# Nerd sniping

1. Dec 26, 2007

### D H

Staff Emeritus
From "xkcd.com"[/URL] :

[ATTACH=full]120592[/ATTACH]

Last edited by a moderator: Apr 23, 2017
2. Dec 26, 2007

### Gokul43201

Staff Emeritus
Is it 0.916333... ohms?

Damn, I missed dinner!

3. Dec 27, 2007

### siddharth

How did you get that? I think that to solve such problems (where a simple superposition doesn't give you the answer), a discrete fourier transform is required, and then the final integral which gives the resistance looks very tricky.

4. Dec 27, 2007

### neutrino

D H: 4 points (or is it 2 and 1, for the engineer? :tongue:)

5. Dec 27, 2007

### D H

Staff Emeritus
At least three points, since I made Gokul miss dinner.

This http://arxiv.org/abs/cond-mat/9909120" [Broken], question 10) uses a different expression and that enables Mathematica to yield $(8-\pi)/(2\pi)\approx0.773$ ohms.

Last edited by a moderator: May 3, 2017
6. Dec 27, 2007

### siddharth

The article uses the same principle of taking the fourier transform and applying ohm's law. The integral one gets after the fourier transform is (eqn 25 in the article)

$$\frac{1}{4 \pi^2} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left( \frac{1-\cos(2x+y)}{2-\cos{x} - \cos{y}} \right) dx dy$$.

I'm at home and don't have access to an integrator. Does anyone have access to a math package that evaluates this integral, or a way to find the value? Does it come out as $(8-\pi)/(2\pi)\approx0.773$?

Mathworld seems to have used some substitution in its expression.

EDIT: Nevermind. They've apparently used a contour integral and then a substitution. It's there in appendix A, which I missed the first time.

Last edited by a moderator: May 3, 2017
7. Dec 27, 2007

### D H

Staff Emeritus
The difference between a normal person and a scientist:

8. Dec 27, 2007

### Staff: Mentor

:rofl:

9. Dec 27, 2007

### neutrino

That's actually a normal person and Homer Simpson => Homer = Scientist?? :surprised

10. Dec 27, 2007

### Gokul43201

Staff Emeritus
I actually did use a superposition - I don't know why it fails! =(

I haven't yet looked at the article posted by D H.

Last edited: Dec 27, 2007
11. Dec 27, 2007

### Oerg

It's funny how this thread turned into a discussion about the problem in the comic strip rather than the humour about it.

12. Dec 27, 2007

### ranger

It just proves the first sketch of the comic strip. Nerd sniping in action.

13. Dec 28, 2007

### mheslep

Last edited by a moderator: Apr 23, 2017
14. Dec 28, 2007

### D H

Staff Emeritus
The resistance betweeb any two horizontally or vertically adjacent nodes is exactly 1/2 ohms. The resistance between nodes further apart than that is not. That 20% fudge on top of 0.5 ohms yields 0.6 ohms, not 0.773 ohms. Its even worse for nodes further apart than that.

Why don't you try working out the answer for nodes separated by two diagonal hops?

15. Dec 28, 2007

### mheslep

Yes, ok, I'll have to make clear I'm using really cheap resistors when I go for the Google interview.

Think I'll code up a quick sim for fun.