Measurement and direction of information - Page 1
Forum section: electronics / general

author: John Heath
author_rank: Frequent Contributor
timestamp: May 14, 2016, 02:35:54 pm
content: I wish to measure the current between one op-amp output at + 1 volt 1 u second pulse impedance 0 and another op-amp at + 1 volt 1 u second pulse impedance 0. On the surface the answer would be 0 current before and after the 1 u second pulse as both op-amps are in agreement voltage wise. However the 1 u second + and + pulses have to travel through 100 feet of coax cable before it reaches the other op-amp. This places the meeting of the + and + 1 volt 1 u second pulse midway in the coax cable not the op-amp outputs. The burning question. When the two pulses meet mid way in the coax cable do they reflect off each other then return to their source or do they pass each other like ships in the night. This is a very tricky question that gets to the heart of the direction of information. Can identical information , pulse , propagating in a coax cable traveling from A to B and B to A make it there? The end result of a reflection , identical information can not travel from A to B , in a coax cable or ships passing in the night , identical information can travel from A to B. The answer has no relevance as either way the end result is the same as I should measure high current on the op-amp outputs when the time displaced pluse arrives. The real question is high current caused by the op-amps own pulse reflected back later in time or high current caused by the pulse from the other op-amp. Put in other words can identical information be conveyed through a coax cable from A to B and B to A or is this physically impossible as they will always reflect off each other in the coax cable?

author: sarepairman2
author_rank: Frequent Contributor
timestamp: May 14, 2016, 03:47:38 pm
content: they had ascii diagrams in 1970But to answer your question, try approaching the phenomenon from a different angle with non electrical waves that have a slower propagation delay, i.e. mechanical shock waves.Reflections happen from impedance mismatch, does the impedance of a transmission line change as a pulse is moving through it? Electrical signals superimpose on each other, if you had a slotted wave guide you can measure this.I think though, on a advanced level, your question reaches into areas of physics that we don't understand too well (if you look at it from a low enough level). i.e. some kind of distortion n energy transfer between waves?

author: edavid
author_rank: Super Contributor
timestamp: May 14, 2016, 03:59:20 pm
content: Quote from: John Heath on May 14, 2016, 02:35:54 pmSure. It doesn't matter if the information is "identical". If the transmission line is linear, it can be used bidirectionally.Reflections only occur at discontinuities.

author: edavid
author_rank: Super Contributor
timestamp: May 14, 2016, 04:51:32 pm
content: Telephone pairs aren't used as transmission lines, so they aren't exactly relevant to the OP's question.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 14, 2016, 04:51:59 pm
content: Quote from: sarepairman2 on May 14, 2016, 03:47:38 pmIn electronics I always think in terms of physics. It just seems easier that way. Inductors are fly wheels , voltage pressure and so on. For this reason when speaking of physics I am really speaking of electronics in the only way my pea brain can understand itTo your thoughts. Say the coax has an impedance of 50 Z ohms. For a pulse the cable feels like 50 Z as it propagates down the cable. If that pulse runs into another identical pulse moving the other way does the cable still feel like 50 Z ??. both pulses go to 1 volt .If they are both 1 volt then there is not a reason for electrons to flow in either direction. This being the case the cable would feel like infinity impedance for the pulses. It would be as if the two pulses ran into a brick wall so they would reflect and reverse direction ,, i think. I have a video of this. Just before 50 percent of the video you can see two pulses colliding into each other. What do you think? Are the pulses passing each other like ships in the night or did they hit a brick wall reflecting off each other reversing direction.To diuligent minds I hope this video meets your requirement for a visualization.

author: edavid
author_rank: Super Contributor
timestamp: May 14, 2016, 04:55:28 pm
content: Quote from: John Heath on May 14, 2016, 04:51:59 pmYes.QuoteThat is not how transmission line propagation works.QuoteNo, absolutely does not happen.

author: Marco
author_rank: Super Contributor
timestamp: May 14, 2016, 05:22:02 pm
content: Quote from: John Heath on May 14, 2016, 04:51:59 pmIt's all abstraction, use the abstraction which isn't silly and will generalize ... so they pass through each other and add amplitude, obviously.https://www.youtube.com/watch?feature=player_detailpage&v=DovunOxlY1k#t=374

author: John Heath
author_rank: Frequent Contributor
timestamp: May 14, 2016, 05:25:26 pm
content: Quote from: DiligentMinds.com on May 14, 2016, 04:10:48 pmInteresting story now that you brought up telephones. I was working on temperature probes in a chemical plant. I idea was to computerize everything with the spiffy new 100 bus computer. This is before personal computers were available. There was a need to communicate while running the temperature cables. In a stroke of genius we used two phones with a 9 volt battery for a current loop. This way by connecting the phones with alligator clips to the temperature cable we could talk back and forth from 100 feet away. The phones used carbon microphones so amplification was not needed. And yes information was traveling in both direction in one pair of wires.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 14, 2016, 05:52:59 pm
content: Quote from: Marco on May 14, 2016, 05:22:02 pmGreat video. The old school videos from the 1950s are always better in my experience as they seem to take the time to cover as much as possible. Possibly the cost of using film leads to better preparation.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 16, 2016, 12:59:28 pm
content: Pulses or waveforms travelling along a coax are mathematically identical to a wave travelling along a long, stretched wire or spring. The waves or pulses travelling in opposite direction do indeed 'pass through' each other, superimposing in amplitude as they pass. There is no more to be said. That is how it works. Does that answer the question? That is a property of waves in general, water waves, sound waves, light waves, radio waves, EM waves in general, etc.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 17, 2016, 04:36:42 am
content: Okay a slinky spring between A and B with a brick wall in the middle. Clearly the longitudinal pules from A and B will just bounce off the wall then return. l will now remove the brick wall and replace it with a tiny string to mark the center point. Again A and B send identical longitudinal pulse. If you look closely the string did not move. Almost as if it were a brick wall ?? If A sends a square pulse and B a sine pulse the string will move but only as much as there is a difference between A and B pulse. This sounds to me as if identical information can not be sent from A to B as this information appears to be bounce back as noted by the tiny string not moving as if it were a brick wall.You did not FEA model my magnet on a TV screen , nag nag.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 17, 2016, 11:17:44 am
content: Quote from: John Heath on May 17, 2016, 04:36:42 amI'm trying hard to understand your question. For any wave scenario you describe, I feel confident that I understand what is happening, and can describe what happens, but am not sure what you are really asking.You start off with 2 identical pulses at A and B, travelling toward each other, so they meet at C, midway between A and B. For convenience, let's talk about electrical pulses travelling along coaxial cables. I will consider the case where the 2 pulses are of the same sign.I think we agree about what actually happens. The 2 pulses meet in the middle, and at one instant in time, you observe a single, double amplitude pulse centred at C. In the normal description, the 2 original pulses continue on their way, with the pulse that started at A eventually arriving at B, and vice-versa.However, in every possible way, the behaviour would be identical if the cable was neatly snipped at C, and the 2 waves simply reflected (in phase) off the two unterminated ends at C. Cute.This leads to an interesting question. When a single pulse travels along a cable, then when the pulse is passing any point P, there is a transfer of energy across the boundary at P.But in the example in question, with 2 counter-propagating pulses, is any energy (or current) transferred across the boundary at C? The answer is NO, no matter how you look at it. If you view the situation as 2 independent pulses passing like ships in the night, then you have equal and opposite transfers of energy travelling L-R and R-L, so the net transfer of energy across C is zero. Alternatively, if you are John Heath, you would say that the 2 pulses 'reflect' in phase at C, so neither pulse transfers energy across C, so the total energy transferred across C is zero.Not sure if I have answered JH's question or not.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 17, 2016, 11:23:47 am
content: For completeness, here is the complementary case.You start off with 2 identical pulses at A and B, travelling toward each other, so they meet at C, midway between A and B. For convenience, let's talk about electrical pulses travelling along coaxial cables. I will consider the case where the 2 pulses are ofoppositesign.I think we agree about what actually happens. The 2 pulses meet in the middle, and at one instant in time, you observe zero amplitude everywhere - total cancellation. In the normal description, the 2 original pulses continue on their way, with the pulse that started at A eventually arriving at B, and vice-versa.However, in every possible way, the behaviour would be identical if the 2 cables wereshortedat C (analog of a brick wall), and the 2 waves simply reflected (out of phase) off the two shorted ends.This leads to an interesting question. When a single pulse travels along a cable, then when the pulse is passing any point P, there is a transfer of energy across the boundary at P.But in the example in question, with 2 counter-propagating pulses, is any energy (or current) transferred across the boundary at C? The answer is NO, no matter how you look at it. If you view the situation as 2 independent pulses passing like ships in the night, then you have equal and opposite transfers of energy travelling L-R and R-L, so the net transfer of energy across C is zero. Alternatively, if you are John Heath, you would say that the 2 pulses 'reflect' in out-of-phase at C, so neither pulse transfers energy across C, so the total energy transferred across C is zero.So if we neatly shorted the coax at C, or snipped the cable at C and shorted the 2 snipped ends, then the behaviour of the 2 pulses in our example would be identical in every way to the case where the cable was continuous. Cute.Not sure if I have answered JH's question or not. I do like the head 'reflecting' off the brick wall, in JH's posting.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 17, 2016, 12:44:14 pm
content: It puts a smile on my face that you to put your finger square on the issue at hand the first time. With a voltage and amp meter at mid point of the cable the voltage doubles with 0 amp for same information. For opposite information the amp double and voltage is zero. In either case energy is 0. At any other point alone the cable the energy is a constant 1 combination of voltage and current but not at the mid point as it will always be 0 energy thus a brick wall for same and opposite information between A and B.The electrons can be modified to parallel movement and magnet made square for the TV if that helps , nuff said.

author: Marco
author_rank: Super Contributor
timestamp: May 17, 2016, 01:48:58 pm
content: Quote from: John Heath on May 17, 2016, 04:36:42 amOnly if the static force on the string is higher than the force of the wave, otherwise it will go slackAs I said before, use the abstraction which generalizes ... it's the more useful kind. They all break down in one corner case or another, but try to minimize it ... creating a new one for a corner case which superposition handles fine is silly.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 18, 2016, 01:23:31 am
content: Quote from: Zeranin on May 16, 2016, 12:59:28 pmThinking further, what I should have said was :-Pulses or waveforms travelling along a coax are mathematicallysimilarto a wave travelling along a long, stretched wire or spring.For the purpose of the discussion, both types of wave propagate at a particular speed, and both exhibit superposition when pulses travelling in opposite direction pass through each other.However, purely for interest, it would appear that although the math describing each is similar, it is not the same. One is not simply a direct analog of the other.Specifically, consider an infinitely long length of 50 ohm coax cable. Connect a 10V source to one end, and leave it connected. The 0V to 10V edge will propagate along the cable and, provided the cable is of infinite length, the power supply will see a 50R load, and will need to provide an input power of 10x10/50 = 2W, for as long as it is connected. This input power is going into ‘charging up’ the coax for an ever-increasing length. Furthermore, the energy that has been supplied will be stored uniformly along the length of ‘charged’ coax.However, if we take an infinitely long length of stretched spring or rubber band, we find a very different behavior. We hold one end, while the other end is attached to a wall at infinity. Then, to generate a long pulse, we rapidly move our hand 200mm (or whatever) upward, and leave our hand in the upward position. The 200mm high square edge thus generated will propagate along the spring, just as with the coax. However, once our hand has generated the edge, and is stationary at the 200mm high position, then unlike with the coax, we are no longer putting any energy into the spring. Furthermore, the only place that the spring is stretched and stores energy is right at the travelling edge of the pulse. This is all quite different to what happens with waves or pulses travelling along coax, and different to any other sort of wave that I can think of as well.Next time I will be more careful before waving my hands and saying that the math behind all waves is the same. It isn’t. The math describing water waves is different yet again, and very complicated. We have an Applied Maths research group here that studies (among other things) ‘rogue ocean waves’. JH would love this. In any ocean, you get random small waves over very large areas. Under certain conditions, these small waves can concentrate into a single, monster ‘rogue wave’ up to 40m high that literally appears out of nowhere, capable of sinking even large ships. Seehttp://www.anu.edu.au/news/all-news/rogue-wave-theory-to-save-shipsJust imagine if the same thing could happen with EM waves, where if you are unlucky you get blinded by a massive ‘rogue wave’ of light!There are similarities between all types of wave, but there are differences also, and it is not just a matter of plugging different physical quantities into the same math.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 18, 2016, 04:22:44 am
content: You should not doubt yourself . It took 2 hours pacing the floor but at the end your original assessment was correct , all waves are the same.The infinitely long coax cable with regulated 10 volt power supply is the same as a infinitely long string for both longitudinal and transverse waves.If I pull on a string with 10 ounces , 10 volts , my hand will move quickly. Quickly means high current smoking your nice 10 volt power supply. What went wrong. The natural series inductance of the coax cable went wrong. Must place lead marbles every inch on the string for the inductance of the cable. Now it works. When I pull on the string with constant 10 ounces , 10 volts , my hand does not move quickly as I now have to drag ever more and more lead balls as the string tightens up. This will lead to constant current , hand velocity , of 200 m amps for 10 volts at 50 ohms Z.As to your transverse wave by pulling the string up with 10 ounces. I have a marked preference for side ways force to take gravity off the table. You can not just move the string sideways. You have to keep moving the string sideways forever with a 10 ounce force for a 10 volt power supply analogy of constant force. Done this way it ends up the same as the coax cable with constant velocity sideways at constant 10 ounce force as you start to accelerate more and more lead marbles on the string.As for water we go from 1 dimension to 2 dimensions. The wave will attenuate faster. This complicates it. I will avoid this by asking another question. How can a wave in water with a parabolic reflector travel in a straight line in one dimension only. What keeps the sides of the wave crest from falling out sideways? And while I am at it why are the little wave crests on the side of a boat always the same angle regardless of the speed of the boat. I think it is 18 degrees but I could be wrong. If you go to a different plant with different gravity it is 12 degrees regardless of how fast their boats are going. In short the angle of the little water crests coming off a boat depends only on the amount of gravity. Why? No answers on my end. I wish there was an electrical analogy for this as I have a feeling something cool would come out of it.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 18, 2016, 01:28:04 pm
content: Quote from: John Heath on May 18, 2016, 04:22:44 amAll along I was consideringtransversewaves/pulses on my stretched spring, so let’s assume transverse waves. I am happy to move the spring sideways, or do the experiment in deep space, so that gravity is not a player. As to the ‘lead marbles’ on the string, you don’t need to do that, because all springs or elastic rubber band already have a characteristic mass-per-length, so you don’t need to add additional masses.What you say is exactly correct. I salute you, Sir.I agree with you that the correct analogy is Force-Voltage and Velocity-Current.Taking things even further, Mass-Inductance and Compliance-Capacitance.A stretched spring is a distributed Mass-Spring system, so our analogy is that an electrical transmission line is a distributed LC system, which is correct.Then it all works, just as you say, and I’ll agree that the math for one applies exactly to the math for then other, by swapping variables as described above.That said, we are a long way from how waves/pulses on stretched springs are usually presented in textbooks. Generating waves/pulses on a slinky spring is a classic high-school experiment. Textbooks and videos love showing creation of a wave or pulse on a long spring. You move your hand back and forth with a displacement of say +- 200mm, and you get a wave with the same amplitude. The peak-peak amplitude of a wave on a long spring is taken as the transverse distance between peaks. I’m sure that’s what all the text books say. You just move the spring sideways to create a wave/pulse, and how far you move it sideways dictates the amplitude of the pulse. Can you show me a textbook or web tutorial that describes transverse waves on a spring differently to this? The tutorial video on this thread certainly shows waves in this way.However, what we are saying is that if you want to treat a pulse propagating along a stretched spring as being analogous to a wave/pulse propagating along a transmission line, then the normal interpretation of the pulse on the spring is not really correct, but a simplistic ‘con job’ that appears OK at first sight, but ultimately fails to be analogous to wave propagation in a coaxial cable, for the reasons I gave in my previous posting.Just as you say, the wave amplitude that is Volts on the electrical transmission line needs to be equated to Force on the stretched spring. This is curious but true. The observed shape of the stretched spring should not be regarded as the amplitude of the propagating wave at all, but actually represents the time integral of the current in the analogous model of a wave propagating along a coaxial cable. To put that another way, when you move your hand to produce a shape on the spring, it is the FORCE that your hand exerts that represents thevoltageamplitude of the equivalent coax wave, and the VELOCITY (dx/dt) of your hand represents thecurrentamplitude of the equivalent coax wave.If you produce a steady sine-wave with the movement of your hand, then the velocity of your hand is a COS wave, which looks identical except for 90 degree phase shift. Thus, when all the textbooks/videos show a lovely sine wave on a stretched spring, then it’s OK to say that the visible sine wave on the spring is analogous to a sine wave propagating along a coax cable, because SIN and COS waves look the same.However, if you want to produce a wave or pulse on your stretched spring that is equivalent to asquarewave in a coaxial cable then it’s a different story. To produce a wave on a stretched springthat is equivalent to a square wave propagating along a coax cable, then your hand needs to produce a constant FORCE in one direction, followed by the same constant force in the other direction, and so on, with the result that your hand will move at constant speed in one direction, followed by constant speed in the other direction etc, so that the shape produced on the spring will actually look like a triangle wave.Interesting.I suppose it’s OK to still describe the moving shapes on a stretched spring as ‘propagating pulses/waves’, just so long as it is realized that such ‘waves’ are not directly analogous with waves/pulses propagating along an electrical transmission line.As for water waves, I might talk about them later, because I'm thinking that the usual hand-waving that says they are 'just the same thing' isn't really right either.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 19, 2016, 04:31:13 am
content: Coax vs slinky or 90 vs 180 degrees for voltage vs current. You do manage to get yourself in these intellectual holes. I can barely see the top of your head with a little dirt being tossed up as you dig yourself deeper and deeper. Digging ones way out of such holes is the only way to truly understand nature. I was in that hole so allow me to hand you a ladder.I have a 300 ohms transmission line running 20 feet along the ground with a standing wave of 100 MHz. If I hold a florescent tube it will light up every 4 feet. At 2 feet the tube is dim but my clamp RF current meter reads high. At 4 feet tube is bright but RF current meter low. At 6 feet and so on and so on. That sounds like 180 not 90 degree difference?? However there are not 1 but 2 bright tube points and 2 high current points within 1 cycle of a sine wave thus 90 vs 180 degrees. You need 2 max stress points and 2 max movement points on a slinky spring to complete 1 cycle of a sine wave . This will return your slinky spring force to movement to 90 degrees just like a coax cable where current to voltage is at 90 degrees. Not sure if that works for you but that was my way to dig out of that hole.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 19, 2016, 08:33:40 am
content: Quote from: John Heath on May 19, 2016, 04:31:13 amHmmm. Actually I stand by all I said, and don't understand which part of what I said you disagree with. I'm happy to comment on your 300 ohm coax example though.You describe the case of astanding wavein your coax. I don't know why you did that, because that has no relevance to any of the examples given previously, that refer to the case of an infinitely long coax or slinky, in which case there can be no standing waves.If you have pure standing waves, then the input to the coax looks purely reactive, so in that case voltage and current are 90 degrees apart, as you say, but that case is not relevant.If you have an infinitely long coax, then the input looks purely resistive, so V and I are in phase.The point I was making in my previous posting, FWIW, is that if you produce an infinitely long transverse square pulse on a spring, apparently equivalent to switching a DC voltage on the input to a coax, then apart from the initial edge, no further power is required to hold your hand at the fixed amplitude level of the spring, thus clearly demonstrating that coax cable and springs are not directly analogous if you equate the voltage amplitude of the wave/pulse on the coax with the transverse displacement amplitude of the wave/pulse on the spring. However, if you instead use a sinusoidal excitation for your infinitely long coax and spring, then they superficially DO appear analogous, because in both cases it DOES require a constant input power for sinusoidal excitation. I say superficially analogous, because for the coax, the required power input is frequency independent (=V/R), whereas for the spring, the required power input scales with frequency, with both cases having sinusoidal, fixed amplitude excitation.Are we completely in agreement?What all of this is saying, really, is that waves/pulses in a stretched spring are NOT directly analogous to waves/pulses in a coaxial cable.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 20, 2016, 06:29:25 am
content: I am going to disagree. Coax cable 90 degrees and you are saying 0 degrees. As this is a disagreement I am not allowed to use a text book reference , computer animation or so and so says this or that. It must be empirical only. To be clear it can not be a standing wave , it can not be a mechanical wave . It must be a electromagnetic wave propagating in one direction where both voltage and current sine waves can be compared for phase difference. A coax cable does not allow measurement from the outside however a 300 ohms TV antenna line does. However a voltage and current probe can not be trusted to have the same phase delay at 100 Mhz , they are different probes. I am running out of options. Any thoughts on your end to resolve this?I pp

author: Zeranin
author_rank: Regular Contributor
timestamp: May 20, 2016, 06:40:10 am
content: Quote from: John Heath on May 20, 2016, 06:29:25 amYou can use any text book you like.If you have an infinitely long coax cable, or a finite length with far end terminated with 50R (for 50R coax), then the input end of the coax looks purely resitive, in every way indistinguishable from a 50R resistor. That is a fact, and a fundamental characteristic of coaxial cables.For a pure resistor, voltage and current are in phase. That's all there is to it, no more to be said. Is there a reason why you object to what I'm saying here? In other words, what is the problem we are trying to resolve?

author: John Heath
author_rank: Frequent Contributor
timestamp: May 20, 2016, 01:35:52 pm
content: The problem is conventional science says it is 90 degrees . You say it is 0 degrees. What some one says or what the majority thinks is not relevant to science. Measurement is the only variable that counts. I only have 1 piece of true empirical evidence and it says you are right 0 degrees. That 1 piece of empirical evidence comes from a noted doctor that would say 90 degrees yet there it is in plain sight 0 degrees. Would like to go over this empirical evidence with you to see if you agree. I will caution ahead of time that this is inconclusive which is why I wanted you to look at it. The picture is from the video farther up this thread. The white dots are the real thing not an animation therefore have to be consider true regardless of what anyone says or a text book says. The X up down difference tells us what the torque is on the tention cable , voltage. The X up down difference also tells us how far the white dot has moved. according to this picture voltage and current are in phase , 0 degrees.When I walk down a transmission line with a standing wave to measure current and voltage I am measuring the vacuum around the white dots not the white dots themselves. The vacuum is what imparts inertia on the electron by counter EMF. I can not the the vacuum in the picture shown therefore there is missing information which is why I cautioned inconclusive.I realize I have the confused the issue beyond any hope of unraveling it by mixing standing wave with a mechanical wave and a coax cable wave.However the picture shown and the transmission line standing wave is the only empirical evidence at hand. Everything else is just hand waving.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 20, 2016, 11:06:48 pm
content: Quote from: John Heath on May 20, 2016, 01:35:52 pmWhich 'conventional science' says that that voltage and current are 90 degrees apart in an infinitely long coax, or one that is terminated with 50R? I have never read or heard of such a thing.It would be easy enough to measure. All you need to do is use a signal generator to send 50 MHz (or whatever) into a longish coax that is properly terminated at the far end, and insert a 1R shunt in the connection between generator and coax braid. Then use a CRO to simultaneously measure the voltage and current at the input to the coax. I'll do the measurement if you like, and there is not the slightest doubt about what the result will be. The voltage and current will be in phase. If you want, you can use the same method to measure V and I anywhere along the coax as well.They MUST be in phase, or to be more precise, theyCANNOTbe 90 degrees out of phase, because if they were, then no power would be going into the coax! Remember, AC power is VI Cos(theta), so if theta is 90 degrees, then the power is zero.There is no mystery to unravel. The only mystery is to figure out why you came to believe the answer is 90 degrees in the first place.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 22, 2016, 04:59:19 pm
content: The 90 degrees comes from time delay in a coax cable . Forward force followed by delay in action as the information propagates down the cable. Forward force would be voltage and inductive effective inertia delaying current. I assumed this would be 90 degrees when impedance is matched? Perhaps I am wrong.Going to switch gears for a moment to an engineering video on origins of a magnetic field using Lorentz Fitzgerald contraction. I always liked this interpretations as it implies no magnetic field , just a Coulomb force. Any attempts to measure this on my part have failed before the attempt is even made. Would like to bounce a few of the ideas I have to see if you agree or have an alternative way to make this measurement. First the video.I have a turn table set up. The thinking was to have 1 amp of current going clock wise around the outside of the turn table , calculate electron drift velocity then then turn the table counter clock wise at the same electron drift velocity. With electrons going clock wise and table turning counter clock wise at the same electron drift velocity the electrons should be standing still from my frame of reference. This being the case I should measure 0 current with my open clamp current meter that you saw for magnet outer edge effective current reading. The problem is the protons in the wire that can not move are now rotating counter clock wise with the table. A counter clock wise positive proton generates the same magnetic field as a clock wise electron drift velocity. You see the problem , its a no win. There has to be a way around this to verify electron drift velocity. I was thinking of doing this backward by electrostatic van de graaff charging the wire to guarantee less negative electrons then protons then rotate the table. However 1 amp is a tall order , 6 X 10^18 electrons , for a charge to be in the range of my current meter. No point in trying. There must be a way to move just electrons but not protons with a turn table to verify electron drift velocity? Frustrated. Anything you can think of would be appreciated.

author: Zeranin
author_rank: Regular Contributor
timestamp: May 23, 2016, 11:54:43 pm
content: I looked at the video, and it seems like nonsense to me. The relativity argument relies on the charges moving at near the speed of light, but in any normal conductor they don’t. The electron drift velocity in a conductor is actually quite slow, mush slower than walking speed, so the whole relativity argument becomes nonsense.It’s very easy to derive and understand the formula for electron drift velocity in a conductor from first principles, and easy to calculate the drift velocity using the said simple formula. An example of drift velocity calculation can be found here :-http://resources.schoolscience.co.uk/cda/16plus/copelech2pg3.htmlThis shows that the drift velocity for a 5A current in a 0.5 mm^2 conductor is a piffling 7.35E-4 m/s, or roughly 0.1 mm/s. That’s at a respectable current density of 10A/mm^2. Even with a water-cooled conductor you are unlikely to exceed 100A/mm^2, so the highest steady drift velocity that you are likely to create is around 1mm/s, or 3.6 meters per hour.The current density in power station alternators, xformers, transmission lines etc will generally be less than 10A/mm2, which as discussed gives an electron drift speed of 0.1mm/sec. At 50Hz powerline frequency, that means that the electrons move for a distance of only 0.01 seconds x 0.1mm/sec = 0.001mm (1 micron) before the direction is reversed. Therefore the electrons in the windings of a power station alternator will never leave the winding, much less the power station, but just jiggle back and forth by 0.001mm.Not sure if the above alters your desire to directly measure electron drift velocity in a conductor, or influences how you might choose to do so.

author: John Heath
author_rank: Frequent Contributor
timestamp: May 24, 2016, 03:10:55 am
content: slower than walking speed, so the whole relativity argument becomes nonsense.Yes the slow walk argument. It is so far from the steep end of the Lorentz curve that the effect would be too small. The counter argument that I googled up is the coulomb force is much stronger therefore that little effect is enough. You could also say electrons are jumping around in all directions close to c with a higher probability of jumping in the drift direction. Both these arguments a bit ,, optimistic to put it in a nice way.This shows that the drift velocity for a 5A current in a 0.5 mm^2 conductor is a piffling 7.35E-4 m/s,I googled 1 mm wire 1 amp 100 u meters per second electron drift velocity which is almost the same as your numbers. That is a very slow walk , ha.Not sure if the above alters your desire to directly measure electron drift velocity in a conductor, or influences how you might choose to do so.I can not live with a knot in a rope. It will haunt for months and months. I suspect you can relate to that. As I said before any attempts to make this measurement with a wire fails as the protons turn the other way when rotating a wire on a turn table. If you stew on it for a bit you will see the problem. If I could just rotate all the electrons in a 1 mm wire 1 revolution per second the magnetic field would be off the scale if 1 u meter per second is 1 amp. 10,000 amps would be the ball park number.One thought was bits of conductive tape on the edge of a disk exposed to a 32 KV TV high voltage line. This should pull a few electrons off the tape. When rotating the disk the missing electrons , protons , on the tape will be moving without electrons to counter the effect. What do you think ?

author: Zeranin
author_rank: Regular Contributor
timestamp: May 25, 2016, 05:21:16 am
content: Quote from: John Heath on May 24, 2016, 03:10:55 amLike you, I cannot sleep if there is something I don't understand. However, where is the 'knot in the rope' in this case? Now that we have that silly relativistic video out of the way, what more is there to know or understand about electron drift velocity in conductors? Why do you still have an apparent desire to measure drift velocity?Your proposed experiment should be a valid way of producing moving charge, though I would need to do some back-of-envelope calcs to know if the equivalent current produced would be enough to measure. At a wild guess, you might get equivalent of a few uA. But what are you even trying to measure in the experiment as described?

author: John Heath
author_rank: Frequent Contributor
timestamp: May 25, 2016, 11:47:10 pm
content: Back of envelope calcs ,, I like it. If you see myself working that expression into another thread you will know where it came from. What am I trying to measure with this experiment ? I would like to see first hand a forced low drift velocity charge by a rotating turn table producing a measurable magnetic field at 5 to 10 revolutions a second. All theories must be held accountable. In short talk is cheap , lets see the beef. Also there are additional concerns. Addressing these other concerns will be a little wordy , apologies in advance.I am going the charge the earth to 10 K volts + by sending a bunch of electrons off to a far far place. The equator is turning at 1000 miles per hour. Where I am around 500 miles per hour. There is no reason for the 10 K volt electric field to turn just because the earth is turning. Its just a 10 K volt field coming from the earth. As luck would have it a gravity field is exactly the same as a electromagnetic field. Both travel at c and both gravity field and a electromagnetic field will will not rotate just because the earth is rotating other than a minor hint of frame dragging. Why complicate things by putting gravity on the table? The reason gravity must be put on the table has to do with an experiment done in the early 1970s by Hafele–Keating .https://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experimentThey flew some atomic clocks around. It turned out east and west bound planes had a difference time dilation. This sounds like a disaster for special relativity however when SR and GR are combined SR is still okay as we are effectively moving through our own gravitational field at 1000 miles per hour when on the equator. As electric field act the same as gravity field then we are moving at 1000 miles per hour through the earth's electric field at the equator. That would be a rather impressive drift velocity for a charge. The probability of the earth having exactly the same amount of electrons as protons is unlikely considering the corona bursts that the sun dumps on us for the northern lights. The earth must have some electric field.With all that has been said a conclusion on my part is a successful test of a magnetic field caused by the rotation of charges on a turn table will confirm that a low electron drift velocity can produce a measurable magnetic field. However the Hafele–Keating says there must also be an oscillation in that magnetic field equal to the frequency of rotation of the turn table as we move faster and slower with every turn of the table through the earth's electric field at the equator. In my mind this oscillation would be check mate to confirm the Lorentz contraction interpretation of a magnetic field in light of the Hafele–Keating experiment.All that is left is the details of isolating + or - charges on bits of aluminum tape then rotate the table. The current probe will do the rest providing enough charges can be put on the tape. Need to do some " back of envelope calcs" for 2 p farad tape to surrounding areas with a 10 K volt source with pointed copper edges facing the tape as it passes by on the turn table. The reason for non contact charging of bits of aluminum tape is to use ionized gas to charge it . Air is the path of least resistance for ions. Once on the tape the ions should stick ,, I think ?

author: RandallMcRee
author_rank: Frequent Contributor
timestamp: May 27, 2016, 12:08:21 am
content: Its all a bit over my head--but can't you charge one of these dissectable capacitors, separate them and do the experiment??Seems like the disassembly accomplishes your main goal of separating the charges quite effectively.Ummm, if you do this--can you make a video?

author: John Heath
author_rank: Frequent Contributor
timestamp: May 27, 2016, 03:40:45 am
content: Thanks for posting the video. I enjoyed it. Odd how it is the jar that holds the charge not the plates? While I have your attention all current cell phone have excellent 3 dimensional hall effect transistors in them. It will display X Y and Z including B for net magnetic field. Perhaps you could start there for your proximity detector. As for my stuff being over your head trust me it is not. If you google to have it explained by someone more qualified than myself it will make more sense.