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A small riddle, just for fun
Yesterday, while repairing my antenna, something came to my mind I had never
focused on before. Let us consider a bipole, that is a "black box" having TWO terminals and including plain passive elements only (like capacitors, inductors, ... , no diodes or other special devices), arranged the way you prefer, it does not matter. In my mind it was quite clear that, when fitting such a bipole into a circuit, the sense makes no difference, i.e. one can reverse the two terminals with no consequence. As a matter of fact, the bipole has an equivalent impedance that remains the same independently of the way it is put in the circuit. Yesterday a case occurred to me in which this is not actually true. Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. Although probably unnecessary, let me recall that a filter is typically a THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one. 73 Tony I0JX Rome, Italy |
A small riddle, just for fun
On 31 ene, 21:20, "Antonio Vernucci" wrote:
Yesterday, while repairing my antenna, something came to my mind I had never focused on before. Let us consider a bipole, that is a "black box" having TWO terminals and including plain passive elements only (like capacitors, inductors, ... , no diodes or other special devices), arranged the way you prefer, it does not matter. In my mind it was quite clear that, when fitting such a bipole into a circuit, the sense makes no difference, i.e. one can reverse the two terminals with no consequence. As a matter of fact, the bipole has an equivalent impedance that remains the same independently of the way it is put in the circuit. Yesterday a case occurred to me in which this is not actually true. Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. Although probably unnecessary, let me recall that a filter is typically a THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one. 73 Tony I0JX Rome, Italy Hello Tony, The answer may be in your own text (the ground issue). There may be a third path via ground (capacitive coupling). You can add a very good common mode choke at the input terminal of you bipole. In that case the path via ground is blocked. Try a very simple bipole: metallic case connected to terminal 1, terminal 2 connected to nothing. When the center conductor of your source is connected to terminal 1, you have the ground path. The effect of reverse connection will be less when you use a very small battery powered source that is completely floating. Measuring data you can get out of it via optical link... Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me. |
A small riddle, just for fun
The answer may be in your own text (the ground issue). There may be a
third path via ground (capacitive coupling). Hi Wim, I acknowledege that your example leads to an asymmetric bipole, that one cannot reverse it without consequences. However, in the case which occurred to me, there is no need to assume the existence of a third path via ground for justifying the asymmetry. Actually the asimmetry remains even if the bipole would be in free space! Forgive me if I do not immediately reveal my case, but I would like to see if there are some more guesses before closing the issue. 73 Tony I0JX |
A small riddle, just for fun
K1TTT wrote:
as long as it is ONLY 2 ports, has ONLY passive linear components, AND is small enough to be considered a lumped element, then you can always reverse the terminals and not know the difference. The most frequently violated condition is the last one, put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off. An electrolytic capacitor acts differently if reverse biased. The results are usually bad. It does qualify as two leaded, passive and linear when used properly. |
A small riddle, just for fun
On Jan 31, 9:38*pm, "Antonio Vernucci" wrote:
The answer may be in your own text (the ground issue). There may be a third path via ground (capacitive coupling). Hi Wim, I acknowledege that your example leads to an asymmetric bipole, that one cannot reverse it without consequences. However, in the case which occurred to me, there is no need to assume the existence of a third path via ground for justifying the asymmetry. Actually the asimmetry remains even if the bipole would be in free space! Forgive me if I do not immediately reveal my case, but I would like to see if there are some more guesses before closing the issue. 73 Tony I0JX as long as it is ONLY 2 ports, has ONLY passive linear components, AND is small enough to be considered a lumped element, then you can always reverse the terminals and not know the difference. The most frequently violated condition is the last one, put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off. |
A small riddle, just for fun
On Mon, 31 Jan 2011 21:20:57 +0100, "Antonio Vernucci"
wrote: Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. Hi Antonio, This would fall into the area of a hidden variable (a ghost terminal, so-to-speak) and is called Common Mode. This occurs in the situation that includes the proximity of ground which is often neglected as an influence. The dipole could be unbalanced forcing currents to be out of balance. This is notorious with cabled down leads which preserves the sense of two terminals, but the imbalance with ground forces a third terminal into the reality of implementation. If we were to divorce ground from this (free space) and simply looked at the two lead impedance at the drive point, then adding a cable will force the same imbalance (albeit to a different degree, as found with the proximity of earth). If the dipole is off-center fed, this will be more profound (even though when it is measured at the feed point, the measurement is immune to pole switching). Reversing the "polarity" of the cable connection will create a new topology (although the Common Mode will persist with new characteristics). Although probably unnecessary, let me recall that a filter is typically a THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one. Well, even though I have anticipated ground, a filter can easily be a two terminal device without need for ground - being a series component. This two terminal description is quite typical too. 73's Richard Clark, KB7QHC |
A small riddle, just for fun
On Mon, 31 Jan 2011 17:23:33 -0600, joe wrote:
An electrolytic capacitor acts differently if reverse biased. The results are usually bad. It does qualify as two leaded, passive and linear when used properly. Hi Joe, The same can be said of a diode. Hence the electrolytic capacitor is non-linear, which violates the premise. 73's Richard Clark, KB7QHC |
A small riddle, just for fun
On Jan 31, 11:23*pm, joe wrote:
K1TTT wrote: as long as it is ONLY 2 ports, has ONLY passive linear components, AND is small enough to be considered a lumped element, then you can always reverse the terminals and not know the difference. *The most frequently violated condition is the last one, put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off. An electrolytic capacitor acts differently if reverse biased. The results are usually bad. It does qualify as two leaded, passive and linear when used properly. an electrolytic that acts that way is not a linear component. |
A small riddle, just for fun
"Antonio Vernucci" ha scritto nel messaggio ... Forgive me if I do not immediately reveal my case, but I would like to see if there are some more guesses before closing the issue. From my very poor knowledge of trasmission lines... ... reversing inner and outer conductor of a coaxial cable ?? :) -.-. --.- |
A small riddle, just for fun
On Jan 31, 12:20*pm, "Antonio Vernucci" wrote:
Yesterday, while repairing my antenna, something came to my mind I had never focused on before. Let us consider a bipole, that is a "black box" having TWO terminals and including plain passive elements only (like capacitors, inductors, ... , no diodes or other special devices), arranged the way you prefer, it does not matter. In my mind it was quite clear that, when fitting such a bipole into a circuit, the sense makes no difference, i.e. one can reverse the two terminals with no consequence. As a matter of fact, the bipole has an equivalent impedance that remains the same independently of the way it is put in the circuit. Yesterday a case occurred to me in which this is not actually true. Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. Although probably unnecessary, let me recall that a filter is typically a THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one. 73 Tony I0JX Rome, Italy I have a circuit I've been working on lately which has a simple series LC in it, no other connection to the node between the inductor and capacitor. It turns out that the order of the inductor and capacitor makes a big difference in the circuit performance. I anticipated that it would, and put them in the intuitively obvious order, only to find out that it was the wrong order! A proper model cleared things up quite nicely. However, in no way would I call that particular part of the circuit a "two terminal" network. The effect is the same as Wim mentioned. Cheers, Tom |
A small riddle, just for fun
As long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always reverse the terminals and not know the difference. The most frequently violated condition is the last one, put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off. Could you please explain me the meaning of your sentence: "put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off" as I have difficulties to interpret it due to my non-mother tongue english. 73 Tony I0JX Rome, Italy |
A small riddle, just for fun
On Feb 1, 6:24*pm, "Antonio Vernucci" wrote:
As long as it is ONLY 2 ports, has ONLY passive linear components, AND is small enough to be considered a lumped element, then you can always reverse the terminals and not know the difference. *The most frequently violated condition is the last one, put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off. Could you please explain me the meaning of your sentence: "put in a piece of coax with that is long enough to be measured at the highest frequency you will use and all bets are off" as I have difficulties to interpret it due to my non-mother tongue english. 73 Tony I0JX Rome, Italy the general case is, if any part of the circuit is more than a small fraction of a wavelength in size you may be able to detect the difference between the ports. one common way to do that is to try to measure a circuit that has a long piece of coax in it, the results may be very different when you reverse the terminals. |
A small riddle, just for fun
On Feb 1, 6:18*pm, K7ITM wrote:
On Jan 31, 12:20*pm, "Antonio Vernucci" wrote: Yesterday, while repairing my antenna, something came to my mind I had never focused on before. Let us consider a bipole, that is a "black box" having TWO terminals and including plain passive elements only (like capacitors, inductors, ... , no diodes or other special devices), arranged the way you prefer, it does not matter. In my mind it was quite clear that, when fitting such a bipole into a circuit, the sense makes no difference, i.e. one can reverse the two terminals with no consequence. As a matter of fact, the bipole has an equivalent impedance that remains the same independently of the way it is put in the circuit. Yesterday a case occurred to me in which this is not actually true. Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. Although probably unnecessary, let me recall that a filter is typically a THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one. 73 Tony I0JX Rome, Italy I have a circuit I've been working on lately which has a simple series LC in it, no other connection to the node between the inductor and capacitor. *It turns out that the order of the inductor and capacitor makes a big difference in the circuit performance. *I anticipated that it would, and put them in the intuitively obvious order, only to find out that it was the wrong order! *A proper model cleared things up quite nicely. *However, in no way would I call that particular part of the circuit a "two terminal" network. *The effect is the same as Wim mentioned. Cheers, Tom then there is either something wrong with your components or you have too much coupling the the surrounding environment... so you really had a 3 terminal network using the stray capacitance or inductance. |
A small riddle, just for fun
On Feb 1, 3:24*pm, K1TTT wrote:
On Feb 1, 6:18*pm, K7ITM wrote: On Jan 31, 12:20*pm, "Antonio Vernucci" wrote: Yesterday, while repairing my antenna, something came to my mind I had never focused on before. Let us consider a bipole, that is a "black box" having TWO terminals and including plain passive elements only (like capacitors, inductors, .... , no diodes or other special devices), arranged the way you prefer, it does not matter. In my mind it was quite clear that, when fitting such a bipole into a circuit, the sense makes no difference, i.e. one can reverse the two terminals with no consequence. As a matter of fact, the bipole has an equivalent impedance that remains the same independently of the way it is put in the circuit. Yesterday a case occurred to me in which this is not actually true. Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. Although probably unnecessary, let me recall that a filter is typically a THREE-terminal device (IN, OUT, GROUND), not a TWO-terminal one. 73 Tony I0JX Rome, Italy I have a circuit I've been working on lately which has a simple series LC in it, no other connection to the node between the inductor and capacitor. *It turns out that the order of the inductor and capacitor makes a big difference in the circuit performance. *I anticipated that it would, and put them in the intuitively obvious order, only to find out that it was the wrong order! *A proper model cleared things up quite nicely. *However, in no way would I call that particular part of the circuit a "two terminal" network. *The effect is the same as Wim mentioned. Cheers, Tom then there is either something wrong with your components or you have too much coupling the the surrounding environment... If by "too much" you mean "more than I wanted," you're absolutely right. But of course with real components, there is no such thing as zero coupling to the surrounding environment. As with all real circuits, there are tradeoffs: high coil Q results in a large coil, which in turn results in more capacitance to the shield than a smaller low-Q coil would have. The shield is required for other reasons... ... so you really had a 3 terminal network using the stray capacitance or inductance. Exactly. Just what I said: the effect is the same as Wim mentioned. Cheers, Tom |
A small riddle, just for fun
On Tue, 1 Feb 2011 15:24:41 -0800 (PST), K1TTT wrote:
then there is either something wrong with your components or you have too much coupling the the surrounding environment... so you really had a 3 terminal network using the stray capacitance or inductance. This can be neutralized employing a "Wagner Earth." For inductors, there is the Modified Owen method. Basically, problems of this sort have been handled with driven shields (a variant of the Wagner Earth/Ground). This exactly century-old topic ( "Wagner Earth") is rarely discussed outside of precision measurement or very high voltage handling circuits, however. 73's Richard Clark, KB7QHC |
A small riddle, just for fun
as long as it is ONLY 2 ports, has ONLY passive linear components, AND
is small enough to be considered a lumped element, then you can always reverse the terminals and not know the difference. You have pinpointed the correct issue: lumped elements. The story began when, a few days ago, I was going to replace a trap of my HF yagi. Not to make mistakes, I consulted the antenna assembly manual where I found a big banner: do not invert traps otherwise the antenna will not work. So, I thought, this is a case in which a bipole cannot be inverted. This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason why it cannot be inverted. 73 Tony I0JX Rome, Italy |
A small riddle, just for fun
Wim,
the story began when, a few days ago, I was going to replace a trap of my HF yagi. Not to make mistakes, I consulted the antenna assembly manual where I found a big banner: do not invert traps otherwise the antenna will not work. So, I thought, this is a case in which a bipole cannot be inverted. This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason, I believe, why it cannot be inverted. 73 Tony I0JX Rome, Italy |
A small riddle, just for fun
Tom,
the story began when, a few days ago, I was going to replace a trap of my HF yagi. Not to make mistakes, I consulted the antenna assembly manual where I found a big banner: do not invert traps otherwise the antenna will not work. So, I thought, this is a case in which a bipole cannot be inverted. This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason why it cannot be inverted. So, as K1TTT has pointed out, a bipole can be inverted without consequences only if it has only 2 ports, has only passive linear components, and is small enough to be considered a lumped element. 73 Tony I0JX Rome, Italy |
A small riddle, just for fun
On 2 feb, 09:37, "Antonio Vernucci" wrote:
Wim, the story began when, a few days ago, I was going to replace a trap of my HF yagi. Not to make mistakes, I consulted the antenna assembly manual where I found a big banner: do not invert traps otherwise the antenna will not work. So, I thought, this is a case in which a bipole cannot be inverted. This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason, I believe, why it cannot be inverted. 73 Tony I0JX Rome, Italy Hello Tony, So it isn't a bipole. The tube containing the trap is a (short) common mode transmission line that has effect on its environment (like a metal case with its capacitive effect on ground). The trap inside the metal tube is in series with the end of the short transmission line. The ground in this case is the boundary where the near field is no longer dominant w.r.t. to the radiating field. Think of a circular ground at 0.16lambda from the structure. So one end of your bipole has more ground to this virtual ground and is therefore a 3-pole device. For the limitation on lumped circuit approach, it is not important whether or not it contributes to the overall radiation of a structure. When you make a floating ground out of 4 quarter-wave radials (monopole?), this structure has minor influence on the far field pattern of the quarter wave (or better half wave) radiator above it. Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me |
A small riddle, just for fun
"Wimpie" napisal w wiadomosci ... On 2 feb, 09:37, "Antonio Vernucci" wrote: For the limitation on lumped circuit approach, it is not important whether or not it contributes to the overall radiation of a structure. When you make a floating ground out of 4 quarter-wave radials (monopole?), this structure has minor influence on the far field pattern of the quarter wave (or better half wave) radiator above it. In http://www.padrak.com/ine/FARADAY1.html Faraday wrote: "The view which I am so bold to put forth considers, therefore, radiation as a kind of species of vibration in the lines of force which are known to connect particles and also masses of matter together. It endeavors to dismiss the aether, but not the vibration. The kind of vibration which, I believe, can alone account for the wonderful, varied, and beautiful phaenomena of polarization, is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, for the vibrations in these cases are direct, or to and from the centre of action, whereas the former are lateral. It seems to me, that the resultant of two or more lines of force is in an apt condition for that action which may be considered as equivalent to a lateral vibration; whereas a uniform medium, like the aether, does not appear apt, or more apt than air or water." Is your 4 quarter-wave radials (monopole?), the same like Faraday's lines? Two or more monopoles produces "polarised waves". Two or more monopoles "is in an apt condition for that action which may be considered as equivalent to a lateral vibration;". Some textbooks say that Faraday suggested the transverse (angular) wibrations. Did he? S* |
A small riddle, just for fun
On Wed, 2 Feb 2011 09:35:38 +0100, "Antonio Vernucci"
wrote: This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason why it cannot be inverted. Hi Antonio, No, in fact. You clearly state in the first sentence the reason why. The external body renders an asymmetry. The two elements it separates are also non-symmetrical (both physically and electrically). This is NOT a lumped circuit exercise that has gone bad. 73's Richard Clark, KB7QHC |
A small riddle, just for fun
On Feb 2, 3:42*am, "Antonio Vernucci" wrote:
Tom, the story began when, a few days ago, I was going to replace a trap of my HF yagi. Not to make mistakes, I consulted the antenna assembly manual where I found a big banner: do not invert traps otherwise the antenna will not work. So, I thought, this is a case in which a bipole cannot be inverted. This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason why it cannot be inverted. So, as K1TTT has pointed out, a bipole can be inverted without consequences only if it has only 2 ports, has only passive linear components, and is small enough to be considered a lumped element. 73 Tony I0JX Rome, Italy A bipole is symmetrical, obviously the internal circuity of the trap is not symmetrical . Jimmie |
A small riddle, just for fun
A bipole is symmetrical, obviously the internal circuity of the trap
is not symmetrical . Jimmie, that is not the point. - if the bipole is made of lumped components, then it can be freely reversed without consequences, independently of whether it is symmetrical or asymmetrical - a trap, which instead contains a distributed element (that is the radiating trap body), can be freely reversed without consequences only if it is symmetrical (which is not my case). 73 Tony I0JX |
A small riddle, just for fun
Hello Tony,
On 2 feb, 19:14, "Antonio Vernucci" wrote: A bipole is symmetrical, obviously the internal circuity of the trap is not symmetrical *. Jimmie, that is not the point. At some frequency, everything becomes a distributed component, it depends on size/lambda ratio and application. Many distributed components can be modelled based on theoretical lumped components (with sufficient accuracy), but a ground for modelling stray capacitance is frequently required. True lumped component behaviour, in my opinion, only applies to networks with size=0. - if the bipole is made of lumped components, then it can be freely reversed without consequences, independently of whether it is symmetrical or asymmetrical What do you define as "lumped component"? - a trap, which instead contains a distributed element (that is the radiating trap body), can be freely reversed without consequences only if it is symmetrical (which is not my case). I agree on the above. 73 Tony I0JX Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me |
A small riddle, just for fun
On Feb 2, 3:53*pm, "Szczepan Bialek" wrote:
"Wimpie" napisal w ... On 2 feb, 09:37, "Antonio Vernucci" wrote: For the limitation on lumped circuit approach, it is not important whether or not it contributes to the overall radiation of a structure. *When you make a floating ground out of 4 quarter-wave radials (monopole?), this structure has minor influence on the far field pattern of the quarter wave (or better half wave) radiator above it. Inhttp://www.padrak.com/ine/FARADAY1.html*Faraday wrote: "The view which I am so bold to put forth considers, therefore, radiation as a kind of species of vibration in the lines of force which are known to connect particles and also masses of matter together. It endeavors to dismiss the aether, but not the vibration. The kind of vibration which, I believe, can alone account for the wonderful, varied, and beautiful phaenomena of polarization, is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, for the vibrations in these cases are direct, or to and from the centre of action, whereas the former are lateral. It seems to me, that the resultant of two or more lines of force is in an apt condition for that action which may be considered as equivalent to a lateral vibration; whereas a uniform medium, like the aether, does not appear apt, or more apt than air or water." I hope you can read what you quoted above and understand it... note specifically: It endeavors to dismiss the aether, but not the vibration. so with this he has decided there is NO aether! is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, so all the comparisons to water waves or sound waves are NOT correct! equivalent to a lateral vibration NOT longitudinal! so by these statements he has said that electromagnetic waves ARE NOT caused by an aether, and ARE lateral, NOT longitudinal... and CAN NOT be compared to sound or water vibrations! |
A small riddle, just for fun
On Feb 2, 7:06*pm, Wimpie wrote:
Hello Tony, On 2 feb, 19:14, "Antonio Vernucci" wrote: A bipole is symmetrical, obviously the internal circuity of the trap is not symmetrical *. Jimmie, that is not the point. At some frequency, everything becomes a distributed component, it depends on size/lambda ratio and application. *Many distributed components can be modelled based on theoretical lumped components (with sufficient accuracy), but a ground for modelling stray capacitance is frequently required. True lumped component behaviour, in my opinion, only applies to networks with size=0. 'True' or theoretical versus 'practical' is a very important distinction that is, or at least was, taught in engineering classes. it is very important to know when you can apply the practical simplifications that allow you to do design work without worrying about insignificant phenomena in the problem domain you are working in... so if i am designing an HF antenna I know that objects below a given size can be safely ignored, but if i'm doing a microwave design i have to take into account much smaller objects. |
A small riddle, just for fun
"K1TTT" napisal w wiadomosci ... On Feb 2, 3:53 pm, "Szczepan Bialek" wrote: Inhttp://www.padrak.com/ine/FARADAY1.html Faraday wrote: "The view which I am so bold to put forth considers, therefore, radiation as a kind of species of vibration in the lines of force which are known to connect particles and also masses of matter together. It endeavors to dismiss the aether, but not the vibration. The kind of vibration which, I believe, can alone account for the wonderful, varied, and beautiful phaenomena of polarization, is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, for the vibrations in these cases are direct, or to and from the centre of action, whereas the former are lateral. It seems to me, that the resultant of two or more lines of force is in an apt condition for that action which may be considered as equivalent to a lateral vibration; whereas a uniform medium, like the aether, does not appear apt, or more apt than air or water." I hope you can read what you quoted above and understand it... note specifically: It endeavors to dismiss the aether, but not the vibration. so with this he has decided there is NO aether! Ik such words: "I suppose we may compare together the matter of the aether and ordinary matter (as, for instance, the copper of the wire through which the electricity is conducted), and consider them as alike in their essential constitution; i.e. either as both composed of little nuclei, considered in the abstract as matter" In your antennas and in the space are free electrons. We do not need a mystery aether. is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, so all the comparisons to water waves or sound waves are NOT correct! You know that the dipole works like the two loudspeakers. equivalent to a lateral vibration NOT longitudinal! And not traversal. The waves radiated from the two ends of a dipole I call "coupled". so by these statements he has said that electromagnetic waves ARE NOT caused by an aether, and ARE lateral, NOT longitudinal... and CAN NOT be compared to sound or water vibrations! Faraday was great. In the Nature no natural acoustic dipoles. For him was obvious that must be something equivalent to a lateral vibration. As you know it is the two or more sources of pressure waves properly phased. Are electrons jumping off from your transmitting antennas? In copper and in space are the same "little nuclei, considered in the abstract as matter". Electrons have mass. Biot-Savart's magnetic whirl is massles. S* |
A small riddle, just for fun
On Feb 2, 7:53*pm, "Szczepan Bialek" wrote:
*"K1TTT" napisal w ... On Feb 2, 3:53 pm, "Szczepan Bialek" wrote: Inhttp://www.padrak.com/ine/FARADAY1.htmlFaraday wrote: "The view which I am so bold to put forth considers, therefore, radiation as a kind of species of vibration in the lines of force which are known to connect particles and also masses of matter together. It endeavors to dismiss the aether, but not the vibration. The kind of vibration which, I believe, can alone account for the wonderful, varied, and beautiful phaenomena of polarization, is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, for the vibrations in these cases are direct, or to and from the centre of action, whereas the former are lateral. It seems to me, that the resultant of two or more lines of force is in an apt condition for that action which may be considered as equivalent to a lateral vibration; whereas a uniform medium, like the aether, does not appear apt, or more apt than air or water." I hope you can read what you quoted above and understand it... note specifically: It endeavors to dismiss the aether, but not the vibration. so with this he has decided there is NO aether! Ik such words: "I suppose we may compare together the matter of the aether and ordinary matter (as, for instance, the copper of the wire through which the electricity is conducted), and consider them as alike in their essential constitution; i.e. either as both composed of little nuclei, considered in the abstract as matter" In your antennas and in the space are free electrons. We do not need a mystery aether. is not the same as that which occurs on the surface of disturbed water, or the waves of sound in gases or liquids, so all the comparisons to water waves or sound waves are NOT correct! You know that the dipole works like the two loudspeakers. equivalent to a lateral vibration NOT longitudinal! And not traversal. The waves radiated from the two ends of a dipole I call "coupled". so by these statements he has said that electromagnetic waves ARE NOT caused by an aether, and ARE lateral, NOT longitudinal... and CAN NOT be compared to sound or water vibrations! Faraday was great. In *the Nature no natural acoustic dipoles. For him was obvious that must be something equivalent to a lateral vibration. As you know it is the two or more sources of pressure waves properly phased. Are electrons jumping off from your transmitting antennas? In copper and in space are the same "little nuclei, considered in the abstract as matter". Electrons have mass. Biot-Savart's magnetic whirl is massles. S* no, electrons don't jump off my antennas, and there can not be free electrons in space or they would all repel each other and fly away. and since they have mass and other detectable properties we would easily be able to measure them if they were conducting electromagnetic waves, such a simple thing as speed of the wave would be VERY different than what we measure now. |
A small riddle, just for fun
On Feb 2, 12:42*am, "Antonio Vernucci" wrote:
Tom, the story began when, a few days ago, I was going to replace a trap of my HF yagi. Not to make mistakes, I consulted the antenna assembly manual where I found a big banner: do not invert traps otherwise the antenna will not work. So, I thought, this is a case in which a bipole cannot be inverted. This is clearly due to the fact that the external body of the trap (an aluminuim can about 2 feet long), which contains two coils resonated at different frequencies by means of built-in capacitors, is effectively part of the antenna radiating element. So, the trap is a bipole not only comprising lumped elements, and that is the reason why it cannot be inverted. So, as K1TTT has pointed out, a bipole can be inverted without consequences only if it has only 2 ports, has only passive linear components, and is small enough to be considered a lumped element. 73 Tony I0JX Rome, Italy As I see it, it has nothing to do with being small enough to be considered a lumped element. Instead, it has to do with coupling between the network elements and the outside world. Clearly your trap is coupled to the outside world. Clearly the series LC in my example is coupled to its surroundings. In such cases, you're NOT dealing with a two-terminal network: there is a path for current other than the two terminals. You can put as many transmission lines in your network as you wish, and as long as they don't have coupling to the rest of the universe except through the two closely-spaced terminals, there will be no difference in behaviour if you reverse the terminals. If you can show a valid counter-example, you've proven a whole lot of textbooks wrong... I suppose equivalently, if you decouple your measurement from the rest of the universe, you'll get the same result. Typical broadband directional couplers (of the sort used in S-parameter test sets) have good decoupling, to be able to separate excitation from return power. Cheers, Tom |
A small riddle, just for fun
In my first EE class, 'way back in the fall of 1962, the instructor
walked in and said "We shall study LLFPB", by which he meant Lumped, Linear, Finite, Passive, and BILATERAL so I'd have to conclude that your bipole is one or more of not lumped and/or not linear and/or not finite and/or not passive. Yesterday, while repairing my antenna, something came to my mind I had never focused on before. Let us consider a bipole, that is a "black box" having TWO terminals and including plain passive elements only (like capacitors, inductors, ... , no diodes or other special devices), arranged the way you prefer, it does not matter. Instead of directly telling which it is, just for fun I wonder whether anyone can figure out a case in which a bipole may not be reversed without consequences. Not difficult, but it anyway requires some thinking. -- -- Myron A. Calhoun. Five boxes preserve our freedoms: soap, ballot, witness, jury, and cartridge NRA Life Member & Certified Instructor for Rifle, Pistol, & Home Firearm Safety Also Certified Instructor for the Kansas Concealed-Carry Handgun (CCH) license |
A small riddle, just for fun
"K1TTT" napisal w wiadomosci ... On Feb 2, 7:53 pm, "Szczepan Bialek" wrote: In such words: "I suppose we may compare together the matter of the aether and ordinary matter (as, for instance, the copper of the wire through which the electricity is conducted), and consider them as alike in their essential constitution; i.e. either as both composed of little nuclei, considered in the abstract as matter" (Faraday). In your antennas and in the space are free electrons. We do not need a mystery aether. Are electrons jumping off from your transmitting antennas? In copper and in space are the same "little nuclei, considered in the abstract as matter". Electrons have mass. no, electrons don't jump off my antennas, Tesla did the electron gun from an antenna. Why yours are different? and there can not be free electrons in space or they would all repel each other and fly away. They do. The rare plasma is produced by the Sun. After condensation they come back to stars. and since they have mass and other detectable properties we would easily be able to measure them if they were conducting electromagnetic waves, The electron gun produces the catode rays. You are able to measure them. Put on a glas bottle with the anticatode on the top of CB radio antenna. Your current meter measures the DC ground current. such a simple thing as speed of the wave would be VERY different than what we measure now. Speed of the electric waves is and will be the same for ever. Of course for the same temperature. Faraday stated that must be: "that action which may be considered as equivalent to a lateral vibration; " The two (or more) sources give the same result like mystery TEM. Are you better than Faraday? S* |
A small riddle, just for fun
On Feb 3, 8:42*am, "Szczepan Bialek" wrote:
*"K1TTT" napisal w ... On Feb 2, 7:53 pm, "Szczepan Bialek" wrote: In such words: "I suppose we may compare together the matter of the aether and ordinary matter (as, for instance, the copper of the wire through which the electricity is conducted), and consider them as alike in their essential constitution; i.e. either as both composed of little nuclei, considered in the abstract as matter" (Faraday). In your antennas and in the space are free electrons. We do not need a mystery aether. Are electrons jumping off from your transmitting antennas? In copper and in space are the same "little nuclei, considered in the abstract as matter". Electrons have mass. no, electrons don't jump off my antennas, Tesla did the electron gun from an antenna. Why yours are different? and there can not be free electrons in space or they would all repel each other and fly away. They do. The rare plasma is produced by the Sun. After condensation they come back to stars. and since they have mass and other detectable properties we would easily be able to measure them if they were conducting electromagnetic waves, The electron gun produces the catode rays. You are able to measure them. Put on a glas bottle with the anticatode on the top of CB radio antenna. Your current meter measures the DC ground current. such a simple thing as speed of the wave would be VERY different than what we measure now. Speed of the electric waves is and will be the same for ever. Of course for the same temperature. Faraday stated that must be: "that action which *may be considered as equivalent to a lateral vibration; " The two (or more) sources give the same result like mystery TEM. Are you better than Faraday? S* no, i am better than you at reading faraday. |
A small riddle, just for fun
Uzytkownik "K1TTT" napisal w wiadomosci ... On Feb 3, 8:42 am, "Szczepan Bialek" wrote: Faraday stated that must be: "that action which may be considered as equivalent to a lateral vibration; " The two (or more) sources give the same result like mystery TEM. Are you better than Faraday? S* no, i am better than you at reading faraday. Do you need " the action which may be considered as equivalent to a lateral vibration; "? Or you are fine with TEM? S* |
A small riddle, just for fun
On Feb 4, 8:04*am, "Szczepan Bialek" wrote:
Uzytkownik "K1TTT" napisal w ... On Feb 3, 8:42 am, "Szczepan Bialek" wrote: Faraday stated that must be: "that action which may be considered as equivalent to a lateral vibration; " The two (or more) sources give the same result like mystery TEM. Are you better than Faraday? S* no, i am better than you at reading faraday. Do you need " the action which may be considered as *equivalent to a lateral vibration; "? Or you are fine with TEM? S* i am fine with the transverse waves described by maxwell's equations that don't require any aether. |
A small riddle, just for fun
Uzytkownik "K1TTT" napisal w wiadomosci ... On Feb 4, 8:04 am, "Szczepan Bialek" wrote: Faraday stated that must be: "that action which may be considered as equivalent to a lateral vibration; " The two (or more) sources give the same result like mystery TEM. Are you better than Faraday? S* no, i am better than you at reading faraday. Do you need " the action which may be considered as equivalent to a lateral vibration; "? Or you are fine with TEM? S* i am fine with the transverse waves described by maxwell's equations that don't require any aether. Maxwell equations are for Heaviside aether: http://www-groups.dcs.st-and.ac.uk/~...Heaviside.html "I saw that it was great, greater, and greatest, with prodigious possibilities in its power. I was determined to master the book... It took me several years before I could understand as much as I possible could. Then I set Maxwell aside and followed my own course. And I progressed much more quickly. Despite this hatred of rigour, Heaviside was able to greatly simplify Maxwell's 20 equations in 20 variables, replacing them by four equations in two variables. Today we call these 'Maxwell's equations' forgetting that they are in fact 'Heaviside's equations'." Heaviside equations have nothing common with Faraday and Maxwell. I prefer the Faraday idea - without any aether. S* |
A small riddle, just for fun
On Feb 4, 6:00*pm, "Szczepan Bialek" wrote:
Uzytkownik "K1TTT" napisal w ... On Feb 4, 8:04 am, "Szczepan Bialek" wrote: Faraday stated that must be: "that action which may be considered as equivalent to a lateral vibration; " The two (or more) sources give the same result like mystery TEM. Are you better than Faraday? S* no, i am better than you at reading faraday. Do you need " the action which may be considered as equivalent to a lateral vibration; "? Or you are fine with TEM? S* i am fine with the transverse waves described by maxwell's equations that don't require any aether. Maxwell equations are for Heaviside aether:http://www-groups.dcs.st-and.ac.uk/~...Heaviside.html "I saw that it was great, greater, and greatest, with prodigious possibilities in its power. I was determined to master the book... It took me several years before I could understand as much as I possible could. Then I set Maxwell aside and followed my own course. And I progressed much more quickly. Despite this hatred of rigour, Heaviside was able to greatly simplify Maxwell's 20 equations in 20 variables, replacing them by four equations in two variables. Today we call these 'Maxwell's equations' forgetting that they are in fact 'Heaviside's equations'." Heaviside equations have nothing common with Faraday and Maxwell. I prefer the Faraday idea - *without any aether. S* it all depends on who is writing history: "The four partial differential equations, now known as Maxwell's equations, first appeared in fully developed form in Electricity and Magnetism (1873). Most of this work was done by Maxwell at Glenlair during the period between holding his London post and his taking up the Cavendish chair. They are one of the great achievements of 19th- century mathematics." http://www-groups.dcs.st-and.ac.uk/~...s/Maxwell.html |
A small riddle, just for fun
On Fri, 4 Feb 2011 12:09:17 -0800 (PST), K1TTT wrote:
it all depends on who is writing history: "The four partial differential equations, now known as Maxwell's equations, first appeared in fully developed form in Electricity and Magnetism (1873). Most of this work was done by Maxwell at Glenlair during the period between holding his London post and his taking up the Cavendish chair. They are one of the great achievements of 19th- century mathematics." http://www-groups.dcs.st-and.ac.uk/~...s/Maxwell.html Yes, indeed it does matter who is "writing history." The quotation above is, in fact, quite wrong. "Maxwell's formulation of electromagnetism consisted of 20 equations in 20 variables. Heaviside employed the curl and divergence operators of the vector calculus to reformulate these 20 equations into four equations in four variables (B, E, J, and rho), the form by which they have been known ever since (see Maxwell's equations)." This "writing of history" above comes from: http://en.wikipedia.org/wiki/Heaviside where the embedded link to (see Maxwell's equations) is: http://en.wikipedia.org/wiki/Maxwell%27s_equations which then "RE-WRITES HISTORY" to state that the forms of the four equations (formulated by Heaviside) were original to Maxwell (whose work of 20 are ignored). This, of course, will have no effect on the Soviet-inspired revisionist polemic that un-informs this side thread from S* our amateur doctrinaire. * * * * * * As an experiment at translate.google.com, I entered that last paragraph above, translated it into Polish, and then took the Polish output and ran it through (in reverse as it were) to see what that looked like in English: "This, of course, will not affect the relationship inspired revisionist polemic that does not inform the thread-by-S * our doctrinaire enthusiasts." Curious how the "Soviet" was airbrushed out of translation. So through carefully crafting the statement to preserve its piquancy, I amended it to: This, of course, will have no effect on the Stalinist-inspired revisionist polemic that un-informs this side thread from S* our amateur doctrinaire. English-Polish-English renders: "This, of course, will have no impact on the Stalinist-inspired revisionist polemic that does not inform the thread-by-S * our doctrinaire enthusiasts." Aside from the last plural in place of singular, quite faithful to the intended irony. * * * * * * * I wasn't going to try to push Maxwell's 20 equations through translate.google.com to see if Heaviside emerged. 73's Richard Clark, KB7QHC |
A small riddle, just for fun
On Feb 4, 9:17*pm, Richard Clark wrote:
On Fri, 4 Feb 2011 12:09:17 -0800 (PST), K1TTT wrote: it all depends on who is writing history: "The four partial differential equations, now known as Maxwell's equations, first appeared in fully developed form in Electricity and Magnetism (1873). Most of this work was done by Maxwell at Glenlair during the period between holding his London post and his taking up the Cavendish chair. They are one of the great achievements of 19th- century mathematics." http://www-groups.dcs.st-and.ac.uk/~...s/Maxwell.html Yes, indeed it does matter who is "writing history." *The quotation above is, in fact, quite wrong. * * * * "Maxwell's formulation of electromagnetism consisted * * * * of 20 equations in 20 variables. Heaviside employed * * * * the curl and divergence operators of the vector * * * * calculus to reformulate these 20 equations into * * * * four equations in four variables (B, E, J, and rho), * * * * the form by which they have been known ever * * * * since (see Maxwell's equations)." This "writing of history" above comes from:http://en.wikipedia.org/wiki/Heaviside where the embedded link to *(see Maxwell's equations) is:http://en.wikipedia.org/wiki/Maxwell%27s_equations which then "RE-WRITES HISTORY" to state that the forms of the four equations (formulated by Heaviside) were original to Maxwell (whose work of 20 are ignored). This, of course, will have no effect on the Soviet-inspired revisionist polemic that un-informs this side thread from S* our amateur doctrinaire. * * * * * * As an experiment at translate.google.com, I entered that last paragraph above, translated it into Polish, and then took the Polish output and ran it through (in reverse as it were) to see what that looked like in English: "This, of course, will not affect the relationship inspired revisionist polemic that does not inform the thread-by-S * our doctrinaire enthusiasts." Curious how the "Soviet" was airbrushed out of translation. So through carefully crafting the statement to preserve its piquancy, I amended it to: This, of course, will have no effect on the Stalinist-inspired revisionist polemic that un-informs this side thread from S* our amateur doctrinaire. English-Polish-English renders: "This, of course, will have no impact on the Stalinist-inspired revisionist polemic that does not inform the thread-by-S * our doctrinaire enthusiasts." Aside from the last plural in place of singular, quite faithful to the intended irony. * * * * * * * I wasn't going to try to push Maxwell's 20 equations through translate.google.com to see if Heaviside emerged. 73's Richard Clark, KB7QHC i thought that rather than going to wikipedia or somewhere else it was more appropriate to quote from the maxwell bio on the same website mr.B used to get his quote. |
A small riddle, just for fun
On Fri, 4 Feb 2011 14:17:00 -0800 (PST), K1TTT wrote:
i thought that rather than going to wikipedia or somewhere else it was more appropriate to quote from the maxwell bio on the same website mr.B used to get his quote. This only works for a rational discussion. S* is only interested in pursuing agitprop, the failure of a decandent counter-cultural ideology. 73's Richard Clark, KB7QHC |
A small riddle, just for fun
Uzytkownik "Richard Clark" napisal w wiadomosci ... On Fri, 4 Feb 2011 14:17:00 -0800 (PST), K1TTT wrote: i thought that rather than going to wikipedia or somewhere else it was more appropriate to quote from the maxwell bio on the same website mr.B used to get his quote. This only works for a rational discussion. S* is only interested in pursuing agitprop, the failure of a decandent counter-cultural ideology. You wrote: "Yes, indeed it does matter who is "writing history." The quotation above is, in fact, quite wrong." So the best approach is to take a glance into the original papers. In Maxwell's model the magnetic lines are like the smoke rings. Nothing flow along them. In Heaviside's model there is the solenoidal flow. It is not simplification. The both models are quite different. But the both are a history. Your radio waves travel in rare plasma. It is interesting that Faraday predicted it. S* |
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