The problem as I see it is that the method of obtaining the solution involves three assertions. The first assertion is that with one ampere fed into into a node, let`s call it node A, of an infinite grid the current divides equally between the four resistors connected to that node, let`s call the resistors a0,a1,a2 and a3. Because of the symmetry of the situation this assertion is certainly true. Similarly for the second assertion a current of one ampere pulled out of the opposite, node B, causes a current of 0.25 amperes to flow out of the four resistors connected to that node. This assertion is also certainly true. The four resistors are named b0,b1,b2and b3. Now a0 is connected to b0 and a1 to b1. The third assertion is that superposition may be invoked when the two currents are applied at the same time. The current in all eight resistors remains 0.25 amperes. Node A will therefore be at 0.5 volts above Node A. If we take the junction of a0 and b0 as the reference node (it is where we attach the negative terminal of our imaginary infinite impedance voltmeter) then node A is at 0.25 volts and node B is at -0.25 volts. Since a2 still has 0.25 amperes flowing through it there must be 0.25 volts across it so the voltage where it joins the rest of the grid must also be zero. So 0.25 amperes flowing into the grid produces zero voltage. The rest of the grid therefore displays zero resistance at that point. But there are only three other one-ohm resistors connected to a2 so the resistance can not be less than one third of an ohm and the voltage can not be less than 1%12 volts. So the third assertion leads to the conclusion that the voltage at this junction is both zero and non-zero, which is incorrect. The conclusion is that superposition fails because of the hidden infinity. The current a0 and a1 is greater than 0.25 amperes and the current through a2 and a3 is less than 0.25 amperes.
On Mon, Jan 28, 2013 at 1:57 PM, Raul Miller <[email protected]> wrote: > 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 > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
