# Bedsheet Problem

(Difference between revisions)
 Revision as of 13:31, 26 May 2011 (edit)← Previous diff Revision as of 17:07, 26 May 2011 (edit) (undo)Next diff → Line 24: Line 24: =====Single Direction===== =====Single Direction===== - {{HideShowThis|ShowMessage=Click to reveal Single Direction|HideMessage=Click to conceal Single Direction|HiddenText=The length of paper required to fold a sheet of paper N times in a single direction, meaning the folding occurs in only one direction, includes how much paper has been lost after a certain number of folds. For a single fold there is a minimum length of the thickness times /pi, the length of the semicircle. + {{HideShowThis|ShowMessage=Click to reveal Single Direction|HideMessage=Click to conceal Single Direction|HiddenText=The length of paper required to fold a sheet of paper N times in a single direction, meaning the folding occurs in only one direction, includes how much paper has been lost after a certain number of folds. For a single fold there is a minimum length of the thickness times $/pi$, the length of the semicircle. + The following definition are used in the equation. + *L=length of material needed + *t=thickness of one sheet of material + *n=number of folds + *k=dummy variable + *i=dummy variable L=$\pi t * \sum_{i=1}^n {\sum_{k=1}^{2^{i-1}} 2^{i-1}-(k-1)}$ L=$\pi t * \sum_{i=1}^n {\sum_{k=1}^{2^{i-1}} 2^{i-1}-(k-1)}$ + + The first summation is for the actual folds. The second is for the amount of sheet loss in a particular fold. The expression $(k-1)$ the $2^{n-1}$ is the radius of the outer folded layer. + + The following are the steps to simplify the equation. + + $\sum_{k=1}^n k-1$=$\sum_ {k=0}^n-1 k$=${(n-1)/n}/2$. + + In our case we get $<$

## Revision as of 17:07, 26 May 2011

Bedsheet Problem
Take a piece of paper. Now try to fold it in half more than 7 times. Is it possible? This leads into the problem: what is the ultimate number of folds a flat piece of material can achieve?
There is an urban legend that a piece of paper cannot fold more than 7 times. All who claimed the myth was valid could only cite empirical evidence, they could not explain or prove it mathematically. The puzzle was both mysterious and inexplicable. The wonder that limits the folding has been discovered. This image shows a sheet folded 12 times.

# Basic Description

This well-known problem was not solved before for many reason. First, many believed the puzzle was true because no one understood or proved mathematically the geometric limitations. Most people folded a sheet of paper and no matter what material was used people usually got around 5, 6, and 7 folds. Because 7 was the highest number it became the folding limit. Also, there was little creativity in disproving the myth. No one realized that the thickness, width, flexibility of the material, strength of the person and crease lines all play a role in how many folds are achievable. Nevertheless, a sheet of paper was folded 12 times.

The thickness of the sheet of paper limits the amount of times the paper can be folded. After each fold the thickness doubles. For example after the third fold the thickness is the same as eight sheets. This brings up the issue with folding a sheet of paper in half: it ends up being very thick really fast. In fact, the limit of the folding occurs when the thickness is more than the width. At this point the paper is bent and has curved ends. It is no longer flat and much harder to fold. If we fold in the same direction, which we would not usually do in reality, a narrow strip with a constant thickness will form. Then after a certain number of folds, the thickness will double each time and the width is halved.

Now consider that we are folding the sheet of paper in alternate directions. Then the width is halved every two folds. The thickness behaves the same: it doubles every fold. Folding a sheet in alternative direction requires that the sheet is wide enough to make a fold in the other direction. From this the sheet would have a greater volume. Folding in alternative direction is a possible advantage because unlike folding in one direction the sheet will not unravel a previous fold.

#### King's Problem

Overall the paper folding connects to exponential growth. This is the essence of exponential growth: very small amounts rapidly become astronomically large through simple doubling. There is a math folktale about a clever merchant who asked the King to pay him with grains of wheat on a chessboard. The merchant asked the King to place one grain on the first square, two on the second, four on the third, eight on the fourth, sixteen on the fifth, and so on until the 64 boxes were filled. The king was too proud to admit that he could not calculate the sum of the grains. He foolishly granted the wish, not knowing that it would wipe out his stock and he would be in debt. If it isn't clear yet imagine the 20th box where the number 2 is multiplied by itself 20 times. This means that just on the 20th box the merchant would have more than a million grains. This is an example of exponential growth.

The bed sheet problem though becomes a little bit more complex. The thickness does not simply double after each fold. When a sheet of paper is folded, one end of the paper is placed onto the opposite end. The paper begins to crumple and it does not fold smoothly. With each fold the additional layers make it hard to lay one end over the opposite side. The paper begins to curve and puff at the ends. After a certain number of folds (this varies based on the thickness and width of the paper) the paper will reach a limit and any additional folds will no longer keep the paper long and flat but rather it will be a semi-circle. In addition to the thickness doubling, the sheet of paper begins to curve. The circumference of the semicircle at the edges are included in the thickness formula.

# A More Mathematical Explanation

#### Derivations

If you keep folding a sheet of paper the thickness doubles after each fold. With [...]

#### Derivations

If you keep folding a sheet of paper the thickness doubles after each fold. With one fold, you have a thickness of two paper, for two folds it has a thickness of 4 sheets of paper and so on. The layers increase by $2^N$ where N is the number of folds. This is somewhat similar to the chessboard example.

However, the sheet does not remain flat and long. It begins to curve at the edge. Each time a fold takes place the edges become round and the radius of the curved section. The radius of each layer is a half of the thickness.
With each layer the radius is different. The curved section of the folded sheet is noticeable when the thickness of the paper is equal or greater than the width. With an increase in folds, the radius section begins to take up a greater percentage of the paper's volume. The volume and thickness in curved section grows exponentially. The volume squares with each fold while the thickness doubles. The folding limit is reached when there is not enough volume or length of the remaining paper to fill the curved section. Any folds pass the absolute limit the entire sheet will become a semi-circle. To be specific we consider the folded section to be the region that has two times the number of folds of the straight layers. The curved ends are not counted as part of the folded section.
##### Single Direction

The length of paper required to fold a sheet of paper N times in a single direction, meaning the folding occurs in only one direction, includes how much paper has been lost after a certain number of folds. For a single fold there is a minimum length of the thickness times $/pi$, the length of the semicircle.

The following definition are used in the equation.

• L=length of material needed
• t=thickness of one sheet of material
• n=number of folds
• k=dummy variable
• i=dummy variable

L=$\pi t * \sum_{i=1}^n {\sum_{k=1}^{2^{i-1}} 2^{i-1}-(k-1)}$

The first summation is for the actual folds. The second is for the amount of sheet loss in a particular fold. The expression $(k-1)$ the $2^{n-1}$ is the radius of the outer folded layer.

The following are the steps to simplify the equation.

$\sum_{k=1}^n k-1$=$\sum_ {k=0}^n-1 k$=${(n-1)/n}/2$.

In our case we get $<$

The following equation gives the loss function for folding a paper in half in one direction.

Eq. 1        $\pi t/6 * (2^n+4)(2^n-1)$

##### Alternative Direction

Alternate direction folding uses both flat directions of the paper, so the width decreases after every fold. The alternative direction derivation is complicated because there are two folding limits. For the last fold to be achieved the previous fold's length has the be $\pi$ times the thickness. This relationship applies for both methods of folding. The sides of the fold are $\pi t 2^{(n-1)}$ and the thickness at this stage is $t 2^{(n-1)}$. The total area of all sheet is the area of a single sheet times the number of sheets.

Area=$\pi^2 t^2 2^{3(n-1)}$.

Taking the square root of the area will give the limiting width of the original sheet.

• W=Limiting width
• t=Thickness of material
• n=Number of folds

Equation 2 refers to the limiting paper width based on the last fold.

Eq. 2        W=$\pi t* 2^{3(n-1)/2}$

The equation though isn't completely accurate because it does not include the materials lost in the radii of previous folds. For odd number of folds there is lost if the paper is folded in the odd fold direction. The old fold do not contribute to the loss in the even fold direction. For large number of folds there is a possible limit that is 60% less than the actual limit. In practice a sheet of material could not fold that tight so equation 2 accurately demonstrates the limiting width.

#### Limitations

In order to compute the length of the paper required and the thickness of each sheet it is important to understand what exactly limits the number of folds.

• Smoothness

In particular, the smoothness of the paper determines if the paper can slide around the curved sections. As the thickness increases so does the stiffness and the resistance to folding. At this point no human or machine effort can produce another fold. The stiffness issue occurs when the folded section is length is less than π times the thickness. In order to do another successful fold the length to thickness ratio has to be greater than π.

Another important factor is the difference in length of layers. At the round ends each layer has a different radius and circumference. This can happen when paper is pushed out the folded section and is not included in the curved section. The radius section then takes the majority of the paper’s volume until it reaches a point from making the folded materials into a semi-circle.

• Folding Technique

After each fold, it becomes more difficult to set each layer flatly on top of each other. This happens when layers are not wrinkle free and lumps begin to form in the inner portions. From this, paper extends beyond the folded section.

• Miscellaneous
1. The strength of a person can affect how many fold are achievable. The stronger the person the flatter the edges.
2. The paper is not uniform.
3. There's a difference in the limited number of folds based on whether the paper is being folded in the single direction or the alternative direction.

#### Related Problem

Rather than how many folds are achievable, a similar problem to the bed sheet problem focuses on the thickness of the sheet of paper. If you were to take a large sheet of paper with thickness 1/400 inch and fold it in half 50 times, how tall would it be (if it was theoretically possible to fold a sheet of paper 50 times)? Most people in their minds would imagine it to be as thick as a phone book. Some might even being daring and say a few feet. After 3 folds the sheet of paper would be as thick as your fingernail. And, if you continue to fold until the paper is at 50 folds then it will be as thick as 40 million miles. The paper would be able to reach the sun and return back to earth. This sounds ridiculous, but it isn't. This is an example in mathematics called geometric progression. Each number in the sequence is multiplied by a fixed number to get the next term.

# Why It's Interesting

This well-known problem is interesting because it was solved by Britney Gallivan, who was at the time a junior in high school. She was asked by her teacher to fold a sheet of paper 12 times and as an incentive she would get extra credit. She failed multiples times. Later she succeeded after using a thin gold sheet and proved the assumption wrong. Gallivan was able to achieve 12 folds by folding a roll of thin toilet paper that stretched over three-fourths of a mile. It took seven hours in a shopping mall with her parents, but Gallivan was able to bust a myth as well as derive a formula relating the width, thickness of a paper and the number of folds achievable. The urban legend of 7 folds was disproved in 2001.

Gallivan showed that any person can do the impossible. She solved a problem that many mathematicians attempted but also failed to solve. It is interesting that a high school student with such fervor demonstrated a person should not accept anything to be true without evidence.

Gallivan has inspired others to break the record for most folds. Late April 2011 some students from Massachusetts claimed the world record in achieving 13 folds. They used the thinnest type of toilet paper that stretched over 2 miles.

# References

http://pomonahistorical.org/12times.htm
http://sciencebits.wordpress.com/2008/09/06/folding-paper/
Gallivan, B. C. "How to Fold Paper in Half Twelve Times: An 'Impossible Challenge' Solved and Explained." Pomona, CA: Historical Society of Pomona Valley, 2002.
http://mathworld.wolfram.com/Folding.html

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