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#81
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On Wed, 03 Aug 2005 00:59:51 GMT, TokaMundo
wrote: Since a thin round copper wire has a very low emissivity it wont give up its heat all that fast. --- Ever hear of convection, dumbass? Got some numbers? Or is this going to be another one of your unsubstantiated claims? --- This will mean that your thermal gradient won't be as prevalent as you suggest. The proof is when one takes a copper wire and places it across a battery's terminals. Notice how the entire wire turns a nice cherry red quite evenly, all the way up to where it is attached to any form of sinking element. --- I see. Just by _looking_ at the glowing wire you can tell what the temperature differential is between the center of the wire and its surface? Amazing!!! --- The current throughout the wire will be even, --- No, it won't. it'll be lower in the parts of the wire which are hotter. --- and it is that current which generates the heat, or more precisely, the resistance to said current flow. --- The current changes the temperature of the wire, but it doesn't generate the resistance of the wire. The wire's resistance and its temperature coefficient of resistance are due to the resistivity and tempco of the material the wire is made of. --- If the wire were giving up its heat real fast, like that of a finned heat sink with air passing over it, I might agree. --- Whether you agree or not is immaterial. The phenomenon occurs with or without your permission. --- In the case of bare copper, however, the temperature throughout the wire is going to be very even. Your gradient will be nearly undetectable. --- Point is, dimwit, it'll still be there. --- For a very large diameter copper bus, it MIGHT have a slight gradient between the center and the outer surface, but not much. For wire, it is as even as even gets. Got some numbers, or is that just some more of your bull**** opinion? -- John Fields Professional Circuit Designer |
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#82
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"Corona discharge" is *NOT* caused by AC or skin effect. Corona discharge is caused by a high voltage gradient in the space around a conductor. This is a combination of the voltage applied to the conductor and the effective radius of the conductor. A high voltage, or very small effective radius can increase the gradient to the point where the air is ionized. Simple proof is that corona discharge is a problem with high DC voltage systems as well as AC. Sometimes hollow tubes are used for high frequency power conductors. This reduces the weight and cost by eliminating the central part of the conductor, where 'skin effect' has rendered the impedence high anyway. So little admittance is lost for a great savings in material/weight. VERY high frequency. NOT AC line frequencies. Not so. I could show you several switchyards within a short drive that use many hollow tube conductors all over the yard. daestrom I thought large diameter conductors in switchyards were for corona reduction. --------- Not directly related, an engineering handbook lists the skin effect for a 500MCM conductor as increasing the resistance by 2% - which I would say is significant. ----------- Since TokaMundo is infallable, consider the possibiltiy that he is actually the pope. He has also been identified as a troll. Bud-- |
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#84
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"TokaMundo" schreef in bericht ... On Wed, 03 Aug 2005 01:24:11 GMT, Repeating Rifle Gave us: Some people say that there is no such thing as a stupid question. Obviously there seems to be no shortage of stupid answers. Bill Yours certainly contributed abso****inglutely nothing, and would certainly fall into the "stupid answer" category. You remind me of a Firesign Theatre quote: "Who wona second world war.. you so smart?" Perhaps, if you are so informed, you should try giving an answer that actually has facts in it that are in sync with the topic of the thread, not merely its title. Someone once tolf me something, I believe it wass: "Don't feed the trolls", it seems to me that they are having a ball with food in abundance. |
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#85
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"TokaMundo" wrote in message ... On Tue, 02 Aug 2005 23:05:04 GMT, "Bob Myers" Gave us: My, again with the personal attacks; I suppose in the absence of practical knowledge, that's about all one is left with. As if declaring that someone has "no practical knowledge" isn't a personal attack. **** off retard. You have social problems. Gee, someone must've received a bad grade last year in their freshman circuit analysis class... Bob M. |
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#86
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On Wed, 03 Aug 2005 17:49:42 GMT, TokaMundo
wrote: On Wed, 03 Aug 2005 05:31:54 -0500, John Fields Gave us: On Wed, 03 Aug 2005 07:19:12 GMT, TokaMundo wrote: Sometimes I find it hard to believe that we actually call ourselves sentient beings with the way some of you act. --- I don't think anyone has ever called you sentient, much to their credit. Your a ****ing kingpin when it comes to being retarded. ^^^^ Tsk, tsk, tsk... Geez, Tokey, I'm not the one making all the spelling, punctuation, and grammar misteaks. Oh, that one was just for fun. -- John Fields Professional Circuit Designer |
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#87
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On Wed, 03 Aug 2005 18:30:47 GMT, "Bob Myers"
wrote: "TokaMundo" wrote in message .. . On Tue, 02 Aug 2005 23:05:04 GMT, "Bob Myers" Gave us: My, again with the personal attacks; I suppose in the absence of practical knowledge, that's about all one is left with. As if declaring that someone has "no practical knowledge" isn't a personal attack. **** off retard. You have social problems. Gee, someone must've received a bad grade last year in their freshman circuit analysis class... --- LOL, as much circuit design and analysis as he's done around here makes me think they kicked his sorry ass out of the class! -- John Fields Professional Circuit Designer |
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#88
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On Wed, 03 Aug 2005 18:48:51 GMT, TokaMundo
wrote: Just as I stated, when the wire turns cherry red from current, it is uniformly heated. --- Sorry, Charlie, there'll still be a temperature gradient across the diameter of the wire. There has to be, since the surface of the wire will be radiating heat and being cooled by convection. --- This will mean that your thermal gradient won't be as prevalent as you suggest. The proof is when one takes a copper wire and places it across a battery's terminals. Notice how the entire wire turns a nice cherry red quite evenly, all the way up to where it is attached to any form of sinking element. --- I see. Just by _looking_ at the glowing wire you can tell what the temperature differential is between the center of the wire and its surface? Amazing!!! --- The current throughout the wire will be even, --- No, it won't. it'll be lower in the parts of the wire which are hotter. The entire wire has the same temp and the same heat. The only exception is the connection points to the current source. --- Nope. I'm sorry that you can't understand why the gradient _has_ to be there, but it does, trust me. Or not, but continue on your present course and all you'll do is further convince everyone that you're as pig-headed now as you ever were. --- and it is that current which generates the heat, or more precisely, the resistance to said current flow. --- The current changes the temperature of the wire, but it doesn't generate the resistance of the wire. I never said it did. --- Sure you did. If you go back and read the sentence more carefully, what you said was: "and it is that current which generates the heat, or more precisely, the resistance to said current flow." Now, if we restructure the sentence in accordance with your instructions about preciseness, it reads: "and it is that current which generates the resistance to said current flow." which is incorrect. If the wire were giving up its heat real fast, like that of a finned heat sink with air passing over it, I might agree. --- Whether you agree or not is immaterial. What you have to say about it is certainly immaterial. The phenomenon occurs with or without your permission. The phenomenon of the entire wire being at the same temperature. --- The only way there would be no resistance variation across the diameter of the wire would be if there was no charge flowing through the wire and it was in an isothermal environment. Period. End of discussion. --- In the case of bare copper, however, the temperature throughout the wire is going to be very even. Your gradient will be nearly undetectable. --- Point is, dimwit, it'll still be there. You have no point. You and your stupidity has been exiled to the pointless forest. --- You always regress to your fourth grade insults when you don't have a leg to stand on and all you want to do is make noise, huh? --- For a very large diameter copper bus, it MIGHT have a slight gradient between the center and the outer surface, but not much. For wire, it is as even as even gets. Got some numbers, or is that just some more of your bull**** opinion? Got something that proves otherwise? --- Sure, but you wouldn't understand it. Even if you did, you'd still continue with your harangue in order to keep from having to admit ignorance like you always do. --- Otherwise your rebuttal is nothing more than bull**** opinion. --- You've got the cart before the horse. I hypothesized that there was a temperature gradient in a current carrying conductor, originally, in a reply to one of John Larkin's posts about skin depth. That concept was elaborated on by daestrom, and then rebutted by you, so unless you can prove that you're right and I'm wrong, my hypothesis stands and the rebuttal, which _you_ made, is nothing more than bull**** opinion. -- John Fields Professional Circuit Designer |
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#89
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On Wed, 03 Aug 2005 18:56:32 GMT, TokaMundo
wrote: On Wed, 03 Aug 2005 09:55:43 -0500, Bud Gave us: Not directly related, an engineering handbook lists the skin effect for a 500MCM conductor as increasing the resistance by 2% - which I would say is significant. I would say that the material the wire is made of would make a difference, and change that figure accordingly so such an arbitrary number is pretty ****ing vague. --- Since very little of what you say is true, a reference would be preferable to an opinion. Something like this: http://en.wikipedia.org/wiki/Skin_effect could even help you to _prove_ your point. -- John Fields Professional Circuit Designer |
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#90
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John Fields wrote:
On Wed, 03 Aug 2005 07:19:12 GMT, TokaMundo wrote: Sometimes I find it hard to believe that we actually call ourselves sentient beings with the way some of you act. --- I don't think anyone has ever called you sentient, much to their credit. -- John Fields Professional Circuit Designer I think "Sediment" would be a better description for TokaMundo, John. At least that's the sound he made when he was tossed in my troll bucket weeks ago. -- Link to my "Computers for disabled Veterans" project website deleted after threats were telephoned to my church. Michael A. Terrell Central Florida |
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#91
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"TokaMundo" wrote in message
... On Tue, 2 Aug 2005 22:58:33 +0000 (UTC), (Don Klipstein) Gave us: Going by "High Frequency Resistance", pages 3323-3325 of the 43rd edition of the "CRC Handbook": They give a formula X=pi*d*SQR((2*u*f)/rho)*SQR(1000) u is magnetic permeability, unity for copper. rho is resistivity in microohm-cm. They simplify this for copper, to x=10*d*.01071SQR(f) d is diameter in centimeters, and f is frequency in Hz. So, for 17 mm diameter copper wire at 60 Hz this "x" is 1.41. Next is a table that gives ratio of AC resistance to DC resistance as a function of this "x". This table has an entry for 1.4, giving AC resistance 1.020 times DC resistance. The table I saw shows the AC and DC resistance as being exactly the same for both. Your flaw is where you failed to note the topic given in the CRC handbook. 60 Hz is NOT high frequency... at all. Try some calculations at 100 kHz and you'll see that those frequencies down near zero (ie 60Hz) yield very nearly nil difference. --------------------- In Engineering Electromagnetics, Hayt points out that in a power station a bus bar for alternating current at 60 Hz much more than 1/3rd of an inch (8 mm) thick is wasteful of copper, and in practice bus bars for heavy AC current are rarely more than 1/2 inch (12 mm) thick except for mechanical reasons. This seems to imply that the bulk of the current is in the outermost 4mm. This does not mean that conductors, at 60 Hz, which are less than 8mm in diameter do not show skin effect. 60 Hz AC resistance/DC resistance for commonly used conductors (say 12 to 6 gauge) may be 1.1 to 1.25 in practice- this includes skin and proximity effects . However, anyone wanting to do the math from scratch better be familiar with Bessel functions. Are you? Skin effect, per se, is not a concern with ACSR power cables as there are a number of other factors which are more important. Possibly the approximations for high frequencies are not valid at 60Hz but this does not mean that skin effect is negligable- except for conductors 000 or higher - provided they are straight. . The point is that there is no hard and fast "rule" covering all situations -- Don Kelly @shawcross.ca remove the X to answer ---------------------------- |
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#92
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"Don Kelly" wrote in message news:dZfIe.110141$s54.2240@pd7tw2no... "TokaMundo" wrote in message ... On Tue, 2 Aug 2005 22:58:33 +0000 (UTC), (Don Klipstein) Gave us: Going by "High Frequency Resistance", pages 3323-3325 of the 43rd edition of the "CRC Handbook": They give a formula X=pi*d*SQR((2*u*f)/rho)*SQR(1000) u is magnetic permeability, unity for copper. rho is resistivity in microohm-cm. They simplify this for copper, to x=10*d*.01071SQR(f) d is diameter in centimeters, and f is frequency in Hz. So, for 17 mm diameter copper wire at 60 Hz this "x" is 1.41. Next is a table that gives ratio of AC resistance to DC resistance as a function of this "x". This table has an entry for 1.4, giving AC resistance 1.020 times DC resistance. The table I saw shows the AC and DC resistance as being exactly the same for both. Your flaw is where you failed to note the topic given in the CRC handbook. 60 Hz is NOT high frequency... at all. Try some calculations at 100 kHz and you'll see that those frequencies down near zero (ie 60Hz) yield very nearly nil difference. --------------------- In Engineering Electromagnetics, Hayt points out that in a power station a bus bar for alternating current at 60 Hz much more than 1/3rd of an inch (8 mm) thick is wasteful of copper, and in practice bus bars for heavy AC current are rarely more than 1/2 inch (12 mm) thick except for mechanical reasons. This seems to imply that the bulk of the current is in the outermost 4mm. This does not mean that conductors, at 60 Hz, which are less than 8mm in diameter do not show skin effect. 60 Hz AC resistance/DC resistance for commonly used conductors (say 12 to 6 gauge) may be 1.1 to 1.25 in practice- this includes skin and proximity effects . However, anyone wanting to do the math from scratch better be familiar with Bessel functions. Are you? Skin effect, per se, is not a concern with ACSR power cables as there are a number of other factors which are more important. Possibly the approximations for high frequencies are not valid at 60Hz but this does not mean that skin effect is negligable- except for conductors 000 or higher - provided they are straight. . The point is that there is no hard and fast "rule" covering all situations -- Don Kelly @shawcross.ca remove the X to answer ---------------------------- Thanks Don! I have been watching this thread for days. I like your answer. It's allways give and take, no hard and fast rules that fit every situation. And even better yet, You made your point without slamming anybody.. Well Done, Hats off.... |
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#93
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"TokaMundo" wrote in message ... On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin Gave us: In big AC transmission lines, there's a complex optimization involving wire weight, tensile strength, ohmic losses, skin effect, corona losses, wire cost, and tower spacing/cost. No ****. For one thing, they are primarily designed for high tensile strength as they have to stay mounted through all weather and environmental conditions. After that, their resistance is an issue as the primary material has to be steel for the tensile forces involved. They usually get clad in Aluminum as copper is too costly for such long runs, and the losses in using aluminum are little in comparison. This is also the reason that high voltages are used in long haul transmission lines. The loss over 2000 feet of line with 120 volts on it is significantly different than the loss over 2000 feet of line with 20,000 volts on it. -------- Gee - I thought I^2R loss depended on the current, not the voltage. For a given power you are right but you didn't state this. ------------- Corona will become a problem as that line voltage is raised. At that time line spacing becomes an issue. Tower spacing is a function of the terrain being traversed. Line spacing ON a given tower design is a function only of the voltage that is proposed to be carried, and the total number of conductors. Skin effect, in these high tension line realms is only an issue if the idiots that made the wire didn't know how deep to make the cladding. If the wire is clad to thinly, there will be more loss as the steel is more resistive, and the wire will heat more as well. If it is clad too thickly, an unnecessary cost is introduced. ------- Right -and I have seen ACSR cable with an aluminum depth that exceeds 2cm. This is unusual and now smaller conductors in bundles (spaced 30-45cm between conductors ) because of lower inductive reactance and surface fields that result-notghing to do with skin effect. This is specifically because the skin depth is so deep at this frequency, NOT due to it being a thin depth! So in power line cases, the effect is an issue of how deep the cladding is, not how thin. ----- Not a big deal. The usual skin depth rules go out the window because of the magnetic core material and the fact that you have strands of aluminum in close proximity. --------- In RF transmission lines, which are typically nickel or silver plated, it becomes a cost issue, and claddings are made as thin as possible for a given application frequency. These cases are where one will see hollow conductors, or plated tube or solids. This is where a Litz configuration or plated conductor will assist one in design of a circuit. At 60Hz, a high voltage step up transformer will have some transfer efficiency number. At switching frequencies, the same transformer design (wire turn count wise) will operate better if the primary, and or secondary have litz wire used in them as the effective resistance of the winding will be reduced at the higher frequencies. -------- Note that the equivalent of Litz wire has been used and is used in 60 Hz generator windings. Wonder why? The individual strands are too small to have an appreciable skin effect but there is also the proximity effect which can be more of a problem. ------------- -- Don Kelly @shawcross.ca remove the X to answer ---------------------------- |
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#94
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Skin effect has nothing to do with voltage. Nor is the 15KV level a
"boundary" . It has long been recognised that larger diameter conductors will reduce surface fields and corona as well as reducing inductance. However, large diameter conductors are heavy. Initially a copper "barrel stave" conductor was used but then the idea of a steel core for strength and aluminum for conductivity replaced this original idea. Skin effect was reduced and was considered in ther design but basically strength without loss of conductivity was the basis for practical large diameter conductors. The bundling of conductors (say 2 to 4 conductors spaced 30-45cm apart) is an extension of this - effective very large diameter and lower surface fields and series inductance at a reasonable price and weight savings. This has nothing to do with skin effect. -- Don Kelly @shawcross.ca remove the X to answer ---------------------------- "Dimitrios Tzortzakakis" wrote in message ... -- Tzortzakakis Dimitrios major in electrical engineering, freelance electrician FH von Iraklion-Kreta, freiberuflicher Elektriker dimtzort AT otenet DOT gr Ï "daestrom" Ýãñáøå óôï ìÞíõìá ... "Dimitrios Tzortzakakis" wrote in message ... -- Tzortzakakis Dimitrios major in electrical engineering, freelance electrician FH von Iraklion-Kreta, freiberuflicher Elektriker dimtzort AT otenet DOT gr Ï "Alexander" Ýãñáøå óôï ìÞíõìá ... "TimPerry" schreef in bericht ... "AllTel - Jim Hubbard" wrote in message ... I am curious about what would happen to an electrical current in 2 situations..... Assume that you have 2 wires that, when joined, complete a closed electrical DC circuit with electrons flowing thusly..... ------------ ============ eeeeeeeeee eeeeeeeeeeeeeee ------------ ============ If you flattened out the end of each wire where they connect , would the resulting electron paths be more like figure A or Figure B? neither ... research "skin effect" Most of the times this just aplies to AC (high frequency) circuits Or of line-to-line voltage equal or above 220 kV.Therefore transmission lines of 400 kV are always designed with a double conductor, thus to reduce the corona discharge due to skin effect. Oh boy, you have a 'couple of crossed wires' there. "Skin effect" is the phenomenon where electric current flow is forced out from the center of a conductor due to the self-inductance in the conductor when carrying AC current. The higher the frequency, the more pronounced the current shift to the exterior. It's mostly a problem with high current situations, even if the voltages are so low that corona discharge is not a problem. "Corona discharge" is *NOT* caused by AC or skin effect. Corona discharge is caused by a high voltage gradient in the space around a conductor. This is a combination of the voltage applied to the conductor and the effective radius of the conductor. A high voltage, or very small effective radius can increase the gradient to the point where the air is ionized. Simple proof is that corona discharge is a problem with high DC voltage systems as well as AC. Sometimes hollow tubes are used for high frequency power conductors. This reduces the weight and cost by eliminating the central part of the conductor, where 'skin effect' has rendered the impedence high anyway. So little admittance is lost for a great savings in material/weight. And for high voltage systems, multiple parallel conductors are used to give a larger 'effective radius', thereby reducing the corona losses. But the two phenomenon are not related, and the two techniques used are not really related. Yes, but also in voltages =15 kV there's a signifigant skin effect, that's why all transmission conductors are constructed with a steel *core* and an *aluminium* outer sheath, because the current tends to flow on the skin of the conductor.I mentioned corona discharge, to bring into evidence the very strong electric field around the conductor in very high voltages. daestrom |
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#95
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On Thu, 04 Aug 2005 06:25:43 GMT, TokaMundo
wrote: On Wed, 03 Aug 2005 15:54:26 -0500, John Fields Gave us: The current changes the temperature of the wire, but it doesn't generate the resistance of the wire. I never said it did. --- Sure you did. If you go back and read the sentence more carefully, what you said was: "and it is that current which generates the heat, or more precisely, the resistance to said current flow." Now, if we restructure the sentence in accordance with your instructions about preciseness, it reads: "and it is that current which generates the resistance to said current flow." Like I said before... I never said that, you ****ing asswipe. You're nothing more than a manipulative lard ass. --- And I've got you dancing to my tune, LOL! --- which is incorrect. Yes, you are. --- Grow up, ya goddam baby. -- John Fields Professional Circuit Designer |
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#96
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John Fields wrote:
Violent little bitch, eh? Everyone is at some time. Why don't you leave a snippet of the post you're replying to intact so that those of us who aren't mind-readers don't have to go searching to find out whom you're threatening? It was a vocatively-ambiguous imperative, duh. BTW, don't be surprised if the law comes knocking at your door... Threats of physical violence are looked on very seriously these days. The law must be rubbed out then. -Aut |
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#97
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John Fields wrote:
And where are you, my dear? there |
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#98
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"TokaMundo" wrote in message ... On Sun, 31 Jul 2005 10:32:46 -0500, "DBLEXPOSURE" Gave us: It is all speculation of course. I have never seen an electron, Have you? You have never seen lightning? It is an entire stream of them, IN MOTION. The visible effects are enough. Ever seen "Ball Lightning"? I suppose ionized air is just the visible result of the electron passing, and it moves so fast as not to be visible with the human eye, so what we see is air turned plasma. Ever had a jolt fire into you? The pain at the entry site tells one actual movement occurs. The jolt through the body and out the exit point also concur. See the photo at alt.binaries.pictures.misc titled "strike" Yes, I have seen lightning, not ball lightning but I have heard of it. "Air turned plasma", something like that. Oh yes, I have felt the effects of the electron. --- The visible effects are enough --- I suppose that depends on how curious your are. |
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#99
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On Wed, 3 Aug 2005 22:37:11 -0500, "DBLEXPOSURE"
wrote: Thanks Don! I have been watching this thread for days. I like your answer. It's allways give and take, no hard and fast rules that fit every situation. And even better yet, You made your point without slamming anybody.. --- Geez, you don't know when to quit, do you, Mr Passive-Aggressive? -- John Fields Professional Circuit Designer |
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#100
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On Thu, 04 Aug 2005 06:41:22 GMT, TokaMundo
wrote: On Sun, 31 Jul 2005 21:09:23 +1200, Jasen Betts Gave us: electrons cannot exceed the speed of light in a vacuum. no physical object can. There was an guy like you saying the same thing about aircraft and the "sound barrier" a little over 57 years ago. --- The difference was that back then bullets were known to be supersonic, so there was no _basic_ prohibition on supersonic flight. Today (actually, 70 years ago...) we have the EPR paradox and quantum entaglement which _hints_ that superluminal velocities are possible, but we also have: m0 mr = -------------------- sqrt (1 - (v²/c²)) which states that anything with a rest mass, m0, will have its relativistic mass, mr, tend toward infinity as its velocity, v, approaches that of light, c. Every experiment ever done to try to refute the veracity of the equation has confirmed that the equation is valid and, consequently, indicates that it is impossible for massive bodies to achieve the speed of light. However, we also have Cerenkov radiation, which is emitted whenever a massive particle exceeds the speed of light in the medium through which the particle is travelling... -- John Fields Professional Circuit Designer |
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#101
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"Don Kelly" wrote in message news:dZfIe.110141$s54.2240@pd7tw2no... However, anyone wanting to do the math from scratch better be familiar with Bessel functions. Are you? Or they need to be in Alameda. After all, Keptin, dis is vere dey keep da nuclear Bessels, no? :-) Bob M. |
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#102
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"John Fields" wrote in message ... On Wed, 3 Aug 2005 22:37:11 -0500, "DBLEXPOSURE" wrote: Thanks Don! I have been watching this thread for days. I like your answer. It's allways give and take, no hard and fast rules that fit every situation. And even better yet, You made your point without slamming anybody.. --- Geez, you don't know when to quit, do you, Mr Passive-Aggressive? -- John Fields Professional Circuit Designer ---- Passive-Aggressive Hmmmm...... |
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#103
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"TokaMundo" wrote in message ... If it were modeled after the earth, with the heat source in the center, I would agree. I feel, however, that it is more closely modeled after Io, which is heated by the magnetic forces of Jupiter, and more closely approximates an evenly heated body. Is there no end of subjects in which you are willing to demonstrate your ignorance? Io is NOT primarily heated by "the magnetic forces of Jupiter," but rather by its gravity - or, more specifically, the tidal forces resulting from the pull of Jupiter and the other major moons. Heating due to internal currents generated by the moon's passage through the Jovian magnetic field occurs, but is small in comparison to the tidal heating. As the specifics of the tidal heating depends on the particular relationship between Io, its parent (Jupiter) and the other satellites at the time, it is by no means even. For details, see: http://www.planetaryexploration.net/...l_heating.html As a result, Io does NOT have a particularly uniform thermal profile, either on its surface or vs. distance from the moon's core; see in particular: http://www.lpi.usra.edu/meetings/lpsc2004/pdf/2048.pdf http://www.lpi.usra.edu/meetings/lpsc2003/pdf/2030.pdf http://astron.berkeley.edu/~fmarchis...rintIcarus.pdf Bob M. |
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#104
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"TokaMundo" wrote in message ... On Sun, 31 Jul 2005 10:32:46 -0500, "DBLEXPOSURE" Gave us: It is all speculation of course. I have never seen an electron, Have you? You have never seen lightning? Hardly the same as seeing an electron itself, any more than observing the glow of the phosphor screen of a CRT is direct observation of an electron. Both, in fact, are examples of materials emitting light because of the absorption of energy from SOME source - "electron" is just a convenient name for the model we use, a particle which transports that energy. The visible appearance of the lightning bolt is also NOT a case of directly observing the "flow" of electric charge or current. It is an entire stream of them, IN MOTION. OK - according to you, how fast are the particles in question - the "stream of them" - moving through the air? Bob M. |
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#105
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"John Fields" wrote in message ... On Thu, 04 Aug 2005 06:41:22 GMT, TokaMundo wrote: On Sun, 31 Jul 2005 21:09:23 +1200, Jasen Betts Gave us: electrons cannot exceed the speed of light in a vacuum. no physical object can. There was an guy like you saying the same thing about aircraft and the "sound barrier" a little over 57 years ago. --- The difference was that back then bullets were known to be supersonic, so there was no _basic_ prohibition on supersonic flight. Today (actually, 70 years ago...) we have the EPR paradox and quantum entaglement which _hints_ that superluminal velocities are possible, but we also have: m0 mr = -------------------- sqrt (1 - (v²/c²)) which states that anything with a rest mass, m0, will have its relativistic mass, mr, tend toward infinity as its velocity, v, approaches that of light, c. Every experiment ever done to try to refute the veracity of the equation has confirmed that the equation is valid and, consequently, indicates that it is impossible for massive bodies to achieve the speed of light. However, we also have Cerenkov radiation, which is emitted whenever a massive particle exceeds the speed of light in the medium through which the particle is travelling... -- John Fields Professional Circuit Designer quantum entaglement, It's like you read my mind. |
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#106
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On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE"
wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer |
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#107
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"TokaMundo" wrote in message ... On Wed, 03 Aug 2005 15:54:26 -0500, John Fields Gave us: --- Sorry, Charlie, there'll still be a temperature gradient across the diameter of the wire. There has to be, since the surface of the wire will be radiating heat and being cooled by convection. --- If it were modeled after the earth, with the heat source in the center, I would agree. I feel, however, that it is more closely modeled after Io, which is heated by the magnetic forces of Jupiter, and more closely approximates an evenly heated body. In the wire, since the heat is generated throughout the medium via current flow, even from low currents on up to my cherry red scenario would show the wire at the same temp from center to outer surface. No, you still don't see it. The heat in the center is conducted outwards through an imaginary cylindrical surface separating the innermost core from the next outer layer. And the heat from the *that* layer, plus the heat from the inner core is transmitted through the next concentric imaginary cylindrical surface separating that layer from the area. The surface area of each imaginary cylindrical surface increases proportionally with the radius out from the center, but the heat that must be conducted through each surface increases proportional to the radius squared. So the temperature gradient across each imaginary cylindrical surface gets stronger and stronger. Thus the *temperature* across the cross-section is parabolic, even though the heat generation is flat/level. This has *nothing* to do with the heat transfer from the outer most surface to it's surroundings. Heat removal from the outer surface by convection, conduction, or radiation will *not* change the shape of the interior temperature gradient. daestrom |
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On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
wrote: "TokaMundo" wrote in message .. . On Wed, 03 Aug 2005 15:54:26 -0500, John Fields Gave us: --- Sorry, Charlie, there'll still be a temperature gradient across the diameter of the wire. There has to be, since the surface of the wire will be radiating heat and being cooled by convection. --- If it were modeled after the earth, with the heat source in the center, I would agree. I feel, however, that it is more closely modeled after Io, which is heated by the magnetic forces of Jupiter, and more closely approximates an evenly heated body. In the wire, since the heat is generated throughout the medium via current flow, even from low currents on up to my cherry red scenario would show the wire at the same temp from center to outer surface. No, you still don't see it. The heat in the center is conducted outwards through an imaginary cylindrical surface separating the innermost core from the next outer layer. And the heat from the *that* layer, plus the heat from the inner core is transmitted through the next concentric imaginary cylindrical surface separating that layer from the area. The surface area of each imaginary cylindrical surface increases proportionally with the radius out from the center, but the heat that must be conducted through each surface increases proportional to the radius squared. So the temperature gradient across each imaginary cylindrical surface gets stronger and stronger. Thus the *temperature* across the cross-section is parabolic, even though the heat generation is flat/level. This has *nothing* to do with the heat transfer from the outer most surface to it's surroundings. Heat removal from the outer surface by convection, conduction, or radiation will *not* change the shape of the interior temperature gradient. That's very, very nearly true. But for any reasonably insulated wire, the thermal conductivity of the insulation (including air) will be minute compared to that of copper, so internal temp gradients will be very low. To force a decent grad, you'd need to run a lot of current through a wire directly in contact with water or something. John |
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"John Fields" wrote in message ... On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE" wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer And how does this hint at super luminal velocities? I am curios. |
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On Thu, 4 Aug 2005 17:10:11 -0500, "DBLEXPOSURE"
wrote: "John Fields" wrote in message .. . On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE" wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer And how does this hint at super luminal velocities? I am curios. --- You have this annoying habit of deleting previously posted material which must then be searched out and reposted in order to properly reply to your queries, which invariably require reference to the previously posted material. I won't play that game. If you're serious, and you'd like to discuss the possibility of massive bodies achieving transluminal or superluminal velocities, aquaint yourself with Einstein, the EPR paradox, quantum entanglement, and then report back with what you've found. Otherwise, well, you know, **** off... -- John Fields Professional Circuit Designer |
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"John Fields" wrote in message ... On Thu, 4 Aug 2005 17:10:11 -0500, "DBLEXPOSURE" wrote: "John Fields" wrote in message . .. On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE" wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer And how does this hint at super luminal velocities? I am curios. --- You have this annoying habit of deleting previously posted material which must then be searched out and reposted in order to properly reply to your queries, which invariably require reference to the previously posted material. I won't play that game. If you're serious, and you'd like to discuss the possibility of massive bodies achieving transluminal or superluminal velocities, aquaint yourself with Einstein, the EPR paradox, quantum entanglement, and then report back with what you've found. Otherwise, well, you know, **** off... -- John Fields Professional Circuit Designer No sir, I replied to your post from which the previously posted material, was missing. But if it helps, here ya go. Today (actually, 70 years ago...) we have the EPR paradox and quantum entaglement which _hints_ that superluminal velocities are possible, but we also have: m0 mr = -------------------- sqrt (1 - (v²/c²)) which states that anything with a rest mass, m0, will have its relativistic mass, mr, tend toward infinity as its velocity, v, approaches that of light, c. Every experiment ever done to try to refute the veracity of the equation has confirmed that the equation is valid and, consequently, indicates that it is impossible for massive bodies to achieve the speed of light. However, we also have Cerenkov radiation, which is emitted whenever a massive particle exceeds the speed of light in the medium through which the particle is travelling... |
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On Thu, 4 Aug 2005 19:20:16 -0500, "DBLEXPOSURE"
wrote: "John Fields" wrote in message .. . On Thu, 4 Aug 2005 17:10:11 -0500, "DBLEXPOSURE" wrote: "John Fields" wrote in message ... On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE" wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer And how does this hint at super luminal velocities? I am curios. --- You have this annoying habit of deleting previously posted material which must then be searched out and reposted in order to properly reply to your queries, which invariably require reference to the previously posted material. I won't play that game. If you're serious, and you'd like to discuss the possibility of massive bodies achieving transluminal or superluminal velocities, aquaint yourself with Einstein, the EPR paradox, quantum entanglement, and then report back with what you've found. Otherwise, well, you know, **** off... -- John Fields Professional Circuit Designer No sir, I replied to your post from which the previously posted material, was missing. But if it helps, here ya go. Today (actually, 70 years ago...) we have the EPR paradox and quantum entaglement which _hints_ that superluminal velocities are possible, but we also have: m0 mr = -------------------- sqrt (1 - (v²/c²)) which states that anything with a rest mass, m0, will have its relativistic mass, mr, tend toward infinity as its velocity, v, approaches that of light, c. Every experiment ever done to try to refute the veracity of the equation has confirmed that the equation is valid and, consequently, indicates that it is impossible for massive bodies to achieve the speed of light. However, we also have Cerenkov radiation, which is emitted whenever a massive particle exceeds the speed of light in the medium through which the particle is travelling... --- OK, so what do you want to know? -- John Fields Professional Circuit Designer |
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"John Fields" wrote in message ... On Thu, 4 Aug 2005 19:20:16 -0500, "DBLEXPOSURE" wrote: "John Fields" wrote in message . .. On Thu, 4 Aug 2005 17:10:11 -0500, "DBLEXPOSURE" wrote: "John Fields" wrote in message m... On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE" wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer And how does this hint at super luminal velocities? I am curios. --- You have this annoying habit of deleting previously posted material which must then be searched out and reposted in order to properly reply to your queries, which invariably require reference to the previously posted material. I won't play that game. If you're serious, and you'd like to discuss the possibility of massive bodies achieving transluminal or superluminal velocities, aquaint yourself with Einstein, the EPR paradox, quantum entanglement, and then report back with what you've found. Otherwise, well, you know, **** off... -- John Fields Professional Circuit Designer No sir, I replied to your post from which the previously posted material, was missing. But if it helps, here ya go. Today (actually, 70 years ago...) we have the EPR paradox and quantum entaglement which _hints_ that superluminal velocities are possible, but we also have: m0 mr = -------------------- sqrt (1 - (v²/c²)) which states that anything with a rest mass, m0, will have its relativistic mass, mr, tend toward infinity as its velocity, v, approaches that of light, c. Every experiment ever done to try to refute the veracity of the equation has confirmed that the equation is valid and, consequently, indicates that it is impossible for massive bodies to achieve the speed of light. However, we also have Cerenkov radiation, which is emitted whenever a massive particle exceeds the speed of light in the medium through which the particle is travelling... --- OK, so what do you want to know? -- John Fields Professional Circuit Designer And how does this, (quantum entaglement) hint at super luminal velocities? I am curios. |
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On Tue, 02 Aug 2005 12:27:47 -0700, John Larkin wrote:
On Tue, 02 Aug 2005 18:28:32 GMT, TokaMundo On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin Before any difference could even be noted, the wired diameter would have to be over 16 mm. Not so. At 0.85 cm depth, current density is down to 1/e (ie, only 0.37 of) the surface density. That's pretty significant. 0.85 cm is pretty thick. 8.5 mm in fact. Double that to get 17mm. Unless the wire is larger than 17mm at 60Hz, the entire wire will carry current. VERY simple math. Current begins to fall off monotonically from the very surface for any wire size at any AC frequency. There's no hard "skin boundary", and the 1/e density is just a handy if arbitrary measurement point. I don't see why this needs arguing over. In a given situation, you just calculate the effects and decide how they affect things. Sometimes a 200% increase in resistance doesn't matter, and sometimes a 1% increase does. But skin effect does often matter in real situations at 60 Hz, and shouldn't be always/automatically discounted. OK, i'ts kinda counter-intuitive to us seat-of-the-pants techies, because we were raised with very small wire. 17 mm is a pretty hefty chunk o' wire! ;-) It seems I've learned something here. With great big huge fat wires, skin effect in significant even at 60 Hz. :-) Thanks! Rich |
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On Wed, 03 Aug 2005 18:31:32 +0200, Alexander wrote:
"TokaMundo" schreef in bericht ... On Wed, 03 Aug 2005 01:24:11 GMT, Repeating Rifle Gave us: Some people say that there is no such thing as a stupid question. Obviously there seems to be no shortage of stupid answers. Bill Yours certainly contributed abso****inglutely nothing, and would certainly fall into the "stupid answer" category. You remind me of a Firesign Theatre quote: "Who wona second world war.. you so smart?" Perhaps, if you are so informed, you should try giving an answer that actually has facts in it that are in sync with the topic of the thread, not merely its title. Someone once tolf me something, I believe it wass: "Don't feed the trolls", it seems to me that they are having a ball with food in abundance. It's a game to them. I don't know if it has a name, but I'd call it something like troll-baiting. I just usually ignore those symbiotic threads. ;-) Cheers! Rich |
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On Tue, 02 Aug 2005 18:40:17 +0000, Bob Myers wrote:
"TokaMundo" wrote in message ... On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin Gave us: Before any difference could even be noted, the wired diameter would have to be over 16 mm. Not so. At 0.85 cm depth, current density is down to 1/e (ie, only 0.37 of) the surface density. That's pretty significant. 0.85cm is 8.5 mm. That means that the wire has to be bigger than that number as a radius before the current flow anywhere else besides the entire wire. Wrong again. You seem to think that the current is uniform down to the "skin depth," and THEN it somehow starts to fall off. As John already pointed out, with seemingly unwarranted patience, that ain't so. Once again: "do the math." And this time, go beyond just using the skin-depth calculator on your favorite web site, and actually figure out what the EFFECTS would be (in terms of resistive loss, heating, whatever) of the skin depth at 60 Hz in a conductor otherwise seemingly-properly-sized for the 800A service that John mentioned as an example. You might be surprised by the result. Once again, STFW to the rescue: http://www.physics.ubc.ca/~mattison/.../lecture17.pdf In aluminum at 60 Hz, the skin depth is about 2 cm. And there's the formula right there. Cheers! Rich |
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On Fri, 05 Aug 2005 04:36:15 GMT, TokaMundo
wrote: John Fields Professional Self Aggrandizer Your lame ass also accuses people of "self aggrandizement". I have yet to see one post from you where you don't do the same. Funny, since you're no more than a fat tub of lard. --- LOL, if I were to write the single line: "Tokamundo is a good guy." and post it, you'd critcise it in your boring, predictable way. Probably by calling me a liar. -- John Fields Professional Circuit Designer |
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On Fri, 05 Aug 2005 04:43:00 GMT, TokaMundo
wrote: On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom" Gave us: No, you still don't see it. No. You don't see "IT". The heat in the center is conducted outwards through an imaginary cylindrical surface separating the innermost core from the next outer layer. No. THAT is for cylinders where the heat source is at the center. This heat source is throughout the medium. --- If the surface of the conductor is cooler than the core, which it _has_ to be by virtue of the fact that heat is radiating and being convected away from it, then there will be a temperature gradient from the conductor's center to its surface. However, since the resistance of the cooler regions of the conductor will be less than that of the hotter regions, the resistance of the cooler regions will be lower, allowing more current to pass through them. This will heat the cooler regions and more nearly make the conductor isothermal. So ya get a choice between a thermal gradient or a current gradient. Or maybe some of both. -- John Fields Professional Circuit Designer |
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"DBLEXPOSURE" schreef in bericht ... "John Fields" wrote in message ... On Thu, 4 Aug 2005 19:20:16 -0500, "DBLEXPOSURE" wrote: "John Fields" wrote in message ... On Thu, 4 Aug 2005 17:10:11 -0500, "DBLEXPOSURE" wrote: "John Fields" wrote in message om... On Thu, 4 Aug 2005 14:40:00 -0500, "DBLEXPOSURE" wrote: quantum entaglement, It's like you read my mind. --- Well, no. I was thinking quantum entanglement. -- John Fields Professional Circuit Designer And how does this hint at super luminal velocities? I am curios. --- You have this annoying habit of deleting previously posted material which must then be searched out and reposted in order to properly reply to your queries, which invariably require reference to the previously posted material. I won't play that game. If you're serious, and you'd like to discuss the possibility of massive bodies achieving transluminal or superluminal velocities, aquaint yourself with Einstein, the EPR paradox, quantum entanglement, and then report back with what you've found. Otherwise, well, you know, **** off... -- John Fields Professional Circuit Designer No sir, I replied to your post from which the previously posted material, was missing. But if it helps, here ya go. Today (actually, 70 years ago...) we have the EPR paradox and quantum entaglement which _hints_ that superluminal velocities are possible, but we also have: m0 mr = -------------------- sqrt (1 - (v²/c²)) which states that anything with a rest mass, m0, will have its relativistic mass, mr, tend toward infinity as its velocity, v, approaches that of light, c. Every experiment ever done to try to refute the veracity of the equation has confirmed that the equation is valid and, consequently, indicates that it is impossible for massive bodies to achieve the speed of light. However, we also have Cerenkov radiation, which is emitted whenever a massive particle exceeds the speed of light in the medium through which the particle is travelling... --- OK, so what do you want to know? -- John Fields Professional Circuit Designer And how does this, (quantum entaglement) hint at super luminal velocities? I am curios. I worked at a university and they achieved a zero speed of light inside a crystal. With this in mind almost everything can achieve super luminal velocities.. http://www.tn.utwente.nl/lf/project....=12&submenu=16 |
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"TokaMundo" schreef in bericht ... On 4 Aug 2005 05:36:24 -0700, "Autymn D. C." Gave us: John Fields wrote: And where are you, my dear? there Over there? I was here, now i'm there. You're not here so are yot there. Wait a moment I will go there Now I'm here which was there a moment ago and first here. Strange a moment ago was there here and here there. |
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