Funny little problem - can you answer? - Page 1
Forum section: electronics / general

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 04, 2022, 01:45:08 pm
content: I have just seen a video and decided to ask you to solve this little interesting question.You have some RG58 coaxial cable and your trusted DMM ohm-meter.What would your meter show if you measured 1 yard of the cable, end to end?Now let's make the cable be insanely long, let's say from earth, around the moon and back again, an about 2 light-second journey.What will your Ohm-meter show if you measured the cable? Remember that your DMM can measure several times before the signal has traveled for the 2 seconds necessary to go all the way. :-)

author: Ian.M
author_rank: Super Contributor
timestamp: May 04, 2022, 02:11:55 pm
content: 'RG58' is more of a general description of the cable's approximate physical size and nominal impedance than an electrical spec. Reputable brand RG58 cables can have center conductor DC resistances from under 10 to over 35 milliohms/meter and an even wider range for shield resistance. However most DMMs are lousy at measuring low (milliohm) resistance values so odds are unless your DMM supports Kelvin (4 wire) resistance measurements, a yard of coax will read pretty much the same as shorting the test leadsIf you could connect a multimeter to the ends of a cable that extends in a loop out to lunar orbit, the most likely result is destruction of the multimeter due to induced currents in the cable!

author: rs20
author_rank: Super Contributor
timestamp: May 04, 2022, 03:20:52 pm
content: Are you connecting to the centre conductors or the shields at each end? Presuming the centre conductors, are the shields floating or grounded at the ends, or grounded regularly? Are the shields at the two ends connected?Assuming the shields are connected at the ends, and we're measuring between the conductors, I'd expect 100 ohms for the first second (assuming a 2 light-second loop), then a bunch of reflection nonsense, eventually settling on 10-35 millohms/meter * c * 2 s (stealing Ian.M's numbers).

author: free_electron
author_rank: Super Contributor
timestamp: May 04, 2022, 03:41:40 pm
content: your question is underdefined. where are the probes connected ?coax cable has 4 terminalsA : left side center conductorB : left side shieldC : right side center conductorD : right side shieldwhere are the probes ?what with the unused terminal ? float ? short ?

author: Benta
author_rank: Super Contributor
timestamp: May 04, 2022, 04:59:17 pm
content: That the meter will measure several times before the signal returns is irrelevant.You'll have some settling time on your result, but normally the current source for resistance measurement is constantly on (=DC), unless it's a special pulse measurement.Stop watching idiotic YT videos.

author: cdev
author_rank: Super Contributor
timestamp: May 04, 2022, 05:13:10 pm
content: I think Ian is probably right, the induced currents would make attempts at measurement useless.Quote from: FriedMule on May 04, 2022, 01:45:08 pm

author: Benta
author_rank: Super Contributor
timestamp: May 04, 2022, 06:23:31 pm
content: Quote from: cdev on May 04, 2022, 05:13:10 pmDC is DC.The bigger problem is, that a trip around the moon with RG-58 would exceed the capabilities of your multimeter.Inner conductor resistance is around 30 ohms/km. Times 750 kkm gives 22.5 Mohms, which is out the resistance range of normal multimeters.

author: SiliconWizard
author_rank: Super Contributor
timestamp: May 04, 2022, 06:23:56 pm
content: Yup for induced current.But we can thank Derek for a potentially endless stream of fun. =)

author: Benta
author_rank: Super Contributor
timestamp: May 04, 2022, 06:49:23 pm
content: Why would a current source (the multimeter) cause induced currents? The logic fails me.

author: Ian.M
author_rank: Super Contributor
timestamp: May 04, 2022, 07:16:21 pm
content: Quote from: Benta on May 04, 2022, 06:49:23 pmhttps://www.nasa.gov/magnetosphere/with a *large* loop of cable moving through it at over 1400 m/s. How can it *NOT* produce a current?You could always postulate that the ridiculously and impossibly long cable is non-inductively wound on a series of cable reels in a magnetically and electrically screened enclosure, which eliminates the orbital loop induced current problem.If the cable was Belden RG-58/U RF200 Riser cable #7807R , its core is 3.3 ohms/1000 ft, so 2x 385,000 km of cable would only have a core resistance of slightly over 8.3 megaohms, in range for most DMMs.However a bigger problem would be that even at $0.20 per foot, the cost of the cable would be over 500 million USD, and the world market for all forms of coax is estimated at only 2841.3 million USD in 2022, so the odds are there isn't enough production capacity to supply that much RG58 in much short of a decade. Then you've got the problem of jointing it as it certainly cant be a single length as it would weigh approx 28.5 kilotonnes for the cable alone!

author: Benta
author_rank: Super Contributor
timestamp: May 04, 2022, 07:57:14 pm
content: Quote from: bdunham7 on May 04, 2022, 07:22:08 pmTotally irrelevant. Do you really think he meant "around the moon" in a literal sense? Which part of "let's say" is misunderstandable?The point was about an extremely long cable. Don't muddy the waters, please.

author: DavidAlfa
author_rank: Super Contributor
timestamp: May 04, 2022, 08:02:06 pm
content: A huge short, at least for a few days/weeks, until the cable charges up?As per 82pF/m, given 2c length (600*10⁹), capacitance would be 49.2F !

author: Benta
author_rank: Super Contributor
timestamp: May 04, 2022, 08:15:30 pm
content: Quote from: DavidAlfa on May 04, 2022, 08:02:06 pmWho said the shield is connected?

author: magic
author_rank: Super Contributor
timestamp: May 04, 2022, 08:32:26 pm
content: Stop being practical.My guess: OP wants to hear 50Ω.

author: FriedMule
author_rank: Frequent Contributor
timestamp: May 04, 2022, 09:22:56 pm
content: I am sorry had to run out the door, and I am sorry for my badly formulated question.Let's say you had an Ohm meter that, let's say measures twice a second, it is able to measure the cables.You measure the center conductor. Your Ohm meter is not particular fantastic, so if you measure 1 inch or 10 inches will just be detected as about zero Ohms.I have deliberately chosen the RG58 because it is a general "standard" without being too technical in frequency, shielding or material.The question is mostly about what you would imagine your Ohm-meter will say when it returns a measuring result before the Ohm-meter's signal can have time to travel all the way. :-)

author: Benta
author_rank: Super Contributor
timestamp: May 04, 2022, 10:01:21 pm
content: Answered in #4.

author: cdev
author_rank: Super Contributor
timestamp: May 04, 2022, 11:31:22 pm
content: So, yes!The Van Allen belt

author: magic
author_rank: Super Contributor
timestamp: May 05, 2022, 05:19:19 am
content: Quote from: FriedMule on May 04, 2022, 09:22:56 pmIf the shield is connected to ground of the DMM on both ends then you will read 50Ω, which is the same thing you will read if the remote end (shield and core) are left floating.With no shield and no externally induced currents in the cable, two ends of the cable coming back to you form a twin-wire transmission line. You will read its wave impedance or something approximating it, because if the distance between the two halves isn't constant (like in a round loop) then the wave impedance isn't too.With a floating shield, not 100% sure, but still something similar.

author: Benta
author_rank: Super Contributor
timestamp: May 05, 2022, 06:19:26 pm
content: Quote from: magic on May 05, 2022, 05:19:19 amWhy would you read 50 ohms? It's a DC measurement. You're talking AC.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 05, 2022, 06:58:28 pm
content: Yes, destruction, shield grounded or no, because it's not nearly good enough of a shield: it's only effective above modest frequencies (10s, 100s kHz?), and even still not very (RG-58 is the cheapest 50 ohm for a reason). Such a condition will generate ~mHz frequencies (corresponding to the relative motion of Earth's field, the loop, and sure let's include solar wind in that too), the shield might as well not exist. Or, the shield is just another conductor on a parallel path, no different from twisted pair if one of the conductors had much higher resistance.Note that Earth's rotation by itself doesn't cause induction, because it's on a similar axis to the magnetic field. There's no such thing as a "rotating magnetic field" on an axis parallel to the lines of flux -- one of the conceits of the image of flux lines that can be overly confusing. But the axes don't match exactly, and there will be orbital or dynamic motion (the loop flopping around in space relative to its attachment point), so it will still move up and down through varying field strength and thus generate voltage.If it were a superconducting shield, including the means to keep it superconducting as it passes through the atmosphere, and enough cross-section to keep it from exceeding critical field strength internally (so, doubly so very much NOT RG-58), the center conductor can be effectively shielded and measure approximately nothing. (Likely it would still measure something, but due to stress/flex of the cable itself.) It would also short-circuit the ionosphere to ground, which must draw... I have no idea how much current, but it's going to get HOT from all that plasma it's literally lightning-rod-ing through.Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 05, 2022, 07:03:25 pm
content: Quote from: Benta on May 05, 2022, 06:19:26 pmIf measuring one end, core to shield, shield grounded, far end flopping in the breeze: hmm probably over that distance, induction still wouldn't quite cancel out and many volts would be measured (but perhaps not destructively so). If we say the cable is well constructed and doesn't accumulate induced voltage in this way, then we're left with the cable itself, and a cable of that length will measure 50 ohms the first time, then increasing amounts once the measurement pulse has had time to propagate the full length. Because the cable is lossy, the pulses won't actually reach the far end, just smearing out, diffusing along its length instead, and so the resistance increases gradually, say some ohms per second.Tim

author: magic
author_rank: Super Contributor
timestamp: May 05, 2022, 08:40:46 pm
content: Ignoring all the magnetic fields, solar wind, aether wind and other drama - just the cable - we have the following:Say that 1V is appiled to the cable, the cable charges to 1V at a speed of 20cm/ns or so.Say it has 100pF per meter, so 20pF/ns are flowing into it for the continuous charging.That happens to be 20mA for 1V applied to the input - 50Ω resistance.Checks outeditWhich is to say, a transmission line really works at all frequencies, down to DC, regardless of its length.Weird things only happen if it isn't terminated in its characteristic impedance and you wait long enough to discover this.Well, OK, I suppose you will also discover at some point that a real world cable is lossy, as T3sl4co1l says.

author: Benta
author_rank: Super Contributor
timestamp: May 05, 2022, 08:59:04 pm
content: Quote from: magic on May 05, 2022, 08:40:46 pmYou're still assuming that shield is grounded and core is connected, which is not given in the original question. Try to make your calculation with a simple conductor instead. (and PLEASE: without Van Allen belts, Earth magnetism, Aliens and other irrelevant voodoo).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 06, 2022, 01:54:25 am
content: The orbital mechanics seem relevant as, why provide such a detail otherwise; but perhaps it's merely in the collective consciousness because of certain recent videos.If we say it's a spool of wire instead, then it all gets much simpler, indeed.For which, testing impedance across far ends of the cable (it doesn't much matter whether either end is connected to the core or shield: the core simply adds Zo to the shield impedance at each end), is just some manner of, somewhere between a wire over ground plane (for whatever length of the wire is laid out on the ground*), to a spooled inductor (all the wire wound up in a more compact package).The smaller it is, of course -- and for that matter, if it's doubled back on itself, in a "noninductive" manner for example -- the less sensitive it will be to ambient fields. So, depending on if you have this laid out maximally, like, literally looping around the Earth multiple times -- you'll have quite a lot of gain to ambient fields, particularly at very low frequencies (mains, submarine communcations, Schumann resonance, even daily ionospheric cycles I suppose), so, still some things to account for in addition to its own self-impedance; or, if in a small package and cancelled out to ambient fields, maybe not much of anything besides what's applied to it.*Not that ground is very well defined, for short lengths; a wire in free space may be a better approximation. Which amounts to a higher Zo for that segment. Of which, a relatively large fraction of that impedance might be due to radiative loss rather than self-impedance. Depends on how good the ground is (soil? metal building?), what all is inbetween (wood or concrete floors?), etc.In any case, that much wire, spooled up, will certainly have quite some inductance. A meter might not notice much impedance at all between the ends, at least until it charges up -- the time constant could be, fairly unsurprisingly: many seconds!Tim