Given an "infinite" 2D square grid of resistors with resistance R, the equivalent resistance across the diagonal of one square is 2R/pi [1,2]. Working out this solution theoretically has sniped many a nerd. Furthermore, it is prohibitively laborious for an amateur to attempt to verify this experimentally -- until now. Using Bare Paint, I drew a square grid of resistors on a page of normal copy paper, and measured resistances with a multimeter. My wishful thinking was that, with a large enough array of resistors (14 x 14), I could approximate "infinity" sufficiently from the perspective of squares in the middle of the page. Drawing the resistors with the Bare Paint seemed easier than gathering and connecting 196 resistors.

While one set of measurements yielded a value of pi as impressive (to me) as 3.38, overall I observed poor consistency of measurements among adjacent squares near the center of the page, and poor repeatability of measurements for the squares tested. Nevertheless, I hope that you, dear reader, will consider trying this experiment for yourself. I will tell you how I did it, and perhaps you will take more care than I to apply the paint accurately and precisely; otherwise, you will too observe a high variance of resistance among your resistors. Note also that the resistance of the Bare Paint decreases as it dries, so be sure to allow ample time for drying (tens of minutes) before measurements.

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Submitted by Ace Monster Toys Hackerspace in Oakland, CA for the Instructables Sponsorship Program

While one set of measurements yielded a value of pi as impressive (to me) as 3.38, overall I observed poor consistency of measurements among adjacent squares near the center of the page, and poor repeatability of measurements for the squares tested. Nevertheless, I hope that you, dear reader, will consider trying this experiment for yourself. I will tell you how I did it, and perhaps you will take more care than I to apply the paint accurately and precisely; otherwise, you will too observe a high variance of resistance among your resistors. Note also that the resistance of the Bare Paint decreases as it dries, so be sure to allow ample time for drying (tens of minutes) before measurements.

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Submitted by Ace Monster Toys Hackerspace in Oakland, CA for the Instructables Sponsorship Program

I used one tube of Bare Paint, a multimeter with leads (pictured in the intro section), and an 8 1/2" x 11" sheet of paper.

I think that your design of the net with conductive ink is a cleaver idea :) <br>The problem is only about the possibility of doing UNIFORM painted lines. Uniform in the amount of ink in each little piece or line (in each differential of line in the mathematic language). <br>May be if you use an old ruling pen and a ruler ...

I am very happy to see this instructable. Google used it as one of their questions for prospective employees on a famous test they gave around the time they went public. someone asked "why pi"? The reason is that the grid is symmetrical to infinity but there are different degrees of infinity so that it is important that the answer contain an element of "equal distance in all directions" which is how a circle is formed. In other words, the solution requires "equal distance to enforce equal infinities". The long derivations to such a simple answer, in my opinion, show that this problem exposes a terrible weakness in today's math. A simple answer from a simple question should have a simple derivation, if the analytical method is in any sense "powerful" for that type of question.

I like that intuition about pi being part of the answer due to symmetry. I also appreciate derivations that avoid long sequences of symbolic manipulations; if you haven't seen Bret Victor's project to "<a href="http://worrydream.com/KillMath/" rel="nofollow">Kill Math</a>", I find it inspiring. I particularly liked the snippet from Strogatz's <u>Nonlinear Dynamics and Chaos</u> he includes at the bottom of that page.

Maybe someone can explain why pi is in the denominator and why 2R is up top. Even a series expansion argument is too complex. I think viewing it as a conductance would help in that effort: Ctotal =C*pi/2. Re-arranging: pi=2*Ctotal/C and since pi = circumference / diameter, C is the diameter of the "simplified (circle) viewpoint" we are looking for and 2*Ctotal is its circumference. Or it might be better to find the viewpoint that says diam=1/2*C and circumference=Ctotal. Yes, the whole intellectual explosion around Chaos in the 1980's was from people finally taking more notice of graphical methods in the back of a few 1960's and 1970's DE books, although the ideas had been around a lot longer. Feynman diagrams are another example of a "break-through" that occurred from drawing pictures instead of equations. Our mind can understand pictures pretty darn good. We need a good picture of how Michael Hudson's economics works and then vote accordingly.

Oh man...I remember seeing this problem in my junior electrical engineering courses. We lucked out - it was given as a final exam question the year before...but as it turns out most aerospace and mechanical engineers are terrible are circuit analysts...so we were spared a repeat of that blunder. <br> <br>Interesting experiment...this would be a good showcase in circuitry courses. Nice approach...I wonder how big an array would actually be needed to reasonably approximate pi? Kudos!

I have not calculated the minimum error associated with approximating pi this way as a function of array dimension; as you mentioned, this could be a useful table to reference.

sounds interesting

Wouldn't a conductive wire mesh like an RF Shield be able to approximate it? If it requires a high resistance for accurate measurement, maybe a type of filtering system that uses tiny meshes may be resistive enough.

Sounds good to me. Try it and report back to us. :)

Keep in mind that a lead (graphite) pencil is conductive and could be used in place of the paint perhaps?

Yes, and it seems user Rogie-Bear has taken this approach.

Brilliant ‘ibble! <br>Certainly the most thought provoking <br>My attempt to emulate your setup using penciled 8X9 grid yielded pi=1.44! <br>Close, but not close enough! <br>

This is a great piece of work. The "final exam" for any EE is the understanding of the infinite grid of resistors. So simple, so majestic and so defining of the difference between an EE and a technician. <br> <br>Fantastic lab exercise for either physics or circuits.

Best instructable I have seen in quite some time.

THIS IS INCREDIBLE! We are completely in awe.

this is the first time I have heard of this problem...

Could you explain why R_diag = R*(2/pi) <br>Thanks a lot in advance.

That result is derived in each of the <a href="https://sites.google.com/site/resistorgrid/" rel="nofollow">two</a> <a href="http://www.mathpages.com/home/kmath668/kmath668.htm" rel="nofollow">references</a> I provided at the end of my first sentence, and the derivation is difficult to summarize in a sentence or two, even with heavy jargon. I recommend printing out one of the derivations and finding a local math junkie with whom to sit and walk through it.

great application for the paint! really cool project, I'd never seen this problem before.