Challenge question

Imagine (or look at a picture of) a convex regular icosahedron. There 12 vertices, 20 faces, 30 edges.

Replace all the edges with 1 ohm resistors. What's the measured resistance between any two distinct vertices of the grid?

You can construct a shape with addition faces by taking each triangle primitive of the icosahedron and subdividing it into four. Doing this recursively several times and the shape approaches a sphere.

What's the resistance between any two distinct vertices of the sphere in the limit as the number of faces goes to infinity?

Reply to
bitrex
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Here's such a sphere I rendered using OpenGL, with perspective and scaling transformations applied, 3 "levels" of recursion:

Reply to
bitrex

On a sunny day (Sat, 3 Jun 2017 12:39:32 -0400) it happened bitrex wrote in :

Imagine an Ohm meter, read it. :-)

Reply to
Jan Panteltje

Where am I going to find an icosahedron with an infinite number of faces, though?

Reply to
bitrex

"Infinite Icosahedrons 'R' Us" of course!

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Rick C
Reply to
rickman

1/2 ohm.
Reply to
edward.ming.lee

1/4 ohm.
Reply to
edward.ming.lee

I see a pattern forming. Will 1/8 ohm be the next answer?

Show your work.

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Rick C
Reply to
rickman

ote:

12

in

No, 1/3 ohm, my final answer. It's symmetrical. So, between any two point s, there is 1 ohm (n=1) and 1/2 ohm (n=3). For n=9 or more, the paral lel resistance is less important. The series will convert to either 1/3 oh m or 1/4 ohm or something close enough.

Reply to
edward.ming.lee

"Any two distinct vertices"? So, anywhere at all? As a function of... what kind of distance? We need axes defined here, because the shape is not perfectly uniform (there is distortion from the original geometric shape).

Or did you mean any pair of vertices joined by an edge? In that case, we still must specify location somehow, because notice the original shape's vertices are always within five triangles, while the others (along edges and in faces) are within six.

Since the surface is mostly regular, the solution can be arrived at using another regular mathematical method: the Fourier transform. For instance, the knights-move on a square grid picks up a factor of pi this way.

In the limit as N vertices --> infty, the 12 original vertices are diluted to ~zero, so that you can instead analyze the resistance of "almost all" edges as that of an infinite (flat) triangle grid. (A sphere of infinite size has zero curvature, as any Flat Earther knows. :-) ) So, at this point, it doesn't matter what the original shape was.

For edges coincidentally very close to an original vertex (a finite number of edges away from the vertex), the five-triangle point looks like a defect, and will have a perturbative effect on the region around it (for which the Fourier method would probably be best).

As it happens, it looks like equivalent methods are needed anyway;

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Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: http://seventransistorlabs.com 

"bitrex"  wrote in message  
news:9DBYA.19448$Qc7.7350@fx36.iad... 
> Imagine (or look at a picture of) a convex regular icosahedron. There 12  
> vertices, 20 faces, 30 edges. 
> 
> Replace all the edges with 1 ohm resistors. What's the measured resistance  
> between any two distinct vertices of the grid? 
> 
> You can construct a shape with addition faces by taking each triangle  
> primitive of the icosahedron and subdividing it into four. Doing this  
> recursively several times and the shape approaches a sphere. 
> 
> What's the resistance between any two distinct vertices of the sphere in  
> the limit as the number of faces goes to infinity? 
> 
>
Reply to
Tim Williams

On a sunny day (Sat, 3 Jun 2017 13:15:50 -0400) it happened rickman wrote in :

I was reading my imagined ohm meter it showed zero Ohm, and noticed my imagined temperature meter indicated 10 K. I am the winner, as OP did not specify temperature.

Now should I google for what an Icosahedron is? I think not :-)

Reply to
Jan Panteltje

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--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

Slowman's are 10 points >:-} ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
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Reply to
Jim Thompson

There's a solution posted on the Web somewhere - that's not an easy problem, either.

You also have to make assumptions about "where does the current at infinity _go_?" which are non-physical...

Reply to
bitrex

Not sure I follow. The current all returns to the other point of contact. No? What does it mean "at infinity"? Wouldn't that be one of those questions that are category errors?

Do you have the same problem calculating a field that extends to infinity?

--

Rick C
Reply to
rickman

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