THE ELECTRONIC BALL

First step to be taken is translating that complicated 3-D thing into something easier to handle and draw, ie we have to flatten the ball. Just have a look to the picture below, you can imagine each branch of our graph -the ball- is made of some ideally elastic resistor, then you stretch one of pentagons so that it can surround all the rest. Of course we loose neat angles and branch lengths you can see on the 3-D ball but this is the only way to flatten it. That neat arrangement of polygons can only exist in 3-D space. Nonetheless topology is the same, in other words connections between resistors in the branches is exactly the same. We still have 20 hexagons and 12 pentagons and connections, common branches and “neighbors” are just the same as in the real 3-D thing. Note that the 12th pentagon is “the biggest one” surrounding the hole graph.























This could be done even using one of the hexagons as graph center taking us to the same conclusions.

Next step is answering the question: Is resistance between any two diametrically opposite nodes the same? This cannot really taken for granted as is. However looking at the picture below we can see circular symmetry as 360/5=72° and this means that circuit and hence resistance between any couple of nodes A-A', B-B', C-C', D-D' and E-E' is the same.





















Of course which pentagon is used as center of the graph has no influence and then the same shall apply to any pentagon in the ball. This will include all the nodes in the ball and gives us opportunity to calculate resistance between any couple of nodes above having the same result.




To do that we shall connect a voltage generator between A and A' and calculate current flowing









This connection will unfortunately invalid the 5-folded symmetry seen above reducing to a simple vertical symmetry. However this 2-symmetry will allow us to reduce circuit to be considered.

In fact due this symmetry voltage on node 1 will be the same as that on node 1', hence voltage across branch 1-1' will be zero and no current will ever flow, so we can even remove this branch without modifying operation of the circuit. The same will naturally apply to 2-2', 3-3', 4-4', 5-5' and 6-6'.


















This is the new equivalent circuit






















We are close to dividing circuit into two halves, we just have to cope with the two branches 7-7' and 8-8'.

We first will replace resistor R with two parallel connected 2R resistors and than we will do the same as before, due symmetry current out of X and X' is the same, then current in the horizontal branch is zero, we then can remove it and finally cut our ball in two halves.













Then the remaining of our ball is
























So resistance seen from any two diametrically opposite nodes will be one half of resistance seen from voltage generator in the equivalent circuit below.





















Despite several tries to simplify the circuit above using common series, parallel and star/delta rules on electric nets I did not manage to achieve any further improvement apart from the obvious series connection of couples of resistors (not shown above to make clear belonging polygon).

So I had to go through the safe and hard way, ie solve the circuit using loop-currents method and Kirchhoff's laws. Let's have a look to diagram below where I replaced series resistors with their equivalent and depicted 12 loop currents and our “most wanted” voltage generator current with their (conventional) directions.























Each loop shall satisfy : sum of voltage generator(s) equals sum of voltage drop across resistors:

V=RI

for instance, looking at loop I1 we see we have no voltage generator ie V=0 and then given different directions of currents involved.

0=7RI1 – RI2 – RI3 – RI4 + 3RI13

This , repeated 13 times will give us a set 13 linear equations in 13 unknowns which can be conveniently described and solved using matrix algebra.



This is the R matrix describing the above circuit:

......................7.-1 . -1. -1. 0. 0. 0. 0. 0. 0. 0. 0. 3
....................-1. .6 . -1 ..0 -1 -1 0 0 0 0 0 0 0
....................-1 -1 5 -1 0 -1 -1 0 0 0 0 0 0
....................-1 . 0 -1 6 0 0 -1 -1 0 0 0 0 2
.....................0 -1 ..0 0 5 -1 0 0 -1 0 0 0 0
.....................0 -1 -1 0 -1 6 -1 0 -1 -1 0 0 0
R =............. 0 0 -1 -1 0 -1 6 -1 0 -1 -1 0 0
.....................0 0 0 -1 0 0 -1 5 0 0 -1 0 2
.....................0 0 0 0 -1 -1 0 0 6 -1 0 -1 0
.....................0 0 0 0 0 -1 -1 0 -1 5 -1 -1 0
.....................0 0 0 0 0 0 -1 -1 0 -1 6 -1 2
.....................0 0 0 0 0 0 0 0 -1 -1 -1 7 1
.....................3 0 0 2 0 0 0 2 0 0 2 1 10

then we need the known terms column ie a vector describing voltage generators connected – only one in loop 13 in our case- and a second vector containing our 13 unknowns.
We shall suppose voltage applied is 1V just to make calculations easier, value is not important at all being the circuit linear. We actually need only I13 but unfortunately this cannot be calculated on its own, we need to solve the whole system instead.


..................0 ............I1
..................0.............I2
..................0............ I3
..................0 ............I4
..................0 ............I5
..................0 ............I6
V =........... 0 .... I = .I7
..................0 ............I8
..................0 ............I9
..................0 ............I10
..................0 ............I11
..................0 ............I12
..................1 .............I13

Once we have these powerful instruments problem becomes logically simple, since

RI=V then I=R-1V

then we “only” have to invert R matrix to get R-1 even if this is logically simple requires quite a huge amount of calculations to be done, hence I got MatLab to do this. I will omit inverted matrix here since we only need one of its elements, in fact given the all-zeroes-and-one-one known term vector we only need R-1(13,13) to calculate I13. So we have:

R-1(13,13)=0.323529412


and then since V=1V and R=1Mohm

I13=R-1(13,13) V13 = 0.323529412 . 1 = 0.323529412 A

and hence resistance of one half of the ball shall be

rh = 1/ 0.323529412=3.0909 Mohm

One interesting consideration is that since all numbers in the R matrix are integers and operations involved in calculating its inverse are sums, differences, products and divisions result must be a rational number ie ratio of two integers. It actually looks like

rh =3 +1/11 =34/11 Mohm

Further investigations with MatLab showed that this is true down to its machine precision so we will assume it for true.
So finally taking into account the second half of the ball parallel connected we get


rball =17/11 Mohm = 1.5454 Mohm






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