# Is there a mathematical way to fold a $20 dollar bill for compactness? I had a strange thought. I used to carry a pill fob on my keys with an emergency$20 bill in it, before the whole thing got stolen.

I always had some trouble fitting the bill inside the fob and barely managed it each time. I would fold it a couple times then roll it really tight and put it into the tube before it expanded.

Imagine that I don't have the restriction of needing a tube. I know to minimize volume a sphere is best, but you can't fold a paper into a sphere, something about Gaussian Curvature. Is there a math formula or technique where a bill can be folded maximally? Specifically to narrow it down I found this formula online: $$L = \frac{\pi t}{6}(2^n + 4)(2^n - 1)$$

Gallivan's formula gives the maximum number of times you can fold a bill going in the same direction. There is also another formula she came up with for an accordion fold.

If I have a finite number of folds that can be done and I know what that number is. Then is there a way to figure out what the most compact method of folding the bill is without having to try all of them?

• I suspect you will get the bill smaller if you just roll into a cylinder without folding. May 14, 2015 at 2:29
• But is there a proof? like I mentioned in the post I did a hybrid fold and roll or roll and fold, but is there a math way to show there isn't anything better?
– Neil
May 14, 2015 at 2:30
• When you fold, unless you crease the fold very tightly, you will get a little airspace inside the hinge. This will make your total diameter larger. If you do a perfect, tight, roll, then there will be no airspace. May 14, 2015 at 2:32
• This has to do more with materials science than with math. This is a good start skeptics.stackexchange.com/questions/5912/… May 14, 2015 at 2:34
• Valentine that isn't very helpful, but good try. I already have the formula from that link in my post. The bill isn't made of a compressible material and it has a standardized thickness, with that information it becomes about getting the smallest volume for a finite set of folds.
– Neil
May 14, 2015 at 2:58

According to this all US bills are the same size: is $$2.62\times 6.14 \times .0043$$ inches. We can call this $$2.62 = h$$, $$6.14 = \ell$$, and $$.0043 = t$$. So it already has a set volume (.0692 cubic inches) that cannot be changed. Minimizing volume, therefore, can only be interpreted as minimizing the amount of air that sneaks in when the folds aren't tight enough. Rolling is probably the easiest way to avoid airspace sneaking it.

So, to fold it compactly, we want it to be as sphere-like as possible. When paper is folded, Gaussian curvature does not change, so if the paper is curved in one direction, it must be flat in the other direction. So the closest we can get to a sphere, therefore, is a cylinder (curved in one direction and flat in the other) where the diameter is as close as possible to the height.

To allow the fold to be as tight as possible, we need to use an accordion fold $$n$$ layers thick rather than folding in half over and over (which is what Gallivan's formula applies to). Since loss in an accordion fold increases linearly, as long as we do nothing ridiculous we can discount it from the calculations. Further, to minimize the folds and prevent loss, we will be folding the height of the bill rather than the length. So will can the formula $$h/n$$ to find the height of the final cylinder and $$nt$$ to find the thickness after the accordion fold.

When we roll it, the area of the cross-section of the cylinder will be $$A = nt\ell,$$ because the thickness $$nt$$ times the length being rolled $$\ell$$ will become the area of the circle assuming it is rolled tightly. Since $$A = \pi r^2$$, we see that $$2r = 2\sqrt{\frac{nt\ell}{\pi}}.$$

Since we want the diameter $$2r$$ to be as close to the height of the cylinder $$H$$ as possible, we set $$H = 2r$$. This gives us \begin{align} 2\sqrt{\frac{nt\ell}{\pi}} =& \frac{h}{n} \\ 2\sqrt{\frac{n\cdot .0043 \cdot 6.14}{\pi}} =& \frac{2.62}{n} \\ \sqrt{.008404 n} =& \frac{1.31}{n} \\ .008404n =& \pm\frac{1.7161}{n^2} \\ .008404n^3 =& \pm 1.7161 \\ n =& \pm 5.889. \end{align} So we should fold the paper into 5.889ths, which we round to 6ths. This is the closest we can get to a cylinder with equal height and diameter.

So the most compact way to fold any US bill (for non-US bills, the numbers would need to be worked out again) is to accordion fold it in 6ths to get a long strip, and then tightly roll the strip.

When actually folding, the rolling portion is fairly difficult and might take several tries, but it is doable. Trying it myself, I ended up with a cylinder with a diameter of 1/2 in and a height of 7/16 inches. This is probably as close to spherical as possible, as shown below.