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Infinite Lattice of 1ohm resistors..?

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St.Steezy
Posts: 4228 - Karma: 20
May as well try here...

I need some help on solving the problem posed in this cartoon, http://www.xkcd.com/356/ . Its a bonus question on my assignment and I may as well give it a shot. Its probably above 99% of the math/physics comprehension of ns.. but you never know.

Cheers.
Mar 11 2008 2:29AM
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St.Steezy
Posts: 4228 - Karma: 20
Click for larger image
Mar 11 2008 2:33AM
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Yauch
Posts: 1685 - Karma: 62
thats a good question,  im not sure, i think there will be some sort of lower limit as the number of resistors approaches infinity. Could be 0. give me props if im right
Mar 11 2008 2:38AM
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coolhandluke
Posts: 4465 - Karma: 153
anything that involves a one-line diagram invokes the black magic of elen's.

this might help you get close:
http://www.geocities.com/frooha/grid/node2.html

looks to be more math than circuitry knowledge
Mar 11 2008 2:39AM
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Pman13a
Posts: 3800 - Karma: 452
i may be completely wrong but it looks like the shortest path is 3 resistors, each having R=1 ohm

R-total = R1+R2+R3 when the resistors are in series

R-total = 1+1+1 = 3 ohms




Mar 11 2008 2:07PM
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Yauch
Posts: 1685 - Karma: 62
yeah but there are an infinite number of other pathways that the electricity could travel that would be in parallel, and thats gonna lower the resistance
Mar 11 2008 2:13PM
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Pman13a
Posts: 3800 - Karma: 452
ya that could be right also, as i said above im not quite sure because of the infinite part
Mar 11 2008 2:33PM
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beer.
Posts: 3096 - Karma: 44
for parellel circuits its 1/r total=1/r1 + 1/r2 + 1/r3 I believe

Id try to help you but I'm pretty sure I am in that bottom 99 percent
Mar 11 2008 2:45PM
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sbark
Posts: 300 - Karma: 11
technically, the resisters are in a grid pattern, they are all the same one-ohm type. and since we cannot see where the power enters or exits the system (considering the amount of resistors is infinite), we can

a) assume there is no power source
b)seeing they are both equally placed and equally identical in resistance we can assume the difference is zero.

keep in mind I am not an electrical engineer, and I finished high-school physics last year, my hypothesis may be wrong.


use this formula,

E=IR

or
1
1/R1+1/R2+1/R3...
Mar 11 2008 3:48PM
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Yauch
Posts: 1685 - Karma: 62
so you think the resistance is 0?
Mar 11 2008 4:23PM
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_dy-lan_
Posts: 206 - Karma: 9
my guess is 1 but i'm not positive. try google, there's a bunch of stuff out there about this question.
Mar 11 2008 6:53PM
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MCIRONCOCK
Posts: 3094 - Karma: 30
1 or 0
Mar 11 2008 7:00PM
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ontwosticks
Posts: 915 - Karma: 95
it would make sense for it to be zero because you're basically doing a limit as the number of resistors approaches infinity. which would be lim(n-->inf) 1/R which would equal zero. that just kind of looking at it by more mathmatical approach. i'm gonna have to say 0.
Mar 11 2008 9:49PM
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BootySpunk.
Posts: 8148 - Karma: 84
my mind just exploded
Mar 11 2008 9:57PM
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samskisalot
Posts: 711 - Karma: 12
im gonna say 3. but im really not sure.
Mar 11 2008 9:57PM
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49566738265
Posts: 0 - Karma: 7,704
None of those are in series OR parallel because they are in a grid and have a resitor in between each one of them.  For example, the horizontal resistors are not in parallel because the veritcal resistors are in between them and they are not in series for the same reason.  So I don't know if that makes the resistance of the system 0 or some infinite number.
Mar 11 2008 10:39PM
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COskier
Posts: 1066 - Karma: 131
Tis an interesting question. My first guess was zero but I'm not sure that is correct because of the whole infinite paths thing. However, when you take limits such as that one you often find that they are not zero because the more roundabout paths for electricity have diminishing effects on the final answer. So, my answer would be just write it out, take the limit as n-> infinity and see what happens
Mar 11 2008 11:32PM
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COskier
Posts: 1066 - Karma: 131
^Not zero, take the limit and you will get it.
Mar 11 2008 11:35PM
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rhcp
Posts: 2598 - Karma: 18
ahahahahahahaha
Mar 12 2008 12:03AM
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COskier
Posts: 1066 - Karma: 131
Final post on this. Way more complicated than I had first thought but in the end it is a converging series, just takes a lot of effort to transform it to that. I got (4/pi)-.5 but I may be way off. Obviously though the answer must be between 0 and three so at least that number is possible.
Mar 12 2008 12:05AM
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Yauch
Posts: 1685 - Karma: 62
that seems reasonable, also it might be that same limit you found no matter where the 2 points are im not sure
Mar 13 2008 1:29AM
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GrassCutter86
Posts: 320 - Karma: 9
Looks like so far you sniped 15 nerds, LOL
Mar 13 2008 2:00AM
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St.Steezy
Posts: 4228 - Karma: 20
I'm pro, atleast 30 points..

I gave up on this problem, I don't have anytime to spare right now... and the assignment was due today.
Mar 13 2008 2:21AM
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[tightpants]
Posts: 1529 - Karma: 17
This sounds right to me. If you use the parallel circuits formula and take the limit as n --> infinity that might be the best way. I really dont know though.
Mar 13 2008 2:31AM
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[tightpants]
Posts: 1529 - Karma: 17
Shit i just realized you would get 0 as an answer then. Maybe it is 0?
Mar 13 2008 2:32AM
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St.Steezy
Posts: 4228 - Karma: 20
^it converges on some value.. for the limit approaching infinity.
Mar 13 2008 3:03AM
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duckets.
Posts: 1168 - Karma: 41
it's either infinity or zero i can't remember if resistors are 1/R or if it's R1 + R2 actually it's one.

it's one.

there's my thought process

there's an infinite number of ways to go and there's an infinite number of both ways to go parallel or in series.

the 1/R as well as R1 + R2 possibilities will add up to 1

i'm almost positive i'm right
Mar 13 2008 11:48AM
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COskier
Posts: 1066 - Karma: 131
no, use a fourier transformation and you'll see it's not gonna converge to one.
Mar 13 2008 1:03PM
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St.Steezy
Posts: 4228 - Karma: 20
Its not one or zero, thats a fact. Those values do not work or make any sense.
Mar 13 2008 2:33PM
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duckets.
Posts: 1168 - Karma: 41
so what was the answer?
Mar 17 2008 4:56PM
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