Keith: Likely you missed the email because of this gmail feature:
http://www.jsoftware.com/jwiki/System/Forums#Gmail Chris On Thu, Mar 7, 2013 at 12:45 AM, Keith Park <[email protected]> wrote: > ---------- Forwarded message ---------- > From: Keith Park <[email protected]> > Date: Mon, Mar 4, 2013 at 7:20 PM > Subject: Re: [Jprogramming] ARE:xkcd356 > To: [email protected] > > > Thanks. I suspect that you were the only person to receive it, possibly > because we had been exchanging messages privately. The mail server seems > to have a mind of its own. I would have expected more noise if it had been > posted to everyone. Thanks again. > > > On Mon, Mar 4, 2013 at 3:11 PM, Don & Cathy Kelly <[email protected]> wrote: > > > I received it normally- in my email from the J forum. > > Don > > > > > > On 04/03/2013 8:28 AM, Keith Park wrote: > > > > My post seems to have gone directly to the archives which is no exactly > > what I wanted. How did you receive it? > > > > On Sun, Mar 3, 2013 at 10:15 PM, Don & Cathy Kelly <[email protected]> wrote: > > > >> > >> On 03/03/2013 1:25 PM, Keith Park wrote: > >> > >>> A long look back, an OOPS!, and a couple of questions. > >>> Sixty years ago when I was a post-graduate student an old professor of > >>> Electrical Engineering asked me this same question about the resistance > >>> between two points in an infinite grid of resistors. In his question > the > >>> resistance was across the diagonal of one small square and the > resistors > >>> were each 2 ohms. He added in his crusty manner, "...and if you write > >>> anything down, Park, you are wrong!". I assumed then that the expected > >>> answer was one ohm which I was never able to demonstrate. I was rather > >>> gruntled therefore to see the elegant proof posted by Kelly. After > >>> mulling > >>> over that result for a few days I came to the conclusion that it was > >>> wrong. > >>> > >> So did I- it was wrong because I overdid symmetry -I assumed that the > >> 1/4A flowing in each branch from A splits equally so I got an > approximation > >> of > >> 2*(1/4)*(1 +1/3)=0.6667 vs 2/pi which is 0.6365. My assumption was > >> wrong. > >> Note that the author in the reference uses an approach which involves > >> integration and trig functions. His second reference uses another > approach > >> but one that converges to values involving pi. > >> > >> Don > >> > >> > >> > >> > >> > >> At first I attributed the failure of the proof due to the > >>> presence > >>> of infinities in the argument, but there I too was wrong. Consider the > >>> simpler case of the resistance between two adjacent points on the grid. > >>> Call these points A and B. There are four resistors connect to each > >>> of A > >>> and B and there is one resistor, call it Rab which is common to both > >>> points. We first inject a current into point A. Because the > resistance > >>> of > >>> the grid to infinity is infinite there must be infinite voltage at > point > >>> A > >>> under this condition. However this infinity does not cause any problem > >>> because we only use this step to determine the current in the resistor > >>> Rab. > >>> The current must be 0.25 amperes because the one-ampere current must > >>> divide equally amongst the four resistors connected to point A. We then > >>> remove the current from point A and pull one ampere out of point B. > >>> Again > >>> there must be 0.25 amperes flowing in resistor Rab. We now apply the > two > >>> currents simultaneously. The principle of superposition allows us to > >>> determine the current in Rab by simply summing the results obtained by > >>> applying each current separately. There must then be 0.5 amperes > flowing > >>> in Rab which will produce a voltage of 0.5 volts. The resistance from > A > >>> to > >>> B is therefore 0.5 ohms. Rab is one ohm so the rest of the infinite > >>> grid > >>> behaves as if it were a one-ohm resistor parallel to it. > >>> Now consider the case where the two points are on opposite > >>> corners > >>> of a square. As before call these points A and B. (If I am belabouring > >>> the > >>> argument it is because there seemed to be so much confusion the last > time > >>> that this subject was discussed.) Call the four resistors connected to > >>> point A a0,a1,a2 and a3. Similarly the four resistors connected to > >>> point > >>> B are b0,b1,b2 and b3. The names are arbitrary so we may call the four > >>> resistors that form the square a0,a1,b0 and b1 with a0 connected to b0 > >>> and > >>> a1 connected b1. We now inject a current of one ampere into point A > with > >>> the current flowing out at infinity. The current flowing in a0 and a1 > >>> must > >>> be 0.25 amperes. With the current removed from A we now pull one > ampere > >>> out of point B. Again it is certainly true that there must be 0.25 > >>> amperes > >>> flowing in each of b0 and b1. By superposition we may find the current > >>> in > >>> each resistor when both currents are applied by adding the results > >>> obtained > >>> from applying each current separately . That is certainly true. So > the > >>> current in a0 and b0 must be 0.25 amperes with both currents applied, > but > >>> this is false, of course. > >>> When we apply a current of one ampere to point A there is 0.25 > >>> amperes flowing in a0 and a1, but there is also a current flowing in b0 > >>> and > >>> b1. Call this current OOPS! because we overlooked it. When we pull one > >>> ampere out of point B there will be, of course, the same current OOPS! > >>> flowing in a0 and a1. So under superposition the current in all four > >>> resistors will be 0.25 + OOPS! amperes, so we may conclude that the > >>> resistance between A and B is 0.5 + 2*OOPS! ohms. The value OOPS! is > not > >>> easy to calculate so superposition is of no help to solve this problem. > >>> Don Kelly provided a reference to a solution to the problem at: > >>> > >>> http://www.mathpages.com/home/kmath668/kmath668.htm > >>> > >>> The unnamed author of the report derives an answer of 2%pi ohms for the > >>> resistance across the diagonal. I do not understand the method or the > >>> math > >>> involved. The answer is almost certainly true. > >>> I have used J to build the admittance matrix for small square > >>> grids. > >>> The resistance calculated for a small grid is too high, but it > approaches > >>> the correct value as the size of the grid is increased. So the result > >>> can > >>> be considered as an upper bound for the true value. If one considers a > >>> small grid for which the outer resistors are all short-circuited > together > >>> then one obtains a resistance that is too small but rises as the grid > >>> size > >>> is increased. These results are then a lower bound for the true > >>> resistance. For the largest grids that my computer could handle the > >>> upper > >>> and lower bounds are 0.636945 and 0.636305 respectively. The mean of > >>> these > >>> two bounds differs from 2 divided by pi by less than 0.001 percent. > This > >>> result may be more of a happy accident than reality as I do not know > how > >>> to > >>> properly compare the results for the "open" and the short-circuited > >>> grids. > >>> If Roger will ever get around to implementing %. for sparse matrices > I > >>> could then perhaps squeeze out a couple more decimal places. > >>> It would be sort of fun if someone would write an explanation > of > >>> the > >>> method used in the mathpages reference. Please keep it > >>> simple....something > >>> even an old engineer could understand. > >>> A couple of questions: > >>> 0) Why does pi appear in a calculation for a SQUARE grid! > >>> 1) Matrix division for this special matrix is therefore a way of > >>> computing pi, albeit a very inefficient one. Is this a new way of > >>> computing pi or is it somehow related to a known sequence? > >>> ---------------------------------------------------------------------- > >>> For information about J forums see http://www.jsoftware.com/forums.htm > >>> > >>> > >> > > > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
