Britney Gallivan's paper folding formulas According to a few YouTube videos and New Scientist, formulas exist (based on algebraic/mathematical premises, thereby making this a valid math question) to describe the limits of paper folding. The following 2 formulas are due to Britney Gallivan (who discovered them while in high school!) and apparently they are accurate:

An upper bound and a close approximation of the actual paper width needed for alternate-direction folding is
  $$W = \pi t 2^{(3/2)\left(n-1\right)}.$$
For single-direction folding (using a long strip of paper), the exact required strip length $L$ is
  $$L = \frac{\pi t}{6}\left(2^{n}+4\right)\left(2^{n}-1\right),$$
  where $t$ represents the thickness of the material to be folded, $W$ is the width of a square piece of paper, $L$ is the length of a paper piece to be folded in only one direction and $n$ represents the number of folds desired.

from Wikipedia.
My question: can anyone explain the derivation and foundational logical steps which give rise to these formulas? I have been unable to find anything on this in the web. 
Thank you :)
 A: I heard about this on the TV show QI. I was intrigued by it, so I looked into the single direction folding case. I found that the derivation of the single direction formula is fairly straightforward.
Every time you fold the paper you "lose" some paper due to the fold. The rest is straight.
The formula gives the total loss after n folds, which is equivalent to the minimum length to make n folds. Folding some paper may help see what is going on.
For paper of length L and thickness t:
On the first fold, you lose a semicircle of radius t. So the length lost is $\pi t.$
On the second fold, you lose a semicircle of radius t and a semicircle of radius 2t. So the length lost is $\pi t + 2\pi t$. Total length lost: $\pi t$ + $(\pi t + 2\pi t)$.
On the third fold, you lose a semicircle of radius t, and a semicircle of radius 2t, and a semicircle of radius 3t and a semicircle of radius 4t. So the length lost is $\pi t + 2\pi t + 3\pi t + 4\pi t$. Total length lost: $\pi t$ + $(\pi t + 2\pi t)$ + $(\pi t + 2\pi t + 3\pi t + 4\pi t)$.
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So after n folds the total length lost is:
$\pi t$ + $(\pi t + 2\pi t)$ + $(\pi t + 2\pi t + 3 \pi t + 4\pi t)$ + ... + $(\pi t + 2\pi t + 3 \pi t + 4\pi t +... + 2^{(n-1)}\pi t)$
Now it's just a matter of cranking the handle a few times to output the formula:
$\pi t$ + $(\pi t + 2\pi t)$ + $(\pi t + 2\pi t + 3 \pi t + 4\pi t)$ + ... + $(\pi t + 2\pi t + 3 \pi t + 4\pi t +... + 2^{(n-1)}\pi t)$ =
$\pi t(1 + (1 + 2) + (1 + 2 + 3 + 4) +...+ (1 +2 + 3 + 4 + ... + 2^{(n-1)})$
Using the formula for the sum of an arithmetic series this is:
$(\pi t/2)((1*2) + (2*3) + (4*5) + (8*9) +...+ (2^{(n-1)}*(2^{(n-1)}+1))$.
The nth term is $2^{(n-1)}*(2^{(n-1)}+1)$. This is $(2^2)^{n-1} + 2^{n-1}$.
So rewrite the formula:
$(\pi t/2)((1*2) + (2*3) + (4*5) + (8*9) +...+ (2^{(n-1)}*(2^{(n-1)}+1))$ = 
$(\pi t/2)((2^2)^0 + (2^2)^1 +...+ (2^2)^{n-1})$ + $(\pi t/2)((2^0 + 2^1 +...+ 2^{n-1})$.
Using the formula for the sum of a geometric series we get:
$(\pi t/2)((2^2)^0 + (2^2)^1 +...+ (2^2)^{n-1})$ + $(\pi t/2)((2^0 + 2^1 +...+ 2^{n-1})$ = 
$(\pi t/2)((2^2)^n-1)/3) + (\pi t/2)(2^n-1)$ =
$(\pi t/6)(2^{2n}-1) + (\pi t/6)(3*2^n-3)$ =
$(\pi t/6)((2^n)^2 + 3*2^n -4)$ = 
$(\pi t/6)(2^n+4)(2^n-1)$.
A: Back in May 2014 I analyzed the paper folding procedure out of curiosity how the published formula had come about. The result, with a detailed description of the derivation process and precise statements of assumptions, including comparisons of different choices for the appearance of the resulting paper stack, is published in a blog here: 
http://paperfolding-analysis.blogspot.com/ 
