Actually, I think I see an inconsistency. Or, at least, an issue which I do not understand.
If I understand your model properly (and, I might not), you have the same potential at all points on a square surrounding A (nodes which have an equal manhattan distance from A). But this suggests that the resistance is the same from all points on that square back to A. Can this be correct? Thanks, -- 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
