0356: "Nerd Sniping"

[unfunk]

I registered just because I wanted to say how much this comic hurt my drunken, nerdy brain

Now, I'm no Math or Physics major - or anything even slightly numberworthy (I majored in music composition), so my initial thought was that the resistance would be 3Ω (three equal length "shortest paths" of 3Ω each, divided by three)

...but then I read what you people have been posting, and now my brain has broken!

/consumes more vodka

[Syphon]

I'm going to go ahead and say it's infinite resistance. Infinite OR zero.

Would this be both parallel and in series?

Code: Select all

. ~ . ~ O
|   |   |
O ~ . ~ .

(+ infinite many other resistors)

-and-

. ~ . ~ O
|
O

(+ infinite many other resistors)

So for parallel...

R_T^-1 = ∞(1/R)
R_T^-1 = ∞
R_T = 0 (Using lim f(x) = 1/x = ∞)
----------------^{x → 0}

And for series

R_T = ∞(R)
R_T = ∞

[d3adf001]

from what i think its 0 but i have to run to an exam. but ill explain when i get back

[jima]

One thing I never liked about this comic is how it excludes "Literary Nerds", if you will.

i.e. "Which is a better literary work, Stephen Crain's "Open Boat" or Ralph Waldo Emerson's "Self Reliance"?"

Go ahead, make a sign and camp out across the street from a library. Keep us posted.

[sillybear25]

I'd like to present a variation on an old adage (I forgot who originally said it...):

If at first you don't succeed, try, try again. Then give up. There's no point being such a damn fool about it.

[schrodingersduck]

I'm going to go ahead and say it's infinite resistance. Infinite OR zero.

A good way to see why it would be neither of these is to imagine a real-world equivalent: suppose you were standing on a planet made of silicon (which can approximated as a large number of silicon resistors joined in a lattice - silicon has a large, but not massive, resistivity, so it can be measured relatively easily), and you jabbed the contacts straight into the ground. The planet can be considered infinite compared to the distance between nodes, yet you would not expect the resistance to be either 0 or infinity. Instead, you would expect it to be almost identical to the resistance found by doing the same experiment with a silicon brick, so the distance between nodes is now comparable to the size of the brick.

[ricree]

I registered just because I wanted to say how much this comic hurt my drunken, nerdy brain

Now, I'm no Math or Physics major - or anything even slightly numberworthy (I majored in music composition), so my initial thought was that the resistance would be 3Ω (three equal length "shortest paths" of 3Ω each, divided by three)

...but then I read what you people have been posting, and now my brain has broken!

/consumes more vodka

A couple things. First, for parallel resistors you get the equivalent resistance with Req=1/(1/R+1/R)
You can actually get a fairly decent approximation by taking just those three paths, but since two of those paths share the first resistor, you can't just throw all three in parallel. Here's what I plugged into google:

(3^-1+(1+(2^-1+2^-1))^-1
which gives 0.833333333

Some earlier pointed out that the answer was

Spoiler:

4/pi-.5 Ohms == .773 Ohms

so while not the world's most precise answer, it is reasonably close (about 7.5% difference, I think) and doesn't involve anything more than really basic rules.

[Tualha]

I haven't had time to read the MathWorld article yet, but it seems to me the answer would be zero.

Consider the simpler case of a "ladder" of 1-ohm resistors, and R_eq across one "rung". We place 1 volt across the rung and find the current. Across that rung, 1 amp. Go up, across, down: 1/3 amp. Ditto for down. Continue, and we end up with the total current being:

1 + 2/3 + 2/5 + 2/7 + ...

This grows faster than the harmonic series, so it diverges. Therefore R_eq is zero, and so in the infinite grid, R_eq between two adjacent points is also zero, and so is R_eq between any two points.

What am I missing?

---------------------------

What I was missing is that the voltage along the ladder drops off faster than it would if each rung were treated independently. So my method is no good.

Surfing around I see the proper treatment involves math that I've either not used in years or have never learned at all, so it's time to drop it.

[schrodingersduck]

I haven't had time to read the MathWorld article yet, but it seems to me the answer would be zero.

Consider the simpler case of a "ladder" of 1-ohm resistors, and R_eq across one "rung". We place 1 volt across the rung and find the current. Across that rung, 1 amp. Go up, across, down: 1/3 amp. Ditto for down. Continue, and we end up with the total current being:

1 + 2/3 + 2/5 + 2/7 + ...

This grows faster than the harmonic series, so it diverges. Therefore R_eq is zero, and so in the infinite grid, R_eq between two adjacent points is also zero, and so is R_eq between any two points.

What am I missing?

Would the potential difference be 1 volt across every rung? I get the feeling that the system would act as a potential divider at each rung, so the voltage will also be tending to zero.

[pieaholicx]

Anybody else have a copy of the Google Labs Aptitude Test? If not, I'll dig up the images when I get home tonight, I've got them on an old laptop of mine. It's got some really interesting questions, my favorite being something to the effect of "The space below has been left blank. Improve upon it."

[jima]

Ah resistors, thy mortal enemy. I always had trouble working out resistance in physics when doing electricity. but what you've really got to ask yourself is: Are they in series or parallel?

I'm just a bit... drugged up on medicine at the moment.. I don't know if I make any sense or not.

Depends. Was that supposed to be a play on Dirty Harry?

[Arancaytar]

Gosh, I want to say that it's infinite. That'd be too easy though, right?

If it were infinite anything, it would be infinitely small. Remember that there are an infinite number of parallel circuits between these two nodes, the minimum resistance of which is 3 Ohm. Since the resistance of parallel circuits is always smaller than the smallest resistance of the individual circuits, the resistance is something smaller than 3 Ohm.

The question is: Do the parallel circuits add up to approach 0 resistance as the grid size approaches infinity? Or is there a finite limit at which the resistances of the longest circuits are so high that they no longer decrease the combined resistance?

I can see how solving this question would fill several blackboards. Ironically, while it is a physics problem, the solution requires at least some mathematics (infinite series). So it's easy to see how the problem, er, flattened the poor physicist.

Edit: The Gedankenexperiment with the brick above made me realize that it would have to be a finite limit. Since the resistance in a brick would be non-zero, and you would not expect a significant difference between the planetary and brick resistance.

So we have determined the solution is a finite positive number smaller than 3. The trivial problem of finding this number is left as an exercise to the reader.

[plasticsuperhero]

You know if xkcd has this kind of motivating power, Randall ought to put it to good use. For example slip in a subtle problem that is equivalent to a proof of NP-completeness or some other intractable task... within hours of being posted someone on the fora might have a solution!

Distributed HUMAN computing?

[Arancaytar]

You know if xkcd has this kind of motivating power, Randall ought to put it to good use. For example slip in a subtle problem that is equivalent to a proof of NP-completeness or some other intractable task... within hours of being posted someone on the fora might have a solution!

Distributed HUMAN computing?

Imagine a Beowulf cluster of nerds, and you have the xkcd fora.

[Sprocket]

Mathematicians are easier (in my experience ) , Physicists should be worth more points. Especially if we group computer scientists with mathematicians.

[karthikp]

The answer is R/9. While I could claim the answer is all mine (solved it on my shower tiles, no less), I can't. Most students in India preparing for the dreaded IIT Joint Entrance Exams will be familiar with the solution because a similar problem shows up in a book of physics problems by Irodov.

Assume you apply a positive voltage +V to point A. We can then imagine the current that leaves A and propagates in all directions. If you refer to the figure I've attached, you'll see that I/9 of it reaches point B.

Solution!

xkcd.png (30.19 KiB) Viewed 4520 times

Superimpose this problem on top of another problem in which the potential farfield is still zero, but you apply a -V voltage at point B. The currents "leaving" A for B will be I/9, too.

Put together, you've applied a voltage difference of 2V between A and B. You got a current of 2I/9. So, the effective resistance between A and B is R/9 where R is the resistance of a single element. So, 1/9 ohms!

I must add two observations. First, this is exam season. Highly inappropriate for nerd sniping exercises of this nature. I myself have an exam in a few hours and ended up getting...sniped. Shame on you, sir! Second, "Send more physics problems!"

Meanwhile, I'll work on extending the solution to getting the resistance between any two points on the grid. After my exam...

[karthikp]

Whoa. That figure came out rather...big! Sorry. Gotta run now...

[DragonHawk]

I'm going to go ahead and say it's infinite resistance. Infinite OR zero.

Would this be both parallel and in series?

This was as far as I got:

Electricity always follows the path of least resistance (regardless of whatever other higher resistance paths might be available). So even without doing anything more than addition, there is obviously a three ohm path between the two points. So it isn't infinite. Indeed, the answer can't be more than three ohms.

The parallel question is trickier, because of the whole infinity thing. You can keep including more parallel paths forever. But for any number of parallel paths you include, you also have to cross some proportionate number of resistors in series. If you add more parallel paths to spread out the resistance more, you also add proportionate series resistance. I suspected one could never get to zero because of that. But I was wondering if this is one of those cases where it approaches zero, because the parallel paths are "growing faster" than the series resistances.

Then some actual EE types came along and ruined everything by posting a method for analysis and math and stuff.

Although I do note that a few impressive-sounding answers have been posted, and they don't all reach the same number. That makes me wonder if at least one is wrong. And if one is wrong, maybe they're all wrong. But the various answers are all relatively close in value, and some state they're approximations or involve irrational numbers like pi, so maybe they're all actually correct within stated parameters. In any event, the people talking about it know more than I do, so I have to bow out.

[trip11]

Another, but much simplier problem that I had some fun with. Imagine a circut like this:

Code: Select all

.   ._R1_._R3_.
.   |    |    |
.---|    R5   |----
.   |    |    |
.   ._R2_._R4_.


So the question is, what is the equivilent resistance of that circut? Don't assume the resistors are equal.

Spoiler:

Unless I missed something obvious, you can't do the simple combinations of resistors into parellel and series sub-sets. You have to use Kirchhoff's rules and solve a bunch of simulatanious equations. But I'd love to hear a simple solution

[Belial]

perhaps he should post a different kind of problem.. something like "what's the minimum set of words needed to make anyone fall in love with you instantly"

now THAT's interesting

14, apparently.

I saw the initial post, and started to answer, then realized you'd beaten me.

[tgape]

Gotta leave for work soon, but just eyeballing it - I'd guess around 3/4 of an ohm. However, I'm pretty sure it's not exactly 0.75 Ohms.

[dangil]

perhaps he should post a different kind of problem.. something like "what's the minimum set of words needed to make anyone fall in love with you instantly"

now THAT's interesting

14, apparently.

but what are those words ?

and you solved part of the problem, since I couldn't remember where I heard about this in the first place..

[d3adf001]

the way i see it is that the trail to the resistors is 2n+1 and if there is only 1 path the resistance would be the sumof 1/(2n+1) from 1 to infinity and that diverges. so since there are an exponential more paths as n goes along. so the resistance should be 0.

[Kin]

/as an EE, this problem warms my heart

I have captured a pun!

[Unfortunately i doubt intended..]

[ricree]

the way i see it is that the trail to the resistors is 2n+1 and if there is only 1 path the resistance would be the sumof 1/(2n+1) from 1 to infinity and that diverges. so since there are an exponential more paths as n goes along. so the resistance should be 0.

The problem, I think, is that those paths aren't simply parallel with one another. Even when you consider paths of length three, there are two paths which share a resistor for part of the way.

PS. Yes, I know I need to make my way over to the intro thread. I'll try to get around to it as soon as possible.

[Kirin42]

The answer is R/9. While I could claim the answer is all mine (solved it on my shower tiles, no less), I can't. Most students in India preparing for the dreaded IIT Joint Entrance Exams will be familiar with the solution because a similar problem shows up in a book of physics problems by Irodov.

Put together, you've applied a voltage difference of 2V between A and B. You got a current of 2I/9. So, the effective resistance between A and B is R/9 where R is the resistance of a single element. So, 1/9 ohms!

I've got to jump in and ask for some further proof that the remaining current is insignificant. What I mean, is, for example: let's say A is the bottom left of the rectangle and B is the top right. Your solution ignores the current that comes from A: right, down, right, up, up to get to B. Some fraction of current (specifically, I/108) will travel this path. Now, while I/108 seems pretty insignificant in the face of I/9, I'm unconvinced that an infinite number of such paths is equally insignificant. Does anyone have a similarly mathy follow-up to this? I'm a computer scientist, not a physicist, but I'm very interested. Anyone with a mathy explanation?

Kirin

(And I, too, created an account on this forum just to talk about this problem. Way to go, Randall...I was hoping to get actual work done today.)

[d3adf001]

the way i see it is that the trail to the resistors is 2n+1 and if there is only 1 path the resistance would be the sumof 1/(2n+1) from 1 to infinity and that diverges. so since there are an exponential more paths as n goes along. so the resistance should be 0.

The problem, I think, is that those paths aren't simply parallel with one another. Even when you consider paths of length three, there are two paths which share a resistor for part of the way.

PS. Yes, I know I need to make my way over to the intro thread. I'll try to get around to it as soon as possible.

that is true but there are an infinite number of parallel paths that dont share that follow the 2n+1. im probably wrong but might not

[ricree]

The answer is R/9. While I could claim the answer is all mine (solved it on my shower tiles, no less), I can't. Most students in India preparing for the dreaded IIT Joint Entrance Exams will be familiar with the solution because a similar problem shows up in a book of physics problems by Irodov.

Assume you apply a positive voltage +V to point A. We can then imagine the current that leaves A and propagates in all directions. If you refer to the figure I've attached, you'll see that I/9 of it reaches point B.

xkcd.png

Superimpose this problem on top of another problem in which the potential farfield is still zero, but you apply a -V voltage at point B. The currents "leaving" A for B will be I/9, too.

You seemed to start with the assumption that current is going to flow equally away from the node in all directions, and that is flat out a wrong assumption.

Edit: Never mind about the KCL part, as you are obviously assuming that the current comes from an leaves through your source lines. Still, I don't see why it is valid to assume that all currents will flow equally from the node, as the voltage difference across the adjectent resistors is going to be different for each resistor.

[Gelsamel]

I have no fucking clue how to solve it....

But I really really want too... I'm trying to resist the urge... I need sleep T_T

Motherfucker! -_-

[LumenPlacidum]

Is there anyone who knows of a standard way of calculating equivalent resistance between two points on an infinite continuous sheet of resistive material? I'm willing to bet that such an answer relies fairly heavily on the notion of distance between the two points. I wonder if you get the correct answer to this problem if you simply replace the notion of Euclidean distance in that solution with the idea of a rectangular distance. Of course, you'd have to go deep with it, replacing the metrics used in definitions of integrals (which I'm sure are used) with the new one.

[DragonHawk]

something like "what's the minimum set of words needed to make anyone fall in love with you instantly"

14, apparently.

but what are those words ?

"What do you get if you multiply six by nine". (Elric the Technomage was using base six when he said "fourteen".)

-----

Is there anyone who knows of a standard way of calculating equivalent resistance between two points on an infinite continuous sheet of resistive material?

I just hook it up to my multimeter using my infinitely long test leads.

[ckt]

I have been successfully sniped (I notice I'm not the only one who posted for the first time in response to this comic).

For those that are obsessed with the problem and need some help, I did a quick literature search on the problem, and I found this:
"Infinite Resistive Lattices" by D. Atkinson and F.J. van Steenwijk (it's in the American Journal of Physics, but a google search also turns up a citeseer copy).

Spoiler:

The poster (floyd4one) who said the answer is 4/pi - 1/2 is correct, I'm just supplying a reference that shows the full solution.

[Robin S]

it just seems like such a convenient but totally hypothetical number and it kind of makes sense in this case.

Yes, it is occasionally convenient. No, it is not totally hypothetical. It is exactly equal to 1. There are plenty of threads on this forum for you to read if you want convincing.

http://www.youtube.com/watch?v=BipvGD-LCjU

OH MY GOD.

[segmentation fault]

judging by the responses in this thread, you all should have been sniped by now.

[fnordboy]

Dug up from two academic journal articles:
http://arxiv.org/abs/cond-mat/9909120v1 and
http://scitation.aip.org/vsearch/servlet/VerityServlet?KEY=AJPIAS&CURRENT=NO&ONLINE=YES&smode=strresults&sort=rel&maxdisp=25&threshold=0&pjournals=AJPIAS&pyears=2001%2C2000%2C1999&possible1=1000&possible1zone=fpage&fromvolume=62&SMODE=strsearch&OUTLOG=NO&viewabs=AJPIAS&key=DISPLAY&docID=1&page=1&chapter=0

For a 2D grid with resistors of value R and points that are (Nx,Ny) nodes apart, the solution for large Nx, Ny is:

R(Nx,Ny) = (R / pi) * (ln(sqrt(Nx^2+Ny^2)) + 0.5772 + (ln8)/2 )

This reduces to approximately the exact solution of 1/2 at for Nx = 1, Ny =0 (or Ny=1,Nx=0) and 2/pi for Nx=Ny=1.

[sab39]

EDIT:

perhaps he should post a different kind of problem.. something like "what's the minimum set of words needed to make anyone fall in love with you instantly"
now THAT's interesting

I think on these fora, it would likely be something horribly nerdy. But then, you'd get different answers for different girls.

"Fruity Oaty Bars" FTW!

[dr_nik]

Ok so first off I have to agree that I have been quite successfully sniped...I have to submit a paper to a conference by the end of the day and I have a ton of work to do on it...but instead I decided to solve this problem.

Second, it seems like everyone working on the problem here and historically has been a mathematician, not an EE. Why? Because an EE would make it into a simpler problem. I'm going to go out on a huge limb here and say that everyone who has posted solutions is wrong since the current will not go through *every* resistor in the plane. Path of least resistance yada yada. So heres my solution:

Spoiler:

The equivalent of an infinite series of resistors is an infinite sheet of metal with resistivity of 1 Ohm.m. Then it is a simple solution of finding the distance between the two points (2.236m assuming each segment is 1m long). That makes it 2.236 ohms

Ok, I wanna hear why I am supremely wrong now =)

Oh, and curse you Randal...

[Murgatroyd]

Re: "Path of Least Resistance":
Yes, current follows the path of least resistance. But it does not follow it exclusively. It actually follows just about all available paths, with more following the paths of less resistance.

That is true but there are an infinite number of parallel paths that dont share that follow the 2n+1. im probably wrong but might not

You're wrong. I guarantee that you cannot find a set of more than four parallel paths of any length that don't share any resistors.

[tetsujin]

You know what this comic reminds me of? Captain Kirk.

Destroying automata with circular logic and contradictions was practically a specialty of his. For instance Nomad ("You must destroy anything flawed: you are flawed, you haven't destroyed yourself, therefore you are flawed."), Norman from Mudd's Planet (Kirk: Anything he says is a lie. Mudd: I am lying.) - I don't remember if there were other cases I've forgotten... (He pretty much just walked the M-5 through a morality problem - not really in the same vein I guess... and Rayna just died of a broken heart...)

[dirtysouthafrican]

This problem piqued my interest also. I dug up a bunch of references (some have already been mentioned here) and consolidated them, so if anyone is interested:
http://www.physics.thetangentbundle.net/wiki/Analog_circuits/infinite_resistive_lattice

Also, a bunch of you might otherwise be interested in the rest of the site (at least the physics/mathematics components), so I invite you to look around.

Cheers