0356: "Nerd Sniping"

[Chalnoth]

So really the point is that the number you compute using Chalnoth's setup might not be the same thing you get if you actually tried to measure the resistance. Now given that we're talking about an infinite grid, perhaps it's silly to worry about this kind of thing. However, a finite grid will behave identically to an infinite grid within the time it takes for current to propagate to the boundary. So if you had an astronomical grid of resistors, maybe you would see some unexpectedly high resistance for a few seconds after you hooked up your multimeter.

Actually, the mere fact that you can take a relatively small finite grid of resistors as a very good approximate solution to the infinite grid indicates that you don't have to wait until the current propagates to infinity before your measurement is accurate: the further the current propagates, the less effect it has.

[DeadCatX2]

If you apply a voltage at one point but let the other float, you do not have a circuit. No circuit, no current flow. Remember, current must flow in loops.

I don't think this is true. To me it is natural to assume that since it's an infinite grid, the current would flow to infinity, so this is your circuit: 1A source at the origin, and then sink at the periphery at infinity.

It is natural for an infinite grid to have a periphery?

Either way, by using the terms "source" and "sink" you are implicitly creating a loop. The periphery is not floating, so it seems that we are in agreement in this regard.

I strongly believe that 1A source at (0,0) is the simplest way to think of the problem. In fact, I would say that the "official" solution is a factor of 2 off. Consider this: with the 1A source at (0,0), obviously, 1/4 A flows out to an adjacent node; the voltage drop between (0,0) and (1,0) is thus 0.25 V, so R_eq = 0.25 ohm.

Two things.

1) Are you aware that the problem is asking for the equivalent resistance between (0,0) and (2,1)?

2) A 1A current source at (0,0) with a sink at the periphery would find the equivalent resistance between (0,0) and the periphery.

[M.qrius]

Not sure if it's relevant to the discussion, but just throwing it out there:
According to wikipedia, the resistance between a point and a point at infinity would be infinity. If you were to do it in a 3d grid, however, the resistance would be zero.

[drwho]

Resistance is Futile! Give up! You will be assimilated! You will be exterminated! Thevenin and Norton will not save you here, boy.....muh ha ha ha ha

[tensorspace]

this is how i want to die. xD

[hnooch]

Not sure if it's relevant to the discussion, but just throwing it out there:
According to wikipedia, the resistance between a point and a point at infinity would be infinity. If you were to do it in a 3d grid, however, the resistance would be zero.

Actually, if you examine one of the earlier cited sources, in a 3d grid the resistance tends to approximately 0.505.

[phlip]

I just had a thought...

The way you find the equivalent resistance of a circuit is to hook up a voltage source, and measure the current that would pass through it (or vice-versa)... but that doesn't work if, when the meter isn't connected, there would be current flowing through the circuit.

For example: take a 1A current source and a 2Ohm resistor in a loop. Hook up a resistance meter that supplies 2V and measures the current. If you hook it up with the same polarity as the current source, it'll get no current and read inf Ohms... hook it up the other way around, it'll get 2A, and read 1Ohm. So at the very least, the equivalent resistance of the circuit is now directional... and it's also dependant on the size of the applied voltage. In short... it's not equivalent to a resistor.

Now, in a finite circuit, if there's current flowing, there has to be a power source... Kirchoff's laws see to that... so the basic rule is "get rid of all the power sources and then hook up meter". However, an infinite circuit can satisfy Kirchoff's laws, and still have current flowing with no power source... say that every horizontal resistor has 1A flowing left-to-right, and the vertical resistors have no current flowing... this satisfies both Kirchoff's current and voltage laws. If the current is flowing in such a manner, then the two nodes in the question are already 2V apart... if you hook up a 2V source, nothing will happen... and you'll get 0A through the meter. Hook it up the other way around, and it'll read something.

So, in short: I'm no longer convinced that this can be accurately measured, even by theoretically ideal equipment.

[Chalnoth]

So, in short: I'm no longer convinced that this can be accurately measured, even by theoretically ideal equipment.

Why not? The contribution of additional resistors drops off very rapidly, such that it's only the relatively nearby lattice that is important.

[silverdevilboy]

I wasn't sure if this has already been posted, but....

If you insert a 1A current at point A, and remove a 1A current at pt B, the current than spreads out evenly throughout the grid in the seperate cases, and you can just overlap the two grids, and look at any direct path from A to B, and find the voltage across the path, which will then be the voltage required to drive a current of 1A. The resistance is then trivial to derive.

[i like pi]

I say again, the fact that people are trying to find the answer to this just give more validity to the comic.

[GCM]

Ahh yes, electricity. I got a C for that.

They're all 1 ohm, right? So for those in series, it will be the sum of the resistance, and for parallel items, it will be 1/number of resistors. Considering that... I got nothing. Anyone else got it?

[IcyBallerina]

They're all 1 ohm, right? So for those in series, it will be the sum of the resistance, and for parallel items, it will be 1/number of resistors. Considering that... I got nothing. Anyone else got it?

We haven't even gotten to electricity in physics yet but this sounds like a harmonic series?

[Chalnoth]

Ahh yes, electricity. I got a C for that.

They're all 1 ohm, right? So for those in series, it will be the sum of the resistance, and for parallel items, it will be 1/number of resistors. Considering that... I got nothing. Anyone else got it?

Yeah, it ends up being solved by Fourier series. I posted an outline of the solution back in post #190:
viewtopic.php?f=7&t=16104&st=0&sk=t&sd=a&start=190

[Stanistani]

What if instead of an infinite grid, the loose ends of an 64-by-64 example were gathered up and linked to form a solid sphere (dropping out unnecessary links and such)?

Would that change the problem?

I'd study this problem, but I'm still working on my proof that all cats are Schrödinger's cat.

[Chalnoth]

What if instead of an infinite grid, the loose ends of an 64-by-64 example were gathered up and linked to form a solid sphere (dropping out unnecessary links and such)?

You could make a torus out of it instead and drop no links, as I don't think you could make a sphere where each node still had four resistors attached to it. But yes, it would change the problem dramatically, in either situation.

[Maseiken]

Screw that, Schrodinger was a jerk! I don't want him taking all the Kitties and putting them in boxes until they die/don't die!

[Startibartfast]

just a shot in the dark here, as electricity has alwats been my worst unit in physics, but could the answer be e?

[M.qrius]

Not sure if it's relevant to the discussion, but just throwing it out there:
According to wikipedia, the resistance between a point and a point at infinity would be infinity. If you were to do it in a 3d grid, however, the resistance would be zero.

Actually, if you examine one of the earlier cited sources, in a 3d grid the resistance tends to approximately 0.505.

Hmmm... Then this must be one of the 3 errors per page of wikipedia. Either that, or the source is wrong.

[Jamaican Castle]

I'd study this problem, but I'm still working on my proof that all cats are Schrödinger's cat.

The key to this proof lies in realizing that there is no cat. It is only the mind that is (not) dead.

Also, Schrödinger's cat tastes like chicken.

[iNap]

Actually, its taste is both chicken-y and non-chicken-y until you take a bite. *fires up the stove*

[eboyce]

The day after this comic posted, I couldn't do anything else for 10-15 minutes while I worked out the fraction of the earth that's directly illuminated by the sun at any instant of time. It's 50.7%, rather than 50.0%, due to atmospheric refraction and the nonzero size of the sun.

I'm assuming you simplified the earth into a smooth sphere for the purposes of this?
To truly find the most accurate answer to this question, you would need a specific moment in time, a topological mapaf at least 2 thirds of the world, and the precise velocity of the earths orbit, spin, and drift. As well as the exact location of each object at the chosen moment in time.

Yes, I assumed a spherical Earth and neglected shadows from mountains. The non-nerd part of my brain kicked in before I tried to take any of that into account.

[CNLohr]

Ok, well, even though the paper discussing one way to discuss it has already been posted by "OP Tipping" I'm a computer engineer, and I've been been one to trust that voodoo the math people or physics people do. One thing I do trust though, is computers, and in paticular, spice. If I would have known that the absolute answer was known, I probably wouldn't have gone forth with this, but I figure, since I already did it and have the results, I should post it. In retrospect, I do trust iNap's readings more than my spice printout. The difference between my answer and his answer of .8, ~.02673 ohms must be floating point error!

I wrote a .c program that produces a spice file that gets compiled, executed, piped into spice, piped out and into a results file. I have my new shiny Pentium Q6600 and I'm not doing anything tonight, so make -j4 seemed suitable. Ignoring the 2 hours and individual processes spiking up to around 400 or 500 MB, I simulated a variety of problems with a variety of # of resistors in my sample set all the way up to 231,200 resistors.

For instance:
make 180.txt (to make a 180x180 grid of resistors, test and run it.)
Direct diagonal, 64,800 resistors, .6366536 Ohms (Pardon the completely incorrect number of sigfigs)
Across one Resistor, 64,800 resistors, .5000169 Ohms (Pardon the completely incorrect number of sigfigs)

I was fairly satisfied with my results. So, I plopped in the coordinates for the Knight's move, and ran the test automatically on a variety of number of resistors. A couple results from the middle are as follows:

Spoiler:

4x4 grid (32 resistors): .9955357 ohms
15x15 grid (450 resistors): .7856096 ohms
40x40 grid (3,200 resistors): .7749523 ohms
120x120 grid (28,800 resistors): .7734296 ohms
weighing in at around 514MB memory used, and 66 minutes run time, 320x320 grid (231,200): .7732632 ohms

So, if anyone is interested in a numerical system for solving arbitrary check points and arbitrary sizes for the field of resistors, feel free to check out my code.

http://userpages.umbc.edu/~lohr1/xkcdresistor/

[iNap]

In retrospect, I do trust iNap's readings more than my spice printout. The difference between my answer and his answer of .8, ~.02673 ohms must be floating point error!

Yay! (As I said earlier, my multimeter's a cheap one from Radio Shack that wouldn't display more than one place after the decimal--and your answer rounds to mine. ^_^)

[DeadCatX2]

Ok, well, even though the paper discussing one way to discuss it has already been posted by "OP Tipping" I'm a computer engineer, and I've been been one to trust that voodoo the math people or physics people do. One thing I do trust though, is computers, and in paticular, spice.

From one computer engineer to another, I salute you! That was absolutely awesome.

The way you find the equivalent resistance of a circuit is to hook up a voltage source, and measure the current that would pass through it (or vice-versa)... but that doesn't work if, when the meter isn't connected, there would be current flowing through the circuit.

You misunderstand how a meter works (in particular, an ohmmeter). Of course you're going to get faulty measurements if current is passing through a resistor as you measure its resistance. The faulty measurement does not mean that resistance is sensitive to polarity, only that your measurement device is sensitive to pre-existing current flow.

Push 1 mA through a resistor and measure it with a voltmeter. Now push 1 mA the other way. You will get a voltage of equal magnitude. Using Ohm's law, you can deduce that the resistance is the same either way.

[phlip]

My point was that you can't control the external currents (that could be potentially running through the infinite grid without locally violating Kirchoff). Push 1A into the grid across that knight's move and measure the voltage... and you'll get a different answer depending on whether there's other current flowing across the grid.

[DeadCatX2]

My point was that you can't control the external currents (that could be potentially running through the infinite grid without locally violating Kirchoff). Push 1A into the grid across that knight's move and measure the voltage... and you'll get a different answer depending on whether there's other current flowing across the grid.

I think I understand what you're saying, but pushing current through the grid isn't going to make the resistors change value. You could still figure out the equivalent resistance by placing 1V across the knight's move and measuring the current used; by superposition, that answer is still valid even in the presence of external currents.

On a side note, when you try to measure ultra-small resistances, you often need to use a Kelvin Connection to supply the current through the resistor. The reason is that you develop a voltage across the power leads to the resistor (proportional to the current you're pushing times the resistance of the leads), and you're measuring this voltage in addition to the drop across the resistor. It's usually negligible, but for large currents or small (measured) resistances, the effect can be pronounced.

A Kelvin connection provides independent paths for the measurement wires and the power delivery wires. Thus, any voltage drop developed across the delivery wires due to their current will not show up on the measurement wires, because there is no juice flowing across those measurement wires.

[phlip]

I think I understand what you're saying, but pushing current through the grid isn't going to make the resistors change value.

Oh, I know that, I'm not claiming that the resistance is changing, I'm just claiming that the result you'll get from measuring it can change (ie, the measurement can be wrong).

You could still figure out the equivalent resistance by placing 1V across the knight's move and measuring the current used; by superposition, that answer is still valid even in the presence of external currents.

Except that you can't... gah, this is hard to word properly... here goes...

Let S1 be the system of the infinite resistors, with no external currents, just the 1V applied across the knight's move. Let S2 being the system with any/all of the external currents.
Now, the voltage in S1 divided by the current in S1 is the answer we need... but what you'll actually measure will be the voltage in S1 and the current in S1+S2... superposition doesn't do what you want it to do here.

I've already given an example of a set of external currents where you can apply 1V across the knight's move and get no current flowing through the voltage source. That fact doesn't change just because it can be considered a superposition of the current flowing one way because of the voltage source, and the current flowing the other way because of the external currents... it's the net of 0 that'll be measured.


I'm not saying the question is absolutely meaningless, or that the answer's wrong... I'm resonably convinced that if the question can be worded such that it has a real answer, the the strange fourier pi thing will be right. I'm just saying that the usual "stick in a voltage/current source, analyse the circuit, calculate the current/voltage, divide" procedure fails for infinite circuits. That's not to say a different definition of "equivalent resistance" won't give a specific answer.

[Chalnoth]

I'm just saying that the usual "stick in a voltage/current source, analyse the circuit, calculate the current/voltage, divide" procedure fails for infinite circuits.

No it doesn't. Why would you think it does?

[phlip]

I'm just saying that the usual "stick in a voltage/current source, analyse the circuit, calculate the current/voltage, divide" procedure fails for infinite circuits.

No it doesn't. Why would you think it does?

You mean aside from the reason I wrote in the post you're quoting?

[Chalnoth]

You mean aside from the reason I wrote in the post you're quoting?

Yes, because that reasoning is invalid. You won't measure any of the "S2" current at all.

[phlip]

Yes, because that reasoning is invalid. You won't measure any of the "S2" current at all.

Why not?

Again I'll bring up my example before... an infinite grid of resistors, all horizontal resistors have 1A going from left-to-right, vertical resistors have 0A. A voltage source over a knight's move, with 2V (with an ammeter in series).
This system:
* Is consistent with Kirchoff's Current Law on any finite set of nodes.
* Is consistent with Kirchoff's Voltage Law on any finite loop.
* Is consistent with the 2V voltage source.
* Has no current running through the voltage source.
Thus the ammeter will read 0A, so it will appear that the circuit has inifinite resistance.

Where's the flaw in this reasoning?

[Fat Tony]

I know nothing about how to go about solving the problem, nor could I solve an incredibly simplified version of it, but I just realized something:
That's the guy from "The Journal" who likes to hurt people!

[Chalnoth]

Yes, because that reasoning is invalid. You won't measure any of the "S2" current at all.

Why not?

Again I'll bring up my example before... an infinite grid of resistors, all horizontal resistors have 1A going from left-to-right, vertical resistors have 0A. A voltage source over a knight's move, with 2V (with an ammeter in series).
This system:
* Is consistent with Kirchoff's Current Law on any finite set of nodes.
* Is consistent with Kirchoff's Voltage Law on any finite loop.
* Is consistent with the 2V voltage source.
* Has no current running through the voltage source.
Thus the ammeter will read 0A, so it will appear that the circuit has inifinite resistance.

Where's the flaw in this reasoning?

What you've neglected is the impact of the measurement apparatus. You could well consider a purely horizontal current in the absence of any measurement apparatus. But once you apply the measurement apparatus, the system changes. The current from the measurement apparatus is added to the background current by superposition, and you measure some current.

[phlip]

What you've neglected is the impact of the measurement apparatus.

How so? The measurement apparatus is a 2V voltage source and an ammeter in series. They are connected across two nodes that are 2V apart. What more needs to be considered?

If you insist on bringing superposition in... the first thing you have to do is replace the voltage source with a short. If you do that, you can't have a purely horizontal current running through the circuit, it'd violate Kirchoff's Voltage Law (you'd have a short between two nodes that're 2V apart). You'd have some current running through the short, and thus through the ammeter. Now get rid of any external currents, and put the voltage source back in... you would of course have current running through it, and the ammeter (but in the other direction). The two currents from these two sub-circuits sum to zero, which is the superposed final answer.

[Chalnoth]

What you've neglected is the impact of the measurement apparatus.

How so? The measurement apparatus is a 2V voltage source and an ammeter in series. They are connected across two nodes that are 2V apart. What more needs to be considered?

If you insist on bringing superposition in... the first thing you have to do is replace the voltage source with a short. If you do that, you can't have a purely horizontal current running through the circuit, it'd violate Kirchoff's Voltage Law (you'd have a short between two nodes that're 2V apart). You'd have some current running through the short, and thus through the ammeter. Now get rid of any external currents, and put the voltage source back in... you would of course have current running through it, and the ammeter (but in the other direction). The two currents from these two sub-circuits sum to zero, which is the superposed final answer.

Yes, I think I see the problem now. The issue is that the way you have set the problem up, each row of resistors is independent, with an arbitrary voltage differential between them. So you'd still be inducing current between the rows. I suspect it'd end up exactly the same value as if the voltage was defined to be zero at infinity (which is what is usually done). I'll have to go back and look at the principle of superposition, but I'm pretty sure that this is correct.

Edit 2:

Yes, the principle of superposition has to hold, as it's a linear system. Thus the system's behavior must be identical to the one with infinite voltage at infinity, as you can decompose that system into the one with zero voltage at infinity with the voltage source, and the one with no source and voltage at infinity.

[phlip]

The issue is that the way you have set the problem up, each row of resistors is independent, with an arbitrary voltage differential between them.

Not so... the vertical resistors have no current flowing through them... so there's 0V across them. All the nodes in a certain column have the same voltage, the column to the right is 1V lower, the column to the left is 1V higher.

...if the voltage was defined to be zero at infinity (which is what is usually done).

Well, if there's a way this is usually done, then that changes matters... we didn't learn anything about infinite circuits in what EE I did at uni... I was under the impression that this was a rather obscure type of puzzle, so there wouldn't be any particular conventions about how to set it up... maybe not the smartest assumption I've made.

[edit]

Edit 2:
Yes, the principle of superposition has to hold, as it's a linear system. Thus the system's behavior must be identical to the one with infinite voltage at infinity, as you can decompose that system into the one with zero voltage at infinity with the voltage source, and the one with no source and voltage at infinity.

Certainly... but I'm still convinced that the measurement on the ammeter will be the superposed sum of the current that would go through it due to the voltage at infinity, and the current that would go though it due to the voltage source... whereas we only want the latter, to measure the equivalent resistance properly.

As you say, fixing the voltage at infinity to be 0 will make this a non-issue, as then it's just the superposition of the current due to the voltage source, and 0. My only concern was that there's nothing in the problem itself that fixes the voltage at infinity as 0...

[Chalnoth]

Hmm, there was an error in my logic. The equivalent system would require that you set the voltage at both infinity and the two test nodes in both situations. You can't superimpose the two systems without them both being the same system, and adding the testing apparatus changes the system.

However, I'm not so sure that there would be zero current flow through the testing apparatus. Consider a different, parallel scenario: that of a single battery connected to a single resistor. If we were to use a testing apparatus with the same voltage as the battery, we wouldn't measure zero resistance: if our testing apparatus had the same electrical properties as the battery, we would actually measure half the resistance, as the current load would be divided equally between our battery and the testing apparatus. I'm not quite sure how this applies to the infinite plane problem, but I suspect that what would be measured would be some function of the equivalent resistance with voltage equal to zero at infinity.

As for this usually being the case, yes, it is. You won't see this in circuits classes, of course, as you usually don't deal with infinite circuits in those classes. But we do infinite electrical systems in physics all the time in electricity and magnetism courses. Makes the math easier.

[DeadCatX2]

Why not?

Again I'll bring up my example before... an infinite grid of resistors, all horizontal resistors have 1A going from left-to-right, vertical resistors have 0A. A voltage source over a knight's move, with 2V (with an ammeter in series).
This system:
* Is consistent with Kirchoff's Current Law on any finite set of nodes.
* Is consistent with Kirchoff's Voltage Law on any finite loop.
* Is consistent with the 2V voltage source.
* Has no current running through the voltage source.
Thus the ammeter will read 0A, so it will appear that the circuit has inifinite resistance.

Where's the flaw in this reasoning?

Well, at first I thought you were saying that the difficulty in measuring the equivalent resistance created by adding a current flow across the grid meant that the resistance would be non-constant. I see now that you do not hold this opinion, but it is why I brought up superposition; the current flowing across the grid is not going to stop other current from flowing.

Besides, it's a grid of ideal resistors and there were no voltage sources specified. This leads one to assume that the periphery of the grid is floating, which is inconsistent with your 1A current, so I will say that it is your current which is inconsistent. But messing up in the presence of mysterious external currents is not a trait solely of infinite circuits, as you pointed out with the multi-meter scenario earlier. So, voltage/current injection and current/voltage reading are just as reasonably applied to infinite circuits as their finite counterparts.

Besides for that, your idea is actually pretty neat; it effectively eliminates the existence of one of the dimensions. But, the circuit only appears to have infinite resistance because of the specific combination of current and voltage sources that you've chosen. You've legislated something like a balanced wheatstone bridge, which is like stacking the cards so we have no choice but to divide by zero. If you were to apply any other voltage or current sources, the ammeter will register a non-zero value.

[ryan_hutchins]

I believe that the equivalent resistance is 13/36 ohms. If anyone knows about the correctness of this answer, please let me know. If it turns out to be right, I'll explain how I got it. I used a similar method to one that I developed as a freshman for doing a certain challenge problem on one of our problem sets.

[Simetrical]

The correct answer was already posted on the first page, twice, and multiple times since then. 4/π − 0.5. References were posted by page three, along with a general solution for the resistance between any two nodes. No offense, but you could at least try to skim the first few pages.