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Loading Coils; was : Vincent antenna
On Sun, 16 Dec 2007 22:14:49 GMT, Cecil Moore
wrote: Richard Clark wrote: On Sun, 16 Dec 2007 19:17:48 GMT, Cecil Moore wrote: Cecil Moore wrote: It just occurred to me that you and I may be talking about two different phases. Continuing: What is the phase shift When you acknowledge there is some confusion as to which phase is being talked about. Do you suppose you know enough to tell us which phase you are talking about? Funny. In the part you deleated, I said it was the phase shift between Vfor1 and Vfor2. Your sneaky underhanded deletion trick is noted. You still show signs of confusion. Phase shift in what? |
Loading Coils; was : Vincent antenna
On Sun, 16 Dec 2007 16:11:05 -0600, Cecil Moore
wrote: Richard Clark wrote: Cecil Moore wrote: I don't think anyone is stupid enough to assert that the phase shift in a capacitor is the same as it is in the absence of any physical impedance discontinuity. Capacitance is not obtained in a physical impedance discontinuity? or is it: You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. More signs of your confusion. How would you know how many terminals if its inside a box? |
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Richard Clark wrote:
You still show signs of confusion. Phase shift in what? I have explained it twice already. Given the following 4-terminal network impedance discontinuity at '+': ---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open Vfor1--|--Vfor2 What is the phase shift in the forward voltage at the impedance discontinuity? The forward voltage on each side of the impedance discontinuity , '+', is not equal in magnitude or phase. We know the stub causes a 90 degree phase shift end to end. Since there is 43.4 deg phase shift in the 600 ohm line and 10 deg phase shift in the 10 ohm line, guess what the phase shift in the forward voltage has to be at the impedance discontinuity? Also previously explained, if you prefer - in s-parameter terms: What is the phase shift between a1 and b2 in the s-parameter equation: b2 = s21*a1 + s22*a2 -- 73, Cecil http://www.w5dxp.com |
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Richard Clark wrote:
Cecil Moore You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. More signs of your confusion. How would you know how many terminals if its inside a box? If you were locked in a black box, you would be ignorant of night and day. Does that mean that night and day would not be happening? Hint: No, it would just mean you are ignorant. Your ignorance changes absolutely nothing outside of the box in which you are locked. Why does my ignorance of what's in the black box change the reality of what's in the black box? Hint: it doesn't. The fact that I am ignorant of the four terminal network in the box doesn't change the fact that it is a four terminal network. The fact that you won't allow me to open the box and measure the phase shift at the impedance discontinuity doesn't change the fact that it is 36.6 degrees. -- 73, Cecil http://www.w5dxp.com |
Loading Coils; was : Vincent antenna
On Mon, 17 Dec 2007 15:22:44 GMT, Cecil Moore
wrote: Richard Clark wrote: Cecil Moore You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. More signs of your confusion. How would you know how many terminals if its inside a box? If you were locked in a black box, you would be ignorant of night and day. Does that mean that night and day would not be happening? Hint: No, it would just mean you are ignorant. Your ignorance changes absolutely nothing outside of the box in which you are locked. Why does my ignorance of what's in the black box change the reality of what's in the black box? Hint: it doesn't. The fact that I am ignorant of the four terminal network in the box doesn't change the fact that it is a four terminal network. The fact that you won't allow me to open the box and measure the phase shift at the impedance discontinuity doesn't change the fact that it is 36.6 degrees. OK, So you are ignorant of what is inside the box. Rather a long ramble to such a simple conclusion. Are you still ignorant of which of two phases? Your rambling confusion has yet to come to terms even to the point of not being able to name both of them in one posting! Like trying to wake a sleep-walker, I hesitate to offer a dangerous suggestion with you in your condition: Could they be "phase" shift and "phase" length? In your somnambulant state of foggy recall It just occurred to me that you and I may be talking about two different phases. can you tell the group what the two different phases are? This could be an important moment in recovery (it may take only 11 steps more). Take your time, we appreciate that your sudden catharsis can jog your mind into curious responses...even though they would be indifferentiable from the several hundred that preceded. ;-) |
Loading Coils; was : Vincent antenna
Richard Clark wrote:
Are you still ignorant of which of two phases? I have spelled out the desired phases three times now. This will be the forth time. --43.4 deg 600 ohm line--+--10 deg 100 ohm line--open Vfor1--|--Vfor2 The phase shift we are looking for is between Vfor1 and Vfor2. I might as well perform the calculations for you since you seen to be incapable of doing so. If we assume a reference of 100 volts at zero degrees incident upon the open end of the stub, then back at the impedance discontinuity: Vfor2 = 100 volts at -10 deg. Vfor1 = 143.33 volts at -46.6 deg The phase shift between Vfor1 and Vfor2 is 36.6 degrees just as it has to be for a 90 degree phase shift to occur end to end in the above 1/4WL stub. Vref2 = 100 volts at 10 deg Vref1 = 143.33 volts at 46.6 deg -- 73, Cecil http://www.w5dxp.com |
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Cecil Moore wrote:
Richard Clark wrote: Cecil Moore wrote: I don't think anyone is stupid enough to assert that the phase shift in a capacitor is the same as it is in the absence of any physical impedance discontinuity. Capacitance is not obtained in a physical impedance discontinuity? or is it: You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. Cecil, This appears to be an unusual definition. How does the "point where two pieces of feedline are connected" become a four-terminal network? One typically thinks of a four-terminal network as having inputs and outputs, with something between. What is that "something between" in the case of two connected feedlines? In your models this "something" seems to have no dimensions and no characteristics other than a phase shift. Are you suggesting that every simple connection is now a four-terminal network? Do all of the textbooks need to be re-written? 73, Gene W4SZ |
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On Mon, 17 Dec 2007 15:11:36 GMT, Cecil Moore
wrote: Richard Clark wrote: You still show signs of confusion. Phase shift in what? I have explained it twice already. You have many explanations that don't actually answer questions. Your confusion, as evidenced in the sudden realization: It just occurred to me that you and I may be talking about two different phases. doesn't really tell us what phase shift. Perhaps if you could state what the two are, and which you are using, the rest of us would be satisified you are no longer confused. Or maybe everyone is actually on the same page with only one phase being mentioned, and some alternate phase expression (yet to be revealed by you) is tormenting your imagination and corrupting your answers with their math errors. |
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Cecil Moore wrote:
I(x,t) = Imax sin(kx) cos(wt) For any point location 'x', it can be seen that the standing wave current is not "flowing" in the ordinary sense of the word but rather, is just oscillating in place at that fixed point. According to the equation you provide above, for any point location 'x', the phase of the current varies continuously with t. Presumably that is what it means to just oscillate in place. 73, ac6xg |
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Gene Fuller wrote:
Cecil Moore wrote: You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. This appears to be an unusual definition. Not unusual at all, Gene. The two input terminals to the black box are on one side. The two output terminals from the black box are on the other side. The impedance discontinuity is inside the box. The black box is extremely small. Give me the four s-parameters, s11, s12, s21, and s22 and I can tell you virtually everything about what is inside the black box without even applying a signal. -- 73, Cecil http://www.w5dxp.com |
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On Mon, 17 Dec 2007 16:12:59 GMT, Cecil Moore
wrote: Richard Clark wrote: Are you still ignorant of which of two phases? I have spelled out the desired phases three times now. This will be the forth time. Fourth. (You suffer from spelling as well as logic, theory, math... errors. Yes, you freely admit to all of this, but stop wringing your hands, we all know it merely exhibits the eccentric's classic signs of that 260 IQ.) However, the assembled group (loitering in the wings, amused at this comic duo) will agree you've arrived at some form of conclusion quicker than in 20 dips into memory's poisoned well. Your netzheimers must be in 16/20ths (as in (20 - 4)/20) remission. Would that be computed to square law to specify your recall at -1.94dB? Perhaps not. Random scraps of error free math does not impart validity. |
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Richard Clark wrote:
Perhaps if you could state what the two are, and which you are using, the rest of us would be satisified you are no longer confused. I've done that four times now, Richard. I even did the math for you. I'm not going to waste any further time on you. -- 73, Cecil http://www.w5dxp.com |
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Richard Clark wrote:
Cecil Moore wrote: I have spelled out the desired phases three times now. This will be the forth time. Fourth. (You suffer from spelling as well as logic, theory, math... errors. My spell checker didn't catch that one. I knew it was only a matter of time until you stooped to complaining about my spelling, close to as low as a person can stoop. Your ulterior motive here is clear to all. -- 73, Cecil http://www.w5dxp.com |
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Cecil Moore wrote:
Gene Fuller wrote: Cecil Moore wrote: You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. This appears to be an unusual definition. Not unusual at all, Gene. The two input terminals to the black box are on one side. The two output terminals from the black box are on the other side. The impedance discontinuity is inside the box. The black box is extremely small. Give me the four s-parameters, s11, s12, s21, and s22 and I can tell you virtually everything about what is inside the black box without even applying a signal. BZZZT! Wrong answer. Nobody ever said anything about the "other side" of the black box. Everything has been referenced to "the point where two pieces of feedline are connected." Nobody is trying to suggest that what is inside one black box is identical to what is inside another black box. The only thing known is that the *two* accessible terminals to each of the black boxes show the same specific impedance at the given frequency. Nothing else. Yet by your models and math the black boxes don't behave the same in your test circuit. You still have not addressed the original question posed by Keith. 73, Gene W4SZ |
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On Mon, 17 Dec 2007 13:22:36 -0600, Cecil Moore
wrote: Richard Clark wrote: Cecil Moore wrote: I have spelled out the desired phases three times now. This will be the forth time. Fourth. (You suffer from spelling as well as logic, theory, math... errors. My spell checker didn't catch that one. Well, for an admitted MENSA member (or lapsed member, which gives the impression of a limp pendant) also having troubles in catching math errors, or confused dualities; then it stands you must also be warned a spell checker doesn't validate syntax, nor the validity of expression. As I know you stumble over English, I will put it in Texican: "Random math does not validate irrational theory; Xeroxed words littered on the page does not sell snake oil." (OK, so it is more like cowboy couplet haiku.) I knew it was only a matter of time until you stooped to complaining about my spelling, Netzheimers remission is clearing the fog, is it? close to as low as a person can stoop. Your ulterior motive here is clear to all. "All" seem to be very few (except for those chuckling in the wings). *** So, moving back to the technical side of the balance sheet. *** No problems with my math assuming the square law computation of your recall at -1.94dB? I thought you might want to boost it to 19/20ths with having to only say it once and suffer a recall of -0.45dB. Was that a new low? This exterior motive was especially formulated for your the sake of your depressed recall response curve. ;-) So that you can preserve your dignity - I will leave you the last word. And remember, always leave them laughing (think of it as another theory you are introducing). |
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On Mon, 17 Dec 2007 13:18:55 -0600, Cecil Moore
wrote: Richard Clark wrote: Perhaps if you could state what the two are, and which you are using, the rest of us would be satisified you are no longer confused. I've done that four times now, Richard. I even did the math for you. I'm not going to waste any further time on you. Retiring from king of the hill, last man standing competition? I Don't Think So! You are forever doomed to fulfill the cliché. :-0 73's Richard Clark, KB7QHC |
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I see that the smudge pots have been lit and the hand waving,
misdirection, and desperate attempts at changing the subject begins. As they have so many times before. What a waste of all that alleged IQ. Roy Lewallen, W7EL |
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"Roy Lewallen" wrote in message ... I see that the smudge pots have been lit and the hand waving, misdirection, and desperate attempts at changing the subject begins. As they have so many times before. What a waste of all that alleged IQ. Roy Lewallen, W7EL I do believe that someone may have inadvertently added 100 to that IQ. Everyone on here with the exception of the gifted one has known for the last 3 days that Richard has been just pulling his bobber under. He finally picked up on it this morning and decided to stop wasting his time. Now if he would only make that a promise! W4ZCB |
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On Mon, 17 Dec 2007 20:33:34 GMT, "Harold E. Johnson"
wrote: I do believe that someone may have inadvertently added 100 to that IQ. Everyone on here with the exception of the gifted one has known for the last 3 days that Richard has been just pulling his bobber under. He finally picked up on it this morning and decided to stop wasting his time. Now if he would only make that a promise! Harold! I resemble that earmark! Add 100 indeed, I thought it was 200. As for your last sentence. The "he" is rather unspecific as two correspondents made a promise (perhaps not so well timed as to fit your observation here). However, I too will sit on the edge of my seat in wonderment to that outcome. 73's Richard Clark, KB7QHC |
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Jim Kelley wrote:
Cecil Moore wrote: I(x,t) = Imax sin(kx) cos(wt) For any point location 'x', it can be seen that the standing wave current is not "flowing" in the ordinary sense of the word but rather, is just oscillating in place at that fixed point. According to the equation you provide above, for any point location 'x', the phase of the current varies continuously with t. Presumably that is what it means to just oscillate in place. Of course, that's what it means. It doesn't move right or left. I(x,t) = Imax sin(kx) cos(-wt) would be just as accurate a description. One cannot even tell which way the standing-wave phasor is rotating. This equates to putting the source on either end of a lossless stub without anything changing. Standing-wave current phase is unchanging up and down a lossless stub. Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole. That current cannot be used to obtain a valid delay through a wire or a coil. -- 73, Cecil http://www.w5dxp.com |
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Gene Fuller wrote:
Cecil Moore wrote: Gene Fuller wrote: Cecil Moore wrote: You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. This appears to be an unusual definition. Not unusual at all, Gene. The two input terminals to the black box are on one side. The two output terminals from the black box are on the other side. The impedance discontinuity is inside the box. The black box is extremely small. Give me the four s-parameters, s11, s12, s21, and s22 and I can tell you virtually everything about what is inside the black box without even applying a signal. BZZZT! Wrong answer. Nobody ever said anything about the "other side" of the black box. That's obviously a lie. I said something about the other side of the black box. Yet by your models and math the black boxes don't behave the same in your test circuit. That's another lie. All my models and math show the black boxes all behaving exactly the same external to the two input terminals. In fact, I have said it is impossible for it to be any other way. Is there no limit to how dishonest you will be? -- 73, Cecil http://www.w5dxp.com |
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As for your last sentence. The "he" is rather unspecific as two correspondents made a promise (perhaps not so well timed as to fit your observation here). However, I too will sit on the edge of my seat in wonderment to that outcome. 73's Richard Clark, KB7QHC Didn't read your last (promise) until after I sent that. You know who I meant. BTW, did you just by any chance ever work for me at the PME Lab in Hawaii ? W4ZCB |
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Roy Lewallen wrote:
I see that the smudge pots have been lit and the hand waving, misdirection, and desperate attempts at changing the subject begins. As they have so many times before. What a waste of all that alleged IQ. Do you think that waving your hands and posting gobblegook like the above is any better? Please, please, make a technical statement with which you know I disagree and let's discuss it like mature individuals. -- 73, Cecil http://www.w5dxp.com |
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Harold E. Johnson wrote:
I do believe that someone may have inadvertently added 100 to that IQ. Everyone on here with the exception of the gifted one has known for the last 3 days that Richard has been just pulling his bobber under. He finally picked up on it this morning and decided to stop wasting his time. Now if he would only make that a promise! I'm a simple trusting person who tries to see the good in everyone, even Richard C. Please pick a technical subject upon which we disagree and let's discuss it like technical gentlemen. -- 73, Cecil http://www.w5dxp.com |
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Roy Lewallen wrote:
I see that the smudge pots have been lit and the hand waving, misdirection, and desperate attempts at changing the subject begins. As they have so many times before. What a waste of all that alleged IQ. Roy, you are always saying I should be using voltages for my calculations and that's exactly what I did. --43.4 deg 600 ohm line--+--10 deg 100 ohm line--open Vfor1--|--Vfor2 With 100 volts at zero degrees incident upon the open end of the stub, I get: Vfor1 = 143.33 volts at -46.6 deg Vfor2 = 110 volts at -10 deg for a phase shift through the impedance discontinuity of 36.6 degrees. What do you get? Don't anybody hold his breath waiting for an answer. The Emperor has no clothes. -- 73, Cecil http://www.w5dxp.com |
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Cecil Moore wrote:
Jim Kelley wrote: Cecil Moore wrote: I(x,t) = Imax sin(kx) cos(wt) Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole. That is correct to the same extent that t in the equation above "changes hardly at all" with time. 73, ac6xg |
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Roy Lewallen wrote:
I see that the smudge pots have been lit and the hand waving, misdirection, and desperate attempts at changing the subject begins. As they have so many times before. What a waste of all that alleged IQ. Please note: This is pure unadulterated political power in action devoid of any technical content. Roy is trying to use his guru status, devoid of any technical argument, to try to discredit someone. If I am so obviously technically wrong, why doesn't he just prove me wrong with mathematics and formulas? (Because he cannot?) -- 73, Cecil http://www.w5dxp.com |
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On Mon, 17 Dec 2007 21:02:16 GMT, "Harold E. Johnson"
wrote: Didn't read your last (promise) until after I sent that. You know who I meant. BTW, did you just by any chance ever work for me at the PME Lab in Hawaii ? Hi Howard, Hmmm, how to answer that. Did that fellow remind you of me? Well, to cut to the chase, I was SUBLANT, USS Holland AS-32, Nuclear Navy; and as far as I know, the closest point of approach would have been Guam (aside from our time in the yards at Bremerton, across the water from where I live now). 73's Richard Clark, KB7QHC |
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Jim Kelley wrote:
Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: I(x,t) = Imax sin(kx) cos(wt) Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole. That is correct to the same extent that t in the equation above "changes hardly at all" with time. t changes hardly at all referenced to the source current phase which is what we are talking about. Please don't try to feign ignorance of that fact. What I don't get is why people like you have to distort the technical facts. What do you possibly have to gain through distortion and diversion? -- 73, Cecil http://www.w5dxp.com |
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Jim Kelley wrote:
Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: I(x,t) = Imax sin(kx) cos(wt) Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole. That is correct to the same extent that t in the equation above "changes hardly at all" with time. The phase is referenced to the source phase, as it is in EZNEC, but you already knew that and just want to perform your usual diversions away from the technical facts. I have said at least a dozen times that the current phase I am talking about is the same as EZNEC reports. If you don't like what EZNEC reports, take it up with Roy. For those who don't understand Jim's diversion above, EZNEC sets t=zero as a reference and then reports the phase. Jim knows that and is just trying to hoodwink the uninitiatated. His motives for such remain a mystery. -- 73, Cecil http://www.w5dxp.com |
Loading Coils; was : Vincent antenna
Didn't read your last (promise) until after I sent that. You know who I
meant. BTW, did you just by any chance ever work for me at the PME Lab in Hawaii ? Hi Howard, Hmmm, how to answer that. Did that fellow remind you of me? Well, to cut to the chase, I was SUBLANT, USS Holland AS-32, Nuclear Navy; and as far as I know, the closest point of approach would have been Guam (aside from our time in the yards at Bremerton, across the water from where I live now). 73's Richard Clark, KB7QHC OK Robert. (Howard was my uncle. He has the orange top). No, I just knew that you were into metrology from your postings, and I had the PMEL at Hickam 1960-63. Now that you've explained that, I'll inquire someday (but not today) as to what the SUBLANT folks were doing in the Pacific. Regards W4ZCB |
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Harold E. Johnson wrote:
Everyone on here with the exception of the gifted one has known for the last 3 days that Richard has been just pulling his bobber under. So now I get blasted for being naive, not for the technical stuff I am presenting? Why am I not surprised? -- 73, Cecil http://www.w5dxp.com |
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"Harold E. Johnson" wrote in message news:H1D9j.253396$Fc.55212@attbi_s21... Didn't read your last (promise) until after I sent that. You know who I meant. BTW, did you just by any chance ever work for me at the PME Lab in Hawaii ? Hi Howard, Hmmm, how to answer that. Did that fellow remind you of me? Well, to cut to the chase, I was SUBLANT, USS Holland AS-32, Nuclear Navy; and as far as I know, the closest point of approach would have been Guam (aside from our time in the yards at Bremerton, across the water from where I live now). 73's Richard Clark, KB7QHC OK Robert. (Howard was my uncle. He has the orange top). No, I just knew that you were into metrology from your postings, and I had the PMEL at Hickam 1960-63. Now that you've explained that, I'll inquire someday (but not today) as to what the SUBLANT folks were doing in the Pacific. Regards W4ZCB the Holland served many areas, quite a history for a sub tender. http://www.tendertale.com/tenders/132/132.html I don't remember being tied up to her, but we were both in Charleston at the same time, though we were sailing out of Kings Bay at the time she was getting overhauled in Charleston. |
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Cecil Moore wrote:
Gene Fuller wrote: Cecil Moore wrote: Gene Fuller wrote: Cecil Moore wrote: You missed the point. A terminating capacitor is a two terminal network. The point where two pieces of feedline are connected is a four-terminal network. A two-terminal network is different from a four-terminal network. This appears to be an unusual definition. Not unusual at all, Gene. The two input terminals to the black box are on one side. The two output terminals from the black box are on the other side. The impedance discontinuity is inside the box. The black box is extremely small. Give me the four s-parameters, s11, s12, s21, and s22 and I can tell you virtually everything about what is inside the black box without even applying a signal. BZZZT! Wrong answer. Nobody ever said anything about the "other side" of the black box. That's obviously a lie. I said something about the other side of the black box. Yet by your models and math the black boxes don't behave the same in your test circuit. That's another lie. All my models and math show the black boxes all behaving exactly the same external to the two input terminals. In fact, I have said it is impossible for it to be any other way. Is there no limit to how dishonest you will be? No lies; just carelessness in a nit-picking contest. As usual you have twisted the question so that you can provide some type of answer. You still have not answered the original question posed by Keith. 73, Gene W4SZ |
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Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: I(x,t) = Imax sin(kx) cos(wt) Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole. That is correct to the same extent that t in the equation above "changes hardly at all" with time. The phase is referenced to the source phase, as it is in EZNEC, but you already knew that and just want to perform your usual diversions away from the technical facts. Actually, I'm trying to figure out what technical meaning there is to be obtained from your repeated observation "Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole." The phase of the standing wave varies with position from one perspective, and with time from another, and with amplitude from yet another. If you hold t fixed, then amplitude and position remain variable. This is a revelation? I have said at least a dozen times that the current phase I am talking about is the same as EZNEC reports. If you don't like what EZNEC reports, take it up with Roy. One should be careful not to invite comparisons to a craftsman holding his tools responsible for poor craftsmanship. ;-) 73, ac6xg |
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Jim Kelley wrote:
Cecil Moore wrote: . . The phase is referenced to the source phase, as it is in EZNEC, but you already knew that and just want to perform your usual diversions away from the technical facts. The current reported by EZNEC isn't referenced to source phase. The phases of all currents and voltages -- wire, source, and load -- are all referenced to the same arbitrary point. Source phase can be assigned by the user to any value relative to this point. Actually, I'm trying to figure out what technical meaning there is to be obtained from your repeated observation "Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole." The phase of the standing wave varies with position from one perspective, and with time from another, and with amplitude from yet another. If you hold t fixed, then amplitude and position remain variable. This is a revelation? I have said at least a dozen times that the current phase I am talking about is the same as EZNEC reports. If you don't like what EZNEC reports, take it up with Roy. As far as I can tell, Cecil has never been quite able to understand just what EZNEC reports. EZNEC doesn't need several different definitions of current to suit the theory du jour or to do its job, so it doesn't report "standing wave current" or other fruits of The Gifted One's overly fertile imagination. It simply reports current. Anyone not acquainted with this concept can refer to any basic text on electricity or physics. What EZNEC reports is exactly what you'll find there. Roy Lewallen, W7EL |
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Gene Fuller wrote:
As usual you have twisted the question so that you can provide some type of answer. You still have not answered the original question posed by Keith. What was that question? I suspect the question was irrelevant because Keith didn't understand what phase shift I was talking about. I also suspect that Keith is beginning to understand what I am talking about. His silence seems a little strange. What are you going to do when your realize you are on the wrong side of the technical argument? Sandbag - like some others have done and try to obscure the technical facts? How about an answer from you? What is the phase shift through the impedance discontinuity between Vfor1 and Vfor2 below? --43.4 deg 600 ohm line--+--10 deg 100 ohm line--open Vfor1--|--Vfor2 I doubt that you even know how to solve the problem. Your lack of an answer will speak volumes. -- 73, Cecil http://www.w5dxp.com |
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Jim Kelley wrote:
Actually, I'm trying to figure out what technical meaning there is to be obtained from your repeated observation "Standing-wave current phase changes hardly at all up and down a stub with losses or a wire 1/2WL dipole." Take a look at the current reported by EZNEC and you will understand. Until you do that, I'm afraid you will just continue to show your ignorance. Hint: Given a reference zero phase for the source signal, EZNEC reports all current phases with respect to that source phase. So either choose to understand or remain ignorant - I just don't care which. -- 73, Cecil http://www.w5dxp.com |
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On Mon, 17 Dec 2007 23:10:24 GMT, "Dave" wrote:
the Holland served many areas, quite a history for a sub tender. http://www.tendertale.com/tenders/132/132.html I don't remember being tied up to her, but we were both in Charleston at the same time, though we were sailing out of Kings Bay at the time she was getting overhauled in Charleston. Hi Dave, Thanx for the link. 73's Richard Clark, KB7QHC |
Loading Coils; was : Vincent antenna
Roy Lewallen wrote:
Jim Kelley wrote: Cecil Moore wrote: . . The phase is referenced to the source phase, as it is in EZNEC, but you already knew that and just want to perform your usual diversions away from the technical facts. The current reported by EZNEC isn't referenced to source phase. The phases of all currents and voltages -- wire, source, and load -- are all referenced to the same arbitrary point. Source phase can be assigned by the user to any value relative to this point. The phase assigned by the user to the source obviously becomes the default reference phase. What on earth do you have to gain by obscuring that technical fact? If the user doesn't do anything, the source phase defaults to zero and all current measurements are referenced to that zero source phase. Are you so determined to discredit me that you are willing to sacrifice your integrity in the process? Do you deny that there is virtually no phase shift in the current up and down a 1/2WL dipole referenced to the source current. If you are up to that denial, please argue with Kraus's graphic. Take a look at the phase angles at: http://www.w5dxp.com/krausdip.jpg Unfortunately, Kraus agrees with EZNEC and that current is exactly what you used to make your meaningless measurements of current phase through a loading coil. So please look at Kraus's graph and tell us again how that current with unchanging phase can be used to measure phase shift through a loading coil. As far as I can tell, Cecil has never been quite able to understand just what EZNEC reports. EZNEC doesn't need several different definitions of current to suit the theory du jour or to do its job, so it doesn't report "standing wave current" or other fruits of The Gifted One's overly fertile imagination. It simply reports current. My point exactly! If the antenna being modeled is a standing-wave antenna, EZNEC reports standing-wave current. If the antenna being modeled is a traveling-wave antenna, EZNEC reports traveling-wave current. Roy apparently doesn't want anyone to know this fact. So Roy, why would the phase shift through a coil be different when it is used in a 1/2WL dipole vs using it in a rhombic at the same frequency??? Magic??? Roy has threatened to take EZNEC away from me for using it to demonstrate traveling-wave current through a loading coil. Here is what EZNEC says about the delay through that loading coil. http://www.w5dxp.com/coil512.ez Click on "Load Dat" and observe what Roy is so afraid that someone besides me is going to discover. -- 73, Cecil http://www.w5dxp.com |
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