An interesting question from Veritasium on YouTube
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Sorry if I didn't describe the setup clearly enough. What I had in mind is using the 2 strands of a coax cable in the same way the 2 simple wires are used in the Veritasium setup:Baluncore said:The problem is very poorly specified, as are the questions.
For example:
- Inner coax strand connects: battery positve to light bulb terminal 1
- Outer coax strand connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart.
- The switch at the battery interrupts both coax strands.
How long will it take for the light bulb to receive any power after the switch is closed?
About 1m/c?
About 1s?
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But there are four simple wires in the symmetrical Veritasium setup. Describing a circuit diagram in words is not a wise move.A.T. said:What I had in mind is using the 2 strands of a coax in the same way the 2 simple wires are used in the Veritasium setup:
Fundamentally, the Veritasium setup uses the near end of a transmission line as a local component in a local circuit. It does that twice, once on the left hand side, and once on the right hand side of the circuit diagram.
What happens at the far end of the two coaxial cables? Are they joined to make the mid-point of one longer cable, or are they separately short circuited, while remaining 1 m apart?A.T. said:- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart.
Since there is no signal on the outside surface of a coaxial cable, the fact that it could be laid out like a hairpin, one metre apart seems irrelevant. It might as well be a circle with a circumference of one second. Maybe I misunderstand you.
If you connect a battery to the near end of a one second long transmission line, the connection transient will take one second to reach a globe at the far end of the line. The geographic path that the line takes is irrelevant. It does not matter how far apart the ends of the delay line are, the signal must follow the line. That is quite different to the Veritasium circuit where there is a reflection from a short circuit at the far end of each transmission line.
That should not be necessary. One switch pole should be sufficient to make or break the circuit at the battery. I wonder why you make that point.A.T. said:- The switch at the battery interrupts both coax strands.
What is a “coax strand”? Maybe it is another opportunity for a miscommunication.
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In my coax setup there is only one coaxial cable. At the far end (1/2 lightsecond from the battery & bulb) it simply bends around. By strands I meant the two coax conductors: outer & inner.Baluncore said:The problem is very poorly specified, as are the questions.
What happens at the far end of the two coaxial cables? What is a “coax strand”?A.T. said:- Inner coax strand connects: battery positve to light bulb terminal 1
- Outer coax strand connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each.
- The switch near at the battery interrupts both coax strands.
How long will it take for the light bulb to receive any power after the switch is closed?
If I understand you correctly it would take 1s in my coax setup?
Of course it is.Baluncore said:That is quite different to the Veritasium circuit ...
For the original Veritasium question it is irrelevant what happens at the far end. The bulb there will light up after ~1m/c, thus long before any information about the switch closing can even reach the far end.Baluncore said:... where there is a reflection from a short circuit at the far end of each transmission line.
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TBH I have no idea what your setup is, precisely, and I am looking through all of your posts.A.T. said:In my coax setup there is only one coaxial cable. At the far end (1/2 lightsecond from the battery & bulb) it simply bends around. By strands I meant the two coax conductors: outer & inner.
Why not draw up an equivalent circuit diagram in spice, like baluncore has done. It resolved, for me, the original issue I had with his explanation. Diagram is needed here, if you really want any proper discuss about it, and if you do put it into spice, then just hit the 'run' button and let spice tell you the answer!
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cmb said:TBH I have no idea what your setup is, precisely, and I am looking through all of your posts.
You only need to read this:
- Inner coax conductor connects: battery positve to light bulb terminal 1
- Outer coax conductor connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
- The switch at the battery interrupts both coax conductors.
How long will it take for the light bulb to receive any power after the switch is closed?
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Are you asking how long a signal takes to travel along a line that you have specified as being one second long, or is this a trick question ?A.T. said:How long will it take for the light bulb to receive any power after the switch is closed?
How do you define "to receive any power" ? Do you mean to receive any energy ?
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I'm supposed to guess what the switch is connected to?A.T. said:You only need to read this:
- Inner coax conductor connects: battery positve to light bulb terminal 1
- Outer coax conductor connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
- The switch at the battery interrupts both coax conductors.
How long will it take for the light bulb to receive any power after the switch is closed?
Just do the diagram in spice and press the 'run' button.
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No, just how long the signal takes, without additional constraints on how it gets there.Baluncore said:Are you asking how long a signal takes to travel along a line ...
What difference does it make whether I ask about "time to receive any power" or "time to receive any energy"?Baluncore said:How do you define "to receive any power" ? Do you mean to receive any energy ?
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The description is complete. (Yikes! get a grip)cmb said:I'm supposed to guess what the switch is connected to?
In this case the bulb will receive a very small signal through the air immediately but will heat up more than one second later, depending upon the coax propagation speed.
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A.T. said:- Inner coax conductor connects: battery positve to light bulb terminal 1
- Outer coax conductor connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
- The switch at the battery interrupts both coax conductors.
How long will it take for the light bulb to receive any power after the switch is closed?
Thanks for the reply. Is the initial very small signal through the air because the fields outside a coax cable are not completely zero in the transient phase, when the current builds up?hutchphd said:In this case the bulb will receive a very small signal through the air immediately but will heat up more than one second later, depending upon the coax propagation speed.
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1/c. The parallel external coaxial braids form a separate external two wire transmission line, shorted at the far end, where the line loops. Call that say Zo = 600 ohms, until the short circuit appears after about 0.7 seconds, because the coaxial external velocity factor will be greater than the inner. That circuit is closed on the centre conductor side by stray inductive and capacitive coupling between battery and bulb, terminals and wires.A.T. said:No, just how long the signal takes, without additional constraints on how it gets there.
Exactly 1 sec, when the signal on the inside of the coax gets to the globe.
Power is the rate of flow of energy. It is not energy.
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No. It is capacitive and inductive coupling of the local wires. The outside of the folded coaxial cable is a quite separate transmission line to the inside of the same coaxial line.A.T. said:Is the initial very small signal through the air because the fields outside a coax cable are not completely zero in the transient phase, when the current builds up?
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So it is a quadruple pole switch, that is on both sides of the battery at the same time?hutchphd said:The description is complete. (Yikes! get a grip)
Somehow?
Really, a diagram would have clarified this.
OK, it's over my head now then!... I'm out.
artis
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Well someone already mentioned this here that the bulb will light up later if the wires to the bulb have insulation around them. This I think is the simplest proof that the energy travels as fields and current is just the result of those fields. Insulation attenuates the field, it's propagation velocity decreases and boom there is your proof.
As for the transmission line and transients if the wire was not superconducting then the bulb would never light up I think simply due to the huge resistance, but a transient would still reach the bulb? I think this because the wires closer to the battery would have a higher voltage than the wire further away and since this is a transmission line the capacitance is equally spaced across the length so there would still be some capacitance in the parts of thew wire that are close to battery between those that are opposite to them and close to bulb.
So the transient would not light up the bulb but a sensitive voltmeter should be able to pick it up across the bulb?
As for the transmission line and transients if the wire was not superconducting then the bulb would never light up I think simply due to the huge resistance, but a transient would still reach the bulb? I think this because the wires closer to the battery would have a higher voltage than the wire further away and since this is a transmission line the capacitance is equally spaced across the length so there would still be some capacitance in the parts of thew wire that are close to battery between those that are opposite to them and close to bulb.
So the transient would not light up the bulb but a sensitive voltmeter should be able to pick it up across the bulb?
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cmb said:So it is a quadruple pole switch, that is on both sides of the battery at the same time?
The technical term would be double pole single throw (DPST) switch.
May I also recommend the Wikipedia article:
https://en.wikipedia.org/wiki/Transmission_line
Really nice succinct rendition (at least for my old brain)
.
cmb
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I took "The switch interrupts/connects both coax conductors simultaneously" to mean it interrupts both coaxial lines.hutchphd said:The technical term would be double pole single throw (DPST) switch.
I can accept the ambiguity may have been intended to imply a double pole on one coax only.
I feel that my proposition for a diagram to aid understanding was very reasonable.
On the subject of a double pole switch, in practice this would create an interesting outcome in that as both could not be timed perfectly, this may also have a consequence.
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Your circuit has internal and external transmission lines. Firstly there is the line inside the coax, say 70 ohm. Then there is the two wire line formed by the outside of the parallel braids, say 300 ohm.A.T. said:The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
Since you are relying on stray capacitance to complete the circuit during the first second, you might copy the Veritasium method by duplicating the TLs on both sides of the circuit.
Notes. DC bias is needed for the model to run, that is applied through nodes a, b, c and d.
The circuit is crossed over at the top to fix the polarity of the two paths so the bulb is driven by both sides in parallel, hence a 35 ohm bulb.
The first energy to reach the globe travels by the Zo of the external braid parallel line, but has high Zo.
Then the reflection of it's short circuit arrives later.
The signal by the internal coax path with the lower velocity arrives at one second. That line is modeled as two halves, joined at the bend. Some of the external braid energy reaching the bulb, heads backwards, down the internal transmission line.
OmCheeto
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How Wrong Is VERITASIUM? A Lamp and Power Line Story
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Yay! I thought I was losing my mind, but Mehdi came to the same conclusion that I did.
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Someone did an actual experiment to settle this. The short answer is that there is a small amount of current through the light bulb almost immediately upon flipping the switch, followed by the circuit ramping up to full current once the signal has time to propagate along the wire in both directions.
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Drakkith said:Someone did an actual experiment to settle this. The short answer is that there is a small amount of current through the light bulb almost immediately upon flipping the switch, followed by the circuit ramping up to full current once the signal has time to propagate along the wire in both directions.
OK, Youtube should be happy now.
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Those who understand transmission lines will agree with that analysis. There is no need for an experiment.Drakkith said:The short answer is that there is a small amount of current through the light bulb almost immediately upon flipping the switch, followed by the circuit ramping up to full current once the signal has time to propagate along the wire in both directions.
Those who do not understand the principles of transmission lines will argue that only their preferred part of TL theory needs to be considered.
OmCheeto
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Derek showed the light coming on instantly. He said it was a lamp that came on regardless of how little current flowed through it. But real lamps don't work like that. According to his definition, his magical lamp should always be on.hutchphd said:Which as near as I can tell differed very little (except for the hype) from Veritasium. Please tell me the substantive difference.
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OmCheeto said:Derek showed the light coming on instantly. He said it was a lamp that came on regardless of how little current flowed through it. But real lamps don't work like that. According to his definition, his magical lamp should always be on.
That's not really what was implied by Derek's statement because it defies engineering logic. It's easy to design systems that can detect a few nA of current from normal background noise or a hand clap with acoustic signals. No, that statement of the lamp being always on is bogus.
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Veritasium
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Thank you for making this video! While in hindsight I can clearly see that I should have gone into more detail with the explanation, I have really enjoyed watching all the response videos. For the record I was not suggesting the lightbulb lights at ANY current value but at some small but significant current value. I tested my LED bulb rated for 12V and found it turns on dimly when I apply 2V. There may yet be a follow up video coming. So thank you for this commentary - I'll incorporate it into any further work I do on this topic.
OmCheeto
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So why did his lamp come on full brightness? He says one thing and displays something which defies engineering logic. Do you not find that bogus?nsaspook said:That's not really what was implied by Derek's statement because it defies engineering logic. It's easy to design systems that can detect a few nA of current from normal background noise or a hand clap with acoustic signals. No, that statement of the lamp being always on is bogus.
Btw, this is starting to sound like science by litigation. Waiting for someone to point out the inflection in Derek's voice as an argument in their favor.
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His main point was that it is the fields that carry the power.OmCheeto said:Do you not find that bogus?
At no point did I think he actually had wires ##10^8## meters long attached to his apparatus. If you thought that he did then in fact it was deceptive (and I have bridge you might be interested in...)
Otherwise give me a break.
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OmCheeto said:So why did his lamp come on full brightness? He says one thing and displays something which defies engineering logic. Do you not find that bogus?
Btw, this is starting to sound like science by litigation. Waiting for someone to point out the inflection in Derek's voice as an argument in their favor.
Because logically (in the eyes of a engineer) he was talking about a small current detector like the one in a garage door opener that can turn on a light or open a heavy door from a tiny detection of current from a remote source. What's bogus is using this lamp detail as a sleight of hand from the facts of the video, "it is the fields that carry the power".
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