## Introduction: Estimate Pi Using Conductive Paint

Third Prize in the

Pocket Sized Electronics

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.

--

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

## Step 1: Materials

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.

## Step 2: Folding the Paper to Make "guide Lines" for Drawing

To make a square grid of lines to guide your eyes and the pen when drawing the resistors, you can first fold the sheet diagonally to be flush with a long edge, as shown. Cut/tear the sheet along the interface between the two-layer part and the still-one-layer part to end up with (after unfolding) a square piece (8 1/2" x 8 1/2") of paper.

Then, for each of the two axes of the sheet, successively fold the sheet in half four times and unfold it completely.

## Step 3: Draw the Grid!

With a calm and steady hand, keeping an eye on the guide lines, and with light pressure on the pen body, draw the grid. Well, to decrease variability, you may try a different method, but drawing by hand is easy. I waited a bit after drawing all lines in one direction before drawing the crossing lines, but I think that was unwise because the drawn lines were able to dry/harden a bit and thus more care was required on my part when drawing across the already-drawn lines -- sometimes the pen seemed to "jump" with some momentum over the already-drawn lines, and thus I had to backtrack to ensure well-connected junctions.

You can clearly see than my drawn grid lacked uniformity, but I didn't know then how much that would matter. Hopefully you will do better than me.

Also, note how I drew "dangling" resistors at the edges of the sheet. These are useful to estimate the resistance of an isolated stretch of the paint equal in length to a side of one square; you need this value to estimate pi.

## Step 4: Measure!

"*To measure is to know*." -Lord Kelvin

There are three simple measurements to do. Using your multimeter in "ohmmeter" mode (the units of resistance are Ohms, symbolized by the Greek capital letter Omega), you can measure (1) the resistance across an isolated "dangling" resistor that is not part of the grid; let's call this resistance R. Next, you can measure (2) the resistance across one side of a square near the center of the paper; the resistance should theoretically be R_side = R / 2. Finally, you can measure (3) the resistance across the diagonal of one square; the resistance should theoretically be R_diag = R * (2 / pi). So, pi should equal 2 * (R / R_diag). Is it close to 3.14? Note that I can list at most three significant digits because my pictured multimeter only provided three. I got as close as 3.38 for one set of measurements.

## Share

## Recommendations

We have a **be nice** policy.

Please be positive and constructive.

## 20 Comments

I think that your design of the net with conductive ink is a cleaver idea :)

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).

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 "Kill Math", I find it inspiring. I particularly liked the snippet from Strogatz's

Nonlinear Dynamics and Chaoshe 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.

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?