This is gorgeous. Thank you.
On 01/10/2013 04:24 AM, Donald Kelly wrote:
This is the resistance between adjacent nodes. Up to today I couldn't handle
non-adjacent nodes but have found what appears to be an elegant way to deal
with it.
This also assumes an infinite grid but would give good results for a grid which
is large compared to the distance between the nodes under consideration.
The first step is to look at the branches from one of the nodes to the adjacent
ones. With a unit injected corrent, and calling on symmetry, the current in
each of the 4 branches is the same so each of the nodes at the ends of these
branches are at the same potential 1/4V (i'm not worrying about sign so I will
call it positive).
These nodes can then be connected into a super node (a diamond)-call it ring 1
without affecting anything else. Each of these nodes has 3 outgoing connections
so that there are 12 connections to the next set of nodes and as the total
current is still 1A the incremental voltage change is 1/12V Now consider the
diamond 'ring' of these nodes ring2 It has 3*4 corner outgoing branches and 4*2
side branches for a total of 20. resultant delta V from here to the next ring
is 1/20V. keep going and the next will have 12corners + 8 side nodes
for a net of 28 so delta V is 1/28
Taking in superposition we can double these to get
2*(1/4)(1+1/3+1/5+1/7+...+1/n) as the voltage and resistance between nodes.
R=0.5*sum from 0 to n of 1/(2n-1)
How do we get n? Simply count the number of resistors in the shortest path
between nodes.
the xkcd 356 problem has n=3
R=:+/\0.5*%1+2*i.3
0.766667
showing part sums below
R=:+/\0.5*%1+2*i.3
0.5 0.666667 0.766667
or
]R=:+/\x:0.5*%1+2*i.3
1r2 2r3 23r30
as to being close together
]R100K=:+/0.5*%1+2*i.100000
3.36911
Note that the series doesn't converge to a finite value but looks logarithmic
----- Original Message -----
From: "Graham Parkhouse" <[email protected]>
To: [email protected]
Sent: Wednesday, January 9, 2013 8:05:57 AM
Subject: Re: [Jprogramming] xkcd 356
-----Original Message-----
Date: Tue, 08 Jan 2013 18:29:12 -0800
From: Don & Cathy Kelly <[email protected]>
To: Raul Miller <[email protected]>
Cc: [email protected]
Subject: Re: [Jprogramming] xkcd 356
Message-ID: <[email protected]>
Content-Type: text/plain; charset=UTF-8; format=flowed
In the case of the resistance between adjacent nodes it is actually
easier than the finite grid case.
Consider 1A injected into one node and drawn out at infinity. At the
injection node 1/4A flows in each of the 4 branches.
Now remove this source and provide on drawing out 1A at the adjacent
node (and current going in at infinity).
Again we have current flowing in at 1/4 ohm in each branch. Now
superimpose to get no current in or out at infinity but 1A going in at
the first node and out at the second node. In the adjoining link the
current will be 1/4 +1/4 =1/2A or half the current injected. So between
the terminals there is 1/2 volt and 1A corresponding to an effective
total resistance of 1/2 ohm.
For a purely resistive grid of any dimension transients can be
neglected.
Don
Beautiful! I think that such an amazingly simple argument only exists for
two nodes this close together. The resistance between adjacent nodes is also
1/2 ohm.
Regards
Graham
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