---------- 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
