## 0356: "Nerd Sniping"

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### Re: "Nerd Sniping" Discussion

J Spade wrote:I came up with 0 < R < 3.

Cuz' resistors in parallel come to less resistance than the least resistant resistor, and the smallest line would have 3 ohms of resistance.

EDIT: But an infinite number of parallel series with potentially infinite resistances would potentially bring R infinitely close to 3?

It seems as though I was unnoticed at the bottom of the page.

Luppoewagan

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### Re: "Nerd Sniping" Discussion

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.
eboyce

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### Re: "Nerd Sniping" Discussion

Chalnoth wrote:
imatrendytotebag wrote:As a pure mathematician, I am interested in solving this rigorously.

As a pure mathematician senior in high school with not much experience in fourier series, I am interested in solving this rigorously in a different manner.

This makes me a masochist (I think) and also throughly sniped, but...

So far I think the most promising stab at a non-fourier-series solution was the method involving keeping a 1 and -1 voltage anchored in the plane and repeatedly changing each point into the average of its neighbors.

This can be solved using combinatorics. I offer no proof here, but consider a function:

f(n) is the number of ways you can get to the point (1,0) starting at (0,0) in exactly n moves (where a move constitutes going from one vertex to an adjacent vertex), such that the path never returns to (0,0) or goes through (2,1).
Similarly, q(n) is the number of ways you can get to the point (1,0) starting at (2,1) in exactly n moves without returning to (2,1) or going through (0,0).

Then the voltage at (1,0) is...

sum(n=0, infinity) (f(n)-q(n))/4^n

Then we use a similar method to find voltages at (0,1), (-1,0), (0,-1) and then subtract 1 from their sum for victory!

Note: This does not seem like a very good method of solution, unless it simplifies very nicely or helpful approximations can be made.
Note #2: The computation is much easier if you have 0 voltage at (2,1), then you don't need to have q(n) but you still need to avoid (2,1) when counting paths for f(n).

I tried thinking about this in terms of paths, but it turned out to be nontrivial for a simple problem.

Consider comparing a series-parallel circuit (one wire splits into two, each has a resistor, then they combine together and pass through one more resistor) to a parallel circuit (one wire splits into two, each has two resistors before coming back together). By a purely path summing formula, both circuits have the same number of paths, and the same number of resistors on each path. But the equivalent resistance is different.

In order to get the path sum approach working, you'd have to take into account how many paths share the same resistor as well. I'm not sure how it would work in general.

I don't think this method works in general. But, due to the symmetric nature of this problem, I believe it DOES work (for mathematical reasons, rather than physical ones). That doesn't make it a good method, though...
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imatrendytotebag

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### Re: "Nerd Sniping" Discussion

My weekend has been sniped. So has my research. I thought I'd be content once I would know the answer, but the solution raises more questions!

The main question is this: do you fix parameters for one node or for both nodes when modeling the measurement?

I wrote a brute-force routine that sets up and solves the system of linear equations for a large but finite grid. I get an answer, but it's a factor of 2 off! Other than that, it matches analytical solution to arbitrary precision (just gotta pick a grid large enough). This was driving me mad. This solution is consistent with all my equations, current conservation, KVL, everything!

Here's the reason the solution is different (this clarifies things somewhat, but not really). How does one model the equivalent R measurement? I am assuming a 1A *current source* at node 1, and then measure the voltage diff. between nodes 1 and 2.

An earlier poster (argonaut) has a solution that actually has the same problem:
Attach the network to a constant voltage source connected at the two marked test nodes 1 and 2. Set the voltage at these points equal to +1 volt and -1 volt. (You can pick any voltage, but these are convenient.)

This is not true! if you pick non-equal voltages, the code argonaut posted gives a completely different R_eq. The reason is this: for non-equal voltages, the net "external" current flows at node 1 and node 2 don't cancel out, and that somehow screws things up.

However, if one runs argonaut's code without fixing voltage at the second node - just letting it float - it spits out the answer that matches mine with the currents source (analytical / 2).

So I guess what I'm hoping to have someone explain to me is -- why does using a current source or floating voltage produce that factor of 2?.. I suppose at the end of the day it all comes down to how analytical papers define the measurement -- and I haven't read them yet because I want to figure that out on my own, now that I got the hint about working with currents and voltages as functions of Fourier-transformed lattice. But if you have insight, pls share.

And for the final question... If you actually figured out how to do this analytically with the Fourier trick all on your own -- what kind of a freak are you?
dnquark

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### Re: "Nerd Sniping" Discussion

dnquark wrote:And for the final question... If you actually figured out how to do this analytically with the Fourier trick all on your own -- what kind of a freak are you?

Oh, I have a Ph.D. in physics

The math I used I learned in solid state physics as an undergrad, though (solid state physics is the physics of solids, where we study stuff like superconductivity). So I probably could have done it back then...
Chalnoth

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### Re: "Nerd Sniping" Discussion

floyd4one wrote:
erl137 wrote:It looks like a preturbative problem to me . . . of course, I'm not as clear on the technical end of that as I'd like, so, somebody please feel free to correct me.

Actually this problem is solved using a 2D Fourier series. I'm guessing the answer is between 2/pi Ohms and 8/3pi Ohms (the solutions between 1 diagonal and 2 diagonal points away respectively)

Runs off to play with Mathematica.... Yes I am a physicist.

Edit: Here's the solution if you want to check yourself
Spoiler:
The answer is 4/pi-.5 Ohms == .773 Ohms.
As I predicted the resistance is between the other two cases which are numerically .637 and .849 Ohms.

This document seems to think the answer is 1/2 ohm.
http://arifatmath.co.uk/Physics_based_P ... ler_v5.pdf

Do you have the mathematica code you used?
omniron

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### Re: "Nerd Sniping" Discussion

omniron wrote:This document seems to think the answer is 1/2 ohm.
http://arifatmath.co.uk/Physics_based_P ... ler_v5.pdf

Do you have the mathematica code you used?

That paper doesn't appear to have any application to the problem at hand. A solution to which has been posted many times over.
Chalnoth

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### Re: "Nerd Sniping" Discussion

eboyce wrote: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.
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The Candylawyer

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### Re: "Nerd Sniping" Discussion

dnquark wrote:M
An earlier poster (argonaut) has a solution that actually has the same problem:
Attach the network to a constant voltage source connected at the two marked test nodes 1 and 2. Set the voltage at these points equal to +1 volt and -1 volt. (You can pick any voltage, but these are convenient.)

This is not true! if you pick non-equal voltages, the code argonaut posted gives a completely different R_eq. The reason is this: for non-equal voltages, the net "external" current flows at node 1 and node 2 don't cancel out, and that somehow screws things up.

Replied in detail by PM, but the reason my code gives the wrong answer when I don't set V1 = -V2 is because I cheated on the boundary conditions. At the outer edge of the finite grid, I set V = 0, but the real boundary condition is no current, I = 0 flowing off the grid. The latter is more complicated to code, you need to explicitly handle the special edge cases. I knew in advance that if I picked V1 = - V2, it wouldn't matter, so I did that instead.

The modification to my code needed to handle the edge cases is left as an exercise to the reader.
argonaut

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### Re: "Nerd Sniping" Discussion

iNap wrote:What about engineers, though? Should I be worried?

Not at all. If we get an approximation that's close enough, then we have a workable solution.
Elfer

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### Re: "Nerd Sniping" Discussion

argonaut wrote:The modification to my code needed to handle the edge cases is left as an exercise to the reader.

Just set it to (V1+V2)/2?
sab39

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### Re: "Nerd Sniping" Discussion

Since I lack the math skills to do something formal... I would just take it out to ten resistor paths. Add them up old school (there are quite a few delta-Y conversion to do). I imagine this would give an answer within 10% or less. Too lazy to actually do it though.
xygonn

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### Re: "Nerd Sniping" Discussion

vwlou89 wrote:Wouldn't the resistance be 0? I actually came up with a number that would satisfy this in my Calculus class: 1-.99999... It comes out to .000...0001 (infinite zeros). It can be represented as "the limit of y such that y=1/x as x goes to infinity". Since it could never really reach zero it would be as close to zero as was possible. Just a thought.

yes, very close to the resistance offered by a wire of the same length. resistors in series (the individual lines of resistors that make the grid) increase resistance, and resistors in parallel decrease resistance. Each row (or column) would have the same resistive value alone (infinite across the entire row/column), but because each row and column is hooked up in series, you'd get 1 over infinite resistance for each one. (1/infinity). Seeing as how there's an infinite number of parallel connections you'd get 1/infinity + 1/infinity ... 1/infinity which would give you = infinity/infinity, which should (but probably doesn't) = 1.

so your load on the circuit would essentially be (ideal resistors, remember) nill. no resistance. (or 1 ohm, depending on who you ask) grats me for figuring it out (i actually knew the answer was "none or so little to not matter" the second i saw it, just takes longer to type out than think!)

The answer could be (and probably is) 1 ohm, but it's the idea that boggles the mind, not the math.
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genewitch

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### Re: "Nerd Sniping" Discussion

genewitch wrote:which would give you = infinity/infinity, which should (but probably doesn't) = 1.

Infinity/infinity is indeterminate: it can equal many different things depending upon where those infinities came from. But this isn't simply a series/parallel circuit, and cannot be solved as one.

genewitch wrote:so your load on the circuit would essentially be (ideal resistors, remember) nill. no resistance. (or 1 ohm, depending on who you ask) grats me for figuring it out (i actually knew the answer was "none or so little to not matter" the second i saw it, just takes longer to type out than think!)

The answer could be (and probably is) 1 ohm, but it's the idea that boggles the mind, not the math.

Incorrect.
Chalnoth

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### Re: "Nerd Sniping" Discussion

argonaut wrote:Replied in detail by PM, but the reason my code gives the wrong answer when I don't set V1 = -V2 is because I cheated on the boundary conditions. At the outer edge of the finite grid, I set V = 0, but the real boundary condition is no current, I = 0 flowing off the grid. The latter is more complicated to code, you need to explicitly handle the special edge cases. I knew in advance that if I picked V1 = - V2, it wouldn't matter, so I did that instead.

The modification to my code needed to handle the edge cases is left as an exercise to the reader.

ok, that makes sense. non-equal voltages imply there must be a net flow of current through the boundary.

I'm still convinced that the simplest approach (and the one that is closer to experimental reality) would be to only fix the voltage (or the current) at only one node, and let the other one float. But doing this results in an answer that is off by exactly a factor of 2 from the alleged "correct" answer.

Can someone shed light on how multimeters actually work? as in, if you were to actually run this experiment, would the multimeter be injecting current or fixing voltage?
dnquark

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### Re: "Nerd Sniping" Discussion

Chalnoth wrote:
dnquark wrote:And for the final question... If you actually figured out how to do this analytically with the Fourier trick all on your own -- what kind of a freak are you?

Oh, I have a Ph.D. in physics

The math I used I learned in solid state physics as an undergrad, though (solid state physics is the physics of solids, where we study stuff like superconductivity). So I probably could have done it back then...

I almost have a Ph.D. in applied physics (well, scratch almost, but I've been in grad school long enough). The question still stands

Granted, solid state has never been my favorite to put it mildly, maybe that's why I haven't seen the connection. Can you point me to somewhere in e.g. Aschroft+Mermin that has the same math?
dnquark

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### Re: "Nerd Sniping" Discussion

dnquark wrote:ok, that makes sense. non-equal voltages imply there must be a net flow of current through the boundary.

I'm still convinced that the simplest approach (and the one that is closer to experimental reality) would be to only fix the voltage (or the current) at only one node, and let the other one float. But doing this results in an answer that is off by exactly a factor of 2 from the alleged "correct" answer.

Can someone shed light on how multimeters actually work? as in, if you were to actually run this experiment, would the multimeter be injecting current or fixing voltage?

You'd be fixing the voltage at the two points.

dnquark wrote:I almost have a Ph.D. in applied physics (well, scratch almost, but I've been in grad school long enough). The question still stands

Granted, solid state has never been my favorite to put it mildly, maybe that's why I haven't seen the connection. Can you point me to somewhere in e.g. Aschroft+Mermin that has the same math?

Well, my undergrad solid state text was Elementary Solid State Physics by Omar, but I think this is pretty standard material. The math comes in when looking at lattice vibrations, which will sometimes be called phonons.
Chalnoth

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### Re: "Nerd Sniping" Discussion

Actually, a brief look at Wikipedia suggests that in at least some instruments current is injected and voltage measured.

I've been holding off on reading the Am. J. Phys. papers... I want to solve this myself, dammit. I did look at the phonon chapters in solid state. Yuck, solid state. No help there. So I started trying to recall my EE-style Fourier analysis... Bingo, DTFT: used to deal with a priori non-periodic but discretized systems.

Ok, so let's use superposition and mentally separate current flowing in and out of each node. Current flowing out of each node will then be the same in all 4 directions due to symmetry. Put a current source on the [0,0] node. Then we simply have:

4 * i_[m,n] = i_[m+1,n] + i_[m-1,n] + i_[m,n+1] + i_[m,n-1] + DeltaFunction[0,0]

Let F[u,v] be the DTFT of i[m,n].

Well, now we have 4* F[u,v] = (2 cos[u] + 2 cos[v]) * F[u,v] + 1 using the old real space space-shift == fourier space phase-shift trick

So, to find i[m,n] we transform back...... except for, we can't, because (1/(2 - cos[u] - cos[v])) du dv doesn't integrate!!! Argh. So close yet so far.

Edit: ok, so what we are *really* after is the actual voltage drop at node (m,n), which in this notation is equal to
i[0,0]-i[m,n] (trust me on this now )

when we now write i[0,0] - i[m,n] all under one integral, the divergence disappears. But my answer is wrong. It is correct for m=0 or n=0 (e.g. points (0,1),(0,2), etc the integral taken numerically matches the analytical solution). No idea where to look for error. Grrr.
dnquark

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### Re: "Nerd Sniping" Discussion

dnquark wrote:Actually, a brief look at Wikipedia suggests that in at least some instruments current is injected and voltage measured.

I've been holding off on reading the Am. J. Phys. papers... I want to solve this myself, dammit. I did look at the phonon chapters in solid state. Yuck, solid state. No help there. So I started trying to recall my EE-style Fourier analysis... Bingo, DTFT: used to deal with a priori non-periodic but discretized systems.

Ok, so let's use superposition and mentally separate current flowing in and out of each node. Current flowing out of each node will then be the same in all 4 directions due to symmetry. Put a current source on the [0,0] node. Then we simply have:

4 * i_[m,n] = i_[m+1,n] + i_[m-1,n] + i_[m,n+1] + i_[m,n-1] + DeltaFunction[0,0]

Let F[u,v] be the DTFT of i[m,n].

Well, now we have 4* F[u,v] = (2 cos[u] + 2 cos[v]) * F[u,v] + 1 using the old real space space-shift == fourier space phase-shift trick

So, to find i[m,n] we transform back...... except for, we can't, because (1/(2 - cos[u] - cos[v])) du dv doesn't integrate!!! Argh. So close yet so far.

Edit: ok, so what we are *really* after is the actual voltage drop at node (m,n), which in this notation is equal to
i[0,0]-i[m,n] (trust me on this now )

when we now write i[0,0] - i[m,n] all under one integral, the divergence disappears. But my answer is wrong. It is correct for m=0 or n=0 (e.g. points (0,1),(0,2), etc the integral taken numerically matches the analytical solution). No idea where to look for error. Grrr.

Integral looks wrong (should have more than just unity in the numerator, but I think the denominator is correct). I think you're not properly dealing with the current conservation at each node (no current inflow/outflow at all nodes except two where you're testing, where the current inflow/outflow is +/- V/R (where V is the difference in voltage applied, and R is the equivalent resistance) The papers posted in this thread also use the Taylor series expansion of the cosines in the denominator to make the integral doable.
Chalnoth

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### Re: "Nerd Sniping" Discussion

Yeah, I'm looking at the paper, http://atkinson.fmns.rug.nl/public_html/resist.pdf, although not in minute detail. It seems to be essentially the same approach. The integrals are 1D with a Fourier series inside, as you said, because it's analytically doable, and this should amount to the 2D transform that I am doing.

Just to clarify: I am only injecting the current in one node, same as what they do in the paper; this is accounted for by the kroenecker delta. The current conservation should hold.

I *still* can't find an error though, and it's driving me nuts. I'm at my wits' end. Someone please help.

The basic equation is this:

4 * i_[m,n] = i_[m+1,n] + i_[m-1,n] + i_[m,n+1] + i_[m,n-1] + DeltaFunction[m,n] (1)

This is the same as the paper above (cf the unnumbered equation before Eq. 1 in the paper. They use V, I use i, but since V = I R, R=1, it's the same thing). Delta function represents their assumption (that they use later) that there's a 1A current source at the (0,0) node.

Finally, I know the equation is correct, because writing it as a system of m*n equations (in actuality less because i used symmetry) and solving it in matlab gives a solution that converges to the analytic one for m*n -> large.

Now, we follow the definitions of the DTFT from here:
http://ccrma.stanford.edu/~jos/mdft/Dis ... sform.html

assuming i[m,n] and F[u,v] are transform pairs. Let j be the imaginary unit.

Now transform (1) and use (a) the fact that xform of the delta fn is 1, and (b) "time shift -> phase shift" property

we get 4 * F[u,v] = 2 cos(u) F[u,v] + 2 cos(v) F[u,v] + 1

Trying to solve for F and transform back yields a diverging integral. But what quantity do we actually want? Well, we assume injecting 1A of current into the (0,0) node, so to find equivalent resistance at node (m,n) we just need to know the voltage drop. So, tracing the route from (0,0) to (m,n) we have

net voltage drop = ( i[0,0] - i[0,1] ) + ( i[0,1] - i[0,2] ) + .... + (i[m-1,n] - i[m,n]) = i[0,0] - i[m,n].

Now, i[0,0] = (1/2) integral 1 / (2 - cos[u] - cos[v]) du dv

and i[m,n] = (1/2) integral exp[j(u m+ v n)] / (2 - cos[u] - cos[v]) du dv

at the end of the day, we have (1/2) integral (1-exp[j(u |m|+ v |n|)]) / (2 - cos[u] - cos[v]) du dv from -pi to pi

Note that absolute values were put in around m and n to enforce symmetry. I'm not very happy that I had to do this by hand.

This integral converges and can be evaluated numerically. So, what does it give?

For m=0, n = anything, (or n=0, m=anything) it gives the answer that is exactly a factor of 4 off from the known analytical answer. For m!=0, it plain doesn't work -- wrong answer.

I cannot find the error here. Tried multiple numerical integration routines, so it's not computational. I'm giving up... someone figure this out pls
dnquark

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### Re: "Nerd Sniping" Discussion

the fact that people are trying to solve this just proves the comic.
Or something to that effect. Hell, I don't know.

i like pi

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### Re: "Nerd Sniping" Discussion

I have read the solutions referenced so far (e.g. http://www.geocities.com/frooha/grid/node2.html), and it looks like they all rely on Kirchoff's Law. I don't see why the assumptions needed to apply Kirchoff's law really apply here. In particular, it's not clear the charge density on our grid is constant.

Please consider that since we have an infinite grid, our grid has an infinite capacity to store charge (Cantor's hotel problem). So as we introduce electrons at the injection point, our grid is free to disperse the extra charge density radially out to the "infinite edge" of the grid, and the charge density at each point will continually oscillate as the extra charge density disperses past it. This is a fundamentally different situation than a finite grid, and models which assume a finite system simply won't work.

Thus not only are solutions based on Kirchoff invalid, neither are solutions based on a series of finite approximations. The solution to this problem is *not* the limit of solutions to finite problems, because no finite grid (no matter how large) reasonably approximates what happens on an infinite grid, at least not without taking into account the actual charge density dispersal over time.

I suggest the grid will act like an "infinite capacitor", accepting as much charge as we load onto it. We will never be able to achieve a sustained charge density any higher than what we started with. Thus actually the effective resistance between any two points is "infinite." To be more precise, no current flow will ever be observed between any two points on the grid, no matter what the applied voltage is.
landijk

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### Re: "Nerd Sniping" Discussion

landijk wrote:Thus actually the effective resistance between any two points is "infinite." To be more precise, no current flow will ever be observed between any two points on the grid, no matter what the applied voltage is.

I agreed with everything you said (zpart from you attributing Hilbert's Grand Hotel to Cantor) up to this point... but I don't see how this follows.

Certainly, given that the plane is infinite, you can spin a Grand-Hotel-style setup where you can hook up a current source to the grid with no sink... and the current will just propagate out across the grid. Each node can individually satisfy Kirchoff's current law without the circuit as a whole doing so.

However, surely that requires current to flow between the nodes? Like, if you injected I into the node (0,0), I/4 would go to each of (1,0),(0,1),(-1,0),(0,-1), and then I/12 would flow out of those into their neighbours, and on it goes... spreading out to the extent of the grid. If the grid was bounded, you'd eventually be unable to satisfy the law at the corners, but since it's not bounded, that won't happen.

Still, I find it hard to believe that, if you put 3V across a knight's-move of the grid, that you wouldn't get at least 1A flowing, though one of the 3ohm paths between the nodes... regardless of how much current flows off to infinity.
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phlip
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### Re: "Nerd Sniping" Discussion

Yup, Kirchoff's law applies just fine, because it's nothing more than a statement of local charge conservation. Local charge conservation doesn't care what happens at infinity.

The only wrinkle that infinity throws into the problem is that in order to solve for the voltage at every point on the grid, you do need to know the potential at infinity. After all, if you had a voltage at infinity that, say, was positive in the +x direction and negative in the -x direction, then you'd have this continuous current flow through the grid.

However, it turns out that this little detail is irrelevant: since the system is linear, and since we are only testing the grid locally, then whatever the potential at infinity is, the full solution of the potential of each grid point will be a superposition of the solution for the actual potential at infinity and the solution where we set the potential at infinity to zero. What this means is that no matter what potential we place at infinity, as far as our finite little voltmeter is concerned, the potential might as well be zero. In essence, asking for the equivalent resistance between two points of finite separation let's us utterly ignore what happens at infinity. You might indeed have some current flow that is not caused by your testing apparatus, but your testing apparatus will only detect the voltages it induces, and won't be able to see this overall current flow (in this situation).
Chalnoth

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### Re: "Nerd Sniping" Discussion

I'm going to suggest that I'm the hardest-sniped of all:

Mmhm. 200 one-ohm resistors. A far cry from infinity, to be sure, and the thread hadn't made it all the way to the posts suggesting that finite representations aren't that useful yet--but I had to give it a shot. Another disclaimer: My multimeter isn't the greatest in the world. (It's a rather basic one from Radio Shack.)

With that said:

Step 1: Cut a hole in the...ahem, starting over.
Step 1: 7 resistors--1.5 ohms.

Step 2: 31 resistors--1.1 ohms.

Step 3: 71 resistors--0.9 ohms.

Step 4: 127 resistors--0.9 ohms.

Step 5: 199 resistors--still 0.9 ohms, though I suspect it was getting closer to 0.8 since I occasionally got that figure instead.

That's about as far as I'm willing to go. (To my credit, I didn't do this in the middle of the road so I could get hit by trucks.) My girlfriend suggested I use it as a wall hanging, which I very well might do. Other suggestions?

iNap

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### Re: "Nerd Sniping" Discussion

Does the law of conservation of charge apply to our infinite grid? In other words, is it reasonable to assume that if we apply a 1A current at one point on the grid we will read a -1A current at any other point?

I argue that conservation does not apply because of the Hilbert's Hotel paradox-- there's always room for more electrons in our grid, and adding them won't raise the overall charge density. Thus the current we apply will simply dissipate over the grid and we won't measure the -1A current anywhere.

Those applying Kirchoff's law to this problem clearly assume that conservation does apply, and I think it would advance everyone's understanding if someone would make the case explicitly. Alternatively, perhaps someone has a clearer case against conservation than my admittedly brief effort. In any case, I think there are subtle problems with the unboundedness of the grid which could use some precise explanation.

p.s. I don't know where I got the idea it was Cantor's hotel earlier. Sorry about that, Hilbert!
landijk

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### Re: "Nerd Sniping" Discussion

I claim that local conservation of charge still applies. That is, if 1A goes into node X, then 1A must come out of node X. Thus Kirchoff's current law still applies, when applied to individual nodes, and finite subcircuits.

Global conservation of charge need not apply over the entire infinite grid. Kirchoff's law doesn't apply when applied to the grid as a whole... it's possible to have a source but no sink, or vice-versa.
While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

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### Re: "Nerd Sniping" Discussion

Yup. All laws of physics are only defined locally, after all. Any global conservation laws that we derive are fundamentally a consequence of local physics. And local physics has no knowledge of what's going on far away, so the size of the grid is irrelevant to how we apply conservation of charge (which is a consequence of fundamental physics).
Chalnoth

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### Re: "Nerd Sniping" Discussion

How does the speed of light and special relativity affect this?
sab39

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### Re: "Nerd Sniping" Discussion

sab39 wrote:How does the speed of light and special relativity affect this?

Since it's a DC problem, it doesn't

One could, of course, solve the problem for a time-varying voltage source, and then the speed of propagation of the current would be taken into account. However, if I remember my electricity and magnetism correctly, different materials have different propagation speeds, and you'd also need the size of the lattice. Oh, and it'd be a horrifically beastly problem to solve. By contrast, the problem as it stands is child's play.

Edit: And no, though I was sniped quite well by the original problem, there's no way in hell I'd attempt the vastly more difficult time varying problem.
Chalnoth

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### Re: "Nerd Sniping" Discussion

Chalnoth wrote:
sab39 wrote:How does the speed of light and special relativity affect this?

Since it's a DC problem, it doesn't

Well, it was claimed that an infinite grid is fundamentally different than any kind of finite grid, because of the lack of boundary conditions. Surely the fact that there's at any given time only a finite area that the voltages could possibly have propagated to in a relativistic universe, means that the grid is effectively finite anyway?
sab39

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### Re: "Nerd Sniping" Discussion

sab39 wrote:Well, it was claimed that an infinite grid is fundamentally different than any kind of finite grid, because of the lack of boundary conditions. Surely the fact that there's at any given time only a finite area that the voltages could possibly have propagated to in a relativistic universe, means that the grid is effectively finite anyway?

Because we assume that the voltage has been there for all time, and isn't applied suddenly, it doesn't make a difference.
Chalnoth

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### Re: "Nerd Sniping" Discussion

Chalnoth wrote:Because we assume that the voltage has been there for all time, and isn't applied suddenly, it doesn't make a difference.

Gotcha.
sab39

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### Re: "Nerd Sniping" Discussion

Earlier I wrote that the infinite grid was like an "infinite capacitor," but really I should have said it's an electrical ground, because that's the right word for what I was trying to say.

Consider this -- if you have an infinite pool of water (with no edges) and you dump in one gallon, do you raise the water level? No! Well, maybe. It depends *where* and *when* you measure, and how you define "water level". What actually happens when you dump in the water is you create a ripple, which is a local oscillation in the height of the water. So you will measure fluctuations in the water level as the ripple spreads. However, the as the ripple spreads the oscillations dampen, and eventually the ripple will become unmeasurable. You will measure the height of the water to be the same everywhere, and that height will be the same height you started with.

Now what if the infinite pool has two edges (so it's like the quadrant x>=0, y>=0), the water is filled exactly to the top of the pool at the edges, and you dump in the gallon of water somewhere in the pool? How much water will spill out? Well, that's a really hard problem to solve. The spillage you get isn't due to the increase in the water level per se, it's due to the ripple effect. I doubt anyone can really solve this one. It certainly depends on exactly *how* you add the water, you have to worry about surface tension, etc. Similarly, what if instead of one gallon, you continuously pump in one gallon per minute at some point? What is the rate at which water overflows the sides of the pool? Again, this is an impossibly difficult problem to solve, and depends on how the water is pumped in.

For the final observation, suppose you replace the pool with a finite pool. Then everyone will agree that if you dump one gallon in, one gallon will overflow (ignoring surface tension). Similarly, if you pump in one gallon per minute, one gallon per minute will overflow. Why do we all agree on this? Because we all know the *global* behavior of the system, namely that "water in == water out," and we ignore the *local* behavior (the ripple effect, surface tension, and so on), because the local behavior is too hard to deal with. Thus if you have a sequence of pools, each one larger than the next, the overflow rate will be the same for each pool-- one gallon per minute. Then if you compute the limit of the overflow rates, that too will be one gallon per minute.

But we already saw that in the case of the infinite pool, the rate is very difficult to compute, but certainly less than one gallon per minute, because some of the added water ripples away from the edge and never spills over. In particular, we can imagine that if we pump in the water far enough from the edges of the infinite pool, the ripple will effectively die out by the time it reaches the edge (i.e. it will become so small that other physical forces become much more important), and actually *no* water will spill over. Thus the limit of the rates of the finite pools is not equal to the rate of the infinite pool. Infinite systems don't work like the "limit" of finite systems!

Similarly, I insist the problem with the infinite electrical grid is either to difficult to solve, or you simply wave your hands and say that the grid is in effect an electrical ground and leave it at that. At the very least we must admit that *some* of the current we add will in effect dissipate over the infinite expanse of the grid (a phenomenon we don't have to worry about with finite grids). The physical model which underlies Kirchoff's law, etc. has to do with how variations in charge density equalize over time (which I admit is a *local* phenomenon), but the assumption that "charge in == charge out" is a *global* phenomenon (just like the water level), and I think that assumption is incorrect. Infinite systems don't work like finite systems. The only option in the case of the infinite grid is to compute the charge "ripple", but I don't think anyone is really prepared to model how an applied charge actually propagates over time on our grid. It certainly depends on how you apply the charge, etc., so you simply cannot say there is some intrinsic effective resistance between the two points.

The problem probably should have been worded differently:

*****

Suppose you have a sequence of grids. The first grid looks like this:

A - B - C
| | |
D - E - F

The capital letters represent nodes, and the lines represent 1-Ohm resistors. To obtain the next grid from some grid, add a band of resistors around the edge to form a new rectangular grid. Now in each grid, we may compute the effective resistance between A and F. Thus the sequence of grids gives us a sequence of resistances, which by Kirchoff's law, etc. is bounded below by zero and decreases monotonically. Thus by the bounded convergence theorem the sequence must converge to some limit. Compute the limit.

*****

p.s. I wonder if it makes a difference whether at each stage we add the resistors in a band around the outside (so that A stays in the "center" of the grid) or just an extra row and column at the bottom and left, so that A always is at the upper-left corner of the grid.
landijk

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### Re: "Nerd Sniping" Discussion

It was silly to argue that no water would overflow the edges of the infinite pool if we added our gallon of water far enough from the edges. Yes, it is true that we could add it so far out that by the time the ripple reached the edge of the pool its amplitude would be less than the diameter of an atom or something, but that is true of large pools bounded on all sides as well. Really we are interested in continuous phenomena, even though strictly speaking the continuous model is incorrect.

I think the right way to look at the infinite pool problem is that you have some initial unbalanced state (there's a "bump" somewhere in the water level), and the water level behaves according to some differential equation. The trouble is the differential equation is different at the edge of the pool than it is away from the edge.

The infinite grid problem could be handled similarly, but I haven't quite figured out how to set it up. Can anyone help? The idea I think is to fix a "high" charge density at the first node to represent your energy source and then look at how the charge density at the second node changes with time. Maybe this problem isn't actually impossible, after all...
landijk

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### Re: "Nerd Sniping" Discussion

Well, no current escapes to infinity in this problem, though. All of the current that enters through one point exits the other. If dV is the difference in the voltages between the two points, and R is the equivalent resistance of the circuit, then at all times dV/R is the current injected into one point, and that exits the other.

It is furthermore useful to recognize that this current inflow is constant. Thus the system is at an equilibrium with the inflow and outflow of the current, resulting in a variation of the voltage across neighboring resistors.
Chalnoth

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### Re: "Nerd Sniping" Discussion

OK, I think I finally understand what Chalnoth is saying. Chalnoth is saying that in this problem we assume there is a 1A current flowing into some node x and out some node y a knight's move away. We also assume the charge density on the grid is uniform, but we don't really care *how* things got this way. Then, given these assumptions, we can use Kirchoff's current and voltage laws to set up the difference equations and use a Fourier transform to solve them. We get the voltages, and from there the resistance. I don't see any problem with that at all.

The problem I had was that I was assuming the grid was at ground state, and then we came along with our multimeter and applied voltage across x and y to find out what the resistance is. In that case, we can't use Kirchoff's current law (or at least if we do all our equations will have an extra time-dependent term which represents the change in charge density at each node), because when we apply our voltage we cause a disturbance in the charge density.

If the grid were finite, we could just wait until the charge density stabilizes and then use Kirchoff's laws as usual. We know the charge density will stabilize because the charge is restricted to a finite area. Indeed, this is why most of the time no one worries about charge density when doing circuit computations. However, with an infinite grid there are two possibilities, depending on whether the charge dissipates over the grid faster than our voltage source can supply it. It seems clear that in the case of fast dissipation, the current we measure at y over time will not approach the current we are supplying at x, and in the slow dissipation case it will as charge builds up around x.

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.

Anyway, I hope this way of stating the issue is helpful. I'm afraid all that nonsense about Hilbert's Hotel, the law of conservation of charge, and infinite swimming pools really sort of missed the point. Sorry for thinking out loud!
landijk

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### Re: "Nerd Sniping" Discussion

dnquark wrote:Ok, so let's use superposition and mentally separate current flowing in and out of each node. Current flowing out of each node will then be the same in all 4 directions due to symmetry.

This can only occur if the periphery of the infinite grid is grounded and you apply a voltage at a single node. Periphery of an infinite grid...hm...something seems paradoxical about that.

dnquark wrote:I'm still convinced that the simplest approach (and the one that is closer to experimental reality) would be to only fix the voltage (or the current) at only one node, and let the other one float. But doing this results in an answer that is off by exactly a factor of 2 from the alleged "correct" answer.

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.

dnquark wrote:Can someone shed light on how multimeters actually work? as in, if you were to actually run this experiment, would the multimeter be injecting current or fixing voltage?

Old meters applied a voltage, new meters supply constant current.

genewitch wrote:yes, very close to the resistance offered by a wire of the same length. resistors in series (the individual lines of resistors that make the grid) increase resistance, and resistors in parallel decrease resistance. Each row (or column) would have the same resistive value alone (infinite across the entire row/column), but because each row and column is hooked up in series, you'd get 1 over infinite resistance for each one. (1/infinity). Seeing as how there's an infinite number of parallel connections you'd get 1/infinity + 1/infinity ... 1/infinity

Er, no. Resistance is proportional to path length, but not all paths are the same length; some are shorter than others, and as you add paths their resistance will go up, but the original shorter paths will maintain their original resistances. So it's more like 1/3 + 1/3 + 1/3 + 1/4 + 1/4 + ...

phlip wrote:Kirchoff's law doesn't apply when applied to the grid as a whole... it's possible to have a source but no sink, or vice-versa.

I have never heard this before in all my years of electrical engineering. Current flows in loops, period. No loop, no current. Could you enlighten me?

landijk wrote:[amazing pool analogy]
Similarly, if you pump in one gallon per minute, one gallon per minute will overflow. Why do we all agree on this? Because we all know the *global* behavior of the system, namely that "water in == water out," and we ignore the *local* behavior (the ripple effect, surface tension, and so on), because the local behavior is too hard to deal with.

Small problem. To measure the equivalent resistance, you need to have a loop. Current must not only go in somewhere, but it must also go out somewhere else. In the infinite resistor grid example, current would go in point A and out point B. Your pool analogy is trying to essentially "ground" the periphery of the grid (because that is the ultimate destination of the water/current), so that all current flows outward with the edge as the destination, but that is not how the circuit would behave because the reference point is located a finite distance away.

landijk wrote:At the very least we must admit that *some* of the current we add will in effect dissipate over the infinite expanse of the grid (a phenomenon we don't have to worry about with finite grids).

The current will dissipate over the entire grid, yes, but remember that all of the current flowing into point A is trying to get back to point B. The longer the path, the less current that flows, but it STILL comes back to point B. As you get farther from the reference points (in any direction), the amount of current flowing through a given path will approach 0, which is why finite grids can indeed approximate the equivalent resistance of an infinite grid, within some finite error value which gets smaller as you make the grid larger. Thus, it would indeed have "limiting" behavior.

landijk wrote:The only option in the case of the infinite grid is to compute the charge "ripple", but I don't think anyone is really prepared to model how an applied charge actually propagates over time on our grid. It certainly depends on how you apply the charge, etc., so you simply cannot say there is some intrinsic effective resistance between the two points.

The latter part of your argument is moot if you specify DC resistance. In such a case, the charge is not applied at some point in time, but rather has existed since the beginning of time.

As far as modeling how an applied charge propagates over time on a grid, there are some fascinating visualizations done when modeling the plane layers of a power distribution system in a printed circuit board that come quite close to doing just that. I posted a link earlier in this discussion about the DC resistance of any two points on an infinite copper plane (quite similar to the infinite grid of ideal resistors). The source is sigcon, which is run by signal integrity guru Dr. Howard Johnson.

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### Re: "Nerd Sniping" Discussion

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. By the way, regarding the above discussion of capacitance -- I agree that the infinite grid has infinite capacitance, but the only quantity of relevance would be capacitance per unit area. In any case, since this is a DC problem, we don't care.

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. The analytical solution gives 1/2.

Oh, and it turns out the setup of my solution above is identical to the one linked on the geocities site... I guess it's a good thing that I didn't see the link before, it makes me proud that I came up with the solution myself.... On the other hand, I wasted countless hours because Mathematica v4 numerical integration was converging to a blatantly wrong value. I am writing this from a different computer with a fresh install of v6, and what do you know, it works. [rant]ARRRGH. Thanks Wolfram Research for wasting more hours of my youth. No, really. I've been using Mathematica as my principal tool in research for the last 4 years, and I've pretty much grown to hate it. I'm slowly migrating to Matlab. [/rant]
dnquark

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### Re: "Nerd Sniping" Discussion

this caused us great amusement at work.

I seem to be at least partially immune to sniping. At least for this particular problem.