom Donaly wrote:
Balanis uses a highly mathematical approach in most of his book, supplemented by many graphs and charts. Cecil's quote, like his quote of Tom Rauch on loading coils is only a very small part of the total. You want me to quote the total Balanis book?????? Why don't you, instead, just pick one subject upon which you think you and I disagree, and discuss it. The only thing I know for sure that you and I disagree on is the current at each end of a bugcatcher coil. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Reg Edwards wrote:
There you (in the plural) go again - using handbooks as bibles. Nope, not a bible, just a quote with which I didn't even say I agreed (or not). But ~4% efficiency sounds about right for an 8 foot center-loaded mobile antenna on 75m. Cec, I'm on deep red, South African Western Cape, Pinotage-Shiraz tonight. You should try some. Makes a change from Californian, Texan and John Wayne, six-shooter politics. I'm chugging Franzia Merlot while cleaning my Colt Python .357 Magnum and Winchester 30-30 Lever-Action Carbine. :-) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Forgotten? How can we forget a "fact" we learned wasn't true in the
first place? According to the many references I have, the equation you quote is a simplified equation that's valid for a single wire over a perfect conducting ground plane, where the height is a very small fraction of a wavelength (i.e., radiation is negligible). Even when you ignore the relatively poor conductivity and the permittivity of real ground, the equation is certainly not valid if the wire is high enough for significant radiation to take place. There are several reasons for this: 1. The field shapes become different from the shapes assumed in deriving the equation. 2. Radiation would make the impedance complex rather than purely real. 3. The voltage between the conductor and ground depends on the path taken to measure it, so "characteristic impedance" takes on a whole different meaning, if it has any at all in this context. There is, of course, also the problem of ignoring the finite conductivity of real ground, which will likewise impact the angle of the impedance. It's surely tempting to take a nice, simplistic equation like this and build from it a whole theory of how things work. The seductive thing about it is that it seems to work, sort of, for some special applications. But it's a house of cards, and is at its root based on invalid assumptions. So all the wonderful theories that follow from it are fatally flawed and not to be trusted. As apparently the only person on this newsgroup to have "learned" this "fact", it would serve you well to un-learn it. That is, if you're really interested in discovering how things really work rather than clinging to possibly mistaken notions about how they do. Roy Lewallen, W7EL Cecil Moore wrote: Apparently, a lot of the otherwise knowledgeable people on this newsgroup have forgotten that the formula for the characteristic impedance of a single-wire transmission line is 138*log(4h/d) where h is the height of the wire above ground and d is the diameter of the wire. There's no difference between that single-wire transmission line and a lot of ham antennas. That single-wire transmission line radiates just like an antenna. 1/2WL of #16 wire 24 feet in the air has a Z0 of 600 ohms. If that center-fed dipole were terminated at each end with a 600 ohm load, it would be a traveling-wave antenna with a feedpoint impedance of 600 ohms. Take away the loads and there's a match to 50 ohm coax at the feedpoint. The only difference in those two antennas is that removing the loads turned the antenna into a standing-wave antenna and reflections are arriving back at the feedpoint, lowering the feedpoint impedance. Any coil installed in a standing wave antenna is going to be subjected to both forward and reflected currents. There is no hope of understanding the current in a loading coil without understanding the component currents flowing both directions through the loading coil. -- 73, Cecil, W5DXP |
Radiation Resistance. Roy, what an excellent, well needed exposition, in plain English, addressed mainly to professionals who should know better, and are guilty (dare I say it) of obtaining money under false pretences. You omitted only "DISTRIBUTED radiation resistance", a term essential to but absent from this newsgroup. It's probably also absent from Terman and Kraus, the latter I have never read. It's an aid to clear logical thought. It should be used whenever radiation resistance is compared with conductor resistance, inductance, capacitance, etc., of elongated wires and loading coils. As an example, it so happens that the distributed, end-to-end, radiation resistance, Rd, of a half-wave diopole is exactly twice (easily proved) the feedpoint resistance of around 72 ohms. The radiating efficiency of a half-wave dipole is then, very simply and accurately - Efficiency = 144 / ( 144 + overall HF conductor resistance ) although some old-wives may wish to argue about it on the grounds that it's far too simple. ---- Reg, G4FGQ |
Sorry, I take issue with this. The radiation resistance, as universally
used in the professional literature, *is* a distributed radiation resistance -- it's the resistance that "consumes" the power radiated from the entire antenna, not just one point on the antenna. But that entire power-consuming property is commonly lumped into a single component -- the "radiation resistance" which can be defined (or "referred to") anywhere on the antenna you'd like, including but not limited to the feedpoint. And when placed at that point, it consumes the amount of power radiated from the entire antenna. It's not one single, absolute value, but a component whose value depends on where you define it on the antenna. A very simple and correct way of looking at it is to realize that if P watts is being radiated from the antenna, the radiation resistance value has to equal P/I^2, where I is the magnitude of the current at the point where you're measuring or defining the radiation resistance. So the radiation resistance always "consumes" P watts. If you want to calculate efficiency, you have to do the same thing with the loss resistance, and make a single R that consumes the same amount of power as the total antenna loss. Again, you can define it anywhere on the antenna including the feedpoint, and it'll have a different value wherever you put it. To calculate efficiency from radiation and loss resistances, both have to be -- correctly -- defined at (or "referred to") the same point. There's no need for additional "essential" fundamental terms -- the simple concept of radiation resistance as I've described it is perfectly adequate to explain and calculate antenna radiation and efficiency. But like other concepts, it does take a little effort to understand it. Roy Lewallen, W7EL Reg Edwards wrote: Radiation Resistance. Roy, what an excellent, well needed exposition, in plain English, addressed mainly to professionals who should know better, and are guilty (dare I say it) of obtaining money under false pretences. You omitted only "DISTRIBUTED radiation resistance", a term essential to but absent from this newsgroup. It's probably also absent from Terman and Kraus, the latter I have never read. It's an aid to clear logical thought. It should be used whenever radiation resistance is compared with conductor resistance, inductance, capacitance, etc., of elongated wires and loading coils. As an example, it so happens that the distributed, end-to-end, radiation resistance, Rd, of a half-wave diopole is exactly twice (easily proved) the feedpoint resistance of around 72 ohms. The radiating efficiency of a half-wave dipole is then, very simply and accurately - Efficiency = 144 / ( 144 + overall HF conductor resistance ) although some old-wives may wish to argue about it on the grounds that it's far too simple. ---- Reg, G4FGQ |
On Thu, 04 Nov 2004 20:38:35 -0700, Wes Stewart
wrote: Or, model a short lossless monopole over perfect ground and determine the feedpoint R. In this case, R is totally due to radiation loss, i.e. "radiation resistance." Add a lossless loading inductance somewhere in the middle and see what happens to R. Hi Wes, The difference between the two (perfect/real) insofar as Z is hardly remarkable. First I will start with a conventionally sized quarterwave and by iteration approach the short antenna and observe effects. I am using the model VERT1.EZ that is in the EZNEC distribution and modifying it by turns. For instance, I immediately turn on the wire loss. 40mm thick radiator 10.3 meters tall: Impedance = 36.68 + J 2.999 ohms which lends every appearance to expectation of Rr that could be expected from a lossless perfect grounded world. Best gain is -0.03dBi next iteration: cut that sucker in half: Impedance = 6.867 - J 301 ohms which, again, conforms to most authorities on the basis of Rr. best gain 0.16dBi How about that! More gain than for the quarterwave (but hardly remarkable). This makes me wonder why any futzing is required except for the tender requirements of the SWR fearing transmitter (which, by the way, could be as easily taken care of with a tuner). next iteration: load that sucker for grins and giggles: load = 605 Ohms Xl up 55% Impedance = 13.43 + J 0.1587 ohms Did I double Rr? (Only my hairdresser knows.) best gain 0.13dBi Hmm, losing ground for our effort, it makes a pretty picture of current distribution that conforms to all the descriptions here (sans the balderdash of curve fitting to a sine wave). I am sure someone will rescue this situation from my ineptitude by a better load placement, so I will leave that unfinished work to the adept practitioners. next iteration: cut that sucker down half again (and remove the load): Impedance = 1.59 - J 624.6 ohms Something tells me that this isn't off the scale of the perfect comparison. best gain: 0.25dBi Hmm, the trend seems to go counter to intuition. next iteration: -sigh- what charms could loading bring us? load = 1220 Ohms Xl up 55% Impedance = 3.791 + J 1.232 ohms more than doubled the Rr? best gain: 0.23dBi Now, all of this is for a source that is a constant current generator; we've monkeyed with the current distribution and put more resistance (Rr?) into the equation with loading; and each time loading craps in the punch bowl. So much for theories of Rr being modified by loading. I would appreciate other effort in kind to correct any oversights I've made (not just the usual palaver of tedious "explanations" - especially those sophmoric studies of current-in/current-out). 73's Richard Clark, KB7QHC |
Roy Lewallen wrote:
According to the many references I have, the equation you quote is a simplified equation that's valid for a single wire over a perfect conducting ground plane, where the height is a very small fraction of a wavelength (i.e., radiation is negligible). Even if everything you say is true, it doesn't nullify the concepts of physics. Even if the Z0 is changing point to point along the length of the wire, as it surely does for a vertical antenna, the idea that standing wave antennas don't possess standing waves is ridiculous. The idea that the net antenna current on a standing wave antenna is not itself a standing wave is ridiculous. The idea that the net antenna current on a standing wave is not the result of the superposition of the forward current and reflected current is ridiculous. The idea that RF waves can stand still is ridiculous. The idea that current flows in only one direction in a standing wave antenna is ridiculous. A Rhombic, for instance, is a traveling wave antenna. Its feedpoint impedance is equivalent to its characteristic impedance which is hundreds of ohms. In a standing wave antenna, it is the reflected wave superposing with the forward wave at the feedpoint that determines the feedpoint impedance, low for 1/2WL center-fed dipoles and high for 1WL center-fed dipoles. The feedpoint impedance depends upon interference between the forward wave and the reflected wave. All you have proven is that this is a difficult subject to quantitize, but we already knew that. It is not a difficult subject to conceptualize. So, Roy, please answer the following true/false questions. Standing wave antennas actually exhibit standing waves as described by Kraus, Balanis, and others. ______ Standing waves are created by the superposition of forward waves and reflected waves. ________ RF waves cannot stand still. _______ If true, it follows that "standing" waves are an artifact of superposition and cannot exist without the two underlying component waves. What is moving at the speed of light is the forward wave and the reflected wave. ________ Hint: RF waves must move at the speed of light. Therefore, RF standing waves have two components, each moving at the speed of light in opposite directions. For what it's worth, here's a quote from The ARRL Antenna Book, 15th edition, page 24-22 under "Single Wire Line": "The characteristic impedance of the single wire line depends on the conductor size and the height of the wire above ground, ranging from 500 to 600 ohms for #12 or #14 conductors at heights of 10 to 30 ft." Nothing said about "perfect ground" or "small fractions of wavelength". Again, the concepts that I discuss fall out perfectly from the laws of physics. That they are difficult to quantify is not a good reason to adopt a closed mind. (This reminds me of my Southern Baptist upbringing where some subjects were forbidden). -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
"Roy Lewallen" wrote - Sorry, I take issue with this. The radiation resistance, as universally used in the professional literature, *is* a distributed radiation resistance -- ==================================== Yes, I know. Didn't I just say so? It's just that nobody ever refers to it as such. It would avoid a lot of misunderstandings and arguments if they did! ---- Reg. |
Roy Lewallen, W7EL wrote:
"Sorry. I take issue with this. The radiation resistance as universally used in the professional literature "is" a distributed radiation resistance---." One definition, but not universal by any means. Terman is as professional as needed for most purposes. Terman defines radiation resistance and the custom for stating it on page 891 of his 1955 edition: "Unless specifically stated to the contrary, it is customary to refer the radiation resistance to a current maximum in the case of an ungrounded antenna, and to the current at the base of the antenna when the antenna is grounded." Terman`s definition is unequivocal and useful. It is echoed by other authors. Kraus says on page 182 of his final 3rd edition: "Ro = 60 times the intergral, zero to pi, of the square of Cos [(beta L/2) cos theta] - cos(beta L/2) / sin theta, d theta Where the radiation resistance Ro is referred to the current maximum. In this case of a 1/2-wave antenna, this is at the center of the antenna or at the terminals of the transmission line (see Fig. 6-7)." The solution for a thin dipole yields 73 ohms. Of course, the impedance of a standing-wave antenna varies continuously along its length due to interaction of waves traveling in opposite directions. For practical purposes, we can define Ro as the resistive part and transform its value to the antenna input terminals if these don`t correspond to a current maximum. Best regards, Richard Harrison, KB5WZI |
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