As far as I can see- YES
It's a matter of circuit analysis as topology/ graph theory- just as
what I think you mean by Manhattan distance.
Note that the path from A to any point on the square has the same
resistance. In the case of the chess knights move of the original
problem (1 over 2 down or 2 over 2 down, the shortest path involves 3 -1
ohm resistors.
Each square ( which is turned 45 degrees to the grid- looking like
diamonds -is connected to the next square by the 1 ohm grid.These
squares don't touch any grid elements except at the nodes. That is good .
Starting from the injection point, the resistances to the adjacent 4
nodes are the same- considering symmetry each of the 4 branches carry
the same current. (1/4A for 1 A injection) The next square is connected
by 1 ohm resistors and there are 12 branches between them- all of 1 ohm.
Again symmetry comes in so the current in each branch is now
1/12A.....etc. one could connect the inner 4 points with a scrap piece
of semiconductor and it would not affect anything inside or outside.
Similarly the next square has a total of 1A coming in and is also an
equipotential but this has 20 outgoing leads---etc etc etc. Recognizing
this along with the equipotential situation spilled the water in my bathtub.
The trouble is that considering only injection in at A and out at
infinity (which can also be considered as a square equipotential) is
insufficient
as the problem involves current in at A and out at B.
Looking at current out at B and in at infinity, the voltage Vab will be
the same (same distance-different square but again symmetry because of
the reference at the end of the universe)
The superposition gives a unit current between A and B and eliminates
the pesky current at infinity,
Don
On 28/01/2013 10:57 AM, Raul Miller 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,
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