I stand corrected -- that's an embarrassing mistake, on my part. And thinking back on what I paid attention to when I read, this both fits what you were describing and also seems [to me] to be a plausible use of infinity.
(That said, I did not try to follow all the details of your post. Instead, I tried a few example cases, to see if I could find any inconsistencies in the numeric results.) -- Raul On Mon, Jan 28, 2013 at 12:11 AM, Don & Cathy Kelly <[email protected]> wrote: > Actually not > inject 1A at A with B open. and draws out 1A at the 'edge" (infinite point > or ring of all points equidistant from both A and B) which you can call > ground as it doesn't matter. > and finds the voltage drop between A and B. > Then inject -1A at B with A open and draws out 1A at the 'edge" > and find the voltage between A and B > Summing the two cases the result is the same as if the current source was > connected between A and B with the 'edge" open > there will be no current drawn out from the edge. It's easy to use infinity > but this is simply a way of stating equidistance from A and B in order to > call on symmetry. Errors will be in the order of 0.25% for distance AB > being about 1% of distances from both to the edge. > > Don Kelly > > > On 27/01/2013 6:34 PM, Raul Miller wrote: >> >> Ok, that clarifies your point. Except, in Kelly's method, it's not >> tied to ground at infinity. >> >> The problem is to determine the resistance between two nodes, A and B, >> which are a finite distance apart. The method injects 1 amp of >> current at A and grounds B. >> > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
