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tuner - feedline - antenna question ?
art wrote:
All those degrees in his possesion and all we get is a belching black smoke column. It's going to be another long night Cecil! Art, I'm still waiting for someone to prove that an EM wave can stand still and continue to exist in violation of Einstein's theory that light (and RF waves) always travel at the speed of light. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Owen Duffy wrote:
- that V/I=Zo for a travelling wave in a transmission line, and that (Vf+Vr)/(If-Ir) at the load end of the line must equal Zload; It seems that we can draw some conclusions from these assumptions. Vf*If*cos(0) = forward joules/sec Vr*Ir*cos(0) = reflected joules/sec The forward and reflected energy waves actually exist and are the building blocks of the standing wave. Vf and Vr are phasors and therefore subject to superposition and interference. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Dave wrote:
How do the physics type adjust their definition to include the Poynting Vector? The same way they handle the Power Flow Vector and the "IEEE Dictionary", like a hot potato. :-) -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Cecil Moore wrote:
art wrote: All those degrees in his possesion and all we get is a belching black smoke column. It's going to be another long night Cecil! Art, I'm still waiting for someone to prove that an EM wave can stand still and continue to exist in violation of Einstein's theory that light (and RF waves) always travel at the speed of light. Not everything about light (or EM waves) travels at the speed of light. Standing waves are simply the RF equivalent of stationary optical interference fringes. -- 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
tuner - feedline - antenna question ?
Cecil Moore wrote:
You apparently still cling to the mistaken notion that standing waves can exist without their forward and reflected wave building blocks. It is the standing wave equation that is the math construct that only exists in the human mind. EM waves cannot stand still. They must necessarily move at the speed of light. Anything else violates the principles of relativity. Cecil, You have fallen into the same trap as those who don't understand the difference between phase velocity and group velocity. Relativity is a physical concept, not a mere result of mathematical manipulation. 73, Gene W4SZ |
tuner - feedline - antenna question ?
Ian White GM3SEK wrote:
Not everything about light (or EM waves) travels at the speed of light. All EM waves (photons) travel at the speed of light. Anything else would violate the principles of relativity. Any detection of light results in the dissipation (absorption) of the associated photons. Standing waves are simply the RF equivalent of stationary optical interference fringes. That they appear to stand still is just an illusion. Photons are continuously traveling at the speed of light to get to your eyes where they are dissipated. When you see an interference ring that appears to stand still, you are actually receiving speed-of- light photons into your eyes. (If the interference rings were truly stationary, you wouldn't be able to see them.) If the interference fringes are being absorbed, photons are being dissipated. If the interference fringes are being scattered, the photons are moving at the speed of light. The photons either disappear or they travel at the speed of light. The photons in a transmission line cannot stand still. Standing waves are caused by superposition and interference between separate and distinct forward and reflected energy waves. And the photons in the standing wave are NOT standing still. Consider two coherent nearly co-linear beams of light. There are bright points of constructive interference and dark points of destructive interference but the interference has no effect on the two waves which emerge from the superposition unaffected and continue on their merry way. Those waves never slow down from their speed-of-light velocity. The "stationary" bright and dark points are an illusion. At every point, photons are either being dissipated or are traveling at the speed of light (including VF of course). Nothing else is possible. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Dave wrote:
Cecil Moore wrote: Dave wrote: Cecil, as an engineer you should stick with standard vocabulary. Just trying to appease the physicists, Dave. They are arguing that it is not power until work is done. A Poynting vector is watts/square angle [watts/degree^2]. It is not being dissipated in free space. It is Diverging [vector relationship]. How do the physics type adjust their definition to include the Poynting Vector? I'll sit back and read the follow up posts for the next few weeks :-) Dave, Physicists are basically reasonable, but clever, folks. ;-) They define power as the time rate at which work is done, OR, the time rate at which energy is transferred. With some imagination applied to the context, the definitions converge. In the case of the Poynting vector, convergence (of the definitions) would require dealing with the ultimate destination of the flowing energy. Might be troublesome in infinite space, but on earth, all of the energy ultimately produces power. The problem seems to be the careless use of "power flow" which, if not meaningless, I find impossible to conceptualize. The flow of the flow of energy? Sort of like common usage of "current flow", which is also the flow of a flow. 73, Chuck ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-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 =---- |
tuner - feedline - antenna question ?
Gene Fuller wrote:
Relativity is a physical concept, not a mere result of mathematical manipulation. My point exactly! All your mathematical manipulation is not going to get a photon to stand still. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Dave wrote:
Cecil Moore wrote: Dave wrote: Cecil, as an engineer you should stick with standard vocabulary. Just trying to appease the physicists, Dave. They are arguing that it is not power until work is done. A Poynting vector is watts/square angle [watts/degree^2]. It is not being dissipated in free space. It is Diverging [vector relationship]. How do the physics type adjust their definition to include the Poynting Vector? I'll sit back and read the follow up posts for the next few weeks :-) And now one for the engineers! How do you interpret a non-zero Poynting vector determined by static E- and H- fields? 73, Chuck ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-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 =---- |
tuner - feedline - antenna question ?
On 2 Mar, 08:45, chuck wrote:
Dave wrote: Cecil Moore wrote: Dave wrote: Cecil, as an engineer you should stick with standard vocabulary. Just trying to appease the physicists, Dave. They are arguing that it is not power until work is done. A Poynting vector is watts/square angle [watts/degree^2]. It is not being dissipated in free space. It is Diverging [vector relationship]. How do the physics type adjust their definition to include the Poynting Vector? I'll sit back and read the follow up posts for the next few weeks :-) And now one for the engineers! How do you interpret a non-zero Poynting vector determined by static E- and H- fields? 73, Chuck ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----http://www.newsfeeds.comThe #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----- Hide quoted text - - Show quoted text - That is a simple question for an engineer! You invest in a antenna computor program that gives you the answer that you expect and move on to the next problem. How would you solve it Chuck? Art |
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