Home 
Search 
Today's Posts 
#1




Conservation of Energy
Cecil, W5DXP wrote:
" ***Before they cancel***, they will burn a hole in the black paper proving that they contain energy." Anyone who can obtain a copy of Arnold B. Bailey`s "TV and Other Receiving Antennas" can turn to page 158 to see a diagram of how waves interfere. Interaction between components of the same signal from two sources are shown in the figure to produce a combined signal which varies regularly with the height above the earth. This has a corresponding effect on the signal received by an antenna placed at a particular altitude. The average signal at all altitudes is the same as if there was no interference. But, that is because the layers of strengthened signal are exactly offset by the layers of reduced signal strength. Best regards, Richard Harrison, KB5WZI 
#2




W5DXP wrote: Jim Kelley wrote: Well, I've already provided the quote from: "Fields and Waves in Communication Electronics", by Ramo, Whinnery, and Van Duzer. The topic of converstaion was: Poynting Vector Energy Anaylsis of An Impedance Discontinuity. Yes it was. So what happens to the reflected power Poynting Vector (that the above reference says exists) when it encounters the impedance discontinuity? 100W source50 ohm feedline+1/2WL 150 ohm feedline50 ohm load The reflected power Poynting vector on the left side of the impedance discontinuity is zero. The reflected power Poynting vector on the right side of the impedance discontinuity is 33.33W pointing toward the left. For about the tenth time, what happens to that reflected power Poynting vector? 
#3




W5DXP wrote: Jim Kelley wrote: Well, I've already provided the quote from: "Fields and Waves in Communication Electronics", by Ramo, Whinnery, and Van Duzer. The topic of converstaion was: Poynting Vector Energy Anaylsis of An Impedance Discontinuity. Yes it was. So what happens to the reflected power Poynting Vector (that the above reference says exists) when it encounters the impedance discontinuity? 100W source50 ohm feedline+1/2WL 150 ohm feedline50 ohm load The reflected power Poynting vector on the left side of the impedance discontinuity is zero. The reflected power Poynting vector on the right side of the impedance discontinuity is 33.33W pointing toward the left. For about the tenth time, what happens to that reflected power Poynting vector? Like I said, you haven't done a Poynting vector energy analysis. If you had, I wouldn't have to solve the problem for you. Why should someone believe what you say when there's no evidence you've even worked the problem yourself? 73, ac6xg 
#4




W5DXP wrote:
I have told you time after time, that the load impedance doesn't matter and the feedline length doesn's matter. It matters in exactly the same way that the index of refraction of the glass substrate matters. 50V at zero degrees and 1A at 180 degrees = 50W 50V at 180 degrees and 1A at zero degrees = 50W Nonsequitur. The topic of conversation had been: Examples where "something is actually destroyed and gives up all it's energy." Dodging another technical question, I see, and we all know why. :) Yes, very technical. If a math question was posed, I must have missed it. You know that the sum of the two above waves is zero volts and zero amps, i.e. destruction of the two waves in their direction of travel. Where does the energy go when the waves go to V=0 and I=0 in their direction of travel. Using the values above, calculate the rate of flow of energy equal to V*I. That's how much energy is involved in your dilema here. It seems that everything you don't know the answer to is a nonsequitur. I fully understand your attitude. Another one of your technical remarks? ac6xg 
#5




Jim Kelley wrote:
W5DXP wrote: I have told you time after time, that the load impedance doesn't matter and the feedline length doesn's matter. It matters in exactly the same way that the index of refraction of the glass substrate matters. Yes, it does, which is not at all if we already know the reflected irradiance which is a given. If the reflected irradiance is a given, you don't need to know the index of refraction of the glass. In fact, if it is unknown, you can calculate it from the given reflected irradiance. 50V at zero degrees and 1A at 180 degrees = 50W 50V at 180 degrees and 1A at zero degrees = 50W :) Yes, very technical. If a math question was posed, I must have missed it. It ain't rocket science. What is the superposed sum of the two above waves? What happens to the intrinsic energy pre existing in those waves before they cancel each other? Using the values above, calculate the rate of flow of energy equal to V*I. That's how much energy is involved in your dilema here. The rate of flow of energy has to be 100 joules/sec since the energy in those two waves cannot stand still and cannot be destroyed. We already know that the energy in those two waves joins the forward traveling power wave toward the load.  73, Cecil, W5DXP 
#6




Jim Kelley wrote:
Why should someone believe what you say when there's no evidence you've even worked the problem yourself? Here's an interesting reply to one of my postings on sci.physics.electromag: Cecil said: Seems to me that destructive interference event occurring in the direction of the source feeds energy to the constructive interference event occurring in the direction of the load. The reply was: Yes, just so. In fact, the phase change/nonchange when the waves are reflected/transmitted by lowtohigh or hightolow interfaces ensures that this will always happen  if in phase on one side, they must be out of phase on the other. Which is exactly what I say in my article.  73, Cecil, W5DXP 
#7




Peter Brackett wrote:
Cecil: It certainly does resolve any problems. All waves contain energy. All waves transfer that energy when they are dissipated or radiated or when they are destroyed by wave cancellation. No they don't, and... Changing the energy in a TEM wave to heat is obviously a transfer of energy. Radiating TEM energy from an antenna is obviously a transfer of energy. When TEM wavefronts are destroyed by wave cancellation in a transmission line, they cease to exist and must therefore transfer their energy somewhere. There is simply no other possibility. At an impedance discontinuity match point with reflections, the destructive interference energy is transferred in phase to the constructive interference event occurring in the other direction. Please see the "Mental Exercise" thread cross posted from sci.physics.electromag What *exactly* is your definition of a "wave"? "traveling wave  The resulting wave when the electric variation in a circuit takes the form of translation of energy along a conductor, such energy being always equally divided between current and potential forms." i.e. P=V*I and Z0=V/I  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 80,000 Newsgroups  16 Different Servers! = 
#8




Cecil, W5DXP wrote:
"energy being always equally divided between current and potential forms." Except at opens and shorts, real and virtual, where amps or volts for an instant however brief are brought down to zero, which is a transfer of all energy to either the electric or magnetic field. Best regards, Richard Harrison, KB5WZI 
#9




"W5DXP" wrote in message ... Jim Kelley wrote: W5DXP wrote: I have told you time after time, that the load impedance doesn't matter and the feedline length doesn's matter. It matters in exactly the same way that the index of refraction of the glass substrate matters. Yes, it does, which is not at all if we already know the reflected irradiance which is a given. Obviously, the load determines the boundary conditions and so it is not irrelevant. You said that it was, and that's not correct. The load impedance is what determines the reflectivity. Go ahead and disagree. :) Yes, very technical. If a math question was posed, I must have missed it. What is the superposed sum of the two above waves? Zero. What happens to the intrinsic energy pre existing in those waves before they cancel each other? The answer the intrinsic energy in the waves where the waves exist is stored in the transmission line, and nothing happens to energy where waves don't exist. The waves in question don't convey energy from the source to the load  obviously because they don't propagate from the source to the load. It ain't rocket science  as you're so fond of saying. Using the values above, calculate the rate of flow of energy equal to V*I. That's how much energy is involved in your dilema here. The rate of flow of energy has to be 100 joules/sec since the energy in those two waves cannot stand still and cannot be destroyed. The rate of flow of energy "in" those two waves is not 100 Joules per second. We already know that the energy in those two waves joins the forward traveling power wave toward the load. The energy does travel forward, but not by way of those two waves. 73, Jim AC6XG 
#10




Richard Harrison wrote:
Cecil, W5DXP wrote: "energy being always equally divided between current and potential forms." Except at opens and shorts, real and virtual, where amps or volts for an instant however brief are brought down to zero, which is a transfer of all energy to either the electric or magnetic field. That was the definition of a traveling wave in a Z0 environment, Richard, and opens and shorts are not a Z0 environment. Can you give an example of a virtual short and explain how it develops?  73, Cecil, W5DXP 
Reply 
Thread Tools  Search this Thread 
Display Modes  

