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Where Does the Power Go?
I've been told that this discussion continues in Letters
to the Editor in QEX magazine, to which I don't subscribe any more. Without knowing the context of those present discussions, I composed a letter to the QEX editors. Here it is: “Where Does the Power Go?” was answered in my magazine article, “An Energy Analysis at an Impedance Discontinuity in a Transmission Line”, published by Worldradio magazine and available on my web page at: http://www.w5dxp.com/energy.htm Here is a thought experiment that should lay the subject to rest. Assume a one-second long lossless 50 ohm open-circuit transmission line being driven by a 141.4 volt, 50 ohm Thevenin source. At time, t = 0, source power is supplied to the transmission line. After one second, 100 joules will have been delivered to the line’s forward wave and the forward wave will have just reached the open-circuit at the other end of the transmission line. What happens next simply follows the laws of physics. The conservation of energy principle tells us that the 100 joules in the forward wave must be conserved. The conservation of momentum principle tells us that the momentum in the forward wave must be conserved. The open-circuit prohibits the forward wave from continuing in the forward direction and the forward wave ceases to exist beyond the open-circuit point. The forward wave ceases to exist so what happens to the energy and momentum in the forward wave? Physically, there is only one possibility. The energy and momentum in the forward electromagnetic wave is moving at the speed of light and is indeed conserved by transferring its energy and momentum to the reflected wave at the open-circuit point in accordance with the rules of the distributed-network/wave-reflection model. This starts to happen at t = 1 second at which time 100 joules of electromagnetic energy exists in the forward wave in the transmission line. (One can actually calculate the force being exerted on the open- circuit by the forward wave. It is akin to the pressure of sunlight.) The transfer of energy and momentum from the forward wave to the reflected wave continues throughout the t =1 to t = 2 second time frame. At the end of two seconds, the forward wave contains 100 joules and the reflected wave contains 100 joules. Thus 200 joules are "stored" in the transmission line during the first two seconds. After two seconds, steady-state has been reached and the Thevenin source ceases to supply any energy. No additional energy is needed since the system is lossless. But please note that 200 joules exist in the transmission line all during steady-state, exactly enough energy to support the 100 joules/second forward watts and the 100 joules/second reflected watts. Any theory of where the energy goes must account for the 200 joules of energy in the transmission line. Those 200 joules cannot be destroyed by sweeping them under the steady-state rug. An ideal (lossless) directional wattmeter will tell us that forward watts = 100 watts and that reflected watts = 100 watts. 200 joules is required to support that number of forward and reflected watts. Is it just a coincidence that the 200 joules existing in the transmission line during steady-state is *exactly* the magnitude of energy required to support 100 watts forward and 100 watts reflected? The number of joules in a transmission line is *always* exactly the number required to support the forward watts and reflected watts. Since EM waves cannot stand still, it seems logical to leave the energy right where it is at the beginning of steady-state - 200 joules per second being exchanged between the forward wave and the reflected wave. The energy exchanges at the two ends of the transmission line are balanced and equal so one might argue that there is no net energy exchange between the forward and reflected waves during steady-state. But since the two energy transfers are occurring thousands of miles apart, that would seem to be a moot point. The transmission line is charged with this energy during the power on transient phase before steady-state is reached. The transmission line is discharged (energy dissipated) during the power down transient phase after steady-state. It is simply conservation of energy at work. It is interesting to note that if the transmission is discharged through a 50 ohm resistor at the far open-end, it will be discharged at a rate of 100 watts for two seconds. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Gosh Cecil... In a lossless line if you put in 100 joules, at the rate
of 100 joules per second, and get 200 joules back, I predict that in less than 30 seconds (30 doublings) there will be a titanic explosion exceeding all the Hydrogen bombs on the planet... Do the math! OK, tongue out of cheek... Assume that we put a 1 second squirt at the rate of 100 Joules per second, into the input end of the line, it is an open line at the far end, and yank the input end of the line out of the transmitter at exactly 1 second... Now we have a lossless line with a wave going forward and a reflected wave coming back... Do you still claim 200 joules? We are going to do this in quarters (25% of time, 25% of 100 joules and 25% of line length) Start pushing energy into the line at T=0 and jump ahead to T = .25 second... The is 25 joules total in the .00 to .25 line section and 0 joules in the other 3 sections... At T=.50 there are 25 joules in each of the first two .25 sections and nothing in the last two .25 sections... Total 50 joules... The next two time periods is more of the same so let's jump to T=1.00 The line is now just 'full', with 4 'quarters' of energy having been inserted into the line, at that instant the leading edge of the wave reaches 1.00 of line (.25 x 4) and reflects off the open terminal, folding back and going towards the input... The total power in the line is 100 joules at T= 1.00.... No surprise there... At 1.25 seconds we have a quarter of the wave reflected back and re-reaching the .75 line point (25 joules in that quarter) laid over top of the quarter of the wave front still going outwards in the .75 to 1.00 section (25 joules in that quarter). So at T = 1.25 in that 0.75 to 1.00 portion of line we have 50 joules of energy (first quarter coming back plus second quarter still going out, passing each other... ) In the .50 to .75 line we have 25 joules of energy... In the .25 to .50 line we have 25 joules of energy... In the .00 to .25 line we have 0 joules of energy... At T = 1.50 we have a wave going out in the .50 to .75 section and also in the .75 to 1.00 section... And reflected wave coming back in both those sections for 50 watts in each quarter section... The .00 to .25, and .25 to .50 sections are empty... For a total of 100 joules... T=1.75 and 2.00 is just more of the same... 4 quarters of 25 joules each lined up a row with 4 quarters of 0 joules each... Chasing each other up and down the line, folding back and passing each other at each ..25 time period... A given quarter line section can have 0 joules, 25 joules or 50 joules, present at the instantaneous T of any given time block.. In a lossless line this power will slosh back and forth forever... Never having more than 50 joules in any quarter section of line and certainly not doubling to 200 joules... If you have ever held a slinky toy between your hands and watched the wave bouncing from hand to hand, you have seen what happens in the 1 second of line section... BTW, this can be visualized by cutting 4 paper strips 3" long and marking them as 25 joules each and pushing them onto your 1 foot ruler (or ugly 250 mm stick) one at a time and just like box cars on a train see what they do... Awww righty, now lets leave the transmitter on the line instead of yanking it... Sort of like non coitus interruptus... At T=1.00 nothing is happening as far as the transmitter knows... The line is still taking power like a hungry calf taking milk from a cow... At T = 1.25 we have put 125 joules into the line... The line section ..75 to 1.00 is doubled up with 25 joules going and 25 joules coming... At T = 1.50 we have put in 150 joules... Line sections 3 and 4 are doubled up to 50 joules each and line sections 1 and 2 have 25 joules each on board... At t = 1.75 sections 2,3,4 are doubled up and only section one has 25 joules currently on board... At T = 2.00 we have 50 joules in each quarter section of line for a total of 200 joules, and Cecil is fat and happy... Thinks he has been vindicated... But wait! There's more... These Ginsu knifes will slice and dice and uuunhhh, - oops, wrong topic... Well, as a thought experiment there is nothing to inhibit me from pumping another 100 joules into the line in the T = 2.00 to 3.00 time block, and again in the 3.00 to 4.00 time block, ad infinitum until we come to the Big Bang... So Cecil, why doesn't this happen? You can't have it both ways, a theoretical thought experiment of putting energy into a lossless line and at T = 2.00 suddenly revert to some mumble jumbo about the energy going away at the same rate it is entering, or the transmitter absorbing the reflected wave (those bottles gonna vaporize on the first long dah)... There is nothing in our theoretical world to differentiate T = 2 from any T = n + 1... Inquiring minds, and all that... But let me suddenly play spoiler here... After 2 seconds, the reflected returning slug of energy present in each quarter line section is 180 degrees out of phase with the out going energy slug in each respective section... The two slugs vectorally neutralize each other... No energy present in that quarter section at any given T... So for each .25 T after T = 2.00 you are merely replacing the energy that has been neutralized by the returning wave.... Remember, energy can neither be created nor destroyed... So, where did the energy go? denny / k8do |
Where Does the Power Go?
Denny wrote:
Assume that we put a 1 second squirt at the rate of 100 Joules per second, into the input end of the line, it is an open line at the far end, and yank the input end of the line out of the transmitter at exactly 1 second... Now we have a lossless line with a wave going forward and a reflected wave coming back... Do you still claim 200 joules? No, 100 joules/second * one second would be 100 joules in your above example. In my previous example the 100 joule/sec "squirt" was two seconds long. Thus, in my example, 200 joules made it into the one-second long line. Never having more than 50 joules in any quarter section of line and certainly not doubling to 200 joules... If you have ever held a slinky toy between your hands and watched the wave bouncing from hand to hand, you have seen what happens in the 1 second of line section... True for your example, but not true for my example. In my example the 100 watt source was delivering 100 watts for two seconds. Last time I checked 100 joules/sec * 2 seconds = 200 joules. At T = 2.00 we have 50 joules in each quarter section of line for a total of 200 joules, and Cecil is fat and happy... Thinks he has been vindicated... But wait! There's more... These Ginsu knifes will slice and dice and uuunhhh, - oops, wrong topic... Well, as a thought experiment there is nothing to inhibit me from pumping another 100 joules into the line in the T = 2.00 to 3.00 time block, ... On the contrary - if the frequency is an integer number of cycles and the feedline is exactly one second long, the source will see an infinite impedance after two seconds and further sourcing of energy will be impossible. So I am ignoring the rest of your posting until you understand that fact. Under the stated the boundary conditions, it is impossible to shove any more than 200 joules into the line. At the end of two seconds, the forward current and the newly arrived reflected current are 180 degrees out of phase and cancel each other, i.e. after 2 seconds, Ifor + Iref = ZERO The reflected voltage arrives back in phase with the forward voltage and so the voltage at that point equals the voltage in the 141.4 volt source and zero current flows between those two equal potentials. There is 200 joules in the transmission line with no possibility of adding any more. The boundary conditions limit the energy content of the transmission line to 200 joules, exactly the number of joules needed to support the steady-state 100 watts forward and 100 watts reflected. The laws of physics strike again. Please rethink your position. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
On Mon, 02 Oct 2006 19:09:56 GMT, Cecil Moore
wrote: So I am ignoring :-) |
Where Does the Power Go?
Richard Clark wrote:
Cecil Moore wrote: So I am ignoring :-) Yep, when someone begins their argument with: Assume 1 = 2, I am inclined to ignore it. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
I've been told that this discussion continues in Letters to the Editor in QEX magazine, to which I don't subscribe any more. Without knowing the context of those present discussions, I composed a letter to the QEX editors. Here it is: “Where Does the Power Go?” was answered in my magazine article, “An Energy Analysis at an Impedance Discontinuity in a Transmission Line”, published by Worldradio magazine and available on my web page at: http://www.w5dxp.com/energy.htm The problem with your paper, Cecil, is the part where you try to invent the "4th Mechanism of Reflection". 73, Jim, AC6XG |
Where Does the Power Go?
On Mon, 02 Oct 2006 20:18:57 GMT, Cecil Moore
wrote: Assume 1 = 2 :-) |
Where Does the Power Go?
the big confusion factor is using power and energy at all. they are both
derived from the much simpler to handle and understand voltage or current waves. the biggest problem is that once you change from voltage or current to power you lose the information necessary to calculate superposition because you no longer have the phase information from the basic wave components. this is partly a result of the common use of the swr meter that measures forward and reflected 'power', everyone thinks they understand how it works, but very few really do. "Jim Kelley" wrote in message ... Cecil Moore wrote: I've been told that this discussion continues in Letters to the Editor in QEX magazine, to which I don't subscribe any more. Without knowing the context of those present discussions, I composed a letter to the QEX editors. Here it is: “Where Does the Power Go?” was answered in my magazine article, “An Energy Analysis at an Impedance Discontinuity in a Transmission Line”, published by Worldradio magazine and available on my web page at: http://www.w5dxp.com/energy.htm The problem with your paper, Cecil, is the part where you try to invent the "4th Mechanism of Reflection". 73, Jim, AC6XG |
Where Does the Power Go?
Jim Kelley wrote:
The problem with your paper, Cecil, is the part where you try to invent the "4th Mechanism of Reflection". I wish I had invented it, Jim, but the mechanism of wave reflection due to interference was well known and under- stood by optical engineers long before I was born. It's how non-reflective glass works. Ideally, interference at the thin-film coating reflects all of the light back toward the picture behind the glass. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Dave wrote:
the big confusion factor is using power and energy at all. they are both derived from the much simpler to handle and understand voltage or current waves. But it is hard to answer the "Where does the power go?" question without using power and energy. BTW, I didn't start that question. Jon Bloom asked that question in a Dec. 1994 QEX article as a rebuttal to the information in "Reflections", by Walter Maxwell. the biggest problem is that once you change from voltage or current to power you lose the information necessary to calculate superposition because you no longer have the phase information from the basic wave components. If one knows the length of the transmission line and the velocity factor, the phases can be deduced. If one is dealing with a Z0-match, which is most common in amateur radio, the phase information is trivial because all the voltages and all the currents are either in-phase or 180 degrees out of phase at the Z0-match point. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote: Jim Kelley wrote: The problem with your paper, Cecil, is the part where you try to invent the "4th Mechanism of Reflection". I wish I had invented it, Jim, but the mechanism of wave reflection due to interference was well known and under- stood by optical engineers long before I was born. It's how non-reflective glass works. Ideally, interference at the thin-film coating reflects all of the light back toward the picture behind the glass. What has been known since long before you were born is that only direct interaction with matter causes EM waves to reflect. You're far too modest Cecil. The "4th mechanism of reflection" is completely your idea! ;-) 73, Jim, AC6XG |
Where Does the Power Go?
Jim Kelley wrote:
What has been known since long before you were born is that only direct interaction with matter causes EM waves to reflect. Would you say the changing characteristic impedance between two waveguides in outer space is a direct interaction with matter? There is no matter inside the waveguide with which to interact. However, I am wondering if I am using the wrong word when I say interference can cause reflections. Since the same thing happens with scattering S-parameters, it may be a 180 degree refraction instead of a 180 degree reflection. I'll have to take a look at the math. But no matter what it is called, the results are the same. "A rose by any other name ..." -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Jim Kelley wrote:
What has been known since long before you were born is that only direct interaction with matter causes EM waves to reflect. You're far too modest Cecil. The "4th mechanism of reflection" is completely your idea! ;-) Nope, it's not. The interference is caused by the interaction of the forward and reflected waves *at a physical impedance discontinuity*. The impedance discontinuity is the primary cause of everything and is certainly a point of "direct interaction with matter". The reflection coefficients based on RF impedance discontinuities are similar in kind to the reflection coefficients based on differing light indices of refraction between materials. The forward reflection of light from a non-reflective thin-film glass surface was well understood before I was born. Optical engineers have no problem with that change of EM wave momentum due to interference between the internal reflection and the external reflection. That's the same mechanism that happens at a Z0-match point in a transmission line. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
[snip] However, I am wondering if I am using the wrong word when I say interference can cause reflections. Since the same thing happens with scattering S-parameters, it may be a 180 degree refraction instead of a 180 degree reflection. I'll have to take a look at the math. Cecil, You may have taken the first step along the path to enlightenment. It has been explained previously on RRAA that interference is a result, not a cause. There are no primary equations related to interference that are useful for analyzing the exact behavior of a system. Sure, there are lots of handwaving explanations, but nothing that can actually give real numbers for fields, currents, or whatever. If you start with the proper equations for the fields, or voltage and current if you desire, add in the correct boundary conditions on the interfaces, and then find the numerical solution, any interference will appear. No need to make a special case. If you do this then two positive things will occur. 1. You will be in accord with virtually every mathematician, physical scientist, optical scientist, and even engineer in applying standard analysis techniques. 2. All of the worry about missing energy, canceling waves at the interfaces, etc. simply disappears. It all pops right out from the proper application of the math, automatically. 73, Gene W4SZ |
Where Does the Power Go?
Gene Fuller wrote:
You may have taken the first step along the path to enlightenment. It has been explained previously on RRAA that interference is a result, not a cause. Of course, the Big Bang is the cause of everything, but what caused the Big Bang? When one says A causes B which causes C which causes D, etc., it is not false to say C causes D. There may be a long line of causes and effects. One step's cause is the previous step's effect. Interference causes visible interference rings in a light experiment. And of course, interference is just one event in a long line of cause and effect. Without interference, there would be no interference rings. Without beams of light, there would be no interference. Without a Big Bang, there would be no light. In a line of events, interference is an event that has a cause and has an effect. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
Gene Fuller wrote: You may have taken the first step along the path to enlightenment. It has been explained previously on RRAA that interference is a result, not a cause. Of course, the Big Bang is the cause of everything, but what caused the Big Bang? When one says A causes B which causes C which causes D, etc., it is not false to say C causes D. There may be a long line of causes and effects. One step's cause is the previous step's effect. Interference causes visible interference rings in a light experiment. And of course, interference is just one event in a long line of cause and effect. Without interference, there would be no interference rings. Without beams of light, there would be no interference. Without a Big Bang, there would be no light. In a line of events, interference is an event that has a cause and has an effect. Cecil, You just proved the point perfectly. I did not say that interference is imaginary or that it is not a useful description. I said that interference is not a primary tool for achieving detailed numerical solutions. It is a result from such calculations. Now put your hands down, solve the real equations, and stop all that mumbo-jumbo about canceling waves and reversing momentum. 8-) 73, Gene W4SZ |
Where Does the Power Go?
On Tue, 03 Oct 2006 16:56:32 GMT, Gene Fuller
wrote: reversing momentum Imagine the G forces of that for an infinitesmal 1/10¹² second or so. |
Where Does the Power Go?
Gene Fuller wrote:
Now put your hands down, solve the real equations, ... Do you mean the mashed-potatoes energy equations where the wave energy components are completely ignored? That's how some arrived at such strange unreal concepts in the first place. When one assumes that standing waves have an existence separate from the necessary components of the standing wave, one loses touch with reality. There should be a rehab clinic for sufferers of the mashed potatoes steady-state addiction. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Do you mean the mashed-potatoes energy equations where
the wave energy components are completely ignored? Calling the steady state solution "mashed-potatoes" is ridiculous. All other terms besides the steady-state ones have left the picture in the steady state. This is true of the waves, their amplitudes, the energy in the line, everything. That's what makes steady-state analysis useful... you can do it and be right. If you want to know what happens during the transient, you MUST start from the original differential equation that describes how the fields were set up in the line in the first place, but it doesn't matter in the steady state. The answer is the same. If you start with the full, time dependent equations for turning the source on into some line, whatever it is that happens with the power and where it goes will become clear from the transient solution, which I know you haven't worked out. Neither have I. I don't want to have to do it but one of these days I think it might be necessary. This is not a problem that will be solved in with Logic and English, Cecil. There is no argument if you start with a full , time dependent mathematical description of the waves. That is what will answer the question "Where Does the Power Go" You'll end up with a time dependent Poynting vector that will tell you. I hypothesize, for no particularly good reason, that such a description will lead to the excess power having been delivered to the load through the interaction of the transient solutions on the line. - - - - - - Copied and pasted from the rest of my post at eHam: In a matched line, none of us would disagree that the power flux out of one end of the line is the power flux into the other end. From what I understand, the electromagnetic energy contained in that line is related to the Poynting vector. The integral of the Poynting vector over the cross sectional area of the line gives you the time averaged power flowing in the line. The power flows at the group velocity of the waves in the line, for normal transmission lines, this is somewhere between 0.6 - 1.0 times the speed of light. The power flow divided by the group velocity gives you the energy density per unit length in the line. The power flow is the Poynting vector integrated over the cross sectional area of the line. The Poynting vector in a mismatched line has been worked out by me, here, and I'd appreciate comments on the MATH. http://en.wikipedia.org/wiki/User:Dan_Zimmerman/Sandbox My claim is that that's it. That's all. No handwaving, no setting up steady state startup and claiming that that energy remains in the line. The real part of the Poynting vector represents real power flux and the imaginary part represents reactive power flux. To get the energy that exists in the line you just add it all up... Take the real part, divide it by the propagation velocity on the line, integrate it over the area of the line (I think for TEM, the fields are uniform and so multiplying by the area of the line is sufficient) This gives you the energy density per unit length for the power that's flowing from source to load. Integrate that over the length of the line. Set aside. Take the imaginary part, do the same thing. This gives you the energy contained in the line of the reactive, circulating power. Add in the previously calculated energy of the flowing power, and you're done. Am I wrong, Cecil et. al? If so, Cecil, could you please write down the description mathematically or get someone to help you out, and could you please look at and check my math? I started with the assumption that there are two counterpropagating waves with some arbitrary electric field amplitudes and an arbitrary phase relationship. I didn't use anything about Thevenin. I didn't use anything about virtual open circuits. The forward and reverse waves have been included, from the beginning. They're both there. When the SWR is 1:1, there is no energy in the line associated with reactive stored power. None at all. The reflection coefficient is zero, and all energy in the line is associated with flowing power. When the SWR is infinite, there is no energy in the line associated with flowing power. None at all. The reflection coefficient is 1, and all energy in the line is associated with circulating power (reactance only, as you would expect from a stub) Dan |
Where Does the Power Go?
Richard Clark wrote:
Gene Fuller wrote: reversing momentum Imagine the G forces of that for an infinitesmal 1/10¹² second or so. The pressure of such a momentum change can be calculated. From "Optics", by Hecht, 4th edition, page 57: "When the surface under illumination is *perfectly reflecting*, the beam that entered with a velocity of +c will emerge with a velocity of -c. This corresponds to twice the change in momentum that occurs on absorption, and hence," Pressure of a absorbed wave = S(t)/c Pressure of a reflected wave = 2*S(t)/c -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote: Richard Clark wrote: Gene Fuller wrote: reversing momentum Imagine the G forces of that for an infinitesmal 1/10¹² second or so. The pressure of such a momentum change can be calculated. From "Optics", by Hecht, 4th edition, page 57: "When the surface under illumination is *perfectly reflecting*, the beam that entered with a velocity of +c will emerge with a velocity of -c. This corresponds to twice the change in momentum that occurs on absorption, and hence," Pressure of a absorbed wave = S(t)/c Pressure of a reflected wave = 2*S(t)/c I'll bet the reason no one can measure the radiation pressure resulting from your "4th mechanism of reflection" is because it cancels out with the radiation pressure from the cancelled reflection in the other direction. Right, Cecil? :-) 73, Jim, AC6XG |
Where Does the Power Go?
On Tue, 03 Oct 2006 10:47:58 -0700, Jim Kelley
wrote: Pressure of a reflected wave = 2*S(t)/c I'll bet the reason no one can measure the radiation pressure resulting from your "4th mechanism of reflection" is because it cancels out with the radiation pressure from the cancelled reflection in the other direction. Right, Cecil? :-) Hi Jim, Who needs reasoning when a Xeroxed formula, like a "4th mechanism of reflection," is simpler to cut and paste than actually offer a practical answer to? This is akin to his sophomoric allusion to anti-glare coating yet again (yawn), he couldn't work the math on that one anywhere closer than pi = 22/7. Soon we will be down the path of 66% allowable error to prove a concept. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Jim Kelley wrote:
I'll bet the reason no one can measure the radiation pressure resulting from your "4th mechanism of reflection" is because it cancels out with the radiation pressure from the cancelled reflection in the other direction. Right, Cecil? :-) There is certainly radiation pressure pushing outward from a 1/4WL thin-film non-reflecting surface even though the reflected waves cancel each other. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
On 2 Oct 2006 11:11:40 -0700, "Denny" wrote:
Assume that we put a 1 second squirt at the rate of 100 Joules per second, into the input end of the line, it is an open line at the far end, and yank the input end of the line out of the transmitter at exactly 1 second... Now we have a lossless line with a wave going forward and a reflected wave coming back... Do you still claim 200 joules? Hi Denny, A question of my own: Did you really expect your question above wouldn't be dodged? I suppose not. Anyway, your observations revealed Cecil's usual lack of rigor. You certainly know how to step back while he juggles un-pinned hand grenades. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Richard Clark wrote:
Who needs reasoning when a Xeroxed formula, like a "4th mechanism of reflection," is simpler to cut and paste than actually offer a practical answer to? This is akin to his sophomoric allusion to anti-glare coating yet again (yawn), he couldn't work the math on that one anywhere closer than pi = 22/7. Richard, I wasn't the one who, through superposition of powers, came up with an irradiance brighter than the surface of the sun at the non-reflective surface interface. Did you ever figure out that superposing powers is a no-no? -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
There is certainly radiation pressure pushing outward from a 1/4WL thin-film non-reflecting surface even though the reflected waves cancel each other. Cecil, Radiation pressure without waves? Isn't that the sort of thing that started the Boston Tea Party? 8-) 73, Gene W4SZ |
Where Does the Power Go?
OK, so we've established you don't know the G forces for changed
momentum, only how to sniff toner at the Xerox. On Tue, 03 Oct 2006 18:34:17 GMT, Cecil Moore wrote: Richard, I wasn't the one who, through superposition of powers, Yes, you did have a problem translating power to energy and back. I could offer any number of common scenarios that would have you gasping for air: There is a common bare light bulb 1 meter away; it illuminates a cm² target with 3µW @ 55nM of POWER; what is: the number of candela per steradians, at the target, from total bandwidth radiation? or: How much power is being supplied to the bulb? came up with an irradiance brighter than the surface of the sun at the non-reflective surface interface. No, true to your form, you rounded errors and fudged numbers to prove light was black. In that regard I will offer you a third choice question from above: Can you see this amount of light on the target? (choose this one, you might guess it right - it doesn't demand any math skill.) :-0 |
Where Does the Power Go?
Gene Fuller wrote:
Cecil Moore wrote: There is certainly radiation pressure pushing outward from a 1/4WL thin-film non-reflecting surface even though the reflected waves cancel each other. Radiation pressure without waves? Please don't deliberately obfuscate things, Gene. There are no reflected waves on the outside of the thin-film and therefore no reflected wave pressure from the outside. All the reflected waves are on the inside of the thin-film outer plane. That's why the reflected wave pressure is pushing outwards. The pressure is where the waves are. There's no wave pressure without waves. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
wrote:
Do you mean the mashed-potatoes energy equations where the wave energy components are completely ignored? Calling the steady state solution "mashed-potatoes" is ridiculous. I'm not calling the solutions ridiculous. I am calling some of the conceptual conclusions ridiculous. For instance, since the *net* Poynting vector equals delivered source watts, there are zero watts in the reflected wave even though there are joules in the reflected wave moving at the speed of light. If you want to know what happens during the transient, you MUST start from the original differential equation that describes how the fields were set up in the line in the first place, but it doesn't matter in the steady state. The answer is the same. Not according to some of the gurus posting here. The joules supplied during the initial transient state are being swept under the steady-state rug. Witness "Food For Thought #1" on www.eznec.com I don't want to have to do it but one of these days I think it might be necessary. What you will find is that there is exactly the energy in the transmission line required to support the real speed-of-light forward traveling wave and the real speed-of-light reflected traveling wave - no more and no less. The argument that there is no more energy in a transmission line with reflections than is being supplied by the source is simply false. The Poynting vector in a mismatched line has been worked out by me, here, and I'd appreciate comments on the MATH. http://en.wikipedia.org/wiki/User:Dan_Zimmerman/Sandbox What about the forward Poynting vector, Pz+, and the reflected Poynting vector, Pz-. Reference page 291, "Fields and Waves ...", Ramo and Whinnery, 2nd edition? My claim is that that's it. That's all. No handwaving, no setting up steady state startup and claiming that that energy remains in the line. The real part of the Poynting vector represents real power flux and the imaginary part represents reactive power flux. To get the energy that exists in the line you just add it all up... The forward wave's average Poynting vector is real. The reflected wave's average Poynting vector is real. Adding two real power flow vectors is a lot easier than integrating a power flow vector with real and imaginary parts. And, bottom line, one obtains exactly the same results with 10% of the work. Subtract the reverse Poynting vector from the forward Poynting vector to determine the net power. Add the two Poynting vectors and multiply by the length in seconds of the feedline to determine the total energy. What could be simpler? Am I wrong, Cecil et. al? No, you are correct but you are going around the world to get from California to New York. My method, supported by Ramo and Whinnery, can be done in much less time and can be more easily understood by people not familiar with your method. By the time you get out your calculator, I can have your answer waiting for you and it will be the same answer you get after wasting a lot of time. When the SWR is infinite, there is no energy in the line associated with flowing power. None at all. The reflection coefficient is 1, and all energy in the line is associated with circulating power (reactance only, as you would expect from a stub) You are, of course, talking about *net* "flowing" power. But your answer is exactly the same as component forward power in real joules/sec plus component reflected power in real joules/sec. And thinking in terms of those real component powers is a lot easier than your imaginary stuff. Pz+ - Pz- = Pz-tot sourced and delivered (Pz+ + Pz-)* feedline length in seconds = total feedline energy There is also a physics problem with the imaginary concept. EM wave energy must necessarily move at the speed of light. A small amount of TV modulation will prove that those forward and reflected waves are still moving end to end at the speed of light and still transferring information. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Cecil Moore wrote:
Gene Fuller wrote: Cecil Moore wrote: There is certainly radiation pressure pushing outward from a 1/4WL thin-film non-reflecting surface even though the reflected waves cancel each other. Radiation pressure without waves? Please don't deliberately obfuscate things, Gene. There are no reflected waves on the outside of the thin-film and therefore no reflected wave pressure from the outside. All the reflected waves are on the inside of the thin-film outer plane. That's why the reflected wave pressure is pushing outwards. The pressure is where the waves are. There's no wave pressure without waves. Cecil, Sorry, I guess my closing was not clear. Try this. 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 73, Gene W4SZ |
Where Does the Power Go?
Gene Fuller wrote:
Sorry, I guess my closing was not clear. Try this. 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-) Sorry, I thought your happy face was a localized variable. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Hi All,
Even armed with college texts on optics, Cecil hasn't got a whisper of a chance in actually answering a direct question such as has been posed below. A Xerox is not, after all, a calculator (nor does it impart knowledge). As such, I will offer direct answers and let it go at that. On Tue, 03 Oct 2006 12:03:34 -0700, Richard Clark wrote: There is a common bare light bulb 1 meter away; it illuminates a cm² target with 3µW @ 55nM of POWER; what is: the number of candela per steradians, at the target, from total bandwidth radiation? 5 or: How much power is being supplied to the bulb? 100W Can you see this amount of light on the target? Depending upon the infirmities claimed: none to some. Aside: only 0.7% as bright as the sun, in the tropics, at noon, on an equinox. 73's Richard Clark, KB7QHC |
Where Does the Power Go?
Cecil Moore wrote: Jim Kelley wrote: I'll bet the reason no one can measure the radiation pressure resulting from your "4th mechanism of reflection" is because it cancels out with the radiation pressure from the cancelled reflection in the other direction. Right, Cecil? :-) There is certainly radiation pressure pushing outward from a 1/4WL thin-film non-reflecting surface even though the reflected waves cancel each other. Actually I agree, but it's all from ordinary reflections, rather than from backscattered interference or anything else from the 'square root of negative one' axis. 73, ac6xg |
Where Does the Power Go?
On Tue, 03 Oct 2006 13:33:26 -0700, Richard Clark
wrote: Aside: only 0.7% as bright as the sun, in the tropics, at noon, on an equinox. Also given that Cecil cannot compute the radiation pressure, the answer with respect to the question offered in the thread above is: 3.2 pico Newtons or 3.2 nano g·m/s² Dare I ask the G forces again? |
Where Does the Power Go?
Jim Kelley wrote:
Actually I agree, but it's all from ordinary reflections, rather than from backscattered interference or anything else from the 'square root of negative one' axis. Jim, the reflected energy and momentum changes direction. Walter Maxwell calls it a virtual short or open with a virtual reflection coefficient of 1.0. I agree with "Optics", by Hecht, that it is total destructive interference in the source direction accompanied by total constructive interference in the load direction. What do you call it? What reflection coefficient does your reflected wave see? Please give it a name. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Richard Clark wrote:
Even armed with college texts on optics, Cecil hasn't got a whisper of a chance in actually answering a direct question such as has been posed below. Actually, I figure someone who superposes power to the extent that non-reflective glass is brighter than the surface of the sun doesn't know enough to ask a decent question. -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
Tom Donaly wrote:
Don't expect much math from Cecil. He always neglects the phase difference of the reflected wave, and won't do the algebra, no matter what. Simply not true, Tom. For a Z0-match, the phase is always zero or 180 degrees. Only addition or subtraction is ever required at a Z0-match. I've got the answer while you guys are still trying to figure out the cosine of zero degrees. :-) -- 73, Cecil http://www.w5dxp.com |
Where Does the Power Go?
On Wed, 04 Oct 2006 00:32:17 GMT, Cecil Moore
wrote: Richard Clark wrote: Even armed with college texts on optics, Cecil hasn't got a whisper of a chance in actually answering a direct question such as has been posed below. Actually, I figure someone who superposes power to the extent that non-reflective glass is brighter than the surface of the sun doesn't know enough to ask a decent question. Sounds like a lame answer from you given you cannot perform ANY of the math. |
Where Does the Power Go?
Richard Clark wrote:
Sounds like a lame answer from you given you cannot perform ANY of the math. Richard, you really need to disguise your snipe hunts a little better. -- 73, Cecil http://www.w5dxp.com |
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