:) another fun problem for ya all.

[Cogman]

I thought it would be fun. From the comic here what do you think the answer will be, and why? I am seeing this as a summation of some sort. You know the answer will be less then 3, in fact with some quick observations you will see that it goes to 1.5 ohms pretty quickly (so anyone that guesses more then that fails )

Post how you did it, or what logic you used to arrive to you answer. My guess is it comes to around .7 ohms

 

[esun]

There are solutions all over the web already (SPOILER: http://www.geocities.com/frooha/grid/node2.html). That would be my answer and justification, accordingly.

 

[TuxDave]

Originally posted by: esun
There are solutions all over the web already (SPOILER: http://www.geocities.com/frooha/grid/node2.html). That would be my answer and justification, accordingly.

That solves a slightly different problem and on top of that..... I have no idea what the hell he's doing.

 

[JohnCU]

i came up with .6

 

[JohnCU]

and yes, that answer posted is slightly different, different nodes

 

[KIAman]

Damnit, why the hell does PI always make it in the answers?!?!

 

[esun]

My bad, here's a better link:

http://answers.google.com/answers/threadview?id=92564

It has the general solution as follows:

R(n,n) = 2/pi * Sum{k=1...n: 1/(2*k-1)}
R(1,1) = 2/pi
R(2,2) = 2/pi * (1 + 1/3)
R(3,3) = 2/pi * (1 + 1/3 + 1/5)

EDIT: Sorry, also appears to be different from the XKCD comic (it wants R(2,1)). I'm guessing the solution follows from the derivations in the links, though.

 

[bobsmith1492]

Haha, that site is hilarious. Oh, I built said infinite network and measured it. It was 1/2 ohms.

 

[Fox5]

Originally posted by: Cogman
I thought it would be fun. From the comic here what do you think the answer will be, and why? I am seeing this as a summation of some sort. You know the answer will be less then 3, in fact with some quick observations you will see that it goes to 1.5 ohms pretty quickly (so anyone that guesses more then that fails )

Post how you did it, or what logic you used to arrive to you answer. My guess is it comes to around .7 ohms

I'd imagine you can use a symmetry argument, and that is the only way it's solvable. It's kind of like the 3 dimensional cube of resistors.

 

[BrownTown]

well when they are 2 apart it is 2/pi, so I would guess that if they are 3 apart it would be 3/pi...

it wouldn't be too hard to construct 3-4 resistor ladders using a computer program with each going out an additional layer and then graph the results and extrapolate out to the final answer. The contribution from the farther out resistors is very small past the first few layers.

 

[BrownTown]

well when they are 2 apart it is 2/pi, so I would guess that if they are 3 apart it would be 3/pi...

 

[Mark R]

Can someone explain why you need to use Fourier transforms to solve this?

It's just that the only general solution I can find involves taking the FT of the positions in the grid in order to describe it as a definite integral. From there, the solution of the integral (although tricky) is merely brute force and ignorance. That said, it's beyond my capabilities.

When algebra fails, at least you can resort to numerical methods:

I get 17/7pi.