Infinite resistor network

That would depend on the configuration. They could be arranged in triangles, or other polygons. It does not have to be a quad array.

Not since you are flipping things around.

IF I were a troll, I would have a very low score. Just ask anyone.

So, since it makes no sense for me to be a troll, much less "SED's finest troll", I would say that you are talking out of your ass...

Again.

Thank you for the simple, straightforward reply. I may have mentioned I'm a pictures kind of guy. I can now visualize the diminishing influence of the farther resistors from the tested central pair as relating to the curvature of a graphed function.

It's nice when someone can give a simple straightforward reply, instead of trying to appear superior without demonstrating any actual knowledge like certain other posters in this thread. Not naming any names...

Mark L. Fergerson

Hi, Mark. There is still the question of the changing signs that nowhere mentioned in his (well, Leibnitz') convergence.

In this problem case, it actually follows out of another way of estimating PI using continued fractions -- most particularly, Brouncker's. The connection here can be seen by taking a close look at the diagonal case (not the knight's move case):

| | | | | | --/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | | | | \\ \\ \\ \\ \\ \\ / / / / / / \\ \\ \\ \\ \\ \\ | | |E1 | | | --/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | | | | \\ \\ \\ \\ \\ \\ / / / / / / \\ \\ \\ \\ \\ \\ |C1 | |D1 | | | --/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | | | | \\ \\ \\ \\ \\ \\ / / / / / / \\ \\ \\ \\ \\ \\ | |C2 |B R2 |D2 |E2 | --/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | | | | \\ \\ \\ \\ \\ \\ / / /R1 /R3 / / \\ \\ \\ \\ \\ \\ | |A |C3 | | | --/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | R4 | | | \\ \\ \\ \\ \\ \\ / / / /R5 / / \\ \\ \\ \\ \\ \\ | | | |C4 | | --/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | | | |

I've placed the points A and B nearer the lower left corner because the focus will be on the right and upper region. Given a current injected at B and withdrawn at A or a voltage source placed across A and B, I think you can see that the diagonal formed along C1, C2, C3, and C4 are all at the same voltage potential, namely (V(B)-V(A))/2. Hopefully, that is clear. So we can ignore everything along that line and below-left and focus on solving the upper-right area, as solving that half solves the other half.

There is another consideration to help simplify. Points D1 and D2 are at equal potentials, as are points E1 and E2. Currents in the square region bounded along the D2 to E2 line and the D1 to E1 line don't modify the resistance because they are bounded by successive points of equipotentials (not that D2 and E2 are equal, but that D1 and D2 are and that E1 and E2 are, etc.)

So this allows a focus on the paths from B to A that lie in the region bounded by the C-line and the B-D2-E2... line.

Our first estimate is that the resistance from B to C3 is just R1. In other words, just R. (Since there is another R on the other side of our dividing line, that would be 2R from B to A, except that all this is duplicated on the other fold along the A-B line, so the 2R becomes just R, again, as a first estimate. (Note that this means solving this part actually answers the larger question.)

However, there are also three resistors providing the closed loop from B to C3 going through R2, R3, and R4, as well. So as a second estimate, we could say that the resistance is R1||(R2+R2+R3). Almost. R4 is also obviously paralleled by R5 to C4, which is at the exact same potential as C3 (a vitual voltage, if you like.) So this really should be R1||(R2+R3+R4||R5). With all R's the same, this works out to R||(2.5R) = (5/7)*R.

Looking back at R3, though, it is also paralleled by another three resistors in a ring to its right. And so on.

These rings of three with one and the fact that a portion of the current in one leg extends into the three, leads to a continued fraction that fits the inverse of Brouncker's continued fraction for PI/2 (all that means is that we insert a zero at the beginning.)

So, it's not quite like Leibnitz' convergence as much as yet another one, Brouncker's.

Another view of the section under study is:

\\ \\ \\ / / / \\ \\ \\ R | R R | R R | R ,--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | \\ \\ \\ / / / \\ \\ \\ (V/2) R | R R | R R | R R ,---+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\--+--/\\/\\-- | | | | | --- \\ \\ \\ \\ - / R / R/2 / R/2 / R/2 --- \\ \\ \\ \\ - | | | | | | | | | | gnd gnd gnd gnd gnd

You can also see the 1,3,1,3,1,3, patterns here.

Jon

f the

I specified wrapp> An obviously (to me) related question; is there an equally "neat"

and you replied:

So, where's your solution to the cylinder-grid problem?

Mark L. Fergerson

Gee, you really cannot read; the proposition was to take the array, snip it and bend it so as to make an infinitely long cylinder. Granted, a few operations on the original infinite matrix were not stated, but...

And like I said... It does NOT assist in visualizing the situation in any way shape or form. You should be able to do that yourself without actually seeing it in that physical arrangement.

If it is an infinite grid, there can be no such cylinder. It would have an ever expanding diameter. Not just length. D'oh!

A flat array arranged such that the finished item looks like a big pizza (an eyelid opening, actually) is all you need.

The two points, regardless of how many nodes there are between them would appear on the edge of the pizza opposite the other. That is all that would need to be considered for the calculation.

The actual influence would look more like an ellipse with pointy-er short radius. Looks like the opening for your eyeball when your eye is open. All resistors outside those will not matter enough to need inclusion.

Currents outside that influence are too small to consider as being significant.

I can still remember my teacher telling me that I was using too many places behind the decimal.

You only need the degree of precision which describes the result well enough to impart the nature of what generated it.

Just because anybody else can easily visualize the cylindrical form as well as solve that problem in less than 10 seconds does not mean you can do either...

Does the word "snip" have a meaning for you or are you that impaired?

I have to presume that, if you are getting your jollies from feeding the troll, you are NOT getting any nooky ;-)

...Jim Thompson

Note that I did not say that I was unable to perform any such visualization. I merely said that it has zero merit. That is what gets your gourd, chump.

I am starting to think that not a day passes where this retarded twit names Thompson doesn't make an absolutely stupid "joke", and actually thinks it is funny.