MORE BALLS PLEASE

Thanks for all your replies on the question of the Electronic Ball. There are more answers then I am able to publish on this blog. Below you will find an answer of Ron from New Zealand and Helmut from France. Ron uses LTSpice to get to the answer - Helmut the 'poor mans approach' Thanks guys!

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Hi Guys,

I have a solution to your resistor ball problem:

I flattened the ball, drew the schematic and loaded it into LTSpice circuit simulator, running on a laptop. I added a 1000 volt battery to the schematic and connected it to two diametrically opposite points. I ran the simulator in DC mode, and it produced a list of all the nodes with their voltages, and all the components (battery and resistors) with the currents through them. Knowing the applied voltage and the battery current, I calculated the resistance for question one.

Helmut from France wrote

Hi,

Here are some thoughts on the calculation of the equivalent resistance of two adjacent nodes, a subject that attracted less attention than the equivalent resistance of the two opposing nodes. Being more a practical kind of person I am not going to do a calculation that belongs in a theoretical treatise on electronics, instead I will try to find some poor man’s approach that will hopefully result in an acceptable approximation.

Some years ago a similar question was raised in Elektor about an infinite flat resistor grid and I was thinking that maybe it would be possible to reuse some of that. My starting point was (and still is) that the further a resistor is away from the two adjacent nodes, the lesser its contribution to the equivalent resistance will be. Intuitively, this value should be somewhere in between two limiting values, depending on the value 0 (short) or infinite (open) attributed to these far away resistors. So why not take the arithmetic average of the two? This approximation should get better when the number of jumps needed to get to the border nodes of the grid increases. These nodes will be given a value of open or shorted. The easiest case is the one with 0 jumps: short equals 0 ohm, open equals 1 Mohm, average equals 0.5 ohm. Averaging several approximate calculations of the equivalent resistance when the circuit gets more complicated gives this table (for the side of a pentagon):

Jumps	0	1	2
Short circuits	0.00	0.27 (3/11)	0.602
Open circuits	1.00	1.00	0.696
Average	0.50	0.64 (8/11)	0.649
Elektor measurement	0.65	0.65	0.65

With more jumps the calculation becomes complicated! But, with resistors of 2%, it seems that even with just one jump (still within reach of my mental calculation skills) we arrive at a good approximation.

Regards,

Helmut Müller