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Number of folds to form a cube, using a square paper?

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7

1

$begingroup$

Take a square paper of size S X S units, and thickness of paper 1 unit.

How many half folds should we do to get form a cube out it?

Consider infinite foldable paper.

mathematics paper-folding

share|improve this question

edited 17 hours ago

Ahmed Ashour

972313

asked 22 hours ago

Amruth AAmruth A

1,57821146

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
  $endgroup$
  – Nick
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Full cube, or hollow cube?
  $endgroup$
  – Laurent LA RIZZA
  18 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
  $endgroup$
  – Nick
  17 hours ago
  

  
  
  
  

add a comment | 

7

1

$begingroup$

Take a square paper of size S X S units, and thickness of paper 1 unit.

How many half folds should we do to get form a cube out it?

Consider infinite foldable paper.

mathematics paper-folding

share|improve this question

edited 17 hours ago

Ahmed Ashour

972313

asked 22 hours ago

Amruth AAmruth A

1,57821146

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
  $endgroup$
  – Nick
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Full cube, or hollow cube?
  $endgroup$
  – Laurent LA RIZZA
  18 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
  $endgroup$
  – Nick
  17 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Take a square paper of size S X S units, and thickness of paper 1 unit.

How many half folds should we do to get form a cube out it?

Consider infinite foldable paper.

mathematics paper-folding

share|improve this question

edited 17 hours ago

Ahmed Ashour

972313

asked 22 hours ago

Amruth AAmruth A

1,57821146

$endgroup$

share|improve this question

edited 17 hours ago

Ahmed Ashour

972313

asked 22 hours ago

Amruth AAmruth A

1,57821146

share|improve this question

edited 17 hours ago

Ahmed Ashour

972313

asked 22 hours ago

Amruth AAmruth A

1,57821146

edited 17 hours ago

Ahmed Ashour

972313

edited 17 hours ago

Ahmed Ashour

972313

asked 22 hours ago

Amruth AAmruth A

1,57821146

asked 22 hours ago

Amruth AAmruth A

1,57821146

3 Answers
3

active

oldest

votes

5

$begingroup$

Firstly there must be an even number of folds for the folded paper to remain square


Consider the volume of the paper which is $S times S times 1 = S^2$


Therefore the resultant cube side is $sqrt[3]{S^2}$


After $N$ pairs of folds the side length is $S / (2^N)$


So $S / (2^N) = sqrt[3]{S^2}$


So $S = 2^{3N}$


So $N = log S / (3 times log 2) $

Or $N = log S / log 8 $


Now suppose $ S = 512$ thickness $T = 1$

The computation gives $ N = 3 $ pairs of folds


Worked example:

After 1st pair of folds $S = 256$ with $T = 4$

After 2nd pair of folds $S = 128$ with $T = 16$

After 3rd pair of folds $S = 64$ with $T = 64$ which is a cube


The question asks how many half folds?

Answer:

Half folds = $ 2 times log S / log 8 $


Obviously the paper can only be folded thus if $N$ is an integer

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Weather VaneWeather Vane

1,49919

$endgroup$

add a comment | 

5

$begingroup$

Consider what happens by doing $2$ folds, one in each direction.

You now have a square that is half the size, but $4$ times the thickness.

So after $2k$ folds

the square has edge length $S/2^k$ and thickness $4^k$.

For this to be a cube we need:

$$frac{S}{2^k} = 4^k\ S=8^k \ k = log_8{S} = frac{log S}{log 8}$$

So the number of folds we need is

$$2k = frac{2log S}{log 8}$$
It will only be a cube if this result is an even whole number, i.e. if S is a power of $8$.

In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which is quite substantial, and the round folded edges also keep it from being in the shape of a cube.

share|improve this answer

edited 14 hours ago

2012rcampion

11.4k14273

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

$endgroup$

add a comment | 

4

$begingroup$

I think it should be...

$N = 2 log_2{sqrt[leftroot{-2}uproot{2}3]{S}}$, where $N$ is the number of folds.

As a caveat, $S$ can only be $2^{3x}$, where $x in mathbb{N}$.


For example:
$S=4096$ ($=2^{12})$
$sqrt[leftroot{-2}uproot{2}3]{S} = 16$; $log_2{16} = 4$; $4 times 2=8$ folds.
$4096 div 2^4$ (I fold each side $4$ times) $= 256$.

Thickness is $2^8$ ($8$ folds) = $256$.

share|improve this answer

edited 16 hours ago

Hugh

2,26811127

answered 22 hours ago

Jan IvanJan Ivan

2,066620

$endgroup$

add a comment | 

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5

$begingroup$

Firstly there must be an even number of folds for the folded paper to remain square


Consider the volume of the paper which is $S times S times 1 = S^2$


Therefore the resultant cube side is $sqrt[3]{S^2}$


After $N$ pairs of folds the side length is $S / (2^N)$


So $S / (2^N) = sqrt[3]{S^2}$


So $S = 2^{3N}$


So $N = log S / (3 times log 2) $

Or $N = log S / log 8 $


Now suppose $ S = 512$ thickness $T = 1$

The computation gives $ N = 3 $ pairs of folds


Worked example:

After 1st pair of folds $S = 256$ with $T = 4$

After 2nd pair of folds $S = 128$ with $T = 16$

After 3rd pair of folds $S = 64$ with $T = 64$ which is a cube


The question asks how many half folds?

Answer:

Half folds = $ 2 times log S / log 8 $


Obviously the paper can only be folded thus if $N$ is an integer

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Weather VaneWeather Vane

1,49919

$endgroup$

add a comment | 

5

$begingroup$

Firstly there must be an even number of folds for the folded paper to remain square


Consider the volume of the paper which is $S times S times 1 = S^2$


Therefore the resultant cube side is $sqrt[3]{S^2}$


After $N$ pairs of folds the side length is $S / (2^N)$


So $S / (2^N) = sqrt[3]{S^2}$


So $S = 2^{3N}$


So $N = log S / (3 times log 2) $

Or $N = log S / log 8 $


Now suppose $ S = 512$ thickness $T = 1$

The computation gives $ N = 3 $ pairs of folds


Worked example:

After 1st pair of folds $S = 256$ with $T = 4$

After 2nd pair of folds $S = 128$ with $T = 16$

After 3rd pair of folds $S = 64$ with $T = 64$ which is a cube


The question asks how many half folds?

Answer:

Half folds = $ 2 times log S / log 8 $


Obviously the paper can only be folded thus if $N$ is an integer

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Weather VaneWeather Vane

1,49919

$endgroup$

add a comment | 

$begingroup$

Firstly there must be an even number of folds for the folded paper to remain square


Consider the volume of the paper which is $S times S times 1 = S^2$


Therefore the resultant cube side is $sqrt[3]{S^2}$


After $N$ pairs of folds the side length is $S / (2^N)$


So $S / (2^N) = sqrt[3]{S^2}$


So $S = 2^{3N}$


So $N = log S / (3 times log 2) $

Or $N = log S / log 8 $


Now suppose $ S = 512$ thickness $T = 1$

The computation gives $ N = 3 $ pairs of folds


Worked example:

After 1st pair of folds $S = 256$ with $T = 4$

After 2nd pair of folds $S = 128$ with $T = 16$

After 3rd pair of folds $S = 64$ with $T = 64$ which is a cube


The question asks how many half folds?

Answer:

Half folds = $ 2 times log S / log 8 $


Obviously the paper can only be folded thus if $N$ is an integer

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Weather VaneWeather Vane

1,49919

$endgroup$

share|improve this answer

edited 20 hours ago

answered 21 hours ago

Weather VaneWeather Vane

1,49919

answered 21 hours ago

Weather VaneWeather Vane

1,49919

answered 21 hours ago

Weather VaneWeather Vane

1,49919

5

$begingroup$

Consider what happens by doing $2$ folds, one in each direction.

You now have a square that is half the size, but $4$ times the thickness.

So after $2k$ folds

the square has edge length $S/2^k$ and thickness $4^k$.

For this to be a cube we need:

$$frac{S}{2^k} = 4^k\ S=8^k \ k = log_8{S} = frac{log S}{log 8}$$

So the number of folds we need is

$$2k = frac{2log S}{log 8}$$
It will only be a cube if this result is an even whole number, i.e. if S is a power of $8$.

In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which is quite substantial, and the round folded edges also keep it from being in the shape of a cube.

share|improve this answer

edited 14 hours ago

2012rcampion

11.4k14273

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

$endgroup$

add a comment | 

5

$begingroup$

Consider what happens by doing $2$ folds, one in each direction.

You now have a square that is half the size, but $4$ times the thickness.

So after $2k$ folds

the square has edge length $S/2^k$ and thickness $4^k$.

For this to be a cube we need:

$$frac{S}{2^k} = 4^k\ S=8^k \ k = log_8{S} = frac{log S}{log 8}$$

So the number of folds we need is

$$2k = frac{2log S}{log 8}$$
It will only be a cube if this result is an even whole number, i.e. if S is a power of $8$.

In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which is quite substantial, and the round folded edges also keep it from being in the shape of a cube.

share|improve this answer

edited 14 hours ago

2012rcampion

11.4k14273

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

$endgroup$

add a comment | 

$begingroup$

Consider what happens by doing $2$ folds, one in each direction.

You now have a square that is half the size, but $4$ times the thickness.

So after $2k$ folds

the square has edge length $S/2^k$ and thickness $4^k$.

For this to be a cube we need:

$$frac{S}{2^k} = 4^k\ S=8^k \ k = log_8{S} = frac{log S}{log 8}$$

So the number of folds we need is

$$2k = frac{2log S}{log 8}$$
It will only be a cube if this result is an even whole number, i.e. if S is a power of $8$.

In reality, this will only work if you cut the paper in half and stack the pieces, instead of folding the paper. I have ignored the amount of paper connecting the different layers of the folded cube, which is quite substantial, and the round folded edges also keep it from being in the shape of a cube.

share|improve this answer

edited 14 hours ago

2012rcampion

11.4k14273

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

$endgroup$

share|improve this answer

edited 14 hours ago

2012rcampion

11.4k14273

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

edited 14 hours ago

2012rcampion

11.4k14273

edited 14 hours ago

2012rcampion

11.4k14273

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

answered 21 hours ago

Jaap ScherphuisJaap Scherphuis

16.1k12772

4

$begingroup$

I think it should be...

$N = 2 log_2{sqrt[leftroot{-2}uproot{2}3]{S}}$, where $N$ is the number of folds.

As a caveat, $S$ can only be $2^{3x}$, where $x in mathbb{N}$.


For example:
$S=4096$ ($=2^{12})$
$sqrt[leftroot{-2}uproot{2}3]{S} = 16$; $log_2{16} = 4$; $4 times 2=8$ folds.
$4096 div 2^4$ (I fold each side $4$ times) $= 256$.

Thickness is $2^8$ ($8$ folds) = $256$.

share|improve this answer

edited 16 hours ago

Hugh

2,26811127

answered 22 hours ago

Jan IvanJan Ivan

2,066620

$endgroup$

add a comment | 

4

$begingroup$

I think it should be...

$N = 2 log_2{sqrt[leftroot{-2}uproot{2}3]{S}}$, where $N$ is the number of folds.

As a caveat, $S$ can only be $2^{3x}$, where $x in mathbb{N}$.


For example:
$S=4096$ ($=2^{12})$
$sqrt[leftroot{-2}uproot{2}3]{S} = 16$; $log_2{16} = 4$; $4 times 2=8$ folds.
$4096 div 2^4$ (I fold each side $4$ times) $= 256$.

Thickness is $2^8$ ($8$ folds) = $256$.

share|improve this answer

edited 16 hours ago

Hugh

2,26811127

answered 22 hours ago

Jan IvanJan Ivan

2,066620

$endgroup$

add a comment | 

$begingroup$

I think it should be...

$N = 2 log_2{sqrt[leftroot{-2}uproot{2}3]{S}}$, where $N$ is the number of folds.

As a caveat, $S$ can only be $2^{3x}$, where $x in mathbb{N}$.


For example:
$S=4096$ ($=2^{12})$
$sqrt[leftroot{-2}uproot{2}3]{S} = 16$; $log_2{16} = 4$; $4 times 2=8$ folds.
$4096 div 2^4$ (I fold each side $4$ times) $= 256$.

Thickness is $2^8$ ($8$ folds) = $256$.

share|improve this answer

edited 16 hours ago

Hugh

2,26811127

answered 22 hours ago

Jan IvanJan Ivan

2,066620

$endgroup$

share|improve this answer

edited 16 hours ago

Hugh

2,26811127

answered 22 hours ago

Jan IvanJan Ivan

2,066620

edited 16 hours ago

Hugh

2,26811127

edited 16 hours ago

Hugh

2,26811127

answered 22 hours ago

Jan IvanJan Ivan

2,066620

answered 22 hours ago

Jan IvanJan Ivan

2,066620