Soccer Ball Resistor Puzzle

[bob012345]

TL;DR

A truncated icosahedron (soccer ball) is made from 90 identical 1 Ω resistors, one on each edge. The exercise is to find the exact equivalent resistances between adjacent vertices.

I worked on this problem during the World Cup. A truncated icosahedron (soccer ball) is made from 90 identical 1 Ω resistors, one on each edge.

Part (a) Determine the exact equivalent resistance between two adjacent vertices across a hexagon–hexagon edge.

Part (b)Without solving a second resistor network from scratch, determine the exact equivalent resistance between two adjacent vertices across a hexagon–pentagon edge.

 

[Baluncore]

I have noticed that many of these problems reduce to a branching ladder network, a continued fraction.

 

[Baluncore]

(a) hex-hex edge R = 0.66897929
(b) hex-pent edge R = 0.6488431
Can you confirm, or what did you get ?

 

[bob012345]

(a) hex-hex edge R = 0.66897929
(b) hex-pent edge R = 0.6488431
Can you confirm, or what did you get ?

Those numbers look numerically close but can you give me the exact values? How did you get these?

 

[bob012345]

Here are my exact values.

$R_{hex-hex}=\frac{16778}{25080}$

$R_{hex-pent}=\frac{16273}{25080}$

 

[Baluncore]

I cheated by avoiding van Steenwijk’s method.

Found the graph of the TI on Wikipedia, numbered all 60 nodes, then numbered all 90 edges.
I wrote out the 90x 1 ohm resistors in SPICE format; R90 node1 node2 1

I pasted the 90 element table into LTspice, then grounded one node, while injecting a current of 1 amp into the other node. The voltage on the injection node gave the resistance of the network to the grounded node.

I checked my results by comparing two similar edges on the network. They were the same, so the graph entered was symmetrical.

 

[bob012345]

I started in a similar vein but a bit different. I drew the 2D projection of the TI modified to fit a flat x-y format. Then I drew it in LTspice also. It looks like this;

The loops numbered 1-12 are the pentagons and the others are the hexagons. This gave me the numerical numbers that matched yours but what I really wanted are the exact fractions. The next step was to use symmetry to reduce this to a much simpler circuit of which I solved the system of equations exactly to get $R_{hex-hex}$.

Once $R_{hex-hex}$ is known, there is a trivial way to get the other nearest neighbor value $R_{hex-pent}$ without having to solve another network. That is part b of this puzzle.

 

[Baluncore]

Two-point resistances in Archimedean resistor networks.
Frédéric Perrier, Frédéric Girault.
Results in Physics 36 (2022) 105443
https://www.sciencedirect.com/science/article/pii/S2211379722001978?via=ihub

 

[bob012345]

Two-point resistances in Archimedean resistor networks.
Frédéric Perrier, Frédéric Girault.
Results in Physics 36 (2022) 105443
https://www.sciencedirect.com/science/article/pii/S2211379722001978?via=ihub

Thanks! There is a lot of interesting data in that paper. I liked the table of resistances between different vertices. You can get all the exact fractions with your simulation by multiplying the numerical answer by that factor 25080 and rounding to the nearest integer but that’s not the method I was thinking of for part b.

 

[bob012345]

To answer part b, I used Foster’s Resistance Theorem. The average equivalent resistance is $\frac{V-1}{E}$ where V and E represent the number of vertices and edges which is 60 and 90 in this case. At any vertex there are three edges to average so we can write $$\frac{1}{3}(2R_{hex-pent} +R_{hex-hex})=\frac{59}{90}$$ so knowing the exact fraction of one gives the other.

 

[Carl Schwent]

Obligatory xkcd reference Nerd Sniping https://xkcd.com/356/

 

[bob012345]

Obligatory xkcd reference Nerd Sniping https://xkcd.com/356/

I see you just joined. Welcome to Physics Forums! One real benefit is that these kinds of problems help keep the mind sharp as one gets older.

P.S. $R_{(2,1)}$≈0.771

 

[Baluncore]

Nerds are people too, but they are non-linear.

In normal times, when a nerd stands near a group of n people, that makes n+1 people.
In difficult times, when a nerd is accepted by a group of n people, they have the power of 10n.

A good example at the moment, is drone and AI development in warfare.