Introduction
Recently, I came across a chapter in the book by Steven Strogatz "The Joy of x: A Guided Tour of Math, from One to Infinity" on the effect of exponential functions. By way of an example the author0 discusses the task of folding a piece of paper with a given length and thickness say 7 or 8 times, or generally n > 7 times. This turns out to be essentially impossible using a commonly available sheet of paper, for instance a large newspaper sheet. The reason is that the thickness of the resulting wad of paper rapidly becomes approximately as thick as the wad is long at which point another meaningful fold is impossible. He cites a formula derived by Britney Gallivan in 2002 that describes the relationship between the paper's geometry, that is, length L and thickness T, and the number n of possible foldings.The formula provided is Gallivan's equation, see also this Wikipedia article
L = (π/6)T·(2n + 4)(2n – 1) . (1)
This expression gives the minimal length L of a paper with thickness T needed to fold it n times in a single direction. The conditions that define "minimal length" exactly, however, were not indicated, nor was it described how the wad of folded paper looks like after the last fold has been made. So in order to better understand the formula and how it was obtained I derived a formula for the folding procedure for which the assumptions are clearly stated.
Folding a Length of Paper
A simplified version of folding paper is to cut the paper in half and stack the two pieces on top of each other. Then the process is repeated by cutting the two pieces in half again and stacking them on top of each other again, and so on. This approach ignores the effect that the folds in a multilayer stack would have on the required paper length and will thus give an upper limit to the number of 'folds' possible; conversely, it determines the minimal length L for paper of thickness T to be folded n times. It will also provide a basis for subsequently discussing the effect of the actual folding process.A) Disregarding the folds: Cutting the paper
Before 'folding' — that is: cutting and stacking the halves — the paper has length L and thickness T; its width W is not relevant.
Counting each cutting by the variable k, after the first cut, that is for k = 1, the length of the resulting stack is l1 = L/2 and the thickness is t1 = 2·T, see the diagram at right. After the next cut, the dimensions are l2 = l1/2 = L/22 and t2 = 2t1 = T·22. In general, after k cuttings and stackings the stack has a thickness of tk = 2·tk–1 = T·2k and its length shrinks to lk = lk–1/2 = L/2k.
For an adequate ending of this process one could arbitrarily require that after the last fold, k = n, there still remains a flat part sufficiently long for the stack to be self-supporting and not topple over, that is, the length ln of the stack is still comparable to the thickness tn. A straightforward condition to fulfill that requirement is that the stack is (approximately) as thick as it is long, or
tn = ln (2)
that is, the stack looks like a square from the side. This condition gives a relationship between paper geometry and cuts as L/2n = T·2n from which the paper's initial length-to-thickness ratio L/T is expressed as a function of the number of achievable cuts
tn = ln (2)
that is, the stack looks like a square from the side. This condition gives a relationship between paper geometry and cuts as L/2n = T·2n from which the paper's initial length-to-thickness ratio L/T is expressed as a function of the number of achievable cuts
L/T = (2n)2 = 22n = 4n (3)
It is readily appreciated that the paper length has to increase very strongly with the number of folds. For each additional fold to be made, the paper has to be four times longer. Even for a very light bond paper of weight 10, that is of size 17" x 22" and thickness of 2 mil (about a hair's diameter), one has a length-to-thickness ratio of L/T = 22 / 0.002 = 11,000. Solving equ. (3) for n in this case gives
n = log(L/T) / log4 = 6.7
of which the nearest integer is taken, n = 6 or n = 7. This means the paper can be cut and stacked either six times with the stack appearing flat, i.e., longer than high, or seven times, then looking higher than long. Conversely, to obtain an exact square cross section after the seventh cut, the length of a paper of 2 mil thickness would have to be 0.002" x 214 = 323/4" long.
B) Accounting for the folds in single-direction folding
The length of paper required to form a fold reduces the length of the stack for which only the flat parts of the folded paper are counted, see the diagram for the details and definitions. The initial length L of the paper is thus split, after the kth fold, between the flat parts of length lk and the semicircular folds of total length Fk asL = 2k·lk + Fk with Fk = ∑j=1k fj . (4)
This amounts to a sequence of odd numbers, multiplied by T/2, that is, the ith layer in the fold is bent with a radius ri = (2i – 1)·T/2 which uses up a semicircle π·ri of paper, cf. the red lines in the diagram. In the jth fold, there are m = 2j–1 layers of paper, that is, each fold contains twice as many layers as the fold before it and, therefore, the amount fj of paper used in that fold is the sum over those semicircles
fj = π·∑1m ri = (π/2)T·∑1m (2i – 1) with m = 2j–1 .
The sum of this sequence of odd numbers, [1+3+5+7+...+(2m–1)], evaluates to m2 so that
fj = (π/2)T·[2(j–1)]2 = (π/2)T·22(j–1)
or
fj = (π/2)T·4(j–1) for the jth fold.
The total amount of paper used in a stack that has been folded k times is the sum of all the folds j, from the first fold, j = 1, to the current fold k. This is described by the term Fk in equ. (4) above which can now be evaluated as
Fk = ∑1k fj = (π/2)T·∑1k 4(j–1) (5)
The sum S = ∑1k a(j–1) in equ. (5) is the power series S = 1 + a + a2 + a3 + ... + ak–1 which can also be written as S = ∑0k–1aj . The sum of this series is S = (ak – 1)/(a – 1) which gives, for a = 4, cf. equ. (5), the expression S = (4k – 1)/3 = (22k – 1)/3.
With this result, equ. (4) can be rewritten as
L = 2k·lk + (π/2)T·(22k – 1)/3 . (6)
The stack height after k foldings has again grown to tk = T·2k. Applying the same criterion for the end of the folding process at k = n, namely that of a square aspect of the flat part of the stack, cf. equ. (2) in section A) above, the stack length is ln = tn = T·2n. This allows one to express the paper's length-to-thickness ratio required to make n folds as
L/T = 22n + (π/2)·(22n – 1)/3 . (7)
This puts about 2/3 of L/T into the flat part (using π/6 ≈ 1/2, and neglecting the –1 term in the parentheses against 22n for n > 3) so the square aspect of the flat stack part (from setting ln = tn) appears to be a waste of paper. Setting instead ln = (π/2)·T in equ. (6) – independent of n – achieves two things at the same time, namely
(i) making the length of the flat part of the stack equal to π/2 (≈1.6) times the thickness T of the paper and
(ii) providing a common multiplier of (π/2) for both terms in equ. (6).
These two features allow equ. (6) to be rewritten as
L/T = (π/2)·[2n + (22n – 1)/3] = (π/6)·[3·2n + 22n – 1]
= (π/6)·[3·2n + (2n + 1)(2n – 1)] (*)
which can be cast into a simpler expression by writing ±3 into the [...] parentheses, forming the term 3·(2n – 1) to get [3·(2n – 1) + (2n + 1)(2n – 1) + 3], neglecting the +3 term against the 2n–containing terms, and then combining it with the (... ± 1) product to give
L/T = (π/6)·(2n + 4)(2n – 1) (8)
which happens to be the Britney Gallivan formula.
The difference in L/T between equ. (*) and equ. (8) is (π/6)·3 = π/2 ≈ 1.57 independent of n which can indeed be neglected for n≥4. This difference between the folding procedures, that is, the effect of the choices for ln on the shape of the folded stack can be visualized with the diagram at right for n = five folds using equations (7) and (8).
With a similar approach, the choice for lk in equ. (6) for k = n could be ln = (π/2)·T·2n instead of just T·2n so that the first term in equ. (7) becomes also multiplied by π/2. This gives a rectangular appearance of the flat part of the stack, longer than high, and requires, of course, a paper even longer than what equ. (7) alone would require. From this choice one obtains a simple expression
fj = (π/2)T·4(j–1) for the jth fold.
The total amount of paper used in a stack that has been folded k times is the sum of all the folds j, from the first fold, j = 1, to the current fold k. This is described by the term Fk in equ. (4) above which can now be evaluated as
Fk = ∑1k fj = (π/2)T·∑1k 4(j–1) (5)
The sum S = ∑1k a(j–1) in equ. (5) is the power series S = 1 + a + a2 + a3 + ... + ak–1 which can also be written as S = ∑0k–1aj . The sum of this series is S = (ak – 1)/(a – 1) which gives, for a = 4, cf. equ. (5), the expression S = (4k – 1)/3 = (22k – 1)/3.
With this result, equ. (4) can be rewritten as
L = 2k·lk + (π/2)T·(22k – 1)/3 . (6)
The stack height after k foldings has again grown to tk = T·2k. Applying the same criterion for the end of the folding process at k = n, namely that of a square aspect of the flat part of the stack, cf. equ. (2) in section A) above, the stack length is ln = tn = T·2n. This allows one to express the paper's length-to-thickness ratio required to make n folds as
L/T = 22n + (π/2)·(22n – 1)/3 . (7)
This puts about 2/3 of L/T into the flat part (using π/6 ≈ 1/2, and neglecting the –1 term in the parentheses against 22n for n > 3) so the square aspect of the flat stack part (from setting ln = tn) appears to be a waste of paper. Setting instead ln = (π/2)·T in equ. (6) – independent of n – achieves two things at the same time, namely
(i) making the length of the flat part of the stack equal to π/2 (≈1.6) times the thickness T of the paper and
(ii) providing a common multiplier of (π/2) for both terms in equ. (6).
These two features allow equ. (6) to be rewritten as
L/T = (π/2)·[2n + (22n – 1)/3] = (π/6)·[3·2n + 22n – 1]
= (π/6)·[3·2n + (2n + 1)(2n – 1)] (*)
which can be cast into a simpler expression by writing ±3 into the [...] parentheses, forming the term 3·(2n – 1) to get [3·(2n – 1) + (2n + 1)(2n – 1) + 3], neglecting the +3 term against the 2n–containing terms, and then combining it with the (... ± 1) product to give
L/T = (π/6)·(2n + 4)(2n – 1) (8)
which happens to be the Britney Gallivan formula.
The difference in L/T between equ. (*) and equ. (8) is (π/6)·3 = π/2 ≈ 1.57 independent of n which can indeed be neglected for n≥4. This difference between the folding procedures, that is, the effect of the choices for ln on the shape of the folded stack can be visualized with the diagram at right for n = five folds using equations (7) and (8).
With a similar approach, the choice for lk in equ. (6) for k = n could be ln = (π/2)·T·2n instead of just T·2n so that the first term in equ. (7) becomes also multiplied by π/2. This gives a rectangular appearance of the flat part of the stack, longer than high, and requires, of course, a paper even longer than what equ. (7) alone would require. From this choice one obtains a simple expression
L/T = (π/6)·(2n+1 + 1)(2n+1 – 1) (9)
as can be best seen by using equ. (6) again for analyzing it. This length-to-thickness value can be considered the upper bound on the paper-folding exercise.
The lower bound would be reached, somewhat unrealistically, by setting ln = 0 for a stack without any flat part at all, the paper being used up completely in the folds; this results in the expression
L/T = (π/6)·(2n + 1)(2n – 1) (10)
which is just one fold less than what is described with equ. (9). Incidentally, equ. (3) and equ. (10) combine to the folding described by equ. (7) which is obvious from the selections: square aspect of the flat part of the stack and no folds, equ. (3), vs. no flat stack part and only folds, equ. (10), add up to the choice ln = tn for equ. (7).
C) Numerical visualization of results
The table below gives paper lengths in units of paper thickness, that is, the ratio L/T, required for the number n of foldings to be possible for the cases discussed above.
It gives an impression of the paper geometries required for n = 4 or more folds. Since for n ≥ 5 the terms –1 or –4 and even the term +3·2n can be neglected for a still fairly good estimate of the situation, that is, with less than 10% error and much less for higher n, the simple case of the 'cut & stack' from Equ. (3) gives a good upper limit, overestimating the minimal paper length solutions of Equ. (8) and (10) by a factor of only ≈ 2 (due to the factor π/6), and underestimating the 'square aspect' case of Equ. (7) by about ~35%. Therefore, for practical purposes, a comfortable estimate of the paper length required to manage n folds is obtained from a length–to–thickness ratio of L/T ≈ 4n, and an absolute lower limit is about 10% more than half of that value as obtained from equ. (10).
Since "thin paper" is about 25 µm thick (or ~1 mil, that is, 1/1000 of an inch) – and "standard paper" about twice as thick – the length of paper needed for n folds is obtained by dividing the numbers in the column under Equ. (3) by 10,000 to convert to cm and multiplying the result by ~25. Thus, for 7 folds, a typical length of "thin paper" of ~40 cm down to ~22 cm will work . To achieve Gallivan's result of 12 folds with that paper it should be a bit longer than ~220 m.
