...can a piece of paper be folded in half?

Well, how many times can you do it? Try!

It is difficult to fold A4 more than 6 times, as in the picture to the left. However, it is not impossible to come beyond the 7-8 times often erroneously cited as the limit, e.g. at Danish Television TV2: TV 2. E.g. I folded a news paper page 9 times.

Why is it so difficult? Because doubling grows extremely fast: 2, 4, 8, 16, 32, 64, 128., 256, 512, 1040, ... We write 2^n for n doublings. Furthermore, in paperfolding it strikes twice: 1) Each fold will halve the length or width, and thus the paper area. The sixth time the 44.6 cm paper strip is folded, the length is 446 / (2^6) = 446 / 64 = 7 mm. 2) The thickness is doubled each time. The same paper strip will be 0.1 * 64 = 6.4 mm thick after 6 foldings. Already now the thickness is as large as the length. As the folding edge each time must "go around" the thickness, there is simply not enough paper left for folding another time.

So the two dominating factors are the paper length (and width), and its thickness. The theoretical connection is described by Britney Gallivan in 2001 while at Pomona High School. The formula is:

L = ((pi * t)/6) * (2^n + 4) * (2^n - 1)

where L is paper length, t paper thickness, and n the number of times to fold over.

Example: Put the Fröbel star strip in, we getthe equation 446 = ((3.1415... * 0.1)/6) * (2^n + 4) * (2^n - 1) = 0,05... * (2^(2*n) + 3 * 2^n - 4), or just about: 27,4 = 2^n * (2^n + 3). (PS: enable javascript to see formula calculator).

Try the calculator. Write the numbers and press "Calculate":

Length (L) in mm:

Number(n):
?
(lenght needed: ? mm, next length: ? mm)

Looking at the A4-folding on the top, it seems likely that another fold is possible. And yes, if you exploit both the length and the width alternatingly, another formula is needed:

W = pi * t * 2^(3*(n-1)/2)

where W is the side length of a square, t paper thickness, and n the number of times to fold.

Put in A4, we see that the paper may be folded 7 times.

Try the formula calculator. Write the numbers and press "Calculate":

Width (W) in mm:

Number of folds (n):
?
(needed width: ? mm, next width: ? mm)

But it doesn't stop here. If instead of alternating, you first fold m times one direction and then n times the other, two equations are needed (this is not proved, but just a "good guess" of mine):

W > ((pi * t)/6) * (2^m + 4) * (2^m - 1)

L > ((pi * (t*2^m))/6) * (2^n + 4) * (2^n - 1)

where W is paper width, L is paper length, t paper thickness, m number of folds in the width direction, and n the number of folds in the length direction. The answer is achieved by maximising (m+n) over these in-equations.

Put in A4, and you get m+n = 5+3 = 8.

Try the calculator yourself. Write the numbers and press "Calculate":

Width (W) in mm:

Number (m+n):
?
(found m+n: ? + ?, using width/length: ? mm / ? mm).

And this yields exactly the 8 times that succeeds for A4, and 9 for tabloid newspaper pages, in pictures to the left.

Please note that we did not consider the difficulty of bending thick paper, nor to lift the paper. E.g. formula (A) tells us that 875 meters of paper 0.1 mm thick must be used to get 12 foldings as Britney Gallivan's record is given for paper. That is 2960 sheets of A4 in the length. As they weigh 5 g a sheet, it is 14,8 kg, and the thickness at the end is 2^12 * 0.1 mm = 409.6 mm, or almost 41 cm. As A4 copy paper is rather stiff, it is probably not possible in the real world. So Britney must have used thinner and possibly more flexible paper, and probably also paper longer than the theoretical minumum given by the formula.

So let me ask again: How many times can you fold a piece of paper?

Until now we have obeyed the "rules of the game" and each time folded all over and through all layers, either in one or the other direction. We change the question into the weaker: "How many layers can a piece of paper be folded into?"

**Other than halfs: **E.g. the tabloid format length (572 mm) is far from the folding limit (281
mm). This may be exploited to fold e.g. in thirds in the final fold, see picture top right. You gain an
extra 256 layers on top of the 512, resulting in 768 layers. But you would not gain more by starting with
thirds, as you then would just eat the paper faster.

**Avoid folding through all layers:** E.g pleat folding always fold only through one layer of
paper, and you avoid wasting as much paper on "going around" the accumulated thickness in each fold. That is
shown for the Fröbel star strip above and to the right. There are 256 layers which would otherwise require 8
foldings in half, that is, 2 more than the Fröbel star strip could otherwise do.

Theoretically the distance between the folding lines in the pleat solution corresponds to taking formula (A) above and insert thickness t, n = 1, and half. This is t * pi/2. The number of layers is then 2 * L / (t * pi). For the Fröbel star strips it should be possible to achieve 2 * 446 / (0.1 * 3.1415...) = 2839 layers, that is 10 times more than my 256. In practice it is hardly possible to do folds that close and actually fold it into the pleated shape. Already the 256 was pretty hard to reach. Maybe someone can go higher, e.g. 512 layers?