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Dave Shrader wrote:
If dV/dt = 0 then I must have maximum voltage. This has meaning. You must be into this new math. What is the slope of a point when dt=0 = the width of a point? :-) Is that point flat, pointing up, or pointing down? :-) -- 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: Cecil seemed to indicate that he thought delta t going to zero meant that t was perpetually zero. If delta-t ever gets to zero, time stands still. All you can allow delta-t to do is to approach zero. Once it reaches zero the ballgame is over. Limit delta-t to a minimum of a yoctosecond and everything will be perfectly OK. -- 73, Cecil http://www.qsl.net/w5dxp Are you trying to start an argument? I wrote "going to zero," not "at zero." Besides, what the hell is a yoctosecond? All this was argued over and discussed in the 18th century. You and Richard Harrison are beginning to sound like Bishop Berkeley. 73, Tom Donaly, KA6RUH |
Richard Harrison wrote:
Keith wrote: "Are you sure you want to discard all thoughts of the instantaneous? Certainly not, but it has little application to power in transmission line problems. Power is the rate of transferring energy or the rate of doing work. Electrical power is measured in joules per seconds or more succinctly in watts. What is the value in watts or joules per second when seconds equal zero? I venture an answer: It is the V x I x cos. theta at that instant, but since work is power x time, it won`t do anything for you in zero seconds. But then instantaneous velocity and instantaneous acceleration won't do anything for you in zero seconds, either. And yet, for example, inertial navigation systems successfully operate by integrating these instantaneous values. Back to your assertion "but it has little application to power in transmission line problems". It is certainly true that for RF, average power is of most interest. It is what gets you communicating. But if you want to understand how things work, exploring the land of the instantaneous is quite valuable. It is instantaneous voltages which make standing waves. It is instantaneous signals which cause distortion in diode demodulators. It is instantaneous voltages and currents which are added and subtracted in Bird wattmeters. It is instantaneous voltages and currents which interact at impedance discontinuities to do all the neat stuff. And it is instantaneous voltages and currents which produce instantaneous power. But I notice an instantaneous willingness to reject the value of instantaneous power. I suspect this is because the conclusions reached when thinking in terms of instantaneous power are inconsistent with some of your long held beliefs and rather than re-examine these beliefs it is simpler to just reject instantaneous power. But to reject instantaneous power in a consistent manner, you need to explain why you do not also reject instantaneous velocity, acceleration, current, flow or any of the many other interesting things which are a derivative with respect to time. For if we accept the argument "in zero time, no energy can flow" then we should also accept: - "in zero time, no charge can flow" -- say bye to instantaneous current - "in zero time, we can move no distance" -- say bye to instantaneous velocity - etc., etc. When you can't find any fundamental reason that the instantaneousness of power is different from the instantaneousness of other common measures like velocity, current, etc., you may wish to return to the original assertion which caused all this fuss: Assertion A: "In a shorted ideal transmission line which has reached steady state, no energy can cross a voltage or current minimum because p(t) = v(t) * i(t) and at a voltage or current minimum, the voltage or current is always zero, so the power is always zero, so there is no energy flow across a voltage or current minimum." This conclusion contradicts a commonly held belief: Belief B: "that in steady-state, energy is flowing along the transmission line to the end where it is reflected and travels back to the beginning." Unless you can find an error in the logic of Assertion A, it would seem reasonable that you re-assess your acceptance of Belief B. Assertion A caused me to reject Belief B and the world did not collapse: - Bird wattmeters still give useful indications - ghosts still occur on TVs - echoes still occur on phone lines - bidirectional communications still occur on two wire lines but the simplistic explanations for these phenomena offered by Belief B need to be re-examined. A better understanding is all that you stand to gain by discarding Belief B. ....Keith ....Keith |
Tom Donaly wrote:
"You and Richard Harrison are beginning to sound lke Bishop Berkeley." We are in good company. Terman says on page 84 of his 1955 opus: "In these equations Zo = sq rt Z/Y is termed the "characteristic impedance" of the line. In the case of radio-frequency lines, Zo can nearly always be assumed to be a pure resistance, as discussed on page 88." When Terman says SWR = Emax / Emin, it makes no difference whether you use instantaneous values or rms values, so long as you are consistent, the ratio is the same. Some complained that nobody provided a trustworthy VSWR related to power. Bird Electronic Corporation does: VSWR = 1+sq rt (reflected pwr/forward pwr) over 1-sq rt (reflected pwr/forward pwr). Millions of conversions have proved this VSWR from measured powers relation. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
For the power sine wave, though the fact that a minus times a minus is a plus results in 2x the voltage frequency, dP/dt=0 at maxima. Alternatively, from amplitude modulation, the product of two sine waves produces a sum and difference frequency. When the two sine waves have the same frequency (as they due for the voltage and current contributing to the power), the result is a double frequency sine wave and a 0 frequency difference which is the DC or average power. A question raised in this thread is, how can energy, which is joules per second times seconds, be zero when the number of seconds is zero? The answer seems obvious. Zero times anything is zero. Exactly. And when the voltage or current is always zero, so must be the power. ....Keith |
Keith wrote:
"---you need to explain why you do not reject instantaneous velocity, acceleration, ---." I don`t reject instantaneous anything including power. Instantaneous power is not particularly useful in working with radio transmission lines. Like infinity, the infinitesimal is unmeasurable. Like infinity, the infinitesimal is useless in calculations. The idea of the infinitesimal is useful in perceiving targeted values approached as a variable approaches a limit. A differential "d" is an infinitesimal smaller than a difference. "dy" is the differential of y. "dx" is the differential of x. "dt" is the differential of t. The ratio "dy/dx" is a slope defined at a point and is equal to the limit as x goes to zero of the ratio delta y over delta x. The basis of differentation is superfluous to this discussion, but Keith asks, why not reject things which are a derivative with respect to time including acceleration. Acceleration may be a good example. Calculus can give the rate at which a variable varies. On the other hand, it can give a function if the rate of change is given. Velocity is the variable in acceleration. Assume velocity is increasing and you have a definition of the function. For a given velocity the acceleration can be determined, and for a given acceleration, the velocity can be determined. My point, repeated again, is that when delta time is zero, no distance is traversed, not that acceleration and velocity are zero. Power x time = energy. Thast`s how the electric power company calculates your bill. If no time elapses during which power is available, no energy is consumed. Best regards, Richard Harrison, KB5WZI |
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Richard Harrison wrote:
Standing waves are perceived only by using a sensor which does not discriminate by direction and accepts power traveling in both directions at the same time. There actually is no abatement of power in either direction. Sadly for ham radio, this fact of physics has been known to man for 300 years. How can a "technical" hobby be 300 years out of date? -- 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! =----- |
wrote:
But if you want to understand how things work, exploring the land of the instantaneous is quite valuable. OTOH, if you want to understand how EM waves work, exploring the discoveries of the past 500 years in the field of optics is quite valuable - discoveries that you have officially discounted when you reject the total knowledge base embodied in the field of optics. Seems my "sin" is pretty small compared to yours. :-) -- 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: Standing waves are perceived only by using a sensor which does not discriminate by direction and accepts power traveling in both directions at the same time. There actually is no abatement of power in either direction. Sadly for ham radio, this fact of physics has been known to man for 300 years. How can a "technical" hobby be 300 years out of date? -- 73, Cecil http://www.qsl.net/w5dxp Larding it on a little thick are we, Cecil? 73, Tom Donaly, KA6RUH |
W5DXP wrote:
wrote: It is instantaneous voltages which make standing waves. That's a no-brainer. It is instantaneous signals which cause distortion in diode demodulators. Please define "instantaneous" for us. I suspect it cannot be what happens in a dt=0 time slot. With a time varying voltage, for each time t there is a particular value of voltage. This voltage can be called the instantaneous voltage to make it clear that it is not the average or RMS or peak or any of the other voltages which may be of interest with regards to the signal. Hence the notation v(t) representing the voltage at any time t. It is instantaneous voltages and currents which are added and subtracted in Bird wattmeters. But the result is rectified and the maximum value is stored on a capacitor. By definition, a capacitor cannot accept an instantaneous value of voltage. If it could, it would be called an inductor. :-) None the less, the original additions and subtractions of voltages and currents must be done with the instantaneous values for the proper results to be obtained. But I notice an instantaneous willingness to reject the value of instantaneous power. Please prove that the human brain is capable of instantaneous willingness. :-) If you will accept proof by example, then simply re-read some of your posts. But to reject instantaneous power in a consistent manner, you need to explain why you do not also reject instantaneous velocity, acceleration, current, flow or any of the many other interesting things which are a derivative with respect to time. Some things logically have an instantaneous value, e.g. voltage. Other instantaneous values do not seem to be related to reality. Let me use your mind-f__king techniques on you. So you are saying that all instantaneous values appearing in any math model anywhere have an associated existence in reality. Absolutely not. But I have asked that if you wish to contend that instantaneous power is of no use while instantaneous current is, that some rationale be provided. The previous arguments were that it made no sense because when dt was 0, there was no power. This seemed a weak argument since a similar argument could be made for the other instantaneous value which do seem to be accepted. Assertion A: "In a shorted ideal transmission line which has reached steady state, no energy can cross a voltage or current minimum because p(t) = v(t) * i(t) and at a voltage or current minimum, the voltage or current is always zero, so the power is always zero, so there is no energy flow across a voltage or current minimum." Assertion B: Since the universe is about 15 billion years older than our solar system, transmission lines have probably been doing their thing for billions of years longer than your above statement has existed. Does this non-sequitor mean that you accept Assertion A and wish to deflect the discussion? This conclusion contradicts a commonly held belief: Yes, and is therefore probably as wrong as can be. Perhaps, but I observe that you have not yet pointed out the flaw. Were you to find the flaw in the logic, I would willingly accept it as false. Merely that the result disagrees with Belief B is not, by itself, a flaw. There is absolutely nothing in physics that prohibits energy from flowing across an area where power is zero. I had always understood that power was defined as the rate of energy flow. If the power is zero, then by definition, there is no flow. Perhaps your definition of power is different and we should resolve that before proceeding. Belief B: "that in steady-state, energy is flowing along the transmission line to the end where it is reflected and travels back to the beginning." Please present a model of how standing waves are possible without forward waves and reflected waves in a single-source, single feedline, single load system. Well, as you agreed near the start of this post (with your statement "That's a no brainer), it is voltage waves which produce standing waves, not waves of energy. So Assertion A is not in conflict at all with standing waves. Until you do that, you are just, IMHO, blowing smoke. It is certainly possible that I have made an error in Assertion A. I observe that you have not yet located that error. But Assertion A is fairly short and based on the most basic of concepts. It should be easy to locate and describe the error, if one exists. Unless you can find an error in the logic of Assertion A, it would seem reasonable that you re-assess your acceptance of Belief B. The error in assertion A is that EM light waves cause power nulls without having any effect on each other whatsoever. And that's exactly what happens in a transmission line. Please go through each of the steps of Assertion A and describe the first one which is in error and why. Assertion A caused me to reject Belief B and the world did not collapse: When you reject the primacy of consciousness in favor of the primacy of existence, you will understand why your thoughts don't effect reality. Your thoughts also do not affect much of reality. What happens when you God-like gurus disagree? - Close to nothing! This, of course, would not assist with the understanding of the nature of 'reflected' power. When you get through with that, consider what happens if time doesn't really exist and is simply a model of change invented by the human mind. What happens when you divorce change from the rotation of the earth? Nor would this. ....Keith |
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Tom Bruhns wrote:
(Agreeing with Keith...) Of course there MUST be instantaneous power, because if there is not, then one cannot find energy by taking the integral of power over time. The instantaneous value may be zero, or positive, or negative, of course. Or perhaps another way to look at it is that if energy, which itself is a function of time, is differentiable, then power must take on instantaneous values. I don't think there is any argument over whether it exists in the math model or not. My argument is that the concept lacks a lot of usefulness. -- 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! =----- |
W5DXP wrote in message ...
Tom Bruhns wrote: (Agreeing with Keith...) Of course there MUST be instantaneous power, because if there is not, then one cannot find energy by taking the integral of power over time. The instantaneous value may be zero, or positive, or negative, of course. Or perhaps another way to look at it is that if energy, which itself is a function of time, is differentiable, then power must take on instantaneous values. I don't think there is any argument over whether it exists in the math model or not. My argument is that the concept lacks a lot of usefulness. Ah, I see. Well, consider this: the concept IS useful to me. That's not an arguable point, as I am the only one who can evaluate it. If it isn't to you, you don't have to use it. I'll not argue with that. Cheers, Tom |
wrote:
Are you saying that there can be power when the voltage or current is always 0? And that this is the error in the sequence of steps which makes Assertion A false? There can and does exist component energy when the NET power is zero. You will understand this when you understand complete destructive interference in light waves represented by the equation I1 + I2 -2*Sqrt(I1*I2) equation 9.16, page 388, in the 4th edition of _Optics_, by Hecht. Two waves are flowing with unchanging energies and producing an absolutely black ring, i.e. zero irradiance equals zero power. Consider again, Assertion A, in more detail this time: It won't do a bit of good to rehash this until you understand complete destructive interference. The two interfering waves flow unabated. You are being fooled by superposition of two waves whose component energies are completely independent of each other. All the steps in Assertion A seem correct to me. Can you help me find the false step which makes Assertion A false? Yes, I already did that more than once. You will not understand until you understand how two light beams can cause zero power without even knowing the other wave is there. There is absolutely no change in the energy levels in the individual waves during complete destructive interference. -- 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! =----- |
Actually, all you've got to do to understand this is to realize that
energy is the time integral of power. (Remember that "useless" quantity of power as a time function?) When you integrate power to get energy, you get a constant. That constant is the energy present when power = 0, and it has to be evaluated by knowing something other than just the power. In a transmission line, that energy is stored in the electric and/or magnetic fields during the times energy isn't moving (i.e., when p(t) = 0). Really, math is cool. Roy Lewallen, W7EL W5DXP wrote: wrote: Are you saying that there can be power when the voltage or current is always 0? And that this is the error in the sequence of steps which makes Assertion A false? There can and does exist component energy when the NET power is zero. You will understand this when you understand complete destructive interference in light waves represented by the equation . . . |
W5DXP wrote:
wrote: All the steps in Assertion A seem correct to me. Can you help me find the false step which makes Assertion A false? Yes, I already did that more than once. You will not understand until you understand how two light beams can cause zero power without even knowing the other wave is there. There is absolutely no change in the energy levels in the individual waves during complete destructive interference. I conclude from this response that you are unable to detect any flaws in the logic of steps 1 to 8. Oh well. ....Keith |
Keith wrote:
"I conclude from this response that you are unable to detect any flaws in the logic of steps 1 to 8." They are contradictory. That`s a fatal flaw. Volts and amps on a transmission line are manifestations of E&H fields guided by the line. These fields are generating the volts and amps seen on the line as the wave travels along its length. Even a hard short or an open circuit doesn`t annihilate the signal on a line. The short or open only produces a reflection. At the open or short, amps or volts are forced to zero. That event forces the energy which had resided in the H-field or E-field to temporarily be accomodated in the field which is its partner in transporting energy. This doubles volts right at the open or doubles amps right at the short.. This single field is only temporary because in a very short travel distance, energy is again balanced between the two fields. Note that the open or short does not stop energy flow. It just turns it around as a reflection. This happens at an actual discontinuity. At an SWR interference point, nothing upsets energy flow at all in a uniform line. SWR nulls are just demonstrations of commingled waves passing through each other without serious consequence. For the nth time, were you to separate the going and coming signals, you would find no variation in average signal power in either direction along the line at any point, including the SWR nulls.. Best regards, Richard Harrison, KB5WZI |
wrote:
6) From 2) and 5), the power (rate of energy flowing) at quarter wave points will be 0 The NET power will be zero. The forward power flow vector is NOT zero and the reflected power flow vector is NOT zero. The NET power of zero is complete destructive interference in action. It happens all the time to two light beams whose energies are not affected by the interference. 7) From 6), the energy crossing quarter wave points is 0 The NET energy crossing quarter wave points is 0. The two component energies crossing quarter wave points are not 0. They are the constant forward power flow vector and the constant reflected power flow vector as explained in Ramo & Whinnery. Their ratio is the power reflection coefficient. Equation 3, page 350, _Fields_ and_Waves_in_Communications_Electronics_, Ramo, Whinnery, & Van Duzer. (Pz-)/(Pz+)=|rho|^2 All the steps in Assertion A seem correct to me. Yes, that is your problem. They are not correct for the two component energies. Can you help me find the false step which makes Assertion A false? Here it is again, for the Nth time. There are two waves flowing in opposite directions, each possessing its own constant energy, each flowing unopposed end to end in the transmission line. -- 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! =----- |
On Mon, 25 Aug 2003 21:19:24 -0500, W5DXP
wrote: Two waves are flowing with unchanging energies and producing an absolutely black ring, i.e. zero irradiance equals zero power. Hi Cecil, Only when you violate conservation of energy, i.e. you have not summed over the complete interval, only that part that satisfies your argument (which allows anyone to perform similar miracles). 73's Richard Clark, KB7QHC |
Richard Clark wrote:
W5DXP wrote: Two waves are flowing with unchanging energies and producing an absolutely black ring, i.e. zero irradiance equals zero power. Only when you violate conservation of energy, ... No violation at all, Richard. The power (irradiance) that is lost in the black rings is gained in the bright rings as explained in _Optics_, by Hecht where he says constructive interference is always equal to destructive interference. -- 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! =----- |
wrote:
Richard Harrison wrote: They are contradictory. That`s a fatal flaw. This may be, but there seems to be some reluctance to point out WHICH step is contradictory. The reluctance seems to be your unwillingness to read the replies which point out exactly that. I'll accept that, but no energy flows past the short or open where the voltage or current is zero which is consistent with p(t) = v(t) * i(t). No NET energy flows past such a point. Plenty of component energy flows past the point in both directions as explained in Ramo, Whinnery, & Van Duzer. I do understand the derivation from this separation of the going and coming signals. But I am unable to rationalize this explanation with the very basic and widely accepted p(t) = v(t) * i(t) and the observed zero voltages and currents. You are talking about NET energy and NET power. Drop down one level to component energy and power and all will become clear - especially if you take a look at light wave interference patterns. Do think of that voltage or current 0 at the shorted or open end of the line. In your model, this voltage or current 0 produces a reflection while other voltage or current zeroes do not. Is this consistent? Yes, perfectly consistent. The short or open end of the line is a physical impedance discontinuity. The other voltage and current zeros are NOT physical impedance discontinuities. They are a result, not a cause of anything. You are simply confusing cause and effect. Consider adding another quarter wave section to the end of the line. The voltage or current 0 remains and to the left of the original end of the line nothing changes, but by your model, suddenly there is no longer a reflection at the original end of the line, but it has moved to the new end. But the voltage and current distributions have remained the same. There is no observable difference. Are you telling us that you cannot locate the physical short or open at the end of the line and don't observe any difference between a physical short or open and a piece of continuous transmission line? All you have to do is open your eyes. -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= 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! =----- |
wrote:
W5DXP wrote: Actually, not. You continue to resist pointing out which step is wrong. You obviously have not read all of my replies. Is it step 2)? "2) At quarter wave points along this line, voltages and currents which are always 0 can be observed -- the standing wave" This seems to be generally accepted. This is where your confusion starts. There are points where the NET voltage and NET current are zero. Those are merely the points where the forward Poynting vector and reflected Poynting vector are out of phase. Is it step 3)? "3) Power is the rate at which energy flows" This is the definition of power. So please apply it to the forward Poynting vector and the reflected Poynting vector as described in Ramo & Whinnery. Is it step 4)? "4) The power (rate of energy flowing) at time t can be computed using p(t) = v(t) * i(t)" This is the well know expression of power in terms of volts and amps. This is NET power. Your sin is not a sin of commission. It is a sin of omission. Is it step 5)? 5) Substituting a voltage or current which is always 0 into the expression above will result in a power which is also always 0 Just normal subsitution of actual values into an equation. Again, you are dealing only with NET power when you need to be dealing with component energies. Is it step 6)? "6) From 2) and 5), the power (rate of energy flowing) at quarter wave points will be 0" The result obtained after substitution. Again, your assertions apply ONLY to NET power which is not what is being discussed. Is it step 7)? "7) From 6), the energy crossing quarter wave points is 0" If steps 1) to 6) are not in error, then step 7) follows. The NET energy is zero. The component energies are not zero. Is it step 8)? "8) From 7), energy can not be flowing down and up the line crossing quarter wave points" Simply follows from 7). And completely false for component energies. You have not yet pointed to any error in the derivation. Yes, I have, numerous times. Let me give you an analogy. To prove my point I say: "My pickup is white. Please prove otherwise." My statement is absolutely true and absolutely irrelevant. So are yours. If your extensive study of optics leads you to believe that the conclusion is incorrect, and, if the conclusion is incorrect, there must be an error in the derivation. What is it? Your error is one of omission. Your assertions are simply not relevant to the discussion. But the derivation is quite simple. Yes, too simple and completely irrelevant. Since you have not yet pointed to an error in the derivation (which would be the obvious way to close the question), I conclude that you have been unable to locate such an error. Seems reasonable, does it not? Since you have not proven that my pickup is not white, you lose the argument. See, I can do the same thing you are attempting. -- 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! =----- |
On Tue, 26 Aug 2003 09:17:33 -0500, W5DXP
wrote: Richard Clark wrote: W5DXP wrote: Two waves are flowing with unchanging energies and producing an absolutely black ring, i.e. zero irradiance equals zero power. Only when you violate conservation of energy, ... No violation at all, Richard. The power (irradiance) that is lost in the black rings is gained in the bright rings as explained in _Optics_, by Hecht where he says constructive interference is always equal to destructive interference. Hi Cecil, So you contradict yourself. Two waves do not produce a black ring, they produce two rings (one dark, one bright) over an interval of one wavelength (the proper bounds for an energy solution) and the sum of them are the same as the applied powers - not zero. Choosing lesser bounds to craft your "proof" constitutes an invalid proof. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
W5DXP wrote: No violation at all, Richard. The power (irradiance) that is lost in the black rings is gained in the bright rings as explained in _Optics_, by Hecht where he says constructive interference is always equal to destructive interference. So you contradict yourself. Two waves do not produce a black ring, they produce two rings (one dark, one bright) ... Don't know what planet you live on, Richard, but on this one if there are two rings, one dark and one bright, there exists a dark ring. No contradiction at all. Please sober up and try to do a better job next time. -- 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 Clark wrote:
W5DXP wrote: That's the flaw in your logic and this is about the 5th time I have explained it to you. And what makes this remarkable? .... his ability to ignore the great body of scientific facts contained in the field of physics from which I have previously quoted. Most of what we are discussing has been known for about 300 years. If he (and you) choose to ignore the known facts of physics, there is absolutely nothing I can do about it. Understanding interference patterns is centuries old but my daughter still cannot understand them. Probably only a small percentage of the human population understand interference patterns - which is a pity. -- 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! =----- |
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On Tue, 26 Aug 2003 11:13:45 -0500, W5DXP
wrote: Richard Clark wrote: W5DXP wrote: That's the flaw in your logic and this is about the 5th time I have explained it to you. And what makes this remarkable? there is absolutely nothing I can do about it. Hi Cecil, And yet you gust on in indifference. Well, I am going to step inside out of the elements and watch the leaves dance. ;-) 73's Richard Clark, KB7QHC |
On Tue, 26 Aug 2003 11:04:48 -0500, W5DXP
wrote: Richard Clark wrote: W5DXP wrote: No violation at all, Richard. The power (irradiance) that is lost in the black rings is gained in the bright rings as explained in _Optics_, by Hecht where he says constructive interference is always equal to destructive interference. So you contradict yourself. Two waves do not produce a black ring, they produce two rings (one dark, one bright) ... Don't know what planet you live on, Richard, but on this one if there are two rings, one dark and one bright, there exists a dark ring. No contradiction at all. Please sober up and try to do a better job next time. Hi Cecil, I live on the blue marble your planet is now approaching as close as it has been in 70,000 years. The sobriety you speak of is a matter of having more oxygen than you. As for jobs? Write when you find work, buckaroo. We have a dead rover somewhere near you that could stand fixing. ;-) 73's Richard Clark, KB7QHC |
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W5DXP wrote:
wrote: W5DXP wrote: Actually, not. You continue to resist pointing out which step is wrong. You obviously have not read all of my replies. On the contrary. But finally this one actually addresses one of the steps rather than just attempting to show that that conclusion is wrong. Is it step 2)? "2) At quarter wave points along this line, voltages and currents which are always 0 can be observed -- the standing wave" This seems to be generally accepted. This is where your confusion starts. There are points where the NET voltage and NET current are zero. Those are merely the points where the forward Poynting vector and reflected Poynting vector are out of phase. This emphasis on NET seems to be the place where the difficulties begin. An (ideal) voltmeter placed at a voltage minima on the line indicates a voltage of 0 volts. You appear to be saying that despite this indication of 0, there is actually voltage present. This contention leads to a number of questions related to voltmeters: - when I use a voltmeter to measure the voltage in a circuit, how do I know when an indication of 0 means 0? - how do I know when an indication of 0 means there are really a number of voltages which sum to 0? - how do I determine what these voltages are which sum to 0? - if it indicates other than 0, is it really indicating a number of voltages which sum to the result? - how do I determine what these voltages actually are? - can voltmeters be trusted at all? - when? In my world, voltmeters indicate volts. There is no need for second guessing. Until this very fundamental difference is settled, there is no value in examining the remaining steps. ....Keith |
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wrote:
This emphasis on NET seems to be the place where the difficulties begin. An (ideal) voltmeter placed at a voltage minima on the line indicates a voltage of 0 volts. You appear to be saying that despite this indication of 0, there is actually voltage present. There are actually two voltages present of equal amplitude and opposite phase. Their phasor sum is zero volts. - how do I determine what these voltages actually are? |Vfwd| = Sqrt(Pfwd*Z0) |Vref| = Sqrt(Pref*Z0) If the voltmeter reads zero, these two voltages are equal in magnitude and opposite in phase. Do you know how to sum two phasor voltages? -- 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! =----- |
Hm, this has me puzzled. Assuming a purely real Z0, you're taking the
square roots of two purely real quantities, each of which can have two values, and speaking of a phase angle between them. Where in the process did they pick up phase information? Or do you just mean when one is the negative of the other? If so, how do you tell -- each has two roots, that is, Sqrt(Pfwd*Z0) can be either positive or negative, and so can Sqrt(Pfref*Z0). How do you know when one is the negative of the other? Roy Lewallen, W7EL W5DXP wrote: Richard Clark wrote: We covered this before. The NET 0 "could be" the result of a bajillion volts and -bajillion volts. It could be but it's not. The NET 0 is the result of Pfwd=Pref Thus, Sqrt(Pfwd*Z0) = Sqrt(Pref*Z0) and when they are 180 deg out of phase, the net voltage is zero. |
Keith wrote:
"-how do I determine what these voltages actually are?" You measure the power in either direction using a wattmeter containing a directional coupler. The volts, amps, and power flowing each direction are the same. That`s why you have zeros. Then, P = Esquared / Zo, so E = sq rt (P)(Zo) Best regards, Richard Harrison, KB5WZI |
I wrote:
"E = sq rt (P)(Zo)" Some may not realize that E is an rms value. Keith early in the thread was inserting instantaneous volts which can be an infinite range of values as they may be taken at any point in a cycle for evaluation. I agree with Cecil`s "tits on a boar hog" characterization of value in transmission line problem utility. Best regards, Richard Harrison, KB5WZI |
Roy Lewallen wrote:
Hm, this has me puzzled. :-) Good one, Roy. :-) Assuming a purely real Z0, you're taking the square roots of two purely real quantities, each of which can have two values, and speaking of a phase angle between them. Where in the process did they pick up phase information? Or do you just mean when one is the negative of the other? If so, how do you tell -- each has two roots, that is, Sqrt(Pfwd*Z0) can be either positive or negative, and so can Sqrt(Pfref*Z0). How do you know when one is the negative of the other? V^2/Z0=P is a well known equation (so is I^2*Z0=P). These are *RMS* values. So the RMS voltage is V = Sqrt(Pfwd*Z0). Root Mean Square AC voltages are equivalent to DC voltages in power dissipation and are generally considered to be positive values because they are the sum of squared terms. We can turn those RMS voltages into phasors by adding the phase angles. When Vfwd+Vref = Vmax, Vfwd and Vref are in phase (at the SWR voltage maximum point). When Vfwd+Vref = Vmin, Vfwd and Vref are 180 degrees out of phase (at the SWR voltage minimum point). Vmax/Vmin = VSWR. Please feel free to pull my leg again anytime. :-) -- 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 Clark wrote:
Certainly we cannot ignore the bajillion volts *RMS*, now, can we? ;-) We can unless Sqrt(P*Z0) equals a bajillion volts which it usually doesn't. These values are all inter-related. A 200W transmitter will pour Sqrt(200*50) = 100 volts RMS into a 50 ohm load. That's a no-brainer. Why do you act like it is a far out big deal? -- 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! =----- |
On Wed, 27 Aug 2003 15:51:20 -0500, W5DXP
wrote: Why do you act like it is a far out big deal? Hi Cecil, Are you now trying to convince us that a bajillion volts *RMS* meeting -bajillion volts *RMS* is NOT NET 0? Those are merely the points where the forward Poynting vector and reflected Poynting vector are out of phase. We are dealing with component energies. The NET energy is zero. The component energies are not zero. This would seem to conflict with much of your wave mechanics for the last 6 months. Shirley you cannot deny the impact of a bajillion volts *RMS* simply because your current argument doesn't need that solution, can you? (I guess you can.) However, that is not to say that we didn't notice that rhetorical slide from "could be" to the firmer "is." We can specify these bajillion volts *RMS* with a phase of a mega-bajillion degrees (with deliberate care to avoid problematic 179 or 181 degrees) to suit any opportunistic need. ;-) 73's Richard Clark, KB7QHC |
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