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On Wed, 10 Mar 2004 19:53:20 GMT, Gene Fuller
wrote: However, don't expect this sort of simple handwaving model to be extendable to all sorts of silliness about energy and momentum transfer. It all goes to the measure of Standard handWaving Ridiculousness. |
Steve Nosko wrote:
"I disagree." (If part of the source resistance were not lossless efficiencies would be limited to 50%.) Before I sent that I asked myself if it were necessary to include the peoviso that the statement only applies to maximum power transfer. Sez to myself, no don`t bother. Any schoolboy knows that when the load resistance is large as compared with the source resistance, you may exceed 50% efficiency. That`s the norm for power distribution. Also, sez to myself, the title of the thread is: " max power transfer theorem" Surely it`s understood the comments refer to maximum power transfer and not to a less demanding condition. Me, myself, and I were wrong. Best regards, Richard Harrison, KB5WZI |
Gene Fuller wrote:
In the perfectly antireflective case all of the waves keep moving in the same direction, from air to thin film to glass. To prove that to be a true statement you must prove that the transistion point between materials of different indices of refraction results in zero reflections. Good luck on that one. For instance, one can change the thin-film thickness from 1/4WL to 1/2WL and cause exactly the opposite effect, i.e. extreme glare. If you are using the quantum electrodynamics model, please let us know. Most of the rest of us are using the EM wave reflection model. -- 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! =----- |
Cecil,
I haven't the foggiest idea what model you might be using. I am using the classical model that is found in virtually any textbook that deals with plane waves in non-conducting media. I don't have to "prove" anything. Just set up the standard wave equations with the standard boundary conditions and the problem practically solves itself. The non-zero remaining waves are all moving in the same direction. I forgot to ask them if they realize that Cecil doesn't approve of such behavior. I suppose this is an prime example of being seduced by "math models", but I believe that is a lesser fault than being seduced by Cecil's imaginary models. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: In the perfectly antireflective case all of the waves keep moving in the same direction, from air to thin film to glass. To prove that to be a true statement you must prove that the transistion point between materials of different indices of refraction results in zero reflections. Good luck on that one. For instance, one can change the thin-film thickness from 1/4WL to 1/2WL and cause exactly the opposite effect, i.e. extreme glare. If you are using the quantum electrodynamics model, please let us know. Most of the rest of us are using the EM wave reflection model. |
Richard Harrison wrote: Cecil, W5DXP wrote: "It`s not magic and is explained on the Melles-Groit web page---. (how the energy magically reverses direction and heads back toward the load.) I agree. It isn`t magic. Optical examples are good because we can see reflections. The phenomenon isn't magical, as can be clearly seen in the optics texts. On the other hand, Cecil's elaborate theories on the subject transcend those described within these treatises. Nowhere other than in Cecil's works will one find a description of waves reversing direction without reflecting from a discontinuity. 73, Jim AC6XG |
Gene Fuller wrote:
In the perfectly antireflective case all of the waves keep moving in the same direction, from air to thin film to glass. I just realized what you are saying. Your above statement is wrong about "all of the waves". Your above statement is correct about "net irradiance". I'm not talking about "net" anything. I am talking about the component forward and reflected waves which are easily proven to exist. -- 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! =----- |
Cecil Moore wrote: Old Ed wrote: "Lossless resistance?" Would that be zero resistance, or perhaps a negative resistance, as in the active part of a tunnel diode's V-I characteristic? I am a career EE, with a couple of graduate EE degrees; and this is something entirely new to me. Could you explain this concept, and/or provide some references? How about an example? If L and C are lossless, then SQRT(L/C) will be lossless with a dimension of ohms, i.e. resistance. Umm, isn't that an example of reactance? I assume you would have us believe they are one in the same. BTW, not all forms of resistive loss are "ohmic". 73, Jim AC6XG 73, Jim AC6XG |
Gene Fuller wrote:
I don't have to "prove" anything. Just set up the standard wave equations with the standard boundary conditions and the problem practically solves itself. The non-zero remaining waves are all moving in the same direction. I forgot to ask them if they realize that Cecil doesn't approve of such behavior. You should have warned us that you were talking about NET waves and NET energy transfer. I'm not discussing that at all. I am talking about component waves and component energy transfer without which standing waves cannot exist. Or maybe you can offer an example of standing waves in the absence of at least two waves traveling in opposite directions. If you can do that, I will admit defeat. I suppose this is an prime example of being seduced by "math models", but I believe that is a lesser fault than being seduced by Cecil's imaginary models. It is indeed an example of being seduced by the NET math model. Please transfer over to the component math model and rejoin the discussion. Lots of interesting things are happening below the threshold of the NET math model. The NET math model doesn't explain anything except the NET results. If your bank account balance doesn't change from one month to another, do you also assume that you have written no checks and have no income for that month? Literally speaking, please get real! -- 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! =----- |
Steve Nosko wrote:
"It is as simple as the fact that it is not internally resistance limited----." We are discussing maximum power transfer which by definition is the condition in which all available power is delivered to the load. This condition requires equal resistances in source and load. Maximum power transfer can be found by varying the resistance used as the load until the load resistance is found which generates the most heat. This assumes there is no reactance or other opposition to power other than that of the resistive type in the source and load. Once you`ve found the load which extracts maximum power from the source, measure its resistance. That is also the resistance of your source. In the Class-C amplifier, some of the source resistance it presents to the load is of the lossless variety. Were it all of the dissipative variety, just as much heat would be generated within the amplifier as within the load, UNDER MAXIMUM POWER TRANSFER CONDITIONS. Some of the internal resistance is the lossy kind. The final amplifying devices have almost full on or off states. There`s little transition, and the saturation voltage is low but not zero. The lossless variety of internal resistance comes from an average of the switched-off time of the amplifier. My example presumed a 50-50 spllit between dissipative and lossless resistances in the Class-C amplifier. That made an efficiency of 66.7%. Not bad and not unusual. That`s the way it works, believe it or not. Best regards, Richard Harrison, KB5WZI |
Jim Kelley wrote:
Nowhere other than in Cecil's works will one find a description of waves reversing direction without reflecting from a discontinuity. You have made that assertion so many times it has turned into a Big Lie. Maybe you should reveal the agenda responsible for such unethical behavoir? What I have said is that the *energy* involved in destructive interference at a Z0-match point reverses direction. The waves are *destroyed* by the destructive interference and the waves therefore cease to exist. So how could destroyed waves possibly reverse direction? They cannot! Your assertions are simply false. Perhaps you are confusing the energy in the cancelled (destroyed) waves with the waves themselves? In any case, unless I accidentally mis-spoke sometime, I have never said that cancelled waves reverse themselves. To say that cancelled waves reverse themselves would be a ridiculous assertion. The *energy* in the destroyed waves cannot be destroyed and we know that it doesn't continue on toward the source. Therefore, the *energy* in the destroyed waves changes directions and becomes constructive interference flowing toward the load. The *energy* in the cancelled (destroyed) reflected waves reverses directions at the Z0-match. There is simply no other possibility since reflected energy toward the source is eliminated by wave cancellation. All this is explained on the Melles-Groit web page if you desire to comprehend. They say the "lost" reflected energy is not lost at all and enhances the forward wave. How can reflected energy enhance the forward wave without changing directions? -- 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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