OK lets put the reference 10^3 or 10^6 nodes away we still end up with
0 current at the reference after superposition.and still retain near
symmetry
even though I have found a major flaw in my simple series model- can't
assume corner nodes at the same potential as side nodes so side nodes
are at the same potential but corner ones at another and this affects
the current distribution.In the second ring out this flaw leads to
having 1/12 =3/20 . Downer-for which I apologize.
Don
On 27/01/2013 3:27 PM, Keith Park wrote:
Sorry about any misunderstanding. The method involves injecting a current
into one node with the rest of the grid tied to ground at infinity. So one
ampere flows out at infinity. The resistance from the point of injection
to infinity is infinity so the voltage at the injection point is infinity.
On Sun, Jan 27, 2013 at 5:09 PM, Raul Miller <[email protected]> wrote:
On Sun, Jan 27, 2013 at 4:53 PM, Keith Park <[email protected]> wrote:
The method of finding the resistance between the two nodes of an infinite
grid of resistances (Don&Kathy Kelly) is erroneous. The method fails
because a one ampere current flowing into the grid produces an infinite
voltage.
What do you mean by this?
Are you claiming that "if one amp were injected into the grid, the
resulting voltage would be infinite"? That can only happen if the
distance is infinite, and is really as much an objection to the
concept of "infinite" as anything else. For a finite separation
between the two significant nodes, the voltage must be finite.
Or, are you instead saying that the proposed method yields infinite
voltages for a finite separation? If so, I must confess that I did
not observe it doing any such thing, and I'd like some explanation
about how you get that result.
Or did you really mean something else?
Thanks,
--
Raul
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