User talk:Constant314/Archive 9

This is an archive of past discussions with User:Constant314. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Hey, I stumbled across this trying to find something else. Excuse me for the late reply lol.

In integration rectangles are often used, and we don't stop using rectangles at the limit of ${\displaystyle f(x)}$ going to zero even if it is only a good approximation for area when ${\displaystyle f(x)>>dx}$ outside of which you maybe expecting the triangle-ish shapes to lead to inaccuracies. It only matters if the upper and lower limit converge to the same value. I won't try to rigorously talk about upper and lower limits in the context of the linked paper but I thought I had to comment since I feel like you left an interesting comment and took your time to go through my linked article. ^^ EditingPencil (talk) 20:31, 3 February 2024 (UTC)

I'm going to wait a while and then make the same improvements to the EM field article again. When I do, you won't revert them. Holographer1 (talk) 19:47, 9 February 2024 (UTC)

My apologies on reverting EM radiation. I had no issue with those edits. I was on the wrong article. Constant314 (talk) 19:49, 9 February 2024 (UTC)

The discuss of physical vs mathematical needs reliable sources and a talk page consensus. When you change a stable article and get reverted, the onus is on you to build a consensus.

Apologize for my reverts on EM radiation. I was on the wrong article. Constant314 (talk) 19:52, 9 February 2024 (UTC)

Regarding your revert https://en.wikipedia.org/w/index.php?title=Characteristic_impedance&oldid=prev&diff=1217577555 : I follow up on your revert comment "Already clarified in the same sentence. I'll be glad to help if you think I missed your intention.".

So the statement in the article as-is is the following:

> The characteristic impedance or surge impedance (usually written Z₀) of a uniform transmission line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that is, a wave travelling in one direction in the absence of reflections in the other direction.

So I assue "already clarified in the same sentence" refers to "a wave travelling in one direction in the absence of reflections in the other directions".

This statement does not make it clear to me what a "single wave" is. The term "single wave" is still vague.

Exemplary suggestive questions that come here up are:

• is a single wave a single period?
• is a single wave a single frequency?
• is it a "packet of energy" / impulse?
• What makes a signal/wave a "travelling wave"?

Thank you very much with providing more insights, e.g. by further extending the explanation, by linking to a definite article, or by creating such article for "single wave". Abdull (talk) 09:35, 9 April 2024 (UTC)

I see what you mean. I believe the intent of "single wave" means no wave traveling in the opposite direction. Constant314 (talk) 14:59, 9 April 2024 (UTC)

I seemed to have welcomed someone at the same time as you - double welcome is not that common, it was accidental - I keep thinking there is lag somewhere to not have seen yours already there - feel free to delete mine if you feel inclined, otherwise... JarrahTree 02:12, 27 April 2024 (UTC)

Nothing wrong with a double welcome. I am sure that there are lags. Nice to meet you. Constant314 (talk) 02:16, 27 April 2024 (UTC)

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Hello. You recently put a tag requesting clarification on the Skin effect article. In your edit summary you said that you could provide illustrations if contacted. I am not terribly skilled in graphics programs, so I was wondering if you could produce an illustration similar to the one on the right. Here the ${\displaystyle \Delta a}$ length is perpendicular to to the long axis of the wire and the ${\displaystyle \ell }$ length is parallel.

Let me know if the description is unclear. Thanks, XabqEfdg (talk) 01:39, 7 May 2024 (UTC)

That shows me what you want. I am not that good either, but this one looks pretty simple. Constant314 (talk) 02:21, 7 May 2024 (UTC)

How is this? Constant314 (talk) 04:25, 7 May 2024 (UTC)

Sorry for the delay in replying. That is excellent! Thank you for making that.

May I ask what used to produce such an image? XabqEfdg (talk) 14:43, 7 May 2024 (UTC)

Thank you for the compliment. I use Microsoft Visio. I do not recommend it, but I have it and I am used to making it do what I want it to do. It is a 2D drawing program. It is tedious to 3D depictions. But I am retired and do not mind. I have posted the image. If you want different symbols or fonts or features, let me know. Constant314 (talk) 15:39, 7 May 2024 (UTC)

Hi there. You reverted a small and useful cross reference I introduced in 1221518613. It looks like you tagged it as REFSPAM. My addition does not meet the definition of citation spamming in any why I can think of. If you think a different reference would be preferable, I'm happy to supply one. But I'd appreciate an explanation of this reversion. Thank you. Shonfeder (talk) 21:50, 8 May 2024 (UTC)

My mistake. I had not noticed that you added a name to list. I apologize. Constant314 (talk) 22:35, 8 May 2024 (UTC)

Ah, that explains it! Thanks for explaining, for undoing the erroneous reversion, and for spending time checking such things to maintain nice articles! Shonfeder (talk) 22:11, 9 May 2024 (UTC)

Hello again! Sorry for messing up the per unit quantities in the derivation, but isn't the equation for voltage ${\displaystyle V=-\int _{d/2}^{D/2}E\cdot {\hat {r}}dr=-\int _{d/2}^{D/2}{\frac {Q}{2\pi \epsilon _{o}r}}dr={\frac {Q}{2\pi \epsilon _{0}}}\ln {\frac {D}{d}}}$ still incorrect? Since ${\displaystyle \int _{r}^{R}{\frac {dr}{r}}=\ln {\frac {R}{r}},}$ this would give a voltage of ${\displaystyle {\frac {Q}{2\pi \epsilon _{0}}}\ln {\frac {d}{D}}}$ since it's negative. It seems like either the bounds of the integral should be swapped or (equivalently) the negative sign should be removed. XabqEfdg (talk) 03:00, 10 May 2024 (UTC)

Yes, it is possible that I have this backward. Let me ponder this for a bit. Constant314 (talk) 03:25, 10 May 2024 (UTC)

@XabqEfdg: Potential increases when you push a positive charge against the E field. Assuming that there is positive charge on the center conductor, then the E field points from the center conductor toward the outer conductor. If you take the path of integration from inner to outer, you are going in the same direction as the field, so the change in potential would be negative. Based on that, it would appear that

${\displaystyle V=-\int _{d/2}^{D/2}E\cdot {\hat {r}}dr}$ is correct.

But, we want a positive number because we are just trying to compute capacitance. So, it would make sense to use either

${\displaystyle V=\int _{d/2}^{D/2}E\cdot {\hat {r}}dr}$ or ${\displaystyle V=-\int _{D/2}^{d/2}E\cdot {\hat {r}}dr}$

So, which is better? People are used to integrating from smaller to larger, but they are also used paths that start at ground which is usually the outer conductor.

I think I would choose ${\displaystyle V=\int _{d/2}^{D/2}E\cdot {\hat {r}}dr}$ as being simpler. Do you concur?

When I make this sort of computation, it usually works better for me to assume the center conductor is ground and the charge is on the outer conductor. Then ${\displaystyle V=-\int _{d/2}^{D/2}E\cdot {\hat {r}}dr}$ is correct, but the expression for E picks up a negative sign because E directed toward the center conductor. Whichever way we fix this, we should explicitly state where the charge is and which conductor is grounded.

I agree that your second option (smaller to larger) is better since the form is more easily understood and it doesn't matter if we integrate from inner to outer or vice versa, as long as the grounding and charge are defined so that voltage is positive. If we let the outer conductor be ground and the inner one be negatively charged, that will give a positive voltage integrating from the inner conductor to the outer one and keep the outer conductor as ground. Would that be acceptable? XabqEfdg (talk) 05:31, 10 May 2024 (UTC)

To be sure, you want the charge on the inside of the outer conductor. If you aren't careful, you do get charge on the outside of the outer conductor, in which case it is a nice antenna. That is why you want the outer conductor to be ground. If you just want the integral to work, it doesn't matter, but if you are building something it does. Gah4 (talk) 10:41, 10 May 2024 (UTC)

@User:Gah4. I just assume a grounded center conductor for analysis. Thanks for joining the discussion.

@XabqEfdg: I am pulled more than one way.

1. We should make it no more complicated than necessary. While I know exactly what you mean, most readers will not.

2. We should not say anything that is incorrect.

Since we are only trying to compute capacitance perhaps we should say that the magnitude of the voltage is given by ${\displaystyle V=|\int _{d/2}^{D/2}E\cdot {\hat {r}}dr|=|\int _{d/2}^{D/2}{\frac {Q}{2\pi \epsilon _{o}r}}dr|={\frac {|Q|}{2\pi \epsilon _{o}}}\ln {\frac {D}{d}}}$ Constant314 (talk) 13:13, 10 May 2024 (UTC)

@XabqEfdg @Gah4 I had a nap! I see a simpler way to do this: just calculate C directly. I you have two concentric closely spaced cylinders of radius r separated by Δr then the capacitance per unit length is 2πrε/Δr. If we had more cylinders radially we just use the formula for capacitors in series. Thus

${\displaystyle {\frac {1}{C}}=\int _{d/2}^{D/2}{\frac {dr}{2\pi \epsilon r}}={\frac {1}{2\pi \epsilon }}\int _{d/2}^{D/2}{\frac {dr}{r}}={\frac {1}{2\pi \epsilon }}\ln {\frac {D}{d}}}$ Constant314 (talk) 16:27, 10 May 2024 (UTC)

The solution with infinitesimal series capacitors would work, and you wouldn't have to talk about charge and grounding, but couldn't you just use that formula for capacitors which are not closely spaced? I.e. "using the formula for the capacitance of coaxial cylinders gives...".

If you were to use voltage and the definition of capacitance, I think the explanation with the magnitude of the voltage is OK, since that is what really matters in this case. XabqEfdg (talk) 16:53, 10 May 2024 (UTC)

Yes, simply using the formula for coaxial cylinders was going to be my next suggestion. :) Of course, it is the same derivation. Likewise, characteristic impedance can be computed directly. Maybe we should just delete the derivation section and replace it with formulas from reliable sources. Constant314 (talk) 17:14, 10 May 2024 (UTC)

That might be best. All the transmission-line specific information is already covered in "Important parameters", so all this is section would be doing is deriving coaxial cable inductance and capacitance. If people are interested, they can check the reference given or a many other physics textbooks covering inductance and capacitance, but I am not sure the derivations belong in the coaxial cable article. XabqEfdg (talk) 18:01, 10 May 2024 (UTC)

Neither do I, per WP:NOTTEXTBOOK. Constant314 (talk) 18:09, 10 May 2024 (UTC)

The formula for capacitance comes from doing the integral, though that might go to the capacitor article. I do remember, so many years ago, doing this for a physics lab. You can also get the propagation velocity from them. The lab I had, had a special cable with spiral wound, high inductance center conductor, giving about 0.1c velocity. It is meant for a delay line. The fun one, though, is that the velocity decreases at about the length of the unwound center conductor. That is, the speed of the wave following along on the helix. Derivation of the propagation velocity might go here. Gah4 (talk) 18:20, 10 May 2024 (UTC)

See the new topic I added below this topic (might take e few minutes). Constant314 (talk) 18:33, 10 May 2024 (UTC)

I used to work for a company that made handheld instruments for the phone company. It included TDR and insulation testing. Bellcore reps came out to review the product. They asked us to have a large spool of new telephone twisted pair cable available. We got a new spool. I think it was 500m of 7 pair cable with an overall shield. The nominal round trip delay is 10ns/m. So, we hooked up the TDR and sure enough there is a pulse at 5000 ns. The try open circuit, short circuit, resistor, etc. That return pulse does behaves exactly as expected. But there was also a pulse exactly at the halfway point (2500 ns). Everybody was going nuts, but since it was exactly halfway, I figured it was an artifact of the wire being on a spool. It is as if the pulse got to the far end and then took a short cut back to the beginning. e unspooled the cable and the mysterious pulse went away. The test was successful. I mentioned it to one of the Bellcore guys and he said that always shows up on new factory wound spools of cable. He said you just have to unwind the spool and then rewind it by hand. The spools coming from the factory are tightly and precision wound. Re respooled the cable and the pulse did not come back. I have not yet found an explanation.

Any guesses? Constant314 (talk) 19:45, 10 May 2024 (UTC)

The physics lab I had, had spools of cable, though not factory wound. We had a pulse generator, series resistor, oscilloscope, spool of wire, and variable terminating resistor. That is, a poor-mans TDR. We measured the impedance by adjusting the terminator, and the attenuation by measuring the size of the reflected pulse. (Or maybe the size of the transmitted pulse.)

Your story reminds me of an Ethernet cable that mostly worked, but not quite as well as it should. So I was supposed to look at it. It seems that one end was mispaired. So one wire of one pair wasn't connected at one end, but one from another pair instead. You would think it wouldn't work at all, but there is enough coupling over the length.

Factory spooled cable has a nice layer wound in one direction, and then the next layer back again. It isn't so obvious, though, how the signal goes in that case. Gah4 (talk) 12:00, 11 May 2024 (UTC)

OK, another coaxial cable physics story. In college, we had a demonstration showing the similarity between resonance in open and closed end air columns, like organ pipes, and open and shorted end coax cables. Sounds nice. In the middle of the demonstration, the demonstrator figured out something was wrong. The open end air column corresponded to the shorted end coax cable, and vice versa. That should have been fine. But the next class, the demonstration was back, this time with a current probe on the oscilloscope. Now it worked out: closed matches closed, open matches open. Most would have ignored the problem, or explained the reason. But we got the second demonstration. And of course, never forget it. Gah4 (talk) 12:08, 12 May 2024 (UTC)

Makes sense. If you equate voltage to pressure and current to displacement, then a closed air column is equivalent an open-ended transmission line. The displacement at the end of a closed column is zero no matter how high the pressure. The current at the open end of a transmission line is zero no matter how high the voltage. Constant314 (talk) 15:01, 12 May 2024 (UTC)

I might be close to answering your question, but another physics demonstration first. This one, as the story goes, by cowboys. First, the wave velocity on a taught string or rope, based on the tension (T) and mass/length (mu) gives velocity sqrt(T/mu). Next, centrifugal force is ${\displaystyle mv^{2}/R}$. If you rotate a ring of rope, as the story goes a cowboy lasso, around its axis, the tension is ${\displaystyle T=\mu Rv^{2}/R}$, or just ${\displaystyle \mu v^{2}}$. If you distort the rope, such as kicking it, the distortion travels as a wave, both directions, with velocity v in the rotating frame, or velocity 0 and 2v in the non-rotating frame. Visually, so the story goes, you see the non-moving kicked dent, and not the fast moving 2v dent. Gah4 (talk) 11:07, 13 May 2024 (UTC)