Hello Friends,
I have come by a paper that might interest some of you.
https://www.jstor.org/stable/25138615#metadata_info_tab_contents
Interesting paper MasterBlaster, thanks for sharing :)
This paper was written right around the time that long-range telegraph was beginning to be developed+implemented, and there was still much confusion and headaches with regards to maintaining consistent phase angle / solving dissipation problems. 'Impedance matching' was a thing that hadn't quite been developed yet (there wasn't even a 'henry' back then :o). So in this paper really gives you an idea what they were trying to map out and solve.
Oliver Heaviside wrote his Telegraph equation / transmission line formulas in 1876 just a few years prior to this paper (1880), but that era I suppose information took quite a while to spread and be tested/adopted among the rest of the community+world.
It is interesting from the historical point of view, not scientific. All the author's astonishments about delays and phases have long been perfectly explained.
If you treat magnetic wave propagation inside a ferromagnetic rod as a transmission line you find that the wave impedance is imaginary (imaginary in the math sense meaning reactive with no real component). Transmission line theory can handle this. When you do the math you find that such a line terminated in a capacitive reactance can offer an input impedance that has a negative resistance. That offers the possibility of a self sustaining resonance. So I think there is still the possibility for astonishment in this world.
Smudge
Quote from: Smudge on 2022.08.30, 14:09:22
If you treat magnetic wave propagation inside a ferromagnetic rod as a transmission line you find that the wave impedance is imaginary (imaginary in the math sense meaning reactive with no real component). Transmission line theory can handle this. When you do the math you find that such a line terminated in a capacitive reactance can offer an input impedance that has a negative resistance. That offers the possibility of a self sustaining resonance. So I think there is still the possibility for astonishment in this world.
Smudge
Hi Smudge,
This is very interesting idea, perhaps you could expand it a little more ?
Thanks,
Vasik
Edit: I see it now, managed to open the link :)
Quote from: Smudge on 2022.08.30, 14:09:22
If you treat magnetic wave propagation inside a ferromagnetic rod as a transmission line you find that the wave impedance is imaginary (imaginary in the math sense meaning reactive with no real component). Transmission line theory can handle this. When you do the math you find that such a line terminated in a capacitive reactance can offer an input impedance that has a negative resistance.
Smudge
That is elegant and insane; I love it ;D
However if this system were symmetric, would the capacitive reactance on the input cancel out the gains? I take it the induction would have to be magnetic and the 'load' capacitive?
Quote from: Hakasays on 2022.08.31, 02:22:15
However if this system were symmetric, would the capacitive reactance on the input cancel out the gains? I take it the induction would have to be magnetic and the 'load' capacitive?
But if you switch capacitor with proper phase ;)
PS I guess now everyone who interested should see how this works... and how similar it is to another device discussed here :)
Quote from: Hakasays on 2022.08.31, 02:22:15
That is elegant and insane; I love it ;D
However if this system were symmetric, would the capacitive reactance on the input cancel out the gains? I take it the induction would have to be magnetic and the 'load' capacitive?
Here is a paper on the magnetic delay line. The arguments about longitudinal waves on your benches might take note that in the air outside the ferromagnetic rods the wave is transverse but inside it is longitudinal. Same goes for the conductor in the twin wire transmission line, inside the conductor there is a longitudinal wave.
Smudge
A calculation made with the equations from the known laws of electromagnetism cannot result in a pure negative resistance, which would be equivalent to saying that energy comes out of nowhere, and the internal coherence of these laws forbids it. We can see that this is not the case, the impedance has a reactive component.
The impedance only makes sense for an established regime. During the establishment of the regime, the elements L and C of the circuit absorb energy as the electric wave progresses in the line, and this energy is then maintained, in the form of reactive energy, or standing wave when the impedance matching generator / line / load is not correct and the length of line is not negligible compared to the wavelengths of the signals.
The reactive component of the impedance absorbs energy and stores it, and the negative resistance allows to recover part of this energy. No free lunch.
Quote from: F6FLT on 2022.09.05, 08:21:41
A calculation made with the equations from the known laws of electromagnetism cannot result in a pure negative resistance, which would be equivalent to saying that energy comes out of nowhere, and the internal coherence of these laws forbids it. We can see that this is not the case, the impedance has a reactive component.
If we have a circuit with both positive and negative reactance (L and C) and negative resistance then it will self oscillate if that negative resistance exceeds any positive resistance in the circuit. The oscillations will build up exponentially. Do the math!!
QuoteThe impedance only makes sense for an established regime. During the establishment of the regime, the elements L and C of the circuit absorb energy as the electric wave progresses in the line, and this energy is then maintained, in the form of reactive energy, or standing wave when the impedance matching generator / line / load is not correct and the length of line is not negligible compared to the wavelengths of the signals.
The reactive store of energy is not maintained in a standing wave, it is continually completely lost and then restored at the cyclic rate (the applied frequency).
QuoteThe reactive component of the impedance absorbs energy and stores it, and the negative resistance allows to recover part of this energy.
Not true, the negative resistance (if it exists) doesn't recover part of that energy, it adds energy as it moves through the resistor from the L to the C or vice versa.
QuoteNo free lunch.
That is the standard view but we are here to challenge that. If you believe that standard science cannot be challenged, why are you here on this forum? Perhaps you could tell us what is wrong with the transmission line equations in my paper that clearly do predict a negative resistance.
Smudge
"Reaction Machines" in Charles Steinmetz' famous book Alternating Current Phenomena covers some of the concept and mathematics as well:
http://www.tuks.nl/pdf/Reference_Material/Steinmetz/Reaction%20Machines%20chapter%20in%20Alternating%20Current%20Phenomona.pdf
@Smudge
A standing wave only attenuates if there are losses in the line.
Now you talk about negative reactance but there is none in your paper, only negative resistance.
Since negative resistance produces energy, if it does not come from the reactive elements, you will have to explain where it comes from, and why your calculation from the equations of standard physics leads to a result contrary to standard physics, which implies an internal inconsistency of the equations of standard physics, which you still have to specify.
If what you said was correct, then the line would produce power even before a generator was connected. And the millions of engineers and technicians who worked on the lines never noticed this, what fools!
Let's be clear: I think that standard science is not complete and that it can be challenged or completed, which is partly why I am here.
Conversely, if you don't consider it solid, why do you use its equations?
First, the coherence of its formalism forbids the creation of energy in a closed system. So a calculation based on the equations of standard physics that leads to a creation of energy ex-nihilo, negative resistance or not, is wrong.
Secundly, to challenge it with simple devices as well known as a line of which all kinds have been manipulated for decades by people in the business, is to believe oneself smarter than them. It is a right to believe so, and it may be true, but then it will be necessary to provide the experimental demonstration.
Quote from: F6FLT on 2022.09.05, 16:25:52
@Smudge
A standing wave only attenuates if there are losses in the line.
Now you talk about negative reactance but there is none in your paper, only negative resistance.
Perhaps I did not make myself clear. In the imaginary plane (multiplied by j or i, sqrt-1) this can be positive or negative, I thought this is well known, hence my reference to L or C reactances.
QuoteSince negative resistance produces energy, if it does not come from the reactive elements, you will have to explain where it comes from, and why your calculation from the equations of standard physics leads to a result contrary to standard physics, which implies an internal inconsistency of the equations of standard physics, which you still have to specify.
If what you said was correct, then the line would produce power even before a generator was connected. And the millions of engineers and technicians who worked on the lines never noticed this, what fools!
How many engineers have worked on lines that have a characteristic impedance that is predominantly imaginary reactive? In my long experience (over 70 years) the impedance of lines has always been real resistive.
QuoteLet's be clear: I think that standard science is not complete and that it can be challenged or completed, which is partly why I am here.
Conversely, if you don't consider it solid, why do you use its equations?
Because in this case it deals with imaginary reactive lines that to my knowledge have not been experimented with.
QuoteFirst, the coherence of its formalism forbids the creation of energy in a closed system. So a calculation based on the equations of standard physics that leads to a creation of energy ex-nihilo, negative resistance or not, is wrong.
This reactive transmission of energy cannot occur in free space. Thus we are dealing with transmission through some material that now involves atoms, the particles that are the atoms and of course the strange quantum world there. So I see the possibility of energy being transduced from that world. Obviously you don't agree.
QuoteSecundly, to challenge it with simple devices as well known as a line of which all kinds have been manipulated for decades by people in the business, is to believe oneself smarter than them. It is a right to believe so, and it may be true, but then it will be necessary to provide the experimental demonstration.
And I did so in my magnetic delay transformer thread, but without success. No reason to give up though.
Smudge
Quote from: Smudge on 2022.09.05, 18:49:32
...
How many engineers have worked on lines that have a characteristic impedance that is predominantly imaginary reactive?
...
The fact that it is a line does not change the principle. A line is a network of L and C elements, it can even be modeled with them, and no passive network provides energy. When a line is not terminated by a resistive load of the same impedance, the impedance seen by the generator is reactive. All engineers have had to deal with open or short-circuited lines, or lines terminated by a reactive load, without ever noticing any anomaly.
Moreover, when the load or generator is not matched, the reactive impedance will depend on the frequency, because of the transmission delays independent of the signal period. The voltage and current nodes along the line do not fall in the same places depending on the frequency. I don't see why this wouldn't be the case with a natively reactive line, and yet your impedance is unique. You should start from the equations of physics, not from those of engineering, which correspond to particular cases.
In your paper you refer to prof. Turtur. Although he is an academic, the only difference between Turtur and a proponent of perpetual motion is that he arbitrarily attributes the cause of perpetual motion to the ZPE. But his engine has never worked with the ZPE. In a vacuum it doesn't work anymore. Turtur is not a serious reference. When what you claim is extraordinary, the experimental demo must be unambiguous.
If you don't experiment with the line yourself, I think you'll have a hard time convincing someone competent in the field to do it.
Quote from: F6FLT on 2022.09.05, 21:32:30
The fact that it is a line does not change the principle. A line is a network of L and C elements, it can even be modeled with them, and no passive network provides energy.
Ferromagnetic transmission lines can be more difficult to model since L changes dynamically with current.
Quote from: Hakasays on 2022.09.06, 04:42:06
Ferromagnetic transmission lines can be more difficult to model since L changes dynamically with current.
Electromagnetic software, such as CST studio, take this into account, you just have to provide them with the B/H curve.
Quote from: F6FLT on 2022.09.05, 21:32:30
The fact that it is a line does not change the principle. A line is a network of L and C elements, it can even be modeled with them, and no passive network provides energy.
There was a time when I would have agreed with you. But that was before I taught myself to solve in the magnetic domain where flux is treated like current (or magnetic displacement current) and mmf is treated like voltage. Most people in this world do not take that beyond magnetic Ohm's Law, dealing with variations of materials or dimensions around a closed magnetic circuit, like air gaps. Few can solve dynamically. Taking the twin wire transmission line shown in my paper it has distributed inductance and capacitance along its length, hence your L and C elements apply. Now take my twin ferrite rod line, what is the distributed L and C there? You can't answer that. It can appear as a ladder network of reluctances, with distributed series reluctance and distributed shunt permeance along the line. That does not tell you the velocity which as you know can depend on many factors like domain wall velocity. And it does not tell you what the effect of those electric field lines are. I can tell you that the dielectric surrounding the ferrite rods will carry displacement current, and that will result in the series reluctance chain having a magnetic "component" in series that behaves nothing like electrical ones. I have used D as the symbol for this component, and it obeys mmf=-D*d
2(flux)/dt
2, the second time differential of the flux. That too has an influence on the propagation velocity. So please don't tell me this line is a network of L and C elements.
QuoteWhen a line is not terminated by a resistive load of the same impedance, the impedance seen by the generator is reactive. All engineers have had to deal with open or short-circuited lines, or lines terminated by a reactive load, without ever noticing any anomaly.
Moreover, when the load or generator is not matched, the reactive impedance will depend on the frequency, because of the transmission delays independent of the signal period. The voltage and current nodes along the line do not fall in the same places depending on the frequency.
I don't need you to teach me classical line stuff.
QuoteI don't see why this wouldn't be the case with a natively reactive line, and yet your impedance is unique.
Yes, unique as I mention above.
QuoteYou should start from the equations of physics, not from those of engineering, which correspond to particular cases.
Oh, I thought I was using equations of physics that covered all cases and not the particular L and C ones you have in mind.
QuoteIf you don't experiment with the line yourself, I think you'll have a hard time convincing someone competent in the field to do it.
Sadly my experimental days are over as I draw ever closer to end-of-life. I am trying hard to convince others but I keep getting negative reactions from certain members of this forum.
Smudge
Quote from: Smudge on 2022.09.06, 08:47:29
There was a time when I would have agreed with you. But that was before I taught myself to solve in the magnetic domain where flux is treated like current (or magnetic displacement current)
Unlike current, which is the flow of charges, a magnetic flux is not the flow of any substance. There is no displacement along the magnetic circuit. The field lines are always looped and expand or contract around the conductor according to the current flowing through it.
I know you know this. While engineering simplifications by treating the flow as a current are sometimes convenient, they too often lead to false reasoning in physics, such as the idea of Bearden's MEG and the "magnetic transistor" in general.
In my opinion the problem should be treated by physics, not by engineering. By characterizing your line by the impedance, you are in fact considering it a priori as a simple RLC network, by definition of what an impedance is.
QuoteI can tell you that the dielectric surrounding the ferrite rods will carry displacement current, and that will result in the series reluctance chain having a magnetic "component" in series that behaves nothing like electrical ones. I have used D as the symbol for this component, and it obeys mmf=-D*d2(flux)/dt2, the second time differential of the flux. That too has an influence on the propagation velocity. So please don't tell me this line is a network of L and C elements.
A line is a quadrupole. The input is a dipole. The impedance of a dipole is expressed as Z = R + j X. When X is positive, the impedance is inductive. When X is negative, the impedance is capacitive. So if your line is something else, it is not by the impedance that you can characterize its specificities.
QuoteYes, unique as I mention above.
Either you place yourself in the case of a quasi-stationary regime, and then the impedance doesn't depend on the frequency but talking about "line" doesn't make sense anymore, it's a simple network, or you place yourself in the usual framework of use of lines, where the length is greater than the wavelength of the signals, so there is propagation. Then the impedance does not depend on the frequency, but only if the load matches the line impedance. In other cases, so in the general case, it depends on it, and even drastically if the line is for example a quarter wave or half wave.
QuoteSadly my experimental days are over as I draw ever closer to end-of-life. I am trying hard to convince others but I keep getting negative reactions from certain members of this forum.
Not from me anyway, when I criticize it's because I see weak points and I don't agree, I say why, it's not to denigrate.
I generally like your ideas and formalizing them as you do with rigor puts you far above the nonsense we usually see in the FE. If a good idea can appear, it's from works like yours.
But I don't see all your work as being of equal value. This one on lines is not based on physics but on engineering formulas that apply to ordinary lines. Moreover it is contrary to the general principle that, except by integrating an exotic energy source, no overunity can come from the equations of classical physics whose coherence forbids the creation of energy ex nihilo.
Your work on the Magnetic Delay Lines seems therefore to me to be questionable, but this does not invalidate my consideration for your work in general, especially on the potential vector.
As I see your brain working well :), I thought maybe the arms too and you could still experiment. But I understand that with age it can become difficult or painful, I feel sorry for you and understand perfectly that you stay in the realm of ideas.
Propagation of Magnetization of Iron as affected hy the Electric Currents in the Iron
This is a rare find but worthy of posting:
Quote from: MasterBlaster on 2022.09.07, 10:06:05
Propagation of Magnetization of Iron as affected hy the Electric Currents in the Iron
...
This is a factor, but not the only cause. The magnetization of an insulating ferrite also has a propagation delay.
The maximum velocity of an electric or magnetic wave in a medium of permittivity ε and permeability µ is v = 1 /√(ε.µ) .
Quote from: F6FLT on 2022.09.06, 13:06:26
Unlike current, which is the flow of charges, a magnetic flux is not the flow of any substance. There is no displacement along the magnetic circuit. The field lines are always looped and expand or contract around the conductor according to the current flowing through it.
I know you know this. While engineering simplifications by treating the flow as a current are sometimes convenient, they too often lead to false reasoning in physics, such as the idea of Bearden's MEG and the "magnetic transistor" in general.
In my opinion the problem should be treated by physics, not by engineering.
I make no apology for treating closed magnetic circuits in a manner identical to electrical ones as the math has the same formalism. There is no false reasoning here.
QuoteBy characterizing your line by the impedance, you are in fact considering it a priori as a simple RLC network, by definition of what an impedance is.
And a simple RLC network can show anomalous behavior if R is negative.
QuoteA line is a quadrupole. The input is a dipole. The impedance of a dipole is expressed as Z = R + j X. When X is positive, the impedance is inductive. When X is negative, the impedance is capacitive.
Clearly what you consider to be a dipole or a quadrupole is not what I am taught. If you consider an inductor to be a magnetic dipole and a capacitor to be an electric dipole then by mixing the two types of pole then perhaps that could be some sort of quadrupole representation for a transmission line, but mixed poles do not a true quadrupole make.
QuoteEither you place yourself in the case of a quasi-stationary regime, and then the impedance doesn't depend on the frequency but talking about "line" doesn't make sense anymore, it's a simple network, or you place yourself in the usual framework of use of lines, where the length is greater than the wavelength of the signals, so there is propagation. Then the impedance does not depend on the frequency, but only if the load matches the line impedance. In other cases, so in the general case, it depends on it, and even drastically if the line is for example a quarter wave or half wave.
I am not sure what you are trying to convey here.
Smudge
Quote from: Smudge on 2022.09.16, 14:42:29
I make no apology for treating closed magnetic circuits in a manner identical to electrical ones as the math has the same formalism. There is no false reasoning here. And a simple RLC network can show anomalous behavior if R is negative.
I did not say it is wrong, I even said "engineering simplifications by treating the flux as a current are sometimes convenient".
So no false reasoning on your part if we stick to engineering calculations as you do, but the idea taken up by others who see the flux as a current leads to misunderstandings, such as the belief that we can modulate the flux as a current, which led to the MEG dead end.
QuoteAnd a simple RLC network can show anomalous behavior if R is negative.
If R is negative, a reasoning strictly limited to RLC network engineering equations, see extra energy. The problem is that engineering equations derive from physics, whose formalism of all theories implies conservation of energy.
Finding R to be negative demonstrates an error somewhere in the use of the formulas. This is why I say that one must go through the basic equations of physics, either to find the error, or to specify the exact source of the extra-energy if one really believes that the equations could show it, which I doubt very much given the internal consistency of the theories of physics.
Quote
Clearly what you consider to be a dipole or a quadrupole is not what I am taught.
A quadrupole is a very simple device with 2 inputs and 2 outputs. The currents and voltages going in and out are enough to characterize it entirely. That's what I'm talking about. The French Wikipedia explains it well, here is the translation:
https://fr-m-wikipedia-org.translate.goog/wiki/Quadrip%C3%B4le?_x_tr_sl=fr&_x_tr_tl=en
It's with a quadrupole that we represent a line, so I don't see why yours couldn't be.
Quote from: F6FLT on 2022.09.16, 21:31:14
A quadrupole is a very simple device with 2 inputs and 2 outputs. The currents and voltages going in and out are enough to characterize it entirely. That's what I'm talking about. The French Wikipedia explains it well, here is the translation:
https://fr-m-wikipedia-org.translate.goog/wiki/Quadrip%C3%B4le?_x_tr_sl=fr&_x_tr_tl=en
OK, I understand what you mean by dipole and quadrupole. I go back a long way and my view of a pole was a point object, in magnetics a N or S pole and in electrics a positive or negative pole. I now see that modern science treats circuit nodes as poles. Going back to your original statement
QuoteA line is a quadrupole. The input is a dipole. The impedance of a dipole is expressed as Z = R + j X. When X is positive, the impedance is inductive. When X is negative, the impedance is capacitive.
That is the impedance expressed entirely within the frequency domain and shows how the impedance (ratio of voltage to current) varies with frequency. It does not show how the impedance varies with time and since time delay along the line is a crucial factor that math is not enough to tell you all you need to know. Yes you can use Fourier transforms to convert from one domain to the other but even that can hide what is really going on.
QuoteSo if your line is something else, it is not by the impedance that you can characterize its specificities.
QuoteIt's with a quadrupole that we represent a line, so I don't see why yours couldn't be.
Your quadrupole transmission line of series inductors and shunt capacitors carries a longitudinal wave of current along the inductors. (Perhaps the arguments going on elsewhere on this forum about longitudinal waves should take note that such waves really do exist.) My transmission line of a series ferromagnetic rod carries a longitudinal wave of flux along the rod. The time varying A field around the rod creating an E field in the dielectric there hence driving displacement current around the rod that in turn induces flux into the rod affects the propagation and can be modeled as a shunt effect. Thus I can create a quadrupole representation for the transmission line but the impedances are NOT your R + jX. IMO transmission along the rod is not modeled by the classical distributed LC transmission line, and I still think there are possibilities to explore here.
Smudge
Quote from: Smudge on 2022.09.20, 10:37:22
...
Yes you can use Fourier transforms to convert from one domain to the other but even that can hide what is really going on.
The frequency and time domains are mathematically equivalent representations, therefore one can use either one, it is not a problem of physics but of interpretation which can be conceptually more or less clear depending on the choice and what one is trying to see.
Quote
I can create a quadrupole representation for the transmission line but the impedances are NOT your R + jX. IMO transmission along the rod is not modeled by the classical distributed LC transmission line, and I still think there are possibilities to explore here.
An impedance is always of the form R+jX, by definition.
And it is from the formula of the impedance of a line that you have developed your paper, to tell us now that it cannot be modelled classically by an LC distribution. Your conclusion is therefore in opposition to the premises. And you don't see any inconsistency in this process?
If a line terminated by a reactive load, a case that occurs all the time in the RF domain, created energy, it would have already been seen. On the other hand, your negative resistance is only negative at a given frequency, the one validating your β parameter. But your negative resistance would produce direct current. And this without even using a signal of the frequency where the resistance is negative! And you don't see any inconsistency?
I'm sorry, but all this seems so implausible that I wouldn't even want to spend time looking for a formal error.
Quote from: F6FLT on 2022.09.21, 11:12:09
An impedance is always of the form R+jX, by definition.
And it is from the formula of the impedance of a line that you have developed your paper, to tell us now that it cannot be modelled classically by an LC distribution. Your conclusion is therefore in opposition to the premises. And you don't see any inconsistency in this process?
If a line terminated by a reactive load, a case that occurs all the time in the RF domain, created energy, it would have already been seen.
It seems you have not quite followed the argument. The classical line that has been extensively examined in the RF domain as you say does not have a reactive Z
0. Such lines would not exhibit negative input resistance under any conditions. Such lines terminated in their characteristic Z
0 have E and H fields that are in phase. The magnetic delay line is different, its E and H fields are at 90 degree phase, the Z
0 is reactive. This I have stated clearly in my paper. And the standing waves brought about by the capacitive termination could lead to anomalous effects since we are dealing with spin waves within the material.
I have a lot of data on measurements taken on a large ferrous ring core with diametrically emplaced small windings so that there is a magnetic delay between input and output. This exhibits the expected resonant peak at the resonant frequency of the terminating capacitor and the inductance of the output coil, but it also exhibits another peak at a higher frequency. This peak coincides exactly with the frequency where the math tells us the input resistance should go negative, but sadly not quite OU. The fact that that non-resonant peak was even there tells me something unusual is going on. Since the time delay is not classical sqrt(LC) but is domain wall movements I am still of the opinion that it is worthy of more examination.
QuoteOn the other hand, your negative resistance is only negative at a given frequency, the one validating your β parameter. But your negative resistance would produce direct current. And this without even using a signal of the frequency where the resistance is negative!
I think you should review your thinking, that is just nonsense. If it can only occur at a given frequency it cannot occur at DC.
QuoteI'm sorry, but all this seems so implausible that I wouldn't even want to spend time looking for a formal error.
There you go again, by your thinking there must be an error. Why don't you open your mind to things being possible, that current science may have overlooked these possibilities?
Smudge
@Smudge
I still don't see any sense in what you are saying. You are comparing a conventional line terminated by a load equal to its characteristic impedance, with a line of your design, which is not.
But a conventional line, terminated by a reactive load, also generates standing waves and also has an input impedance that can be equal to anything, depending on the impedance of the terminal load and the frequency, exactly like your line.
Then a negative resistor is not an LC circuit with negative L or C, which could generate an AC signal. A negative resistor produces only DC. This is inconsistent with the fact that the negative resistance seen at the input would depend on the frequency. On the frequency of what? Of a signal that we don't even have to generate?
Finally, multiple resonances are commonplace in classical lines when they are loaded by a reactive load. So it is not a specificity of your line. You seem to be unaware of how classical lines without a suitable load work and you attribute new effects to your line when, apart from negative R, it is commonplace in all lines.
Quote from: F6FLT on 2022.09.23, 17:01:40You seem to be unaware of how classical lines without a suitable load work and you attribute new effects to your line when, apart from negative R, it is commonplace in all lines.
I left school at age 16 and went straight into work as a Scientific Assistance at a research establishment. My entire working life of 49 years was in electronics and electromagnetics during which I schooled myself on General Relativity in order to understand some of the weirder aspects of EM. I rose to the giddy heights of Chief Systems Engineer in the division I worked for. So I guess my knowledge of how classical lines work is at least as good as yours and possible better. I was deeply involved in near-field radar as proximity detectors for shells, bombs, missiles, torpedoes and mines, so I understand the difference between far-field (where the E and H vectors are in space quadrature but in time phase) and near-field (where the E and H vectors ore not in phase). The far-field has the so-called wave impedance of 377 Ohms and that is a real resistance, not a reactance. The near-field has an impedance that varies with distance from the transmitting antenna; for the classical small electric dipole the impedance rises as you get nearer to the antenna and for the small magnetic antenna it falls (small meaning antenna dimensions smaller then a wavelength). In both cases the resistance value changes while the reactance value rises so that close to the antenna the wave impedance is either capacitive reactance of inductive reactance. The classical delay line that you and most people are familiar with has a characteristic impedance that is dominantly resistive. I can find no reference to anyone with a knowledge and experience of a delay line that has a reactive impedance. I am sure that you have no such experience.
Transmission line theory readily accepts a reactive line impedance and when you look into this it predicts a negative input resistance when the line is terminated with a capacitance. Now you might argue that such a line with its mathematically imaginary impedance is actually imaginary, it doesn't exist in real life, it is only in one's mind. You have not presented that argument, you have continually claimed that any line must be modelled by actual L, C and R components and these cannot result in a negative resistance. I have argued that magnetic field propagation along a core or a pair of cores IS a transmission line having reactive line impedance, and that is something that has been overlooked in science. It can NOT be modelled by the usual LC network. With my self-taught methods of solving problems in the magnetic domain I have perhaps muddied the waters by using an LC network where the Ls and Cs are magnetic ones relating flux to mmf in the same manner that actual Ls and Cs relate current to voltage. If that has created confusion I apologise.
Wave propagation along a core has been looked at and it is known as magnetic viscosity. Its usual effect is to create losses, and this is easily modelled by introducing a time delay between input and output that converts the linear BH characteristic into a hysteresis loop that is traversed CCW. A CCW loop represents a loss, and it is assumed that this loss must go as heat in the core. I can find no evidence of calorific measurements that verify this input loss is actually converted into heat. I can hear you challenging this saying where can the energy go except as radiation or heat? We know that the electron dipoles are very active, they precess at the Larmor rate, free electrons are whizzing about, orbital electrons are hopping into and out of their stationary orbits, there is a lot of activity going on. If we can somehow connect into that activity, is it not possible that we can both source and sink energy there. If the activity really is a connection to an active aether why can't that supply or sink energy?
Within the core material magnetic wave propagation is very complex with electron dipoles obeying quantum rules involving their orbital or inherent spin, dipoles flipping, domain walls moving etc etc. Barkhausen jumps are a known effect; before the jump, as an externally applied field increases value the total field obeys dB/dH = μ
0, there is no relative permeability until a dipole flips, then there is the field jump to an increased value. What happens if the H field is pulsed yielding a dH over a time dt that takes B to the point of initiation of a Barkhausen jump, then the dH is suddenly reversed, do we get a jump in B while H returns to its original value? Of course in normal transformers these many jumps are not seen, the material behaves as though the accepted permeability rules apply dB/dH = μ
R μ
0. But within the material there could be small discrete volumes where B changes by μ
0dH in synch with H, then the jump by μ
Rμ
0dH occurs. Who knows what happens when there are H waves traveling both ways within the core influencing those small regions, what is the overall effect? Has this been studied? Shouldn't we be studying these possibilities? If transmission line theory predicts an anomalous effect for a line with reactive impedance, and our only known line of this type is a transformer core operating in an unusual manner, and that core's characteristics come from active elements within the core, isn't there a possibility, however slight, that we could make it behave in an active manner. I certainly believe so.
Smudge
Edit, changed CW to CCW to correct an error brought about by a senior moment.
@Smudge
You go into digressions. I don't doubt your technical skills regarding engineering, but I don't doubt your way of embroidering around it either. The fact that you have experience in a field where I too have experience without ever having noticed the slightest anomaly, and where, I suppose, no OU of any kind has ever been noticed nor used by the companies you worked for, is therefore of little relevance when it comes to talking about new phenomena. More than ever, the argument of authority does not hold.
A line being made of real resistances, real inductances and real capacities, it can be modelled by these elements. The difference with a simple network is the propagation time which plays a role when it is not negligible compared to the period of the signals, which is the general case of the use of lines. For this reason, the input impedance of a line depends on the frequency. This is the same problem we have in designing broadband antennas to ensure a constant impedance throughout its frequency range.
The only case where the input impedance of a line does not depend on frequency is when the resistive terminal load is equal to its characteristic impedance, which is not the case here since the terminal load is assumed to be reactive.
So there is no sense in talking about negative input resistance if you don't specify at what frequency. I think I found the question of frequency in your text, you see I looked for it and I am of good will, it is according to me in the phase constant β and thus θ too: the length of line θ=β*x being expressed in radians, it is relative to the period of the signal.
Now according to the line impedance given in your equation (5), R is not necessarily negative when X0>Xload, but only when tan(θ) * (X0-Xload) is negative because it also depends on tan(θ), so at certain frequencies relative to the line length. Is it possible to choose Xload such that tan(θ) * (X0-Xload) < 0? Of course, since Xload can be chosen arbitrarily, so the error is not there.
So where does the error come from?
It is in the sentence "We are interested in the case where Z0 is purely reactive (say jX0)". In the absence of losses, a characteristic impedance is always resistive since it depends on the ratio of the impedance of an inductor to that of a capacitor (or the reverse if C and L are swapped), so omega disappears. With losses (R series in L and R // in C) the characteristic impedance can be reactive, therefore frequency dependent, but never "purely reactive" and it will always remain to be demonstrated that R<0 at the input will not be cancelled out by this resistive part of the characteristic impedance, due to losses.
Concerning the physics part of your development, I have no opinion for the moment, it is a lot of ideas to see but without experimental proposal, difficult to know if the tracks are valid.
Quote from: F6FLT on 2022.09.25, 11:25:37
..........So where does the error come from?
It is in the sentence "We are interested in the case where Z0 is purely reactive (say jX0)". In the absence of losses, a characteristic impedance is always resistive since it depends on the ratio of the impedance of an inductor to that of a capacitor (or the reverse if C and L are swapped), so omega disappears. With losses (R series in L and R // in C) the characteristic impedance can be reactive, therefore frequency dependent, but never "purely reactive" and it will always remain to be demonstrated that R<0 at the input will not be cancelled out by this resistive part of the characteristic impedance, due to losses.
There you go again, you are fixated on the magnetic transmission line acting like a classical transmission line where the characteristic impedance "depends on the ratio of the impedance of an inductor to that of a capacitor". Were it so I would agree with your statements.
The magnetic line is not classical. When you compare the twin wire line and its distributed L and C with the twin ferrous rod line there are significant differences. Both have E and H fields in the space outside the "conductors" but is I show they are swapped over. And as I stated the phase between the E and H fields is different. In the twin wire case not only is Z
0 related to the ratio of L and C but also the propagation velocity is related to the product of L and C, and these also apply to the transverse E.H wave travelling along the line with E and H in phase as in far field radiation. In the twin ferrous rod case the propagation velocity is primarily determined by the "conductor" properties, its magnetic viscosity. And with the E and H fields in phase quadrature the wave is like near-field radiation. In the twin wire case the E x H Poynting vector carries the energy along the line. In the twin rod line with E and H in phase quadrature what does the Poynting vector do? It is certainly different. I am sorry to labor this point but that 90-degree phase between E and H tells me the line has a reactive Z
0, it is not a classical transmission line.
QuoteConcerning the physics part of your development, I have no opinion for the moment, it is a lot of ideas to see but without experimental proposal, difficult to know if the tracks are valid.
But surely worth investigating just in case it leads somewhere. Below is a short paper showing the positive reluctance of a core as the slope of flux v. mmf, where introducing a time delay between the application of mmf and the flux creates a CCW loop as an energy loss. Also shown is the negative reluctance induced into the magnetic circuit by a capacitively loaded coil. Of interest is a time delay between the mmf (current) of that coil and flux which creates a CW loop, indicating an energy gain. If the area of the CW loop could be made larger than the area of the CCW loop then there would be an overall gain. Since the magnetic viscosity plays its part in both loops then this possibility may seem improbable. But the magnetic wave propagation can also be influenced by surrounding the ferrous rods with high K dielectric, which increases the magnetic delay. A series of coils along the rods each one shunted by a capacitor also increases delay.
This argument may also play into the Holcomb claims. They may have got their arguments wrong with regard to how electron spin accounts for their results, but if they really do have positive results the energy comes from somewhere, and it seems the magnetic steel is the important part. The Manelos and Sweet devices also obtain their gains from the ferrous material.
Smudge
Quote from: Smudge on 2022.09.25, 15:44:51
...
The magnetic line is not classical. When you compare the twin wire line and its distributed L and C with the twin ferrous rod line there are significant differences.
...
I remind you of what you wrote: "The impedance Zin looking into a line of characteristic impedance Z0, of length x(m) and terminated in an impedance Zload is given by ..."
All your development is based on the classical equation of the impedance of a line. This leads to inconsistencies that I have pointed out. You only answer them by claiming that it is not a classical line. Then the use of your starting equation is not justified, which invalidates your conclusions of negative R.
Quote from: F6FLT on 2022.09.25, 16:07:48
I remind you of what you wrote: "The impedance Zin looking into a line of characteristic impedance Z0, of length x(m) and terminated in an impedance Zload is given by ..."
All your development is based on the classical equation of the impedance of a line. This leads to inconsistencies that I have pointed out. You only answer them by claiming that it is not a classical line. Then the use of your starting equation is not justified, which invalidates your conclusions of negative R.
You have made the assumption that the classical transmission-line equations I have used can only apply to classical lines. That is not the case, they are much broader than that. You and I and the rest of the world have extensive experience of classical transmission lines where the characteristic impedance is non frequency dependent and is resistive. The line equations of course cover those classical lines,
but they also cover non-classical lines where the characteristic impedance is non frequency dependent and is reactive. I suspect that you and the rest of the world do not knowingly have experience of such non-classical lines and would find it hard to describe a line having that characteristic. Your criticism is unfounded.
Smudge
@Smudge
The only thing I'm saying is that if you use the classical line equation to describe your own, and it results in inconsistencies, then your choice to use it is not justified.
But you don't answer the objections.
The main inconsistencies are :
Negative R only appears in the equations at certain frequencies, but negative R does not allow the generation of an AC current.
The equations of electromagnetism, including those of lines, guarantee the conservation of energy.
Quote from: F6FLT on 2022.09.26, 09:44:14
@Smudge
The only thing I'm saying is that if you use the classical line equation to describe your own, and it results in inconsistencies, then your choice to use it is not justified.
Why not justified? Because of the inconsistencies? Shouldn't we be looking into what might be causing those inconsistencies?
QuoteBut you don't answer the objections.
I have even offered an alternative method of looking into the effect of the inconsistency.
QuoteThe main inconsistencies are :
Negative R only appears in the equations at certain frequencies, but negative R does not allow the generation of an AC current.
Oh really, have you tried solving the time evolution of current in a L C series negative R circuit? You would find it instructive.
QuoteThe equations of electromagnetism, including those of lines, guarantee the conservation of energy.
So you keep saying, and if you believe that there is no way of going beyond those equations why are you on this forum?
Smudge
Quote from: Smudge on 2022.09.26, 10:34:57
...
Oh really, have you tried solving the time evolution of current in a L C series negative R circuit? You would find it instructive.
I would be curious to see your equations that would show the maintenance of an LC oscillation with only negative R!
No more need for a transmitter, no more need for electronics, because R negative being like a DC generator with R positive, you would just have to connect an antenna via an inductance and a capacitor, to the car battery...
Quote
So you keep saying, and if you believe that there is no way of going beyond those equations why are you on this forum?
Seriously, do you believe that in the equations that guarantee the conservation of energy you will find a non-conservation?
You think that by staying in your country you will see what is beyond?
I remind you that science models what we observe. The models guarantee the conservation of energy of a closed system because we have never observed anything else. No matter how you twist the models, since they have an internal mathematical consistency, if you see OU in them, you can be sure that the error is yours.
In case of real OU, you will either have to use new models, or you will have to integrate the discovered energy source into the old models.
The OU will be in the discovery, not in the known equations. That's why I'm here. The known equations are a starting point, to be completed, they do not contain their own beyond.
Quote from: F6FLT on 2022.09.26, 11:58:32
I would be curious to see your equations that would show the maintenance of an LC oscillation with only negative R!
Not maintenance of an LC oscillation but an exponential build-up of the LC oscillation (following an e
+at rise in the envelope) rising to infinity. If there is positive R there as well the oscillations build up to a fixed level. Is done in any RF oscillator where the negative R is created by positive feedback but the build up is often not of interest. It is of interest in super-regenerative amplifiers where the oscillations are built up then allowed to decay and this is repeated at the quenching frequency.
QuoteSeriously, do you believe that in the equations that guarantee the conservation of energy you will find a non-conservation?
Depends on what equations you are talking about. In the case of the transmission line equations, if they apply to non-classical lines that involve propagation that is not related to EM wave equations but to other dynamics within materials such as atomic dipole flips, quantum rules, fermi velocity of conduction electrons etc. etc., then yes I think there is a possibility that there is a connection to an active aether.
QuoteI remind you that science models what we observe. The models guarantee the conservation of energy of a closed system because we have never observed anything else.
If an active aether is the source of the energy then our experiment is not a closed system. But of course you don't believe in an active aether.
QuoteIn case of real OU, you will either have to use new models, or you will have to integrate the discovered energy source into the old models.
The OU will be in the discovery, not in the known equations. That's why I'm here. The known equations are a starting point, to be completed, they do not contain their own beyond.
I could not agree more. Why then do you pour cold water on any attempt to discover an energy source?
Smudge
Quote from: Smudge on 2022.09.26, 15:36:11
Not maintenance of an LC oscillation but an exponential build-up of the LC oscillation (following an e+at rise in the envelope) rising to infinity.
This is a triviality, like applying a voltage swing to an LC circuit. But this is not the case for the line, where the supposedly negative resistance does not appear instantaneously.
Quote
If an active aether is the source of the energy then our experiment is not a closed system. But of course you don't believe in an active aether.
Your OU, you don't get it from the ether but from a conventional equation that makes no assumption of ether.
Digressions and non sequitur, it becomes nonsense because it has no logical relation with your initial paper, the one I am contesting.
Quote from: F6FLT on 2022.09.27, 08:00:15
Your OU, you don't get it from the ether but from a conventional equation that makes no assumption of ether.
But it does make an assumption of a line that can't be made from L and C and R. So it is possible that another form of line that has the "impossible" characteristic could yield that "impossible" result. I even demonstrated how the capacitively loaded coil produces a CW flux v. mmf loop in the presence of a time delay and that did not use any equations. As you know a CW loop represents an energy source that can be modelled as a negative R. So the magnetic delay line with that "impossible" characteristic is worth investigating and if it did offer OU then the energy source could well be the active aether.
QuoteDigressions and non sequitur, it becomes nonsense because it has no logical relation with your initial paper, the one I am contesting.
But it could clarify why the conventional equations yield the "impossible" result, and surely that is worthwhile?
Smudge
Quote from: Smudge on 2022.09.27, 09:54:33
But it does make an assumption of a line that can't be made from L and C and R.
...
I never said it was impossible. But what do you answer except to your own subjects?!
Your fallacy: https://en.wikipedia.org/wiki/Straw_man
I said that
maintaining an oscillation was impossible with negative resistance.
Once again what you answer is irrelevant, the rhetorical pirouettes are getting to be too much.