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Richard Clark wrote:
Cecil Moore wrote: You furnished less than none. Short memory in long supply. The complete treatment in math was offered successfully rebutting your proposition and you have shown nothing new. The negation stands. There is zero net refraction, given by definition. So all your refraction math was irrelevant and negated nothing. All that exists in the example is forward energy and reflected energy which you chose not to deal with at all. The reason that optical engineers know so much more about power and energy in EM waves is because that's about all they could measure for 100 years. They don't have the luxury of measuring the voltage in an EM light wave. And using voltage to analyze photonic EM energy waves doesn't reveal the whole story. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
On Sat, 16 Jul 2005 17:15:58 -0500, Cecil Moore
wrote: There is zero net refraction, given by definition. So all your refraction math was irrelevant and negated nothing. The math remains inviolate, exhibits the laws of conservation, and negate your premise. So far you have added nothing to offset this. The reason that optical engineers know so much more about power and energy in EM waves is because Uh-huh. Against myself, a practicing and successful optical engineer with optical patents; you, a xerox jockey, are comparing yourself? :-) Bubba, you can't even answer simple power questions like: You have a one square cm target that is irradiated with 64 microWatts of 660nM radiation at a distance of 1M from a light bulb. (which is less than the power reflected from one of your example interfaces and still visibly quite bright.) How much power is found in the 555nM spectrum expressed in Lux? Both are power spectrums within a 30nM BW. How much total power is the light bulb radiating? Optical engineers can answer this, and I will by midnight. ;-) No one expects a binary engineer can (example of a simple 1 or 0). |
Richard Clark wrote:
Cecil Moore wrote: There is zero net refraction, given by definition. So all your refraction math was irrelevant and negated nothing. The math remains inviolate, exhibits the laws of conservation, and negate your premise. So far you have added nothing to offset this. Most of the reflection examples in _Optics_, by Hecht, assume zero net refraction and I will continue to follow Hecht's lead. (Complicating examples beyond what is needed for understanding the principles involved is a form of obfuscation.) -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Richard Clark, KB7GHQ wrote:
"No one expects a binary engineer can (answer that)." I have a different take on reflection of the reflected wave at the generator. I think current results from potential difference. When there is no difference, there is no current. Suppose you have an ideal transformer with a perfectly centertapped secondary. Volts are the same on both sides of the centettap but 180-degrees out of phase with respect to the centertap. Connect the extremes together and a high current results from the short circuit. Suppose you modify the secondary by untwisting the sentertap and this results in two identical secondary coils. Connect these in the same phase in parallel. No current flows between the coils because no potential difference ever exists between the coils. An ideal transmission line with a complete reflection at its far end is much like the transformer with two identical secondaries. The incident and reflected voltages are equal and at some points can completely add or subtract. Back at the generator connected to a line with a complete reflection at the load end, there are line lengths which produces a reflection arriving back at the generator 180-degrees out of phase with the generator voltage. This produces the equivalent of a short circuit on the generator. If line length results in reflected volts exactly in-phase with generated volts, no current flows into the line once steady-state conditions are established. This is what happens in the short-circuited 1/4-wave stub (metal insulator). It is the standoff between identical voltages which produces the extremely high impedance at the input to the stub. I have not tried my Bird wattmeter on this. Maybe Cecil has and will say I am wrong. Best regards, Richard Harrison, KB5WZI |
On Sat, 16 Jul 2005 22:00:47 -0500, Cecil Moore
wrote: Most of the reflection examples in _Optics_, by Hecht Dear Readers, it isn't midnight and there is no chance that our scribbler has actually had the experience to express any solution bearing on power but his own cooked theory. After-all if we can see the reflections from these anti-reflective layers, and that is sufficient proof to invalidate this folderol; well, experience must be a lie compared to tarted up references (which is so much sacred hamburger). So, for the sake of those following this (and sufficiently wise enough not to jump into this sewer without a snorkel), I posed a simple question as to the amount of power in the 555nM band (within a 30nM BW) given a known power of 64microWatts illuminating a square cM target. I further asked that this be expressed in Lux. Well, this problem is no more difficult than being able to simply take the one power already known, the characteristic of a tungsten lamp and transforming it into the other wavelength. This is a commonplace of optical engineering unknown to the binary engineer who finds the sum total of his entire instruction in two pages xeroxed from a library book. I will skip the expression in Lux simply because that would be showing off, and cut to the chase of power expressed in conventional terms (the binary engineer will be wholly lost in the arcana of practical measures of light and couldn't be trusted to answer if any power were actually visible - commonplace experience is a mystery) = 220 nW. This answer reveals there is more to the characteristic of the tungsten lamp, than meets the eye. 220 nW is actually quite bright, and yet we are being handed sloppy work that is acknowledged to dismiss nearly a thousand times as much power as a trivial difference that doesn't invalidate a claim of "total" cancellation. I wouldn't trust such a personality as a bank teller, nor a goldsmith, nor a surgeon, bridge builder, ... in short, no one serious about the subject. This kind of slop is what CFA and EH antennas are built from. So, this practiced optical engineer has delivered what the binary engineer could not. Nothing amazing about that - experience clearly differentiates knowledge from wishing. I could continue with turning this into Lux/Lumens/Foot-Candles/Candelas (terms of confusion to our scribbler), shifting the wavelength again, expressing the total power radiated, expressing the total light seen (or power in the BW selected light), and so on that are complete mysteries that utterly wipe out these facades breathlessly offered as compelling proofs. The single most embarrassing question I've offered to this correlation of "glare-proof" optics was to ask the obvious: What wavelength is Glare? Even here there is a practical answer that stumps the binary mentality limited by the lack of experience, and the dearth of exposition from two pages of thumb worn xeroxes. So, we have these "Can you solve this?" howlers where the author is so utterly unversed on the topic that he cannot describe power; fails to acknowledge if that glare's reflection could be seen at the typical values found in perfect math solutions; what wavelength we are talking about; why the problem is posed out to 5 places of precision and with only 1 place of accuracy results; why a light level of 1/1000th of the typical perfect math solution is still visible but is dismissed as a correct result. It has been amusing nonetheless. 73's Richard Clark, KB7QHC |
Richard Harrison wrote:
I have not tried my Bird wattmeter on this. Maybe Cecil has and will say I am wrong. No, I agree with your analysis, Richard, as far as it goes but an expanded look will reveal some interesting facts. Concerning open-circuited or short-circuited transmission lines, if the voltages add up to zero, that looks like a short, and current is at a maximum. If the currents add up to zero, that looks like an open and the voltage is at a maximum. If these conditions occur at the source, 100% re-reflection of the incident reflected wave energy is guaranteed. The Bird reads equal forward and reflected power all up and down the feedline. If an open-circuited 1/2WL of 50 ohm lossless feedline is connected to a simple class-A source having a series source impedance of 50 ohms and a voltage source of 141.4v (as in W7EL's "Food For Thought #1) the steady-state impedance seen by the source is an open-circuit so the source voltage is 141.4v and the source current is 0 amps. The implication is that it is like having nothing connected at all but that's not correct. Measuring the current at the mid-point of that 1/2WL of feedline will prove the feedline is filled with EM wave energy which must travel at the speed of light. The Bird will read 100w forward and 100 reflected on the feedline. An RF current meter at the center of that 1/2WL of feedline will read 2.828 amps. That's the sum of the forward current and reflected current in phase, 1.414 amps in the forward direction and 1.414 amps in the rearward direction. 1.414^2*50 = 100w for both forward and reflected powers. The Bird is right. Those powers are really there supporting the forward and reflected waves and cannot be used for any other purpose. Since the feedline is lossless, there's no lost energy to replace so the source power output is zero. Anything else would violate the conservation of energy principle. Note that at the above current maximum point, the net flow of energy is zero since the Poynting vectors for forward and reflected power add up to zero. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Richard Clark wrote:
Cecil Moore wrote: Most of the reflection examples in _Optics_, by Hecht, assume zero net refraction and I will continue to follow Hecht's lead. After-all if we can see the reflections from these anti-reflective layers, and that is sufficient proof to invalidate this folderol; Following exactly the same reasoning, if we can measure loss in all transmission lines, the concept of a lossless transmission line is also folderol. The example I gave was designed for clarity of understanding. If you choose to obfuscate clarity, don't expect me to join you. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
On Sun, 17 Jul 2005 07:55:32 -0500, Cecil Moore
wrote: The example I gave was designed for clarity of understanding. It is a poor example of understanding when you purposely inject error. There is nothing clear about intentional mistakes. Any rejection of a complete solution is a suspect agenda from the beginning. Understanding does not come of clouded data and murky results. What has been offered is evidence of poor knowledge, and no experience. That poor foundation built on sand has translated into outrageous conclusions that are nothing more than castles in the sky. I can demonstrate this in that when posed with a REAL power application, silence typically falls for the simplest of computations. Like how much power does a light bulb radiate to illuminate in the 660nM region a target of 1 cm square, at 1 M with 64 microWatts/30 nM of bandwidth? The inability to do such trivial power models reveals every thing else offered has the makings of superstition that builds CFAs and their ilk. Dear Readers, Knowing the binary result beforehand (0) to this simple appeal, I will render that answer before the day is out. It will exhibit how little optical knowledge is contained in this "Can you solve this" banality. You may all note that my embarrassing question: What is the wavelength of Glare? remains without comment or response, even though there is a practical answer and a perfectly reasonable explanation. The gulf of silence that attends this remarks how complete the void of experience is. Can you imagine basing an entire exposition around such a commonplace problem and not knowing the basics? In short, this "Can you solve this?" is more an appeal for knowledge than a demonstration of skill. For one, you need to know WHO needs this glare cancellation capacity, and then you would ask WHY; and it then follows that the wavelength falls within these particular aspects. Fairly simple stuff for the optical engineer, but wholly outside of the binary engineer's experience and education. The greater embarrassment is that apparently it is outside of the skill of performing a simple Google search to fill that gap of knowledge. As I offered, once that knowledge came to mind, the WHY and WHO would explain the WHEREFORE instead of this rummaging through text to xerox formulas to force-fit a presumed theory of "total" cancellation. It is still entertaining tho', as a burlesque of engineering. ;-) 73's Richard Clark, KB7QHC |
On Sun, 17 Jul 2005 11:07:12 -0700, Richard Clark
wrote: I can demonstrate this in that when posed with a REAL power application, silence typically falls for the simplest of computations. Like how much power does a light bulb radiate to illuminate in the 660nM region a target of 1 cm square, at 1 M with 64 microWatts/30 nM of bandwidth? Dear Readers, It comes as no great surprise that this simple example of optical power is so powerfully baffling to a neophyte. Optical studies are not nearly as simple as some might believe, and on the other hand, the field yields answers as easily as any if you simply observe first principles. Those principles are as simple as knowing the area of the total surface illuminated by the standard light bulb (we will ignore the shadow cast by the base). This is quite easily determined as I already offered the distance of 1M and the equivalent power within a 1cM area of that total surface. The math is hardly as arcane as "interference math," but as I have demonstrated how poorly that was performed, perhaps a little walk through here is in order. The area of sphere, at 1M radius quickly resolves to 4·pi M². Conversion to cM² "should" be a knock off, but when you use 5 place math to perform 1 place accurate results.... Well, let's just cut to the chase and skip all the so obviously difficult parts and just answer the question above. Our source is a 10W bulb. Common, mud-ordinary situation offering smaller values of power than supposedly encountered in the "reflections" of this "Can you solve this?" drek. And as I offered, such reflections as you would observe (much like the mathematical impossibility of a bee's flight) - quite bright. However, this was on scale with 1cM² and when the original question posed a 1W laser - well you can appreciate that I do not hesitate to point out that the power of those "totally" cancelled reflections are easily 10,000 time brighter than my insipid Christmas tree bulb - and AT LEAST 10 TIMES BRIGHTER THAN THE SUN! One must truly imagine quite hard to dismiss such brilliance as being "totally" canceled out to the point of invisibility. :-) It has been enjoyable, and yet there are questions remaining that relate to the point of this origin in "Glare." I notice that this too draws a vacuum of response and yet it was so central to "illustrating" a thesis. No wonder such papers find their way to the round file at the editor's desk. The only chance for publication, however, is strictly meeting the academic strictures of a vanity press - its vaudeville quality. In this regard, the opus of "total cancellation" is destined for serialization and possibly a Hollywood movie for next summer. Myself, I enjoyed the new "Hitchhiker's Guide to the Galaxy," but I found the new Ford Prefect rather lame. Oh well, I went to the matinee showing to save a buck or two - so let's call it break-even. 73's Richard Clark, KB7QHC |
Cecil, W5DXP wrote:
"Concerning open-circuited or short-circuited transmission lines, if the voltages add up to zero, that looks like a short, and current is at a maximum. If these conditions occur at the source, 100% re-reflection of the incident reflected wave energy is guaranteed. The Bird reads equal forward and reflected power all up and down the feedline." Yes. A bird is indicatihg power in one direction only and this is the same along a low-loss line everywhere along the line regardless of SWR. A short does not kill line energy. It merely transfers electric field energy into the magnetic field for an instant producing an SWR in which voltage and current patterns are 90-degrees apart. When the load end of a transmission line is shorted, 100% of the incident wave is reflected but the reflected wave has a complete reversal in phase of its voltage with no change in the phase of its associated current. When the load end of a transmission line is open-circuited, 100% of the incident wave is reflected but the reflected wave has a complete reversal in the phase of its current with no change in the phase of its associated voltage. When the source end of a transmission line is effectively a short or open circuit, ir re-reflects the reflected wave. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote: When the source end of a transmission line is effectively a short or open circuit, ir re-reflects the reflected wave. Are we then supposed to infer that it [the source] doesn't re-reflect the wave if anything other than a short or open circuit appears there? 73, ac6xg |
Cecil Moore wrote:
Measuring the current at the mid-point of that 1/2WL of feedline will prove the feedline is filled with EM wave energy which must travel at the speed of light. So it follows that measuring voltage at the wall outlet proves there's energy filling the wall. As always, it's important to remember that any such energy would of course be traveling at the speed of light - and no faster. :-) The Bird will read 100w forward and 100 reflected on the feedline. An RF current meter at the center of that 1/2WL of feedline will read 2.828 amps. That's the sum of the forward current and reflected current in phase, 1.414 amps in the forward direction and 1.414 amps in the rearward direction. 1.414^2*50 = 100w for both forward and reflected powers. The Bird is right. Those powers are really there supporting the forward and reflected waves and cannot be used for any other purpose. A veritable black hole of logic: it's inescapable! :-) Since the feedline is lossless, there's no lost energy to replace so the source power output is zero. Anything else would violate the conservation of energy principle. Note that at the above current maximum point, the net flow of energy is zero since the Poynting vectors for forward and reflected power add up to zero. Noting of course that EM energy can't normally put itself back into the source after it's done bouncing around. So, since there's no load and the system is lossless, no energy is produced or transferred, which means zero power. In this instance, the readings on the Bird wattmeter are not at all helpful toward understanding the flow of EM energy - which as Cecil is always kind enough to remind us, travels not just at any speed, but at the speed of light. 73, ac6xg |
Jim Kelley wrote:
"Are we then supposed to infer that it (the source) doesn`t re-reflect the wave if anything other than a short or open circuit appears there?" The reflection may be incomplete unless either a hard short or complete open-circuit appears at the source where the generator meets the transmission line. If the source appears as a complete short or open the reflection is total. Best regards, Richard Harrison, KB5WZI. |
Cecil Moore wrote:
Jim Kelley wrote: Richard Clark wrote: Jim Kelley wrote: Born and Wolf has an interesting comment in the section on total reflection. "...the electromagnetic field in the second medium does not disappear, only there is no longer a flow of energy across the boundary." your source, and yet unable or unwilling to confront this single observation. Apparently that would mean the waves aren't traveling at the speed of light and it would violate his "waves cannot exist without energy" law of physics, so therefore the book is wrong. Wrong. All it means is that reflected energy doesn't make it across the match point. It says, and means more than just that, obviously. "... when two waves of equal amplitude and wavelength that are 180-degrees out of phase with each other meet, they are (canceled but) not actually annihilated. All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation. (There are only two directions available in a transmission line.) Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy (back toward the load) rather than the spontaneous construction or destruction of light." (Words in parentheses are mine added for clarity.) Interesting to note that the interference phenomenon is often described as a redistribution, but is never described in any reference as a reflection, or re-reflection as you have done. The reason is that when light destructively interferes in some direction, it simply does not go in that direction. It doesn't, contrary to your assertion, first go in the reflected direction, say oooops, then turn around in mid-air and go in the other direction. The reflection is prevented. Comprende senor? The Bird wattmeter can be misleading in this regard. It measures the effect of a field (sometimes like the one in Born and Wolf that doesn't have transfer of energy associated with it), and in every case assumes energy and power. But it's simple minded so it has an excuse. 73, ac6xg |
On Sun, 17 Jul 2005 21:37:50 -0700, Richard Clark
wrote: there are questions remaining that relate to the point of this origin in "Glare." I notice that this too draws a vacuum of response and yet it was so central to "illustrating" a thesis. Dear Readers, Taking a deep breath and wading into the hits found at Google comes one very typical observation: " The methods discussed above will work with varying degrees of success, depending on the details of the manufacturer's process. The glare reduction factors claimed by many manufacturers should not be taken too seriously. Some claim glare reduction factors of up to 250 to 1. Such numbers are either dreamed up in the advertising department or are the result of unrealistic test conditions. A good quality anti-glare filter with a multi-layer optical coating will reduce glare and reflections by about a factor of 20. Visually compare glare reductions before you buy anything." I leave the WHO undisclosed (being it is such a tantalizing overture to yet another, unanswerable question). We may all note that what is described here is "multi-layer" which suggests more than one layer for more than one BW of glare components (yet another tantalizing, unanswerable question: What wavelength is Glare?). I will pose that the necessity for multiple layers is derived from the same need for one (which is actually quite useless in ordinary life). As I offered, this is all a very common arena for the optical engineer, and thin-films are tools for SPECIFIC needs (the WHEREFOR, or yet another tantalizing, unanswerable question whose origin is one of practicality) rather than as sacred cows for the theory of moonbeams. There are correlatives in RF and antennas such that this is not an alien discussion. However, it takes little more effort to be correct instead of simply clowning through the math, because these side-show results we have been subjected with can be stretched to cover any kind of bizarre theory. As the final line of the quote above suggests: "Visually compare glare reductions before you buy anything." rather ordinary advice that is so aggressively shunned by, instead, shoveling formulas at us instead. 73's Richard Clark, KB7QHC |
Richard Harrison wrote: Jim Kelley wrote: "Are we then supposed to infer that it (the source) doesn`t re-reflect the wave if anything other than a short or open circuit appears there?" The reflection may be incomplete unless either a hard short or complete open-circuit appears at the source where the generator meets the transmission line. If the source appears as a complete short or open the reflection is total. Best regards, Richard Harrison, KB5WZI. Hmmm. I wonder if the phase change on re-reflection the same as it does on reflection. If it does, and the transmission line is a half wave long, the Bird wattmeter readings would be real hard to explain. And isn't it true that if there were actually a real hard short or a complete open circuit at the source, there wouldn't even be a signal on the transmission line? 73, ac6xg |
11001000 nW
"Richard Clark" wrote in message ... On Sat, 16 Jul 2005 17:15:58 -0500, Cecil Moore wrote: Bubba, you can't even answer simple power questions like: You have a one square cm target that is irradiated with 64 microWatts of 660nM radiation at a distance of 1M from a light bulb. (which is less than the power reflected from one of your example interfaces and still visibly quite bright.) How much power is found in the 555nM spectrum expressed in Lux? Both are power spectrums within a 30nM BW. How much total power is the light bulb radiating? Optical engineers can answer this, and I will by midnight. ;-) No one expects a binary engineer can (example of a simple 1 or 0). |
On Mon, 18 Jul 2005 16:53:37 -0400, "Fred W4JLE"
wrote: 11001000 nW Hi Fred, Well, actually 11011100 nW, but as you are within 1010% you have certainly come closer than any other binary engineer. ;-) 73's Richard Clark, KB7QHC |
On Mon, 18 Jul 2005 13:46:02 -0700, Jim Kelley
wrote: Hmmm. I wonder if the phase change on re-reflection the same as it does on reflection. If it does, and the transmission line is a half wave long, the Bird wattmeter readings would be real hard to explain. Hi Jim, In fact it is explainable (done it several times here in fact) and is called mismatch uncertainty which is quantifiable error derived from attempting to measure power between two mismatches. When the generator mismatches the line by as little as 2:1 and so does the load, you are already pushing 20% error. However, the quantification is resolved through interference math. No, not like the stuff presented in this "Can you solve this," but close enough (sans the howling errors of commission). And isn't it true that if there were actually a real hard short or a complete open circuit at the source, there wouldn't even be a signal on the transmission line? Ah, Reciprocity! The first law ditched over the side when a new "theory" hits the boards. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
It is a poor example of understanding when you purposely inject error. There is nothing clear about intentional mistakes. Any rejection of a complete solution is a suspect agenda from the beginning. Richard, since you seem to be incapable of understanding the simplest lossless, refractionless, laser example, it makes no sense to complicate things with additional details. -- 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 =--- |
Richard Clark wrote:
It comes as no great surprise that this simple example of optical power is so powerfully baffling to a neophyte. Who said it is baffling? The problem is that if you cannot understand the simplest of examples involving lossless, refractionless, laser systems, you cannot possibly understand anything more complicated. I haven't even read past your inability to understand that simple example. -- 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 =--- |
Richard Harrison wrote:
When the source end of a transmission line is effectively a short or open circuit, ir re-reflects the reflected wave. Yep, which is exactly what happens with W7EL's "Food For Thought #1". The fact that Roy didn't choose to mention the 2.828 amps of current flowing in the middle that 1/2WL of transmission line is interesting. -- 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 =--- |
Jim Kelley wrote:
Richard Harrison wrote: When the source end of a transmission line is effectively a short or open circuit, ir re-reflects the reflected wave. Are we then supposed to infer that it [the source] doesn't re-reflect the wave if anything other than a short or open circuit appears there? To that list of two, we can add two more. If the source end of a transmission line is terminated in a pure reactance, it re-reflects the reflected wave. When 100% wave cancellation of reflected waves occurs at the source due to total destructive interference, it re-reflects the reflected wave energy components as constructive interference energy in the opposite direction. Anything else would violate the conservation of energy principle. -- 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 =--- |
Jim Kelley wrote:
Cecil Moore wrote: Measuring the current at the mid-point of that 1/2WL of feedline will prove the feedline is filled with EM wave energy which must travel at the speed of light. So it follows that measuring voltage at the wall outlet proves there's energy filling the wall. As always, it's important to remember that any such energy would of course be traveling at the speed of light - and no faster. :-) There are thousands of unterminated wall outlets and hundreds of terminated wall outlets spaced only a fraction of a wavelength from each other. By all means, if you cannot understand the simplest of examples, create an example that is so complicated that nobody can understand. This is an example of someone trying to obfuscate things in order to reduce everyone down to his/her low level of understanding. Noting of course that EM energy can't normally put itself back into the source after it's done bouncing around. So, since there's no load and the system is lossless, no energy is produced or transferred, which means zero power. Zero *NET* power has nothing to do with the component powers. All it means is that the forward power and reflected power are equal. The losses in a real world transmission line depend upon the magnitude of the forward and reflected powers which pretty much shoots your illogical argument in the foot. The higher the voltage applied to a real-world 1/2WL transmission line, the greater the losses absorbed by the feedline. How the heck do you explain that one by neglecting forward and reflected energy? -- 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 =--- |
Cecil Moore wrote: Jim Kelley wrote: Richard Harrison wrote: When the source end of a transmission line is effectively a short or open circuit, ir re-reflects the reflected wave. Are we then supposed to infer that it [the source] doesn't re-reflect the wave if anything other than a short or open circuit appears there? To that list of two, we can add two more. If the source end of a transmission line is terminated in a pure reactance, it re-reflects the reflected wave. When 100% wave cancellation of reflected waves occurs at the source due to total destructive interference, it re-reflects the reflected wave energy components as constructive interference energy in the opposite direction. Anything else would violate the conservation of energy principle. In other words a system in which all of the power from the source reaches the load and none is reflected back to the source without first reflecting then re-reflecting would violate conservation of energy. It's like saying a ball violates conservation of energy if it rolls down hill without first passing through a Rube Goldberg contraption. jeez, ac6xg |
Jim Kelley wrote:
Interesting to note that the interference phenomenon is often described as a redistribution, but is never described in any reference as a reflection, or re-reflection as you have done. That's easy to understand in the context of light which can take a vector in any direction in 3D space. However, a transmission line is essentially an one-dimensional world. If something happens and it doesn't happen in one direction, there is only one other direction available. In other words, a redistribution of energy in a transmission line is, by definition, a reflection (or re-reflection). If things continue on in the original direction, it is not a redistribution. If it is a redistribution, it must change direction by 180 degrees. That's so simple a concept even you should be able to understand :-) but you obviously haven't admitted such so far. It doesn't, contrary to your assertion, first go in the reflected direction, say oooops, then turn around in mid-air and go in the other direction. I never said it did, Jim. That is at worst a straw man on your part or at least an extreme lack of understanding of what I said. Because of your semantic objection, I changed interference events from happening on either side of the match point to happening *AT* the dimisionless match point in my article. In my article, everything happens *AT* the match point, NOT on either side of the match point. The reflection is prevented. Comprende senor? Yes, I agree 100% and always have. The reflection is prevented by wave cancellation of two reflected waves. We have been over this item a hundred times and you still think your straw men will work? Instead of introducing every diversion known to man, why don't you just discuss the technical details? We agree 100% except for the most minute of details and you have been aware of that for months now. You have even said that in private emails to me. The Bird wattmeter can be misleading in this regard. It measures the effect of a field (sometimes like the one in Born and Wolf that doesn't have transfer of energy associated with it), and in every case assumes energy and power. But it's simple minded so it has an excuse. It certainly depends on your definition of power. Yours is not the same as the IEEE Dictionary so you have an excuse. That excuse is obviously: "Physicists are superior to engineers in every way and are allowed special sacred cow privileges when defining words!" :-) Engineering power does NOT require that it be dissipated, it only requires that energy be flowing. Joules/Sec flowing past a point satisfies the definition of power as defined by the IEEE. The Bird Wattmeter measures IEEE power. -- 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 =--- |
Richard Clark wrote:
Cecil Moore wrote: Most of your past objections are personal opinions Curious how you flail at these imaginary demons and blow off his actual quote from your source negating your premise. Curious how you delete "his actual quote ... negating my premises". -- 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 =--- |
Jim Kelley wrote:
And isn't it true that if there were actually a real hard short or a complete open circuit at the source, there wouldn't even be a signal on the transmission line? Yep, that's why one cannot use circuit analysis on distributed network problems. -- 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 =--- |
On Mon, 18 Jul 2005 16:35:16 -0500, Cecil Moore
wrote: The problem is that if you cannot understand the simplest of examples involving lossless, refractionless, laser systems, you cannot possibly understand anything more complicated. Um, yes. Can tell us why your example exhibits a reflection product TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled completely? :-) This question, like others, is likely to suffer the fate of you whining on about "understanding." Go ahead anyway, it establishes you as an academy of one - just don't chew the scenery. |
Jim Kelley wrote:
In other words a system in which all of the power from the source reaches the load and none is reflected back to the source without first reflecting then re-reflecting would violate conservation of energy. Everything I said is 100% consistant with the laws of physics. Your "in other words" statements violate the laws of physics. Every "in other words" statement that you have ever made, Jim, has been 100% incorrect. Maybe you should try to understand the concept before offering an "in other words" statement? -- 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 =--- |
On Mon, 18 Jul 2005 16:25:56 -0500, Cecil Moore
wrote: it makes no sense to complicate things with additional details They are your details and you can't do the math. You instead beg it doesn't make sense to? Certainly not from the perspective of your theories. ;-) |
There are 10 types of people, those that understand binary and those that
don't... "Richard Clark" wrote in message ... On Mon, 18 Jul 2005 16:53:37 -0400, "Fred W4JLE" wrote: 11001000 nW Hi Fred, Well, actually 11011100 nW, but as you are within 1010% you have certainly come closer than any other binary engineer. ;-) 73's Richard Clark, KB7QHC |
Jim Kelley wrote:
"In other words a system in which all of the power from the source reaches the load and none is reflected back to the source without first reflecting then re-reflecting would violate conservation of energy." Conservation of energy means that energy is neither created nor destroyed, but that heat and other forms of energy are quantitifiable and convertable in their equivalence. The total amount of mechanical, thermal, chemical, electrical, and other forms of energy in any isolated system remains constant. A century ago, Einstein broadened the law to include equivalence of mass and energy. Regardless of reflections and re-reflections, all the energy sourced into a transmission line ends up in the load if it isn`t lost in transmission by radiation or conversion into heat. There`s no place else for it to go. Best regards, Richard Harrison, KB5WZI |
Richard Clark wrote:
Cecil Moore wrote: The problem is that if you cannot understand the simplest of examples involving lossless, refractionless, laser systems, you cannot possibly understand anything more complicated. Um, yes. Can tell us why your example exhibits a reflection product TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled completely? :-) It doesn't. All reflections are eliminated by wave cancellation. That is a given boundary condition for the simple example. You can argue that there's no such thing in reality as a dimensionless point or a line of only one dimension or a plane of only two dimensions. That doesn't keep such from being taught as concepts in every plane geometry class. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Richard Clark wrote:
Cecil Moore wrote: it makes no sense to complicate things with additional details They are your details and you can't do the math. No, they are your details and I choose not to waste my time with your logical diversion of the basic issue. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Richard Harrison wrote:
Regardless of reflections and re-reflections, all the energy sourced into a transmission line ends up in the load if it isn`t lost in transmission by radiation or conversion into heat. There`s no place else for it to go. Hi Richard, does that statement assume that all energy dissipated as heat in the source was never sourced? -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
On Tue, 19 Jul 2005 01:45:26 -0500, Cecil Moore
wrote: Um, yes. Can tell us why your example exhibits a reflection product TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled completely? :-) It doesn't. All reflections are eliminated by wave cancellation. That is a given boundary condition for the simple example. What a larf. :-) If you can't explain it, you simply re-write the math to suit the outcome? This is neither demonstrable nor provable, unless, of course you simply fudge the numbers until they agree with your result. Given that every reader here has had the opposite experience as your prediction, this is classic thumb on the scale work - quite shabby. |
Richard Clark wrote:
Cecil Moore wrote: it makes no sense to complicate things with additional details They are your details and you can't do the math. You instead beg it doesn't make sense to? Certainly not from the perspective of your theories. ;-) It seems you have forgotten the purpose of the laser example. It was supposed to be as much like a transmission line example as is possible. Refraction is irrelevant in a transmission line so there's no logical reason to introduce refraction into the simple laser example. A laser was chosen to make the example single frequency and coherent like an RF source. Occam's Razor is used to trim off all irrelevant stuff. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
On Tue, 19 Jul 2005 01:48:00 -0500, Cecil Moore
wrote: No, they are your details All the math was per your explicit design. If that embarrasses you, fine, I will accept that as "my" details. and I choose not to waste my time with your logical diversion of the basic issue. No one is forcing you to explain what you cannot. What you choose to respond to is strictly your choice and wholly your responsibility. If you don't care to answer for why the reflections of your model are TEN TIMES BRIGHTER THAN THE SUN, and call this power cancellation instead, no one really expects you to anyway. Why don't you give us another hymn instead about understanding and such? Is your paper in the hands of a devotional publication? Do they cater to a herd of sacred cows? Would this make you a holy cowboy? :-) |
Richard Clark wrote:
Cecil Moore wrote: Um, yes. Can tell us why your example exhibits a reflection product TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled completely? :-) It doesn't. All reflections are eliminated by wave cancellation. That is a given boundary condition for the simple example. If you can't explain it, you simply re-write the math to suit the outcome? The purpose of the laser example is to make it as much like a transmission line example as possible. A Bird wattmeter indicates that all reflections are eliminated in the T-line example so the laser example assumes that as a boundary condition. Your insistance on rewriting the math to make it as different from a T-line example as possible is simply a diversion away from the original purpose. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
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