Help writing area function for folding paper

[jwpaine]

Just for fun:

I just got thinking about folding an 8*11 piece of paper: and when I do, what the 2d area and how many stacks of paper it would be:

I wanted to write a function so that I could plug in large numbers and see how small the piece of paper would theroticly be.

for page thickness p(f) in pages in terms of number of folds, f, I have: p(f) = 2^f
So, with 4 folds, I have p(4) = 2^4 = 16 pages


Now I need some help on the area: This is what I have so far.....

FOR ALL NUMBER OF FOLDS divisible by 2:

area(f) = (length*width)/(f)

So, say length = 11 and width = 8
Folding the paper 2 times would produce an area of (11*8)/(2)


Now how would I write this area function of mine without having it divisible by 2? (because for every 2 even folds) area = (currentlength*currentWidth)/2

How would I write the function expecting to put in folds = 15 etc.


EDIT: Hmmm... well I guess depending on what side I started folding first so for either H or W... 15 folds = 15-1

 

[Denis]

So, say length = 11 and width = 8
Folding the paper 2 times would produce an area of (11*8)/(2)

No: folding once produces that area.
Folding twice produces area of (11*8)/4.

 

[jwpaine]

thanks.

 

[tkhunny]

You can't use no paper while folding. You have to cut it. Even then, you'll lose some paper unless you have a "Subtle Knife".

 

[Denis]

FOR ALL NUMBER OF FOLDS divisible by 2.....

Ahhh...I see; but why complicate it?

Take a 8 by 12 sheet instead; area = 96 (f = folds):

f   size_after_folding    number     area
0    8x12                 2^0=1    96/1 = 96
1    4x12 or 6x8          2^1=2    96/2 = 48
2    4x6                  2^2=4    96/4 = 24
3    2x6 or 3x4           2^3=8    96/8 = 12
4    2x3                  2^4=16   96/16 = 6
5    1x3 or 1.5x2         2^5=32   96/32 = 3
6    1x1.5                2^6=64   96/64 = 1.5

By the way, if you could fold a sheet of paper 50 times,
the resulting height (200 thicknesses = 1 inch) would be:
2^50 / (200*12*5280) = 88,849,424.47.... MILES :shock:

 

[jwpaine]

Thanks Denis.

Sorry I questioned you. I did my math wrong.


So, Jwpaine summarizes:

Thickness of stack in pages, p can be defined with the function p(folds) = 2^folds
Surface area (of top or bottom) can be defined by area(folds)=[(8*11)/(2^folds)]

So total volume = V(folds) = ((8*11)/(2^folds))*(2^folds(SinglePageThinkness)))

so v(folds) should always stay the same, because as the top and bottom of the cube decrease in surface area, the length of the cube increases in length. (The decrease in surface area of the two sides is directly proportional to the increase in length).

Thanks, John.