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Nerd sniping 
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#1
Dec2607, 08:37 PM

Mentor
P: 15,065




#2
Dec2607, 09:36 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

Is it 0.916333... ohms?
Damn, I missed dinner! 


#3
Dec2707, 06:38 AM

HW Helper
PF Gold
P: 1,198




#4
Dec2707, 07:12 AM

P: 2,047

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



#5
Dec2707, 09:45 AM

Mentor
P: 15,065

At least three points, since I made Gokul miss dinner.
This arXiv article (see appendix A) gives an absolute mess. Mathworld (http://mathworld.wolfram.com/news/20041013/google, question 10) uses a different expression and that enables Mathematica to yield [itex](8\pi)/(2\pi)\approx0.773[/itex] ohms. 


#6
Dec2707, 10:12 AM

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PF Gold
P: 1,198

[tex] \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 [/tex]. 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 [itex](8\pi)/(2\pi)\approx0.773[/itex]? 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. 


#7
Dec2707, 10:29 AM

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P: 15,065

The difference between a normal person and a scientist:



#9
Dec2707, 10:37 AM

P: 2,047

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



#10
Dec2707, 12:01 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

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


#11
Dec2707, 01:03 PM

P: 364

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



#12
Dec2707, 01:08 PM

PF Gold
P: 1,685




#13
Dec2807, 09:09 PM

PF Gold
P: 3,081

BTW, engineer's solution: assume the infinite grid looks like a short to any one node except for the four resistors directly attached into the node. The parallel resistance into the node then is 1/4 ohm and therefore the resistance between any two, nonadjacent nodes is 1/2 ohm. With actual (cheap) 20% resistors thats close enough to the actual (.7xx is it?) 


#14
Dec2807, 09:28 PM

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P: 15,065

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
Dec2807, 09:44 PM

PF Gold
P: 3,081




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