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Number of folds to form a cube, using a square paper?
Folding a hexagon to a rectangle or square, with uniform overlapWhich 3D shape can you make out of this?The square and the compass II - Midpoints1 Cube 1 Square 1 LinePFG: A pretty spiral!Fold a standard paper to get a RhombusA new PSE member with square reputationMaximal-volume cube net from unit square paperWhat is the largest number of cubes that can be cut?Folding a paper wouldn't be this hard
$begingroup$
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
edited 4 hours ago
Ahmed Ashour
972313
asked 10 hours ago
Amruth AAmruth A
1,57321145
$endgroup$
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
edited 4 hours ago
Ahmed Ashour
972313
asked 10 hours ago
Amruth AAmruth A
1,57321145
$endgroup$
1
$begingroup$
Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
$endgroup$
– Nick
8 hours ago
2
$begingroup$
Full cube, or hollow cube?
$endgroup$
– Laurent LA RIZZA
6 hours ago
$begingroup$
Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
$endgroup$
– Nick
4 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
edited 4 hours ago
Ahmed Ashour
972313
asked 10 hours ago
Amruth AAmruth A
1,57321145
$endgroup$
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
mathematics paper-folding
edited 4 hours ago
Ahmed Ashour
972313
asked 10 hours ago
Amruth AAmruth A
1,57321145
edited 4 hours ago
Ahmed Ashour
972313
asked 10 hours ago
Amruth AAmruth A
1,57321145
edited 4 hours ago
Ahmed Ashour
972313
edited 4 hours ago
Ahmed Ashour
972313
edited 4 hours ago
Ahmed Ashour
972313
972313
asked 10 hours ago
Amruth AAmruth A
1,57321145
asked 10 hours ago
Amruth AAmruth A
1,57321145
asked 10 hours ago
Amruth AAmruth A
1,57321145
1,57321145
1
$begingroup$
Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
$endgroup$
– Nick
8 hours ago
2
$begingroup$
Full cube, or hollow cube?
$endgroup$
– Laurent LA RIZZA
6 hours ago
$begingroup$
Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
$endgroup$
– Nick
4 hours ago
add a comment |
1
$begingroup$
Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
$endgroup$
– Nick
8 hours ago
2
$begingroup$
Full cube, or hollow cube?
$endgroup$
– Laurent LA RIZZA
6 hours ago
$begingroup$
Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
$endgroup$
– Nick
4 hours ago
1
1
$begingroup$
Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
$endgroup$
– Nick
8 hours ago
$begingroup$
Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
$endgroup$
– Nick
8 hours ago
2
2
$begingroup$
Full cube, or hollow cube?
$endgroup$
– Laurent LA RIZZA
6 hours ago
$begingroup$
Full cube, or hollow cube?
$endgroup$
– Laurent LA RIZZA
6 hours ago
$begingroup$
Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
$endgroup$
– Nick
4 hours ago
$begingroup$
Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
$endgroup$
– Nick
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
answered 8 hours ago
Weather VaneWeather Vane
1,49919
$endgroup$
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.
edited 2 hours ago
2012rcampion
11.4k14273
answered 9 hours ago
Jaap ScherphuisJaap Scherphuis
16.1k12772
$endgroup$
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) = $256$.
edited 3 hours ago
Hugh
2,26811127
answered 9 hours ago
Jan IvanJan Ivan
2,066620
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
answered 8 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
answered 8 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
answered 8 hours ago
Weather VaneWeather Vane
1,49919
$endgroup$
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
answered 8 hours ago
Weather VaneWeather Vane
1,49919
edited 7 hours ago
answered 8 hours ago
Weather VaneWeather Vane
1,49919
answered 8 hours ago
Weather VaneWeather Vane
1,49919
answered 8 hours ago
Weather VaneWeather Vane
1,49919
1,49919
add a comment |
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.
edited 2 hours ago
2012rcampion
11.4k14273
answered 9 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.
edited 2 hours ago
2012rcampion
11.4k14273
answered 9 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.
edited 2 hours ago
2012rcampion
11.4k14273
answered 9 hours ago
Jaap ScherphuisJaap Scherphuis
16.1k12772
$endgroup$
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.
edited 2 hours ago
2012rcampion
11.4k14273
answered 9 hours ago
Jaap ScherphuisJaap Scherphuis
16.1k12772
edited 2 hours ago
2012rcampion
11.4k14273
edited 2 hours ago
2012rcampion
11.4k14273
edited 2 hours ago
2012rcampion
11.4k14273
11.4k14273
answered 9 hours ago
Jaap ScherphuisJaap Scherphuis
16.1k12772
answered 9 hours ago
Jaap ScherphuisJaap Scherphuis
16.1k12772
answered 9 hours ago
Jaap ScherphuisJaap Scherphuis
16.1k12772
16.1k12772
add a comment |
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) = $256$.
edited 3 hours ago
Hugh
2,26811127
answered 9 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) = $256$.
edited 3 hours ago
Hugh
2,26811127
answered 9 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) = $256$.
edited 3 hours ago
Hugh
2,26811127
answered 9 hours ago
Jan IvanJan Ivan
2,066620
$endgroup$
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) = $256$.
edited 3 hours ago
Hugh
2,26811127
answered 9 hours ago
Jan IvanJan Ivan
2,066620
edited 3 hours ago
Hugh
2,26811127
edited 3 hours ago
Hugh
2,26811127
edited 3 hours ago
Hugh
2,26811127
2,26811127
answered 9 hours ago
Jan IvanJan Ivan
2,066620
answered 9 hours ago
Jan IvanJan Ivan
2,066620
answered 9 hours ago
Jan IvanJan Ivan
2,066620
2,066620
add a comment |
add a comment |
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1
$begingroup$
Fun question! Brought me to try folding an origami cube :D ... (there are various manuals online of someone)
$endgroup$
– Nick
8 hours ago
2
$begingroup$
Full cube, or hollow cube?
$endgroup$
– Laurent LA RIZZA
6 hours ago
$begingroup$
Actually it was a diamond window, hollow cube. I should have counted the number of foldings though..
$endgroup$
– Nick
4 hours ago