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Subject: "Question on the Limits on Paper Folding"     Previous Topic | Next Topic
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Cliff P
guest
Aug-13-07, 05:21 PM (EST)
 
"Question on the Limits on Paper Folding"
 
   I'm having a little trouble understanding the exact meaning of "L" in the
formula at the following web page, which relates to paper folding. I'd like to explain L to a broad audience, and don't quite understand it myself. Can anyone help?

https://www.sciencenews.org/articles/20040124/mathtrek.asp

I also am unsure of the meaning of this sentence: "This means that for the 12th folding of paper in half, 2798250 times as much material has been lost to potential folding as was lost on the first fold."

What does it mean to "lose paper to potential folding"?

Thanks. Cliff


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alexb
Charter Member
2228 posts
Aug-13-07, 11:26 PM (EST)
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1. "RE: Question on the Limits on Paper Folding"
In response to message #0
 
   >I'm having a little trouble understanding the exact meaning
>of "L" in the
>formula at the following web page, which relates to paper
>folding. I'd like to explain L to a broad audience, and
>don't quite understand it myself. Can anyone help?
>
>https://www.sciencenews.org/articles/20040124/mathtrek.asp
>
>I also am unsure of the meaning of this sentence: "This
>means that for the 12th folding of paper in half, 2798250
>times as much material has been lost to potential folding as
>was lost on the first fold."
>
>What does it mean to "lose paper to potential folding"?
>

I am at a loss. There's Pi in the formula, so I guess something must be right, but what limit'she's talking about I could not fathom.

But you should buy her book, you know. Just to have a chance to learn from the source. May it be that I. Peterson got the formula wrong.

Learn from the masters: I just got a copy of de Morgan's A Budget of Paradoxes - a 400 page book. In the introduction he mentions that he only used the material from his own library.


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Mcc
guest
Aug-14-07, 08:18 AM (EST)
 
2. "RE: Question on the Limits on Paper Folding"
In response to message #1
 
   "Lost to paper folding" appears to be how they define the amount of paper that has been folded up to leave you with one unit of paper width.

It'stops being possible to fold something when it is higher than it is wide, simply because there is not enough material to close the fold. Ignoring the amount that is lost on the corners, you can think of the increase in height per decrease in width as a doubling in height for every fold, resulting in a pile that is now half as wide but twice as thick. Of course there are corners to include too, but this will do for now. If you know what the amount of paper is that you would need to start with to leave you with a m metre wide pile after n folds, and you know what the height of the pile after n-folds is, then you can calculate whether or not it will be possible to fold it again (is it'still wide enough to be able to close the fold one more time, or is there no longer enough material)

In the formula, L is the minimum length of material that is needed to enable that number of folds with that thickness of material... any shorter and it will not have enough material to close the fold - it will not be wide enough to close over the height of the pile. The formula looks the way it does because folding the material involves "loosing" some in the corners (which are assumed circular). Actually I am surprised that this paper-folding thing was touted as an impossibility, because it truely is dependent on the material used, though the fact you can not keep folding indefinitely is obviously true.


Hope that helps


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Cliff P
guest
Aug-14-07, 10:24 AM (EST)
 
3. "RE: Question on the Limits on Paper Folding"
In response to message #2
 
   MCC, thanks! It'seems that you would agree with this following:

In 2001, Gallivan determined equations that characterize the limit on the number times we can fold a sheet of paper of a given size. For the case of folding in a single direction of a paper with thickness t, we can estimate the initial minimal length L of a paper that is required in order to achieve n folds: L = <(PI*t)/6>*(2^n+4) * (2^n-1). We may study the behavior of (2n+4) × (2n-1). Starting with n=0, we have 0, 1, 4, 14, 50, 186, 714, 2794, 11050, 43946, 175274, 700074…, This means that for the 11th folding of the paper in half, 700,074 times as much material has been lost to folding, at the curved edges along the fold, as was lost on the first fold.


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phrend
guest
May-23-08, 08:48 PM (EST)
 
4. "RE: Question on the Limits on Paper Folding"
In response to message #3
 
   Gallivan showed when paper is folded, each time some of the paper can not be considered folded as it is not on top of the rest of the material. Surprisingly the amount lost, according to her formula goes up by a factor of about 4 for each fold. When the amount lost due to folding for a particular number of folds is larger than the size of the available paper it can not be fold that number of time, and thus the limit. She was also cleaverly showed how to remove wadding and include it into the maximum number of folds.


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