The conventional wisdom:

It is impossible to fold any piece of paper, no matter how big, small, thin, or thick, more than 8 times.

It has been cited by many sources and quoted often in kids' programs. For example, Math@Home and PBS Kids each talked about such.

Well, it is false.

Britney Gallivan, now a student at U.C. Berkeley, successively folded a piece of paper twelve times when she was still a junior in high school. Her achievements made MathWorld and is reported by the Historical Society of Pomona Valley, who also sells a forty-page booklet about her achievement (why the historical society is beyond me).

But she not only folded a piece of paper 12 times: she also derived equations for how big a piece of paper you need to start with in order to fold a certain number of times.

For a long, skinny piece of paper, which is only folded in one direction, her calculations give the following equation: where L is the minimal length of the paper (width doesn't matter, since you are folding parallel to it), d is the thickness of the paper, and k is the number of folds. This is also known as the Loss Function (I'll show why in a bit).

For a square piece of paper folded in alternating directions, she gives the minimal width W by

---

How did she come up with the equations? Imagine we started with a sponge, instead of paper. Suppose we fold it in half: it is easy to see that the face on the inside gets compressed a bit and the face on the outside gets stretched a bit. We can assume that at the center of the sponge, it is neither stretched nore compressed. The criterion Britney defined for a successive fold gives some hint:

For a sheet to be considered folded n times it must be convincingly documented and independently verified that (2ⁿ) unique layers lie in at least one straight line. Sections that do not meet this criteria are not counted as a part of the folded section.

What does this tell us? Going back to our sponge (which I conveniently pictured below, having been folded twice), only the parts shaded in blue can count toward a fold. In other words, the parts not shaded are lost. That is why it is called a Loss Function! In the case that the paper started out with the minimal length required, we will end up with no shaded region at all: all the paper would be in the fold and we will have one vertical line down the middle where the paper is "folded".

So let's calculate. After the first fold, the center of the the sponge/paper forms an arc/semi-circle with diameter equal to the thickness of the paper. The second fold produces two arcs (you can see it in the picture above), one with diameter equal to the thickness of the paper, the other one with thickness of thrice the diameter. By induction, we can show simply that the n-th fold will produce 2^(n-1) arcs, with diameters ranging from 1, 3, 5, ..., (2^n-1) times the thickness of the paper.

Given that the arc length of a semicircle is

Arc = π Diameter / 2

We can add up the total length of the various arcs after k number of folds: The inner sum can be calculated as a simple arithmetic sequence

Finite sum of arithmetic series = (Top + Bottom) Height / 2

and evaluates to (1 + (2*2^(i-1) - 1))* 2^(i-1) / 2 = 2^(2i-2). We rewrite that as 4^i / 4 and sum as a geometric sequence by

Finite sum of geometric series = (Ratio Top - Bottom) / (Ratio - 1)

which evaluates to (4^(k+1) - 4)/(4-1) / 4 = (4^k - 1) /3.

Putting it all together, we get

Loss = π/6 d (2^2k-1)

Wait, you say, it doesn't agree with Britney's results!! What's wrong?

Well, For my calculation, I made the assumption that a paper is like a sponge: when you fold it, the inner side shrinks while the outerside stretches, and so I measured the arcs in the center of the paper. If you factor (2^(2k) - 1) you get (2^k+1)(2^k-1), which is slightly different from Britney's formula by one number. My guess is that she made the (perhaps more physical) assumption that the outer side of the paper won't stretch, while the innerside compresses. The point is: to the highest order of calculation, our results agree (since this is only a model anyway, we don't have absolute precision in our calculations).

Well, let's calculater some real-world numbers. According to Paper-Paper, every-day 20lb paper has a thickness just shy of 0.1 miillimeter. Letting k = 12, 2^(2k) = 2^24 is 16777216, which means that if Britney were to start with common paper, she would need a length of 879 meters to boot... that is a LOT of paper.

(Take a look at the Pomona historical society's webpage. It has a picture of Britney with 11 folds done... the left over thing looks just about 2 - 3 feet. 11 folds gives 2048 layers, so that means she started with a piece of paper somewhere between 1.2 and 1.8 kilometers; for those metric impaired, that is a 1-mile-long piece of paper she started with...)

How about copy-machine paper at letter size? 11 inches is 28 centimeters, divide that by 0.1 millimeters and multiply by 6/π, we get roughtly 5350. So roughly speaking, by the estimate, if you only fold along the long edge of a normal, letter-sized, xerox-paper, you should be able to fold it 6 times max.

The derivation for the square paper, then, as always, is left as an exercise for the reader.