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"Waves of Average Power"
Tom Donaly wrote:
... it can all be explained neatly by superposition. There's no reason to make up any crackpot theories, or magical, mystical stories. It was all understood long before you were born. No one is making up "crackpot theories or magical mystical stories". I am reporting the laws of physics by physicists who understood them before the first man- made RF antenna or transmission line ever existed. Some people on this newsgroup are trying to sweep those laws of optical physics under the rug and ignore them. I am merely attempting to lift the rug and expose what has been known since long before you were born. What happens to the ExH joules/sec in two coherent collinear EM waves that are superposed such that the resultant wave contains ExH=0 joules/sec forever in the original direction of travel? None of you "experts" have ever answered that question. You guys have tried your best to completely ignore the fact of physics that EM waves contain energy and cannot exist without energy. Some of you have gone so far as to assert that reflected waves exist without energy and just slosh around in violation of the laws of physics for EM waves. There is your magical, mystical story but it is not coming from me. The advice from the gurus here is to use the voltages in the waves and completely ignore the necessary energy in the waves along with the conservation of energy principle. Ignore things that you can see with your own eyes? Now *that* is your "crackpot theory". Optical physicists who do not have the crutch of voltage upon which to lean, understood the energy content of EM waves a century ago. All they could measure was the power density and as a result, the field of optics understands EM waves while the field of RF remains ignorant of such. As a result, we are fed old wives' tales about reflected waves containing zero energy and just "sloshing" around when there is nothing to cause them to slosh. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
On 5 Nov, 04:57, Cecil Moore wrote:
Tom Donaly wrote: ... it can all be explained neatly by superposition. There's no reason to make up any crackpot theories, or magical, mystical stories. It was all understood long before you were born. No one is making up "crackpot theories or magical mystical stories". I am reporting the laws of physics by physicists who understood them before the first man- made RF antenna or transmission line ever existed. Some people on this newsgroup are trying to sweep those laws of optical physics under the rug and ignore them. I am merely attempting to lift the rug and expose what has been known since long before you were born. What happens to the ExH joules/sec in two coherent collinear EM waves that are superposed such that the resultant wave contains ExH=0 joules/sec forever in the original direction of travel? None of you "experts" have ever answered that question. You guys have tried your best to completely ignore the fact of physics that EM waves contain energy and cannot exist without energy. Some of you have gone so far as to assert that reflected waves exist without energy and just slosh around in violation of the laws of physics for EM waves. There is your magical, mystical story but it is not coming from me. The advice from the gurus here is to use the voltages in the waves and completely ignore the necessary energy in the waves along with the conservation of energy principle. Ignore things that you can see with your own eyes? Now *that* is your "crackpot theory". Optical physicists who do not have the crutch of voltage upon which to lean, understood the energy content of EM waves a century ago. All they could measure was the power density and as a result, the field of optics understands EM waves while the field of RF remains ignorant of such. As a result, we are fed old wives' tales about reflected waves containing zero energy and just "sloshing" around when there is nothing to cause them to slosh. -- 73, Cecil http://www.w5dxp.com From a optical physicists stand point I can understand your dilema but I still don't understand how it applies to antennas, what problems it creats and how can we fix it. Obviously there are very few people in ham radio, unless there are walk overs from CB, who are sufficiently educated in your field and would be very unlikely to delve into the mysteries of same unless they could see the future, or want to combat the problems shown today. A case in point is the wave versus particle theory, and it is just a theory.I have found that there is little or no interest in that at all in ham radio as it is a hobby and thus the notion holds true that all is known. Fortunately there is a world of science out there that pursue science for its interest regardless where it leads but hams view the hobby of radio as a means to talk, endlessly in some cases, about things that are already known and to defend such notions. Thus the presence of somebody who wants to delve into the unknown properties of radio communication is obviously on the other side of the fence as ham radio goes. Yes, I am like you who does not believe that all is known tho lacking in education in the field that you are in. But the days have gone where ham radio was populated by those interested in science and/or do not posses the powers of logic required. Frankly this newsgroup is for regurguration only and to argue not about the pursuit of science but as a means to spend your time in retirement, a fact that is reflected often in the form of senior moments that occur so often. Cecil, we can't win for losing if success can not be recognised by ignorance. Your only avenue is to write a book that people will then use as a datum and regurgitate "facts" that you provide. That is the only way with respect to ham radio can you project your self as an "expert", remember what is written and be consistent with the responses given and jeer at those who will not concurr Best regards Art Unwin KB9MZ....XG (uk) |
"Waves of Average Power"
art wrote:
From a optical physicists stand point I can understand your dilema but I still don't understand how it applies to antennas, EM waves obey the same rules, no matter what the frequency. Some people would have us believe that RF waves don't obey the same rules as light waves, that the ability to measure the voltage associated with an EM wave somehow changes its nature - that RF waves are capable of sloshing back and forth in a transmission line at sub-light speeds - that some RF waves are completely devoid of an ExH power density. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
On 5 Nov, 07:18, Cecil Moore wrote:
art wrote: From a optical physicists stand point I can understand your dilema but I still don't understand how it applies to antennas, EM waves obey the same rules, no matter what the frequency. Some people would have us believe that RF waves don't obey the same rules as light waves, that the ability to measure the voltage associated with an EM wave somehow changes its nature - that RF waves are capable of sloshing back and forth in a transmission line at sub-light speeds - that some RF waves are completely devoid of an ExH power density. -- 73, Cecil http://www.w5dxp.com You make a good point. Many scientists have tried to connect chemical, electrical and mechanical laws into one since all possess a quantised measure of energy. But this can be very difficult if one believes that a electron can behave both as a particle and a wave when the definition of both of these words are not chiselled in stone. Read "Secrets of the atom", a new unified field theory, by Dr Weldon Vlasak for some new analysis of the day. Should be of interest to you as it evolves around shell energy as in electricity no less Art |
"Waves of Average Power"
On Nov 4, 3:08 pm, "Stefan Wolfe" wrote:
Do you agree that a series of ground radials for a vertical antenna is a true "ground" in the sense that a 1/4 wave antenna radiator is "grounded"? Or do you believe that the only true ground for a 1/4 wave antenna radiator is true earth ground (or as close to that as you can get). If you are writing about the buried radials typically used by commercial AM broadcast stations, then they provide a much better "ground" in terms of conductivity than the earth itself. Without those buried radials, r-f losses in the earth within ~1/2-wavelength of the vertical greatly reduce the radiated fields it will produce, as those losses are in series with the r-f current flowing on the vertical. The radial system actually radiates as an antenna element and that gives the perception that the radial system is acting as true "ground" (but only at a specific frequency. Whether buried or elevated, r-f currents flow in opposite directions on opposing pairs of radials. So if all such pairs of radials are installed orthogonal to the vertical radiator, the useful far-field radiation from the radials themselves essentially is zero. // |
"Waves of Average Power"
On Nov 5, 4:57 am, Cecil Moore wrote:
Tom Donaly wrote: ... it can all be explained neatly by superposition. There's no reason to make up any crackpot theories, or magical, mystical stories. It was all understood long before you were born. No one is making up "crackpot theories or magical mystical stories". I am reporting the laws of physics by physicists who understood them before the first man- made RF antenna or transmission line ever existed. Some people on this newsgroup are trying to sweep those laws of optical physics under the rug and ignore them. I am merely attempting to lift the rug and expose what has been known since long before you were born. What happens to the ExH joules/sec in two coherent collinear EM waves that are superposed such that the resultant wave contains ExH=0 joules/sec forever in the original direction of travel? None of you "experts" have ever answered that question. You guys have tried your best to completely ignore the fact of physics that EM waves contain energy and cannot exist without energy. Some of you have gone so far as to assert that reflected waves exist without energy and just slosh around in violation of the laws of physics for EM waves. There is your magical, mystical story but it is not coming from me. The advice from the gurus here is to use the voltages in the waves and completely ignore the necessary energy in the waves along with the conservation of energy principle. Ignore things that you can see with your own eyes? Now *that* is your "crackpot theory". Optical physicists who do not have the crutch of voltage upon which to lean, understood the energy content of EM waves a century ago. All they could measure was the power density and as a result, the field of optics understands EM waves while the field of RF remains ignorant of such. As a result, we are fed old wives' tales about reflected waves containing zero energy and just "sloshing" around when there is nothing to cause them to slosh. -- 73, Cecil http://www.w5dxp.com Gee, Cecil, I'm sorry you've so completely misunderstood what I've posted on the subject. I've tried to be very clear about it, but I've apparently failed with respect to communicating with you. I'm sorry that's the way it is. Of course I don't consider myself an expert, so perhaps I'm excused from your catch-phrase classification of others who've accurately described the situation here. |
"Waves of Average Power"
K7ITM wrote:
Gee, Cecil, I'm sorry you've so completely misunderstood what I've posted on the subject. Could you be a little clearer? Does a reflected EM wave have an associated ExH power density? Does a reflected wave obey the principles of conservation of energy and momentum? Is there exactly the amount of energy in a transmission line needed to support the measured forward power and reflected power? -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
On Nov 5, 11:07 am, Cecil Moore wrote:
K7ITM wrote: Gee, Cecil, I'm sorry you've so completely misunderstood what I've posted on the subject. Could you be a little clearer? No, I'm sorry, Cecil. I've tried, but I'm afraid I'm just inept at communicating. I take all the blame for it. However, life's too short to spend a lot of time worrying about it, and I need to get back to calculating the even and odd mode impedances of some coupled line structures now. I just hope I don't have too much trouble applying the results to some real-world problems. |
"Waves of Average Power"
K7ITM wrote:
On Nov 5, 11:07 am, Cecil Moore wrote: Could you be a little clearer? No, I'm sorry, Cecil. I'm sorry you trimmed out and ignored all the yes/no questions that I posted. One wonders why you guys refuse to answer simple yes/no questions. Here one again: Does a reflected wave possess energy proportional to ExH, i.e. the cross product of the E-field and the H-field? Given the magical thinking on this newsgroup, it is pretty obvious why you guys cannot afford to answer that question. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
Cecil Moore wrote:
OK, the graphic is at http://www.w5dxp.com/thinfilm.gif It's a decent start, however you must remember that power doesn't reflect or propagate. You can't add power algebraically, so you won't be able to take phase into consideration (sort of crucial if you want to show cancellation). You have to use a vector quantity. Once you change to more sensible units, you can produce a sum at each reflection and show how the total changes as a function of time. Remember also that the front surface continues to be irradiated at each t sub n, so the amplitude of the signal penetrating the front surface will be different at each subsequent t sub n. 73, ac6xg |
"Waves of Average Power"
Jim Kelley wrote:
Cecil Moore wrote: OK, the graphic is at http://www.w5dxp.com/thinfilm.gif It's a decent start, however you must remember that power doesn't reflect or propagate. There's no power shown propagating. The quantities in watts are the irradiance values in joules/sec/unit-area existing within a unit-area on the surface of the thin-film. You can't add power algebraically, so you won't be able to take phase into consideration (sort of crucial if you want to show cancellation). Any physicist should be able to solve this problem using the data given. The reason you cannot solve it is that you have apparently never understood the irradiance equation. It is true that one cannot add powers algebraically but powers (irradiances) are added all the time in optical physics using the irradiance equation. (Powers do not superpose. There is special equation governing the addition of powers.) You have to use a vector quantity. That's where you are wrong. One doesn't need to use vectors. The only thing needed is the phase angle between the external reflection and the 1st internal reflection. You can figure out that phase angle by knowing that the thin film is 1/4 wavelength thick. If you cannot, I'll just tell you that the phase angle between the external reflection and the internal reflections is 180 degrees. You have all you need to know to answer the question: What happens to the total reflected power (irradiance) in the air back toward the source when the first internal reflection arrives assuming the reflections are coherent and collinear? I hope you will not go silent on this thread just because you don't like the emerging technical facts. If you pursue the discussion to its logical end, either you or I will learn something new. I am eager to discover any flaws in my technical logic. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
Cecil Moore wrote: I am eager to discover any flaws in my technical logic. Then, instead of flapping your lips and waving your hands, consider doing what I suggested. ac6xg |
"Waves of Average Power"
Jim Kelley wrote:
Cecil Moore wrote: I am eager to discover any flaws in my technical logic. Then, instead of flapping your lips and waving your hands, consider doing what I suggested. I will save us some time and concede that a vector analysis, which I already know how to do, would work and yield valid results. You and I would not discover anything new in the process. We would agree on the results and not discover any points of disagreement. If you want to divert the issue in that manner, I will agree with any vector analysis that you perform that doesn't contain errors. Now it's your turn to do a power-density analysis using the data given. Please use the irradiance equation to determine the reflected power toward the source after the external reflection and 1st internal reflection encounter each other. All the data that you need for such an analysis is contained in the example which is the only data that optical physicists had to work with 100 years ago when they couldn't directly measure any phase angles. It is in the area of the irradiance equation that I might discover any flaws in my technical logic. If you refuse to use the irradiance equation, I will understand completely. Everyone else is afraid of what they will discover. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
Cecil Moore wrote:
I will save us some time and concede that a vector analysis, which I already know how to do, would work and yield valid results. You and I would not discover anything new in the process. I think you might - if you worked it all the way from first transient through to steady state. On the other hand the irradiance equation can apparently lead some people to believe things about the nature of the phenomenon which are not precisely correct. 73, ac6xg |
"Waves of Average Power"
Jim Kelley wrote:
Cecil Moore wrote: I will save us some time and concede that a vector analysis, which I already know how to do, would work and yield valid results. You and I would not discover anything new in the process. I think you might ... I will agree with everything you say about a vector analysis (unless you make a mistake). Please let's proceed to the crux of our argument. Please apply the same irradiance equation that optical physicists were using before you were born. Any physics professor should be able to accomplish that simple task. If optical physicists were in error in using that equation 100 years ago, please explain their error. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
In article , Jim Kelley
wrote: Cecil Moore wrote: OK, the graphic is at http://www.w5dxp.com/thinfilm.gif It's a decent start, however you must remember that power doesn't reflect or propagate. Hello, Jim , and power (energy per unit time) doesn't propagate? Where do we get all that radiant heat from 93 million miles away? Sincerely, and 73s from N4GGO, John Wood (Code 5550) e-mail: Naval Research Laboratory 4555 Overlook Avenue, SW Washington, DC 20375-5337 |
"Waves of Average Power"
J. B. Wood wrote:
Where do we get all that radiant heat from 93 million miles away? When the infrared EM energy encounters the earth, it is transformed from EM energy to heat energy. Before that transformation, it is just another set of EM waves differing only in frequency from any other EM wave, e.g. RF and light. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
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"Waves of Average Power"
In article , Richard Clark
wrote: Energy propagates much as we expect it does; power - well, not always (hardly ever). 73's Richard Clark, KB7QHC Hello, and the above statement is simply not true as any undergraduate textbook in electromagnetics will point out. Transmission lines, for example, be they 60 Hz or at RF due in fact transmit power from source to load. At any point along the line the average power is given by 1/2 the real part of the product of the voltage and complex conjugate of the current. Voltage and current contain their respective components of travelling waves in both directions (source-to-load and load-to-source). Of course the transport medium doesn't have to be a transmission line - it can be free space, say from a transmitting antenna to a receiving antenna. I have no idea how many of those in our ham hobby have taken any courses in electromagnetics (traditionally called "fields" by undergrad EE students). Such courses are part of an undergrad EE program and if you major in electrophysics (as I did) at the grad level you delve much deeper into the subject. Sincerely, John Wood (Code 5550) e-mail: Naval Research Laboratory 4555 Overlook Avenue, SW Washington, DC 20375-5337 |
"Waves of Average Power"
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"Waves of Average Power"
On 7 Nov, 08:41, (J. B. Wood) wrote:
In article , Richard Clark wrote: Energy propagates much as we expect it does; power - well, not always (hardly ever). 73's Richard Clark, KB7QHC Hello, and the above statement is simply not true as any undergraduate textbook in electromagnetics will point out. Transmission lines, for example, be they 60 Hz or at RF due in fact transmit power from source to load. At any point along the line the average power is given by 1/2 the real part of the product of the voltage and complex conjugate of the current. Voltage and current contain their respective components of travelling waves in both directions (source-to-load and load-to-source). Of course the transport medium doesn't have to be a transmission line - it can be free space, say from a transmitting antenna to a receiving antenna. I have no idea how many of those in our ham hobby have taken any courses in electromagnetics (traditionally called "fields" by undergrad EE students). Oh I believe that there are quite a few who memorised what was taught and what was in the books but now they are getting older and the memory is failing where they didn't understand the basics. Since you may be younger and have taken the EE course. Try adding a time varient to Gaussian statics law and thus show how it equals Maxwell's law. But them you may be relying on memory as well and bypassed mathematics. Richard has come a long way by doing that and nobody is equiped in mathematics to call his bluff. Things were like that when I was in the military. Don't ask why just put it into the memory box and follow orders. If you can't remember then follow the rest of the squad. Doesn't that sound like ham radio? Art Such courses are part of an undergrad EE program and if you major in electrophysics (as I did) at the grad level you delve much deeper into the subject. Sincerely, John Wood (Code 5550) e-mail: Naval Research Laboratory 4555 Overlook Avenue, SW Washington, DC 20375-5337 |
"Waves of Average Power"
Richard Clark wrote:
On Wed, 07 Nov 2007 11:41:29 -0500, (J. B. Wood) Hi John, I majored in English. Yet and all that intellectually crippled, I found you couldn't respond to heat (the expression of power by your own choice) being the dissipation of energy in a load, and that energy is the entity that propagates - not power. As the topic has migrated away from your original thesis of heat and the sun towards fields; I, as an English major, am at a loss to discover the correlation that resolves how fields can resurrect the missing heat in an Earth equivalent volume of space, or provide the surplus heat found in a smaller volume of the moon - all from the same source of those fields, the sun. Are fields selectively powerful (like a god)? Do the fields of Hello, Richard and all. With regard to power (be it from the sun or some other source of energy), strictly speaking I would agree that what is transported is energy. Physicists and engineers usually include the ability to do work (as well as the work itself) as part of the concept of power. For example, we can have "available power" at the terminals of a receiving antenna even though the antenna is not connected to a dissipating termination. Radio/antenna engineers and physicists also use the concept of "power density" (usually in units of watts per square meter) at some distance from a transmitting antenna. We then speak of an antenna having an effective "capture area" that can extract a portion of this power from incident electromagnetic waves. In this regard we speak of power being transmitted through a medium. If we clock energy flow across a boundary for a specified time then the quantity of energy divided by that time represents power flow across the boundary. With regard to the sun/volume question I'm not qualified to answer that. The earth still retains its primordial inner sources of heat and other forms of energy as manifiested in erupting volcanoes, geysers, and deep sea thermal vents. Of course these are in addition to radiant energy from the sun. Sincerely, |
"Waves of Average Power"
On Wed, 07 Nov 2007 13:23:12 -0500, "J.B. Wood"
wrote: In this regard we speak of power being transmitted through a medium. Hi John, You might be interested in the work of Debra R. Rolison, Ph.D. Naval Research Laboratory -- Surface Chemistry Branch. The lead-in would be the medium of aerogels where her presentation that I attended nearly 5 years ago was called "The Importance of Nothing in Nanostructured Materials." You will learn from her that power in the form of Heat (not many other forms left) does not transmit through aerogels. However, for her research, that is one of those "People" magazine kind of qwik-facts where her work investigates to a greater depth. She is quite conversant and has a vice-like grip on the topic. Ask about Phonons insofar as how Power moves through a medium. 73's Richard Clark, KB7QHC |
"Waves of Average Power"
J. B. Wood wrote:
Transmission lines, for example, be they 60 Hz or at RF due in fact transmit power from source to load. The unit of power is the watt. If power is being transmitted, the transmitted power would have units of watts/sec or joules/sec/sec. What would be the physical meaning of joules per second squared? -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
Cecil Moore wrote:
J. B. Wood wrote: Transmission lines, for example, be they 60 Hz or at RF due in fact transmit power from source to load. The unit of power is the watt. If power is being transmitted, the transmitted power would have units of watts/sec or joules/sec/sec. What would be the physical meaning of joules per second squared? I have no idea what you're talking about, Cecil. (But then I haven't read every post in this thread.) Sounds like you just coined a term for energy acceleration, whatever that might apply to. If you don't object I think I'll stick with what I've gleaned over the years as an engineer and going back to my university days. You are free of course to provide your own interpretation of electrical phenomena. Sincerely, and 73s, |
"Waves of Average Power"
J.B. Wood wrote:
Cecil Moore wrote: The unit of power is the watt. If power is being transmitted, the transmitted power would have units of watts/sec or joules/sec/sec. What would be the physical meaning of joules per second squared? I have no idea what you're talking about, Cecil. I know and you are not alone. I was taught about power transmission in college but it was actually energy transmission they were talking about. The power company doesn't charge me for power - they charge me for KWH, i.e. energy. Consider a Bird wattmeter in a flat transmission line. It is at a fixed point reading watts. Are the watts moving? The Bird is displaying joules/sec passing a fixed point. The joules are certainly moving but are the joules/sec moving? As you noted, it seems that if the joules/sec are moving then the joules must necessarily be accelerating. You can measure the number of cars passing over a bridge in one hour and write that number on your notepad. The cars are moving but is that cars/hour measurement written on your notepad moving? If so, where is it going? -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
On 7 Nov, 14:40, Cecil Moore wrote:
J.B. Wood wrote: Cecil Moore wrote: The unit of power is the watt. If power is being transmitted, the transmitted power would have units of watts/sec or joules/sec/sec. What would be the physical meaning of joules per second squared? I have no idea what you're talking about, Cecil. I know and you are not alone. I was taught about power transmission in college but it was actually energy transmission they were talking about. The power company doesn't charge me for power - they charge me for KWH, i.e. energy. Consider a Bird wattmeter in a flat transmission line. It is at a fixed point reading watts. Are the watts moving? The Bird is displaying joules/sec passing a fixed point. The joules are certainly moving but are the joules/sec moving? As you noted, it seems that if the joules/sec are moving then the joules must necessarily be accelerating. You can measure the number of cars passing over a bridge in one hour and write that number on your notepad. The cars are moving but is that cars/hour measurement written on your notepad moving? If so, where is it going? -- 73, Cecil http://www.w5dxp.com Cecil, Velocity and acceleration are two different things Velocity is the amount of movement and accelleration is the rate of change of movement dependent on the number of samples taken over a period of time Joules are not necessarily accelerating. Think Newton....ut + 1/2 f(t sqd) Art Art |
"Waves of Average Power"
Cecil Moore wrote:
Please apply the same irradiance equation that optical physicists were using before you were born. Any physics professor should be able to accomplish that simple task. If optical physicists were in error in using that equation 100 years ago, please explain their error. So far no response to this simple request. The graphic of the non-reflected glass example is at http://www.w5dxp.com/thinfilm.gif We know that the reflections are 100% canceled during steady-state. The problem is: With the given data, calculate the magnitude of the total reflection back toward the source immediately after the first internal reflection arrives back at the thin-film to air surface at time t3 and is superposed with the external reflection. The two superposed waves are 180 degrees apart. The external reflection is 0.01 watts at a reference angle of zero deg. This is the normal reflection from the thin-film surface The first internal reflection is 0.009801 watts at 180 degrees. This is the first reflection from the glass. These two waves superpose at t3. The irradiance equation, using P for power density, is: Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A) where (A) is the angle between the two waves and Ptotal is the total power density of the reflections toward the source after the two waves are superposed. Ptotal = 0.01 + 0.009801 + 2*SQRT(0.01*0.009801)(-1) Ptotal = 0.019801 - 0.0198 = 0.000001 watt There is 0.0198 watts of destructive interference which, according to the conservation of energy principle, must result in 0.0198 watts of constructive interference in the opposite direction. The magnitude of the reflection toward the source drops from 0.01 watt to 0.000001 watt at the arrival of the first internal reflection from the glass. That's a five magnitude reduction in the reflections at the time of the arrival of the first internal reflection. The reflections back toward the source are eventually completely eliminated during steady-state. There was no vector math and no need to calculate the magnitudes and phases of the electric and magnetic fields - just a straight forward calculation to get the answer. As long as the source remains in place, the steady- state cancellation of the reflections toward the source is permanent. It is difficult to analyze how that could happen unless the internal reflections interact with the external reflection resulting in wave cancellation. Everyone is invited to prove the above calculation to be incorrect. Hint: if one actually performs a vector analysis, the magnitude of reflection result will be exactly the same as above. The above method does not invalidate or replace a vector analysis. Unlike a vector analysis, it simply shows where the energy goes. Optical physicists knew this a century ago - most RF engineers have yet to learn it. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
Cecil, W5DXP wrote:
"The cars are moving but is the cars/hour measurement written on your notepad moving?" A simple question deserves a simple answer. Yes it is moving downstream to the next tallyman`s position. Best regards, Richard Harrison, KB5WZI |
"Waves of Average Power"
Art wrote:
"---Joules are not necessarily accelerating." True. Terman says: "Radio waves are produced to some extent whenever a wire in open space carries a high-frequency current." Kraus says: "Thus, time-changing current radiates and acceleerated charge radiates." A joule is a unit of work or energy in the MKS system. A joule is the quantity of energy needed to transport one coulomb of charge between two points of one volt potential difference. Incidentally, "average power" is the kind of power we ordinarily use. Radio waves as Terman uses above are of the energy which escapes from the near field of the wire and is unlikekly to be recaptured by the wire. Best regards, Richard Harrison, KB5WZI |
"Waves of Average Power"
K7ITM wrote:
On Nov 2, 9:54 pm, K7ITM wrote: On Nov 2, 3:58 pm, Cecil Moore wrote: Do you know of any way to achieve wave cancellation without any interaction between the waves? It's called "vector addition," not "interaction." A bit more on the lack of "interaction" between two waves... A bit more on the "interaction" between two waves... From "Optics", by Hecht: "Briefly then, optical interference corresponds to the *interaction* of two or more light waves yielding a resultant irradiance that deviates from the sum of the component irradiances." (emphasis mine) Since the "interaction" between two waves is good enough for Hecht, it is good enough for me. -- 73, Cecil http://www.w5dxp.com |
"Waves of Average Power"
Cecil Moore wrote:
K7ITM wrote: On Nov 2, 9:54 pm, K7ITM wrote: On Nov 2, 3:58 pm, Cecil Moore wrote: Do you know of any way to achieve wave cancellation without any interaction between the waves? It's called "vector addition," not "interaction." A bit more on the lack of "interaction" between two waves... A bit more on the "interaction" between two waves... From "Optics", by Hecht: "Briefly then, optical interference corresponds to the *interaction* of two or more light waves yielding a resultant irradiance that deviates from the sum of the component irradiances." (emphasis mine) Since the "interaction" between two waves is good enough for Hecht, it is good enough for me. Gimme that old time religion... 73, Tom Donaly, KA6RUH |
"Waves of Average Power"
Tom Donaly wrote:
Since the "interaction" between two waves is good enough for Hecht, it is good enough for me. Gimme that old time religion... Please take your objection up with Eugene Hecht. -- 73, Cecil http://www.w5dxp.com |
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