Folding paper, again I was looking at the earlier topic about folding paper, and I found something interesting. Clearly, for a given thickness of paper, the larger it is, the more times you can fold it. So what is the relationship between the size of the paper, and the number of times you can fold it before it becomes impossible to fold any more? If I can fold a Post-It note six times, how big a piece of paper would I have to have to be able to fold it seven times? How about eight times? The result was kind of surprising to me. Try working out the answer "theoretically," then try folding some differently sized pieces of paper and see if your prediction holds out. Ham Fisted Wednesday, April 21, 2004 I think 7 times is the max for any size paper Just a guess Monday, April 26, 2004 I've seen a large sheet of paper folded 11 times... so no. Ham Fisted Tuesday, April 27, 2004 we have to stop folding, when both our dimensions of the piece we hold are less than the twice the height of the folded thing - there won't be enough paper to wrap around the edge. So our width not only halves, but there's some more gone around the edge. Unfortunately, the simple model developed that way - ie, from X, Y, Z go to X/2 - Z, Y, 2Z, doesn't really preserve volume, so it seems quite unrealistic. Nevertheless, it's amusing to play with - if we only fold along one side, thus having 1D paper, maximum folds is something like 1/2 * log2 ( 3X/2Z +1 ), where X and Z are the original dimensions. So in the case of X = 0.1m, Z = 0.0001m, this comes to around 5... Then we try folding both ways, alternating, and we get 2/3 * log2 ( 7X/2Z +1 ), which arrives at the 7 mentioned. There might be some +/- 1, so if you're really strong, you might get 8 from the equation. First time in my experience where physical strength can give you a different solution of a maths problem. I have no idea why alternating would be the best sequence, though it certainly seems that way. If we now try to preserve volume, the model becomes: X, Y, Z goes to X/2 - Z, Y, Z1, where Z1 is 2Z adjusted so the volume stays the same ( some justification for that is that there will be some creasing in the inner folds, in-creasing the height ) This one, unfortunately, leads to a badly non-linear equation, so I haven't found a decent closed form. It seems that here the subtraction mostly contributes some lower order term, so all that matters is the halving/doubling - ie we get log4 ( X/Z ) phew... if only I was so productive at work... Ivan Wednesday, May 12, 2004   Fog Creek Home