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?

  • $\begingroup$ I suspect you will get the bill smaller if you just roll into a cylinder without folding. $\endgroup$ – vadim123 May 14 '15 at 2:29
  • $\begingroup$ 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? $\endgroup$ – Neil May 14 '15 at 2:30
  • $\begingroup$ 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. $\endgroup$ – vadim123 May 14 '15 at 2:32
  • $\begingroup$ This has to do more with materials science than with math. This is a good start skeptics.stackexchange.com/questions/5912/… $\endgroup$ – Valentin May 14 '15 at 2:34
  • $\begingroup$ 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. $\endgroup$ – Neil May 14 '15 at 2:58

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